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# How to write an equation for an exponential graph
R squared is equal to one over nine. Uses function notation incorrectly or not at all. See—this is the same form as the inverse variation function formula. When you do have a coefficient in front of the exponent, you can always get rid of it by manipulating the index. Instructional Procedures View This lesson can be fun for students because it illustrates how exponential and logarithmic functions are used in the real world.
What happens if you substitute one for x in your function. Every time you increase your Now let's figure out the exponential function. Okaloosa Is this Resource freely Available. Did you write the function. Consider providing a more student friendly definition with examples.
But either way, we have figured out the linear function. Writing exponential functions Video transcript - [Instructor] The graphs of the linear function f of x is equal to mx plus b and the exponential function g of x is equal to a times r to the x where r is greater than zero pass through the points negative one comma nine, so this is negative one comma nine right over here, and one comma one.
When x is negative one, y is nine. And then we could substitute this back into either one of these other two to figure out what a is.
It gets very hard to plot the values accurately because they are so small — for instance is 0. We will see some examples of exponential functions shortly.
The fact that g of x keeps approaching, it's getting closer and closer and closer to zero as x increases. But if you can take 40 hours to get there, you only have to average 25 miles per hour, since.
One minus nine is negative eight. Or we could write, let me scroll down a little bit. Maybe this was 5. This means you have to be very careful when you plan your graphs of exponential relations.
Discuss with the student differences in the rates of increase of the two functions and relate these differences to the forms of the equations.
Our change in x, we're going from x equals negative one to x equals one.
You can also have a coefficient in front of the exponent itself, like this: In fact this is so special that for many people this is THE exponential function. Provide frequent opportunities to use function notation, so the student can become familiar and comfortable with its use.
We started at negative one. This is a typical line shape for an exponential graph. And here on the x axis, we're marking off every half. Another way of looking at the expression "loga x" is "to what power exponent must a be raised to get x?. a. 3x + 1 = 35 Write the equation.
x + 1 = 5 Equate the exponents. − 1 Subtract 1 from each side. − 1 x = 4 Simplify. b. Solving Exponential Equations by Graphing Sometimes, it is impossible to rewrite each side of an exponential equation using the same base. You can solve these types of equations by graphing each side and fi nding.
Graphing Exponential Functions Worksheets This Algebra 1 Graphing Exponential Functions worksheet will give you exponent functions to graph.
You may choose to graph an equation or write an equation from a graph. • Graph exponential functions and use the One-to-One Property. • Recognize, evaluate, and graph exponential functions with base e.
• Use exponential functions to model and solve real-life problems. What You Should Learn. 3 Exponential Functions. 4 Exponential Functions. A pair of an input value and its corresponding output value is called an ordered pair and can be written as (a, b).
In an ordered pair the first number, the input a, corresponds to the horizontal axis and the second number, the output b, corresponds to the vertical axis. The graph of an exponential function can also be reflected over the x-axis or the y-axis, and rotated around the origin, as in Heading.
The general form of an exponential function is f (x) = c ƒ a x-h + k, where a is a positive constant and a ≠1. The letters a and b in the exponential function equation are coefficients that form the base of the exponent x. This interactive graph lets you manipulate these values on the graph of the equation y .
How to write an equation for an exponential graph
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Writing an explicit formula for the graph of an exponential relationship | LearnZillion
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# 10.3 Solving ratio problems where the total of one part of the ratio is given
Take a look at the worked example below:
You are growing tomatoes. The instructions on the tomato feed say:
• Use 1 part feed to 4 parts water
If you use 600 ml of water, how much tomato feed should you use?
These questions make much more sense if you look at them visually:
Figure 15 Solving ratio problems to grow tomatoes
You can now see clearly that 600 ml of water is worth 4 parts of the ratio. To find one part of the ratio you need to do:
• 600 ml ÷ 4 = 150 ml
Since the feed is only 1 part, feed must be 150 ml. If feed was more than one part you would multiply 150 ml by the number of parts.
Just as with the previous type of question, you need to try to work out the value of 1 part. The value of any other number of parts can be worked out from this.
## Activity 26: Ratio problems with one part given
Practise your skills by tackling the ratio problems below:
1. A recipe requires flour and butter to be used in the ratio 3:5. The amount of butter used is 700 g.
How much flour will be needed?
1. Flour:Butter
Figure 16 Using ratios in recipes
To find one part you do 700 g ÷ 5 = 140 g
To find the amount of flour needed you then do 140 g × 3 = 420 g flour.
1. When looking after children aged between 7 and 10, the ratio of adults to children must be 1:8.
• a.For a group of 32 children, how many adults must there be?
• b.If there was one more child in the group, how would this affect the number of adults required?
• a.
Figure 17 Working out the ratio of adults to children
To find one part you do 32 ÷ 8 = 4.
• b.If there were 33 children, one part would be 33 ÷ 8 = 4.125.
Since you cannot have 4.125 adults, you need to round up to 5 adults so you would need one more adult for 33 children.
1. A shop mixes bags of muesli using oats, sultanas and nuts in the ratio 6:3:1.
If the amount of sultanas used is 210 g, how heavy will the bag of muesli be?
1. Oats:Sultanas:Nuts
Figure 18 Working out the ratio of oats, sultanas and nuts
Sultanas are 3 parts so to find 1 part you do 210 g ÷ 3 = 70 g.
Oats are 6 parts so 6 × 70 = 420 g.
Nuts are only 1 part so they are 70 g.
The total weight of the bag would be 210 g + 420 g + 70 g = 700 g.
Next you’ll look at ratio problems where only the difference in amounts is given.
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# Riemann Sum & Table Problems (Type 5)
### AP Questions Type 5: Riemann Sum & Table Problems
Information given in tables may be used to test a variety of ideas in calculus including analysis of functions, accumulation, theory and theorems, and position-velocity-acceleration, among others. Numbers and working with numbers are part of the Rule of Four and table problems are one way they are tested. This question often includes an equation in a latter part of the problem that refers to the same situation.
What students should be able to do:
• Find the average rate of change over an interval
• Approximate the derivative using a difference quotient. Use the two values closest to the number at which you are approximating. This amounts to finding the slope or rate of change. Show the quotient even if you can do the arithmetic in your head and even if the denominator is 1.
• Use a left-, right-, or midpoint- Riemann sums or a trapezoidal approximation to approximate the value of a definite integral using values in the table (typically with uneven subintervals). The Trapezoidal Rule, per se, is not required; it is expected that students will add the areas of a small number of trapezoids without reference to a formula.
• Average value, average rate of change, Rolle’s theorem, the Mean Value Theorem, and the Intermediate Value Theorem. (See 2007 AB 3 – four simple parts that could be multiple-choice questions; the mean on this question was 0.96 out of a possible nine points.)
• These questions are usually presented in context and answers should be in that context. The context may be something growing (changing over time) or linear motion.
• Use the table to find a value based on the Mean Value Theorem (2018 AB 4(b)) or Intermediate Value Theorem. Also, 2018 AB 4 (d) asked a related question based on a function given by an equation.
• Unit analysis.
Dos and Don’ts
Do: Remember that you do not know what happens between the values in the table unless additional information is given. For example, do not assume that the largest number in the table is the maximum value of the function, or that the function is decreasing between two values just because a value is less than the preceding value.
Do: Show what you are doing even if you can do it in your head. If you’re finding a slope, show the quotient even if the denominator is 1.
Do Not do arithmetic: A long expression consisting entirely of numbers such as you get when doing a Riemann sum, does not need to be simplified in any way. If you simplify a correct answer incorrectly, you will lose credit.
Do Not leave expression such as R(3) – pull its numerical value from the table.
Do Not: Find a regression equation and then use that to answer parts of the question. While regression is perfectly good mathematics, regression equations are not one of the four things students may do with their calculator. Regression gives only an approximation of our function. The exam is testing whether students can work with numbers.
This question typically covers topics from Unit 6 of the CED but may include topics from Units 2, 3, and 4 as well.
Free-response examples:
• 2007 AB 3 (4 simple parts on various theorems, yet the mean score was 0.96 out of 9),
• 2017 AB 1/BC 1, and AB 6,
• 2016 AB 1/BC 1
• 2018 AB 4
• 2021 AB 1/ BC 1
• 2022 AB4/BC4 – average rate of change, IVT, Rieman sum, Related Rate (part (d) good question)
Multiple-choice questions from non-secure exams:
• 2012 AB 8, 86, 91
• 2012 BC 8, 81, 86 (81 and 86 are the same on both the AB and BC exams)
Revised March 12, 2021, March 25, 2022
The Mean Value Theorem (MVT) is proved by writing the equation of a function giving the (directed) length of a segment from the given function to the line between the endpoints as you can see here. Since the function and the line intersect at the endpoints of the interval this function satisfies the hypotheses of Rolle’s theorem and so the MVT follows directly. This means that the derivative of the distance function is zero at the points guaranteed by the MVT. Therefore, these values must also be the location of the local extreme values (maximums and minimums) of the distance function on the open interval. *
Here is an exploration in three similar examples that use this idea to foreshadow the MVT. You, of course, can use your own favorite function. Any differentiable function may be used, in which case a CAS calculator may be helpful. Answers are at the end.
First example:
Consider the function $\displaystyle f\left( x \right)=x+2\sin \left( {\pi x} \right)$ defined on the closed interval [–1,3]
1. Write the equation of the line thru the endpoints of the function.
2. Write an expression for h(x) the vertical distance between f(x) and the line found in part 1.
3. Find the x-coordinates of the local extreme values of h(x) on the open interval (–1,3).
4. Find the slope of f(x) at the values found in part 3.
5. Compare your answer to part 4 with the slope of the line. Is this a coincidence?
Second example: slightly more difficult than the first.
Consider the function $\displaystyle f\left( x \right)=1+x+2\cos (x)$ defined on the closed interval $\displaystyle [\tfrac{\pi }{2},\tfrac{{9\pi }}{2}]$.
1. Write the equation of the line thru the endpoints of the function.
2. Write an expression for h(x) the vertical distance between f(x) and the line found in part 1.
3. Find the x-coordinates of the local extreme values of h(x) on the open interval $\displaystyle \left( {\tfrac{\pi }{2},\tfrac{{9\pi }}{2}} \right)$.
4. Find the slope of f(x) at the values found in part 3.
5. Compare your answer to part 4 with the slope of the line. Is this a coincidence?
Third example: In case you think I cooked the numbers. You may want to use a CAS for this one.
Consider the function $\displaystyle f(x)={{x}^{3}}$ defined on the closed interval $\displaystyle [-4,5]$.
1. Write the equation of the line thru the endpoints of the function.
2. Write an expression for h(x) the vertical distance between f(x) and the line found in part 1.
3. Find the x-coordinates of the local extreme values of h(x) on the open interval $\displaystyle \left( {-4,5} \right)$.
4. Find the slope of f(x) at the values found in part 3.
5. Compare your answer to part 4 with the slope of the line. Is this a coincidence?
First example:
1. y = x
2. $\displaystyle h(x)=f(x)-y(x)=x-\left( {x+2\sin (\pi x} \right)=\left( {2\sin (\pi x} \right)$
3. $\displaystyle {h}'\left( x \right)=2\pi \cos \left( {\pi x} \right)=0$ when x = –1/2, ½, 3/2 and 5/2
4. $\displaystyle {f}'\left( x \right)=1+2\pi \cos \left( {\pi x} \right)=1$, the slope = 1 at all four points
5. They are the same. Not a coincidence.
Second example:
1. The endpoints are $\displaystyle \left( {\tfrac{\pi }{2},1+\tfrac{\pi }{2}} \right)$ and $\displaystyle \left( {\tfrac{9\pi }{2},1+\tfrac{{9\pi }}{2}} \right)$; the line is $\displaystyle y=x+1$
2. $\displaystyle h(x)=f(x)-y(x)=\left( {1+x+2\cos (x)} \right)-(x+1)=2\cos (x)$
3. $\displaystyle {h}'\left( x \right)=-2\sin (x)=0$ when $\displaystyle x=\pi ,2\pi ,3\pi ,\text{ and }4\pi$
4. $\displaystyle {f}'\left( x \right)=1-2\sin \left( x \right)=1$; at the points above the slope is 1.
5. They are the same. Not a coincidence.
Third example:
1. The endpoints are (-4, -64) and (5, 125), the line is $\displaystyle y=125+21\left( {x-5} \right)=21x+20$.
2. $\displaystyle h\left( x \right)={{x}^{3}}-21x-20$
3. $\displaystyle {h}'\left( x \right)=3{{x}^{2}}-21=0$ when $\displaystyle x=\sqrt{7},-\sqrt{7}$
4. $\displaystyle {f}'\left( {\pm \sqrt{7}} \right)=3{{\left( {\pm \sqrt{7}} \right)}^{2}}=21$
5. They are the same. Not a coincidence.
See this post for links to other posts discussing the full development of the MVT
* It is possible that the derivative is zero and the point is not an extreme value. This is like the situation with a point of inflection when the first derivative is zero but does not change sign.
This post was originally published on October 19, 2018.
# Unit 5 – Analytical Applications of Differentiation
Unit 5 covers the application of derivatives to the analysis of functions and graphs. Reasoning and justification of results are also important themes in this unit. (CED – 2019 p. 92 – 107). These topics account for about 15 – 18% of questions on the AB exam and 8 – 11% of the BC questions.
You may want to consider teaching Unit 4 after Unit 5. Notes on Unit 4 are here.
Reasoning and writing justification of results are mentioned and stressed in the introduction to the topic (p. 93) and for most of the individual topics. See Learning Objective FUN-A.4 “Justify conclusions about the behavior of a function based on the behavior of its derivatives,” and likewise in FUN-1.C for the Extreme value theorem, and FUN-4.E for implicitly defined functions. Be sure to include writing justifications as you go through this topic. Use past free-response questions as exercises and also as guide as to what constitutes a good justification. Links in the margins of the CED are also helpful and give hints on writing justifications and what is required to earn credit. See the presentation
### Topics 5.1
Topic 5.1 Using the Mean Value Theorem While not specifically named in the CED, Rolle’s Theorem is a lemma for the Mean Value Theorem (MVT). The MVT states that for a function that is continuous on the closed interval and differentiable over the corresponding open interval, there is at least one place in the open interval where the average rate of change equals the instantaneous rate of change (derivative). This is a very important existence theorem that is used to prove other important ideas in calculus. Students often confuse the average rate of change, the mean value, and the average value of a function – See What’s a Mean Old Average Anyway?
### Topics 5.2 – 5.9
Topic 5.2 Extreme Value Theorem, Global Verses Local Extrema, and Critical Points An existence theorem for continuous functions on closed intervals
Topic 5.3 Determining Intervals on Which a Function is Increasing or Decreasing Using the first derivative to determine where a function is increasing and decreasing.
Topic 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema Using the first derivative to determine local extreme values of a function
Topic 5.5 Using the Candidates’ Test to Determine Absolute (Global) Extrema The Candidates’ test can be used to find all extreme values of a function on a closed interval
Topic 5.6 Determining Concavity of Functions on Their Domains FUN-4.A.4 defines (at least for AP Calculus) When a function is concave up and down based on the behavior of the first derivative. (Some textbooks may use different equivalent definitions.) Points of inflection are also included under this topic.
Topic 5.7 Using the Second Derivative Test to Determine Extrema Using the Second Derivative Test to determine if a critical point is a maximum or minimum point. If a continuous function has only one critical point on an interval then it is the absolute (global) maximum or minimum for the function on that interval.
Topic 5.8 Sketching Graphs of Functions and Their Derivatives First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.
Topic 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.
### Topics 5.10 – 5.11
Optimization is an important application of derivatives. Optimization problems as presented in most textbooks, begin with writing the model or equation that describes the situation to be optimized. This proves difficult for students, and is not “calculus” per se. Therefore, writing the equation has not been asked on AP exams in recent years (since 1983). Questions give the expression to be optimized and students do the “calculus” to find the maximum or minimum values. To save time, my suggestion is to not spend too much time writing the equations; rather concentrate on finding the extreme values.
Topic 5.10 Introduction to Optimization Problems
Topic 5.11 Solving Optimization Problems
### Topics 5.12
Topic 5.12 Exploring Behaviors of Implicit Relations Critical points of implicitly defined relations can be found using the technique of implicit differentiation. This is an AB and BC topic. For BC students the techniques are applied later to parametric and vector functions.
### Timing
Topic 5.1 is important and may take more than one day. Topics 5.2 – 5.9 flow together and for graphing they are used together; after presenting topics 5.2 – 5.7 spend the time in topics 5.8 and 5.9 spiraling and connecting the previous topics. Topics 5.10 and 5.11 – see note above and spend minimum time here. Topic 5.12 may take 2 days.
The suggested time for Unit 5 is 15 – 16 classes for AB and 10 – 11 for BC of 40 – 50-minute class periods, this includes time for testing etc.
Finally, were I still teaching, I would teach this unit before Unit 4. The linear motion topic (in Unit 4) are a special case of the graphing ideas in Unit 5, so it seems reasonable to teach this unit first. See Motion Problems: Same thing, Different Context
This is a re-post and update of the third in a series of posts from last year. It contains links to posts on this blog about the differentiation of composite, implicit, and inverse functions for your reference in planning. Other updated post on the 2019 CED will come throughout the year, hopefully, a few weeks before you get to the topic.
Previous posts on these topics include:
Then There Is This – Existence Theorems
What’s a Mean Old Average Anyway
Did He, or Didn’t He? History: how to find extreme values without calculus
Mean Value Theorem
Fermat’s Penultimate Theorem
Rolle’s theorem
The Mean Value Theorem I
The Mean Value Theorem II
Graphing
Concepts Related to Graphs
The Shapes of a Graph
Joining the Pieces of a Graph
Extreme Values
Extremes without Calculus
Concavity
Far Out! An exploration
Open or closed Should intervals of increasing, decreasing, or concavity be open or closed?
Others
Lin McMullin’s Theorem and More Gold The Golden Ratio in polynomials
Soda Cans Optimization video
Optimization – Reflections
Curves with Extrema?
Good Question 10 – The Cone Problem
Here are links to the full list of posts discussing the ten units in the 2019 Course and Exam Description.
Limits and Continuity – Unit 1 (8-11-2020)
Definition of t he Derivative – Unit 2 (8-25-2020)
Contextual Applications of the Derivative – Unit 4 (9-22-2002) Consider teaching Unit 5 before Unit 4
Analytical Applications of Differentiation – Unit 5 (9-29-2020) Consider teaching Unit 5 before Unit 4 THIS POST
LAST YEAR’S POSTS – These will be updated in coming weeks
2019 – CED Unit 6 Integration and Accumulation of Change
2019 – CED Unit 7 Differential Equations Consider teaching after Unit 8
2019 – CED Unit 8 Applications of Integration Consider teaching after Unit 6, before Unit 7
2019 – CED Unit 9 Parametric Equations, Polar Coordinates, and Vector-Values Functions
2019 CED Unit 10 Infinite Sequences and Series
# Discovering the MVT
Today’s Blog is an exploration that will lead up to the Mean Value Theorem (MVT) and, I hope, help your students better understand the MVT and why it is true.
While you may do this by hand, using a graphing calculator will make things way easier. This is good calculator practice and can be done on a graphing, non-CAS, calculator without writing anything down. Try it that way.
Here are the steps to follow. My solution with screen pictures is below.
1. Choose your favorite differentiable function. Call it f(x) and enter it in your calculator as Y1.
2. Choose two values, a and b, in the domain of your function. Save (store) these on your calculator as a and b.
3. Find the slope of the line (a, f(a)) and (b, f(b)). It would be best, but not necessary, that the line intersects the function only at (a, f(a)) and (b, f(b)) not between them, and not be horizontal. Store this in your calculator as m.
4. Write the equation of the line through (a, f(a)) and (b, f(b)) and enter it as Y2.
5. Write a function, h(x), that gives the vertical distance between f(x) and the line found in step 3. (Hint: upper curve minus the lower.) Enter this as Y3
6. Find the x-coordinate local extreme value of h(x). Store this number to c.
7. Find the slope of the tangent line to f(x) at the value found in step 6.
8. What do you notice? Compare your result and conclusion with the other in your class. Discuss.
My solution.
Step 1: I choose $f\left( x \right)=x+2\sin \left( x \right)$ and entered this in my calculator as Y1
Step 2: I choose a = 1 and b = 3 and stored them in my calculator.
Step 3: I calculated the slope in my calculator – see first figure.
Step 4: The equation of the line is $y=f\left( a \right)+m\left( {x-a} \right)$. I entered this as Y2 in my calculator.
Step 5 $h\left( x \right)=Y1(x)-Y2(x)=\left( {x+2\sin (x)} \right)-\left( {f\left( a \right)+m\left( {x-a} \right)} \right)$
Step 6: ${h}'\left( x \right)=1+2\cos (x)-m$
Solve ${h}'\left( x \right)=0$ for the value between a and b on your calculator. See second figure.
Step 6 and 7: I stored this value to C in my calculator and computed ${f}'(c)$ on the home screen. See third figure.
Step 8: It is no coincidence that ${f}'\left( c \right)=m$.
The Mean Value Theorem states that for a function that is continuous on the interval [ab] and differentiable on the open interval (ab) there exists a number c in (a, b) such that
$\displaystyle {f}'\left( c \right)=\frac{{f\left( b \right)-f\left( a \right)}}{{b-a}}$
1. Can you show why ${f}'\left( c \right)=m$. ? Hint: Look at the expression for ${h}'\left( c \right)$ in step 5; set it equal to zero. Why must the solution be the value that makes ${f}'\left( x \right)=m$?
2. What does this mean graphically?
1. Pick a different value for a and/or b so that the line between (a, f(a)) and (b, f(b)) intersects f(x) two (or more) times. The derivative will now have two (or more) zeros. Find them and calculate the slope at each one. What do you notice?
Students often confuse the Mean Value Theorem, the Average Rate of Change of a function on an interval, and the Average Value of a function on an interval. This is understandable because of the similarity in their names and the similarity of their results. Be sure to point this out as you teach them and help them learn the meanings of each.
Other posts related to the Mean Value Theorem
Foreshadowing the MVT Other examples using this technique
Existence Theorems
Fermat’s Penultimate Theorem A lemma for Rolle’s Theorem: Any function extreme value(s) on an open interval must occur where the derivative is zero or undefined.
Rolle’s Theorem A lemma for the MVT: On an interval if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b) and f(a) = f(b), there must exist a number in the open interval (a, b) where ‘(c) = 0.
Mean Value Theorem I Proof
Mean Value Theorem II Graphical Considerations
Darboux’s Theorem the Intermediate Value Theorem for derivatives.
Mean Tables
The Definite Integral and the FTC
# Analytical Applications of Differentiation – Unit 5
Unit 5 covers the application of derivatives to the analysis of functions and graphs. Reasoning and justification of results are also important themes in this unit. (CED – 2019 p. 92 – 107). These topics account for about 15 – 18% of questions on the AB exam and 8 – 11% of the BC questions.
You may want to consider teaching Unit 4 after Unit 5. Notes on Unit 4 are here.
Reasoning and writing justification of results are mentioned and stressed in the introduction to the topic (p. 93) and for most of the individual topics. See Learning Objective FUN-A.4 “Justify conclusions about the behavior of a function based on the behavior of its derivatives,” and likewise in FUN-1.C for the Extreme value theorem, and FUN-4.E for implicitly defined functions. Be sure to include writing justifications as you go through this topic. Use past free-response questions as exercises and also as guide as to what constitutes a good justification. Links in the margins of the CED are also helpful and give hints on writing justifications and what is required to earn credit. See the presentation
### Topics 5.1
Topic 5.1 Using the Mean Value Theorem While not specifically named in the CED, Rolle’s Theorem is a lemma for the Mean Value Theorem (MVT). The MVT states that for a function that is continuous on the closed interval and differentiable over the corresponding open interval, there is at least one place in the open interval where the average rate of change equals the instantaneous rate of change (derivative). This is a very important existence theorem that is used to prove other important ideas in calculus. Students often confuse the average rate of change, the mean value, and the average value of a function – See What’s a Mean Old Average Anyway?
### Topics 5.2 – 5.9
Topic 5.2 Extreme Value Theorem, Global Verses Local Extrema, and Critical Points An existence theorem for continuous functions on closed intervals
Topic 5.3 Determining Intervals on Which a Function is Increasing or Decreasing Using the first derivative to determine where a function is increasing and decreasing.
Topic 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema Using the first derivative to determine local extreme values of a function
Topic 5.5 Using the Candidates’ Test to Determine Absolute (Global) Extrema The Candidates’ test can be used to find all extreme values of a function on a closed interval
Topic 5.6 Determining Concavity of Functions on Their Domains FUN-4.A.4 defines (at least for AP Calculus) When a function is concave up and down based on the behavior of the first derivative. (Some textbooks may use different equivalent definitions.) Points of inflection are also included under this topic.
Topic 5.7 Using the Second Derivative Test to Determine Extrema Using the Second Derivative Test to determine if a critical point is a maximum or minimum point. If a continuous function has only one critical point on an interval then it is the absolute (global) maximum or minimum for the function on that interval.
Topic 5.8 Sketching Graphs of Functions and Their Derivatives First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.
Topic 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.
### Topics 5.10 – 5.11
Optimization is important application of derivatives. Optimization problems as presented in most text books, begin with writing the model or equation that describes the situation to be optimized. This proves difficult for students, and is not “calculus” per se. Therefore, writing the equation has not be asked on AP exams in recent years (since 1983). Questions give the expression to be optimized and students do the “calculus” to find the maximum or minimum values. To save time, my suggestion is to not spend too much time writing the equations; rather concentrate on finding the extreme values.
Topic 5.10 Introduction to Optimization Problems
Topic 5.11 Solving Optimization Problems
### Topics 5.12
Topic 5.12 Exploring Behaviors of Implicit Relations Critical points of implicitly defined relations can be found using the technique of implicit differentiation. This is an AB and BC topic. For BC students the techniques are applied later to parametric and vector functions.
### Timing
Topic 5.1 is important and may take more than one day. Topics 5.2 – 5.9 flow together and for graphing they are used together; after presenting topics 5.2 – 5.7 spend the time in topics 5.8 and 5.9 spiraling and connecting the previous topics. Topics 5.10 and 5.11 – see note above and spend minimum time here. Topic 5.12 may take 2 days.
The suggested time for Unit 5 is 15 – 16 classes for AB and 10 – 11 for BC of 40 – 50-minute class periods, this includes time for testing etc.
Finally, were I still teaching, I would teach this unit before Unit 4. The linear motion topic (in Unit 4) are a special case of the graphing ideas in Unit 5, so it seems reasonable to teach this unit first. See Motion Problems: Same thing, Different Context
This is a re-post and update of the third in a series of posts from last year. It contains links to posts on this blog about the differentiation of composite, implicit, and inverse functions for your reference in planning. Other updated post on the 2019 CED will come throughout the year, hopefully, a few weeks before you get to the topic.
Previous posts on these topics include:
Then There Is This – Existence Theorems
What’s a Mean Old Average Anyway
Did He, or Didn’t He? History: how to find extreme values without calculus
Mean Value Theorem
Fermat’s Penultimate Theorem
Rolle’s theorem
The Mean Value Theorem I
The Mean Value Theorem II
Graphing
Concepts Related to Graphs
The Shapes of a Graph
Joining the Pieces of a Graph
Extreme Values
Extremes without Calculus
Concavity
Far Out! An exploration
Open or Closed Should intervals of increasing, decreasing, or concavity be open or closed?
Others
Lin McMullin’s Theorem and More Gold The Golden Ratio in polynomials
Soda Cans Optimization video
Optimization – Reflections
Curves with Extrema?
Good Question 10 – The Cone Problem
Here are links to the full list of posts discussing the ten units in the 2019 Course and Exam Description.
Limits and Continuity – Unit 1 (8-11-2020)
Definition of t he Derivative – Unit 2 (8-25-2020)
Contextual Applications of the Derivative – Unit 4 (9-22-2002) Consider teaching Unit 5 before Unit 4
Analytical Applications of Differentiation – Unit 5 (9-29-2020) Consider teaching Unit 5 before Unit 4 THIS POST
LAST YEAR’S POSTS – These will be updated in coming weeks
2019 – CED Unit 6 Integration and Accumulation of Change
2019 – CED Unit 7 Differential Equations Consider teaching after Unit 8
2019 – CED Unit 8 Applications of Integration Consider teaching after Unit 6, before Unit 7
2019 – CED Unit 9 Parametric Equations, Polar Coordinates, and Vector-Values Functions
2019 CED Unit 10 Infinite Sequences and Series
# Riemann Sum & Table Problems (Type 5)
### AP Questions Type 5: Riemann Sum & Table Problems
Tables may be used to test a variety of ideas in calculus including analysis of functions, accumulation, theory and theorems, and position-velocity-acceleration, among others. Numbers and working with numbers is part of the Rule of Four and table problems are one way this is tested. This question often includes an equation in a latter part of the problem that refers to the same situation.
What students should be able to do:
• Find the average rate of change over an interval
• Approximate the derivative using a difference quotient. Use the two values closest to the number at which you are approximating. This amounts to finding the slope or rate of change. Show the quotient even if you can do the arithmetic in your head and even if the denominator is 1.
• Use a left-, right-, or midpoint- Riemann sums or a trapezoidal approximation to approximate the value of a definite integral using values in the table (typically with uneven subintervals). The Trapezoidal Rule, per se, is not required; it is expected that students will add the areas of a small number of trapezoids without reference to a formula.
• Average value, average rate of change, Rolle’s theorem, the Mean Value Theorem and the Intermediate Value Theorem. (See 2007 AB 3 – four simple parts that could be multiple-choice questions; the mean on this question was 0.96 out of a possible 9.)
• These questions are usually presented in some context and answers should be in that context. The context may be something growing (changing over time) or linear motion.
• Use the table to find a value based on the Mean Value Theorem (2018 AB 4(b)) or Intermediate Value Theorem.
• One of the parts of this question asks a related question based on a function given by an equation.
• Unit analysis.
Do’s and Don’ts
Do: Remember that you do not know what happens between the values in the table unless some other information is given. For example, do not assume that the largest number in the table is the maximum value of the function, or that the function is decreasing between two values just because a value is less than the preceding value.
Do: Show what you are doing even if you can do it in your head. If you’re finding a slope, show the quotient even if the denominator is 1.
Do Not do arithmetic: A long expression consisting entirely of numbers such as you get when doing a Riemann sum, does not need to be simplified in any way. If you a simplify correct answer incorrectly, you will lose credit.
Do Not leave expression such as R(3) – pull its numerical value from the table.
Do Not: Find a regression equation and then use that to answer parts of the question. While regression is perfectly good mathematics, regression equations are not one of the four things students may do with their calculator. Regression gives only an approximation of our function. The exam is testing whether students can work with numbers.
This question typically covers topics from Unit 6 of the 2019 CED but may include topics from Units 2, 3, and 4 as well.
Free-response examples:
• 2007 AB 3 (4 simple parts on various theorems, yet the mean score was 0.96 out of 9),
• 2017 AB 1/BC 1, and AB 6,
• 2016 AB 1/BC 1
• 2018 AB 4
Multiple-choice questions from non-secure exams:
• 2012 AB 8, 86, 91
• 2012 BC 8, 81, 86 (81 and 86 are the same on both the AB and BC exams)
Revised March 12, 2021
# 2019 CED Unit 5 Analytical Applications of Differentiation
Unit 5 covers the application of derivatives to the analysis of functions and graphs. Reasoning and justification of results are also important themes in this unit. (CED – 2019 p. 92 – 107). These topics account for about 15 – 18% of questions on the AB exam and 8 – 11% of the BC questions.
Reasoning and writing justification of results are mentioned and stressed in the introduction to the topic (p. 93) and for most of the individual topics. See Learning Objective FUN-A.4 “Justify conclusions about the behavior of a function based on the behavior of its derivatives,” and likewise in FUN-1.C for the Extreme value theorem, and FUN-4.E for implicitly defined functions. Be sure to include writing justifications as you go through this topic. Use past free-response questions as exercises and also as guide as to what constitutes a good justification. Links in the margins of the CED are also helpful and give hints on writing justifications and what is required to earn credit. See the presentation
### Topics 5.1
Topic 5.1 Using the Mean Value Theorem While not specifically named in the CED, Rolle’s Theorem is a lemma for the Mean Value Theorem (MVT). The MVT states that for a function that is continuous on the closed interval and differentiable over the corresponding open interval, there is at least one place in the open interval where the average rate of change equals the instantaneous rate of change (derivative). This is a very important existence theorem that is used to prove other important ideas in calculus. Students often confuse the average rate of change, the mean value, and the average value of a function – See What’s a Mean Old Average Anyway?
### Topics 5.2 – 5.9
Topic 5.2 Extreme Value Theorem, Global Verses Local Extrema, and Critical Points An existence theorem for continuous functions on closed intervals
Topic 5.3 Determining Intervals on Which a Function is Increasing or Decreasing Using the first derivative to determine where a function is increasing and decreasing.
Topic 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema Using the first derivative to determine local extreme values of a function
Topic 5.5 Using the Candidates’ Test to Determine Absolute (Global) Extrema The Candidates’ test can be used to find all extreme values of a function on a closed interval
Topic 5.6 Determining Concavity of Functions on Their Domains FUN-4.A.4 defines (at least for AP Calculus) When a function is concave up and down based on the behavior of the first derivative. (Some textbooks may use different equivalent definitions.) Points of inflection are also included under this topic.
Topic 5.7 Using the Second Derivative Test to Determine Extrema Using the Second Derivative Test to determine if a critical point is a maximum or minimum point. If a continuous function has only one critical point on an interval, then it is the absolute (global) maximum or minimum for the function on that interval.
Topic 5.8 Sketching Graphs of Functions and Their Derivatives. First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.
Topic 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative. First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.
### Topics 5.10 – 5.11
Optimization is important application of derivatives. Optimization problems as presented in most text books, begin with writing the model or equation that describes the situation to be optimized. This proves difficult for students, and is not “calculus” per se. Therefore, writing the equation has not be asked on AP exams in recent years (since 1983). Questions give the expression to be optimized and students do the “calculus” to find the maximum or minimum values. To save time, my suggestion is to not spend too much time writing the equations; rather concentrate on finding the extreme values.
Topic 5.10 Introduction to Optimization Problems
Topic 5.11 Solving Optimization Problems
### Topics 5.12
Topic 5.12 Exploring Behaviors of Implicit Relations Critical points of implicitly defined relations can be found using the technique of implicit differentiation. This is an AB and BC topic. For BC students the techniques are applied later to parametric and vector functions.
### Timing
Topic 5.1 is important and may take more than one day. Topics 5.2 – 5.9 flow together and for graphing they are used together; after presenting topics 5.2 – 5.7 spend the time in topics 5.8 and 5.9 spiraling and connecting the previous topics. Topics 5.10 and 5.11 – see note above and spend minimum time here. Topic 5.12 may take 2 days.
The suggested time for Unit 5 is 15 – 16 classes for AB and 10 – 11 for BC of 40 – 50-minute class periods, this includes time for testing etc.
Finally, were I still teaching, I would teach this unit before Unit 4. The linear motion topic (in Unit 4) are a special case of the graphing ideas in Unit 5, so it seems reasonable to teach this unit first. See Motion Problems: Same thing, Different Context
Previous posts on these topics include:
Then There Is This – Existence Theorems
What’s a Mean Old Average Anyway
Did He, or Didn’t He? History: how to find extreme values without calculus
Mean Value Theorem
Fermat’s Penultimate Theorem
Rolle’s theorem
The Mean Value Theorem I
The Mean Value Theorem II
Graphing
Concepts Related to Graphs
The Shapes of a Graph
Joining the Pieces of a Graph
Extreme Values
Extremes without Calculus
Concavity
Far Out! An exploration
Open or Closed Should intervals of increasing, decreasing, or concavity be open or closed?
Others
Lin McMullin’s Theorem and More Gold The Golden Ratio in polynomials
Soda Cans Optimization video
Optimization – Reflections
Curves with Extrema?
Good Question 10 – The Cone Problem
Here are links to the full list of posts discussing the ten units in the 2019 Course and Exam Description.
2019 CED – Unit 1: Limits and Continuity
2019 CED – Unit 2: Differentiation: Definition and Fundamental Properties.
2019 CED – Unit 3: Differentiation: Composite , Implicit, and Inverse Functions
2019 CED – Unit 4 Contextual Applications of the Derivative Consider teaching Unit 5 before Unit 4
2019 – CED Unit 5 Analytical Applications of Differentiation Consider teaching Unit 5 before Unit 4
2019 – CED Unit 6 Integration and Accumulation of Change
2019 – CED Unit 7 Differential Equations Consider teaching after Unit 8
2019 – CED Unit 8 Applications of Integration Consider teaching after Unit 6, before Unit 7
2019 – CED Unit 9 Parametric Equations, Polar Coordinates, and Vector-Values Functions
2019 CED Unit 10 Infinite Sequences and Series
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# How do you find the length of the curve y^2 = 16(x+1)^3 where x is between [0,3] and y>0?
Feb 24, 2015
The answer is: $L = \frac{1}{54} \left(\sqrt{145} - 37 \sqrt{37}\right)$.
The curve can be written:
$y = \pm 4 \sqrt{{\left(x + 1\right)}^{3}}$, but we need only the branch with $y > 0$, so:
$y = 4 \sqrt{{\left(x + 1\right)}^{3}}$.
To find the lenght of a curve written as a function and in cartesian coordinates we have to remember this formula:
$L = {\int}_{a}^{b} \sqrt{1 + {\left[f ' \left(x\right)\right]}^{2}} \mathrm{dx}$.
So, we have first of all calculate the derivative of:
$y = 4 {\left(x + 1\right)}^{\frac{3}{2}} \Rightarrow y ' = 4 \cdot \frac{3}{2} {\left(x + 1\right)}^{\frac{1}{2}} = 6 \sqrt{x + 1}$
$L = {\int}_{0}^{3} \sqrt{1 + 36 \left(x + 1\right)} \mathrm{dx} = {\int}_{0}^{3} \sqrt{36 x + 37} \mathrm{dx} =$
$= \frac{1}{36} {\int}_{0}^{3} 36 \sqrt{36 x + 37} \mathrm{dx} = \frac{1}{36} {\left[{\left(36 x + 37\right)}^{\frac{1}{2} + 1} / \left(\frac{1}{2} + 1\right)\right]}_{0}^{3} =$
$= \frac{1}{36} \cdot \frac{2}{3} {\left[\sqrt{{\left(36 x + 37\right)}^{3}}\right]}_{0}^{3} = \frac{1}{54} \left(\sqrt{145} - \sqrt{{37}^{3}}\right) = \frac{1}{54} \left(\sqrt{145} - 37 \sqrt{37}\right)$.
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# Find all the zeros of
Question:
Find all the zeros of $2 x^{4}-13 x^{3}+19 x^{2}+7 x-3$ if two of its zeros are $(2+\sqrt{3})$ and $(2-\sqrt{3})$.
Solution:
Let $f(x)=2 x^{4}-13 x^{3}+19 x^{2}+7 x-3$
It is given that $(2+\sqrt{3})$ and $(2-\sqrt{3})$ are two zeroes of $f(x)$
Thus, $f(x)$ is completely divisible by $(x-2-\sqrt{3})$ and $(x-2+\sqrt{3})$.
Therefore, one factor of $f(x)$ is $\left((x-2)^{2}-3\right)$
$\Rightarrow$ one factor of $f(x)$ is $\left(x^{2}-4 x+1\right)$
We get another factor of $f(x)$ by dividing it with $\left(x^{2}-4 x+1\right)$.
On division, we get the quotient $2 x^{2}-5 x-3$.
$\Rightarrow f(x)=\left(x^{2}-4 x+1\right)\left(2 x^{2}-5 x-3\right)$
$=\left(x^{2}-4 x+1\right)\left(2 x^{2}-6 x+x-3\right)$
$=\left(x^{2}-4 x+1\right)(2 x(x-3)+1(x-3))$
$=(x-2-\sqrt{3})(x-2+\sqrt{3})(2 x+1)(x-3)$
To find the zeroes, we put $f(x)=0$
$\Rightarrow(x-2-\sqrt{3})(x-2+\sqrt{3})(2 x+1)(x-3)=0$
$\Rightarrow(x-2-\sqrt{3})=0$ or $(x-2+\sqrt{3})=0$ or $(2 x+1)=0$ or $(x-3)=0$
$\Rightarrow x=2+\sqrt{3}, 2-\sqrt{3},-\frac{1}{2}, 3$
Hence, all the zeroes of the polynomial $f(x)$ are $2+\sqrt{3}, 2-\sqrt{3},-\frac{1}{2}$ and 3 .
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Home/Archive/Archive - 2015-16/Std-11, Science, Mathematics, Application of Derivatives
## Std-11, Science, Mathematics, Application of Derivatives
Good afternoon students, today we will study maxima and minima increasing decreasing
functions and other applications of Derivatives. Let us start.
Derivatives are very widely used and they have many different applications. As we studied in
the last lecture, firstly they are used in finding equation of tangents, equation of normals.
Secondly, they are also used for checking whether the function is increasing or decreasing.
Their third application is maxima and minima and mean value of theorems and many other
different applications. We have talked about tangents and normals in last class, today we
will talk about increasing, decreasing functions, maxima and minima and mean value
theorems.
Function is increasing when derivative is positive. Function is decreasing when derivative is
negative. When derivative is positive function is increasing function and when derivative is
negative function is decreasing function. If derivative is positive function will increase and
derivative is negative function will decrease. Increasing function will look like this, going
from left to right, value of function increases, graph goes upwards and decreasing function is
like this, going from left to right and graph goes downwards. Decreasing function and
increasing function. This graph is made when derivative is negative and this graph is made
when derivative is positive.
Maxima and Minima – I hope everybody had studied maxima and minima for school, yes,
very good. This is exactly same as what you studied for boards. Exactly, same types of
questions are asked. Firstly we have to find critical points. Critical points are the points
where f dash x is zero or not defined. First of all we will find critical points and after that we
will check if critical point is maxima or minima. Second point is we check whether critical
point is maxima or minima. Critical point is that point where derivative is zero. Now, how to
check whether critical point is maxima point or minima point. Say, x equals to a is critical
point, now we find f double dash a. We will find double derivative, if it is negative then it is
maxima or f double dash a is positive then x equals to a is minima. I hope everybody got it.
Yes, very good, to check maxima and minima of functions, we find critical points. Critical
points are the points whether f dash is zero or not defined. After that we can apply two ways
firstly what I have taught you is second derivative test. You find double derivative and put
critical point if the answer is negative then critical point is maxima point or if it is positive
then it is minima point. But double derivative point many a times fail, when double
derivative doesn’t exists only or double derivative instead of being positive or negative is
zero.
So, we have one more test which is first derivative test. What do first derivative test says? It
says if sign of derivative changes from negative to positive, what is this – sign of f dash x, on
the both sides of critical point, we have to check the sign of f dash x, if it goes from negative
to positive, then critical point is, function has increased instead of decreasing. As soon as the
function increases instead of decreasing, critical point is point of minima. You can very
clearly say function is increasing instead of decreasing. Means the value decrease at a
minimum value and then increases, so the critical point is of minima. And if sign of f dash
becomes turns positive to negative on the both sides of critical point. Means f dash should
change from positive to negative, after the functions increasing it decreases so the critical
point is point of maxima. To find maxima and minima functions, we need to do two things,
first is we need to critical points, second thing is we should check whether critical point is of
maxima or minima, for this we have two tests. Second derivative test and first derivative
test. In second derivative test, in double derivative value we will put critical value and if it is
positive then minima and if it is negative then maxima. In first derivative test, on both the
sides of critical points, we will check sign of f dash, if first derivative sign changes from
negative to positive, so the function is from decreasing to increasing, means this point will
be of minima. And if first derivative sign changes from positive to negative, then the function
will go from increasing to decreasing. So this point will be of maxima. And third thing which
may happen is.
2016-03-17T10:46:27+00:00 Categories: Archive - 2015-16||0 Comments
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# Perpendicular Bisectors
## Intersect line segments at their midpoints and form 90 degree angles with them.
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Perpendicular Bisectors
What if you were given FGH\begin{align*}\triangle FGH \end{align*} and told that GJ\begin{align*} \overleftrightarrow{GJ}\end{align*} was the perpendicular bisector of FH¯¯¯¯¯¯¯¯\begin{align*}\overline{FH}\end{align*}? How could you find the length of FG\begin{align*}FG\end{align*} given the length of GH\begin{align*}GH\end{align*}? After completing this Concept, you'll be able to use the Perpendicular Bisector Theorem to solve problems like this one.
### Watch This
First watch this video.
Next watch this video.
Then watch this video.
Finally, watch this video.
### Guidance
A perpendicular bisector is a line that intersects a line segment at its midpoint and is perpendicular to that line segment, as shown in the construction below.
One important property related to perpendicular bisectors is that if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. This is called the Perpendicular Bisector Theorem.
If CDAB¯¯¯¯¯¯¯¯\begin{align*}\overleftrightarrow{CD} \perp \overline{AB}\end{align*} and AD=DB\begin{align*}AD = DB\end{align*}, then AC=CB\begin{align*}AC = CB\end{align*}.
In addition to the Perpendicular Bisector Theorem, the converse is also true.
Perpendicular Bisector Theorem Converse: If a point is equidistant from the endpoints of a segment, then the point is on the perpendicular bisector of the segment.
Using the picture above: If AC=CB\begin{align*}AC = CB\end{align*}, then CDAB¯¯¯¯¯¯¯¯\begin{align*}\overleftrightarrow{CD} \perp \overline{AB}\end{align*} and AD=DB\begin{align*}AD = DB\end{align*}.
When we construct perpendicular bisectors for the sides of a triangle, they meet in one point. This point is called the circumcenter of the triangle.
#### Example A
If MO\begin{align*}\overleftrightarrow{MO}\end{align*} is the perpendicular bisector of LN¯¯¯¯¯¯¯¯\begin{align*}\overline{LN}\end{align*} and LO=8\begin{align*}LO = 8\end{align*}, what is ON\begin{align*}ON\end{align*}?
By the Perpendicular Bisector Theorem, LO=ON\begin{align*}LO = ON\end{align*}. So, ON=8\begin{align*}ON = 8\end{align*}.
#### Example B
Find x\begin{align*}x\end{align*} and the length of each segment.
WX\begin{align*}\overleftrightarrow{WX}\end{align*} is the perpendicular bisector of XZ¯¯¯¯¯¯¯¯\begin{align*}\overline{XZ}\end{align*} and from the Perpendicular Bisector Theorem WZ=WY\begin{align*}WZ = WY\end{align*}.
2x+11168=4x5=2x=x\begin{align*}2x + 11 &= 4x - 5\\ 16 &= 2x\\ 8 &= x\end{align*}
WZ=WY=2(8)+11=16+11=27\begin{align*}WZ = WY = 2(8) + 11 = 16 + 11 = 27\end{align*}.
#### Example C
Find the value of x\begin{align*}x\end{align*}. m\begin{align*}m\end{align*} is the perpendicular bisector of AB\begin{align*}AB\end{align*}.
By the Perpendicular Bisector Theorem, both segments are equal. Set up and solve an equation.
3x8x=2x=8\begin{align*}3x-8 &=2x\\ x &=8\end{align*}
### Guided Practice
1.OQ\begin{align*}\overleftrightarrow{OQ}\end{align*} is the perpendicular bisector of MP¯¯¯¯¯¯¯¯¯\begin{align*}\overline{MP}\end{align*}.
a) Which line segments are equal?
b) Find x\begin{align*}x\end{align*}.
c) Is L\begin{align*}L\end{align*} on OQ\begin{align*}\overleftrightarrow{OQ}\end{align*}? How do you know?
2. Find the value of x\begin{align*}x\end{align*}. m\begin{align*}m\end{align*} is the perpendicular bisector of AB\begin{align*}AB\end{align*}.
3. Determine if ST\begin{align*}\overleftrightarrow{S T}\end{align*} is the perpendicular bisector of XY¯¯¯¯¯¯¯¯\begin{align*}\overline{XY}\end{align*}. Explain why or why not.
1. a) ML=LP, MO=OP\begin{align*}ML = LP, \ MO = OP\end{align*}, and MQ=QP\begin{align*}MQ = QP\end{align*}.
b) 4x+34xx=11=8=2\begin{align*}4x + 3 & = 11\\ 4x & = 8\\ x & = 2\end{align*}
c) Yes, L\begin{align*}L\end{align*} is on OQ\begin{align*}\overleftrightarrow{OQ}\end{align*} because ML=LP\begin{align*}ML = LP\end{align*} (the Perpendicular Bisector Theorem Converse).
2. By the Perpendicular Bisector Theorem, both segments are equal. Set up and solve an equation.
x+6x=22=16\begin{align*}x+6 &=22\\ x &=16\end{align*}
3. ST\begin{align*}\overleftrightarrow{S T}\end{align*} is not necessarily the perpendicular bisector of XY¯¯¯¯¯¯¯¯\begin{align*}\overline{XY}\end{align*} because not enough information is given in the diagram. There is no way to know from the diagram if ST\begin{align*}\overleftrightarrow{S T}\end{align*} will extend to make a right angle with XY¯¯¯¯¯¯¯¯\begin{align*}\overline{XY}\end{align*}.
### Practice
For questions 1-4, find the value of x\begin{align*}x\end{align*}. m\begin{align*}m\end{align*} is the perpendicular bisector of \begin{align*}AB\end{align*}.
\begin{align*}m\end{align*} is the perpendicular bisector of \begin{align*}\overline{AB}\end{align*}.
1. List all the congruent segments.
2. Is \begin{align*}C\end{align*} on \begin{align*}m\end{align*}? Why or why not?
3. Is \begin{align*}D\end{align*} on \begin{align*}m\end{align*}? Why or why not?
For Question 8, determine if \begin{align*}\overleftrightarrow{S T}\end{align*} is the perpendicular bisector of \begin{align*}\overline{XY}\end{align*}. Explain why or why not.
1. In what type of triangle will all perpendicular bisectors pass through vertices of the triangle?
2. Fill in the blanks of the proof of the Perpendicular Bisector Theorem.
Given: \begin{align*}\overleftrightarrow{C D}\end{align*} is the perpendicular bisector of \begin{align*}\overline{AB}\end{align*}
Prove: \begin{align*}\overline{AC} \cong \overline{CB}\end{align*}
Statement Reason
1. 1.
2. \begin{align*}D\end{align*} is the midpoint of \begin{align*}\overline{AB}\end{align*} 2.
3. 3. Definition of a midpoint
4. \begin{align*}\angle CDA\end{align*} and \begin{align*}\angle CDB\end{align*} are right angles 4.
5. \begin{align*}\angle CDA \cong \angle CDB\end{align*} 5.
6. 6. Reflexive PoC
7. \begin{align*}\triangle CDA \cong \triangle CDB\end{align*} 7.
8. \begin{align*}\overline{AC} \cong \overline{CB}\end{align*} 8.
### Vocabulary Language: English
circumcenter
circumcenter
The circumcenter is the point of intersection of the perpendicular bisectors of the sides in a triangle.
perpendicular bisector
perpendicular bisector
A perpendicular bisector of a line segment passes through the midpoint of the line segment and intersects the line segment at $90^\circ$.
Perpendicular Bisector Theorem Converse
Perpendicular Bisector Theorem Converse
If a point is equidistant from the endpoints of a segment, then the point is on the perpendicular bisector of the segment.
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# ISEE Middle Level Math : How to divide fractions
## Example Questions
### Example Question #11 : How To Divide Fractions
Simplify:
Explanation:
Rewrite as a division, then solve:
### Example Question #12 : How To Divide Fractions
Evaluate:
Explanation:
To divide any number by a fraction, we can multiply that number by the reciprocal of the fraction. That means:
So we have:
### Example Question #13 : How To Divide Fractions
Simplify:
Explanation:
Multiply each term by 6:
### Example Question #14 : How To Divide Fractions
Evaluate:
Explanation:
A mixed number represents the sum of an integer and a fraction. In order to evaluate this problem first we need to change the mixed number ( ) to an improper fraction. Improper fractions are fractions whose numerator is greater than the denominator. So we can write:
Now we should evaluate and we have:
### Example Question #15 : How To Divide Fractions
Simplify:
Explanation:
We can first find a common denominator for the expression in the numerator, which is . This gives us:
### Example Question #531 : Numbers And Operations
Which of the following is the reciprocal of ?
Explanation:
First, rewrite this as an improper fraction:
The reciprocal of an improper fraction can be found by switching its numerator and denominator, retaining the negative sign, so the reciprocal is .
### Example Question #532 : Numbers And Operations
Which of the following is the reciprocal of 2.8?
Explanation:
First, rewrite this as an improper fraction:
The reciprocal of an improper fraction can be found by switching its numerator and denominator, so the reciprocal is .
Solve:
Explanation:
### Example Question #534 : Numbers And Operations
Which of the following is the reciprocal of 31.25?
Explanation:
Rewrite 31.25 as a fraction:
Exchange the positions of the numerator and the denominator to get . Now divide 4 by 125:
Solve:
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Seven Minute Trigonometry
Although it’s not possible to learn everything you need to know to quiz out of college trigonometry in 7 minutes, it is possible for you to understand the basics of trigonometry, sine, cosine, tangent, and theta, in just that short a time.
Yes, thanks to the power of the Mark Clarkson method, anyone can master a basic understanding of trigonometry in just 7 minutes. The first thing to know about trigonometry is that, despite its intimidating name, trigonometry is the study of right triangles. Period. That’s it. I swear. Right triangles. How do the parts of a right triangle relate to each other?
The second thing to know about trigonometry is that there are only two important numbers – sine and cosine. Everything else follows from sine and cosine. What are sine and cosine? The answer’s simpler than you might suspect.
Let’s say you take a small stick and poke it in the ground so it stands up on its own. You might want to measure the stick, and see how tall it is.
You notice that the stick casts a shadow on the ground. You measure the shadow to see how long it is.
That’s your sine and cosine. Sine is how tall the stick is.
Cosine is how long the shadow is.Why are they called sine and cosine instead of, say, length and width? For arcane historical reasons and because it makes the uninitiated feel stupid. But that’s all they are. Here’s a mnemonic: sine and ‘high’ both have a long ‘i’ sound; ‘co’ and ‘low’ both have a long ‘o’ sound.
If you connect the open ends of your two lines (or your stick and its shadow), you’ll make a right triangle. As you remember, a right triangle is any triangle with a 90 degree angle in it somewhere, even if it faces left. Your stick and its shadow (the sine and cosine) are at right angles to each other, hence, a right triangle.
The line that closes the triangle is called the hypotenuse. I have no idea why.
The angle at the bottom left of the triangle is called ‘theta’. Theta is the only angle trig cares about.
Theta derives from the sine and cosine. That is, if you know the sine and cosine, you can figure theta. Likewise, if you know theta, you can compute the sine and cosine. Every combination of sine and cosine results in a unique angle, theta (and vice versa). Put in a taller stick, or wait until the sun sinks lower in the sky, and theta changes.
Note: You can think about sine and cosine as arising from theta, or you can think about theta as arising from a particular sine and cosine, whichever you prefer. It doesn’t make any difference. Trig is about the relationship between these three numbers, sine, cosine, and theta.
Since theta is the only angle trigonometry cares about, the other two angles don’t have a name. The one is always 90 degrees. The other is always 90 degrees minus theta.
Let’s say you measure the three sides of a triangle, and the measurements are 3, 4, and 5. It doesn’t matter 3,4, and 5 of what. Call it inches if that makes you more comfortable.
Now, you may remember that I said there were really only two numbers in trig. Here, however, you’ll notice that there are three numbers, A, B, C. That’s okay, because we’re going to throw one of those numbers — C, the hypotenuse — away. We don’t need it. Trig never uses it. How can that be? I’ll tell ya.
Those crafty Pythagoreans realized that they could simplify things considerably if they always assumed that the hypotenuse was equal to 1. Remember that: in trig, the hypotenuse is always equal to 1. Always.
But you measured your triangle and the hypotenuse wasn’t equal to 1; it was equal to 5. What to do? Simple, divide by 5 to make the hypotenuse equal to 1. Pretend that your triangle was 1/5th the size you measured.
If you divide all three sides by 5, the hypotenuse becomes 1; the sine becomes 0.6 (3/5); and the cosine becomes 0.8 (4/5).
If your hypotenuse had been 8, or 108,123, then you’d divide all your sides by 8, or by 108,123. If your measured hypotenuse had been 1/2, then you would have multiplied the sides by 2. Whatever it takes to make the hypotenuse equal one because, in trig, the hypotenuse is always 1.
You may have noticed that trigonometry is often applied to problems of circles and arcs and ellipses. Yet I insisted that it’s just about triangles, and only right triangles at that. Remember the old adage: “If your only tool is a hammer, then everything looks like a nail?” That applies to trigonometry very well. In trig, that hammer is the right triangle. Advanced trigonometry is about restating questions about circles and lines and curves and ellipses and whatever else mathematicians may think of, as questions about right triangles.
Let’s look at an example. Let’s superimpose our triangle onto a circle, so that the point of the angle, theta — the only angle we care about — falls on the origin, that is, the center point of the circle. Give this circle a radius of 1, the same as the hypotenuse of your triangle. The far end of the hypotenuse will just touch the perimeter of the circle. The X and Y coordinates of the point where the triangle touches the circle are 0.8 and 0.6 … your cosine and sine! Remember, cosine measures the length of the triangle, along the X axis. Sine measures the height of the triangle along the Y axis. So the cosine and sine correspond to the X,Y coordinates where the triangle touches the circle.
You can think about sine and cosine either as the coordinates of the point where the hypotenuse touches the circle, or as the lengths of the other two sides of the triangle, whichever seems more natural to you. It’s the same thing.
Look at the triangle above. Imagine that the hypotenuse, labeled ‘C’, is a hand on a clock. Right now it’s pointing to about 2 o’clock. As it moves toward 3 o’clock, imagine how the cosine, or ‘A’, will lengthen. Similarly, imagine how the sine, or ‘B’, will grow shorter.
When the hypotenuse ‘hand’ reaches 3 o’clock, cosine will be 1, and sine will be 0. The triangle will have collapsed into a flat line.
As the hand continues to sweep around the clock, cosine will begin to shrink again, and sine to lengthen. (Notice that as the hand passes below 3 o’clock, the sine becomes negative; it’s now measuring the distance beneath the X axis.)
Both sine and cosine will alternate from 1 to 0 to -1 and back again. If you graph the sine (or cosine) over time, as the ‘hand’ sweeps around the ‘clock’, you’ll get – surprise! – a ‘sine curve.’
Start at the center of the circle and draw a straight line coming out from it at any angle, 0 to 360. The point where that line crosses the circle is the sine and cosine of that angle. The x coordinate of that point is the cosine of that angle. The y coordinate of that point is the sine of the angle.
The right triangle may not always be shown, but it is always really there. It has to be, because trigonometry is only about right triangles.
Tangent
There’s another number that occurs almost as often as sine and cosine: tangent. The tangent is the slope of the hypotenuse. Whether we’re talking about roofs, or stairs, or right triangles, slope is rise over run. A roof which rises 3 feet for every 2 feet inwards has a steeper slope than a roof which rises 1 foot for every 2 feet inwards.
In the case of a right triangle, rise would be the same as sine. Remember, sine is the height of the triangle. The run would be the same as the cosine. Cosine is the length of the triangle.
So the slope of the hypotenuse is defined by the sine (the rise) over the cosine (the run). In the case of your 3 by 4 by 5 triangle – or your 0.6 by 0.8 by 1 triangle, whichever – the tangent (or slope) would be 3 over 4 (3/4), or 0.75.
That’s it. That’s all there is. Everything else in trig is just increasingly abstruse ways of restating these relationships, and ways to restate different classes of problems as problems about right triangles.
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2 Responses to Seven Minute Trigonometry
1. deesh says:
love your 7 minute trig! hope you write a follow-up ie how to use these ratios to find theta or what is inverse cos?
• Rahul says:
This has got to be symptomatic of what it takes to suececd in school. From K-16, quantitative tests are aced by knowing how to apply formulas.Students adapt to the goals they’re asked to achieve. Maybe we should be giving them half the day to build stuff instead.
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# My Personal Pizza
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## Objective
SWBAT create a pizza and name the different fractions for the slices by using fraction pieces.
#### Big Idea
Different fractions can be used to make up a whole.
## Whole Class Discussion
15 minutes
In today's lesson, the students learn to compare fractions by using fration squares and fraction circles to create a pizza with different size slices. This aligns with 4.NF.A2 because the students are comparing fractions with different numerators and denominators and using fraction circles and squares to justify their answers.
Because we have been working on fractions for a while, I wanted to give the students the opportunity to be creative with fractions. Before we begin, I ask the students to tell me what they have learned about fractions: Some student responses: The numerator is at the top and the denominator is at the botttom; the numerator tells you how many pieces; the denominator tells you how many to cut it into; and the top number is sometimes bigger than the bottom number.
I let them know that they are correct. For this lesson, I want to challenge the students' thinking in a creative way to compare fractions. I am requiring the students to create their own pizza the way that they want. The slices does not have to be the same size. (I realize that this will puzzle some of them, but struggle is good.)
Scenario:
My mom baked a large pizza for the family. My dad said that he wanted 1/3 of the pizza. My sister said that she was not that hungry, so she only wants 1/6 of the pizza. I only want 1/4 of the pizza. My mom said that she will eat 1/4 of the pizza as well. My mom said okay, she will cut the pizza the size that we want. How is this possible? Let's find out.
I point out to the students that when a whole is cut into different size pieces, then that creates fractions with different denominators. If they are cut into pieces that are the same size, then the denominator is the same (common).
Look at the pizza. Which piece is larger, the 1/6 or the 1/3? I give the students a few minutes to think about the question. The students respond that 1/3 is larger than 1/6. How many 1/6 pieces will it take to equal 1/3? Most students yell out 2, but a few students say 3. I take my marker and divide the 1/3 piece in half. The students can clearly see that it takes two 1/6 pieces to equal 1/3.
We can compare fractions within the whole. In this pizza, the largest piece is the 1/3. The smallest piece is the 1/6. Remember that models are excellent to help you compare fractions.
Let's practice comparing fractions in an activity.
## Skill Building/Exploration
20 minutes
For this activity, I give each student an activity sheet (My Personal Pizza.docx). The students will all complete the activity independently, but they will share resources, such as the fraction squares and circles (MP5). The students must create their own pizza cut into any sizes that equal 1 whole, therefore, the students must be accurate with the fraction sizes in order to come up with a whole pizza (MP6).
The students are provided with the necessary materials: construction paper, scissors, glue, fraction squares or circles, and an activity sheet. The students use the fraction squares or circles to help them create a proportional pizza that equals 1 whole. First, the students take the 1 whole fraction square or circle and trace it on construction paper. The students cut out the 1 whole piece. The students determine what toppings they want, as well as how much of it they want. As they create their pizza, they must identify the name of the fractions. As they use the fraction squares or circles to lay out for the different pieces, this is allowing the students to see the different fractions and see how they compare to each other in size. For example, if they create their pizza using 1/2, and 2/3, the students will need to determine what other fraction can be used. As they do this, they see that it must be a smaller than 1/2 and 1/3. They then see that 1/6 is a fraction that they can use. Once the students have the desired pizza made up of fractions pieces that equal 1 whole, they must trace the fraction pieces onto construction paper and then cut out the pieces. The students glue the pieces onto the 1 whole to make their pizza (Student Working2). Last, the students glue the pizza onto their activity sheet and label the fractions.
As the students work side by side, they are guided to a conceptual understanding through questioning by their classmates, as well as by me. Even though they are working on individual products, the students always talk and discuss what they are doing. (Especially, when it is an activity that they enjoy.)
As they work, I monitor and assess their progression of understanding through questioning.
1. Which toppings covers the largest amount of the pizza? What fraction is it?
2. Do the fraction squares validate that your pizza equals 1 whole?
3. What is the smallest fraction on your pizza? How do you know?
Any groups that finish the assignment early, can go to the computer to practice the skill at the following site until we are ready for the whole group sharing: http://www.math-play.com/math-fractions-games.html
## Closure
15 minutes
To close the lesson, I have one or two students share their answers. This gives those students who still do not understand another opportunity to learn it. I like to use my document camera to show the students' work during this time. Some students do not understand what is being said, but understand clearly when the work is put up for them to see.
I feel that by closing each of my lessons by having students share their work is very important to the success of the lesson. Students need to see good work samples, as well as work that may have incorrect information. Samples of student work can be found here, as well as in the resources (Student Work - Pizza.jpg were the students used 1/6 cheese, 1/6 sausage, 1/4 peppers, 1/6 hamburger, and 1/4 pepperoni) and (Student Work - Pizza 2.jpg were the students used 2/6 cheese, 1/4 hamburger, 1/3 pepperoni, and 1/12 other.)
At the end of the lesson, I display the homework assignment on the board for the students to copy.
Assignment:
Write to Explain: How can you use fraction circles to help build a pizza that equals 1 whole?
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# 2.6: Chapter 2 Review
Difficulty Level: At Grade Created by: CK-12
## Symbol Toolbox
\begin{align*}\rightarrow\end{align*} if-then
\begin{align*}\land\end{align*} and
\begin{align*}\therefore\end{align*} therefore
\begin{align*}\sim\end{align*} not
\begin{align*}\lor\end{align*} or
Keywords
Inductive Reasoning
The study of patterns and relationships is a part of mathematics. The conclusions made from looking at patterns are called conjectures. Looking for patterns and making conjectures is a part of inductive reasoning, where a rule or statement is assumed true because specific cases or examples are true.
Conjecture
The study of patterns and relationships is a part of mathematics. The conclusions made from looking at patterns are called conjectures.
Counterexample
We can disprove a conjecture or theory by coming up with a counterexample. Called proof by contradiction, only one counterexample is needed to disprove a conjecture or theory (no number of examples will prove a conjecture). The counterexample can be a drawing, statement, or number.
Conditional Statement (If-Then Statement)
Geometry uses conditional statements that can be symbolically written as \begin{align*}p \rightarrow q\end{align*} (read as “if \begin{align*}p\end{align*}, then \begin{align*}q\end{align*}”). “If” is the hypothesis, and “then” is the conclusion.
Hypothesis
The conditional statement is false when the hypothesis is true and the conclusion is false.
Conclusion
The second, or “then,” part of a conditional statement. The conclusion is the result of a hypothesis.
Converse
A statement where the hypothesis and conclusion of a conditional statement are switched.
Inverse
A statement where the hypothesis and conclusion of a conditional statement are negated.
Contrapositive
A statement where the hypothesis and conclusion of a conditional statement are exchanged and negated.
Biconditional Statement
If \begin{align*}p \rightarrow \ q\end{align*} is true and \begin{align*}q \rightarrow \ p\end{align*} is true, it can be written as \begin{align*}p \rightarrow \ q\end{align*}.
If \begin{align*}p\end{align*} is not true, then we cannot conclude \begin{align*}q\end{align*} is true.
If we are given \begin{align*}q\end{align*}, we cannot make a conculsion. We cannot conclude \begin{align*}p\end{align*} is true.
Logic
The study of reasoning.
Deductive Reasoning
Uses logic and facts to prove the relationship is always true.
Law of Detachment
The Law of Detachment states: If \begin{align*}p \ q\end{align*} is true and \begin{align*}p\end{align*} is true, then \begin{align*}q\end{align*} is true.
If \begin{align*}p\end{align*} is not true, then we cannot conclude \begin{align*}q\end{align*} is true
If we are given \begin{align*}q\end{align*}, we cannot make a conclusion. We cannot conclude \begin{align*}p\end{align*} is true.
Law of Contrapositive
If the conditional statement is true, the converse and inverse may or may not be true. However, the contrapositive of a true statement is always true. The contrapositive is logically equivalent to the original conditional statement.
Law of Syllogism
The Law of Syllogism states: If \begin{align*}p \rightarrow \ q\end{align*} and \begin{align*}q \rightarrow \ r\end{align*} are true, then \begin{align*}p \rightarrow \ r\end{align*} is true.
Right Angle Theorem
If two angles are right angles, then the angles are congruent.
Same Angle Supplements Theorem
If two angles are supplementary to the same angle (or to congruent angles), then the angles are congruent.
Same Angle Complements Theorem
If two angles are complementary to the same angle (or to congruent angles), then the angles are congruent.
Reflexive Property of Equality
\begin{align*}a = a\end{align*}.
Symmetric Property of Equality
\begin{align*}a = b\end{align*} and \begin{align*}b = a\end{align*}.
Transitive Property of Equality
\begin{align*}a = b\end{align*} and \begin{align*}b = c\end{align*}, then \begin{align*}a = c\end{align*}.
Substitution Property of Equality
If \begin{align*}a = b\end{align*}, then \begin{align*}b\end{align*} can be used in place of \begin{align*}a\end{align*} and vise versa.
If \begin{align*}a = b\end{align*}, then \begin{align*}a + c = b + c\end{align*}.
Subtraction Property of Equality
If \begin{align*}a = b\end{align*}, then \begin{align*}a - c = b - c\end{align*}.
Multiplication Property of Equality
If \begin{align*}a = b\end{align*}, then \begin{align*}ac = bc\end{align*}.
Division Property of Equality
If \begin{align*}a = b\end{align*}, then \begin{align*}a \div \ c = b \div \ c\end{align*}.
Distributive Property
\begin{align*}a(b + c) = ab + ac\end{align*}.
Reflexive Property of Congruence
For Line Segments \begin{align*}\overline{AB} \cong \overline{AB}\end{align*} For Angles \begin{align*}\overline{AB} \cong \angle ABC \cong \angle CBA\end{align*}
Symmetric Property of Congruence
For Line Segments If \begin{align*}\overline{AB} \cong CD\end{align*}, then \begin{align*}\overline{CD} \cong \overline{AB}\end{align*} For Angles \begin{align*}\overline{CD} \cong \overline{AB}\end{align*} If \begin{align*}\angle ABC \cong \angle DFF \cong \angle ABC\end{align*}
Transitive Property of Congruence
For Line Segments If \begin{align*}\overline{AB} \cong \overline{CD}\end{align*} and \begin{align*}\overline{CD} \cong \overline{EF}\end{align*}, then \begin{align*}\overline{AB} \cong \overline{EF}\end{align*} For Angles If \begin{align*}\angle ABC \cong \angle DEF\end{align*} and \begin{align*}\angle DEF \cong \angle GHI\end{align*}, then \begin{align*}\angle ABC \cong \angle GHI\end{align*}
## Review
Match the definition or description with the correct word.
1. \begin{align*}5 = x\end{align*} and \begin{align*}y + 4 = x\end{align*}, then \begin{align*}5 = y +4\end{align*} — A. Law of Contrapositive
2. An educated guess — B. Inductive Reasoning
3. \begin{align*}6(2a + 1) = 12a +12\end{align*} — C. Inverse
4. 2, 4, 8, 16, 32,... — D. Transitive Property of Equality
5. \begin{align*}\overline{AB} \cong \overline{CD}\end{align*} and \begin{align*}\overline{CD} \cong \overline{AB}\end{align*} — E. Counterexample
6. \begin{align*}\sim p \rightarrow \sim q\end{align*} — F. Conjecture
7. Conclusions drawn from facts. — G. Deductive Reasoning
8. If I study, I will get an “\begin{align*}A\end{align*}” on the test. I did not get an \begin{align*}A\end{align*}. Therefore, I didn’t study. — H. Distributive Property
9. \begin{align*}\angle A\end{align*} and \begin{align*}\angle B\end{align*} are right angles, therefore \begin{align*}\angle A \cong \angle B\end{align*}. — I. Symmetric Property of Congruence
10. 2 disproves the statement: “All prime numbers are odd.” — J. Right Angle Theorem — K. Definition of Right Angles
## Texas Instruments Resources
In the CK-12 Texas Instruments Geometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9687.
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# Sum of Sequence of Squares
## Theorem
$\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6$
## Proof 1
Proof by induction:
For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
$\displaystyle \sum_{i \mathop = 1}^n i^2 = \frac{n \left({n+1}\right)\left({2n+1}\right)} 6$
When $n = 0$, we see from the definition of vacuous sum that:
$0 = \displaystyle \sum_{i \mathop = 1}^0 i^2 = \frac{0 \left({1}\right)\left({1}\right)} 6 = 0$
and so $P(0)$ holds.
### Base Case
When $n=1$, we have $\displaystyle \sum_{i \mathop = 1}^1 i^2 = 1^2 = 1$.
Now, we have:
$\displaystyle \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6 = \frac {1 \left({1 + 1}\right) \left({2 \cdot 1 + 1}\right)} 6 = \frac 6 6 = 1$
So $P(1)$ is true. This is our base case.
### Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.
So this is our induction hypothesis:
$\displaystyle \sum_{i \mathop = 1}^k i^2 = \frac {k \left({k + 1}\right) \left({2 k + 1}\right)} 6$
Then we need to show:
$\displaystyle \sum_{i \mathop = 1}^{k+1} i^2 = \frac{\left({k+1}\right) \left({k+2}\right) \left({2 \left({k+1}\right) + 1}\right)} 6$
### Induction Step
This is our induction step:
Using the properties of summation, we have:
$\displaystyle \sum_{i \mathop = 1}^{k+1} i^2 = \sum_{i \mathop = 1}^k i^2 + \left({k+1}\right)^2$
We can now apply our induction hypothesis, obtaining:
$$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^{k+1} i^2$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac{k \left({k + 1}\right) \left({2 k + 1}\right)} 6 + \left({k + 1}\right)^2$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac{k \left({k + 1}\right) \left({2 k + 1}\right) + 6 \left({k + 1}\right)^2} 6$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac{\left({k + 1}\right) \left({k \left({2 k + 1}\right) + 6 \left({k + 1}\right)}\right)} 6$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac{\left({k + 1}\right) \left({2 k^2 + 7 k + 6}\right)} 6$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac{\left({k + 1}\right) \left({k + 2}\right) \left({2 \left({k + 1}\right) + 1}\right)} 6$$ $$\displaystyle$$ $$\displaystyle$$
So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
$\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6$
$\blacksquare$
## Proof 2
$$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n 3 i \left({i + 1}\right)$$ $$=$$ $$\displaystyle$$ $$\displaystyle n \left({n + 1}\right) \left({n + 2}\right)$$ $$\displaystyle$$ $$\displaystyle$$ Sum of Sequence of Products of Consecutive Integers $$\displaystyle$$ $$\displaystyle \implies$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n 3 i^2 + \sum_{i \mathop = 1}^n 3 i$$ $$=$$ $$\displaystyle$$ $$\displaystyle n \left({n + 1}\right) \left({n + 2}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \implies$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n 3 i^2$$ $$=$$ $$\displaystyle$$ $$\displaystyle n \left({n + 1}\right) \left({n + 2}\right) - 3 \frac {n \left({n + 1}\right) } 2$$ $$\displaystyle$$ $$\displaystyle$$ Closed Form for Triangular Numbers $$\displaystyle$$ $$\displaystyle \implies$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n i^2$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {n \left({n + 1}\right) \left({n + 2}\right)} 3 - \frac {n \left({n + 1}\right) } 2$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \implies$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n i^2$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {2 n \left({n + 1}\right) \left({n + 2}\right) - 3 n \left({n + 1}\right) } 6$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \implies$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n i^2$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac{n \left({n + 1}\right) \left({2n + 1}\right)} 6$$ $$\displaystyle$$ $$\displaystyle$$
$\blacksquare$
## Proof 3
Observe that:
$3 i \left({i + 1}\right) = i \left({i + 1}\right) \left({i + 2}\right) - i \left({i + 1}\right) \left({i - 1}\right)$
That is:
$(1): \quad 6 T_i = \left({i + 1}\right) \left({\left({i + 1}\right) + 1}\right) \left({\left({i + 1}\right) - 1}\right) - i \left({i + 1}\right) \left({i - 1}\right)$
Then:
$$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle n^2$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {n^2 + n + n^2 - n} 2$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {n \left({n+1}\right) + n \left({n-1}\right)} 2$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {n \left({n+1}\right)} 2 + \frac{n \left({n-1}\right)} 2$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle T_n + T_{n-1}$$ $$\displaystyle$$ $$\displaystyle$$
where $T_n$ is the $n$th Triangular number.
Then:
$$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n i^2$$ $$=$$ $$\displaystyle$$ $$\displaystyle 1 + \left({T_1 + T_2}\right) + \left({T_2 + T_3}\right) + \left({T_3 + T_4}\right) + \cdots + \left({T_{n-1} + T_n}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle 1 + 2T_2 + 2T_3 + 2T_4 + \cdots + 2T_{n-1} + T_n$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle 1 - T_1 - T_n + 2 \left({T_1 + T_2 + T_3 + T_4 + \cdots + T_n}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle 2 \left({\sum_{i \mathop = 1}^n T_i} \right) - \frac{n \left({n+1}\right)} 2$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle 2 \left({\frac{n \left({n+1}\right)\left({n+2}\right)} 6}\right) - \frac{n \left({n+1}\right)} 2$$ $$\displaystyle$$ $$\displaystyle$$ Telescoping Series from $(1)$ above $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac{2n \left({n^2 + 3n+2}\right) - \left({3n^2 + 3n}\right)} 6$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac{2n^3 + 3n^2 + n} 6$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac{n \left({n + 1}\right) \left({2n + 1}\right)} 6$$ $$\displaystyle$$ $$\displaystyle$$
$\blacksquare$
## Proof 4
We can observe from the above diagram that:
$\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \sum_{i \mathop = 1}^n \left({\sum_{j \mathop = i}^n j}\right)$
Therefore we have:
$$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n i^2$$ $$=$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n \left({\sum_{j \mathop = i}^n j}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n \left( {\sum_{j \mathop = 1}^n j - \sum_{j \mathop = 1}^{i-1} j} \right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n \left({\frac {n \left({n+1}\right)} 2 - \frac {i \left({i-1}\right)} 2}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \implies$$ $$\displaystyle$$ $$\displaystyle 2 \sum_{i \mathop = 1}^n i^2$$ $$=$$ $$\displaystyle$$ $$\displaystyle n^2 \left({n+1}\right) - \sum_{i \mathop = 1}^n i^2 + \sum_{i \mathop = 1}^n i$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \implies$$ $$\displaystyle$$ $$\displaystyle 3 \sum_{i \mathop = 1}^n i^2$$ $$=$$ $$\displaystyle$$ $$\displaystyle n^2 \left({n+1}\right) + \sum_{i \mathop = 1}^n i$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \implies$$ $$\displaystyle$$ $$\displaystyle 3 \sum_{i \mathop = 1}^n i^2$$ $$=$$ $$\displaystyle$$ $$\displaystyle n^2 \left({n+1}\right) + \frac {n \left({n+1}\right)} 2$$ $$\displaystyle$$ $$\displaystyle$$ Closed Form for Triangular Numbers $$\displaystyle$$ $$\displaystyle \implies$$ $$\displaystyle$$ $$\displaystyle 6 \sum_{i \mathop = 1}^n i^2$$ $$=$$ $$\displaystyle$$ $$\displaystyle 2 n^2 \left({n+1}\right) + n \left({n+1}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle n \left({n+1}\right) \left({2n+1}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \implies$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n i^2$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {n \left({n+1}\right) \left({2n+1}\right)} 6$$ $$\displaystyle$$ $$\displaystyle$$
$\blacksquare$
## Proof 5
$$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n \left({\left({i + 1}\right)^3 - i^3}\right)$$ $$=$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n \left({i^3 + 3i^2 + 3i + 1 - i^3}\right)$$ $$\displaystyle$$ $$\displaystyle$$ Binomial Theorem $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n \left({3i^2 + 3i + 1}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle 3 \sum_{i \mathop = 1}^n i^2 + 3 \sum_{i \mathop = 1}^n i + \sum_{i \mathop = 1}^n 1$$ $$\displaystyle$$ $$\displaystyle$$ Summation is Linear $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle 3\sum_{i \mathop = 1}^n i^2 + 3 \frac {n \left({n + 1}\right)} 2 + n$$ $$\displaystyle$$ $$\displaystyle$$ Closed Form for Triangular Numbers
On the other hand:
$$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n \left({\left({i + 1}\right)^3 - i^3}\right)$$ $$=$$ $$\displaystyle$$ $$\displaystyle \left({n + 1}\right)^3 - n^3 + n^3 - \left({n - 1}\right)^3 + \left({n - 1}\right)^3 - \cdots + 2^3 - 1^3$$ $$\displaystyle$$ $$\displaystyle$$ Definition of Summation $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \left({n + 1}\right)^3 - 1^3$$ $$\displaystyle$$ $$\displaystyle$$ Telescoping Series $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle n^3 + 3n^2 + 3n + 1 - 1$$ $$\displaystyle$$ $$\displaystyle$$ Binomial Theorem $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle n^3 + 3n^2 + 3n$$ $$\displaystyle$$ $$\displaystyle$$
Therefore:
$$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle 3\sum_{i \mathop = 1}^n i^2 + 3 \frac {n \left({n + 1}\right)} 2 + n$$ $$=$$ $$\displaystyle$$ $$\displaystyle n^3+3n^2+3n$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \implies$$ $$\displaystyle$$ $$\displaystyle 3 \sum_{i \mathop = 1}^n i^2$$ $$=$$ $$\displaystyle$$ $$\displaystyle n^3 + 3n^2 + 3n - 3 \frac {n \left({n + 1}\right)} 2 - n$$ $$\displaystyle$$ $$\displaystyle$$
Therefore:
$$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^n i^2$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac 1 3 \left({n^3 + 3n^2 + 3n-3 \frac {n \left({n + 1}\right)}2 - n}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac 1 3 \left({n^3 + 3n^2 + 3n - \frac {3 n^2} 2 - \frac {3 n} 2 - n}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac 1 3 \left({n^3 + \frac {3 n^2} 2 + \frac n 2}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac 1 6 n \left({2n^2 + 3n + 1}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac 1 6 n \left({n + 1}\right) \left({2n+1}\right)$$ $$\displaystyle$$ $$\displaystyle$$
$\blacksquare$
## Historical Note
This result was documented by Āryabhaṭa in his work Āryabhaṭīya of 499 CE.
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# Find the equation of the tangent and normal to the curve y=cosx at x=π/3?
Jun 22, 2018
Tangent's equation is $y = - \frac{\sqrt{3}}{2} x + \frac{\pi}{2 \sqrt{3}} + \frac{1}{2}$
Normal's equation is $y = \frac{2}{\sqrt{3}} x - \frac{2 \pi}{3 \sqrt{3}} + \frac{1}{2}$
#### Explanation:
First, you find the first derivative of the function then substitute with the given point $\left({x}_{1} , {y}_{1}\right)$ to get the slope$\left(m\right)$ and then substitute in this function
color(green)((y-y_1)=m(x-x_1)
apply for the given function
$y = \cos x$
At $x = \frac{\pi}{3}$ $\rightarrow$ $y = \cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$
$\left({x}_{1} , {y}_{1}\right) =$$\left(\frac{\pi}{3} , \frac{1}{2}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \sin x$
Substitute with the point $\left(\frac{\pi}{3} , \frac{1}{2}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = {m}_{1} = - \sin \left(\frac{\pi}{3}\right) = - \frac{\sqrt{3}}{2}$
Now Substitute in the formula in green with $\left({x}_{1} , {y}_{1}\right) , {m}_{1}$
to get the equation of the tangent to the curve at the given point
$y - \frac{1}{2} = - \frac{\sqrt{3}}{2} \left(x - \frac{\pi}{3}\right)$
$y = - \frac{\sqrt{3}}{2} x + \frac{\pi}{2 \sqrt{3}} + \frac{1}{2}$
the normal to the curve would be at the same point but its slope is different and in order to get its slope ${m}_{2}$
We use this
${m}_{1} \cdot {m}_{2} = - 1$
${m}_{2} = \frac{2}{\sqrt{3}}$
Now Substitute with $\left({x}_{1} , {y}_{1}\right) , {m}_{2}$ in the equation in green to find the equation of the normal
$y - \frac{1}{2} = \frac{2}{\sqrt{3}} \left(x - \frac{\pi}{3}\right)$
$y = \frac{2}{\sqrt{3}} x - \frac{2 \pi}{3 \sqrt{3}} + \frac{1}{2}$
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# Focus on Math
## Helping children become mathematicians!
### How Many More to Make 30? February 12, 2013
This is an activity I created to use with two grade 2 classes that I work with at Charlie Lake Elementary. In BC, grade two students work extensively with numbers to 100. The activity, like “How many more to make 20?” (see post from Feb. 5, 2013), is based on one of the foundational number relationships which is, for numbers 1 to 10, anchoring each number to 10.
30 was chosen as the focal point for this activity since multiples of 10 are also important anchoring numbers.
Once again I was delighted to put some special dice to use, in this case 30-sided dice.** Each child rolled the dice and then, using a set of 10 frames, created the number rolled at the top of the sheet, right over the blank ten frames there. Thus, if 14 were rolled, the child placed a full ten frame and a one showing four on the paper, and then recorded the number 14 in the roll column of the T-chart. Then he looked to see how many would be needed to make 30, in this case 6 to fill the partial ten frame and one more full ten. 16 was recorded beside 14 on the T-chart (see picture).
Similarly, if 7 were rolled, the child placed a ten frame showing seven on the paper, and then recorded the number 7 in the roll column of the T-chart. He could see that to make 30 he would need 3 more to fill the partial ten frame and two more full 10 frames, and thus 23 was recorded on the T-chart (see picture).
As in the “How many more to make 20?” activity, some of the children stopped making the numbers with their ten frames soon into the activity. Clearly they could imagine the anchoring relationship in their minds and did not need to manipulate the cards to “see” the numbers. Other children needed the support for every roll, but they were still able to be successful because of the scaffolding the ten frames provided.
I hope you will try the activity with your students!
Mathematically yours,
Carollee
Download the recording page here.
**If you do not have 30-sided dice, having students draw numbers from a bag or spinning numbers on a spinner will do nicely. You could even give students the page with the first column already filled in with numbers of your choice.
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# Binomial theorem
### Binomial calculator raised to some exponent
The structure of the calculator is as follows:
\left( \text{Monomial 1} + \text{Monomial 2} \right)^{\text{Exponent}} = \text{Result}
• You can enter decimal values, but it is best to only enter whole values
• Do not enter letters or other characters. Except for the character “-” that indicates that an integer is negative
Binomial theorem program Monomial 1:
Monomial 2:
Exponent:
### Let’s go with the theory of the binomial theorem
The same binomial theorem is known as the binomial formula because, that is, a formula. The binomial theorem is only valid in terms of an integer and positive power of a binomial. Let’s see the first five values of the power:
\begin{array}{c c l} (x + y)^{1} & = & x + y \\ (x + y)^{2} & = & x^{2} + 2xy + y^{2} \\ (x + y)^{3} & = & x^{3} + 3x^{2}y + 3xy^{2} + y^{3} \\ (x + y)^{4} & = & x^{4} + 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + y^{4}\\ (x + y)^{5} & = & x^{5} + 5x^{4}y + 10x^{3}y^{2} + 10x^{2}y^{3} + 5xy^{4} + y^{5} \end{array}
There is a lot of theory that proves the binomial theorem, but we are going to go directly with the way in which you can write binomials with larger exponents, we need the pascal triangle and understand something very simple. The first thing we need to understand is that when we raise a binomial to a certain power, we will observe the following behavior:
(x+y)^{n} = x^{n} + a\ x^{n-1}y + b \ x^{n-2} y^{2} + \dots + b \ x^{2}y^{n-2} + a \ x y^{n-1} + y^{n}
Assuming that we do not know the coefficients that each monomial will have, what I want you to observe is that the exponent of the first term, x, starts with the value of the binomial power and is subtracted one by one until that its exponent reaches zero. The opposite occurs with the second term of the binomial, y, its exponent starts at zero (because a number raised to zero is equal to one) and the exponent is increased one by one until it reaches the same value as the exponent of the binomial.
Well, since we have very clear the aforementioned, let’s go to the pascal triangle, we are going to place the values that the coefficients of the development of a certain binomial would take depending on its exponent, let’s see:
\begin{array}{c c c} n = 0 & \quad & 1 \\ n = 1 & & 1 \quad 1\\ n = 2 & & 1 \quad 2 \quad 1 \\ n = 3 & & 1 \quad 3 \quad 3 \quad 1 \\ n = 4 & & 1 \quad 4 \quad 6 \quad 4 \quad 1 \\ n = 5 & & 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1 \\ n = 6 & & 1 \quad 6 \quad 15 \quad 20 \quad 15 \quad 6 \quad 1 \end{array}
Great, once we have the pascal triangle, let’s go with our example, develop the following binomial:
(x+y)^{6}
To do this the first thing we are going to do is to place the ascending and descending order of the exponents and then we are going to add the coefficients that are obtained from the pascal triangle when n = 6:
\begin{array}{c c c c c c c c c c c c c} x^{6} & \ & x^{5} y & \ & x^{4} y^{2} & \ & x^{3} y^{3} & \ & x^{2} y^{4} & \ & x y^{5} & \ & y^{6} \\ \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow \\ 1 & & 6 & & 15 & & 20 & & 15 & & 6 & & 1 \\ \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow \\ x^{6} & \ & 6 x^{5} y & \ & 15 x^{4} y^{2} & \ & 20 x^{3} y^{3} & \ & 15 x^{2} y^{4} & \ & 6 x y^{5} & \ & y^{6} \end{array}
\begin{array}{c} \Downarrow \\ Finally \\ \Downarrow \\ x^{6} + 6 x^{5} y + 15 x^{4} y^{2} + 20 x^{3} y^{3} + 15 x^{2} y^{4} + 6 x y^{5} + y^{6} \end{array}
With that you can develop binomials with larger exponents!
#### What happens if a term in my binomial is negative?
Quiet, it is much easier than it sounds, what is going to be done is to encapsulate the negative in a parenthesis and we will place a positive sign. Let’s solve the following binomial:
(x-y)^{6}
It is easier to group the negative with a parenthesis and place a positive sign because the brain is easier. Let’s see how it looks:
(x+ (-y))^{6}
Great, our second term is now (-y). It only remains to write how our developed binomial will be, so let’s write almost the same as we wrote above:
\begin{array}{c c c c c c c c c c c c c} x^{6} & \ & x^{5} (-y) & \ & x^{4} (-y)^{2} & \ & x^{3} (-y)^{3} & \ & x^{2} (-y)^{4} & \ & x (-y)^{5} & \ & (-y)^{6} \\ \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow \\ 1 & & 6 & & 15 & & 20 & & 15 & & 6 & & 1 \\ \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow \\ x^{6} & \ & 6 x^{5} (-y) & \ & 15 x^{4} (-y)^{2} & \ & 20 x^{3} (-y)^{3} & \ & 15 x^{2} (-y)^{4} & \ & 6 x (-y)^{5} & \ & (-y)^{6} \end{array}
\begin{array}{c} \Downarrow \\ Almost \quad finally \\ \Downarrow \\ x^{6} + 6 x^{5} (-y) + 15 x^{4} (-y)^{2} + 20 x^{3} (-y)^{3} + 15 x^{2} (-y)^{4} + 6 x (-y)^{5} + (-y)^{6} \end{array}
The only thing missing is to develop the parentheses of each (-y). Remember that a negative raised to an even power will always be positive, and if it is raised to an odd power it will always be negative.
x^{6} - 6 x^{5} y + 15 x^{4} y^{2} - 20 x^{3} y^{3} + 15 x^{2} y^{4} - 6 x y^{5} + y^{6}
Now, you know how to develop binomials with very large exponents.
#### Before you leave for the time being
I leave you a PDF where you can find the binomials from n = 0 to n = 20 and you will also find the formula of the binomial theorem, just click here \Rightarrow Binomial theorem and formulas
Thank you for being at this moment with us : )
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Is 13212 a prime number? What are the divisors of 13212?
## Is 13212 a prime number?
No, 13212 is not a prime number.
For example, 13212 can be divided by 2: 13212 / 2 = 6 606.
To be 13212 a prime number, it would have been required that 13212 has only two divisors, i.e., itself and 1.
## Parity of 13212
13212 is an even number, because it is evenly divisible by 2: 13212 / 2 = 6 606.
## Is 13212 a perfect square number?
A number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 13212 is about 114.943.
Thus, the square root of 13212 is not an integer, and therefore 13212 is not a square number.
## What is the square number of 13212?
The square of a number (here 13212) is the result of the product of this number (13212) by itself (i.e., 13212 × 13212); the square of 13212 is sometimes called "raising 13212 to the power 2", or "13212 squared".
The square of 13212 is 174 556 944 because 13212 × 13212 = 132122 = 174 556 944.
As a consequence, 13212 is the square root of 174 556 944.
## Number of digits of 13212
13212 is a number with 5 digits.
## What are the multiples of 13212?
The multiples of 13212 are all integers evenly divisible by 13212, that is all numbers such that the remainder of the division by 13212 is zero. There are infinitely many multiples of 13212. The smallest multiples of 13212 are:
• 0: indeed, 0 is divisible by any natural number, and it is thus a multiple of 13212 too, since 0 × 13212 = 0
• 13212: indeed, 13212 is a multiple of itself, since 13212 is evenly divisible by 13212 (we have 13212 / 13212 = 1, so the remainder of this division is indeed zero)
• 26 424: indeed, 26 424 = 13212 × 2
• 39 636: indeed, 39 636 = 13212 × 3
• 52 848: indeed, 52 848 = 13212 × 4
• 66 060: indeed, 66 060 = 13212 × 5
• etc.
## Nearest numbers from 13212
Find out whether some integer is a prime number
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# Algebraic Expressions
In these lessons, we will learn what variables, constants, terms, expressions and coefficients are in algebra.
The following diagram gives an example of algebraic expression. Scroll down the page for more examples.
### Algebraic Terms & Expressions
In mathematics, often the value of a certain number may be unknown. A variable is a symbol, usually a letter, which is used to represent an unknown number.
Some examples of variables are:
x, a, t, y, b
A term can be a number, a variable, or a number and variable combined by multiplication or division.
Some examples of terms are:
x, 8, 4y,
An expression can be term or a collection of terms separated by addition or subtraction operators. Some examples of expressions, with the numbers of terms, are listed below:
Expression Number of Terms Description 6x 1 A number multiplied by a variable. The number is always written first followed by the variable(s). 3w – 8 2 Terms separated by – 7b + 5t – 6 3 Terms separated by + and – 1 All multiplication and division, no + or – symbol
Example:
Determine the number of terms in the following expressions:
a) 5xyz
b) 3x + 2y – 2x + 6
Solution:
a) 5xyz has one term
b) 3x + 2y – 2x + 6 has four terms
### Coefficients Of Algebraic Terms
The number (positive or negative) in the algebraic term is called the coefficient.
For example:
For the term 4x, 4 is the coefficient
For the term –7y, –7 is the coefficient
The coefficient of 1 in an algebraic term is usually not written.
For example:
1y can be written as simply y. So, 1y and y are the same.
Example:
Find the coefficient and variable of each of the following terms:
a) 0.2m
b) –3p
c) q
d) z
e)
Solution:
a) 0.2m (The coefficient is 0.2 and the variable is m)
b) –3p (The coefficient is –3 and the variable is p)
c) q (The coefficient is and the variable is q)
d) z (The coefficient is 1 and the variable is z)
e) (The coefficient is and the variable is x)
### Terms Of An Algebraic Expression
A presentation of the terms of an algebraic expression.
The number and types of terms are presented.
Constants, variables and numerical coefficients, terms and expressions are discussed.
### Parts Of An Expression And How To Evaluate Expressions
This video shows the parts of an expression (terms, coefficients and variables) then explains about evaluating expressions using some examples.
What is a Term?
A term is any coefficient and its variable.
Identify the term in the expression:
4x - 2y + 5
Identify the coefficient and variable in the term:
7xy
Example:
1. If x = 4 and y = 2
Evaluate 3x + 5y
2. Evaluate the expression if a = 7 and b = 2
(3 + a) - 2b
3. Jim and Jane like to go running in the morning. Jim runs twice as far as Jane. Let J be the distance Jane runs. What is expression for the combined distance they travel?
### Simplifying Expressions And Combining Like Terms
Expressions are put into their simplest form so as not to be confusing or too complex. One way of simplifying expressions is to combine like terms. By combining terms we can shorten and simplify our expressions, making them easier to read. Like terms often contain the same variable or variables.
This video shows how to simplify algebraic expressions by using the distributive property and combine like terms.
Example:
1. Simplify 5x - 8 + 3(2x - 5)
2. Simplify 5y2 - y - (3y2 - 6y) + 8y
3. Simplify 9x3 + 5 - [5(x3 - 3) + 12]
### Algebraic Expression, Terms, And Coefficients
In this video, you’ll learn how to identify the parts of an expression in Algebra. You’ll learn about expressions, terms, coefficients, and more.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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## The Decimal Point: How a 7-year-old Invented Something Better
By Raphael Kohlberg
The decimal point baffles many children. They often don’t know what it’s for and what the numbers that follow it represent. It doesn’t have to be this way! The best success I’ve had teaching the decimal point has come from guiding my students to essentially inventing it themselves.
When we’re first teaching kids numbers, we’re only concerned with whole numbers (1, 2, 3, 4, etc.) so we can simply say that the number, or digit, that’s furthest to the right corresponds to the ones, which with Digi-Block is the smallest block. But what if we want to be able to express numbers that aren’t whole numbers, such as 20.75? Or 0.3? I can introduce the decimal blocks, but now there is no longer a “smallest” block, because the blocks could in theory always get smaller and smaller, so we can no longer use the simple rule that “the smallest block is the number at the right end.” We need something new!
The way I teach explores these ideas and shows the need for a new device for writing numbers. This engages and excites students. They might come up with a solution that works. Even if they don’t, they’ll now appreciate the solution and have a deeper understanding of why we’ve introduced something new.
First, you need to explore the idea that there’s stuff out there smaller than 1, but bigger than 0. When I teach with Digi-Block blocks, I ask kids, what do I find if I open up a block-of-10? They quickly respond, “10 ones!” I then hold up a one, or single, and ask, what do you think I’d find if I could open this block up too? They think about it and say, “10 smaller blocks inside?” And I say, “Yes, you’re right!” And then I show them the tenths blocks.
Start by writing a multi digit number down, such has “572”. Ask your student to read you the number. Ask him “How do you know which digit is the hundreds? Which digit is the tens? Which digit is the ones?” This question might leave him staring blankly back at you or you might have an interesting discussion about place value.
Regardless, the next step is to open up your student’s thinking. You’ll use different color ink for each power of ten. The idea here is that we can use color, not position, to determine the value of each digit. Tell him “I’m going to use purple for hundreds, blue for tens, and green for ones.” Now write the same digits on a piece of paper, but use green for the “2,” purple for the “5,” and blue for the “7.” Now ask your student to read you this number. If he deciphers the code correctly, he should realize it’s the same number as before: 572! Try a few more examples until he gets the hang of it.
Now ask him if he can think of a different way, maybe without color. I had one student who wrote numbers in different sizes to represent powers of ten. It looked like this:
As he tried to write more and more numbers, and longer numbers that required more than 3 digits, he decided this wasn’t very practical.
I then presented him with our problem: we have things smaller than 1. I bring the tenths blocks back out to show him what I mean. (It’s not important to name these yet, you’re just using them to show a unit smaller than the singles.)
I put out several blocks (a few hundreds, tens, ones, and tenths) and said, how do I write this down? I then took them away and said, or if you write a number down for me, a number that should use those little blocks (the tenths blocks), how will I know what to build?
After some time to think about it. I came back to the “572” we had written before. I asked which way do the digits mean bigger blocks? Which way do they mean smaller blocks? He quickly saw that the digits to the left were for larger blocks and the digits to the right were for smaller blocks. He jumped up and said that the digit for the smaller block must to the right of the one!
He wrote a 4 digit number down, “4823,” but realized it looked like four thousand eight hundred twenty three.
I asked him, what can we do to let me know which digit is for which size block? (Like how we used color before…)
He thought about it and then showed me his idea: he would circle the digit in the ones place. I thought this was fantastic! After some more thinking, I think this is actually better than the decimal point.
Circling the digit in the one’s place shows the relationship between the digits to the left and to the right of the one’s place much more clearly. Ten is one power of ten greater than 1, and one tenth is one power of ten less than 1. 100 is two powers of ten greater than 1 and one hundredth is two powers of ten less than 1, and so on. The decimal point obscures this and makes it look like the relationship between 1 and one tenth, 10 and one hundredth, and so on, should be emphasized…
Back to my student! I told him I thought his idea was great, but a long time ago, when people had to solve this problem, they came up with something a little different: the decimal point. I showed him how to write a number with a decimal point and then asked him to build it in blocks.
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# How do you solve the following system: 7x - y = 10 , 8x - y = 13 ?
Jan 6, 2016
#### Answer:
$x = 3$ and $y = 11$
#### Explanation:
$7 x - y = 10$$\ldots \ldots \ldots \ldots \ldots . . \left(i\right)$
$8 x - y = 13$$\ldots \ldots \ldots \ldots \ldots \ldots \left(i i\right)$
Subtract $\left(i\right)$ from $\left(i i\right)$
$\implies x = 3$
Put $x = 3$ in $\left(i\right)$
$\implies 7 \left(3\right) - y = 10$
$\implies 21 - y = 10$
$\implies y = 11$
$\implies y = 11$
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# Solving Simple Inequalities
## Presentation on theme: "Solving Simple Inequalities"— Presentation transcript:
Solving Simple Inequalities
1-5 Solving Simple Inequalities Pre-Algebra Warm Up Solve. 1. x + 6 = 13 2. 8n = 48 3. t 2 = 56 4. 6 = x = 7 n = 6 t = 58 z 6 z = 36 Homework out to check!
Learn to solve and graph inequalities.
An inequality compares two quantities and typically uses one of these symbols:
< is less than is greater than is less than or equal to is greater than or equal to
Additional Example 1: Completing an Inequality
Compare. Write < or >. A. 23 – > B. 5(12) <
Try This: Example 1 Compare. Write < or >. A. 19 – < B. 4(15) >
An inequality that contains a variable is an algebraic inequality.
A number that makes an inequality true is a solution of the inequality. The set of all solutions is called the solution set. The solution set can be shown by graphing it on a number line.
Word Phrase Sample Solutions
Inequality Sample Solutions Solution Set x is less than 5 x < 5 x = 4 4 < 5 x = 2.1 2.1 < 5
Word Phrase Sample Solutions
Inequality Sample Solutions Solution Set y is less than or equal to 2 y is at most 2 y 2 y = 0 0 2 y = 1.5 1.5 2 –3 –2 –
Solving inequalities is the same as solving an equation, but with 1 exception – when you multiply or divide by a negative #.
Joke of the day A new teacher was trying to make use of her psychology courses. She started her class by saying, "Everyone who thinks you're stupid, stand up!“ After a few seconds, Little Johnny stood up. The teacher said, "Do you think you're stupid, Little Johnny?" "No, ma'am, but I hate to see you standing there all by yourself!"
Solve and graph the inequality.
A. x 8 –2.5 –2.5 Subtract 2.5 from both sides. x 5.5
Solve and graph the inequality.
B. 5t > 15 5t > 15 Divide both sides by 5. 5 5 t > 3
Solve and graph the inequality.
C. w – 1 < 8 Add 1 to both sides. w < 9
Solve and graph the inequality.
4 D. 3 3 p 4 4 • 4 • Multiply both sides by 4. 12 p
Lesson Quiz Use < or > to compare each inequality.
(2) – Solve and graph each inequality. 3. k + 9 < 12 4. 3 5. A school bus can hold 64 passengers. Three classes would like to use the bus for a field trip. Each class has 21 students. Write and solve an inequality to determine whether all three classes will fit on the bus. > > k< 3 –5 –4–3–2– m 2 6 m –4 –3–2– 3(21) 64; 63 64; yes ?
Write an equation & solve
Homework out Warm-up Write an equation & solve The average weight of an elephant is 5,450 kg. This is 4,270 kg more than the average weight of a giraffe. What is the average weight of the giraffe?
Positive choices Respect Integrity Determination Excellence P.R.I.D.E.
Altadena Middle School P.R.I.D.E. P Positive choices R Respect I Integrity D Determination E Excellence Monday
Mid-chp 1 quiz You need 1 pieces of paper out
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# 4 Examples Of Rational Numbers
In the context of mathematics, a rational number is a number that can be expressed as the ratio of two integers. A rational number is a number that is equal to the quotient of two integers p and q. In other words, a rational number can be expressed as some fraction where the numerator and denominator are integers. Some examples of rational numbers include:
1. The number 8 is rational because it can be expressed as the fraction 8/1 (or the fraction 16/2)
2. the fraction 5/7 is a rational number because it is the quotient of two integers 5 and 7
3. the decimal number 1.5 is rational because it can be expressed as the fraction 3/2
4. the repeating decimal 0.333… is equivalent to the rational number 1/3
Traditionally, the set of all rational numbers is denoted by a bold-faced Q. Rational numbers are distinguished from the natural number, integers, and real numbers, being a superset of the former 2 and a subset of the latter. There also exist irrational numbers; numbers that cannot be expressed as a ratio of two integers. An example of an irrational number is √2. √2 cannot be written as the quotient of two integers.
Let’s take a step back and talk about the different kinds of numbers.
## Kinds Of Numbers
It may come as a surprise to some that there exist different classes of numbers. After all, a number is a number, so how can some numbers be fundamentally different than other numbers? In a nutshell, numbers can be differentiated by how they behave when being added, subtracted, multiplied, or divided.
### Natural Numbers
Credit: Good Free Photos CC0 1.0
Let’s start with the most basic group of numbers, the natural numbers. The set of natural numbers (denoted with N) consists of the set of all ordinary whole numbers {1, 2, 3, 4,…} The natural numbers are also sometimes called the counting numbers because they are the numbers we use to count discrete quantities of things. A key feature of natural numbers is that they can be represented without some fractional or decimal component. There are an infinite amount of natural numbers stretching from 1 to infinity.
The natural numbers are considered the most basic kind of number because all other kinds of numbers can be defined as extensions of the natural numbers. Traditionally, the natural numbers do not contain the number zero (0), though some mathematicians consider 0 to be a natural number.
The natural numbers are closed under addition and multiplication. This means that if you add or multiply any two natural numbers, your answer will be another natural number. Adding 4 and 4 gives equals the natural number 8 and multiplying 5 by 1,000,000 equals the natural number 5,000,000. Adding or multiplying two natural numbers will always give you another natural number, no exceptions.
What about subtraction though? The natural numbers are not closed under subtraction. This means that if you subtract two natural numbers, your answer may not always be a natural number, which leads us to…
### Integers
Next up are the integers. The integers (denoted with Z) consists of all natural numbers and all negative whole numbers (…-4, -3, -2, -1) The set of integers is constructed by adding the additive inverse of every natural number, so it contains all positive and negative whole numbers {…-4, -3, -2, -1, 0, 1, 2, 3, 4,…}. As a consequence, all natural numbers are also integers. Like the naturals, there are an infinite amount of integers spanning from negative infinity to positive infinity.
Credit: WikiCommons CC0 1.0
Like the natural numbers, the integers are closed under addition and subtraction. Adding or multiplying any two integers will always give you another integer. Introducing negatives into our number systems makes it so that the integers are also closed under subtraction. Subtracting any two integers will always give you another integer. 6−3 = -3 and 12−40 = -28.
Now we have a set of numbers that is closed under addition, multiplication, and subtraction. What about division though? A moment’s thinking should tell you that no, the integers are not closed under division. Dividing two integers may not always result in another integer. This realization leads us to the next set of numbers…
### Rational Numbers
Credit: Pixabay CC0 1.0
Enter the rational numbers. The addition of rational numbers (denoted Q) allows us to express numbers as the quotient of two integers. Every rational number can be uniquely represented by some irreducible fraction. The number 3/2 is a rational number because it is expressed as a fraction in simplest form. Consequently, the rational number 6/4 is also equal to 3/2, because 6/4 can be simplified to 3/2. All integers (and so all natural numbers) can be expressed as an irreducible fraction (8 = 8/1 and -5 = -5/1), so all integers and natural numbers are also rational numbers.
Rational numbers are added to the number system to allow that numbers also be closed under division (with the lone exception of division by 0). The quotient of any two rational numbers can always be expressed as another rational number. This insight can be seen in the general rule for dividing fractions (i.e. rational numbers).
a/b ÷ c/d = ad/bc when d and b ≠ 0
a/b and c/d are rational numbers, meaning that by definition a, b, c, and d are all integers. Since the integers are closed under multiplication, ad and bc are also integers. ad/bc is represented as a ratio of two integers, which is the exact definition of a rational number. Therefore, the rational numbers are closed under division.
Rational numbers can also be expressed as decimals. Converting from fraction to decimal notation is easy: all you have to do is set up a long division problem and divide the numerator by the denominator. Dividing out an irreducible fraction will give you one of two results: either (i) long division will terminate in some finite decimal sequence or (ii) long division will produce an infinitely repeating sequence of decimals (e.g. 1/3 = 0.333… and 6/11 = 0.5454…).
Converting from a decimal to a fraction is likewise easy. All you have to do is multiply the decimal by some power of 10 to get rid of the decimal point and simplify the resulting fraction.
eg) 0.25 × 100/100 = 25/100 = 1/4
The rational numbers are the simplest set of numbers that is closed under the 4 cardinal arithmetic operations, addition, subtraction, multiplication, and division. This property makes them extremely useful to work with in everyday life. Rational numbers are not the end of the story though, as there is a very important class of numbers that cannot be expressed as a ratio of two integers.
### Irrational Numbers
An irrational number is a number that cannot be expressed as a ratio of two integers. Irrational numbers cannot be represented as a fraction in lowest form. The two sets of rational and irrational numbers are mutually exclusive; no rational number is irrational and no irrational number is rational. Common examples of irrational numbers include π, Euler’s number e, and the golden ratio φ. None of these three numbers can be expressed as the quotient of two integers.
How do we even know irrational number exist? Here is a simple proof by contradiction which shows that √2 is an irrational number:
Assume √2 is a rational number. If √2 is a rational number, then that means it can be expressed as an irreducible fraction of two integers. Let’s call those two integers p and q.
(1.) √2 = p/q
Since p/q is an irreducible fraction (per the definition of a rational number) they do not have any factors in common. Squaring both sides to get rid of the left hand radical gives us:
(2.) 2 = p2/q2
which we can rearrange into:
(3.) 2q2 = p2
This result implies that p2 is an even number because 2 is one of its factors. The only way p2 could be even is if p itself is even. If p is even, then there is some number k such that p = 2k. Substituting 2k for p in equation (3.) gives us:
(4.) 2q2 = (2k)2
(5.) 2q2 = 4k2
Equation (5.) can be rewritten as:
(6.) q2 = 2k2
By similar reasoning, q2 and must be even. Therefore, both p and q are even numbers. However, this contradicts our requirement from (1.) that p and q do not share any factors. Since we derived a contradiction, our initial assumption (that √2 is rational) must be false. Therefore, √2 is an irrational number and cannot be expressed as the quotient of two integers.
Irrational numbers rear their head all over the place. As it turns out, the square roots of most natural numbers are irrational. Many commonly seen numbers in mathematics are irrational. For example, the number π which is the ratio of the diameter of a circle to its circumference is irrational Additionally, Euler’s number e, the unique number whose natural logarithm is 1, is also irrational.
For some time, it was thought that all numbers were rational numbers. The preoccupation with rational numbers stems back to ancient Greece with the teaching of the Pythagoreans. The Pythagoreans were a quasi-religious sect who believed that numbers are the basic constituents of the universe. Central to their beliefs was the idea that all quantities could be expressed as rational numbers. The legend goes that the Pythagorean Hippasus first discovered the existence of irrational numbers when trying to solve for the hypotenuse of a right triangle with sides of equal length. Hippasus discovered that the length of the hypotenuse could not be understood as proportional to the lengths of its sides, and in doing so discovered irrational numbers. Reportedly, his discovery so greatly distressed the other Pythagoreans that they had Hippasus drowned as punishment for sacrilege.
Nowadays, we understand that not only do irrational numbers exist but that the vast majority of numbers are actually irrational. Comparatively, the set of rational numbers (which includes the integers and natural numbers) is incomprehensibly dwarfed by the size of the set of irrational numbers.
To sum up, rational numbers are numbers that can be expressed as the quotient of two integers. Rational numbers form an important class of numbers and are the simplest set of numbers that is closed under the 4 cardinal arithmetic operations of addition, subtraction, multiplication, and division. Rational numbers are distinguished from irrational numbers; numbers that cannot be written as some fraction.
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Permutation and Combination
# Permutation and Combination
## Concept of Permutation and Combination
Permutation is an arrangement of objects in a specific order. There are n! ways to permute n objects, where n! is the product of all the different permutations of n objects.
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For example, there are six different permutations of the letters A, B, and C:
ABC
ABD
ACD
BCD
BDA
CDA
## Concept of Permutation And Combination
Permutation and combination are two different concepts in mathematics. Permutation is the arrangement of objects in a particular order. Combination is the selection of objects from a particular set, without regard to the order.
## What is Factorial?
The factorial of a number is the product of all the integers up to and including that number. For example, the factorial of 4 is 4! = 4 × 3 × 2 × 1 = 24.
## What is Permutation?
A permutation is an arrangement of a set of objects in a particular order.
## What is Combination?
A combination is a set of things that are put together.
## Difference Between Permutation and Combination
The key difference between permutation and combination is that permutation is a selection of items in a specific order, whereas combination is a selection of items without any order.
A permutation is an ordered selection of a certain number of items from a given set. The order matters, so for example, the permutation (1,2,3) is different from the permutation (3,1,2). The number of permutations is calculated by multiplying the number of items in the set by itself the number of times the selection can be made (the order matters). For example, if there are three items in the set, then there are 3! or six permutations (1,2,3,3,2,1; 2,1,3,3,1,2; 3,2,1,1,2,3; 1,3,2,2,1,1; etc).
A combination is a selection of items without any order. The order doesn’t matter, so for example, the combination (1,2,3) is the same as the combination (3,1,2). The number of combinations is calculated by multiplying the number of items in the set by itself the number of times the selection can be made (the order doesn’t matter). For example, if there are three items in the set, then there are 3! or six combinations (1,2,3,2,3,1
## :
The amount of the subsidy for a student who is a dependent of a veteran and is enrolled in a public institution of higher education in this state is the lesser of the amount of the tuition and fees for the student or the amount of the tuition and fees for the most expensive public institution of higher education in this state.
The amount of the subsidy for a student who is a dependent of a veteran and is enrolled in a private institution of higher education in this state is the lesser of the amount of the tuition and fees for the student or the amount of the tuition and fees for the most expensive private institution of higher education in this state.
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Review question
# Can we find the sum of the integers from $2k$ to $4k$ inclusive? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R5849
## Solution
1. The positive integer $k$ is given.
1. Find, in terms of $k$, an expression for $S_1$, the sum of the integers from $2k$ to $4k$ inclusive.
#### Approach 1
We could add together the integers $2k$, $2k+1$, $\dotsc$, $4k$ excluding $3k$ in pairs as follows: \begin{align*} 2k+4k &= 6k \\ (2k+1)+(4k-1) &= 6k \\ \vdots \quad & \\ (3k-1)+(3k+1) &= 6k. \end{align*}
And there are $k$ of these pairs in total.
So to add the numbers between $2k$ and $4k$ inclusive, we can add $k$ pairs that add to $6k$, together with $3k$, that is, $S_1=6k^2+3k.$
#### Approach 2
We could use the fact that the sum of the integers from $1$ to $n$ is $P_n=\dfrac{n(n+1)}{2}$.
So we want to find $P_{4k}-P_{2k-1}$.
We have \begin{align*} S_1 &= P_{4k}-P_{2k-1} \\ &=\frac{1}{2} (4k)(4k+1) - \frac{1}{2} (2k-1)(2k)\\ &=6k^2+3k. \end{align*}
1. Find, in terms of $k$, an expression for $S_2$, the sum of the odd integers lying between $2k$ and $4k$.
The idea behind the first approach above can be used again for (a)(ii).
### Approach 1
If $k$ is even, we can add the odd integers between $2k$ and $4k$ in $k/2$ pairs that add to $6k$: \begin{align*} (2k+1)+(4k-1) &= 6k \\ (2k+3)+(4k-3) &= 6k \\ \vdots \quad & \\ (3k-1)+(3k+1) &= 6k. \end{align*} So \begin{align*} S_2 &= 6k\times\frac{k}{2}\\ &=3k^2. \end{align*}
We can use exactly the same approach if $k$ is odd, only this time the odd integers between $2k$ and $4k$ consist of $(k-1)/2$ pairs that add to $6k$, together with $3k=6k/2.$
So once again $S_2=3k^2.$
### Approach 2
We need $(2k+1) + (2k+3) + \cdots + (4k-1)$, or $(2k+1) + (2k+3) + \cdots + (2k+(2k-1))$.
This is an arithmetic sequence with $k$ terms, so for the formula, $n = k$, $d = 2$ and $a = 2k+1$.
The sum is therefore $\dfrac{n}{2}(2a+(n-1)d) = \dfrac{k}{2}(4k+2+2(k-1)) = k(3k ) =3k^2$.
1. Show that $\dfrac{S_1}{S_2} = 2+\dfrac{1}{k}$.
We have \begin{align*} \frac{S_1}{S_2} &= \frac{6k^2+3k}{3k^2}\\ &=2+\frac{1}{k}. \end{align*}
1. Prove that the sum of the first $n$ terms of the geometric progression having first term $a$ and common ratio $r$ ($r \neq 1$) is $a \left( \frac{1-r^n}{1-r} \right).$
The sum of the first $n$ terms of a geometric progression is $T_n=a+ar+ar^2+\cdots+ar^{n-1}.$
Now notice that $r\times T_n = ar+ar^2+\cdots+ar^{n-1}+ar^n.$
Subtracting, most of the terms cancel out so that \begin{align*} (1-r)T_n &= a-ar^n\\ \implies\quad T_n &= \frac{a(1-r^n)}{(1-r)}. \end{align*}
By regarding the recurring decimal $0.\dot{0}7\dot{5}$ ($=0.075075\ldots$, where the figures $075$ repeat) as an infinite geometric progression, or otherwise, obtain the value of the decimal as a fraction in its lowest terms.
Now $0.075075\ldots$ can be written as the sum of a geometric progression with first term $0.075$ and common ratio $0.001$.
Using the expression above for the sum of the first $n$ terms, and taking the limit as $n\to \infty$, we have $T_n\to a \left( \frac{1}{1-r} \right)$ (since $|r|<1$, and $r^n \to 0$ as $n\to\infty$).
Substituting in our values of $a$ and $r$ gives $0.\dot{0}7\dot{5}=\frac{0.075}{1-0.001}=\frac{0.075}{0.999}=\frac{25}{333}.$
Alternatively, let $x = 0.075075\ldots$, so $1000x = 75.075075\ldots$.
Subtracting, we get $999x = 75$, so $x=\dfrac{75}{999} = \dfrac{25}{333}$.
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Lesson Plan
## Midpoints, Distance, and Slope
• Assessment Anchors
• Eligible Content
• Big Ideas
• Mathematical statements can be justified through deductive and inductive reasoning and proof.
• Numbers, measures, expressions, equations, and inequalities can represent mathematical situations and structures in many equivalent forms.
• Objects can be transformed in an infinite number of ways. Transformations can be described and analyzed mathematically.
• Patterns exhibit relationships that can be extended, described, and generalized.
• Relations and functions are mathematical relationships that can be represented and analyzed using words, tables, graphs, and equations.
• Similarity relationships between objects are a form of proportional relationships. Congruence describes a special similarity relationship between objects and is a form of equivalence.
• Spatial reasoning and visualization are ways to orient thinking about the physical world.
• There are some mathematical relationships that are always true and these relationships are used as the rules of arithmetic and algebra and are useful for writing equivalent forms of expressions and solving equations and inequalities.
• Concepts
• 2- and 3-dimensional figures
• Geometric Relations: Congruence and Similarity
• Geometric Representations
• Reasoning and Proof
• Competencies
• Define and describe types of geometrical reasoning and proof, using them to verify valid conjectures as they surface in the study of geometry; develop a counter example to refute an invalid conjecture.
• Define, describe, and analyze 2- and 3-dimensional figures, their properties and relationships, including how a change in one measurement will affect other measurements of that figure.
### Objectives
In this lesson, students will use geometric figures in the coordinate plane to find slopes of lines, distances between two points, and the midpoints between two points. Students will:
• learn how the slopes of parallel and perpendicular lines are related.
• use the distance formula to determine the length of a side in a geometric shape in the coordinate plane.
• use the midpoint formula to determine the midpoint of the segment, given its endpoints in the coordinate plane.
#### Essential Questions
• How can you use coordinates and algebraic techniques to represent, interpret, and verify geometric relationships?
### Vocabulary
• Concave: Curving inward; a curve is concave toward a point if it bulges away from the point; a polygon is concave if it is not convex, i.e., if at least one of its interior angles is greater than 180 degrees. [IS.1 - Preparation] [IS.2 - Struggling Learners and ELL Students]
• Convex: Curving outward; a curve such that any straight line cutting the curve cuts it in just two points; a polygon is convex if it lies on one side of any one of its sides extended, i.e., if each interior angle is less than or equal to 180 degrees.
• Coordinate Plane: A surface for which any set of numbers locate a point, line, or any geometry element in space; for Cartesian coordinates, the point, can be located by its distances from two intersecting straight lines, the distance from one line being measured along a parallel to the other line.
• Distance Formula: The formula that represents the length of the line segment joining two points. In analytic geometry, it is found by taking the square root of the sum of the squares of the differences of the corresponding rectangular Cartesian coordinates of the two points, .
• Equation: A statement of equality between two quantities, generally divided into two types, identities and conditional equations. A conditional equation is true only for certain values of the unknown; an identity is true for all values of the variables.
• Formula: A general answer, rule, or principle stated in mathematical language.
• Geometry: The science that treats the shape and size of things; the study of invariant properties of given elements under specified groups of transformations.
• Midpoint: The point that divides a line segment into two equal parts; the point that bisects the line.
• Parallel Lines: Equidistant, apart; if two lines are cut by a transversal, and the sum of the interior angles on one side of the transversal is less than a straight angle, the two lines will meet if produced, and will meet on that side of the transversal. Only one line can be drawn parallel to a given line through a given point not on the line.
• Perpendicular Lines: Two lines are perpendicular to each other if, in a plane, the slope of one of the lines is the negative reciprocal of the other; two straight lines that intersect such that they form a pair of equal adjacent angles.
• Polygon: A closed-plane figure consisting of points called vertices and lines called sides, which have no common point except for end points. A polygon is convex if each interior angle is less than or equal to 180 degrees. A polygon is concave if it is not convex.
• Slope: The angle of inclination; for a straight line, the tangent of the angle that the line makes with the positive x-axis.
### Duration
90–120 minutes/1–2 class periods [IS.3 - All Students]
### Prerequisite Skills
Prerequisite Skills haven't been entered into the lesson plan.
### Related Materials & Resources
The possible inclusion of commercial websites below is not an implied endorsement of their products, which are not free, and are not required for this lesson plan.
### Formative Assessment
• View
• The Introductory Activity uses knowledge acquired from Algebra 1 on graphing linear equations to revisit the concepts of parallel and perpendicular. Students make the link between the geometry (parallel and perpendicular) and the congruent slopes and negative reciprocal associated with algebra.
• Lesson 1 Exit Ticket activity evaluates students’ understanding of correspondence between ordered pair and location on the coordinate grid, knowledge that slope is the ratio of the change in y (vertical distance) to change in x (horizontal distance), and skill in using the midpoint algorithm. The evaluation is done by requiring students to plot points, calculate slope, and calculate midpoint.
### Suggested Instructional Supports
• View
Active Engagement, Explicit Instruction
[IS.9 - All Students]
W: This lesson begins with a pairs activity of reviewing linear functions. Students observe how the slopes of parallel lines and perpendicular lines are related. They need to know this in order to verify properties of geometric figures in the coordinate plane. They learn the distance formula to find lengths of sides of geometric figures in the coordinate plane, as well as the midpoint figure to determine the midpoint of a line segment. H: This lesson begins with an investigation that incorporates a real-world application: money. Students visualize what it means to owe money (a negative sloped line) and what it means to earn money (a positive sloped line). [IS.10 - All Students] They also design a dart board using the concepts of slope, distance, and midpoint. E: The opening investigation is a good way to gauge where students’ abilities are in terms of finding the slope of a line. This topic should be review for students, but then they will apply the slopes of lines to a geometric topic. The graphic organizer equips students with the formulas they need to solve the dart board problem. They will be interested to see how these formulas can be used in the real world and how creative they can be with math. R: Students have many opportunities to check their work and revise anything they make mistakes on. Students complete the dart board problem in pairs and then get together with another group to check work. Collaboration is a good way for students to check their thought processes and to see how other students solve problems. When some students present how they completed the dart board problem, students can refine their work before moving on to the exit ticket. E: Students can evaluate their own work when they get together with another pair. They can find other ways to solve the same problems or share with one another how they completed the task at hand. T: If necessary, the pace of the lesson can be slowed down if students need a little extra practice at finding the slope between two points. The main activity of this lesson is great for all learners. The visual learner can see where slope, distance, and midpoint can be used in a real-world application. The auditory learner is paired up with someone who can describe the process of finding slopes, distances, and midpoints. The kinesthetic learner can actually create the dart board if given the opportunity. O: This lesson begins with an investigation that ties a topic students have seen before to a new concept: slopes in geometric figures. Students then fill out a graphic organizer which they use for the main activity of the lesson: creating a dart board. Once the activity is done and the class has discussed it, students are asked to complete an exit ticket to see if they have really understood the topics of the day.
IS.1 - Preparation In addition to the vocabulary document that follows in the materials section, consider a review of specific vocabulary terms prior to this lesson. Some of these words may need to be taught explicitly. Use a graphic organizer such as the Frayer Model to review words such as concave, convex,parallel lines, perpendicular lines etc. These may also be defined on a word wall in the classroom, or a student’s vocabulary notebook IS.2 - Struggling Learners and ELL Students A mini-graphic with key word, definition, and example/related symbols, pictures, expression may be helpful for ELL students as well as struggling learners. IS.3 - All Students Consider pre-teaching the concepts critical to this lesson, including the use of hands-on materials . IS.4 - Struggling Learners and ELL Students Struggling students and ELL students may have difficulty with the amount of information o n the page. Consider cutting up the sheet and dissecting it into clusters of 2 or 3 problems that the students can complete IS.5 - Struggling Learners Provide physical assistance for those students who may have difficulty filling out this worksheet due to visual or physical disabilities. IS.6 - All Students Consider providing a visual of a dart board for those students who do not have the background knowledge. IS.7 - All Students Consider using this as formative assessment and using the information gained to correct any misconceptions. IS.8 - All Students Consider this a great example of minute-by-minute and/or day-by-day formative assessment that may need to be acted upon prior to students leaving the classroom. IS.9 - All Students Consider imbedding these practices throughout the lesson as frequently as possible to assure retention of new concepts. IS.10 - All Students Consider showing the students “Slope Dude” on Youtube to lock in the concept of positive and negative slope. This may help clarify the concepts for ELLs and struggling students.
### Instructional Procedures
• View
Part 1: Introductory Activity
Hand out the Intro Worksheet (M-G-5-1_Intro Worksheet.doc and M-G-5-1_Intro Worksheet KEY.doc). Have students work on it in pairs. “What can we conclude about the slopes of lines that are perpendicular?” (Their slopes multiply to negative 1.) “What can we conclude about the slopes of lines that are parallel?” (Their slopes are the same.) [IS.5 - Struggling Learners]
Hand out the Graphic Organizer (M-G-5-1_Lesson 1 Graphic Organizer.doc and M-G-5-1_Lesson 1 Graphic Organizer KEY.doc) and fill it out with students.
Part 2 (Think-Pair-Share)
“How many of you like games, such as arcade games or board games?” Allow time for students to respond. “Today we are going to help a young man create his own dart board by using what we know about slopes of lines, squares, and finding distances and midpoints.” [IS.6 - All Students] Hand out copies of Mason’s Dart Board Activity (M-G-5-1_Mason's Dart Board Activity.doc and M-G-5-1_Mason's Dart Board KEY.doc) and the Dart Board Graph and Table Worksheet (M-G-5-1_Dart Board Graph and Table.doc and M-G-5-1_Dart Board Graph and Table KEY.doc).
Students should work on this in pairs using their graphic organizer. Walk around while students are working. [IS.7 - All Students] You may have to hint to them on number 2 that they need to determine if the side lengths are the same size and if the sides make a right angle (by checking their slopes). After the pairs have completed the activity, group them with another pair to check one another’s work. Groups can then present how they came up with the last two squares, as well as the slopes of their sides and lengths of side. When the dart board has been created, wrap up this activity with a discussion of how to number the different sections or how many points certain sections should be worth.
Part 3
Hand out the Lesson 1 Exit Ticket (M-G-5-1_Lesson 1 Exit Ticket.doc and M-G-5-1_Lesson 1 Exit Ticket KEY.doc) to evaluate whether students understand the concepts. [IS.8 - All Students]
Extension:
• Hand out the Lesson 1 Extension Activity (M-G-5-1_Lesson 1 Extension Activity.docx). Student responses will vary for the Extension Activity. Rectangles will vary in side length and shape. The line adjoining the midpoints of two adjacent sides will form the hypotenuse of a right triangle and the length of that hypotenuse must correspond with the distance formula calculation. Verify that students have used the correct ordered pairs in the distance formula.
### Related Instructional Videos
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Instructional videos haven't been assigned to the lesson plan.
DRAFT 10/13/2011
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## Least Squares Linear Predictor¶
In this section we are going to step away from the bivariate normal distribution and see if we can identify the best among all linear predictors of one numerical variable based on another, regardless of the joint distribution of the two variables.
For jointly distributed random variables $X$ and $Y$, you know that $E(Y \mid X)$ is the least squares predictor of $Y$ based on functions of $X$. We will now restrict the allowed functions to linear functions and see if we can find the best among those. In the next section we will see the connection between this best linear predictor, the best among all predictors, and the bivariate normal distribution.
### Minimizing Mean Squared Error¶
Let $h(X) = aX + b$ for constants $a$ and $b$, and let $MSE(a, b)$ denote $MSE(h)$.
$$MSE(a, b) ~ = ~ E\big{(} (Y - (aX + b))^2 \big{)} ~ = ~ E(Y^2) + a^2E(X^2) + b^2 -2aE(XY) - 2bE(Y) + 2abE(X)$$
To find the least squares linear predictor, we have to minimize this MSE over all $a$ and $b$. We will do this using calculus, in two steps:
• Fix $a$ and find the value $b_a^*$ that minimizes $MSE(a, b)$ for that fixed value of $a$.
• Then plug in the minimizing value $b_a^*$ in place of $b$ and minimize $MSE(a, b_a^*)$ with respect to $a$.
#### Step 1¶
Fix $a$ and minimize $MSE(a, b)$ with respect to $b$.
$$\frac{d}{db} MSE(a, b) ~ = ~ 2b - 2E(Y) + 2aE(X)$$
Set this equal to 0 and solve to see that the minimizing value of $b$ for the fixed value of $a$ is
$$b_a^* ~ = ~ E(Y) - aE(X)$$
#### Step 2¶
Now we have to minimize the following function with respect to $a$:
\begin{align*} E\big{(} (Y - (aX + b_a^*))^2 \big{)} ~ &= ~ E\big{(} (Y - (aX + E(Y) - aE(X)))^2 \big{)} \\ &= ~ E\Big{(} \big{(} (Y - E(Y)) - a(X - E(X))\big{)}^2 \Big{)} \\ &= ~ E\big{(} (Y - E(Y))^2 \big{)} - 2aE\big{(} (Y - E(Y))(X - E(X)) \big{)} + a^2E\big{(} (X - E(X))^2 \big{)} \\ &= ~ Var(Y) - 2aCov(X, Y) + a^2Var(X) \end{align*}
The derivative with respect to $a$ is $-2Cov(X, Y) + 2aVar(X)$. Thus the minimizing value of $a$ is
$$a^* ~ = ~ \frac{Cov(X, Y)}{Var(X)}$$
At this point we should check that what we have is a minimum, not a maximum, but based on your experience with prediction you might just be willing to accept that we have a minimum. If you're not, then differentiate again and look at the sign of the resulting function.
### Slope and Intercept of the Regression Line¶
The least squares straight line is called the regression line.You now have a proof of its equation, familiar to you from Data 8. The slope and intercept are given by
\begin{align*} \text{slope of regression line} ~ &= ~ \frac{Cov(X,Y)}{Var(X)} ~ = ~ r_{X,Y} \frac{\sigma_Y}{\sigma_X} \\ \\ \text{intercept of regression line} ~ &= ~ E(Y) - \text{slope} \cdot E(X) \end{align*}
### Regression in Standard Units¶
If both $X$ and $Y$ are measured in standard units, then the slope of the regression line is the correlation $r_{X,Y}$ and the intercept is 0.
In other words, given that $X = x$ standard units, the predicted value of $Y$ is $r_{X,Y}x$ standard units. When $r_{X,Y}$ is positive but not 1, this result is called the regression effect: the predicted value of $Y$ is closer to 0 than the given value of $X$.
### The Line and the Shape of the Scatter Diagram¶
The calculations above show that:
• The regression line goes through the point $(E(X), E(Y))$.
• The equation of the regression line holds regardless of the joint distribution of $X$ and $Y$.
• There is always a best straight line predictor among all straight lines, regardless of the relation between $X$ and $Y$. If the relation isn't roughly linear you won't want to use the best straight line for predictions, because the best straight line is only best among a bad class of predictors. But it exists.
|
# Absolutely
##### Absolute Value
The majority of students learn about absolute value long before high school. That is, they learn a lot of wrong things about absolute value.
• They learn that “the absolute value of a number is the number without its sign” or some such nonsense. All numbers, except zero have a sign! This sort of works with numbers, but becomes a problem when variables appear. True or false | x | = x? True or false | –x | = x? Most kids will say they are both true; in fact, as you know, they are both false.
• They also learn that “the absolute value of a number is its distance from zero on the number line.” True and works for numbers, but what about variables?
• They learn that “the absolute value of a number is the larger of the number and its opposite.” True again. How do you use it with variables?
• They learn $\left| x \right|=\sqrt{{{x}^{2}}}$ which is correct, useful for order-of-operation practice, and useful in other ways later, But they still compute $\sqrt{{{\left( -3 \right)}^{2}}}=-3$ and $\sqrt{{{x}^{2}}}=x$ since the square and square “cancel each other out.”
So here is a good vertical team topic. Get to those teachers in elementary and middle school and be sure they are not doing any of the above. They should start with the correct definition in words:
• The absolute value of a negative number is its opposite.
• The absolute value of a positive number (or zero) number is the same number.
This works all the time and will continue to work all the time. Teaching anything else will eventually require unlearning what they are using, and unlearning is far more difficult than learning.
When they start using variables and reading symbols translated into English, then the definition becomes their first piecewise define function:
• $\text{ If }x\ge 0,\text{ then }\ \left| x \right|=x;$ and if $x<0,\text{ then }\left| x \right|=-x$
• $\left| x \right|=\left\{ \begin{matrix} x & \text{ if }x\ge 0 \\ -x & \text{ if }x<0 \\ \end{matrix} \right.$
When reading this definition be sure to say “the opposite of the number” not “negative x” which in this case is probably a positive number.
Give variations of the two True-False questions above on every quiz and test until everyone gets it right!
When you see absolute value bars and want to be rid of them the first question to ask is, “Is the argument positive or negative? “Any time there is an absolute value situation, this is the way to proceed.
And yes, this does show up on the AP Calculus exams. Consider $\int_{0}^{1}{\left| x-1 \right|dx}$ which appeared as a multiple-choice question a few years ago. Give it a try before reading on.
On the interval of integration, [0,1], $\left( x-1 \right)\le 0$ so $\left| x-1 \right|=-\left( x-1 \right)$
$\displaystyle \int_{0}^{1}{\left| x-1 \right|dx}=\int_{0}^{1}{-\left( x-1 \right)}dx=\left. -\tfrac{1}{2}{{x}^{2}}+x \right|_{0}^{1}=-\tfrac{1}{2}+1-0=\tfrac{1}{2}$
Now try $\displaystyle \int_{0}^{1}{\sqrt{{{x}^{2}}-2x+1}\,dx}$, or did we do this one already?
|
# Making sense of irrational numbers
Hello mathematicians!
Why do I call you a mathematician? Yes, you all are excellent. Since junior and senior KG, you have met with the numbers. White pencil and black slate both were your friend to draw every number and loudly say, “it’s a one, two, …”. Later we learned tables which were again game of numbers. Still, we are playing with the numbers. However, as each standard has changed, new concepts have added to the study.
Let’s take an example of irrational numbers. Remember? Quickly recall it.
Contents
### What Is an Irrational Number?
An irrational number is a real number that is can not be expressed as a ratio of integers means fraction.
Rational numbers are the opposite and can be expressed as a fraction.
Who discover the irrational numbers?
Hippasus of Metapontum, in the 5th century, B.C.He was the ancient Greek philosopher and mathematician. He proved the existence of irrational numbers first.
He found the problems with Root 2.
Later, he decided to go with the help of a square shape where he used Pythagoras theorem.
a^2 + b^2 = c^2.
Now, the problem has arrived.
The diagonal length would be the square root of 2 that means c^2 =2 and c=√2
However, he could not prove that as the ratio of two integers. Hippasus did not get the expected result. Without wasting time and never give up thinking he started again in a different way.
Step 1
He assumed that Root 2 is a rational number. As you know, every rational number can be expressed as a ratio or ratio of two integers.
He supposed the two integers namely p and q. We write a ratio in the form of p/q, it could not have common factors that mean example, 2 can be expressed as 8/4 or 10/5, we get the answer 2 after division. So, 2 can be represented as 2/1, 1 is invisible.
Step 2
Hippasus assumed that p/q can not exist. Multiplied by q on both sides and squared both sides.
The original picture is,
√2 = p / q
After multiplying by q, you get
√2 q = p
Square of both sides give the equation,
2 (q)^2 = p^2
Step 3
Square of an odd number is always odd
3^2 = 9
and the square of an even number is even only.
4^2 = 16
Step 4
So, p would have to be an even integer and expressed or replaced by 2x where x is an integer.
The equation has changed,
2 (q)^2 = (2x)^2
2 (q)^2 = 4x^2
This implies that q is also even.
Initially, p and q had no common factors. However, it seems both are even now and 2 is the common factor for both. The initial assumption of Root 2 being a rational number goes false. Contradiction had found. This is how making sense of irrational numbers.
### Conclusion
Alike scientist and their experiment, mathematician Hippasus experimented and solved the contradictory examples. We never think while learning how each statement and concept arise with strong proof.
” Assumption helped to solve the confusion“!!!
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Class 10
### Topic Covered
♦ Elimination Method
### Elimination Method
=> Now let us consider another method of eliminating (i.e., removing) one variable. This is sometimes more convenient than the substitution method. Let us see how this method works.
Q 3139578412
The ratio of incomes of two persons is 9 : 7 and the ratio of their
expenditures is 4 : 3. If each of them manages to save 2000 per month, find their
monthly incomes.
Class 10 Chapter 3 Example 11
Solution:
Let us denote the incomes of the two person by 9x and 7x and their expenditures by 4y and 3y respectively. Then the equations formed in the situation is given by :
9x – 4y = 2000 \ \ \ \ (1)
and 7x – 3y = 2000 \ \ \ \ (2)
Step 1 : Multiply Equation (1) by 3 and Equation (2) by 4 to make the coefficients of
y equal. Then we get the equations:
27x – 12y = 6000.............. (3)
28x – 12y = 8000............. (4)
Step 2 : Subtract Equation (3) from Equation (4) to eliminate y, because the coefficients
of y are the same. So, we get
(28x – 27x) – (12y – 12y) = 8000 – 6000
i.e., x = 2000
Step 3 : Substituting this value of x in (1), we get
9(2000) – 4y = 2000
i.e., y = 4000
So, the solution of the equations is x = 2000, y = 4000. Therefore, the monthly incomes
of the persons are 18,000 and 14,000, respectively.
Verification : 18000 : 14000 = 9 : 7.` Also, the ratio of their expenditures =
18000 – 2000 : 14000 – 2000 = 16000 : 12000 = 4 : 3
Remarks :
1. The method used in solving the example above is called the elimination method,
because we eliminate one variable first, to get a linear equation in one variable.
In the example above, we eliminated y. We could also have eliminated x. Try
doing it that way.
2. You could also have used the substitution, or graphical method, to solve this
problem. Try doing so, and see which method is more convenient.
Let us now note down these steps in the elimination method :
Step 1 : First multiply both the equations by some suitable non-zero constants to make
the coefficients of one variable (either x or y) numerically equal.
Step 2 : Then add or subtract one equation from the other so that one variable gets
eliminated. If you get an equation in one variable, go to Step 3.
If in Step 2, we obtain a true statement involving no variable, then the original
pair of equations has infinitely many solutions.
If in Step 2, we obtain a false statement involving no variable, then the original
pair of equations has no solution, i.e., it is inconsistent.
Step 3 : Solve the equation in one variable (x or y) so obtained to get its value.
Step 4 : Substitute this value of x (or y) in either of the original equations to get the
value of the other variable.
Now to illustrate it, we shall solve few more examples.
Q 3169578415
Use elimination method to find all possible solutions of the following
pair of linear equations :
2x + 3y = 8 (1)
4x + 6y = 7 (2)
Class 10 Chapter 3 Example 12
Solution:
Multiply Equation (1) by 2 and Equation (2) by 1 to make the
coefficients of x equal. Then we get the equations as :
4x + 6y = 16 (3)
4x + 6y = 7 (4)
Step 2 : Subtracting Equation (4) from Equation (3),
(4x – 4x) + (6y – 6y) = 16 – 7
i.e., 0 = 9, which is a false statement.
Therefore, the pair of equations has no solution.
Q 3179578416
The sum of a two-digit number and the number obtained by reversing
the digits is 66. If the digits of the number differ by 2, find the number. How many such
numbers are there?
Class 10 Chapter 3 Example 13
Solution:
Let the ten’s and the unit’s digits in the first number be x and y, respectively.
So, the first number may be written as 10x + y in the expanded form (for example,
56 = 10(5) + 6).
When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s
digit. This number, in the expanded notation is 10y + x (for example, when 56 is
reversed, we get 65 = 10(6) + 5).
According to the given condition.
(10x + y) + (10y + x) = 66
i.e., 11(x + y) = 66
i.e., x + y = 6 (1)
We are also given that the digits differ by 2, therefore,
either x – y = 2 (2)
or y – x = 2 (3)
If x – y = 2, then solving (1) and (2) by elimination, we get x = 4 and y = 2.
In this case, we get the number 42.
If y – x = 2, then solving (1) and (3) by elimination, we get x = 2 and y = 4.
In this case, we get the number 24.
Thus, there are two such numbers 42 and 24.
Verification : Here 42 + 24 = 66 and 4 – 2 = 2. Also 24 + 42 = 66 and 4 – 2 = 2.
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At a Glance - Compound Events
In the world of probability, compound events are probabilities of two or more things happening at once. For example, what is the probability that you forgot to do your homework and there will be a pop quiz in class?
We'll go over three different ways to compute these probabilities (organized lists, tree diagrams, and area models) and show you examples using each method.
Measuring Compound Events Using Organized Lists
Using the organized list method, you would list all the different possible outcomes that could occur. This can be difficult, because there's a high probability that we will forget one or two options.
For example, if you flip a coin and roll a die, what is the probability of getting tails and an even number?
First, we need to start by listing all the possible outcomes we could get. (H1 means flipping heads and rolling a 1.)
H1 T1 H2 T2 H3 T3 H4 T4 H5 T5 H6 T6
There are twelve possible outcomes, and three of these outcomes give a desired outcome (tails plus an even number). These are T2, T4, and T6. So the probability is:
Example 1
In Clarajean's closet are four pairs of pants (black, white, grey, and brown), and five different shirts (blue, white, red, yellow, and purple). How many different outfits can she make with these options?
And, if she was to blindly pick a pair of pants and a shirt, what is the probability that the pants and shirt will be the same color?
First we need to list all the different outfits she can make, like white pants with a red shirt. The first color will be for the pants, and the second will be for the shirts.
Black-Blue Black-White Black-Red Black-Yellow Black-Purple White-Blue White-White White-Red White-Yellow White-Purple Grey-Blue Grey-White Grey-Red Grey-Yellow Grey-Purple Brown-Blue Brown-White Brown-Red Brown-Yellow Brown-Purple
Whew, that was long and tedious, but it got the job done. There are 20 different outfits.
Only one of the outfits has matching pants and shirt (white-white). So, the probability of blindly picking pants and shirts of the same color is:
Example 2
If you toss a coin three times, what is the probability of flipping at least 2 heads?
Now, we are working with three different events (each flip counts as an individual event). To help keep this organized, we can list these in three columns.
Flip 1Flip 2Flip 3
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
There are 8 different outcomes. These are the favorable ones: HHT, HTH, THH, and HHH (at least 2 heads includes flipping three). So the probability is:
Measuring Compound Events Using Tree Diagrams
Tree diagrams will give you the same answer as lists
Let's look at the coin and die example again: if you flip a coin and roll a die, what is the probability of getting tails and an even number?
We can chart all the possible outcomes by making a tree. The first set of "branches" will be all the possible outcomes of the first event. (It doesn't matter which event we put first, the total outcomes will be the same.) From each of those outcomes, draw branches for all the possibilities of the second event.
It could look like this:
Or this:
By counting the smallest branches, we see that there are 12 possibilities. So the probability of flipping a tail and rolling an even number is:
Example 1
In Clarajean's closet are four pairs of pants (black, white, grey, and brown), and five different shirts (blue, white, red, yellow, and purple). How many different outfits can she make with these options?
And, if she was to blindly pick a pair of pants and a shirt, what is the probability that the pants and shirt will be the same color?
If we start with the shirts, then add the pants, our tree diagram should look something like this:
Only one branch has matching pants and shirt (white pants and white shirt). So, the probability of blindly picking pants and shirts of the same color is:
Example 2
If you toss a coin three times, what is the probability of flipping at least 2 heads?
With three events, we will have three sets of branches on our tree.
There are eight different possibilities, four of which give at least two heads. So the probability is:
Measuring Compound Events Using Area Models
The final way we will chart these problems is the area model.We're smitten with this method because it's hard to forget an outcome using it.
Example: if you flip a coin and roll a die, what is the probability of getting tails and an even number?
Start by making a table with the outcomes of one event listed on the top and the outcomes of the second event listed on the side. Fill in the cells of the table with the corresponding outcomes for each event. We can shade the cells that fit our probability.
There are twelve cells, of which three are shaded. So the probability is:
Example 1
Going back to Clarajean's closet, she has four pairs of pants (black, white, grey, and brown), and five different shirts (blue, white, red, yellow, and purple). How many different outfits can she make with these options?
And, again, if she was to blindly pick a pair of pants and a shirt, what is the probability that the pants and shirt will be the same color?
Here's our table. We shaded the favorable outcome.
There are 20 different cells, which means 20 different options (4x5=20). One cell is shaded to represent the one time when she could pick the same color for both shirt and pants. Based on the calculation below, the probability that she'll pick the same color for both is 5%.
Example 2
If you toss a coin three times, what is the probability of flipping at least 2 heads?
Three events make the area model a bit more complicated since there are only two places to list the events (top and side). To work around this, draw one table for the first two events,
Now use those outcomes (HH, HT, TH, TT) for the rows or columns of a second table.
In the final table, there are eight different cells, four of which are shaded.
Exercise 1
Use this tree diagram of drawing a card from a deck of cards, then flipping a coin, then rolling a die, to answer the questions.
Exercise 2
How many different outcomes could happen?
Exercise 3
What is the probability drawing a black card, flipping tails, and rolling a 6?
Exercise 4
What is the probability of drawing a red card, flipping heads and rolling an odd number?
Exercise 5
Make an area model to illustrate the different possible outcomes for drawing a certain suit from a deck of cards and rolling a die.
Exercise 6
Using the area model above, what is the probability of drawing a spade and rolling a number less than three?
|
# Exercise 6.3
### 1. 5x2 × 4x3
#### Solution:
To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices. However, use of these laws is subject to their applicability in the given expressions.
In the present problem, to perform the multiplication, we can proceed as follows:
5x2 × 4x3
= (5×4)×(x2×x3)
= 20×5 (∵am × an = am+n)
### 2. −3a2×4b4
#### Solution:
To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, am×an=am+n, wherever applicable.
We have:
−3a2×4b4
= (−3×4) × (a2×b4)
= −12a2b4
### 3. (−5xy) × (−3x2yz)
#### Solution:
To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, am×an=am+n, wherever applicable.
We have:
(−5xy) × (−3x2yz)
= [(−5) × (−3)] × (x×x2) × (y×y)×z
= 15×(x1+2) × (y1+1)×z
= 15x3y2z
#### Solution:
To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, that is, am×an=am+n
We have:
#### Solution:
To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, that is, am×an=am+n
We have:
#### Solution:
To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, that is, am×an=am+n
We have:
#### Solution:
To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, that is, am×an=am+n
We have:
Solution:
To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, that is, am×an=am+n
## 9. Find the products
### (7ab) × (− 5ab2c) × (6abc2)
#### Solution:
To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, a× an = am + n, wherever applicable.
We have:
### (−5a) × (−10a2) × (−2a3
#### Solution:
To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, a× an = am + n, wherever applicable.
We have:
## 11. Find the products
### Solution:
To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, a× an = am + n, wherever applicable.
We have:
### (2.3xy) × (0.1x) × (0.16)
#### Solution:
To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, a× an = am + n, wherever applicable.
We have:
(2.3xy) × (0.1x) × (0.16)
= (2.3 × 0.1 × 0.16) × (x × x) × y
= (2.3 × 0.1 × 0.16) × (x1+1) × y
= 0.0368x2y
### 19. Express each of the following products as a monomials and verify the result in each case for x = 1:
#### Solution:
To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, am × an = am + n, wherever applicable.
We have:
Updated: November 7, 2019 — 4:31 pm
|
# Solving Rational Equations in Algebra
Many word problems about proportions and ratios require you to solve a rational equation. (SEE BOTTOM OF ARTICLE FOR SOLUTION)
A rational function is a fraction where the numerator and denominator are both polynomials. In the context of algebraic equation solving, a rational equation is an equation of the form
a(x)/b(x) = c(x)/d(x),
where a(x), b(x), c(x), and d(x) are all polynomials. To solve such an equation for x, you must first cross multiply the denominators and transform the original equation into the equivalent equation
a(x)*d(x) = c(x)*b(x), or
a(x)*d(x) - c(x)*b(x) = 0.
Since the left hand side is just another polynomial, solving the original problem just amounts to finding the roots of the polynomial a(x)*d(x) - c(x)*b(x) = 0. Here are some examples.
## Example Problem
Rational equations in which the numerators and denominators of both sides are linear functions are the easiest to solve, since the degree of a(x)*d(x) - c(x)*b(x) is at most 2 (i.e. quadratic).
Example: Alloy A is 10% copper and 90% zinc by weight, Alloy B is 25% copper and 75% zinc, and Alloy C is 50% and 50% zinc. Mary has 100 grams of Alloy A and 200 grams of Alloy B. If she adds equal amounts of Alloy C to both metals, how much of Alloy C does she have to add so that the new metals have the same ratio of copper to zinc?
Solution: Before any of Alloy C is added, the ratio of copper to zinc in Alloy A is 10g/90g, and the ratio of copper to zinc in Alloy B is 50g/150g. (The numerator is the amount of copper and the denominator is the amount of zinc.) If she adds x grams of Alloy C to each metal, the new ratios will be
(10+0.5x)/(90+0.5x) {Alloy A}
(50+0.5x)/(150+0.5x) {Alloy B}
Thus, we need to solve
(10+0.5x)/(90+0.5x) = (50+0.5x)/(150+0.5x), or
(10+0.5x)(150+0.5x) = (50+0.5x)(90+0.5x)
1500 + 80x + 0.25x^2 = 4500 + 70x + 0.25x^2
10x = 3000
x = 300.
So, if Mary adds 300 grams of Alloy C to both metals then they will have the same composition.
## Solution to Problem Shown in Image
(3x + 5) / (x + 1) = (2x - 7) / (4 - x)
(3x + 5)*(4 - x) = (2x - 7)*(x + 1) [cross multiplying to clear denominators]
-3x^2 + 7x + 20 = 2x^2 - 5x - 7 [expanding into quadratic]
0 = 5x^2 - 12x - 27 [combining like terms]
x = 3.815339, -1.415339 [solving with quadratic formula]
24
4
1
1
0
## Popular
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4
• ### How to Solve for X Using a Casio Calculator
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0 of 8192 characters used
• TR Smith 5 years ago from Eastern Europe
Nice explanation and example.
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# Elements of the circumference
🏆Practice circle
## What is circumference?
This question is not easy to answer and even more complicated to understand. If you imagine any point on a flat surface and a series of points whose distance from that point is identical, then you are looking at a circle.
## Test yourself on circle!
$$r=11$$
Calculate the circumference.
## Circumference characteristics
### Circumference perimeter
The perimeter of any circumference can be calculated. Generally, we can say that to calculate it, we must multiply by $2$ the value of $π$ (pi) and the length of the radius.
Click to access the article on the perimeter of the circumference.
## Circle area
Another equally important data that we can obtain with respect to any circumference is the area of the circle. To find it, we must raise the length of the radius squared and then multiply the result obtained by π. Click here to access the article on the area of the circle.
Join Over 30,000 Students Excelling in Math!
Endless Practice, Expert Guidance - Elevate Your Math Skills Today
## Examples and practice
### Exercise 1
Let's consider the following circle.
The radius of the circle is equal to $7 cm$.
Use the image and the data provided to calculate:
1. the diameter of the circumference
2. the perimeter of the circle
3. the area of the circle
Solution:
$P=2\times R\timesπ=2\times7\times3,14=43,96$
The perimeter of the circumference equals $43,96 cm$.
3. To calculate the area of the space inside the circumference, i.e., the circle, we must square the length of the radius of the circumference and then multiply the result obtained by the value of $π$.
Thus, we obtain:
$S=π\times R\times R=3,14X7\times7=153,86$
The area of the circle is $153,86cm²$
The diameter of the circle equals $14 cm$.
### Exercise 2
Let's consider the following circle.
We know that its diameter is $20cm$.
Use the image and the data provided to calculate:
1. the radius of the circle
2. the perimeter of the circle
3. the area of the circle
Solution:
1. The diameter of the circumference is actually the length of the radius multiplied by $2$. In our case, we already know what the diameter is, so all we have to do to find the length of the radius is to divide the diameter by $2$. By dividing it, we get that the radius of the circumference equals $10 cm$ $(20/2)$ .
2. As we have already said, to calculate the perimeter of the circumference we must multiply the value of $π$ l the length of the radius by $2$ (or use the value of the diameter directly instead of multiplying the length of the radius by $2$). The value of $π$ is $3,14$.
We obtain that:
$P=2\times R\timesπ=2\times10\times3,14=62,8$
The perimeter of the circumference is $62,8 cm$.To calculate the area of the circle, we must raise the length of the radius (obtained in a. above) of the circumference to the square and then multiply the result obtained by the value of $π$.
Thus, we obtain:
$S=π\times R\times R=3,14\times10\times10=314$
1. The area of the circle is $314 cm²$.
Do you know what the answer is?
## Circumference exercises:
### Exercise 1
Assignment:
Given the circumference of the figure
The diameter of the circle is $13$,
Solution
It is known that the diameter of the circle is twice its radius, i.e. it is possible to know the radius of the circle in the figure.
$13:2=6.5$
Tofind the area of the circle, we replace the data we have in the formula for the calculation of the circle
$A=\pi\times R²$
We replace the data we have:
$A=\pi\times6.5²$
$A=\pi42.25$
$pi42.25$
### Exercise 2
Request
Given an equilateral triangle in a circle
What is the areaof the circle?
Solution
Recall first the theorem that a circumferential angle that is inclined about the diameter is equal to 90 degrees.
That is, the triangle inside the circle is a right triangle and isosceles, so we can use the Pythagorean formula.
We replace the data we have with the Pythagorean formula
$X=Diámetro$
$(\sqrt{2})²+(\sqrt{2})²=X²$
The root cancels the power and therefore we obtain that
$2+2=X²$
That is
$X²=4$
The root of $4$ is $2$ and therefore the diameter is equal to $2$ and the radius is equal to half of the diameter and therefore it is equal to $1$
Therefore we obtain that the diameter is equal to $2$
And the radius is $1$
Then we add the formula for the area of the circle
$π$
### Exercise 3
Given the semicircle:
Consigna
Calculate its area
Solution
Since we know that it is a semicircle we can conclude that the base of the semicircle is the diameter.
We know that the diameter is twice the radius and therefore we can know the radius of the circle.
$Diámetro = 14$
$14:2=7$
$Radio = 7$
The formula for calculating the area of the circumference is
$A=\pi\times R²$
We replace the data in the formula
$A=\pi\times 7²$
$A=\pi\times 49$
Since the formula was the area of the semicircle we divide the area of the circle by $2$ and get the answer.
$\pi49:2=24.5\pi$
$\pi49:2=24.5\pi$
### Exercise 4
Request
Given the circumference of the figure
Given that the radius is equal to $6$,
What is the circumference?
Solution
The radius of the circle is $r=6$
We use the formula of the circumference $2\pi r$
Replace by $r=6$
We obtain
$2\cdot\pi\cdot6$
$12\pi$
$12\pi$
Do you think you will be able to solve it?
### Exercise 5
Request
Given the circumference of the figure.
Given that the radius is equal to $\frac{2}{3}$,
What is the circumference?
Solution
The radius of the circle $r=\frac{2}{3}$
We use the formula of the circumference $2\pi r$
Replace by $r=\frac{2}{3}$
We obtain
$2\cdot\pi\cdot\frac{2}{3}=$
$\frac{4}{3}\pi$
$\frac{4}{3}\pi$
### Exercise 6
Request
Given the circumference of the figure.
Given that the radius is equal to $5$,
What is the circumference?
Solution
Diameter of the circumference
$2r=5$
We divide by $2$
$r=\frac{5}{2}=2\frac{1}{2}$
Therefore the radius of the circle is $r=2\frac{1}{2}$
We use the formula of the circumference $2\pi r$
Replace by $r=2\frac{1}{2}$
We obtain
$2\cdot\pi\cdot2\frac{1}{2}=$
$2\cdot2\frac{1}{2}\cdot\pi=$
$5\pi$
$5\pi$
If this article interests you, you may also be interested in the following articles:
On the Tutorela blog you will find a variety of articles about mathematics.
## examples with solutions for circle
### Exercise #1
O is the center of the circle in the figure below.
What is its circumference?
### Step-by-Step Solution
We use the formula:$P=2\pi r$
We replace the data in the formula:$P=2\times8\pi$
$P=16\pi$
$16\pi$ cm
### Exercise #2
Look at the circle in the figure:
Its radius is equal to 4.
What is its circumference?
### Step-by-Step Solution
The formula for the circumference is equal to:
$2\pi r$
### Exercise #3
Look at the circle in the figure:
The radius is equal to 7.
What is the area of the circle?
### Step-by-Step Solution
Remember that the formula for the area of a circle is
πR²
We replace the data we know:
π7²
π49
49π
### Exercise #4
Given the circle whose diameter is 7 cm
### Step-by-Step Solution
First, let's remember the formula for the area of a circle:
$\pi r^2$
In the question, we are given the diameter of the circle, but we need the radius.
It is known that the radius is actually half of the diameter, therefore:
$r=7:2=3.5$
We replace in the formula
$\pi3.5^2=12.25\pi$
$12.25\pi$ cm².
### Exercise #5
O is the center of the circle in the diagram below.
What is its area?
### Step-by-Step Solution
Remember that the formula for the area of a circle is
πR²
We replace the data we know:
π3²
π9
$9\pi$ cm²
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# Experimental probability for AQA GCSE Maths
1. Understanding randomness
2. Understanding fairness
3. Understanding equally likely events
4. Calculating outcomes of multiple experiments
5. Expected frequencies and theoretical probability
When a random process is occuring, it is rare that the same results are repeated. The results of a random process are not affected by previous results.
A fair process is one that doesn't favour a particular outcome. A process that has a higher probability of giving a specific outcome is called biased.
When events are equally likely, the probability of an event occuring can be calculated by dividing the number of ways an outcome can happen by the total number of possible outcomes.
Probability outcomes can be calculated from experiments.
Relative frequency, also known as experimental probability, is a probability calculated using the results of an experiment by dividing the number of times an event happens by the total number of trials. Experimental probability may be different to theoretical probability. The relative frequency is made more accurate to the theoretical probability by repeating the experiment. Probabilities lie on the 0 to 1 probability scale, meaning that a probability value cannot be negative nor greater than 1. A probability of 0 means that an event is impossible and a probability of 1 means that an event is certain.
# 1
A fair coin is flipped 10 times and the results are recorded: H, T, T, T, H, H, T, H, H, H. State the theoretical probability of getting a Heads and calculate the experimental probability.
Since the coin is fair, the theoretical probability is 1/2.
Experimental probability = 6/10 = 0.6.
# 2
What is meant by a fair process?
A fair process is one that doesn't favour a particular outcome.
# 3
A deck of 15 cards is made up of red, green and blue cards. Two cards are randomly drawn from the deck without replacement. Calculate the probability of drawing two red cards, given that all three colours are initially equally likely events.
P(two red cards) = 5/15 × 4/14 = 2/21
2/21
# 4
A random card is being drawn by a common deck of 52 cards that has half of the cards blue and and half of them black. Calculate the probability that a red card is drawn.
There are 26 equally-likely possible red cards that can be drawn.
There is a total number of 52 cards.
P(red) = 26/52 = 1/2
1/2
# 5
A fair die is rolled 12 times. The following results are recorded: 1, 5, 5, 2, 3, 6, 3, 3, 2, 1, 4, 6. State the theoretical probability of rolling a 3 and alculate the experimental probability of rolling a 3.
Since the die is fair, the theoretical probability is 1/6.
Experimental probability = 3/12 = 0.25.
End of page
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# Fractions of an Amount - All Types
## Fractions of an Amount - How it Works - Video with Diagram
### Example 1a
Example 1a:
For this example, the fraction, 2/3, tells us so much. The 2 (numerator) tells us how many equal sections we need and the 3 (denominator) tells us how many total sections we need. The amount in questioned is 24, so we draw 24 circles to represent that. So we have 3 total sections with 8 circles in each section. We need 2 equal sections and in this case, there are 16 circles. So the answer is 16.
### Example 1b
Example 1b:
For this example, the fraction, 2/3, tells us so much. The 2 (numerator) tells us how many equal sections we need and the 3 (denominator) tells us how many total sections we need. Now we going to use multiplication and division to solve the problem. We use the amount, 24, and divide by the denominator, 3, then multiply by the numerator, 2. So we have 24 divided by 3 which is 8, then we multiply 8 by 2 to get 16.
### Example 2a
Example 2a:
For this example, the fraction, 3/4, tells us so much. The 3 (numerator) tells us how many equal sections we need and the 4 (denominator) tells us how many total sections we need. The amount in questioned is 20, so we draw 20 circles to represent that. So we have 3 total sections with 5 circles in each section. We need 3 equal sections and in this case, there are 15 circles. So the answer is 15.
### Example 2b
Example 2b:
For this example, the fraction, 3/4, tells us so much. The 3 (numerator) tells us how many equal sections we need and the 4 (denominator) tells us how many total sections we need. Now we going to use multiplication and division to solve the problem. We use the amount, 20, and divide by the denominator, 4, then multiply by the numerator, 3. So we have 20 divided by 4 which is 5, then we multiply 5 by 3 to get 15.
### Example 3a
Example 2a:
For this example, the fraction, 3/5, tells us so much. The 3 (numerator) tells us how many equal sections we need and the 5 (denominator) tells us how many total sections we need. The amount in questioned is 30, so we draw 30 circles to represent that. So we have 3 total sections with 6 circles in each section. We need 3 equal sections and in this case, there are 18 circles. So the answer is 18.
### Example 3b
Example 3b:
For this example, the fraction, 3/5, tells us so much. The 3 (numerator) tells us how many equal sections we need and the 5 (denominator) tells us how many total sections we need. Now we going to use multiplication and division to solve the problem. We use the amount, 30, and divide by the denominator, 5, then multiply by the numerator, 3. So we have 30 divided by 5 which is 6, then we multiply 6 by 3 to get 18.
## Fractions of an Amount - How it Works - Video with Multiplication/Division
### Example 1
Example 1:
For this example, the fraction, 3/4, tells us so much. The 3 (numerator) tells us how many equal sections we need and the 4 (denominator) tells us how many total sections we need. We have to split the 8 into 4 equal sections since the denominator in 3/4 is 4. Now, we divided the amount or 8 by the denominator, 4, and that give us 2, because 8 divided by 4 is 2. Since we need 3 equal sections to determine 3/4 of 8, we multiply that 2 by the numerator or 3 in this case. 2 times 3 is 6, so 3/4 of 8 is 6.
### Example 2
Example 2:
For this example, the fraction, 2/3, tells us so much. The 2 (numerator) tells us how many equal sections we need and the 3 (denominator) tells us how many total sections we need. We have to split the 24 into 3 equal sections since the denominator in 2/3 is 3. Now, we divided the amount or24 by the denominator, 3, and that give us 8, because 24 divided by 3 is 8. Since we need 2 equal sections to determine 2/3 of24, we multiply that 8 by the numerator or 2 in this case. 8 times 2 is 16, so 2/3 of 24 is 16.
### Example 3
Example 3:
For this example, the fraction, 2/5, tells us so much. The 2 (numerator) tells us how many equal sections we need and the 5 (denominator) tells us how many total sections we need. We have to split the 35 into 5 equal sections since the denominator in 2/5 is 5. Now, we divided the amount or 35 by the denominator, 5, and that give us 7, because 35 divided by 5 is 7. Since we need 2 equal sections to determine 2/5 of 35, we multiply that 7 by the numerator or 2 in this case. 7 times 2 is 14, so 2/5 of 35 is 14.
## Live Worksheet
Here is the link if you prefer.
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# Addition And Subtraction In Vedic Maths
Addition and subtraction in vedic maths
Vedic maths Addition
Usually if we want to add to numbers say 52 and 66 we would add the unit digits. and if there is any remainder we will bring it to the tens digit and atlast we will add the tens digit.
Now we will do it in the different way.
Let us take the same number 52 and 66.
The first step is to add the first number and the unit digit of the second number
ie. 52 + 6 = 58
The second step is to jump tens digit time.
our tens digit is 6.
so the answer is 118
Vedic Maths Subtraction
This Vedic Maths Subtraction method found as sutra in ancient vedas, is given below is very useful for such subtractions.
For example 1000 – 357 = ?
We simply take each figure in 357 from 9 and the last figure from 10.
Step 1. 9-3 = 6
Step 2. 9-5 = 4
Step 3. 10-7 = 3
So the answer is 1000 – 357 = 643
This always works for subtractions from numbers consisting of a 1 followed by
noughts: 100; 1000; 10,000 etc.
Similarly 10,000 – 1049 = 8951 (subtraction from 10000)
9-1 = 8
9-0 = 9
9-4 = 5
10-9 = 1
So answer is 8951
For 1000 – 83, in which we have more zeros than figures in the numbers being subtracted, we simply suppose 83 is 083.
So 1000 – 83 becomes 1000 – 083 = 917
Vedic maths multiplication
The Vedic method (the general method, at least) is based on the Urdhva-Tiryagbhyam sutra. A very terse sutra, it simply translates in English to say “vertically and crosswise”. The sutra is rather vague, so the technique, as well as an algebraic analysis of the technique, is presented in the following steps:
An Algebraic Perspective
All numbers n the base 10 number system (and number systems of any other base, for that matter) consist of a number of digits. Each digit represents a multiple times a power of 10 (or whatever the number system’s base is). So, for example, given a number like 52, we could rewrite it as 5*10+2.
Algebraically speaking, we can express any 3-digit number as: ax+b (where a, and b are integers).
So, suppose we wanted to multiply 2 2-digit numbers. We can express them in polynomial form. Then, by foiling:
(ax+b)(cx+d) = acx2+(ad+bc)x+bd
Algebraic Multiplication for Higher Numbers of Digits
So, for brief review, 2-digit by 2-digit algebraic multiplication goes as follows (x in all of the following examples is the base of the number system being used, which is usually 10):
(ax+b)(cx+d) = acx2+(ad+bc)x+bd
Expanding to 3-digit by 3-digit algebraic multiplication:
(ax2+bx+c)(dx2+ex+f) = adx4+(ae+bd)x3+(af+be+cd)x2+(bf+ec)x+cf
Now, 4-digitby 4-digit algebraic multiplication (ax3+bx2+cx+d)(ex3+fx2+gx+h)=aex6+(af+be)x5+(ag+bf+ce)x4+(ah+bg+cf+de)x3+(bh+cg+df)x2+(ch+dg)x+dh
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# Lesson 4: Calculating Probabilities for Chance Experiments with Equally Likely Outcomes
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1 Lesson : Calculating Probabilities for Chance Experiments with Equally Likely Outcomes Classwork Example : heoretical Probability In a previous lesson, you saw that to find an estimate of the probability of an event for a chance experiment you divide: P(event) = Number of observed occurrences of the event otal number of observations Your teacher has a bag with some cubes colored yellow, green, blue, and red. he cubes are identical except for their color. Your teacher will conduct a chance experiment by randomly drawing a cube with replacement from the bag. Record the outcome of each draw in the table below. rial Outcome
2 Exercises 6. Based on the 20 trials, estimate for the probability of a. choosing a yellow cube. b. choosing a green cube. c. choosing a red cube. d. choosing a blue cube. 2. If there are 0 cubes in the bag, how many cubes of each color are in the bag? Explain. 3. If your teacher were to randomly draw another 20 cubes one at a time and with replacement from the bag, would you see exactly the same results? Explain.
3 . Find the fraction of each color of cubes in the bag. Yellow Green Blue Each fraction is the theoretical probability of choosing a particular color of cube when a cube is randomly drawn from the bag. When all the possible outcomes of an experiment are equally likely, the probability of each outcome is P(outcome) = Number of possible outcomes. An event is a collection of outcomes, and when the outcomes are equally likely, the theoretical probability of an event can be expressed as P(event) = Number of favorable outcomes Number of possible outcomes. he theoretical probability of drawing a blue cube is P(blue) = Number of blue cubes otal number of cubes = Is each color equally likely to be chosen? Explain your answer. 6. ow do the theoretical probabilities of choosing each color from Exercise compare to the experimental probabilities you found in Exercise?
4 Example 2 An experiment consisted of flipping a nickel and a dime. he first step in finding the theoretical probability of obtaining a head on the nickel and a head on the dime is to list the sample space. For this experiment, the sample space is shown below. Nickel Dime If the counts are fair, these outcomes are equally likely, so the probability of each outcome is. Nickel Dime Probability he probability of two heads is or P(two heads) =. Exercises Consider a chance experiment of rolling a number cube. a. What is the sample space? List the probability of each outcome in the sample space. b. What is the probability of rolling an odd number? c. What is the probability of rolling a number less than 5?
5 8. Consider an experiment of randomly selecting a letter from the word: number. a. What is the sample space? List the probability of each outcome in the sample space. b. What is the probability of selecting a vowel? c. What is the probability of selecting the letter z? 9. Consider an experiment of randomly selecting a cube from a bag of 0 cubes. a. Color the cubes below so that the probability of selecting a blue cube is 2.
6 b. Color the cubes below so that the probability of selecting a blue cube is Students are playing a game that requires spinning the two spinners shown below. A student wins the game if both spins land on. What is the probability of winning the game? Remember to first list the sample space and the probability of each outcome in the sample space. here are eight possible outcomes to this chance experiment. Blue Blue Green Yellow
7 Lesson Summary When all the possible outcomes of an experiment are equally likely, the probability of each outcome is P(outcome) = Number of possible outcomes. An event is a collection of outcomes, and when all outcomes are equally likely, the theoretical probability of an event can be expressed as P(event) = Number of favorable outcomes Number of possible outcomes. Problem Set. In a seventh grade class of 28 students, there are 6 girls and 2 boys. If one student is randomly chosen to win a prize, what is the probability that a girl is chosen? 2. An experiment consists of spinning the spinner once. a. Find the probability of landing on a 2. b. Find the probability of landing on a. c. Is landing in each section of the spinner equally likely to occur? Explain An experiment consists of randomly picking a square section from the board shown below. a. Find the probability of choosing a triangle. b. Find the probability of choosing a star. c. Find the probability of choosing an empty square. d. Find the probability of choosing a circle.
8 . Seventh graders are playing a game where they randomly select two integers from 0 9, inclusive, to form a twodigit number. he same integer might be selected twice. a. List the sample space for this chance experiment. List the probability of each outcome in the sample space. b. What is the probability that the number formed is between 90 and 99, inclusive? c. What is the probability that the number formed is evenly divisible by 5? d. What is the probability that the number formed is a factor of 6? 5. A chance experiment consists of flipping a coin and rolling a number cube with the numbers 6 on the faces of the cube. a. List the sample space of this chance experiment. List the probability of each outcome in the sample space. b. What is the probability of getting a head on the coin and the number 3 on the number cube? c. What is the probability of getting a tail on the coin and an even number on the number cube? 6. A chance experiment consists of spinning the two spinners below. Blue Green Yellow a. List the sample space and the probability of each outcome. b. Find the probability of the event of getting a red on the first spinner and a red on the second spinner. c. Find the probability of a red on at least one of the spinners.
### Lesson 4: Calculating Probabilities for Chance Experiments with Equally Likely Outcomes
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Assertion and Reason Questions for Class 9 Maths Chapter 8 Quadrilaterals
Home » CBSE Class 9 Maths » Assertion Reason Questions for Class 9 Maths » Assertion and Reason Questions for Class 9 Maths Chapter 8 Quadrilaterals
Assertion and Reason Questions for Class 9 Maths Chapter Quadrilaterals
Directions:
(a) Assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Assertion is false but reason is true.
Q.1. Assertion : The angles of a quadrilateral are x0, (x – 10)0, (x + 30)0 and (2x)0, the smallest angle is equal to 580.
Reason : Sum of the angles of a quadrilateral is 3600.
Q.2. Assertion : If the diagonals of a parallelogram ABCD are equal, then ∠ABC = 900.
Reason : If the diagonals of a parallelogram are equal, it becomes a rectangle.
Q.3. Assertion : A parallelogram consists of two congruent triangles.
Reason : Diagonal of a parallelogram divides it into two congruent triangles.
Q.4. Assertion : ABCD is a square. AC and BD intersect at O. The measure of ∠ABC = 900.
Reason : Diagonals of a square bisect each other at right angles.
Answer Answer: (a) Since, diagonals of a square bisect each other at right angles.
Q.5. Assertion : ABCD and PQRC are rectangles and Q is a midpoint of AC . Then DP = PC.
Reason : The line segment joining the midpoint of any two sides of a triangle is parallel to the third side and equal to half of it.
Q.6. Assertion : In ΔABC , median AD is produced to X such that AD DX = . Then ABXC is a parallelogram.
Reason : Diagonals AX and BC bisect each other at right angles
Q.7. Assertion : The consecutive sides of a quadrilateral have one common point.
Reason : The opposite sides of a quadrilateral have two common point
Q.8. Assertion : Two opposite angles of a parallelogram are (3x-2)0 and (50-x)0. The measure of one of the angle is 370.
Reason : Opposite angles of a parallelogram are equal.
Answer Answer: (a) Hint: Since, opposite angles of a parallelogram are equal.
Q.9. Assertion : If the angles of a quadrilateral are in the ratio 2 : 3 : 7 : 6, then the measure of angles are 400, 600, 1400 1200, respectively.
Reason : The sum of the angles of a quadrilateral is 3600.
|
V.T.U.
Web-Based Education
Engineering Mathematics – I (MAT-11)
By Dr. K.S. Chandrashekhar
Dept. of Mathematics, N.I.E.,
Mysore-8
Lesson 1 Differential Equations
Session – 1
Introduction
Many problems in all branches of science and engineering when analysed for putting in a
mathematical form assumes the form of a differential equation.
An engineer or an applied mathematician will be mostly interested in obtaining a solution
for the associated equation without bothering much on the rigorous aspects. Accordingly
the study of differential equations at various levels is focused on the methods of solving
the equations.
Preliminaries
Ordinary Differential Equation (O.D.E)
If y = f (x) is an unknown function, an equation which involves atleast one derivative of
y, w.r.t. x is called an ordinary differential equation which in future will be simply
referred to as Differential Equation (D.E).
The order of D.E is the order of the highest derivative present in the equation and the
degree of the D.E. is the degree of the highest order derivative after clearing the
fractional powers.
Finding y as a function of x explicitly [y = f (x)] or a relationship in x and y satisfying the
D.E. [f (x, y)= c] constitutes the solution of the D.E.
Observe the following equations along with their order and degree.
1 2
dy
dx
x ·
[order = 1, degree = 1]
2 3 2 0
2
dy
dx
dy
dx
F
H
G
I
K
J
+
F
H
G
I
K
J
+ ·
[order = 1, degree = 2]
1
General solution and particular solution
A solution of a D.E. is a relation between the dependent and independent variables
satisfying the given equation identically.
The general solution will involve arbitrary constants equal to the order of the D.E.
If the arbitrary constants present in the solution are evaluated by using a set of given
conditions then the solution so obtained is called a particular solution. In many physical
problems these conditions can be formulated from the problem itself.
Note : Basic integration and integration methods are essential prerequisites for this
chapter.
Solution of differential equations of first order and first degree
Recollecting the definition of the order and the degree of a D.E., a first order and first
degree equation will be the form
dy
dx
f x, y M x, y dx +N x, y dy = · b g b g b g or 0
We discuss mainly classified four types of differential equations of first order and first
degree. They as are as follows :
• Variables separable equations
• Homogenous equations
• Exact equations
• Linear equations
Variables separable Equations
If the given D.E. can be put in the form such that the coefficient of dx is a function of the
variable x only and the coefficient of dy is a function of y only then the given equation is
said to be in the separable form.
The modified form of such an equation will be,
P (x) dx + Q (y) dy = 0
Integrating we have
P x dx + Q y dy = c bg bg z z
This is the general solution of the equation.
2
Example –1
Solve :
e y dx e + dy=
x x
− + 1 2 4 0 b g
>> e y dx e + dy=
x x
− + 1 2 4 0 b g
Dividing throughout by (y-1) (e
x
+ 4) we get,
e
e
dx
dy
y
Variables are separated
x
x
+
+
·
4
2
1
0
+
+
z
·
z
e
e
dx
dy
y
c
x
x
4
2
1
i. e. e y c
x
log log + + − · 4 2 1
or log log e y c
x
+ + − · 4 1
2
i. e. e y k say
x
log log + − · 4 1
2
b g
or is the required solution e y k
x
+ − · 4 1
2
Example 2
Solve given that
or
i. e. by separating the variables.*
i. e.
Put or
Hence we have,
i e
or is the general solution.
dy
dx
x e y
dy
dx
x e
dy
dx
x e e
dy
e
x e dx
e dy x e dx= c
e dy x e dx= c
x t x dx =dt x dx = dt
e e dt c
e
e
c
y x
y x y x
y
x
y x
y x
y t
y
t
· ·
>> · ·
·
z
z
− −
z
− · ∴− −
− + ·
z
− + ·
− ·
− −
− −
− −
2
2 2
2
2
2
0 0
2 2
2
2
2
bg
e
e c
x
y
2
2
Now we consider y (0) = 0 That is y = 0 when x = 0,
* For animation
3
The general solution becomes
1
2
1
1
2
− · − c c = or
Now the general solution becomes
e
e
x
y
− · −
2
2
1
2
or is the particular solution. e e
x y − −
+ ·
2
1 2
e j
Example –3
Solve :
i. e.,
i. e.,
or by separating the variables.*
or
i. e., 1 log
i. e., log log
or
x y
dy
dx
x +y +xy
>> x y
dy
dx
x +y +xy
x y
dy
dx
x y +x
x y
dy
dx
x +y
ydy
y
x
x
dx
y
+y
dy
x
x
dx c
y
+y
dy
x
dx dx c
dy
+y
dy x x = c
y y x x = c
· +
· +
· + +
· +
+
·
+
⇒ −
+
·
z z
+ −
− −
z
·
z z
− −
z
z
− + − −
− − +
1
1
1 1
1 1
1
1
1
1
1 1
1
1
1
1
1
1
y x x y b glog 1 · c is the required solution.
Example – 4 :
Solve : y x
dy
dx
y
dy
dx
− · +
2
>> Rearranging the given equation we have,
4
y y
dy
dx
x +
dx
x
dy
y y
dx
x
dy
y y
c
x
dy
y y
c
− ·
+
·
+
z
·
·
z
+ −
·
z
2
2
1
1
1 1
1
1
1
or by separating the variables.*
i. e. logb g
We have to employ the method of partial fractions for the second term of the above.
Let
1
(1 )
Put
Put
or
log log log
1
y y
A
y
B
y
A y B y
y = A
y = B
dy
y y y
dy +
y
dy
dy
y y y
dy +
y
dy
dy
y y
y + y
y
y
+ +
⇒ · − +
·
·
z
·
z z
z
· −
z z
· − − ·
− F
H
G
I
K
J z
1
1 1
0 1
1 1
1
1 1
1
1
1 1
1
1
1
b g
b g
b g
b g
b g
i e
Using this result in (1) we get,
log log
1
or log
( ) ( )
log say
is the required solution.
x +
y
y
c
x y
y
k
1
1 1
b g+
− F
H
G
I
K
J
·
+ − L
N
M
O
Q
P
·
∴ −
x + y ky, 1) (1 )=
Example – 5 :
Solve : tan cos( ) +cos( )
>> The given equation on expanding terms in the R. H. S. becomes
tan cos cos sin sin cos cos sin sin
i e tan cos cos
y
dy
dx
x y x y
y
dy
dx
x y x y x y x y
y
dy
dx
x y
· + −
· − + +
·
2
5
or
tan
cos
cos by separating the variables.*
tan sec cos
is the required solution.
y
y
dy= x dx
y. y dy x dx c
2
2 ⇒ −
z z
·
∴ − sec 2sin y x =c
Session – 2
Equations reducible to the variables separable form
Some differential equations can be reduced to the variables separable from by taking a
suitable substitution. We present a few examples.
Example – 6
Solve : cos (x + y + 1) dx – dy = 0
>> We have = cos ( )
dy
dx
x +y +1
… (1)
Put or
Now (1) becomes cos or cos
i. e.
cos cos
t = x +y +
dt
dx
dy
dx
dt
dx
dy
dx
dt
dx
t
dt
dx
t
dt
t
dx
dt
t
dx=c
1 1 1
1 1
1 1
∴ · + − ·
− · · +
+
· ⇒
+
z
z
i. e.
cos
or
1
2
sec
2
2
dt
t
x =c t dt x =c
2 2
2
−
z z
i. e. tan ( / 2) t x =c −
i. e. is the required solution. tan
+1
2
x + y
x = c
F
H
G
I
K
J
Example – 7
Solve tan
tan( ) +1 .... (1)
Put or
:
dy
dx
x y x
dy
dx
x y x
t = y x
dt
dx
dy
dx
dt
dx
dy
dx
· − +
>> · −
− ∴ · − + ·
1
1 1
6
Now (1) becomes tan or tan
Hence
tan
by separating the variables.*
cot
i. e. log(sin )
Thus ( ) is the required solution.
dt
dx
x t
dt
dx
x t.
dt
t
x dx
t dt x dx=c
t
x
c
+ · + ·
·
∴ −
z z
− ·
− −
1 1
2
2
log sin
2
2
y x
x
Example – 8
Solve : x y dx dy dx +dy + − · 2
>> Rearranging the terms in the given equation we have,
x + y dx= x + y + dy
dy
dx
x + y
x + y
2 1 2 1
2 1
2 1
1
·
+
or
Put
or
1
2
The becomes
1
2
i. e.,
t = x + y
dt
dx
dy
dx
dt
dx
dy
dx
d. e
dt
dx
t
t +
dt
dx
t
t
dt
dx
t
t +
2
1 2 1
1
1
1
2
1
1
1
3 1
1
∴ · + −
F
H
G
I
K
J
·
F
H
G
I
K
J
·
∴ ·
+
+
·
or
i. e.,
t
t
dt =dx
t
t
dt dx =c
t
t
dt x =c
+
+
z
z
+
z
1
3 1
1
3 1
1
3 1
… (2)
Let t + 1 = l(3t−1) + m
or t + 1 = ( 3l ) t + (−l+m)
⇒ −
− ·
∴ + · − +
3 1 1
1 3 1 3 1 4 3
1 1 3 3 1 4 3
l = l +m =
l = +m= m
t . t
and
Hence and or
Thus (2) can be written as
7
i. e.
1 3 3 1 4 3
3 1
t
t
dt x c
− +
− ·
z
i. e.
1
3
1
4
3 3 1
dt +
dt
t
x =c
z z
i. e. log(3 1)
t
t x =c
3
4
9
+ − −
i. e. log(3 6 )
x + y
x y x c
2
3
4
9
1 + + − − ·
i. e. is the required solution.
2
3
4
9
log 6 y x + x + y c − − · 3 1
Example – 9 :
Solve :
2
dy
dx
x +y +
x + y
·
+
1
2 3
>> ·
+
We have
2(
dy
dx
x +y +
x +y
1
3
Put or t =x +y
dt
dx
dy
dx
dt
dx
dy
dx
∴ · + − · 1 1
The given equation becomes,
dt
dx
t
t
dt
dx
t
t +
− ·
+
+
· +
+
1
1
2 3
1
1
2 3
or
i. e. or
dt
dx
t t
t +
dt
dx
t
t
·
+ + +
·
+
+
2 3 1
2 3
3 4
2 3
i. e.,
2 3
3 4
t
t +
dt = dx
+
+
z
z
2 3
3 4
t
t +
dt dx = c
Let 2t + 3 = l (3t + 4) + m
or 2t + 3 = (3l) t + (4l + m)
⇒ 3 2 4 3 l = l +m= and
∴ l = m= m = 2 3 8 3 1 3 and / 3 + or
Thus 2 t + 3 = 2/3 . (3t+ 4) + 1/3
Hence (1) becomes
2 / 3.(3 + 4) +1 / 3
3 + 4
t
t
dt dx =c
z
z
i. e.,
2
3
1
1
3 3 4
dt +
dt
t
x =c
+
z
z
8
i. e.,
2
3
log where
t
t x =c, t = x +y + + −
1
9
3 4
∴ + + −
2
3
1
9
3 3 4
x +y
x y + x =c log
i. e., is the required solution.
1
3
(2 ) +
1
9
log(3 + 3 + 4) = y x x y c −
Example – 10
Solve : sec
2
x
dy
dx
x y + xy
3 2
0 + ·
The given d.e. can be written as,
x x
dy
dx
y xy
2
0 1 +
F
H
G
I
K
J
+ · sec
2
Put t = xy
dt
dx
x
dy
dx
y ∴ · +
Now (1) becomes, sec
2
x
dt
dx
t =
2
0 +
i. e., sec or
sec
2
2
x
dt
dx
t
dt
t
dx
x
2
2
· − · −
z z
· cos
2
t dt +
dx
x
c
2
i. e.
cos 1 2
2
1 + t
dt
x
c −
z
·
i. e.,
1
2
sin2
where t +
t
x
c, t = xy.
2
1 F
H
G
I
K
J
− ·
i. e., + is the required solution.
1
4
2 sin(2 )
1
= xy xy
x
c, −
9
EXERCISES :
Solve the following differential equations
1 0
2 2
xy x dx + x y +y dy = +
2 0
2 2 2 2 2
x yx dy + y +x y dx = −
3
2
y x
dy
dx
a y
dy
dx
− · +
F
H
G
I
K
J
4 x +y dx dy dx +dy − ·
5
4 3
x
dy
dx
x y+ xy + cosec( ) = 0
1 1 1
2 2
+ + · x y c
2 1 1 log y + y + x x c − ·
3 1 − · ay a+x cy
4y x + x +y c − · log
5
2
cos( ) +1 / 2 xy x c ·
Session –3
A function u = f (x, y) is said to be homogeneous function of degree n if
u = x g y / x u = y g x / y
n n
b g b g or
A D.E. of the form M (x, y) dx+ N (x, y) dy = 0 is said to be a homogeneous differential
equation if both M (x, y) and N (x, y) are homogeneous functions of the same degree.
Solution of a homogeneous equation
We prefer to have the differential equation in the form
dy
dx
f x, y
g x, y
·
b g
b g
after recognizing that the D.E. is a homogeneous one.
We take the substitution y = v x so that,
dy
dx
v. x
dv
dx
· + 1 by product rule.
With these the given d.e. can be solved by separating the variables.
10
Example –1
Solve : x y dx x y dy =
2 3 3
0 − +
>> (Observe that the coefficient of dx and dy are homogeneous functions of degree 3)
The given equation can be written as,
dy
dx
x y
x y
y = v x
dy
dx
v +x
dv
dx
v +x
dv
dx
x . vx
x +v x
·
+
∴ ·
·
Put
Now (1) becomes,
2
3 3
2
3 3 3
1
i. e., or
i. e., or
by separating the variables.*
v +x
dv
dx
x v
x v
x
dv
dx
v
v
v
x
dv
dx
v v v
v
x
dv
dx
v
v
v
v
dv =
dx
x
·
+
·
+
·
− −
+
·
+
+ −
3
3 3 3
4
3
4
3
3
4
1 1
1 1
1
(* for animation)
z z
·
z
+
− +
1 1
3
1
3
3
4
3
3
3
3
v
dv +
v
dv +
dx
x
c
v
v + x =c
v
vx c, v =
y
x
x
y
y =c,
i. e., log log
i. e., + log ( ) = where
Thus log is the required solution.
Example –2
Solve :
>> We have
or
x dy y dx = x +y dx
x dy = y + x +y dx
dy
dx
y + x +y
x
L
N
M
O
Q
P
·
2 2
2 2
2 2
1
11
Put
Now (1) becomes
i. e., or
by separating the variables. *
i. e., sin log where
Thus sin log is the required solution.
y =vx
dy
dx
v +x
dv
dx
v +x
dv
dx
vx x v x
x
v +x
dv
dx
x v + v
x
x
dv
dx
v
dv
v
dx
x
dv
v
dx
x
c
h v x =c, v = y / x
h y / x x =c,
∴ ·
·
+ +
·
+
F
H
I
K
· +
+
·
+
z
z
·
2 2 2
2
2
2
2
1
1
1
1
1
1
Example –3
Solve :
We have
or
x
dy
dx
y
x
y
>> x
dy
dx
y
y
x
dy
dx
xy y
x
+ ·
· −
·
2
2
2
2
1
Put
Now (1) becomes
i. e., or
y = v x
dy
dx
v +x
dv
dx
v +x
dv
dx
x . v x v x
x
v +x
dv
dx
x v v
x
x
dv
dx
v
∴ ·
·
·
· −
2 2
2
2 2
2
2
∴ · −
z
+
z
· − +
− +
dv
v
dx
x
dv
v
dx
x
c
v
x =c, v =
y
x
x
y
x = c,
2
2
1
by separating the variables.*
Hence i. e., log where
Thus log is the required solution.
12
Example – 4
Solve : tan ( ) sec sec
>> The given equation can be written as,
sec tan( )
sec
2 2
2
2
x y / x y y / x dx +x y / x dy =
dy
dx
y y / x x y / x
x y / x
·
0
1
Put or
Now (1) becomes,
sec tan
sec
i. e.,
sec tan
sec
2
2
2
2
y
x
v v = v x
dy
dx
v +x
dv
dx
v +x
dv
dx
v x v x v
x v
v +x
dv
dx
x v v v
x v
· ∴ ·
·
·
−
i. e.,
sec tan
sec
i. e.,
sec tan sec
sec
i. e.,
tan
sec
or
sec
tan
2
2
2 2
2
2
2
x
dv
dx
v v v
v
v
x
dv
dx
v v v v v
v
x
dv
dx
v
v
v
v
dv =
dx
x
·
·
− −
·
Hence
sec
tan
i. e., log (tan ) + log
or log (tan ) = = log (say),
2
v
v
dv+
dx
x
c
v x =c
v. x c k
z
·
z
⇒x v = k v = y / x
x y / x k,
tan where
Thus tan ( ) = is the required solution.
Example – 5
Solve : log log
The given equation can be written in the form
log( ) +1
x
dy
dx
y y x +
>>
dy
dx
y
x
y / x
· −
·
1
1
13
Put
Now (1) becomes log
i. e., log or
log
Hence
1/
log
y = v x
dy
dx
v +x
dv
dx
v +x
dv
dx
v v
x
dv
dx
v v
dv
v v
dx
x
v
v
dv
dx
x
=c
∴ ·
· +
· ·
z
z
1
i. e., log log log log say
i. e., log log log log
i. e., log log log
log where
Thus l g( ) = is the required solution.
v x =c = k
v k + x
v kx
v kx v = y / x
y / x k x,
·
·
⇒ ·
Example – 6
Solve : 1+
(As we observe terms with , we need to express the
equation relating to and the terms are homogeneous functions of degree 0.
We have 1+
e dx +e
x
y
dy =
x / y
dx / dy
e dx =e
x
y
dy
x/ y x/ y
x/ y x/ y
e j
e j
1 0
1
F
H
G
I
K
J
>>
F
H
G
I
K
J
or
1+
Put or
Now (1) becomes,
dx
dy
e
x
y
e
x / y = v x = v y
dx
dy
v +y
dv
dy
v +y
dv
dy
e v
e
x/ y
x/ y
v
v
·
F
H
G
I
K
J
∴ ·
·
+
1
1
1
1
e j
i. e., i. e.,
i. e., or
(1+
Hence
(1+ )
y
dv
dy
e v
e
v y
dv
dy
e v e v e v
e
y
dv
dy
e +v
e
e dv
e +v
dy
y
e dv
e v
dy
y
c
v
v
v v v
v
v
v
v
v
v
v
·
+
− ·
− − −
+
·−
+
· −
+
+
z z
·
1
1 1
1
14
i. e., log ( ) log
or log log (say)
where
Thus is the required solution.
e +v y =c
e +v y k
e +v y =k, v = x / y.
y e x =k,
v
v
v
x/ y
+
·
+
Session – 4
Equations reducible to the homogeneous form
Consider the differential equation in the form :
ax +by +c dx a' x +b' y +c' dy = t 0
We first express the equation in respect of dy / dx and the procedure is narrated by taking
dy
dx
a x + b y + c
a' x +b' y +c'
a
a'
b
b'
· ≠ where 1
This condition implies that there are no common factors for the x and y terms in the
numerator as well as in the denominator.
Put x = X + h, y = Y + k where h and k are constants to be chosen appropriately later.
Now
Hence
dy
dx
dy
dY
dY
dX
dX
dx
dY
dX
.
dy
dx
dY
dX
·
· · 1 1
As a consequence of these (1) becomes
dY
dX
a X +h b Y +k +c
a' X +h b' Y +k +c'
dY
dX
aX +bY ah+bk +c
a' X +b' Y a' h+b' k +c'
·
+
+
·
+
+
b g b g
b g b g
b gb g
b gb g
i. e., 2
Now, let us choose h and k such that :
ah+bk +c a' h+b' k +c' = · 0 0 and
Solving these equations we get the value for h and k.
Thus (2) now assumes the form
dY
dX
aX +bY
a' X + b' Y
· 3
It is evident that (3) is a homogeneous equation in the variables X and Y. This equation
can be solved by putting Y = VX as discussed already. Finally we substitute for X and
Y where X = x −h, Y = y−k.
15
Example –1
Solve : ( )
We have
Put and where and are constants to be choosen suitably later.
x y dx + x +y dy =
x +y dy = x y dx
dy
dx
x y
x +y
x = X +h y =Y +k h k
− − −
>> − − − −
∴ ·
− + +
4 9 4 2 0
4 2 4 9
4 9
4 2
Now
Hence
dy
dx
dy
dY
dY
dX
dX
dx
dY
dX
dy
dx
dY
dX
·
· · 1 1
Thus (1) becomes
dY
dX
X h Y +k
X +h Y +k
dY
dX
X Y h+ k
X +Y h+k
h k
h+ k h+k
h = k
·
− + + +
+ −
·
− + + − +
+ −
− + · − ·
· −
4 9
4 2
4 4 9
4 4 2
2
4 9 0 4 2 0
1 2
i. e.,
Let us choose and such that
and
Solving these equations we get, and
Thus (2) becomes
Put
Now (3) becomes,
i. e., or
i. e., i. e.,
by separating the variables.*
dY
dX
X Y
X +Y
Y = VX
dY
dX
V + X
dV
dx
V + X
dV
dX
X VX
X +VX
V + X
dV
dX
X V
X +V
X
dV
dX
V
+V
V
X
dV
dX
V V V
+V
X
dV
dX
V
+V
V dV
V
dX
X
·
− +
∴ ·
·
− +
·
− +
·
− +
·
− + − −
·
− +
+
+
· −
4
4
3
4
4
1 4
4
1 4
4
1 4 4
4
1
4
4
1
2 2
2
16
+
+
z
+
+
z
·
z
+
+
4
1 1
1
2
2 2
2
2 2
1 2
1 2
1 2 2
dV
V
V dV
V
dX
X
c
V + V X =c
V + V X = c
V + V X = c, V =Y / X
i. e., 4 tan log(1+ log
i. e., 8 tan log (1+ log
i. e., 8 tan log (1+ where
∴ + ·
− − −
F
H
G
I
K
J
+ − + + ·
8 tan log ( + say
But and
8 tan log
This is the required solution.
1 2 2
1 2 2
2
1 2
2
1
1 2
Y / X X Y c =k
X = x h = x Y = y k = y +
y +
x
x y k
Example – 2
Solve :
dy
dx
x + y
x +y
·
2 3
2 3
>> ·
We have
dy
dx
x + y
x +y
2 3
2 3
1
Put x = X + h and y = Y + k, where h and k are constants to be chosen suitably later.
∴ ·
· ·
·
+ −
+ −
·
+ −
+ −
dy
dx
dy
dY
dY
dX
dX
dx
dY
dX
dy
dx
dY
dX
dY
dX
X +h Y +k
X +h Y +k
dY
dX
X + Y h+ k
X +Y h+k
1 1
2 3
2 3
2 2 3
2 2 3
Hence
Thus(1) becomes,
i. e.,
……(2)
Let us choose h and k such that,
h+ k h+k
h k
dY
dX
X + Y
X +Y
2 3 0 3 0
1 1
2
2
3
− · − ·
· ·
·
and 2
Solving these equations we get and
Thus (2) becomes
17
Put
Now (3) becomes
i. e.,
Y = VX
dY
dX
V + X
dV
dX
V + X
dV
dX
X + VX
X +VX
V + X
dV
dX
X V
X V
∴ ·
·
·
+
+
2
2
1 2
2
or or
by separating the variables.*
2
i. e., log
1+
log( log
i. e., log
1+
log( log
X
dV
dX
V
V
V X
dV
dX
V
V
V
V
dv =
dX
X
dV
V
+
V dV
V
dX
X
c
V
V
V X =c
V
V
V X = c
·
+
+
− ·
+
+
z
− ·
z z
F
H
G
I
K
J
− − −
F
H
G
I
K
J
− − −
1 2
2
1
2
2
1
1 1
2
1
2 1
1
2
1
2
1
1 2 2
2
2
2 2
2
2
i. e., log
1+
log (
or log
(
( (
log say
(
(
or
( )
since
ut and
Thus is the required solution.
V
V
V X = c
V
V V X
k
V
V X
k
X +Y
X Y
k V =
Y
X
X = x h = x Y = y k y
x y k x y
1
1 2
1
1 1
1
1
1 1
2
2
2 2
2
2 2 2
3 2 3
3
F
H
G
I
K
J
− −
+
− −
L
N
M
O
Q
P
·
+
·
·
− − − · −
+ − · −
b g
Exercises :
Solve the following equations:
1
2 0
3
4 4 2
5 3 1 2 0
3 3
2
x +y dx xy x +y dy
x +y x y y dx x x y dy
x / y y dx x dy y dy
y x dx= y +x dy
x y dy x y dx =
·
− − − ·
− ·
− − −
− − + −
log log log log
Sin ( )
18
1
2 1
3
2
1
4
4 1 3 2 3 1
5 1 4 2 1
2 1
4
y / x + y x c
x+ y / x y / x c
.
x
y
x y y =c
x + y y x + c
x +y c x y
log
log log( )
sin(2 ) log
log ( ) tan
2
− ·
− ·
− −
+ − + − ·
− · − −
Session - 5
We say that M (x, y) dx+N (x, y) dy = 0 is an exact differential equation if there exists a
function f (x, y) such that
df = M (x, y) dx+N (x, y) dy.
The necessary and the sufficient condition for the D.E. M (x, y) dx + N (x, y) dy = 0 to be
an exact equation is
M
y
N
x
·
Further the solution of the exact equation is given by
Mdx + N y dy = c
z zbg
where, in the first term we integrate M (x, y) w.r.t x keeping y fixed and N (y) indicate
the terms in N not containing x.
Example –1
Solve : [ cos log sin
Let cos and log sin
y x y dx + x + x x y dy
M = y x y N = x + x x y
1 1 0
1 1
+ + − ·
>> + + −
b g
b g
∴ · + − · + −
·
z z
M
y x
y
N
x x
y
M
y
N
x
M dx+ N y dy =c
1
1
1
1
sin and sin
Since the given equation is exact.
The solution is
bg
i. e., cos
Thus log cos is the required solution.
y
x
y dx + dy= c
y x + x x y =c,
1
1
0 +
F
H
G
I
K
J
+
L
N
M
O
Q
P
z z
+ b g
19
Example –2
Solve :
cos sin
sin cos
The given equation is put in the form,
cos sin sin cos
Let cos sin and sin cos
dy
dx
y x+ y+ y
x +x y +x
y x+ y+ y dx + x +x y +x dy =
M = y x+ y+ y N = x +x y +x
+ ·
>>
0
0 b g b g
∴ · ·
·
z z
z z
M
y
x+ y+
N
x
x+ y+
M
y
N
x
Mdx+ N y dy =c
y x+ y+ y dx = dy= c
y x + x y + xy = c,
cos cos and cos cos
Since the given equation is exact.
The solution is
i. e., cos sin
Thus sin sin is the required solution.
1 1
0
bg
b g
Example – 3
Solve :
Let and
y e x dx + xye y dy
M= y e x N = x y e y
M
y
y e xy +e y,
N
x
x y e y ye
xy xy
xy xy
xy xy xy xy
2 3 2
2 3 2
2 2
2 2
2 2
2 2 2 2
4 2 3 0
4 2 3
2 2 2 2
+ − ·
>> + −
∴ · · +
e j e j
i. e., and
Since the given equation is exact.
The solution is
M
y
xy e ye
N
x
xy e ye
M
y
N
x
M dx + N y dy= c
xy xy xy xy
· + · +
·
z z
2 2 2 2
3 3
2 2 2 2
i. e., y e x dx + y dy =c
xy 2 3 2
2
4 3 +
z
z e j
i. e.,
Thus is the required solution.
y
e
y
x y c
e x y c,
xy
xy
2
2
4 3
4 3
2
2
+ − ·
+ − ·
Equations reducible to the exact form
Sometimes the given differential equation which is not an exact equation can be
transformed into an exact equation by multiplying with some function (factor) known as
the integrating factor (I.F.)
20
The procedure to find such a factor is as follows.
Suppose that, for the equation M dx + N dy = 0
M
y
N
x
≠ , then we take their difference.
If
1
or
1
N
M
y
N
x
f x
M
M
y
N
x
g y
F
H
G
I
K
J
· −
F
H
G
I
K
J
· bg bg
then or is an integrating factor. e e
f x dx g y dy bg bg z
z
Example – 4
Solve : (4 ) xy + y x dx +x x + y dy = 3 2 0
2
− b g
Let 4 and
and (The equation is not exact. )
Consider close to
M= xy + y x N = x x + y x xy
M
y
x + y
N
x
x + y.
M
y
N
x
x y = x + y N.
3 2 2
4 6 2 2
2 4 2 2
2 2
− · +
· ·
− · +
Now
1
Hence is an integrating factor.
Multiplying the given equation by we now have,
and
Solution is given by
Thus ( )
log log
N
M
y
N
x
x + y
x x + y x
f x
e
e e e e x
x
M= x y + x y x N = x x y
M dx + N y dy =c
x y + x y x dx + dy =c
f x dx
f x dx
x
dx
x x
F
H
G
I
K
J
· · ·
· · · ·
− +
z z
z z
z
z
z
z z
2 2
2
2
4 3 2
4 3 0
2
2 2
2
3 2 2 3 4 3
3 2 2 3
2
bg
bg
bg
i. e., is the required solution. x y +x y
x
c,
4 3 2
4
4
− ·
21
Example – 5
Solve :
Let and
i. e., and
y x y + dx +x x y + dy =
M= y x y + N = x x y +
M= xy y y N = x xy + x
2 1 3 4 3 0
2 1 3 4 3
2 3 4 3
2 2
− −
>> − −
− + −
b g b g
b g b g
M
y
x y
N
x
x y
M
y
N
x
x y x y + M.
M
M
y
N
x
x y +
y x y + y
g y
· − + · − +
− · − + − · − −
F
H
G
I
K
J
·
− −
· − ·
2 2 1 6 4 3
4 2 2 2 2 1
2 2 1
2 1
2
b g
b g
b g
bg
near to
Now,
1
Hence
Multiplying the given equation with we now have,
and
log log
I. F =e e =e e y
y
M= x y y y N = x y xy xy
g y dy y
dy
y y −
z
z
· ·
− + − +
2
2 2
2
3 4 3 2 2 3 2
2
2 3 4 3
The solution is
i. e.,
Thus is the required solution.
M dx + N y dy = c
x y y y dx + dy = c
x y xy xy c,
bg
e j
z z
− +
z z
− + ·
2 0
3 4 3
2 3 4 3
EXERCISES:
Solve the following differential equations
1. cos x (e
y
+1)dx + sin x e
y
dy = 0
2. [4x
3
y
2
+ y cos (xy)] dx+[2x
4
y+x cos (xy)] dy =0
3 2 3 2 1 1
2 2
xy x y dx +x y dy = x +y dy ; y + − + ·
4 1 3 2 0 y x +y dx +x x + y dy + + · b g
5 2 3 2 6 3 0
2 3 2
x + y y dx + x +xy dy y + · given that (1) = 2
22
1 1
2
3
2 2
2 3 1
4
2
5 3 15
4 2
2 2 2
2
2 2
3 2
6 4 3 4
sin
sin
x e c
x y xy c
x y x
xy + x y
x y
xy xy c
x x y x y
y
+ ·
+ ·
+ − − ·
+ + ·
+ + ·
bg
Session – 6
Linear Equations
A differential equation of the form
dy
dx
Py =Q +
…(1)
where P and Q are functions of x only is called a linear equation in ‘y’.
y e Q e dx +c
dx
dy
Px =Q ...
P dx P dx
z z
·
z
+
is the solution of the linear equation (1).
An equation of the form :
2
where P and Q are functions of y is called a linear equation in x.
The solution can simply be written by interchanging the role of x and y.
i. e.. x e Q e dy +c
P dy P dy
z z
·
z
is the solution for the linear equation (2)
Working procedure for problems
• The given equation must be first put in the form conformal to the standard
form of the linear equation in x or y.
• The expression for P and Q is to be written by simple comparison.
We equip with the I. F. or e e
P dx P dy
z z
• We assume the associated solution and we only need to tackle the R.H.S. part
of the solution to finally arrive at the required solution.
23
Example – 1
Solve : cot cos
cot cos is of the form
where cot and cos
sin
cot log (sin
dy
dx
y x = x
>>
dy
dx
y x = x
dy
dx
Py =Q, P = x Q = x
e e e x
P dx x dx x
+
+
+
∴ · · ·
z z z
The solution is
i. e., sin cos sin
i. e., sin
sin 2
Thus sin
cos 2
4
, is the required solution.
y e Qe dx +c
y x = x. x dx +c
y x =
x
dx +c
y x =
x
c
P dx P dx
z z
·
z
z
z
+
2
Example – 2
Solve : 2 cos sin sin given that
Dividing the given equation by cos we have
tan = sin sin 2 cos sin
This is of the form where tan and sin
y' x+ y x = x y
>> x,
dy
dx
x y x x= x x
dy
dx
Py =Q, P x Q x
4 2 3 0
2
2 2
2
b g·
+
+ · ·
e e e x
pdx x dx x
z z z
· · ·
2 tan 2 log (sec ) 2
sec
The solution is y e Qe dx +c
P dx P dx
z z
·
z
i. e., sec sin sec
i. e., sec tan sec
Thus sec sec , is the general solution. ..... (1)
Consider That is when
Hence (1) becomes 0 = 2 + or
Thus sec sec is the required particular solution.
y x = x. x dx +c
y x = x x c
y x = x dx +c
y y = x =
c c =
y x= x ,
2 2
2
2
2
3 0 0 3
2
2
z
z
+
·
b g
24
Example – 3
Solve : (1+ ) tan
We have
tan
or
tan
This is of the form
Here and
tan
;
tan
y dx + x y dy =
dx
dy
y x
y
dx
dy
x
y
y
y
dx
dy
Px =Q
P=
y
Q=
y
y
e e
P dy y
2 1
1
2
2
1
2
2
1
2
0
1
1 1
1
1 1
1
−
>> ·
+
+
+
·
+
+
+ +
·
z
The solution is given by
i. e.,
tan
By putting tan we obtain
1
1+
i. e., by parts.
Thus tan is the required solution.
tan tan
tan
tan
tan tan
x e Q e dy+c
x e
y
y
e dy+c
y =t,
y
dy = dt
x e t e dt + c
x e t e e + c,
x e e y +c,
P dy P dy
y y
y t
y t t
y y
z z
·
z
·
+
z
∴ ·
z
· −
· −
− −
− −
1 1
1
1
1 1
1
2
1
2
1
1
1
Equations reducible to linear form
Form (i): where and are function of
We put
The given equation becomes which is a linear equation in
Similarly, where and are function of can be
reduced to the linear form by putting
f ' y
dy
dx
P f y Q, P Q x.
f y t f ' y
dy
dx
dt
dx
dt
dx
Pt =Q t.
f ' x
dx
dy
P f x Q, P Q y
f x t
+ ·
· ∴ ·
+
+ ·
·
bg
25
Form (ii) : where and are functions of
This equation is called as Bernoulli' s equation
We first divide the equation throughout by to obtain
1
Put or
1
dy
dx
+P y =Q y P Q x.
y.
y
y
dy
dx
P y =Q .....
y t n y
dy
dx
dt
dx y
dy
dx n
dt
dx
n
n
n
n
n n
n
+
· ∴ − · ·
− −
1
1
1
1
1
1
Hence (1) becomes,
or which is a linear equation in
Similarly , where and are functions of is called
Bernoulli' s equation . We first divide by and later put =
to obtain a linear equation in .
1
1
1 1
1
+
+ − −
+
n
dt
dx
P t =Q
dt
dx
n P . t = n Q t.
dx
dy
P x =Q x P Q y
x x x t
t
n
n n
Example - 4
Solve : tan tan cos cos
>> Dividing the given equation by cos we have
sec tan sec tan cos
2
2
y
dy
dx
x = y x
y
y y
dy
dx
y x = x
+
+ 1
Now, put sec sec tan
Hence (1) becomes tan cos
This equation is of the form + where we have,
tan and cos
2
2
y = t y y
dy
dx
dt
dx
dt
dx
t x = x
dt
dx
Pt =Q,
P= x Q= x
∴ ·
+
∴ · ·
·
z
z
z z z
z z
e e e x
t e Qe dx +c
y x = x . x dx +c
P dx x dx x
P dx P dx
tan log (sec
2
= sec
The solution is
i. e., sec sec cos sec
i. e., sec sec cos
Thus sec sec sin is the required solution.
y . x = x dx +c
y x = x +c,
z
26
Example – 5
Solve :
Dividing the given equation by we have,
1
Put
Hence (1) becomes
or
dy
dx
y
x
y x
>> y
y
dy
dx y x
x
y
t
y
dy
dx
dt
dx
dt
dx
t
x
x
dt
dx
t
x
x
+ ·
+ ·
· ∴
·
+ ·
− · −
2
2
2
2
1
1
1 1
This equation is a linear equation of the
where and
The solution is
i. e.,
Thus is the required solution.
log
dt
dx
Pt =Q,
P=
x
Q= x
e e =e
x
t e Q e dx +c
t .
x
x.
x
dx + c
xy
x +c,
P dx
x
dx
x
P dx P dx
+
∴ · ·
·
z
· −
z
· −
z
z
z z
1
1
1 1
1
1
Example – 6
Solve :
Consider
or Dividing by we get,
xy xy
dy
dx
dx
dy
x y +x y
dx
dy
xy = x y x
1 1
2
2 3
2 3 2
+ ·
>> ·
1
Put
x
dx
dy x
y = y
x
t
x
dx
dy
dt
dy
2
3
2
1
1
1 1
· ∴
·
27
Hence (1) becomes or
This equation is of the form where,
and
The solution is
− − −
+
− ∴ · ·
·
z
z z
z z
dt
dy
t y = y
dt
dy
+t y = y
dt
dy
P t = Q
P = y Q = y e e e
t e Q e dy +c
P dy y dy y
P dy P dy
3 3
3 2
2
i. e.,
Put
Also or
i. e., , on integration by parts.
Thus is the required solution.
t e y e dy +c
y u y dy = du
y y dy = y du y dy = u du
t e u e du+c
t e u e e +c
e
x
e
y
c ,
y
y
y u
y u u
y
y
2
2
2
2
2
2
2 3
2
2
2 2 3
2
2
2
2
2
2
2
2
2
2 1
2
· −
z
· ∴
∴ · −
z
· − −
· −
F
H
G
I
K
J
+
EXERCISES
Solve the following equations
1 4 2 0
2 2
dy
dx
y x = x x ; y
x
dy
dx
y = x
+ ·
cot cosec
sin 2 tan
3 0 x dy + y x xy x dx − − · tan b g
4
5 2 3 0
2 6
4 2
dy
dx
y
x
x y
y xy dx + x dy =
+ ·
1 2 2
2
3
4
1 5
2
5 2
2 2
5 5 2
2 3
y x = x
y x = x c
x y x x+ x x +c
x y x
c
x y x
sin
cot log tan
cos =cos sin
+
− ·
· +
28
Session – 7
Orthogonal Trajectories
Definition : If two family of curves are such that every member of one family intersects
every member of the other family at right angles then they are said to be orthogonal
trajectories of each other.
Working procedure for problems
Case – i : (Cartesian family)
* Given f (x, y, c) = 0, differentiate w.r.t x and eliminate c.
* Replace by and solve the equation.
dy
dx
dx
dy
Case – ii : (Polar family)
* Given an equation in r and θ , we prefer to take logarithms first and then
differentiate w.r.t θ .
* fter ensuring that the given parameter is elimiated we replace by
and solve the equation.
dr
d
r
d
d r
2
Example –1
Find the Orthogonal trajectories of the family of parabolas y
2
= 4 a x.
>> · Consider
y
x
a
2
4
Now differentiating (1) w.r.t x we have,
x. y
dy
dx
y
x
x y
dy
dx
y
2 1
0 0
2
2
2
· − ·
or 2
i. e., 2 is the D. E. of the given family.
Replacing by we have, 2
or
x
dy
dx
y =
dy
dx
dx
dy
x
dx
dy
y =
x dx + y dy =
x dx+ y dy = c
− −
F
H
G
I
K
J
z z
0
0
2 0
2
i. e., or 2 say
Thus is the required O. T.
x
y
c x y c =k
x y k
2
2
2 2
2 2
2
2
2
+ · + ·
+ ·
29
Example – 2
Find the orthogonal trajectories of the family of curves
where is the parameter.
x
a
y
b
2
2
2
2
1 +
+
·
>> We have
Differentiating w. r. t we have,
where
i. e.,
x
a
y
b
x
x
a
y y
b
y
dy
dx
x
a
y y
b
2
2
2
2
2
1
2
1
2
1
2
1 1
2 2
0
2
+
+
·
+
+
· ·
·
+
Also from (1)
or
Now, dividing (2) by (3) we get,
or
Now, let us replace by
x
a
y
b
x a
a
y
b
x
x a
y y
y
x
x a
y
y
y
dy
dx
dx
dy
2
2
2
2
2 2
2
2
2
2 2
1
2 2 2
1
1
1
3
− ·
+
·
+
·
·
·
· −
F
H
G
I
K
J
− −
⇒ − +
z z z
x
x a y
dx
dy
y dy =
x a
x
y dy = x dx +a
dx
x
c
2 2
2 2
2
1
or by separating the variables.
i. e., log
or log = 0 is the required orthogonal trajectories. Here
y x
+a x +c
x +y a x b b = c
2 2
2
2 2 2
2 2
2 2
·
− −
Example-3
Show that the family of parabolas y
2
= 4a (x+a) is a self orthogonal.
>> Consider y
2
= 4a (x+a) ….(1)
Differentiating w.r.t x, we have
2 4
2
1
1
y
dy
dx
a a =
yy
y
dy
dx
· ∴ · where
Substituting this value of ‘a’ in (1) we have,
30
y yy x +
yy
y xy yy
y xy yy
2
1
1
1 1
2
1 1
2
2
2
2
2 2
·
F
H
G
I
K
J
· +
· +
or
Thus we have,
This is the D.E. of the given family.
Now replacing y
1
by −1 / y
1
(2) becomes
y = x
y
y
y
y =
x
y
y
y
yy x y y .......
2
1 1 2
2 3
1 1
2
1 1
2
1
2
1
− F
H
G
I
K
J
+
− F
H
G
I
K
J
+
+ ·
or
i. e.,
(3) the D.E. of the orthogonal family which is same as (2) being the D.E. of the given
family.
Thus the family of parabolas y
2
= 4a(x+a) is self orthogonal.
Example –4
Find the O.T. of the family r =a(1+sin θ )
>> We have r = a (1 + sin θ )
⇒log log + log (1+ sin r = a
Differentiating w.r.t θ we have,
1
0
1
2
2
r
dr
d
dr
d
r
d
dr
r
r
d
dr
· +
F
H
G
I
K
J
·
cos
1+ sin
Replacing by we get,
cos
1+ sin
i. e.,
cos
1+ sin
or
1+ sin
cos
by separating the variables.
1+ sin
cos
i. e., log + sec tan
− ·
·
z
+
z
·
z
+ ·
z
r
d
dr
d
dr
r
dr
r
d c
r d d c
i. e., log + log(sec tan ) + log(sec
i. e., log sec tan sec log (say)
cos
sin
cos cos
i. e.,
1+ sin
cos
or
1+ sin
sin
2
r c
r b
r b
r
b
r
b
+ ·
+ ·
⇒ +
F
H
G
I
K
J
·
·
·
b g
1 1
1
2
Thus sin is the required O. T. r =b 1−
31
Example – 5
Find the O.T. of the family r
n
cos nθ = a
n
>> We have r
n
cos n θ = a
n
⇒n r + n n a log log(cos ) = log
Differentiating w.r.t θ we have,
n
r
dr
d
n n
n r
dr
d
n
dr
d
r
d
dr
r
r
d
dr
n r
d
dr
n
d
n
dr
r
+
− F
H
G
I
K
J
· ·
F
H
G
I
K
J
· − ·
∴ ·
sin
cos
i. e.
1
tan
Replacing by we have,
1
tan or tan
tan
by separating the variables.
0
2
2
z
+ ·
z
dr
r
n d c cot
i. e., log log(sin ) =
or log log (sin ) =
r +
n
n c
n r + n nc
1
i. e., log( sin ) = log (say)
sin = is the required O. T.
r n b
r n b,
n
n
∴
EXERCISES
Find the orthogonal trajectories of the following family of curves
1
2 2 0
4
2
2 2
2
y = ax
x y y +c =
x a y +a
+ +
·
being the parameter.
3. Show that the family of parabolas is a self orthogonal.
4. sec
cos
2
r = a
r=a
2
5 1+
1 2
2 2 0
2
5 1
2 2
2 2
2
x y c
. x y kx c = k
r =b
r =b
+ ·
+ + +
is arbitrary.
4. cosec
cos
32
LESSON – 2 Part – C Integral Calculus
Session - 1
Introduction :
We are familiar with various methods of integration, definite integrals and the associated
application of finding the area under a curve.
In this chapter we first discuss reduction formulae and later discuss the method of tracing
cartesian and polar curves. By knowing the shape of a given curve we disucss application
of definite integrals such as area, length or perimeter, surface area of plane curves and
volume of solids.
In all these applications reduction formulae plays a vital role in the evaluation of definite
integrals.
Reduction Formulae
Reduction formulae is basically a recurrence relation which reduces integral of functions
of higher degree in the form of (where m and n are non negative f x dx, f x g x dx
n m n
z z
Reduction formulae is basically a recurrence relation which reduces integral of functions
of higher degree in the form of (where m and n are non negative f x dx, f x g x dx
n m n
z z
integers) to lower degree. The successive application of the recurrence relation finally
end up with a function of degree 0 or 1 so that we can easily complete the integration
process.
We discuss certain standard reduction formulae in the form of indefinite integrals and the
evaluation of these with standard limits of integration.
Reduction formula for sin and sin where is a positive integer
0
/2
Let sin
sin sin say
n n
x dx x dx dx n
I x dx
x . x dx = u v dx
n
n
n
z z
·
z
·
z z
π
1
We have the rule of integration by parts,
u v dx = u vdx v dx.u' dx −
zz z z
∴ · − − − −
z
− −
z
·− − −
z
· − −
z
− −
z
− −
− −
− −
− −
I x x x n x x dx
= x x + n x x dx
x x + n x x dx
x x + n x dx n x dx
n
n n
n n
n n
n n n
sin cos cos sin cos
sin cos sin cos
sin cos sin sin )
sin cos sin sin
1 2
1 2 2
1 2 2
1 2
1
1
1 1
1 1
33
i. e. sin cos
i. e. sin cos
sin
sin cos
I x x + n I n I
I n x . x + n I
I x dx =
x . x
n
n
n
I
n
n
n n
n
n
n
n
n
n
n
· − − − −
+ − · − −
∴ ·
z
+
1
2
1
2
1
2
1 1
1 1 1
1
1
This is the required reduction formula.
Illustration
i To find sin
4
x dx
z
>> Comparing with L.H.S. of (1), we need to take n = 4 and use the established result.
∴ ·
z
·
+ I x dx
x x
I
4 2
4
3
4
sin
sin cos
4
3
We need to apply the result (1) again by taking n = 2
i. e.,
sin cos sin cos
3
I
x x x x
I
4 0
4
3
4 2
1
2
·
+
+
R
S
T
U
V
W
We cannot find I
0
from (1). But basically we have
I x dx= dx = x
0
1 ·
z z
sin
0
Thus sin
sin cos
sin cos
4
I x dx=
x x
x x +
x
c
4
3
4
3
8
3
8
·
z
− +
Corollary :
To evaluate sin
n
x dx
0
2
z
>> ·
z
Let sin I x dx
n
n
0
2
Equation (1) must be established first.
∴ · −
L
N
M
O
Q
P
+
from (1)
sin cos
I
x. x
n
n
n
I
n
n
n
0
2
1
2
1
But cos (π / 2) = 0 = sin 0
Thus I
n
n
I .
n n
·
1
2
2
We use this recurrence relation to find I
n
2
by simply replacing n by (n− 2).
34
i. e.,
Hence
Similarly from (2)
Hence again by back substitution.
Continuing like this the reduction process will end up as below.
if is odd
= if is even
But sin
I
n
n
I
I
n
n
n
n
I
I
n
n
I
I
n
n
n
n
n
n
I
I
n
n
n
n
n
n
I n
n
n
n
n
n
n
I n
I x dx =
n n
n n
n n
n n
n
− −
− −
·
·
− −
·
·
− −
·
− −
− −
· −
2 4
4
4 6
6
1
0
1 0
3
2
1 3
2
5
4
1 3
2
5
4
1 3
2
5
4
2
3
1 3
2
5
4
1
2
2
0
2
0 0
2
0
2
0
2
0 1 1
2
cos
and sin
0
x
I x dx = dx = x
z
· − − ·
·
z
·
z
Thus we have,
sin
if is even
if is odd
n
x dx =
n
n
n
n
n
n
n
n
n
n
n
n
n
n
− −
− −
R
S
|
T
|
z
1 3
2
5
4
1
2 2
1 3
2
5
4
2
3
0
2
Note :
The reduction formula for cos can be established in a similar way.
n
x dx
z
The result is as follows.
cos
cos sin
n
n
n
x dx =
x . x
n
n
n
I
+
z
1
2
1
sin cos by the property
n n
a a
x dx = x dx f x dx = f a x dx
0
2
0 0 0
2
z z
z z
35
Session – 2
Reduction formula for tan and tan where is a positive integer.
0
n n
x dx x dx n
Let tan
tan tan tan sec
tan sec tan
tan sec
For the first term, put tan sec
Now
Thus
tan
2 2
2
2
,
I x dx
x. x dx x x dx
x x dx x dx
x x dx I
x = t x dx = dt
I t dt I
t
n
I
I
x
n
I
n
n
n n
n n
n
n
n
n
n
n
n
n
n
n
z z
·
z
·
z
· −
z
·
z
z
·
z
· −
z
·
·
− −
− −
2 2
2 2
2
2
2
2
2
1
2
1
2
1
1
1
1
Next, let tan
tan
by using (1)
I x dx
I
x
n
I
n
n
n
n
n
·
z
∴ ·
L
N
M
O
Q
P
0
4
1
0
4
2
1
But tan( / 4) =1 and tan 0 = 0
Thus tan
or or
I x dx
n
I
I I
n
n I I
n
n
n
n n n n
·
z
·
+ ·
+ ·
− + −
0
4
2
2 1 1
1
1
2
1
1
1
Illustration
(i) To find tan
5
x dx
z
>> ·
z
· −
− −
R
S
T
U
V
W
·
z
·
z
·
z
· − +
I x dx
x
I
x x
I
I x dx x dx x
x dx
x x
x c
5 3
1
1
tan
tan
4
by using (1).
i. e., =
tan
4
tan
2
But tan tan log (sec )
tan
tan
4
tan
2
log (sec ) +
5
4
4 2
1
5
4 2
Note : The reduction formula for cot can be established in a similar way
and the result is as follows
n
x dx
.
z
36
I x dx=
x
n
I
n
n
n
n
·
z
cot
cot
1
2
1
3. Reduction formula for sec where is a positive integer
n
x dx n
z
>> ·
z
·
z
Let sec
sec sec
2
I x dx
x dx
n
n
n 2
Integrating by parts we have,
I x x x. n x x x dx
= x x n x x dx
= x x n x x dx
= x x n x dx + n x dx
n
n n
n n
n n
n n n
· −
z
− −
z
− −
z
− −
z
z
− −
− −
− −
− −
sec tan tan sec sec tan
sec tan sec tan
sec tan sec sec
sec tan sec sec
2
2
2
2 3
2 2
2 2
2
2
2
2 1
2 2
I x x n I n I
I n
n
x x n I
I x dx =
n
x x
n
n
n
I
I x dx
I
n
x x
n
n
n
I
n
n
n n
n n
n
n
n
n
n
n n
· − − + −
+ − ·
+ −
·
+
F
H
G
I
K
J z
·
z
∴ ·
L
N
M
O
Q
P
+
F
H
G
I
K
J
sec tan
i. e., sec tan
Thus sec
sec tan
Next let sec
sec tan
2
2
2
2
2
2
0
4
2
0
4
2
2 2
1 2 2
1
2
1
1
2
1
…(1)
But sec( / 4) = 2 tan ( / 4) =1, sec0 =1, tan 0 = 0
Thus sec
/ 4
n
n
n
x dx =
n
n
n
I
2
1
2
1
2
2
0
2
z
+
F
H
G
I
K
J
Illustrations
(i) To find sec
sec
sec tan
by using (1).
i. e., =
sec tan sec tan
But sec log(sec tan )
5
5
3
3
x dx
I x dx
x x
I
x x x x
I
I x dx = x x
z
>> ·
z
· +
+ +
R
S
T
U
V
W
· +
z
5 3
1
1
4
3
4
4
3
4 2
1
2
37
z
· + +
z
· + · + +
R
S
|
T
|
U
V
|
W
|
· + ·
z
·
z
·
+
F
H
G
I
K
J
sec
sec tan
sec tan log(sec tan ) +
(ii) To evaluate sec
The reduction formula for cosec can be established in
a similar way and the result is as follows.
cosec
cosec cot
5
3
4
x dx
x x
x x + x x c
x dx
>>I I
x dx
I x dx
x x
n
n
n
I
n
n
n
n
n
4
3
8
3
8
2
3
2
3
2
3
2
3
2
1
0
2
3
2
3
4
3
1
2
1
0
4
4
2
2
0
2
2
e j e j
Note :
Session – 3
Reduction formula for sin cos and sin cos
where and are positive integers.
0
/ 2
m n m n
x x dx x x dx
m n
z z
Let sin cos
sin sin cos say
I x x dx
x x x dx= uv dx
m, n
m n
m n
·
z
·
z z
1
We have
Here sin cos
cos
by putting cos
uv dx = u v dx v dx. u' dx
v dx = x dx
x
n
x = t
n
n
zz z z
z z
·
+
+1
1
Now (sin
cos cos
sin cos
i. e., =
sin cos
sin cos
I x
x
n
x
n
m x x dx
x x
n
m
n
x x dx
m, n
m
n n
m
m n
m n
·
+
F
H
G
I
K
J
+
z
+
+
+
z
+ +
− +
− +
1
1 1
2
1 1
2 2
1 1
1
1
1
1
=
x x
n
m
n
x x x dx
=
x x
n
m
n
x x x dx
m n
m n
m n
m n
+
+
+
z
+
+
+
z
− +
− +
sin cos
sin cos cos
sin cos
sin cos ( sin
2
2
1 1
2
1 1
2
1
1
1
1
1
1
1
=
x x
n
m
n
x x dx
m
n
x x dx
m n
m n m n
+
+
+
z
+
z
− +
sin cos
sin cos sin cos
1 1
2
1
1
1
1
1
38
I
x x
n
m
n
I
m
n
I
I
m
n n
x x + m I
I
m n
n n
x x + m I
I x x dx =
x x
m+n
m
m+n
I
m, n
m n
m n m, n
m, n
m n
m n
m, n
m n
m n
m, n
m n
m n
m n
·
+
+
+
+
+
+
L
N
M
O
Q
P
·
+
− −
+
+
L
N
M
O
Q
P
·
+
− −
∴ · − +
z
− +
− +
− +
− +
sin cos
i. e., sin cos
sin cos
sin cos
sin cos
1 1
2
1 1
2
1 1
2
1 1
2
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
…(1)
Note : If we decompose sin cos sin cos cos and integrate by parts taking
cos sin cos we can obtain
sin . cos
m n m n
n m
m, n
m n
m, n
x x= x x x
u = x, v = x x
I
x x
m+n
n
m+n
I
+ −
·
+
1
1
1 1
2
1
2
Note:
sin cos
m n
x x dx =
m m n n
m n m n m n
k
− − − −
+ + − + −
×
z
1 3 1 3
2 4
0
2
where k = π / 2 when m and n are even. This is known as Walli’s rule.
Illustrations
i sin cos
ii sin cos
iii sin cos
iv sin cos
4
5
5
6
5
0
2
7
0
2
6
0
2
8
4 2 3 1
9 7 5 3 1
8
315
6 4 2 4 2
12 10 8 6 4 2
1
120
5 3 1 4 2
11 9 7 5 3 1
8
693
7 5 3 1 5 3 1
14 12 10 8 6 4 2
x x dx =
x x dx =
x x dx =
x x dx =
× × × ×
·
z
× × × × ×
·
z
× × × × ×
·
z
× × × × × ×
2
5
4096
0
2
·
z
Session – 4
Note for problems : Basic properties of definite integrals are the prerequisites for
working problems on reduction formulae.
Example –1
Evaluate sin
Let sin
8
8
x x dx
I = x x dx
0
0
z
z
39
We have the property
sin sin
8 8
f x dx = f a x dx
I = x x dx = x x dx
a a
0 0
0 0
z
z
∴ − −
z
z
· −
z z
z z
x dx x x dx
I = x dx I I = x dx
I =
I =
sin sin
sin or 2 sin by a property.
Hence by reduction formula.
Thus
8 8
8 8
0 0
0
2
0
2
2
7
8
5
6
3
4
1
2 2
35
256
Example –2
Evaluate cos sin using reduction formula.
Let cos sin
sin 6 sin 3 cos 3 sin sin cos (
cos sin cos 3
4 2
/ 6
4 2
4
3 6
3 6
2 2 2 2
3 2 3
0
0
6
2
0
6
x x dx
I = x x dx
x = x x
I = x x x dx
/
/
z
>>
z
·
z
i. e., sin cos 3
Put 3
If If
sin cos =
4
3
sin cos
by reduction formula.
Thus
2 6
2 6 2 6
I = x x dx
x = y dx =dy
x y x = y =
I = y y
dy
y y dy
I =
I =
/
/ /
4 3
3
0 0 6 2
4
3
4
3
1 5 3 1
8 6 4 2 2
5
192
0
6
0
2
0
2
z
· ·
z z
× × ×
L
N
M
O
Q
P
40
Example – 3
Using reduction formula find the value of
Let
Put sin cos and varies from 0 to
cos cos
sin cos cos sin cos
Hence by reduction formula.
Thus
0
1
2 3
3 4
x x dx
>> I = x x dx
x = dx = d
x
I = d d
I =
I =
2 2 3 2
2 2 3 2
0
1
2 3 2 3 2
2
0
2
2
0
2
1
1
2
1
1 3 1
6 4 2 2
32
z
z
− · ·
z
·
z
× ×
Example – 4
Evaluate (i) (ii)
Let
Put sin sin cos
varies from 0 to
Also sin sin
i. e., = sin sin sin cos sin cos
0
2
0
2
2
x ax x dx
x dx
ax x
I x ax x dx
x = a dx = a d
ax x a a
a a a
a a
a
2 2
2
2
0
2
1
2 2
2
2 2 2 2 4
2 2 2 2 2
2
2
2
2 4
2
2 4 4
4 1 4 2
z
z
>> · −
z
− · −
− · ·
∴ ·
z
·
z
·
× × ×
·
·
I a a a d
a d
a
I
a
1
2
0
2
4
0
2
4
1
4
4 2 4
32
32
5 3 1
8 6 4 2 2
5
8
sin sin cos sin cos
sin cos
by reduction formula.
Thus
4
6 2
41
ii)
sin
sin cos
sin cos
sin by reduction formula.
Thus
4
4
I
a
a
a d
a d a
I
a
2
2
0
2
2
0
2
2
2
2
4
2
4
8 8
3
4
1
2 2
3
2
·
z
·
z
·
·
Example – 5
Evaluate
>> Let
Put tan sec
If If
Also (1+ ) tan sec sec
0
0
2
4 2 2 8
x
x
dx
I =
x
x
dx
x = dx = d
x = x = ,
x
4
2 4
4
2 4
2 4 4
1
1
0 0 2
1
+
z
+
z
· ∞ ·
· + · ·
z
·
z
·
z
·
z
× ×
· ·
I = d d
d d
I =
I =
/ /
/
tan
sec
sec
tan
sec
cos
sin
cos
sin cos
Hence by reduction formula.
Thus
4
8
2
4
6
6
4
4
4 2
0
2
0
2
0
2
0
2
3 1 1
6 4 2 2
32
Exercises :
Evaluate the following integrals
1
2
3
0
4
0
4
2 2
0
sin 4 cos 2
sin cos
4 3
7 2
x x dx
x x x dx
x
a x
dx
a
z
z
z
4
1
3
2 4
0
1
x
x
dx
+
z
5
1
6
2 9 2
0
x
x
dx
+
z
1 128 1155 2 16 315 3 3 16 4 1 24 5 1 7
4
a
42
Session –5
Tracing of Curves
Introduction :
This topic gives an insight to the process of finding the shape of a plane curve based on
its equation by examining certain features. Based on these features we can draw a rough
sketch of the curve. It is highly essential to known the shape of the curve to find its
area, length, surface area and volume of solids.
List of important points to be examined for tracing a cartesian curve f (x, y) = 0
1. Symmetry : If the given equation has even powers of x only then the curve is
symmetrical about the y axis and if the given equation has even powers of y only
then the curve is symmetrical about the x-axis.
If f (x, y) = f (y, x) then the curve is symmetrical about the line y = x. Also
if f (x, y) = f (−x, − y) then the curve is symmetrical about the
origin.
2. Special points on the curve : If f (0, 0) = 0 then the curve passes through the
origin. In such a case we can find the equations of tangents at the origin by
equating the groups of lowest degree terms in x and y to zero.
The points of intersection of the curve with the x-axis is got by putting y = 0 and
that with the y-axis is got by putting x = 0.
3. Asymptotes : Asymptote of a given curve is defined to be the tangent to the given
curve at infinity. In otherwords these are lines touching the curve at infinity.
Equating the coefficient of highest degree terms in x to zero we get asymptotes
parallel to the x-axis and equating the coefficient of highest degree terms in y to
zero we get asymptotes parallel to the y-axis.
4. Region of existence : Region of existence can be determined by finding out the set
of permissible (real) values x and y. The curve does not lie in the region whenever
x or y is imaginary.
By examining these features we can draw a rough sketch of the curve.
Note : In the case of a parametric curve : x = x (t) and y = y (t), we need to vary the
parameter t suitably to take a note of the variations in x and y so that the curve can be
drawn accordingly.
List of important points to be examined for tracing a polar curve f ( r, θ ) = 0
1. Symmetry : f ( r, θ ) = f ( r, −θ ) then the curve is symmetrical about the initial
line θ = 0 and θ · π .
If f ( r, θ ) = f ( r, π −θ ) then the curve is symmetrical about the line
θ · π / 2 (positive y-axis)
43
If f ( r, θ ) = f ( r, π / 2 −θ ) then the curve is symmetrical about the line
θ · π / 4 (the line y = x)
If f ( r, θ ) = f ( r, 3 π / 2 −θ ) then the curve is symmetrical about the line
θ · 3 π / 4 (the line y = −x)
If f ( r, θ ) = f (−r, θ ) then the curve is symmetrical about the pole. (origin)
2. Curve passing through the pole : If r = 0 gives a single value of θ say θ
1
between 0 and 2π then the curve passes through the pole once. θ = θ
1
is a
tangent to the curve at the pole.
If it gives two values then the curve passes through the pole twice.
3
0 0
Asymptote : If as then the line is an asymptote. r → ∞ → ·
4. Region of existence : If r is imaginary for θ ∈ (α, β ) i.e, α< θ < β then the
curve does
not exist in the region between θ · α and θ · β .
5. Special points : We can tabulate a set of values of r for convenient values of θ .
These give some specific points through which the curve passes.
By examining these features we can draw a rough sketch of the curve
Example – 1
Trace the curve y
2
(a−x) = x
3
, a > 0
>> We have y
2
(a−x) = x
3
. (This curve is known as cissoid)
We observe the following features of the curve.
1. Symmetry : The equation contain even powers of y.
the curve is symmetrical about x- axis.
2. Special points : The curve passes through (0, 0).
The given equation is ay
2
−xy
2
=x
3
.
The lowest degree term is ay
2
and ay
2
= 0
y = 0 which is the equation of x-axis. Hence
x-axis is the tangent to the curve at the origin.
Putting y = 0 we get x = 0 and vice versa. This means that the curve meets the x-axis and
y-axis at the origin.
3. Asymptotes : Equating the co-efficient of the highest degree term in y i.e. coefficient
of y
2
being a−x to zero we get x=a which is a line parallel to the y-axis. Hence x=a is an
asymptote. Also coefficient of the highest degree term in x is x
3
whose coefficient is 1
0. This implies that there is no asymptote parallel to the x-axis.
4. Region of existence : y
2
= x
3
/ (a−x)
∴ − y = x a x.
3
44
This is positive if x > 0, a −x > 0 or x < 0, a −x < 0 i.e. x > 0, x < a, x < 0, x > a. Since
a > 0 the second case is not possible. Hence y is real if x > 0 and x < a which implies that
the curve lies in the interval 0 < x < a. Further as x increases y also increases.
The shape of the curve is as follows
Note : Since the curve meets the coordinate axes at the origin only, the origin called a
‘cusp’ with x-axis as the common tangent.
Example – 2
Trace the curve y
2
(a−x) = x
2
(a + x), a > 0
>> y
2
(a−x) = x
2
(a + x) (This curve is known as ‘Strophoid’.)
We observe the following features of the curve.
1. Symmetry : The equation contain even powers of y.
the curve is symmetrical about x- axis.
2. Special points : The curve passes through the origin. The equation of the curve
can be put in the form
a (y
2
−x
2
) −x y
2
−x
3
= 0
Equating the lowest degree terms to zero we have a (y
2
−x
2
) = 0
Hence y = + x which are the tangents to the curve at the origin. Since there are two
tangents, the origin is called a ‘node’.
Next putting y =0 we get x
2
(a + x)= 0 x=0, x = − a.
The points are (0, 0) and (−a, 0)
Also putting x = 0 we get ay
2
= 0 or y = 0 and the point is (0, 0)
Hence we say that the curve intersects x-axis at (0,0) (−a, 0) and intersects the y-axis at
(0, 0) only.
3. Asymptotes : The co-efficient of the highest degree term in x being x
3
is −1 and
hence there is no asymptote parallel to the x-axis. Also the coefficient of the highest
45
degree in y being a−x, a−x = 0 gives x = a. Hence x = a is the only asymptote which is
a line parallel to the y-axis.
4
2
Region of existence : y = x a+x a x. −
When a+x < 0 and also when a−x < 0 y is imaginary.
Hence we can say that the curve lies between the lines x = −a and x = + a
The shape of the curve is as follows.
Example – 3
Trace the curve a y
2
= x
2
(a−x), a > 0
We observe the following features of the curve.
1. Symmetry : The equation contain even powers of y and hence the curve is
2. Special points : The curve passes through (0, 0). The equation of the curve is
ay
2
−ax
2
+ x
3
= 0. Equating the lowest degree terms to zero we have
a (y
2
−x
2
) = 0 y = + x
These are the equations of the tangents to the curve at the origin.
If y = 0 then x
2
(a − x) = 0, or x = 0, x = a and if x = 0, y = 0. Hence the curve meets
the x-axis at (a, 0) and meets the y-axis at (0,0) only.
3. Asymptotes : The coefficient of the highest degree terms in x and y are
respectively 1 and a. Since these are constants, there are no asymptotes.
4. Region of existence : y = x a x a −
y is positive if x > 0 and a−x > 0. x > 0 and x < a or 0 < x < a.
The curve lies between x = 0 and x = a.
46
The shape of the curve is as follows.
Session – 6
Example – 4
Trace the curve r = a sin 3 (Three leaved rose)
We observe the following features of the curve.
f (r, ) f (r,− ) the curve is not symmetrical about the initial line.
f (r, ) f (−r, ) the curve is not symmetrical about the pole.
f (r, ) f (r, − ) the curve is symmetrical about the line
r = 0 gives sin 3 0 n or n
Taking values for n = 0, 1, 2, …..6 we get the corresponding values of
and the curve passes through the pole for these values.
If 0 < r is positive and r = a if
If < r is positive and r = 0 if
If < r is negative and r = −a if
These observations implies that r increases from 0 to a as varies from 0 to
r decreases from a to 0 as varies from to
.
r increases numerically from 0 to a as varies from to
47
Further f(r −) = f(r,) implies that the curve is
symmetrical about the line = / 6 so that we conclude that there is a loop
between the lines and
Similarly we can examine the path of the curve as moves from to
and also from to 2.
Let us tabulate a set of values of r corresponding to some values of
r 0 a 0 −a 0 a 0 −a 0 a 0 −a
The curve is symmetrical about and 32
The shape of the curve is as follows.
Example – 5
Trace the curve r
2
= a
2
cos 2 (Lemniscate of Bernoulli)
>> We observe the following features of the curve.
f (r, ) = f (r, −) the curve is symmetrical about the initial line.
f (r, ) = f (−r, ) the curve is symmetrical about the pole.
r = 0 gives a
2
cos 2 = 0 i.e., cos2 = 0 2and
andare the tangents to
the curve at the pole.
When = 0, r
2
= a
2
or r = + a.
Hence the curve meets the initial line at the points (+a, 0) and (− a, 0).
48
Since the curve is symmetrical about the initial line it is composed of two loops. r is real
for and . Also
r doesnot tend to infinity for any and hence there are no asymptotes.
The shape of the curve is as follows.
Application to find area, length and volume of solids of revolution
The relevent formulae for finding these are as follows.
1. Area : The area (A) bounded by a curve y = f (x) , the x-axis and the ordinates x = a
and x = b is given by
A= y dx
x=a
b
z
The area (A) between the curves y = f (x) and y = g (x) between x = a and x = b is given
by
A= f x dx g x dx.
a
b
a
b
−
z z
The area (A) called the sectorial area, bounded by a polar curve r = f ( and the lines
and is given by
A= r d
1
2
2
1
2
z
2. Length : The length (s) of the arc of a curve between two specified points on it for
various types of the curves are given by the following formulae.
Such a process is called rectification and the entire length of the curve is called as the
perimeter of the curve.
(i) Cartesian curve y = f (x) or x = f (y)
s =
dy
dx
dx
dx
dy
dy
x=a
b
y=c
d
1 1
2 2
+
F
H
G
I
K
J z
+
F
H
G
I
K
J z
or
49
(ii) Parametric curve x = x (t), y = y (t)
s =
dx
dt
dy
dt
dt
t
t
F
H
G
I
K
J
+
F
H
G
I
K
J z
2 2
1
2
(iii) Polar curve r = f (
s = r
dr
d
d s = r
d
d r
dr
r r
r
2
2
2
2
1
2
1
2
1 +
F
H
G
I
K
J z
+
F
H
G
I
K
J z
· ·
or
3. Surface area : When a curve revolves about the x-axis a surface is generated and
the same is called a surface of revolution. If a curve is bounded by the ordinates
x = a and x = b revolves once completely about the x-axis, the area of the surface
(S) generated is given by
S = y ds= y
ds
dx
dx
x=a
b
a
b
2 2
z z
where
ds
dx
dy
dx
· +
F
H
G
I
K
J
1
2
Similarly the surface area of revolution about the y-axis is given by
S = x ds = x
ds
dy
dy,
c
d
y=c
d
2 2
z z
where
ds
dy
dx
dy
· +
F
H
G
I
K
J
1
2
In the case of a polar curve the surface area of revolution about the initial line is given by
S = r ds= r
ds
d
d 2 2
1
2
1
2
sin sin
z z
·
where
ds
d
r
dr
d
· +
F
H
G
I
K
J
2
2
4. Volume of revolution : The volume (V) of the solid generated by the revolution of the
curve y = f (x) between the ordinates x = a and x = b, about the x-axis is given by
V = y dx
x=a
b
2
z
Similarly if the axis of revolution is the y-axis, the volume of the solid is given by
V = x dy
y=c
d
2
z
50
Also in the case of a polar curve r = f () the volume (V) of the solid generated is given
by
V = r d
V = r d
2
3
2
3
2
3
3
z
z
·
sin (revolution about the initial line)
cos (revolution about the line )
Session – 7
Application related to standard curves
Example –1
The Astroid : Astroid is the curve represented by the equation
x
2/3
+ y
2/3
= a
2/3
Its parametric equation is x = a cos
3
and y = a sin
3
.
We shall find its shape first and then determine the associated area, perimeter, surface
area and the volume.
(a) Trace curve x
2/3
+ y
2/3
= a
2/3
>> Let us consider its parametric equation :
x = a cos
3
y = a sin
3
.
We tabulate x, y corresponding to certain angles of in the interval [0, 2
x a 0 −a 0 a
y 0 a 0 −a 0
From the table we conclude that the curve meets the x-axis at the points (a, 0) and (
− a, 0). Also it meets the y-axis at the points (0, a) and (0, − a). Since | cos | < | and
| sin | < 1, we have | x | < a and | y | < a. Hence we infer that the entire curve lies within
a circle of radius ‘a’ having origin as the centre.
51
Also we have from the cartesian equation of the curve,
f (x, y) = f (−x, y) ; f (x, y) = f (x, − y) ; f (x, y) = f (y, x)
Hence the curve is symmetrical about the coordinate axes and also about the line y = x.
Taking a note of the values of x and y as advances from one quadrant to the other
the shape of the curve is as follows.
(b) Find the area enclosed by astroid x
2/3
+ y
2/3
= a
2/3
>> Note : In any problem on applications we need to draw the curve first by briefly
examining the important features.
The curve astroid is symmetrical about the coordinate axes and hence the required area
(A) is equal to four times the area in the first quadrant.
i. e., A= y dx= y
dx
d
d
a a
4 4
0 0
z z
We have cos sin cos sin x =a y =a
dx
d
a
3 3 2
3
∴ · −
When x = 0 : a cos
3
= 0 or cos
3
= 0
x = 1 : a cos
3
= a or cos
3
= 1
∴ −
z
·
A= a a d 4 3
0 2
0
sin cos sin
3 2
·
z
12
2
0
2
a d sin cos
4 2
·
× ×
12
3 1 1
6 4 2 2
3 8
2
2
a
A a
by reduction formula.
Thus the area ( ) enclosed is sq. units.
(c) Find the parameter of the astroid x
2/3
+ y
2/3
= a
2/3
52
>> Since the curve is symmetrical about the coordinate axes, the perimeter (entire length)
of the curve is four times its length in the first quadrant.
∴ ·
F
H
G
I
K
J
+
F
H
G
I
K
J z
·
l
dx
d
dy
d
d 4
2 2
0
2
· +
z
4 9 9
2 2
0
2
a a d cos sin sin cos
4 2 4 2
· +
z
4 9
2
0
2
a d cos sin cos sin
2 2 2 2
·
z
4 3
0
2
a d cos sin
·
z
·
− L
N
M
O
Q
P
· − − · − − − ·
6
6
2
3 3 1 1 6
0
2
0
2
a d
a a a a
a
/
sin 2
cos 2
cos cos 0
Thus the perimeter of the curve is 6 units.
(d) Find the surface area of revolution of the astroid x
2/3
+ y
2/3
= a
2/3
>> Because of symmetry the required surface area is equal to twice the surface area by
the revolution of the first quadrant of the curve.
∴ · ×
z z
·
S y ds = y
ds
d
d
a
2 2 4
0
2
0
But cos sin
Hence sin cos sin
3
ds
d
dx
d
dy
d
a
S = a a d
·
F
H
G
I
K
J
+
F
H
G
I
K
J
·
z
2 2
0
2
3
4 3
= a d
= a
a
12
12
3
5 3 1
2
0
2
2
2
sin cos
by reduction formula.
Thus the required surface area = 12 / 5 sq. units.
4
z
× ×
(e) Find the volume of the solid generated by the revolution of the astroid x
2/3
+y
2/3
= a
2/3
>> Because of the symmetry the required volume (V) is equal to twice the volume of the
solid generated by the curve in the first quadrant about the x-axis.
53
∴ · ×
z z
· −
z
·
z
V y dx = y
dx
d
d
a a d
a d
a a
2 2
2 3
6
2 2
0 0
2
2
0
3
0
2
sin cos sin
sin cos
6 2
7 2
·
× × × ×
6
6 4 2 1
9 7 5 3 1
3
3
a
a
by reduction formula.
Thus the required volume of the solid is 32 / 105 cubic units.
Example –2
Cycloid : Cycloid is a curve generated by a point on the circumference of a circle which
rolls on a fixed straight line known as the base. Imagine a wheel rolling on a straight line
without slipping. A fixed point on the rim of the wheel traces the cycloid. The parametric
equation of the cycloid can be in the following forms :
i sin cos
ii sin cos
iii sin cos
iv sin cos
x =a y =a
x =a y =a
x =a y =a
x =a y =a
− −
+ −
− +
+ +
1
1
1
1
(a) Trace the cycloid :
x =a y =a − − sin cos 1
>> Let us tabulate x, y for certain values of in the interval [0, 2] where is
x 0 a
( )
a
a(3 ) 2a
y 0 a 2a A 0
From the table we can conclude that the curve intersects the x-axis at x = 0 and 2a.
Also, we have y = a (1cos) and since | cos | < 1 y is non negative. Hence the
curve lies above the x-axis.
Taking a note of the values of x and y as advances in the interval [0, 2] the shape
of the curve is as follows. It is called an arch of the curve.
54
(b) Find the area of an arch of the cycloid :
x =a y =a − − sin cos 1
>>
z
− −
z
·
Area
i. e., cos cos
A = y
dx
d
d
A= a a d
0
2
0
2
1 1
· −
z
·
z
a d
a d
2 2
0
2
2
0
2
1
4 2
cos
sin (
4
Put varies from 0 to
sin sin
4 4
2 2
8 8 2
2
0
2
0
2
· ∴ ·
z z
· ·
t d dt, t
A= a t dt = a t dt
t t
i. e., by reduction formula. A= a 16
3
4
1
2 2
2
Thus the area enclosed by an arch of the curve on its base is 3 a
2
sq.units.
(c) Find the length of an arch the cycloid :
x =a y =a − − sin cos 1
55
>>
F
H
G
I
K
J
+
F
H
G
I
K
J z
− +
z
· − + +
z
· − ·
z z
·
z
· −
L
N
M
O
Q
P
· − − · − − − ·
Length ( ) =
i. e., cos sin
cos cos sin
cos sin (
sin (
cos (
cos cos 0
=0
2
2
2
2
l
dx
d
dy
d
d
l = a a d
a d
a d a d
a d
a
a a
2 2
2 2 2
0
2
2
0
2
0
2
0
2
0
2
0
2
1
1 2
2 1 2 2 2
2 2
2 2
1 2
4 4 1 1 8
a
Thus the required length is 8a
(d) Find the surface area generated by the revolution of an arch of the cycloid
x = a (−sin ), y = a (1−cos ) about the x-axis.
>>
z
·
F
H
G
I
K
J
+
F
H
G
I
K
J
·
∴ −
z
Surface area (
But sin ( / 2)
cos sin ( / 2)
S = y
ds
d
d
ds
d
dx
d
dy
d
a
S = a a d
2
2
2 1 2
0
2
2 2
0
2
· ·
z z
· ∴ ·
·
z z
4 2 8
2 2
8 2 16 2
32
2
3
3
2
0
2
2
0
2
2
0
2
0
2
2
2
a d a d
t d dt t
S = a t . dt a t dt
i. e., S = a
a
t= t=
sin ( / 2) sin ( / 2)
Put and varies from 0 to
Hence sin sin
by applying reduction formula.
Thus the required surface area is 64 sq. units.
3 3
3 3
(e) Find the volume of the solid generated by the revolution of the cycloid
x =a y =a − − sin cos 1
56
>>
z
· − −
z
· ·
z z
· ·
z z
V = y
dx
d
d
a a d
a d a d
a d a d
2
0
2
2 2
0
2
3
3
3
0
2
0
2
3 3
0 0
1 1
2 2 8 2
8 2 2 16 2
cos cos
sin sin
sin sin
2 6
6 6
o t
Put and varies from 0 to
sin
by reduction formula.
Thus the required volume is 5 cubic units.
6
2 2 2
16 2
32
5
6
3
4
1
2 2
3
0
2
3
2 3
· ∴ ·
z
·
t d dt t
V = a t . dt
a ,
a
Session –8
Example –3
Cardiode : r = a (1+cos )
(a) Trace the curve r = a (1+cos )
>> r = a (1+cos )
We observe the following features of the curve.
(i) f (r, ) = f (r, −) and hence the curve is symmetrical about the initial line.
(ii) r = 0 when = and hence the curve passes through the pole. =
is a tangent to the curve at the pole.
(iii) Since | cos | < |, | r |< 2a and hence the curve lies with in the circle of radius
2a having its centre at the pole.
Let us tabulate r for certain angles of .
r 2a 3a/2 a a/2 0
57
It is evident thatas increases from 0 to , r decreases from 2a to 0. The shape
of the curve is as follows.
(b) Find the area of the cardioide r = a (1 + cos )
>> Since the curve is symmetrical about the initial line, the total area (A) is twice the area
above the initial line.
i. e., cos
cos ( )] cos ( )
2
2 2 4
A= r d a d
=a d a d
2
1
2
1
2 2 4 2
2 2
0 0
2
0
2
0
· +
z z
z
·
z
Put and varies from 0 to
cos cos
by reduction formula.
Thus the area enclosed is 3 / 2 sq. units.
4 2 4
2
2 2 2
4 2 8
8
3
4
1
2 2
2
0
2
0
2
2
· ∴ ·
z z
·
t d dt t
A= a t. dt = a t dt
a
a
t= t=
(c) Find the perimeter of cardioide r = a (1 + cos )
>> Perimeter (length) = 2 (length of the upper half of the curve)
i. e., = 2 where
Now cos sin cos
2
ds
d
ds
d
r
dr
d
ds
d
a a a
0
2
2
2 2 2
1 2 1
z
· +
F
H
G
I
K
J
· + + · +
·
z
·
L
N
M
O
Q
P
·
2 2
2 4
1 2
8
0
0
a
a d a a
a
cos (
perimeter = 2 cos( / 2)
sin( / 2)
Thus the perimeter of the curve is 8 units.
58
(d) Find the surface area of the revolution of the curve about the initial line.
>>
z
· +
F
H
G
I
K
J
·
∴ +
z
·
z
·
·
Surface area sin
But cos ( )
cos sin cos ( )
cos sin ( cos( ) cos( )
S = r
ds
d
d
ds
d
r
dr
d
a
S = a a d
a d
2
2 2
2 1 2 2
4 2 2 2 2 2 2
2
2
0
2 2
0
·
z
· ∴ ·
z
×
16 2 2
2 2 2
16 2
32
3 1
5 3
5
2 4
0
2 4
0
2
2
2
a d
t d dt t
S = a t t . dt
= a
a
t=
cos sin (
Put and varies from 0 to .
Hence cos sin
by reduction formula.
Thus the required surface area is 32 sq. units.
(e) Find the volume generated by the revolution of the curve r = a (1 + cos )
the initial line.
>> ·
z
+
z
+ ∴ −
· ·
∴ − ·
z z
·
L
N
M
O
Q
P
· − ·
V = r d a d
t = dt = d
t = t =
V =
a
t dt
a
t dt
a t a a
a
2
3
2
3
1
1
0 2 0
2
3
2
3
2
3 4
2
3
4 0
8
3
3
3
0
3 3
0
3
3
3
2
0
3
0
2
3 4
0
2
3 3
3
sin cos sin
Put cos sin
If and if
Thus the required volume is 8 cubic units.
Example – 4
Regarding sphere as the solid generated by revolving a circle about a diameter, find the
volume of a sphere of radius ‘a’
59
>> Let x
2
+ y
2
= a
2
be the equation of the circle and when the semicircle revolves about
the diameter (x-axis) a sphere of radius ‘a’ is generated.
Volume
Thus the required volume is cubic units.
V = y dx
V y dx a x dx
a x x a
a
a
a
a
a
a
a
2
2 2 2
2 3 3
3
3 4 3
4 3
z
·
z
· −
z
· − ·
− −
Example – 5
Find the volume generated by the parabola y
2
= 4 ax when revolved about the y-axis
between y = 0 and y = 2a
>> Required volume (
i. e.,
Thus the required volume is cubic units.
V = x dy
V y a dy
a
y
a
a
a
a
a
y=
a
a
2
2 2
0
2
2
5
0
2
2
5
5
2
3
4
16 5
80
2
32
80
2 5
z
·
z
·
L
N
M
O
Q
P
· ·
Example – 6
Find the perimeter of the curve r = a sin
3
(/ 3)
>> Let us tabulate r for certain angles of where r = a sin
3
(/ 3)
r 0 a/8
3 3 8 a
A
3 3 8 a
a/8 0
60
∴ +
z
+
F
H
G
I
K
J
· +
· +
· +
·
∴ + ·
·
perimeter
sin sin cos
sin sin cos
sin sin cos
sin
sin
6 2
6 4 2
4 2 2
4
2
l = r dr / d d
r
dr
d
a a.
a a
a
a
r dr / d a
2 2
0
3
2
2
2
2
2 2
2
2
2 2
3 3 3 3 1 3
3 3 3
3 3 3
3
3
o t
Hence, perimeter sin
Put and varies from 0 to
sin sin
i. e., by reduction formula.
Thus the required perimeter is 3 / 2 units.
2
2 2
l = a d
d d
l =a d a d
l = a .
a
3
3 3
3 3 2
6
1
2 2
0
3
0 0
2
z
· ∴ ·
∴ ·
z z
·
Exercises
1. Show that the length of the arc from = 0 to = along the curve
x = a (cos + sin ), y = = a (sin − cos ) is a
/ 2
2. Find the surface area of a sphere of radius a.
3. Show that the surface area generated by the revolution of the loops of the curve
r
2
= a
2
cos 2 about the initial line 2 a
2
(
2 2 −
).
4. Find the volume of the sphere of radius ‘a’.
5. Find the volume of the solid generated by the revolution of a loop of the curve
r
2
= a
2
cos 2 about the initial line.
2. 4a
2
4. 4a
3
/3 5.
2 15
3
a
61
Infinite Series
Session - 1
Introduction:
Infinite series is basically a summation of infinite number of terms. The summation will
be meaningful if there is no significant change in the sum as more and more terms gets
added up which is the concept of convergence. In this topic we discuss the aspect of
convergence, divergence and oscillation of an infinite series which are attributed as the
nature or the behaviour of an infinite series. In this connection we make use of well
established tests so as to draw a valid conclusion on the nature of an infinite series.
‘Progression’ is the pre-requisite for this topic
Definitions
If u
n
is a function of n defined for all integral values of n, an expression of the form
u + u + u +... + u infinite series
n 1 2 3
containing infinite number of terms is called an
usually denoted by or simply u u
n
n
n
·
∑ ∑
1
u
n
is called the n
th
term or the general term of the infinite series. The sum of the first n
terms of the series is denoted by s
n
i. e., s =u u u u
n n 1 2 3
+ + + +
Convergence, Divergence and Oscillation
A series is said to be if
lim
where is a finite quantity.
is said to be if
lim
A series is said to be if tends to more thanone limit as
u s l l
u s
u s n
n
n
n
n
n
n
n n
∑ ·
∑ · t∞
∑ → ∞
→∞
→∞
convergent
divergent
Oscillatory
Geometric series as an example for convergence, divergence and oscillation.
Let us consider the geometric series a + ar + ar
2
+ …to discuss its nature for various
values of r. The sum to n terms of this series is given by
62
s
a r
r
r
s
a r
r
r
r| < r n
s
a
r
a
r
n
n
n
n
n
n
n
·
<
·
>
→ → ∞
·
− ·
− →∞
1
1
1 1
1
1
1 2
1 0
1
1 0
1
if
and if
Now if as and from (1), we have
lim
which is a finite quantity.
Hence we conclude that the geometric series is convergent for | r |< 1.
Next if from (2)
lim lim
If as
r s
a r
r
r r n
n
n
n
n
n
> ·
· ∞
> → ∞ → ∞
→∞ →∞
1
1
1
1
Hence we conclude that the geometric series is divergent if r > 1
Also if r = 1 the series becomes a + a + a +…
s
n
= a + a + a… to n terms = na
∴ · ∞
→∞ →∞ n
n
n
s n a =
lim lim
Hence the series is divergent if r = 1.
When r = −1, the series become a −a + a … where the sum of the first n terms s
n
is 0 or
' a' n s a s
n
n n
according as is even or odd and we can say that
lim
= 0 or . tends to more
→∞
than one limit as n ∞
Hence we conclude that the series is oscillatory if r = −1.
Further if r <−1, r
n
∞ or −
∞ according as n is even or odd. s
n
tends to more
than one limt as n ∞
Hence we conclude that the series is oscillatory if r < −1.
Note : An infinite series of positive terms is either convergent or divergent.
Property : If is convergent, then
lim
but not conversely. u u = ,
n
n
n
→∞
0
An example to show that the converse is not true.
63
Consider the series
Here
i. e.,
n+ n+
u n n+
n+ n+ n+ n+
n+ n+
u
n+ n
n+ n+ n+ n+
n=
n
n
2 1
2 1
2 1 2 1
2 1
2 1
2 1
1
2 1
1
− ∑
· + − ·
− +
+
·
− +
+
·
+
e je j
e j
e j
Clearly
lim
Next
i. e.,
i. e.,
lim lim
n
n
n n n
n
n
n
n
n
u
s u u u u u
s n+ n n+ n+
s n+
s n+
→∞
→∞ →∞
·
· + + + +
· − + − + − + + − + −
·− +
∴ · − + · ∞
0
3 2 4 3 5 4 1 2 1
2 2
2 2
1 2 3 1
e j e j e j e j e j
Hence we conclude that u
n
is divergent.
This example shows that the converse of the property is not true.
Remarks :
1. The contrapositive of the statement of this theorem provides a fundamental test
for the divergence of an infinite series which can be stated as follows :
If u
n
does not tend to zero as n tends to infinity, then u
n
is not convergent.
Thus
lim
is divergent.
n
n n
u u
→∞
≠ ⇒ ∑ 0
2. Addition or deletion or multiplication of each term of infinite series by a nonzero
constant doesnot alter the nature of infinite series.
Tests for convergence for series of positive terms
Comparison Test
If and be any two series of positive terms such that
lim
is a non zero finite quantity. Then and behave alike.
That is and both will converge or both will diverge.
u v
u
v
u v
u v
n n
n
n
n
n n
n n
∑ ∑
∑ ∑
∑ ∑
→∞
64
Remark : In this test the given series u
n
is to be compared with another series v
n
whose nature is to be known to us. We usually compare with the harmonic series
1
1 n
p p p
p
n
( series) which is convergent for >1 and divergent for 1 − ≤ ∑
·
(The proof is established later)
Working procedure for problems
• Given an infinite series, we first write the n
th
term (general term) u
n
.
[The n
th
term of the A.P; a, a+ d, a+2 d, … is a+(n− 1)d ]
• Comparison test is usually applicable when u
n
involves expression of n in the
numerator and the denominator like (n+1), (2n+1), (n
2
+1),
n
, n
3
+ 4 etc.
• ·
We choose where and are the highest power in the denominator v
n
p q n
n
p q
1
and numerator of u
n
respectively.
• − −
+
+
However if is of the form or irrational factors connected
by a negative sign), we must rationalize by multiplying with or
as the case may be and then chose as stated.
u f n g n f n g n
f n g n
f n g n v
n
n
Order Test
We say that
1
is the order of . If ( ) = (say) then we say that is of the order
1
and conclude that is converngent if and divergent if .
Sometimes this is referred to as the order test.
n
u p q k u
n
u k k
p q
n n
k
n
∑ > ≤ 1 1
Session – 2
Example – 1
65
Examine the series
1
1.3.5
for convegence.
>> Numerators are 1, 2, 3, .... Hence the general term is .
In the denominator we have,
First factors : 1, 3, 5, ... term = 1+
Second factors : 3, 5, 7, ... term = 3 +
Third factors : 5, 7, 9 ... term = 5 +
Thus term of the given series is given by
th
th
th
th
+ + +
∴ − · −
∴ − · +
∴ − · +
2
3 5 7
3
5 7 9
1 2 2 1
1 2 2 1
1 2 2 3
n
n n n
n n n
n n n
n
u
n
n n n
v
n
n n
u
v
n
n n n
n
u
v
n
n n n n
u v
n n
n
n
n
n
n
n
n n
·
− + +
· ·
·
− + +
×
·
− + +
·
∑ ∑
→∞ →∞
2 1 2 1 2 3
1
2 1 2 1 2 3 1
2 1 2 1 2 3
1
8
3 2
2
3
3
Choose
Thus
Now
lim lim
By comparision test, and behave alike.
Here is of the form is convergent.
Hence is also convergent.
v
n n
p
u
n
p
n
· · > ∑ ∑ ∑
1 1
2 1
2
Example – 2
Find the nature of the series
1
1
1 2
1 2
1 2 3
1 2 3
2 2 2 2 2 2
+
+
+
+
+ +
+ +
+
>> It should be observed that there will be n terms in the numerator and the
denominator of the general term u
n
.
66
∴ ·
+ + + +
+ + + +
·
·
+ + +
·
+
+ +
·
+
·
·
+
·
+
F
H
G
I
K
J
·
∑ ∑
· · · ∑ ∑ ∑
→∞ →∞
→∞
u
n
n
n
n
u
n n n n n
u
n n
n n n n
v
n
u
v n
n
n
u v
v
n n
p
n
n
n n
n
n
n
n
n
n n
n
1 2 3
1 2 3
1
2
1 2 1
6
1
2
6
1 2 1
3
2 1
1
3
2 1 1
3
2 1
3
2
1 1
1
2 2 2 2 2
1
i. e.,
i. e., Choose
Now
lim lim
lim
By comparison test, and behave alike.
But is divergent.
Hence is also divergent. u
n
Example – 3
Test for convegence
By data Choose
Now
lim lim
lim lim
By comparison test and behave alike.
But = ( =1 / 2 <1) is divergent.
Hence is also divergent.
Alternative version of the problem.
1
1
1
1
1
1
1 1
1 1 1 1 1
1
2
1
1
1 2
n n
u
n n
v
n
u
v n n
n
n
n n n
n
n n
u v
v
n
p
u
n
n n
n
n
n
n
n n
n n
n
/
n
+ +
>> ·
+ +
·
·
+ +
·
+ +
·
+ +
·
∑ ∑
∑ ∑
·
→∞ →∞
→∞ →∞
Note :
67
Discuss the convergence of the infinite series
>> and on rationalizing we have
i. e., which is the same as the example discussed.
n n
u n n
u n n
n n
n n
n n
n n
u
n n
n
n
n
n
+ − ∑
· + −
· + −
+ +
+ +
·
+ −
+ +
·
+ +
·
1
1
1
1
1
1
1
1
1
1
e j
e j
e j
Example – 4
Test for convergence
By data,
We have an elementary formula
or
If we take and assumes the form
by above.
i. e., Choose
n n
u n n
a b a b a +ab+b a b
a b
a +ab+b
a = n b =n, u a b
u
n n
n n n+n
u
n n n+n
v
n
n
n
n
n
n n
3 3
1
3 3
3 3 2 2
3 3
2 2
3 3
3 3
3 2 3 3 3 2
3 2 3 3 3 2
1
1
1
1
1 1
1
1 1
1
+ −
L
N
M
O
Q
P
>> · + −
L
N
M
O
Q
P
− · − − ·
+ −
∴ ·
+ −
+ + +
·
+ + +
·
·
2
Now
lim lim
lim
lim
lim
By comparison test and behave alike.
Hence is also convergent.
n
n
n
n
n
n
n
n n
n
u
v
n n n+n
n
n
n n n n n+n
n
n n n n n
n
n n n
u v
u
→∞ →∞
→∞
→∞
→∞
·
+ + +
×
·
+ + +
·
+ + + +
·
+ + + +
L
N
M
O
Q
P
·
+ +
·
∑ ∑
1
1 1
1
1 1 1 1
1 1 1 1
1 1 1 1 1
1
1 1 1
1
3
3 2 3 3 3 2
2
2
3 3
2 3
3 3 3 2
2
2 3 2 3 2 3 3 2
2
2 3 2 3 3 3
68
Exercises :
Find the nature of the following series
1
1
1 2 3
4
2 5 7
7
3 811
2
1
1
1 2
1 2
1 2 3
1 2 3
3
1
3 5
1
4 6
1
5 7
4 3
2
3
2 2
3 3
2 2 2
3 3 3
4 2
+ + +
+
+
+
+
+ +
+ +
+
+
+
+
+
+
+ −
L
N
M
O
Q
P
∑ n n
1. Convergent 2. Divergent 3. Divergent 4. Convergent
Session – 3
D’Alembert’s ratio test and Raabe’s test
If u
n
is a series of positive terms and if
n
n
n
u
u
l
→∞
+
·
lim
(finite qty)
1
then u
n
is convergent if l < 1, divergent if l > 1 and the test fails if l = 1.
When the ratio test fails, we can apply the comparison test (order test) if possible or
Raabe’s test stated as follows.
If u
n
is a series of positive terms and if
n
n
n
n
u
u
l
→∞
+
F
H
G
I
K
J
·
lim
(finite qty)
1
1
then u
n
is convergent if l > 1, divergent if l < 1 and the test fails if l = 1.
Remark :
Usually we try ratio test in the following cases.
(i) n
th
term of the series is not of the order 1/n
k
.
(ii) n
th
term of the series involves variables like x
n
, x
2n
etc.,
69
(iii) n
th
term of the series involves n !, (n+1) !, (n !)
2
etc.
(iv) The number of factors in the numerator and denominator increase steadily from term
to term like
1
3
(carry over series) + + +
12
3 5
1 2 3
3 5 7
Working procedure for problems
We first write the general term u
n
for the given series and find u
n+1
by replacing n
by (n+1)
• → ∞
+
We find the ratio simplify and find its limit as to decide the nature of the series.
u
u
n
n
n
1
When the ratio test fails we can try to apply the order test in the first
instance or
apply Raabe’s test.
Raabe’s test becomes inevitable when the ratio test fails in the case of carry over
series.
Example – 1
x
.
x x
u
x
n n
u
x
n n
u
u
x
n n
n n
x
n
n
x
u
u
n
n
x =
n
x = x
n
n
n
n
n
n
n
n
n
n
n
n n
12 2 3 3 4
1
1 2
1 2
1
2
2
1
1 2
2 3
1
1
1
1
1
+ + +
>> ·
+
∴ ·
+ +
·
+ +
+
·
+
∴ ·
+ +
+
+
+
+
→∞
+
→∞ →∞
Now
lim lim lim
>
R
S
T
·
·
+
·
+
·
+
by D' Alembert' s ratio test is
divergent if
convergent if
and the test fails if
But when
u
x
x<
x
x = u
n n n n n n
n
n
n
1
1
1
1
1
1
1
1
1
2
70
u n p u
u x x
n n
n
is of order 1 / ( = 2 >1) and hence is convergent.
Hence we conclude that
is convergent if and divergent if
2
1 1
≤ ∑ >
Example – 2
Find the nature of the series
1
2
13
2 4
13 5
2 4 6
13 5 2 1
2 4 6 2
1
13 5 2 1 1
2 4 6 2 1
13 5 2 1
2 4 6 2 2
13 5 2 1 2 1
2 4 6 2 2 2
2 3
1
1 1
1
x + x x
u =
n
n
x
u
n
n
x
n
n
x
u
n n
n n
x
n
n
n
n n
n
+ +
>>
·
+ −
+
·
+
+
·
− +
+
+
+ +
+
i. e.,
n+1
2
From (1) and (2) we have,
u
u
n n
n n
x
n
n x
u
u
n
n
x ...
n
n
n
n
n
n
+ +
+
·
− +
+
×
·
+
+
1 1
1
13 5 2 1 2 1
2 4 6 2 2 2
2 4 6 2
13 5 2 1
2 1
2 2
3
i. e.,
∴ ·
+
+
+
+
>
< R
S
T
·
→∞
+
→∞ →∞ n
n
n
n n
n
u
u
n
n
x=
n n
n n
x = x
u
x
x
x
x
lim lim lim
Hence by ratio test
is
divergent if
convergent if
and the test fails if =1
Now when we shall apply Raabe' s test.
1
2 1
2 2
2 1
2 2
1
1
1
When from (3)
Now
lim
lim lim
lim
is divergent by Raabe' s test Hence we conclude that the given series
is convergent if and divergent if
x
u
u
n
n
u
u
n
n
n
u
u
n
n
n
n
n n
n
n
n n
u
u x < x
n
n
n
n
n
n
n
n n
n
n
n
· ·
+
+
∴ ·
+
+
F
H
G
I
K
J
·
+
+
F
H
G
I
K
J
·
+ − −
+
F
H
G
I
K
J
·
+
· <
∴∑
∑ ≥
+
+
→∞
+
→∞ →∞
→∞
1
2 1
2 2
2 2
2 1
1
2 2
2 1
1
2 2 2 1
2 1
2 1
1
2
1
1 1
1
1
1
71
Example –3
Test for convergence
By data
i. e.,
3 6 9 3
4 710 3 1
5
3 2
3 6 9 3
4 710 3 1
5
3 2
3 6 9 3 1
4 710 3 1 1
5
3 1 2
3 6 9 3 3 3
4 710 3 1 3 4
5
3 5
1
1
1
1
1
n
n n+
u
n
n n+
u
n
n n +
u
n n
n n n
n
n
n
n
n
n
n
n
+
>> ·
+
∴ ·
+
+ + +
·
+
+ + +
·
+
+
+
+
Now
i. e.,
lim lim
lim
u
u
n n
n n n
n
n
n
u
u
n n
n n
u
u
n n
n n
n
n
n
n
n
n
n
n
n
n
n
n
+
+
+
→∞
+
→∞
→∞
·
+
+ + +
×
+ +
·
+ +
+ +
∴ ·
+ +
+ +
·
+
1
1
1
1
3 6 9 3 3 3
4 710 3 1 3 4
5
3 5
4 710 3 1
3 6 9 3
3 2
5
3 3 3 2
3 4 3 5
5
3 3 3 2
3 4 3 5
5
3 3
3 2
3 4 3 5
5 5 1
+
+ +
· >
n
n n
u
n
Hence by the ratio test is divergent.
Example – 4
72
Find the nature of the series
By data,
i. e.,
From (1) and (2) we have,
4 710 3 1
4 710 3 1
1
4 710 3 1 1
1
4 710 3 4
1
4 710 3 1 3 4
1
2
4
1
1
1 1
1
1
1
n
n
x
u
n
n
x
u
n
n
x
n
n
x
u
n n
n n
x
u
u
n
n
n
n
n
n n
n
n
n
n
+
>> ·
+
·
+ +
+
·
+
+
·
+ +
+
·
·
+
+ +
+
+
+
710 3 1 3 4
1 4 710 3 1
3 4
1
3 4
1
3 4
1 1
3
1
1
1
n n
n n
x
n
n x
u
u
n
n
x
u
u
n
n
x=
n
n
x = x
n
n
n
n
n
n
n
n n
+ +
+
×
+
·
+
+
∴ ·
+
+
+
+
+
+
→∞
+
→∞ →∞
i. e.,
lim lim lim
Hence by ratio test is
divergent if 3 or
convergent if 3 or
and the test fails if 3 or =1 / 3
If we have =
1
3
or
Now
lim lim lim lim
Hence is diverge
u
x x
x x
x = x
x =
u
u
n
n
u
u
n
n
n
u
u
n
n
n
n
n n
u
n
n
n
n
n
n
n
n
n n n
n
> >
< < R
S
T
+
+
·
+
+
F
H
G
I
K
J
·
+
+
F
H
G
I
K
J
·
+
F
H
G
I
K
J
·
+
·
<
+
+
→∞
+
→∞ →∞ →∞
1 1 3
1 1 3
1
1 3
3 4
1
3 3
3 4
1
3 3
3 4
1
1
3 4
1
3 4
1
3
1
1
1
1
nt when 3 =1 by Raabe' s test.
Thus is convergent if <1 / 3 and divergent if 1 / 3.
x
u x x
n
∑ ≥
Example – 5
Test for convergence of the infinite series
1
2
2
3
3
4
4
2 3 4
+ + + +
>> The first term of the given series can be written as 1!/1
1
so that we have,
u
n
n
u
n
n
n n
n
n
n
n
n
n
n n n
· ∴ ·
+
+
·
+
+
·
+
+
+ +
1
1 1
1
1
1
1 1
73
Now
lim lim
Hence by ratio test is convergent.
u
u
n
n
n
n
n
n n
u
u n e
u
n
n
n
n n
n n
n
n
n
n
n
n
+
→∞
+
→∞
·
+
·
+
∴ ·
+
· <
1
1
1 1 1
1
1 1
1
1
Exercises
1 1
3
1
3
2
3
3
2
2
3 4
3
4 5
4
5 6
3
1
3
1 2
3 5
12 3
3 5 7
4
3
4 5
3 6
4 7 8
3 6 9
4 710 11
5
3 5 7 2 1
3 6 9 3
2 3
2 3
2 3
2 3
1
+ + + +
+ +
+ +
+ + +
+
·
x + x x
x + x x
x x x
n
n
x
n
n
1. Convergent
2. Convergent if x < 1, divergent if x > 1
1. Convergent if x < 2, divergent if x > 2
2. Convergent if x < 1, divergent if x > 1
3. Convergent if x < 3/2, divergent if x > 3/2
Session - 4
Cauchy’s root test
If u
n
is a series of positive terms and if
n
n
n
u l
→∞
·
lim
(finite)
1
then the series is convergent if l < 1, divergent if l > 1 and the test fails if l = 1.
When the test fails we can try the order test or the fundamental test.
Note : 1. We try to apply this test if u
n
is of the form [f (n)]
g (n)
2. The following standard limits will be useful
74
(i)
lim
ii
lim
iii
lim
n
n
n
n
n
n
x
n
n
e;
x
n
e
→∞ →∞ →∞
· +
F
H
G
I
K
J
· +
F
H
G
I
K
J
·
1
1 1
1
1
Example – 1
Test for convergence
by data.
1
1
1
1
1
1
1
1
1
1
3 2
3 2
3 2
1 2
1
1
1
1
+
L
N
M
O
Q
P
>> · +
L
N
M
O
Q
P
∴ · +
L
N
M
O
Q
P
R
S
|
T
|
U
V
|
W
|
· +
L
N
M
O
Q
P
· +
L
N
M
O
Q
P
·
− −
n
u
n
u
n
u
n n
n
n
n
n
n
n
n
n
n
n
n n
Now
lim lim
lim
as also
Hence by Cauchy' s root test is convergent.
n
n
n
n
n
n
n
n
u
n
n
e
n , n
u
→∞ →∞
→∞
· +
L
N
M
O
Q
P
·
+
L
N
M
O
Q
P
· <
→ ∞ → ∞
1
1
1
1
1
1
1
1
Example – 2
Discuss the convergence of
by data.
( )
lim
( )
lim lim lim
1/
1/
n x
n
u =
n x
n
u
n x
n
n x
n
n x
n n
u
n x
n n
n n x
n n
n x
n
n n
n
n
n
n n
n
n
n
n
n
n n
n+ n n n
n
n
n
n
n
n
n
n
n
+
>>
+
∴ ·
+
·
+
·
+
·
+
·
+
·
+
+
·
+
+
→∞ →∞ →∞ →∞
1
1
1
1 1
1 1 1 1 1
1
1
1
1
1
1 1 1 1 1
1 1 1
o t
75
n
n
n n
n
n
n
n
n
n
n
n
n
n n
n
n
u
n x
n
x
x, n
u
x>
x
x
x = , u
n
n
u n n n u
u x < x
→∞
→∞
→∞
→∞
+
+
·
+
·
+
· ·
< R
S
T
·
·
+
∑ ≥
lim
( )
lim
lim
lim
Hence by Cauchy' s root test is
divergent if 1
convergent if
and the test fails if
When
is of order / =1 / and hence is divergent.
Thus is convergent if 1 and divergent if 1.
1/
1 1
1 0
1
1
1
1
1
1
1
1
1
1
Example – 3
Find the nature of the series
1
2
3
3
4
0
1
2
1
2
1
2
1 1
1 2
1
1
2
2
1
1
1
1
+
F
H
G
I
K
J
+
>> ·
+
+
F
H
G
I
K
J
∴ ·
+
+
F
H
G
I
K
J
R
S
|
T
|
U
V
|
W
|
·
+
+
F
H
G
I
K
J
·
+
+
>
< R
S
T
→∞ →∞
x + x x >
u
n
n
x
u
n
n
x
n
n
x
u
n
n
x = x
u
x
x
n
n
n
n
n
n
n
n n
n
n
n
n
n
Omitting the first term,
Now
lim lim
Hence by Cauchy' s root test is
divergent if
convergent if
and the test fails if =1 x
76
But when
lim lim
using
lim
Since
lim
is divergent when
Thus is convergent if <1 and divergent if 1.
x = u
n
n
n
n
n
n
u
n
n
e
e e
x
n
e
u
e
u x
u x
n
n
n
n n
n
n
n
n
n
n
n
x
n
n n
n
1
1
2
1
1
2
1 1
1 2
1 1
1 2
1
1
1
0 1
2
·
+
+
F
H
G
I
K
J
·
+
+
F
H
G
I
K
J
·
+
+
L
N
M
O
Q
P
·
+
+
· · +
F
H
G
I
K
J
·
· ≠ ∑ ·
∑ ≥
→∞ →∞ →∞
→∞
Example – 4
Test for convergence
1
2
The first term of the series can be put in the form 1 / 1 so that we have,
lim lim lim
Hence by Cauchy' s root test is convergent.
1
+ + +
>>
·
∴ ·
F
H
G
I
K
J
· · <
→∞ →∞ →∞
1
2
1
3
1
1 1
0 1
2 2
1
1
u
n
u
n n
u
n
n
n
n
n
n
n
n
n
n
Exercises :
Tests for convergence the following series
1 1
3
2
1 3
4
3
4
3
7
5
10
7
13
9
4 1
2 3 4
0
2
2
2 3 4
2
2
3
3
+
F
H
G
I
K
J ∑
+ F
H
G
I
K
J
F
H
G
I
K
J ∑
+
F
H
G
I
K
J
+
F
H
G
I
K
J
+
F
H
G
I
K
J
+
+ + + + >
n
n
n
x x x
x
n
n n
77
(1) Convergent (2) Divergent (3) Divergent (4) Convergent
Cauchy’s integral test
If is a series of positive terms and if = ( ) be such that u u f x
n x
(i) ( ) is continuous in 1 <
(ii) ( ) decreases as increases [ ( ) is monotonically decreasing] then the series is
convergent or divergent according as the integral ( ) is finite or infinite.
The condition ( ) < 0 ensures that ( ) decreases as increases where ( )
is the derivative of ( ).
1
f x x <
f x x f x u
f x dx
f ' x f x x f ' x
f x
n
z
Note :
Example :
Apply Cauchy’s integral test to discuss the nature of the harmonic series (p- series)
1
1
1
0
1
1
1
n
u
n
f n
f x
x
x f ' x px
p
x
f x
p
n
n
p
p
p p
p
·
− − −
+
>> · ·
· · ∴ · − ·
<
By data
Hence is decreasing.
Now if
: Let or ( ) > 0
Then as
which is finite.
Thus is convergent if >1.
1 1 1
1
f x dx =
x
dx = x dx =
x
p
p
p > p
x
x
x p
f x dx =
x
p p p
u p
p
p
p
p
p
p
n
∞ ∞
z z
L
N
M
O
Q
P
z
· → → ∞ − >
z
L
N
M
O
Q
P
· −
·
1
1
1
1 1
1
0 1 0
1
0
1
1
1
1
1
1
1
1
1
1
Case - i
78
Case - ii : Let or 1 > i. e., (1 ) > 0
As above
But as (1 ) > 0
Hence
Thus is divergent if <1.
1
1
p < p p
f x dx =
x
p
x x . p
f x dx =
p
u p
p
p
n
1
1
1
1
1
1
1
z
L
N
M
O
Q
P
→ ∞ → ∞ −
z
∞ −
· ∞
Case - iii
Remark :
: Let
log
Thus is divergent if =1.
Conclusion : is convergent if >1 and divergent if 1.
We have used the behaviour of this series while solving problems on comparison test
to discuss the nature of a givenseries in comparison with a series of the form
1 1
1
1
p =
f x dx =
x
dx = x
u p
n
p p
n
n
p
n=
p
n=
1
1
0
1
1
1
1
z
· ∞ − · ∞
z
≤ ∑
Session – 5
Alternating series
If , , , ... are all positive, then a series of the form is called an
alternating series represented by
1 2 3 4 1 2 3 4
u u u u u u + u u + ...
u
n
n
n
− −
− ∑
·
1
1
1
Leibnitz test for convergence of an alternating series
If , , , ... are all positive and if
1 2 3 4
u u u u
(i) , for all i. e., > ...
1 2 3 4
u >u n u >u >u u
n n+1
(ii)
lim
n
n
u
→∞
· 0
then the alternating series + +... is convergent.
1 2 3 4
u u u u − −
Remark : If
lim
then the series is oscillatory.
n
n
u
→∞
≠ 0
79
Absolute convergence and conditional convergence
The alternating series + +... is said to be absolutely convergent if the series of
positive terms + +... is convergent.
1 2 3 4
1 2 3 4
u u u u
u u u u
− −
+ +
If the given alternating series is convergent and the absolute series (series of positive
terms) is divergent then the alternating series is said to be conditionally convergent.
Property : Every absolutely convergent series is convergent but the converse is not true.
Generalised D’Alembert’s test
If is a general series and if
lim
=| | then is absolutely convergent if | |
not convergent if | | and the test fails if | | =1.
u
u
u
l u l
l l
n
n
n
n
n
∑ ∑ <
>
→∞
+1
1
1
Example – 1
Test for convergence the series, 1
Here =
1
We shall apply Leibnitz test.
Now
1 1 1
i. e.,
− + − +
>>
− · −
+
·
+ −
+
·
+
>
− > ⇒ >
+
+ +
1
2
1
3
1
4
1 1
1
1
0
0
1
1 1
u
n
u u
n n
n n
n n n n
u u u u
n
n n
n n n n
Also
lim lim
1
n
n
n
u
n →∞ →∞
· · 0
Hence the given series is convergent by Leibnitz test.
Remark : We have mentioned a property that “every absolutely convergent series is
convergent”, but the converse is not true. The series 1 serves as an − + − +
1
2
1
3
1
4
example for the converse of the statement not being true. That is a convergent series not
being absolutely convergent.
80
The alternating series is convergent by Leibnitz test, but the absolute series 1
being is divergent . ( series where
+ + +
·
1
2
1
3
1
1
1
n
p - p =
n
Thus the alternating series is not absolutely convergent.
Example –2
Test the series + +... for
(a) convergence (b) absolute convergence (c) conditional convergence
>> We have
Now
since ( +1) >
i. e.,
Also
lim lim
1
1
2 2
1
3 3
1
4 4
1 1 1
1
1 1
1
1
1
0
0
1
0
3 2
1
3 2
1
3 2 3 2
3 2 3 2
3 2
1 1
− −
· · ∴ ·
+
− · −
+
·
+ −
+
>
− > ⇒ >
· ·
+
+
+ +
→∞ →∞
u
n n n
u
n
u u
n n
n n
n n
n n.
u u u u
u
n n
n n
n n
n n n n
n
n
n
Hence the given series is convergent by Leibnitz test.
The absolute series is given by + + +...
We have
is a harmonic series of the type is convergent.
1
1
2 2
1
3 3
1
4 4
1 1
1 1 3 2 1
3 2
3 2
1 1
+
· ·
∑ ∑ · >
·
·
u
n n n
n n p
n
n
p
n
Thus the alternating series is absolutely convergent and it is not conditionally
convergent.
Example – 3
81
Find the nature of the series
3
4
5
7
7
10
9
13
>> We have
Now
i. e.,
Also
lim lim
i. e.,
lim
n n n
− + − +
·
+
+
∴ ·
+ +
+ +
·
+
+
− ·
+
+
+
+
·
+ +
>
>
·
+
+
· ≠
+
+
+
→∞ →∞ →∞
u
n
n
u
n
n
n
n
u u
n
n
n
n n n
u u
u
/ n
n
u
n n
n n
n n
n n
2 1
3 1
2 1 1
3 1 1
2 3
3 4
2 1
3 1
2 3
3 4
1
3 1 3 4
0
2 1
3 1
2
3
0
1
1
1
Hence the given series is oscillatory.
Example – 4
Show that the series
1
1
2
1
3
1
4
1
5
1
6
1
7
1
8
1
1
2
1
3
1
4
1
5
1
6
1
7
1
8
1
1
2 2 2 2 2 2 2
2 2 2 2 2 2 2
2
2
+ − − + + − − +
+ + + + + + + +
·
is convergent.
>> Consider the absolute series,
The general term of this series is given by
is a harmonic series ( = 2 >1) and it is convergent.
u
n
n
p
n
Hence we conclude that the given series is absolutely convergent.
But every absolutely convergent series is convergent.
Hence the given alternating series is convergent.
Remark : The problem can also be done by applying Leibnitz test.
Example – 5
82
Discuss the nature of the series
x
x x x
u
x
n
u
x
n
n
n n
n
n n
− + − +
>>
·
∴ ·
+
+
+
2 3 4
1
1
1
2 3 4
1 1
1
The general term of the series is given by
We shall apply generalised D' Alembert' s test.
Now
lim lim
lim
n
n
n
n
n n
n n
n
u
u
x
n
n
x
n
n n
x x
→∞
+
→∞
+
→∞
·
+ −
· −
+
·
1
1
1
1
1 1
1
1 1
Hence the series is absolutely convergent if | x | < 1, not convergent if | x | > 1 and the test
fails if | x | = 1.
Since absolute convergence implies convergence, the given series is convergent if |x|<1,
that is –1< x <1,not convergent when | x | > 1 and the test fails if | x | = 1.
If | x | = 1 then x = 1 or –1.
If =1, the series becomes
We have = = Clearly
Also
lim
=
lim
Hence the series is convergent by Leibnitz test
If the series becomes
Here the series of positive terms = is divergent.
the given series is divergent (to ) when
x
u
n
u
n
u u
u
n
x
n n
x
n n n n
n
n
n
1
1
2
1
3
1
4
1 1
1
1
0
1
1
1
2
1
3
1
1
2
1
3
1 1
1
1 1
1 2
− + − +
+
>
·
· −
− − − − · − + + +
F
H
G
I
K
J
∑ ∑
∴ − ∞ · −
− ≤ · −
+ +
→∞ →∞
Thus the given series is convergent if 1 < 1 divergent when 1 x x
83
Exercises :
Find the nature of the following series :
1 1
1
4
1
7
1
10
2
1
7
1
12
1
17
1
22
− + − +
− + − + Examine the series for
(a) convergence for (b) absolute convergence
(c) conditional convergence
(3) Show that the seies
1+
is convergent.
(4) Show the series
is absolutely convergent if | | < 1, conditionally convergent if =1 and divergent
if =1.
1
2 2
1
3 3
1
4 4
1
5 5
1
6 6
1
7 7
1
8 8
1
1
4
1
7
1
10
2 3
− − + + − − +
− − +
x + x x
x x
x
1. Convergent
2. Convergent, conditionally convergent, not absolutely convergent.
84
V.T.U. Web-Based Education
Engineering Mathematics – I (MAT-11)
By Prof.M.G.Geetha
Dept. of Mathematics, PESIT
Bangalore
ANALYTICAL GEOMETRY IN 3
DIMENSIONS
Introduction
Any point in a plane can be located by means of two real numbers, called the
coordinates. These numbers could either be distances from two fixed straight lines or a
distance form a fixed point and an angle with reference to a straight line.
The former is called the Cartesian System of coordinates and the latter the polar system
of coordinates. If the two reference lines in the Cartesian system are perpendicular to
each other, the system is called the rectangular Cartesian coordinate system.
Any point in a plane is represented by P(x, y) in Cartesian system and P(r, θ ) in polar
system of coordinates.
In space, we need three real numbers to locate a point and the reference lines, called the
coordinate axes, which are three concurrent mutually perpendicular lines. The system of
coordinates is called the rectangular Cartesian coordinate system in three dimension.
F I G 1
In the above figure, the lines OX,OY and OZ are the three axes with O as the
origin of the coordinate system.
85
If P(x,y,z) is any point in this system, then ON = x, NM = y and MP = z where
M is the foot of the perpendicular from P to the XOY plane and N is the foot of
the perpendicular from M to the axis OX.
Distance formula :
Then from the right angled triangle OMP,
OP
2
= OM
2
+ MP
2
= ON
2
+ MN
2
+ MP
2
, because ONM is a right angled triangle.
= x
2
+ y
2
+ z
2
Or, the distance OP =
2 2 2
z y x + +
Distance between two given points :
Let P (x
1
, y
1
, z
1
) and Q (x
2
, y
2
, z
2
) be the two given points.
Shift the origin to P keeping the axes parallel to the original axes.
Then A≡ (x
2
- x
1
, y
2
- y
1
, z
2
- z
1
) with reference to the new axes.
∴Distance
2
1 2
2
1 2
2
1 2
) ( ) ( ) ( z z y y x x PQ − + − + − ·
Section Formula
If a point R divides the line segment joining the points P(x
1
, y
1
, z
1
) and Q(x
2
, y
2
, z
2
)
in the ratio p : q , then
]
]
]
+
+
+
+
+
+
q p
qz pz
q p
qy py
q p
qx px
R
1 2 1 2 1 2
, ,
If R is the middle point of PQ, then
]
]
]
+ + +
2
,
2
,
2
1 2 1 2 1 2
z z y y x x
R
Example:
Find the coordinates of the centroid of the triangle whose vertices are
A (x
1
, y
1
, z
1
), B (x
2
, y
2
, z
2
) and C (x
3
, y
3
, z
3
).
If G is the centroid of the triangle ABC, then it divides AD, the median from A,
in the ratio 2:1.
86
Since D is the mid point of the side BC, D ≡
]
]
]
+ + +
2
,
2
,
2
3 2 3 2 3 2
z z y y x x
≡ ∴ G ]
]
]
+ + + + + +
3
,
3
,
3
3 2 1 3 2 1 3 2 1
z z z y y y x x x
Direction Cosines:
If a straight line makes angles α , β and γ with the coordinate axes, then cosα , cosβ
and cosγ are defined as the Direction Cosines (D.C.s) of the straight line. They are
denoted by l, m, n.
Numbers proportional to the D.C.s are called Direction Ratios (D.R.s), denoted by a. b.
c.
For instance,
1, 0, 0 are the D.C.s of the X-axis and s,0,0 are its D.R.s for any real number s.
Note
1. Two parallel lines have the same D.C.s
2. If A(x, y, z) is any point on the line OA that makes angles α , β and γ with the
three axes, then
3. α cos ·
r
x
, β cos ·
r
y
, cos ·
r
z
γ where OA = r
∴cos
2
α + cos
2
β + cos
2
γ = 1, using the distance formula r =
2 2 2
z y x + +
4. If P is any point at a distance r from the origin, then P is given by P ≡ (r l, r
m, r n) where l, m, n are the D.C.s of the line OP.
5. l
2
+ m
2
+ n
2
=1 for any line whose D.C.s are l, m, n
6. If a, b, c are D.R.s of a line, then a, b, c are proportional to the D.C.s l, m, n
87
i.e.,
2 2 2
2 2 2
2 2 2
c b a
n m l
c b a
n
c
m
b
l
a
+ + ·
+ +
+ +
· · ·
∴The D.C.s of the line are
2 2 2 2 2 2 2 2 2
, ,
c b a
c
c b a
b
c b a
a
+ + + + + +
D.C.s of the line joining two points: fig 1.3
Let A (x
1
, y
1
, z
1
) and B (x
2
, y
2
, z
2
) be the two points.
Shifting the origin to A (x
1
, y
1
, z
1
) with the axes parallel to the original axes,
We get B ≡ (x
2
- x
1
, y
2
- y
1
, z
2
- z
1
)
If l, m, n are the D.C.s of the line AB with length r units, we have B ≡ (r l, r m, r
n) since A is the origin in the new system of coordinates.
1 2 1 2 1 2
z z
n
y y
m
x x
l
·
·
∴ The D.C.s of AB are proportional to x
2
– x
1
, y
2
- y
1
, z
2
– z
1
.
or one set of D.R.s can be taken as x
2
– x
1
, y
2
- y
1
, z
2
– z
1
.
Angle between two lines with known D.C.s :
Fig 1.4
Let l
1
, m
1
, n
1
and l
2
, m
2
, n
2
be the D.C.s of two given straight lines.
Draw OP and OQ parallel to the given lines passing through the origin.
Let OP = r
1
and OQ = r
2.
Then P ≡ (r
1
l
1
, r
1
m
1
, r
1
n
1
) and Q ≡ (r
2
l
2
, r
2
m
2
, r
2
n
2
)
∴ PQ
2
= (r
2
l
2
– r
1
l
1
)
2
+ (r
2
m
2
– r
1
m
1
)
2
+ (r
2
n
2
– r
1
n
1
)
2
= r
1
2
+ r
2
2
– 2r
1
r
2
(l
1
l
2
+ m
1
m
2
+ n
1
n
2
), using the distance formula
Remember : l
1
2
+ m
1
2
+ n
1
2
= 1 and l
2
2
+ m
2
2
+ n
2
2
= 1 .
Also, using Cosine formula for triangle OPQ,
We have PQ
2
= OP
2
+ OQ
2
– 2OP.OQ cos θ , where θ =
Q O P
.
Equating the two expressions for PQ
2
, we get cos θ = l
1
l
2
+ m
1
m
2
+ n
1
n
2
Note:
88
1. The expression for the angle between two lines in terms of their D.R.s (a
1
, b
1,
c
1
and
a
2
, b
2
, c
2
)
is
2
2
2
2
2
2
2
1
2
1
2
1
2 1 2 1 2 1
cos
c b a c b a
c c b b a a
+ + + +
+ +
· θ
2. The condition for perpendicularity of two lines is
l
1
l
2
+ m
1
m
2
+ n
1
n
2
= 0 or a
1
a
2
+ b
1
b
2
+ c
1
c
2
= 0.
Projection of a line joining two points on a straight line with known D.C.s :
Fig. 1.5
Let A ≡ (x
1
, y
1
, z
1
), B ≡ (x
2
, y
2
, z
2
) and let l, m, n be the D.C.s of the line MN.
MN is the projection of AB on MN.
If θ is the angle between AB and MN, then MN = AB cos θ .
∴MN
2
1 2
2
1 2
2
1 2
1 2 1 2 1 2 2
1 2
2
1 2
2
1 2
) ( ) ( ) (
) ( ) ( ) ((
) ( ) ( ) (
z z y y x x
z z n y y m x x l
z z y y x x
− + − + −
− + − + −
− + − + − ·
∴MN = l(x
2
– x
1
) + m(y
2
- y
1
) +n(z
2
- z
1
).
Remember: The D.C.s of a line joining the points (x
1
, y
1
, z
1
) and (x
2
, y
2
, z
2
) are
2
1 2
2
1 2
2
1 2
1 2
) ( ) ( ) ( z z y y x x
x x
− + − + −
etc.
Exercise:
1. The distance between the points (0,0,3) and (-4, 0, 0) is
a)* 5 unit b) √5 units c) 25 units d) 4 units
2. If A ≡ (-1, 0, 4) and B ≡ (1, -6, 2), then the mid-point of AB is
a) (1, -3, 1) b)* (0, -3, 3) c) (1, -3, -1) d) (0, 3, -1)
3. The line which is perpendicular to the line with D.R.s 1,2,3 has D.C.s proportional to
a) 3, 2, 1 b) *0, 3, -2 c) 1, 3, -2 d) 1, -2, 3.
4. The vertices (5, -1, 1), (7, -4, 7), (1, -6, 10) and (-1, -3, 4) form a
a) rectangle b) Square c)* Rhombus d) None of the others
89
5. The point A (5,8,-1) is joined to the mid-point M of the line joining P(1,2,-3) and
Q(3,4,5). The angle between these two lines is
a)* 90
o
b) 30
o
c) 60
o
d) 45
o
Planes:
A Plane is a surface such that the line joining any two points on the surface lies entirely
on the surface.
Theorem:
A linear equation ax + by + cz +d = 0 in x, y and z represents a plane.
Proof :
If A(x
1
, y
1
, z
1
) and B(x
2
, y
2
, z
2
) are two points on the surface represented by the above
equation, then the point
]
]
]
+
+
+
+
+
+
q p
qz pz
q p
qy py
q p
qx px
C
1 2 1 2 1 2
, ,
is a point lying on the
line AB dividing it in the ratio p : q.
Since A and B are points lying on the surface ax + by + cz +d = 0
------------------------- (1)
we have ax
1
+ by
1
+ cz
1
+ d = 0 and ax
2
+ by
2
+ cz
2
+ d = 0
0 ) ( ) ( ) ( ) (
1 2 1 2 1 2
· + + + + + + + ∴ q p d qz pz c qy py b qx px a
Or
0
1 2 1 2 1 2
· +
,
`
.
|
+
+
+
,
`
.
|
+
+
+
,
`
.
|
+
+
d
q p
qz pz
c
q p
qy py
b
q p
qx px
a
which implies that the point C lies on the surface given by equation (1).
Since C is any arbitrary point lying on the line AB, the entire line AB lies on the surface
(1).
∴By the definition of a plane, equation (1) represents a plane.
Normal Form of equation to a plane :
Fig. 1.6
Let OL be drawn perpendicular (normal) to the plane from the origin.
90
Let its length be p and the D.C.s be l, m, n.
If P (x, y, z) is any point on the plane, then OL is the projection of OP on the normal
OL.
By the projection formula, we have p = lx + my + nz.
or lx + my + nz = p is the normal form of equation to a plane.
Note: Comparing the general and the normal forms of equations of a plane, we get
p
d
n
c
m
b
l
a
· · ·
from which we observe the following:
1. a, b and c, the coefficients of x, y and z in the equation of a plane are
proportional to the D.C.s of the normal to the plane.
In other words, the coefficients of x, y, z are a set of D.R.s of the normal.
2. Since each ratio is equal to
2 2 2
2 2 2
2 2 2
c b a
n m l
c b a
+ + ·
+ +
+ +
,
we get
2 2 2
c b a
d
p
+ +
·
∴ Length of the perpendicular from the origin to the plane is
2 2 2
c b a
d
+ +
3. To find the length of the perpendicular from any point P ≡ (x
1
, y
1
, z
1
) to the plane
ax + by + cz + d = 0, shift the origin to the point P keeping the directions of the axes
unchanged.
Then any general point (X, Y, Z) in the new system of coordinates changes to
(x – x
1
, y – y
1
, z –z
1
)
∴The equation of the plane in the new system of coordinates is
a (x + x
1
) + b (y + y
1
) + c (z + z
1
) + d = 0
or ax + by + cz + (ax
1
+ by
1
+ cz
1
+ d) = 0
so that the perpendicular distance of the plane from P, the origin in the new system, is
2 2 2
1 1 1
c b a
d cz by ax
+ +
+ + +
Three Point form of Equation:
91
Given the three points (x
i
, y
i
, z
i
); i =1, 2, 3.
If the plane given by ax + by + cz +_d = 0 passes through these three points, then they
satisfy the equation of the plane.
Eliminating ‘d’, we get
a(x - x
1
) + b(y – y
1
) + c(z – z
1
) = 0
a(x
1
– x
2
) + b(y
1
– y
2
) + c( z
1
– z
2
) = 0
and a(x
2
– x
3
) + b(y
2
– y
3
) + c(z
2
– z
3
) = 0
This system of three equations has a non-trivial solution if
3 2 3 2 3 2
2 1 2 1 2 1
1 1 1
z z y y x x
z z y y x x
z z y y x x
− − −
− − −
− − −
= 0.
Note:
The three-point form of equation to a plane gives the condition for coplanarity of four
points
(x
i
, y
i
, z
i
) ; i = 1, 2, 3, 4 as
4 3 4 3 4 3
3 2 3 2 3 2
2 1 2 1 2 1
z z y y x x
z z y y x x
z z y y x x
− − −
− − −
− − −
= 0.
Intercept form of equation:
If a plane makes intercepts A, B, and C with the coordinate axes, then the points
(A,0,0), (0,B,0) and (0,0,C) lie on the plane.
If ax + by + cz + d = 0 is the equation of the plane, then
C
d
c
B
d
b
A
d
a
·
·
· & ,
∴The equation of the plane reduces to
1 · + +
C
z
B
y
A
x
.
This is known as the intercept form of equation to the plane.
Note:
92
1. The angle between the two planes a
i
x + b
i
y + c
i
z + d
i
= 0 ; i = 1, 2 ; is equal to the
angle between their normals and hence is
]
]
]
]
+ + + +
+ +
2
2
2
2
2
2
2
1
2
1
2
1
2 1 2 1 2 1 1
cos
c b a c b a
c c b b a a
2. If the two planes are perpendicular, then a
1
a
2
+ b
1
b
2
+ c
1
c
2
= 0.
3. Equation of a plane passing through the line of intersection of two planes
a
i
x + b
i
y + c
i
z + d
i
= 0, i =1,2 ; is
(a
1
x + b
1
y + c
1
z +d
1
) + k(a
2
x + b
2
y + c
2
z +d
2
) = 0 where k is a constant.
Exercise:
1. The planes 2x + 3y - z = 5 and 6x +9y – 3z -7 = 0 are
a) Perpendicular b)* Parallel c) coinciding d) inclined at an angle of
45
o
2. The distance of the point (1,1,1) from the plane 2x – y +2z =3 is
a)* 0 b) 1/3 c) 1 d) 2
3. The D.C.s of the normal to the XY-plane are
a) 1,1,1 b) 0,1,0 c)* 0,0,1 d) 1,0,0
4. Which of the following points lie on the same side of the plane 3x – y + 2x – 8 = 0
a) (3,0,0) b) (3,0,2) c) (2,0,1) d)* (2,0,0)
5. One of the following planes is perpendicular to the plane 2x – 3y +5z = 1
a) x – y – z = 4 b) 5y - 3z = 1 c) x + y – z = 14 d) 3x – 2y + 5 = 0
True of False
1. The plane 4x – 3y +5 = 12 passes through the intersection of the planes
2x - 5y + z = 6 and x + y + 2z = 3. (T)
2. The point (1, -5,2) lies on the plane 3x + y + z = 1. (F)
3. The Plane 2x + 3y – z + 6 = 0 and 10x - 15y - 5z + 24 = 0 are parallel (F)
93
Straight Lines:
A curve in space is defined as the set of points common to two intersecting surfaces
and it is represented by two equations - equations of the intersecting surfaces.
A straight line is the curve of intersection of two planes.
Two plane form of equation to a straight line:
Two planes interest in a straight line and therefore the equations of the intersecting
planes together represent a straight line.
If a
1
x + b
1
y + c
1
z + d
1
= 0 and a
2
x + b
2
y + c
2
z + d
2
= 0 represent two intersecting
planes, then the equations of the straight line of intersection are
a
1
x + b
1
y + c
1
z + d
1
= 0 = a
2
x + b
2
y + c
2
z + d
2
This is called two-plane form of equations .
Symmetric form of equations:
If P(x, y, z) is any point on a straight line that passes through the point (x
1
, y
1
, z
1
),
then D.C.s of the straight line l, m, n are proportional to x - x
1
, y - y
1
, z - z
1.
∴The equations of the straight line are ,
1 1 1
n
z z
m
y y
l
x x −
·
·
called the symmetric form of equations.
Note:
1. Any point lying on the straight line whose equations are
,
1 1 1
n
z z
m
y y
l
x x −
·
·
is given by (lt + x
1
, mt + y
1
, nt + z
1
) where t is a parameter .
2. The symmetric form of equations to a straight line can be converted to its two-
plane form of equations and vice versa.
3. When the two-plane form of equations to a straight line are given, the D.C.s of the
straight line are proportional to b
1
c
2
– b
2
c
1
, c
1
a
2
– c
2
a
1
and a
1
b
2
– a
2
b
1
,
since the straight line is perpendicular to the normal to the two intersecting planes.
94
Coplanarity of two straight lines:
Consider two straight lines
2
2
2
2
2
2
1
1
1
1
1
1
n
z z
m
y y
l
x x
and
n
z z
m
y y
l
x x −
·
·
− −
·
·
.
If these lines are coplanar, let the equation of the plane containing them be
ax + by + cz + d = 0.
Then the two points (x
1
, y
1
, z
1
) and (x
2
, y
2
, z
2
) lie on this plane. Also the normal to the
plane is perpendicular to the two lines.
∴we have the following conditions :
ax
1
+ by
1
+ cz
1
+ d = 0 --------------------------- (1)
ax
2
+ by
2
+ cz
2
+ d = 0 --------------------------- (2)
al
1
+ bm
1
+ cn
1
= 0 --------------------------- (3)
al
2
+ bm
2
+ cn
1
= 0 --------------------------- (4)
Eliminating d from equations (1), (2) and ax + by + cz + d = 0, we get
a (x – x
1
) + b ( y – y
1
) + c (z –z
1
) = 0 and
a (x
1
– x
2
) + b ( y
1
– y
2
) + c (z
1
–z
2
) = 0
Eliminating a, b, c from this equation, (3) and (4) , we get the condition for
coplanarity of straight lines in the form
2 2 2
1 1 1
2 1 2 1 2 1
n m l
n m l
z z y y x x − − −
= 0
and the equation of the plane containing them is given by
2 2 2
1 1 1
1 1 1
n m l
n m l
z z y y x x − − −
= 0.
Note:
If two lines are coplanar, then the lines are either parallel or intersecting. When they
intersect, the point of intersection can be found by solving the equations of the two lines.
95
Skew Lines:
Two lines are said to be skew when they are non-coplanar. This is so, when they are
neither parallel nor intersecting.
Shortest distance between skew lines:
When two lines are skew, there exists a line perpendicular to both called the line of
shortest distance. The length of the common perpendicular intercepted between the
skew lines is the shortest distance between them.
Fig. 1.7
Let the two skew lines be given by
1
1
1
1
1
1
n
z z
m
y y
l
x x −
·
·
-------------- (1)
and
2
2
2
2
2
2
n
z z
m
y y
l
x x −
·
·
-------------(2)
with P ≡ (x
1
, y
1
, z
1
) and Q ≡ (x
2
, y
2
, z
2
) .
AB, the line of shortest distance is perpendicular to both the lines.
Then the D.C.s λ , µ , ν of the line of shortest distance are proportional to
m
1
n
2
– m
2
n
1
, n
1
l
2
– n
2
l
1
, l
1
m
2
– l
2
m
1
We observe that the length AB is the projection of the line PQ on the line of shortest
distance.
∴ By the projection formula,
AB = (x
2
– x
1
)λ + (y
2
– y
1
)µ + (z
2
– z
1
2
1 2 2 1
2
1 2 2 1
2
1 2 2 1
1 2 2 1 1 2 1 2 1 1 1 2 1 2 2 1 1 2
) ( ) ( ) (
) )( ( ) )( ( ) )( (
m l m l l n l n n m n m
m l m l z z l n l n y y n m n m x x
− + − + −
− − + − − + − −
·
This gives the shortest distance between the two skew lines.
Equations of the line of shortest distance:
The shortest distance line is coplanar with line (1) since it intersects line (1) at the point
A. Point B is the point of intersection of the line of shortest distance and line (2).
∴ We have
96
0
1 1 1
1 1 1
·
− − −
n m l
z z y y x x
υ µ λ
and
0
2 2 2
2 2 2
·
− − −
n m l
z z y y x x
υ µ λ
These two equations together represent the line of shortest distance between the skew
lines given by (1) and (2).
Exercise:
1. The straight line
c
z z
b
y y
a
x x
1 1 1
·
·
is
a)* Perpendicular b) Parallel to c) lies entirely on d) at an angle of 45
o
with
the plane ax + by + cz + d = 0 .
2. The point of intersection of the lines
2 3 7 −
· ·
z y x
and
z
y x
·
·
4 3
is
a) (-5, -13, -37) b) (4, 7, -3) c)* (0, 0, 0) d) (1, -1, 2)
3. The lines
2
3
4
5
3
2 +
·
·
− z y x
and
4
3 2
4
6
3
4 +
·
·
− z y x
are
a) Perpendicular b)* Parallel c) coincident d) Skew
4. The image of the point (1, 1, 1) in the plane x + y + z = 0 is
a) (0, 0, 0) b) (1, -1, -1) c)* (-1, -1, -1) d) (
3
1
,
3
1
,
3
1
)
True or False
1. The lines
2
3
2
2
1
·
· −
z y
x
and
2
4
5
3
4
2 z y x −
·
·
are
perpendicular .
2. The point (1, 2, 3) lies on the line
z
y x
− ·
·
2
4
6
3
4
.
3. The line
0
2
0
1 z
y
x
· − ·
and the z – axis are skew lines.
4. The equations
4
3
2
2
1
− ·
·
z
y x
and 3x –2y +1 = 0 = y – 3z +10
represent the same straight line.
5. The plane 2x + 3y + 3z –1 = 0 contains the line
4
1
2
1
3
2
·
+
·
− z y x
.
97
6. The shortest distance between the lines
4
4
3
2
2
1
·
·
− z y x
and
5
4
4
2
3
1 −
·
·
− z y x
is 0.
7. The straight lines which do not intersect are called skew lines.
Right Circular cone and Right Circular Cylinder:
The solid generated when a straight line rotates about a fixed line is either a right
circular cone or a right circular cylinder. The rotating line is a generator of the solid.
The fixed line is called the axis.
In the case of a cone, the generators are inclined at an angle α with the axis while in a
cylinder, the generators are parallel to the axis.
Note:
The angle α is called the semi-vertical angle of the cone and the point of
intersection of the generators with the axis is called the vertex of the cone.
Equation of a right circular Cone:
Fig. 1.8
Let the axis of the cone be given by
n
z z
m
y y
l
x x
1 1 1
·
·
with its vertex as V≡ (x
1
, y
1
, z
1
).
If P (x, y, z) be any point on the surface of the Cone, then the D.C.s of the generator
passing through V are proportional to x – x
1
, y – y
1
, z – z
1.
If α is the semi-vertical angle of the cone, then
2 2 2 2
1
2
1
2
1
1 1 1
) ( ) ( ) (
) ( ) ( ) (
cos
n m l z z y y x x
n z z m y y l x x
+ + − + − + −
− + − + −
· α
or ( ) ( ) ( ) [ ]
2
1 1 1
n z z m y y l x x − + − + − ( ) [ ]
2
1
2
1
2 2 2 2
) ( ) ( cos z z x x n m l − + − + + · α
is the equation of a right circular cone with its vertex at (x
1
, y
1
, z
1
), semi-vertical angle
as α and the D.C.s of its axis proportional to l, m, n.
98
Equation of a right circular cylinder:
Fig/ 1.8
n
z z
m
y y
l
x x
1 1 1
·
·
be the axis of the right circular cylinder with A ≡ (x
1
, y
1
,
z
1
) and r as its radius.
If P ≡ (x, y, z) is any point on the surface of the cylinder, then AM is the projection of
AP on the axis with M as the foot of the perpendicular from P to the axis.
or AM = AP cos θ , where θ =
P A M
2 2 2 2
1
1 1 1 2
1
2
1
2
1
) (
) ( ) ( ) (
. ) ( ) ( ) (
n m l x x
z z n y y m x x l
z z y y x x AM
+ + −
− + − + −
− + − + − · ∴
Also AM
2
= AP
2
– MP
2
with MP = r
[ ]
2 2 2
2
1 1 1 2 2
1
2
1
2
1
) ( ) ( ) (
) ( ) ( ) (
n m l
z z n y y m x x l
r z z y y x x
+ +
− + − + −
+ · − + − + − ∴
is the equation of the right circular cylinder.
Exercise
Fill in the blanks :
1. The semi-vertical angle of the right circular cone passing through the origin, vertex at
(1,2,1) with its axis parallel to the line
1
1
2
1
2 −
+
·
·
z y x
is ____________. π /3
2. The vertex of the right circular cone whose axis is the line
2
7
3
7 +
·
·
z y
x with the line
2
2
3
3
1
+ ·
·
+
z
y x
as one of its generators is _______________ (2,1,-3)
3. The radius of the right circular cylinder passing through the origin with its axis lying
along the line
3
5
2
4
10
9
·
·
+ z y x
is _______ units.
3.
99
Questions
1. Find the distance of the point (1, 2, -3) from the middle point of the line segment
joining (-3, 4, -2) and (1, 2, 6)
The mid-point of the given line segment is (-1, 3, 2).
∴ The distance of (1, 2, -3) from this point is 30 25 1 4 · + + units
2. Show that the points A (2, 3, 4), B (3, 4, 2) and C(-1, 0, 10) are collinear.
D.R.s of the line AB are 1, 1, -2 and those of the line BC are 4, 4, -8 or 1, 1, -2
∴ The lines AB and BC have proportional D.R.s. The lines have a common point B.
Hence the points A, B and C are collinear.
3. Find the centroid of the triangle ABC where
A ≡ (-1, 2, 1), B ≡ (-3, 4, -4) and C ≡ (-1, 0, 3).
The centroid G is given by
,
`
.
| + − + + + + −
3
3 4 1
,
3
0 4 2
,
3
1 3 1
or (1, 2, 0).
4. Find the angles made by a line with the coordinate axes if it is equally inclined with
the axes.
We know that cos
2
α + cos
2
β + cos
2
γ = 1
But α = β = γ is given . ∴ angle =
3
1
cos
1 −
· α
5. Find the angles of the triangle ABC formed by the vertices
A (3, 0, -1), B(-1, 4, 1) and C(3, 4, 2) .
D.R.s of AB are –4, 4, 2 and D.R.s. of AC are 0, 4, 3.
15
11
25 36
6 16 0
cos ·
+ +
· ∴ A
D.R.s of BA are 4, -4, -2 and D.R.s of BC are 4, 0, 1
17 3
7
cos · ∴ B
D.R.s of CA are 0, -4, -3 and D.R.s of CB are –4, 0, -1
100
17 5
3
cos · ∴ C
∴The angles are
,
15
11
cos
1
·
17 3
7
cos
1
·
and
17 5
3
cos
1
·
6. Find the angle between any two diagonals of a cube.
Fig. 1.10
Let the cube be as shown in the figure with one of its vertices O as the origin of the
coordinate system and the axes along the three sides OA, OB, OC of the cube
respectively.
If d is the length of each side of the cube, then its vertices are given by
O ≡ (0, 0, 0), O
≡ (d, d, d),
A ≡ (d, 0, 0), A
≡ (0, d, d),
B ≡ (0, d, 0), B’ ≡ (d, 0, d),
C ≡ (0, 0, d), C
≡ (d, d, 0),
We can see that the four diagonals of the cube are OO’, AA’, BB’, and CC’.
D.R.s of OO’ and AA’ are d, d, d and –d, d, d respectively
∴ Angle between these two diagonals is given by
3
1
cos
3 3
cos
1
2 2
2 2 2
1 − −
·
+ +
·
d d
d d d
7. Find the angle between a diagonal of a cube and
(i) an intersecting diagonal of a face (ii) a non-intersecting diagonal of a face.
(i) The diagonal AA
of the cube and the diagonal O’A of a face of the cube
intersect.
D.R.s of these two lines are respectively –d, d, d, and 0, d, d .
∴Angle between them is given by
3
2
cos
2 3
2
cos
1
2 2
2
1 − −
· ·
d d
d
(ii) The diagonal AA’ of the cube and the diagonal B’C’
of a face of the cube do not
intersect. Their D.C.s are respectively proportional to –d, d, d and 0, d, -d. .
101
∴Angle between them is given by
2
2 3
0
cos
2 2
2 2
1
π
·
− +
·
d d
d d
8. Given a cube as in Fig. 1.10, the diagonals passing through O, of the faces meeting at
O, make angles α , β , γ with the diagonal OO’ of the cube, prove that
cos
2
α + cos
2
β + cos
2
γ = 2.
The diagonals of the faces meeting at O and passing through O are OA
, OB
, OC
’.
whose D.R.s are 0, d, d; d, 0, d and d, d, 0.
If these lines are inclined at angles α , β , γ with OO
, whose D.R.s are d, d, d ;
then the angles are given by
6
2
6
2
cos
2
2
· ·
d
d
α
or
3
2
cos
2
· α
Also
3
2
cos
2
· β
and
3
2
cos
2
· γ
. Hence the required result.
9. Find the angle between the two lines whose D.C.s are given by the relations
l + m + n = 0 and 2lm + 2nl – mn = 0.
Eliminating l from the given relations, we get 0 1 2 2
2
· + +
,
`
.
|
n
m
n
m
. 0 1 2 2 ·
,
`
.
|
+ +
,
`
.
|
+ ⇒
n
m
n
m
0 2 · + ⇒ n m
or
0 2 · +n m
These with l + m + n = 0 give us two sets of l,m,n proportional to –1, -1, 2 and 1,
-2, 1.
The angle between these two lines is
3 6
2 2 1
cos
1
π
·
+ + −
.
10. Show that the straight liens whose D.C.s are given by ul + vm + wn = 0 and
pl
2
+qm
2
+rn
2
= 0
are (i) perpendicular if u
2
(q+r) + v
2
(r+p) + w
2
(p+q) = 0
and (ii) parallel if
0
2 2 2
· + +
r
w
q
v
p
u
102
Eliminating l from the two given relations,
We get 0 ) ( 2 ) (
2 2
2
2
2 2
· + + + + ru pw
n
m
pvw
n
m
qu pv -----------------( 1)
(i) Equation (1) is quadratic in
n
m
and has two roots, say,
1
1
n
m
and
2
2
n
m
Then the product of the 2 roots is given by
2 2
2 2
2
2
1
1
pv qu
ru pw
n
m
n
m
+
+
·
or
2 2
2 1
2 2
2 1
pv qu
n n
ru pw
m m
+
·
+
Similarly, by eliminating m and n from the given relations and then getting the
product of the roots and simplifying, we get
2 2
2 1
2 2
2 1
qw rv
l l
pv qu
n n
+
·
+
and
2 2
2 1
2 2
2 1
ru pw
m m
qw rv
l l
+
·
+
2 2
2 1
2 2
2 1
2 2
2 1
qw rv
l l
pv qu
n n
ru pw
m m
+
·
+
·
+
---------------(2)
where l
i
, m
i,
n
i ;
i= 1, 2 are the two straight lines given by the relations.
When these lines are perpendicular, we have l
1
l
2
+ m
1
m
2
+ n
1
n
2
= 0
⇒ pw
2
+ ru
2
+ qu
2
+ pv
2
+ rv
2
+ qw
2
= 0 from (2)
⇒ u
2
(q+r) + v
2
(r+p) + w
2
(p+q) = 0
(ii) When the two lines are parallel, they have the same D.C.s.
This is true when equation (1) has two equal roots.
⇒ 4p
2
v
2
w
2
- 4(pv
2
+ qu
2
) (pw
2
+ ru
2
) = 0
⇒ qru
2
+ rpv
2
+ pqw
2
= 0
0
2 2 2
· + +
r
w
q
v
p
u
Planes:
11. Find the equation of the plane passing through the points (1, 2, 3), (2, 0, -1) and (-
1, 4, 5).
Also find the distance of the point (0, 2, -4) from the plane.
103
What is the angle made by this plane with 2x-y+z = 6 ?
The plane through the three given points is
2 2 2
4 2 1
3 2 1
− −
− − − z y x
= 0 or 2x+3y-z
= 5.
Perpendicular distance of (0,2,-4) from this plane is
14
5 4 6 0 − + +
=
14
5
.
The angle between this plane and 2x-y+z = 6 is given by cosθ = 0.
or the two planes are perpendicular.
12. Show that the points (3, 4, 1), (-1, -2, 5), (1, 7, 1) and (1, 10, 0) are coplanar.
Find the equation of the plane containing them and also the intercept made by it on the
x-axis.
The plane containing (3,4,1), (-1, -2, 5) and (1,7,1) is
4 9 2
2 3 2
1 4 3
− − − z y x
= 0 or 3x +2y + 6z - 23= 0.
Since the point (1, 10, 0) lies on this plane, the four given points are coplanar.
The equation of the plane in intercept form is
1
6
23
2
23
3
23
· + +
z y x
∴The x-intercept by the plane is
3
23
units.
13. Find the equation of the plane bisecting perpendicularly the join of the points (2,1,3)
and (4, 3, -1).
Let the equation of the required plane be ax+by+cz + d = 0
Then the mid-point (3,2,1) lies on it. Also a, b, c, are proportional to 1, 1, -2.
∴The equation is x + y – 2z – 3 = 0.
104
14. Find the equation to the plane perpendicular to x + y – 4z = 0 and through the
intersection of the planes x +3y –z = 4 and 2x + 2y +2z = 1.
The equation of the plane through the intersection of the given planes is
x + 3y – z – 4 + k(2x + 2y + 2z – 1) = 0
or (1 + 2k) x + (3 + 2k) y + (-1 + 2k) z + (-4 – k ) =0
But this is given to be perpendicular to x + y – 4z = 0
∴ 1 + 2k + 3 + 2k + 4 –8k =0 ⇒k = 2.
∴ The equation of the required plane is 5x + 7y + 3z =6.
15. The plane 4x + 7y + 4z +81 = 0 is rotated through a right angle about its line of
intersection with the plane 5x + 2y + 5z = 25. Find the equation of the plane in its
new position.
The plane in its new position passes through the line of intersection of the two given
planes.
Fig. 1.11
∴Its equation is given by 4x + 7y + 4z +81 + k (5x + 2y +5z – 25) = 0
or (4 + 5k) x + (7 + 2k) y + (4 + 5k) z + (81 –25k) = 0
But this is perpendicular to the plane 4x + 7y +4z +81 =0
∴We have (4 + 5k) 4 + (7 + 2k) 7 + (4 + 5k) 4 = 0. ∴
2
3
− · k
.
∴The equation of the required plane is 7x-8y+7z-237 =0.
Straight lines:
16. Find the equations of the straight line of intersection of the planes 2x – y + z = 5
and
x +3y – 2z +1=0
The line of intersection is perpendicular to both the planes.
If l, m, n are its D.C.s, we have 2l – m + n = 0 and l + 3m – 2n = 0 .
105
∴l : m : n = -1 : 5 : 7
Choosing z = 0 in the equations of the planes and solving the resulting equations,
we get (2, -1, 0) as a point lying on both the planes.
∴ Equations of the straight line are
7 5
1
1
2 z y x
·
+
·
17. Find the image of the point P(1, 3, 4) in the plane 2x – y +z + 3 = 0 .
Fig. 1.12
If P
is the image of the point P in the given plane, then the line PP
is perpendicular
to the plane and the plane bisects the line at the point, say, M .
∴ the equations of PP
are
k
z y x
·
·
·
1
4
1
3
2
2
(say)
M is given by M ≡ (2k+1, -k+3, k+4) and lies on the given plane. ∴k = 1.
∴M ≡ (-1, 4, 3)
∴ P’≡ (-2, 5, 2)
18. Find the image of the point (1, -1, 2) in the line
2
3
3
1
· · −
z y
x
.
Let P ≡ (1, -1, 2) and P’ be its image.
Then PP’ is perpendicular to the given line and its mid-point, say, M lies on the
given line.
∴M can be written as (k + 1, 3k, -2k + 3) .
∴D.R.s if PP
are k + 1 –1, 3k+1, -2k +3 –2 or k, 3k + 1, -2k+1.
Since PP’ is perpendicular to the line
2
3
3
1
· · −
z y
x
,
we have k + 3 (3k + 1) –2(-2k + 1) = 0 ⇒
14
1
− · k
.
14
44
,
14
3
,
14
13
M
,
`
.
| −
≡ ∴
.
7
30
,
7
4
,
7
6
P'
,
`
.
|
≡ ∴
106
19. Find the image of the line
4
3
3
2
2
1 −
·
·
− z y x
in the plane x + y + z + 3 =0 .
The given line is not parallel to the given plane.
∴It must intersect the plane.
Any point on the given line is (2k + 1, 3k +2, 4k + 3). If this lies on the plane, then k
= -1.
∴(-1, -1, -1) is the point of intersection of the given line and the given plane and
hence is a point lying on the image of the line in the plane.
(1,2, 3) is a point on the given line. Its image in the given plane is (-5, -4, -3) and is
another point on the image of the line.
∴D.R.s of the image line are 4, 3, 2 .
∴Its equations are
2
3
3
4
4
5 +
·
+
·
+ z y x
.
20. Find the equation of the plane through the points (1, 0, -1) and (3, 2, 2,) and parallel
to the line
3
2
2
1
1
·
· −
z y
x
.
The equation of the required plane is of the form a(x-1) + by + c (z+1) =0 --------
(1)
This plane also passes through the point (3, 2, 2).
∴We have 2a + 2b + 3c = 0 ----------- --- (2)
The plane is parallel to
3
2
2
1
1
·
· −
z y
x
.
∴ The normal to the plane with D.R.s a, b, c is perpendicular to the line.
∴ a – 2b + 3c = 0 ---------------------(3)
Evaluating a, b, c from the three equations, we get
0
3 2 1
3 2 2
1 1
·
+ − z y x
or 4x-y-2z-6 = 0 is the equation of the plane.
107
21. Find the angle between the lines
2 2
z
y
x
· ·
and 4x – y + z –5 =0 = 14x – 3y +
4z – 20.
D.R.s of the first line are 2, 1, 2 .
D.R.s of the second line are got by solving the equations 4a – b + c = 0 & 14a
– 3b + 4c=0
and they are 1, 2, -2 .
∴ Angle between the lines is
2 9
4 2 2
cos
1
π
·
− +
22. Find the angle that the line
4
3 2
2
2
1 +
· − ·
− z
y
x
makes with the plane x – y +
z = 0.
The required angle is θ
π
cos
2
− , where θ is the angle between the line and
the normal to the plane.
Here
3
1
cos
3 3
2 1 2
cos
1 1 − −
·
+ −
· θ
∴ Required angle is
3
1
sin
1 −
.
23. Find the distance of the point (1, -2, 3) from the plane x – y + z =13
measured
parallel to the line
6 3 2 −
· ·
z y x
.
Fig. 1.13
Let P ≡ (1, -2, 3). The equations of the line through P parallel to the given line are
6
3
3
2
2
1
·
+
·
− z y x
. Let this line intersect the given plane at M.
Then M ≡ (2k + 1, 3k - 2, -6k + 3).
108
Since M lies on the plane, we have 2k + 1 –3k + 2 – 6k + 3 = 13 ⇒k = -1.
∴ M ≡ (-1, -5, 9).
∴ The required distance PM = 7 units .
24. Find the distance of the point (-1, -5, -10) from the point of intersection of the line
12
2
4
1
3
2 −
·
+
·
− z y x
and the plane x - y + z =5.
Any point on the given line is (3k + 2, 4k –1, 12k+2).
If this lies on the given plane, then
3k + 2 – 4k +1 +12k +2 = 5 ⇒k =0.
∴(2, -1, 2) is the point of intersection of the given line with the given plane.
∴ The required distance is 13 ) 10 2 ( ) 1 5 ( ) 1 2 (
2 2 2
· + + − + + units
25. Find the line through (3, -4, 1), parallel to the plane 2x + y –z +8 = 0 and
intersecting the line 2
3
1
2
3
− ·
+
·
z
y x
.
The equations of the required line are
n
z
m
y
l
x 1 4 3 −
·
+
·
, where l, m, n are
its D.C.s.
This line is parallel to 2x + y – z +8 = 0.
∴We have 2l + m – n = 0 -----------------(1)
The line intersects
2
3
1
2
3
− ·
+
·
z
y x
.
∴They are coplanar.
0 3 3 0
1 3 0
1 3 2 · − + ⇒ ·
− −
− n m l
n m l
-----------------------(2)
Solving equations (1) and (2) , we get l : m : n are 2 :-3 :1 .
∴ Equations of the required line are
1
1
3
4
2
3 −
·
+
·
− z y x
.
109
26. Find the equations of the line through (2, 1, 3) and perpendicular to the line
3
2
2
6
5
·
+
· −
z y
x
.
Also find the length of the perpendicular from (2, 1, 3).
If M is the foot of the perpendicular from (2, 1, 3) to the given line, M ≡ (k+5, 2k-
6,3k+2).
∴ D.R.s of PM are k + 3, 2k – 7, 3k –1 with P ≡ (2, 1, 3) .
But PM is perpendicular to the given line.
∴ k + 3 + 2 (2k – 7) + 3 ( 3k –1) = 0 ⇒k =1
∴ M ≡ (6, -4, 5) ∴ Perpendicular distance = 3 3
Equations of the perpendicular line are
2
3
5
1
4
2 −
·
·
− z y x
.
27. A line with D.C.s proportional to 7, 4, -1 is drawn to intersect the lines
2
1
7
3
1
+ ·
·
z
y x
and
4
5
2
3
3
1 −
·
·
+ z y x
. Find the points of intersection and
also the length intercepted.
Let P and Q be the points of intersection of the line with D.R.s 7, 4, -1 with the two
given lines.
Then P ≡ (3k
1
+ 1, k
1
+ 7, k
1
- 2)
and Q ≡ (-3k
2
– 3, 2k
2
+ 3, 4k
2
+ 5) for some k
1
& k
2
∴ D.R.s of PQ are 3k
1
+ 3k
2
+4, -k
1
–2k
2
+ 4, k
1
–4k
2
– 7
They are also given by 7, 4, -1.
∴These two sets of D.R.s are proportional .
We get the equations 19k
1
+ 26k
2
= 12 and k
1
– 6k
2
= 8,
which when solved, yield k
1
=2, k
2
= -1 .
∴Points of intersection are P ≡ (7, 5, 0) and Q ≡ (0, 1, 1) .
∴Length intercepted 66 units
110
28. Find the shortest distance between the lines
4
3
3
2
2
4 −
·
·
− z y x
and
5x +3y +7z +6 = 0 = 6x +7y -5z +14. Also find the equations of the shortest
distance line.
Line 1 has D.R.s 2, -3, 4 and (4, 2, 3) lying on it.
Line 2 has D.R.s –64, 67, 17 and (0, -2, 0) lying on it.
∴The shortest distance line has D.R.s 11, 10, 2.
But the shortest distance is the projection of the join of (4, 2, 3) and (0, -2, 0) on the
line of shortest distance.
∴ Shortest distance length
6 ) 0 3 (
15
2
) 2 2 (
15
10
) 0 4 (
15
11
· − + + + −
units
Equations of the shortest distance line are given by
2 10 11
17 67 64
2
0
2 10 11
4 3 2
3 4 4
+
· · −
− − − z y x z y x
or 46x – 40y – 53z + 55 = 0 = 4x – 35y + 153z – 70.
29. Find the shortest distance between the line
7
3
4
2
2
1 −
·
·
− z y x
and
6
2
5
1
2
2
·
·
− z y x
.
What do you conclude here?
The D.C.s of the shortest distance line are proportional to 11, 9, 2 .
Shortest distance =
( ) ( )
0
206
) 3 2 ( 2 1 9 1 2 11
·
− + − + −
.
∴The lines are coplanar. We can see that they are not parallel. Hence they intersect.
111
30. Find the plane through the line 5x + 3y + 7z +6 = 0 = 6x + 7y – 5z +2 and parallel
to the line
4
3
3
1
2
1 −
·
·
− z y x
. Hence find the length of the shortest distance
between the given lines.
Any plane through the given line is (5 + 6k) x + (3 + 7k) y + (7 – 5k) z + (6 +2k) =
0.
This is parallel to the line
4
3
3
2
2
1 −
·
·
− z y x
.
∴(5 + 6k) 2 + (3+7k) (-3) + (7-5k) 4 = 0 ⇒k = 1.
∴ The equation of the required plane is 11x +10y + 2z + 8 = 0.
Now the length of the perpendicular from the point (1, 2, 3) on to this plane gives
the shortest distance between the two given lines.
∴It is
3
15
45
·
units.
Cones and Cylinders :
31. Find the equation of the right circular cone with vertex at (1, 2, -3), 30
o
as its semi-
vertical angle and its axis along the line
1
3
3
1
2
1 +
·
·
− z y x
.
( ) ( ) ( ) [ ]
( ) ( ) ( ) [ ]
2 2 2
2
2
3 2 1 14
3 1 3 1 2
4
3
30 cos
+ + − + −
+ + − + −
· ·
z y x
z y x
o
or ( ) ( ) ( ) [ ] [ ]
2 2 2 2
5 3 2 2 3 2 1 21 − + + · + + − + − z y x z y x is the equation.
32. Show that the equation of the right circular cone with semi-vertical angle as α and its
axis along the z –axis, is given by x
2
+ y
2
= (z-3)
2
tan
2
α given that the vertex is at
(0, 0, 3).
[ ]
2 2 2
2
2
) 3 (
) 3 ( ) 0 ( 0 ) 0 ( 0
cos
− + +
− + − + −
·
z y x
z y x
α
( ) [ ]
2 2 2 2 2
) 3 ( cos 3 − · − + + ⇒ z z y x α
α
2 2 2 2
tan ) 3 ( − · + ⇒ z y x
112
33. Given that the vertex of a right circular cone passing through the origin is at the point
(1,2,1) whose axis is along the line
2
1
1
2
2
1
·
·
− z y x
. Find its equation.
[ ]
27
2
6 9
) 1 0 ( 2 ) 2 0 ( 1 ) 1 0 ( 2
cos
2
2
·
− − − + −
·
x
θ
If (x, y, z) is any general point on the surface of the cone, then
[ ]
( ) ( ) ( ) [ ]
2 2 2
2
1 2 1 9
) 1 ( 2 ) 2 ( 1 ) 1 ( 2
27
2
− + − + −
− − − + −
·
z y x
z y x
or ( ) ( ) ( ) [ ] [ ]
2 2 2 2
2 2 2 3 1 2 1 2 − − + · + + − + − z y x z y x is the equation.
34. Find the equation of the right circular cylinder whose axis is along the line
2
1
3
1
2
·
+
· −
z y
x
with the radius of its base as 3 units.
Let A ≡ (2,-1, 1), then
( ) ( ) ( ) [ ]
14
1 2 1 3 2
2
2
− + + − −
·
z y y
AM . (check fig. 1.9 )
But AM
2
= AP
2
– PM
2
, by Pythagoras Theorem.
= (x – 2)
2
+ (y +1)
2
+ (y - 1)
2
–9 .
∴(x –3y + 2z -7)
2
= 14[(x-2)
2
+ (y + 1)
2
+ (z - 1)
2
] –126 is the equation of the
right circular cylinder.
35. Show that the equation of a right circular cylinder whose axis is along x –2 = 0 = z
–1
2
+ y
2
– 4x – 2z + 4 = 0.
The equations of the axis are
0
1
1
0
0
2 −
·
·
− z y x
.
Here A = (2, 0, 1) and r = 1.
∴We have x
2
+ z
2
– 4x – 2z +4 = 0 as the equation.
113
36. Find the equation of a right circular cylinder whose axis is
5
4
4
3
3
2 −
·
·
− z y x
and
a generator is
5
2
4
3
3
4 −
·
·
− z y x
A ≡ (2, 3, 4) and Q ≡ (4, 3, 2) is a point on the surface of the cylinder.
If QM perpendicular to the axis, QM = r, radius and r
2
= AQ
2
– AM
2
[ ]
25
192
50
) 2 ( 5 ) 0 ( 4 ) 2 ( 3
4 0 4
2
·
− + +
− + + · .
If P ≡ (x, y, z) is any general point on the surface of the cylinder, then
AL
2
= AP
2
– PL
2
, PL perpendicular to the axis.
[ ]
. 25 / 192 ) 4 ( ) 3 ( ) 2 (
50
) 4 ( 5 ) 3 ( 4 ) 2 ( 3
2 2 2
2
− − + − + − ·
− + − + −
z y x
z y x
or ( ) ( ) ( ) [ ] 384 ) 38 5 4 3 ( 4 3 2 50
2 2 2 2
· − + + − − + − + − z y x z y x
is the equation of the right circular cylinder.
TEST :
1. If α , β , γ are the angles made by a line with the coordinate axes, the value of
sin
2
α + sin
2
β + sin
2
γ is
a) 1 b) 2 c) 3 d) 4
2. A straight line is inclined at angles 45
o
and 60
o
with the y and the z-axes respectively.
Its inclination with x-axes is
a) 60
o
b) 45
o
c) 75
o
d) 90
o
3. The D.C.s of the line perpendicular to the plane ABC formed by A≡ (1, 1, 1), B≡ (3,
4, -1) and C≡ (1, 4, 0) are
a) 3, 2, 6 b)
7
6
,
7
2
,
7
3
, c)
7
6
,
7
2
,
7
3
d)
7
6
,
7
2
,
7
3 −
4. The foot of the perpendicular from the origin to the line joining (1, 2, 3) and (2, 3, 4)
is
a) (1, 0,1) b) (-1, 0, 1) c) (1, 0, -1) d) (-1, 0, -1)
114
5. The points A(3, 4, 2), B(6, -1, 10) and C(9, 1, 4) are the vertices of a triangle.
Traingle ABC is
a) Isosceles b) Right angled c) equilateral d) Right angled isosceles
6. The perpendicular distance between the planes 2x – 2y + z + 3 = 0 and 4x – 4y +
2y +9 = 0 is
a)
2
1
b)
6
1
c)
3
1
d)
3
2
7. The equation of the plane through the points (2, 3, 4), (-3, 5, 1) and (4, -1, 2) is
a) x-y+z1=0 b) x+y-z+1=0 c) x+y-z-1=0 d) x-y-z-1 =0
8. If the points (1, 1, -1), (7, 1, 2), (λ , -1, 2) and the origin are coplanar, the value of λ
is
a) 1 b) 2 c) 3 4) 4
9. The equation of the plane through (1, -2, 4) and parallel to the plane x+2y-3z-7 = 0
is
a) x+2y-2z=0 b) x+2y-3z+7=0 c) x+2y-3z-15=0 d) x+2y-3z+15=0
10. The acute angle between the plane 2x-y+z = 7 and x+y+2z-11 = 0 is
a)
3
π
b)
3
c)
6
5
cos
1 −
d)
6
1
cos
1 −
11. The equation of the plane through the line of intersection of the planes 2x-
3y+4z+1=0 and x+y+z-5=0 and perpendicular to the latter plane is
a) 3x-2y+5z-4=0 b) x-4y+3z+6=0 c) 11x+6y+13z-44=0 d) 12x+7y+14z-
49=0
12. The foot of the perpendicular from (3, 2, 1) to x+3y-2z+7 = 0 is
a) (2, -2, 3) b) (4, 5, -1) c) (2, -1, -1) d) (3, 2, 1)
13. The equations of the straight line joining the points (3, 2, 4) and (4, 5, 2) are
a)
2
4
5
2
4
3 −
·
·
− z y x
b)
2
4
3
2
3
·
· −
z y
x
c)
4
2
2
5
3
4 −
·
·
− z y x
d)
2
2
3
3
1
+
·
· −
z y
x
14. The equation of the plane through the line of intersection of the planes 3x-4y+5z-10
= 0
and 2x+2y-3z-4 = 0 and parallel to the straight line x = 2y = 3z is
115
a) x+20y –27z=14 b) x-20y-27z+14=0 c) x-20y +27z=14 d) x +20y – 27z
+14=0
15. Equations of the straight line through the point (2, 3,1) and parallel to the line
x-2y-z +5 = 0 = x+y+3z-6 are
a)
3
1
4
3
5
2
·
·
− z y x
b)
3
1
4
3
5
2
·
·
− z y x
c)
3
1
4
3
5
2 −
·
·
− z y x
d)
3
1
4
3
5
2 −
·
·
− z y x
16. Equations of the straight line through the point (2, 1, -3) and perpendicular to the
plane 3x+2y-4y-7=0 are
a)
4
3
2
1
3
2 +
·
·
− z y x
b)
4
3
2
1
3
2 −
·
·
− z y x
c)
4
3
2
1
3
2
+
·
·
− z y x
d)
4
1
2
1
3
2
·
·
− z y x
17. Equations of the straight line through the point (3, 4, -3) and perpendicular to two
lines whose D.C.s are
1 : 2 : 2 and 2 : 1 : 3 are
a)
3
3
1
4
4
3
+
·
·
− z y x
b)
3
3
1
4
4
3
·
·
− z y x
c)
3
3
1
4
4
3
·
·
− z y x
d)
3
3
1
4
4
3
+
·
·
− z y x
18. A(1, 2, 3), B(2, 1, -5) and C(4, -1, 3) are the vertices of a triangle.
The equations of the median through A are
a)
2
3
1
2
1
1 −
·
·
− z y x
b)
2
5
1
1
1
2 +
·
·
− z y x
c)
4
3
1
1
1
4 −
·
+
·
− z y x
d)
4
1
1
2
1
1 −
·
·
− z y x
19. The equation of the plane through the point (2, -1, 0) & (3, -4, 5) and parallel to the
line 2x=3y=4z is
a) 29x-27y –22z=85 b) 29x+27y-22z-85=0 c) 29x-27y-22z+85=0 d) 29x-27y+22z-
85=0
20. The plane 4x+5y-z+15=0 ______________ the line
3
4
1
3
2
1 −
·
+
·
− z y x
a) Perpendicular b) Contains c) Parallel to d) is at 45
o
with
116
21. The point of intersection of the straight lines
3
1
2
1
1
1 +
·
·
− z y x
and
2
2
1
5
2
2 +
·
·
+ z y x
is
a) (2, 3, 1) b) (2, 3, 2) c) (3, 3, 1) d) (3, 3, 2)
22. The straight line
4
3
3
2
2
1 −
·
·
− z y x
is non-coplanar with one of the following
straight lines.
a)
2
11
1
8
2
2 −
·
·
− z y x
b)
4
13 3
1
3
2
5 3 −
·
·
− z y x
c)
5
5
4
4
3
2 −
·
·
− z y x
d)
2
1
3
1 −
·
+
·
z y
x
23. The shortest distance between the straight lines x+y = 0 = y+z and x+y = 0 =
x+y+z-a is
a)
6
a
b)
3
2a
c)
3
2a
d) a
3
2
24. The equation x
2
+y
2
-z
2
=0 represents a
a) Sphere b) Cone c) Cylinder d) Two planes
25. The equation y
2
+z
2
=1, in space, represents a
a) Cylinder b) cone c) Sphere d) none of the three
117
Notes by Prof. A.T.Eshwar, PES, Mandya
Visvesvaraya Technological University, Belgaum
WEB-BASED EDUCATION
Engineering Mathematics – I (I Semester)
(VTU Sub.Code:MAT -11)
Syllabus
PART –A
1. Analytical geometry in 3 dimensions
Direction cosines and direction ratios - planes - straight lines -Angle between
planes/straight lines - coplanar lines-shortest distance between skew lines, right circular
cone and right circular cylinder.
PART - B
2. Differential calculus
Determination of n
th
derivatives of standard functions, Leibnitz's theorem (without
proof) - problems only. Polar curves - Angle between the radius vector and the tangent -
pedal equations of polar curves only. Partial differentiation: Euler's Theorem. Total
differentiation.
Differentiation of Composite and implicit functions - Jacobins - Errors and
approximations. -Illustrative Engg. Oriented problems.
PART – C
3. Integral calculus
Reduction formulae for the functions sin
n
x, cos
n
x, tan
n
x, cot
n
x, sec
n
x, cosec
n
x and
sin
m
x cos
n
x - Evaluation of these integrals with standard limits - problems. Tracing of
standard curves in Cartesian form, parametric form and Polar form,
Applications to find area, length, volume and surface area.
PART - D
4. Differential equations
Solutions of 1
st
order & 1
st
degree equation- variables separable - Homogeneous and
Non- Homogeneous, Exact equations and reducible to exact form, Linear and Bernoulli’s
equations. Orthogonal trajectories of Cartesian and polar forms. (Use of initial condition
should be emphasized).-Illustrative examples from engineering field.
5. Infinite Series
Convergence, divergence and oscillation of an infinite series, Comparison Test, P-series,
D'Alembert's ratio test, Raabe's test, Cauchy's root test, Cauchy's integral test (All tests
without proof) for series of positive terms. Alternating series. Absolute and conditional
convergence, Leibnitz's test (without proof)
118
Text Book:
1. Higher Engineering Mathematics by B.S.Grewal, 36
th
Edition - July 2001.
Reference book:
2. Advanced Engineering Mathematics: - by E. Kreyszig, John Wiley & Sons, 6
th
Ed.
• Engineering Applications of MAT-11 at glance
• Contents of Part-B
A.T.Eswara, PESCE, Mandya
Sl.
No
Engg. Maths -I
(Code:MAT 11)
Applications
1
PART –A
Analytical
Geometry
in 3 dimensions
Wave Theory in E&C; Filed Theory in E&E;
Drawings in Mech. Engg.; Applied Mechanics in Civil
Engg.,
2 PART - B
Differential calculus
Basic of Applied Mechanics, Bending, deflection torsion,
in Civil & Mech. Engg.; Spreading the functions &
Electron beam deflections in E&C and E&E
3
Part – C
Integral Calculus
Basic Thermodynamics for Mech.Engg., Area, Volume
and Surface area, Tracing of Curves for all branches of
Engg.,
4
PART - D
Differential
equations
Mathematical Modeling in circuits theory for E&E,
E&C, CS&E, IS&E, Mechanical Vibrations, Deflection
of beams,Whilring of shafts in Civil/ Mech.Engg
5
PART - D
Infinite
Series
Theory of signals in E&E, E&C; Series solutions of
Differential Equations in all branches; Description of
Periodic phenomena
119
Differential Calculus
(Part-B)
Chapter – 1
LESSON -1 : Successive Differentiation
• In this lesson, the idea of differential coefficient of a function and its successive
derivatives will be discussed. Also, the computation of n
th
derivatives of some
standard
functions is presented through typical worked examples.
1.0 Introduction:- Differential calculus (DC) deals with problem of calculating
rates of
change.When we have a formula for the distance that a moving body covers as a
function of
time, DC gives us the formulas for calculating the body’s velocity and
acceleration at any
instant.
• Definition of derivative of a function y = f(x):-
Fig.1. Slope of the line PQ is
x
x f x x f
− ∆ + ) ( ) (
120
Chapter 1
Successive
Differentiation
Chapter 3
Partial
Differentiation
Chapter 2
Polar Curves
Lesson-1
n
th
derivative
of standard
functions
Lesson-2
Leibnitz`
s theorem
Lesson-1
Angle
between
polar curves
Lesson-2
Pedal
equations
Lesson-1
Euler’s
theorem
Lesson-2
Total derivative,
differentiation of
composite &
implicit functions
Lesson-3
Applications to
Jacobians, Errors
& Approximations
The derivative of a function y = f(x) is the function
) (x f ′
whose value at each x is
defined as
dx
dy
=
) (x f ′
= Slope of the line PQ (See Fig.1)
=
0
lim
→ ∆x
x
x f x x f
− ∆ + ) ( ) (
-------- (1)
=
0
lim
→ ∆x
(Average rate change)
= Instantaneous rate of change of f at x provided the limit exists.
The instantaneous velocity and acceleration of a body (moving along a line) at any instant
x is the derivative of its position co-ordinate y = f(x) w.r.t x, i.e.,
Velocity =
dx
dy
=
) (x f ′
--------- (2)
And the corresponding acceleration is given by
Acceleration ) (
2
2
x f
dx
y d
′ ′ · · ---------- (3)
• Session - 1
1.1 Successive Differentiation:-
The process of differentiating a given function again and again is called as
Successive differentiation and the results of such differentiation are called
successive derivatives.
• The higher order differential coefficients will occur more frequently in spreading
a function all fields of scientific and engineering applications.
• Notations:
i.
dx
dy
,
2
2
dx
y d
,
3
3
dx
y d
,…….., n
th
order derivative:
n
n
dx
y d
ii
) (x f ′
,
) (x f ′ ′
,
) (x f ′ ′ ′
,…..., n
th
order derivative: ) (x f
n
iii Dy, y D
2
, y D
3
,………..., n
th
order derivative: y D
n
iv
y′
,
y ′ ′
,
y ′ ′ ′
,……, n
th
order derivative:
) (n
y
v.
1
y ,
2
y , 3
y
…, n
th
order derivative: n
y
• Successive differentiation – A flow diagram
Input function:
) (x f y ·
Output function
) (x f
dx
df
y ′ · · ′
(first order
derivative)
Input function
) (x f y ′ · ′
Output function ) (
2
2
x f
dx
f d
y ′ ′ · · ′ ′ (second
order derivative)
121
Operation
dx
d
Operation
d x
d
Input function
) (x f y ′ ′ · ′ ′
Output function ) (
3
3
x f
dx
f d
y ′ ′ ′ · · ′ ′ ′ (third order
derivative)
-----------------------------------------------------------------------------------------------------------
------
Input function ) (
1 1
x f y
n n − −
· Output function ) (x f
dx
f d
y
n
n
n
n
· · (nth
order
derivative)
Animation Instruction
(Successive Differention-A flow diagram)
Output functions are to appear after operating
on Input functions, successively.
• 1.1 Solved Examples :
1. If
) sin(sin x y ·
, prove that 0 cos tan
2
2
2
· + + x y
dx
dy
x
dx
y d
Solution: Differentiating
) sin(sin x y ·
--------- (1) w.r.t.x, we get
x x
dx
dy
y cos ). cos(sin
1
· ·
-------------- (2)
Again differentiating
dx
dy
y ·
1
w.r.t.x gives
[ ] x x x x x
dx
y d
y cos ) sin(sin ( cos ) sin )( cos(sin
2
2
2
− + − · · Using product rule
[ ] ) sin(sin cos ) cos(sin sin
2
2
2
2
x x x x
dx
y d
y + − · ·
i.e.
]
]
]
+ − · ) sin(sin cos ) cos(sin cos
cos
sin
2
2
x x x x
x
x
y
[ ] xy xy y
2
1 2
cos tan + − · , using Eqs. (1) and (2)
or 0 cos tan
2
1 2
· + + xy xy y
or 0 cos tan
2
2
2
· + + x y
dx
dy
x
dx
y d
122
Operation
dx
d
Operation
d x
d
Operation
dx
d
2. If
) (
) (
d cx
b ax
y
+
+
·
, show that
2
2 3 1
3 2 y y y ·
Solution: We rewrite
) (
) (
d cx
b ax
y
+
+
·
, by actual division of ax+b by c x+d, as
( )
1
1
+ + ·
+
,
`
.
|
− + · d cx k
c
a
d cx c
b
c
a
y
---------- (1)where
,
`
.
|
− ·
c
b k
Differentiating (1) successively thrice, we get
( )
2
1
+ − · · d cx kc y
dx
dy
---------- (2)
( )
3 2
2
2
2
2
+ − · · d cx kc y
dx
y d
---------- (3)
( )
4 3
3
3
3
6
+ − · · d cx kc y
dx
y d
---------- (4) From (2), (3) and (4) we get
{ ¦{ ¦ [ ]
4 3 2
3 1
) ( 6 ) ( 2 2
− −
+ − + − · d cx kc d cx kc y y
6 4 2
3 1
) ( 12 2
+ · d cx c k y y
[ ]
2
3 2
3 1
) ( 2 3 2
+ − · d cx kc y y
Therefore
2
2 3 1
3 2 y y y · , as desired.
3. If t x sin · ,
pt y sin ·
, Prove that 0 ) 1 (
2
2
2
2
· + − − y p
dx
dy
x
dx
y d
x
Solution: Note that the function is given in terms a parameter t. So we find,
t
dt
dy
cos ·
and
pt p
dt
dy
cos ·
, so that
t
p t p
d x
d y
y
d t
d x
d t
d y
c o s
c o s
1
· · · . Squaring on both sides
( )
2
2 2
2
2 2
2
2 2
2
1
1
) 1 (
sin 1
) sin 1 (
cos
cos
x
y p
t
pt p
t
pt p
y
·
· · (by data)
∴( )( ) ( )
2 2 2
1
2
1 1 y p y x − · − .
Differentiating this equation w.r.t x, we get
( ) ) 2 ( ) 2 ( ) ( 2 1
1
2 2
1 2 1
2
yy p x y y y x − · − + − .
Canceling
1
2y throughout, this becomes
( ) y p xy y x
2
1 2
2
1 − · − −
or ( ) 0 1
2
1 2
2
· + − − y p xy y x
i.e. 0 ) 1 (
2
2
2
2
· + − − y p
dx
dy
x
dx
y d
x
4. If
) sin (cos t t t a x + ·
,
) cos (sin t t t a y − ·
, find
2
2
dx
y d
123
Solution:
( ) t at t t t t a
dt
dy
cos sin cos sin · + + − ·
( ) t at t t t t a
dt
dy
sin cos sin cos · − + ·
∴ t
t a t
t a t
d x
d y
d t
d x
d t
d y
t a n
c o s
s i n
· · ·
,
`
.
|
Hence,
t at t at
t
dx
dt
t
dx
y d
3
2 2
2
2
cos
1
cos
1
sec sec ·
,
`
.
|
·
,
`
.
|
·
5. If ( )
a
x
a y cosh · , prove that
2
1
2
2
2
1 y y a + ·
Solution: ( )( ) ( ), sinh
1
sinh
1
a
x
a a
x
a
dx
dy
y · · · and
( )( ),
1
cosh
2
2
2
a a
x
dx
y d
y · ·
∴ ( )
a
x
ay cosh
2
· , so that ( )
a
x
y a
2 2
2
2
cosh ·
i.e. ( )
2
1
2 2
2
2
1 sinh 1 y
a
x
y a + · + · , as desired.
• Problem Set No. 1.1 for practice.
1. If bx e y
ax
sin · , prove that ( ) 0 2
2 2
1 2
· + + − y b a ay y
2. If 1 2
2 2
· + + by hxy ax , prove that
( )
3
2
2
2
by hx
ab h
dx
y d
+
·
3. If ) cos( c t l Ae y
kt
+ ·
, show that ( ) 0 2
2 2
2
2
· + + + y l k
dx
dy
k
dx
y d
4. If ( )
2
1 log x x y + + · , prove that ( ) 0 1
2
2
2
· + +
dx
dy
x
dx
y d
x
5. If ( ) x y sinh tan
1 −
· , prove that 0 tan
2
2
2
·
,
`
.
|
+
dx
dy
y
dx
y d
6. If
( )
2
t a n l o g c o s
t
t a x + ·
,
t a y sin ·
, find
2
2
dx
y d
Ans:
t a
t
4
cos
sin
7. Find
2
2
dx
y d
, when θ
3
cos a x · , θ
3
sin b y · Ans:
2
4
3
sec cos
a
ec b θ θ
124
8. Find
3
3
dx
y d
, where
,
`
.
|
·
2
1
1
tan
x
x
y
Ans:
( )
2
5
2
2
1
2 1
x
x
+
9. If t t x 2 cos cos 2 − · , t t x 2 sin sin 2 − · , Find
2
2
2
π
·
,
`
.
|
x
dx
y d
Ans: -3/2
10. If
x x
be e xy
+ · , prove that 0 2
2
2
· − + xy
dx
dy
dx
y d
x
• Session -2
1.2 Calculation of n
th
derivatives of some standard functions
• Below, we present a table of n
th
order derivatives of some standard functions for
Table : 1
125
• We proceed to illustrate the proof of some of the above results, as only the above
functions are able to produce a sequential change from one derivative to the
other. Hence, in general we cannot obtain readymade formula for nth derivative
of functions other than the above.
1. Consider
mx
e . Let
mx
e y · . Differentiating w.r.t x, we get
·
1
y
mx
me . Again differentiating w.r.t x, we get
( )
mx
me m y ·
2
=
mx
e m
2
Similarly, we get
3
y
=
mx
e m
3
Sl.
No
y = f(x)
y D
dx
y d
y
n
n
n
n
· ·
1
mx
e
mx n
e m
2
mx
a ( )
mx n n
a a m log
3
( )
m
b ax + i. ( )( ) ( ) ( )
n m n
b ax a n m m m m
+ + − − − 1 .... 2 1 for all
m
.
ii. 0 if
n m <
iii.
( ) ! n
a
n
if
n m ·
iv.
( )
n m
x
n m
m
− !
!
if
n m <
4
( ) b ax +
1
n
n
n
a
b ax
n
1
) (
! ) 1 (
+
+
5.
( )
m
b ax +
1
n
n m
n
a
b ax m
n m
+
+ −
− + −
) ( ! ) 1 (
! ) 1 ( ) 1 (
6.
) log( b ax +
n
n
n
a
b ax
n
) (
! ) 1 ( ) 1 (
1
+
− −
7.
) sin( b ax +
)
2
sin(
π
n b ax a
n
+ +
8.
) cos( b ax +
)
2
cos(
π
n b ax a
n
+ +
9. ) sin( c bx e
ax
+
) sin( θ n c bx e r
ax n
+ + ,
) ( t a n
1 2 2
a
b
b a r
· + · θ
10. ) cos( c bx e
ax
+
) cos( θ n c bx e r
ax n
+ + ,
) ( t a n
1 2 2
a
b
b a r
· + · θ
126
4
y
=
mx
e m
4
…………….
And hence we get
n
y
=
mx n
e m ∴ [ ] ·
mx
n
n
e
dx
d
mx n
e m .
2. ( )
m
b ax + (See Sl. No-3 of Table-1 )
let
· y
( )
m
b ax + Differentiating w.r.t x,
1
y =
m ( ) a b ax
m 1 −
+ . Again differentiating w.r.t x, we get
2
y =
m
( ) 1 − m
( )
2 2
a b ax
m−
+
Similarly, we get
3
y
=
m
( ) 1 − m
( ) 2 − m
( )
3 3
a b ax
m−
+
………………………………….
And hence we get
n
y
=
m
( ) 1 − m
( ) 2 − m
……….
( ) 1 + −n m
( )
n n m
a b ax
+ for all m.
Case (i) If
n m ·
(m-positive integer),then the above expression becomes
n
y
=
n ( ) 1 − n ( ) 2 − n
……….3.2.1( )
n n n
a b ax
+
i.e. ( )
n
n
a n y ! ·
Case (ii) If m<n,(i.e. if n>m) which means if we further differentiate the above
expression, the
right hand site yields zero. Thus ( ) [ ] ( ) n m if b ax D
m n
< · + 0
Case (iii) If m>n, then ( )( ) ( )( )
n n m
n
a b ax n m m m m y
+ + − − − · 1 ...... 2 1 becomes
( )( ) ( )( )
( )
( )
n n m
a b ax
n m
n m n m m m m
+
− + − − −
·
!
! 1 ...... 2 1
. i.e
( )
( )
n n m
n
a b ax
n m
m
y
+
·
!
!
3.
( )
m
b ax +
1
(See Sl. No-5 of Table-1 )
( )
( )
m
m
b ax
b ax
y Let
+ ·
+
·
1
Differentiating w.r.t x
( ) ( ) ( )
( )
a b ax m a b ax m y
m m 1 1
1
1
+ − − −
+ − · + − ·
( ) ( ) ( ) ( )
( )
[ ] ( ) ( ) ( )
( ) 2 2 2 1 1
2
1 1 1 1 a b ax m m a b ax m m y
m m + − − + −
+ + − · + + − − ·
Similarly, we get ( ) ( )( )( )
( ) 3 3 3
3
2 1 1 a b ax m m m y
m+ −
+ + + − ·
( ) ( ) ( ) ( ) ( )
( ) 4 4 4
4
3 2 1 1 a b ax m m m m y
m+ −
+ + + + − ·
……………………………
( ) ( )( ) ( )( )
( ) n n m n
n
a b ax n m m m m y
+ −
+ − + + + − · 1 ..... 2 1 1
This may be rewritten as
127
( ) ( )( ) ( ) ( )
( )
( )
( ) n n m
n
n
a b ax
m
m m m n m n m
y
+ −
+
− + − + − + −
·
! 1
! 1 1 ..... 2 1 1
or
( ) ( )
( ) ( )
n
n m
n
n
a
b ax m
n m
y
+
+ −
− + −
·
! 1
! 1 1
4.
( ) b ax +
1
(See Sl. No-4 of Table-1 )
Putting
, 1 · m
in the result
n
n m
n
m
n
a
b ax m
n m
b ax
D
+
+ −
− + −
·
]
]
]
+ ) ( ! ) 1 (
! ) 1 ( ) 1 (
) (
1
we get
n
n
n
n
a
b ax
n
b ax
D
+
+ −
− + −
·
]
]
]
+
1
) ( ! ) 1 1 (
! ) 1 1 ( ) 1 (
) (
1
or
n
n
n
n
a
b ax
n
b ax
D
+
+
·
]
]
]
+
1
) (
! ) 1 (
) (
1
• 1.2.1. Worked Examples:-
In each of the following Questions find the n
th
derivative after reducing them into
standard functions given in the table 1.2.1
1. (a) ) 1 9 log(
2
− x (b) [ ]
7 5
) 3 4 ( log
+
+
x
e x (c)
6
2
10
) 1 (
) 3 2 ( ) 5 3 (
log
+
− +
x
x x
Solution: (a) Let { ¦ ) 1 3 )( 1 3 ( log ) 1 9 log(
2
− + · − · x x x y
) 1 3 log( ) 1 3 log( − + + · x x y
(
B A AB log log ) log( + ·
)
∴ { ¦ { ¦ ) 1 3 log( ) 1 3 log( − + + · x
dx
dn
x
dx
dn
y
n n
n
i.e
n
n
n
n
n
n
n
x
n
x
n
y ) 3 (
) 1 3 (
! ) 1 ( ) 1 (
) 3 (
) 1 3 (
! ) 1 ( ) 1 (
1 1
− −
+
+
− −
·
− −
(b) Let [ ]
7 5 7 5
log ) 3 4 log( ) 3 4 ( log
+ +
+ + · + ·
x x
e x e x y
e x x
e
log ) 7 5 ( ) 3 4 log( + + + ·
( A B A
B
log log · )
) 7 5 ( ) 3 4 log( + + + · x x y
(
1 log · e
e )
0 ) 4 (
) 3 4 (
! ) 1 ( ) 1 (
1
+
+
− −
·
n
n
n
n
x
n
y
5 ) 6 5 ( · + x D
0 ) 6 5 (
2
· + x D
) 1 ( 0 ) 1 5 ( > · + n x D
n
(c) Let
6
2
10
) 1 (
) 3 2 ( ) 5 3 (
log
+
− +
·
x
x x
y
128
¹
¹
¹
'
¹
¹
¹
¹
'
¹
+
− +
·
6
2
) 1 (
) 3 2 ( ) 5 3 (
10 log
1
x
x x
e
10 log
log
log
10
e
e
X
X ·
¹
'
¹
¹
'
¹
¹
'
¹
¹
'
¹
+
− +
·
6
2
) 1 (
) 3 2 ( ) 5 3 (
log
2
1
10 log
1
x
x x
e
A B A
B
log log ·
B A
B
A
log log log − ·
,
`
.
|
{ ¦
6 2
) 1 log( ) 3 2 log( ) 5 3 log(
10 log 2
1
+ − − + + · x x x
e
{ ¦ ) 1 log( 6 ) 3 2 log( ) 5 3 log( 2
10 log 2
1
+ − − + + · x x x y
e
Hence,
¹
'
¹
¹
'
¹
+
− −
− −
− −
+
+
− −
·
− − −
n
n
n
n
n
n
n
n
n
e
n
x
n
x
n
x
n
y ) 1 (
) 1 (
! ) 1 ( ) 1 (
. 6 ) 3 (
) 3 2 (
! ) 1 ( ) 1 (
) 3 (
) 5 3 (
! ) 1 ( ) 1 (
. 2
10 log 2
1
1 1 1
2. (a)
4 2 4 2
6
+ +
+
x x
e (b) x x 4 cosh 4 cosh
2
+
(c) x x e
x
2 cosh 3 sinh
(d)
5
4
) 8 6 (
) 4 5 (
1
) 5 4 (
1
+ +
+
+
+
x
x x
Solution: (a) Let
4 2 4 2
6
+ +
+ ·
x x
e y
4 2 4 2
6 6
x x
e e + ·
∴ ) 6 ( 1296 ) (
2 2 4 x x
e e y + ·
hence ) 6 ( 1296 ) (
2 2 4 x
n
x
n
n
dx
dn
e
dx
dn
e y + ·
{ ¦ { ¦
x n n x n
e e
2 2 4
6 ) 6 (log 2 1296 2 + ·
(b) Let x x y 4 cosh 4 cosh
2
+ ·
2
4 4 4 4
2 2
,
`
.
| +
+
,
`
.
| +
·
− − x x x x
e e e e
( ) { ¦ ) )( ( 2 ) ( ) (
4
1
2
1
4 4 2 4 2 4 4 4 x x x x x x
e e e e e e
− − −
+ + + + ·
( ) { ¦ 2
4
1
2
1
8 8 4 4
+ + + + ·
− − x x x x
e e e e y
hence,
[ ] [ ] 0 ) 8 ( 8
4
1
) 4 ( 4
2
1
8 8 4 4
+ − + + − + ·
− − n n n n x n x n
n
e e e e y
(c) Let x x e y
x
2 cosh 3 sinh
·
¹
'
¹
¹
'
¹ +
¹
'
¹
¹
'
¹ −
·
− −
2 2
2 2 3 3 x x x x
x
e e e e
e
129
{ ¦ ) )( (
4
2 2 3 3 x x x x
x
e e e e
e
− −
+ − ·
{ ¦
x x x x
x
e e e e
e
5 5
4
− −
− + − ·
{ ¦
x x x
e e e
6 2 4
1
4
1
− −
− + − ·
{ ¦
x x x
e e e y
6 2 4
1
4
1
− −
− − + ·
Hence,
{ ¦
x n x n x n
n
e e e y
6 2 4
) 6 ( ) 2 ( ) 4 ( 0
4
1
− −
− − − − + ·
(d) Let
5
4
) 8 6 (
) 4 5 (
1
) 5 4 (
1
+ +
+
+
+
· x
x x
y
Hence,
( ) { ¦
5
4
8 6
) 4 5 (
1
) 5 4 (
1
+ +
¹
'
¹
¹
'
¹
+
+
¹
'
¹
¹
'
¹
+
· x
dx
dn
x dx
dn
x dx
dn
y
n n n
n
0 ) 5 (
) 4 5 ( ! ) 1 4 (
! ) 1 4 ( ) 1 (
) 4 (
) 5 4 (
! ) 1 (
4 1
+
+ −
− + −
+
+
·
+ +
n
n
n
n
n
n
x
n
x
n
i.e
n
n
n
n
n
n
n
x
n
x
n
y ) 5 (
) 4 5 ( ! 3
! ) 3 ( ) 1 (
) 4 (
) 5 4 (
! ) 1 (
4 1 + +
+
+ −
+
+
·
• Session - 3
• 1.2.2 Worked examples:-
1. (i)
8 6
1
2
+ − x x
(ii)
3 2
1
1
x x x + − −
(iii)
6 7 2
2
2
+ + x x
x
(iv)
9 12 4
1
1
2
2
+ +
+
,
`
.
|
+
+
x x x
x
(v) ( )
a
x
1
tan
(vi) x
1
tan
(vii)
,
`
.
|
+
x
x
1
1
tan
1
• In all the above problems, we use the method of partial fractions to reduce
them into standard forms.
130
Solutions: (i) Let
8 6
1
2
+ −
·
x x
y . The function can be rewritten as
) 2 )( 4 (
1
− −
·
x x
y
• This is proper fraction containing two distinct linear factors in the denominator.
So, it can be split into partial fractions as
) 2 ( ) 4 ( ) 2 )( 4 (
1
+
·
− −
·
x
B
x
A
x x
y
Where the constant A and B are found
as given below.
) 2 )( 4 (
) 4 ( ) 2 (
) 2 )( 4 (
1
− −
− + −
·
− − x x
x B x A
x x
) 4 ( ) 2 ( 1 − + − · x B x A
-------------(*)
Putting x = 2 in (*), we get the value of B as
2
1
− · B
Similarly putting x = 4 in(*), we get the value of A as
2
1
· A
2
) 2 / 1 (
4
) 2 / 1 (
) 2 )( 4 (
1
+
·
− −
·
x x x x
y
Hence
,
`
.
|
,
`
.
|
·
2
1
2
1
4
1
2
1
x dx
d
x dx
d
y
n
n
n
n
n
]
]
]
]
]
]
·
+ +
n
n
n
n
n
n
x
n
x
n
) 1 (
) 2 (
! ) 1 (
2
1
) 1 (
) 4 (
! ) 1 (
2
1
1 1
]
]
]
− ·
+ + 1 1
) 2 (
1
) 4 (
1
! ) 1 (
2
1
n n
n
x x
n
(ii) Let
) 1 )( 1 (
1
) 1 ( ) 1 (
1
1
1
2 2 3 2
x x x x x x x x
y
− −
·
− − −
·
+ − −
·
ie
) 1 ( ) 1 (
1
) 1 )( 1 )( 1 (
1
2
x x x x x
y
+ −
·
+ − −
·
Though y is a proper fraction, it contains a repeated linear factor
2
) 1 ( x − in its
denominator. Hence, we write the function as
x
C
x
B
x
A
y
+
+
+
·
1 ) 1 ( ) 1 (
2
in terms of partial fractions. The
constants A, B, C
are found as follows:
131
x
C
x
B
x
A
x x
y
+
+
+
·
+ −
·
1 ) 1 ( ) 1 ( ) 1 ( ) 1 (
1
2 2
ie
2
) 1 ( ) 1 ( ) 1 )( 1 ( 1 x C x B x x A − + + + + − · -------------(**)
Putting x = 1 in (**), we get B as
2
1
· B
Putting x = -1 in (**), we get C as
4
1
· C
Putting x = 0 in (**), we get C B A + + · 1
4
1
4
1
2
1
1 1 · − − · − − · C B A
4
1
· A
Hence,
) 1 (
) 4 / 1 (
) 1 (
) 2 / 1 (
) 1 (
) 4 / 1 (
2
x x x
y
+
+
+
·
]
]
]
+
+
]
]
]
− −
− + −
+
]
]
]
·
+ + +
n
n
n
n
n
n
n
n
n
n
x
n
x
n
x
n
y ) 1 (
) 1 (
! ) 1 (
4
1
) 1 (
) 1 ( ! ) 1 2 (
! ) 1 2 ( ) 1 (
2
1
) 1 (
) 1 (
! ) 1 (
4
1
1 2 1
]
]
]
+ −
+ −
+
]
]
]
+
+
− ·
+ +
2 ) 1 (
! ) 1 ( ) 1 (
2
1
) 1 (
1
) 1 (
1
! ) 1 (
4
1
1 1
n x
n
x x
n
n
n n
n
(iii) Let
6 7 2
2
2
+ +
·
x x
x
y (VTU July-05)
This is an improper function. We make it proper fraction by actual division
and later
spilt that into partial fractions.
i.e
6 7 2
) 3 (
2
1
) 6 7 2 (
2
2
7
2 2
+ −
− −
+ · + + ÷
x x
x
x x x
) 2 )( 3 2 (
3
2
1
2
7
+ +
+ ·
x x
x
y Resolving this proper fraction into partial
fractions, we get
]
]
]
+
+
+
+ ·
) 2 ( ) 3 2 ( 2
1
x
B
x
A
y
. Following the above examples for finding A &
B, we get
]
]
]
+
+
+
+ ·
2
) 4 (
3 2 2
1
2
9
x x
y
Hence,
]
]
]
+
]
]
]
+
+ ·
+ +
n
n
n
n
n
n
n
x
n
x
n
y ) 1 (
) 2 (
! ) 1 (
4 ) 2 (
) 3 2 (
! ) 1 (
2
9
0
1 1
132
i.e
]
]
]
+
+
− ·
+ + 1 1
2
9
) 2 (
4
) 3 2 (
) 2 (
! ) 1 (
n n
n
n
n
x x
n y
(iv) Let
9 12 4 ) 1 (
) 2 (
2
+ +
+
+
+
·
x x
x
x
x
y
(i) (ii)
Here (i) is improper & (ii) is proper function. So, by actual division (i)
becomes
,
`
.
|
+
+ ·
,
`
.
|
+
+
1
1
1
1
2
x x
x
. Hence, y is given by
2
) 3 2 (
1
1
1
1
+
+
,
`
.
|
+
+ ·
x x
y
[ 9 12 4 ) 3 2 (
2 2
+ + · + x x x ]
Resolving the last proper fraction into partial fractions, we get
2 2
) 3 2 ( ) 3 2 ( ) 3 2 ( +
+
+
·
+ x
B
x
A
x
x
. Solving we get
2
1
· A and
2
3
− · B
]
]
]
+
+
+
+
,
`
.
|
+
+ ·
2
2
3
2
1
) 3 2 ( ) 3 2 ( 1
1
1
x x x
y
]
]
]
+ +
+ −
]
]
]
+
+
]
]
]
+
+ ·
+
n
n
n
n
n
n
n
n
n
n x
n
x
n
x
n
y ) 2 (
2 ) 3 2 (
! ) 1 ( ) 1 (
2
3
) 2 (
) 3 2 (
! ) 1 (
2
1
) 1 (
) 1 (
! ) 1 (
0
1
(v) ( )
a
x
1
tan
Let ( )
a
x
y
1
tan
·
( )
2 2 2
1
1
1
1
a x
a
a
a
x
y
+
·
,
`
.
|
+
·
,
`
.
|
+
· · ·
− −
2 2
1
1
1
) (
a x
a
D y D y D y
n n n
n
Consider
) )( (
2 2
ai x ai x
a
a x
a
− +
·
+
) ( ) ( ai x
B
ai x
A
+
+
·
, on resolving into partial fractions.
( ) ( )
) (
2
1
) (
2
1
ai x
i
ai x
i
+
+
·
, on solving for A & B.
133
,
`
.
|
+
,
`
.
|
+
·
,
`
.
|
+
− − −
a i x
D
a i x
D
a x
a
D
i n i n n 2
1
1 2
1
1
2 2
1
]
]
]
− −
,
`
.
|
+
]
]
]
+
− −
,
`
.
|
− ·
− −
n
n
n
n
ai x
n
i ai x
n
i ) (
! ) 1 ( ) 1 (
2
1
) (
! ) 1 ( ) 1 (
2
1
1 1
-----------(*)
• Since above answer containing complex quantity i we rewrite the answer in terms of
real quantity, We take transformation
θ θ sin cos r a r x · ·
where
2 2
a x r + · ,
,
`
.
|
·
x
a
1
tan θ
( )
θ
θ θ
i
re i r ai x · + · + sin cos
( )
θ
θ θ
i
re i r ai x
· − · − sin cos
( )
n
in
in n n
r
e
e r ai x
θ
θ
· ·
1 1
,
( )
n
in
n
r
e
ai x
θ −
·
+
1
now(*) is
( ) ( )
[ ]
θ θ in in
n
n
n
e e
r i
n
y
− −
·
2
! 1 1
1
( )
( )
( ) ( )
θ θ n
r
n
n i
r i
y
n
n
n
n
n
sin
! 1 1
sin 2
2
1
1 1
− −
·
− −
(vi) Let x y
1
tan
· .Putting a = 1 in Ex.(v) we get
n
y
which is same as above with ( )
x
x r
1
tan 1
1 2 −
· + · θ
( ) θ θ cot cot
1
· ·
x or x
θ
θ
θ θ
n
n n
ec r
ec r sin
cos
1 1
cos 1 cot
2
· · ⇒ · + · ∴
( ) ( ) ( ) x where n n x D
n n n 1 1 1
cot sin sin ! 1 1 tan
− − −
· − − · θ θ θ
(vii) Let
,
`
.
|
+
·
x
x
y
1
1
tan
1
put θ tan · x x
1
tan
· θ
]
]
]
+
·
θ
θ
tan 1
tan 1
tan
1
y
[ ] ) t a n ( t a n
4
1
θ
π
+ ·
( )
,
`
.
|
+
· +
θ
θ
θ
π
tan 1
tan 1
4
tan
) ( tan
4
4
1
x
+ · + ·
π
θ
π
) ( tan
4
1
x y
+ ·
π
) (tan 0
1
x D y
n
n
+ ·
134
]
]
]
− −
,
`
.
|
+
]
]
]
+
− −
,
`
.
|
− ·
− −
n
n
n
n
ai x
n
i ai x
n
i ) (
! ) 1 ( ) 1 (
2
1
) (
! ) 1 ( ) 1 (
2
1
1 1
Problem set No. 1.2.1 for practice
Find the n
th
derivative of the following functions:
1.
) 4 )( 1 (
6
2
− − x x
x
2.
) 1 2 )( 2 (
2
+ − + x x x
x
3.
2 2
1 4
2 3
2
− − +
+ +
x x x
x x
4.
3 4
2
− − x x
x
5.
2 3
2
3
+ − x x
x
6.
,
`
.
|
2
1
1
2
tan
x
x
7.
,
`
.
|
− +
x
x 1 1
tan
2
1
• Session 4
1.
) sin( b ax +
.(See Sl. No-7 of Table-1 )
Let
) sin( b ax y + ·
. Differentiating w.r.t x,
a b ax y ). cos(
1
+ · As
X X cos )
2
sin( · +
π
135
We can write
). 2 / sin(
1
π + + · b ax a y
Again Differentiating w.r.t x,
a b ax a y ). 2 / cos(
2
π + + · Again using
X X cos )
2
sin( · +
π
,we get
2
y as
a b ax a y ). 2 / 2 / sin(
2
π π + + + ·
i.e. ). 2 / 2 sin(
2
2
π + + · b ax a y
Similarly, we get
). 2 / 3 sin(
3
3
π + + · b ax a y
). 2 / 4 sin(
4
4
π + + · b ax a y
……………………………
). 2 / sin( π n b ax a y
n
n
+ + ·
2. ( ). sin c bx e
ax
+ (See Sl. No-9 of Table-1 )
( ) ) 1 .....( sin c bx e y Let
ax
+ ·
Differentiating using product rule ,we get
·
1
y ( ) ( )
ax ax
ae c bx b c bx e + + + sin cos
i.e. ·
1
y ( ) ( ) [ ] c bx b c bx a e
ax
+ + + cos sin . For computation of higher order
derivatives
it is convenient to express the constants ‘a’ and ‘b’ in terms of the
constants r and
θ defined by
θ cos r a ·
&
θ sin r b ·
,so t
2 2
b a r + · and
( )
a
b
1
tan
· θ .thus,
1
y can be rewritten as
( ) ( ) ( ) ( ) [ ] c bx r c bx r e y
ax
+ + + · cos sin sin cos
1
θ θ
or ( ) ( ) [ ] } cos cos cos {sin
1
θ θ c bx c bx r e y
ax
+ + + ·
i.e. ( ) [ ] ) 2 .( .......... sin
1
θ + + · c bx re y
ax
Comparing expressions (1) and (2), we write
2
y as
( ) θ 2 sin
2
2
+ + · c bx e r y
ax
( ) θ 3 sin
3
3
+ + · c bx e r y
ax
Continuing in this way, we get
( ) θ 4 sin
4
4
+ + · c bx e r y
ax
( ) θ 5 sin
5
5
+ + · c bx e r y
ax
…………………………….
( ) θ n c bx e r y
ax n
n
+ + · sin
( ) [ ] ( ), sin sin θ n c bx e r c bx e D
ax n ax n
+ + · + where
2 2
b a r + · &
( )
a
b
1
tan
· θ
• 1.2.3 Worked examples
136
1. (i) x x
3 2
cos sin + (ii) x
3 3
cos sin (iii) x x x 3 cos 2 cos cos
(iv) x x x 3 sin 2 sin sin (v) x e
x
2 cos
3
(vi) ( ) x x e
x 3 2 2
cos sin +
• The following formulae are useful in solving some of the above problems.
(i)
2
2 cos 1
cos ) (
2
2 cos 1
sin
2 2
x
x ii
x
x
+
·
·
(iii) x x x iv x x x cos 3 cos 4 3 cos ) ( sin 4 sin 3 3 sin
3 3
− · − ·
(v)
( ) ( ) B A B A B A − + + · sin sin cos sin 2
(vi)
( ) ( ) B A B A B A − − + · sin sin sin cos 2
(vii)
( ) ( ) B A B A B A − + + · cos cos cos cos 2
(viii)
( ) ( ) B A B A B A + − − · cos cos sin sin 2
Solutions: (i) Let
( ) x x
x
x x y cos 3 3 cos
4
1
2
2 cos 1
cos sin
3 2
+ +
,
`
.
| −
· + ·
( ) ( ) [ ] ( ) ( ) ( ) [ ]
2
cos 3
2
3 cos 3
4
1
2
2 cos 2 0
2
1
π π π n
x
n
x
n
x y
n n
n
+ + + + + − · ∴
(ii)Let y =
]
]
]
+ −
· ·
,
`
.
|
·
4
2 sin 3 6 sin
8
1
8
2 sin
2
2 sin
cos sin
3
3
3 3
x x x x
x x
[ ] x x 6 sin 2 sin 3
32
1
− ·
]
]
]
,
`
.
|
+ −
,
`
.
|
+ ·
2
6 sin 6
2
2 sin 2 . 3
32
1 π π n
x
n
x y
n n
n
(iii) )Let y = x x x 2 cos cos 3 cos
=
( ) [ ] x x x x x x 2 cos 2 cos 4 cos
2
1
2 cos 2 cos 4 cos
2
1
2
+ · +
=
( )
]
]
]
+ +
2
4 cos 1
2 cos 6 cos
2
1
2
1 x
x x
( ) x
x
x 4 cos 1
4
1
4
2 cos
6 cos
4
1
− + + ·
4
2
4 cos 4
4
2
2 cos 2
2
6 cos 6
4
1
,
`
.
|
+
,
`
.
|
+
+
,
`
.
|
+ · ∴
π π
π
n
x
n
x
n
x y
n n
n
n
(iv) )Let y =
x sn x x 2 sin 3 sin
( ) [ ] x x x 2 sin 4 sin 2 sin
2
1
− ·
137
[ ] x x x 2 sin 4 sin 2 sin
2
1
2
− ·
=
( )
]
]
]
− −
x x
x
6 sin 2 sin
2
1
2
4 cos 1
2
1
( )
]
]
]
− −
,
`
.
| −
· x x
x
6 sin 2 sin
4
1
4
4 cos 1
]
]
]
,
`
.
|
+ +
,
`
.
|
+ −
,
`
.
|
+ ·
2
6 sin 6
2
2 sin 2
2
4 cos 4
4
1 π π π n
x
n
x
n
x y
n n n
n
(v) Let x e y
x
2 cos
3
· (Refer Sl.No. 10 of Table 1)
( ) θ n x re y
x
n
+ · ∴ 2 cos
3
where
2 2
2 3 + · r 13 · &
,
`
.
|
·
3
2
tan
1
θ
(vi) Let y = ( ) x x e
x 3 2 2
cos sin +
We know that
[ ] [ ] x x
x
x x cos 3 3 cos
4
1
2
2 cos 1
cos sin
3 2
+ +
· +
∴y =
[ ] [ ] x x
e x
e x x e
x
x x
cos 3 3 cos
4 2
2 cos 1
cos sin
2
2 3 2 2
+ +
]
]
]
· +
[ ] [ ] x e x e x e e y
x x x x
cos 3 3 cos
4
1
2 cos
2
1
2 2 2 2
+ + − · ∴
Hence,
( ) [ ] ( ) ( ) [ ]
3
2
3 2
2
2 1
2
1
2
cos 3 3 cos
4
1
2 cos 2
2
1
θ θ θ n x e r n x e r n x e r e y
x n x n x n x n
n
+ + + + + − ·
where
2 2
1
2 2 + · r 8 · ;
2 2
2
3 2 + · r 13 · ;
2 2
3
1 2 + · r 5 ·
;
2
1
tan ;
2
3
tan ;
2
2
tan
1
3
1
2
1
1
,
`
.
|
·
,
`
.
|
·
,
`
.
|
·
− − −
θ θ θ
Problem set No. 1.2.2 for practice
Find n
th
derivative of the following functions:
1. ) cos (sin
2 3
x x + 2. x x 3 cos 2 sin 3. x x 3 sin . 2 cos 4. x x 2 cos cos
5. x x 2 s i n s i n 6. ) cos (sin
2 3 3
x x e
x
+ 7. x x e
x
4 cos 2 cos 8. x x e
x
2 cos sin
2 −
.9. x e
x 3 3
cos
(VTU Jan-04)
LESSON -2 : Leibnitz’s Theorem
• Session - 1
138
• Leibnitz’s theorem is useful in the calculation of n
th
derivatives of product of two
functions.
• Statement of the theorem:
If u and v are functions of x, then
v uD v uD D C v uD D C uDv D C uv D uv D
n r r n
r
n n n n n n n
... .... ) (
2 2
2
1
1
+ + + + + ·
− − −
,
where
dx
d
D ·
, n C
n
·
1
,
( )
( )! !
!
,........,
2
1
2
r n r
n
C
n n
C
r
n n
·
·
• Worked Examples
1. If
that prove pt y t x sin , sin · ·
( ) ( ) ( ) 0 1 2 1
2 2
1 2
2
· − + + − −
+ + n n n
y n p xy n y x (VTU July-05)
Solution: Note that the function
) (x f y ·
is given in the parametric form with a
parameter t.
So, we consider
t
p t p
d x
d y
d t
d x
d t
d y
c o s
c o s
· · (p – constant)
or
2
2 2
2
2 2
2
2 2
2
1
) 1 (
sin 1
) sin 1 (
cos
cos
x
y p
t
pt p
t
pt p
dx
dy
·
· ·
,
`
.
|
or ( ) ( )
2 2 2
1
2
1 1 y p y x − · −
So that ( ) ( )
2 2 2
1
2
1 1 y p y x − − − Differentiating w.r.t. x,
( ) ( ) ( ) [ ] ( ) 0 2 2 2 1
1
2 2
1 2 1
2
· − − − + − yy p x y y y x
( ) 0 1
2
1 2
2
· + − − y p xy y x --------------- (1) [
1
2y ÷ , throughout]
Equation (1) has second order derivative
2
y in it. We differentiate (1), n times,
term wise,
using Leibnitz’s theorem as follows.
( ) [ ] 0 1
2
1 2
2
· − − − y p xy y x D
n
i.e { ¦ { ¦ ( ) 0 ) 1 (
2
1 2
2
· − − − y p D xy D y x D
n n n
---------- (2)
(a) (b) (c)
Consider the term (a):
( ) [ ]
2
2
1 y x D
n
− . Taking
2
y u · and ) 1 (
2
x v − · and applying Leibnitz’s theorem
we get
[ ] ...
3 3
3
2 2
2
1
1
+ + + + ·
− − −
v uD D C v D D C uDv D C uv D uv D
n n n n n n n n
i.e
[ ] ... ) 1 ( ) ( ) 1 ( ) ( ) 1 ( ). ( ) 1 ).( ( ) 1 (
2 3
2
3
3
2 2
2
2
2
2
2
1
1
2
2
2
2
+ − + − + − + − · −
− − −
x D y D C x D y D C x D y D C x y D x y D
n n n n n n n n
... ) 0 .( .
! 3
) 2 )( 1 (
) 2 .(
! 2
) 1 (
) 2 .( )
2 ) 3 ( 2 ) 2 ( 2 ) 1 (
2
2 ) (
+
− −
+ −
+ − + − ·
+ − + − + − + n n n n
y
n n n
y
n n
x ny x y
( ) [ ] ( )
n n n
n
y n n nxy y x y x D ) 1 ( 2 1 1
1 2
2
2
2
− − − − · −
+ +
----------- (3)
139
Consider the term (b):
[ ]
1
xy D
n
. Taking
1
y u · and
x v ·
and applying Leibnitz’s
theorem, we get
[ ] .... ) ( ). ( ) ( . ) ).( ( ) (
2
1
2
2 1
1
1 1 1
+ + + ·
− −
x D y D C x D y D C x y D x y D
n n n n n n
.... ) 0 (
! 2
) 1 (
.
2 ) 2 ( 1 ) 1 ( 1 ) (
+
+ + ·
+ − + − + n n n
y
n n
ny x y
[ ]
n n
n
ny xy xy D + ·
+1 1
---------- (4)
Consider the term (c):
n
n n
y p y D p y p D
2 2 2
) ( ) ( · · --------- (5)
Substituting these values (3), (4) and (5) in Eq (2) we get
( ) { ¦ { ¦ { ¦ 0 ) 1 ( 2 1
2
1 1 2
2
· + + − − − − −
+ + + n n n n n n
y p ny xy y n n nxy y x
ie ( ) 0 ) 1 2 ( 1
2 2
1 2
2
· + − + − + − −
+ + n n n n n n
y p ny ny y n xy n y x
∴( ) ( ) 0 ) 1 2 ( 1
2 2
1 2
2
· − + + − −
+ + n n n
y n p xy n y x as desired.
2. If ) 1 log( 2 sin
1
+ ·
x y or
[ ] ) 1 log( 2 sin + · x y
or [ ]
2
) 1 log( sin + · x y
or ) 1 2 log( sin
2
+ + · x x y , show that
( ) ( )( ) ( ) 0 4 1 1 2 1
2
1 2
2
· + + + + + +
+ + n n n
y n y x n y x (VTU Jan-03)
Out of the above four versions, we consider the function as
) 1 log( 2 ) ( sin
1
+ ·
x y
Differentiating w.r.t x, we get
,
`
.
|
+
·
1
2
) (
1
1
1
2
x
y
y
ie
2
1
1 2 ) 1 ( y y x − · +
Squaring on both sides
( ) ) 1 ( 4 1
2 2
1
2
y y x − · +
Again differentiating w.r.t x,
( ) ( ) ( ) ) 2 ( 4 ) 1 ( 2 2 1
1
2
1 2 1
2
yy x y y y x − · + + +
or ( ) ) 2 ( 4 ) 1 ( 1
1 1 2
2
y y y x y x ÷ − · + + +
or ( ) 0 4 ) 1 ( 1
1 2
2
· + + + + y y x y x -----------*
Differentiating * w.r.t x, n-times, using Leibnitz’s theorem,
{ ¦ 0 4 ) 1 ( ) 1 )( ( ) 2 )( (
! 2
) 1 (
) 1 ( 2 ) ( ) 1 (
1
1
1 2
2
2
1 2
2
· + + + +
¹
'
¹
¹
'
¹ −
+ + + +
− − −
y D y nD x g D y D
n n
x y nD x y D
n n n n n n
On simplification, we get
( ) ( )( ) ( ) 0 4 1 1 2 1
2
1 2
2
· + + + + + +
+ + n n n
y n y x n y x
140
3. If
) tan(log y x ·
, then find the value of
( ) ( )
1 1
2
) 1 ( 1 2 1
− +
− + − + +
n n n
y n n y nx y x (VTU July-04)
Consider
) tan(log y x ·
i.e. y x log tan
1
·
or
x
e y
1
tan
·
Differentiating w.r.t x,
2 2
tan
1
1 1
1
.
1
x
y
x
e y
x
+
·
+
·
∴( ) ( ) 0 1 1
1
2
1
2
· − + · + y y x ie y y x -----------*
We differentiate * n-times using Leibnitz’s theorem,
We get
( ) [ ] 0 ) ( 1
1
2
· − + y D y x D
n n
ie.
{ ¦ { ¦ 0 .... ) 1 ( ) ( ) 1 ( ) ( ) 1 )( (
2 2
1
2
2
2
1
1
1
2
1
· − + + + + + +
− −
y D x D y D C x D y D C x y D
n n n n n n
ie
0 .... 0 ) 2 (
! 2
) 1 (
) 2 ( ) 1 (
1
2
1
· −
¹
'
¹
¹
'
¹
+ +
+ + +
− + n n n n
y y
n n
x ny x y
( ) ( ) 0 ) 1 ( 1 2 1
1 1
2
· − + − + +
− + n n n
y n n y nx y x
4. If
x y y
m m
2
1 1
· +
, or
[ ]
m
x x y 1
2
− + ·
or
[ ]
m
x x y 1
2
− − ·
Show that ( ) ( ) 0 ) 1 2 ( 1
2 2
1 2
2
· − + + + −
+ + n n n
y m n xy n y x (VTU Feb-02)
Consider
x
y
y x y y
m
m m m
2
1
2
1
1 1 1
· + ⇒ · +
( ) ( ) 0 1 2
1 1
2
· + − ⇒
m m
y x y
m
y
1
2
4 4 2
) 1 ( 2
) 1 )( 1 ( 4 ) 2 ( ) 2 (
2 2
1 − t
·
− − t − −
·
x x x x
y
m
( ) ( ) 1 1
2
1 2 2
2 2
2
1
− t · ⇒ − t ·
− t
· x x y x x
x x
m
( )
m
x x y 1
2
− t ·
. So, we can consider
[ ]
m
x x y 1
2
− + ·
or
[ ]
m
x x y 1
2
− − ·
Let us take
[ ]
m
x x y 1
2
− + ·
( )
,
`
.
|
+ − + ·
) 2 (
1 2
1
1 1
2
1
2
1
x
x
x x m y
m
( )
,
`
.
|
+ −
− + ·
1
1
1
2
2
1
2
1
x
x x
x x m y
m
or
( ) my y x · −
1
2
1 . On squaring
141
( )
2 2 2
1
2
1 y m y x · − . Again differentiating w.r.t x,
( ) ) 2 ( ) 2 ( 2 1
1
2 2
1 2 1
2
yy m x y y y x · + −
or
( ) ) 2 ( 1
1
2
1 2
2
y y m xy y x ÷ · + −
or
( ) 0 1
2
1 2
2
· − + − y m xy y x ------------(*)
Differentiating (*) n- times using Leibnitz’s theorem and simplifying, we get
( ) ( ) 0 ) 1 2 ( 1
2 2
1 2
2
· − + + + −
+ + n n n
y m n xy n y x
Problem set 1.3.1
In each of the following, apply Leibnitz’s theorem to get the results.
1. show that
]
]
]
− − − −
·
]
]
]
+
n
x
x
n
x
x
dx
d
n
n
n
n
1
.........
3
1
2
1
1 log
! ) 1 ( log
1
Hint: Take
x
u x v
1
; log · ·
2. If , ) 1 (
2 n
x y − · Show that n
y
satisfies the equation
( ) 0 ) 1 ( 2 1
2
2
2
· + + − − y n n
dx
dy
x
dx
y d
x Hint : It is required to show that
( ) ( ) 0 1 2 1
1 2
2
· + + − −
+ + n n n
y n n xy y x
3. If
), sin(log ) cos(log x b x a y + ·
( ) ( ) 0 1 1 2
2
1 2
2
· + + + +
+ + n n n
y n xy n y x
4. If ,
1
sin x m
e y
· Prove That
( ) ( ) ( ) 0 1 2 1
2 2
1 2
2
· + − + − −
+ + n n n
y m n xy n y x
5. If , log cos
1
n
n
x
b
y
,
`
.
|
·
,
`
.
|
Show that
( ) 0 2 1 2
2
1 2
2
· + + +
+ + n n n
y n xy n y x
6. ( ) That ove x m y Pr , sin sin
1 −
·
( ) ( ) ( ) 0 1 2 1
2 2
1 2
2
· + − + − −
+ + n n n
y m n xy n y x
7. If ( ), log x x D y
n n
n
· Prove That
(i)
)! 1 (
1
− + ·
n ny y
n n (ii)
! n y
n
·
]
]
]
+ + + + +
n
x
1
.........
3
1
2
1
1 log
8. If , log x x y
n
· Show that
x
n
y
n
!
1
·
+
• Summary :- The idea of successive differentiation was presented. The computation
of n
th
derivatives of a few standard functions and relevant problems were discussed.
Also, the concept of successive differention was extended for special type of
functions using Liebnitz’s theorem.
142
r = f(ө)
A
P(r, ө)
O
r
Ө
Chapter – 2 : POLAR CURVES
LESSON -1 : Angle between Polar Curves
• In this chapter we introduce a new coordinate system, where we can understand
the idea of polar curves and their properties.
• Session-1
2.1.0 Introduction:- We are familiar with Cartesian coordinate system for specifying a
point in the xy – plane. Another useful system for similar purpose is Polar coordinate
system, and the curves specified by these coordinates are referred to as polar curves.
• A polar curve by name “three-leaved rose” is displayed below:
• Any point P can be located on a plane with co-ordinates
( ) θ , r
called polar
co-ordinates of P where r = radius vector OP,(with pole ‘O’) ;θ = projection
of OP
on the
initial axis OA.(See Fig.1)
• The equation
( ) θ f r ·
is known as a polar curve.
• Polar coordinates
( ) θ , r
can be related with Cartesian coordinates
( ) y x,
through the relations Fig.1. Polar
coordinate
θ cos r x · &
θ sin r y ·
. system
ө = 0
ө =
ө = ө =
ө = π
ө =
143
T
2.1.1 Important results
• Theorem 1 : Angle between the radius vector and the tangent.:
i.e. With usual notation prove that
dr
d
r
θ
φ · tan
• Proof:- Let “
φ
” be the angle between the radius vector OPL
and the tangent
1
TPT
at the point `P` on the polar
curve
( ) θ f r ·
. (See fig.2)
From Fig.2, Fig.2. Angle between radius
φ θ ψ + ·
vector and the tangent
tan
( )
φ θ
φ θ
φ θ ψ
tan tan 1
tan tan
tan
+
· + ·
i.e.
....... ..........
tan tan 1
tan tan
φ θ
φ θ
+
·
dx
dy
(1)
On the other hand, we have θ cos r x · ;
θ sin r y ·
differentiating these,
w.r.tθ,
( )
,
`
.
|
+ − ·
θ
θ θ
θ d
dr
r
d
dx
cos sin
&
( )
,
`
.
|
+ ·
θ
θ θ
θ d
dr
r
d
dy
sin cos
//NOTE//
( )
( )
,
`
.
|
+ −
,
`
.
|
+
· ·
θ
θ θ
θ
θ θ
θ
θ
d
dr
r
d
dr
r
d
dx
d
dy
dx
dy
cos sin
sin cos
θ
θ
cos &
d
dr
by Dr Nr the dividing
( )
( ) 1 tan
tan
+ −
+
·
θ θ
θ
θ
dr rd
dr
d
r
dx
dy
i.e.
( )
( )
dr
rd
dr
d
r
dx
dy
θ
θ
θ
θ
tan 1
tan
+
· ………………….(2)
Comparing equations (1) and (2)
we get
dr
d
r
θ
φ · tan
• Note that
,
`
.
|
·
θ
φ
d
dr
r
1
cot
• A Note on Angle of intersection of two polar curves :-
If
1
φ and
2
φ are the angles between the common radius vector and the
tangents
at the point of intersection of two curves ( ) θ
1
f r · and ( ) θ
2
f r · then the
angle intersection of the curves is given by 2 1
φ φ −
144
r = f(ө)
φ
φ
T
1
1
1
P(r, ө)
O
r
Ө
ψ
L
A
Y
φ
O
P(r, ө)
P
N
r = f (ө)
φ
r
Ө
Ψ
• Theorem 2 : The length “p” of perpendicular from pole to the tangent in a
polar curve i.e.
(i)
φ sin r p ·
or (ii)
,
`
.
|
· ·
θ d
dr
r r p
4 2 2
1 1 1
2
• Proof:- In the Fig.3, note that ON = p, the length of the perpendicular from the
pole to
the tangent at p on
( ) θ f r ·
.from the right angled triangle OPN,
( ) φ φ sin sin OP ON
OP
ON
· ⇒ ·
i.e.
) ....( .......... sin i r p φ ·
Consider
φ
φ
ec
r r p
cos
1
sin
1 1
· ·
( ) φ φ 2 cot 1
1
cos
1 1
2
2
2 2
+ · · ∴
r
ec
r p
Fig.3 Length of the perpendicular
from the pole to the tangent
]
]
]
]
,
`
.
|
+ ·
2
2 2
1
1
1 1
θ d
dr
r r p
) ..( ..........
1 1 1
2
4 2 2
ii
d
dr
r r p
,
`
.
|
+ · ∴
θ
• Note:-If
,
1
r
u ·
we get
2
2
2
1
,
`
.
|
+ ·
θ d
du
u
p
• Session-2
• In this session, we solve few problems on angle of intersection of polar curves
and pedal equations.
2.1.2 Worked examples:-
• Find the acute angle between the following polar curves
1.
( ) θ cos 1+ ·a r
and
( ) θ cos 1− ·b r
(VTU-July-2003)
2
( ) θ θ cos sin + · r
and θ sin 2 · r (VTU-July-2004)
3. ( )
2
sec 16
2
θ
· r and ( )
2
cos 25
2
θ
ec r ·
4.
θ log a r ·
and
θ log
a
r ·
(VTU-July-2005)
5.
θ
θ
+
·
1
a
r and
2
1 θ +
·
a
r
145
Solutions:
1. Consider Consider
( ) θ cos 1+ ·a r ( ) θ cos 1− ·b r
Diff w.r.t θ Diff w.r.t θ
θ
θ
sin a
d
dr
− · θ
θ
sin b
d
dr
·
( )
θ
θ θ
sin
cos 1
a
a
dr
d
r
+
·
( )
θ
θ θ
sin
cos 1
b
b
dr
d
r
·
( )
( ) ( )
2
cos
2
sin 2
2
cos 2
tan
2
1
θ θ
θ
φ − ·
( )
( ) ( )
2
c o s
2
s i n 2
2
s i n 2
t a n
2
1
θ θ
θ
φ − ·
2
cot
θ
− ·
2
tan
θ
·
i.e ( ) ( )
2 2 2 2
tan ta n
1 1
θ π
φ
θ π
φ + · ⇒ + ·
2 1 1
2
tan tan φ φ
θ
φ · ⇒ ·
Angle between the curves
( )
2 2 2 2
2 1
π θ θ π
φ φ · − + · −
Hence ,the given curves intersect orthogonally
2. Consider Consider
( ) θ θ cos sin + · r
θ sin 2 · r
Diff w.r.t θ Diff w.r.t θ
θ θ
θ
sin cos − ·
d
dr
θ
θ
cos 2 ·
d
dr
θ θ
θ θ θ
sin cos
cos sin
+
·
dr
d
r
θ
θ θ
cos 2
sin 2
·
dr
d
r
θ
θ
φ
tan 1
1 tan
tan
1
+
·
(÷ Nr & Dr θ cos ) θ φ tan tan
2
·
i.e ( ) θ
π
θ
θ
φ + ·
+
·
4
tan
tan 1
1 tan
tan
1
θ φ · ⇒
2
θ
π
φ + · ⇒
4
1
∴Angle between the curves = ( )
4 4
2 1
π
θ θ
π
φ φ · − + · −
3. Consider Consider
( )
2
sec 16
2
θ
· r ( )
2
cos 25
2
θ
ec r ·
146
Diff w.r.t θ Diff w.r.t θ
( ) ( )
2
1
.
2
tan
2
sec 32
2
θ θ
θ
·
d
dr
( ) ( )
2
1
.
2
cot
2
cos 50
2
θ θ
θ
ec
d
dr
− ·
( ) ( )
2
tan
2
sec 16
θ θ
·
( ) ( )
2
cot
2
cos 25
2
θ θ
ec − ·
( )
( ) ( )
2
ta n
2
se c 16
2
s e c 1 6
2
2
θ θ
θ
θ
·
dr
d
r
( )
( ) ( )
2
co t
2
c o s 2 5
2
c os 25
2
2
θ θ
θ
θ
ec
e c
dr
d
r
·
( )
2 2
tan
2
cot tan
1
θ π θ
φ − · ·
( )
2
tan
2
tan tan
2
θ θ
φ − · − ·
( )
2 2
1
θ π
φ − · ⇒
2
2
θ
φ − · ⇒
Angle of intersection of the curves = ( ) ( )
2 2 2
2 1
θ θ π
φ φ − − · −
2
π
·
4. Consider Consider
θ log a r ·
θ log
a
r ·
Diff w.r.t θ Diff w.r.t θ
θ
θ
a
d
dr
· ( )
θ
θ
θ
1
. log
2
a
d
dr
− ·
( )
a
a
dr
d
r
θ
θ
θ
log ·
( )
,
`
.
|
,
`
.
|
− ·
a
a
dr
d
r
θ θ
θ
θ
2
log
log
) ( .......... log tan
1
i θ θ φ ·
) ( .......... log tan
2
ii θ θ φ − ·
We know that
( )
2 1
2 1
2 1
tan tan 1
tan tan
tan
φ φ
φ φ
φ φ
+
· −
147
( )
( )( ) θ θ θ ϑ
θ θ θ θ
log log 1
log log
− +
− −
·
i.e
( )
( )
) ..( ..........
log 1
log 2
tan
2
2 1
iii
θ θ
θ θ
φ φ
· −
From the data:
( ) 1 log 1 log
log
log
2
t · · ⇒ · · θ θ
θ
θ or
a
r a
As θ is acute, we take by θ=1
NOTE e · ⇒θ
Substituting e · θ in (iii), we get
( )
( )
,
`
.
|
·
· −
2 2
2 1
1
2
log 1
log 2
tan
e
e
e e
e e
φ φ
( ) 1 log ·
e
e
,
`
.
|
· − ∴
2
1
2 1
1
2
tan
e
e
φ φ
5. Consider Consider
θ
θ
+
·
1
a
r as
2
1 θ
θ
+
·
a
r
( ) 1
1
1 1 1
+ ·
+
·
θ
θ
θ
a a r
( )
r
a
· + ∴
2
1 θ
Diff w.r.t θ Diff w.r.t θ
( )
2
2
1
1 1
θ
θ
− · −
a d
dr
r
θ
θ
d
dr
r
a
2
2
·
2
1
θ θ a
r
d
dr
r
·
θ
θ
d
dr
r a
r 1 2
·
r
a
dr
d
r
2
θ θ
· i.e
θ
θ
r
a
dr
d
r
2
·
( ) θ
θ
θ
φ
+
·
1
tan
2
1
a
a
,
`
.
| +
− ·
a
a
2
2
1
2
tan
θ
θ
φ
( ) θ θ φ + · ∴ 1 tan
1
( )
2
2
1
2
1
tan θ
θ
φ + − ·
Now, we have
( ) ( ) θ θ θ
θ θ
θ
+ · + ⇒
+
· ·
+
1 1
1 1
2
2
a a
a
r
a
or 1 1
3 3
· ⇒ + · + θ θ θ θ or 1 · θ
2 tan
1
· ∴ φ
& ( ) 1 tan
2
− · φ
Consider
( )
( )( )
2 1
2 1
2 1
tan tan 1
tan tan
tan
φ φ
φ φ
φ φ
+
· −
( )
( )( )
3 3
1 2 1
1 2
· − ·
− +
− −
·
148
( ) 3 tan
1
2 1
· − ∴ φ φ
Problem Set No. 2.1.1 for practice.
• Find the acute angle between the curves
1. ( ) θ θ n n a r
n n
sin cos + · and θ n a r
n n
sin · (ans:
4
π
)
2.
n n
a n r · θ cos and
n n
b n r · θ sin (ans:
2
π
)
3. θ a r · and
θ
a
r · (ans:
2
π
)
4. θ cos a r · and
2
a
r · (ans:5
6
π
)
5. θ m a r
m m
cos · and θ m b r
m m
sin · (ans:
2
π
)
LESSON -2 : Pedal Equations
• Session - 1
• 2.2.0 Pedal equations (p-r equations):- Any equation containing only p & r
is
known as pedal equation of a polar curve.
• Working rules to find pedal equations :-
(i) Eliminate r and
φ
from the Eqs.: (i)
( ) θ f r ·
&
φ sin r p ·
(ii) Eliminate only θ from the Eqs.: (i)
( ) θ f r ·
&
2
4 2 2
1 1 1
,
`
.
|
+ · ∴
θ d
dr
r r p
• 2.2.1 Worked Examples on pedal equations:-
• Find the pedal equations for the polar curves:-
1.
θ cos 1
2
− ·
r
a
2.
α θ cot c
e r ·
3. θ θ m b m a r
m m m
cos sin + · (VTU-Jan-2005)
4. θ cos 1 e
r
l
+ ·
Solutions:
1. Consider
θ cos 1
2
− ·
r
a
……….(i)
Diff. w.r.t θ
( ) θ
θ
sin
1
2
2
·
d
dr
r
a
a
r
d
dr
r 2
sin 1 θ
θ
·
θ
θ
sin
1 2
r
a
dr
d
r − ·
149
( )
( )
2
t a n
2
c o s
2
s i n 2
2
s i n 2
s i n
c o s 1
t a n
2
θ
θ θ
θ
θ
θ
φ − · − ·
− ·
( )
2 2
t a n t a n
θ
φ
θ
φ
· ⇒
·
Using the value of
φ
is
, sin φ r p ·
we get
( ) ) ...( ..........
2
sin
2
sin ii r r p
θ θ
− ·
·
Eliminating “θ” between (i) and (ii)
,
`
.
|
·
,
`
.
| −
· ·
r
a r
r r p
2
2 2
cos 1
2
sin
2
2 2 2 2
θ
θ
[See eg: - (i)]
.
2
ar p ·
This eqn. is only in terms of p and r and hence it is the pedal equation of the polar
curve.
2. Consider
α θ cot
e r ·
Diff. w.r.t θ
( ) ( )
α θ α θ
α α
θ
cot cot
cot cot e r r e
d
dr
· · ·
We use the equation
2
4 2 2
1 1 1
,
`
.
|
+ ·
θ d
dr
r r p
( )
2
4 2
cot
1 1
α r
r r
+ ·
( ) ( ) α α α
2
2
2
2
2
4 2
cos
1
cot 1
1
cot
1 1
ec
r r r r
· + · + ·
α
2
2 2
cos
1 1
ec
r p
·
α
2
2
2
cosec
r
p · or α
2 2 2
cos ec p r · is the required pedal equation
3.Consider θ θ m b m a r
m m m
cos sin + ·
Diff. w.r.t θ
( ) ) sin ( cos
1
θ θ
θ
m m b m m a
d
dr
mr
m m m
− + ·
θ θ
θ
m b m a
d
dr
r
r
m m
m
sin cos − ·
θ θ
θ θ
θ m b m a
m b m a
d
dr
r
m m
m m
cos sin
sin cos 1
+
·
θ θ
θ θ
φ
m b m a
m b m a
m m
m m
cos sin
sin cos
cot
+
·
150
Consider
φ sin r p ·
,
φ ec
r p
cos
1 1
·
φ
2
2 2
cos
1 1
ec
r p
·
( ) φ
2
2
cot 1
1
+ ·
r
]
]
]
]
,
`
.
|
+
+ ·
2
2
cos sin
sin cos
1
1
θ θ
θ θ
m b m a
m b m a
r
m m
m m
( ) ( )
( ) ]
]
]
]
+
− + +
·
2
2
2
2
cos sin
sin cos cos sin 1
θ θ
θ θ θ θ
m b m a
m b m a m b m a
r
m m
m m m m
Note
r
b a
r p
m
m m
]
]
]
+
·
2
2 2
2 2
1 1
( )
m m
m
b a
r
p
2 2
1 2
2
+
· ⇒
+
is the required p-r equation
4. Consider ( ) θ cos 1+ ·
r
l
Diff w.r.t θ
θ
θ
θ
θ
sin
1
sin
1
2
e
d
dr
r r
l
e
d
dr
r
l ·
,
`
.
|
⇒ − ·
,
`
.
|
( ) θ φ sin cot e
r
l
·
( ) θ φ sin cot e
l
r
· ∴
We have
( ) φ
2
2 2
cot 1
1 1
+ ·
r p
(see eg: 3 above) Now
]
]
]
+
·
2
2 2 2 2
2 2
sin 1 1
l
r e l
r p
θ
( ) θ
2
2
2 2
2
sin 1
1
l
r e
r
+ ·
r
l
e · + θ cos 1
r
r l
e
· θ cos
]
]
]
]
]
]
]
¹
¹
¹
'
¹
¹
¹
¹
'
¹
,
`
.
| −
− +
·
2
2
2 2 2
2 2
1
1 1
l
re
r l
r e l
r p
,
`
.
| −
·
re
r l
θ cos
θ θ
2 2
cos 1 sin − ·
On simplification
lr e
e
p
2 1 1
2
2
2
+
,
`
.
| −
·
2
1
,
`
.
| −
− ·
re
r l
151
• Problem Set No. 2.2.1 for practice.
Find the pedal equations of the following polar curves
1.
n n
a n r · θ cos and
n n
b n r · θ sin
2. θ a r · and
θ
a
r ·
3. θ cos a r · and
2
a
r ·
4 . θ m a r
m m
cos · and θ m b r
m m
sin ·
Chapter-3: PARTIAL DIFFERENTIATION
Lesson-1: Euler’s Theorem
Session-1
3.1.0 Introduction : We often come across qualities which depend on two or more
variables. For e.g. the area of a rectangle of length x and breadth y is given by
Area = A(x,y) = xy. The area A(x, y) is, obliviously, a function of two variables.
Similarly, the distances of the point (x, y, z) from the origin in three-dimensional space is
an example of a function of three variables x, y, z.
• 3.1.1 Partial derivatives: Let z = f(x, y) be a function of two variables x and y.
The first order partial derivative of z w.r.t. x, denoted by
x
z
or
x
f
or z
x
or f
x
is
defined as
( ) ( )
x
y x f y x x f
x
z
x
δ
δ
δ
, ,
lim
0
− +
·
• From the above definition, we understand that
,
`
.
|
x
z
is the ordinary derivative of
z w.r.t x, treating y as constant.
The first order partial derivative of z w.r.t y, denoted by
y
z
or
y
f
or z
y
or f
y
is defined
as
( ) ( )
y
y x f y y x f
y
z
y
δ
δ
δ
, ,
lim
0
− +
·
• From the above definition, we understand that
,
`
.
|
y
z
is the ordinary derivative of
z w.r.t y, treating x as constant
• The partial derivatives
2
2
x
z
x
z
x ∂
·
,
`
.
|
or
2
2
x
f
or z
xx
or f
xx
;
2
2
y
z
y
z
y ∂
·
,
`
.
|
or
2
2
y
f
or z
yy
or f
yy
;
y x
z
y
z
y ∂ ∂
·
,
`
.
|
2
or z
yx
or f
yx
152
and
y x
z
y
z
y ∂ ∂
·
,
`
.
|
2
or z
xy
or f
xy
are known as second order Partial derivatives.
In all ordinary cases, it can be verified that
x y
z
y x
z
∂ ∂
·
∂ ∂
2 2
• The third and higher order partial derivatives of f(x,y) are defined in an analogous
way Also, the second and higher order partial derivatives of more than two
independent variables are defined similarly.
• A note on rules of partial differentiation :-
All the rules of differentiation applicable to functions of a single independent
variable are applicable for partial differentiation also; the only difference is that
while differentiating partially w.r.t one independent variable all other independent
variables are treated as constants.
• 3.1.2 Worked examples : -
In solving the following examples, we learn the partial differentiation
of so called explicit functions of more than one variables.
1. Evaluate ( )
x
z
and
( ) y x ∂ ∂
, if
(a)
xy x y x z sin
2
− ·
(b)
z z y x log · + +
Solution: - (a) Consider
xy x y x z sin
2
− ·
Differentiating
z w.r.t x, keeping y as a
constant, we get the partial derivative,
( ) ( )( ) ( )( ) { ¦ 1 sin cos 2 xy y xy x y x
x
z
+ − ·
i.e.
( ) xy xy xy xy xy xy xy z
x
sin cos 2 sin cos 2 − − · − − ·
Similarly, differentiating z w.r.t y keeping x as a constant
,
we get the partial derivative
( ) ( ) ( )( ) { ¦ 0 sin 1
2
xy x xcoxy x
y
z
+ − ·
i.e
( ) xy x xy x x z
y
cos 1 cos
2 2 2
− · − ·
(b) Consider
z z y x log · + +
i.e.
* log − − − − − + · − y x z z
Differentiating * partially w.r.t x treating y as
Constant,
1 1
1
0 1
1
·
,
`
.
|
,
`
.
|
⇒ + ·
z x
z
x
z
x
z
z
i.e.
1
1
·
,
`
.
| −
z
z
x
z
or
,
`
.
|
·
z
z
x
z
1
153
Similarly, differentiating * partially w.r.t y, treating x as constant,
1 1
1
0 1
1
·
,
`
.
|
,
`
.
|
+ ·
z x
z
or
x
z
x
z
z
,
`
.
|
·
z
z
x
z
1
,
`
.
|
·
·
z
z
y
z
x
z
1
2. .If
t
r
n
e t
4
2
· θ
,
what value of n will make
?
1
2
2
t r
r
r r ∂
·
,
`
.
|
∂ θ θ
Solution: Consider
t
r
n
e t
4
2
· θ
------------------------------------------------- (*)
Differentiating * w.r.t. r, treating t as constant, we get
·
r
θ
t
r
n
t
r
n
e t
r
t
r
e t
4
1
4
2 2
2 4
2
·
,
`
.
| −
t
r
n
e t
r
r
r
4
1
3
2
2
2
− ·
θ
and hence
,
`
.
|
,
`
.
|
·
,
`
.
|
t
r
e t
r
e t
r
r
r
r
t
r
n
t
r
n
4
2
2 2
3
4
1
3
4
1
2
2
2 2
θ
t
r
n n
e t
r
t
r
r
r r
4
2
2
1 2
2
2
4 2
3 1
− −
,
`
.
|
+ − ·
,
`
.
|
θ
------------------- (1)
Also, differentiating (*)
w.r.t t, treating r as constant, we get
,
`
.
|
+ ·
− −
2
2
4 4
1
4
2
2
t
r
e t e nt
t
t
r
n
t
r
n
θ
t
r
n n
e t r nt
4
2 2 1
2
4
1
− −
,
`
.
|
+ ·
------------------------ (2)
Since
,
,
1
2
2
t r
r
r r ∂
·
,
`
.
|
∂ θ θ
equation (1) and (2) yield,
t
r
n n
t
r
n n
e t
r
nt e t
r
t
4
2
2
1
4
2
2
1
2
2
4 4
2
3
− −
− −
,
`
.
|
+ ·
,
`
.
|
+
i.e.
( ) ( )
2
3
0
2
3
0
2
3 4
1
2
· · + ⇒ · +
n or n e t n
t
r
n
3. If
( ), by ax f e z
by ax
− ·
+
prove that
abz
y
z
a
x
z
b 2 ·
+
(VTU-Jan-2004)
Solution: Consider
( ), by ax f e z
by ax
− ·
+
------------------- (*)
Differentiating (*)
w.r.t x using product and chain rules, we get
( )( ) { ¦ ( ) ( ) { ¦ b e by ax f b by ax f e
y
z
by ax by ax + +
− + − − ·
1
154
( ) ( ) { ¦ by ax f by ax f be
y
z
by ax
− − − ·
+ 1
------------------- (2)
Multiplying eq (1) by b and eq(2) by a, and adding we get
( ) ( ) ( ) ( ) [ ] by ax f by ax f by ax f by ax f abe
y
z
a
x
z
b
by ax
− − − + − + + ·
+
+ 1 1
( ) [ ] by ax f abe
by ax
− ·
+
2
( ) [ ] by ax f e ab
by ax
− ·
+
2
abz 2 ·
,
as desired
4.Given
( ), sin cos
cos
θ
θ
r e u
r
· ( ) θ
θ
sin sin
cos
r e v
r
·
,
prove that
θ ∂
·
∂ v
r r
u 1
and
θ ∂
− ·
∂ u
r r
v 1
Solution: Consider
( ), sin cos
cos
θ
θ
r e u
r
·
Differentiating u w.r.t. r and
θ
partially, we
get
,
( )( ) { ¦ ( ) ( ) { ¦ θ θ θ θ
θ θ
cos sin cos sin sin sin
cos cos r r
e r r e
r
u
+ − ·
i.e.
( ) ( ) { ¦ θ θ θ θ
θ
sin sin sin cos sin cos
cos
r r e
r
u
r
or
( ) { ¦ θ θ
θ
+ ·
sin cos
cos
r e
r
u
r
--------------(1)
and
( )( ) { ¦ ( ) ( ) { ¦ θ θ θ θ
θ
θ θ
sin sin cos cos sin sin
cos cos
r e r r r e
u
r r
− + − ·
( ) ( ) { ¦ θ θ θ θ
θ
sin sin cos cos sin sin
cos
r r re
r
+ − ·
i.e.
( ) { ¦ θ θ
θ
θ
+ ·
− sin sin
1
cos
r e
u
r
r
---------------------(2)
Next consider
,
) sin sin(
cos
θ
θ
r e v
r
·
Differentiating v w.r.t r &
θ
,
Partially we get
( ) { ¦ ( ) ( ) { ¦ θ θ θ θ
θ θ
cos sin sin sin sin cos
cos cos r r
e r r e
r
v
+ ·
( ) ( ) { ¦ θ θ θ θ
θ
sin sin cos cos sin sin
cos
r r e
r
+ ·
i.e
( ) { ¦ θ θ
θ
+ ·
sin sin
cos
r e
r
v
r
--------------------(3);
( )( ) { ¦ ( ) ( ) { ¦ θ θ θ θ
θ
θ θ
sin sin sin cos sin cos
cos cos
r e r r r e
v
r r
− + ·
( ) ( ) { ¦ θ θ θ θ
θ
sin sin sin cos sin cos
cos
r r re
r
− ·
i.e.
( ) { ¦ θ θ
θ
θ
+ ·
sin cos
1
cos
r e
v
r
r
----------------------------- (4)
155
Thus, from eqs (1) and (4),
we obtain
θ ∂
·
∂ v
r r
u 1
, and from eqs (2) and (3) we obtain
θ ∂
− ·
∂ u
r r
v 1
.
Session-2
• 3.1.3 Worked Examples:- In each of following examples, we try to
show that
x y
u
y x
u
∂ ∂
·
∂ ∂
2 2
, where
( ) y x f u , ·
is a function of two variables.
1.
y
x u · ; 2. ( ) y y y x e u
x
sin sin − ·
3.
,
`
.
|
·
x
y
u
1
sin
Solution: - 1. Differentiating
y
x u · partially w.r.t y, we get
x x
y
u
y
log ·
( )
,
`
.
|
· x a a
dx
d
x x
log
differentiating the
above partially w.r.t x, we get
( ) x x
x y
u
x
y
log
·
,
`
.
|
i.e
( )
1
2
log
1
+
,
`
.
|
·
∂ ∂
y y
yx x
x
x
y x
u
or
( ) x y x
y x
u
y
log 1
1
2
+ ·
∂ ∂
---------------(1)
Next, differentiating
y
x u · partially w.r.t. x, we get
1 −
·
y
yx
x
u
so that
( )
1 −
·
,
`
.
|
y
yx
y x
u
y
i.e
( ) ( ) 1 log
1 1
2
− −
+ ·
∂ ∂
y y
x x x y
x y
u
or
( ) x y x
x y
u
y
log 1
1
2
+ ·
∂ ∂
---------------(2)
from (1) and (2), we get
x y
u
y x
u
∂ ∂
·
∂ ∂
2 2
2. Differentiating
,
`
.
|
·
x
y
u
1
sin
partially w.r.t. x, we get
156
2 2
2
2
1
1
y x x
y
x
y
x
y
x
u
·
,
`
.
|
,
`
.
|
·
( ) ( )
( )
2 2 2
2 2
2 2
2
2
1
1
y x x
y
y x
x y y x x
x
u
y −
¹
¹
¹
'
¹
¹
¹
¹
'
¹
,
`
.
|
− −
− ·
,
`
.
|
i.e.
( )
( ) ( )
2
3
2 2
2
3
2 2 2
2 2 2 2
y x
x
y x x
xy y x x
x y
u
− ·
¹
¹
¹
'
¹
¹
¹
¹
'
¹
+ −
− ·
∂ ∂
----------------(1)
Next, differentiating
,
`
.
|
·
x
y
u
1
sin
partially w.r.t. y, we get
( )
2 2
2
1
1
1
1
y x
x
x
y
y
u
·
,
`
.
|
·
( ) ( ) ( )
2 2
2 2
2 2
2
2
1
1 0
y x
x
y x
y x
y
u
x −
− −
− ·
,
`
.
|
i.e.
( )
2
3
2 2
2
y x
x
y x
u
− ·
∂ ∂
------------- (2)
From (1)
and (2), we get
x y
u
y x
u
∂ ∂
·
∂ ∂
2 2
3. Differentiating
( ) y y y x e u
x
sin sin − ·
w.r.t y
,
we get
( ) y y y y x e
y
u
x
sin cos cos − − ·
Differentiating w.r.t x, we get
( ) ( )
x x
e y y y y x y y x e
y
u
x
sin cos cos 0 0 cos sin − − + − − + − ·
,
`
.
|
i.e.
( ) y y y y x y y x e
y x
u
x
sin cos cos cos sin
2
− − + + − ·
∂ ∂
Or
( ) ( ) { ¦ y x y y x e
y x
u
x
sin 1 cos 1
2
+ − − + ·
∂ ∂
----------------- (1)
Next, differentiating
w.r.t x, we get
{ ¦ ( )
x x
e y y y x y y x e
x
u
sin sin sin cos − + + ·
( ) { ¦ y y x y x e
x
u
x
sin 1 cos − + + ·
Differentiating this w.r.t. y, we get
157
( ) ( ) ( ) { ¦ 1 sin cos 1 0 sin − + − + + + − ·
,
`
.
|
y y y x y x e
x
u
y
x
( ) ( ) { ¦ y x y y x e
x
sin 1 cos 1 + − − + ·
----------------- (2)
from (1)and (2),we
get
x y
u
y x
u
∂ ∂
·
∂ ∂
2 2
3.1.4 In the following few examples involve equations where partial derivatives of
higher order occur. These equations frequently appear is engineering applications.
1. If
,
`
.
|
·
2 2
1
2
tan
y x
xy
u
, prove that
0
2
2
2
2
·
+
y
u
x
u
Solution:- Differentiating
,
`
.
|
·
2 2
1
2
tan
y x
xy
u
partially w.r.t.x, we get
( )( ) ( )
( ) ¹
¹
¹
'
¹
¹
¹
¹
'
¹
− −
,
`
.
|
+
·
2
2 2
2 2
2
2 2
2 2 2
2
1
1
y x
x xy y y x
y x
xy
x
u
i.e.
( )
2 2
2
y x
y
x
u
+
·
Differentiating Partially this w.r.t. x, we get
( )( ) ( ) ( )
( )
2
2 2
2 2
2 2 2
2
2 2 0 2
y x
x y y x
y x
y
x x
u
+
− −
·
,
`
.
|
+
·
i.e.
( )
2
2 2
2
2
4
y x
xy
x
u
+
·
-------------------------- (1)
Next differentiating
,
`
.
|
·
2 2
1
2
tan
y x
xy
u
partially w.r.t. y, we get
( )( ) ( )
( ) ¹
¹
¹
'
¹
¹
¹
¹
'
¹
− − −
,
`
.
|
+
·
2
2 2
2 2
2
2 2
2 2 2
2
1
1
y x
y xy x y x
y x
xy
y
u
( )
2 2
2
y x
x
y
u
+
·
Differentiating partially this w.r.t.y, we get
( )( ) ( ) ( )
( )
2
2 2
2 2
2 2 2
2
2 2 0 2
y x
y x y x
y x
x
x x
u
+
− −
·
,
`
.
|
+ ∂
·
i.e.
( )
2
2 2
2
2
4
y x
xy
y
u
+
·
-----------------------(2)
Adding eqs (1) and (2) we get
0
2
2
2
2
·
+
y
u
x
u
, as desired
158
Note:
(a) The equation
0
2
2
2
2
·
+
y
u
x
u
is known as Laplace’s equation in two dimension
which has variety of applications in potential theory.
(b) A similar equation viz,
0
2
2
2
2
2
2
·
+
+
z
u
y
u
x
u
is known as Laplace’s equation in
three- dimensions where
( ) z y x u u , , ·
2. If
( ) ( ), ay x g ay x f u − + + ·
show that
2
2
2
2
2
x
u
a
y
u
·
Solution:- Differentiating
( ) ( ), ay x g ay x f u − + + ·
partially w.r.t.x, we get
( )( ) ( )( ) 1 1
1 1
ay x g ay x f
x
u
− + + ·
Again differentiating partially w.r.t.x, we see that
( )( ) ( )( ) 1 1
11 11
2
2
ay x g ay x f
x
u
− + + ·
( ) ( ) [ ] ay x g ay x f a
x
u
a − + + ·
11 11 2
2
2
2
-----------(1)
Next, differentiating
( ) ( ), ay x g ay x f u − + + ·
partially w.r.t.y
( )( ) ( )( )
2 1 2 1
. a ay x g a ay x f
y
u
− + + ·
Again differentiating partially w.r.t.y, we see that
( )( ) ( )( )
2 11 2 11
2
2
a ay x g a ay x f
x
u
− + + ·
i.e
( ) ( ) [ ] ay x g ay x f a
y
u
− + + ·
11 11 2
2
2
------------------(2)
from eqs (1) and (2), we see that
2
2
2
2
2
x
u
a
y
u
·
Note:- (a). The equation
2
2
2
2
2
x
u
a
y
u
·
or the equation
2
2
2
2
2
x
u
a
t
u
·
is known as
one-dimensional wave equation
(b). A similar equation viz,
,
`
.
|
+
+
·
2
2
2
2
2
2
2
2
2
z
u
y
u
x
u
a
t
u
is known as three- dimensional wave equation
159
3. If x e u
t
3 cos
2 −
· , find the value of ‘c’ such that
2
2
2
x
u
c
t
u
·
Solution:- Differentiating x e u
t
3 cos
2 −
· , partially w.r.t. t we get
x e
t
u
t
3 cos 2
2 −
− ·
--------------- (1)
Next, differentiating x e u
t
3 cos
2 −
· partially w.r.t.x,
x e
x
u
t
3 sin 3
2 −
− ·
Again differentiating partially w.r.t x,
x e
x
u
t
3 cos 9
2
2
2 −
− ·
( )
2
9
3 cos 9 3 cos 2
2 2 2
2
2
2
· ⇒ · − ⇒
·
− −
c x e x e
x
u
c
t
u
t t
or
2
9
· c
Note (a)
The equation
2
2
2
x
u
c
t
u
·
is called one-dimensional heat equation
(b) Similarly, the equation viz,
,
`
.
|
+
+
·
2
2
2
2
2
2
2
z
u
y
u
x
u
c
t
u
is known as
three-dimensional heat equation
4. If ( ) xyz z y x u 3 log
3 3 3
− + + · , prove that
(a)
,
`
.
|
+ +
·
+
+
z y x z
u
y
u
x
u 3
(b)
( )
2
2
9
z y x
u
z y x + +
− ·
,
`
.
|
+
+
(VTU Feb-2005)
Solution:- (a) Differentiating u partially w.r.t. x,y,z we get respect or
xyz z y x
yz x
x
u
3
3 3
3 3 3
2
− + +
·
xyz z y x
zx y
y
u
3
3 3
3 3 3
2
− + +
·
And
xyz z y x
xy z
z
u
3
3 3
3 3 3
2
− + +
·
adding these partial derivatives we get
160
( ) ( )
( ) ( ) z x y z x y z y x z y x
z x y z x y z y x
x y z z y x
z x y z x y z y x
z
u
y
u
x
u
− − − + + + +
− − − + +
·
− + +
− − − + +
·
+
+
2 2 2
2 2 2
3 3 3
2 2 2
3
3
3
,
`
.
|
+ +
·
z y x
3
(b) By definition, we have
·
,
`
.
|
+
+
u
z y x
2
,
`
.
|
+
+
z y x
u
z y x
,
`
.
|
+
+
=
,
`
.
|
+
+
z y x
,
`
.
|
+
+
z
u
y
u
x
u
=
,
`
.
|
+
+
z y x
,
`
.
|
+ + z y x
3
=
,
`
.
|
+ + ∂
z y x x
3
+
,
`
.
|
+ + ∂
z y x y
3
+
,
`
.
|
+ + ∂
z y x z
3
=
( )
,
`
.
|
+ +
2
3
z y x
+
( )
,
`
.
|
+ +
2
3
z y x
+
( )
,
`
.
|
+ +
2
3
z y x
=
( )
2
9
z y x + +
5. If
( ) r f u ·
, where
2 2 2
z y x r + + · , show that
( ) ( ) r f
r
r f
z
u
y
u
x
u
1 11
2
2
2
2
2
2
2
+ ·
+
+
Solution:- By data
2 2 2 2
z y x r + + · Differentiating this partially w.r.t x, we get
x
x
r
r 2 2 ·
or
,
`
.
|
·
r
x
x
r
Similarly, we get
,
`
.
|
·
r
y
y
r
and
,
`
.
|
·
r
z
z
r
Now, differentiating
( ) r f u ·
, partially w.r.t.x,
( )
x
r
r f
x
u
·
1
As
,
`
.
|
·
r
x
x
r
, we get
( )
,
`
.
|
·
r
x
r f
x
u
1
Differentiating
( )
¹
'
¹
¹
'
¹
·
r
r f
x
x
u
1
partially w.r.t.x, we get
( ) ( )
( )
( ) 1
1 1
1
2
1 11
2
2
r
r f
x
r
r
r f
x
r
r f
r
x
x
u
+
¹
'
¹
¹
'
¹
,
`
.
|
− +
·
i.e.
( ) ( )
( )
r
r f
r
x
r
r f
r
x
r f
r
x
x
u
1
2
1 11
2
2
1 1
+
¹
'
¹
¹
'
¹
,
`
.
|
,
`
.
|
− +
,
`
.
|
·
or
( ) ( ) ( )
r
r f
r
r f
r
r f
r
x
x
x
u
1
2
1 11
2
2
+
¹
'
¹
¹
'
¹
,
`
.
|
·
…………. (1)
161
( )
( )
¹
'
¹
¹
'
¹
− + ·
2
2 1
11
2
2
2
2
1
r
x
r
r f
r f
r
x
x
u
Similarly, we get
and
( )
( )
( ) 3 . . . . . 1
2
2 1
1 1
2
2
2
2
¹
'
¹
¹
'
¹
− + ·
r
z
r
r f
r f
r
z
z
u
Adding eqs (1) ,(2) and (3), we get
( )
( )
( )
¹
'
¹
¹
'
¹ + +
− + + + ·
+
+
2
2 2 2 1
2 2 2
2
11
2
2
2
2
2
2
3
r
z y x
r
r f
z y x
r
r f
z
u
y
u
x
u
( )
( )
{ ¦ 1 3
1
11
− + ·
r
r f
r f
( )
( )
r
r f
r f
1
11
2 + ·
Problem Set No.3.1.1 for practice
1. Verify
x y
z
y x
z
∂ ∂
·
∂ ∂
2 2
where
(i) axy y x z 3
3 3
− + · (ii) ( )
2 2 1
tan y x z + ·
(iii)
( ) x y y x z sin sin log + ·
2. If
y
x
y
x
y
x u
1 2 1 2
tan tan
− −
,
`
.
|
·
, prove that
2 2
2 2 2
y x
y x
x y
u
+
·
∂ ∂
Hint
,
`
.
|
·
∂ ∂
x y x y
u
2
3. If,
y x
y x
z
+
+
·
2 2
, show that
,
`
.
|
− ·
,
`
.
|
y
z
x
z
y
z
x
z
1 4
2
(VTU-Jan-2005)
4. If
( )( )( ) x z z y y x u − − − ·
, show that
0 ·
+
+
z
u
y
u
x
u
5. If
xy
e z · , prove that
]
]
]
]
,
`
.
|
+
,
`
.
|
·
+
2
2
2
2
2
2
2
1
y
z
x
z
z y
z
x
z
6. If
x
z
z
y
u + ·
, prove that
0 ·
+
+
z
u
z
y
u
y
x
u
x
7. If
, sin , cos θ θ r y r x · ·
show that
]
]
]
]
,
`
.
|
+
,
`
.
|
·
+
2 2
2
2
2
2
1
y
r
x
r
r y
r
x
r
(Hint: By data
2 2 2
r y x · + )
162
( )
( )
( ) 2 .... 1
2
2 1
11
2
2
2
2
¹
'
¹
¹
'
¹
− + ·
r
y
r
r f
r f
r
y
y
u
Session –3
• 3.1.5: Introduction to homogenous functions:-
An expression of the form
n
n
n n n
y a y x a y x a x a + + + +
− −
.. ..........
2 2
2
1
1 0
In which every term is of the n
th
degree, is called homogenous function of order n. This
can be rewritten as
]
]
]
,
`
.
|
+ +
,
`
.
|
+
,
`
.
|
+
n
n
n
x
y
a
x
y
a
x
y
a a x ... ..........
2
2 1 0
,
`
.
|
·
x
y
x
n
φ
.
Thus any function
( ) y x f ,
which can be expressed in the below
,
`
.
|
x
y
x
n
φ
, is called
homogenous function of order n (or derivative n) in x and y
For example: (i)
( )
]
]
]
,
`
.
|
,
`
.
|
,
`
.
|
,
`
.
|
+ · + − + ·
3 2
3 3 2 2 3
1 ,
x
y
x
y
x
y
x y xy y x x y x f
,
`
.
|
·
x
y
x
n
φ
is a homogenous function of degree 3.
(ii)
( )
,
`
.
|
·
,
`
.
|
+
,
`
.
|
+
·
+
+
·
x
y
x
x
y
x
x
y
x
y x
y x
y x f φ
2
1
1
1
,
is a homogenous function
of degree ½
(iii)
( )
,
`
.
|
·
,
`
.
|
·
,
`
.
|
·
x
y
x
x
y
x
x
y
y x f φ
0 0
sin sin ,
is a homogenous function of degree 0
Note: - (1) Any function, which can be expressed as,
,
`
.
|
y
x
y
n
φ
is also a homogenous
function of degree n.(or order n)
(2) In general, a function
( ) ,......... , , , t z y x f
is said to be a homogenous
function
of order n in x, y, z, t… if it can be expressed as
,
`
.
|
....... ,......... , ,
x
t
x
z
x
y
x
n
φ
3.1.6: Euler’s Theorem:-
The following theorem, named in honour of the great Swiss mathematician
L. Euler (1707-1783) gives a very useful formula for a particular combination of
partial derivatives of homogenous functions.
• Statement of Euler’s theorem:- If u is a homogenous function of x and y with
degree n, then
nu
y
u
y
x
u
x ·
+
(VTU Feb-2005)
Proof: - Since u is a homogenous function of degree n, we can put u is the form
163
,
`
.
|
·
x
y
x u
n
φ
.
Differentiating this w.r.t x partially, we get
{ ¦
1
2
1 −
,
`
.
|
+
,
`
.
|
,
`
.
|
·
n n
nx
x
y
x
y
x
y
x
x
u
φ φ
or
,
`
.
|
+
,
`
.
|
− ·
− −
x
y
nx
x
y
y x
x
u
n n
φ φ
1 1 2
----------(1)
Again differentiating
,
`
.
|
·
x
y
x u
n
φ
partially w.r.t y, we get
( ) 0
1
1
+
,
`
.
|
·
x
x
y
x
y
u
n
φ
,
`
.
|
·
x
y
x
y
u
n 1 1
φ
…………. (2)
Consider
¹
'
¹
¹
'
¹
,
`
.
|
+
¹
'
¹
¹
'
¹
,
`
.
|
+
,
`
.
|
− ·
+
− − −
x
y
x y
x
y
nx
x
y
y x x
y
u
y
x
u
x
n n n 1 1 1 1 2
φ φ φ
,
`
.
|
+
,
`
.
|
+
,
`
.
|
− ·
− −
x
y
y x
x
y
nx
x
y
y x
n n n 1 1 1 1
φ φ φ
=
,
`
.
|
x
y
nx
n
φ
nu ·
. Since
,
`
.
|
·
x
y
x u
n
φ
Note: - In general, if u is a homogenous function of order x, y, z, t… then
nu
t
u
t
z
u
z
y
u
y
x
u
x · +
+
+
+
.. ..........
• Corollary to Euler’s theorem :- If u is homogenous function of x and y with
degree n, then
( )u n n
y
u
y
y x
u
xy
x
u
x 1 2
2
2
2
2
2
2
2
− ·
+
∂ ∂
+
(VTU Feb-2005)
Proof: - As u is a homogenous function of degree, we have, by Euler’s theorem
nu
y
u
y
x
u
x ·
+
-----------(*)
Differentiating (*) partially w.r.t x,
x
u
n
y x
u
y
x
u
x
u
x
·
∂ ∂
+
+
2
2
2
( )
x
u
n
y x
u
y
x
u
x
− ·
∂ ∂
+
1
2
2
2
………….(1)
Differentiating (*) Partially w.r.t y
164
y
u
n
y
u
y
u
y
x y
u
x
·
+
+
∂ ∂
2
2 2
or
( )
y
u
n
x y
u
x
y
u
y
− ·
∂ ∂
+
1
2
2
2
………….. (2)
Multiplying esq. (1) by x and esq. (2) by y we get
( )
¹
'
¹
¹
'
¹
+
− ·
¹
'
¹
¹
'
¹
∂ ∂
+
+
¹
'
¹
¹
'
¹
∂ ∂
+
y
u
y
x
u
x n
x y
u
xy
y
u
y
y x
u
xy
x
u
x 1
2
2
2
2
2
2
2
2
or
( )( )
( )u n n
y
u
y
y x
u
xy
x
u
x
nu n
y
u
y
y x
u
xy
x
u
x
1 2
1 2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
− ·
+
∂ ∂
+
− ·
+
∂ ∂
+
,
`
.
|
∂ ∂
·
∂ ∂
y x
u
y x
u
2 2
• Session - 4
• 3.1.7: Worked Examples
1. Verify Euler’s theorem for
i).
,
`
.
|
+
·
y x
xy
u
ii)
2 2
2 by hxy ax u + + ·
Solution:- (i) Consider
( )
( )
,
`
.
|
·
,
`
.
|
·
,
`
.
|
+
·
+
·
+
·
x
y
x
x
y
x
x y
x y
x
x y x
x y x
y x
xy
u φ φ
1
2
1 1
This shows that
y x
xy
u
+
·
is a homogenous function of degree1 (1=n).
Hence, Euler’s theorem when applied to u becomes
u
y
u
y
x
u
x ) 1 ( ·
+
………….(*) (n=1)
we verify the equation (*) by showing actually LHS=RHS
now as
,
`
.
|
+
·
y x
xy
u
,
( )
2
2
y x
y
x
u
+
·
&
( )
2
2
y x
x
y
u
+
·
∴ LHS of (*)
y
u
y
x
u
x
+
·
( ) ( )
( )
( ) ( ) y x
xy
y x
y x xy
y x
x
y
y x
y
x
+
·
+
+
·
,
`
.
|
+
+
,
`
.
|
+
·
2 2
2
2
2
=RHS
This verifies the result(*)
(ii) Consider
,
`
.
|
+
,
`
.
|
+ · + + ·
2
2
2
2 2 2
2 2
x
y
b
x
y
h
x
a
x by hxy ax u
,
`
.
|
·
x
y
x φ
2
,which means u is a
homogenous function of degree 2. Hence, on applying Euler’s theorem
to u, we get
u
y
u
y
x
u
x 2 ·
+
………….(*)
165
Now, as
2 2
2 by hxy ax u + + · , we see that
( ) hy ax
x
u
+ ·
2
&
( ) hx ay
y
u
+ ·
2
Consider, LHS of (*)
y
u
y
x
u
x
+
( ) ( ) ( ) ( ) hx ay y hy ax x + + + · 2 2
( )
2 2 2 2
2 2 2 2 2 2 by hxy ax ay hxy hxy ax + + · + + + ·
u 2 · = RHS
This verifies the result (*)
(2) If
,
`
.
|
+
·
y x
y x
u
2 2
sin
, show that
u
y
u
y
x
u
x tan 3 ·
+
(VTU July-2005)
Solution:- Consider
f
y x
y x
u ·
,
`
.
|
+
·
2 2
sin
,say We note that
,
`
.
|
·
,
`
.
|
+
,
`
.
|
·
,
`
.
|
+
·
x
y
x
x
y
x
x
y
x
y x
y x
f φ
3
2
4
2 2
1
.Thus, `f` is a homogenous function of degree 3.
Applying Euler’s theorem to f, we get
f
y
f
y
x
f
x 3 ·
+
As
u f sin ·
,we see that
( ) ( ) ( ) u u
y
y u
x
x sin 3 sin sin ·
+
i.e.
( ) ( ) u
y
u
u y
x
u
u x sin 3 cos cos ·
¹
'
¹
¹
'
¹
+
¹
'
¹
¹
'
¹
or we get
u
y
u
y
x
u
x tan 3 ·
+
as desired
2. If
,
`
.
|
+
+
·
y x
y x
e u
4 3
3 3
, show that
u u
y
u
y
x
u
x log 2 ·
+
(VTU july-2004)
Solution: - Consider
,
`
.
|
+
+
·
y x
y x
e u
4 3
3 3
f
y x
y x
u ·
,
`
.
|
+
+
· ∴
4 3
log
3 3
,say
,
`
.
|
·
,
`
.
|
,
`
.
|
+
,
`
.
|
+
·
+
+
· ∴
x
y
x
x
y
x
x
y
x
y x
y x
f φ
2
3
3
3
3 3
4 3
1
4 3
.Thus, f is a homogenous function of degree 2.
Applying Euler's theorem to f, we get
f
y
f
y
x
f
x 2 ·
+
i.e
( ) ( ) ( ) u u
y
y u
x
x log 2 log log ·
+
or
u
y
u
u
y
x
u
u
x log 2
1 1
·
,
`
.
|
+
,
`
.
|
166
or
u u
y
u
y
x
u
x log 2 ·
+
4. If
y x
y x
u
+
+
· cos
Prove that
u
y
u
y
x
u
x cot
2
1
− ·
+
(VTU- Jan-2004)
Solution:- Let
f
x
y
x
x
y
x
x
y
x
y x
y x
u ·
,
`
.
|
·
,
`
.
|
+
,
`
.
|
+
·
+
+
· φ
2
1
1
1
cos
,say Hence, f is a
homogenous function of degree ½
On applying Euler's theorem to f, we get
f
y
f
y
x
f
x
2
1
·
+
Since
u f cos ·
( ) ( ) u u
y
y u
x
x cos
2
1
cos cos ·
+
or
u
y
u
u y
x
u
u x cos
2
1
sin sin ·
¹
'
¹
¹
'
¹
− +
¹
'
¹
¹
'
¹
or
u
u
y
u
y
x
u
x
sin
cos
2
1
− ·
+
or
u
y
u
y
x
u
x cot
2
1
− ·
+
5. If
,
`
.
|
+
+
·
y x
y x
u
3 3
1
tan
, show that
u yu xu
y x
2 sin · +
.Hence to evaluate
yy xy xx
u y xyu u x
2 2
2 + +
(VTU July-2005)
Solution: - Consider
,
`
.
|
+
+
·
y x
y x
u
3 3
1
tan
f
y x
y x
u ·
+
+
· ∴
3 3
tan
, say
Now,
,
`
.
|
·
,
`
.
|
+
,
`
.
|
+
·
+
+
· ∴
x
y
x
x
y
x
x
y
x
y x
y x
f φ
2
3
3
3
3 3
1
1
.So, f is a homogenous function of
degree2. Applying Euler’s theorem to f we get
f
y
f
y
x
f
x 2 ·
+
As
u f tan ·
, we see that
( ) ( ) u u
y
y u
x
x tan 2 tan tan ·
+
u
y
u
u y
x
u
u x tan 2 sec sec
2 2
·
¹
'
¹
¹
'
¹
+
¹
'
¹
¹
'
¹
u u
u
u
y
u
y
x
u
x cos sin 2
sec
tan 2
2
· ·
+
u
y
u
y
x
u
x 2 sin ·
+
…………….(*),as required.
167
To get the value of yy xy xx
u y xyu u x
2 2
2 + +
we proceed as follows Differentiating (*)
partially w.r.t.x & y we get, respectively
( )
x
u
u
y x
u
y
x
u
x
u
x
·
∂ ∂
+
+
2 cos 2
2
2
2
…………..(1)
( )
y
u
u
y
u
y
u
y
x y
u
x
·
+
+
∂ ∂
2 cos 2
2
2 2
…………….(2)
Multiplying eq (1) by x, and eq(2) by y and adding thereafter
we get
( )
,
`
.
|
+
− ·
+
∂ ∂
+
y
u
y
x
u
x u
y
u
y
y x
u
xy
x
u
x 1 2 cos 2 2
2
2
2
2 2
2
( )( ) u u 2 sin 1 2 cos 2 − ·
u u u y xyu u x
yy xy xx
2 sin 4 sin 2
2 2
− · + +
6. If
,
`
.
|
,
`
.
|
·
− −
y
x
y
x
y
x u
1 2 1 2
tan tan
, show that
u
y
u
y
y x
u
xy
x
u
x 2 2
2
2
2
2 2
2
·
+
∂ ∂
+
Solution:- We note that
w v u − ·
where
,
`
.
|
·
x
y
x v
1 2
tan
&
,
`
.
|
·
y
x
y w
1 2
tan
So that v and w are homogenous functions of degree 2. Applying the corollary to the
Euler’s theorem to v and w, we obtain
( ) v v
y
v
y
y x
v
xy
x
v
x 2 1 2 2 2
2
2
2
2 2
2
· − ·
+
∂ ∂
+
( ) w w
y
w
y
y x
w
xy
x
w
y 2 1 2 2 2
2
2
2
2 2
2
· − ·
+
∂ ∂
+
Taking the difference of these two expressions, we get
u
y
u
y
y x
u
xy
x
u
x 2 2
2
2
2
2 2
2
·
+
∂ ∂
+
Problem set 3.1.2 for practice
1. If
,
`
.
|
+
·
y x
y x
u
2 2
1
sin
,show that
u
y
u
y
x
u
x tan ·
+
2. If
,
`
.
|
+
·
y x
y x
z
2 2
log
,show that
1 ·
+
y
z
y
x
z
x
168
3. If
y x
y x
u
+
+
·
−1
cot
,prove that
u
y
z
y
x
z
x 2 sin
4
1
− ·
+
4. If
,
`
.
|
+
·
3 3
3 3
1
tan
y x
y x
z
,prove that
u
y
u
y
x
u
x 2 sin
2
3
·
+
169
Chapter - 3 : PARIAL DIFFERENTIATION
LESSON-2: Total derivatives, Differentiation of Composite
and Implicit functions
• In this lesson we learn the concept of total derivatives of functions of two
or more
Variables and, also rules for differentiation of composite and implicit
functions.
Session-1
3.2.0: a) Total differential and Total derivative:-
For a function
) , ( y x f z ·
of two variables, x and y the total differential
(or
exact differential ) dz is defined by:
dy
x
f
dx
x
f
dz
+
·
--------------------------(1)
Further, if
) , ( y x f z ·
where
) (t x x ·
,
) (t y y ·
i.e. x and y are themselves
functions of an independent variable t, then total derivative of z is given by
dt
dy
y
f
dt
dx
x
f
dt
dz
+
·
-------------------------(2)
Similarly, the total differential of a function
) , , ( z y x f u ·
is defined by
dz
z
f
dy
y
f
dx
x
f
du
+
+
·
-----------------(3)
Further, if
) , , ( z y x f u ·
and if
) (t x x ·
,
) (t y y ·
,
) (t z z ·
, then the total
derivative of u is given by
dt
dz
z
f
dt
dy
y
f
dt
dx
x
f
dt
du
+
+
·
---------------(4)
(b) Differentiation of implicit functions:-
An implicit function with x as an independent variable and y as the dependent
variable is generally of the form
0 ) , ( · · y x f z
. This gives
0 ·
,
`
.
|
·
,
`
.
|
dx
df
dx
dz
.Then, by virtue of expression (2) above, we get
dx
dy
y
f
x
f
dx
dz
+
·
or
dx
dy
y
f
x
f
dx
df
+
·
, and hence
170
dx
dy
y
f
x
f
+
· 0
, so that we get
,
`
.
|
,
`
.
|
− ·
y
f
x
f
dx
dy
----------- (5)
(c) Differentiation of composite functions:-
Let z be an function of x and y and that
) , ( v u x φ ·
and
) , ( v u y ϕ ·
are
functions of u and
v then,
&
v
y
y
f
v
x
x
f
v
z
+
·
-------------------(6)
Similarly, if
) , ( v u f z ·
are functions of u and v and if
) , ( y x u φ ·
and
) , ( y x v ϕ ·
are functions of x and y then,
&
y
v
v
f
y
u
u
f
y
z
+
·
--------------------------
(7)
Note:-1) The above formulae can be extended to functions of three are more
variables and formulas (6) and(7) are called Chain rule for partial differentiation.
2) The second and higher order partial derivatives of
) , ( y x f z ·
can be
obtained by repeated applications of the above formulas
3.2.1: Worked examples:-
1. Find the total differential of
(i) [ ] y y y x e
x
cos sin + (ii)
xyz
e
Ans:- (i) Let ( ) [ ] y y y x e y x f z
x
cos sin , + · · Then
( ) [ ] y y y x e
x
z
x
cos sin 1 + + ·
and
( ) [ ] y y y x e
y
z
x
sin cos 1 − + ·
Hence, using formula (1), we get
dy
y
z
dx
x
z
dz
+
·
i.e ( ) [ ] ( ) [ ]dy y y y x e dx y y y x e dz
x x
sin cos 1 cos sin 1 − + + + + ·
(ii) Let
xyz
e z y x f z · · ) , , ( Then
171
u
y
y
f
u
x
x
f
u
z
+
·
x
v
v
f
x
u
u
f
x
z
+
·
( )
xyz
e yz
x
u
·
;
( )
xyz
e xz
y
u
·
;
( )
xyz
e xy
z
u
·
∴ Total differential of
) , , ( z y x f z ·
is (see formula (3) above)
dz
z
u
dy
y
u
dx
x
u
du
+
+
·
( ) xydz zxdy yzdx e
xyz
+ + ·
2. Find
dt
dz
if
(i) y x xy z
2 2
+ · ,where
2
at x · ,
at y 2 ·
(ii)
,
`
.
|
·
x
y
u
1
tan
,where
t t
e e x
− · ,
t t
e e y
+ · (VTU-Jan 2003)
Ans:- (i) Consider y x xy z
2 2
+ ·
xy y
x
z
2
2
+ ·
&
2
2 x xy
y
z
+ ·
Since
2
at x · &
at y 2 ·
, We have
at
dt
dx
2 ·
,
a
dt
dy
2 ·
Hence, using formula (2), we get
dt
dy
y
z
dt
dx
x
z
dt
dz
+
·
( )( ) ( )( ) a x xy at xy y 2 2 2 2
2 2
+ + + ·
( ) ( )
2 2
2 2 2 x xy a y xy y + + + · , Using
at y 2 ·
2 2 3
2 4 2 ax axy xy y
dt
dz
+ + + ·
To get
,
`
.
|
dt
dz
explicitly in terms of t, we substitute
2
at x · &
at y 2 ·
, to get
( )
4 3 3
5 8 2 t t a
dt
dz
+ ·
,
`
.
|
(ii) Consider
,
`
.
|
·
x
y
u
1
tan
2 2
y x
y
x
u
+
·
,
2 2
y x
x
y
u
+
·
Since
t t
e e x
− · &
t t
e e y
+ · ,we have
y e e
dt
dx
t t
· + ·
x e e
dt
dy
t t
· − ·
Hence
dt
dy
y
u
dt
dx
x
u
dt
du
+
·
( see Eqn (2))
,
`
.
|
+
·
,
`
.
|
+
+
,
`
.
|
+
·
2 2
2 2
2 2 2 2
y x
y x
x
y x
x
y
y x
y
Substituting
t t
e e x
− · &
t t
e e y
+ · , we get
172
t t
e e dt
du
2 2
2
+
·
Session-2
3. Find
,
`
.
|
dx
dy
if (i)
x y
y x + =Constant
(ii) xy e x
y
2 · +
Ans: - (i) Let
x y
y x y x f z + · · ) , ( =Constant. Using formula (5)
y
f
x
f
dx
dy
− ·
-----------------(*)
But
y y yx
x
f
x y
log
1
+ ·
and
1
log
+ ·
x y
xy x x
y
f
Putting those in(*), we get
¹
'
¹
¹
'
¹
+
+
− ·
1
1
log
log
x y
x y
xy x x
y y yx
dx
dy
(ii) Let xy e e y x f z
y x
2 ) , ( − + · · =Constant
Now,
y e
x
f
x
2 − ·
;
x e
y
f
y
2 − ·
Using this in (8),
¹
'
¹
¹
'
¹
− ·
¹
¹
¹
'
¹
¹
¹
¹
'
¹
− ·
x e
y e
y
f
x
f
dx
dy
y
x
2
2
4. (i) If
) , ( y x f z ·
,where θ cos r x · ,
θ sin r y ·
show that
2
2
2
2
2
1
,
`
.
|
+
,
`
.
|
·
,
`
.
|
+
,
`
.
|
θ
z
r r
z
y
z
x
z
(VTU July-2005)
(ii) If
) , ( y x f z ·
,where
v u
e e x
+ · &
v u
e e y − ·
,Show that
y
z
y
x
z
x
v
z
u
z
·
Ans: - As θ cos r x · and
θ sin r y ·
, we have
θ cos ·
r
x
,
θ
θ
sin r
x
− ·
;
θ sin ·
r
y
&
θ
θ
cos r
y
·
. Using Chain rule (6) & (7)
we have
( ) ( ) θ θ sin cos
y
z
x
z
r
y
y
z
r
x
x
z
r
z
+
·
+
·
,
`
.
|
Squaring on both sides, the
above equations, we get
θ θ θ θ cos sin 2 sin cos
2
2
2
2 2
,
`
.
|
,
`
.
|
+
,
`
.
|
+
,
`
.
|
·
,
`
.
|
y
z
x
z
y
z
x
z
r
z
173
( ) ( ) θ θ
θ θ θ
cos sin r
y
z
r
x
z y
y
z x
x
z z
+ −
·
+
·
,
`
.
|
θ θ θ θ
θ
cos sin 2 sin cos
1
2
2
2
2 2
2
,
`
.
|
,
`
.
|
,
`
.
|
+
,
`
.
|
·
,
`
.
|
y
z
x
z
y
z
x
z z
r
Adding the above equations , we get
( ) θ θ
θ
2 2
2 2 2
2
2
sin cos
1
+
¹
¹
¹
'
¹
¹
¹
¹
'
¹
,
`
.
|
+
,
`
.
|
·
,
`
.
|
+
,
`
.
|
y
z
x
z z
r r
z
=
2
2
,
`
.
|
+
,
`
.
|
y
z
x
z
as desired.
(ii) As
v u
e e x
+ · &
v u
e e y − ·
, We have
u
e
u
x
·
,
v
e
v
x
− ·
,
u
e
u
y
− ·
&
v
e
v
y
− ·
Using Chain rule (6) we get
( ) ( )
v u
e
y
z
e
x
z
u
y
y
z
u
x
x
z
u
z
·
+
·
,
`
.
|
( ) ( )
v v
e
y
z
e
x
z
v
y
y
z
v
x
x
z
v
z
− −
·
+
·
,
`
.
|
( ) ( )
v u v u
e e
y
z
e e
x
z
v
z
u
z
− +
·
,
`
.
|
,
`
.
|
− −
y
y
z
x
x
z
·
5. (i) If
) / , , ( z y z x f u ·
Then show that
0 ·
z
u
z
y
u
y
x
u
x
(VTU-July-2004)
(ii) If
) , , ( x z z y y x f H − − − ·
, show that
0 ·
+
+
z
H
y
H
x
H
(VTU-July-2003)
Ans: - (i) Let
) , ( w v f u ·
, where
xz v ·
and
z
y
w ·
z
x
v
·
,
0 ·
y
v
,
x
z
v
·
&
0 ·
x
w
,
z
y
w
1
·
,
2
z
y
z
w
·
Using Chain rule,
( ) ( )
v
u
z
w
u
z
v
u
x
w
w
u
x
v
v
u
x
u
·
+
·
+
·
0
( ) ( )
w
u
z
z
w
u
v
u
y
w
w
u
y
v
v
u
y
u
·
+
·
+
·
∂ 1
1
0
( )
w
u
z
y
v
u
x
z
y
w
u
x
v
u
z
w
w
u
z
v
v
u
z
u
·
,
`
.
|−
+
·
+
·
2
2
From these, we get
,
`
.
|
·
w
u
z
y
v
u
x z
w
u
z
y
v
u
xz
z
u
z
y
u
y
x
u
x
2
= 0
(ii) Let
) , , ( w v u f H ·
Where
x z w z y v y x u − · − · − · , ,
174
Now,
0 , 1 , 1 ·
− ·
·
z
u
y
u
x
u
1 , 1 , 0 − ·
·
·
z
v
y
v
x
v
1 , 0 , 1 ·
·
− ·
z
u
y
w
x
w
Using Chain rule,
( ) ( ) ( ) 1 0 1 −
+
+
·
+
+
·
w
H
v
H
u
H
x
w
w
H
x
v
v
H
x
u
u
H
x
H
( ) ( ) ( ) 0 1 1
w
v
v
H
u
H
y
w
w
H
y
v
v
H
y
u
u
H
y
H
+
+ −
·
+
+
·
( ) ( ) ( ) 1 1 0
w
H
v
H
u
H
z
w
w
H
z
v
v
H
z
u
u
H
z
H
+ −
+
·
+
+
·
Adding the above equations, we get
0 ·
+
+
z
H
y
H
x
H
, as desired.
3.2.2: Problem Set No: 3.2.1
1. Find the total differentials of
(i) ( )
1 −
+ xyz xyz (ii) x z z y y x
2 2 2
+ +
2. Find
,
`
.
|
dt
du
If
(i)
2 2
y x u − · , , cos t e x
t
· t e y
t
sin ·
(ii)
t
e y t x xy u · · · , log , sin
2
(iii)
t t
e z e y
t
x zx yz xy u
· · · + + · , ,
1
,
(iv) ( ) t z t y e x z y x u
t
cos , sin , , log · · · + + ·
3. Find
,
`
.
|
dx
dy
is each of the following cases:-
(i)
( ) ( ) y x y x x + · − sin
(ii) ( ) ( )
x y
y x sin cos ·
4. If
) , ( y x f z ·
and
uv y v u x · − · ,
Prove that
(i)
( )
v
z
v
x
z
u
x
z
v u
·
+
(ii)
( )
v
z
u
z
y
z
v u
+
·
+
5. If , 5 8 , 4 3 2 ,
2 2
− + − · + − · − · s r y s r z y x u Prove that
y x
r
u
2 4 + ·
6. If
) , ( s r f z ·
, where
bt y y at x r + · + · ,
Prove that
y
u
b
x
u
a
t
u
+
·
Chapter - 3 : PARIAL DIFFERENTIATION
175
LESSON-3: Applications to Jacobians, Errors and
Approximations
In this lesson, we study Jacobians, errors and approximations using the
concept of
partial differentiation.
Session - 1
3.3.0 Jacobians:-
Jacobians were invented by German mathematician C.G. Jacob Jacobi (1804-
1851),
who made significant contributions to mechanics, Partial differential equations and
calculus
of variations.
Definition:- Let u and v are functions of x and y, then Jacobian of u and v w.r.t x and
y,
denoted by
J or
,
`
.
|
y x
v u
J
,
,
or
( )
( ) y x
v u
,
,
is defined by
y
v
x
v
y
u
x
u
y x
v u
J
·
,
`
.
|
,
,
Similarly, if u, v, w are functions of three independent variables of x, y, z, then
176
z
w
y
w
x
w
z
u
y
v
x
v
z
u
y
u
x
u
z y x
w v u
J J
·
,
`
.
|
·
, ,
, ,
Remark:- In a similar way, Jacobian of n functions in n-variables can be defined
Note:- (i) If
( )
( ) y x
v u
J
,
,
·
, then the "inverse Jacobian" of the Jacobian J,
denoted by J ′ ,is defined as
( )
( ) v u
y x
J
,
,
·
(ii) Similarly, "inverse Jacobian" of
( )
( ) z y x
w v u
J
, ,
, ,
·
is defined as
( )
( ) w v u
z y x
J
, ,
, ,
· ′
3.3.1: Properties of Jacobians :-
Property 1:- If
( )
( ) y x
v u
J
,
,
·
and
( )
( ) v u
y x
J
,
,
· ′
then 1 · ′ J J
Proof:- Consider
177
( )
( )
( )
( )
v
y
u
y
v
x
u
x
y
v
x
v
y
u
x
u
v u
y x
y x
v u
J J
×
·
×
·
,
,
,
,
1
1
0 1
0 1
· ·
+
+
+
+
·
v
y
y
v
v
x
x
v
u
y
y
v
u
x
x
v
v
y
y
u
v
x
x
u
u
y
y
u
u
x
x
u
Property 2:- (Chain rule for Jacobians):- If u and v are functions of r&s and r,s
are functions x&y,then
,
`
.
|
×
,
`
.
|
·
,
`
.
|
·
y x
s r
J
s r
v u
J
y x
v u
J
,
,
,
,
,
,
Proof:- Consider
178
y
s
x
s
y
r
x
r
s
v
r
v
s
u
r
u
y x
s r
J
s r
v u
J
×
·
,
`
.
|
×
,
`
.
|
,
,
,
,
y
s
s
v
y
r
r
v
x
s
s
v
x
r
r
v
y
s
s
u
y
r
r
u
x
s
s
u
x
r
r
u
+
+
+
+
·
,
`
.
|
·
·
y x
v u
J
y
v
x
v
y
u
x
u
,
,
3.3.2:-Jacobians in various co-ordinate systems:-
179
1. In Polar co-ordinates, θ cos r x · ,
θ sin r y ·
we have
( )
( )
r
r
v u
·
θ ,
,
2. In spherical coordinates,
, , sin , cos z z y x · · · φ ρ φ ρ
we have
( )
( )
ρ
φ ρ
·
z
z y x
, ,
, ,
3. In spherical polar co-ordinates,
φ θcos sin r x ·
,
φ θsin sin r y ·
, θ cos r z ·
Proof of 1:- we have,
θ cos ·
r
x
and
θ sin ·
r
y
θ
θ
sin r
x
− ·
and
θ
θ
cos r
y
·
( )
( ) θ θ
θ θ
θ
θ
θ c o s s i n
s i n c o s
,
,
r
r
y
r
y
x
r
x
r
y x
·
·
( ) r r r r · + · + · θ θ θ θ
2 2 2 2
sin cos sin cos
Proof of 2 :-we have
φ
ρ
cos ·
∂x
,
φ
ρ
sin ·
∂y
,
0 ·
ρ
z
φ ρ
φ
sin − ·
∂x
,
φ ρ
φ
cos ·
∂y
,
0 ·
φ
z
0 ·
ρ
z
,
0 ·
φ
z
,
0 ·
z
z
180
( )
( )
ρ φ ρ φ
φ ρ φ
φ ρ
φ ρ
φ ρ
φ ρ
·
·
·
1 0 0
0 c o s s i n
0 s i n c o s
, ,
, ,
z
z z z
z
y y y
z
x x x
z
z y x
Proof of 3:- We have
θ θcos sin ·
r
x
,
φ θ
θ
cos cos r
x
·
,
φ θ
φ
sin sin r
x
− ·
φ θsin sin ·
r
y
,
φ θ
θ
sin cos r
y
·
,
φ θ
φ
cos sin r
y
·
θ cos ·
r
z
,
θ
θ
sin r
z
− ·
,
0 ·
φ
z
( )
( )
0 s i n c o s
c o s s i n s i n c o s s i n s i n
s i n s i n c o s c o s c o s s i n
, ,
, ,
θ θ
φ θ φ θ φ θ
φ θ φ θ θ θ
φ θ
r
r r
r r
r
z y x
·
θ sin
2
r ·
Session – 2
181
3.3.3:-Worked Examples:-
1. If
2 2
2y x u − · ,
2 2
2 y x v − · , where
θ cos r x · ,
θ sin r y ·
show that
( )
( )
θ
θ
2 sin 6
,
,
3
r
r
v u
·
(VTU-Jan-2006)
Consider θ θ
2 2 2 2 2 2
sin 2 cos 2 r r y x u − · − ·
θ θ
2 2 2 2 2 2
sin cos 2 2 r r y x v − · − ·
θ θ
2 2
sin 4 cos 2 r r
r
u
− ·
,
θ θ
2 2
sin 2 cos 4 r r
r
v
− ·
θ θ θ θ
θ
cos sin 4 sin cos 2
2 2
r r
u
− − ·
θ θ θ θ
θ
cos sin 2 sin cos 4
2 2
r r
v
− − ·
( )
( )
θ θ θ θ θ θ
θ θ θ θ θ θ
θ
θ
θ
c o s s i n 2 s i n c o s 4 s i n 2 c o s 4
c o s s i n 4 s i n c o s 2 s i n 4 c o s 2
,
,
2 2 2 2
2 2 2 2
r r r r
r r r r
v
r
v
u
r
u
r
v u
− − −
− − −
·
·
( )( ) θ θ θ θ θ θ cos sin 2 sin cos 4 sin 4 cos 2
2 2 2 2
r r r r − − − ·
( )( ) θ θ θ θ θ θ
2 2 2
sin 2 cos 4 cos sin 4 sin cos 2 r r r r − − − ·
θ 2 sin 6
3
r ·
2. If
( ) , , 1 uv y v u x · − ·
Prove that
1
,
,
,
,
1
·
,
`
.
|
×
,
`
.
|
v u
y x
J
v u
y x
J
(VTU-2001)
182
Consider
v
u
x
− ·
1
,
u
v
x
− ·
v
u
y
·
,
u
v
y
·
u v
u v
v
y
u
y
v
x
u
x
v u
y x
J
− −
·
·
,
`
.
|
1
,
,
( ) ( ) u uv uv u uv u v · + − · − − − · 1
) 1 (
,
,
− − − ·
,
`
.
|
∴ u
v u
y x
J
Further, as
( ) , , 1 uv y v u x · − ·
uv u − ·
We write,
y u x − ·
y x u + · ∴
and
,
`
.
|
+
· ·
y x
y
u
y
v
,
`
.
|
+
· ∴
y x
y
v
, 1 ·
x
u
1 ·
y
u
and
( )
,
2
y x
y
x
v
+
− ·
( )
2
y x
x
y
v
+
·
183
( ) ( )
2 2
1
1 1
,
,
y x
x
y x
y
y
v
x
v
y
u
x
u
y x
v u
J
+ +
·
·
,
`
.
|
( ) ( ) u y x y x
y
y x
x 1 1
2 2
·
,
`
.
|
+
·
+
+
+
·
1
1
1
· · ∴
u
u JJ
3. If , cos v e x
u
· , sin v e y
u
· Prove that
( )
( )
( )
( )
1
,
,
,
,
·
×
y x
v u
v u
y x
Consider v e x
u
cos · v e y
u
sin ·
v e
u
x
u
cos ·
v e
u
y
u
sin ·
v e
v
x
u
sin − ·
v e
v
y
u
cos ·
184
( )
( )
v e v e
v e v e
v
y
u
y
v
x
u
x
v u
y x
u u
u u
c o s s i n
s i n c o s
,
,
·
·
( )
( )
) 1 (
,
,
.
2
− − − − − ·
u
e
v u
y x
e i
Again Consider , cos v e x
u
· , sin v e y
u
·
u
e y x
2 2 2
· + ∴ or
( )
2 2
log
2
1
y x u + ·
&
v
x
y
tan ·
or
,
`
.
|
·
x
y
v
1
tan
Hence
; ,
2 2
y x
x
x
u
+
·
; ,
2 2
y x
y
y
u
+
·
& 2 2 2 2
;
y x
x
y
v
y x
y
x
v
+
·
+
·
185
( )
( )
( )
1
2 2
2 2
2 2 2 2
2 2 2 2
1
,
, −
+ ·
+
·
+ +
+ +
·
·
∴ y x
y x
y x
x
y x
y
y x
y
y x
x
y
v
x
v
y
u
x
u
y x
v u
i.e
( )
( )
) 2 (
,
,
2
− − − − − − − ·
− u
e
y x
v u
( )
( )
( )
( )
1
,
,
,
,
2 2
· × ·
×
− u u
e e
y x
v u
v u
y x
4. If
, , ,
z
xy
w
y
zx
v
x
yz
u · · ·
Show that
( )
( )
4
, ,
, ,
·
z y x
w v u
186
Now,
( )
( )
2
2
2
, ,
, ,
z
x y
z
x
z
y
y
x
y
z x
y
z
x
y
x
z
x
y z
z
w
y
w
x
w
z
v
y
v
x
v
z
u
y
u
x
u
z y x
w v u
·
·
i.e
( )
( )
¹
'
¹
¹
'
¹
,
`
.
| −
,
`
.
|
,
`
.
|
,
`
.
|
,
`
.
|
+
¹
'
¹
¹
'
¹
,
`
.
|
,
`
.
|
,
`
.
| −
,
`
.
|
,
`
.
|
¹
'
¹
¹
'
¹
,
`
.
| −
,
`
.
| −
,
`
.
|
− ·
2
2 2 2 2
, ,
, ,
y
zx
z
y
z
x
y
z
x
y
y
x
z
y
z
xy
y
z
x
z
z
xy
y
zx
x
yz
z y x
w v u
= 4, as desired.
5. If
φ θcos sin r x ·
,
φ θsin sin r y ·
, θ cos r z · ,show that
( )
( )
θ
φ θ
sin
, ,
, ,
2
r
r
z y x
·
Now, by definition
187
( )
( )
φ θ
φ θ
φ θ
φ θ
·
z z
r
z
y y
r
y
x x
r
x
r
z y x
, ,
, ,
i.e
( )
( )
0 s i n c o s
c o s s i n s i n c o s s i n s i n
s i n s i n c o s c o s c o s s i n
, ,
, ,
θ θ
φ θ φ θ φ θ
φ θ φ θ φ θ
φ θ
r
r r
r r
r
z y x
·
( ) { ¦ φ θ φ θ cos sin 0 cos sin
2 2
r − ·
( ) { ¦ φ θ θ φ θ cos cos sin 0 cos cos r r − −
{ ¦ φ θ φ θ φ θ sin cos sin sin sin sin
2 2
r r r − − −
( ) φ θ θ φ θ θ
2 2 2 2 2 2
cos cos sin cos sin sin r r + ·
( ) φ θ φ θ θ θ
2 2 2 2 2 2
sin sin cos cos sin sin r r + + ·
( ) φ φ θ
2 2 2
sin cos sin + · r
θ sin
2
r · , as required
3.3.4:-Problem set No: 3.3.1
1. θ cos r x · ,
θ sin r y ·
show that
1
,
,
,
,
1
·
,
`
.
|
×
,
`
.
|
y x
r
J
r
y x
J
θ
θ
2. If , sec v e x
u
· , tan v e y
u
· Show that
( )
( )
( )
( )
1
,
,
,
,
·
×
y x
v u
v u
y x
3.If
z y x w zx yz xy v xyz u + + · + + · · , ,
show that
188
( )
( )
( )( )( ) x z z y y x
z y x
w v u
− − − ·
, ,
, ,
4.If
, , , uvw z v z y u z y x · · + · + +
find the value of
( )
( ) w v u
z y x
, ,
, ,
5.If y e v u y e v u
x x
sin , cos · − · + find
,
`
.
|
y x
v u
J
,
,
[Hint: Start from ( ) y y
e
u
x
sin cos
2
+ · and ( ) y y
e
v
x
sin cos
2
− ·
3.3.5:-Errors and Approximations
Session-3:-
While doing scientific or engineering computations there is always scope for
computation errors. Obviously, it is going to have an impact on the result which may
be negligible or significant too. In such situations, an account of various types of
errors is necessary to present approximate value of the function under consideration.
3.3.6:-Definition:-
If
) , ( y x f z ·
is a function of two variables x and y then the error or absolute error
z ∂ is given by
y
y
z
x
x
z
z δ δ δ
,
`
.
|
+
,
`
.
|
·
-------------(1)
Similarly, if
) , , ( z y x f u ·
is a function of three variables x,y,z then the error or
absolute error u δ is given by
z
z
u
y
y
u
x
x
u
u δ δ δ δ
,
`
.
|
+
,
`
.
|
+
,
`
.
|
·
----------- (2)
Remark:- The quantity o is called relative error in z or u respectively. Similarly, the
quantity
,
`
.
|
×
,
`
.
|
× 100 100
u
u
or
z
z δ δ
is called percentage error in z or u respectively
Note:- The following results similar to that of differentials will also be useful.
(i)
( ) u c cu δ δ ·
(ii)
( ) v u v u δ δ δ t · t
(iii)
( ) u v v u uv δ δ δ + ·
(iv)
2
v
v u u v
v
u δ δ
δ
·
,
`
.
|
3.3.7:-Worked Examples:-
1. Find the percentage error in measuring the area of a traingle if 1% error (each) is
made in measuring its base as well as height.
Area A of a traingle is given by
189
ah A
2
1
·
---------- (1)
where `a`and `h` are base and height of the triangle respectively. Therefore, the
change A ∂ due to a change a δ in a, and a change h δ in h, is given by
( ) [ ] a h h a ah A δ δ δ δ + · ·
2
1
2
1
[ ]
ah
a h h a
A
A
2
1
2
1
δ δ
δ
+
· ∴
a
a
h
h
A
A δ δ δ
+ · ∴
,
`
.
|
× +
,
`
.
|
× · × ∴ 100 100 100
a
a
h
h
A
A δ δ δ
=1 % +1% (by data)
=2%
The percentage error in A is 2%
2. The focal length f of a lens is given by
q p f
1 1 1
+ ·
where p and q are the
distances of the lens from the object and the image respectively. For a certain lens
p and q are each 25cms with a possible error of almost 0.5cm. Find the
approximate value of the maximum error in f.
Consider
q p f
1 1 1
+ ·
. By data
25 · ·q p
and
5 . 0 · · q p δ δ
Now,
,
`
.
|
+ ·
,
`
.
|
q p f
1 1 1
δ δ
i.e
,
`
.
|
+
,
`
.
|
·
,
`
.
|
q p f
1 1 1
δ δ δ
q
q
p
p
f
f
δ δ δ
2 2 2
1 1 1
− − · −
i.e
]
]
]
+ ·
2 2
2
q
q
p
p
f f
δ δ
δ
As
, q p ·
and
, q p δ δ ·
we have
]
]
]
·
2
2
2
p
p
f f
δ
δ
Since
08 . 0
25
1
25
1 1
· + ·
f
we have f=12.5. Hence,
( )
( )
25 . 0
25
5 . 0
2 5 . 12
2
2
·
]
]
]
]
,
`
.
|
· f δ
∴ The maximum error in f=0.25
190
3. The current measured by a tangent galvanometer is given by the relation
θ tan k c · Where θ is the angle of deflection .Show that the relative error in c
due to a given error in θ is maximum where
0
45 · θ .
Consider θ tan k c · (k=constant)
∴ ( ) θ δ δ tan k c ·
( ) 0 sec
2
+ · δ θ θ k
∴ δ θ
θ
θ δ
tan
sec
2
k
k
c
c
·
or
δ θ
θ
δ
,
`
.
|
·
2 sin
2
c
c
-----------(*)
The relation error in c being
c
c δ
is maximum when the denominator of RHS of
(*) is maximum and the maximum of sine function is unity.
0 0
45 90 2 1 2 sin · ⇒ · ⇒ · ∴ θ θ θ
Thus, the relative error in c is minimum when
0
45 · θ .
4. If
2
2
1
mv T ·
is the kinetic energy, find approximately the change in T as m
changes From 49 to 49.5 and V changes from 1600 to 1590.
By data,
5 . 0 5 . 49 , 49 · ∂ ⇒ · ∂ + · m m m m
10 1590 , 1600 − · ∂ ⇒ · ∂ + · v v v v
As
2
2
1
mv T ·
,
( )
2
2
1
mv T δ δ ·
( ) ( ) [ ] m v v v m δ δ
2
2
2
1
+ ·
i.e
( )( )( )( ) ( )( ) [ ]
2
1600 5 . 0 10 1600 2 49
2
1
+ − · T δ
=-1,44,000
5. If the sides of a triangle ABC vary in such a way that its circum radius remains
constant, Prove that
0
cos cos cos
· + +
c
c
B
b
A
a δ δ δ
If the triangle ABC is inscribed in a circle of radius R and if a,b,c respectively
denotes the sides opposite to the angles A,B,C we have the sine rule given by
R
c
c
B
b
A
a
2
sin sin sin
· + ·
( ) ( ) A R A R a A R a sin 2 sin 2 sin 2 δ δ δ · · ⇒ · ⇒
; cos 2 A A R a δ δ · ∴
( ) ( ) B R B R b B R b sin 2 sin 2 sin 2 δ δ δ · · ⇒ · ⇒
; cos 2 B B R b δ δ · ∴
191
( ) ( ) C R C R c C R c sin 2 sin 2 sin 2 δ δ δ · · ⇒ · ⇒
C C R C δ δ cos 2 · ∴
From the above, we write
C R
C
c
B R
B
b
A R
A
a
δ
δ
δ
δ
δ
δ
2
cos
& 2
cos
; 2
cos
· · ·
C R B R A R
C
c
B
b
A
a
δ δ δ
δ δ δ
2 2 2
cos cos cos
+ + · + +
( ) C B A R δ δ δ + + ·2
( ) t Cons R tan 2 δ ·
t Cons radians C B A tan · · + + π
( ) 0 2 δ R ·
=0
3.3.8:-Problem set No: 3.3.2
1. Find the percentage error in the evaluation of the area of an ellipse if 1% error is
made while measuring the major and minor axis
.
2. The time T of a Complete oscillation of a simple pendulum is given by the
formula
( ) π 2
g
l
T ·
, where g is constant .If there is an error of 3% in the value of
l, Find the percentage value of T.
3. The radius of a Sphere is found to be 10cms with possible error of 0.02cm. What
is the relative error in computing the volume?
( Hint: Volume of the Sphere=
3
3
4
r π
)
4. The pressure p and the volume v of a gas are concentrated by the relation
pv
1.4
=constant. find the percentage increase in pressure corresponding to a
dimension of
2
1
% in volume
192
193
General solution and particular solution A solution of a D.E. is a relation between the dependent and independent variables satisfying the given equation identically. The general solution will involve arbitrary constants equal to the order of the D.E. If the arbitrary constants present in the solution are evaluated by using a set of given conditions then the solution so obtained is called a particular solution. In many physical problems these conditions can be formulated from the problem itself. Note : Basic integration and integration methods are essential prerequisites for this chapter. Solution of differential equations of first order and first degree Recollecting the definition of the order and the degree of a D.E., a first order and first degree equation will be the form dy = f x, y or M x, y dx + N x, y dy = 0 dx We discuss mainly classified four types of differential equations of first order and first degree. They as are as follows : • • • • Variables separable equations Homogenous equations Exact equations Linear equations
bg bg bg
Variables separable Equations If the given D.E. can be put in the form such that the coefficient of dx is a function of the variable x only and the coefficient of dy is a function of y only then the given equation is said to be in the separable form. The modified form of such an equation will be, P (x) dx + Q (y) dy = 0 Integrating we have P x dx Q yg z bg + z bdy = c
This is the general solution of the equation.
2
Example –1 Solve : e x y − 1 dx + 2 e x + 4dy= 0
x x
b g >> e b 1g + 2 + 4 y − dx e dy= 0
Variables are separated
Dividing throughout by (y-1) (ex + 4) we get, ex dy dx + 2 =0 x − 1 y e +4 ⇒
z
ex dy dx + 2 =c x y −1 e +4
z
i.e. log x + 4 2 log − 1 c e + y =
2 or log x + 4 log − 1= c e + y
2 i.e. log x + 4 y − 1 = log k say e
bg
2 or x + 4 y − 1 = k is the required solution e
Example 2 Solve
2 dy = x e y − x given that y 0 = 0 dx 2 2 dy dy >> = x e y − x or = x e y − x e dx dx 2 dy i. e. y = x e − x dx by separating the variables.* e
bg
⇒ e − y dy − x e − x dx = c
2
z
z
i. e. − e − y dy − x e − x dx = c
2
z
Put − x 2 = t ∴− 2 x dx = dt or − x dx = dt 2 Hence we have, − e − y + e t dt = c 2 i e
2
z
− e− y +
et =c 2
or
e−x − e − y = c is the general solution. 2
Now we consider y (0) = 0 That is y = 0 when x = 0, * For animation
3
4 . i. e.The general solution becomes 1 1 − 1 = c or c = − 2 2 Now the general solution becomes e− x 1 − e− y = − 2 2 2 or e − x + 1 = 2e − y is the particular solution.. or ⇒ or i. 2 e j Example –3 Solve : x y >> x y i. e. dy = 1 + x + y + xy dx or y − x − log x+ y= c is the required solution. i..* 1+ y x y 1+ x dy − dx = c 1+ y x + y 1 1 − 1 dy − dx − 1 dx = c 1+ y x 1 1 dy − dy − log x − x = c 1+ y y − log y log x − x = c 1+ − z z z z z z z >> Rearranging the given equation we have. 1 Example – 4 : Solve : y − x dy dy = y2 + dx dx b g dy = 1 + x + y + xy dx dy xy = + x y+ x 1 + 1 dx dy xy = + x 1 + y 1 dx ydy 1 + x = dx by separating the variables. e... e.
* x + 1 y − y2 dx dy ⇒ = =c x + 1 y− y 1 dy i. log x + 1 − =c 1 y− y 1 z z b gz b g We have to employ the method of partial fractions for the second term of the above.y − y 2 = dy + 1 x dx dx dy or = by separating the variables. x Example – 5 : Solve : tan y dy = cos( x + y ) + cos ( x − y ) dx >> The given equation on expanding terms in the R. log x + 1 + log ∴ + 1) (1 − y )= ky. is the required solution.H.S. Let 1 A B + + y(1 − y ) y 1 − y ⇒ 1= A 1 − y + B y Put y = 0 1 = A Put y =1 = B 1 dy 1 1 ∴ = dy + dy y 1− y y 1− y zb g z z dy 1 −1 or − z = − z dy + z dy yb y g y 1− 1− y i − e dy 1 = 1 = zb− yg −log y + log b− yglog F−y y I G J y1 H K Using this result in (1) we get. becomes dy tan y = x cos y − sin x sin y x cos y + sin x sin y cos + cos dx dy i tan y e = 2 cos x cos y dx 1 b g F−y y I = c G J H K Lx + 1) (1 − y) Olog k ( or log M = N y P say Q 5 . e.
G2 J H K Example – 7 Solve : >> dy = x tan − x 1 y + dx . e... z z Session – 2 Equations reducible to the variables separable form Some differential equations can be reduced to the variables separable from by taking a suitable substitution. e. We present a few examples. z cosdt2− x = c 2 t 2 or 1 t 2 dt z sec − x = c 2 2 i.tan y dy= 2 cos x dx by separating the variables. sec y dy − 2 cos x dx = c or ∴ sec y − 2 sin x = c is the required solution. tan x F + y + 1I − x = c is the required solution.* cos y ⇒ tan y. tan (t / 2) − x = c i. Example – 6 Solve : cos (x + y + 1) dx – dy = 0 dy = cos ( x + y +1) dx >> Put We have … (1) dt dy dt dy =1+ or 1= − dx dx dx dx dt dt Now (1) becomes − 1 = cos t or = 1 + cos t dx dx dt dt i. = dx ⇒ − dx =c 1 + cos t 1 + cos t t = x + y +1 ∴ z z i. e.. e. (1) dy = x tan ( y − x ) + 1 dx dt dy dt dy Put t = y − x ∴ = − 1 or = +1 dx dx dx dx 6 .
e.e.* tan t ∴ cot t dt − x dx =c z z i.Now (1) becomes Hence dt dt = x tan t + 1 or +1 x tan t.e becomes G − 1J = 2 H K t +1 dx dt dy 1 dt dy =1 + 2 or −1 = dx dx 2 dx dx dt − 1 t =2 +1 dx + 1 t dt 3t − 1 i. + 2 y − 1 = + 2 y + 1 x dx x dy dy + 2 y 1 x − or = 1 dx + 2 y 1 x + ∴ F I G J H K 1F dt I t − 1 The d.. 2 Example – 8 Solve : + 2 ydx − dy + dy x = dx >> Rearranging the terms in the given equation we have. dt − x = c 3t − 1 … (2) Put t = x + 2 y ∴ or z z z Let t + 1 = l(3t −1) + m or t + 1 = ( 3l ) t + ( −l+m) ⇒ 3 l =1 and − l+ m = 1 Hence l = 1 and − 1 + m = 1 or m = 4 3 3 3 ∴t + 1 = 1 . = dx t + 1 t +1 t +1 dt =dx ⇒ dt − dx = c 3t − 1 3t − 1 t +1 i.. log (sin t ) − x2 =c 2 x2 Thus log sin ( y − x ) − is the required solution. e. = dx dx dt = x dx by separating the variables. t − 1 4 3 3 + 3 Thus (2) can be written as 7 .
i. e. e.. ⇒ 2 +3 dx ztt + 4 dt − z = c 3 Let 2t + 3 = l (3t + 4) + m or 2t + 3 = (3l) t + (4l + m) ⇒ 3 l = 2 and 4 l+m= 3 ∴ l= 2 3 and 8 / 3 + m= 3 or m = 1 3 Thus 2 t + 3 = 2/3 . (3t+ 4) + 1/3 Hence (1) becomes i.1 3 3 z33t t−−11 + 4 dt − x = c 1 4 dt i. dt t +1 dt t +1 −1= or =1 + dx 2t + 3 dx 2t + 3 i. e. zdt + z 1 − x=c 3 3 3t − 1 i. e. dt 2t + 3 + t + 1 dt 3t + 4 = or = dx 2t + 3 dx 2t + 3 2t + 3 dt = dx 3t + 4 i. 2 dx z / 3. y + 3 = 3 9 dy + y 1 x + = dx 2 x + 2 y + 3 dy + y 1 x + = dx 2( x + y 3 + dt dy dt dy =1 + or −1 = dx dx dx dx Example – 9 : Solve : >> We have Put t = x + y ∴ The given equation becomes.(33tt ++4) + 1 / 3 dt − z = c 4 2 1 dt 1 dt + − x=c 3 3 3t + 4 8 z z . e.. e. e. i. t 4 + log (3t − 1) − x = c 3 9 x + 2y 4 + log (3 x + 6 y − 1) − x = c 3 9 2 4 − x log x + 6 y − 1 c is the required solution. i. e.
e. where t = xy. where t = x + y 3 − 3 9 2 + y 1 x + log x + 3 y + 4 x = c 3 − 3 9 1 1 (2 y − x ) + log (3x + 3 y + 4) = c is the required solution. G t+ J− = c. can be written as. 4 x 9 . 2 H 2 Kx i. e. x 2 i.e. 1 1 2xy + sin (2 xy ) − = c... e. ∴ 2t 1 + log t + 4 x = c. e. 3 9 i. e. Example – 10 Solve : x 3 dy + x 2 y + sec 2 0 xy = dx The given d.. e.. z 1 z+ cos 2t dt − 1 = c 2 x 1 F sin 2t I 1 i. is the required solution.i. x2 x F dy + yI + sec 0 Gdx J H K xy = 2 1 Put t = xy ∴ dt dy =x +y dx dx dt + sec 2 t = 0 dx Now (1) becomes. x 2 dt dt dx = − sec 2 t or =− 2 2 dx sec t x dx z =c x 2 ∴ cos2 t dt + i..
EXERCISES : Solve the following differential equations 1xy 2 + x dx + 2 y + y = 0 x dy 2 2 2 2 2 2x − yx + + x y = 0 dy y dx dy dy 3 − x y = a y2 + dx dx 4 + y − dy dx + dy x dx = dy 5 4 x + x 3 y+ cosec ( xy) = 0 dx F G H I J K ANSWERS : 2 2 1 + x + y = c 1 1 2 log y + 1 y + 1 − x = c x 3 − ay + x cy 1 a = 4 − x + log x + y = c y 5 ( xy) + 1 / 2 x 2 = c cos Session –3 A function u = f (x. y = dx g x. can be solved by separating the variables. Solution of a homogeneous equation We prefer to have the differential equation in the form after recognizing that the D. dy f x. We take the substitution y = v x so that. y bg bg 10 . y) is said to be homogeneous function of degree n if u = x n g y / x or u = y n g x / y b g b g A D. of the form M (x. dy dv = v. dx dx With these the given d. y) dy = 0 is said to be a homogeneous differential equation if both M (x.E. y) and N (x.e.E. is a homogeneous one. y) are homogeneous functions of the same degree.1 + x by product rule. y) dx+ N (x.
e. e. v + x = dx x 3 + v 3 x 3 i. x = or x = dx dx 1 + v 3 1 + v3 ∴ 1 + v3 − dx dv = separating the variables. is the required solution. e. + log v + log x = c −3 1 y i. e.. where v = x 3v 3 x Thus − 3 + log y = c..* by 4 x v 1 dx z1 dv +z dv + zx = c v v 4 (* for animation) ⇒ v −3 i. vx Now (1) becomes. v + x dv x3 v dv v = 3 or x = −v 3 dx x + v dx 1 + v 3 1 dv v − v − v 4 dv − v 4 i. 3y Example –2 Solve : x dy − y dx = >> We have x dy = y + x 2 + y 2 dx or dy y + x 2 + y 2 = dx x L M N x 2 + y 2 dx O P Q 1 11 ..Example –1 Solve : x 2 y dx − 3 + y 3 = 0 x dy >> (Observe that the coefficient of dx and dy are homogeneous functions of degree 3) The given equation can be written as. − 3 + log (vx ) = c. dy x2 y = 3 dx x + y 3 Put y = v x ∴ 1 dy dv =v+ x dx dx dv x 2 ..
. y − Example –3 dy y 2 + =y dx x dy y2 >> We have x y − = dx x 2 dy xy − y or = dx x2 dy dv Put y = v x ∴ = v + x dx dx dv x . v + x = or x = −v 2 2 dx dx x Solve : x ∴ 1 dv dx =− by separating the variables. − + log x = c. is the required solution..* 2 x v dv dx 1 y Hence 2 + = c i. sin h −1 v − log x = c. v + x = or x = 1 + v2 dx x dx dv dx ∴ = by separating the variables.v x − v 2 x 2 Now (1) becomes v + x = dx x2 dv x 2 − v 2 v dv i. y z z 12 . e.. e. e.. is the required solution. where v = y / x Thus sin h −1 / x log x = c.e. where v = x v x v x Thus − + log x = c. * x 1 + v2 F H I K ⇒ dx z 1dvv − zx = c + 2 i.Put y = vx ∴ dy dv = v+ x dx dx dv vx + x 2 + v 2 x 2 Now (1) becomes v + x = dx x 2 dv x v + 1 + v dv i.
x dv x sec 2 v − tan v v = dx x sec 2 v 1 dv v sec 2 v − tan v = −v dx sec 2 v dv v sec 2 v − tan v − v sec 2 v = dx sec 2 v dv − tan v sec 2 v dx = or dv = − 2 dx sec v tan v x sec 2 v dx Hence dv+ =c tan v x i.Example – 4 Solve : x tan ( y / x ) − y sec 2 / x + x sec 2 / x = 0 y dx y dy >> The given equation can be written as.. v + x = dx x sec 2 v Put i.. e. dy y sec 2 / x x tan ( y / x ) y − = 2 dx x sec / x y y dy dv = v or v = v x ∴ = v + x x dx dx 2 dv v x sec v − x tan v Now (1) becomes. v + x i. e. Example – 5 dy = y y − log x + 1 log dx >> The given equation can be written in the form dy y = log ( y / x ) + 1 1 dx x Solve : x z z 13 . e. ⇒ x tan v = k where v = y / x Thus x tan ( y / x ) = k.. is the required solution.. x i. e. log (tan v ) + log x = c or log (tan v. x ) = c = log k (say).. e. x i.
we need to express the equation relating to dx / dy and the terms are homogeneous functions of degree 0... x = v log v or = dx v log v x 1/ v dx Hence dv − =c log v x Put y = v x ∴ z z i. e. y dv e v − 1 v = dy + e v 1 dv e v − 1 v dv e v v − e v − v − e v v = − v i. y = dy + e v dy 1 + e v 1 dv v + v e (1 + e v dv dy i. Example – 6 Solve : 1 + e x/y dx + e x/y 1 − e j F x I dy = 0 G yJ H K F − 1I dy x G J H K y >> (As we observe terms with x / y .. y =− or =− v v dy y + e 1 e +v Hence (1 dy z e+ e+ )vdv + zy = c v v 14 . v + y i. e.. e. log v log x = c = log k log − say i. is the required solution. We have 1 + e x/y dx = e x/y x/y or Put e j F I x e G− 1J H K y dx = dy 1 e+ e j x /y 1 dx dv = v+y dy dy x / y = v or x = v y ∴ Now (1) becomes.. e. e..dy dv = v+ x dx dx dv Now (1) becomes v + x = v v + 1 log dx dv dv dx i. log v log log = kx ⇒ log v = kx where v = y / x Thus l g ( y / x ) = k x.. log v log k + log x log = i. e. e.
e. is the required solution. Finally we substitute for X and Y where X = x −h. let us choose h and k such that : Solving these equations we get the value for h and k.i. Put x = X + h.. e y Thus y e x/y + x = k. Now dy dy dY dX = dx dY dX dx dY dy dY = 1 . b gb g b g b g aX + ah b + bY gb + bk + cg dY = dX b X + b' Y gbh + b' k + c' g a' + a' ah + bk + c = 0 and a' h + b' k + c' = 0 2 Now. Y = y −k. e. Session – 4 Equations reducible to the homogeneous form Consider the differential equation in the form : + by + c ± x + b' y + c' = 0 ax dx a' dy We first express the equation in respect of dy / dx and the procedure is narrated by taking dy a x + b y + c a b = where ≠ dx a' x +b' y + c' a' b' 1 This condition implies that there are no common factors for the x and y terms in the numerator as well as in the denominator.1 Hence = dX dx dX As a consequence of these (1) becomes a X +h + b Y +k +c dY = dX a' X + h + b' Y + k + c' i. where v = x / y. Thus (2) now assumes the form dY aX + bY = dX a' X + b' Y 3 It is evident that (3) is a homogeneous equation in the variables X and Y. 15 .. This equation can be solved by putting Y = VX as discussed already. y = Y + k where h and k are constants to be chosen appropriately later. log (e v + v ) + log y = c or log v + v = log k (say) e y ⇒ v + v = k.
dy dy dY dX Now = dx dY dX dx dY dy dY = 1 Hence 1 = dX dx dX Thus (1) becomes dY − + h 4 + k 9 X + Y + = dX 4 + h + k 2 X + Y − dY X + 4Y h + 4 k + 9 − + − i.Example –1 Solve : ( x − 4 y − 9) dx + x + y − 2 = 0 4 dy >> We have x + y − 2 = − − 4 y − 9 4 dy x dx dy − x + 4 y + 9 ∴ = dx 4 x + y − 2 Put x = X + h and y = Y + k where h and k are constants to be choosen suitably later. X = dX 4+V dX + V 4 + V 4 dV dX ∴ =− by separating the variables. e.. V + X = or X = −V dX X+ V 4 dX 4+V dV −1 + 4V − 4V − V 2 dV −+ V 2 1 i. e.* 2 X + V 1 16 . V + X = dX 4 X + VX dV X 1 + 4V − dV −1 + 4V i. X = i.. e. e.. h = 1 and k = −2 Thus (2) becomes 2 dY − X + 4Y = 3 dX 4 X + Y dY dV Put Y = VX ∴ =V + X dX dx dV − X + 4VX Now (3) becomes. = dX X + Y h + k − 2 4 + 4 Let us choose h and k such that − h + 4 k + 9 = 0 and 4 h + k − 2 = 0 Solving these equations we get..
e. e. h+ 2 k − 3 = 0 and 2 h + k − 3 = 0 Solving these equations we get h = 1 and k = 1 dY X + 2Y Thus (2) becomes = dX 2 X + Y 3 17 . = dX X + Y h + k − 3 2 + 2 ……(2) Let us choose h and k such that.⇒4 dV dX z+ V + zV+dV + zX = c 1 1 V 2 2 1 i. 8 tan −1 V + log (1 + V 2 2 log X = 2c + i. where h and k are constants to be chosen suitably later. where V = Y / X X ∴ 8 tan −1 / X log ( X 2 + Y 2 2c = k Y + = say But X = x − h = x − 1 and Y = y − k = y + 2 8 tan −1 y F+ 2 I+ log − 1+ + 2 = k y G− 1J H K x x 2 2 This is the required solution. = dX 2 + h + k 3 X + Y − dY + 2Y + 2 k − 3 X + h i. 4 tan −1 V + log (1 + V 2 log X = c + 2 i. 8 tan −1 V + log (1 + V 2 2 = 2c. ∴ dy dy dY dX = dx dY dX dx dY dy dY = 1 Hence 1 = dX dx dX dY + h 2 + k 3 X + Y − Thus(1) becomes.. Example – 2 Solve : dy x + 2 y − 3 = dx 2 x + y − 3 dy x + 2 y − 3 = dx 2 x + y − 3 1 >> We have Put x = X + h and y = Y + k. e....e.
.. log G J − log (1 − V = 2c X H− V K 1 L (1 + V Olog k or log M = P say (1 − V (1 − V Q X N 2 2 2 2 2 2 2 2 z ⇒ (1 + V (X +Y) Y = k or = k since V = 3 2 3 X (1 − V X X −Y b g ut X = x − h = x − 1 and Y = y − k = y − 1 3 Thus + y − 2 k − y is the required solution. e. 2 log − log (1 − V 2 log X = c − 2 1− V 2 z z F I G J H K 1 F+ V I i.* 2 X 1− V dV V dV dX ⇒2 + − =c 2 2 X 1− V 1− V 1 1+ V 1 i. x = x Exercises : Solve the following equations: 1x 3 + y 3 = xy + y dx x dy 2x + y log x − y log y − x x − log y = 0 dx log dy 3 2 ( x / y) y dx − x dy= y dy Sin 4 − x − 4 = + x − 2 y dx y dy 53 x − y − 1 + 2 − y = 0 dy x dx 18 .. V + X = dX X + V 2 Put Y = VX ∴ dV 1 + 2V dV 1 − V 2 or X = − V or X = dX 2 + V dX 2 + V 2+V dX ∴ dv = by separating the variables. e. e. 2 log G J − log (1 − V 2 log X = 2c − H− V K 1 1 F+ V I i.. e.dY dV =V+X dX dX dV X + 2VX Now (3) becomes V + X = dX 2 X + VX dV X + 2V 1 i.
y) dx + N (x.5 We say that M (x. in the first term we integrate M (x. y) dy = 0 to be an exact equation is M N = y x Further the solution of the exact equation is given by Mdx N y dy z + z bg = c where..E. y) w. y) dy = 0 is an exact differential equation if there exists a function f (x. y) such that df = M (x. y) dy. y) dx+N (x. y) dx+N (x. The necessary and the sufficient condition for the D. is the required solution. the y x The solution is M dx + N y dy = c b g >> Let M = y b 1 gcos y and N = x + log x − x sin y 1+ x + ∴ z z bg LF+ 1 I dx i. e.Answers : 1 / x + log − x c y y = 2 x + / x log ( y / x ) − 1 c log y = x 1 3. Example –1 Solve : [ y 1 + 1 + cos y + + log x − x sin y = 0 x dx x dy M 1 N 1 = 1 + − sin y and = 1 + − sin y y x x x M N Since = given equation is exact. M (x. x+ + 19 .r. − sin (2 x y) − log y = c 2y 4 2 4 ( x + 1)2 + − 3 + 2 tan −1 − 3 + 1= c log y y x 4 5 x + y − 1 = c x − 2 y − 1 4 Session .t x keeping y fixed and N (y) indicate the terms in N not containing x. zyG J + cos y P + z dy= c M xK O 0 H N1 Q Thus y b log x g x cos y = c.
.. the y x The solution is M dx + N y dy = c i. is the required solution. z bg dx 0 b zy cos x +sin y+ yg = z dy= c z e j e Thus y sin x + x sin y + xy = c. e. e. y dx sin dy bcos x +sin y+ yg + b x + x cos y + xg = 0 Let M = y cos x + sin y+ y and N = sin x + x cos y + x ∴ M N = cos x + cos y+ 1 and = cos x + cos y+ 1 y x M N Since = given equation is exact.. e 2 2 = 2 x y e xy 2 + 2 ye xy y y x 2 2 2 2 N M = 2 xy 3 e xy + 2 ye xy and = 2 xy 3 e xy + 2 ye xy y x N M Since = given equation is exact. e.F. is the required solution. i. z − zy e + 4 x jdx + z3y dy = c e 2 xy 2 2 z 3 2 e xy i.Example –2 Solve : dy y cos x + sin y+ y + =0 dx sin x + x cos y + x >> The given equation is put in the form. Example – 3 Solve : y 2 e xy + 4 x 3 dx + 2 xye xy − 3 y 2 dy = 0 >> Let M = y 2 e xy + 4 x 3 and N = 2 x y e xy − 3 y 2 2 2 2 2 N M ∴ = y 2 xy xy +e xy y . the y x The solution is M dx + N y dy= c 2 2 2 2 j i.) 2 20 . e. y 2 + x 4 − y 3 = c y 2 Thus e xy + x 4 − y 3 = c.. Equations reducible to the exact form Sometimes the given differential equation which is not an exact equation can be transformed into an exact equation by multiplying with some function (factor) known as the integrating factor (I.
Suppose that..The procedure to find such a factor is as follows. for the equation M dx + N dy = 0 M N ≠ . 4 21 . then we take their difference. y x If 1 M N 1 M N − = f x or − =g y N y x M y x f x dx − g y dy then e z bg or e F G H I bg J K zbg F G H I b J g K is an integrating factor. (The equation is not exact. 2 F G H I J K bg z dx 2 f x dx log e z bg = e x = e z log x = e z x = x 2 Multiplying the given equation by x 2 we now have.e. M = 4 x 3 y + 3 x 2 y 2 − x 3 and N = x 4 + 2 x 3 y Solution is given by M dx + N = c y dy Thus ( z4 x y + 3x y 3 4 3 2 z 2 2 z − x ) dx + z dy = c 0 3 x4 i. Example – 4 Solve : (4 xy + 3 y 2 − x ) dx + x x + 2 y dy = 0 Let M = 4 xy + 3 y 2 − x and N = x + 2 y x 2 + 2 xy x = M N = 4 x + 6 y and = 2 x + 2 y. is the required solution. x y x Now 1 M N 2 + 2 y 2 x − = = = f x N y x x + 2 y x x f x dx 2 b g Hence e z bg is an integrating factor.) y x M N Consider − = 2 x + 4 y = 2 + 2 y close to N. x y + x y − = c.
.. EXERCISES: Solve the following differential equations 1. z2 x y e z 3 − y4 z bg + y jdx + z dy = c 0 3 Thus x 2 y 3 − xy 4 + xy 3 = c. M = 2 xy − y 2 + y and N = 3 x 2 − 4 xy + 3 x M N = 2 x − 2 y + 1 = 6x − 4y + 3 y x M N − = −4 x + 2 y − 2 = −2 2 x − y + 1 near to M. b b g 2 1 FM I −2 b − y + 1g 2 N y Gy − J = y bxx− y +1g= − y = gbg MH xK 2 − g y dy Hence I. M = 2 x y 3 − y 4 + y 3 and N = 3 x 2 y 2 − 4 xy 3 + 3 xy 2 The solution is M dx + N y dy = c i. y x Now.Example – 5 Solve : y 2 x − y + 1 dx + x 3 x − 4 y + 3 dy = 0 g b g >> Let M = y b − y + 1g N = xb − 4 y + 3g 2x and 3x i. [4x3 y2+ y cos (xy)] dx+[2x4y+x cos (xy)] dy =0 3xy 2 + x − 2 y + 3dx + x 2 y dy = 2 + y . F = e z e 2 z dy y y = e 2 log y = e log = y 2 2 Multiplying the given equation with y 2 we now have. e. e. is the required solution. y = 1 x dy 1 4 x + y + 1 dx + x + 3 y + 2 = 0 y x dy 5x 2 + 2 y 3 + 6 y + 3 + xy 2 = 0 given that y (1) = 2 23 dx x dy b g ANSWERS: 22 . cos x (ey +1)dx + sin x eydy = 0 2.
S. F. part of the solution to finally arrive at the required solution. z is the solution of the linear equation (1). P P We equip with the I. The solution can simply be written by interchanging the role of x and y. The expression for P and Q is to be written by simple comparison. P P i. 2 where P and Q are functions of y is called a linear equation in x.1 xy + 1 c sin e = 2 4 y 2 + sin xy = c x x 2 y2 x 2 3 + − 2 xy + 3 x − y 2 = 1 2 2 2 2 x y 4 + xy 3 + xy 2 = c 2 5 6 + x 4 y 3 + 3 x 4 y = 15 x Session – 6 Linear Equations A differential equation of the form dy + Py = Q dx P P y e z dx = Q e z dx dx + c bg …(1) where P and Q are functions of x only is called a linear equation in ‘y’. e z dx or e z dy We assume the associated solution and we only need to tackle the R. An equation of the form : dx + Px = Q dy .H.. x e z dy = Q e z dy dy + c z is the solution for the linear equation (2) Working procedure for problems • • • • The given equation must be first put in the form conformal to the standard form of the linear equation in x or y. 23 . e...
4 Solve : 2 y' cos x + 4 y sin x = sin 2 x given that y 3 = 0 >> Dividing the given equation by 2cos x. (1) z x = z x sec x + c tan z bg 24 .. sin x dx + c i.. where P = cot x and Q = cos x dx P cot log ∴ e z dx = e z x dx = e z (sin x sin x = P P The solution is y e z dx = Q e z dx dx + c i.. y sec 2 x = sin x. e. we have dy + tan x = sin x sin 2 x = 2 cos x sin x 2 y dx dy This is of the form + Py = Q. y sin x = cos x. is the required solution. z sin 2 x y sin x = z dx + c 2 z Thus y sin x = Example – 2 − cos 2 x + c .Example – 1 Solve : >> dy + y cot x = cos x dx dy + y cot x = cos x is of the form dx dy + Py = Q. e. sec 2 x dx + c i.. y sec 2 Thus y sec 2 x = sec x dx + c. ... e. is the general solution. Consider y 3 = 0 That is y = 0 when x = 3 Hence (1) becomes 0 = 2 + c or c = −2 Thus y sec 2 x = sec x − 2. where P = 2 tan x and Q = sin x dx pdx 2 2 e z = e z tan x dx = e z log (sec x ) = sec 2 x P P The solution is y e z dx = Q e z dx dx + c bg i. e.. is the required particular solution..
tan + Equations reducible to linear form dy Form (i): f ' + P f Q.. f ' + P f x = Q.Example – 3 Solve : (1 + y 2 )dx + − tan −1 y = 0 x dy >> We have or dx tan −1 y − x = dy 1 + y2 dx x tan −1 y + = dy 1 + y 2 1 + y 2 dx This is of the form + Px = Q dy Here P = −1 1 tan −1 y P and Q = . e z dy = e tan y 1 + y2 1 + y2 P P The solution is given by x e z dy = Q e z dy dy + c i. = e tan −1 Thus x e tan −1 y −1 y − 1 c. where P and Q are function of x. x e tan −1 y = tan z+ yy e 1 2 −1 z tan −1 y dy + c 1 dy = dt 1 + y2 By putting tan −1 y = t. we obtain ∴ x e tan −1 y = t e t dt + c −1 z y y i. dx dx Similarly. where P and Q are function of y can be x dy reduced to the linear form by putting f t x = bg 25 .. y y = dx dy dt We put f t ∴ f ' = y = y dx dx dt The given equation becomes + Pt = Q which is a linear equation in t. e. e. is the required solution. by parts. x e tan = t e t − e t + c.
t = − n which is a linear equation in t. 1 P 1 Q dx dx Similarly + P x = Q x n . sec y . or dy dt 1 dy 1 dt = or n = dx dx 1 dx y dx − n 1 dt + P t=Q − n 1 dx dt + − n . sec x = z x dx + c cos 2 z i.e. sec x dx + c Thus sec y sec x = sin x + c. 26 ... Example . is the required solution. e.4 dy + tan x = cos y cos2 x dx >> Dividing the given equation by cos y we have dy sec y tan y + sec y tan x = cos2 x 1 dx Solve : tan y Now... where we have.. where P and Q are functions of y is called dy Bernoulli' s equation x.. 1 y n dx Put y1− n = t ∴ − n− n 1 y Hence (1) becomes. We first divide by x n and later put x 1− n = t to obtain a linear equation in t. put sec y = t ∴ sec y tan y Hence (1) becomes dy dt = dx dx dt + t tan x = cos2 x dx dt This equation is of the form + Pt = Q.dy + P y = Q yn where P and Q are functions of x. sec y sec x = cos x . dx P = tan x and Q = cos 2 x P tan log ∴ e z dx = e z x dx = e z (sec x = sec x P P The solution is t e z dx = Q e z dx dx + c z i. dx This equation is called as Bernoulli' s equation y. Form (ii) : We first divide the equation throughout by y n to obtain 1 dy + P y1− n = Q .
Example – 5 dy y + = y2 x dx x >> Dividing the given equation by y 2 we have. dx P ∴ e z dx = e = e − log x = P The solution is t e z dx 1 x P = Q e z dx dx + c i.. Thus 1 1 = − x . 1 dy 1 + =x 2 y dx y x Solve : 1 −1 dy dt =t ∴ 2 = y y dx dx − dt t Hence (1) becomes + =x dx x dt t or − = −x dx x Put This equation is a linear equation of the where P = −1 and Q = − x x − 1 z dx x 1 dt + Pt = Q. dx + c x x z z 1 = − x + c. is the required solution.e. dy 1 dx 1 − y = y3 2 x dy x 1 −1 dx dt Put =t ∴ 2 = x x dy dy 1 27 . t . xy Example – 6 dy Solve : xy + xy 2 = 1 1 dx dx >> Consider = x y + x 2 y3 dy dx or − xy = x 2 y 3 Dividing by x 2 we get.
x 2 F I G J H K EXERCISES Solve the following equations dy 1 + y cot x = 4 x cosec x . e. dy Hence (1) becomes − P P The solution is t e z dy = Q e z dy dy + c P y P = y and Q = − y 3 ∴ e z dy = e z dy = e y 2 2 i. t e y2 2 2 2 ey y2 2 Thus = 2 ey 1 − + c . t e y 2 2 2 = − y3 e z z y2 2 dy + c Put y 2 = u ∴ y dy = du Also y 2 dy = y 2 du or y 3 dy = 2u du y ∴ t ey 2 2 = −2 u e u du+c = −2 e u − e u c . u + 2 z i. e. is the required solution. on integration by parts.. y 0 2 = dx dy 2 2 x sin − 2 y = tan x dx 3 dy + y − x − xy tan x dx = 0 x dy y 4 + = x 2 y 6 dx x 5 y 4 − 2 xy + 3 x 2 dy = 0 dx Answers: 1y sin x = 2 x 2 − 2 2 2 cot x = log tan x + c y 3 y cos x = cos x + x sin x + c x 1 5 4 5 5 − 2 = c x y 2x 5 2 = y 3 + 2 x x b g 28 ..dt dt − t y = y 3 or + t y = − y3 dy dy dt This equation is of the form + P t = Q where.
t x and eliminate c. E.Session – 7 Orthogonal Trajectories Definition : If two family of curves are such that every member of one family intersects every member of the other family at right angles then they are said to be orthogonal trajectories of each other.T. 2 x − − y=0 dx dy dy * F I G J H K or 2 x dx + y dy = 0 ⇒ 2 x dx + y dy = c y2 2 i. c) = 0. we prefer to take logarithms first and then differentiate w. z z 29 . 2 x − y = 0 is the D.. y. of the given family.t θ .t x we have. x + = c or 2 x 2 + y 2 = 2c = k say 2 Thus 2 x 2 + y 2 = k is the required O.2y − y2 1 dy dx = 0 or 2 x y − y2 = 0 2 dx x dy i. dy dx * Replace by − and solve the equation. dr d * fter ensuring that the given parameter is elimiated we replace by − r 2 d dr and solve the equation. dy x . differentiate w..r. Working procedure for problems Case – i : (Cartesian family) Given f (x.r. e. y2 >> Consider = 4a x Now differentiating (1) w. Example –1 Find the Orthogonal trajectories of the family of parabolas y2 = 4 a x.r.e. dx dy dx dx Replacing by − we have. dx dy Case – ii : (Polar family) * Given an equation in r and θ .
Here b = 2c z z z Example-3 Show that the family of parabolas y2 = 4a (x+a) is a self orthogonal.. e. >> Consider y2 = 4a (x+a) …. 2 x 2 y y1 dy + 2 = 0 where y1 = 2 dx a b + x −y y i. r.(1) Differentiating w. = + a 2 log x + c 2 2 2 2 or x + y − 2 a 2 log x − b = 0 is the required orthogonal trajectories. x dx ⇒ y dy = − x dx + a 2 +c x y2 − x 2 i. 30 . dividing (2) by (3) we get. x yy x y = 21 or 2 = 1 2 2 2 y x −a y x −a Now. 2 = 2 1 a b + x2 − y2 Also from (1) 2 − 1 = 2 a b + 2 2 2 x −a −y or = 2 2 a b + Now. let us replace y1 = ∴ x 1 dx = − 2 y dy x −a 2 1 2 3 F I G J H K dy − dx by dx dy −2 − a 2 x or y dy = by separating the variables. a2 b + x2 y2 >> We have 2 + 2 =1 a b + Differentiating w.Example – 2 Find the orthogonal trajectories of the family of curves x2 y2 + 2 = 1 where is the parameter. we have dy yy dy 2y = 4 a ∴ a = 1 where y1 = dx 2 dx Substituting this value of ‘a’ in (1) we have..t x. t x we have. e.r.
of the orthogonal family which is same as (2) being the D.. e. of the given family. cos r dr 1 + sin ⇒ + d c = r cos i. e. 1 31 .T. 3 (3) the D...y 2 = 2 yy1 x + F yy I or y = 2 xy + yy G 2J H K 1 1 2 2 1 2 Thus we have. d dr 1 d cos −r 2 = r dr 1 + sin d cos i.r. Now replacing y1 by −1 / y1 (2) becomes F1I F1I − − y = 2 x G J + yG J or y = y y HK HK 1 1 2 yy1 −2 x y + 2 y1 y1 i.. z r+ sin 1 r+ sin 1 = b or =b cos2 − sin 2 1 Thus r = b− sin is the required O.E. Example –4 Find the O. log r + sec tan c d + d = i. − r = dr 1 + sin 1 + sin − dr or d = by separating the variables.. log r + log (sec tan + log (sec = c + ) F I G J H K z z z i. e.E. + 2 x y1 = y . e. e. of the family r =a(1+sin θ ) >> We have r = a (1 + sin θ ) ⇒ log r = log a + log (1 + sin Differentiating w.t θ we have. e. y = 2 xy1 + yy1 2 This is the D.. 1 dr cos =0+ r d 1 + sin dr d Replacing by − r 2 we get. log r b tan sec = log b (say) sec + g F 1 sin I 1 = b ⇒ rG + J H cos cos K cos i. of the given family.E.. Thus the family of parabolas y2 = 4a(x+a) is self orthogonal.....T..
T.t θ we have..e. y is 2 4. 3.Example – 5 Find the O. log( r sin n = log b (say) ) ∴ r n sin n b. k 4. T. r = a sec 2 5r =a + cos 1 ANSWERS 1x 2 + 2 y 2 = c 2. of the family rn cos nθ = an >> We have rn cos n θ = an ⇒ n log r + log (cos n = n log a ) Differentiating w. r = b cosec 2 2 5r = b − cos 1 32 .. e. log r + log (sin n = c ) n or n log r + log (sin n = nc ) n i. d dr F G H I J K 1 d −r 2 = tan n r dr ∴ F I G J H K z z or − r d = tan n dr d dr = by separating the variables. e. = EXERCISES Find the orthogonal trajectories of the following family of curves 1y = ax 2 2x 2 + y 2 + 2 + c = 0 y being the parameter. = tan n r d cos n r d dr d Replacing by − r 2 we have. x 2 + y 2 + 2 kx + c = 0 is arbitrary. n dr − n sin n 1 dr + = 0 i. Show that the family of parabolas x 2 = 4 a + a a self orthogonal.r. tan n − r dr ⇒ + cot n c d = r 1 i. is the required O.
In this chapter we first discuss reduction formulae and later discuss the method of tracing cartesian and polar curves. Reduction formula for sin n x dx and Let In = sin x dx n −1 z π /2 0 zsin x dx dx where n is a positive integer n z =z sin n x . 33 . We discuss certain standard reduction formulae in the form of indefinite integrals and the evaluation of these with standard limits of integration. In all these applications reduction formulae plays a vital role in the evaluation of definite integrals. length or perimeter.LESSON – 2 Session . dx (where m and n are non negative f x g x z n z m z n z m integers) to lower degree.1 Introduction : Part – C Integral Calculus We are familiar with various methods of integration. Reduction Formulae Reduction formulae is basically a recurrence relation which reduces integral of functions n of higher degree in the form of f x dx. surface area of plane curves and volume of solids. sin x dx = u v dx say z u vdx z z v dx = uz − zv dx. dx (where m and n are non negative f x g x Reduction formulae is basically a recurrence relation which reduces integral of functions n of higher degree in the form of f x dx. definite integrals and the associated application of finding the area under a curve. The successive application of the recurrence relation finally end up with a function of degree 0 or 1 so that we can easily complete the integration process. By knowing the shape of a given curve we disucss application of definite integrals such as area.u' dx ∴ I = sin xcos x z cos x − 1 − − − n sin x cos x dx = − sin x x + − 1sin x cos n cos x dx z = − sin x x + − 1sin x − sin x ) dx cos n z 1 = −sin x x + − 1sin x dx − − 1 cos n n sin z z x dx n n −1 n−2 n −1 n−2 2 n −1 n−2 2 n −1 n−2 n We have the rule of integration by parts.
of (1). In = − sin n −1 x cos x + n − 1In − 2 − − 1n n I i. e. cos x + − 1n − 2 n n I ∴ In = sin n x dx = z −sin n −1 x . cos x n − 1 + In − 2 n n Illustration 1 This is the required reduction formula. I4 = + + I0 4 4 2 2 I0 = sin 0 x dx = 1 dx = x Thus I4 4 3 R S T U V W We cannot find I0 from (1). 34 .S.i. In 1 + − 1= − sin n −1 x . 2 We use this recurrence relation to find In − 2 by simply replacing n by (n − 2). cos x n O+ n −1 I P n Q 2 0 n−2 But cos (π / 2) = 0 = sin 0 Thus In = n −1 In − 2 n . But basically we have z z −sin x cos x 3 3x = z x dx = sin − sin x cos x + +c 4 8 8 2 0 Corollary : To evaluate >> Let In = zsin n n x dx 2 0 zsin x dx Equation (1) must be established first.H..e. ∴ I4 = sin 4 x dx = z z −sin 3 x cos x 3 + I2 4 4 We need to apply the result (1) again by taking n = 2 −sin 3 x cos x 3 −sin x cos x 1 i. To find sin 4 x dx i >> Comparing with L. we need to take n = 4 and use the established result.e. ∴ from (1) In L sin = −M N n −1 x .
n−3 In − 4 n−2 n −1 n − 3 Hence In = In − 4 n n−2 n−5 Similarly from (2) In − 4 = In −6 n−4 n −1 n − 3 n − 5 Hence In = In −6 again by back substitution. The result is as follows. n n−2 n−4 Continuing like this the reduction process will end up as below. n −1 n − 3 n − 5 2 In = I1 if n is odd n n−2 n−4 3 n −1 n − 3 n − 5 1 = I0 if n is even n n−2 n−4 2 i. sin x + n n 1 I n f x dx f a dx zsin x dx = zcos x dx by the property z =z − x n n−2 2 0 n 2 0 n a 0 a 0 n −1 z 35 . e.. In − 2 = But I1 = and I0 0 = zsin x dx = − cos x = − − 1 1 = z sin x dx = zdx = = x 2 2 0 2 0 2 0 0 2 0 2 0 Thus we have. − cos z x dx = cos x . 2 0 n n n 2 R− 1 − 3 n − 5 if n is odd | n n − 2 n − 4 3 zsin x dx = S − 1 n − 3 n − 5 1 |n n − 2 n − 4 if n is even 2 2 T n Note : The reduction formula for cos n x dx can be established in a similar way.
= − − I1 4 2 1 1 4 2 5 n z and the result is as follows. put tan x = t ∴ sec 2 x dx = dt Now In = t n − 2 dt −In − 2 = tan n −1 x Thus In = − In − 2 n −1 Next. 36 . z =z tan =z tan =z tan z x . where n is a positive integer.Session – 2 Reduction formula for tan n x dx and Let In = tan n x dx n−2 n−2 z 0 ztan n x dx. let In = ∴ In 4 0 n But tan ( / 4) = 1 and tan 0 = 0 Thus In = 4 0 ztan x dx L x O − I by using (1) tan =M Nn − 1 P Q n −1 4 n−2 0 ztan n x dx = 1 − In − 2 n −1 2 or In + In − 2 = 1 or nn +1 + In −1 1 I = n −1 Illustration (i) To find tan 5 x dx tan 4 x >> I5 = tan x dx = − I3 using (1). by 4 5 z R U S V T W But I = z x dx = z x dx = log (sec x ) tan tan tan x tan x ∴z x dx = tan − + log (sec x ) + c 4 2 Note : The reduction formula for z x dx can be established in a similar way cot tan 4 x tan 2 x i.. tan 2 x dx = tan n − 2 x sec 2 x − 1dx x sec 2 n−2 n−2 z x dx − z tan x dx x sec 2 x dx − In − 2 t n −1 − In − 2 n −1 1 For the first term. e.
sec 0 = 1..In = cot n x dx = z − cot n −1 x − In − 2 n −1 3.. e. tan 0 = 0 / tan / Thus z n− 2 2 n−2 sec x dx = + In − 2 n −1 n −1 n F I G J H K 2 Illustrations (i) To find sec 5 x dx >> I5 = sec 5 x dx = sec 3 x tan x 3 + I3 using (1). In = sec n −2 x x − tan x . − 2 n −3 x sec x tan x dx tan n sec = sec = sec n−2 n−2 = sec n − 2 z x tan x − − 2sec n z x tan x − − 2sec n z x tan x − − 2sec n − 2 x tan 2 x dx n n−2 n z x 2 x − 1 sec dx x dx + − 2sec n − 2 x dx n z In = sec n −2 x tan x − − 2n + − 2n − 2 n I n I n−2 x tan x + − 2 i. = + + I1 4 4 2 2 But I1 = sec x dx = log (sec x + tan x ) z z z R S T U V W 37 . e. Reduction formula for sec n x dx where n is a positive integer >> Let In = sec n x dx n−2 z =z sec z x 2 dx sec Integrating by parts we have. I + − 2 = sec 1 n n I n sec n− 2 x tan x n−2 Thus In = sec n x dx = + In − 2 n −1 n −1 z F I G J H K n−2 Next let In = ∴ In 4 0 2 zsec n x dx 4 L n− x tan x O + F − 2 I I sec n =M P G− 1 J n N n −1 Q H K 0 4 / 0 n−2 …(1) But sec ( 4) = 2 ( 4) = 1. by 4 4 sec 3 x tan x 3 sec x tan x 1 i.
Let I m. − cosec n − 2 x cot x n−2 In = cosec x dx = + In − 2 n −1 n −1 z n F I G J H K Reduction formula for sin m x cos n x dx and where m and n are positive integers. = − + sin m − 2 x cos n + 2 x dx n +1 n +1 m −1 n +1 sin x cos x m − 1 =− + sin m − 2 x cos n x cos 2 x dx n +1 n +1 m −1 n +1 sin x cos x m − 1 =− + sin m − 2 x cos n x (1 − sin 2 x dx n +1 n +1 sin m −1 x cos n +1 x m − 1 m −1 =− + sin m − 2 x cos n x dx − sin m x cos n x dx n +1 n +1 n +1 z z z z z 38 . n m −1 n +1 n +1 m−2 sin m −1 x cos n +1 x m − 1 i..u' dx uv v v z − cos x Here z dx = z x cos dx = v sin by putting cos x = t n +1 Fcos x I− z cos x − 1 x cos x dx − Now I = (sin x Gn + 1 J − n + 1 m sin H K n n +1 m.sec 3 x tan x 3 3 ∴ sec x dx = + sec x tan x + log (sec x + tan x ) + c 4 8 8 5 z (ii) To evaluate 2 4 0 R 2j U2 2 4 |e + 0|= + = >> I S1 V3 3 3 3 3 | | T W Note : The reduction formula for z cosec x dx can be established in 4 zsec 2 4 x dx 0 e2 j + 2 I = 2 2 = + 3 3 n a similar way and the result is as follows. n = sin m x n x dx cos m −1 n z Session – 3 /2 0 zsin m x cos n x dx z =z sin x x cos x = z dx sin dx uv say We have z dx = u z dx − z dx. e.
n = + Im − 2n − Im. n = −sin m +1 x . n ∴ Im. e. n L O M P N Q m L + n O 1 −sin x cos x + − 1 = m I M+ 1 P n + 1 n N Q sin x cos x m − 1 = z x cos x dx = − sin + I m+n m+n m −1 n +1 m − 2 n m n m −1 n +1 m n m m −2 n n −1 Note : If we decompose sin x cos x = sin x cos x cos u = cos n −1 x. v = sin m x cos x we can obtain Im. 2 zsin m x cos n x dx = i 2 0 z 0 Illustrations 4 1 8 2 3 sin 5 x cos 4 x dx = = 9 × 7 × 5 × 3 × 1 315 7 ii 2 zsin 0 x cos 5 x dx = 2 6 4 2 4 12 × 10 × 8 × 6 × 4 × 2 = 1 120 iii iv 2 zsin 6 x cos 5 x dx = x cos6 x dx = 4 8 5 2 3 1 = 11 × 9 × 7 × 5 × 3 × 1 693 2 0 zsin 8 3 7 5 5 3 1 1 5 = 14 × 12 × 10 × 8 × 6 × 4 × 2 2 4096 Session – 4 Note for problems : Basic properties of definite integrals are the prerequisites for working problems on reduction formulae. Im. n n +1 n +1 n +1 m −1 1 i. This is known as Walli’s rule. n −2 m+n m+n …(1) x and integrate by parts taking 2 Note: m − 1 − 3 n − 1n − 3 m ×k + n + n − 2 + n − 4 m m m 0 where k = π / 2 when m and n are even.. n 1 + = −sin m −1 x cosn +1 x + − 1m − 2 m I n n +1 n +1 Im. Example –1 Evaluate 0 0 x z sin 8 x dx Let I = x sin 8 x dx z 39 . cos n −1 x n − 1 + Im.−sin m −1 x cosn +1 x m − 1 m −1 Im.
z We have the property f = f − x x dx a dx 0 0 8 0 8 a a 0 0 8 0 8 0 8 2 0 8 z 7 5 3 1 Hence I = by reduction formula. >> Let I = z cos 3 x sin 6 x dx Evaluate /6 0 4 2 6 / 0 4 2 sin 6 x = 2 sin 3 x cos 3 x sin = 2 sin 2 ( 2 cos ∴I= 2 zcos 3x sin 3x cos 3xdx i. 8 6 4 2 2 2 35 Thus I = 256 Example –2 zcos 3x sin 6 x dx using reduction formula.. 3 N 6× 4 ×2 2P 8× Q 2 / 0 2 6 2 / 0 2 6 Thus I = 5 192 40 . I = 4 z 3 x cos 3 x dx sin 6 / 0 4 2 6 / 0 2 6 Put 3x = y ∴ dx = dy 3 If x = 0y = 0 If x = 6 y = 2 ∴I=4 zsin y cos y dy = 4 zsin y cos y dy 3 3 4 L 5 3 O 1 1 I= M by reduction formula. e.z ∴ I = z − x − x =z − x x dx sin dx sin = sin x dx − z sin x dx x z I = sin x dx − I or 2 I = 2 z sin x dx by a property.
8×6×4×2 2 5 4 a Thus I1 = 8 41 .. 2 ax − x 2 = 4 a 2 sin 2 4 a 2 sin 4 − = 4 a 2 sin 2 1 − sin 2 = 4 a 2 sin 2 cos2 = 2 a sin cos 2 =0 ∴ I1 = z4a 2 sin 4 2 a sin 4 a sin cos cos d 6 = 32 a 4 2 0 zsin cos2 d 5 3 1 = 32 a 4 by reduction formula. by 6×4×2 2 Thus I = 32 Hence I = 2a 0 Evaluate (i) >> Let I1 = zx 2 2 ax − x 2 dx (ii) 2 ax − x 2 dx 2a 0 z 2ax − x Example – 4 x 2 dx 2 2a 0 zx 2 Put x = 2 a sin 2 ∴dx = 4 a sin cos d varies from 0 to 2 Also i. e.Example – 3 Using reduction formula find the value of x 1 z − x dx >> Let I = z − x x 1 1 2 2 3 2 0 1 0 2 2 3 2 dx Put x = sin ∴ dx = cos d and varies from 0 to 2 3 − x 2 2 = 2 3 = cos3 1 cos 2 2 0 ∴ I= zsin 2 cos3 cos d = 2 0 zsin 2 cos 4 d 1 3 1 reduction formula.
ii) I2 = 2 0 z 4 a 2 sin 4 a sin 4 cos d 2 a sin cos 2 0 = 8 a2 zsin 4 4 3 1 8a 2 by reduction formula. = 2 If Also (1 + x 2 ) 4 = + tan 2 4 = 2 4 = sec8 1 sec 4 2 tan / 2 tan 4 / ∴ I= sec 2 d = d 8 6 0 sec = 0 sec = z z 2 / sin 4 4 d = sin 4 2 cos d cos 0 0 1 3 1 Hence I = by reduction formula. 6×4×2 2 Thus I = 32 Exercises : = 2 zcos 6 z 1 zsin 2 x sin z 4 0 0 a 0 Evaluate the following integrals 4 4 x cos3 2 x dx x cos 2 x dx dx 7 3 za x4 2 − x2 1 x3 4 dx 1 2 4 0 + x ∞ x6 5 dx 2 1 2 9 0 + x z z Answers 1 128 1155 2 315 3 4 4 16 3 a 16 1 24 5 1 7 42 . d = 4 2 2 3 2 a Thus I2 = 2 Evaluate ∞ 0 Example – 5 z1 +xx dx 2 4 ∞ >> Let I = x4 dx 1 2 4 0 + x z Put x = tan dx = sec 2 ∴ d If x = 0 = 0 x = ∞ .
List of important points to be examined for tracing a cartesian curve f (x. The curve does not lie in the region whenever x or y is imaginary. θ ) = f ( r. 2. In such a case we can find the equations of tangents at the origin by equating the groups of lowest degree terms in x and y to zero. θ ) = 0 1. It is highly essential to known the shape of the curve to find its area. Also if f (x. Based on these features we can draw a rough sketch of the curve. 3. − y) then the curve is symmetrical about the origin. θ ) = f ( r.Session –5 Tracing of Curves Introduction : This topic gives an insight to the process of finding the shape of a plane curve based on its equation by examining certain features. The points of intersection of the curve with the x-axis is got by putting y = 0 and that with the y-axis is got by putting x = 0. Asymptotes : Asymptote of a given curve is defined to be the tangent to the given curve at infinity. If f ( r. 0) = 0 then the curve passes through the origin. List of important points to be examined for tracing a polar curve f ( r. 4. we need to vary the parameter t suitably to take a note of the variations in x and y so that the curve can be drawn accordingly. Equating the coefficient of highest degree terms in x to zero we get asymptotes parallel to the x-axis and equating the coefficient of highest degree terms in y to zero we get asymptotes parallel to the y-axis. Symmetry : If the given equation has even powers of x only then the curve is symmetrical about the y axis and if the given equation has even powers of y only then the curve is symmetrical about the x-axis. Symmetry : f ( r. Region of existence : Region of existence can be determined by finding out the set of permissible (real) values x and y. π −θ ) then the curve is symmetrical about the line θ = π / 2 (positive y-axis) 43 . y) = f (y. −θ ) then the curve is symmetrical about the initial line θ = 0 and θ = π . Note : In the case of a parametric curve : x = x (t) and y = y (t). If f (x. x) then the curve is symmetrical about the line y = x. surface area and volume of solids. y) = f ( −x. In otherwords these are lines touching the curve at infinity. y) = 0 1. length. By examining these features we can draw a rough sketch of the curve. Special points on the curve : If f (0.
Also coefficient of the highest degree term in x is x3 whose coefficient is 1 ≠ 0. 4. Hence x=a is an asymptote. This implies that there is no asymptote parallel to the x-axis. If it gives two values then the curve passes through the pole twice.If f ( r.axis. 1. 2. (This curve is known as cissoid) We observe the following features of the curve. (origin) 2. π / 2 −θ ) then the curve is symmetrical about the line θ = π / 4 (the line y = x) If f ( r. By examining these features we can draw a rough sketch of the curve Example – 1 Trace the curve y2 (a −x) = x3. Special points : The curve passes through (0. Special points : We can tabulate a set of values of r for convenient values of θ . Curve passing through the pole : If r = 0 gives a single value of θ say θ 1 between 0 and 2π then the curve passes through the pole once. The given equation is ay2 −xy2=x3. θ = θ 1 is a tangent to the curve at the pole. θ ) then the curve is symmetrical about the pole. 3 Asymptote : If r → ∞ as → then the line is an asymptote. a > 0 >> We have y2 (a −x) = x3. 3. The lowest degree term is ay2 and ay2 = 0 ⇒ y = 0 which is the equation of x-axis. θ ) = f ( r. Region of existence : y2 = x3 / (a −x) ∴ y = x 3 a − x. These give some specific points through which the curve passes. θ ) = f ( r. ⇒ the curve is symmetrical about x. Symmetry : The equation contain even powers of y. 44 . = 0 0 4. then the 5. β ) i. 0). Hence x-axis is the tangent to the curve at the origin. θ ) = f ( −r.e. 3 π / 2 −θ ) then the curve is symmetrical about the line θ = 3 π / 4 (the line y = −x) If f ( r. coefficient of y2 being a −x to zero we get x=a which is a line parallel to the y-axis. α < θ < β curve does not exist in the region between θ = α and θ = β . Asymptotes : Equating the co-efficient of the highest degree term in y i.e. Region of existence : If r is imaginary for θ ∈ (α. This means that the curve meets the x-axis and y-axis at the origin. Putting y = 0 we get x = 0 and vice versa.
Hence y is real if x > 0 and x < a which implies that the curve lies in the interval 0 < x < a. x > a. 0) Hence we say that the curve intersects x-axis at (0. x = a. x > 0. Since there are two tangents. x < a. Further as x increases y also increases. Asymptotes : The co-efficient of the highest degree term in x being x3 is −1 and hence there is no asymptote parallel to the x-axis.This is positive if x > 0. a −x > 0 or x < 0. Next putting y =0 we get x2 (a + x)= 0 x=0. The shape of the curve is as follows Note : Since the curve meets the coordinate axes at the origin only. Special points : The curve passes through the origin. a −x < 0 i. Also the coefficient of the highest 45 .axis. Since a > 0 the second case is not possible. (This curve is known as ‘Strophoid’. x < 0. ⇒ 2. 0) and ( −a. − The points are (0.) We observe the following features of the curve. Symmetry : The equation contain even powers of y. Example – 2 Trace the curve y2 (a −x) = x2 (a + x).e. 0) only. the origin is called a ‘node’. 0) Also putting x = 0 we get ay2 = 0 or y = 0 and the point is (0.0) ( −a. 0) and intersects the y-axis at (0. a > 0 >> y2 (a −x) = x2 (a + x) 1. The equation of the curve can be put in the form a (y2 −x2) −x y2 −x3 = 0 Equating the lowest degree terms to zero we have a (y2 −x2) = 0 Hence y = + x which are the tangents to the curve at the origin. the curve is symmetrical about x. the origin called a ‘cusp’ with x-axis as the common tangent. 3.
Asymptotes : The coefficient of the highest degree terms in x and y are respectively 1 and a.axis. a > 0 We observe the following features of the curve. The equation of the curve is ay2 −ax2 + x3 = 0. Since these are constants. Hence the curve meets the x-axis at (a. Symmetry : The equation contain even powers of y and hence the curve is symmetrical about x. 0) and meets the y-axis at (0. 4.0) only. If y = 0 then x2 (a − x) = 0. The curve lies between x = 0 and x = a. there are no asymptotes. x > 0 and x < a or 0 < x < a. a When a+x < 0 and also when a −x < 0 y is imaginary. y = 0. a −x = 0 gives x = a. x = a and if x = 0. 2.degree in y being a −x. Example – 3 Trace the curve a y2 = x2 (a −x). 0). Hence we can say that the curve lies between the lines x = −a and x = + a The shape of the curve is as follows. Special points : The curve passes through (0. Hence x = a is the only asymptote which is a line parallel to the y-axis. 3. 4 Region of existence : y = x 2 a + x − x. Equating the lowest degree terms to zero we have a (y2 −x2) = 0 y = + x These are the equations of the tangents to the curve at the origin. Region of existence : y = x − x a a y is positive if x > 0 and a −x > 0. 46 . 1. or x = 0.
f (r. ) f (r. 1. ) f ( −r. f (r. ) f (r. 2. − ) the curve is symmetrical about the line r = 0 gives sin 3 0 n or n Taking values for n = 0. ) the curve is not symmetrical about the pole. r increases numerically from 0 to a as varies from to 47 . Session – 6 Example – 4 Trace the curve r = a sin 3 (Three leaved rose) We observe the following features of the curve. ) the curve is not symmetrical about the initial line.6 we get the corresponding values of and the curve passes through the pole for these values. − f (r.The shape of the curve is as follows.. If 0 < r is positive and r = a if r = 0 if r = −a if If < If < r is positive and r is negative and These observations implies that r increases from 0 to a as varies from 0 to r decreases from a to 0 as varies from to . ….
−) the curve is symmetrical about the initial line. f (r. Example – 5 Trace the curve r2 = a2 cos 2 (Lemniscate of Bernoulli) >> We observe the following features of the curve.e.. When = 0. r2 = a2 or r = + a. r = 0 gives a2 cos 2 = 0 i. cos2 = 0 2and andare the tangents to the curve at the pole. ) the curve is symmetrical about the pole. 48 .) implies that the curve is symmetrical about the line = / 6 so that we conclude that there is a loop between the lines and Similarly we can examine the path of the curve as moves from to and also from to 2. Let us tabulate a set of values of r corresponding to some values of r 0 a 0 −a 0 a 0 −a 0 a 0 −a The curve is symmetrical about and 32 The shape of the curve is as follows.Further f(r −) = f(r. ) = f (r. ) = f ( −r. f (r. 0). 0) and ( − a. Hence the curve meets the initial line at the points (+a.
Length : The length (s) of the arc of a curve between two specified points on it for various types of the curves are given by the following formulae. 1. Such a process is called rectification and the entire length of the curve is called as the perimeter of the curve.Since the curve is symmetrical about the initial line it is composed of two loops. bounded by a polar curve r = f ( and is given by A= 2 1 and the lines r z d 2 2. Area : The area (A) bounded by a curve y = f (x) . Also r doesnot tend to infinity for any and hence there are no asymptotes. The shape of the curve is as follows. the x-axis and the ordinates x = a and x = b is given by A= b x=a y zdx The area (A) between the curves y = f (x) and y = g (x) between x = a and x = b is given by A = f − g x dx x dx. r is real for and . a a b z 1 2 b z The area (A) called the sectorial area. length and volume of solids of revolution The relevent formulae for finding these are as follows. (i) Cartesian curve y = f (x) or x = f (y) FI dy dx FI s = z 1 + G J dx or z 1 + G J dy HK dx HK dy b 2 d 2 x=a y=c 49 . Application to find area.
the volume of the solid is given by V = x 2 dy y=c d z 50 .(ii) Parametric curve x = x (t). y = y (t) dx dy FI FI s = z G J + G J dt H K HK dt dt t2 t1 2 2 (iii) Polar curve r = f ( s= 2 = 1 z r2 + dr d F I d or s = z 1 + r F I dr GJ Gr J HK HK d d 2 r2 2 2 r = r1 3. Volume of revolution : The volume (V) of the solid generated by the revolution of the curve y = f (x) between the ordinates x = a and x = b. d c ds dx where = 1+ dy dy 2 FI GJ HK 2 In the case of a polar curve the surface area of revolution about the initial line is given by S= 2 ds 2 sin 2 r sin d r ds= d = 1 1 z z ds dr where = r2 + d d FI GJ HK 2 4. the area of the surface (S) generated is given by S = 2 y ds= 2 y x=a a b z ds where = dx d z dy FI 1 + GJ HK dx b 2 ds dx dx Similarly the surface area of revolution about the y-axis is given by S= y=c ds 2 z2x ds = z x dy dy . If a curve is bounded by the ordinates x = a and x = b revolves once completely about the x-axis. Surface area : When a curve revolves about the x-axis a surface is generated and the same is called a surface of revolution. about the x-axis is given by V = y 2 dx x=a b z Similarly if the axis of revolution is the y-axis.
We tabulate x. a) and (0. 0). 51 . surface area and the volume. Since | cos | < | and | sin | < 1. 0) and ( − a. y corresponding to certain angles of in the interval [0. Also it meets the y-axis at the points (0. perimeter. We shall find its shape first and then determine the associated area. 2 x y a 0 0 a −a 0 0 −a a 0 >> Let us consider its parametric equation : From the table we conclude that the curve meets the x-axis at the points (a. (a) Trace curve x 2/3 + y 2/3 = a 2/3 x = a cos3 y = a sin3 . − a).Also in the case of a polar curve r = f () the volume (V) of the solid generated is given by 2 3 r sin d (revolution about the initial line) 3 2 3 V= r cos d (revolution about the line 2) = 3 V= z z Session – 7 Application related to standard curves Example –1 The Astroid : Astroid is the curve represented by the equation x 2/3 + y 2/3 = a 2/3 Its parametric equation is x = a cos3 and y = a sin3 . we have | x | < a and | y | < a. Hence we infer that the entire curve lies within a circle of radius ‘a’ having origin as the centre.
i. a (c) Find the parameter of the astroid x 2/3 + y 2/3 = a 2/3 52 .. f (x. The curve astroid is symmetrical about the coordinate axes and hence the required area (A) is equal to four times the area in the first quadrant. y) . A = 4 y dx = 4 y 0 0 a z a z ddx d dx = −3a cos 2 sin d We have x = a cos3 y = a sin 3 ∴ When x = 0 : a cos3 = 0 or cos3 = 0 x = 1 : a cos3 = a or cos3 = 1 ∴ A=4 0 0 = 2 a z sin 2 0 3 −3 a cos2 d sin 4 = 12 a 2 zsin 2 d cos 1 3 1 = 12 a 2 by reduction formula. y) = f (y. Taking a note of the values of x and y as advances from one quadrant to the other the shape of the curve is as follows. − y) . f (x. x) Hence the curve is symmetrical about the coordinate axes and also about the line y = x. y) = f (x. units.Also we have from the cartesian equation of the curve. f (x. 6×4×2 2 Thus the area ( A) enclosed is 3 2 8 sq.e. y) = f ( −x. (b) Find the area enclosed by astroid x 2/3 + y 2/3 = a 2/3 >> Note : In any problem on applications we need to draw the curve first by briefly examining the important features.
units. (d) Find the surface area of revolution of the astroid x 2/3 + y 2/3 = a 2/3 >> Because of symmetry the required surface area is equal to twice the surface area by the revolution of the first quadrant of the curve. 53 . ∴ FI FI dx dy l = 4 z G J +G J d d K d K H H = 4 z 9a cos 9a sin sin + cos 2 2 2 =0 2 0 2 4 2 2 4 2 d =4 2 0 sin z 9a cos 2 2 2 cos 2 sin 2 d + = 4 3a cos sin d 0 2 z 0 z − Lcos 2O = − 3a − cos 0 −3a1 − 1 6a = 6a M cos = − = N2 P Q = 6 a sin 2 d 2 / 2 0 Thus the perimeter of the curve is 6 a units. the perimeter (entire length) of the curve is four times its length in the first quadrant. 5 × 3×1 Thus the required surface area = 12 2 / 5 sq. a (e) Find the volume of the solid generated by the revolution of the astroid x 2/3+y 2/3= a 2/3 >> Because of the symmetry the required volume (V) is equal to twice the volume of the solid generated by the curve in the first quadrant about the x-axis.>> Since the curve is symmetrical about the coordinate axes. ∴ S = 2 × 2 ds = 4 y y 0 0 = 2 2 z z ddsd FI FI ds dx dy But = G J + G J = 3a cos sin d H K H K d d Hence S = 4 za sin 3a cos sin d a 2 2 0 3 = 12 2 a 2 0 zsin 4 cos d 3 = 12 2 a by reduction formula.
Hence the curve lies above the x-axis. x y 0 0 a ( ) a a 2a a(3 ) A 2a 0 From the table we can conclude that the curve intersects the x-axis at x = 0 and 2a. Taking a note of the values of x and y as advances in the interval [0. Also. 9 × 7 × 5 × 3 ×1 Thus the required volume of the solid is 32 3 / 105 cubic units. y for certain values of in the interval [0. Imagine a wheel rolling on a straight line without slipping. The parametric equation of the cycloid can be in the following forms : x = a − sin = a − cos i y 1 x = a + sin = a − cos ii y 1 = a − sin = a + cos iii x y 1 = a + sin = a + cos iv x y 1 (a) Trace the cycloid : x = a − sin = a − cos y 1 >> Let us tabulate x. a = 6 3 a Example –2 Cycloid : Cycloid is a curve generated by a point on the circumference of a circle which rolls on a fixed straight line known as the base. 2] where is inradians. we have y = a (1cos) and since | cos | < 1 y is non negative. It is called an arch of the curve. 2] the shape of the curve is as follows.∴ V = 2 × 2 dx = 2 y 2 y 0 0 0 2 6 z z ddxd = 2 z sin −3a cos d a sin = 6 z a sin cos d a a 2 2 3 2 0 7 2 6 4 2 1 by reduction formula. 54 . A fixed point on the rim of the wheel traces the cycloid.
A = 16 a 2 by reduction formula. e.. (c) Find the length of an arch the cycloid : x = a − sin y = a − cos 1 55 . t varies from 0 to = ∴ A = 8a 2 sin 4 t dt = 8a 2 2 t =0 z zsin 4 t dt 3 1 i. e.. A = a− cos a − cos d 1 1 = a2 = a2 2 0 z 2 0 1 z− cos d d z4 sin ( 2 4 2 t =0 Put 2 = t ∴ d 2 dt.(b) Find the area of an arch of the cycloid : x = a − sin y = a − cos 1 >> Area A = 2 0 2 =0 zy ddxd 2 i. 4 2 2 Thus the area enclosed by an arch of the curve on its base is 3 a2 sq.units.
e. l = 2 0 2 2 =0 dx dy GI GI J HJ z F +F d HK d K d 2 2 2 2 2 1 z a − cos + a sin d = z − 2 cos cos sin d a 1 + + = a z 2 − cos d a z 2 sin ( 2 d 1 = 2 2 2 O La cos ( = 2a z ( 2 d − M sin = 2 N 1 P Q 2 0 2 2 2 0 2 0 2 2 0 2 0 = −4 a cos 0 −4 a 1 − 1 8a cos − = − = Thus the required length is 8a (d) Find the surface area generated by the revolution of an arch of the cycloid x = a ( −sin ). e. y = a (1 −cos ) about the x-axis. a 3 (e) Find the volume of the solid generated by the revolution of the cycloid x = a − sin = a − cos y 1 56 . a 3 Thus the required surface area is 64 2 sq.>> Length (l ) = i. 2 dt = 16 2 a 2 2 t=0 zsin 3 t dt 2 i... S = 32 2 by applying reduction formula. units. z ddsd ds dx dy FI FI But = G J + G J = 2 a sin ( 2) / d H H d K d K ∴ S = 2 z− cos a sin ( 2) d a1 2 / >> Surface area ( S 2 y = 0 2 2 2 2 0 =4 2 a 2 0 z2 sin t=0 3 ( 2) d 8 2 / = a 2 0 zsin 3 ( 2) d / Put 2 = t ∴ d 2 dt and t varies from 0 to = Hence S = 8 2 a sin z 3 t .
>> V = zy ddx d = z − cos a − cos d a 1 1 = z 2 sin 2 d 8 z a o t = a sin 2 d = 8 z 16 z a 2 sin 2 d = a sin 2 d 2 0 2 2 0 2 2 3 2 0 2 3 3 2 0 6 3 0 6 3 0 6 2 0 Put 2 = t ∴ d 2 dt and t varies from 0 to 2 = ∴ V = 16 3 a zsin 6 t . = is a tangent to the curve at the pole. r 57 . 2a 3a/2 a a/2 0 Let us tabulate r for certain angles of . | r |< 2a and hence the curve lies with in the circle of radius 2a having its centre at the pole. r = 0 when = and hence the curve passes through the pole. Session –8 Example –3 Cardiode : r = a (1+cos ) (a) Trace the curve r = a (1+cos ) >> r = a (1+cos ) We observe the following features of the curve. ) = f (r. a 6 4 2 2 2 Thus the required volume is 5 a 3 cubic units. by reduction formula. (i) (ii) (iii) f (r. 2 dt 5 3 1 = 32 3 . −) and hence the curve is symmetrical about the initial line. Since | cos | < |.
. (b) Find the area of the cardioide r = a (1 + cos ) >> Since the curve is symmetrical about the initial line. 1 i. 4 2 2 Thus the area enclosed is 3 2 / 2 sq. a (c) Find the perimeter of cardioide r = a (1 + cos ) >> Perimeter (length) = 2 (length of the upper half of the curve) i. 2 dt = 8 a 2 2 t=0 zcos 4 t dt 3 1 = 8 a 2 by reduction formula.It is evident thatas increases from 0 to . r decreases from 2a to 0. 58 . The shape of the curve is as follows. the total area (A) is twice the area above the initial line. e.. Now =2 ds ds dr where = r2 + d d 0d z FI GJ HK 2 ds = a 2 + cos 2 + a 2 sin 2 a 2 + cos 1 = 1 d = 2 a cos ( 2 0 ∴ perimeter = 2 2 a cos( 2) d = 4 a / z sin L (/ 2) O 8a M1 P= N 2 Q 0 Thus the perimeter of the curve is 8a units. e. A = 2 r 2 d = a 2 + cos 2 d 1 20 0 = a2 0 z z 2 cos z2 cos ()] d = 4a z 2 2 2 0 4 ( 2) d Put 2 = t ∴ d 2 dt and t varies from 0 to 2 = ∴ A = 4a 2 2 t=0 zcos 4 t. units.
units. 3 Example – 4 Regarding sphere as the solid generated by revolving a circle about a diameter. 2 dt 3 1 = 32 2 a by reduction formula. 5×3 Thus the required surface area is 32 2 sq. a 5 (e) Find the volume generated by the revolution of the curve r = a (1 + cos ) about the initial line. ds >> Surface area 2 sin d S = r d ds dr But = r2 + = 2 a cos ( 2) d d ∴ S = 2 a+ cos sin 2 a cos ( 2) d 1 =0 2 2 z FI GJ HK 2 z = 4 z cos sin ( 2 cos( 2) cos( 2) d a 2 2 2 0 = 0 = 16 2 cos 4 ( 2 d a 2 sin Put 2 = t ∴ d 2 dt and t varies from 0 to 2. find the volume of a sphere of radius ‘a’ 59 . 2 2 3 >> V = r 3 sin d = a + cos 3 sin 1 d 3 0 3 0 Put t = 1 + cos ∴ dt = −sin d If 0 t = 2 and if t = 0 = = 2 3 0 3 a 2 3 2 3 a ∴V = t dt − = t dt 3 2 3 0 2 3 a = 3 4 2 0 3 z z z z L O= 2 a − 0 8a t MP 3 4 = 3 4 NQ 3 Thus the required volume is 8a 3 cubic units. = 2 t=0 z Hence S = 16 2 a zcos 4 t sin t .(d) Find the surface area of the revolution of the curve about the initial line.
a 3 Example – 5 Find the volume generated by the parabola y2 = 4 ax when revolved about the y-axis between y = 0 and y = 2a >> Required volume (V x 2 dy = i.. z V = z dx = z − x y a dx Volume V = y 2 dx a 2 a 2 −a −a 2 a −a = a 2 x − 3 = 4 3 x 3 a 3 Thus the required volume is 4 3 cubic units. a 5 Example – 6 Find the perimeter of the curve r = a sin3 (/ 3) >> Let us tabulate r for certain angles of where r = a sin3 (/ 3) r 0 a/8 3 3a 8 A 3 3a 8 a/8 0 60 . e. V = 2 ady y 4 2 y=0 5 2a 0 z L O y = MP 5 16 a N Q 2a 2 z = 32 5 a a= 2 5 2 80 a 80 a 2 Thus the required volume is 2 3 cubic units.>> Let x2 + y2 = a2 be the equation of the circle and when the semicircle revolves about the diameter (x-axis) a sphere of radius ‘a’ is generated.
perimeter = l za sin 2 3 d Put 3 = ∴ d 3d and = varies from 0 to ∴ l =a sin 2 d 3 a 3 = 2 =0 z 2 0 zsin 2 d 1 by reduction formula. Exercises 1. 3. 4a2 4. 5.. Show that the surface area generated by the revolution of the loops of the curve r2 = a2 cos 2 about the initial line 2 a2 ( 2 − 2 ). Find the volume of the sphere of radius ‘a’. y = = a (sin − cos ) is a / 2 2. 2 61 .∴ perimeter = l 2 3 0 = 6 zr 2 + / d 2 d dr 2 r2 + F I = a sin o 3 sin dr 3 + a. l = 6 a . Show that the length of the arc from = 0 to = along the curve x = a (cos + sin ).e. 2 2 Thus the required perimeter is 3a 2 units. / i. G d K HJ 2 3 3 3 cos 1 t 2 = a 2 sin 6 a 2 sin 4 2 3 + 3 cos 3 = a 2 sin 4 sin 2 cos 2 3 3 + 3 = a 2 sin 4 3 ∴ r 2 + / d 2 = a sin 2 dr 3 3 0 Hence. 4. Find the volume of the solid generated by the revolution of a loop of the curve r2= a2 cos 2 about the initial line. Answers 2. 4a3/3 a 3 15 5. Find the surface area of a sphere of radius a.
. + un containing infinite number of terms is called an infinite series usually denoted by ∑ un or simply ∑ un n =1 ∞ un is called the nth term or the general term of the infinite series. In this connection we make use of well established tests so as to draw a valid conclusion on the nature of an infinite series. divergence and oscillation of an infinite series which are attributed as the nature or the behaviour of an infinite series. n→∞ ∑ un is said to be divergent if lim sn = ± ∞ n→∞ A series ∑ un is said to be Oscillatory if sn tends to more than one limit as n → ∞ Geometric series as an example for convergence.1 Introduction: Infinite series is basically a summation of infinite number of terms. e. divergence and oscillation. Divergence and Oscillation A series ∑ un is said to be convergent if lim sn = l where l is a finite quantity. The summation will be meaningful if there is no significant change in the sum as more and more terms gets added up which is the concept of convergence. Let us consider the geometric series a + ar + ar2 + …to discuss its nature for various values of r. sn = u1 + u2 + u3 +u n + Convergence.Infinite Series Session .. an expression of the form u1 + u2 + u3 +. The sum to n terms of this series is given by 62 .. The sum of the first n terms of the series is denoted by sn i. In this topic we discuss the aspect of convergence. ‘Progression’ is the pre-requisite for this topic Definitions If un is a function of n defined for all integral values of n.
but not conversely. then lim un = 0. sn tends to more Hence we conclude that the series is oscillatory if r < −1. Property : If ∑ un is convergent. 63 . we have r a a − 0 1 = which is a finite quantity.a− r n 1 if r < 1 1 1− r an − 1 r and sn = if r > 1 2 r −1 Now if r|< 1 n → 0 as n → ∞ and from (1). Further if r < −1. lim sn = 1− r 1− r n→∞ sn = Hence we conclude that the geometric series is convergent for | r |< 1. n→∞ An example to show that the converse is not true. the series become a −a + a … where the sum of the first n terms sn is 0 or ' a' according as n is even or odd and we can say that lim sn = 0 or a. Note : An infinite series of positive terms is either convergent or divergent. rn than one limt as n ∞ ∞ or − ∞ according as n is even or odd. When r = −1. Next if r > 1 from (2) lim sn = lim n→∞ ar n − 1 =∞ r −1 n→∞ If r > 1 → ∞ as n → ∞ r Hence we conclude that the geometric series is divergent if r > 1 Also if r = 1 the series becomes a + a + a +… sn = a + a + a… to n terms = na ∴ lim sn = lim n a = ∞ n→∞ n→∞ n Hence the series is divergent if r = 1. sn tends to more n→∞ than one limit as n ∞ Hence we conclude that the series is oscillatory if r = −1.
Then ∑ un and ∑ vn behave alike. This example shows that the converse of the property is not true. 64 . un = + 2 + 1 n − n = n + 2 + n +1 en + 2 − n +1jen + 2 + en + 2 + n +1j 1 n +1 n +1 j en + 2 + j n +1 Clearly lim un = 0 n→∞ Next sn = u1 + u2 + u3 +un −1 + un i. n→∞ 2. e. n→∞ v n That is ∑ un and ∑ vn both will converge or both will diverge..Consider the series ∑ ∞ n=1 n + 2 − n +1 Here un = n + 2 − n + 1 = i.. e. Tests for convergence for series of positive terms Comparison Test If ∑ un and ∑ vn be any two series of positive terms such that un lim is a non zero finite quantity. sn = i. e.. Remarks : 1. sn = − 2 + n + 2 e3 − 2 j+ e4 − 3 j+ e5 − 4 j++ en + 1 − n j+ en + 2 − j ∴ lim sn = lim − 2 + n + 2 = ∞ n→∞ n→∞ Hence we conclude that un is divergent. Thus lim un ≠ 0 ⇒ ∑ un is divergent. The contrapositive of the statement of this theorem provides a fundamental test for the divergence of an infinite series which can be stated as follows : If un does not tend to zero as n tends to infinity. then un is not convergent. Addition or deletion or multiplication of each term of infinite series by a nonzero constant doesnot alter the nature of infinite series.
1 We choose vn = p − q where p and q are the highest power n in the denominator n and numerator of un respectively. [The nth term of the A. we must rationalize by multiplying with f g the case may be and then chose vn as stated. … is a+(n − 1)d ] Comparison test is usually applicable when un involves expression of n in the numerator and the denominator like (n+1). If ( p − q ) = k (say) then we say that un is of the order 1 and conclude that ∑ un is converngent if k > 1 and divergent if k ≤ 1. n + n as Order Test We say that 1 n p− q is the order of un . We usually compare with the harmonic series ∑ ∞ n =1 1 ( p − series) which is convergent for p > 1 and divergent for p ≤ 1 np (The proof is established later) Working procedure for problems • • • Given an infinite series. (n2+1).P. • However if un is of the form f n − g or n f girrational factors connected n − n f g n + n or by a negative sign). we first write the nth term (general term) un.Remark : In this test the given series un is to be compared with another series vn whose nature is to be known to us. a+ d. a. n . a+2 d. n3 + 4 etc. Session – 2 Example – 1 65 . (2n+1). nk Sometimes this is referred to as the order test.
. 5. . 5. 1. .. Examine the series First factors : 1. 2.5 3 59 57 7 >> Numerators are 1. 7. Here ∑ vn = ∑ Example – 2 Find the nature of the series 1 1+ 2 1+ 2 + 3 + 2 + 2 + 2 2 1 1 +2 1 + 22 + 32 >> It should be observed that there will be n terms in the numerator and the denominator of the general term un. 9 . 7. In the denominator we have. Hence the general term is n.. .. Second factors : 3. 66 .3.1 2 3 + + + for convegence. Now lim 1 1 is of the form ∑ p = 2 > 1 convergent.. 3. ∴ n th term = 1 + n − 12 = 2 n − 1 ∴ n th term = 3 + − 1= 2 n + 1 n 2 ∴ n th term = 5 + − 1= 2 n + 3 n 2 Thus n th term of the given series is given by un = Thus n n 1 Choose vn = 3 = 2 n − 1 n + 1 n + 3 2 2 2 n n un n n2 = × vn n − 1 n + 1 n + 3 1 2 2 2 un n3 1 = lim 3 = n→∞ vn n→∞ n − 1 n + 1 + 3 8 2 2 n 2 n By comparision test.. ∑ un and ∑ vn behave alike.. 3... Third factors : 5. p is 2 n n Hence ∑ un is also convergent.
.∴ un = 1 + 2 + 3++ n ∑n = 2 2 2 2 1 + 2 + 3 ++ n ∑n n + 1 n + 12n + 1 n n i. Note : Alternative version of the problem. ∑ un and ∑ vn behave alike.. un = = Choose vn = 2 n + 12n + 1 2n + 1 n n u 3 n Now lim n = lim n→∞ vn n→∞ 2 n + 1 1 2 = lim n→∞ F 3 I= 3 G+ 1 J 2 H nK 2 By comparison test. n1/2 Hence ∑ un is also divergent.e. p is n n Hence ∑ un is also divergent. 1 ( p = 1 / 2 < 1) is divergent. Example – 3 Test for convegence ∑ >> By data un = Now lim ∞ n =1 1 n +1+ n 1 1 Choose vn = n +1 + n n 1 n 1 n +1+ n n = lim n + 1 n + n n→∞ n 1 n 1 + 1 + 1 n = 1 2 n→∞ un = lim vn n→∞ = lim n→∞ By comparison test ∑ un and ∑ vn behave alike. 1 1 But ∑ vn = ∑ = ∑ 1 = 1 divergent.e. But ∑ vn = ∑ 67 . un = 2 6 n + 1 n 6 3 1 i.
e. un = en + 1 + n en + 1 − n j n + 1 + n j = n + 1− n n +1+ n 1 which is the same as the example discussed. 2 3 + 13 + 3 n 3 + 1 + n 2 n n 1 1 i.Discuss the convergence of the infinite series ∑ ∞ n =1 en + 1 − n j >> un = n + 1 − n and on rationalizing we have un = i. un assumes the form − b a ∴ un = by above.. L M N n2 O P Q = 1 1 = 1+1+1 3 68 . n +1+ n Example – 4 Test for convergence ∑ >> By data. e.. un = Choose vn = 2 3 2 3 3 3 2 n + 1 + n + 1 + n n n Now lim n→∞ 3 + 1 n 3 n − un 1 n2 = lim × 2 vn n→∞ 3 + 1 3 + 3 n 3 + 1 + n 2 1 n n n2 n 3 + 1 3 + 3 n 3 + 1 3 n + n 2 1 n 1 n n2 n 2 + 1 3 3 + n 2 1 n 2 3 2 3 = lim = lim = lim n→∞ n→∞ 1 + 1 3 + n 2 n n→∞ n 2 + 1 3 3 + 3 1 + 1 3 + 1 1 n 2 n By comparison test ∑ un and ∑ vn behave alike. un 3 3 We have an elementary formula Ln + 1 − n O M P N Q = Ln + 1 − n O M P N Q ∞ 3 3 n =1 a 3 − b 3 = − b 2 + ab + b 2 − b a a or a = a3 − b3 a 2 + ab + b 2 If we take a = 3 n 3 + 1 and b = n. Hence ∑ un is also convergent.
Remark : Usually we try ratio test in the following cases.. we can apply the comparison test (order test) if possible or Raabe’s test stated as follows. Convergent Session – 3 D’Alembert’s ratio test and Raabe’s test If un is a series of positive terms and if n→∞ 2. If un is a series of positive terms and if n→∞ lim n Fu − 1I = l (finite qty) G J u H K n n +1 then un is convergent if l > 1. When the ratio test fails. Convergent lim un +1 = l (finite qty) un then un is convergent if l < 1. x2n etc. (i) (ii) nth term of the series is not of the order 1/nk. divergent if l > 1 and the test fails if l = 1. nth term of the series involves variables like xn. divergent if l < 1 and the test fails if l = 1. 69 . Divergent 3.Exercises : Find the nature of the following series 1 4 7 1 + + + 13 27 311 2 5 8 12 12 + 22 12 + 2 2 + 32 2 3 + 3 + + 1 1 + 2 3 13 + 2 3 + 33 1 1 1 3 + + 3+ 5 4+ 6 5+ 7 4∑ Ln + 3 − n O M P N Q 4 2 Answers : 1. Divergent 4.
(n !)2 etc. Raabe’s test becomes inevitable when the ratio test fails in the case of carry over series.(iii) nth term of the series involves n !. (iv) The number of factors in the numerator and denominator increase steadily from term to term like 1 1 13 2 2 + + +(carry over series) 3 3 37 5 5 Working procedure for problems We first write the general term un for the given series and find un+1 by replacing n by (n+1) • We find the ratio un+1 simplify and find its limit as n → ∞ to decide the nature of the series.2 2 3 3 4 >> un = xn nn + 1 x n +1 ∴ un +1 = + 1 + 2 n n Now ∴ lim un +1 x n +1 n + 1 n n = = x un + 1 + 2 x n n n n+2 un +1 n 1 = lim x = lim x=x un n→∞ n + 2 1 n→∞ + 2 n n→∞ convergent if x <1 ∴ by D' Alembert' s ratio test ∑ un is divergent if x > 1 and the test fails if x = 1 1n 1 1 But when x = 1un = = = 2 n + 1 n + 1 n + n n n R S T 70 . (n+1) !. un instance or When the ratio test fails we can try to apply the order test in the first apply Raabe’s test. Example – 1 x x2 x3 + + + 1.
un +1 = x 2 26n n + 2 42 2 From (1) and (2) we have.un is of order 1 / n 2 ( p = 2 > 1) and hence ∑ un is convergent. e. un +1 2 n + 1 = x un 2n + 2 un +1 2n + 1 n + 1 2 n = lim = lim x = x x un n→∞ 2 n + 2 2 n n→∞ n + 2 . we When x = 1 from (3) Now lim n n→∞ R S T Fu G u H n − 1 = lim n n→∞ n +1 I J K un +1 2 n + 1 u 2n + 2 = ∴ n = un 2n + 2 un +1 2 n + 1 n→∞ 2 2 Fn + 2 − 1I = lim n Fn + 2 − 2n − 1I Gn + 1 J G 2n + 1 J H K H K 2 = lim n 1 = <1 2 n n→∞ n + 1 2 ∴ ∑ un is divergent by Raabe' s test Hence we conclude that the given series ∑ un is convergent if x < 1 and divergent if x ≥ 1 71 . e.. un +1 15n − 1 n + 1 n +1 32 2 262 n 4 = x × un 26n n + 2 42 2 15n − 1n 32 x i.. Hence we conclude that ∑ un is convergent if x ≤ 1 and divergent if x > 1 Example – 2 Find the nature of the series 1 1 2 15 3 3 3 x+ x + x + 2 2 4 26 4 15n − 1 n 32 >> un = x 262n 4 un +1 = 15 2 + 1 1 3 n − x n +1 = 1 152 n + 1 n +1 3 x 262 n + 1 4 262 n + 2 4 15n − 1 n + 1 n+1 32 2 i... 3 ∴ lim n→∞ Hence by ratio test convergent if x < 1 ∑ un is divergent if x > 1 and the test fails if x = 1 Now when x = 1 shall apply Raabe' s test.
un +1 39 n n + 3 5n +1 6 3 3 410n + 13n + 2 7 3 = × un 410n + 1 n + 43n + 5 7 3 3 39 3n 6 5n un +1 n + 33n + 2 3 = 5 un n + 43n + 5 3 un +1 n + 33n + 2 3 = lim 5 un n→∞ n + 43n + 5 3 = lim ∴ lim n→∞ + 3 3 + 2 3 n n = 5 >1 5 3 n n n→∞ + 4 3 + 5 Hence by the ratio test ∑ un is divergent. Example – 4 72 . e.Example –3 Test for convergence ∑ >> By data un = ∴ un +1 = ∞ n =1 39 3n 6 5n 410n + 13n + 2 7 3 39 3n 6 5n 410n + 1 3n + 2 7 3 39 3 + 1 6 n 5n +1 4103 + 1 1 3 + 1 2 7 n + n + 39 3nn + 3 5n +1 6 3 i. un +1 = 410n + 1 n + 43n + 5 7 3 3 Now i... e.
un = x 1 n 410 n + 1 1 n +1 410n + 4 n +1 7 3 + 7 3 un +1 = x = x + 1 n + 1 n 410n + 13n + 4 n +1 7 3 i. e. un = 73 .. Example – 5 Test for convergence of the infinite series 1+ 2 3 4 + + + 2 2 33 4 4 n + 1 + 1 n n n n ∴ un +1 = = = n n +1 n +1 n n + 1 n + 1 n + 1 n >> The first term of the given series can be written as 1!/11 so that we have..410n + 1 n 7 3 x n n =1 410n + 1 n 7 3 >> By data. Thus ∑ un is convergent if x < 1 / 3 and divergent if x ≥ 1 / 3. un +1 410n + 1 n + 4 n +1 7 3 3 n = × x un + 1 n n 410n + 1 n 7 3 x u 3n + 4 i. un +1 = x 2 + 1 n n From (1) and (2) we have. e. n +1 = x un n +1 Find the nature of the series ∑ ∞ ∴ lim n→∞ un +1 3n + 4 3+ 4 n = lim = lim x = 3x x un n n→∞ n + 1 n→∞ 1 + 1 convergent if 3 x < 1 or x < 1 3 Hence by ratio test ∑ un is divergent if 3 x > 1 or x > 1 3 and the test fails if 3 x = 1 or x = 1 / 3 u n + 4 3 1 u 3n + 3 If x = 1 we have n +1 = 3 or n = un + 1 3 n un+1 3n + 4 Now lim n n→∞ n n+1 n→∞ n→∞ R S T Fu − 1I= lim n Fn + 3 − 1I = lim n F −1 I = lim 3 G+4 J G J H K u 3n 3n H K H K G + 4J n→∞ −1 −1 = <1 + 4 3 3 n Hence ∑ un is divergent when 3 x = 1 by Raabe' s test.
n n n nn un +1 Now = = n un + 1 n n n + 1 n 1 nn 1 1 u ∴ lim n +1 = lim = <1 n n→∞ un n→∞ + 1 e 1 n Hence by ratio test ∑ un is convergent. divergent if x > 2 2. Convergent if x < 3/2. Convergent if x < 1.4 Cauchy’s root test If un is a series of positive terms and if n→∞ u 1 n lim n = l (finite) then the series is convergent if l < 1. divergent if l > 1 and the test fails if l = 1. We try to apply this test if un is of the form [f (n)]g (n) 2. divergent if x > 1 1. Note : 1. Convergent 2. Exercises 1 1 + 3 32 33 + + + 1 2 3 2 3 2 4 3 2 x+ x + x + 3 4 4 5 5 6 1 1 2 13 3 2 2 3 x + x + x + 3 3 5 37 5 3 x 3 x 2 39 x 3 6 6 4 + + + 4 5 4 8 410 11 7 7 ∞ 37n + 1 52 5 ∑ xn 393n 6 n =1 Answers 1. Convergent if x < 2. When the test fails we can try the order test or the fundamental test. The following standard limits will be useful 74 . divergent if x > 3/2 Session . divergent if x > 1 3. Convergent if x < 1.
1 N nQ R 1OU |L P | ∴ = S+ u 1 N |M n Q V | T W L+ 1 O L+ 1 O = M P =M P u 1 1 N nQ N nQ L+ 1 O Now lim = lim M P u 1 N nQ 2 − n3 n =1 L+ 1 O 1 ∑ M P N nQ ∞ 1 n n n 1 n 2 − n3 n 1 n 2 − n1 − n n→∞ n 1 n − n n→∞ = lim n→∞ L+ 1 O 1 M nP N Q n 1 = 1 < 1 e n → ∞. G nJ H K n − n3 2 iii lim 1 + n→∞ F xI = e G nJ H K n x Example – 1 Test for convergence L+ 1 O >> u = M P by data. n also → ∞ as Hence by Cauchy' s root test ∑ un is convergent. Example – 2 n + 1x n n Discuss the convergence of ∑ n =1 n n +1 n + 1x n n >> un = by data. n n +1 ∞ ∴ (un ) 1/ n n o + 1t = n→∞ 1/ n lim (un ) + 1 + 1 n x n x = 1+1 n 1 n n n nn + 1 n x n+ 1 n 1 x + 1 1 n x = lim = lim = lim 1 n 1 n 1 n n→∞ n→∞ n→∞ nn nn n n+1 1 n n n 1 n n x 1 = 75 .n (i) lim n1 = 1 n→∞ ii lim 1 + n→∞ F 1 I = e.
un = n + 1 n n n +1 un is of order n n / n n +1 = 1 / n and hence ∑ un is divergent. u = G J x H+ 2 K n Rn + 1 I U = F + 1 Ix n |F J | x G J ∴ =S u G+ 2 KV H+ 2 K n |H | Tn W n n n n 1 n n n 1 n 1 n FI x GJ HK 2 2 + x > 0 Now lim n n = lim u 1 n→∞ + 1 1 n x=x 1 n n→∞ + 2 convergent if x < 1 Hence by Cauchy' s root test ∑ un is divergent if x > 1 and the test fails if x = 1 R S T 76 . Example – 3 Find the nature of the series 1+ 2 3 x+ 3 4 n F + 1I >> Omitting the first term. lim n1 = 1 1 n→∞ convergent if x < 1 Hence by Cauchy' s root test ∑ un is divergent if x >1 and the test fails if x = 1 When x = 1. R S T Thus ∑ un is convergent if x < 1 and divergent if x ≥ 1.n→∞ lim (un ) 1/ n = n→∞ 1 n x lim + 1 n→∞ lim n 1 n = + 0 1 x n = x.
1 un = n n Test for convergence ∴ lim n = u n→∞ 1 n n→∞ 1 F1 I lim G J = lim = 0 < 1 HK n n 1 n n n→∞ Hence by Cauchy' s root test ∑ un is convergent. Exercises : Tests for convergence the following series F+ 3 I 1∑ G J 1 H nK n 3 F + 1I F I 2∑ G J G Hn K HJ 4K 4 FI F I F I 7 10 13 3 + G + GJ + GJ + J H K HK 3 H 5K 7 9 − n2 n2 n 2 3 4 41 + x x2 x3 + + + > 0 x 2 32 4 3 Answers : 77 .n n n O F + 1 I F + 1 I L1 + 1 But when x = 1 = G J 1 = G J = M u H+ 2 K H+ 2 K N+ 2 n n 1 n P Q + 1 e 1 1 n F+ x I = = using lim G J = e 1 lim u = lim H nK e + 2 e 1 n n n n n n n n n→∞ n x n→∞ n 2 n→∞ 1 Since lim un = ≠ 0 un is divergent when x = 1 ∑ e n→∞ Thus ∑ un is convergent if x < 1 and divergent if ≥ 1. Example – 4 1 1 1 + 2 + 2 + 2 2 3 >> The first term of the series can be put in the form 1 / 11 so that we have.
Example : Apply Cauchy’s integral test to discuss the nature of the harmonic series (p.i : Then x 1− p ∞ 1 L O 1 = 1 which is finite. x is Now ∞ 1 z ∞ 1 x1− p f = p dx = x − p dx = x dx 1− p 1 x 1 ∞ z z Case . Note : The condition f ' ( x ) < 0 ensures that f ( x ) decreases as x increases where f ' ( x ) is the derivative of f ( x ). 78 .(1) Convergent (2) Divergent (3) Divergent (4) Convergent Cauchy’s integral test If ∑ un is a series of positive terms and if ux = f ( x ) be such that (i) f ( x ) is continuous in 1 < x < ∞ (ii) f ( x ) decreases as x increases [ f ( x ) is monotonically decreasing] then the series ∑ un is convergent or divergent according as the integral ∞ 1 f z( x) dx is finite or infinite. x ∴ z =M P= 0 − f x dx 1− N pQ 1− p p −1 1− p ∞ 1 Let p > 1 or ( p − 1) > 0 1 = p −1 → 0 as x → ∞ − 1 0 p > x L O p ≠1 M Pif N Q ∞ 1 Thus ∑ un is convergent if p > 1.series) n =1 ∞ ∑ 1 np 1 = f n np >> By data un = f x = 1 −p = x − p ∴ f ' − px − p −1 = p +1 < 0 x = p x x Hence f decreasing.
Case . Remark : If lim un ≠ 0 then the series is oscillatory. u2 .. (1 − p) > 0 But x 1− p ∞ 1 → ∞ as x → ∞. (1 − p) > 0 Hence x dx zf = ∞ − 1 −1 p = ∞ Let p = 1 ∞ 1 1 ∞ 1 Thus ∑ un is divergent if p < 1. (ii) lim un = 0 n→∞ then the alternating series u1 − u2 + u3 − u4 +. Case ..iii : ∞ 1 zf xdx = zx1 dx =log x =∞−0=∞ Thus ∑ un is divergent if p = 1... u1 > u2 > u3 > u4 .. is called an alternating series represented by ∑ −1n −1 un n =1 ∞ Leibnitz test for convergence of an alternating series If u1 .. n→∞ 79 . u3 . u2 . e.e. is convergent. for all n i. then a series of the form u1 − u2 + u3 − u4 + . u4 . ∞ 1 Conclusion : ∑ p is convergent if p > 1 and divergent if p ≤ 1....are all positive and if (i) un > un+1 ... n=1 n Remark : We have used the behaviour of this series while solving problems on comparison test ∞ 1 to discuss the nature of a given series in comparison with a series of the form ∑ p n=1 n Session – 5 Alternating series If u1 . u3 .are all positive. u4 .ii : L O x As above z =M P f x dx 1− N pQ ∞ 1 1− p ∞ 1 Let p < 1 or 1 > p i..
un − un +1 > 0 ⇒ un > un +1 Also lim un = lim n→∞ n→∞ 1 =0 n Hence the given series is convergent by Leibnitz test.. Example – 1 Test for convergence the series. Property : Every absolutely convergent series is convergent but the converse is not true. but the converse is not true. n 1 1 n +1− n 1 Now un − un +1 = − = = >0 n n + 1 nn + 1 n + 1 n i. Remark : We have mentioned a property that “every absolutely convergent series is convergent”. is convergent. 80 . 1 − >> Here un = 1 1 1 + − + 2 3 4 1 We shall apply Leibnitz test.... is said to be absolutely convergent if the series of positive terms u1 + u2 + u3 + u4 +. If the given alternating series is convergent and the absolute series (series of positive terms) is divergent then the alternating series is said to be conditionally convergent. e.Absolute convergence and conditional convergence The alternating series u1 − u2 + u3 − u4 +.. That is a convergent series not being absolutely convergent. The series 1 − 1 1 1 + − +serves as an 2 3 4 example for the converse of the statement not being true. Generalised D’Alembert’s test If ∑ un is a general series and if lim n→∞ un +1 =| l | then ∑ un is absolutely convergent if | l | < 1 un not convergent if | l | > 1 and the test fails if | l | = 1.
. Example – 3 81 . 1 n n = 1 n 3 2 ∞ n =1 3 2 np p 2 is ∑ 1 n is a harmonic series of the type ∑ 1 = 3 > 1 convergent.The alternating series is convergent by Leibnitz test. Example –2 +. but the absolute series 1 + being ∑ ∞ 1 1 + + 2 3 n =1 1 is divergent ..series where p = 1 n Thus the alternating series is not absolutely convergent. for 2 2 3 3 4 4 (a) convergence (b) absolute convergence (c) conditional convergence 1 1 1 >> We have un = = 3 2 ∴ un +1 = 3 + 12 n n n n 1 1 Now un − un +1 = 3 − 2 3 n + 1 2 n 3 2 + 1 2 − n 3 n > 0 since (n + 1) > n. n→∞ n→∞ Also lim un = lim 1 The absolute series is given by 1 + We have un = ∞ 1 2 2 + 1 3 3 + 1 4 4 +.. un − un +1 > 0 ⇒ un > un +1 Test the series 1 − 1 + 1 − 1 = =0 n n Hence the given series is convergent by Leibnitz test. n =1 Thus the alternating series is absolutely convergent and it is not conditionally convergent. 2 n + 13 n i.. ( p . e..
. Example – 5 82 . Remark : The problem can also be done by applying Leibnitz test. n 1+ Hence we conclude that the given series is absolutely convergent. lim un ≠ 0 n 3 n →∞ 3 + 1 n →∞ Hence the given series is oscillatory. un > un +1 Find the nature of the series Also lim un = lim n→∞ 2 +1/ n 2 = i.. e. 1 1 1 1 1 1 1 1 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 3 4 5 6 7 8 1 The general term of this series is given by un = 2 n 1 ∑ 2 is a harmonic series ( p = 2 > 1) and it is convergent. Hence the given alternating series is convergent. 22 3 4 5 6 7 8 >> Consider the absolute series. But every absolutely convergent series is convergent.e. Example – 4 Show that the series 1 1 1 1 1 1 1 − 2 − 2 + 2 + 2 − 2 − 2 + is convergent.3 5 7 9 − + − + 4 7 10 13 2n + 1 2 + 1 1 2 n + 3 n + >> We have un = ∴ un +1 = = 3n + 1 3 + 1 1 3n + 4 n + 2n + 1 2n + 3 1 Now un − un +1 = − = >0 3n + 1 3n + 4 n + 13n + 4 3 i.
Discuss the nature of the series x2 x3 x4 + − + 2 3 4 >> The general term of the series is given by x− 1 1 x n − n− 1x n +1 − n un = ∴ un +1 = n n +1 We shall apply generalised D' Alembert' s test. the series becomes 1 − We have un = 1 1 1 + − + 2 3 4 1 1 ∴ un +1 = Clearly un > un +1 n n +1 1 Also lim un = lim =0 n→∞ n→∞ n Hence the series is convergent by Leibnitz test If x = −1 the series becomes 1 1 1 1 − −= − 1 + + + 2 3 2 3 1 1 = ∑ 1 is divergent. Here the series of positive terms ∑ n 2 n ∴ the given series is divergent (to − ∞) when x = −1 Thus the given series is convergent if − 1 < x ≤ 1 divergent when x = −1 −1− F G H I J K 83 . If x = 1.not convergent when | x | > 1 and the test fails if | x | = 1. Since absolute convergence implies convergence. that is –1< x <1. not convergent if | x | > 1 and the test fails if | x | = 1. If | x | = 1 then x = 1 or –1. the given series is convergent if |x|<1. un +1 1x n +1 − n n Now lim = lim n −1 n n→∞ un n→∞ − n + 1 1 x = lim −1 n→∞ n x = x n + 1 1 n Hence the series is absolutely convergent if | x | < 1.
Exercises : Find the nature of the following series : 1− 1 1 1 1 + − + 4 7 10 2 Examine the series 1 1 1 1 − + − +for 7 12 17 22 (a) convergence for (b) absolute convergence (c) conditional convergence (3) Show that the seies 1 1 1 1 1 1 1 1+ − − + + − − + 2 2 3 3 4 4 5 5 6 6 7 7 8 8 is convergent. Answers : 1. conditionally convergent. Convergent. 84 . not absolutely convergent. conditionally convergent if x = 1 and divergent if x = 1. Convergent 2. 1 1 1 (4) Show the series 1 − x + x 2 − x 3 + 4 7 10 is absolutely convergent if | x | < 1.
If the two reference lines in the Cartesian system are perpendicular to each other. PESIT Bangalore ANALYTICAL GEOMETRY IN 3 DIMENSIONS Introduction Any point in a plane can be located by means of two real numbers. called the coordinates.T.G.V. the system is called the rectangular Cartesian coordinate system. called the coordinate axes. The former is called the Cartesian System of coordinates and the latter the polar system of coordinates. the lines OX. F I G 1 In the above figure. y) in Cartesian system and P(r.Geetha Dept.U. of Mathematics. In space. Any point in a plane is represented by P(x. The system of coordinates is called the rectangular Cartesian coordinate system in three dimension. 85 .OY and OZ are the three axes with O as the origin of the coordinate system.M. Web-Based Education Engineering Mathematics – I (MAT-11) By Prof. θ ) in polar system of coordinates. we need three real numbers to locate a point and the reference lines. These numbers could either be distances from two fixed straight lines or a distance form a fixed point and an angle with reference to a straight line. which are three concurrent mutually perpendicular lines.
NM = y and MP = z where M is the foot of the perpendicular from P to the XOY plane and N is the foot of the perpendicular from M to the axis OX.z1) with reference to the new axes. y1.If P(x. z1) and Q(x2. then R ≡ 2 2 2 2 Example: Find the coordinates of the centroid of the triangle whose vertices are A (x1. 1. OP2 = OM2 + MP2 = ON2 + MN2 + MP2. Fig.2 If G is the centroid of the triangle ABC. the median from A. y1. B (x2. Shift the origin to P keeping the axes parallel to the original axes. the distance OP = Distance between two given points : Let P (x1.y1. = x2 + y2 + z2 x 2 + y 2 +z 2 Or. then ON = x.y. . z2) in the ratio p : q . If R is the middle point of PQ. Answer : in the ratio 2:1. y2. Distance formula : Then from the right angled triangle OMP. y3. z1). y2. y1. Then A≡ (x2. z2. z3). then it divides AD.x1. 86 . because ONM is a right angled triangle. z1) and Q (x2. z2) and C (x3. ∴ Distance PQ = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 + ( z 2 − z1 ) 2 Section Formula If a point R divides the line segment joining the points P(x1. px + qx1 py 2 + qy1 pz 2 + qz 1 . . y2. y2. then R ≡ 2 p+q p +q p +q x + x1 y 2 + y1 z 2 + z1 . z2) be the two given points.z) is any point in this system.
r m. 5. b.C. m.C.C.s for any real number s. x y = cos α .R. . Numbers proportional to the D.s) of the straight line. Note 1. then P is given by P ≡ (r l. c.R.s l. then a. = cos β .s are l.s 2.0. denoted by a. n 6.R. n are the D. l2 + m2 + n2 =1 for any line whose D. 1. For instance. m. 3 3 3 Direction Cosines: If a straight line makes angles α .s). Two parallel lines have the same D.s of a line.C. 0. n 87 . b.C. z) is any point on the line OA that makes angles α . m. If A(x.s of the line OP.s of the X-axis and s.C.C. 4. then cosα . c are D. r n) where l. r r z = cos γ r where OA = r using the distance formula r = ∴ cos2 α + cos2 β x 2 + y 2 +z 2 + cos2 γ = 1. b. y. m. x 2 + x3 y 2 + y 3 z 2 + z 3 . D ≡ 2 2 2 ∴ G ≡ x1 + x 2 + x3 y1 + y 2 + y 3 z1 + z 2 + z 3 .s are called Direction Ratios (D. cosβ and cosγ are defined as the Direction Cosines (D. They are denoted by l. β and γ with the three axes. Since D is the mid point of the side BC. n. then 3. 0 are the D. c are proportional to the D. If P is any point at a distance r from the origin. . β and γ with the coordinate axes.0 are its D. If a.
y1.s of the line are a a +b +c 2 2 2 .y1.s of AB are proportional to or one set of D.s of two given straight lines.x1. m. m2. z2) be the two points.4 Let l1. using Cosine formula for triangle OPQ. We have PQ2 = OP2 + OQ2 – 2OP. b a +b +c 2 2 2 .R. n2 be the D. Equating the two expressions for PQ2. r m.OQ cos θ . y1. n are the D. y2 . y2 . z2 . Let OP = r1 Then and OQ = r2. P ≡ (r1l1.i. r n) since A is the origin in the new system of coordinates.C.C. x2 – x1. z2 – z1.C.e. r2n2) = r12 + r22 – 2r1r2 (l1l2 + m1m2 + n1n2).3 Let A (x1. using the distance formula and l22 + m22 + n22 = 1 .z1) If l. r2m2. z1) with the axes parallel to the original axes.C. r1n1) and Q ≡ (r2l2.C. ∧ ∴ PQ2 = (r2l2 – r1l1)2 + (r2m2 – r1m1)2 + (r2n2 – r1n1)2 Remember : l12 + m12 + n12 = 1 Also.s : Fig 1. m1. Note: where θ = P O Q . Shifting the origin to A (x1.s of the line AB with length r units. z1) and B (x2. z2 – z1. y2.s can be taken as x2 – x1. ∴ l m n = = x 2 − x1 y 2 − y1 z 2 − z1 ∴ The D.y1. we get cos θ = l1l2 + m1m2 + n1n2 88 . r1m1. y1.s of the line joining two points: fig 1. y2 . Angle between two lines with known D. n1 and l2. we have B ≡ (r l. We get B ≡ (x2 . c a + b2 + c2 2 D. a b c a2 + b2 + c2 = = = = a2 + b2 + c2 2 2 2 l m n l +m +n ∴ The D.C. Draw OP and OQ parallel to the given lines passing through the origin..
c1 and a2.R. The expression for the angle between two lines in terms of their D. n be the D. 3.3) and (-4.s a) 3. 3. (7.C. 3. y1. b2. MN is the projection of AB on MN. -6. c2 ) is cosθ = a1 a 2 + b1b2 + c1 c 2 2 2 a + b12 + c12 a 2 + b22 + c 2 2 1 2. 1 2.s (a1. 1. 0) is a)* 5 unit b) √5 units c) 25 units d) 4 units If A ≡ (-1. y1.s of the line MN. -1) b) *0. 4) form a a) rectangle 89 . then MN = AB cos θ . z1) and (x2. 7).y1) +n(z2 . y2. d) None of the others The line which is perpendicular to the line with D.5 Let A ≡ (x1. y2. If θ is the angle between AB and MN. b1. z1). 10) and (-1. The condition for perpendicularity of two lines is l1l2 + m1m2 + n1n2 = 0 or a1a2 + b1b2 + c1c2 = 0.s : Fig. -2 b) Square c) 1.0. 4.z1).R.1.s of a line joining the points (x1. B ≡ (x2. -3. 1). -2 c)* Rhombus d) (0. Remember: The D. 1) b)* (0. 0.2. Projection of a line joining two points on a straight line with known D. z2) and let l. etc. 4) and B ≡ (1. 3) c) (1.s proportional to d) 1. 3.C. -6. -1.3 has D. -3. The vertices (5. z2) are x 2 − x1 ( x2 − x1 ) 2 + ( y 2 − y1 ) 2 + ( z 2 − z1 ) 2 Exercise: 1. 3. 2). m. The distance between the points (0. -4. -3.C. 0. (1.C. -3. -1) 1. -2. 2. then the mid-point of AB is a) (1. 2 2 2 ∴ MN = ( x2 − x1 ) + ( y 2 − y1 ) + ( z 2 − z1 ) l (( x2 − x1 ) + m( y 2 − y1 ) + n( z 2 − z1 ) ( x2 − x1 ) 2 + ( y 2 − y1 ) 2 + ( z 2 − z1 ) 2 ∴ MN = l(x2 – x1) + m(y2 .
y2. the entire line AB lies on the surface (1).5).(1) ax1 + by1 + cz1 + d = 0 and ∴a ( px 2 + qx 1 ) + b ( py 2 + qy 1 ) + c ( pz 2 + qz 1 ) + d ( p + q ) = 0 Or px + qx1 py 2 + qy1 pz 2 + qz1 a 2 p + q + b p + q + c p + q + d = 0 which implies that the point C lies on the surface given by equation (1). 1. equation. z1) and B(x2. Theorem: A linear equation Proof : If A(x1.8.-1) is joined to the mid-point M of the line joining P(1. y and z represents a plane.-3) and Q(3.2. line AB dividing it in the ratio p : q. then the point C ≡ 2 is a point lying on the p+q p+q p+q ax + by + cz +d = 0 in x. The point A (5. Normal Form of equation to a plane : Fig. ∴ By the definition of a plane.6 Let OL be drawn perpendicular (normal) to the plane from the origin. y1. z2) are two points on the surface represented by the above px + qx1 py 2 + qy 1 pz 2 + qz 1 . Since C is any arbitrary point lying on the line AB.5. . The angle between these two lines is a)* 90o b) 30o c) 60o d) 45o Planes: A Plane is a surface such that the line joining any two points on the surface lies entirely on the surface.4. 90 . equation (1) represents a plane. Since we have A and B are points lying on the surface ax2 + by2 + cz2 + d = 0 ax + by + cz +d = 0 ------------------------.
z –z1) ∴ The equation of the plane in the new system of coordinates is a (x + x1) + b (y + y1) + c (z + z1) + d = 0 or ax + by + cz + (ax1 + by1 + cz1+ d) = 0 ax 1 + by 1 + cz 1 + d a2 + b2 + c2 so that the perpendicular distance of the plane from P. or lx + my + nz = p is the normal form of equation to a plane.R. we have p = lx + my + nz.C. Since each ratio is equal to we get p= −d a + b2 + c2 2 a2 + b2 + c2 l +m +n 2 2 2 = a2 + b2 + c2 . then OL is the projection of OP on the normal Note: Comparing the general and the normal forms of equations of a plane. is Three Point form of Equation: 91 .C. z1) to the plane ax + by + cz + d = 0. ∴ Length of the perpendicular from the origin to the plane is d a +b2 +c2 2 3. In other words. y1. n. By the projection formula. m. y and z in the equation of a plane are proportional to the D. y.s of the normal to the plane. a. shift the origin to the point P keeping the directions of the axes unchanged. y. we get a b c d = = = l m n −p from which we observe the following: 1. z are a set of D. 2. b and c. the coefficients of x. z) is any point on the plane. To find the length of the perpendicular from any point P ≡ (x1.s of the normal. y – y1. P (x. Then any general point (X. If OL. the coefficients of x. the origin in the new system.s be l.Let its length be p and the D. Z) in the new system of coordinates changes to (x – x1. Y.
Note: The three-point form of equation to a plane gives the condition for coplanarity of four points (xi. If ax + by + cz + d = 0 is the equation of the plane. 2. we get a(x .x1) + b(y – y1) + c(z – z1) = 0 a(x1 – x2) + b(y1 – y2) + c( z1 – z2) = 0 and a(x2 – x3) + b(y2 – y3) + c(z2 – z3) = 0 x −x 1 y −y 1 y1 − y 2 y 2 −y 3 z −z 1 z1 −z 2 z 2 −z 3 This system of three equations has a non-trivial solution if x1 −x 2 x 2 −x 3 = 0.Given the three points If the plane given by (xi. then a= −d −d −d . yi. 3. Intercept form of equation: If a plane makes intercepts A. 3. i = 1. then they satisfy the equation of the plane. B.0. (0. Note: 92 .0). 4 as x1 −x 2 x 2 −x 3 x3 −x 4 y1 −y 2 y 2 −y 3 y 3 −y 4 z1 −z 2 z 2 −z 3 z 3 −z 4 = 0. and C with the coordinate axes.0) and (0. Eliminating ‘d’. b= & c= A B C x y z + + =1 . i =1. zi) . yi. A B C ∴ The equation of the plane reduces to This is known as the intercept form of equation to the plane. then the points (A. ax + by + cz +_d = 0 passes through these three points.0.C) lie on the plane. 2. zi).B.
0.C.0) b) 1/3 b) 0.1 a) (3.0.0 b) (3.0.1.1) from the plane 2x – y +2z =3 is a)* 0 a) 1. and (T) (F) (F) 10x . then aix + biy + ciz + di = 0.2) c) 1 c)* 0. -5.3z = 1 c) x + y – z = 14 b)* Parallel are c) coinciding d) inclined at an angle of True of False 1.2 . Equation of a plane passing through the line of intersection of two planes (a1x + b1y + c1z +d1) + k(a2x + b2y + c2z +d2) = 0 Exercise: 1. 2 . 3. 3.1) d) 2 d) 1. The point (1. The distance of the point (1.z = 5 and 6x +9y – 3z -7 = 0 a) Perpendicular 45o 2. is equal to the a1 a 2 + b1b2 + c1c 2 −1 angle between their normals and hence is cos 2 2 2 2 2 2 a1 + b1 + c1 a 2 + b2 + c2 2. is where k is a constant.0 d)* (2.0) d) 3x – 2y + 5 = 0 The D. i =1. 4. The Plane 2x + 3y – z + 6 = 0 93 .1. If the two planes are perpendicular.1. The planes 2x + 3y .1 c) (2.s of the normal to the XY-plane are Which of the following points lie on the same side of the plane 3x – y + 2x – 8 = 0 One of the following planes is perpendicular to the plane 2x – 3y +5z = 1 a) x – y – z = 4 b) 5y . 3.0.0. i = 1.15y . The angle between the two planes aix + biy + ciz + di = 0 . a1a2 + b1b2 + c1c2 = 0.5z + 24 = 0 are parallel 2x . 5.1.2) lies on the plane 3x + y + z = 1. The plane 4x – 3y +5 = 12 passes through the intersection of the planes x + y + 2z = 3.5y + z = 6 and 2.0.
nt + z1) where t plane form of equations and vice versa. n are proportional to ∴ The equations of the straight line are called the symmetric form of equations. c1a2 – c2a1 and a1b2 – a2b1.x1.z1. If a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 represent two intersecting a1x + b1y + c1z + d1 = 0 = a2x + b2y + c2z + d2 This is called two-plane form of equations .C. mt + y1. Any point lying on the straight line whose equations are x .s of the straight line l.C. z .s of the straight line are proportional to b1c2 – b2c1.y1. z1). 94 . When the two-plane form of equations to a straight line are given. y. x − x1 y − y1 z − z1 = = . l m n planes. The symmetric form of equations to a straight line can be converted to its two3. m. Note: 1. then the equations of the straight line of intersection are x − x1 y − y1 z − z1 = = . since the straight line is perpendicular to the normal to the two intersecting planes.Straight Lines: A curve in space is defined as the set of points common to two intersecting surfaces and it is represented by two equations . Two plane form of equation to a straight line: Two planes interest in a straight line and therefore the equations of the intersecting planes together represent a straight line.equations of the intersecting surfaces. y . the D. then D. z) is any point on a straight line that passes through the point (x 1. Symmetric form of equations: If P(x. A straight line is the curve of intersection of two planes. is a parameter . y1. 2. l m n is given by (lt + x1.
Then the two points (x1.(4) and ax + by + cz + d = 0.(1) ax2 + by2 + cz2 + d = 0 --------------------------. (2) a (x – x1) + b ( y – y1) + c (z –z1) = 0 a (x1 – x2) + b ( y1 – y2) + c (z1 –z2) = 0 Eliminating coplanarity x1 −x 2 l1 l2 y1 − y 2 m1 m2 a. ∴ we have the following conditions : ax1 + by1 + cz1 + d = 0 --------------------------. Also the normal to the plane is perpendicular to the two lines. let the equation of the plane containing them be ax + by + cz + d = 0.(3) = 0 --------------------------. (3) of z1 −z 2 n1 n2 and (4) . 95 . and we get the condition for in the form we get Eliminating d from equations (1). z1) and (x2. lines straight = 0 x −x 1 l1 l2 y −y1 m1 m2 z −z 1 n1 n2 and the equation of the plane containing them is given by = 0. the point of intersection can be found by solving the equations of the two lines. y1.Coplanarity of two straight lines: Consider two straight lines x − x1 y − y1 z − z1 = = and l1 m1 n1 x − x2 y − y 2 z − z 2 = = . c from this equation. Note: If two lines are coplanar. z2) lie on this plane. l2 m2 n2 If these lines are coplanar. b.(2) al1 + bm1 + cn1 al2 + bm2 + cn1 = 0 --------------------------. then the lines are either parallel or intersecting. y2. When they intersect.
Shortest distance between skew lines: When two lines are skew. the line of shortest distance is perpendicular to both the lines. ∴ By the projection formula.C. y2. 1. there exists a line perpendicular to both called the line of shortest distance. of the line of shortest distance are proportional to l1m2 – l2m1 m 1n2 – m2n1.Skew Lines: Two lines are said to be skew when they are non-coplanar. AB = (x2 – x1)λ + (y2 – y1)µ + (z2 – z1)ν We observe that the length AB is the projection of the line PQ on the line of shortest = ( x 2 − x1 )( m1 n 2 − m 2 n1 ) + ( y 2 − y1 )( n1l1 − n 2 l1 ) + ( z 2 − z1 )(l1 m 2 − l 2 m1 ) (m1 n 2 − m 2 n1 ) 2 + ( n1l 2 − n 2 l1 ) 2 + (l1 m 2 − l 2 m1 ) 2 This gives the shortest distance between the two skew lines. µ . n1l2 – n2l1. z1) Then the D. This is so. y1.7 x − x1 y − y1 z − z1 = = -------------. Equations of the line of shortest distance: The shortest distance line is coplanar with line (1) since it intersects line (1) at the point A. Point B is the point of intersection of the line of shortest distance and line (2). ν and Q ≡ (x2.s λ . when they are neither parallel nor intersecting. z2) . The length of the common perpendicular intercepted between the skew lines is the shortest distance between them. distance. AB. Fig.(1) l1 m1 n1 x − x2 y − y2 z − z 2 = = -------------(2) l2 m2 n2 Let the two skew lines be given by and with P ≡ (x1. ∴ We have 96 .
-13. 0) and d) (1. The lines x −1 = 2 − y z −3 = 2 2 and x −2 y −3 4 − z = = 4 5 2 are perpendicular . and 3x –2y +1 = 0 = y – 3z +10 The equations x −1 y − 2 = =z −4 2 3 represent the same straight line. -1) c)* (-1. 3 2 −4 97 . The straight line a)* Perpendicular with the plane 2. . The lines x −2 y −5 z +3 = = 3 −4 2 x − 4 6 − y 2z + 3 = = 3 4 4 a) Perpendicular 4. 0) b) (1. 2. The plane 2x + 3y + 3z –1 = 0 contains the line x − 2 y +1 z −1 = = . 0. -3) and is c)* (0. 0. 3 4 and the z – axis are skew lines. 4. 7. 3. The point The line (1. -1. 1. -1. -1) d) ( . ax + by + cz + d = 0 .x −x 1 y −y 1 z −z 1 x −x 2 =0 y −y 2 z −z 2 λ l1 µ m1 υ n1 and λ l2 µ m2 υ n2 =0 These two equations together represent the line of shortest distance between the skew lines given by (1) and (2). -1. b)* Parallel c) coincident d) Skew 1 3 1 3 1 3 The image of the point (1. Exercise: 1. 5. x y z = = 7 3 −2 y x = =z 3 −4 x − x1 y − y1 z − z1 = = a b c is d) at an angle of 45o b) Parallel to c) lies entirely on The point of intersection of the lines a) (-5. 1) in the plane x + y + z = 0 is a) (0. ) True or False 1. 2. 2) are 3. -37) b) (4. 3) lies on the line x −1 z = y −2= 0 0 x −4 y −6 = =2−z.
If P (x. The fixed line is called the axis. m. 1.6. then the D.C. Note: The angle α is called the semi-vertical angle of the cone and the point of intersection of the generators with the axis is called the vertex of the cone. Right Circular cone and Right Circular Cylinder: The solid generated when a straight line rotates about a fixed line is either a right circular cone or a right circular cylinder. y1. The shortest distance between the lines x −1 y − 2 z − 4 = = 2 3 −4 and x −1 y − 2 z − 4 = = −3 4 5 is 0. 98 .8 Let the axis of the cone be given by x − x1 y − y1 z − z1 = = l m n with its vertex as V≡ (x1. y1.s of the generator passing through V are proportional to x – x1.C. If α is the semi-vertical angle of the cone. the generators are inclined at an angle α with the axis while in a cylinder. The straight lines which do not intersect are called skew lines. z) be any point on the surface of the Cone. Equation of a right circular Cone: Fig. z – z1. 7. The rotating line is a generator of the solid. then cosα = or ( x − x1 )l + ( y − y1 )m + ( z − z1 )n ( x − x1 ) 2 + ( y − y1 ) 2 + ( z − z1 ) 2 l 2 + m 2 + n 2 2 1 1 [ ( x − x ) l + ( y − y ) m + ( z − z ) n] 1 cos2 α = (l 2 + m 2 + n 2 ) ( x − x1 ) + ( z − z1 ) 2 2 [ ] is the equation of a right circular cone with its vertex at (x1. n.s of its axis proportional to l. y. semi-vertical angle as α and the D. In the case of a cone. y – y1. z1). z1). the generators are parallel to the axis.
where θ = ∧ M AP l ( x − x1 ) + m( y − y 1 ) + n( z − z1 ) ∴ AM = ( x − x1 ) 2 + ( y − y1 ) 2 + ( z − z1 ) 2 . y. then AM is the projection of AP on the axis with M as the foot of the perpendicular from P to the axis. y1.1) with its axis parallel to the line x y −1 z + 1 = = is ____________. The semi-vertical angle of the right circular cone passing through the origin. Also AM 2 = AP2 – MP2 with MP = r ∑(x − x ) l + m + n 2 2 2 1 2 ∴ ( x − x1 ) 2 + ( y − y 1 ) 2 + ( z − z1 ) 2 = r 2 + [ l ( x − x ) + m( y − y ) + n( z − z )] 1 1 1 2 l 2 + m2 + n2 is the equation of the right circular cylinder.-3) 3. vertex at (1.Equation of a right circular cylinder: Fig/ 1. or AM = AP cos θ . z1) and r as its radius. Exercise Fill in the blanks : 1.8 x − x1 y − y1 z − z1 = = l m n be the axis of the right circular cylinder with A ≡ (x1. x +9 y −4 z −5 = = 10 −2 −3 is _______ units. 99 . π /3 2 2 −1 2. z) is any point on the surface of the cylinder.2. The radius of the right circular cylinder passing through the origin with its axis lying along the line 3. The vertex of the right circular cone whose axis is the line x +1 y − 3 = =z +2 −3 2 x= y −7 z +7 = −3 2 with the line as one of its generators is _______________ (2.1. If P ≡ (x.
4). 3.R.R.s of CA are 0. -2 ∴ The lines AB and BC have proportional D. 3. 2. 11 15 ∴cos A = D. -1 100 . -2) and (1. 0. B(-1. 4. -4. Find the centroid of the triangle ABC where A ≡ (-1. 3). 1).s of AB are –4. 1 7 3 17 ∴cos B = D. 2) . 3.R. -3 and D. 1. -4) and C ≡ (-1. −1 + 3 +1 2 + 4 + 0 1 − 4 + 3 . D. 0. 2. =1 1 3 α =β =γ −1 ∴ angle = α = cos 5.R. Hence the points A. . 2. The lines have a common point B.s. 2). -4. -2 and those of the line BC are 4. 4. 3. 2 0 + 16 + 6 36 25 = and D. 0. 4. 2.s of the line AB are 1. 0.R. 2. D. 0. 3 3 3 The centroid G is given by 4. 0). 4. -1). 1. Show that the points A (2. 1) and C(3.R. of AC are 0.Questions 1.s of BA are 4. -8 or 1. ∴ The distance of (1. 10) are collinear.s. Find the angles made by a line with the coordinate axes if it is equally inclined with the axes. -2 and D. B ≡ (-3. Find the distance of the point (1.R. 4. -3) from the middle point of the line segment joining (-3. 4. B and C are collinear. 6) The mid-point of the given line segment is (-1. B (3. -3) from this point is 4 +1 + 25 = 30 units 2. Find the angles of the triangle ABC formed by the vertices A (3.s of BC are 4. 4. 4. 2) and C(-1. or (1.R.s of CB are –4. We know that But cos2 α + cos2 β + cos2 γ is given .
d). d). d). 0). A’ ≡ (0. 0.R.10 Let the cube be as shown in the figure with one of its vertices O as the origin of the coordinate system and the axes along the three sides OA. Fig. (i) The diagonal AA’ of the cube and the diagonal O’A of a face of the cube intersect. AA’.C. 101 . d.s of OO’ and AA’ are d. B’ ≡ (d. A ≡ (d. d. d. d and 0. D. 0). cos −1 = 2d 2 3d 2 2d 2 = cos −1 2 3 (ii) The diagonal AA’ of the cube and the diagonal B’C’ of a face of the cube do not intersect. d. 15 cos −1 = 7 3 17 −1 and cos = 3 5 17 6. OC of the cube respectively. then its vertices are given by O ≡ (0. C ≡ (0. -d. d . d. d. d. d. . 0. d respectively cos −1 = d2 +d2 +d2 3d 2 We can see that the four diagonals of the cube are OO’. BB’. Find the angle between a diagonal of a cube and (i) an intersecting diagonal of a face (ii) a non-intersecting diagonal of a face. and 0. and –d. 0). Their D. d. d. C’ ≡ (d.s are respectively proportional to –d.∴cos C = 3 5 17 ∴ The angles are cos −1 = 11 . d ∴ Angle between these two diagonals is given by 3d 2 = cos −1 1 3 7. D. d).s of these two lines are respectively ∴ Angle between them is given by –d. 0). O’ ≡ (d. If d is the length of each side of the cube. and CC’. 1. 0. d. OB. B ≡ (0.R. Find the angle between any two diagonals of a cube. 0.
0. β .10. 2 6 If these lines are inclined at angles α . n n ⇒ m + 2n = 0 or 2m + n = 0 These with l + m + n = 0 give us two sets of l.s are 0. β . 1. d . we get 2 m m ⇒ + 2 + 2 + 1 = 0.R. m m + 2 +1 = 0 n n 2 Eliminating l from the given relations. -1. d. prove that cos2 α + cos2 β + cos2 γ = 2. d. OC’. d. d and d. whose D. with OO’.C.s are given by the relations l+m+n=0 and 2lm + 2nl – mn = 0.s are given by pl2+qm2+rn2 = 0 are and ul + vm + wn = 0 and (i) perpendicular if u2(q+r) + v2(r+p) + w2(p+q) = 0 (ii) parallel if u2 v2 w2 + + =0 p q r 102 . of the faces meeting at O. OB’.n proportional to –1. d. 3 Hence the required result. The angle between these two lines is cos −1 −1 + 2 + 2 π = .∴ Angle between them is given by cos −1 = 0+d2 −d2 3d 2 2d 2 = π 2 8.m. γ with the diagonal OO’ of the cube. whose D. γ then the angles are given by cos α = Also cos 2 β = 2 3 2d 2 6d 2 = or cos 2 α = 2 3 and cos 2 γ = 2 . Show that the straight liens whose D. d. -2. Given a cube as in Fig.R. 1. 6 3 10. 2 and 1. The diagonals of the faces meeting at O and passing through O are OA’. 9.s are d. make angles α .C. Find the angle between the two lines whose D. the diagonals passing through O. 0.
3). say. mi. they have the same D. we have l1l2 + m1m2 + n1n2 = 0 ⇒ pw2 + ru2 + qu2 + pv2 + rv2 + qw2 = 0 ⇒ u2 (q+r) + v2(r+p) + w2(p+q) = 0 (ii) When the two lines are parallel. When these lines are perpendicular. i= 1. ni . -1) and (1. 0. ⇒ 4p2v2w2 . Also find the distance of the point (0. We get (i) ( pv 2 + qu 2 ) m2 m + 2 pvw + ( pw 2 + ru 2 ) = 0 -----------------( 1) 2 n n m and has two roots. 2. 103 . 2. This is true when equation (1) has two equal roots.s.4(pv2 + qu2) (pw2 + ru2 ) = 0 ⇒ qru2 + rpv2 + pqw2 = 0 ⇒ u2 v2 w2 + + =0 p q r from (2) Planes: 11. 5).Eliminating l from the two given relations. -4) from the plane.C. 2 are the two straight lines given by the relations. we get n1 n 2 ll = 212 2 2 qu + pv rv + qw 2 and l1l 2 m1 m 2 = 2 rv + qw pw 2 + ru 2 2 ∴ m1 m 2 nn ll = 2 1 2 2 = 2 1 2 2 ---------------(2) 2 2 pw + ru qu + pv rv + qw where li. by eliminating m and n from the given relations and then getting the product of the roots and simplifying. 4. Find the equation of the plane passing through the points (1. n Equation (1) is quadratic in m1 m2 and n1 n2 m1 m2 pw 2 + ru 2 = Then the product of the 2 roots is given by n1 n2 qu 2 + pv 2 or m1 m 2 pw + ru 2 2 = n1 n 2 qu + pv 2 2 Similarly. (2.
-2. ∴ The equation is x + y – 2z – 3 = 0. 4. or the two planes are perpendicular. 10. Find the equation of the plane bisecting perpendicularly the join of the points (2.-4) from this plane is 0 +6 + 4 −5 14 = 5 14 .2.1). (-1.23= 0. 12. 10. Show that the points (3. -1). 5) and (1. The angle between this plane and 2x-y+z = 6 is given by cosθ = 0. (1. 3. -2. 1. Since the point (1.7.What is the angle made by this plane with 2x-y+z = 6 ? x −1 1 −2 y −2 −2 2 z −3 −4 2 The plane through the three given points is = 5. 104 . b. -2.4.1) lies on it. x The equation of the plane in intercept form is 23 3 + y z + =1 23 23 2 6 ∴ The x-intercept by the plane is 2 3 units. c. Find the equation of the plane containing them and also the intercept made by it on the x-axis. 7. 5). Also a. 1) and (1. 0) lies on this plane.3) and (4. (-1.1) is x −3 2 2 y −4 3 9 z −1 −2 −4 = 0 or 3x +2y + 6z . 3 13.1.2. Let the equation of the required plane be ax+by+cz + d = 0 Then the mid-point (3. the four given points are coplanar. 0) are coplanar. The plane containing (3. 1). = 0 or 2x+3y-z Perpendicular distance of (0. are proportional to 1.
we have 2l – m + n = 0 and l + 3m – 2n = 0 . 2 But this is perpendicular to the plane 4x + 7y +4z +81 =0 ∴ We have (4 + 5k) 4 + (7 + 2k) 7 + (4 + 5k) 4 = 0. ∴ k =− 7x-8y+7z-237 =0. If l. ∴ The equation of the required plane is Straight lines: 16.C. Fig.14. ∴ The equation of the required plane is 5x + 7y + 3z =6. Find the equations of the straight line of intersection of the planes 2x – y + z = 5 and x +3y – 2z +1=0 The line of intersection is perpendicular to both the planes. The plane 4x + 7y + 4z +81 = 0 is rotated through a right angle about its line of intersection with the plane 5x + 2y + 5z = 25. The equation of the plane through the intersection of the given planes is x + 3y – z – 4 + k(2x + 2y + 2z – 1) = 0 or (1 + 2k) x + (3 + 2k) y + (-1 + 2k) z + (-4 – k ) =0 But this is given to be perpendicular to x + y – 4z = 0 ∴ 1 + 2k + 3 + 2k + 4 –8k =0 ⇒ k = 2. n are its D. 15.s. 1. Find the equation of the plane in its new position. Find the equation to the plane perpendicular to x + y – 4z = 0 and through the intersection of the planes x +3y –z = 4 and 2x + 2y +2z = 1. The plane in its new position passes through the line of intersection of the two given planes.11 ∴ Its equation is given by or 4x + 7y + 4z +81 + k (5x + 2y +5z – 25) = 0 (4 + 5k) x + (7 + 2k) y + (4 + 5k) z + (81 –25k) = 0 3 . m. 105 .
∴ the equations of PP’ are x −2 y −3 z −4 = = = k (say) 2 −1 1 M is given by M ≡ (2k+1. 18. say. 2) and P’ be its image. M lies on the given line. 3k + 1.R. we get (2. ∴ M ≡ (-1. . 2) y z −3 = . 3k. Find the image of the point P(1. 3 −2 we have k + 3 (3k + 1) –2(-2k + 1) = 0 ⇒ k = − 13 −3 44 ∴M ≡ . 3 −2 ∴ k = 1. then the line PP’ is perpendicular to the plane and the plane bisects the line at the point. -2k + 3) . 3) ∴ P’≡ (-2.∴ l : m : n = -1 : 5 : 7 Choosing z = 0 in the equations of the planes and solving the resulting equations. 2) in the line x −1 = Let P ≡ (1. . -1. ∴ D. Then PP’ is perpendicular to the given line and its mid-point. k+4) and lies on the given plane. ∴ Equations of the straight line are x − 2 y +1 z = = −1 5 7 17. 3. . -2k+1. y z −3 = . -2k +3 –2 Since PP’ is perpendicular to the line x −1 = or k. 3k+1. 14 14 14 6 4 30 ∴P' ≡ . 0) as a point lying on both the planes. 4. -k+3. say. M . -1. . 7 7 7 1 14 106 . 4) in the plane 2x – y +z + 3 = 0 . Find the image of the point (1.12 If P’ is the image of the point P in the given plane. -1. 1. ∴ M can be written as (k + 1.s if PP’ are k + 1 –1. 5. Fig.
Its image in the given plane is (-5. ∴ Its equations are x +5 y +4 z +3 = = . (1. 2. 2). -1. 107 .s of the image line are 4. ∴ It must intersect the plane. The given line is not parallel to the given plane. c from the three equations.19. -3) and is another point on the image of the line. c is perpendicular to the line. Find the equation of the plane through the points (1.s a. 4 3 2 20. -1) is the point of intersection of the given line and the given plane and hence is a point lying on the image of the line in the plane. 3. -4.(2) x −1 = y −1 z − 2 = . b. Any point on the given line is (2k + 1. -1) and (3.) and parallel to the line x −1 = 1− y z −2 = . ∴ (-1. we get x −1 2 1 y 2 −2 z +1 3 =0 3 or 4x-y-2z-6 = 0 is the equation of the plane. then k = -1. ∴ D. 3) is a point on the given line.--. If this lies on the plane. 3k +2. ∴ We have 2a + 2b + 3c = 0 ----------. 0.R. 2 3 The equation of the required plane is of the form a(x-1) + by + c (z+1) =0 -------(1) This plane also passes through the point (3. b. ∴ a – 2b + 3c = 0 ---------------------(3) Evaluating a.R. 4k + 3). Find the image of the line x −1 y − 2 z − 3 = = 2 3 4 in the plane x + y + z + 3 =0 . 2 . 2. 2.2. −2 3 The plane is parallel to ∴ The normal to the plane with D.
R. Find the angle that the line z = 0. 1. 2 3 −6 Then M ≡ (2k + 1. x −1 2z +3 = y −2 = makes with the plane x – y + 2 4 The required angle is the normal to the plane. Find the distance of the point (1.s of the first line are – 3b + 4c=0 and they are 1. 2 3 −6 Fig. 23.21. The equations of the line through P parallel to the given line are x −1 y + 2 z − 3 = = . 2.s of the second line are got by solving the equations 4a – b + c = 0 & 14a −1 ∴ Angle between the lines is cos 2+2−4 π = 9 2 22. -6k + 3). -2 . x z =y= and 4x – y + z –5 =0 = 14x – 3y + 2 2 2. −1 Here θ = cos π 2 −cos θ . D. Find the angle between the lines 4z – 20. 2 . 3) from the plane x – y + z =13 measured parallel to the line x y z = = . 3k .2. 3). D. 1. 108 . where θ is the angle between the line and 2 −1 + 2 3 3 = cos −1 1 3 1 3 −1 ∴ Required angle is sin . Let this line intersect the given plane at M. -2. -2.R.13 Let P ≡ (1.
PM = 7 units . ∴ l 2 0 m −3 −3 n 1 = 0 ⇒3l + m − 3n = 0 -----------------------(2) −1 Solving equations (1) and (2) . ∴ The required distance 2k + 1 –3k + 2 – 6k + 3 = 13 ⇒ k = -1. 1). 12k+2). n are 2x + y – z +8 = 0. we have ∴ M ≡ (-1. ∴ Equations of the required line are x − 3 y + 4 z −1 = = . ∴ The required distance is ( 2 +1) 2 + (5 −1) 2 + ( 2 +10 ) 2 =13 units 25. If this lies on the given plane. 3 4 12 Any point on the given line is (3k + 2. -5. m. then 3k + 2 – 4k +1 +12k +2 = 5 ⇒ k =0.s. Find the distance of the point (-1. 2 −3 x − 3 y + 4 z −1 = = . 2) is the point of intersection of the given line with the given plane.C. -1. l m n The equations of the required line are its D. parallel to the plane 2x + y –z +8 = 0 and intersecting the line x − 3 y +1 = = z −2. 24. Find the line through (3. -10) from the point of intersection of the line x − 2 y +1 z − 2 = = and the plane x .Since M lies on the plane. -4. This line is parallel to where l. 2 −3 ∴ They are coplanar. 4k –1. ∴ (2. -5. 2 −3 1 109 . ∴ We have 2l + m – n = 0 -----------------(1) The line intersects x − 3 y +1 = = z −2 . we get l : m : n are 2 :-3 :1 .y + z =5. 9).
which when solved. 5.s are proportional . k1 –4k2 – 7 units 110 . 4k2 + 5) They are also given by 7. Find the points of intersection and 3 −1 −3 2 4 also the length intercepted. -1 with the two given lines. Let P and Q be the points of intersection of the line with D. 3) to the given line. Find the equations of the line through (2. 4 −5 2 27. 1. 1.C. ∴ k + 3 + 2 (2k – 7) + 3 ( 3k –1) = 0 ⇒ k =1 ∴ M ≡ (6.2) and Q ≡ (-3k2 – 3. ∴ D. k1 + 7. k2 = -1 . 2 3 and perpendicular to the line Also find the length of the perpendicular from (2. A line with D. We get the equations 19k1 + 26k2 = 12 and k1 – 6k2 = 8. 1. If M is the foot of the perpendicular from (2.R. -k1 –2k2 + 4. ∴ Length intercepted 66 for some k1 & k2 ∴ D. ∴ Points of intersection are P ≡ (7. 0) and Q ≡ (0. -1 is drawn to intersect the lines x −1 y − 7 x +1 y − 3 z − 5 = = z + 2 and = = . 3) x −5= y +6 z −2 = .R. 3k –1 with P ≡ (2. 2k6.s of PM are k + 3.26. 2k2 + 3. 1) .s proportional to 7. 4. 1.R. -4. 4. ∴ These two sets of D. M ≡ (k+5. Then P ≡ (3k1 + 1.s 7. 4. 3) . 5) ∴ Perpendicular distance = Equations of the perpendicular line are 3 3 x − 2 y −1 z − 3 = = .R. k1. 3). -1.s of PQ are 3k1 + 3k2 +4. yield k1 =2.3k+2). 1. 2k – 7. But PM is perpendicular to the given line.
Line 2 has D. x −1 y − 2 z − 3 = = and 2 −4 7 29. 2 5 −6 What do you conclude here? The D.28.s –64. x −4 y −2 z −3 = = 2 −3 4 and Also find the equations of the shortest Line 1 has D. -2. -2. 111 . Find the shortest distance between the lines 5x +3y +7z +6 = 0 = 6x +7y -5z +14.s 11. ∴ Shortest distance length 11 10 2 (4 − 0) + ( 2 + 2) + (3 − 0) = 6 units 15 15 15 Equations of the shortest distance line are given by x −4 2 11 y −4 −3 10 z −3 x 4 = 0 = −64 2 11 y +2 67 10 z 17 2 or 46x – 40y – 53z + 55 = 0 = 4x – 35y + 153z – 70. 3) lying on it. 10. ∴ The lines are coplanar. -3. 0) on the line of shortest distance.R. ∴ The shortest distance line has D. 2. 2 .C. Hence they intersect. 17 and (0. Shortest distance = 11( 2 − 1) + 9(1 − 2 ) + ( 2 − 3) 206 =0.s 2. 2. 0) lying on it.s of the shortest distance line are proportional to 11. But the shortest distance is the projection of the join of (4. distance line.R. 2. Find the shortest distance between the line x − 2 y −1 z − 2 = = . 67. We can see that they are not parallel. 4 and (4. 9.R. 3) and (0.
This is parallel to the line x −1 y − 2 z − 3 = = .30. 3). 15 Cones and Cylinders : 31. Find the plane through the line to the line 5x + 3y + 7z +6 = 0 = 6x + 7y – 5z +2 and parallel x −1 y −1 z − 3 = = . 2. ∴ The equation of the required plane is 11x +10y + 2z + 8 = 0. 2 −3 4 ∴ (5 + 6k) 2 + (3+7k) (-3) + (7-5k) 4 = 0 ⇒ k = 1. ∴ It is 45 = 3 units. cos 2 α = x2 + y2 = (z-3)2 tan2 α given that the vertex is at [ 0( x − 0) + 0( y − 0) + ( z − 3)] 2 x 2 + y 2 + ( z − 3) 2 2 ⇒ x 2 + y 2 + ( z − 3) cos2 α = ( z − 3) 2 ⇒ x 2 + y 2 = ( z − 3) 2 tan 2 α [ ] 112 . is given by (0. 2. Hence find the length of the shortest distance 2 −3 4 between the given lines. 0. 2 3 1 2 [ 2( x − 1) + 3( y − 1) + ( z + 3) ] 3 cos 30 = = 4 14 ( x − 1) 2 + ( y − 2) 2 + ( z + 3) 2 or 21 ( x − 1) + ( y − 2 ) + ( z + 3) = 2[ 2 x + 3 y + z − 5] 2 2 2 [ [ ] ] 2 is the equation. 30o as its semivertical angle and its axis along the line 2 o x −1 y −1 z + 3 = = . 32. Any plane through the given line is (5 + 6k) x + (3 + 7k) y + (7 – 5k) z + (6 +2k) = 0. 3) on to this plane gives the shortest distance between the two given lines. Show that the equation of the right circular cone with semi-vertical angle as α and its axis along the z –axis. Find the equation of the right circular cone with vertex at (1. -3). Now the length of the perpendicular from the point (1.
1) whose axis is along the line cos 2 x −1 y − 2 z −1 = = . 0 1 0 x –2 = 0 = z The equations of the axis are Here A = (2. = (x – 2)2 + (y +1)2 + (y . [ 2(0 − 1) + 1(0 − 2) − 2(0 − 1)] 2 θ= 9 x6 = 2 27 If (x. ∴ (x –3y + 2z -7)2 = 14[(x-2)2 + (y + 1)2 + (z . then AM 2 [ ( y − 2) − 3( y + 1) + 2( z − 1) ] 2 = 14 . then 2 [ 2( x − 1) + 1( y − 2) − 2( z − 1)] = 27 9 ( x − 1) 2 + ( y − 2 ) 2 + ( z − 1) 2 2 [ or 2 ( x − 1) + ( y − 2) 2 [ 2 ] + ( z + 1) ] = 3[ 2 x + y − 2 z − 2] 2 2 is the equation. is x2 + y2 – 4x – 2z + 4 = 0. by Pythagoras Theorem. Find the equation of the right circular cylinder whose axis is along the line x −2 = y +1 z −1 = −3 2 with the radius of its base as 3 units. 1. Show that the equation of a right circular cylinder whose axis is along –1 of radius unity. 0. 113 . (check fig. x − 2 y − 0 z −1 = = .1)2] –126 is the equation of the right circular cylinder. z) is any general point on the surface of the cone.9 ) But AM2 = AP2 – PM2 . 34. 35. ∴ We have x2 + z2 – 4x – 2z +4 = 0 as the equation. y.1)2 –9 . Given that the vertex of a right circular cone passing through the origin is at the point (1.33. Let A ≡ (2. 1) and r = 1. 1).2.-1. 2 1 −2 Find its equation.
-1) 114 . 2.36. PL perpendicular to the axis. . 0. QM = r.1) b) (-1. 0. 7 7 7 d) 3 −2 6 . 2 7 . 4) is a) (1. 1. 0) are a) 3. 3.s of the line perpendicular to the plane ABC formed by A≡ (1. 4. 6 b) 3 7 . γ are the angles made by a line with the coordinate axes. c) 3 2 6 . the value of sin2α + sin2 β + sin2 γ is a) 1 b) 2 c) 3 d) 4 2. then AL2 = AP2 – PL2. 6 7 . The foot of the perpendicular from the origin to the line joining (1. If QM perpendicular to the axis. B≡ (3. If α . ] is the equation of the right circular cylinder. radius and r2 = AQ2 – AM2 =4+0+4− [ 3(2) + 4(0) + 5(−2)] 2 50 = 192 . β . 2. 3. 2 [ 3( x − 2) + 4( y − 3) + 5( z − 4)] or 50 ( x − 2 ) + ( y − 3) + ( z − 4 ) − (3 x + 4 y + 5 z − 38) 2 = 384 2 2 2 [ 50 = ( x − 2) 2 + ( y − 3) 2 + ( z − 4) 2 − 192 / 25. z) is any general point on the surface of the cylinder. -1) and C≡ (1. 2) is a point on the surface of the cylinder. 0. y. The D. A straight line is inclined at angles 45o and 60o with the y and the z-axes respectively. 3.C. . 4) and Q ≡ (4. Find the equation of a right circular cylinder whose axis is a generator is x − 4 y −3 z − 2 = = 3 4 5 x −2 y −3 z −4 = = and 3 4 5 A ≡ (2. 7 7 7 4. 0. 4. 1). TEST : 1. 25 If P ≡ (x. 1) c) (1. Its inclination with x-axes is a) 60o b) 45o c) 75o d) 90o 3. 3) and (2. -1) d) (-1.
-1. 3) b) (4. 2. 10) and C(9. the value of λ b) 2 c) 3 4) 4 9. -1. -1. -2. 2) is a) x-y+z1=0 is a) 1 is a) x+2y-2z=0 b) x+2y-3z+7=0 2π 3 b) x+y-z+1=0 c) x+y-z-1=0 d) x-y-z-1 =0 8. (-3. 2). 2) and the origin are coplanar. 4). Traingle ABC is a) Isosceles 2y +9 = 0 is a) 1 2 b) Right angled c) equilateral d) Right angled isosceles 6. 4) and parallel to the plane x+2y-3z-7 = 0 c) x+2y-3z-15=0 5 6 d) x+2y-3z+15=0 1 6 10. The foot of the perpendicular from (3. 1) 13. 2.5. -1). 1. 5. The equation of the plane through (1. 4) and (4. -1. B(6. 2) are a) c) x −3 y −2 z −4 = = 4 5 2 b) x-4y+3z+6=0 c) 11x+6y+13z-44=0 d) 12x+7y+14z- b) x − 3 = d) x − 1 = y −2 z −4 = 3 2 x −4 y −5 z −2 = = 3 2 4 y −3 z +2 = 3 −2 14. 1) and (4. The equations of the straight line joining the points (3. 4) are the vertices of a triangle. 2. The points A(3. (7. 5. 5. -2. The equation of the plane through the line of intersection of the planes =0 and 2x+2y-3z-4 = 0 and parallel to the straight line x = 2y = 3z is 3x-4y+5z-10 115 . The perpendicular distance between the planes 2x – 2y + z + 3 = 0 and 4x – 4y + 1 6 1 3 2 3 b) c) d) 7. The acute angle between the plane 2x-y+z = 7 a) and x+y+2z-11 = 0 is d) cos −1 π 3 b) c) cos −1 11. 4. 1. If the points (1. The equation of the plane through the points (2. -1) d) (3. The equation of the plane through the line of intersection of the planes 2x3y+4z+1=0 and x+y+z-5=0 and perpendicular to the latter plane is a) 3x-2y+5z-4=0 49=0 12. 3. 2). 1. (λ . 1) to x+3y-2z+7 = 0 is a) (2. -1) c) (2.
A(1. The equation of the plane through the point (2. -4.1) and parallel to the line x-2y-z +5 = 0 = x+y+3z-6 are a) c) x − 2 y − 3 z −1 = = 5 −4 −3 x − 2 y − 3 z −1 = = −5 4 3 b) d) x − 2 y − 3 z −1 = = 5 4 −3 x − 2 y − 3 z −1 = = 5 −4 3 16. The plane 4x+5y-z+15=0 ______________ the line a) Perpendicular b) Contains x −1 y + 3 z − 4 = = 2 −1 3 c) Parallel to d) is at 45o with 116 . 3. Equations of the straight line through the point (3.a) x+20y –27z=14 b) x-20y-27z+14=0 c) x-20y +27z=14 +14=0 d) x +20y – 27z 15. -3) and perpendicular to two lines whose D. 4. -1. 3). 5) and parallel to the line 2x=3y=4z is a) 29x-27y –22z=85 b) 29x+27y-22z-85=0 c) 29x-27y-22z+85=0 d) 29x-27y+22z85=0 20. The equations of the median through A are a) c) x −1 y − 2 z − 3 = = −1 1 2 x − 4 y +1 z − 3 = = 1 −1 4 b) d) x − 2 y −1 z + 5 = = −1 −1 2 x −1 y − 2 z −1 = = 1 −1 4 19. Equations of the straight line through the point (2.s are 1:2:2 a) c) and 2 : 1 : 3 are b) d) x −3 y −4 z −3 = = 4 1 −3 x −3 y −4 z +3 = = 4 1 −3 x −3 y −4 z +3 = = 4 −1 −3 x −3 y −4 z −3 = = −4 1 −3 18. B(2. 2. 1. 3) are the vertices of a triangle. 1. -5) and C(4. -3) and perpendicular to the plane 3x+2y-4y-7=0 are a) c) x − 2 y −1 z + 3 = = 3 2 4 x − 2 y −1 z + 3 = = 3 2 −4 b) d) x − 2 y −1 z − 3 = = 3 2 4 x − 2 y −1 z −1 = = 3 2 −4 17.C. Equations of the straight line through the point (2. 0) & (3. -1.
21. 2) c) (3. The equation x2+y2-z2=0 represents a a) Sphere b) Cone a) Cylinder b) cone c) Cylinder d) Two planes d) none of the three 25. The straight line straight lines. a) c) b) (2. represents a c) Sphere 117 . 3. 3. The shortest distance between the straight lines x+y = 0 = y+z and x+y = 0 = x+y+z-a is a) a 6 b) 2a 3 c) 2a 3 d) 2 a 3 24. 3. 2) x −1 y − 2 z − 3 = = is non-coplanar with one of the following 2 3 4 3x − 5 y − 3 3 z − 13 = = 2 1 4 x − 2 y − 8 z − 11 = = 2 −1 2 b) x −2 y −4 z −5 = = 3 4 5 d) x = y + 1 z −1 = 3 2 23. The point of intersection of the straight lines x +2 y −5 z +2 = = is 2 −1 2 x −1 y −1 z + 1 = = and 1 2 3 a) (2. in space. 1) 22. 3. The equation y2+z2=1. 1) d) (3.
Angle between the radius vector and the tangent pedal equations of polar curves only. P-series. parametric form and Polar form. Tracing of standard curves in Cartesian form. PART . 5.coplanar lines-shortest distance between skew lines. PES. Polar curves .planes . Cauchy's integral test (All tests without proof) for series of positive terms. Infinite Series Convergence.Homogeneous and Non. A. length.Eshwar. Leibnitz's theorem (without proof) . secn x. Differential calculus Determination of nth derivatives of standard functions. Oriented problems. Applications to find area. tann x.T. cosec n x and sinm x cosn x .Homogeneous. cosn x. Alternating series.D 4.-Illustrative examples from engineering field. volume and surface area. Belgaum WEB-BASED EDUCATION Engineering Mathematics – I (I Semester) (VTU Sub.Code:MAT -11) Syllabus PART –A 1. divergence and oscillation of an infinite series. Cauchy's root test. Analytical geometry in 3 dimensions Direction cosines and direction ratios . (Use of initial condition should be emphasized). right circular cone and right circular cylinder. Comparison Test.variables separable . Absolute and conditional convergence. Exact equations and reducible to exact form.Jacobins . Leibnitz's test (without proof) 118 .Evaluation of these integrals with standard limits . Differential equations Solutions of 1st order & 1st degree equation. cotn x. Partial differentiation: Euler's Theorem. Differentiation of Composite and implicit functions .problems only. Integral calculus Reduction formulae for the functions sinn x.B 2. Raabe's test. -Illustrative Engg. Linear and Bernoulli’s equations.problems. Total differentiation. D'Alembert's ratio test.straight lines -Angle between planes/straight lines . PART – C 3. PART .Notes by Prof. Orthogonal trajectories of Cartesian and polar forms. Mandya Visvesvaraya Technological University.Errors and approximations.
. Series solutions of Differential Equations in all branches. E&C. Engg. Volume and Surface area. John Wiley & Sons. Filed Theory in E&E. Mandya 119 .D Differential equations 4 5 PART . PESCE. Description of Periodic phenomena • Contents of Part-B Differential Calculus (Part-B) A. Mathematical Modeling in circuits theory for E&E. Basic of Applied Mechanics.. Deflection of beams. Maths -I No (Code:MAT 11) 1 PART –A Analytical Geometry in 3 dimensions 2 PART .. Tracing of Curves for all branches of Engg.D Infinite Series Theory of signals in E&E. E&C. IS&E.Grewal. Higher Engineering Mathematics by B. Area. Reference book: 2.Whilring of shafts in Civil/ Mech.Engg Sl. in Civil & Mech. Engg. Engg.T. • Engineering Applications of MAT-11 at glance Applications Wave Theory in E&C. Spreading the functions & Electron beam deflections in E&C and E&E Basic Thermodynamics for Mech. CS&E. Applied Mechanics in Civil Engg. Advanced Engineering Mathematics: .. deflection torsion.July 2001.. Drawings in Mech.Engg.B Differential calculus 3 Part – C Integral Calculus PART .by E.S. Mechanical Vibrations.Eswara. Bending. Kreyszig. 36th Edition . 6th Ed.Text Book: 1.
1. Also. the idea of differential coefficient of a function and its successive derivatives will be discussed.Chapter 1 Successive Differentiation Chapter 2 Polar Curves Chapter 3 Partial Differentiation Lesson-1 n derivative of standard functions th Lesson-2 Leibnitz` s theorem Lesson-1 Angle between polar curves Lesson-2 Pedal equations Lesson-1 Euler’s theorem Lesson-2 Lesson-3 Total derivative. the computation of nth derivatives of some standard functions is presented through typical worked examples.When we have a formula for the distance that a moving body covers as a function of time. Definition of derivative of a function y = f(x):- • Fig.1.Differential calculus (DC) deals with problem of calculating rates of change. Applications to differentiation of Jacobians. Slope of the line PQ is f ( x + ∆x ) − f ( x) ∆x 120 . DC gives us the formulas for calculating the body’s velocity and acceleration at any instant.0 Introduction:. Errors composite & & Approximations implicit functions Chapter – 1 LESSON -1 : Successive Differentiation • In this lesson.
y 2 . y1 . The instantaneous velocity and acceleration of a body (moving along a line) at any instant x is the derivative of its position co-ordinate y = f(x) w. nth order derivative: D n y iv y ′ . D 2 y . y 3 …...1 1. • Notations: dy d 2 y d 3 y dny i.e. Velocity = dy = f ′(x) dx --------.1 Successive Differentiation:The process of differentiating a given function again and again is called as Successive differentiation and the results of such differentiation are called successive derivatives.... f ′′(x ) .….………..1) dx f ( x + ∆x ) − f ( x) lim = ∆→ -------..(1) x 0 ∆x = lim (Average rate change) ∆ → x 0 = Instantaneous rate of change of f at x provided the limit exists. y ′′ . nth order derivative: y n • Successive differentiation – A flow diagram Input function: y = f (x) derivative) Input function y ′ = f ′(x) order derivative) Operation Output function y′ = d df = f ′(x ) dx (first order dx Output function y ′′ = d2 f = f ′′( x) (second dx 2 Operation d dx 121 .……. .. nth order derivative: y ( n ) v. . nth order derivative: f n (x ) iii Dy.The derivative of a function y = f(x) is the function f ′(x) whose value at each x is defined as dy = f ′(x) = Slope of the line PQ (See Fig.t x.r. i. f ′′′(x) . .. • The higher order differential coefficients will occur more frequently in spreading a function all fields of scientific and engineering applications. D 3 y . y ′′′ .(2) And the corresponding acceleration is given by Acceleration = d2y = f ′′( x) dx 2 ---------.(3) • Session . nth order derivative: dx dx 2 dx 3 dx n ii f ′(x) .…….
y 2 = − cos x sin x y 2 = − tan xy1 + cos2 xy .(1) w.t.x dx Again differentiating y1 = gives d2y y2 = = [ cos(sin x)( − sin x) + cos x(− sin(sin x) cos x ] Using product rule dx 2 d2y y2 = = − sin x cos(sin x) + cos 2 x sin(sin x) 2 dx [ ] cos x cos(sin x) + cos 2 x sin(sin x) i.t. prove that + tan x + y cos 2 x = 0 2 dx dx Solution: Differentiating y = sin(sin x) --------. • 1.x. (1) and (2) or or y 2 + tan xy1 + cos xy = 0 2 [ ] d2y dy + tan x + y cos 2 x = 0 2 dx dx 122 .(2) dy w.r.1 Solved Examples : d2y dy 1. using Eqs. cos x dx -------------. we get y1 = dy = cos(sin x).r. successively.Input function y ′′ = f ′′(x ) derivative) Operation Output function y ′′′ = d d3 f = f ′′′( x ) (third order dx 3 dx ---------------------------------------------------------------------------------------------------------------dn f Input function y n −1 = f n −1 ( x) Operation Output function y n = n = f n (x) (nth order dx d derivative) dx Animation Instruction (Successive Differention-A flow diagram) Output functions are to appear after operating Operation d dx on Input functions. If y = sin(sin x ) .e.
(4) From (2). by actual division of ax+b by c x+d. dy dy = cos t and = p cos pt . we get 1 − x 2 2 y1 y 2 + ( y1 ) 2 ( −2 x ) = p 2 ( −2 yy1 ) .r. y = sin pt . [ ] 2 d2y dy 3.(ax + b) 2 .(2) ---------. If y = Differentiating (1) successively thrice.e. So we find.(1)where k = b − c c cx + d c c 2. If x = a (cos t + t sin t ) . as (cx + d ) ad 1 a ad −1 y = a + b − = + k ( cx + d ) ---------. y = a (sin t − t cos t ) . Differentiating this equation w. Squaring on both sides c o ts (by data) ) ( ) ( ) Canceling 2 y1 throughout. show that 2 y1 y 3 = 3 y 2 (cx + d ) (ax + b) Solution: We rewrite y = .(3) ---------. (3) and (4) we get d2y −3 = y 2 = −2kc 2 ( cx + d ) 2 dx d3y −4 = y 3 = −6kc 3 ( cx + d ) 3 dx 2 y1 y 3 = 2 {− kc (cx + d ) −2 }{− 6kc 3 (cx + d ) −4 } 2 y1 y 3 = 12 k c (cx + d ) 2 4 −6 [ ] 2 y1 y 3 = 3 −2kc 2 (cx +d ) −3 2 Therefore 2 y1 y 3 = 3 y 2 . find dx 2 ( ) ) 123 . Prove that (1 − x 2 ) 2 − x + p2 y = 0 dx dx Solution: Note that the function is given in terms a parameter t. this becomes 1 − x 2 y 2 − xy1 = − p 2 y ( or 1 − x 2 y 2 − xy1 + p 2 y = 0 d2y dy i. dy y1 = = dx ( dy dt = pc o s t p . If x = sin t . we get dy −2 = y1 = −kc ( cx + d ) dx ---------.t x. (1 − x 2 ) 2 − x + p2 y = 0 dx dx d2y 4. as desired. so that dt dt dx dt 2 2 2 2 2 2 ( y1 ) 2 = p cos2 pt = p (1 − sin2 pt ) = p (1 − 2y ) cos t 1 − sin t 1− x 2 2 2 2 ∴1 − x ( y1 ) = p 1 − y .
a 2 2 1 ( )( ) ( )( ) ( ) • Problem Set No. a 2 y 22 ( ) . If −1 ( ) d y h − ab = 2 dx ( hx + by ) 3 2 2 ( ) d2y dy + 2k + k2 +l2 y = 0 2 dx dx 2 d y dy . If ax 2 + 2hxy + by 2 = 1 . prove that 1 + x 2 +x =0 2 dx dx ( ) ( ) x = a( c o ts+ l o ga nt 2 ) . y = a sin t . prove that d 2 + tan y dy = 0 dx dx 2 Ans: sin t a cos4 t Ans: d2y . when x = a cos 3 θ . If y = tan 6. If y = e ax sin bx .1 for practice. If y = Ae −kt cos( l t + c) . so that a y = cosh ( x a) = 1 + sin h ( x ) = 1 + y . show that 4. If y = a cosh x ( a ) . prove that a y 2 2 2 = 1 + y12 Solution: y1 = dy = a sinh x 1 = sinh x . Find 124 . a a dx 2 2 2 2 2 ay ∴ 2 = cosh x a i. 1. dy dt dx dt = a ts i nt = ta n t a tc o ts d2y 1 dt 1 = sec 2 t = sec 2 t = 2 3 dx dx at cos t at cos t 5. 1. prove that 3. find d y t dx 2 2 2 y ( sinh x ) . y = b sin 3 θ 2 dx b cos ecθ sec 4 θ 3a 2 7. as desired.e. prove that y 2 − 2ay1 + a 2 + b 2 y = 0 2. If y = log x + 1 + x 2 5.Solution: dy = a ( − sin t + t cos t + sin t ) = at cos t dt dy = a ( cos t + t sin t − cos t ) = at sin t dt d y ∴ = d x Hence. and a a a dx d2y y2 = = cosh x 1 .
x d3y −1 8. prove that x Ans: -3/2 d2y dy +2 − xy = 0 2 dx dx • Session -2 1. If xy = e x + be −x . Table : 1 125 . If x = 2 cos t − cos 2t .2 Calculation of nth derivatives of some standard functions • Below. we present a table of nth order derivatives of some standard functions for ready reference. where y = tan 2 3 dx 1− x Ans: d2y π x = 2 2 dx (1 − x ) 2 1 + 2x 2 5 2 9. Find 10. Find . x = 2 sin t − sin 2t .
iii. we get y 2 = m me mx = m 2 e mx Similarly.Sl. 7. ii. Consider e mx . m( m −1)( m − 2 ).t x. No 1 2 3 y = f(x) yn = dny = Dn y dx n e mx a mx ( ax + b ) m m n e mx n m n ( log a ) a mx i.. r n e ax cos( bx + c + nθ ) .t x. 10. we get y1 = me mx . we get y3 = m 3 e mx ( ) 126 . for all 0 if (n!) a n if m = n m! x m −n if ( m − n )! m< n m< n 4 5. Hence. Again differentiating w. 6.r.r. 8. as only the above functions are able to produce a sequential change from one derivative to the other. iv. ( m − n +1) a n ( ax + b ) m −n m.. Let y = e mx . Differentiating w. r = a2 + b2 r = a2 + b2 θ = ta n− 1 (b a) θ = ta n− 1 ( b a) We proceed to illustrate the proof of some of the above results. 9.. • 1 ( ax + b ) (−1) n n ! an n +1 (ax + b) (−1) n (m + n −1) ! n a (m −1) !(ax + b) m +n ( − ) n −1 ( n −1) ! n 1 a (ax +b) n 1 ( ax + b ) m log( ax + b) sin( ax + b) cos( ax + b) e ax sin( bx + c ) e ax cos( bx + c) a n sin(ax + b + n π ) 2 a n cos(ax + b + n π ) 2 r n e ax sin( bx + c + nθ ) . in general we cannot obtain readymade formula for nth derivative of functions other than the above. 1.
.e. ( m −n +1) ( ax + b ) m −n a n for all m. No-5 of Table-1 ) y= 1 = ( ax + b ) −m ( ax + b ) m y 2 = ( − 1)( m ) − ( m + 1)( ax + b ) Similarly. And hence we get y n = m ( m −1) ( m − 2 ) ………... Case (i) If m = n (m-positive integer)..e. ( ax + b ) m (See Sl.(i.y 4 = m 4 e mx …………….r.t x.e m( m −1)( m − 2 )..t x − m −1 − ( m +1) y1 = −m( ax + b ) a = ( − 1) m( ax + b ) a [ − ( m +1) −1 3 y 4 = ( − 1) m( m + 1)( m + 2)( m + 3)( ax + b ) …………………………… n y 3 = ( − 1) m( m + 1)( m + 2 )( ax + b ) − ( m+ 4 ) a = ( − 1) m( m + 1)( ax + b ) 2 −( m+3 ) ] − ( m+ 2 ) a2 a3 a4 −( m + n ) y n = ( − 1) m( m + 1)( m + 2 ). 1 ( ax + b ) m Let i. ( m + n − 1)( ax + b ) an This may be rewritten as 127 . ( m − n + 1)( ax + b ) a n becomes [ ] ..2. Thus D n ( ax + b ) m = 0 if ( m < n ) Case (iii) m −n If m>n. y n = ( n!) a n Case (ii) If m<n. we get y 2 = m ( m −1) ( ax + b ) m−2 a 2 Similarly.3.. the right hand site yields zero. y1 = m ( ax +b ) m −1 a . (See Sl. we get y3 = m ( m −1) ( m − 2 ) ( ax + b ) m−3 a 3 …………………………………. And hence we get d n mx y n = m n e mx ∴ [e ] = m ne mx . Again differentiating w... ( m − n +1)( m − n ) ! ( ax + b ) m −n a n ( m − n) ! m! yn = ( ax + b ) m−n a n ( m − n) ! = 3. if n>m) which means if we further differentiate the above expression..r. we get 4 Differentiating w. No-3 of Table-1 ) let y = ( ax + b ) m Differentiating w.. n dx 2.. then y n = m( m − 1)( m − 2 )..t x.1 ( ax + b ) n −n a n i..then the above expression becomes y n = n ( n −1) ( n − 2 ) ……….r.
.1.. ( m +1) m( m −1) ! ( ax + b ) −( m +n ) a n ( m −1) ! n ( −1) ( m + n − 1) ! a n yn = or ( m −1) !( ax + b ) m +n 4. (a) log( 9 x 2 −1) (b) log [(4 x + 3)e 5 x +7 ] (c) log 10 (3 x + 5) 2 ( 2 − 3 x ) ( x + 1) 6 Solution: (a) Let y = log( 9 x 2 −1) = log {(3 x +1)( 3 x −1)} y = log( 3 x +1) + log( 3 x −1) ( log( AB ) = log A + log B ) ∴y n = dnn { log( 3x + 1)} + dnn { log(3x − 1)} dx dx i..e y n = (−1) n −1 ( n −1) ! (−1) n −1 (n −1) ! (3) n + (3) n (3x +1) n (3 x −1) n (b) Let y = log [(4 x + 3)e 5 x +7 ] = log( 4 x + 3) + log e 5 x +7 = log( 4 x + 3) + (5 x + 7) log e e ( log A B = B log A ) ∴y = log( 4 x + 3) + (5 x + 7) ( log e e = 1 ) ∴y n = (−1) n −1 (n −1) ! ( 4) n + 0 (4 x + 3) n D (5 x + 6) = 5 D 2 (5 x + 6) = 0 D n (5 x +1) = 0 (n > 1) (c) Let y = log 10 (3 x + 5) 2 ( 2 − 3x) ( x + 1) 6 128 .2. 1 ( ax + b ) (See Sl.2. Worked Examples:In each of the following Questions find the nth derivative after reducing them into standard functions given in the table 1.1 1. No-4 of Table-1 ) Putting m =1.yn = ( −1) n ( m + n −1)( m + n − 2 ).. in the result (−1) n (m + n −1) ! n 1 Dn = a m m +n ( ax + b) ( m −1) !(ax + b) (−1) n (1 + n −1) ! n 1 Dn = a we get 1+n ( ax + b) (1 −1) !(ax + b) n n or D = (ax + b)1+n a (ax + b) 1 (−1) n n ! • 1.
(3) n + ( −3) n − 6. [ ] [ ] (c) Let y = e − x sinh 3 x cosh 2 x e 3 x − e −3 x e 2 x + e −2 x = e −x 2 2 129 . (a) e 2 x +4 + 6 2 x +4 (b) cosh 4 x + cosh 2 4 x 1 1 + (6 x + 8) 5 (c) e − x sinh 3 x cosh 2 x (d) (4 x + 5) + (5 x + 4) 4 Solution: (a) Let y = e 2 x +4 + 6 2 x +4 ∴ hence = e2xe4 + 62x64 y = e 4 (e 2 x ) +1296 (6 2 x ) yn = e 4 dn 2 x dn (e ) + 1296 n (6 2 x ) n dx dx 4 n 2x n = e {2 e } + 1296 {2 (log 6) n 6 2 x } 2 (b) Let y = cosh 4 x + cosh 2 4 x e 4 x + e −4 x e 4 x + e −4 x + = 2 2 1 4x 1 4x 2 = e + e −4 x + (e ) + (e −4 x ) 2 + 2(e 4 x )( e −4 x ) 2 4 1 4x 1 8x y = e + e −4 x + e + e −8 x + 2 2 4 1 n 4x 1 y n = 4 e + ( −4) n e −4 x + 8 n e 8 n + (−8) n e −8 n + 0 2 4 ( ) { } ( ) { } hence. yn = 1 2 log e 10 ( −1) n −1 ( n −1) ! ( −1) n −1 ( n −1) ! ( −1) n −1 ( n −1) ! n 2. (1) (3 x + 5) n (2 − 3x) n ( x +1) n 2.= 1 (3 x + 5) 2 (2 − 3 x) log e 10 ( x + 1) 6 log 10 X = log e X log e 10 = (3 x + 5) 2 ( 2 − 3 x) 1 1 log log e 10 2 ( x + 1) 6 log A B = B log A A log = log A − log B B = 1 log( 3 x + 5) 2 + log( 2 − 3 x) − log( x +1) 6 2 log e 10 1 ∴ y = 2 log 10 {2 log( 3x + 5) + log( 2 − 3x) − 6 log( x +1)} e { } Hence.
Hence,
e −x (e 3 x − e −3 x )( e 2 x + e −2 x ) 4 e −x 5x = e − e − x + e x − e −5 x 4 1 = e 4 x − e −2 x + 1 − e −6 x 4 1 y = 1 + e 4 x − e −2 x − e −6 x 4 1 y n = 0 + (4) n e 4 x − (−2) n e −2 x − (−6) n e −6 x 4 1 1 + (6 x + 8) 5 (d) Let y = (4 x + 5) + (5 x + 4) 4 =
{
}
{
}
{
}
{
}
{
}
Hence, y n =
dn 1 dn 1 + ( 6 x + 8) 5 + n 4 (4 x + 5) dx (5 x + 4) dx n ( −1) n n ! ( −1) n ( 4 + n −1) ! = ( 4) n + (5) n + 0 n +1 4 +n ( 4 x + 5) ( 4 −1) !(5 x + 4) dn dx n
{
}
i.e y n =
(−1) n n ! ( −1) n (3 + n) ! ( 4) n + (5) n ( 4 x + 5) n +1 3!(5 x + 4) n +4
• Session - 3
• 1.2.2 Worked examples:1 1 x2 1. (i) 2 (ii) (iii) x − 6x + 8 1 − x − x2 + x3 2x 2 + 7x + 6 (iv) (v)
1 x +2 + x + 1 4 x 2 + 12 x + 9 1+ x tan −1 1 − x
tan −1 x
( a)
(vi) tan −1 x (vii)
In all the above problems, we use the method of partial fractions to reduce them into standard forms.
130
Solutions: (i) Let y = •
1 1 . The function can be rewritten as y = ( x − 4)( x − 2) x − 6x + 8 This is proper fraction containing two distinct linear factors in the denominator. So, it can be split into partial fractions as
2
y=
1 A B = + Where the constant A and B are found ( x − 4)( x − 2) ( x − 4) ( x − 2)
as given below.
1 A( x − 2) + B ( x − 4) = ( x − 4)( x − 2) ( x − 4)( x − 2)
∴ 1 = A( x − 2) + B( x − 4)
-------------(*)
Putting x = 2 in (*), we get the value of B as B = − 1
2
Similarly putting x = 4 in(*), we get the value of A as A = 1 2
∴y = ( x − 4)( x − 2) = x − 4 +
yn = 1 dn 2 dx n 1 1 dn − n x − 4 2 dx 1 x −2
1
(1 / 2)
( −1 / 2) x −2
Hence
=
1 (−1) n n ! 1 (−1) n n ! (1) n − (1) n n +1 n +1 2 ( x − 4) 2 ( x − 2)
1 1 1 ( −1) n n ! − 2 ( x − 4) n +1 ( x − 2) n +1
=
(ii) Let y =
1 1 1 = = 2 3 2 1− x − x + x (1 − x) − x (1 − x) (1 − x )(1 − x 2 ) 1 1 ie y = (1 − x)(1 − x)(1 + x ) = (1 − x) 2 (1 + x)
Though y is a proper fraction, it contains a repeated linear factor
(1 − x ) 2 in its
denominator. Hence, we write the function as
y= A B C + + 2 (1 − x) (1 − x ) 1+ x
in terms of partial fractions. The
constants A, B, C are found as follows:
131
y=
1 (1 − x) (1 + x)
2
=
A B C + + 2 (1 − x) (1 − x) 1+ x
ie 1 = A(1 − x)(1 + x) + B (1 + x) + C (1 − x) 2 -------------(**) Putting x = 1 in (**), we get B as B = 1
2 Putting x = -1 in (**), we get C as C = 1 4
Putting x = 0 in (**), we get 1 = A + B + C
∴A = 1 − B − C = 1 − 1 2 − 1 4 = 1 4 A ∴ = 14
Hence, y =
yn =
1 (−1) n ( 2 + n − 1) ! n 1 ( −1) n n ! 1 ( −1) n ! (1) n + (1) + (1) n n +1 2 +n n +1 4 (1 − x) 2 ( 2 − 1) !(1 − x) 4 (1 + x)
n
(1 / 4) (1 / 2) (1 / 4) + + 2 (1 − x ) (1 − x ) (1 + x)
=
1 (−1) n ( n +1) ! 1 1 1 ( −1) n n ! + + n +1 4 (1 + x ) n +1 2 (1 − x ) n + 2 (1 − x )
x2 (iii) Let y = 2 (VTU July-05) 2x + 7x + 6 This is an improper function. We make it proper fraction by actual division and later spilt that into partial fractions.
i.e
−7 x − 3 1 2 ∴y = + 2 (2 x + 3)(x + 2)
1 (− 7 2 x − 3) x ÷ (2 x + 7 x + 6) = + 2 2 2x − 7x + 6
2 2
Resolving this proper fraction into partial
fractions, we get
y= 1 A B + + . Following the above examples for finding A & 2 ( 2 x + 3) ( x + 2)
B, we get
1 92 ( − 4) y= + + 2 2 x + 3 x + 2
Hence, y n = 0 +
(−1) n n ! 9 (−1) n n ! ( 2) n − 4 (1) n n +1 n +1 2 ( 2 x + 3) ( x + 2)
132
Hence. y is given by x +1 x +1 1 1 y =1+ + 2 x + 1 ( 2 x + 3) [ (2 x + 3) 2 = 4 x 2 +12 x + 9 ] Resolving the last proper fraction into partial fractions.e 9 2 (2) n 4 y n = (− 1) n ! − n+1 ( x + 2) n + 1 (2 x + 3) n (iv) Let y = ( x + 2) x + 2 ( x + 1) 4 x + 12 x + 9 (i) (ii) Here (i) is improper & (ii) is proper function. Solving we get (2 x + 3) (2 x + 3) 2 (2 x + 3) 2 A= 1 B=−3 2 and 2 − 32 1 12 + ∴y = 1 + + 2 1 + x (2 x + 3) (2 x + 3) ( −1) n n ! n 1 (−1) n n ! 3 (−1) n ( n + 1) ! yn = 0 + (1) + ( 2) n − ( 2) n ∴ n n +1 (1 + x) 2 ( 2 x + 3) 2 ( 2 x + 3) n + 2 (v) tan −1 x Let ( a) 1 1+ x y = tan−1 x ( a) a 1 = 2 2 2 a x + a ∴y1 = ( a) 2 a y n = D n y = D n −1 ( y1 ) = D n −1 2 2 x +a Consider a a = 2 ( x + ai )( x − ai ) x +a A B = + .i. ( x + ai ) ( x − ai ) = ( x + ai) (− 12i ) + ( 12i ) ( x − ai) . by actual division (i) becomes x +2 1 =1+ . So. 133 . on resolving into partial fractions. we get x A B = + . on solving for A & B.
θ = tan −1 a x x + ai = r ( cos θ + i sin θ ) = re iθ x − ai = r ( cos θ − i sin θ ) = re −iθ 1 1 e inθ 1 e −inθ = n −inθ = n . We take transformation x = r cos θ a = r sin θ where r = x 2 + a 2 .(v) we get y n which is same as above with or x = cot θ r = x 2 + 1 θ = tan−1 1 ( x) ∴ r = cot 2 θ + 1 = cos ecθ ⇒ 1 1 = = sin n θ n r cos ec nθ n −1 D n ( tan −1 x ) = ( − 1) ( n − 1) ! sin n θ sin nθ where θ = cot −1 x −1 (vii) Let y = tan 1 + x 1 − x put x = tan θ θ = tan −1 x 1 + tan θ ∴ y = tan −1 1 − tan θ = ta n− 1 [ ta nπ (4 + θ )] π tan π ( 4 1 + tan θ +θ = 1 − tan θ ) + θ = + tan −1 ( x) 4 4 π y = + tan −1 ( x) 4 y n = 0 + D n (tan −1 x ) =π 134 . a n − 1 − 12i n − 1 12i ∴ 2 2= D D + D x +a x + a i x − a i n− 1 n −1 n −1 1 ( −1) ( n −1) ! 1 ( −1) ( n −1) ! = − + -----------(*) n n 2i ( x + ai ) 2i ( x − ai ) • Since above answer containing complex quantity i we rewrite the answer in terms of real quantity. = n r ( x − ai ) n r e ( x + ai ) n r now(*) is y n yn = 2 i rn ( −1) n −1 ( n −1)! [e inθ = 2i r n − e −inθ ] ( −1) n −1 ( 2 i sin nθ ) ⇒ ( −1) n −1 ( n −1)! sin nθ rn (vi) Let y = tan −1 x θ = cot −1 ( x ) .Putting a = 1 in Ex.
3 4.a 2 • Session 4 π As sin( X + ) = cos X 135 . 2 2 2 2 ( x −1)( x − 4) ( x + 2)( x − 2 x +1) 4x − x − 3 x + 2x − x − 2 x3 x 2 − 3x + 2 −1 2 x 6.2. tan 2 1 − x − 1 7. sin( ax + b) .r. Differentiating w. tan 1 + x 2 −1 x 1.n −1 n −1 1 ( −1) ( n −1) ! 1 ( −1) ( n −1) ! = − + n n 2i ( x + ai ) 2i ( x − ai ) Problem set No. y1 = cos( ax + b). 1.t x. 5. No-7 of Table-1 ) Let y = sin( ax + b) .1 for practice Find the nth derivative of the following functions: 1. 6x x x x 2 + 4x + 1 2. 3.(See Sl.
. y n = r n e ax sin ( bx + c + nθ ) y1 can be rewritten as 1.e.r. 2. (See Sl.. we write y 2 as y 2 = r 2 e ax sin ( bx + c + 2θ ) y 3 = r 3 e ax sin ( bx + c + 3θ ) Continuing in this way. No-9 of Table-1 ) Let y = e ax sin ( bx + c )...a y 2 as π Again using sin( X + ) = cos X . θ = tan−1 b • ( a) n ax n ax ∴ D [e sin ( bx + c ) ] = r e sin ( bx + c + nθ ) .we get i. we get y 4 = r 4 e ax sin ( bx + c + 4θ ) y 5 = r 5 e ax sin ( bx + c + 5θ ) …………………………….e.( 2) Comparing expressions (1) and (2). ..2..We can write y1 = a sin( ax + b + π / 2). Again Differentiating w. y 2 = a 2 sin( ax + b + 2π / 2).t x.. where r = a 2 + b 2 & y1 = e ax [ ( r cos θ ) sin ( bx + c ) + ( r sin θ ) cos( bx + c ) ] y1 = e ax [ r{sin ( bx + c ) cos θ + cos( bx + c ) cos θ }] y1 = re ax [ sin ( bx + c + θ ) ]. …………………………… y n = a n sin( ax + b + nπ / 2). For computation of higher order it is convenient to express the constants ‘a’ and ‘b’ in terms of the constants r and θ defined by a = r cos θ & b = r sin θ . y 2 = a cos( ax + b + π / 2)....a i. derivatives y1 = e ax cos ( bx + c )b + sin ( bx + c ) ae ax y1 = e ax [ a sin ( bx + c ) + b cos ( bx + c ) ] .so t r = a 2 + b 2 and θ = tan−1 b ( a ) .e. e ax sin ( bx + c ). or i.. we get y 3 = a 3 sin( ax + b + 3π / 2). Similarly. y 4 = a 4 sin( ax + b + 4π / 2).we get 2 y 2 = a sin( ax + b + π / 2 + π / 2).thus..( 1) Differentiating using product rule .3 Worked examples 136 ..
(i) sin 2 x + cos 3 x (ii) sin 3 cos 3 x (iii) cos x cos 2 x cos 3 x (iv) sin x sin 2 x sin 3 x (v) e 3 x cos 2 x (vi) e 2 x (sin 2 x + cos 3 x ) • The following formulae are useful in solving some of the above problems.2 sin 2 x + 2 − 6 sin 6 x + 2 32 (iii) )Let y = cos 3x cos x cos 2 x = 1 ( cos 4 x + cos 2 x ) cos 2 x = 1 cos 4 x cos 2 x + cos 2 2 x 2 2 1 1 1 − cos 4 x = ( cos 6 x + cos 2 x ) + 2 2 2 1 cos 2 x 1 = cos 6 x + + (1 − cos 4 x ) 4 4 4 nπ nπ n 2 n cos 2 x + 4 cos 4 x + 1 n nπ 2 2 ∴ y n = 6 cos 6 x + − + 4 2 4 4 [ ] (iv) )Let y = sin 3 x sin x sn 2 x = 1 [sin ( 2 x ) − sin 4 x] sin 2 x 2 137 .1. (i) sin 2 x = 1 − cos 2 x 2 (ii ) cos 2 x = 1 + cos 2 x 2 (iii) sin 3x = 3 sin x − 4 sin 3 x (iv ) cos 3 x = 4 cos 3 x − 3 cos x (v) 2 sin A cos B = sin ( A + B ) + sin ( A − B ) (vi) 2 cos A sin B = sin ( A + B ) − sin ( A − B ) (vii) 2 cos A cos B = cos ( A + B ) + cos ( A − B ) (viii) 2 sin A sin B = cos ( A − B ) − cos ( A + B ) 1 − cos 2 x 1 + ( cos 3 x + 3 cos x ) 2 4 2 3 Solutions: (i) Let y = sin x + cos x = ∴ yn = 1 1 n n 0 − ( 2 ) cos 2 x + nπ + ( 3) cos 3 x + nπ + 3 cos x + nπ 2 4 2 2 2 3 [ ( )] [ ( ) ( )] (ii)Let y = sin 3 x cos 3 x = = sin 3 2 x 1 − sin 6 x + 3 sin 2 x sin 2 x = = 8 8 4 2 1 [3 sin 2 x − sin 6 x ] 32 1 n nπ nπ n yn = 3.
1.2.No.1 138 . r2 = 2 2 + 3 2 = 13 . s i xns i 2 x 6.9. 10 of Table 1) ∴ y n = re cos ( 2 x + nθ ) where 3x (v) Let y = e 3 x cos 2 x r = 3 2 + 2 2 = 13 −1 & θ = tan (vi) Let y = e 2 x (sin 2 x + cos 3 x ) 2 3 1 − cos 2 x 1 + [ cos 3 x + 3 cos x ] 2 4 2x ∴ = e 2 x sin 2 x + cos 3 x = e 2 x 1 − cos 2 x + e4 [ cos 3x + 3 cos x] y 2 1 2x 1 2x ∴y = e − e 2 x cos 2 x + e cos 3 x + 3e 2 x cos x 2 4 We know that [sin 2 x + cos 3 x ] = [ ] [ ] [ ] Hence. yn = 1 n 2x 1 2 e − r1n e 2 x cos ( 2 x + nθ1 ) + r2n e 2 x cos ( 3 x + nθ 2 ) + 3r3n e 2 x cos ( x + nθ 3 ) 2 4 = 8 [ ] [ ] where r1 = 2 2 + 2 2 . sin 3 x 4.2 for practice Find nth derivative of the following functions: 1.θ 3 = tan −1 . cos 2 x. e (sin x + cos x) 7. r3 = 2 2 + 12 = 5 2 3 1 θ1 = tan −1 . (sin 3 x + cos 2 x) 2.= 1 sin 2 2 x − sin 4 x sin 2 x 2 1 1 − cos 4 x 1 − ( sin 2 x − sin 6 x ) = 2 2 2 [ ] 1 − cos 4 x 1 = − ( sin 2 x − sin 6 x ) 4 4 1 nπ n nπ nπ n y n = 4 n cos 4 x + − 2 sin 2 x + + 6 sin 6 x + 4 2 2 2 (Refer Sl.θ 2 = tan −1 . e cos 3 x (VTU Jan-04) LESSON -2 : Leibnitz’s Theorem • Session . sin 2 x cos 3x 3. e cos 2 x cos 4 x 8. e sin x cos 2 x −3 x . cos x cos 2 x 3x 3 2 x −x 2 n 5. 2 2 2 Problem set No.
. term wise.uD n v .• • Leibnitz’s theorem is useful in the calculation of nth derivatives of product of two functions.D(1 − x 2 )+ n C 2 D n −2 ( y 2 ) D 2 (1 − x 2 )+ n C 3 D n −3 ( y 2 ) D 3 ( [ ] = y ( n ) +2 − x 2 ) + ny ( n −1) +2 .(1 − x 2 ) + n C1 D n −1 ( y 2 ). where D = n( n −1) n! d n n ... n times..( 0) +. throughout] Equation (1) has second order derivative y 2 in it..(1) [ ÷ 2 y1 ...(3) 139 ...r.( − ) + 2 .e D n y 2 (1 − x 2 ) = D n ( y 2 ). We differentiate (1)...t.. y ( n −3) +2 . Taking u = y 2 and v = (1 − x 2 ) and applying Leibnitz’s theorem we get [( ) ] D n [uv ] = D n uv +n C1 D n −1uDv +n C 2 D n −2 D 2 v +n C 3 D n −3 uD 3 v + .(2) (a) (b) (c) Consider the term (a): D n 1 − x 2 y 2 . then n D (uv ) = D n uv +n C1 D n −1uDv +n C 2 D n −2 uD 2 v + .e D n { (1 − x 2 ) y 2 } − D n { xy1 } − D n ( p 2 y ) = 0 [( ) ] ---------... So. Statement of the theorem: If u and v are functions of x. using Leibnitz’s theorem as follows.. 1 1 2 2 1 2 − xy1 + p y = 0 --------------. x. i. C 2 = 2 r!( n − r )! dx • Worked Examples 1.. C1 = n . y = sin pt prove that (1 − x 2 ) y n+2 − ( 2n + 1) xy n+1 + ( p 2 − n 2 ) y n = 0 (VTU July-05) Solution: Note that the function y = f (x) is given in the parametric form with a parameter t.. +n C r D n −r uD r v + ..( − x) + 2 D n 1 − x 2 y 2 = 1 − x 2 y n+ 2 − 2nxy n+1 − n(n − 1) y n [( ) ] ( ) n(n −1) n(n −1)( n −2) y ( n −2 ) +2 . we consider dy dt (p – constant) dx dt dy = dx ( pc o s t p = c o ts ) ( p 2 cos 2 pt p 2 (1 − sin 2 pt ) p 2 (1 − y 2 ) dy = = or = cos 2 t 1 − sin 2 t 1− x2 dx 2 or 1 − x 2 y12 = p 2 1 − y 2 So that 1 − x 2 y12 − p 2 1 − y 2 2 [ (1 − x )( 2 y y ) + y ( − 2 x) ] − p ( − 2 yy ) = 0 (1 − x ) y 2 ( ) ( ) ) Differentiating w. n C r = . If x = sin t . 2 2 D n 1 − x 2 y 2 − xy1 − p 2 y = 0 i. 2! 3! ----------.
r. n n( n −1) n −2 2 n− 1 ( y 2 ) 2( x +1) + D ( y 2 )( 2) + D n ( g 1 )( x +1) + nD n −1 y1 (1) + 4 D n y = 0 D y 2 ( x +1) + nD 2! { } On simplification. = y ( n ) +1 .x +ny ( n −1) +1 + D n [ xy1 ] = xy n +1 + ny n n(n −1) y ( n −2 ) +2 (0) +. 2! ---------. using Leibnitz’s theorem.( x) + n C1 D n −1 y1 .r. (4) and (5) in Eq (2) we get D n ( p 2 y) = p 2 D n ( y) = p 2 y n 2 n+ 2 2 2 2 ie (1 − x ) y n + 2 − (2n + 1) xy n +1 − n y n + ny n − ny n + p y n = 0 ( ∴1 − x 2 ) y n+2 − (2n + 1) xy n +1 + ( p 2 − n 2 ) y n = 0 as desired.t x. If sin −1 y = 2 log( x +1) or ( x + 1) 2 y n +2 + ( 2n + 1)( x + 1) y n +1 + ( n 2 + 4) y n sin −1 y = sin [ 2 log( x +1) ] or y = sin log( x + 1) 2 or y = sin log( x 2 + 2 x +1) . we get 2 ( y1 ) = ie ( x +1) y = 2 1 − y 2 1 x +1 1− y 1 2 Squaring on both sides ( x + 1) 2 y12 = 4(1 − y 2 ) Again differentiating w.. n-times.(4) Consider the term (c): --------.t x. Taking u = y1 and theorem. we get v = x and applying Leibnitz’s D n [ y1 ( x )] = D n ( y1 ).t x.(5) Substituting these values (3). we get ( x + 1) 2 y n +2 + ( 2n + 1)( x + 1) y n +1 + ( n 2 + 4) y n = 0 140 .. D 2 ( x) + .. we consider the function as ( y ) = 2 log( x +1) Differentiating w. {(1 − x ) y − 2nxyn+1 − n(n − 1) y n } − { xyn +1 + ny n } + { p 2 y n } = 0 2.Consider the term (b): D n [ xy1 ] . show that [ ] = 0 (VTU Jan-03) Out of the above four versions..r.. ( x + 1) 2 ( 2 y1 y 2 ) + y12 ( 2( x + 1) ) = 4(−2 yy1 ) or or ( x + 1) 2 y 2 + ( x + 1) y1 = −4 y (÷2 y1 ) -----------* ( x + 1) 2 y 2 + ( x + 1) y1 + 4 y = 0 Differentiating * w.D( x )+ n C 2 D n −2 ( y1 )..
On squaring ) 141 . If y m + y − m = 2x .} − { D n y} = 0 2 ie y n +1 (1 + x ) + ny n (2 x) + ie. we can consider or y = x − x 2 −1 m [ ] m ) m −1 y1 = m x + x 2 − 1 ( ) m −1 1 1 + (2 x) 2 x 2 −1 2 x −1 + x x 2 −1 or ( x 2 − 1 y1 = my . [( ) ] (1 + x ) y 2 1 n +1 + ( 2nx − 1) y n + n( n − 1) y n −1 = 0 n( n −1) y n −1 ( 2) + 0 +. So.e. −1 − 1 ( ) ( ) We get D n 1 + x 2 y1 − D n ( y ) = 0 {D n ( y1 )(1 + x 2 ) + n C1 D n −1 ( y1 ) D(1 + x 2 )+ n C 2 D n −2 ( y1 ) D 2 (1 + x 2 ) + . If x = tan(log y ) . 1 y y1 = e tan x . then find the value of (1 + x 2 ) y n+1 + ( 2nx −1) y n + n(n −1) y n−1 (VTU July-04) Consider x = tan(log y ) i.r...t x. = 2 1+ x 1+ x2 ∴1 + x 2 y1 = y ie 1 + x 2 y1 − y = 0 -----------* We differentiate * n-times using Leibnitz’s theorem.3. − y n = 0 2! 2 2 2 Show that ( x − 1) y n + 2 + ( 2n + 1) xy n +1 + ( n − m ) y n = 0 (VTU Feb-02) Consider 4... tan −1 x = log y or y = e tan x Differentiating w.. 1 or y = x + x 2 −1 [ ] m or y = x − x 2 −1 [ ] m y m + y − m = 2x 1 1 ⇒ ym+ 1 ⇒ (y ) 1 m 1 m 2 − 2 x y m + 1 = 0 Which is quadratic equation in y 1 ( ) 1 = 2x 1 ym 1 m ∴y = − ( −2 x) ± ( −2 x) 2 − 4(1)(1) 2x ± 4x 2 − 4 = 2(1) 2 = [ Let us take y = [x + ∴y 1 ∴y = ( x ± 2x ± 2 x 2 − 1 1 = x ± x2 −1 ⇒ y m = x ± x2 −1 2 x 2 −1 y = x + x 2 −1 x2 = m x + x 2 −1 ( ] −1 ] m ) ( ) ( ) m ..
.3... If y = x n log x. The computation of nth derivatives of a few standard functions and relevant problems were discussed. show that dn dx n 3. u = x 2. we get ( x 2 − 1) y n+2 + (2n + 1) xy n+1 + ( n 2 − m 2 ) y n = 0 2 ) Problem set 1. + 2 3 n 8.. If y = ( x 2 −1) n . Pr ove That (1 − x 2 ) y n+2 − ( 2n + 1) xy n+1 − ( n 2 + m 2 ) y n = 0 n n 7. If cos −1 = log . − 1 (1 − x ( 2 ) )y n +2 − 2 xy n +1 + n( n + 1) y n = 0 x 2 y n +2 + ( 2n + 1) xy n +1 + ( n 2 + 1) y n = 0 4. y = sin ( m sin −1 x )..t x. apply Leibnitz’s theorem to get the results. If y = e m sin x ...r.. 142 .1 In each of the following.The idea of successive differentiation was presented.times using Leibnitz’s theorem and simplifying. If y n = D ( x log x ). Show that y n satisfies the equation d2y dy 1− x2 − 2x + n(n + 1) y = 0 Hint : It is required to show that 2 dx dx 1.. Prove That (1 − x 2 ) y n+2 − ( 2n + 1) xy n+1 − ( n 2 + m 2 ) y n = 0 y x 5.. Prove That x y n +2 2 b n + ( 2n + 1) xy n +1 + 2n 2 y n = 0 (i) y n = ny n −1 + ( n −1)! (ii) y n = n! log x +1 + 1 1 1 + + . − 1 2 y1 y 2 + y12 (2 x) = m 2 (2 yy1 ) − 1 y 2 + xy1 = m 2 y (÷2 y1 ) ) (x (x (x 2 ) 2 ) − 1 y 2 + xy1 − m 2 y = 0 ------------(*) Differentiating (*) n. n Hint: Take x 1 v = log x. Show that n 6. Also... n 1 1 1 log x ( −1) n! = n +1 x log x −1 − 2 − 3 − ... the concept of successive differention was extended for special type of functions using Liebnitz’s theorem... Again differentiating w. If y = a cos(log x) + b sin(log x). Show that y n +1 • n! = x Summary:.(x or or 2 − 1 y12 = m 2 y 2 .
and the curves specified by these coordinates are referred to as polar curves. θ) can be related with Cartesian coordinates ( x. Polar O coordinate x = r cos θ & y = r sin θ . Another useful system for similar purpose is Polar coordinate system. y ) through the relations Fig.(with pole ‘O’) .We are familiar with Cartesian coordinate system for specifying a point in the xy – plane. ө) r Ө r = f(ө) A 143 . Session-1 2.1) • The equation r = f (θ ) is known as a polar curve.1.Chapter – 2 : POLAR CURVES LESSON -1 : Angle between Polar Curves • • In this chapter we introduce a new coordinate system. system P(r.(See Fig.0 Introduction:. θ = projection of OP on the initial axis OA. θ) called polar co-ordinates of P where r = radius vector OP.1. • A polar curve by name “three-leaved rose” is displayed below: ө= ө= ө=π ө= ө=0 ө= • Any point P can be located on a plane with co-ordinates ( r . • Polar coordinates ( r . where we can understand the idea of polar curves and their properties.
Angle between radius ψ = θ +φ vector and the tangent dy tan θ + tan φ i... With usual notation prove that tan φ = r • dθ dr Y T L φ φ P(r. dθ d y tanθ + r d r = d x 1 − tanθ rdθ dr ( ( ) ) ………………….t θ ...2. dx = 1 −tan θ tan φ .. ө) T ψ 1 1 1 Proof:.(2) Comparing equations (1) and (2) we get tanφ = r dθ • • Note that cot φ = 1 dr r dθ dr A Note on Angle of intersection of two polar curves:If φ and φ are the angles between the common radius vector and the 1 2 tangents at the point of intersection of two curves r = f 1 (θ ) and r = f 2 (θ ) then the 1 2 angle intersection of the curves is given by φ −φ 144 .2.e. (1) On the other hand. Fig.: i. y = r sin θ differentiating these..1 Important results • Theorem 1: Angle between the radius vector and the tangent..r. ...e. dy dx dr dr = r ( − sin θ ) + cos θ = r ( cos θ ) + sin θ //NOTE// & dθ dθ dθ dθ dr dy r ( cos θ ) + sin θ dy dr dθ dθ = = dividing the Nr & Dr by cos θ dx dx dθ dr dθ r ( − sin θ ) + cos θ dθ r dθ + tanθ dy dr = dx − ( rdθ dr) tanθ + 1 ( ) i.. A tanψ = tan (θ + φ ) = 1 − tan θ tan φ tan θ + tan φ w..2) curve From Fig..e.. we have x = r cos θ .2.. (See fig.1..Let “ φ” be the angle between the radius vector OPL r 1 Ө and the tangent TPT at the point `P` on the polar O r = f(ө) r = f (θ ) .
r = and r = 1 +θ 1+θ2 145 .. we solve few problems on angle of intersection of polar curves and pedal equations..• Theorem 2: The length “p” of perpendicular from pole to the tangent in a polar curve i.( ii ) 2 p r r dθ 2 2 • • • Note:-If u = 1 1 du .( i ) 1 1 1 Consider p = r sin φ = r cos ec φ 1 1 1 ∴ 2 = 2 cos ec 2φ = 2 (1 + cot 2φ ) p r r φ r Ө φ r = f (ө) P(r.. we get 2 = u 2 + r p dθ Session-2 In this session.e.... 2. . 4.. r = a (1 + cos θ ) and r = b(1 − cos θ ) (VTU-July-2003) 2 3....In the Fig..e. ... sin φ = ON OP ⇒ ON = ( OP ) sin φ i. the length of the perpendicular from the pole to the tangent at p on r = f (θ ) .2 Worked examples:• Find the acute angle between the following polar curves 1.. p = r sin φ.3 Length of the perpendicular from the pole to the tangent 2 1 1 1 dr = 2 1 + p 2 r r dθ N ∴ 1 1 1 dr = 2 + 4 . ө) O Ψ P Fig.. r = ( sin θ + cos θ ) and r = 2 sin θ (VTU-July-2004) 2 r = 16sec2 θ r = a log θ ( 2 ) and r = 25cosec (θ 2 ) (VTU-July-2005) and r = a logθ a aθ 5.1...3.... (i) p = r sin φ or (ii) 1 1 1 dr = = p 2 r 2 r 4 dθ 2 • Proof:..from the right angled triangle OPN. note that ON = p..
t θ r = ( sin θ + cos θ ) Diff w.r.t θ dr = 2 cos θ dθ dθ 2 sin θ r = dr 2 cos θ r = 2 sin θ dr = cos θ − sin θ dθ dθ sin θ + cos θ r = dr cos θ − sin θ tan θ + 1 tan φ1 = (÷ Nr & Dr cos θ ) 1 − tan θ tan φ2 = tan θ ⇒ φ2 = θ i. ( ) r = 16sec2 θ Consider ( 2) Consider r = 25cosec2 θ ( 2) 146 .Solutions: 1. Consider r = a (1 + cos θ ) Consider Diff w.t θ Diff w. Consider Consider ( ) Diff w.e ta nφ 1 = ta nπ + θ ⇒ φ 1 = π + θ 2 2 2 2 ( 2 ) ( ) = tanθ ( 2) ta nφ = − (2 2 sin(θ ) c o sθ ) 2 2 s in2 θ 1 2 tanφ1 = tanθ ⇒ φ1 = φ 2 2 Angle between the curves φ1 − φ 2 = π + θ − θ = π 2 2 2 2 Hence .r.r.r.the given curves intersect orthogonally 2.e tan φ1 = ⇒ φ1 = π + θ 4 tan θ + 1 = tan π + θ 4 1 − tan θ ( ) ∴ Angle between the curves = φ1 − φ 2 = π + θ − θ = π 4 4 3.t θ r = b(1 − cos θ ) dr = −a sin θ dθ dθ a (1 + cos θ ) r = dr − a sin θ ( 2) ta nφ = − 2 sin(θ ) c o s(θ ) 2 2 2 c o s2 θ 1 dr = b sin θ dθ dθ b(1 − cos θ ) r = dr b sin θ = − cotθ i.
r. 1 θ dθ ( log θ ) 2 θ dθ r = − a log θ dr a 147 ...... 1 2 2 2 dθ dr = −50 cos ec 2 θ cot θ . (ii ) We know that tan φ1 − tan φ2 tan (φ1 − φ2 ) = 1 + tan φ1 tan φ2 r ( ) dr 2 = − a ( log θ ) ... Consider r = a log θ Consider r= a Diff w.( 2 ) tan(θ 2 ) = − 25cosec (θ ) cot(θ ) 2 2 = 16secθ 2 Diff w..t θ dr a = θ dθ Diff w.t θ logθ dθ = a log θ θ a dr tan φ1 = θ log θ...r.t θ dr = 32 sec 2 θ tan θ .. 1 2 2 2 dθ ( ) ( ) ( ) ( ) Diff w.t θ 1 6se c2 θ dθ 2 r = 2θ d r 1 6se c ta nθ 2 2 tan φ1 = cot θ 2 2 = tan π ( ) ( ) ( ) ( 2 −θ 2 = tan − θ ) 2 5c o se c2 θ dθ 2 r = 2θ d r − 2 5c o se c co tθ 2 2 ( ) ( ) ( ) tan φ2 = − tan θ ( 2 ) φ1 − φ 2 = π − θ 2 2 =π 2 ⇒ φ1 = π − θ 2 2 ( ) Angle of intersection of the curves = ( ⇒ φ2 = −θ ) ( 2) 2 −θ 4...r....r... (i ) tan φ2 = −θ log θ.
. we take by θ =1 ⇒ =e NO E Substituting θ = e in (iii). Consider 1 1+θ 1 1 = = +1 r aθ a θ Diff w.e tan (φ1 −φ2 ) = 2 1 − (θ log θ ) = 2 From the data: a logθ = r = a logθ ⇒ ( logθ ) = 1 or logθ = ± 1 θ T As θ is acute.( iii ) i...e r dr 2 rθ a 1 +θ 2 2θ a tanφ1 = aθ aθ 2 (1 + θ ) tan φ2 = − ∴tan φ1 = θ (1 + θ ) tan φ2 = − Now..θ log θ − ( −θ log θ ) 1 + (ϑ log θ )( −θ log θ ) 2θ log θ .. we have aθ a =r= ⇒ aθ (1 + θ 2 ) = a (1 + θ ) 2 1+θ 1+θ or θ + θ 3 = 1 + θ ⇒ θ 3 = 1 or θ = 1 ∴tan φ1 = 2 & tan φ2 = ( − 1) Consider tan (φ1 − φ2 ) = = 1 (1 + θ 2 ) 2θ 2 − ( −1) = −3 = 3 1 + ( 2 )( −1) tan φ1 − tan φ2 1 + ( tan φ1 )( tan φ2 ) 148 .r.r..t θ 2θ = − a aθ 1+θ 2 r − r 1 dr r = r dθ aθ 2 dθ aθ 2 dr = r 1 dr 1 1 = − θ2 r 2 dθ a ( ) dr r dθ 2 − 2 rθ 1 dr = a r dθ dθ −a = i....t θ aθ r= as 1 +θ Consider r= ( ) ∴ (1 + θ 2 ) = a Diff w. . we get tan ( φ1 − φ2 ) = 1 − ( e log e ) 2e log e 2 2e = 2 1− e ( log e = 1) e 2e ∴ φ1 − φ2 = tan −1 2 1− e 5..
2. 2. Consider Diff.1 θ 4.: (i) r = f (θ ) & ∴ • 1 1 1 dr = 2 + 4 2 p r r dθ 2 • • 2.(i) r ( ) 149 . r n cos nθ = a n and r n sin nθ = b n 3. r = a cos θ and r = a 2 • 2. • Find the acute angle between the curves (ans: π (ans: π (ans: π 1.2.∴φ −φ2 = tan −1 ( 3) 1 Problem Set No. 4.: (i) r = f (θ ) & p = r sin φ (ii) Eliminate only θ from the Eqs.t θ dr 2a − 1 2 = sin θ r dθ 1 dr − r sin θ = r dθ 2a dθ 2a 1 r =− dr r sin θ 2a = 1 − cos θ ………. Working rules to find pedal equations:(i) Eliminate r and φ from the Eqs. w.Any equation containing only p & r is known as pedal equation of a polar curve. l = 1 + e cosθ r Solutions: 1.0 Pedal equations (p-r equations):. r = aθ and r = a 4) 2) 2) (ans:5 π ) 6 5.1 for practice. r = e θc cot α 3. r m = a m cos mθ and r m = b m sin mθ (ans: π ) 2 LESSON -2 : Pedal Equations • Session .r. r n = a n ( cos nθ + sin nθ ) and r n = a n sin nθ 2.1. r m = a m sin mθ + b m cos mθ (VTU-Jan-2005) 1.1 Worked Examples on pedal equations:Find the pedal equations for the polar curves:2a = 1 − cos θ r 2.
.(ii) 2 1 − cos θ r 2a = r2 = 2 2 2 r [See eg: . 2.(i)] This eqn..t θ We use the equation 1 1 1 dr = 2 + 4 2 p r r dθ 2 dr = eθ cot α ( cot α ) = r cot α dθ ( r = eθ cot α ) 1 1 2 + 4 ( r cot α ) 2 r r 1 1 1 1 = 2 + 4 ( cot 2 α ) = 2 (1 + cot 2 α ) = 2 cos ec 2α r r r r = 1 1 = 2 cos ec 2α 2 p r 2 p2 = r cosec2α or r 2 = p 2 cos ec 2α is the required pedal equation 3.. w....r.Using the value of φis p = r sin φ.. is only in terms of p and r and hence it is the pedal equation of the polar curve.Consider r m = a m sin mθ + b m cos mθ Diff. Consider r = eθ cot α Diff..r....t θ mr m −1 dr = a m ( m cos mθ ) + b m ( −m sin mθ ) dθ r m dr = a m cos mθ − b m sin mθ r dθ 1 dr a m cos mθ − bm sin mθ = r dθ a m sin mθ + b m cos mθ cot φ = a m cos mθ − b m sin mθ a m sin mθ + b m cos mθ 150 . ( 2 ) = − r sinθ 2 . w.. we get 2 = − ta nθ 2 θ sinθ 2 sin θ c o s 2 2 ta nφ = ta n− θ ⇒ φ = − θ 2 2 ( 1 − c o sθ ) = − ta nφ = − 2 sin2 θ ( ) ( ) p = r sin − θ Eliminating “ θ ” between (i) and (ii) p 2 = r 2 sin 2 θ p 2 = ar .
Consider l = ( 1 + cosθ ) Diff w. p = r cos ec φ 1 1 = 2 cos ec 2φ 2 p r 1 1 = = 1 (1 + cot 2 φ ) 2 r 2 1 a m cos mθ − bm sin mθ 1 + m r 2 a sin mθ + b m cos mθ r 2 ( m +1) is the required p-r equation ⇒ p = 2m a + b2m 2 2 2 1 ( a m sin mθ + bm cos mθ ) + ( a m cos mθ − bm sin mθ ) = 2 r (a m sin mθ + bm cos mθ )2 2m 2m 1 1 a + b = 2 Note 2 p r r 2m 4.r.t θ dr 1 dr l − 2 = − e sinθ ⇒ l r 1 r = e sinθ r dθ dθ r l ( cotφ ) = e sinθ r ∴ cotφ = r e sinθ l 1 1 ( ) 2 We have p 2 = r 2 (1 + cot φ ) (see eg: 3 above) Now 1 1 l 2 + e 2 r 2 sin 2 θ = 2 p2 r l2 = 2 2 1 1 + e r 2 sin2 θ 2 l r ( ) 1 + e cosθ = l e cos θ = l −r r r 2 l − r 2 1 l + e 2 r 2 − re 1 1 = 2 p2 r l2 l −r cos θ = re sin 2 θ = 1 − cos 2 θ l −r =1− re 2 e2 − 1 2 1 = 2 + On simplification 2 p e lr 151 .Consider p = r sin φ .
y. z) from the origin in three-dimensional space is an example of a function of three variables x. r = a cos θ and r = a 2 4 . r m = a m cos mθ and r m = b m sin mθ Session-1 3. • 3.1. 2 or ∂x ∂x ∂x ∂x 2 ∂ ∂y ∂z ∂2 z ∂2 f ∂y = ∂y 2 or ∂ 2 or zyy or fyy. denoted by ∂ or ∂ or zy or fy is defined y y as ∂z f ( x. ∂z is the ordinary derivative of ∂x ∂ z ∂ f The first order partial derivative of z w.t y.t.g.0 Introduction: We often come across qualities which depend on two or more variables. For e. ∂z f ( x + δx.r. we understand that z w. y ) = lim ∂x δx →0 δx ∂ z ∂ f or or zx or fx is ∂ x ∂ x Chapter-3: PARTIAL DIFFERENTIATION Lesson-1: Euler’s Theorem The first order partial derivative of z w.t y. the distances of the point (x. The area A(x. y. y ) = lim ∂y δy →0 δy • • From the above definition.1.y) = xy. treating x as constant ∂ ∂z ∂2 z ∂2 f = The partial derivatives or zxx or fxx. 2. a function of two variables.• Problem Set No. denoted by defined as • From the above definition.1 Partial derivatives: Let z = f(x. r = aθ and r = a θ 3. Similarly. we understand that ∂y is the ordinary derivative of z w. y ) − f ( x.r. y 2 ∂ ∂ z ∂ z = or zyx or fyx ∂ ∂ ∂ ∂ y y x y ∂ z 152 .r. y + δy ) − f ( x.1 for practice.t x. Find the pedal equations of the following polar curves 1. obliviously. z. the area of a rectangle of length x and breadth y is given by Area = A(x. r n cos nθ = a n and r n sin nθ = b n 2. treating y as constant.r. y) is.2. x. y) be a function of two variables x and y.
• • 1. we get the partial derivative z y = x 2 − x 2 cos xy = x 2 (1 − cos xy ) i. In all ordinary cases.r. Evaluate (a) (b) z = x 2 y − x sin xy x + y + z = log z (∂z ∂x ) and ( 2 ∂ ∂ ) .r. z ∂x ∂x ∂z 1 − z ∂z z = =1 ∂x z ∂x 1 − z or 153 . 3. ∂z = ( 2 x ) y − { x cos ( xy )( y ) + sin ( xy )(1)} ∂x z x = 2 xy − xy cos xy − sin xy = xy ( 2 − cos xy ) − sin xy ∂z = x 2 (1) − { xcoxy ( x ) + sin ( xy )( 0 )} ∂y i.t x. the second and higher order partial derivatives of more than two independent variables are defined similarly. we learn the partial differentiation of so called explicit functions of more than one variables.(a) Consider z = x y − x sin xy Differentiating z w.r. A note on rules of partial differentiation:All the rules of differentiation applicable to functions of a single independent variable are applicable for partial differentiation also. keeping y as a constant.e. the only difference is that while differentiating partially w. log z − z = x + y − − − − − * Differentiating * partially w.e (b) Consider x + y + z = log z i.1.e.t y keeping x as a constant.t x treating y as i. it can be verified that 2 2 ∂z ∂z = ∂∂ x y ∂∂ y x • The third and higher order partial derivatives of f(x.∂ and ∂ y ∂ z ∂2 z = ∂ ∂ ∂ or zxy or fxy x y y are known as second order Partial derivatives.2 Worked examples: In solving the following examples.y) are defined in an analogous way Also.t one independent variable all other independent variables are treated as constants. if x y Solution: . we get the partial derivative.r. differentiating z w. 1 ∂z ∂z ∂z 1 − = 1 + 0 ⇒ − 1 = 1 ∂x z Constant.e. Similarly.
Similarly, differentiating * partially w.r.t y, treating x as constant,
1 ∂z ∂z ∂z 1 − = 1 + 0 or −1 = 1 z ∂x ∂x ∂x z ∂z z = ⇒ ∂x 1 − z ∂z ∂z z ∴ = = ∂x ∂y 1 − z 1 ∂ 2 ∂θ ∂θ r2 − ? r = 2 n 4t 2. .If θ = t e , what value of n will make r ∂r ∂r ∂t
Solution: Consider θ = t e ------------------------------------------------- (*) Differentiating * w.r.t. r, treating t as constant, we get r2 −r 2 − − 2r − r n −1 4t ∂θ n 4t =t e t e = ∂r 2 4t
n
r2 4t
∂θ r3 ∴r = − t n −1e ∂r 2
2
2
−r 2 4t
∂ 2 ∂θ − 3r n −1 −r t e r = ∂r ∂r 2
2
and hence r 3 n −1 −r 4t − 2 t e
2
4t
2r − 4t
1 ∂ 2 ∂θ 3 n −1 r 2 n −2 −r 2 4t r = − t + t e 4 r 2 ∂r ∂r 2
2 2
------------------- (1)
Also, differentiating (*) w.r.t t, treating r as constant, we get r − −r r2 ∂θ n −1 n 4t 4t = nt e +t e 4t 2 ∂t
1 ∂ 2 ∂θ ∂θ , r = 2 Since, r ∂r ∂r ∂t equation (1) and (2) yield, 2 − 3 n −1 r 2 n −2 −4rt −r 2 r2 t + t e = nt n −1 + t n −2 e 4 t 2 4 4
n −1 1 2 n − 2 − r 2 4t = nt + r t e 4 ------------------------ (2)
i.e.
(
−r n + 3 t n − 1e 2
)
2
4t
= 0 ⇒ n + 3 = 0 orn = − 3 2 2
(
)
∂z ∂z b +a = 2abz z = e ax +by f ( ax − by ), prove that ∂x ∂y 3. If (VTU-Jan-2004) ax +by f ( ax − by ), Solution: Consider z = e
------------------- (*)
Differentiating (*) w.r.t x using product and chain rules, we get
∂z = {e ax +by f 1 ( ax − by )( − b )} + { f ( ax − by )e ax +by ( b )} ∂y
154
∂z = be ax +by { f ( ax − by ) − f 1 ( ax − by )} ∂y ------------------- (2)
Multiplying eq (1) by b and eq(2) by a, and adding we get
b ∂z ∂z +a = abe ax +by f 1 ( ax + by ) + f ( ax − by ) + f ( ax − by ) − f 1 ( ax − by ) ∂x ∂y
[
]
= abe ax +by [ 2 f ( ax − by ) ]
∂u 1 ∂v = u = e r cos θ cos ( r sin θ ), v = e r cos θ sin ( r sin θ ) prove that ∂r r ∂θ and 4.Given , ∂v 1 ∂u =− ∂r r ∂θ r cos θ cos ( r sin θ ), Differentiating u w.r.t. r and θ partially, we Solution: Consider u = e
= 2ab e ax +by f ( ax − by ) = 2abz ,as desired
[
]
get,
∂u = e r cos θ {− sin ( r sin θ )( sin θ )} + cos ( r sin θ ){e r cos θ ( cos θ )} ∂r ∂u r cos θ {cos ( r sin θ ) cos θ − sin ( r sin θ ) sin θ } e i.e. ∂r ∂u = e r cos θ {cos ( r sin θ + θ )} ∂r or --------------(1) ∂u r cos θ {− sin ( r sin θ )( r cos θ )} + cos ( r sin θ ){e r cos θ ( − r sin θ )} =e and ∂θ = −re r cos θ {sin ( r sin θ ) cos θ + cos ( r sin θ ) sin θ } 1 ∂u − = e r cos θ {sin ( r sin θ + θ )} i.e. r ∂θ ---------------------(2)
Next consider,
∂v = e r cos θ { cos ( r sin θ ) sin θ } + sin ( r sin θ ){e r cos θ ( cos θ )} ∂r = e r cos θ {sin ( r sin θ ) cos θ + cos ( r sin θ ) sin θ } ∂v = e r cos θ {sin ( r sin θ + θ )} ∂r i.e --------------------(3); ∂v r cos θ =e {cos ( r sin θ )( r cos θ )} + sin ( r sin θ ){e r cos θ ( − r sin θ )} ∂θ = re r cos θ {cos ( r sin θ ) cos θ − sin ( r sin θ ) sin θ } 1 ∂v = e r cos θ { cos ( r sin θ + θ )} r ∂θ i.e. ----------------------------- (4)
Differentiating v w.r.t r & θ , Partially we get
v = e r cos θ sin( r sin θ)
155
Thus, from eqs (1) and (4), we obtain
∂v 1 ∂u =− . ∂r r ∂θ
∂u 1 ∂v = , and from eqs (2) and (3) we obtain ∂r r ∂θ
Session-2 • 3.1.3 Worked Examples:- In each of following examples, we try to
2 2 ∂u ∂u = , where u = f ( x, y ) is a function of two variables. ∂∂ x y ∂∂ y x 1. u = x y ; 2. u = e x ( x sin y − y sin y )
show that
−1 y 3. u = sin x Solution: - 1. Differentiating u = x y partially w.r.t y, we get
∂u = x y log x ∂y ∂ ∂x
d (a x ) = a x log x dx
differentiating the above partially w.r.t x, we get
∂u ∂ y ∂y = ∂x ( x log x ) ∂2 u 1 = x y + log x yx y −1 x i.e ∂x∂y
(
)
∂u = yx ∂x
∂2 u = x y −1 (1 + y log x ) ∂x∂y or ---------------(1)
Next, differentiating u = x y partially w.r.t. x, we get
∂ ∂u ∂ ( yx y −1 ) = ∂y ∂x ∂y
∂2 u = y ( x y −1 log x ) + x y −1 (1) ∂∂ y x i.e ∂2 u = x y −1 (1 + y log x ) ∂y∂x or ---------------(2)
2 2 ∂u ∂u = x y ∂∂ y x from (1) and (2), we get ∂ ∂
y− 1
so that
−1 y 2. Differentiating u = sin x partially w.r.t. x, we get
156
∂u = ∂x
−y − y = 2 x x x2 − y2 1− y x 1
2
x ( x 2 − y 2 ) + xy2 ∂ 2u x = − =− 2 3 ∂ y∂ x x2 ( x2 − y2 ) 2 ( x − y 2 ) 32 i.e.
( − 2 y ) x x 2 − y 2 (1) − y x 1 2 2 2 x −y ∂ ∂u ∴ =− 2 2 2 ∂y ∂x x (x − y )
----------------(1)
−1 y Next, differentiating u = sin x partially w.r.t. y, we get ∂u 1 1 1 = = 2 x ∂y x2 − y2 1− y x
( )
∂ ∂u ∴ =− ∂x ∂y
x 2 − y 2 ( 0) − (1)
1
2
2 x − y2 x2 − y2
(2x)
∂ 2u x =− ∂ x∂ y ( x 2 − y 2 ) 32 i.e.
------------- (2)
2 2 ∂u ∂u = x y ∂∂ y x From (1) and (2), we get ∂ ∂
x 3. Differentiating u = e ( x sin y − y sin y ) w.r.t y, we get
∂u = e x ( x cos y − y cos y − sin y ) ∂y
Differentiating w.r.t x, we get
∂ ∂u = e x ( − x sin y + cos y − 0 − 0 ) + ( x cos y − y cos y − sin y ) e x ∂x ∂y
∂2 u = e x ( − x sin y + cos y + x cos y − y cos y − sin y ) ∂x∂y i.e. ∂2 u = e x {(1 + x − y ) cos y − (1 + x ) sin y} ∂x∂y Or ----------------- (1)
Next, differentiating w.r.t x, we get
∂u = e x { x cos y + sin y} + ( x sin y − y sin y ) e x ∂x ∂u = e x { x cos y + (1 + x − y ) sin y} ∂x
Differentiating this w.r.t. y, we get
157
∂ ∂u x = e { x ( − sin y ) + 0 + (1 + x − y ) cos y + sin y ( − 1)} ∂y ∂x = e x {(1 + x − y ) cos y − (1 + x ) sin y}
----------------- (2)
from (1)and (2),we get
2 2 ∂u ∂u = ∂∂ x y ∂∂ y x
3.1.4 In the following few examples involve equations where partial derivatives of higher order occur. These equations frequently appear is engineering applications.
∂2 u ∂2 u , prove that + =0 ∂x 2 ∂y 2 2 xy −1 Solution:- Differentiating u = tan 2 x − y2
−1 1. If u = tan
2 xy 2 2 x −y
partially w.r.t.x, we get
∂u = ∂x 1
2
( x 2 − y 2 )( 2 y ) − 2 xy ( 2 x ) 2 2 2 (x − y ) 2 xy 1+ 2 x − y2 ∂u − 2y = 2 ∂x x + y 2 Differentiating Partially this w.r.t. x, we get i.e.
∂ 2 u ∂ − 2 y ( x 2 − y 2 )( 0) − ( 2 y )( 2 x ) = = ∂x 2 ∂x x 2 + y 2 ( x2 + y2 )2
(
)
∂2u 4 xy = 2 2 ∂x ( x + y 2 )2 i.e.
-------------------------- (1)
−1 Next differentiating u = tan
2 xy 2 2 x −y
∂ u ∂ 2 x ( x 2 − y 2 )( 0) − ( 2 x )( 2 y ) = = ∂x 2 ∂x x 2 + y 2 ( x2 + y2 )2
2
( x 2 − y 2 )( 2 x ) − 2 xy ( − 2 y ) 2 2 2 (x − y ) 2 xy 1+ 2 x − y2 ∂u 2x = 2 ∂y x + y 2 Differentiating partially this w.r.t.y, we get ∂u = ∂y 1
2
partially w.r.t. y, we get
(
)
∂ 2u − 4 xy = 2 2 ∂y ( x + y 2 )2 i.e.
-----------------------(2)
Adding eqs (1) and (2) we get
∂2 u ∂2 u + = 0 , as desired ∂x 2 ∂y 2
158
Note: (a) The equation
∂2 u ∂2 u + = 0 is known as Laplace’s equation in two dimension ∂x 2 ∂y 2
which has variety of applications in potential theory.
∂2 u ∂2 u ∂2 u + + = 0 is known as Laplace’s equation in (b) A similar equation viz, ∂x 2 ∂y 2 ∂z 2
three- dimensions where u = u ( x, y , z )
2. If u = f ( x + ay ) + g ( x − ay ), show that
Solution:- Differentiating u = f ( x + ay ) + g ( x − ay ), partially w.r.t.x, we get Again differentiating partially w.r.t.x, we see that ∂2u = f 11 ( x + ay )(1) + g 11 ( x − ay )(1) 2 ∂x ∂ 2u a 2 2 = a 2 [ f 11 ( x + ay ) + g 11 ( x − ay ) ] -----------(1) ∂x Next, differentiating u = f ( x + ay ) + g ( x − ay ), partially w.r.t.y
∂u = f 1 ( x + ay ) ( a 2 ) + g .1 ( x − ay ) ( a 2 ) ∂y
∂2 u ∂2 u = a2 ∂y 2 ∂x 2
∂u = f 1 ( x + ay )(1) + g 1 ( x − ay )(1) ∂x
Again differentiating partially w.r.t.y, we see that ∂2u = f 11 ( x + ay ) ( a 2 ) + g 11 ( x − ay ) ( a 2 ) 2 ∂x i.e
∂2 u = a 2 f 11 ( x + ay ) + g 11 ( x − ay ) ------------------(2) ∂y 2
[
]
from eqs (1) and (2), we see that
∂2 u ∂2 u = a2 ∂y 2 ∂x 2
Note:- (a). The equation
one-dimensional wave equation
∂2 u ∂2 u ∂2u ∂2u = a2 = a 2 2 is known as 2 2 or the equation ∂y ∂x ∂t 2 ∂x
∂2 u ∂2 u ∂2 u ∂2 u = a2 2 + 2 + 2 (b). A similar equation viz, 2 ∂x ∂t ∂y ∂z
is known as three- dimensional wave equation
159
r. If u = log ( x 3 + y 3 + z 3 − 3xyz ) .z we get respect or ∂u 3x 2 − 3 yz = 3 ∂x x + y 3 + z 3 − 3 xyz ∂u 3 y 2 − 3zx = 3 ∂y x + y 3 + z 3 − 3xyz ∂u 3z 2 − 3xy = 3 3 3 And ∂z x + y + z − 3xyz adding these partial derivatives we get 160 . the equation viz.t x.t.x.t.t.(a) Differentiating u partially w. ∂u = −3e −2 t sin 3 x ∂x Again differentiating partially w.(1) ∂t Next.r. partially w.r.r. If u = e −2 t cos 3x .y.∂u ∂2 u = c2 ∂t ∂x 2 Solution:.Differentiating u = e −2 t cos 3x . differentiating u = e −2 t cos 3x partially w. x. prove that (a) ∂u ∂u ∂u 3 + + = x + y +z ∂x ∂y ∂z 2 ∂ ∂ ∂ 9 (b) + (VTU Feb-2005) ∂x ∂y + ∂z u = − ( x + y + z) 2 Solution:. t we get 3. find the value of ‘c’ such that ∂u = −2e −2 t cos 3 x --------------. ∂x ∂t ∂y ∂z three-dimensional heat equation 4. 2 − 2t ∴ ∂u ∂ 2u = c 2 2 ⇒ −2e −2 t cos 3 x = 9e −2 t cos 3x ⇒ c 2 = 9 or c = 2 ∂t ∂x ∂ u = − 9e cos3x 2 ∂x ( ) 9 2 Note (a) The equation ∂u ∂2u = c 2 2 is called one-dimensional heat equation ∂t ∂x ∂2u ∂2u ∂ 2u ∂u = c 2 2 + 2 + 2 is known as (b) Similarly.
t x. partially w. show that ∂2 u ∂2 u ∂2 u 2 + 2 + 2 = f 11 ( r ) + f 1 ( r ) 2 ∂x ∂y ∂z r Solution:.t. 2 = x f ( r ) + f ( r ) − 2 + ∂x r r r r r ∂r x ∂u ∂r = . (1) 161 . we have ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + ∂ + ∂ u = ∂ + ∂ + ∂ ∂ + ∂ + ∂ u x y z x y z y z x 2 = ∂ ∂ ∂ ∂ + ∂ + ∂ y z x ∂u ∂u ∂u ∂x + ∂y + ∂z 3 x + y +z ∂ ∂ ∂ 3 3 3 = x + y + z + ∂y x + y + z + ∂z x + y + z ∂x = ∂ ∂ ∂ ∂ + ∂ + ∂ x y z = 5.r.e.∂ u ∂ u ∂ u 3( x 2 + y 2 + z 2 − x y− y −z z )x 3( x 2 + y 2 + z 2 − x y− y −z z )x + + = 3 3 3 = ∂ x ∂ y ∂ z x + y + z − 3x y z ( x + y + z) ( x 2 + y 2 + z 2 − x y− y −z z )x 3 = x + y +z (b) By definition. we get = and = = 2 x or ∂x r ∂y r ∂z r ∂x 1 1 ∂ 2u ∂r 1 ∂r f ( r ) (1) = x f 11 ( r ) + f 1 ( r ) − 2 + ∂x 2 ∂x r r ∂x r 1 1 11 x ∂ 2u 1 x f ( r ) 1 i.By data r 2 = x 2 + y 2 + z 2 Differentiating this partially w. differentiating u = f ( r ) .x. where r = 9 −3 −3 −3 − 2 + 2 + 2 = ( x + y + z)2 ( x + y + z) ( x + y + z) ( x + y + z) x2 + y2 +z2 . we get = f 1 (r) As ∂x r ∂x ∂x f 1 (r) ∂u ∂u x = x = f 1 ( r ) Differentiating partially w.r.r. If u = f ( r ) .x. we get 2r Now. we get ∂x ∂x r r or 11 1 1 ∂ 2u x f ( r ) f ( r ) f ( r ) = x − 2 + ∂x 2 r r r r ∂ 2 u x 2 11 f 1 ( r) x 2 = 2 f ( r) + 1 − ∂x 2 r r r2 ………….t. ∂r x ∂r y ∂r z ∂r = Similarly.
.1... If u = x tan x − y tan x y . r r ∂ 2u z 2 1 1 f 1( r) z2 ( = 2 f ( r) + 1 − 2 . prove that x ∂x + y ∂y + z ∂z = 0 z x x = r cos θ. If z = e .Similarly. we get f 1( r ) {3 −1} r f 1(r) = f 11 ( r ) + 2 r = f 11 ( r ) + ∂ 2u ∂ 2u ∂ 2u f 11 ( r ) 2 f 1( r ) x2 + y2 + z 2 + 2 + 2 = x + y2 + z2 + 3 − ∂x 2 ∂y ∂z r2 r r2 ( ) Problem Set No. y = r sin θ. If 2 xy 2 2 2 ∂r ∂2 r ∂2 r 1 ∂r (Hint: + 2 = + ∂y ∂x 2 ∂y r ∂x By data x 2 + y 2 = r 2 ) 162 . If u = + ... show that ∂x + ∂y + ∂z = 0 2 ∂z ∂2 z ∂2 z 1 ∂z + 2 = + 2 5.. Verify 2 2 ∂z ∂z = where ∂∂ x y ∂∂ y x (i) z = x 3 + y 3 − 3axy (ii) z = tan −1 ( x 2 + y 2 ) (iii) z = log ( x sin y + y sin x ) 2 −1 y 2 −1 2. we get 2 ∂u y2 = 2 f ∂ 2 y r 1 1 1 2 (r ) + f (r ) − y2 (2 ) and 1 .. show that x +y ∂z ∂z ∂z ∂z ∂x − ∂y = 41 − ∂x − ∂y (VTU-Jan-2005) ∂u ∂u ∂u 4. z = .(2) and (3). show that 7. If. prove that ∂2u x2 − y2 ∂2u ∂∂ = 2 = Hint 2 ∂∂ y x ∂ ∂ y x ∂y∂x x +y x2 + y2 3.3) 2 r r ∂z r Adding eqs (1) . If u = ( x − y )( y − z )( z − x ) .1 for practice 1.3. prove that 2 ∂y ∂x ∂y z ∂x ∂u ∂u ∂u y z 6.
. . is called homogenous function of order n.. is called homogenous function of order n (or derivative n) in x and y 2 3 f ( x.. Euler (1707-1783) gives a very useful formula for a particular combination of partial derivatives of homogenous functions.6: Euler’s Theorem:The following theorem.. . y.. z ... . + a n y x x x = x nφ y ..1.... ) is said to be a homogenous function of order n in x...If u is a homogenous function of x and y with ∂u ∂u degree n. z...... + a n y n In which every term is of the nth degree. named in honour of the great Swiss mathematician L.Since u is a homogenous function of degree n. y ) = x 3 + x 2 y − xy 2 + y 3 = x 3 1 + y − y − y For example: (i) x x x (ii) f ( x. x n y Thus any function f ( x. y ) = sin x = x sin x = x φ x is a homogenous function of degree 0 n Note: ....1. t ......Session –3 • 3. x x 1 + y 1 x = x 2φ y x is a homogenous function y x 1 + x of degree ½ 0 0 y y y (iii) f ( x.. y . which can be expressed as.....(1) Any function.. z . then x ∂x + y ∂y = nu (VTU Feb-2005) Proof: . y ) which can be expressed in the below x φ x ........ x x x 3.. t ..5: Introduction to homogenous functions:An expression of the form a 0 x n + a1 x n −1 y + a 2 x n −2 y 2 + . a function f ( x. t… if it can be expressed as x nφ y . y φ x y is also a homogenous function of degree n... • Statement of Euler’s theorem:.(or order n) (2) In general. we can put u is the form 163 ... y ) = x+ y x+ y = = x nφ y is a homogenous function of degree 3. This can be rewritten as 2 n x n a 0 + a1 y + a 2 y + .
t y 164 .. . z. we get ∂u = x nφ 1 y − y 2 + φ y nx n −1 x x x ∂x ∂u = − x n −2 yφ 1 y + nx n −1φ y ----------(1) or x x ∂x n y Again differentiating u = x φ x partially w.r..In general.. y.r.. x ∂ 2 u ∂u ∂2 u ∂u + +y =n 2 ∂x ∂x ∂x ∂x∂y x ∂2 u ∂2 u ∂u +y = ( n −1) ………….t x partially. then x x 2 ∂2 u ∂2 u ∂2 u + 2 xy + y2 = n( n −1)u ∂x∂y ∂x 2 ∂y 2 (VTU Feb-2005) Proof: .(1) 2 ∂x∂y ∂x ∂x Differentiating (*) Partially w. x Differentiating this w. we have.. = nu ∂ x ∂ y ∂ z ∂ t • Corollary to Euler’s theorem:.t y. t… then x ∂ u ∂ u ∂ u ∂ u +y +z +t +..t x. (2) x ∂y ∂u ∂u +y = x − x n −2 yφ 1 y x ∂x ∂y ( ) Consider x + nx n −1φ y x + y x n −1φ 1 y x = −x n −1 yφ 1 y + nx nφ y + x n −1 yφ 1 y x x x n y = nx φ x = nu .If u is homogenous function of x and y with degree n.r..As u is a homogenous function of degree.u = x nφ y . we get { } ∂u = x nφ 1 y 1 +0 x x ∂y ∂u = x n −1φ1 y ………….. y n Since u = x φ x Note: ...r. if u is a homogenous function of order x. by Euler’s theorem ∂u ∂u +y = nu -----------(*) ∂x ∂y Differentiating (*) partially w.
Verify Euler’s theorem for i). x+ y Hence.which means u is a x homogenous function of degree 2.(*) (n=1) ∂ x ∂ y 2 we verify the equation (*) by showing actually LHS=RHS now as u = y ∂u x xy ∂u = . Hence. 2 & ∂x = ∂y ( x + y ) 2 ( x + y) x+y ∂ u ∂ u LHS of (*) = x ∂ + y ∂y x 2 ∴ y2 = x ( x + y) 2 + x2 y ( x + y) 2 xy ( x + y ) xy = ( x + y ) 2 = ( x + y ) =RHS This verifies the result(*) y2 2 2 2 y (ii) Consider u = ax + 2hxy + by = x a 2 + 2h x + b x2 x = x 2φ y . (2) by y we get 2 ∂2u ∂u ∂2u 2 ∂2u ∂2u ∂u + xy + xy +y x + y = ( n −1) x 2 2 ∂x∂y ∂y ∂y∂x ∂y ∂x ∂x x2 ∂2 u ∂2 u ∂2 u + 2 xy + y2 = ( n −1)( nu ) ∂x∂y ∂x 2 ∂y 2 ∂2 u ∂2 u or 2 ∂2 u x + 2 xy + y2 = n( n −1)u ∂x∂y ∂x 2 ∂y 2 ∂2 u ∂2 u = ∂∂ x y ∂∂ x y Session .(*) ∂x ∂y 165 .4 • 3..(i) Consider u = xy is a homogenous function of degree1 (1=n). u = xy x+y • ii) u = ax 2 + 2hxy + by 2 y x x 2 ( y x) xy 1 y y = = x 1 + y x = xφ x = x φ x x+y x(1 + y x ) Solution:.x ∂2 u ∂2 u ∂u ∂u +y + =n 2 ∂y∂x ∂y ∂y ∂y ∂2 u ∂2 u ∂u +x = ( n −1) …………. we get x ∂u ∂u +y = 2u …………. on applying Euler’s theorem to u. (1) by x and esq.7: Worked Examples 1. (2) 2 ∂y∂x ∂y ∂y or y Multiplying esq. Euler’s theorem when applied to u becomes This shows that u = x ∂ u ∂ u +y = (1)u ………….1.
say We note that x+y ∂f ∂f +y = 3 f As f = sin u . we get x ∂f ∂f +y =2 f ∂x ∂y ∂ ∂ i. as u = ax 2 + 2hxy + by 2 .e x ∂x ( log u ) + y ∂y ( log u ) = 2( log u ) 1 ∂u 1 ∂u + y or x u ∂y = 2 log u u ∂x 166 . LHS of (*) x ∂x + y ∂ y = x( 2( ax + hy ) ) + y ( 2( ay + hx ) ) ∂ u ∂ u = 2ax 2 + 2hxy + 2hxy + 2ay 2 = 2( ax 2 + 2hxy + by 2 ) = 2u = RHS This verifies the result (*) (2) If sin u = x y ∂u ∂u .say 3 x 3 1 + y 3 3 3 x x +y ∴f = = = x 2φ y . f is a homogenous function of degree 2. we see that ∂u = 2( ay + hx ) ∂y ∂u = 2( ax + hy ) & ∂x Consider. f = x x+ y x 1 + y x Applying Euler’s theorem to f.Consider sin u = 4 2 x2 y2 = f . x 3x + 4 y x 3 + 4 y x Applying Euler's theorem to f. we get Solution: . show that x 2. show that x +y = 3 tan u (VTU July-2005) ∂x ∂y x+y 2 2 Solution:. `f` is a homogenous function of degree 3.Consider u =e x3 + y 3 3 x +4 y x3 + y3 ∴ log u = 3 x + 4 y = f .Thus.we see that ∂x ∂y ∂ x ( sin u ) + y ∂ ( sin u ) = 3( sin u ) ∂x ∂y ∂u ∂u i.Thus. x ( cos u ) + y ( cos u ) = 3 sin u ∂x ∂y ∂u ∂u or we get x ∂x + y ∂y = 3 tan u as desired x3 + y3 ∂u ∂u +y = 2u log u (VTU july-2004) 3 x +4 y .Now. If ∂x ∂y u = e x x y x x2 y2 = = x 3φ y .e.
f is a homogenous function of Now. Applying Euler’s theorem to f we get ∂f ∂f +y = 2 f As f = tan u .or x ∂x + y ∂y = 2u log u 4. If u = tan ∂u ∂u x+y ∂u ∂u 1 x3 + y3 x+y . ∂x ∂y 167 . say x+y y3 x 3 1 + 3 x x3 + y3 = x 2φ y .Hence to evaluate x 2 u xx + 2 xyu xy + y 2 u yy (VTU July-2005) x3 + y3 u = tan −1 Solution: .So.Let cosu = x+ y y x 1 + x homogenous function of degree ½ On applying Euler's theorem to f. ∴ f = = x y x+ y x 1 + x degree2.say Hence. we get ∂ f ∂ f 1 +y = f Since f = cos u ∂ x ∂ y 2 ∂ ( cos u ) + y ∂ ( cos u ) = 1 cos u x ∂x ∂y 2 ∂u ∂u 1 cos u ∂u ∂u 1 or x sin u + y − sin u = cos u or x ∂x + y ∂y = − 2 sin u ∂x ∂y 2 ∂u ∂u 1 or x ∂x + y ∂y = − 2 cot u x −1 5. show that xu x + yu y = sin 2u . f is a x Solution:.(*). we see that ∂x ∂y ∂ x ( tan u ) + y ∂ ( tan u ) = 2 tan u ∂x ∂y 2 ∂u ∂u x sec 2 u + y sec u = 2 tan u ∂x ∂y x x ∂u ∂u 2 tan u +y = = 2 sin u cos u ∂x ∂y sec 2 u ∂u ∂u x +y = sin 2u ……………. If cos u = x + y Prove that x ∂x + y ∂y = − 2 cot u (VTU.as required.Jan-2004) x 1 + y 1 x x+ y = = x 2 φ y = f .Consider x+y 3 3 x +y ∴ tan u = = f .
we obtain x2 y2 ∂2 v ∂2 v ∂2 v + 2 xy + y2 = 2( 2 −1)v = 2v ∂x ∂x∂y ∂y 2 ∂2 w ∂2 w ∂2 w + 2 xy + y2 = 2( 2 −1) w = 2 w ∂x ∂x∂y ∂y 2 ∂2 u ∂2 u ∂2 u + 2 xy + y2 = 2u ∂x ∂x∂y ∂y 2 u = v − w where v = x 2 tan −1 y x & w = y 2 tan −1 x y Taking the difference of these two expressions. we get x2 Problem set 3.(2) 2 ∂y∂x ∂y ∂y ∂y x Multiplying eq (1) by x.r.(1) 2 ∂x ∂x∂y ∂x ∂x ∂2 u ∂2 u ∂u ∂u x +y + = ( 2 cos 2u ) …………….show that x +y = tan u ∂x ∂y x+y x2 y2 ∂ z ∂ z z = log 2.2 for practice −1 1. and eq(2) by y and adding thereafter we get ∂u ∂2 u ∂2 u ∂2 u ∂u + 2 xy + y2 = ( 2 cos 2u −1) x 2 ∂x + y ∂y ∂x ∂x∂y ∂y = ( 2 cos 2u −1)( sin 2u ) x 2 u xx + 2 xyu xy + y 2 u yy = sin 4u − sin 2u x2 2 −1 y 2 −1 6. show that x2 ∂2 u ∂2 u ∂2 u + 2 xy + y2 = 2u ∂x ∂x∂y ∂y 2 Solution:. Applying the corollary to the Euler’s theorem to v and w.We note that So that v and w are homogenous functions of degree 2.t. If u = sin x2 y2 ∂u ∂u . respectively 2 2 ∂2 u ∂u ∂2 u ∂u + +y = ( 2 cos 2u ) …………. If u = x tan x − y tan x y . If x + y .show that x ∂ + y ∂ =1 x y 168 .x & y we get..To get the value of x u xx + 2 xyu xy + y u yy we proceed as follows Differentiating (*) partially w.1.
prove that x ∂x + y ∂y = − 4 sin 2u ∂u ∂u 3 +y = sin 2u .prove that x ∂x ∂y 2 ∂z ∂z 1 4. −1 If z = tan x3 y3 3 3 x +y 169 . If u = cot x+ y x+y .−1 3.
y . if z = f ( x.Then. then total derivative of z is given by dz ∂ dx f ∂ dy f = + -------------------------(2) dt ∂ dt x ∂ dt y Similarly. the total differential of a function u = f ( x.2. then the total derivative of u is given by du ∂ dx f ∂ dy f ∂ dz f = + + ---------------(4) dt ∂ dt x ∂ dt y ∂ dt z (b) Differentiation of implicit functions:An implicit function with x as an independent variable and y as the dependent dz df variable is generally of the form z = f ( x. y ) where x = x (t ) . y ) = 0 . if u = f ( x.3 : PARIAL DIFFERENTIATION LESSON-2: Total derivatives. z ) and if x = x (t ) . y ) of two variables. z = z (t ) . Session-1 3. also rules for differentiation of composite and implicit functions. x and y are themselves functions of an independent variable t. y = y (t ) i. and hence df ∂ f ∂ dy f 170 . by virtue of expression (2) above. y = y (t ) . y . x and y the total differential ∂f ∂f dx + dy --------------------------(1) ∂x ∂x Further. z ) is defined by du = ∂ f ∂ f ∂ f dx + dy + dz -----------------(3) ∂ x ∂ y ∂ z Further.0: a) Total differential and Total derivative:(or exact differential ) dz is defined by: dz = For a function z = f ( x. Differentiation of Composite and Implicit functions • In this lesson we learn the concept of total derivatives of functions of two or more Variables and. This gives = = 0 dx dx .e.Chapter . we get dz ∂ f ∂ dy f = + dx ∂ x ∂ dx y or dx = ∂x + ∂y dx .
∂ z ∂ ∂ f u ∂ ∂ f v = + ∂ x ∂ ∂ u x ∂ ∂ v x & ∂ = ∂ ∂ +∂ ∂ y u y v y (7) ∂ z ∂ ∂ f u ∂ ∂ f v -------------------------- Note:-1) The above formulae can be extended to functions of three are more variables and formulas (6) and(7) are called Chain rule for partial differentiation. so that we get ∂ x ∂ dx y dx Let z be an function of x and y and that x =φ(u . y ) and v =ϕ( x.e dz = e x [ (1 + x ) sin y + y cos y ]dx + e x [ (1 + x ) cos y − y sin y ]dy (ii) Let z = f ( x.(i) Let z = f ( x. y. z ) = e xyz Then 171 . 2) The second and higher order partial derivatives of z = f ( x. if z = f (u .1: Worked examples:1. y ) can be obtained by repeated applications of the above formulas 3. y ) = e x [ x sin y + y cos y ] Then ∂z = e x [ (1 + x ) sin y + y cos y ] ∂x ∂z ∂z ∂z dx + dy ∂x ∂y x and ∂y = e [(1 + x ) cos y − y sin y ] Hence. v) and y =ϕ(u . we get dz = i. using formula (1). ∂ z ∂ ∂ f x ∂ ∂ f y = + ∂ u ∂ ∂ x u ∂ ∂ y u & ∂ = ∂ ∂ +∂ ∂ v x v y v ∂ z ∂ ∂ f x ∂ ∂ f y -------------------(6) Similarly. Find the total differential of (i) e x [ x sin y + y cos y ] (ii) e xyz Ans:. ∂f ∂x = − ----------. v ) are functions of u and v then. v) are functions of u and v and if u =φ( x.2.(5) ∂f ∂y (c) Differentiation of composite functions:dy ∂ f ∂ dy f 0= + . y ) are functions of x and y then.
∂u ∂u ∂u = ( xz )e xyz ; = ( yz )e xyz ; = ( xy ) e xyz ∂y ∂x ∂z du = ∂u ∂u ∂u dx + dy + dz ∂x ∂y ∂z = e xyz ( yzdx + zxdy + xydz
Total differential of z = f ( x, y , z ) is (see formula (3) above)
)
2.
Find
(i) z = xy 2 + x 2 y ,where x = at 2 , y = 2at −1 y (ii) u = tan x ,where x = e t − e −t , y = e t + e −t (VTU-Jan 2003) Ans:- (i) Consider z = xy 2 + x 2 y
∂z ∂z = 2 xy + x 2 = y 2 + 2 xy & ∂y ∂x
dz if dt
Since x = at 2 & y = 2at , We have
dz ∂z dx ∂ dy z = + dt ∂x dt ∂y dt
dx dy = 2at , = 2a dt dt
Hence, using formula (2), we get
= ( y 2 + 2 xy )( 2at ) + ( 2 xy + x 2 )( 2a ) = ( y 2 + 2 xy ) y + 2a ( 2 xy + x 2 ) , Using y = 2at
dz = y 3 + 2 xy 2 + 4axy + 2ax 2 dt dz To get explicitly in terms of t, we substitute dt x = at 2 & y = 2at , to get dz 3 3 4 = 2a (8t + 5t ) dt
(ii) Consider
∂u −y ∂u x = 2 = 2 2 , ∂x x + y ∂y x + y2
u = tan −1 y x
Since x = e t − e −t & y = e t + e −t ,we have
dx dy = e t + e −t = y = e t − e −t = x dt dt
du ∂u dx ∂u dy
Hence dt = ∂x dt + ∂y dt
−y = 2 x + y2
( see Eqn (2))
x2 − y2 x y + 2 x = 2 x + y2 x + y2
Substituting x = e t − e −t & y = e t + e −t , we get
172
Session-2 3. Find
du −2 = 2t dt e + e −2t
dy if (i) x y + y x =Constant dx
(ii) x + e y = 2 xy Ans: - (i) Let z = f ( x, y ) = x y + y x =Constant. Using formula (5)
∂f dy = − ∂x -----------------(*) ∂f dx ∂y
But
∂ f ∂f = x y log x + xy x −1 Putting those in(*), we get = yx y −1 + y x log y and ∂ y ∂x yx y −1 + y x log y dy = − y x −1 dx x log x + xy
(ii) Let z = f ( x, y ) = e x + e y − 2 xy =Constant Now,
∂f ∂f = e y − 2 x Using this in (8), = ex −2y ; ∂y ∂x
4. (i) If z = f ( x, y ) ,where x = r cos θ , y = r sin θ show that
∂z 1 ∂z ∂z ∂z + = + 2 (VTU July-2005) ∂y r ∂θ ∂x ∂r
2 2 2 2
∂f ex − 2y dy ∂x = − = − y ∂f dx e − 2x ∂y
(ii) If z = f ( x, y ) ,where x = e u + e −v & y = e −u − e v ,Show that
∂ z ∂ z ∂ z ∂ z − =x −y ∂ u ∂ v ∂ x ∂ y y = r sin θ , we have Ans: - As x = r cos θ and
we have
∂x ∂x ∂y ∂y = cos θ , = −r sin θ ; = sin θ & = r cos θ . Using Chain rule (6) & (7) ∂r ∂θ ∂r ∂θ
∂z ∂z ∂x ∂z ∂y ∂z ( cos θ ) + ∂z ( sin θ ) + = = ∂x ∂y ∂r ∂x ∂r ∂y ∂r
∂z ∂z ∂z ∂x ∂z ∂y + = ( − r sin θ ) + ∂z ( r cos θ ) = ∂y ∂θ ∂x ∂θ ∂y ∂θ ∂x
Squaring on both sides, the
above equations, we get
∂z ∂z ∂z ∂z ∂z 2 2 = cos θ + sin θ + 2 sin θ cos θ ∂y ∂r ∂x ∂x ∂y
2 2 2
173
∂z 1 ∂z ∂z ∂z ∂z 2 2 = cos θ + sin θ − 2 sin θ cos θ 2 ∂y r ∂θ ∂x ∂x ∂y
2
2
2
Adding the above equations , we get
2 2 2 2 ∂z 1 ∂z ∂z ∂z 2 2 + 2 = + cos θ + sin θ ∂y r ∂θ ∂r ∂x
(
)
∂z ∂z = + as desired. ∂x ∂y (ii) As x = e u + e −v & y = e −u − e v , We have Using Chain rule (6) we get
2
2
∂x ∂x ∂y ∂y = eu , = −e −v , = −e −u & = −e v ∂u ∂v ∂u ∂v
∂z u ∂z −v ∂z ∂z ∂x ∂z ∂y + = e − e = ∂x ∂y ∂u ∂x ∂u ∂y ∂u ∂z ∂z ∂z ∂z ∂x ∂z ∂y + = − e −v − −ev = ∂v ∂x ∂v ∂y ∂v ∂x ∂y ∂z −u ∂z ∂z ∂z u e + e −v − e − ev − = ∴ ∂y ∂u ∂v ∂x
( ) (
( )
(
)
(
)
)
(
)
∂u
=
∂ z ∂ z x− y ∂ x ∂ y
5. (i) If u = f ( x, z , y / z ) Then show that x ∂x − y ∂y − z ∂z = 0 (VTU-July-2004) (ii) If H = f ( x − y , y − z , z − x) , show that
∂H ∂H ∂H + + =0 ∂x ∂y ∂z
∂u
∂u
(VTU-July-2003)
Ans: - (i) Let u = f (v, w) , where
∂ v ∂v ∂v =0 , =z , =x ∂ y ∂x ∂z
&
Using Chain rule,
∂w ∂w − y ∂w =1 , =0 , z ∂z = ∂y z2 ∂x
v = xz and w = y z
∂u ∂u ∂v ∂u ∂w ∂u = + = ( z ) + ∂u ( 0) = z ∂u ∂x ∂v ∂x ∂w ∂x ∂v ∂w ∂v
∂u ∂u ∂v ∂u ∂w ∂u = + = ( 0) + ∂u 1 z = 1 ∂u ∂y ∂v ∂y ∂w ∂y ∂v ∂w z ∂w
( )
From these, we get ∂u ∂u ∂u ∂u y ∂u y ∂u ∂u x −y −z = xz − − z x − 2 ∂x ∂y ∂z ∂v z ∂w ∂v z ∂w =0 (ii) Let H = f (u , v, w) Where u = x − y, v = y − z, w = z − x
∂u ∂u ∂v ∂u ∂w ∂u = + = ( x ) + ∂u − y z 2 = x ∂u − y2 ∂u ∂z ∂v ∂z ∂w ∂z ∂v ∂w ∂v z ∂w
174
y = uv Prove that ∂z ∂z ∂z ∂z ∂z ∂z =u −v (i) ( u + v ) (ii) ( u + v ) ∂y = ∂u + ∂v ∂x ∂x ∂v t 3.3 : PARIAL DIFFERENTIATION 175 . ∂z = 0 ∂ v ∂v ∂ v = 0.1 1. Find du If dt (i) u = x 2 − y 2 .2: Problem Set No: 3.2. y = y + bt Prove that ∂u ∂u ∂u =a +b ∂t ∂x ∂y Chapter . as desired. y = e . where r = x + at . z = 2r − 3s + 4. =1. z = cos t dy is each of the following cases: dx (i) x sin ( x − y ) = ( x + y ) (ii) ( cos x ) y = ( sin y ) x 4. Find 5. x = 1 . z = e (iv) u = log ( x + y + z ). x = e −t . Find the total differentials of (i) xyz + ( xyz ) −1 (ii) x 2 y + y 2 z + z 2 x 2. =− 1 ∂ x ∂y ∂ z ∂w ∂w ∂u = −1. y = e t sin t (ii) u = sin xy 2 . If z = f ( x. If u = x 2 − y 2 . y = sin t . y = −r + 8s − 5. x = log t . s ) . y = e t t −t (iii) u = xy + yz + zx. y ) and x = u − v. = 0. we get ∂H ∂H ∂H + + = 0 . ∂x ∂y ∂z ∂u ∂u ∂u ∂H ∂H ∂u ∂H ∂v ∂H ∂w ∂H (1) + ∂H ( 0) + ∂H ( −1) = + + = ∂x ∂u ∂x ∂v ∂x ∂w ∂x ∂u ∂v ∂w ∂ H ∂ ∂ H u ∂ ∂ H v ∂ ∂ H w ∂ H ( −1) + ∂H (1) + ∂v ( 0) = + + = ∂ y ∂ ∂ u y ∂ ∂ v y ∂ ∂ w y ∂ u ∂ v ∂ w Adding the above equations. ∂x =1. =1 Using Chain rule. x = e t cos t . ∂x ∂y ∂z ∂H ∂H ∂u ∂H ∂v ∂H ∂w ∂H ( 0) + ∂H ( −1) + ∂H (1) = + + = ∂z ∂u ∂z ∂v ∂z ∂w ∂z ∂u ∂v ∂w 3. Prove that ∂u = 4x + 2 y ∂r 6.2. If z = f ( r .Now. ∂y = −1.
y ) is defined by ∂u u. y. y ∂ v ∂x ∂u ∂y ∂v ∂y Similarly. Errors and Approximations In this lesson.t x and y. y or ∂( x. v J x.r. v. denoted by J or ∂( u .LESSON-3: Applications to Jacobians. w are functions of three independent variables of x. then 176 . z. v ) u.G. then Jacobian of u and v w. errors and approximations using the concept of partial differentiation. if u. who made significant contributions to mechanics. v ∂ x J = x. Definition:. Partial differential equations and calculus of variations.3. we study Jacobians.0 Jacobians:Jacobians were invented by German mathematician C.Let u and v are functions of x and y. Jacob Jacobi (18041851). Session .1 3.
v.In a similar way. y ) ∂( u . z ∂ x ∂ y ∂ z ∂w ∂w ∂w ∂x ∂y ∂z Remark:. y. v. y ) and J ′ = then JJ ′ = 1 ∂( x. z ) ∂( u .(i) If J = denoted by J ′ . w) ∂( x.1: Properties of Jacobians :Property 1:. "inverse Jacobian" of J = 3.If J = Proof:. v ) 177 . z ) is defined as J ′ = ∂( x. v ) ∂( u .3. w) (ii) Similarly. v ) ∂( x.Consider ∂( u .is defined as J = ∂( x. then the "inverse Jacobian" of the Jacobian J. w ∂ v ∂ v ∂ u J = J = x. Jacobian of n functions in n-variables can be defined ∂( u . v ) . v. y ) ∂( u . y ) Note:. ∂( x. y .∂u ∂u ∂u ∂x ∂y ∂z u . y.
v r.Consider 178 .If u and v are functions of r&s and r. s u.(Chain rule for Jacobians):. v ) ∂ v ∂ v ∂ y ∂ y ∂x ∂y ∂u ∂v ∂u∂x ∂u∂y ∂u∂x ∂u∂y + + ∂x∂u ∂y∂u ∂x∂v ∂y∂v 1 0 = = =1 ∂v∂x ∂v∂y ∂v∂x ∂v∂y 1 0 + + ∂x∂u ∂y∂u ∂x∂v ∂y∂v Property 2:.s are functions x&y. y Proof:. y ) ∂ ( u .then u. y) ∂ x ∂ y ∂ u ∂ v 1 J=J × = × ∂ ( x.∂u ∂u ∂x ∂x ∂ ( u. y = J r . v J = x . s × J x.v) ∂ ( x.
sr ∂ r ∂ s ∂ x ∂ y J × J = × . y ∂ v ∂ v ∂ s ∂ s ∂r ∂s ∂x ∂y ∂u ∂r ∂u ∂s + ∂r ∂x ∂s ∂x = ∂v∂r ∂v ∂s + ∂r ∂x ∂s ∂x ∂u ∂u ∂ x ∂ y u. v = = J ∂ v ∂ v x. y ∂x ∂y 3.∂u ∂u ∂r ∂r .3.sr x.vu .2:-Jacobians in various co-ordinate systems:- ∂u ∂r ∂u ∂s + ∂r ∂y ∂s ∂y ∂v∂r ∂v ∂s + ∂r ∂y ∂s ∂y 179 .
∂ρ = 0 ∂x ∂y ∂z = −ρ sin φ .yx ) ∂ r ∂ θ c θ − ros θ i s n ∴ = = ∂ ( r. x = ρ cos φ. z = r cos θ Proof of 1:. =0 ∂ρ ∂φ ∂z 180 . =0 ∂φ ∂φ ∂φ ∂z ∂z ∂z =0 .θ ) ∂ y ∂ y s θ rci θ n o s ∂r ∂θ = r cos 2 θ + r sin 2 θ = r ( cos 2 θ + sin 2 θ ) = r ∂x ∂y ∂z Proof of 2 :-we have ∂ρ = cos φ . = ρ cos φ . In spherical polar co-ordinates. z ) =ρ ∂( ρ. we have ∂( x. ∂x ∂y = cos θ and = sin θ ∂r ∂r ∂x ∂y = −r sin θ and = r cos θ ∂θ ∂θ ∂x ∂x ∂ ( . θ ) 3.we have. In spherical coordinates.1. x = r sin θ cos φ . v ) =r ∂( r . y = r sin θ sin φ . y. =0 . In Polar co-ordinates. z ) ∂( u . x = r cos θ . ∂ρ = sin φ . y = r sin θ we have 2. y = ρ sin φ. φ. z = z.
z ) ∴ = s θ i s n φi nr c θ os φis nr s θ i c n φ o s ∂ ( r . = r cos θ sin φ .We have ∂x ∂x ∂x = −r sin θ sin φ = sin θ cos θ . ∂φ ∂r ∂θ s θ i c n θ o r cs θ oc φso − sr s θ i s n φi n ∂ ( x. ∂φ ∂r ∂θ ∂z ∂z ∂z =0 = cos θ .∂x ∂x ∂x ∂ρ ∂φ ∂z c φ − oρ s φ 0is n ∂ ( . ∂φ ∂r ∂θ ∂y ∂y ∂y = r sin θ cos φ = sin θ sin φ . = r cos θ cos φ .. φ ) c θ o s− r s θ i n 0 = r 2 sin θ Session – 2 181 . zyx ) ∂ y ∂ y ∂ y ∴ = = s φ ρic φ n 0o = ρ s ∂ ( φρ ..θ . y . = −r sin θ . z) ∂ ρ ∂ φ ∂ z 001 ∂z ∂z ∂z ∂ρ ∂φ ∂z Proof of 3:.
y = r sin θ show that ∂( u . If u = x 2 − 2 y 2 . v 182 . v u. v ) = 6r 3 sin 2θ ∂( r .θ ) ∂ v ∂ v 4r c θ o− 2rs s θ i − n4r c θ so θ −i 2rs ns θ ci θ n o s ∂r ∂θ = 2r cos 2 θ − 4r sin 2 θ − 4r 2 cos θ sin θ − 2r 2 sin θ cos θ = ( − 2r cos θ sin θ − 4r 2 sin θ cos θ )( 4r cos 2 θ − 2r sin 2 θ ) ( )( ) = 6r 3 sin 2θ 2. v = 2 x 2 − y 2 . θ ) (VTU-Jan-2006) Consider u = x 2 − 2 y 2 = r 2 cos 2 θ − 2r 2 sin 2 θ v = 2 x 2 − y 2 = 2r 2 cos 2 θ − r 2 sin 2 θ ∴ ∂v ∂u = 2r cos 2 θ − 4r sin 2 θ . y x.3. y × J 1 = 1 (VTU-2001) u. Prove that J x.3:-Worked Examples:1. If x = u (1 − v ). y = uv . where x = r cos θ .v) ∂ r ∂ θ 2r c θ o− 4rs s θ i − n2r c θ so θ −i 4rs ns θ ci θ n o s = = 2 2 2 2 ∂ ( r. = 4r cos 2 θ − 2r sin 2 θ ∂r ∂r ∂u = −2r 2 cos θ sin θ − 4r 2 sin θ cos θ ∂θ ∂v = −4r 2 cos θ sin θ − 2r 2 sin θ cos θ ∂θ ∂u ∂u 2 2 2 2 ∂ ( u.3.
v Further.Consider ∂x ∂x =1 − v .yx ∂ u ∂ v 1− v − u ∴ J = = . ∂ y ∂ x ∂v x ∂v y = =− . 2 ∂x ( x + y ) ∂y ( x + y ) 2 = u − uv 183 . =u ∂u ∂v ∂x ∂x . = −u ∂u ∂v ∂y ∂y =v . x = u − y ∴ = x + y and y y y = ∴v = x+y x +y u ∂ u ∂ u =1 and ∴ =1. as x = u (1 − v ). u We write. y = uv .vu ∂ y ∂ y v u ∂u ∂v v= = (1 − v )u − ( − uv ) = u − uv + uv = u x. y ∴ J = u − − −(1) u.
y ) Consider ∂x ∂u ∂x ∂v y = e sin v x = e u cos v ∂y = e u cos v = e u sin v ∂u ∂y u = −e sin v = e u cos v ∂v u 184 .∂u ∂u 1 1 1 u. y = e u sin v. v ) × =1 ∂( u . v ∂ x ∂ y ∴ J = = y x x . v ) ∂( x. y ) ∂( u . If x = e u cos v. Prove that ∂( x.y ∂ v ∂ v − 2 2 ( x + y) ( x + y) ∂x ∂y = ∴JJ 1 = u 1 x y + = 2 2 ( x + y) ( x + y) x + 1 = y u 1 =1 u 3.
or u = log ( x 2 + y 2 ) 1 2 y −1 y = tan v or v = tan x x ∂u x ∂u y Hence ∂x = x 2 + y 2 . v ) Again Consider x = e u cos v. . y ) = e 2u − − − − − (1) ∂( u .e ∴x 2 + y 2 = e 2u ∂( x.yx ) ∂ u ∂ v e c v − eos v is n ∴ = =u u ∂ ( u. ∂v −y ∂v x & ∂x = x 2 + y 2 . .∂x ∂x u u ∂ ( . ∂y = x 2 + y 2 . v ) ∂ y ∂ y e s v e i c v n o s ∂u ∂v i. ∂y = x 2 + y 2 & 185 . y = e u sin v.
z ) yz zx xy ∂( u . y ) ∂( x. w = z .∂u ∂u x y 2222 ∂ (u. v ) ∂( x. v ) ∴ × = e 2u × e −2u = 1 ∂( u . Show that ∂( x.v) ∂x ∂ y + yx + yx 1 2 2−1 ∴ = = = 2 2 += yx ∂ (x.y) ∂v ∂ v − y x + yx 2222 ∂x ∂ y + yx + yx () ∂( u . If u = x . v. v = y . w) 186 . v ) = e −2u − − − − − − − ( 2) ∂( x. y ) i. y ) ∂( u .e =4 4. y .
z ) x 2 y 2 z 2 x y z 2 z y y z x y − zx + − 2 x y z z y = 4. 5. as desired.e ∂( u .Now. w) yz − zx − xy z z − xy y x = − − − ∂( x. θ . v w) ∂ v ∂ v ∂ v z − z x x = = 2 ∂ ( x y. If x = r sin θ cos φ . z ) = r 2 sin θ ∂( r .. z) ∂ x ∂ y ∂ z y y y ∂w ∂w ∂w y x − x y 2 ∂x ∂y ∂z z z z i. ∂ u ∂ u ∂ u y z yz −2 ∂x ∂y ∂z x x x ∂ ( u . y = r sin θ sin φ . z = r cos θ .show that Now. v. y. by definition ∂( x. φ ) 187 . y.
y ) ∂( u . v ) ∂( x. v = xy + yz + zx.1 1.∂x ∂r ∂ ( x. θ x.3. y = eu tan v. as required 2 ( ( ) ) ) 3.4:-Problem set No: 3.If u = xyz .e s θi cn φ o r cs θ oc sφ o − sr s θi sn φi n ∂ ( x. θ . y = r sin θ show that r. w = x + y + z show that that 188 . z ) = s θi sn φi nr c θ os φsi n r s θi cn φ o s ∂ ( r. φ ) c θ o s − r s θi n 0 = r 2 sin 2 θ sin θ cos 2 φ + r 2 sin θ cos 2 θ cos 2 φ = r 2 sin θ sin 2 θ + cos 2 θ cos 2 φ + r 2 sin θ sin 2 φ = r 2 sin θ cos 2 φ + sin 2 φ ( = sin θ cos φ {0 − ( r 2 sin 2 θ cos φ )} − r cos θ cos φ{ 0 − ( r sin θ cos θ cos φ )} − r sin θ sin φ{− r sin 2 θ sin φ − r cos 2 θ sin φ} = r sin θ . y =1 r. y J × J 1 x. θ x = eu sec v.3. z ) ∂ y = ∂ ( r. x = r cos θ . If ∂( x. y ) 3. Show 2. φ ) ∂ r ∂z ∂r ∂x ∂θ ∂y ∂θ ∂z ∂θ ∂x ∂φ ∂y ∂φ ∂z ∂φ i. v ) × =1 ∂( u . y . y . θ .
Area A of a traingle is given by (ii) δ ( u ± v ) = δu ± δv u vδu − uδv (iv) δ = 2 v v 189 . it is going to have an impact on the result which may be negligible or significant too. find the value of ∂( x. Similarly. the δz δu quantity ×100 or ×100 is called percentage error in z or u respectively z u Note:. z ) 4. w ) 5.5:-Errors and Approximations Session-3:While doing scientific or engineering computations there is always scope for computation errors.∂( u . z ) is a function of three variables x.3. u − v = e x sin y find J x. v. an account of various types of errors is necessary to present approximate value of the function under consideration. y [Hint: Start from u = x ex ( cos y + sin y ) and v = e ( cos y − sin y ) 2 2 u.If x + y + z = u.If u + v = e x cos y. y + z = v.z then the error or u absolute error δ is given by δu = ∂u ∂u ∂u δx + ∂y δy + ∂z δz ----------. In such situations. v 3.The quantity o is called relative error in z or u respectively. Obviously.3. if u = f ( x.3.y.(2) ∂x Remark:.7:-Worked Examples:1. v.The following results similar to that of differentials will also be useful. z = uvw . y. 3. Find the percentage error in measuring the area of a traingle if 1% error (each) is made in measuring its base as well as height. w) = ( x − y )( y − z )( z − x ) ∂( x. (i) δ ( cu ) = cδu (iii) δ ( uv ) = uδv + vδu 3. z ) ∂( u . y ) is a function of two variables x and y then the error or absolute error ∂ is given by z δz = ∂z ∂z δx + ∂y δy -------------(1) ∂x Similarly. y. y .6:-Definition:If z = f ( x.
Hence.e f 1 1 1 1 1 1 − 1 1 1 δf = − 2 δp − 2 δq f2 p q δp 2 i. 0.A= 1 ah ---------. Therefore. is given by 1 1 δA = δ ( ah ) = [ aδh + hδa ] 2 2 1 [ aδh + hδa ] δA ∴ = 2 1 A ah 2 ∴ δh δa = + A h a δA δh δa ∴ ×100 = ×100 + ×100 A h a δA =1 % +1% (by data) =2% The percentage error in A is 2% 2.25 190 .5 2 = 0.5) 2 2 ( 25) ∴ The maximum error in f=0. the A a h change ∂ due to a change δ in a. By data p = q = 25 and δp = δq = 0. and a change δ in h.5.e δf = f 2 + 2 q p As p = q.5cm. we have δq 1 1 1 2δp = + = 0.25 δf = (12. and δp =δq. For a certain lens p and q are each 25cms with a possible error of almost 0. The focal length f of a lens is given by f = p + q where p and q are the distances of the lens from the object and the image respectively. Consider f = p + q .08 δf = f 2 2 Since f 25 25 p we have f=12. Find the approximate value of the maximum error in f.(1) 2 where `a`and `h` are base and height of the triangle respectively.5 Now. δ = δ + f p q 1 1 1 1 1 1 δ = δ +δ p q i.
3. δT = δ mv 2 2 2 1 = m( 2vδv ) + v 2 ( δm ) 2 1 2 i. The current measured by a tangent galvanometer is given by the relation c = k tan θ Where θ is the angle of deflection .44. the relative error in c is minimum when θ = 45 0 . Prove that δa δb δc cos A + cos B + cos c =0 If the triangle ABC is inscribed in a circle of radius R and if a.000 5. find approximately the change in T as m 2 changes From 49 to 49.5)(1600 ) 2 v =1600 . δ B 191 .C we have the sine rule given by a b c = + = 2R sin A sin B sin c ⇒ a = 2 R sin A ⇒ δa = δ ( 2 R sin A) = 2 Rδ ( sin A) ∴ a = 2 R cos Aδ . By data. 2 ∴sin 2θ = 1 ⇒ 2θ = 90 0 ⇒ θ = 45 0 Thus. m = 49 . v + ∂v =1590 ⇒∂v = −10 As T = ( ) [ ] [ ] =-1. m + ∂m = 49 . δ A ⇒ b = 2 R sin B ⇒ δb = δ ( 2 R sin B ) = 2 R δ ( sin B ) ∴ b = 2 R cos Bδ . Consider c = k tan θ (k=constant) ∴ δc = δ ( k tan θ ) = k sec 2 θ δ θ+ 0 ( ) ∴ δc k sec θ = δθ c k tan θ δc 2 = δ θ-----------(*) or c sin 2θ δ c The relation error in c being is maximum when the denominator of RHS of c (*) is maximum and the maximum of sine function is unity. If T = 1 mv 2 is the kinetic energy.b.Show that the relative error in c due to a given error in θ is maximum where θ = 45 0 .5 1 1 mv 2 . If the sides of a triangle ABC vary in such a way that its circum radius remains constant.B.5 ⇒ ∂m = 0.e δT = ( 49 )( 2 )(1600 )( −10 ) + ( 0.c respectively denotes the sides opposite to the angles A. 4.5 and V changes from 1600 to 1590.
we write δa δb δc = 2 RδA.If there is an error of 3% in the value of l. where g is constant . Find the percentage value of T. 3. What is the relative error in computing the volume? ( Hint: Volume of the Sphere= 4 π r3) 3 4.3.8:-Problem set No: 3.3. The radius of a Sphere is found to be 10cms with possible error of 0. The pressure p and the volume v of a gas are concentrated by the relation pv1. 2. δa δb cos A + cos B + δc = 2 RδA + 2 RδB + 2 RδC cos C = 2 R(δA + δB + δC ) = 2 Rδ ( Cons tan t ) A + B + C = π radians = 2 R δ( 0 ) = Cons tan t =0 3.2 1. = 2 RδB & = 2 RδC cos A cos B cos C Adding. The time T of a Complete oscillation of a simple pendulum is given by the formula T = ( 2π ) l g .4=constant. Find the percentage error in the evaluation of the area of an ellipse if 1% error is made while measuring the major and minor axis . find the percentage increase in pressure corresponding to a dimension of 1 % in volume 2 192 .02cm.⇒ c = 2 R sin C ⇒ δc = δ ( 2 R sin C ) = 2 R δ ( sin C ) ∴ C = 2 R cos Cδ δ C From the above.
193 .
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### Ratio and Proportion
Ratio and Proportion are explained majorly based on fractions. When a fraction is represented in the form of a:b, then it is a ratio whereas a proportion states that two ratios are equal. Here, a and b are any two integers. The ratio and proportion are the two important concepts, and it is the foundation to understand the various concepts in mathematics as well as in science.
In our daily life, we use the concept of ratio and proportion such as in business while dealing with money or while cooking any dish, etc. Sometimes, students get confused with the concept of ratio and proportion. In this article, the students get a clear vision of these two concepts with more solved examples and problems.
Examples
Word Problems
For example, ⅘ is a ratio and the proportion statement is 20/25 = ⅘. If we solve this proportional statement, we get:
20/25 = ⅘
20 x 5 = 25 x 4
100 = 100
Therefore, the ratio defines the relationship between two quantities such as a:b, where b is not equal to 0. Example: The ratio of 2 to 4 is represented as 2:4 = 1:2. And the statement is said to proportion here. The application of proportion can be seen in direct proportion.
What is Ratio and Proportion in Maths?
The definition of ratio and proportion is described here in this section. Both concepts are an important part of Mathematics. In real life also, you may find a lot of examples such as the rate of speed (distance/time) or price (rupees/meter) of a material, etc, where the concept of the ratio is highlighted.
Proportion is an equation which defines that the two given ratios are equivalent to each other. For example, the time taken by train to cover 100km per hour is equal to the time taken by it to cover the distance of 500km for 5 hours. Such as 100km/hr = 500km/5hrs.
Ratios and Proportion
Let us now learn Maths ratio and proportion concept one by one.
Definition of Ratio
In certain situations, the comparison of two quantities by the method of division is very efficient. We can say that the comparison or simplified form of two quantities of the same kind is referred to as ratio. This relation gives us how many times one quantity is equal to the other quantity. In simple words, the ratio is the number that can be used to express one quantity as a fraction of the other ones.
The two numbers in a ratio can only be compared when they have the same unit. We make use of ratios to compare two things. The sign used to denote a ratio is ‘:’.
A ratio can be written as a fraction, say 2/5. We happen to see various comparisons or say ratios in our daily life.
Key Points to Remember:
The ratio should exist between the quantities of the same kind
While comparing two things, the units should be similar
There should be significant order of terms
The comparison of two ratios can be performed, if the ratios are equivalent like the fractions
Definition of Proportion
Proportion is an equation which defines that the two given ratios are equivalent to each other. In other words, the proportion states the equality of the two fractions or the ratios. In proportion, if two sets of given numbers are increasing or decreasing in the same ratio, then the ratios are said to be directly proportional to each other.
For example, the time taken by train to cover 100km per hour is equal to the time taken by it to cover the distance of 500km for 5 hours. Such as 100km/hr = 500km/5hrs.
Ratio and proportions are said to be faces of the same coin. When two ratios are equal in value, then they are said to be in proportion. In simple words, it compares two ratios. Proportions are denoted by the symbol ‘::’ or ‘=’.
Continued Proportion
Consider two ratios to be a: b and c: d.
Then in order to find the continued proportion for the two given ratio terms, we convert the means to a single term/number. This would, in general, be the LCM of means.
For the given ratio, the LCM of b & c will be bc.
Thus, multiplying the first ratio by c and second ratio by b, we have
First ratio- ca:bc
Second ratio- bc: bd
Thus, the continued proportion can be written in the form of ca: bc: bd
Direct And Inverse Proportion
Ratio To Percentage
Ratio and Proportion Formula
Now, let us learn the Maths ratio and proportion formulas here.
Ratio Formula
Assume that, we have two quantities (or two numbers or two entities) and we have to find the ratio of these two, then the formula for ratio is defined as;
a: b ⇒ a/b
where a and b could be any two quantities.
Here, “a” is called the first term or antecedent, and “b” is called the second term or consequent.
Example: In ratio 4:9, is represented by 4/9, where 4 is antecedent and 9 is consequent.
If we multiply and divide each term of ratio by the same number (non-zero), it doesn’t affect the ratio.
Example: 4:9 = 8:18 = 12:27
Proportion Formula
Now, let us assume that, in proportion, the two ratios are a:b & c:d. The two terms ‘b’ and ‘c’ are called ‘means or mean term,’ whereas the terms ‘a’ and ‘d’ are known as ‘extremes or extreme terms.’
a/b = c/d or a : b :: c : d
Example: Let us consider one more example of a number of students in a classroom. Our first ratio of the number of girls to boys is 3:5 and that of the other is 4:8, then the proportion can be written as:
3 : 5 :: 4 : 8 or 3/5 = 4/8
Here, 3 & 8 are the extremes, while 5 & 4 are the means.
Note: The ratio value does not affect when the same non-zero number is multiplied or divided on each term.
Important Properties of Proportion
The following are the important properties of proportion:
Addendo – If a : b = c : d, then a + c : b + d
Subtrahendo – If a : b = c : d, then a – c : b – d
Dividendo – If a : b = c : d, then a – b : b = c – d : d
Componendo – If a : b = c : d, then a + b : b = c+d : d
Alternendo – If a : b = c : d, then a : c = b: d
Invertendo – If a : b = c : d, then b : a = d : c
Componendo and dividendo – If a : b = c : d, then a + b : a – b = c + d : c – d
Difference Between Ratio and Proportion
To understand the concept of ratio and proportion, go through the difference between ratio and proportion given here.
### Difference Between Ratio and Proportion
To understand the concept of ratio and proportion, go through the difference between ratio and proportion given here.
S.No Ratio Proportion 1 The ratio is used to compare the size of two things with the same unit The proportion is used to express the relation of two ratios 2 It is expressed using a colon (:), slash (/) It is expressed using the double colon (::) or equal to the symbol (=) 3 It is an expression It is an equation 4 Keyword to identify ratio in a problem is “to every” Keyword to identify proportion in a problem is “out of”
Fourth, Third and Mean Proportional
If a : b = c : d, then:
d is called the fourth proportional to a, b, c.
c is called the third proportion to a and b.
Mean proportional between a and b is √(ab).
Comparison of Ratios
If (a:b)>(c:d) = (a/b>c/d)
The compounded ratio of the ratios: (a : b), (c : d), (e : f) is (ace : bdf).
Duplicate Ratios
If a:b is a ratio, then:
a2:b2 is a duplicate ratio
√a:√b is the sub-duplicate ratio
a3:b3 is a triplicate ratio
Ratio and Proportion Tricks
Let us learn here some rules and tricks to solve problems based on ratio and proportion topic.
If u/v = x/y, then uy = vx
If u/v = x/y, then u/x = v/y
If u/v = x/y, then v/u = y/x
If u/v = x/y, then (u+v)/v = (x+y)/y
If u/v = x/y, then (u-v)/v = (x-y)/y
If u/v = x/y, then (u+v)/ (u-v) = (x+y)/(x-y), which is known as componendo -Dividendo Rule
If a/(b+c) = b/(c+a) = c/(a+b) and a+b+ c ≠0, then a =b = c
Solved Questions
Question 1: Are the ratios 4:5 and 8:10 said to be in Proportion?
Solution:
4:5= 4/5 = 0.8 and 8: 10= 8/10= 0.8
Since both the ratios are equal, they are said to be in proportion.
Question 2: Are the two ratios 8:10 and 7:10 in proportion?
Solution:
8:10= 8/10= 0.8 and 7:10= 7/10= 0.7
Since both the ratios are not equal, they are not in proportion.
Question 3: Given ratio are-
a:b = 2:3
b:c = 5:2
c:d = 1:4
Find a: b: c.
Solution:
Multiplying the first ratio by 5, second by 3 and third by 6, we have
a:b = 10: 15
b:c = 15 : 6
c:d = 6 : 24
In the ratio’s above, all the mean terms are equal, thus
a:b:c:d = 10:15:6:24
Word Problems
Question 4: Out of the total students in a class, if the number of boys is 5 and the number of girls being 3, then find the ratio between girls and boys.
Solution: The ratio between girls and boys can be written as 3:5( Girls: Boys). The ratio can also be written in the form of factor like 3/5.
Question 5: Two numbers are in the ratio 2 : 3. If the sum of numbers is 60, find the numbers.
Solution: Given, 2/3 is the ratio of any two numbers.
Let the two numbers be 2x and 3x.
As per the given question, the sum of these two numbers = 60
So, 2x + 3x = 60
5x = 60
x = 12
Hence, the two numbers are;
2x = 2 x 12 = 24
and
3x = 3 x 12 = 36
24 and 36 are the required numbers.
What is the ratio with an example?
A ratio is a mathematical expression written in the form of a:b, where a and b are any integers. It expresses a fraction. For example. 2:3 = ⅔.
What is a proportion with example?
A proportion is a statement where two or more ratios are equivalent. For example, ⅔ = 4/6 = 6/9.
How to solve proportions with examples?
If a:b::c:d is a proportion, then;
a/b=c/d
|
# Simple Interest Questions
FACTS AND FORMULAE FOR SIMPLE INTEREST QUESTIONS
1. Principal: The money borrowed or lent out for a certain period is called the principal or the sum.
2. Interest: Extra money paid for using other's money is called interest
3. Simple Interest (S.I.) : If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.
Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then,
(i) $S.I=\left(\frac{P×T×R}{100}\right)$
(ii)
Q:
At what rate percent per annum will a sum of money double in 8 years.
A) 12.5% B) 13.5% C) 11.5% D) 14.5%
Explanation:
Let principal = P, Then, S.I.= P and Time = 8 years
We know that S.I. = PTR/100
Rate = [(100 x P)/ (P x 8)]% = 12.5% per annum.
453 74776
Q:
A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was:
A) Rs.2000 B) Rs.10000 C) Rs.15000 D) Rs.20000
Explanation:
Principal =
160 67772
Q:
A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
A) 650 B) 690 C) 698 D) 700
Explanation:
S.I. for 1 year = Rs. (854 - 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (815 - 117) = Rs. 698
245 49774
Q:
How much time will take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest ?
A) 4 years B) 2.5 years C) 3.5 years D) 2 years
Answer & Explanation Answer: A) 4 years
Explanation:
Time = (100 x 81) / (450 x 4.5) = 4 years.
102 49106
Q:
What annual instalment will discharge a debt of Rs 1092 due in 3 years at 12% simple interest?
A) Rs.325 B) Rs.545 C) Rs.560 D) Rs.550
Explanation:
Let each instalment be Rs.x .
1st year = [x + (x * 12 * 2)/100]
2nd year = [ x + (x *12 * 1)/100]
3rd year = x
Then, [x + (x * 12 * 2)/100] + [ x + (x *12 * 1)/100] + x =1092
3x + ( 24x/100 ) + ( 12x/100 ) = 1092
336x =109200
Therefore, x = 325
Each instalment = Rs. 325
194 46880
Q:
A certain sum of money amounts to Rs 1008 in 2 years and to Rs 1164 in 3 ½ years. Find the sum and the rate of interest.
A) 800, 14% B) 800, 13% C) 800, 12% D) 800, 19%
Answer & Explanation Answer: B) 800, 13%
Explanation:
S.I. for 1 ½ years = Rs (1164 - 1008) = Rs 156 .
S.I. for 2 years = Rs (156 x $23$ x 2)= Rs 208.
Therefore, Principal = Rs (1008 - 208) = Rs 800.
Now, P = 800, T= 2 and S.I. = 208.
Therefore, Rate = (100 x S.I.) / (P x T) = [ (100 x 208)/(800 x 2)]% = 13%
135 46042
Q:
A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at $614$ p.a for 2 years. Find his gain in the transaction per year
A) Rs. 112.50 B) Rs.150.25 C) Rs.167.50 D) Rs.170
Answer & Explanation Answer: A) Rs. 112.50
Explanation:
= Rs. (625 - 400)
= Rs. 225
108 40664
Q:
A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched Rs 360 more. Find the sum.
A) Rs.4000 B) Rs.9000 C) Rs.5000 D) Rs.6000
Explanation:
Let sum = P and original rate = R. Then
[(P * (R+2) * 3)/100] - [ (P * R * 3)/100] = 360
3P*(R+2) - 3PR = 36000
3PR + 6P - 3PR = 36000
6P = 36000
P = 6000
|
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# Solving Linear Equations
To solve an equation means to find its solution set. It is easy to determine whether a given number is in the solution set of an equation, but we want to have a method for solving equations. The most basic method for solving equations involves the properties of equality.
## Properties of Equality
Adding the same number to both sides of an equation does not change the solution set to an equation. In symbols, if a = b then a + c = b + c
Multiplication Property of Equality
Multiplying both sides of an equation by the same nonzero number does not change the solution set to the equation. In symbols, if a = b and c ≠ 0, then ca = cb.
Because subtraction is defined in terms of addition, the addition property of equality also allows us to subtract the same number from both sides. For example, subtracting 3 from both sides is equivalent to adding -3 to both sides. Because division is defined in terms of multiplication, the multiplication property of equality also allows us to divide both sides by the same nonzero number. For example, dividing both sides by 2 is equivalent to multiplying both sides by
Equations that have the same solution set are called equivalent equations. In the next example we use the properties of equality to solve an equation by writing an equivalent with x isolated on one side of the equation.
Example 1
Using the properties of equality
Solve the equation 6 - 3x = 8 - 2x.
Solution
We want to obtain an equivalent equation with only a single x on the left-hand side and a number on the other side.
6 -3x = 8 -2x 6 -3x - 6 = 8 -2x - 6 Subtract 6 from each side. -3x = 2 - 2x Simplify. -3x + 2x = 2 - 2x + 2x Add 2x to each side. -x = 2 Combine like terms -1 · (-x) = -1 · 2 Multiply each side by -1. x = -2
Replacing x by -2 in the original equation gives us
6 -3(-2) = 8 - 2(-2),
which is correct. So the solution set to the original equation is {-2}.
The addition property of equality allows us to add 2x to each side of the equation in Example 1 because 2x represents a real number.
Caution
If you add an expression to each side that does not always represent a real number, then the equations might not be equivalent. For example
are not equivalent because 0 satisfies the first equation but not the second one. (The expression is not defined if x is 0. )
|
## Sunday, April 27, 2014
### Combinations: Arranging Permutations with Three Ascending Consecutive Numbers
Question: How many ways can a digits of a number be arranged so that at least three digits are in (i) ascending order and (ii) consecutive positions?
The Number 1234 (4 digit numbers)
Question: How many ways can I arrange the digits of the number 1234 so that each permutation has at least three digits are in (i) ascending order and (ii) consecutive positions?
For example, desired permutations are 1234, 2134, and 4123.
Calculation:
The requirements that three digits are in ascending and consecutive order are be satisfied if the permutation contains any of the following: 123, 124, 134, and 234.
Treat the mentioned permutations as one object and the remaining number as one. Three slots has the group 123, 124, 134, and 234. The corresponding last digit for each of the group is 4, 3, 2, and 1, respectively. The number of arrangements so far is 4 * 2 = 8. We are not done though. Here are the eight arrangements calculated:
1234
1243
1342
2341
4123
3124
2134
1234
Note that 1234 appears twice. Let's remove the duplicate. We are left with 4 * 2 - 1 = 7. A table of all possible arrangements of 1234 (with the desired permutations highlighted).
The Number 12345 (5 Digits)
Let's address the same question, this time with the arranging the digits of 12345.
Calculation:
The requirements that three digits are in ascending and consecutive order are be satisfied if the permutation contains any of the following: 123, 124, 125, 134, 135, 145, 234, 235, 245, and 345. There are 10 three-digit combinations.
Like before, treat the three digit combinations as "one object" and the two remaining digits separately. For example, the combo 124 will fill three digits, 3 and 5 will complete the other two digits. The gross number of permutations are (10 * 3) * 2 * 1 = 60. Like the last problem, we have to account for duplicates. In those 60 permutations counted, 12345, 12354, 12453, 13452, 23451, 21345, 51234, 41235, 31245 are counted twice, and 12345 counted thrice. Removing 10 duplicate permutations, we arrive at our final answer: 60 - 10 = 50.
The complete calculation is: (10 * 3) * 2 * 1 - 10 = 50.
A table of all possible arrangements of 12345 (with the desired permutations highlighted) is shown below:
I am not 100% sure if there was a formula for answering this question - but here is a way to find such arrangements using brute force. What inspired me to do pursue this question was this video published by Numberphile:
In this video, Simon Pampena arranges nine numbered cards and addressed how often those cards are arranged with at least four cards are in ascending or descending order. However, Pampena does not the requirement that the ordered cards are arranged in consecutive slots.
BTW, I am back from Seattle in Southern California.
Have a great weekend - the rest of it - and I'll talk to you next time!
Eddie
## Thursday, April 24, 2014
### Greetings from Seattle - and a short review of the PCalc App. (Updated 4/28/2014)
I am in at the 1st Ave and Pike St Starbucks: the original Starbucks. Actually, the first one opened that first opened in 1971 moved to this location in 1976. The original Pike Place Brew is strong! Thankfully the coffee mellows after a while.
It is a dream of mine to blog from here. Thank you Starbucks! (Twitter: @starbucks)
Short Review of PCalc
This was recommend to me by bb010g. Thanks for the recommendation!
Prices:
PCalc Lite: Free (basic scientific calculator - Algebraic and RPN modes)
PCalc Full: \$9.99 on iOS. (Includes Engineering functions, additional themes, conversions, programmer pack - base conversions and Boolean logic - each can be purchased separately)
Also available on Mac, but not available on Android devices.
I have the full version on both my iPad and iPod Touch. I like the intelligent layout of the keyboard, especially on the iPod Touch. The keys are big but you can still access all the major functions without much trouble. The choice of settings are plenty: everything from calculator settings to whether sound the keys make, if at all.
The features such as conversions, constants, and custom functions are accessed through the A>B, 42, and f(x) keys respectively. Each of the menus offers its options in a style consistent to standard iOS devices.
The calculator app has 10 memory registers and 16 other temporary registers that are used for programming. You can program custom functions. While the language does not contain loops, it does include relational testing (if true then skip n steps). Instead of working with the stack, you work with the memory and temporary registers, which takes a little getting used to. I hope to publish future posts explaining PCalc programming language in detail in the near future.
Here is a little sample of snippets I learned with the PCalc Programming:
The commands are constructed using proper English. (e.g. "Multiply M3 by M1", "Set R0 to 22/15")
Arithmetic Operators: Execute (add, subtract, multiply, divide) on a designated register with a certain value. The result is stored in the designated register.
Register X is the "display". Use this register to display your final answer. (Assuming your function has only one output).
To take the absolute value (on register X for example), execute the following steps:
Power X by 2
Power X by 0.5
PCalc is a great calculator app worth looking into. Website: http://www.pcalc.com
Update 4/28/2014:
Thanks to Terry for alerting me to this: In Radians mode, cos(1.57079632) returned an answer of 6.7948967066 x 10^-9, which is not accurate. Checking with Wolfram Alpha and with several calculators (HP 32Sii for example), cos(1.57079632) returns the correct answer of 6.7948966... x 10^-9. Hopefully, this gets corrected in the next update.
I checked cos(pi/2) and PCalc was accurate with answer of 0.
This ends my blog entry for now - off to see Seattle! Talk to you all soon! Thanks for the comments, recommendations, corrections, and compliments. As always, they are much appreciated.
Eddie
This blog is property of Edward Shore. 2014
## Sunday, April 20, 2014
### HP Prime: EC and BLOTCH - two programs using MOUSE and DRAWMENU
Happy Easter!
Happy 4/20 Day!
Happy Sunday!
These two programs for the HP Prime illustrate the use of DRAWMENU and MOUSE.
EC
Functions featured:
head: 1st element of a list
tail: all elements of a list except the 1st
l2norm: L-2 norm of a vector
ker: kernel of a matrix
even: is the number even?
Input: EC(argument)
The argument needs to be appropriate type of what you want to do.
For head and tail: the argument needs to be a list.
For l2norm: the argument needs to be a vector.
For ker and SPECRAD: the argument needs to be matrix
For even: we need an integer
Examples:
EC({7,8,9}), choosing head will return {7} while tail returns {8,9}.
EC([7,2,6,9]) while choosing l2norm returns √170.
EC([[1,4],[-3,-12]]) while choosing ker returns [[4, -1]].
EC([[1, 4],[-3, -12]]) while choosing SPECRAD returns 13.0384048104 (approximately)
For even, a result of 1 indicates that the number is even and 0 if the number is odd.
Program:
EXPORT EC(x)
BEGIN
// EWS 2014-04-20
LOCAL m,m1,mx,my;
WHILE MOUSE(1)≥0 DO END;
RECT;
TEXTOUT_P("Choose the function.",1,1,4);
TEXTOUT_P("head: 1st element of a list",1,18,4);
TEXTOUT_P("tail: all elements of a list except the 1st",1,35,4);
TEXTOUT_P("l2norm: L-2 norm of a vector",1,52,4);
TEXTOUT_P("ker: kernel of a matrix",1,69,4);
TEXTOUT_P("even: is the number even?",1,103,4);
REPEAT
m:=MOUSE;
m1:=m(1);
UNTIL SIZE(m1)>0;
mx:=m1(1);
my:=m1(2);
IF my≥220 AND my≤239 THEN
IF mx≥0 AND mx≤51 THEN
RETURN SUB(x,1,1);
END;
IF mx≥53 AND mx≤104 THEN
RETURN SUB(x,2,SIZE(x));
END;
IF mx≥106 AND mx≤157 THEN
RETURN exact(ABS(x));
END;
IF mx≥159 AND mx≤210 THEN
RETURN ker(x);
END;
IF mx≥212 AND mx≤263 THEN
RETURN CAS.SPECNORM(x);
END;
IF mx≥265 AND mx≤319 THEN
RETURN even(x);
END;
END;
END;
BLOTCH
Blotch Drawing Program
S = size of the square blotch
D = size of each box in the blotch. Each box is a randomized color.
To draw a square blotch, just touch the screen outside the menu.
Input: BLOTCH( )
Program:
EXPORT BLOTCH( )
BEGIN
// EWS 04-20-2014
// Initialize
LOCAL m,m1,mx,my,j,k,r;
WHILE MOUSE(1)≥0 DO END;
// Clear Canvas
RECT;
LOCAL s:=50, d:=4;
// Menu - to be redrawn
// Start main loop
REPEAT
// Get mouse data
REPEAT
m:=MOUSE; m1:=m(1);
UNTIL SIZE(m1)>0;
mx:=m1(1); my:=m1(2);
// Clear Screen
IF (my≥220 AND my≤319) AND (mx≥0 AND mx≤51) THEN
RECT;
END;
// Change Size
IF (my≥220 AND my≤319) AND (mx≥53 AND mx≤104) THEN
IF s<80 THEN s:=s+5; END;
END;
IF (my≥220 AND my≤319) AND (mx≥106 AND mx≤157) THEN
IF s>5 THEN s:=s-5; END;
END;
// Change Depth
IF (my≥220 AND my≤319) AND (mx≥159 AND mx≤210) THEN
IF d<8 THEN d:=d+2; END;
END;
IF (my≥220 AND my≤319) AND (mx≥212 AND mx≤263) THEN
IF d>2 THEN d:=d-2; END;
END;
// Exit Key
IF (my≥220 AND my≤319) AND (mx≥256 AND mx≤319) THEN
BREAK;
END;
// Draw Blotch
FOR j FROM mx-s/2 TO mx+s/2 STEP d DO
FOR k FROM my-s/2 TO my+s/2 STEP d DO
r:=RANDINT(1677215);
RECT_P(j,k,j+d-1,k+d-1,r);
END; END;
// Close main loop
UNTIL (my≥220 AND my≤319) AND (mx≥256 AND mx≤319);
RECT_P(0,220,319,239);
TEXTOUT_P("DONE!",146,220,4,#FF0000h);
WAIT(-1);
END;
As always, thank you for your comments, compliments, and questions. Happy Sunday everyone!
I will be heading to Seattle later this week - hope to blog from the Original Starbucks when I am there.
This blog is property of Edward Shore. 2014
## Thursday, April 17, 2014
### MAA Southern California-Nevada Section Spring Meeting: Highlights
Introduction
On April 12, 2014, I went to the Southern California-Nevada Section of the MAA Spring Meeting, held at Concordia University in Irvine, CA. It has been more than ten years since I last went to a meeting. It feels good to go back.
Link on to their website: http://sections.maa.org/socalnv/
Highlights
Hal Stern, University of Redlands talked about how statistics play a significant role in sports, and how it can be useful in measuring and predicting performance.
A student poster session, which lasted an hour. Given how excellent and engaging the student's posters were, I was only able to look at four in the hour given for the poster sessions, and I only wished that I looked at more.
After the poster session, Rachel Levy of Harvey Mudd College, spoke about her journey as a mathematician deals with the media. She states that good incidental communication is important because it form an impression on anyone who is listening.
Levy starts by emphasizing the need for mathematicians to communicate and be aware of how they communicate. Harvey Mudd requires all math majors take a public speaking class. The class has shown to have positive affects on her students. Levy also emphasize the use affirmations, such as saying to students "You are thinking like a math major," leading students to believe that they can join the mathematics community.
Levy stresses the need for positive communication. She challenges the often used saying "So easy even your grandmother can do it," implying the referred groups are seen as novices. This inspired her to start her blog, Grandma got STEM, which highlights grandmothers and their mathematical and scientific accomplishments. After communication with a librarian, her blog gained a significant increase in readers, leading to radio interviews world-wide.
Link to Grandma got STEM: ggstem.wordpress.com
Levy talked about how Twitter can be used to advertise positive messages regarding mathematics, advertising math blogs and events, and send thank you notes.
In the final part of the presentation, Levy describes her dealings with the general press, stating it is a risky proposition, as the press can easily distort the intended message (either intentionally or unintentionally). She gives tips include having your talking points prepared, thinking about the audience, having photographs and videos ready, and making sure the one takeaway point is said during the interview.
This is my favorite part of the spring meeting.
The next talk was given by Perla Myers, University of San Diego. Myers describe her mission to change the prevailing feelings of fear and distraught when people think of mathematics. She specializes in training future teachers to enhance mathematical understanding and introduce activities designed to make learning math enjoyable, such as the use of origami.
Jamie Pommersheim, Reed College, gave he final talk of the day. The topic: dissecting squares into triangles of equal areas.
It is possible to accomplish this task by using an even number of triangles, but what about odd number of triangles? This question was first addressed by Fred Richman, who at first posed this questions to his students. After finding out the difficulty of this task, he turned the question to American Mathematical Monthly publication.
It was later proved by Paul Monsky that splitting the square into an odd number of triangles of equal area was impossible. Pommersheim devoted the rest of his talk to describe why, using two approaches.
The first approach describes Monsky's proof. Pommersheim starts by describing the 2-adic norm which is described by:
|| n || = || 2^t * r/s || = 2^(-t)
Where n is a rational number, and r and s are odd integers. The 2-adic norm of 0 is defined to be 0.
Examples of calculating the 2-adic norms:
|| 6 || = || 2^1 * 3 || = 1/2 (t = 1)
|| 16 || = || 2^4 || = 1/16 (t = 4)
|| 5/8 || = || 1/8 * 5/1 || = || 2^(-3) * 5 || = 8 (t = -3)
Consider a square with corner points (0,0), (1,0), (1,1), and (0,1). Each corner point and any point that helps form triangles within that square is assigned a "color". For each point (x,y), the color is assigned as follows:
The color A is assigned if:
* x has the largest 2-adic norm or
* x at least as big of 2-adic norm of either y or 1.
The color B is assigned if:
* y has the biggest 2-adic norm or
* the 2-adic norm of y is 1 and x has a 2-adic norm is less than 1.
The color C is assigned if both x and y have 2-adic norms less than 1.
For the corner points, the following colors are assigned:
(0,0) has the color C
(1,0) has the color A
(1,1) has the color A
(0,1) has the color B
It is next shown that three collinear points cannot have all three colors A, B, and C. Consider the three points (0,0), (x1, y1), and (x2, y2). Point (0,0) is assigned the color C.
The area of a general triangle can be calculated by:
Area = 1/2 * det([[x1, y1, 1],[x2, y2, 1],[x3, y3, 1]])
Using this formula above, the "area" is x1*y2 - x2*y1. We know the area of any line is 0. Hence, x1*y2 - x2*y1 = 0. And:
Show that a straight line can contain points of only two colours.
x1*y2 = x2*y1
Taking the two 2-adic norms of both sides to get:
|| x1*y2 || = || x2*y1 ||
|| x1 || * || y2 || = || x2 || * || y1||
This implies that both points must be assigned the same color which contradicts the assumption that a line made of three collinear points can have three different colors.
The proof goes on to use Sperner's Lemma, which states (briefly) given any dissection of square there exists of a tricolored triangle. Also, the 2-adic norm of an area of tricolored triangle is greater than 1. However, if the square is divided into an odd number of triangles, the 2-adic norm of each triangle is 1.
Pommersheim shows a second way to demonstrate that squares cannot be cut into an odd number of equal area triangles. He uses finds a polynomial of areas that is associated with each dissection.
For a square dissected into four triangles, the associated polynomial is
D + B - (A + C), which A, B, C, and D represent the areas for each triangle. In this case each area of the triangle is n/4 where n is the area of the square. Clearly, n/4 + n/4 - (n/4 + n/4) = 0, which is the desired result.
For a square dissected into six triangles the polynomial becomes:
(A + C + E)^2 - 4*A*B - (B + D + F)^2 + 4*D*F.
Substituting n/6, the area of each triangle in this case, and the value of the polynomial is 0.
Pommersheim eliminates triangle B. Now we have five triangles, each with area n/5. The polynomial becomes:
(A + C + E)^2 - 4*A*C - (D + F)^2 + 4*D*F
The trouble comes when we evaluate the polynomial with each area n/5, which leaves the value n^2/5. This shows that it is impossible to divide a square into an odd number of triangles of equal area.
There it is. I hope you find this enjoying, informational, and inspiring. I look forward to going to the next one. Until next time,
Eddie
This blog is property of Edward Shore. 2014
### HP Prime Video: Scatterplots
A simple video on how to plot scatter plots on the HP Prime:
http://youtu.be/Q9E0ovCRMQc
This blog is property of Edward Shore. 2014
## Monday, April 14, 2014
### Program - HP 32SII: Stopping Sight Distance
Background: The stoping sight distance is the distance traveled when a person operating perceives the need to the stop and stops the vehicle. The velocity used in calculating stopping sight distance is the referred to as the design speed. The stopping sight distance is the sum of two parts:
1. Reaction Distance, which is the distance while the operator perceives the need to stop, and
2. Breaking Distance, which is the distance traveled while the operator puts the breaks on the vehicle, slowing the vehicle to a stop.
The general formula is:
SSD = v * t_r + v^2/(2*(a + g*G))
where:
v = the design speed of the vehicle
t_r = perception-reaction time
a = deceleration rate of the vehicle
g = gravity constant (32.174 ft/s^2 or 9.80665 m/s^2)
G = grade of the road. Grade is positive for uphill roads. If a road has a grade of 1%, it means for every 100 ft travelled horizontally, the road has risen 1 ft. In this formula, grade is given as a percentage (i.e. 1%, G = 1)
The AASHTO (American Association of State Highway and Transportation Officials) recommends t_r = 2.5 seconds and a = 11.2 ft/s^2.
The formula for SSD using U.S. units and recommended constants is given as:
SSD = 55/15 * v + (1.075*v^2)/(11.2 + 0.32*G)
Where velocity is given in mph. The HP 32sII program given below has the U.S. Formula.
Source: Goswami, Indramil Ph.D. P.E. "All In One Civil Engineering PE Breadth and Depth Exam Guide" 2nd Edition. McGraw Hill: 2012
PROGRAM
Input:
Y: speed of vehicle or design speed (mi/hr)
Output:
X: Stopping Sight Distance (in feet)
Formula used:
SSD = 55/15 * V + 1.075*V^2/(11+.32*G)
Assumptions:
* Total reaction time is 2.5 seconds. The deceleration rate of the vehicle is 11 ft/s^2. Both values are recommended by the AASHTO (American Association of State Highway and Transportation Officials).
Program:
S01 LBL S
S02 0.32
S03 *
S04 11.2
S05 +
S06 1/x
S07 1.075
S08 *
S09 x<>y
S10 ENTER
S11 R-down
S12 x^2
S13 *
S14 R-up
S15 55
S16 *
S17 15
S18 ÷
S19 +
S20 RTN
Examples:
Input:
Y: 65 (V)
X: 2 (G)
Result (Fix 4): 621.9376 ft
Input:
Y: 35 (V)
X: -4 (G)
Result: 261.0828
This blog is property of Edward Shore. 2014
## Thursday, April 10, 2014
### BASIC Programming Language Turns 50 on May 1
Thanks to Don Shepherd sharing this at the MoHPC website, here is the information regarding BASIC's 50th birthday! The programming language was born at Dartmouth College in Hanover, NH. John Kemeny and Thomas Kurtz were the original developers.
https://www.dartmouth.edu/basicfifty/
#programming #math
This blog is property of Edward Shore. 2014
## Wednesday, April 9, 2014
### Video: Calculating NPV, NFV, NUS using (most) any financial calculator
In this video I demonstrate on how to calculate net future value (NFV) and net utility stream (NUS) using any financial calculator with NPV (net present value) and TVM (time value of money) worksheets.
Eddie
This blog is property of Edward Shore. 2014
## Saturday, April 5, 2014
### Video: FILLPOLY command for the HP Prime
A new video has been posted!
http://youtu.be/ArdXnE2b6RQ
Syntax for the FILLPOLY command:
Cartesian:
FILLPOLY(points, color, alpha*)
Pixels:
FILLPOLY_P(points, color, alpha*)
points: a list of vertices {x_1, y_1, x_2, y_2...}
color: either a hexadecimal integer #RRGGBBh or use the RGB command
alpha: transparency factor, 0 to 255 (optional)
Eddie
This blog is property of Edward Shore. 2014
## Friday, April 4, 2014
### Civil Engineering Notes: Transportation
Source: Goswami, Indramil Ph.D. P.E. "All In One Civil Engineering PE Breadth and Depth Exam Guide" 2nd Edition. McGraw Hill: 2012
Intro:
Over the last year or so, the topic of civil engineering has fascinated me. While I don't expect to study for the Civil Engineering exam I wanted to get some sort of an encompassing resource that I learn a few things from in a desire to better understand some facets of engineering. I ended up purchasing a Kindle Edition of the All In One Civil Engineering book primarily because I felt it provided me with the best bang for the buck. Plus it was a better alternative to get a newer edition electronically for half of what I would have paid for a first edition of physical book. From time to time, I plan to post some notes of what I read.
This post is regarding transportation.
Terms:
Trip Generation: the number of trips made on a network. The network is implied to be made of zones. Zones are small areas that are determined by population, employment, or other given measurement, which are said to be anywhere from 1/4 square mile to a full mile. Trips are said to original from one designated zone (trip producer) to a destination zone (trip attractor).
Time Mean Speed (S_T): the arithmetic mean of speeds of vehicles passing over a given point. Let s_1, s_2, ... , s_n be the speeds of passing vehicles. Then the time mean speed is:
S_T = (s_1 + s_2 + ... + s_n) / n
Space Mean Speed (S_R): the average mean speeds with delays taken into account. This is theorized by taking the harmonic mean of speeds of vehicles passing over a given point. A mathematical definition is:
S_R = n / (1/s_1 + 1/s_2 + ... + 1/s_n)
Note that this is not the only way to calculate space mean speed. Using the strict definition:
S_T = total distance ÷ total time including stops and delays
In general, S_R ≤ S_T.
Consider two speeds A and B, assuming A and B are positive. Then:
S_T = (A+B)/2 and
S_R = 2/(1/A +1/B)
= 2 * (A+B)/(A*B)
Setting S_T and S_R equal to each other yields:
1/2 * (A+B) = 2 * (A+B)/(A*B)
A * B = 4
Hence if A * B < 4, S_T < S_T. Otherwise, S_T > S_R.
Speed-Volume-Density
q = s * d
Where:
q = flow rate of vehicles per hour
s = average vehicle per hour
d = density, measured in vehicle per mile
Spacing per foot is calculated by: (5280 ft/mi)
d_ a = 5280/d
Where vehicle headway is calculated by: (3600 sec/hr)
h_a = 3600/q
Example: Observing a part of a highway for one hour, an engineer observes 890 vehicles pass at an average of 55.5 miles per hour in one direction.
The density is: d = 890/55.5 ≈ 16.0360 vehicles per mile
This leads to a spacing per foot as: d_a ≈ 5280/16.0360 ≈ 329.2584 feet per vehicle
And the headway per vehicle is: h_ a = 3600/890 ≈ 4.0449 seconds per vehicle
There are several models regarding traffic flow and density. One is the Greenfield's Linear Model, a general model that is used in most traffic conditions.
S_R = S_ f * (1 - D/D_j)
S_R = average space mean speed (harmonic average of speeds)
S_f = free flow speed. This is the speed one would travel if there was no traffic, the weather was near perfect, and there are no obvious road obstacles.
D = actual density
D_j = jam density. The jam density is achieved when traffic is stopped. If you live in the Greater Los Angeles area, you can easily observe freeways at jam density. And yes, it is a pain to drive on the I-210 Parking Lot.
There are other models that the boom describes, such as the Greenberg Density model, which is used for dense traffic situations only.
The Space Mean Speed of a Train
The book provides a very interesting example of using simple velocity and acceleration equations from physics to help calculate S_R. Here is an example (not from the book, but one I made up):
A train, the Star Train, travels from Mario Station to Luigi Station, which is a 14 mile (73920 ft) trip. The train takes one minute to load and drop passengers at each station. Information for the Star Train is as follows:
Acceleration: 5 ft/s^2
Deceleration: 4.5 ft/s^2
Peak Speed: 84 mph ≈ 123.48 ft/s
What is the space mean speed?
We have the first find out the total time, including delays.
Accelerating phase:
Time it takes to accelerate to full speed: t _a = v_f/a = 123.48/5 ≈ 24.696 s
Distance traveled: s_a = a*t_a^2/2 = 5 * 24.696^2 / 2 ≈ 1524.731 ft
Deceleration phase:
Time it takes to decelerate from full speed: t_d = -v_f/-a_d = -123.48/-4.5 ≈ 27.44 s
Distance traveled:
s_d = -a_d*t_d^2/2 + t*v_f = -4.5*27.44^2/2 + 27.44*123.48 ≈ 1694.1456 ft
Constant Speed Phase:
Here we reverse the order of calculations: distance, the time.
Distance:
s_c = L - s_a - s_d = 73920 - 1524.731 - 1694.1456 = 70701.123
Time:
t_c = s_c/v_f = 70701.123/123.48 ≈ 572.5715 ft/s
Total time:
T = t_a + t_c + t_d + t_delays
= 24.696 + 572.5715 + 27.44 + 60 * 2
= 744.7075 s (about 12 min 24 s)
Finally calculate space mean speed:
S_R = L/T = 73920/744.7075 ≈ 99.2604 ft/s ≈ 67.5241 mph
So the space mean speed of the Star Train from Mario Station to Luigi Station is about 67.5241 miles and hour.
(1 mi/hr ≈ 1.47 ft/s, actually 1.4666666666666... ft/s = 22/15 ft/s)
Hopefully this helps and you enjoyed this post as I did. As always comments and questions are welcome and appreciated. Until next time, take care!
Eddie
This blog is property of Edward Shore. 2014
### Three Generations of the BA II Plus
Here is my collection of the BA II Plus calculators I have with the year I bought them typed into each one.
Eddie
|
##### Math Word Problems For Dummies
If you need to calculate someone’s salary increase or raise based on the percentage of the raise, you can do this using the distributive property. Here’s an example:
Alison’s salary was \$40,000 last year, and at the end of the year she received a 5% raise. What will she earn this year?
To solve this problem, first realize that Alison got a raise. So whatever she makes this year, it will be more than she made last year. The key to setting up this type of problem is to think of percent increase as “100% of last year’s salary plus 5% of last year’s salary.” Here’s the word equation:
This year’s salary = 100% of last year’s salary + 5% of last year’s salary
Before you continue, you should be familiar with the distributive property. It says that multiplying a number by the sum of two other numbers in parentheses is the same thing as multiplying by the numbers in parentheses individually and then adding their products. In other words,
3(1 + 5) = 3(1) + 3(5)
If you evaluate both sides of the equation, here’s what you get. You can see that they’re equal:
3(6) = 3 + 15
18 = 18
The property also works for subtraction:
5(6 – 4) = 5(6) – 5(4)
5(2) = 30 – 20
10 = 10
In percent terms, you can say that the following statements are true:
Salary (100% + 5%) = Salary (100%) + Salary (5%)
Salary (100% – 5%) = Salary (100%) – Salary (5%)
What’s the point? Well, you can solve percent problems either way, but adjusting the percents first is often easier. You may be able to do the addition or subtraction in your head, and as soon as you do the multiplication, you have your answer.
Now, returning to the above problem, you can just add the percentages:
This year’s salary = (100% + 5%) of last year’s salary = 105% of last year’s salary
Change the percent to a decimal and the word of to a multiplication sign; then fill in the amount of last year’s salary:
This year’s salary = 1.05 \$40,000
Now you’re ready to multiply:
This year’s salary = \$42,000
So Alison’s new salary is \$42,000.
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### Counting Factors
Is there an efficient way to work out how many factors a large number has?
### Repeaters
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### Oh! Hidden Inside?
Find the number which has 8 divisors, such that the product of the divisors is 331776.
# Eminit
##### Stage: 3 Challenge Level:
Correct solutions were received form Robert of Forres Academy, Andrei of School NO. 205, Bucharest and Chong, Chen and Teo of Secondary 1B, River Valley High School, Singapore.
Each solution identified that groups of six 8s and six 9s are divisible by seven.
(Is it the case that all six figure numbers made by repeating the same digit are divisible by 7?). This meant that because the original number had 50 digits on each side of M: there are eight groups of six and two digits left.
Which puts the problem into the form:
What is M if 88M99 is divisible by 7?
Robert and Andrei found M by trial and improvement methods and the full solution from the Valley High School Singapore is given below. Well done
The value of the digit M is 5.
When 888888 is divided by 7, we will get 126984. Besides, 8, 88, 888, 8888 and 88888 cannot be divided 7 without any remainder and 888888 is the smallest number made up of eights which can be divided by 7. Thus, when the digit 8 is repeated 50 times (88888888888888 888888888888888888888888888888888888) is divided by 7, we will get the number 126984 repeated 8 times and 12 behind the number, which is 12698412698412698412698412698412698412698412698412 and has a remainder 4.
On the other hand, when 999999 is divided by 7, we will get 142857. Besides, 9, 99, 999, 9999 and 99999 cannot be divided 7 without any remainder and 999999 is the smallest number made up of nines which can be divided by 7. Thus when the digit 9 is repeated 50 times (99999999 999999999999999999999999999999999999999999) is divided by 7, we will get the number 142857 repeated 8 times and 14 behind the number, which is 142857142857142857142857142857142 85714285714285714 and has a remainder 1.
Thus, when the digit 8 is repeated 50 times and divided by 7, the remainder 4 is carried forward and in order to avoid getting any remainder when the digit 9 is repeated 50 times and is divided by 7, we have to take out the first two 9s. Therefore, 4M99 must be able to be divided by 7, so after many trials and errors, we managed to find out that the number that can be divided by 7, is 4599. Therefore, we are able to conclude that the value of the digit M is 5.
Well done to all of you.
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# Law of Sines
## Definition
The Law of Sines is a relationship linking the sides of a triangle with the sine of their corresponding angles. The statement is as follows: Given triangle $$ABC$$, with corresponding side lengths $$a, b$$ and $$c$$ and corresponding angles $$\alpha, \beta$$ and $$\gamma$$, and $$R$$ as the radius of the circumcircle of triangle $$ABC$$, we have the following:
$\frac { a}{ \sin \alpha} = \frac {b}{\sin \beta} = \frac {c} {\sin \gamma} = 2R.$
Note: The statement without the third equality is often referred to as the Sine Rule. The third equality is often referred to as the Extended Sine Rule.
### Proof of the Extended Sine Rule
Let $$O$$ be the center of the circumcircle. Let $$D$$ be the midpoint of $$BC$$, then $$OD$$ is perpendicular to $$BC$$. Note that $$\angle BOC$$ is equal to $$2\alpha$$ or $$360^\circ - 2\alpha$$, depending on whether $$O$$ is in the triangle or not. This gives $$\angle BOD = \alpha$$ or $$180^\circ - \alpha$$, and thus $$\sin BOD = \sin \alpha$$. As such, $\frac {BD}{OB} = \sin \alpha \Rightarrow \frac {a} {\sin \alpha } = 2R.\, _\square$
## Technique
One real-life application of the sine rule is the Sine Bar, which is used to measure the angle of a tilt in engineering. Other common examples include measurement of distances in navigation, and the measurement of distance between two stars in astronomy.
A common textbook application of the sine rule, is to determine the triangle $$ABC$$ given some of its sides and angles. It is worthwhile to mention the 'ambiguous case': Given $$BC=a, AB=c, \angle BAC = \alpha$$ such that $$c \sin \alpha < a < c$$, then there are 2 distinct triangles that satisfy such a configuration.
## Application and Extensions
### Show that the area of triangle $$ABC$$ is equal to $$\frac {abc} {4R}.$$
Let $$D$$ be the foot of the perpendicular from $$A$$ to $$BC$$. Using $$BC$$ as the base and $$AD$$ as the height, the area of the triangle is $$\frac {1}{2} a \cdot |AD|$$. From right triangle $$CAD$$, $$\sin \gamma = \frac {|AD|} {b}$$. Thus, the area of the triangle is $$\frac {1}{2} a \cdot |AD| = \frac {1}{2} a b \sin \gamma$$, which is often quoted. Now, using the Extended sine rule, we have, $$\frac {c}{\sin \gamma} = 2R$$, thus the area of the triangle is $$\frac {1}{2} a b \sin \gamma = \frac {1}{2} a b \frac {c}{2R} = \frac {abc} {4R}$$. $$_\square$$
### In a circle of radius 5, two perpendicular chords $$AB$$ and $$CD$$ are drawn, such that they intersect at $$P$$ within the circle. What is the value of $$AP^2 + BP^2 + CP^2 +DP^2?$$
Applying the Pythagorean theorem to triangles $$APC$$ and $$BPD$$, we get $$AP^2 + CP^2 = AC^2$$, $$BP^2 + PD^2 = BD^2$$. Applying the Extended Sin Rule to triangles $$ABC$$ and $$BCD$$, we get $$AC = 2R \sin \angle ABC$$ and $$BD = 2R \sin \angle DCB$$, where $$R=5$$ is the radius of the circle. Applying the trigonometic form of the Pythagorean theorem to triangle $$PBC$$, we get $$\sin^2 \angle PCB + \sin ^2 \angle PBC = 1$$. Putting this all together, we have
\begin{align} AP^2 + BP^2 + CP^2 + DP^2 &= AC^2 + BD^2 \\ &= 100 \sin ^2 \angle ABC + 100 \sin^2 \angle BCD \\ &= 100 (\sin^2 \angle PBC + \sin^2 \angle PCB) \\ &=100. _\square \end{align}
Note that this sum is independent of the point $$P$$.
### [Angle Bisector Theorem] In triangle $$ABC$$, let $$D$$ be a point on $$BC$$ such that $$AD$$ is the angle bisector of $$\angle BAC$$. Show that $$\frac {AB}{BD} = \frac {AC}{CD}$$.
Applying the sine rule to triangle $$ABD$$, we get:
$\frac {AB}{\sin \angle ADB} = \frac {BD} {\sin \angle BAD} \Rightarrow \frac {AB}{BD} = \frac {\sin \angle ADB}{\sin \angle BAD}.$
Applying the sine rule to triangle $$ACD$$, we get:
$\frac {AC}{\sin \angle ADC} = \frac {CD}{\sin\angle DAC} \Rightarrow \frac {AC}{CD} = \frac {\sin \angle ADC} {\sin \angle DAC}.$
Since $$AD$$ is the angle bisector of $$\angle BAC$$, $$\angle BAD = \angle DAC$$ implies that $$\sin \angle BAD = \sin \angle DAC$$. Since $$BDC$$ is a straight line, $$\angle ADB = 180^\circ - \angle ADC \Rightarrow \sin \angle ADB = \sin \angle ADC$$. Thus, $\frac {AB}{BD} =\frac {\sin \angle ADB}{\sin \angle BAD} = \frac {\sin \angle ADC} {\sin \angle DAC} = \frac {AC}{CD}. _\square$
Note by Arron Kau
4 years, 6 months ago
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we are presently studying this topic in the school...I proved this while doing vectors
- 4 years, 2 months ago
A regular pentagon is inscribed in a circle whose radius is 5 cm. Find the one of the sides of the pentagon by using law of sines.
- 3 years, 7 months ago
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## Modeling and Finding Equivalent Fractions
### Learning Outcomes
• Model equivalent fractions
• Find equivalent fractions
Let’s think about Andy and Bobby and their favorite food again. If Andy eats ${\Large\frac{1}{2}}$ of a pizza and Bobby eats ${\Large\frac{2}{4}}$ of the pizza, have they eaten the same amount of pizza? In other words, does ${\Large\frac{1}{2}}={\Large\frac{2}{4}}?$ We can use fraction tiles to find out whether Andy and Bobby have eaten equivalent parts of the pizza.
### Equivalent Fractions
Equivalent fractions are fractions that have the same value.
Fraction tiles serve as a useful model of equivalent fractions. You may want to use fraction tiles to do the following activity. Or you might make a copy of the fraction tiles shown earlier and extend it to include eighths, tenths, and twelfths.
Start with a ${\Large\frac{1}{2}}$ tile. How many fourths equal one-half? How many of the ${\Large\frac{1}{4}}$ tiles exactly cover the ${\Large\frac{1}{2}}$ tile?
Since two ${\Large\frac{1}{4}}$ tiles cover the ${\Large\frac{1}{2}}$ tile, we see that ${\Large\frac{2}{4}}$ is the same as ${\Large\frac{1}{2}}$, or ${\Large\frac{2}{4}}={\Large\frac{1}{2}}$.
How many of the ${\Large\frac{1}{6}}$ tiles cover the ${\Large\frac{1}{2}}$ tile?
Since three ${\Large\frac{1}{6}}$ tiles cover the ${\Large\frac{1}{2}}$ tile, we see that ${\Large\frac{3}{6}}$ is the same as ${\Large\frac{1}{2}}$.
So, ${\Large\frac{3}{6}}={\Large\frac{1}{2}}$. The fractions are equivalent fractions.
### Example
Use fraction tiles to find equivalent fractions. Show your result with a figure.
1. How many eighths (${\Large\frac{1}{8}}$) equal one-half (${\Large\frac{1}{2}}$)?
2. How many tenths (${\Large\frac{1}{10}}$) equal one-half (${\Large\frac{1}{2}}$)?
3. How many twelfths (${\Large\frac{1}{12}}$) equal one-half (${\Large\frac{1}{2}}$)?
Solution
1. It takes four ${\Large\frac{1}{8}}$ tiles to exactly cover the ${\Large\frac{1}{2}}$ tile, so ${\Large\frac{4}{8}}={\Large\frac{1}{2}}$.
2. It takes five ${\Large\frac{1}{10}}$ tiles to exactly cover the ${\Large\frac{1}{2}}$ tile, so ${\Large\frac{5}{10}}={\Large\frac{1}{2}}$.
3. It takes six ${\Large\frac{1}{12}}$ tiles to exactly cover the ${\Large\frac{1}{2}}$ tile, so ${\Large\frac{6}{12}}={\Large\frac{1}{2}}$.
Suppose you had tiles marked ${\Large\frac{1}{20}}$. How many of them would it take to equal ${\Large\frac{1}{2}}$? Are you thinking ten tiles? If you are, you’re right, because ${\Large\frac{10}{20}}={\Large\frac{1}{2}}$.
We have shown that ${\Large\frac{1}{2},\frac{2}{4},\frac{3}{6},\frac{4}{8},\frac{5}{10},\frac{6}{12}}$, and ${\Large\frac{10}{20}}$ are all equivalent fractions.
## Find Equivalent Fractions
We used fraction tiles to show that there are many fractions equivalent to ${\Large\frac{1}{2}}$. For example, ${\Large\frac{2}{4},\frac{3}{6}}$, and ${\Large\frac{4}{8}}$ are all equivalent to ${\Large\frac{1}{2}}$. When we lined up the fraction tiles, it took four of the ${\Large\frac{1}{8}}$ tiles to make the same length as a ${\Large\frac{1}{2}}$ tile. This showed that ${\Large\frac{4}{8}}={\Large\frac{1}{2}}$. See the previous example.
We can show this with pizzas, too. Image (a) shows a single pizza, cut into two equal pieces with ${\Large\frac{1}{2}}$ shaded. Image (b) shows a second pizza of the same size, cut into eight pieces with ${\Large\frac{4}{8}}$ shaded.
This is another way to show that ${\Large\frac{1}{2}}$ is equivalent to ${\Large\frac{4}{8}}$.
How can we use mathematics to change ${\Large\frac{1}{2}}$ into ${\Large\frac{4}{8}}$? How could you take a pizza that is cut into two pieces and cut it into eight pieces? You could cut each of the two larger pieces into four smaller pieces! The whole pizza would then be cut into eight pieces instead of just two. Mathematically, what we’ve described could be written as:
${\Large\frac{1\cdot\color{blue}{4}}{2\cdot\color{blue}{4}}}={\Large\frac{4}{8}}$
These models lead to the Equivalent Fractions Property, which states that if we multiply the numerator and denominator of a fraction by the same number, the value of the fraction does not change.
### Equivalent Fractions Property
If $a,b$, and $c$ are numbers where $b\ne 0$ and $c\ne 0$, then
${\Large\frac{a}{b}}={\Large\frac{a\cdot c}{b\cdot c}}$
When working with fractions, it is often necessary to express the same fraction in different forms. To find equivalent forms of a fraction, we can use the Equivalent Fractions Property. For example, consider the fraction one-half.
${\Large\frac{1\cdot\color{blue}{3}}{2\cdot\color{blue}{3}}}={\Large\frac{3}{6}}$ so ${\Large\frac{1}{2}}={\Large\frac{3}{6}}$
${\Large\frac{1\cdot\color{blue}{2}}{2\cdot\color{blue}{2}}}={\Large\frac{2}{4}}$ so ${\Large\frac{1}{2}}={\Large\frac{2}{4}}$
${\Large\frac{1\cdot\color{blue}{10}}{2\cdot\color{blue}{10}}}={\Large\frac{10}{20}}$ so ${\Large\frac{1}{2}}={\Large\frac{10}{20}}$
So, we say that ${\Large\frac{1}{2},\frac{2}{4},\frac{3}{6}}$, and ${\Large\frac{10}{20}}$ are equivalent fractions.
### Example
Find three fractions equivalent to ${\Large\frac{2}{5}}$.
### Try it
Find three fractions equivalent to ${\Large\frac{3}{5}}$.
Find three fractions equivalent to ${\Large\frac{4}{5}}$.
### Example
Find a fraction with a denominator of $21$ that is equivalent to ${\Large\frac{2}{7}}$.
### Try it
In the following video we show more examples of how to find an equivalent fraction given a specific denominator.
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# Completing the Square: Two Geometrical Proofs
On this website I previously showed you why the formulas used to complete the square work – and how they can be used to derive formulas such as the quadratic equation. Now, I’ll be doing something different, but related… On this post, I’ll be showing you how to come up with the formulas (2 in total) used to complete the square – geometrically.
#### Completing the Square Formula (Derivation 1):
In the diagram above, what we can see is that:
${ x }^{ 2 }+2\cdot \frac { b }{ 2 } x+{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ \left( x+\frac { b }{ 2 } \right) }^{ 2 }$
This means that:
${ x }^{ 2 }+bx+{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ \left( x+\frac { b }{ 2 } \right) }^{ 2 }\\ \\ \therefore \quad { x }^{ 2 }+bx={ \left( x+\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }$
#### Completing the Square Formula (Derivation 2):
In the diagram above, what we can see is that:
${ \left( x-\frac { b }{ 2 } \right) }^{ 2 }+2\cdot \frac { b }{ 2 } \left( x-\frac { b }{ 2 } \right) +{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ x }^{ 2 }$
This means that:
${ \left( x-\frac { b }{ 2 } \right) }^{ 2 }+b\left( x-\frac { b }{ 2 } \right) +{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ x }^{ 2 }\\ \\ { \left( x-\frac { b }{ 2 } \right) }^{ 2 }+bx-\frac { { b }^{ 2 } }{ 2 } +\frac { { b }^{ 2 } }{ 4 } ={ x }^{ 2 }\\ \\ { \left( x-\frac { b }{ 2 } \right) }^{ 2 }+bx-\frac { 2{ b }^{ 2 } }{ 4 } +\frac { { b }^{ 2 } }{ 4 } ={ x }^{ 2 }\\ \\ { \left( x-\frac { b }{ 2 } \right) }^{ 2 }+bx-\frac { { b }^{ 2 } }{ 4 } ={ x }^{ 2 }\\ \\ { \left( x-\frac { b }{ 2 } \right) }^{ 2 }+bx-{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ x }^{ 2 }\\ \\ \therefore \quad { x }^{ 2 }-bx={ \left( x-\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }$
I hope these geometrical proofs have helped you better understand why the formulas we use to complete the square are in existence. Thanks for reading! 😀
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Serviceline Industrial Sensors
Serviceline Explosion Protection
# Did You Find Out? Get the Answer to the Riddle of E-News Issue No. 6, November 2019
2019-11-13
## Field Trip to Industry 4.0
How old is the chemical plant and the automation technology it uses?
A group of students goes on a field trip to a chemical plant to experience Industry 4.0 in practice. One student recognizes that the automation technology is modern, but the plant as a whole must be significantly older. The student asks how old the plant and its installed automation technology are. Knowing that the "intelligent" automation technology can answer this question, the chemical plant supervisor challenges the students with the following puzzle:
The chemical plant is seven times as old as the automation technology was, when the chemical plant was as old as the automation technology is today. When the automation technology is as old as the plant is today, the sum of the age of the chemical plant and the automation technology will be 51 years.
How old are the chemical plant and the installed automation technology right now?
## Solution
If you translate the student's question into a mathematical expression, the resulting equations solve for the following two unknown quantities:
c = Age of chemical plant
a = Age of automation technology
The equation referring to the statement “The chemical plant is seven times as old as the automation technology was when the chemical plant was as old as the automation technology is today” is:
c = 7 * ( a – ( c – a))
c = 7 * ( a – c + a)
c = 7 * (2a – c)
c = 14a – 7c
8c = 14a
c = 14a/8
c = (7/4)a
The equation referring to the statement “When the automation technology will be as old as the plant is today, the sum of the age of the chemical plant and the automation technology will be 51 years” is:
c + (c – a) + a + (c – a) = 51
2c – a + a + c – a = 51
3c – a = 51
Plug first equation into second equation:
3 * (7/4)a – a = 51
21/4 a – a = 51
17/4 a = 51 / (: 17/4)
a = 12
According to this, the automation technology is 12 years old today
Plugging this result (a = 12) into the first equation yields the age of the chemical plant.
c = (7/4)a
c = (7/4) * 12 = 21
According to this, the chemical plant is currently 21 years old.
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# Medians
## Line segment that joins a vertex and the midpoint of the opposite side of a triangle.
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Medians
What if your art teacher assigned an art project involving triangles? You decide to make a series of hanging triangles of all different sizes from one long piece of wire. Where should you hang the triangles from so that they balance horizontally?
You decide to plot one triangle on the coordinate plane to find the location of this point. The coordinates of the vertices are (0, 0), (6, 12) and (18, 0). What is the coordinate of this point? After completing this Concept, you'll be able to use medians to help you answer these questions.
### Guidance
A median is the line segment that joins a vertex and the midpoint of the opposite side (of a triangle). The three medians of a triangle intersect at one point, just like the perpendicular bisectors and angle bisectors. This point is called the centroid , and is the point of concurrency for the medians of a triangle. Unlike the circumcenter and incenter, the centroid does not have anything to do with circles. It has a different property.
##### Investigation: Properties of the Centroid
Tools Needed: pencil, paper, ruler, compass
1. Construct a scalene triangle with sides of length 6 cm, 10 cm, and 12 cm (Investigation 4-2). Use the ruler to measure each side and mark the midpoint.
2. Draw in the medians and mark the centroid.
Measure the length of each median. Then, measure the length from each vertex to the centroid and from the centroid to the midpoint. Do you notice anything?
3. Cut out the triangle. Place the centroid on either the tip of the pencil or the pointer of the compass. What happens?
From this investigation, we have discovered the properties of the centroid. They are summarized below.
Concurrency of Medians Theorem: The medians of a triangle intersect in a point that is two-thirds of the distance from the vertices to the midpoint of the opposite side. The centroid is also the “balancing point” of a triangle.
If $G$ is the centroid, then we can conclude:
$AG &= \frac{2}{3} AD, CG=\frac{2}{3} CF, EG=\frac{2}{3} BE\\DG &= \frac{1}{3} AD, FG=\frac{1}{3} CF, BG=\frac{1}{3} BE$
And, combining these equations, we can also conclude:
$DG=\frac{1}{2} AG, FG=\frac{1}{2} CG, BG=\frac{1}{2} EG$
In addition to these ratios, $G$ is also the balance point of $\triangle ACE$ . This means that the triangle will balance when placed on a pencil at this point.
#### Example A
Draw the median $\overline{LO}$ for $\triangle LMN$ below.
From the definition, we need to locate the midpoint of $\overline{NM}$ . We were told that the median is $\overline{LO}$ , which means that it will connect the vertex $L$ and the midpoint of $\overline{NM}$ , to be labeled $O$ . Measure $NM$ and make a point halfway between $N$ and $M$ . Then, connect $O$ to $L$ .
#### Example B
Find the other two medians of $\triangle LMN$ .
Repeat the process from Example A for sides $\overline{LN}$ and $\overline{LM}$ . Be sure to always include the appropriate tick marks to indicate midpoints.
#### Example C
$I, K$ , and $M$ are midpoints of the sides of $\triangle HJL$ .
a) If $JM = 18$ , find $JN$ and $NM$ .
b) If $HN = 14$ , find $NK$ and $HK$ .
a) $JN$ is two-thirds of $JM$ . So, $JN =\frac{2}{3} \cdot 18=12$ . $NM$ is either half of 12, a third of 18 or $18 - 12$ . $NM = 6$ .
b) $HN$ is two-thirds of $HK$ . So, $14 =\frac{2}{3} \cdot HK$ and $HK = 14 \cdot \frac{3}{2} = 21$ . $NK$ is a third of 21, half of 14, or $21-14$ . $NK = 7$ .
Watch this video for help with the Examples above.
#### Concept Problem Revisited
The point that you should put the wire through is the centroid. That way, each triangle will balance on the wire.
The triangle that we wanted to plot on the $x-y$ plane is to the right. Drawing all the medians, it looks like the centroid is (8, 4). To verify this, you could find the equation of two medians and set them equal to each other and solve for $x$ . Two equations are $y=\frac{1}{2} x$ and $y=-4x+36$ . Setting them equal to each other, we find that $x = 8$ and then $y = 4$ .
### Guided Practice
1. Find the equation of the median from $B$ to the midpoint of $\overline{AC}$ for the triangle in the $x-y$ plane below.
2. $H$ is the centroid of $\triangle ABC$ and $DC = 5y - 16$ . Find $x$ and $y$ .
3. True or false: The median bisects the side it intersects.
1. To find the equation of the median, first we need to find the midpoint of $\overline{AC}$ , using the Midpoint Formula.
$\left(\frac{-6+6}{2}, \frac{-4+(-4)}{2}\right)=\left(\frac{0}{2}, \frac{-8}{2}\right)=(0,-4)$
Now, we have two points that make a line, $B$ and the midpoint. Find the slope and $y-$ intercept.
$m &= \frac{-4-4}{0-(-2)}=\frac{-8}{2}=-4\\y &= -4x+b\\-4 &= -4(0)+b\\-4 &= b$
The equation of the median is $y=-4x-4$
2. $HF$ is half of $BH$ . Use this information to solve for $x$ . For $y$ , $HC$ is two-thirds of $DC$ . Set up an equation for both.
$\frac{1}{2} BH &= HF \ \text{or} \ BH=2HF && HC=\frac{2}{3} DC \ \text{or} \ \frac{3}{2} HC=DC\\3x+6 &= 2(2x-1) && \frac{3}{2} (2y+8)=5y-16\\ 3x+6 &= 4x-2 && \quad 3y+12=5y-16\\8 &= x && \qquad \quad \ 28=2y$
3. This statement is true. By definition, a median intersects a side of a triangle at its midpoint. Midpoints divide segments into two equal parts.
### Explore More
For questions 1-4, find the equation of each median, from vertex $A$ to the opposite side, $\overline{BC}$ .
1. $A(9, 5), B(2, 5), C(4,1)$
2. $A(-2, 3), B(-3, -7), C(5, -5)$
3. $A(-1, 5), B(0, -1), C(6, 3)$
4. $A(6, -3), B(-5, -4), C(-1, -8)$
For questions 5-9, $B, D$ , and $F$ are the midpoints of each side and $G$ is the centroid. Find the following lengths.
1. If $BG = 5$ , find $GE$ and $BE$
2. If $CG = 16$ , find $GF$ and $CF$
3. If $AD = 30$ , find $AG$ and $GD$
4. If $GF = x$ , find $GC$ and $CF$
5. If $AG = 9x$ and $GD = 5x - 1$ , find $x$ and $AD$ .
Use $\triangle ABC$ with $A(-2, 9), B(6, 1)$ and $C(-4, -7)$ for questions 10-15.
1. Find the midpoint of $\overline{AB}$ and label it $M$ .
2. Write the equation of $\overleftrightarrow{CM}$ .
3. Find the midpoint of $\overline{BC}$ and label it $N$ .
4. Write the equation of $\overleftrightarrow{AN}$ .
5. Find the intersection of $\overleftrightarrow{CM}$ and $\overleftrightarrow{AN}$ .
6. What is this point called?
Another way to find the centroid of a triangle in the coordinate plane is to find the midpoint of one side and then find the point two thirds of the way from the third vertex to this point. To find the point two thirds of the way from point $A(x_1, y_1)$ to $B(x_2, y_2)$ use the formula: $\left(\frac{x_1+2x_2}{3}, \frac{y_1+2y_2}{3}\right)$ . Use this method to find the centroid in the following problems.
1. (-1, 3), (5, -2) and (-1, -4)
2. (1, -2), (-5, 4) and (7, 7)
3. (2, -7), (-5, 1) and (6, -9)
### Vocabulary Language: English
centroid
centroid
The centroid is the point of intersection of the medians in a triangle.
Median
Median
The median of a triangle is the line segment that connects a vertex to the opposite side's midpoint.
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# 10.3 | Finding and Using Taylor Series
Find the Taylor series representation for $$\displaystyle f(x) = \frac{1}{1+x}$$ at $$x=2$$.
\begin{aligned} f(x) & = \frac{1}{1+x} = \frac{1}{3+(x-2)} \\ & = \frac{1}{3 \left[ 1 + \left( \frac{x-2}{3} \right) \right]} = \frac{1}{3} \frac{1}{1-\left(-\frac{x-2}{3} \right)} \\ & = \sum_{n=0}^{\infty} (-1)^n \frac{(x-2)^n}{3^{n+1}}, \; \; \left|-\frac{x-2}{3} \right|<1 \end{aligned} where
• we change the form of $$f(x)$$ so that it includes the expression $$(x-2)$$,
• and then apply the formula
$$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n, |x| <1$$ replacing $$x$$ with $$-\frac{x-2}{3}$$.
So the Taylor series representation for $$\displaystyle f(x) = \frac{1}{1+x}$$ at $$x=2$$ is
$$\sum_{n=0}^{\infty} (-1)^n \frac{(x-2)^n}{3^{n+1}},$$
and the interval of convergence is $$-1 < x < 5$$.
Find the Maclaurin series for $$\displaystyle f(x) = x^2 \sin 2x$$.
\begin{aligned} f(x) & = x^2 \sin 2x = x^2 \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n+1}}{(2n+1)!} \\ & = \sum_{n=0}^{\infty} (-1)^n \frac{2^{2n+1} x^{2n+3}}{(2n+1)!} \end{aligned}
where
• we apply the formula
$$\sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$
replacing $$x$$ with $$2x$$,
• distribute the exponent $$2n+1$$ over $$2x$$,
• and multiply in $$x^2$$
So the Maclaurin series for $$\displaystyle f(x) = x^2 \sin 2x$$ is
$$\sum_{n=0}^{\infty} (-1)^n \frac{2^{2n+1} x^{2n+3}}{(2n+1)!},$$
and the interval of convergence is $$-\infty < x < \infty$$.
Find the Taylor series for $$\displaystyle f(x) = e^{-2x}$$ centered at $$3$$.
Observe that,
\begin{aligned} f(x) & = e^{-2x} = e^{-2(x-3) -6} = e^{-6}e^{-2(x-3)} \\ & = e^{-6}\sum_{n=0}^{\infty} \frac{(-2(x-3))^n}{n!} \\ & = \sum_{n=0}^{\infty} \frac{(-1)^n2^n(x-3)^n}{e^6 n!} \end{aligned}
where
• rewrite $$f(x)$$ so that it includes the expression $$(x-3)$$
• apply the formula
$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$
replacing $$x$$ with $$2(x-3)$$,
• distribute the exponent $$n$$ over $$2(x-3)$$,
• and multiply in $$e^{-6}$$
So the Taylor series for $$\displaystyle f(x) = e^{-2x}$$ centered at $$3$$ is
$$\sum_{n=0}^{\infty} \frac{(-1)^n2^n(x-3)^n}{e^6 n!},$$
and the interval of convergence is $$-\infty < x < \infty$$.
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# NCERT Solutions For Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.2
Here, Below you all know about NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.2 Question Answer. I know many of you confuse about finding Chapter 14 Mathematical Reasoning Ex 14.2 Of Class 11 NCERT Solutions. So, Read the full post below and get your solutions.
## NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.2
NCERT TEXTBOOK EXERCISES
Ex 14.2 Class 11 Maths Question 1.
Write the negation of the following statements:
(i) Chennai is the capital of Tamil Nadu.
(ii) √2 is not a complex number.
(iii) All triangles are not equilateral triangle.
(iv) The number 2 is greater than 7.
(v) Every natural number is an integer.
Solution:
(i) Negation of statement is: Chennai is not the capital of Tamil Nadu.
(ii) Negation of statement is: √2 is a complex number.
(iii) Negation of statement is: All triangles are equilateral triangles.
(iv) Negation of statement is: The number 2 is not greater than 7.
(v) Negation of statement is: Every natural number is not an integer.
Ex 14.2 Class 11 Maths Question 2.
Are the following pairs of statements negations of each other:
(i) The number x is not a rational number.
The number x is not an irrational number.
(ii) The number x is a rational number.
The number x is an irrational number.
Solution:
(i) Let p: The number x is not a rational number.
q: The number x is not an irrational number.
Now, ~p: The number x is a rational number. ~q: The number x is an irrational number.
∴ ~p = q and ~q = p
Thus, p and q are negations of each other.
(ii) Let p: The number x is a rational number.
q: The number x is an irrational number.
Now, ~p: The number x is not a rational number.
~q: The number x is not an irrational number.
∴ ~p = q and ~q = p
Thus, p and q are negations of each other.
Ex 14.2 Class 11 Maths Question 3.
Find the component statements of the following compound statements and check whether they are true or false.
(i) Number 3 is prime or it is odd.
(ii) All integers are positive or negative.
(iii) 100 is divisible by 3,11 and 5.
Solution:
(i) The component statements are:
p: Number 3 is prime
q: Number 3 is odd.
Both the component statements p and q are true.
(ii) The component statements are:
p: All integers are positive.
q: All integers are negative.
Both the component statements p and q are false.
(iii) The component statements are:
p: 100 is divisible by 3.
q: 100 is divisible by 11.
r: 100 is divisible by 5.
The component statements p and q are false whereas r is true.
## NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Exercise 14.2 PDF
For NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.2, you may click on the link below and get your NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Exercise pdf file.
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# A positive integer is of the form $3q+1$, $q$ being a natural number. Can you write its square in any form other than $3m+1$, $3m$ or $3m+2$ for some integer $m$? Justify your answer.
Given: A positive integer of the form $3q\ +\ 1$.
To prove: Here we have to check if the square of $3q\ +\ 1$ is in any form other than $3m+1$, $3m$ or $3m+2$ for some integer $m$.
Solution:
According to Euclid's lemma,
If $a$ and $b$ are two positive integers;
• $a\ =\ bq\ +\ r$, where $0\ \underline{< }\ r\ <\ b$.
If $b\ =\ 3$, then;
• $a\ =\ 3q\ +\ r$, where $0\ \underline{< }\ r\ <\ 3$.
• So, $r\ =\ 0,\ 1,\ 2$
When, $r\ =\ 0$:
$a\ =\ 3q$
Squaring on both sides, we get:
$a^2\ = (3q)^2$
$a^2\ = 9q^2$
$a^2\ = 3(3q^2)$
$a^2\ = 3m$, where $m\ =\ 3q^2$
When, $r\ =\ 1$:
$a\ =\ 3q\ +\ 1$
Squaring on both sides, we get:
$a^2\ = (3q\ +\ 1)^2$
$a^2\ = 9q^2\ +\ 6q\ + 1$
$a^2\ = 3(3q^2\ +\ 2q)\ +\ 1$
$a^2\ = 3m\ +\ 1$, where $m\ =\ 3q^2\ +\ 2q$
When, $r\ =\ 2$:
$a\ =\ 3q\ +\ 2$
Squaring on both sides, we get:
$a^2\ = (3q\ +\ 2)^2$
$a^2\ = 9q^2\ +\ 12q\ + 4$
$a^2\ = 9q^2\ +\ 12q\ + 3\ +\ 1$
$a^2\ = 3(3q^2\ +\ 4q\ +\ 1)\ +\ 1$
$a^2\ = 3m\ +\ 1$, where $m\ =\ 3q^2\ +\ 4q\ +\ 1$
Hence, square of a positive integer of the form $3q\ +\ 1$ is always of the form $3m$ or $3m\ +\ 1$ for some integer $m$.
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Updated on: 10-Oct-2022
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# How To Calculate The Mean, Median, And Mode
How To Calculate The Mean, Median, And Mode? Mean, Median, and Mode are 03 types of averages. There are many averages in statistics, but Mean, Median, and Mode are the most common. Use the below given Formula and Shortcut Tricks to calculate the Mean, Median, And Mode in a simple manner.
Without Mean, Median, And Mode, it will be impossible to understand much of the data used in our daily routine or life. The Mean, Median, And Mode are used to search the arithmetical medium in a group of numbers, but they all do in another way.
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How To Find Mean, Median, And Mode!!!
With the help of below provided Formulas, you can easily find the Mean, Median, And Mode. Once you have mastered the basics of mean, median, and mode, you can start to learn about more and more statistical concepts.
Just scroll down this page of recruitmentresult.com to know more about How to Calculate the Mean, Median, And Mode. Have a look for Mean, Median, And Mode Formula, Shortcut tricks etc.
## How To Calculate The Mean, Median, And Mode
Mean, Median, And Mode Formula, Shortcuts Tricks:
Mean:
Mean Definition: The mean is the arithmetic average of a set of given numbers.
Mean Formula:
How to Calculate the Mean? Shortcut Tricks
• The mean is measured by adding up the scores and dividing the total by number of scores.
• Consider the following number set: 3, 4, 6, 6, 8, 9, 11.
• The mean is calculated in the following manner:
• 3 + 4 + 6 + 6 + 8 + 9 + 11 = 47
• 47 / 7 = 6.7
• The mean (average) of the number set is 6.7.
Get Best Answer: How to Prepare for Maths
Median:
Median Definition: The median is middle score in a set of given numbers.
Median Formula:
If the total number of observation given is odd, then the formula to calculate the median is:
Median = {(n+1)/2}thterm
If the total number of observation is even, then the median formula is:
Median = [(n/2)th term + {(n/2)+1}th]/2
How to Calculate the Mean?
• To measure the mean, simply add all of your numbers together.
• Next, divide the sum by however many numbers you added.
• The result is your mean or average score.
• For example, let’s say you have four test scores: 15, 18, 22, and 20.
• To search the average, you would first add all four scores together, then divide the sum by four. The average would be 19
(15 + 18 + 22 + 20) / 4 = 75 / 4 = 18.75
Get Here: Maths Formulas
Mode:
Mode Definition: The mode is the most normally occurring data point in a dataset. It is useful when there are a lot of repeated values in a dataset.
Mode Formula: The mode is the most frequently occurring score or values.
How to Calculate the Mode?
• Mode is the most frequently occurring score in a distribution; simply select the most common score as your mode.
• Consider the following number distribution of 2, 3, 6, 3, 7, 5, 1, 2, 3, 9.
• The mode of these numbers will be 3 since three is the most frequently occurring number.
• For instance, consider the following set of numbers: 13, 17, 20, 20, 21, 23, 23, 26, 29, 30. In this set, both 20 and 23 occur twice.
Start Now: Online Quiz for Competitive Exams
Uses of Mean, Median, And Mode Formula:
Mean:
• Takes into account every number in the data set.
• That means all numbers are added in calculating the mean.
• Simple and quick way to signify the entire data values by a single or unique number due to its straightforward technique of calculation.
• Each set has a unique mean value.
Median:
• Not affected by the outliers in the data set.
• An outlier is a data point that is radically “distant” or “away” from common trends of values in a given set.
• It does not represent a typical number in the set.
• Each set has a unique median value.
Mode:
• Just like the median, the mode is not affected by outliers.
• Useful to find the most “popular” or common item. This includes data sets that do not involve numbers.
Read Also: How to Prepare for Written Exam
Each dimension is an attempt to detain the essence of how a typical entry or number in the data set may look like. If you have any query regarding How to Calculate the Mean, Median, And Mode, then drop the comment below provided comment box.
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# Video: AQA GCSE Mathematics Higher Tier Pack 5 • Paper 2 • Question 13
Here is some information about the time spent in a week reading fiction books by students in a school. Shortest time: 15 minutes, longest time: 8 hours, lower quartile: 1 hour 15 minutes, interquartile range: 3 hours 30 minutes, median time: 3 hours. Draw a box plot to show this information.
03:35
### Video Transcript
Here is some information about the time spent in a week reading fiction books by students in a school. Shortest time: 15 minutes, longest time: eight hours, lower quartile: one hour 15 minutes, interquartile range: three hours 30 minutes, median time: three hours. Draw a box plot to show this information.
A box plot — sometimes called a box and whisker plot — is a diagram that can be used to display information about the range, the median, and the quartiles of a dataset. We can read several pieces of information from the box plot. The smallest and largest values are represented by the end of each whisker. The box itself represents the middle 50 percent of the data. The left-hand side of the box tells us the lower quartile. Remember that’s just one-quarter of the way through the dataset. And the right-hand side of the box tells us the upper quartile. That’s three-quarters of the way through the dataset. The line inside the box is the median average of the data.
Let’s use this information to draw a box plot representing our dataset. Before we do though, let’s double-check the scale. We can see that four small squares represent one hour. We can work out the value of one small square by dividing by four. When we do though, it’s sensible to work out the time represented by one small square in minutes. There are 60 minutes in one hour. And when we divide this by four, we can halve it and halve it again to do that quickly, we get 15 minutes. One small square on our scale represents 15 minutes.
We can now plot the shortest and longest time. The shortest time is 15 minutes. That’s one small square above zero. The longest time is eight hours. That’s here. The lower quartile is the left-hand side of the box and it’s one hour 15 minutes. That’s one small square above one. We don’t actually know the value of the upper quartile. Instead, we’re told that the interquartile range is three hours and 30 minutes. Remember the interquartile range — I’ve written IQR — is the difference between the upper quartile and the lower quartile.
If we rearrange this formula by adding the lower quartile to both sides, we can see that to work out the upper quartile, we need to add the interquartile range to the lower quartile. That’s three hours 30 minutes plus one hour 15 minutes. Three hours plus one hour is four hours. And 30 minutes plus 15 minutes is 45 minutes. So the upper quartile is four hours 45 minutes. That’s one square below the number five.
At this point, it is important to remember that we can’t easily use our calculator to work this out. We don’t want to add 3.3 and 1.15. Time is a tricky one and we’re best to avoid using the calculator for this type of problem wherever possible. Finally, let’s add the median in at three. Remember we need to join the edges of our box together as shown. And finally, it wouldn’t be complete without adding in the whiskers.
And we’re done. This is the box plot representing the data about the time spent in a week reading fiction books by students in a school.
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# Applications of Differentiation
A LevelAQAEdexcelOCR
## Applications of Differentiation
In this section, we’ll look at how to use differentiation in mechanics and practical problems.
Make sure you are happy with the following topics before continuing.
A LevelAQAEdexcelOCR
## Finding Maximum and Minimum Values of Volume and Area
We can use differentiation to find optimal values of dimensions of objects.
So, let’s say wish to create a hollow box of length $2\textcolor{blue}{x}$, width $\textcolor{blue}{x}$ and height $\textcolor{purple}{h}$, and we have a limit of $600\text{ cm}^2$ of wood to use.
To find the maximum volume possible, we must find two equations in volume and area:
$\textcolor{red}{V} = 2\textcolor{blue}{x} \times \textcolor{blue}{x} \times \textcolor{purple}{h} = 2\textcolor{blue}{x}^2\textcolor{purple}{h}$
and
$\textcolor{limegreen}{A} = 2(2\textcolor{blue}{x} \times \textcolor{blue}{x}) + 2(2\textcolor{blue}{x} \times \textcolor{purple}{h}) + 2(\textcolor{blue}{x} \times \textcolor{purple}{h}) = 4\textcolor{blue}{x}^2 + 6\textcolor{blue}{x}\textcolor{purple}{h} = 600$
Of course, we only want to differentiate in terms of one variable, so we’ll transform the second equation to $\textcolor{purple}{h} = \dfrac{300 - 2\textcolor{blue}{x}^2}{3\textcolor{blue}{x}}$
Plugging this back into our equation for $\textcolor{red}{V}$, we have
$\textcolor{red}{V} = \dfrac{2\textcolor{blue}{x}^2(300 - 2\textcolor{blue}{x}^2)}{3\textcolor{blue}{x}} = 200\textcolor{blue}{x} - \dfrac{4\textcolor{blue}{x}^3}{3}$
We can then differentiate with respect to $\textcolor{blue}{x}$ to find the maximum volume of the box:
$\dfrac{d\textcolor{red}{V}}{d\textcolor{blue}{x}} = 200 - 4\textcolor{blue}{x}^2 = 0$, giving $\textcolor{blue}{x} = \sqrt{50}$
We must also verify that this is a maximum, rather than a minimum, so we find $\dfrac{d^2\textcolor{red}{V}}{d\textcolor{blue}{x}^2}$:
$\dfrac{d^2\textcolor{red}{V}}{d\textcolor{blue}{x}^2} = - 8\textcolor{blue}{x} = -8\sqrt{50} < 0$, so this is definitely a maximum point.
Therefore, we have $\textcolor{blue}{x} = \sqrt{50}$ and $\textcolor{purple}{h} = \dfrac{200}{3\sqrt{50}}$, giving $\textcolor{red}{V}_{max} = \dfrac{20000}{3\sqrt{50}}$.
A LevelAQAEdexcelOCR
@mmerevise
## Finding Rates of Change in Mechanics
In Mechanics, we’ll talk about the derivation of acceleration and velocity from displacement.
From the displacement $\textcolor{limegreen}{s}$, we can differentiate with respect to $\textcolor{purple}{t}$ to find $\textcolor{red}{v}$, and differentiate again to find $\textcolor{blue}{a}$.
So,
$\dfrac{d\textcolor{limegreen}{s}}{d\textcolor{purple}{t}} = \textcolor{red}{v}$
and
$\dfrac{d^2\textcolor{limegreen}{s}}{d\textcolor{purple}{t}^2} = \dfrac{d\textcolor{red}{v}}{d\textcolor{purple}{t}} = \textcolor{blue}{a}$
So, for example, let’s say we have an equation for the displacement $\textcolor{limegreen}{s} = 2\textcolor{purple}{t}^3 - 3\textcolor{purple}{t}^2 -4\textcolor{purple}{t} + 1$.
We can find an equation for the velocity:
$\textcolor{red}{v} = 6\textcolor{purple}{t}^2 - 6\textcolor{purple}{t} - 4$
and an equation for the acceleration:
$\textcolor{blue}{a} = 12\textcolor{purple}{t} - 6$
Therefore, we can find the values of $\textcolor{purple}{t}$ such that $\textcolor{red}{v} = 0$:
$6\textcolor{purple}{t}^2 - 6\textcolor{purple}{t} - 4 = 0$ has solutions at $\textcolor{purple}{t} = \dfrac{1}{2} \pm \sqrt{\dfrac{11}{12}}$, but we cannot have a negative time, so we have $\textcolor{purple}{t} = \dfrac{1}{2} + \sqrt{\dfrac{11}{12}}$.
$\textcolor{blue}{a} = 12\left( \dfrac{1}{2} + \sqrt{\dfrac{11}{12}}\right) - 6 = 11.49$, so we can conclude that this is a point of minimum displacement.
A LevelAQAEdexcelOCR
## Applications of Differentiation Example Questions
To find the maximum volume possible, we must find two equations in volume and area:
$V = 5x \times 3x \times h = 15x^2h$
and
$A = 2(5x \times 3x) + 2(5x \times h) + 2(3x \times h) = 30x^2 + 16xh = 30000$
Of course, we only want to differentiate in terms of one variable, so we’ll transform the second equation to
$h = \dfrac{30000 - 30x^2}{16x}$
Plugging this back into our equation for $V$, we have
$V = \dfrac{15x^2(30000 - 30x^2)}{16x} = 28125x - 28.125x^3$
We can then differentiate with respect to $x$ to find the maximum volume of the box:
$\dfrac{dV}{dx} = 28125 - 84.375x^2 = 0$, giving $x = \sqrt{\dfrac{1000}{3}}$
We must also verify that this is a maximum, rather than a minimum, so we find $\dfrac{d^2V}{dx^2}$:
$\dfrac{d^2V}{dx^2} = - 168.75x = -168.75\sqrt{\dfrac{1000}{3}} < 0$, so this is a maximum point.
Therefore, we have $x = \sqrt{\dfrac{1000}{3}}$ and $h = \dfrac{20000}{16\sqrt{\dfrac{1000}{3}}}$, giving $V_{max} = \dfrac{6250000\sqrt{3}}{\sqrt{1000}}$.
$h = 1000\left( \dfrac{t^2}{40} - \dfrac{t^3}{600}\right)$ gives
$\dfrac{dh}{dt} = 1000 \left( \dfrac{t}{20} - \dfrac{t^2}{200}\right)$
and
$\dfrac{d^2h}{dt^2} = 1000 \left( \dfrac{1}{20} - \dfrac{t}{100}\right)$
Setting $\dfrac{dh}{dt} = 0$ gives $1000\left( \dfrac{t}{20}\left( 1 - \dfrac{t}{10}\right) \right) = 0$, meaning that $t = 0$ or $t = 10$.
Using $t=10$ gives $h=833\text{ m}$.
Volume of bath, $V$:
$V = \dfrac{1}{2}(2.5 + 1.5) \times 0.5 \times 1 = 1\text{ m}^3 = 1 \times 10^6 \text{ cm}^3$
Then, we must find $t$ such that $\dfrac{1}{2}t^2 + 4t - 1000000 = 0$.
Using the quadratic formula, we have
$t = \dfrac{-4 \pm \sqrt{16 - \left( 4 \times \dfrac{1}{2} \times -1000000 \right)}}{1} = -1418.22, 1410.22\text{ s}$
So, we know that the bath takes approximately $23.5\text{ minutes}$ to fill.
When $t = 1410.22$, $\dfrac{dV_{filled}}{dt} = t + 4$, so the water is filling the bath at a rate of $1414.22\text{ cm}^3\text{ s}^{-1}$.
A Level
A Level
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Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.3 Text Book Back Questions and Answers, Notes.
## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.3
Question 1.
Expand
(i) (3m + 5)2
(ii) (5p – 1)2
(iii) (2n – 1)(2n + 3)
(iv) 4p2 – 25q2
(i) (3m + 5)2
Comparing (3m + 5)2 with (a + b)2 we have a = 3m and b = 5
(a + b)2 = a2 + 2ab + b2
(3m + 5)2 = (3m)2 + 2(3m) (5) + 52
= 32 m2 + 30 m + 25
= 9m2 + 30 m + 25
(ii) (5p – 1)2
Comparing (5p – 1)2 with (a – b)2 we have a = 5p and b = 1
(a – b)2 = a2 – 2ab + b2
(5p – 1)2 = (5p)2 – 2(5p)(1) + 12
= 52p2 – 10 p + 1
= 25p2 – 10p + 1
(iii) (2n – 1)(2n + 3)
Comparing (2n – 1) (2n + 3) with (x + a)(x + b) we have a = -1; b = 3
(x + a) (x + b) = x2 + (a + b)x + ab
(2n + (-1)) (2n + 3) = (2n)2 + (-1 + 3)2n + (-1) (3)
= 22 n2 + 2(2n) – 3 = 4n2 + 4n – 3
(iv) 4p2 – 25q2 = (2p)2 – (5q)2
Comparing (2p)2 – (5q)2 with a2 – b2 we have a = 2p and b = 5q
(a2 – b2) = (a + b)(a – b)
= (2p + 5q)(2p – 5q)
Question 2.
Expand
(i) (3 + m)3
(ii) (2a + 5)3
(iii) (3p + 4q)3
(iv) (52)3
(v) (104)3
(i) (3 + m)3
Cornparing (3 + m)3 with (a + b)3 we have a = 3 ; b = m
(a + b)3 = a2 + 3a2b + 3ab2 + b3
(3 + m)3 = 33 + 3(3)2 (m) + 3(3)m2 + m3
= 27 + 27m + 9m2 + m3
= m3 + 9m2 + 27m + 27
(ii) (2a + 5)3 =
Comparing (2a + 5)3 with (a + b)3 we have a = 2a, b = 5
(a + b)3 = a3 + 3a2b + 3ab2 + b3
= (2a)3 + 3(2a)2 5 + 3 (2a) 52 + 53
= 23a3 + 3(22a2)5 + 6a(25) + 125
= 8a3 + 60a2 + 150a + 125
(iii) (3p + 4q)3
Comparing (3p + 4q)3 with (a + b)3 we have a = 3p and b = 4q
(a + b) = a3 + 3a2b + 3ab2 + b2
(3p + 4q)3 = (3p)3 + 3(3p)2 (4q) + 3(3p)(4q)2 + (4q)3
= 33p3 + 3(9p2)(4q) + 9p(16q2) – 43q3
= 27p3 + 108p2q + 144pq3 + 64q3
(iv) (52)3
(52)3 = (50 + 2)3
Comparing (50 + 2)3 with (a + b)3 we have a = 50 and b =2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(50 + 2)3 = 503 + 3 (50)2 2 + 3 (50)(2)2 + 23
523 = 125000 + 6(2,500) + 150(4) + 8
= 1,25,000 + 15,000 + 600 + 8
523 = 1,40,608
(v) (104)3
(104)3 = (100 + 4)3
Comparing (100 + 4)3 with (a + b)3 we have a = 100 and b = 4
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(100 + 4)3 = (100)3 + 3 (100)2 (4) + 3 (100) (4)2 + 43
= 10,00,000 + 3(10000)4 + 300(16) + 64
= 10,00,000 + 1,20,000 + 4,800 + 64= 11,24,864
Question 3.
Expand
(i) (5- x)3
(ii) (2x – 4y)3
(iii) (ab – c)3
(iv) (48)3
(v) (97xy)3
(i) (5- x)3
Comparing (5 – x)3 with (a – b)3 we have a = 5 and b = x
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(5 – x)3 = 53 – 3(5)2(x) + 3(5)(x2) – x3
= 125 – 3(25)(x) + 15x2 – x3
= 125 – 75x + 15 x2 – x3
(ii) (2x – 4y)3
Comparing (2x – 4y)3 with (a – b)3 we have a = 2x and b = 4y
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(2x – 4y)3 = (2x)3 – 3(2x)2 (4y) + 3(2x) (4y)2 – (4y)3
= 23x3 – 3(22x2)(4y) + 3(2x) (42y2) – (43y3)
= 8x3 – 48x2y + 96xy2 – 64y3
(iii) (ab – c)3
Comparing (ab – c)3 with (a – b)3 we have a = ab and b = c
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(ab – c)3 = (ab)3 – 3(ab)2 c + 3ab (c)2 – c3
= a3b3 – 3(a2b2) c + 3abc2 – c3
= a3b3 – 3a2b2c + 3abc2 – c3
(iv) (48)3
(48)3 = (50 – 2)3
Comparing (50- 2)3 with (a – b)3 we have a = 50 and b = 2
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(50 – 2)3 = (50)3 – 3(50)2 (2) + 3 (50)(2)2 – 23
= 1,25,000 – 15000 + 600 – 8
= 1,10,000 + 592
= 1,10,592
(v) (97xy)3
(97xy)3 = 973 x3 y3 = (100 – 3) x3y3 … (1)
Comparing(100 – 3)3 with (a – b)3 we have a = 100, b = 3
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(100 – 3)3 = (100)3 – 3(100)2 (3) + 3 (100)(3)2 – 33
973 = 10,00,000 – 90000 + 2700 – 27
973 = 910000 + 2673
973 = 912673
97x3y3 = 912673x3y3
Question 4.
Simplify (p – 2)(p + 1)(p – 4)
(p – 2)(p + 1)(p – 4) = (p + (-2)) (p + 1) (p + (-4))
Comparing (p – 2) (p + 1) (p – 4) with (x + a) (x + b) (x + c) we have x = p ; a = -2;
b = 1 ; c = -4.
(x + a)(x + b)(x + c) = x2 + (a + b + c) x2 + (ab + bc+ ca)x + abc
= p3 + (-2 + 1 + (-4)) p2 + (-2)( 1) + (1)(-4) + (-4) (-2))p + (-2) (1) (-4)
= p3 +(-5)p2 + (-2 + (-4) + 8)p + 8
= p2 – 5p2 + 2p + 8
Question 5.
Find the volume of the cube whose side is (x + 1) cm
Given side of the cube = (x + 1) cm
Volume of the cube = (side)3 cubic units = (x + 1)3 cm3
We have (a + b)3 = (a33 + 3a2b + 3ab2 + b3) cm3
(x + 1)3 = (x3 + 3x2(1) + 3x(1)2 + 13)cm3
Volume = (x3 + 3x2 + 3x + 1) cm3
Question 6.
Find the volume of the cuboid whose dimensions are (x + 2),(x – 1) and (x – 3)
Given the dimensions of the cuboid as (x + 2), (x – 1) and (x – 3)
∴ Volume of the cuboid = (l × b × h) units3
= (x + 2) (x – 1) (x – 3) units3
We have (x + a)(x + b) (x+c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
∴ (x+2) (x- 1) (x-3) = x3 + (2 – 1 – 3)x2 + (2 (-1) + (-1) (-3) + (-3) (2)) x + (2) (-1) (-3)
= x3 – 2x2 + (-2 + 3 – 6)x + 6
Volume = x3 – 2x3 – 5x + 6 units3
Objective Type Questions
Question 7.
If x2 – y2 = 16 and (x + y) = 8 then (x – y) is ________
(A) 8
(B) 3
(C) 2
(D) 1
(C) 2
Hint:
x2 – y2 = 16
(x + y) (x – y) = 16
8 (x – y) = 16
(x – y) = $$\frac { 16 }{ 8 }$$ = 2
Question 8.
$$\frac{(a+b)\left(a^{3}-b^{3}\right)}{\left(a^{2}-b^{2}\right)}$$ = _________
(A) a2 – ab + b2
(B) a2 + ab + b2
(C) a2 + 2ab + b2
(D) a22 – 2ab + b2
(B) a2 + ab + b2
Hint:
= a2 + ab + b2
Question 9.
(p + q)(p2 – pq + q2) is equal to __________
(A) p3 + q3
(B) (p + q)3
(C) p3 – q3
(D) (p – q)3
(A) p3 + q3
Hint:
a3 + b3 = (a + b) (a2 – ab + b2)
Question 10.
(a – b) = 3 and ab = 5 then a3 – b3 = __________
(A) 15
(B) 18
(C) 62
(D) 72
(D) 72
Hint:
(a – b) = 3
(a – b)2 = 32
a2 + b2 – 2ab = 9
a2 + b2 – 2(5) = 9
a2 + b2 = 9 + 10
a2 + b2 = 19
a3 – b3 = (a – b)(a2 + ab + b2) = 3(19 + 5)
= 3(24) = 72
Question 11.
a3 + b3 = (a + b)3 _________
(A) 3a(a + b)
(B) 3ab(a – b)
(C) -3ab(a + b)
(D) 3ab(a + b)
|
# Quadrilateral with equal opposite sides is a parallelogram
In this chapter we will prove that “any quadrilateral with equal opposite sides is actually a parallelogram.
Given:
Consider quadrilateral ABCD in which opposite sides are equal.
AB = CD and AD = BC
To Prove:
Prove that the quadrilateral is parallelogram.
i.e. AB || CD and AD || CB
Proof:
Consider triangle ABC and CDA;
AB = CD ( given )
AC = CA ( common side )
AD = CB ( given )
By SSS congruency, both the triangles are congruent.
i.e. \mathtt{\triangle ABC\ \cong \triangle CDA}
Since both triangles are congruent, we can write;
∠BAC = ∠DCA and ∠BCA = ∠DAC
Since ∠BAC = ∠DCA;
This is possible when line AB || CD and intersected by AC as transversal.
Also as ∠BCA = ∠DAC;
This is possible when line AD || BC and intersected by AC as transversal.
Since both the opposite sides are parallel to each other, this means that the given figure is of parallelogram.
Hence Proved.
|
# Difference between revisions of "2004 AMC 12B Problems/Problem 14"
## Problem
In $\triangle ABC$, $AB=13$, $AC=5$, and $BC=12$. Points $M$ and $N$ lie on $AC$ and $BC$, respectively, with $CM=CN=4$. Points $J$ and $K$ are on $AB$ so that $MJ$ and $NK$ are perpendicular to $AB$. What is the area of pentagon $CMJKN$?
$[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0); pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) ); draw( A--B--C--cycle ); draw( M--J ); draw( N--K ); label("A",A,NW); label("B",B,SE); label("C",C,SW); label("M",M,SW); label("N",N,S); label("J",J,NE); label("K",K,NE); [/asy]$
$\mathrm{(A)}\ 15 \qquad \mathrm{(B)}\ \frac{81}{5} \qquad \mathrm{(C)}\ \frac{205}{12} \qquad \mathrm{(D)}\ \frac{240}{13} \qquad \mathrm{(E)}\ 20$
## Solution
### Solution 1
The triangle $ABC$ is clearly a right triangle, its area is $\frac{5\cdot 12}2 = 30$. If we knew the areas of triangles $AMJ$ and $BNK$, we could subtract them to get the area of the pentagon.
Draw the height $CL$ from $C$ onto $AB$. As $AB=13$ and the area is $30$, we get $CL=\frac{60}{13}$. The situation is shown in the picture below:
$[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0); pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) ); pair L=intersectionpoint(A--B, C--(C+rotate(90)*(B-A)) ); draw( A--B--C--cycle ); draw( M--J ); draw( N--K ); draw( C--L, dashed ); label("A",A,NW); label("B",B,SE); label("C",C,SW); label("M",M,SW); label("N",N,S); label("J",J,NE); label("K",K,NE); label("L",L,NE); [/asy]$
Now note that the triangles $ABC$, $AMJ$, $ACL$, $CBL$ and $NBK$ all have the same angles and therefore they are similar. We already know some of their sides, and we will use this information to compute their areas. Note that if two polygons are similar with ratio $k$, their areas have ratio $k^2$. We will use this fact repeatedly. Below we will use $[XYZ]$ to denote the area of the triangle $XYZ$.
We have $\frac{CL}{BC} = \frac{60/13}{12} = \frac 5{13}$, hence $[ACL] = \frac{ 25[ABC] }{169} = \frac{750}{169}$.
Also, $\frac{CL}{AC} = \frac{60/13}5 = \frac{12}{13}$, hence $[CBL] = \frac{ 144[ABC] }{169} = \frac{4320}{169}$.
Now for the smaller triangles:
We know that $\frac{AM}{AC}=\frac 15$, hence $[AMJ] = \frac{[ACL]}{25} = \frac{30}{169}$.
Similarly, $\frac{BN}{BC}=\frac 8{12} = \frac 23$, hence $[NBK] = \frac{4[CBL]}9 = \frac{1920}{169}$.
Finally, the area of the pentagon is $30 - \frac{30}{169} - \frac{1920}{169} = \boxed{\frac{240}{13}}$.
### Solution 2
Split the pentagon along a different diagonal as follows:
$[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0); pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) ); pair L=intersectionpoint(A--B, C--(C+rotate(90)*(B-A)) ); draw( A--B--C--cycle ); draw( M--J ); draw( N--K ); draw( M--N, dashed ); label("A",A,NW); label("B",B,SE); label("C",C,SW); label("M",M,SW); label("N",N,S); label("J",J,NE); label("K",K,NE); label("L",L,NE); [/asy]$ The area of the pentagon is then the sum of the areas of the resulting right triangle and trapezoid. As before, triangles $ABC$, $AMJ$, and $NBK$ are all similar.
Since $BN=12-4=8$, $NK=\frac{5}{13}(8)=\frac{40}{13}$ and $BK=\frac{12}{13}(8)=\frac{96}{13}$. Since $AM=5-4=1$, $JM=\frac{12}{13}$ and $AJ=\frac{5}{13}$.
The trapezoid's height is therefore $13-\frac{5}{13}-\frac{96}{13}=\frac{68}{13}$, and its area is $\frac{1}{2}\left(\frac{68}{13}\right)\left(\frac{12}{13}+\frac{40}{13}\right)=\frac{34}{13}(4)=\frac{136}{13}$.
Triangle $MCN$ has area $\frac{1}{2}(4)(4)=8$, and the total area is $\frac{104+136}{13}=\boxed{\frac{240}{13}}$.
|
TOPICS
Exercise - 12.1
Question-1 :- A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
Solution :-
``` Side of traffic signal board = a
Perimeter of traffic signal board = 3 × a
2s = 3a
s = 3a/2
(s - a) = (s - b) = (s - c) = 3a/2 - a = a/2
By Heron's formula,
area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √3a/2 x a/2 x a/2 x a/2
= a²√3/4 square unit
Perimeter of traffic signal board = 180 cm
Side of traffic signal board (a) = 180/3 cm = 60 cm
Using equation (1), area of traffic signal board = √3/4 x (60)² cm²
= √3/4 x 3600 cm²
= √3 x 900 cm²
= 900√3 cm²
```
Question-2 :- The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m. The advertisements yield an earning of ` 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?
Solution :-
``` The sides of the triangle are of 122 m, 22 m, and 120 m respectively.
Perimeter of triangle = (122 + 22 + 120) m
2s = 264 m
s = 132 m
s - a = 132 - 122 = 10m
s - b = 132 - 22 = 110m
s - c = 132 - 120 = 12m
area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √132 x 10 x 110 x 12
= 1320 m²
Rent of 1 m² area per year = Rs 5000
Rent of 1 m² area per month = Rs 5000/12
Rent of 1320 m² area for 3 months = 5000/12 x 3 x 1320
= Rs (5000 × 330) = Rs 1650000
Therefore, the company had to pay Rs 1650000.
```
Question-3 :- There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.
Solution :-
``` Let length, breadth, and height of the rectangular hall be l m, b m, and h m respectively.
Area of four walls = 2lh + 2bh = 2(l + b) h
Perimeter of the floor of hall = 2(l + b) = 250 m
Area of four walls = 2(l + b) h = 250h m²
Cost of painting per m² area = Rs 10
Cost of painting 250h m² area = Rs (250h × 10) = Rs 2500h
However, it is given that the cost of paining the walls is Rs 15000.
15000 = 2500h
h = 6 m
Therefore, the height of the hall is 6 m.
```
Question-4 :- Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.
Solution :-
``` Let the third side of the triangle be x.
Perimeter of the given triangle = 42 cm
18 cm + 10 cm + x = 42
x + 28 = 42
x = 42 - 28
x = 14 cm
s = (a + b + c)/2
s = (18 + 10 + 14)/2
s = 42/2
s = 21cm
s - a = 21 - 18 = 3cm
s - b = 21 - 10 = 11cm
s - c = 21 - 14 = 7cm
area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √21 x 3 x 11 x 7
= 21√11 cm²
```
Question-5 :- Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.
Solution :-
``` Let the common ratio between the sides of the given triangle be x.
Therefore, the side of the triangle will be 12x, 17x, and 25x.
Perimeter of this triangle = 540 cm
12x + 17x + 25x = 540 cm
54x = 540 cm
x = 10 cm
a = 12x = 12 x 10 = 120 cm
b = 17x = 17 x 10 = 170 cm
c = 25x = 25 x 10 = 250 cm
s = (a + b + c)/2
s = (120 + 170 + 250)/2
s = 540/2
s = 270cm
s - a = 270 - 120 = 150cm
s - b = 270 - 170 = 100cm
s - c = 270 - 250 = 20cm
area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √270 x 150 x 100 x 20
= 9000 cm²
```
Question-6 :- An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution :-
``` Let the third side of this triangle be x.
Perimeter of triangle = 30 cm
12 cm + 12 cm + x = 30 cm
x = 6 cm
s = (a + b + c)/2
s = (12 + 12 + 6)/2
s = 30/2
s = 15cm
s - a = 15 - 12 = 3cm
s - b = 15 - 12 = 3cm
s - c = 15 - 6 = 9cm
area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √15 x 3 x 3 x 9
= 9√15 cm²
```
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# Let the integral $I = \int\limits_0^1 {\dfrac{{{x^3}\cos 3x}}{{2 + {x^2}}}dx}$. Then which condition on $I$ holds true$\left( a \right){\text{ }}\dfrac{{ - 1}}{2} < I < \dfrac{1}{2} \\ \left( b \right){\text{ }}\dfrac{{ - 1}}{3} < I < \dfrac{1}{3} \\ \left( c \right){\text{ - 1}} < I < 1 \\ \left( d \right){\text{ }}\dfrac{{ - 3}}{2} < I < \dfrac{3}{2} \\$
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Hint: In this question we have to find the range of the integral $I$ and not the exact value so direct solving of integral using definite integral property won’t help, we have to think in such a way that what is the range of the trigonometric entity defined inside the integral and then form the integral’s range using this concept.
Given integral
$I = \int\limits_0^1 {\dfrac{{{x^3}\cos 3x}}{{2 + {x^2}}}dx}$………… (1)
Now as we know that cos x always lies between -1 to 1 therefore cos 3x also lies between -1 to 1.
$\Rightarrow - 1 \leqslant \cos 3x \leqslant 1$
Now multiply by ${x^3}$ in above equation we have,
$\Rightarrow - {x^3} \leqslant {x^3}.\cos 3x \leqslant {x^3}$
Now divide by $\left( {2 + {x^2}} \right)$ in above equation we have,
$\Rightarrow - \dfrac{{{x^3}}}{{2 + {x^2}}} \leqslant \dfrac{{{x^3}.\cos 3x}}{{2 + {x^2}}} \leqslant \dfrac{{{x^3}}}{{2 + {x^2}}}$
Now apply integration from 0 to 1 we have,
$\Rightarrow - \int\limits_0^1 {\dfrac{{{x^3}}}{{2 + {x^2}}}dx} \leqslant \int\limits_0^1 {\dfrac{{{x^3}.\cos 3x}}{{2 + {x^2}}}dx} \leqslant \int\limits_0^1 {\dfrac{{{x^3}}}{{2 + {x^2}}}} dx$
Now from equation (1)
$I = \int\limits_0^1 {\dfrac{{{x^3}\cos 3x}}{{2 + {x^2}}}dx}$
Therefore above equation becomes
$\Rightarrow - \int\limits_0^1 {\dfrac{{{x^3}}}{{2 + {x^2}}}dx} \leqslant I \leqslant \int\limits_0^1 {\dfrac{{{x^3}}}{{2 + {x^2}}}} dx$
$\Rightarrow - \int\limits_0^1 {\dfrac{{x.{x^2}}}{{2 + {x^2}}}dx} \leqslant I \leqslant \int\limits_0^1 {\dfrac{{x.{x^2}}}{{2 + {x^2}}}} dx$………. (2)
Now let $\left( {2 + {x^2}} \right) = t$…………………. (3)
Differentiate this equation w.r.t. x we have
$\left( {0 + 2x} \right)dx = dt$
$\therefore xdx = \dfrac{{dt}}{2}$………. (4)
Now from equation (3)
If x = 0, $\Rightarrow t = 0 + 2 = 2$
And If x = 1, $\Rightarrow t = 1 + 2 = 3$
So the integration limit is changed from 2 to 3.
Now from equation (2), (3) and (4) we have
$\Rightarrow - \int\limits_2^3 {\dfrac{{t - 2}}{t}\left( {\dfrac{{dt}}{2}} \right)} \leqslant I \leqslant \int\limits_2^3 {\dfrac{{t - 2}}{t}} \left( {\dfrac{{dt}}{2}} \right)$
$\Rightarrow - \int\limits_2^3 {\left( {1 - \dfrac{2}{t}} \right)\left( {\dfrac{{dt}}{2}} \right)} \leqslant I \leqslant \int\limits_2^3 {\left( {1 - \dfrac{2}{t}} \right)} \left( {\dfrac{{dt}}{2}} \right)$
Now as we know integration of 1 is t w.r.t. t and integration of $\dfrac{2}{t}$ is 2logt so, apply this in above equation we have,
$\Rightarrow - \left[ {t - 2\log t} \right]_2^3 \leqslant I \leqslant \left[ {t - 2\log t} \right]_2^3$
Now apply integral limit we have
$\Rightarrow - \dfrac{1}{2}\left[ {3 - 2\log 3 - 2 + 2\log 2} \right] \leqslant I \leqslant \dfrac{1}{2}\left[ {3 - 2\log 3 - 2 + 2\log 2} \right]$
As we know $\left( {\log a - \log b = \log \dfrac{a}{b}} \right)$ so, use this property we have,
$\Rightarrow - \dfrac{1}{2}\left[ {1 - 2\log \dfrac{3}{2}} \right] \leqslant I \leqslant \dfrac{1}{2}\left[ {1 - 2\log \dfrac{3}{2}} \right]$
Now as we know$\left( {\log \dfrac{3}{2} = 0.1761} \right)$, so substitute this value we have,
$\Rightarrow - \dfrac{1}{2}\left[ {1 - 2 \times 0.1761} \right] \leqslant I \leqslant \dfrac{1}{2}\left[ {1 - 2 \times 0.1761} \right]$
Now on simplifying we have,
$\Rightarrow - 0.3239 \leqslant I \leqslant 0.3239$
As 0.3239 is approximately equal to$\dfrac{1}{3}$.
$\Rightarrow - \dfrac{1}{3} \leqslant I \leqslant \dfrac{1}{3}$
Hence, option (b) is correct.
Note: Whenever we face such types of problems the key concept is the basic understanding of the range of the trigonometric entities present inside the integral. Sometimes we also may need to simplify our integral so using proper substitution methods along with basic logarithm properties, we can easily evaluate the integral and reach the right answer.
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# Linear Algebra
```A LEVEL PURE MATHS REVISON NOTES
1 ALGEBRA AND FUNCTIONS
a) INDICES
Rules to learn :
π₯ π × π₯ π = π₯ π+π
π₯ π ÷ π₯ π = π₯ π−π
3
(π₯ π )π = π₯ ππ
1
π₯ −π =
π
1
π₯π
π
π
π
π₯ π = √π₯ π = ( √π₯ )
Solve 32π₯ × 25π₯ = 15
Simplify 2π₯(π₯ − π¦)2 + 3(π₯ − π¦)2
1
2
1
2
(3 × 5)2π₯ = 151
= (π₯ − π¦) (2π₯(π₯ − π¦) + 3))
= (π₯ − π¦) (2π₯ 2 − 2π₯π¦ + 3)
2π₯ = 1
π₯=
1
2
b) SURDS
• A root such as √3 that cannot be written as a fraction is IRRATIONAL
• An expression that involves irrational roots is in SURD FORM
• RATIONALISING THE DENOMINATOR is removing the surd from the denominator (multiply by the conjugate)
Simplify
Rationalise the denominator
√75 − √12
= √5 × 5 × 3 − √2 × 2 × 3
= 5√3 − 2√3
= 3√3
=
2
2−√3
2
2−√3
The conjugate of the denominator
2 - √3 is 2 + √3 so that
(2 - √3)( 2 + √3)
= 22 - √32
=1
2+√3
√3
×2+
= 4 + 2√3
Factorising – identifying the roots of the equation ax2 + bx + c = 0
• Look for the difference of 2 squares x2 – a2 = (x + a)(x – a) or (ax)2 - b2 = (ax + b)( ax – b)
• Look for the perfect square x2 + 2ax + a2 = (x + a)2
• Look out for equations which can be transformed into quadratic equations
12
Solve π₯ + 1 − π₯ = 0
π₯ 2 + π₯ − 12 = 0
(π₯ + 4)(π₯ − 3) = 0
Solve 6π₯ 4 − 7π₯ 2 + 2 = 0
Let z = x2
6π§ 2 − 7π§ + 2 = 0
(2π§ − 1)(3π§ − 2) = 0
x = -4 x = 3
π§=
1
2
1
π₯ = ±√2
π§=
2
3
Completing the square – identifying the vertex and line of symmetry
y = (x + a)2 + b
vertex at (-a , b) line of symmetry as equation x = -a
Line of symmetry
x=2
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2
π₯ = ±√3
π₯=
−π±√π2 −4ππ
2π
for solving ax2 + bx + c = 0
The DISCRIMINANT b2 – 4ac can be used to identify the number of roots
b2 – 4ac > 0 there are 2 real distinct roots (graph crosses the x-axis twice)
b2 – 4ac = 0 there is a single repeated root (the x-axis is a tangent)
b2 – 4ac < 0 there are no real roots (the graph does not touch the x-axis)
d) SIMULTANEOUS EQUATIONS
Solving by elimination
3x – 2y = 19 × 3
2x – 3y = 21 × 2
9x – 6y = 57
4x – 6y = 42
5x – 0y =15
x = 3 ( 9 – 2y = 19)
y = -5
Solving by substitution
x + y = 1 (y = 1 – x)
x2 + y2 = 25
x2 + (1 – x)2 = 25
2x2 – 2x – 24 = 0
2(x – 4)(x + 3) = 0
x = 4 y = -3
x=-3 y=4
If you end up with a quadratic equation when solving simultaneously the discriminant can be used to determine the
relationship between the graphs
If b2 – 4ac > 0 the graphs intersect at 2 distinct points
b2 – 4ac = 0 the graphs intersect at 1 point (or tangent)
b2 – 4ac < 0 the graphs do not intersect
e) INQUALITIES
Linear Inequality - solve using the same method as solving a linear equation but remember to reverse the
inequality if you multiply or divide by a negative number
Quadratic Inequality – always a good idea to sketch a graph
≤ ≥ plot the graph as a solid line or curve
< > plot as a dotted/dashed line or curve
If you are unsure of which area to shade pick a point in one of the regions and check the inequalities using the
coordinates of the point
f) POLYNOMIALS
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•
•
•
•
A polynomial is an expression which can be written in the form axn + bxn-1 + cxn-2 + … where a,b, c are
constants and n is a positive integer.
The order of the polynomial is the highest power of x in the polynomial
Polynomials can be divided to give a Quotient and Remainder
Factor Theorem – If (x – a) is a factor of f(x) then f(a) = 0 and is root of the equation f(x) = 0
Show that (x – 3) is a factor of x3 – 19x + 30 = 0
f(x) = x3 – 19x + 30
f(3) = 33 -19 × 3 + 20
=0
f(3) = 0 so x – 3 is a factor of f(x)
g) GRAPHS OF FUNCTIONS
Sketching Graphs
• Identify where the graph crossed the y-axis (x = 0)
• Identify where the graph crossed the x-axis (y = 0)
• Identify any asymptotes and plot with a dashed line
y=mx + c
y = kx2
y=kx3
y is proportional to x2
y is proportional to x2
y=
π
π₯
Asymptotes at
x = 0 and y = 0
y is proportional to
π
π₯
y=
π
π₯2
Asymptotes at
x = 0 and y = 0
y is proportional to
π
π₯2
Modulus Graphs
• |x| is the ‘modulus of x’ or the absolute value |2|=2 |-2|= 2
• To sketch the graph of y = |f(x)| sketch y = f(x) and take any part of the graph which is below the x-axis and
reflect it in the x-axis
Solve |2x - 4|<|x|
2x - 4 = x
x =4
2x – 4 = -x
3x = 4
4
x=3
4
3
<x<4
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h) FUNCTIONS
• A function is a rule which generates exactly ONE OUTPUT for EVERY INPUT
• DOMAIN – defines the set of the values that can be ‘put into’ the function f(x) = √π₯ domain x ≥ 0
• RANGE – defines the set of values ‘output’ by the function – make sure it is defined in terms of f(x) and not x
π βΆ π₯ β¦ π₯2 π₯ ∈ β
means an input a is converted to a2 where the input ‘a’ can be any real number
π
ππππ π(π₯) ≥ 0
3
π(π₯) = π₯+2 find π −1 (π₯)
-1
• INVERSE FUNCTION denoted by f (x)
3
The domain of f-1(x) is the range of f(x)
π¦ = π₯+2
-1
The range of f (x) is the domain of f(x)
3
Using the same scale on the x and y axis the graphs of
a function and it’s inverse have reflection symmetry
in the line y = x
•
π₯ =π¦−2
3
π −1 (π₯) = π₯ − 2
COMPOSITE FUNCTIONS
The function gf(x) is a composite function which tells you ‘to do’ f first and then use the output in g
f(x) = 4x
g(x) = x2 – 1
fg(x) = 4(x2 -1)
= 4x2 - 4
gf(x) = (4x)2 - 1
= 16x2 - 1
i) TRANSFORMING GRAPHS
Translation
π
To find the equation of a graph after a translation of [ ] replace x by (x – a) and y by (y – b)
π
y = f(x -a) + b
3
The graph of y = x2 -1 is translated by [ ]
−2
Find the equation of the resulting graph.
(y + 2) = (x – 3)2 - 1
y = x2 – 6x + 6
Reflection
Reflection in the x-axis replace y with -y y = -f(x)
Reflection in the y-axis replace x with -x y = f(-x)
Stretch
Stretch in the y-direction by scale factor a y = af(x)
1
Stretch on the x-direction by scale factor π y = f(ax)
Combining Transformations
Take care with the order in which the transformations are carried out.
π
The graph of y = x2 is translated by [ ] and then
π
reflected in the y axis. Find the equation of the
resulting graph
Translation y = (x – 3)2
= x2 -6x + 9
The graph of y = x2 is reflected in the y axis and
π
then translated by [ ]. Find the equation of the
π
resulting graph
Reflection y = (-x)2
= x2
Reflection y = (-x)2 -6(-x) + 9
= x2 + 6x + 9
Translation y = (x – 3)2
= x2 - 6x + 9
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j) PARTIAL FRACTIONS
Any proper algebraic fractions with a denominator that is a product of linear factors can be written as partial fractions
• Useful for integrating a rational function
• Useful for finding binomial approximations
ππ₯+π
(ππ₯+π)(ππ₯+π)(ππ₯+π)
=
π΄
ππ₯+π
π΅
πΆ
ππ₯+π
(ππ₯+π)(ππ₯+π)2
+ ππ₯+π + ππ₯+π
5
π΄
=
π΄
ππ₯+π
π΅
π΅
Express (π₯−2)(π₯+3) in the form π₯−2 + π₯+3
π΄
π₯−2
π΅
+ π₯+3 ≡
π΄(π₯+3)+π΅(π₯−2)
(π₯+3)(π₯−2)
A(x + 3) + B(x - 2) = 5
x = 2 5A = 5
A=1
5
(π₯−2)(π₯+3)
=
1
π₯−2
x = -3 -5B = 5
B = -1
−
1
π₯+3
2 COORDINATE GEOMETRY
a) Graphs of linear functions
y = mx + c
the line intercepts the y axis at (0, c)
πβππππ ππ π¦
πβππππ ππ π₯
Finding the equation of a line with gradient m through point (x1, y1)
Use the equation (y – y1) = m(x – x1)
If necessary rearrange to the required form (ax + by = c or y = mx + c)
Parallel and Perpendicular Lines
y = m1x + c1
y = m2x + c2
If m1 = m2 then the lines are PARALLEL
If m1 × m2 = -1 then the lines are PERPENDICULAR
Find the equation of the line perpendicular to the line y – 2x = 7 passing
through point (4, -6)
Gradient of y – 2x = 7 is 2 (y = 2x + 7)
Gradient of the perpendicular line = - ½
(2 × -½ = -1)
Equation of the line with gradient –½ passing through (4, -6)
(y + 6) = -½(x – 4)
2y + 12 = 4 – x
x + 2y = - 8
Finding the mid-point of the line segment joining (a,b) and (c,d)
Mid-point = (
π+π π+π
2
,
2
)
Calculating the length of a line segment joining (a,b) and (c,d)
Length = √(π − π)2 + (π − π)2
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πΆ
+ ππ₯+π + (ππ₯+π)2
b) Circles
A circle with centre (0,0) and radius r has the equations x2 + y2 = r2
A circle with centre (a,b) and radius r is given by (x - a)2 + (y - b)2 = r2
Finding the centre and the radius (completing the square for x and y)
Find the centre and radius of the circle x2 + y2 + 2x – 4y – 4 = 0
x2 + 2x + y2 – 4y – 4 = 0
(x + 1)2 – 1 + (y – 2)2 – 4 – 4 = 0
(x + 1)2 + (y – 2)2 = 32
Centre ( -1, 2)
The following circle properties might be useful
Angle in a semi-circle
The perpendicular from the centre
is a right angle
to a chord bisects the chord
The tangent to a circle is
Finding the equation of a tangent to a circle at point (a,b)
The gradient of the tangent at (a,b) is perpendicular to the gradient of the radius which meets the
circumference at (a, b)
Find equation of the tangent to the circle x2 + y2 - 2x - 2y – 23 = 0 at the point (5,4)
(x - 1)2 + (y – 1)2 – 25 = 0
Centre of the circle (1,1)
4−1
3
4
4
Gradient of tangent = - 3
4
Equation of the tangent (y – 4) = - 3(x – 5)
3y – 12 = 20 - 4x
4x + 3y = 32
Lines and circles Solving simultaneously to investigate the relationship between a line and a circle will result in
a quadratic equation. Use the discriminant to determine the relationship between the line and the circle
b2 – 4ac > 0
b2 – 4ac = 0 (tangent)
b2 – 4ac < 0
c) Parametric Equations
• Two equations that separately define the x and y coordinates of a graph in terms of a third variable
• The third variable is called the parameter
• To convert a pair of parametric equations to a cartesian equation you need to eliminate the parameter
(you may need to use trig identities if the parametric equations involve trig functions)
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Find the cartesian equation of the curve given by the parametric equations given by π₯ = πππ π π¦ = π ππ2π
π¦ = π ππ2π
π¦ = 2π ππππππ π
π¦ 2 = 4π ππ2 ππππ 2 π
= 4(1 − πππ 2 π)πππ 2 π
π¦ 2 = 4(1 − π₯ 2 )π₯ 2
3. SEQUENCES AND SERIES
a) Binomial
Expansion of (π + π)π |π| < π π ∈ β
π(π − 1) 2 π(π − 1)(π − 2) 3
(1 + π₯)π = 1 + ππ₯ +
π₯ +
π₯ … … … … + ππ₯ π−1 + π₯ π
1×2
1×2×3
1
Use the binomial expansion to write down the first four terms of (2−3π₯)2
3
2
−2
2−2 (1 − π₯)
3
2
= 2−2 (1 + −2 × (− π₯) +
=
1
4
1
(1 + 3π₯ +
3
= 4+ 4π₯ +
27
4
27 2
π₯
16
π₯2 +
+
2
−2×−3
3
(− π₯)
1×2
2
27
2
27 3
π₯
8
+
−2×−3×−4
3
(− π₯)3
1×2×3
2
π₯3)
Expansion of (π + π)π π ∈ β€+
π(π−1) π−2 2
π(π−1)(π−2) π−3 3
(π + π)π = ππ + πππ−1 π +
π
π +
π
π … … … … + πππ π−1 + π π
1×2
1×2×3
Find the coefficient of the x3 term in the expansion of (2 + 3x)9
(3x)3 must have 26 as part of the coefficient (3 + 6 = 9)
9×8×7
1×2×3
× 26 × (3π₯)3 = 145152 (x3)
b) Sequences
• An inductive definition defines a sequence by giving the first term and a rule to find the next term(s)
π’π+1 = π(π’π )
π’1 = π
Find the first 3 terms of a sequence defined by π’π+1 = 2π’π + 1
π’1 = 2 π’2 = 2 × 2 + 1
π’3 = 2 × 5 + 1
= 5
= 11
•
•
•
An increasing sequence is one where π’π+1 > π’π for all n
An decreasing sequence is one where π’π+1 < π’π for all n
A sequence may converge to a limit L π’π+1 = π(π’π ) as π → ∞
The sequence defined by π’π+1 = 0.2π’π + 2
L = 0.2L + 2
0.8L = 2 L = 2.5
•
π’1 = 2
π’π+1 = π’π = πΏ
π’1 = 3 converges to a limit L . Find L
A periodic sequence repeats itself over a fixed interval
π’π+π = π’π for all n for a constant a which is the period of the sequence
c) Sigma Notation – sum of
6
∑(π 2 + 1) = (12+1) + (22+1) + (32+1) + (42+1) + (52+1) + (62+1)
r=1
= 2 + 5 + 10 + 17 + 26 + 37
= 97
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Staring with the 1st term r = 1
Ending with the 6th term r = 6
d) Arithmetic sequences and series
• Each term is found by adding a fixed constant (common difference d) to the previous term
• The first term is a giving the sequence a , a + d, a + 2d, a + 3d…… where π’π = π + (π − 1)π
• The sum of the first n terms can be found using:
π
π
ππ = 2 (2π + (π − 1)π) or ππ = 2 (π + π) where l is the last term
e) Geometric sequence and series
• Each term is found by multiplying the previous term by a fixed constant (common ratio r)
• The first term is a giving the sequence a, ar, ar2, ar3, ar4…
• The sum of the first n terms can be found using
ππ =
π(1−π π )
1−π
or
ππ =
π(π π −1)
π −1
π
π∞ = 1 −π
|r| < 1
4. TRIGONOMETRY
MAKE SURE YOU KNOW AND CAN USE THE FOLLOWING FROM GCSE
π΄πππ =
π
ππππ΄
1
πππ πππΆ
2
π
π
= ππππ΅ = ππππΆ
or
ππππ΄
π
=
ππππ΅
π
=
ππππΆ
π
π2 = π 2 + π 2 − 2πππΆππ π΄
0
Sin
0
Cos
1
Tan
0
π
6
1
2
√3
2
√3
3
y = sin x
π
4
√2
2
√2
2
π
3
√3
2
1
2
π
2
π
1
0
0
1
1
√3
---
0
y = tan x
y = cos x
−
•
You MUST work in radians if you are integrating or differentiating trig functions
•
For an angle at the centre of a sector of θ radians
Arc Length = rθ
Area of the sector = ½r2θ
θ
b) Small angle approximations (θ in radians)
π πππ ≈ π
πππ π ≈ 1 −
π2
2
When π is small show that
π2
)÷π
2
2−π2
÷π
2
2−π2
2π
πππ π
π πππ
π‘πππ ≈ π
can be written as
2−π2
2π
(1 −
=
=
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π
2
−
π
2
c) Inverse Functions (sin-1x, cos-1x, tan-1 x)
By definition a function must be one-to-one which leads to restricted domains for the inverse trig functions
y = sin-1x (arcsin x)
Domain : -1 ≤ x ≤ 1
y = cos-1x (arccos x)
Domain : -1 ≤ x ≤ 1
π
π
2
-1
y = tan-1x (arctan x)
Domain x∈ β
π
2
1
π
-2
-1
π
-2
0
1
d) Reciprocal Trig Functions and identities (derived from π ππ2 π₯ + πππ 2 π₯ = 1)
1
1
sec π₯ = cos π₯
πππ ππ π₯ = sin π₯
1 + π‘ππ2 π₯ = π ππ 2 π₯
1
cos π₯
πππ‘ π₯ = tan π₯ ( sin π₯ )
1 + πππ‘ 2 π₯ = πππ ππ 2 π₯
Solve for 0° < θ < 360° the equation π ππ 4 π − π‘ππ4 π = 2
π ππ 2 π₯ = 1 + π‘ππ2 π₯
π ππ 4 π₯ = 1 + 2π‘ππ2 π₯ + π‘ππ4 π₯
1 + 2π‘ππ2 π₯ + π‘ππ4 π₯ − π‘ππ4 π₯ = 2
2π‘ππ2 π₯ = 1
e) Double angle and addition formulae
The addition formulae are given the formula booklet
Make sure you can use these to derive :
DOUBLE ANGLE FORMULAE
sin 2A = 2 sinAcosA
cos 2A = cos2A – sin2A
= 2cos2A – 1
= 1 - 2sin2A
Tan 2A =
•
•
1
2
tan π₯ = ±√
x = 35.3°, 145°, 215°,325°
3 s.f.
sin(π΄ ± π΅) = π πππ΄ πππ π΅ ± πππ π΄π πππ΅
cos(π΄ ± π΅) = πππ π΄ πππ π΅ β π πππ΄π πππ΅
tan(π΄ ± π΅) =
π‘πππ΄ ± π‘πππ΅
1 β π‘πππ΄π‘πππ΅
2 tan π΄
1−π‘ππ2 π΄
Useful to solve equations
cos 2A often used to integrate trig functions involving sin2x or cos2x
EXPRESSING IN THE FORM ππππ(π½ ± πΆ) and ππππ(π½ ± πΆ)
• Useful in solving equations asin θ + bcos θ = 0
• Useful in finding minimum/maximum values of ππππ π ± ππ πππ
and
ππ πππ ± ππππ π
Find the maximum value of the expression 2sinx + 3cosx and the value of x where this occurs (x < 180°)
2sinx + 3cosx = Rsin(x + α)
(Rsin x cos α + Rcos x sin α) rcos α = 2 rsin α = 3
3
2
2
R = √2 + 3
tanα = 2
= √13
α = 56.3°
2sinx + 3cosx = √13 sin(π₯ + 56.3°)
Max value = √13 occurs when sin(π₯ + 56.3°) = 1
x = 33.7°
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5
LOGARITHMS AND EXPONENTIALS
•
•
•
y = a-x
y = ax
A function of the form y = ax is an exponential function
The graph of y = ax is positive for all values of x
and passes through (0,1)
A logarithm is the inverse of an exponential function
y = ax
x = loga y
Logarithms – rules to learn
loga a = 1
loga ax = x
loga 1 = 0
loga m + loga n = loga mn
π
loga m - loga n = loga ( π )
πlogπ π₯ = x
kloga m = loga mk
Write the following in the form alog 2 where a is an integer 3log 2 + 2log 4 – ½log16
8×16
)=
4
Method 1 : log 8 + log 16 – log 4 = log (
log 32 = 5log 2
Method 2 : 3log 2 + 4log 2 – 2log 2 = 5log 2
An equation of the form ax = b can be solved by taking logs of both sides
a) MODELLING CURVES
Exponential relationships can be changed to a linear form y = mx + c allowing the constants m and c
to be ‘estimated’ from a graph of plotted data
Plot log y against log x. n is the
y = Axn log y = log (Axn) log y = n log x + log A
gradient of the line and log A is
y = mx + c
the y axis intercept
y = Abx
log y = log (Abx)
log y = x log b + log A
y = mx + c
Plot log y against x. log b is the
gradient of the line and log A is
the y axis intercept
V and x are connected by the equation V = axb
The equation is reduced to linear form by taking logs
log V = b log x + log a
Intercept = 3
(y = mx + c) (log V plotted against log x)
From the graph b = 2
log a = 3 a = 103
V = 1000x2
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b) The exponential function y = ex
Exponential Growth
y = ex
Exponential Decay y = e-x
Solve 2ex-2 = 6 leaving your
π π₯−2 = 3
ln(π π₯−2 ) = ln 3
π₯ − 2 = ln 3
π₯ = ln 3 + 2
The inverse of y = ex is the natural logarithm denoted by ln x
The rate of growth/decay to find the ‘rate of change’ you need to differentiate to find the gradient
LEARN THIS
ππ¦
π¦ = π΄π ππ₯
The number of bacteria P in a culture is modelled by
P = 600 + 5e0.2t where t is the time in hours from the start of the
experiment. Calculate the rate of growth after 5 hours
= π΄ππ ππ₯
ππ₯
ππ
ππ‘
P = 600 + 15e0.2t
t=5
ππ
ππ‘
= 3π 0.2π‘
= 3π 0.2×5
= 8.2 bacteria per hour
6
DIFFERENTIATION
•
•
ππ¦
The gradient is denoted by ππ₯ if y is given as a function of x
The gradient is denoted by f’(x) is the function is given as f(x)
LEARN THESE
π¦ = π₯π
π¦ = π ππ₯
π¦ = sin ππ₯
ππ¦
ππ₯
= ππ₯ π−1
ππ¦
ππ₯
= ππ kx
ππ¦
ππ₯
= ππππ ππ₯
ππ¦
π¦ = ππ₯ π
ππ₯
ππ¦
π¦ = πππ₯
π¦ = cos ππ₯
ππ₯
ππ¦
ππ₯
= πππ₯ π−1
1
ππ¦
PRODUCT RULE for differentiating y = f(x)g(x)
π(π)
QUOTIENT RULE for differentiating y = π(π)
PARAMETRIC EQUATIONS y = f(t)
x = g(t)
ππ₯
π
π
=
π
π
ππ¦
ππ₯
=
π¦ = πππ₯
=π₯
= −ππ ππ ππ₯
a) Methods of differentiation
CHAIN RULE for differentiating y = fg(x) y = f(u) where u = g(x)
π¦= π
ππ¦
ππ₯
ππ¦
ππ₯
π¦ = tan ππ₯
=
ππ¦
ππ’
ππ’
× ππ₯
= π ′ (π₯)π(π₯) + π(π₯)π′(π₯)
π′ (π)π(π)−π(π)π′(π)
[π(π)]π
ππ¦
ππ‘
ππ₯
÷ ππ‘
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ππ¦
ππ₯
=0
= (ππππ)πππ₯
ππ¦
ππ₯
= ππ ππ 2 ππ₯
IMPLICIT DIFFERENTIATION- take care as you may need to use the product rule too (xy2, xy, ysinx)
π[π(π¦)]
ππ₯
=
π[π(π¦)]
ππ¦
ππ¦
× ππ₯
b) Stationary (Turning) Points
• The points where
•
ππ¦
ππ₯
= 0 are stationary points (turning points/points of inflection) of a graph
The nature of the turning points can be found by:
Maximum point
Minimum Point
ππ¦
>0
ππ₯
ππ¦
<0
ππ₯
ππ¦
=0
ππ₯
ππ¦
=0
ππ₯
ππ¦
>0
ππ₯
ππ¦
<0
ππ₯
π2 π¦
Maximum if
ππ₯ 2
Minimum if
π2 π¦
ππ₯ 2
>0
<0
Find and determine the nature of the stationary points of the curve y = 2x3 - 3x2 + 18
ππ¦
ππ₯
ππ¦
ππ₯
= 6π₯ 2 − 6π₯
= 0 at a stationary point
6x(x - 1) = 0 Turning points at (0, 18) and (1,17)
π2 π¦
ππ₯ 2
= 12x – 6 x = 0
x=1
Points of inflection occur when
Convex curve :
π2 π¦
Concave curve :
ππ₯ 2
π2 π¦
ππ₯ 2
π2 π¦
ππ₯ 2
π2 π¦
ππ₯ 2
< 0 (0,18) is a maximum
> 0 (1,17) is a minimum
= 0 (π ′′ (π₯) = 0)
π2 π¦
………but ππ₯ 2 = 0 could also indicate a min or max point
> 0 for all values of x in the ‘convex section of the curve’
π2 π¦
ππ₯ 2
< 0 for all values of x in the ‘concave section of the curve’
c) Using Differentiation
Tangents and Normals
The gradient of a curve at a given point = gradient of the tangent to the curve at that point
The gradient of the normal is perpendicular to the gradient of the tangent that point
Find the equation of the normal to the curve y = 8x – x2 at the point (2,12)
ππ¦
ππ₯
= 8 − 2π₯ Gradient of tangent at (2,12) = 8 – 4 = 4
Gradient of the normal = - ¼
(y - 12) = -¼ (x – 2)
4y + x = 50
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d) Differentiation from first principles
As h approaches zero the gradient of the chord gets
closer to being the gradient of the tangent at the point
π(π₯+β)−π(π₯)
π ′ (π₯) = lim (
(x+h,f(x+h))
β→0
β
)
Find from first principles the derivative of x3 – 2x + 3
π(π₯+β)−π(π₯)
)
β
π ′ (π₯) = lim (
(x,f(x))
β→0
(π₯+β)3 −2(π₯+β)+3 –(π₯ 3 −2π₯+3)
h
= lim (
β
β→0
)
π₯ 3 +3π₯ 2 β+3π₯β 2 +β3 −2π₯−2β +3 −π₯ 3 + 2π₯−3)
)
β
β→0
3π₯ 2 β+3π₯β 2 +β3 −2β
lim (
)
β
β→0
lim (3π₯ 2 + 3π₯β + β2 − 2)
β→0
2
= lim (
=
=
= 3π₯ − 2
7
INTEGRATION
Integration is the reverse of differentiation
LEARN THESE
∫ π₯ π ππ₯ =
∫ π ππ₯ ππ₯ =
π₯ π+1
π+1
1
π
+π
(c is the constant of integration)
1
π ππ₯ + π
∫ π₯ ππ₯ = πππ₯ + π
1
∫ sin ππ₯ ππ₯ = − π cos ππ₯ + π
∫ cos ππ₯ ππ₯ =
1
π
sin ππ₯ + π
a) Methods of Integration
INTEGRATION BY SUBSTITUTION
Transforming a complex integral into a simpler integral using ‘u = ‘ and integrating with respect to u
∫ π₯√1 − π₯ 2 ππ₯
ππ’
ππ’
Let u = 1 - x2 ππ₯ = -2x so dx = −2π₯
ππ’
∫ π₯ √1 − π₯ 2 ππ₯ = ∫ π₯√π’
−2π₯
1
1
= - 2 ∫ π’2 ππ’
1
= − 3 (1 − π₯
1 3
= − 3 π’2
3
2 )2
If it is a definite integral it is often easier to
calculate the limits in terms of u and
substitute these in after integrating
+c
Look for integrals of the form
∫ π ππ₯+π ππ₯
+π
Look out for integrals of the form
1
∫ π ′ (π₯)[π(π₯)]π = π+1 [π(π₯)]π+1 + π
π′(π₯)
∫ π(π₯) ππ₯ = ln|π(π₯)| + π
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∫ cos(ππ₯ + π) ππ₯
1
∫ ππ₯+π ππ₯
INTEGRATION BY PARTS
π
π
π
π
∫ π π
π π
π = ππ − ∫ π π
π π
π
∫ πππ₯ ππ₯
ππ’
ππ₯
ππ£
Take care in defining u and ππ₯
ππ£
∫ π₯π 2π₯ ππ₯
π’=π₯
∫ π₯ππ ππ₯
π’ = πππ₯
= π 2π₯
ππ₯
ππ£
ππ₯
=
1
π₯
ππ£
ππ₯
π’ = ln π₯
=1
π£=π₯
1
∫ πππ₯ ππ₯ = π₯πππ₯ − ∫ π₯ π₯ ππ₯
=π₯
= π₯πππ₯ − π₯ + c
PARAMETRIC INTEGRATION
ππ₯
To find the area under a curve defined parametrically use area = ∫ π¦ ππ‘ ππ‘
Remember that the limits of the integral must be in terms of t
4
A curve is defined parametrically by π₯ = π‘ − 1 π¦ = π‘ Calculate the area of the region
included by the line x= 2, the x-axis and the y-axis.
x=2 t=3
x=0 t=1
34
ππ₯
=1
∫1 ππ‘ = [4πππ‘] 13
ππ‘
π‘
= 4ππ3 − 4ππ1
= 4ln 3
b) AREA UNDER A CURVE
The area under a graph can be approximated using rectangle of height y and width dx. The limit as the
number of rectangles increases is equal to the definite integral
π
π
lim ∑ π¦π πΏπ₯ = ∫ π¦ ππ₯
π→∞
π=1
π
Calculate the area under the graph y = 4x – x3 between x = 0 and x = 2
2
∫0 4π₯ − π₯ 3 ππ₯
2
π₯4
= [2π₯ − ]
4
0
2
= (8 − 4) − (0 − 0)
= 4
For an area below the x-axis the integral will
result in a negative value
c) AREA BETWEEN 2 CURVES
If no limits are given you need to identify the x coordinates of the points where the curve intersect
Determine which function is ‘above’ the other
π₯
2
∫π₯ [π(π₯) − π(π₯)]ππ₯
g(x)
1
f(x)
π₯1
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π₯2
d) SOLUTION OF DIFFERENTIAL EQUATIONS
Separating the variables
If you are given the coordinates of a point on the curve a particular solution
can be found if not a general solution is needed
Find the general solution for the differential
equation
ππ¦
π¦ ππ₯ = π₯π¦ 2 + 3π₯
ππ¦
π¦ ππ₯ = π₯(π¦ 2 + 3)
π¦
∫ π¦2 +3 ππ¦ = ∫ π₯ ππ₯
1
ln|π¦ 2
2
8
1
+ 3| = 2 π₯ 2 + π
NUMERICAL METHODS
a) CHANGE OF SIGN – locating a root
For an equations f(x) = 0 , if f(x1) and f(x2) have opposite signs and f(x) is a continuous function between x1 and
x2 then a root of the equation lies in the interval x1< x < x2
b) STAIRCASE and COWBEB DIAGRAMS
If an iterative formula (recurrence relation) of the form xn+1=f(xn) converges to a limit, the value of
the limit is the x-coordinate of the point of intersection of the graphs y = f(x) and y = x
The limit is the solution of the equation f(x) = x
A staircase or cobweb diagram based on the graphs y = f(x) and y = x shows the convergence
Use you ANS
button on your
calculator
c) NEWTON-RAPHSON iteration π(π₯) = 0
π(π₯ )
π₯π+1 = π₯π − π′(π₯π )
π
The equation e-2x – 0.5x = 0 has a root close to 0.5. Using 0.5
as the first approximation use the Newton-Raphson you find
the next approximation
x1 = 0.5 f(0.5) = e-1 – 0.25
f’(x) = -2e-2x – 0.5 f’(0.5) = -2e-1 – 0.5
x2 = 0.5 −
π −1 – 0.25
−2π −1 −0.5
x2 = 0.595
Limitations of the Newton-Raphson method
As the method uses the tangent to the curve, if the
starting value is a stationary point or close to a
stationary point (min, max or inflection) the
method does not work
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d) APPROXIMATING THE AREA UNDER A CURVE
TRAPEZIUM RULE – given in the formula book but make sure you know how to use it!
The trapezium rule gives an approximation of the area under a graph
π
1
∫π π¦ ππ₯ ≈ 2 β[(π¦0+ π¦π ) + 2(π¦1 + π¦2 +. . . π¦π−1 )] where β =
π−π
π
An easy way to calculate the y values is to use the TABLE function on a calculator – make sure you list the
values in the formula (or a table) to show your method
• The rule will underestimate the area when the curve is concave
• The rule will overestimate the area when the curve is convex
UPPER and LOWER bounds - Area estimated using the area of rectangles
For the function shown below if the left hand ‘heights’ are used the total area is a Lower Bound – the
rectangles calculated using the right hand heights the area results in the Upper Bound
Upper Bound
Lower Bound
9
VECTORS
A vector has two properties magnitude (size) and direction
a) NOTATION
Vectors can be written as
3
a=( )
4
a = 3i + 4j where i and j perpendicular unit vectors (magnitude 1)
j
i
Magnitude-direction form (5, 53.1β°) also known as polar form
The direction is the angle the vector makes with the positive x axis
Express the vector p = 3i – 6j in polar form
|p| = √32 + (−6)2
= 3√5
p = ( 3√5, 63.4β°)
The Magnitude of vector a is denoted by |a| and can be found using Pythagoras |a| = √32 + 42
A Unit Vector is a vector which has magnitude 1
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A position vector is a vector that starts at the origin (it has a fixed position)
2
βββββ
ππ΄ = ( )
4
2π + 4π
b) ARITHMETIC WITH VECTORS
Multiplying by a scalar (number)
3
a = ( ) 3i + 2j
2
3
6
2a = 2 ( ) = ( ) 6i + 4j
2
4
a and 2a are parallel vectors
Multiplying by -1 reverses the direction of the vector
2
3
a =( ) b =( )
3
1
Subtraction of vectors
2
a =( )
3
2
3
5
a + b = ( ) + ( ) =( )
3
1
4
resultant
3
b =( )
1
2
3
−1
a - b = ( ) - ( ) =( )
3
1
2
This is really a + -b
A and B have the coordinates (1,5) and (-2,4).
a)
Write down the position vectors of A and B
1
βββββ
ππ΄ = ( )
5
b)
−2
βββββ
ππ΅ = ( )
4
Write down the vector of the line segment joining A to B
ββββββ
ββββββ = −πΆπ¨
ββββββ − πΆπ¨
ββββββ
−2
1
−3
βββββ
π΄π΅ = ( ) − ( ) = ( )
4
5
−1
Collinear - vectors in 2D and 3D can be used to show that 3 or more points are collinear
(lie on a straight line)
Show that A(3,1,2) B(7,4,5) and C(19,13,14)
βββββ = 4π + 3π + 3π π΅πΆ
βββββ = 12π + 9π + 9π
π΄π΅
βββββ
ββββββββ
βββββ
π΅πΆ = 3π΄π΅
π΄π΅ and βββββ
π΅πΆ are parallel vectors sharing a common point B
and are therefore collinear
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10 PROOF
Notation
⇒
If x = 3 then x2 = 9
x = 3 ⇒ x2 = 9
x = 3 is a condition for x2 = 9
βΈ
x = 3 βΈ x2 = 9 is not true as x could = - 3
βΊ
x+1=3βΊx=2
a) Proof by deduction – statement proved using known mathematical principles
Useful expressions : 2n (an even number)
2n + 1 (an odd number)
Prove that the difference between the squares of any consecutive
even numbers is a multiple of 4
Consecutive even numbers 2n, 2n + 2
(2n + 2)2 – (2n)2
4n2 + 8n + 4 – 4n2
=8n + 4
=4(2n +1) a multiple of 4
b) Proof by exhaustion – showing that a statement is true for every possible case or value
Prove that (π + 2)3 ≥ 3π−1 for ππΝ, n<4
We need to show it is true for 1,2 and 3
n = 1 27 ≥ 1
n = 2 64 ≥ 3
n = 3 125 ≥ 9
True for all possible values hence proof that the
statement is true by exhaustion
c) Disproof by counter example – finding an example that shows the statement is false.
Find a counter example for the statement
‘2n + 4 is a multiple of 4’
n = 2 4 + 4 = 8 a multiple of 4
n = 3 6 + 4 = 10 NOT a multiple of 4
d) Proof by contradiction - assume first that the statement is not true and then show that this is not possible
Prove that for all integers n, if n3 + 5 is odd then n is even
Assume that n3 + 5 is odd and n is odd
Let n3 + 5 = 2k + 1 and let n = 2m + 1 (k and m integers)
2k + 1 = (2m + 1)3 + 5
2k + 1 = 8m3 + 12m2 + 6m + 6
2k = 8m3 + 12m2 + 6m + 5
2k = 2(4m3 + 6m2 + 3m) + 5
5
k = (4m3 + 6m2 + 3m) + 2
4m3 + 6m2 + 3m has an integer value leaving k as a non-integer value
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# rational number
The set of rational numbers, usually denoted by $$\mathbb{Q}$$, is the subset of real numbers that can be expressed as a ratio of integers, that is, as ratios of positive or negative whole numbers, like $$\frac{2}{3}$$ or $$\frac{-13}{5}$$. Real numbers that cannot be so expressed are called irrational numbers (e.g., $$\pi$$, $$\sqrt{2}$$, etc.)
The arithmetic of rational numbers takes some getting used to. For any rational numbers $$\frac{a}{b}$$ and $$\frac{c}{d}$$ the operations of addition, multiplication, and division work as follows:
$\begin{eqnarray*} \frac{a}{b} + \frac{c}d{} & = & \frac{ad+bc}{bd} \\ & & \\ \frac{a}{b}\cdot\frac{c}{d} & = & \frac{ac}{bd} \\ & & \\ \frac {a}{b}\div \frac{c}{d} & = & \frac{ad}{bc} \\ \end{eqnarray*}$
These rules are most commonly remembered by saying that to add fractions we need a common denominator, to multiply them we just multiply straight across the numerators and straight across the denominators, and to divide them we simply turn the divisor over and then multiply. So for example we have:
$\begin{eqnarray*} \frac{1}{2} + \frac{3}{5} & = & \frac{1\times 5+3\times 2}{2\times 5} = \frac{11}{10} \\ & & \\ \frac{1}{2}\cdot\frac{3}{5} & = & \frac{1\times 3}{2\times 5} = \frac{3}{10} \\ & & \\ \frac {1}{2}\div \frac{3}{5} & = & \frac{1\times 5}{2\times 3} = \frac{5}{6} \end{eqnarray*}$
Formally, the rational numbers are defined as a set of equivalence classes of ordered pairs of integers, where the first component of the ordered pair is the numerator and the second is the denominator. The equivalence classes arise from the fact that a rational number may be represented in any number of ways by introducing common factors to the numerator and denominator. For instance, $$\displaystyle \frac{2}{3}$$ and $$\displaystyle \frac{6}{9}$$ are the same number:
$\begin{eqnarray*} \frac{6}{9} & = & \frac{2\times 3}{3\times 3} \\ & & \\ & = & \frac{2}{3}\times \frac{3}{3} \\ & & \\ & = & \frac{2}{3} \times 1 \\ & & \\ & = & \frac{2}{3} \end{eqnarray*}$
The equivalence class $$\sim$$ between ordered pairs of integers that determines when two fractions represent the same rational number is given by:
$\frac{a}{b}\sim\frac{c}{d} \Leftrightarrow ad = bc$
Thus for instance, we see that $$\displaystyle \frac{2}{3}=\frac{6}{9}$$ because $$2\times 9=3\times 6$$.
Notice that this similarity definition breaks down if either of the denominators is zero, so this must be outlawed.
So, finally, the complete set-theoretic definition of the rational numbers is as follows:
$\mathbb{Q}= \left. \left\{ \left. \frac{p}{q} \, \right| \,\, p,q \in \mathbb{Z}, q\neq 0 \right\} \right/ \sim$
It is common for students to find fractions to be “hard,” and there is a good reason for this. It is because they are hard. When working with a ratio of integers one is working not with a number per se, but with a representative of an equivalence class of ordered pairs. Of course it’s hard. But once you get the knack, a facility with fractions is a great boon because from then on you can frequently avoid frittering away your time with decimals and calculators. (And you can impress the socks off your friends, too.)
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Distance Point Construction
Historical records show that besides Alberti’s construction, there were other methods for constructing pavimenti. One of them was known as the distance point construction, and was found in the treatise of Jean Pelerin (1445 – 1522), also known as the Viator. Entitled De Artificiali Perspectiva, it was first published in Toul in 1505 and later pirated at Nuremberg in 1509. It produces the same results as Alberti’s construction, but constructs the pavimenti differently. Fig. 8 – Choosing the distance point D. As before, the ground line AB is divided equally, and each of these division points are joined to the centric point C. Next, the distance point D is chosen. The distance CD is the viewing distance. Fig. 9 – The distance point construction. The line AD will intersect all the orthogonals. These intersection points are used to draw the transversals. Fig. 10 – The plan and vertical section corresponding to the distance point construction. We will next formulate a geometrical proof for the distance point construction. Fig. 10 shows the floor plan for the distance point construction. This is similar to Fig. 7. For Alberti’s construction, it is the distance from the viewer to the last transversal which is of great importance. We can picture the line ER as the line HP rotated 90° anticlockwise about the point X. Hence, we show that MP is equal to NR. For the distance point construction, we can picture the line MD as the line MP rotated 90° anticlockwise about the point M. D is the distance point. If one was to stand at D instead of P, one can easily see that the distances MP and MD are equal. It follows that MD is also the viewing distance. We have shown that MD and NR are the correct viewing distances. Hence, MD and NR must be the same length. It follows that Alberti’s and the distance point construction are equivalent. To obtain a three-dimensional “proof” of the distance point construction, consider a square tile shown in the diagram below. Fig. 11 – Three-dimensional setup of the distance point construction. Suppose we have a “cardboard peephole”, which is a piece of cardboard with the trapezium cut out. Now we wish to position the square grid behind the cardboard peephole, in such a way that the entire grid can be seen through the trapezium hole. Fig.12 – Viewing the grid through the cardboard peephole. The viewing distance is x. Now imagine that the cardboard peephole is rotated about its axis line, as shown in Fig. 13. Fig. 13 – Notice that the viewing distance x is due to the distance point construction. Now imagine that the cardboard peephole is displaced to the right, such that it sits exactly on top of the grid (Fig. 14). Fig. 14 – Notice that the viewing distance x is due to Alberti’s construction. After the displacement, notice that x is now the viewing distance obtained from Alberti’s construction. Though hardly a rigorous proof, it suggests a way of obtaining viewing distances from both methods using physical models, and provides an intuitive “feel” for the distance point construction. Most importantly, the model shows that Alberti’s and the distance point construction are equivalent, as they yield the same viewing distances.
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# 5+3/4n+8=-2/4n+8
## Simple and best practice solution for 5+3/4n+8=-2/4n+8 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.
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## Solution for 5+3/4n+8=-2/4n+8 equation:
5+3/4n+8=-2/4n+8
We move all terms to the left:
5+3/4n+8-(-2/4n+8)=0
Domain of the equation: 4n!=0
n!=0/4
n!=0
n∈R
Domain of the equation: 4n+8)!=0
n∈R
We add all the numbers together, and all the variables
3/4n-(-2/4n+8)+13=0
We get rid of parentheses
3/4n+2/4n-8+13=0
We multiply all the terms by the denominator
-8*4n+13*4n+3+2=0
We add all the numbers together, and all the variables
-8*4n+13*4n+5=0
Wy multiply elements
-32n+52n+5=0
We add all the numbers together, and all the variables
20n+5=0
We move all terms containing n to the left, all other terms to the right
20n=-5
n=-5/20
n=-1/4
`
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# Signals & Systems Questions and Answers – Basics of Linear Algebra
This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Basics of Linear Algebra”.
1. Find the values of x, y, z and w from the below condition.
$$5\begin{bmatrix} x & z \\ y & w \\ \end{bmatrix} = \begin{bmatrix} 2 & 10 \\ 3 & 2x+y \\ \end{bmatrix} + \begin{bmatrix} z & 5 \\ 7 & w \\ \end{bmatrix}$$.
a) x=1, y=3, z=4, w=0
b) x=2, y=3, z=8, w=1
c) x=1, y=2, z=3, w=1
d) x=1, y=2, z=4, w=1
View Answer
Answer: d
Explanation: 5z=10+5 => 5z=15 => z=3
5x=2+z => 5x=5 => x=1
5y=3+7 => 5y=10 => y=2
5w=2+2+w => 4w=4 => w=1.
2. The matrix A is represented as $$\begin{bmatrix} 1 & 4 \\ -2 & 9 \\ -3 & -8 \\ \end{bmatrix}$$. The transpose of the matrix of this matrix is represented as?
a) $$\begin{bmatrix} 1 & 4 \\ -2 & 9 \\ \end{bmatrix}$$
b) $$\begin{bmatrix} 1 & 4 \\ -2 & 9 \\ -3 & 8 \\ \end{bmatrix}$$
c) $$\begin{bmatrix} 1 & -2 & -3\\ 4 & 9 & 8\\ \end{bmatrix}$$
d) $$\begin{bmatrix} -1 & 2 & 3\\ -4 & -9 & 8\\ \end{bmatrix}$$
View Answer
Answer: c
Explanation: Given matrix is a 3×2 matrix and the transpose of the matrix is 3×2 matrix.
The values of matrix are not changed but, the elements are interchanged, as row elements of a given matrix to the column elements of the transpose matrix and vice versa but the polarities of the elements remains same.
3. Find the inverse of the matrix $$A = \begin{bmatrix} 8 & 5 & 2\\ 4 & 6 & 3\\ 7 & 4 & 2\\ \end{bmatrix}$$.
a) $$\frac{1}{13}*\begin{bmatrix} 90 & 65 & 80\\ 65 & 61 & 54\\ 80 & 58 & 69\\ \end{bmatrix}$$
b) $$\frac{1}{14}*\begin{bmatrix} 93 & 68 & 80\\ 68 & 61 & 58\\ 80 & 58 & 69\\ \end{bmatrix}$$
c) $$\frac{1}{13}*\begin{bmatrix} 94 & 67 & 80\\ 67 & 60 & 56\\ 80 & 58 & 69\\ \end{bmatrix}$$
d) $$\frac{1}{13}*\begin{bmatrix} 93 & 68 & 80\\ 68 & 61 & 58\\ 80 & 58 & 69\\ \end{bmatrix}$$
View Answer
Answer: d
Explanation: The inverse of matrix A = $$\frac{adjA}{|A|}$$,
adjA=AA-1,
adjA = $$\frac{1}{13}*\begin{bmatrix} 93 & 68 & 80\\ 68 & 61 & 58\\ 80 & 58 & 69\\ \end{bmatrix}$$, |A|=13.
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4. Given the equations are 4x+2y+z=8, x+ y+ z=3, 3x+y+3z=9. Find the values of x, y and z.
a) 5/3, 0, 2/3
b) 1, 2, 3
c) 4/3, 1/3, 5/3
d) 2, 3, 4
View Answer
Answer: a
Explanation: The matrix from the equations is represented as M=$$\begin{bmatrix} 4 & 2 & 1\\ 1 & 1 & 1\\ 3 & 1 & 3\\ \end{bmatrix}$$
The another matrix is X = $$\begin{bmatrix} 8\\ 3\\ 9\\ \end{bmatrix}$$
Then |M| = 6
For x=$$\begin{bmatrix} 8 & 2 & 1\\ 3 & 1 & 1\\ 9 & 1 & 3\\ \end{bmatrix}$$ = 5/3
Similarly, y=0, z=-2/3.
5. Find the adjacent A as A=$$\begin{bmatrix} 1 & 7 & -3\\ 5 & 4 & -2\\ 6 & 8 & -6\\ \end{bmatrix}$$.
a) $$\begin{bmatrix} 1 & 1 & 1\\ 1 & 2 & 3\\ 2 & 3 & 4\\ \end{bmatrix}$$
b) $$\begin{bmatrix} 31 & 39 & 80\\ 39 & 45 & 74\\ 80 & 74 & 136\\ \end{bmatrix}$$
c) $$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{bmatrix}$$
d) $$\begin{bmatrix} 35 & 34 & 32\\ 56 & 67 & 48\\ 98 & 74 & 52\\ \end{bmatrix}$$
View Answer
Answer: b
Explanation: The adjacency of A is given by AAT
AT = $$\begin{bmatrix} 1 & 5 & 6\\ 7 & 4 & 8\\ -3 & -2 & -6\\ \end{bmatrix}$$,
AAT = $$\begin{bmatrix} 1 & 7 & -3\\ 5 & 4 & -2\\ 6 & 8 & -6\\ \end{bmatrix} × \begin{bmatrix} 1 & 5 & 6\\ 7 & 4 & 8\\ -3 & -2 & -6\\ \end{bmatrix}$$
adjA=$$\begin{bmatrix} 31 & 39 & 80\\ 39 & 45 & 74\\ 80 & 74 & 136\\ \end{bmatrix}$$.
6. Find the rank of the matrix A=$$\begin{bmatrix} 1 & 3 & 5\\ 4 & 6 & 7\\ 1 & 2 & 2\\ \end{bmatrix}$$.
a) 3
b) 2
c) 1
d) 0
View Answer
Answer: a
Explanation: To find out the rank of the matrix first find the |A|
If the value of the|A| = 0 then the matrix is said to be reduced
But, as the determinant of A has some finite value then, the rank of the matrix is 3.
7. The rank of the matrix (m × n) where m<n cannot be more than?
a) m
b) n
c) m*n
d) m-n
View Answer
Answer: a
Explanation: let us consider a 2×3 matrix $$\begin{bmatrix} 1 & 1 & 1\\ 4 & 5 & 6\\ \end{bmatrix}$$
Where R1≠R2 rank is 2
Another 2×3 matrix $$\begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ \end{bmatrix}$$
Here, R1=R2 rank is 1
And the rank of these two matrices is 1, 2
So rank is cannot be more than m.
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8. Given A=$$\begin{bmatrix} 2 & -0.1 \\ 0 & 3 \\ \end{bmatrix} A^{-1} = \begin{bmatrix} 1/2 & a \\ 0 & b \\ \end{bmatrix}$$ then find a + b.
a) $$\frac{6}{20}$$
b) $$\frac{7}{20}$$
c) $$\frac{8}{20}$$
d) $$\frac{5}{20}$$
View Answer
Answer: b
Explanation: AA-1 = I = $$\begin{bmatrix} 1 & 2-0.1b \\ 0 & 3b \\ \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$$
Therefore, a = $$\frac{1}{60}$$ and b = $$\frac{1}{3}$$ and a + b = $$\frac{7}{20}$$.
9. If a square matrix B is skew symmetric then.
a) BT = -B
b) BT = B
c) B-1 = B
d) B-1 = BT
View Answer
Answer: a
Explanation: The transpose of a skew symmetric matrix should be equal to the negative of the matrix
Example: let us consider a matrix B = $$\begin{bmatrix} a & e & d\\ -e & b & f\\ -d & -f & c\\ \end{bmatrix}$$, BT = $$\begin{bmatrix} a & -e & -d\\ e & b & -f\\ d & f & c\\ \end{bmatrix}$$.
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10. For the following set of simultaneous equations 1.5x-0.5y=2, 4x+2y+3z=9, 7x+y+5=10.
a) The solution is unique
b) Infinitely many solutions exist
c) The equations are incompatible
d) Finite number of multiple solutions exist
View Answer
Answer: a
Explanation: The equations can be written as $$\begin{bmatrix} 1.5 & -0.5 & 0\\ 4 & 2 & 3\\ 7 & 1 & 5\\ \end{bmatrix}$$
It can also be written as A = $$\begin{bmatrix} 3 & -2 & 0\\ 4 & 2 & 3\\ 7 & 1 & 5\\ \end{bmatrix}$$, |A|=19
Hence, it has a unique solution.
Sanfoundry Global Education & Learning Series – Signals & Systems.
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If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
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# Tutor profile: Dylan H.
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Dylan H.
Tutor for 5 years
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## Questions
### Subject:SAT
TutorMe
Question:
$$a^2–26a+69=0$$. If $$a–5>0$$, what is the value of a? A. 3 B. 5 C. 18 D. 23
Inactive
Dylan H.
There are two methods for solving this problem. First there is the "plug-and-chug" method, which involves testing the answer choices. From $$a–5>0$$, we have that $$a>5$$ so answers A and B are not correct. So now we plug $$18$$ and $$23$$ into $$a^2–26a+69=0$$. By calculation, $$a=23$$ so D is the answer. The second method involves solving the polynomial $$a^2–26a+69=0$$ for $$a$$. There are several methods to factor this including using the quadratic equation or completing the square. We will use completing the square because it involves little computation: $(a^2-26a=-69$) We add $$(26/2)^2=169$$ to both sides then factor the left-hand side of the equation: $(a^2-26a+169=-69+169$) $((a-13)^2=100$) Taking the square root, we have $$a-13=\pm10$$. Thus, $$a=3$$ and $$a=23$$. The second equation tells us that $$a>5$$ so $$a=23$$ (option D) is the solution.
### Subject:Discrete Math
TutorMe
Question:
Show that $$1+2+3+...+n=\frac{n(n+1)}{2}$$.
Inactive
Dylan H.
We prove this by induction. First, the base case of $$n=1$$ has that $$1=\frac{1(1+1)}{2}$$. This is true, so we assume $$1+2+3+...+n=\frac{n(n+1)}{2}$$ for the purpose of induction. We need to show that $$1+2+3+...+n+(n+1)=\frac{(n+1)(n+2)}{2}$$. By our inductive assumption, we have that $$1+2+3+...+n+(n+1)=\frac{n(n+1)}{2}+(n+1)$$. Using algebra, we simplify: $(=\frac{n^2+n}{2}+\frac{2n+2}{2}=\frac{n^2+3n+2}{2}=\frac{(n+1)(n+2)}{2}$) This closes the induction and we have shown what was needed.
### Subject:Calculus
TutorMe
Question:
Derive the derivative for $$y = x^3$$.
Inactive
Dylan H.
The derivative is the instantaneous change in a function at a point. The change in a function is the estimated by the linear change between two nearby points on that line: $(\frac{(x+x_0)^3-x^3}{(x+x_0)-x}$) where $$x_0$$ is the difference between our initial value of $$x$$ and our other value for $$x$$ on the function. It can be visually shown that as $$x_0$$ becomes smaller, the slope is closer to the slope at $$x$$ itself. (I can't show this here because the website doesn't integrate graphing.) We now simplify the equation from above using algebra: $(=\frac{3x^2x_0+3xx_0^2+x_0^3}{x_0}=3x^2+3xx_0+x_0^2$) This allows us to take the limit as $$x_0$$ approaches $$0$$. (Notice that this is different from dividing by $$0$$ because $$x_0$$ only approaches $$0$$ but never IS $$00$$.) So, we have: $(=3x^2+3x(0)+0^2=3x^2$) Thus, the derivative of $$y=x^3$$ is $$\frac{dy}{dx}=3x^2$$.
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# The Method of Partial Fractions
## Introduction
Early in Algebra you learn how to combine "simple'' fractions into a "more complicated'' one. Here is a typical example:
The Method of Partial Fractions does the opposite: It dissects a complicated fraction into a sum of simple fractions. While this is a little more complicated than going the other direction, it is also more useful. Major applications of the method of partial fractions include:
## How simple can it get?
A simple fraction is a fraction with a simple denominator. The first step consists of detecting the factors (the building blocks) of the given denominator. The Fundamental Theorem of Algebra tells us what is possible: Every polynomial can be factored into linear factors (degree 1 polynomials) and irreducible polynomials of degree 2.
#### Some Examples.
• The polynomial can be factored into three linear factors as follows (Do you remember how to do this?):
• The polynomial has a linear factor and an irreducible factor of degree 2:
• The polynomial has a twice repeated irreducible factor of degree 2:
#### How can you tell whether a degree 2 polynomial is irreducible (over the field of real numbers), or can be factored further into two linear factors?
There are different methods to decide:
• Graphically: A reducible quadratic polynomial has 2 zeros or one repeated zero, an irreducible quadratic polynomial has no zeros!
This quadratic polynomial can be factored
An irreducible quadratic polynomial
• Algebraically: If the quadratic formula results in a negative expression under the radical (the discriminant), the associated polynomial is irreducible: Consider the polynomial : Using the quadratic formula for yields:
Since the discriminant (the expression under the radical) is negative, the polynomial is irreducible!
#### Try it yourself!
Is the polynomial irreducible or reducible?
Helmut Knaust
Fri Jul 5 13:54:22 MDT 1996
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# 2013 AMC 10A Problems/Problem 7
## Problem
A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16$
## Solution 1
Let us split this up into two cases.
Case $1$: The student chooses both algebra and geometry.
This means that $3$ courses have already been chosen. We have $3$ more options for the last course, so there are $3$ possibilities here.
Case $2$: The student chooses one or the other.
Here, we simply count how many ways we can do one, multiply by $2$, and then add to the previous.
WLOG assume the mathematics course is algebra. This means that we can choose $2$ of History, Art, and Latin, which is simply $\dbinom{3}{2} = 3$. If it is geometry, we have another $3$ options, so we have a total of $6$ options if only one mathematics course is chosen.
Thus, overall, we can choose a program in $6 + 3 = \boxed{\textbf{(C) }9}$ ways
## Solution 2
We can use complementary counting. Since there must be an English class, we will add that to our list of classes for $3$ remaining spots for the classes. We are also told that there needs to be at least one math class. This calls for complementary counting. The total number of ways of choosing $3$ classes out of the $5$ is $\binom{5}3$. The total number of ways of choosing only non-mathematical classes is $\binom{3}3$. Therefore the amount of ways you can pick classes with at least one math class is $\binom{5}3-\binom{3}3=10-1=\boxed{\textbf{(C) }9}$ ways.
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#### xaktly | Mathematics
Trigonometric functions
### From a black box to a function
So far we've looked at trig functions as ratios of sides, and where those are known, that's easy to calculate. The other way we've used them is as "black boxes" — we put a number into a calculator and we get one back out. Here's a triangle for which we know the lengths of all sides:
\begin{align} sin(\theta) &= \frac{o}{h} = \frac{5}{9.434} = 0.530 \\[5pt] cos(\theta) &= \frac{a}{h} = \frac{8}{9.434} = 0.056 \\[5pt] tan(\theta) &= \frac{o}{a} = \frac{5}{8} = 0.625 \end{align}
Later we'll learn how to recover angles when the side lengths are shown, but let's move on for now.
The other way we've used trigonometry is to enter angle measures (in degrees or radians) into a calculator or computer, and it just gives us the value of the trig function for that angle — magically.
Here's an example of how the trig-function keys will look on your scientific calculator:
It's far from magic, as we will see below, and you'll learn more about that if you study infinite series as part of a calculus course.
### The unit circle – the fundamental basis for trigonometry
#### Units of Angle Measure
In trigonometry and many subfields of math, we use the unit radians for angle measure. The radian arises more or less naturally from the geometry of the circle. There are 2π (yes, 6.283 ..., but we usually just say 2π) radians in a circle, so 180˚ = π radians. The figure on the left shows the most frequently-used radian measures. They are generally multiples of π/6 (30˚) and π/4 (45˚).
You should memorize this unit circle, including the radian measurements. It will grease the wheels for what's ahead. The tool below might help. It reduces the task to simple counting by multiples of 30˚ and 45˚. If you come back to it once in a while, a little more will sink in each time and soon you'll have it.
Note: Unit circle angles begin with zero (degrees or radians) on the far right, and increase in the counterclockwise direction.
There are $2 \pi = 6.28$ radians and 360˚ in a circle, $\pi$ radians and 180˚ in a half circle.
Usually, we don't multiply the $\pi$ when reporting radian measurements. We just say 1.1$\pi$ radians, and so forth.
### Unit circle learning tool
Press the forward arrow as many times as you want to advance the arrow and find the angle, in degrees and radians, on the unit circle.
The circle is "counted around" in 180˚ increments, then in 90˚, 45˚ and 30˚ increments. Go around the circle enough times that you can correctly anticipate the next angle, both in degrees and radians.
Press the back button at any time to reset the circle and start over.
Pro tip: I can't emphasize enough how important it is that you know your way around the unit circle. Now that that's (hopefully) done, we can take a deeper look at the trigonometric functions.
### The special triangles
Two common triangles you should memorize.
These two essential right triangles, their angle measures and the ratios of their side lengths will pop up time and again throughout your studies and work in math and science. Knowing how to jot them down and use them, and from where the measurements came, will be invaluable to you. Study them carefully and memorize their proportions.
### The 45-45-90 triangle
The 45-45-90 triangle is just what it sounds like, a right triangle that has two 45˚ angles, and is thus an isosceles triangle. We'll work with just symbols for the lengths of the sides, like this:
Now the Pythagorean theorem has to hold for this right triangle, so
$$^2 + x^2 = r^2$$
Now combine the like terms on the left,
$$2x^2 = r^2$$
and let's solve for $x$ (you'll see why in a minute), by first dividing both sides by 2.
$$x^2 = \frac{r^2}{2}$$
Now to solve for $x$, take the square root of both sides:
$$x = \sqrt{\frac{r^2}{2}}$$
Taking the square root of the numerator gives r, with $\sqrt{2}$ in the denominator:
$$x = \frac{r}{\sqrt{2}}$$
Finally, just for completeness, not that it really matters mathematically, the denominator of such an equation is usually "rationalized" by multiplying by $\sqrt{2}/\sqrt{2}:$
$$x = \frac{r \sqrt{2}}{2}$$
Now imagine if we let r, the length of the hypotenuse of our triangle, be 1. Then we get this triangle, complete with side measures:
What's great about such a triangle, is that it will give us the lengths of the sides of a 45-45-90 triangle with any hypotenuse length because we have the ratios of the side lengths, $1:\sqrt{2}/2.$
### The 30-60-90 triangle
The 30-60-90 (degrees) right triangle is another frequently-encountered triangle that's worth knowing a lot about.
It isn't quite as easy to find an expression for the side lengths x and y, unless we combine two of these triangles to form an equilateral triangle like this:
Notice here that I've replaced x from the original figure with r/2. We know that the gray vertical line both bisects the top 60˚ angle (by construction) and bisects the base of the triangle. For a proof the the angle bisector of an isosceles triangle also bisects the opposite side, click here.
Now we can set up the Pythagorean theorem for one of the two 30-60-90 triangles shown:
$$r^2 = y^2 + \left( \frac{r}{2} \right)^2$$
Squaring the last term gives
$$r^2 = y^2 + \frac{r^2}{4}$$
Let's solve for y in terms of r by separating terms across the = sign:
$$y^2 = r^2 - \frac{r^2}{4}$$
The difference on the right is easy to calculate; just use 4 as a common denominator:
$$y^2 = \frac{3r^2}{4}$$
Now taking the square root of both sides gives
$$y = r \frac{\sqrt{3}}{2}$$
Now if we take r = 1, we arrive at the prototype 30-60-90 triangle:
So for a 30-60-90 triangle, the shorter leg (recall that the sides that form the right angle of a right triangle are called "legs") has a length that is half the length of the hypotenuse, and the longer leg is $\frac{\sqrt{3}}{2}$ of the hypotenuse.
### Trig ratios in the special triangles
##### $$30^{\circ}, \; \frac{\pi}{6} \text{ rad}$$
\require{cancel} \begin{align} sin(30^{\circ}) &= \frac{1}{2} \\[8pt] cos(30^{\circ}) &= \frac{\sqrt{3}}{2} \\[5pt] tan(30^{\circ}) &= \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\cancel{2}} \frac{\cancel{2}}{\sqrt{3}} = \frac{\sqrt{3}}{3} \end{align}
##### $$45^{\circ}, \; \frac{\pi}{4} \text{ rad}$$
\begin{align} sin(45^{\circ}) &= \frac{\sqrt{2}}{2} \\[8pt] cos(45^{\circ}) &= \frac{\sqrt{2}}{2} \\[5pt] tan(45^{\circ}) &= \frac{\cancel{\frac{\sqrt{2}}{2}}}{\cancel{\frac{\sqrt{2}}{2}}} = 1 \\[5pt] \end{align}
##### $$60^{\circ}, \; \frac{\pi}{3} \text{ rad}$$
\begin{align} sin(60^{\circ}) &= \frac{\sqrt{3}}{2} \\[8pt] cos(60^{\circ}) &= \frac{1}{2} \\[5pt] tan(60^{\circ}) &= \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{\sqrt{3}}{\cancel{2}} \frac{\cancel{2}}{1} = \sqrt{3} \\[5pt] \end{align}
### Trigonometric functions
Now we want to use the unit circle to create graphs of the three major trigonometric functions,
\begin{align} f(x) &= sin(x) \\[5pt] f(x) &= cos(x) \\[5pt] f(x) &= tan(x) \end{align}
We'll do that by drawing a series of triangles, both 45-45-90 and 30-60-90, inside of the unit circle, and calculating the trig function values.
Let's start at 30˚ = θ/6. Notice that we can use that triangle to calculate all three trig functions. We'll set those aside as we work around the circle with triangles, and look at all of the values together at the end of the trip.
Now we really should work our way around the circle in increments of 30˚, but I'm going to skip around a bit and hope you'll get the idea. Next is a 135˚ angle, or 45˚ past 90˚, working counterclockwise (as usual) around the unit circle.
One quirk of the unit circle is that the length of the radius (the hypotenuses of all of our triangles) is always 1, but the other triangle sides can have negative or positive values that match with the x,y axis directions. That's how certain sine, cosine or tangent values end up negative.
You might have noticed that the 90˚ (which we passed up) angle doesn't really define a triangle. The key to understanding what happens there is to consider an 89˚ angle, in which the opposite side is very small and the adjacent side is nearly equal to the hypotenuse, making the sine function small (close to zero) and the cosine function close to one. Well, at 90˚, those values are zero and one, respectively, and we make similar arguments at all of the multiples of 90˚ around the circle.
Now let's do a 240˚ = 4θ/3 angle. Notice that because the x-and y-coordinates of the tip of the radius are negative, we treat the sides of our triangle as negative, and that affects the values of the trig functions.
Finally, let's look at an angle in the 4th quadrant, 315˚ = 7θ/4, which is seven increments of 45˚ around the unit circle, and produces a 45-45-90 triangle, just with some "negative" sides.
Notice that all of this would just repeat as we moved our radius vector around the unit circle again and again. Next we'll gather all of this data in a table and determine some patterns in the sine, cosine and tangent functions.
### Trig functions are periodic
Here's a table of trig function values at key angles in one round trip of the unit circle.
The trigonometric functions are repetitive, or cyclic, and we usually refer to them as periodic, meaning that they repeat the same basic pattern predictably.
$\theta$(deg.) $\theta$(rad) sin $(\theta)$ cos $(\theta)$ tan $(\theta)$ 0˚ 0 0 $\frac{\sqrt{0}}{2}$ 1 $\frac{\sqrt{4}}{2}$ 0 30˚ $\frac{\pi}{6}$ $\frac{1}{2}$ $\frac{\sqrt{1}}{2}$ $\frac{\sqrt{3}}{2}$ $\frac{\sqrt{3}}{2}$ $\frac{\sqrt{3}}{3}$ 45˚ $\frac{\pi}{4}$ $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{2}}{2}$ $1$ 60˚ $\frac{\pi}{3}$ $\frac{\sqrt{3}}{2}$ $\frac{\sqrt{3}}{2}$ $\frac{1}{2}$ $\frac{\sqrt{1}}{2}$ $\sqrt{3}$ 90˚ $\frac{\pi}{2}$ $1$ $\frac{\sqrt{4}}{2}$ $0$ $\frac{\sqrt{0}}{2}$ $\infty$ 120˚ $\frac{2\pi}{3}$ $\frac{\sqrt{3}}{2}$ $\frac{\sqrt{3}}{2}$ $-\frac{1}{2}$ $-\frac{\sqrt{1}}{2}$ $-\sqrt{3}$ 135˚ $\frac{3\pi}{4}$ $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{2}}{2}$ $-\frac{\sqrt{2}}{2}$ $-\frac{\sqrt{2}}{2}$ $-1$ 150˚ $\frac{5\pi}{6}$ $\frac{1}{2}$ $\frac{\sqrt{1}}{2}$ $-\frac{\sqrt{3}}{2}$ $-\frac{\sqrt{3}}{2}$ $-\frac{\sqrt{3}}{3}$ 180˚ $\pi$ $0$ $\frac{\sqrt{0}}{2}$ $-1$ $-\frac{\sqrt{4}}{2}$ $0$
Forget about the gray columns for now. You can pick up a bunch of patterns as we walk around the circle and look at sin(θ), cos(θ) and tan(θ). Notice that the tangent function is a different beast, and we'll get to that later. Notice also that both the sine and cosine functions oscillate between ±1 and pass periodically through y = 0.
The gray columns are just re-expressions of the sine and cosine values to the left. Each is just re-stated with a square root in the numerator so that the increasing-decreasing pattern is more obvious.
The signs of the trig functions change according to which quadrant the tip of radius vector is in (UL = upper left, LR = lower right, and so on). This table shows how the sign of each function depends on the quadrant of the angle:
Signs of the Trig. Functions
UR (I)
+
+
+
UL (II)
+
LL (III)
+
LR (IV)
+
### Trig function graphs
Graphs of the sine (black) and cosine (magenta) functions are drawn below as $f(\theta)$ vs.$\theta$, for $\theta$ between $0$ and a littl more than $4\pi$ - twice around the unit circle plus a little more. Notice that these are periodic functions. Every $2\pi$ radians, they repeat, and that goes on infinitely in both positive and negative directions. The negative direction just means traversing the unit circle in the opposite (clockwise) direction.
All of the points in the table above are plotted on this graph. Be sure to think about these graphs for a while.
Notice a few patterns: They oscillate between ±1; they cross or are farthest apart (vertically) at multiples of $\pi/4$ and $\text{sin}(\theta)$ is the same curve as $\text{cos}(\theta)$, but shifted to the right by $\pi/2$ radians or 90˚. Notice also that $\text{sin}(x) = \text{cos}(x)$ when $x$ is $\frac{\pi}{4}$ and then every $\pi$ radians from there on.
These curves are also referred to as "sine waves". We usually don't say "cosine waves" because the cosine function is the same as the sine, just shifted by $\pi/2$. It is common to represent many periodic natural phenomena using sine and cosine curves: Waves, the motion of a pendulum, and the motion of a bouncing spring, for example.
### Transformations
In order to model periodic behavior using sines and cosines, we need, as usual, some transformations to translate and scale our function at will. In our typical way (see functions section), the transformations of $f(x) = sin(x)$ are shown on the right.
When we use trig. functions to model real data later, we'll make a slight modification of this prototype equation in order to keep the units straight.
The transformations of the other trig. functions are analogous.
### Horizontal stretching
Move the slider to change $c$ in the function $f(x) = sin(f/c)$ to get an idea of the effect of horizontal stretching on a periodic function. The function is written both as $f(x) = sin(x/c)$ and as $f(x) = sin[(1/c)·x]$.
If this function represented a wave (e.g. a sound, light or water wave) then the horizontal stretching parameter, $c$, would change the wavelength, the domain distance between successive peaks or troughs. The parameter $f = 1/c$ is called the frequency factor.
Notice that for small c, the frequency (number of cycles on the graph) is large, and for small $c$, it's large. Because the sine function is odd, when $c$ is negative, the sine curve is inverted across the x-axis.
(Notice also that the graph disappears when we try to divide by zero!)
### Vertical stretching
Move the slider to see the effect of the vertical-stretching parameter, $A$, on our sine function.
If this function represented a wave like a sound or light wave, the vertical distance between $f(x) = 0$ and the maximum (or minimum) value of the sine function would be the amplitude, which would be proportional to volume (loudness) of sound or brightness (intensity) of light.
Notice that when $A \lt 0$, the function is reflected across the x-axis, as we would expect for any function.
### Vertical and horizontal translation
Vertical translation of a periodic function is performed just like on any other function: Simply add a constant number (k) to the value of the function for each point in the domain. Vertical translation is used in modeling to move the average value of the function up or down. For example, to model a tide that fluctuates by ± 3 feet from an average depth of 10 feet, we'd want a sine or cosine function (turns out it doesn't matter which), that oscillates between 7 and 13 feet. The function would read f(some stuff) + 10.
Horizontal translation is very important when we model waves because it represents a property called phase. In our tide example, we might need to shift our model wave over to the right a bit to get the timing of the tides just right (and we'd also have to adjust the wavelength, which would represent the time between high or low tides). Phase is extremely important in many areas of science like x-ray crystallography (used to study molecular structure) and medical imaging.
### The tangent function
Now let's think about a class of trig functions called the reciprocal trig functions. The tangent is the most prominent of these, but there are three more. We can look at the tangent using SOH-CAH-TOA; we know that
$$sin(\theta) = \frac{o}{h}, \phantom{00} cos(\theta) = \frac{a}{h}, \phantom{00} tan(\theta) = \frac{o}{a}.$$
Now notice that
$$\frac{sin(\theta)}{cos(\theta)} = \frac{\frac{o}{h}}{\frac{a}{h}} = \frac{o}{\cancel{h}} \frac{\cancel{h}}{a} = \frac{o}{a}$$
This means that the tangent function is the ratio of the sine and cosine functions:
$$tan(\theta) = \frac{sin(\theta)}{cos(\theta)}$$
Now because $cos(\theta)$ — the denominator of the tangent expression — can be zero, the graph of the tangent function will have vertical asymptotes, as shown in the graph. The tangent of odd multiples of $\pi/2$ is infinite because $cos(\pi/2)$ (and odd multiples of $\pi/2$) is zero.
This has an interesting consequence if you are a rock climber or a lineperson (person who strings cable, e.g. electric wires, from pole to pole. You'll find an explanation here.
There are three other trigonometric functions left to explore, and like the tangent, all are ratios of other trig. functions, so we can expect asymptotic behavior there, too.
$$tan(\theta) = \frac{sin(\theta)}{cos(\theta)}$$
### All six trigonometric functions
Finally, let's define the three other trig. functions. The reciprocal of the sine function (not to be confused with the inverse) is called the cosecant (csc) function. The reciprocal of the cosine is (paradoxically, I know) the secant (sec) and the reciprocal of the tangent is the cotangent (cot). csc(θ), sec(θ) and cot(θ) are defined below.
#### Basic
\begin{align} \text{sine} &\phantom{000} sin(\theta) = \frac{o}{h} \\[5pt] \text{cosine} &\phantom{000} cos(\theta) = \frac{a}{h} \\[5pt] \text{tangent} &\phantom{000} tan(\theta) = \frac{o}{a} \tag{*} \end{align}
#### Reciprocal
\begin{align} \text{cosecant} &\phantom{000} \text{csc} \phantom{00} csc(\theta) = \frac{1}{sin(\theta)} = \frac{h}{o} \\[5pt] \text{secant} &\phantom{000} \text{sec} \phantom{00} sec(\theta) = \frac{1}{cos(\theta)} = \frac{h}{a} \\[5pt] \text{cotangent} &\phantom{000} \text{cot} \phantom{00} cot(\theta) = \frac{cos(\theta)}{sim(\theta)} = \frac{a}{o} \end{align}
(*) Strictly speaking, the tangent is a reciprocal trig function, too, but we'll consider it here to be one of the "big three" most-frequently-used trig functions.
#### Pro tip: Remembering sec, csc and tan
It should be easier to remember which of the main trig. functions are connected to sec, csc and tan, but due to an accident of history, it's not. Sine should go with secant, and cosine should go with cosecant. Unfortunately, those are reversed, and you'll have to remember that. Tangent goes with cotangent, though!
### Graphs of all six trig functions
Several cycles of each of the six trig functions are plotted below. Note that the sin(x) and cos(x) graphs are vertically expanded by a factor of ten compared to the other graphs. This means that the sine graph would fit in between the U-shaped graphs (they are not parabolas!) in the csc(x) graph, same for the cos(x) and sec(x) graphs. Except for sine and cosine,
### Example 1
Determine the values (to the thousandths place) of all six trigonometric functions of the angle $\theta$.
Solution: First find the length of the missing side using the Pythagorean theorem:
\begin{align} o &= \sqrt{21^2 - 17^2} \\[5pt] &= 12.329 \text{ cm} \end{align}
Now that all of the sides are known, we can calculate all of the ratios:
\begin{align} sin(\theta) &= \frac{12.329}{} = 0.587 \\[5pt] cos(\theta) &= \frac{17}{21} = 0.810 \end{align}
\begin{align} tan(\theta) &= \frac{12.329}{17} = 0.725 \\[5pt] csc(\theta) &= \frac{21}{12.329} = 1.703 \\[5pt] sec(\theta) &= \frac{21}{17} = 1.235 \\[5pt] cot(\theta) &= \frac{17}{12.329} = 1.379 \\[5pt] \end{align}
There is much more to trigonometry than can fit into one section like this. You'll need to know how to use inverse trig. functions, how to relate trig functions to one another (analytic trig), and how to use trigonometry on non-right triangles. Here are links to other trigonometry-related pages:
### Practice problems
For problems 1 and 2, calculate the measure of any missing angles and the lengths of any missing sides of the triangle.
1.
Solution
First, the length of the hypotenuse is
$$h = \sqrt{63^2 + 14^2} = \sqrt{4165} = 64.537$$
Now the trig functions are:
\begin{align} sin(\theta) &= \frac{14}{64.537} = 0.217 \\[5pt] cos(\theta) &= \frac{63}{64.537} = 0.976 \\[5pt] tan(\theta) &= \frac{14}{63} = 0.222 \\[5pt] csc(\theta) &= \frac{64.537}{14} = 4.61 \\[5pt] sec(\theta) &= \frac{64.537}{63} = 1.024 \\[5pt] cot(\theta) &= \frac{63}{14} = 4.50 \end{align}
2.
Solution
First, the length of the adjacent side is
$$h = \sqrt{21^2 + 8.5^2} = \sqrt{368.75} = 19.203$$
Now the trig functions are:
\begin{align} sin(\theta) &= \frac{8.5}{21} = 0.405 \\[5pt] cos(\theta) &= \frac{19.203}{21} = 0.914 \\[5pt] tan(\theta) &= \frac{8.5}{19.203} = 0.443 \\[5pt] csc(\theta) &= \frac{21}{8.5} = 2.47 \\[5pt] sec(\theta) &= \frac{21}{19.203} = 1.094 \\[5pt] cot(\theta) &= \frac{19.203}{14} = 2.259 \end{align}
1. The cosine of an angle of a right triangle is $\frac{3}{11}$. Calculate the values of the other five trigonometric functions of this angle, and sketch the triangle.
Solution
Here's a diagram of the triangle, including the length of the missing side:
The missing side is $o = \sqrt{11^2 - 3^2} = 10.583$.
Now the values of the six trig functions are:
\begin{align} sin(x) &= \frac{10.583}{11} = 0.962 \\[5pt] cos(x) &= \frac{3}{11} = 0.273 \\[5pt] tan(x) &= \frac{10.583}{3} = 3.528 \\[5pt] csc(x) &= \frac{11}{10.583} = 1.039 \\[5pt] sec(x) &= \frac{11}{3} = 3.667 \\[5pt] cot(x) &= \frac{3}{10.583} = 0.283 \end{align}
2. The tangent of an angle of a right triangle is $\frac{7}{5}$. Calculate the values of the other five trigonometric functions of this angle and sketch the triangle.
Solution
Here's a diagram of the triangle, including the length of the missing side:
The missing side is $o = \sqrt{11^2 - 3^2} = 10.583$.
Now the values of the six trig functions are:
\begin{align} sin(x) &= \frac{10.583}{11} = 0.962 \\[5pt] cos(x) &= \frac{3}{11} = 0.273 \\[5pt] tan(x) &= \frac{10.583}{3} = 3.528 \\[5pt] csc(x) &= \frac{11}{10.583} = 1.039 \\[5pt] sec(x) &= \frac{11}{3} = 3.667 \\[5pt] cot(x) &= \frac{3}{10.583} = 0.283 \end{align}
3. A certain point on the unit circle is $\left(\frac{7}{25}, \; y \right)$ with $y \gt 0$. Determine the exact value of $y$.
Solution
Here's a diagram of the unit circle with our triangle roughly placed in it:
Now it's clear that our task is just to find the missing side of the triangle, the $y$ coordinate of the point on the circle. That is
\begin{align} y &= \sqrt{1^2 - \left( \frac{7}{25} \right)^2} \\[5pt] &= \sqrt{\frac{625-49}{625}} \\[5pt] &= \sqrt{\frac{576}{625}} \\[5pt] &= \frac{24}{25} \end{align}
4. A certain point on the unit circle is $\left(\frac{15}{17}, \; y \right)$ with $y \lt 0$. Determine the exact value of $y$.
Solution
Here's a diagram of the unit circle with our triangle roughly placed in it:
Now it's clear that our task is just to find the missing side of the triangle, the $y$ coordinate of the point on the circle. That is
\begin{align} y &= -\sqrt{1^2 - \left( \frac{15}{17} \right)^2} \\[5pt] &= -\sqrt{\frac{289-225}{289}} \\[5pt] &= -\sqrt{\frac{64}{289}} \\[5pt] &= -\frac{8}{17} \end{align}
5. Sketch a graph of $f(x) = sin(x) + cos(x)$ over the domain $[0, \, 2\pi]$.
Solution
Let's make a quick table of sums of sin(x) and cos(x) at a few select points:
$\theta$ $sin(\theta)$ $cos(\theta)$ $s(\theta) + c(\theta)$ $0^{\circ}$ $0$ $1$ $1$ $45^{\circ}$ $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{2}}{2}$ $\sqrt{2}$ $90^{\circ}$ $1$ $0$ $1$ $135^{\circ}$ $\frac{\sqrt{2}}{2}$ $-\frac{\sqrt{2}}{2}$ $0$ $180^{\circ}$ $0$ $-1$ $-1$ $225^{\circ}$ $-\frac{\sqrt{2}}{2}$ $-\frac{\sqrt{2}}{2}$ $-\sqrt{2}$ $270^{\circ}$ $\frac{\sqrt{2}}{2}$ $-\frac{\sqrt{2}}{2}$ $0$ $305^{\circ}$ $-\frac{\sqrt{2}}{2}$ $-\frac{\sqrt{2}}{2}$ $-\sqrt{2}$
Here is what the graph looks like:
It's still a periodic function with a period of $2 \pi$ radians or 360˚, but it has been shifted by -45˚.
6. Sketch a graph of $g(x) = sin^2(x) + cos^2(x)$. Note that $sin^(x)$ is just a commonly-used shorthand notation for $[sin(x)]^2$, just the square of the result of $sin(x)$.
Solution
Let's make a quick table of sums of sin2(x) and cos2(x) at a few select points:
$\theta$ $sin^2(\theta)$ $cos^(\theta)$ $s^2(\theta) + c^2(\theta)$ $0^{\circ}$ $0$ $1$ $1$ $45^{\circ}$ $\frac{1}{2}$ $\frac{1}{2}$ $1$ $90^{\circ}$ $1$ $0$ $1$ $135^{\circ}$ $\frac{1}{2}$ $\frac{1}{2}$ $1$ $180^{\circ}$ $0$ $1$ $1$ $225^{\circ}$ $\frac{1}{2}$ $\frac{1}{2}$ $1$ $270^{\circ}$ $\frac{1}{2}$ $\frac{1}{2}$ $1$ $305^{\circ}$ $\frac{1}{2}$ $\frac{1}{2}$ $1$
Notice that the sum $sin^2(\theta) + cos^2(\theta) = 1$ for all values of $\theta$. We haven't calculated any values in between, but it's true. This is a relationship that we'll use a lot going into analytic trigonometry and it's called the Pythagorean Identity:
$$sin^2(\theta) + cos^2(\theta) = 1$$
For any angle, $\theta$.
### Video examples
#### 1. Counting around the unit circle
You should know how to label the most-frequently-used angles around the unit circle (circle of radius r = 1), both in degrees and radians. It's not that difficult if you just think of it as counting around the circle in different increments.
#### Triangles
Back up and refresh your memory about the properties of triangles, key to understanding trigonometry.
#### Trigonometric functions
Learn about the properties and graphs of the trigonometric functions, sin(θ), cos(θ), tan(θ) and three others.
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### mnemonic
A mnemonic (nee·mon'·ick) is a word or phrase designed to help a person remember something. An example would be the pseudo-word "ROYGBIV" or the phrase "Rogers of York Gave Battle in Vain." Both are designed to help us remember the colors of the visible spectrum: red, orange, yellow, green, blue, indigo & violet.
X
#### The Greek alphabet
alpha Α α beta Β β gamma Γ γ delta Δ δ epsilon Ε ε zeta Ζ ζ eta Η η theta Θ θ iota Ι ι kappa Κ κ lambda Λ λ mu Μ μ nu Ν ν xi Ξ ξ omicron Ο ο pi Π π rho Ρ ρ sigma Σ σ tau Τ τ upsilon Υ υ phi Φ φ chi Χ χ psi Ψ ψ omega Ω ω
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# How to convert Binary to Octal?
To convert a binary number to octal, start by multiplying each digit of the binary number by 2 raised to the nth power, where n is the position of the digit from the right.
For example, if the binary number is 1011, then you would multiply 1 by 2^3, 0 by 2^2, 1 by 2^1, and 1 by 2^0. The resultant decimal number should be 11, which is the equivalent octal number for 1011.
To get the remainder when dividing 11 by 8, divide 11 by 2 three times. The remainders will be 3, 1, and 1. These numbers correspond to the octal digit 321.
## How do you convert decimal numbers to octal numbers?
Decimal to octal: We can convert a decimal number to its equivalent octal form by dividing the number by 8 and writing down the remainders in reverse order.
For example, let’s convert the decimal number 234 to octal. We divide 234 by 8 and write down the remainders in reverse order:
234 ÷ 8 = 29 with a remainder of 2 29 ÷ 8 = 3 with a remainder of 5 3 ÷ 8 = 0 with a remainder of 3 Therefore, the octal form of 234 is 3530.
## How do you convert binary to octal and hexadecimal?
Binary to octal conversion is quite simple. You just need to remember that each binary digit corresponds to a power of two, starting with the rightmost (least significant) digit and working left.
So, if you have a binary number “1010”, this can be expressed as 10 = 1*2^1 + 0*2^2 + 1*2^3 This can also be represented in exponential form as 10 = 1*10^1 + 0*10^2 + 1*10^3 To convert from binary to octal, you simply group the digits in threes, starting at the rightmost (least significant) digit.
So, our example “1010” would be grouped like this: 101 010, And then you replace each group with the corresponding octal digit. In this case: 101 = 5 010 = 2 So, our original binary number “1010” converts to the octal number “52”.
Converting from binary to hexadecimal is a little more complicated than converting to octal, but it’s still pretty straightforward. Similar to converting from binary to octal, you start by breaking the number up into groups of four bits, starting at the rightmost (least significant) bit and working left.
So, if we take our example binary number “1010” again, we would group it like this: 1010 Then, we replace each group with the corresponding hexadecimal digit. In this case: 1010 = A So, our original binary number “1010” converts to the hexadecimal number “A”.
## What is the octal equivalent of decimal number 8?
The octal equivalent of decimal number 8 is 10. Octal numbers are base-8 numbers, and they use a combination of the digits 0 to 7. The first digit in an octal number can be any number from 0 to 7, but the second digit can only be 0 to 3 because 8 multiplied by 2 equals 16, which is greater than 10.
## How can we find the octal number?
We can find the octal numbers by converting the decimal number into a binary number first and then grouping the binary digits in threes starting from the right. The octal digit for each group will be the equivalent decimal number of that binary group.
## How do you convert binary to octal?
To convert a binary number to an octal, follow these steps:
1. Multiply each digit of the binary number by 2n-1, where n is the position of the digit from the decimal point. The resultant is the equivalent decimal number for the given binary number.
2. Divide the decimal number by 8.
3. Note the remainder. This will be the first digit of the octal number.
4. Repeat steps 2 and 3 until all digits of the octal number have been determined.
## How do you convert to octal?
There are a few steps in converting decimal to octal numbers. In decimal, we use the base ten because there are ten digits, 0-9, that we can use. When we convert to octal, we are using the base eight because there are only eight digits, 0-7.
To convert from decimal to octal, we divide the number by 8 and write the remainders in reverse order to get the equivalent octal number.
Let’s look at an example: Converting the decimal number 25 to octal: We start by dividing 25 by 8: 25/8 = 3 remainder 1 3/8 = 0 remainder 3 0/8 = 0 remainder 0 We take the remainders in reverse order to get the final octal number: 130
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# 7.08 Division where answers are decimals
Lesson
We've previously learned how to use a division algorithm with short division . Let's practice this concept.
### Examples
#### Example 1
Find the value of 856\div8.
Worked Solution
Create a strategy
Use short division.
Apply the idea
Set up the short division.
8 goes into 8 once, so we put a 1 in the hundreds place at the top.
8 goes into 5 zero times with 5 remaining, so we put a 0 in the tens place at the top and carry the 5 to the units column.
8 goes into 56 seven times, so we put a 7 in the units column.
856 \div 8 = 107
Idea summary
We can use short division to divide one number by another. We start dividing at the highest place value and carry any remainders to the next place value.
## Divide a 3 digit number by a 1 digit number
Sometimes when we divide whole numbers there is a remainder, this remainder can be expressed as a decimal. This video will show us how.
### Examples
#### Example 2
Find the value of 460\div8.
Worked Solution
Create a strategy
Use short division.
Apply the idea
Set up the short division.
8 can fit into 4 zero times with 4 remaining, so we put 0 on the hundreds place at the top and carry the 4 to the tens column.
8 can fit into 46 five times with 6 remaining, so we put 5 in the tens place at the top and carry the 6 to the units column.
8 can fit into 60 seven times with 4 remaining, so we put 7 in the units place at the top.
We have run our of digits to carry the remainder to, so we need to a decimal point and a 0 in the tenths place. Now we can carry the 4 to the tenths column.
8 can fit into 40 five times with 0 remaining, so we put 5 in the tenths place at the top.
There are no more digits to divide by 8, so we know that we are finished.
460\div8=57.5
Idea summary
When dividing a whole number using short division, if we have a remainder after dividing the units place we will end up with a decimal answer.
## Short division
This video looks at short division and how to use short division when there is a remainder.
### Examples
#### Example 3
Find the value of 3653\div4.
Worked Solution
Create a strategy
Use short division.
Apply the idea
Set up the short division.
4 fits into 36 nine times with 0 remaining, so we put 9 in the hundreds place at the top.
4 fits into 5 once with 1 remaining, so we put 1 in the tens place at the top and carry 1 to the units column.
4 fits into 13 three times with 1 remaining, so we put 3 in the units place at the top.
Since we have a remainder, we add a decimal point and a 0 in the tenths place and carry 1 to the tenths column.
4 fits into 10 two times with 2 remaining, so we put 2 in the tenths place at the top.
We add a 0 in the hundredths place and carry the 2 to the hundredths column.
4 fits into 20 five times with 0 remaining, so we put 5 in the hundredths place at the top.
There are no more digits to divide by 4, so we know that we are finished.
3653\div4=913.25
Idea summary
When dividing a whole number using short division, if we have a remainder, we can add a decimal point and add 0s after the decimal point to carry the remainders to until there are no more remainders.
## Divide a 4 digit number by a 2 digit number
This video looks at how to divide by a two digit number using short division. By remembering place value, we can divide each digit, working from left to right. We move from units to tenths and so on if necessary.
### Examples
#### Example 4
Find the value of 460\div25.
Worked Solution
Create a strategy
Use short division.
Apply the idea
Set up the short division.
25 fits into 4 zero times so we put a 0 above the 4. \,\, 25 fits into 46 once, so we put 1 in the tens place at the top.
25\times 1=25 and 46-25=21 so we have a remainder of 21 which we carry to the units column.
25 fits into 210 eight times since 25\times 8 =200. So we put 8 in the units place at the top and carry 10 to the tenths column by adding a decimal point and a 0.
25 fits into 100 four times, so we put 4 in the units place at the top.
There are no more digits to divide by 25, so we know that we are finished.
460\div25=18.4
Idea summary
When you are dividing, you always start with the digit that is farthest to the left. If you get to a digit that you can't divide into, make sure you put a placeholder zero in the answer, before moving to the next digit.
### Outcomes
#### VCMNA215
Multiply decimals by whole numbers and perform divisions by non-zero whole numbers where the results are terminating decimals, with and without digital technologies
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# Systems Using Substitution
## Solve for one variable, substitute the value in the other equation
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Practice Systems Using Substitution
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Solving Linear Systems by Substitution
Have you ever been to the Omni theater? Take a look at this dilemma.
Kelly loved the Omni presentation on the rainforest so much that she decided to go and see it again. She asked Tyler if he wanted to go with her on Saturday afternoon.
“Do you want to go with me?” Kelly asked.
“Sure, but I have karate first, so I will have to meet you there. What time is the show?” Tyler asked.
“The show starts at 2:30 pm. I’m going to leave at one o’clock so I can look around,” Kelly said.
“Well, I don’t get done karate until then, so I probably won’t leave until 2:00 pm,” Tyler said.
On Saturday, the two went about their day and both left for the museum. Kelly’s Mom tends to drive slowly and cautiously, so she was traveling on city streets at 45 mph. Tyler’s karate class is downtown, so he could take the highway to get to the Omni theater and his Dad drove at an average of 55 mph. Will the two catch up to each other?
This is a problem about equations and systems of equations. You will need to solve a system of equations to figure this one out. To figure this out, you would need to find a solution that will work for both equations. You will learn all about it in this Concept.
### Guidance
Let’s recall that in a system of equations, we are looking for the same x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} values that make all equations true. So, the solution to the system
3yyx=x2=4
is the ordered pair (x,y)\begin{align*}(x, y)\end{align*} that makes the first and the second equation true. In other words, x\begin{align*}x\end{align*} in the first equation equals x\begin{align*}x\end{align*} in the second equation and y\begin{align*}y\end{align*} in the first equation equals y\begin{align*}y\end{align*} in the second equation. Now, look at the second equation, yx=4\begin{align*}y-x=4\end{align*}. It is simple to solve for the y\begin{align*}y\end{align*} value—add x\begin{align*}x\end{align*} to both sides. So y=x+4\begin{align*}y=x+4\end{align*}. Well, if y\begin{align*}y\end{align*} is the same in both equations and y=x+4\begin{align*}y=x+4\end{align*}, then we can substitute x+4\begin{align*}x + 4\end{align*} in the place of y\begin{align*}y\end{align*} in the first equation. That is why this is called the substitution method.
Now that we have substituted, we can solve this equation because it has a single variable.
3(x+4)3x+122x+122xx=x2=x2=2=14=7
If x=7\begin{align*}x = -7\end{align*}, substitute again to find y\begin{align*}y\end{align*}:
yyy=x+4=7+4=3
Our solution, then, is (-7, -3).
Solve each system by using substitution.
#### Example A
2yy=x+4=3x
Solution: x=45,y=225\begin{align*}x = \frac{4}{5}, y= 2 \frac{2}{5}\end{align*}
#### Example B
3yy=x22=4x
Solution: x=2,y=8\begin{align*}x = -2,y = -8\end{align*}
#### Example C
6yy=x34=3x
Solution: x=2,y=6\begin{align*}x = 2, y = 6\end{align*}
Now let's go back to the dilemma from the beginning of the Concept.
The first equation we can write is to represent Tyler’s time. His Dad is traveling 55 mph. Therefore his distance is a function of speed and time.
d=55t\begin{align*}d = 55t\end{align*}
Kelly left one hour before Tyler did. She is traveling 45 miles per hour. Therefore, the speed times Tyler’s time plus one hour equals Kelly’s time.
d=45(t+1)\begin{align*}d = 45(t + 1)\end{align*}
Now see if there is a solution that will work for both equations. We can try to solve this by using substitution.
55t55t55t45t10tt=45(t+1)=45t+1=1=1=110
Now we go to Tyler.
dd=55(110)=5.5
The solution could be the following values for d\begin{align*}d\end{align*} and t\begin{align*}t\end{align*}. However, when you substitute those values into both equations, the solution does not work. Therefore there isn’t a solution for this system, and the Kelly and Tyler won’t meet up while driving.
### Vocabulary
System of Equations
Two or more equations at the same time. The solution will be the ordered pair that works for both equations.
### Guided Practice
Here is one for you to try on your own.
Anglica’s mother leaves to visit her grandmother for her birthday. Her grandmother lives 450 miles away and her mother drives at an average of 60mph. Three hours later, Angelica’s step-father notices that her mother forgot the gift for her grandmother so he decides to try to catch up to her. If he drives at an average of 50mph, will he catch up to her before she gets to her grandmother’s house?
Write a system of equations to model the situation and solve using the substitution method.
Solution
First, we can figure out how long it will take Anglica's mother to reacher her grandmother's house given the distance.
d=rt\begin{align*}d = rt\end{align*}
She is driving 50 mph, and her grandmother lives 450 miles away.
4808=60t=t
It will take her eight hours.
Now let's write and solve a system of equations to see if her step-father will catch up before her mother arrives.
d=60t\begin{align*}d = 60t\end{align*} This is the mother's driving.
d=50t+3\begin{align*}d = 50t + 3\end{align*} This is the step-father's driving.
60t=50t+3\begin{align*}60t=50t+3\end{align*}
Now we solve the equation for t, the time.
60t50t10tt=3=3=310
Now substitute this back into the mother's time.
60×310\begin{align*}60 \times \frac{3}{10}\end{align*}
18\begin{align*}18\end{align*}
It will take the step-father 18 hours to catch up. He will not arrive before Anglica's mother.
### Practice
Directions: Solve each linear system by using substitution.
1. .
y2y+2x=8=2
1. .
x2y+2x=6=2
1. .
2x2y+x=8=10
1. .
3yy+2x=9=11
1. .
y82y+x=8=20
1. .
4yy2x=8=8
1. .
x24x+y=4=12
1. .
yy+2x=x+8=11
1. .
2y+8y+2x=12=20
1. .
y33y+3x=6=9
1. .
4y1y4x=11=5
1. .
2y82y+2x=8=2
1. .
4x+y2x3y=2=1
1. .
y6xy=2x=8
1. .
x+4y2x+10y=6=6
### Vocabulary Language: English
Consistent
Consistent
A system of equations is consistent if it has at least one solution.
distributive property
distributive property
The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.
linear equation
linear equation
A linear equation is an equation between two variables that produces a straight line when graphed.
substitute
substitute
In algebra, to substitute means to replace a variable or term with a specific value.
system of equations
system of equations
A system of equations is a set of two or more equations.
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## How do you get 30% of 80?
Percent means per one-hundred and hence 30% is 30/100. Thus 30% of 80 is (30/100) x 80 = 24.
## What number is 35% of 80?
28
How do you find 70 percent of a number?
Example 1. Find 70% of 80. Following the shortcut, we write this as 0.7 × 80. Remember that in decimal multiplication, you multiply as if there were no decimal points, and the answer will have as many “decimal digits” to the right of the decimal point as the total number of decimal digits of all of the factors.
21
### How do you find percentages without a calculator?
If you need to find a percentage of a number, here’s what you do – for example, to find 35% of 240: Divide the number by 10 to find 10%. In this case, 10% is 24. Multiply this number by how many tens are in the percentage you’re looking for – in this case, that’s 3, so you work out 30% to be 24 x 3 = 72.
How do you calculate percentages quickly?
To calculate 10 percent of a number, simply divide it by 10 or move the decimal point one place to the left. For example, 10 percent of 230 is 230 divided by 10, or 23. 5 percent is one half of 10 percent. To calculate 5 percent of a number, simply divide 10 percent of the number by 2.
52
## What number is 20% of 70?
14
What number is 40% of 80?
32
### What number is 35% of 70?
Latest calculated numbers percentages
35% of 70 = 24.5 Mar UTC (GMT)
8.32% of 100,000 = 8,320 Mar UTC (GMT)
186% of 30.5 = 56.73 Mar UTC (GMT)
1.8% of 100 = 1.8 Mar UTC (GMT)
74% of 2,000 = 1,480 Mar UTC (GMT)
### What grade is a 60 out of 80?
75%
What number is 60% of 70?
42
30
## What number is 20% of 75?
15
What number is 10% of 80?
8
7
### What number is 30% of 80%?
24
How do you find 40% 70?
Percentage Calculator: What is 40 percent of 70? = 28.
## How do you find 15% of a number?
15% is 10% + 5% (or 0.15 = 0.1 + 0.05, dividing each percent by 100). Thinking about it this way is useful for two reasons. First, it’s easy to multiply any number by 0.1; just move the decimal point left one digit. For example, 75.00 x 0.1 = 7.50, or 346.43 x 0.1 = 34.64 (close enough).
## What number is 20% of 80?
16
How do you find 75%?
How much is 15 percent off?
1. Divide your original number by 20 (halve it then divide by 10).
2. Multiply this new number by 3.
3. Subtract the number from step 2 off of your original number.
4. You’ve just found your percentage off!
### What is the formula for calculating percentage?
To calculate the percentage, multiply this fraction by 100 and add a percent sign. 100 * numerator / denominator = percentage . In our example it’s 100 * 2/5 = 100 * 0.4 = 40 . Forty percent of the group are girls.
### How do I figure out percentages?
How to calculate percentage
1. Determine the whole or total amount of what you want to find a percentage for.
2. Divide the number that you wish to determine the percentage for.
3. Multiply the value from step two by 100.
What number is 25 percent of 80?
20
50
## How do I calculate percentage on a test?
You can find your test score as a percentage by dividing your score by the total number of points, then multiplying by 100.
What number is 85% of 20?
17
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225-Notes02
225-Notes02 - Stat 225 Lecture Notes "Counting Techniques"...
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Stat 225 Lecture Notes “Counting Techniques” Ryan Martin Spring 2008 1 Basic Counting Rule Recall that in the classical probability model, the probability of an event E is P ( E ) = N ( E ) N (Ω) , (1.1) the number of outcomes in E divided by the total number of outcomes. Counting the number of outcomes that make up a given event can be very diFcult and impractical since, even for relatively simple situations, there can be a tremendous number of such outcomes. ±or example: There are 2,598,960 ways to deal a poker hand. A sequence of 10 tosses of a coin has 1024 possible outcomes. There are 13,983,816 to select 6 out of 49 lottery numbers. Let’s consider an easier example ²rst. Example 1.1. Suppose we ³ip a coin and then roll a 6-sided die. Let E be the event that we get a “Tail” on the coin and an even number on the die. To use formula (1.1) to ²nd P ( E ), we need to know the total number of outcomes. ±rom a tree diagram (see ±igure 1) it is easy to see that there are 12 possible outcomes. Three of the outcomes result in a “Tail” and an even roll, therefore, P ( E ) = 3 / 12 = 0 . 25. Exercise 1.2. I own 5 shirts, 2 pair of pants and two pair of shoes. How many di´erent out²ts could I wear? Draw a tree diagram. Mathematicians are lazy and did not want to work this hard. They developed a set of counting rules, which are included in what is called combinatorics . The ²rst is the Basic Counting Rule (BCR), often called the multiplication rule for obvious reasons. Proposition 1.3 (BCR) . Suppose that r actions are to be performed in a deFnite order. Suppose further that there are m k possibilities for action k , for k = 1 , . . ., r . Then there are m 1 m 2 ··· m r possibilities altogether for the r actions. 1
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H , 1 H , 2 H , 3 H , 4 H , 5 H , 6 T , 1 T , 2 T , 3 T , 4 T , 5 T , 6 Fig 1: A tree diagram for Exercise 1.1. The ±rst set of branches denote the outcome of the coin toss (H or T) and the second set denote the outcome of the die roll (1–6). Exercise 1.4. If you rolled four fair six-sided dice, what is the probability that you get at least one 3? Exercise 1.5. Prove that a set containing n elements has 2 n possible subsets. How many di²erent pizzas can be made with 10 distinct toppings? Exercise 1.6. Let A and B be ±nite sets, each with n elements. How many one-to-one functions are there from A to B ? Relate your answer to the number of tries it would take to successfully decode a simple substitution cipher by guessing. (Note: a substitution cipher is the coding scheme used in a standard cryptogram.) 2 Sampling With and Without Replacement Defnition 2.1. Consider a population with N members. Sampling with replacement is where n members of the population are selected, one at a time, and after the observation is made the member is returned to the population for possible re-selection.
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This note was uploaded on 06/15/2008 for the course STAT 225 taught by Professor Martin during the Spring '08 term at Purdue University.
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225-Notes02 - Stat 225 Lecture Notes "Counting Techniques"...
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1.4: Calculating Limits with Limit Laws
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Think back to the functions you know and the sorts of things you have been asked to draw, factor and so on. Then they are all constructed from simple pieces, such as
• constants — $$c$$
• monomials — $$x^n$$
• trigonometric functions — $$\sin(x), \cos(x)$$ and $$\tan(x)$$
These are the building blocks from which we construct functions. Soon we will add a few more functions to this list, especially the exponential function and various inverse functions.
We then take these building blocks and piece them together using arithmetic
• addition and subtraction — $$f(x) = g(x) + h(x)$$ and $$f(x) = g(x) - h(x)$$
• multiplication — $$f(x) = g(x) \cdot h(x)$$
• division — $$f(x) = \frac{g(x)}{h(x)}$$
• substitution — $$f(x) = g( h(x) )$$ — this is also called the composition of $$g$$ with $$h\text{.}$$
The idea of building up complicated functions from simpler pieces was discussed in Section 0.5.
What we will learn in this section is how to compute the limits of the basic building blocks and then how we can compute limits of sums, products and so forth using “limit laws”. This process allows us to compute limits of complicated functions, using very simple tools and without having to resort to “plugging in numbers” or “closer and closer” or “$$\epsilon-\delta$$ arguments”.
In the examples we saw above, almost all the interesting limits happened at points where the underlying function was badly behaved — where it jumped, was not defined or blew up to infinity. In those cases we had to be careful and think about what was happening. Thankfully most functions we will see do not have too many points at which these sorts of things happen.
For example, polynomials do not have any nasty jumps and are defined everywhere and do not “blow up”. If you plot them, they look smooth 1. Polynomials and limits behave very nicely together, and for any polynomial $$P(x)$$ and any real number $$a$$ we have that
\begin{align*} \lim_{x \to a} P(x) &= P(a) \end{align*}
That is — to evaluate the limit we just plug in the number. We will build up to this result over the next few pages.
Theorem 1.4.1 Easiest limits.
Let $$a,c \in \mathbb{R}\text{.}$$ The following two limits hold
\begin{align*} \lim_{x \to a} c & = c & \text{ and }&& \lim_{x \to a} x &= a. \end{align*}
Since we have not seen too many theorems yet, let us examine it carefully piece by piece.
• Let $$a,c \in \mathbb{R}$$ — just as was the case for definitions, we start a theorem by defining terms and setting the scene. There is not too much scene to set: the symbols $$a$$ and $$c$$ are real numbers.
• The following two limits hold — this doesn't really contribute much to the statement of the theorem, it just makes it easier to read.
• $$\mathbf{\lim_{x \to a} c = c}$$ — when we take the limit of a constant function (for example think of $$c=3$$), the limit is (unsurprisingly) just that same constant.
• $$\mathbf{\lim_{x \to a} x = a}$$ — as we noted above for general polynomials, the limit of the function $$f(x) = x$$ as $$x$$ approaches a given point $$a\text{,}$$ is just $$a\text{.}$$ This says something quite obvious — as $$x$$ approaches $$a\text{,}$$ $$x$$ approaches $$a$$ (if you are not convinced then sketch the graph).
Armed with only these two limits, we cannot do very much. But combining these limits with some arithmetic we can do quite a lot. For a moment, take a step back from limits for a moment and think about how we construct functions. To make the discussion a little more precise think about how we might construct the function
\begin{align*} h(x) &= \frac{2x-3}{x^2+5x-6} \end{align*}
If we want to compute the value of the function at $$x=2\text{,}$$ then we would
• compute the numerator at $$x=2$$
• compute the denominator at $$x=2$$
• compute the ratio
Now to compute the numerator we
• take $$x$$ and multiply it by 2
• subtract 3 to the result
While for the denominator
• multiply $$x$$ by $$x$$
• multiply $$x$$ by 5
• add these two numbers and subtract $$6$$
This sequence of operations can be represented pictorially as the tree shown in Figure 1.4.2 below.
Such trees were discussed in Section 0.5 (now is not a bad time to quickly review that section before proceeding). The point here is that in order to compute the value of the function we just repeatedly add, subtract, multiply and divide constants and $$x\text{.}$$
To compute the limit of the above function at $$x=2$$ we can do something very similar. From the previous theorem we know how to compute
\begin{align*} \lim_{x \to 2} c &= c && \text{and} & \lim_{x \to 2} x &= 2 \end{align*}
and the next theorem will tell us how to stitch together these two limits using the arithmetic we used to construct the function.
Theorem 1.4.3 Arithmetic of limits.
Let $$a,c \in \mathbb{R}\text{,}$$ let $$f(x)$$ and $$g(x)$$ be defined for all $$x$$'s that lie in some interval about $$a$$ (but $$f,g$$ need not be defined exactly at $$a$$).
\begin{align*} \lim_{x \to a} f(x)&=F & \lim_{x \to a} g(x) &=G \end{align*}
exist with $$F,G \in \mathbb{R}\text{.}$$ Then the following limits hold
• $$\lim_{x \to a} ( f(x) + g(x) ) = F+G$$ — limit of the sum is the sum of the limits.
• $$\lim_{x \to a} ( f(x) - g(x) ) = F - G$$ — limit of the difference is the difference of the limits.
• $$\lim_{x \to a} c f(x) = c F\text{.}$$
• $$\lim_{x \to a} ( f(x) \cdot g(x) ) = F \cdot G$$ — limit of the product is the product of limits.
• If $$G \neq 0$$ then $$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{F}{G}$$
Note — be careful with this last one — the denominator cannot be zero.
The above theorem shows that limits interact very simply with arithmetic. If you are asked to find the limit of a sum then the answer is just the sum of the limits. Similarly the limit of a product is just the product of the limits.
How do we apply the above theorem to the rational function $$h(x)$$ we defined above? Here is a warm-up example:
Example 1.4.4 Using limit laws.
You are given two functions $$f,g$$ (not explicitly) which have the following limits as $$x$$ approaches 1:
\begin{align*} \lim_{x \to 1} f(x)&=3 && \text{and} & \lim_{x\to 1} g(x)&=2 \end{align*}
Using the above theorem we can compute
\begin{align*} \lim_{x \to 1} 3f(x) &= 3 \times 3 = 9\\ \lim_{x \to 1} 3f(x) -g(x) &= 3\times 3 -2 = 7\\ \lim_{x \to 1} f(x) g(x) &= 3\times 2 = 6\\ \lim_{x \to 1} \frac{f(x)}{f(x) -g(x)} &= \frac{3}{3-2} = 3 \end{align*}
Another simple example
Example 1.4.5 More using limit laws.
Find $$\lim_{x \to 3} 4x^2-1$$
We use the arithmetic of limits:
\begin{align*} \lim_{x \to 3} 4x^2-1 &= \left( \lim_{x \to 3} 4x^2 \right) - \lim_{x \to 3} 1 & \text{difference of limits}\\ &= \left( \lim_{x \to 3} 4 \cdot \lim_{x \to 3} x^2 \right) - \lim_{x \to 3} 1 & \text{product of limits}\\ &= 4 \cdot \left( \lim_{x \to 3} x^2 \right) - 1 & \text{limit of constant}\\ &= 4 \cdot \left( \lim_{x \to 3} x \right) \cdot \left( \lim_{x \to 3} x \right)-1 & \text{product of limits}\\ &= 4 \cdot 3 \cdot 3 - 1 & \text{limit of $x$}\\ &= 36 - 1\\ &= 35 \end{align*}
This is an excruciating level of detail, but when you first use this theorem and try some examples it is a good idea to do things step by step by step until you are comfortable with it.
Example 1.4.6 Yet more using limit laws.
Yet another limit — compute $$\lim_{x\to 2} \frac{x}{x-1}\text{.}$$
To apply the arithmetic of limits, we need to examine numerator and denominator separately and make sure the limit of the denominator is non-zero. Numerator first:
\begin{align*} \lim_{x \to 2} x & = 2 & \text{limit of $x$}\\ \end{align*}
and now the denominator:
\begin{align*} \lim_{x \to 2} x-1 & = \left( \lim_{x \to 2} x \right) - \left( \lim_{x \to 2} 1 \right) & \text{difference of limits}\\ & = 2 - 1 & \text{limit of $x$ and limit of constant} & = 1 \end{align*}
Since the limit of the denominator is non-zero we can put it back together to get
\begin{align*} \lim_{x\to 2} \frac{x}{x-1} &= \frac{\lim_{x\to 2} x}{ \lim_{x \to 2}(x-1)}\\ &= \frac{2}{1}\\ &= 2 \end{align*}
In the next example we show that many different things can happen if the limit of the denominator is zero.
Example 1.4.7 Be careful with limits of ratios.
We must be careful when computing the limit of a ratio — it is the ratio of the limits except when the limit of the denominator is zero. When the limit of the denominator is zero Theorem 1.4.3 does not apply and a few interesting things can happen
• If the limit of the numerator is non-zero then the limit of the ratio does not exist
\begin{align*} \lim_{x \to a} \frac{f(x)}{g(x)} &= DNE & \text{when $\lim_{x\to a} f(x) \neq 0$ and $\lim_{x \to a} g(x)=0$} \end{align*}
For example, $$\lim_{x \to 0} \frac{1}{x^2} = DNE\text{.}$$
• If the limit of the numerator is zero then the above theorem does not give us enough information to decide whether or not the limit exists. It is possible that
• the limit does not exist, eg. $$\lim_{x \to 0} \frac{x}{x^2} = \lim_{x \to 0} \frac{1}{x} = DNE$$
• the limit is $$\pm \infty\text{,}$$ eg. $$\lim_{x \to 0} \frac{x^2}{x^4} = \lim_{x \to 0} \frac{1}{x^2} = +\infty$$ or $$\lim_{x \to 0} \frac{-x^2}{x^4} = \lim_{x\to 0} \frac{-1}{x^2} = -\infty\text{.}$$
• the limit is zero, eg. $$\lim_{x \to 0} \frac{x^2}{x} = 0$$
• the limit exists and is non-zero, eg. $$\lim_{x \to 0} \frac{x}{x} = 1$$
Now while the above examples are very simple and a little contrived they serve to illustrate the point we are trying to make — be careful if the limit of the denominator is zero.
We now have enough theory to return to our rational function and compute its limit as $$x$$ approaches 2.
Example 1.4.8 More on limits of ratios
Let $$h(x) = \frac{2x-3}{x^2+5x-6}$$ and find its limit as $$x$$ approaches $$2\text{.}$$
Since this is the limit of a ratio, we compute the limit of the numerator and denominator separately. Numerator first:
\begin{align*} \lim_{x \to 2} 2x-3 &= \left( \lim_{x \to 2} 2x \right) - \left( \lim_{x \to 2} 3 \right) & \text{difference of limits}\\ &= 2 \cdot \left( \lim_{x \to 2} x \right) -3 & \text{product of limits and limit of constant}\\ &= 2 \cdot 2 -3 & \text{limits of $x$}\\ &= 1 \end{align*}
Denominator next:
\begin{align*} \lim_{x \to 2} x^2+5x-6 &= \left( \lim_{x \to 2} x^2 \right) + \left( \lim_{x \to 2} 5x \right) - \left( \lim_{x \to 2} 6 \right) & \!\!\!\!\!\!\text{sum of limits}\\ &= \left( \lim_{x \to 2} x \right)\cdot \left( \lim_{x \to 2} x \right) + 5 \cdot \left( \lim_{x \to 2} x \right) - 6\\ &\hskip1.0in \text{product of limits and limit of constant}\\ &= 2 \cdot 2 + 5 \cdot 2 - 6 & \text{limits of $x$}\\ &= 8 \end{align*}
Since the limit of the denominator is non-zero, we can obtain our result by taking the ratio of the separate limits.
\begin{align*} \lim_{x \to 2} \frac{2x-3}{x^2+5x-6} &= \frac{\lim_{x \to 2} 2x-3}{ \lim_{x \to 2} x^2+5x-6} = \frac{1}{8} \end{align*}
The above works out quite simply. However, if we were to take the limit as $$x \to 1$$ then things are a bit harder. The limit of the numerator is:
\begin{align*} \lim_{x \to 1} 2x-3 &= 2 \cdot 1 - 3 = -1 \end{align*}
(we have not listed all the steps). And the limit of the denominator is
\begin{align*} \lim_{x \to 1} x^2 +5x-6 &= 1 \cdot 1 + 5 - 6 = 0 \end{align*}
Since the limit of the numerator is non-zero, while the limit of the denominator is zero, the limit of the ratio does not exist.
\begin{align*} \lim_{x \to 1} \frac{2x-3}{x^2+5x-6} &= DNE \end{align*}
It is IMPORTANT TO NOTE that it is not correct to write
\begin{align*} \lim_{x \to 1} \frac{2x-3}{x^2+5x-6} &= \frac{-1}{0} = DNE \end{align*}
Because we can only write
\begin{align*} \lim_{x \to a} \frac{f(x)}{g(x)} &= \frac{ \lim_{x \to a} f(x)}{\lim_{x \to a} g(x) } = \text{something} \end{align*}
when the limit of the denominator is non-zero (see Example 1.4.7 above).
With a little care you can use the arithmetic of limits to obtain the following rules for limits of powers of functions and limits of roots of functions:
Theorem 1.4.9 More arithmetic of limits — powers and roots.
Let $$n$$ be a positive integer, let $$a \in \mathbb{R}$$ and let $$f$$ be a function so that
\begin{align*} \lim_{x \to a} f(x) &= F \end{align*}
for some real number $$F\text{.}$$ Then the following holds
\begin{align*} \lim_{x \to a} \left( f(x) \right)^n &= \left(\lim_{x \to a} f(x) \right)^n = F^n \end{align*}
so that the limit of a power is the power of the limit. Similarly, if
• $$n$$ is an even number and $$F \gt 0\text{,}$$ or
• $$n$$ is an odd number and $$F$$ is any real number
then
\begin{align*} \lim_{x \to a} \left( f(x) \right)^{1/n} &= \left(\lim_{x \to a} f(x) \right)^{1/n} = F^{1/n} \end{align*}
More generally 3, if $$F>0$$ and $$p$$ is any real number,
$\lim_{x \to a} \left( f(x) \right)^p = \left(\lim_{x \to a} f(x) \right)^p = F^p \nonumber$
Notice that we have to be careful when taking roots of limits that might be negative numbers. To see why, consider the case $$n=2\text{,}$$ the limit
\begin{align*} \lim_{x \to 4} x^{1/2} &= 4^{1/2} = 2\\ \lim_{x \to 4} (-x)^{1/2} &= (-4)^{1/2} = \text{not a real number} \end{align*}
In order to evaluate such limits properly we need to use complex numbers which are beyond the scope of this text.
Also note that the notation $$x^{1/2}$$ refers to the positive square root of $$x\text{.}$$ While $$2$$ and $$(-2)$$ are both square-roots of $$4\text{,}$$ the notation $$4^{1/2}$$ means $$2\text{.}$$ This is something we must be careful of 4.
So again — let us do a few examples and carefully note what we are doing.
Example 1.4.10 $$\lim_{x \to 2} (4x^2-3)^{1/3}$$.
\begin{align*} \lim_{x \to 2} (4x^2-3)^{1/3} &= \left( (\lim_{x\to 2} 4x^2) - (\lim_{x \to 2} 3) \right)^{1/3}\\ &= \left( 4 \cdot 2^2 - 3 \right)^{1/3}\\ &= \left( 16-3 \right)^{1/3}\\ &= 13^{1/3} \end{align*}
By combining the last few theorems we can make the evaluation of limits of polynomials and rational functions much easier:
Theorem 1.4.11 Limits of polynomials and rational functions.
Let $$a \in \mathbb{R}\text{,}$$ let $$P(x)$$ be a polynomial and let $$R(x)$$ be a rational function. Then
\begin{align*} \lim_{x \to a} P(x) &= P(a) \end{align*}
and provided $$R(x)$$ is defined at $$x=a$$ then
\begin{align*} \lim_{x \to a} R(x) &= R(a) \end{align*}
If $$R(x)$$ is not defined at $$x=a$$ then we are not able to apply this result.
So the previous examples are now much easier to compute:
\begin{align*} \lim_{x \to 2} \frac{2x-3}{x^2+5x-6} &=& \frac{4-3}{4+10-6} &=& \frac{1}{8}\\ \lim_{x\to 2} (4x^2-1) &=& 16-1 &=& 15\\ \lim_{x\to 2} \frac{x}{x-1} &=& \frac{2}{2-1} &=& 2 \end{align*}
It is clear that limits of polynomials are very easy, while those of rational functions are easy except when the denominator might go to zero. We have seen examples where the resulting limit does not exist, and some where it does. We now work to explain this more systematically. The following example demonstrates that it is sometimes possible to take the limit of a rational function to a point at which the denominator is zero. Indeed we must be able to do exactly this in order to be able to define derivatives in the next chapter.
Example 1.4.12 Numerator and denominator both go to 0.
Consider the limit
\begin{gather*} \lim_{x \to 1} \frac{x^3-x^2}{x-1}. \end{gather*}
If we try to apply the arithmetic of limits then we compute the limits of the numerator and denominator separately
\begin{align*} \lim_{x \to 1} x^3-x^2 &= 1-1 = 0\\ \lim_{x \to 1} x-1 &= 1-1 = 0 \end{align*}
Since the denominator is zero, we cannot apply our theorem and we are, for the moment, stuck. However, there is more that we can do here — the hint is that the numerator and denominator both approach zero as $$x$$ approaches 1. This means that there might be something we can cancel.
So let us play with the expression a little more before we take the limit:
\begin{align*} \frac{x^3-x^2}{x-1} &= \frac{x^2(x-1)}{x-1} = x^2 & \text{ provided } x \neq 1. \end{align*}
So what we really have here is the following function
\begin{align*} \frac{x^3-x^2}{x-1} &= \begin{cases} x^2 & x \neq 1\\ \text{undefined } & x = 1 \end{cases} \end{align*}
If we plot the above function the graph looks exactly the same as $$y=x^2$$ except that the function is not defined at $$x=1$$ (since at $$x=1$$ both numerator and denominator are zero)
When we compute a limit as $$x \to a\text{,}$$ the value of the function exactly at $$x=a$$ is irrelevant. We only care what happens to the function as we bring $$x$$ very close to $$a\text{.}$$ So for the above problem we can write
\begin{align*} \frac{x^3-x^2}{x-1} &= x^2 & \text{when $x$ is close to $1$ but not at $x=1$} \end{align*}
So the limit as $$x \to 1$$ of the function is the same as the limit $$\lim_{x\to 1} x^2$$ since the functions are the same except exactly at $$x=1\text{.}$$ By this reasoning we get
\begin{align*} \lim_{x \to 1} \frac{x^3-x^2}{x-1} &= \lim_{x \to 1} x^2 = 1 \end{align*}
The reasoning in the above example can be made more general:
Theorem 1.4.13.
If $$f(x) = g(x)$$ except when $$x=a$$ then $$\lim_{x\to a} f(x) = \lim_{x\to a} g(x)$$ provided the limit of $$g$$ exists.
How do we know when to use this theorem? The big clue is that when we try to compute the limit in a naive way, we end up with $$\frac{0}{0}\text{.}$$ We know that $$\frac{0}{0}$$ does not make sense, but it is an indication that there might be a common factor between numerator and denominator that can be cancelled. In the previous example, this common factor was $$(x-1)\text{.}$$
Example 1.4.14 Another numerator and denominator both go to 0 limit.
Using this idea compute
\begin{gather*} \lim_{h \to 0} \frac{(1+h)^2-1}{h} \end{gather*}
• First we should check that we cannot just substitute $$h=0$$ into this — clearly we cannot because the denominator would be $$0\text{.}$$
• But we should also check the numerator to see if we have $$\frac{0}{0}\text{,}$$ and we see that the numerator gives us $$1-1 = 0\text{.}$$
• Thus we have a hint that there is a common factor that we might be able to cancel. So now we look for the common factor and try to cancel it.
\begin{align*} \frac{(1+h)^2-1}{h} &= \frac{1+2h+h^2-1}{h} & \text{expand}\\ &= \frac{2h+h^2}{h} = \frac{h(2+h)}{h} & \text{factor and then cancel}\\ &= 2+h \end{align*}
• Thus we really have that
\begin{align*} \frac{(1+h)^2-1}{h}&= \begin{cases} 2+h & h \neq 0 \\ \text{undefined} & h=0 \end{cases} \end{align*}
and because of this
\begin{align*} \lim_{h \to 0} \frac{(1+h)^2-1}{h} &= \lim_{h \to 0} 2+h\\ &= 2 \end{align*}
Of course — we have written everything out in great detail here and that is way more than is required for a solution to such a problem. Let us do it again a little more succinctly.
Example 1.4.15 $$\lim_{h \to 0} \frac{(1+h)^2-1}{h}$$.
Compute the following limit:
\begin{gather*} \lim_{h \to 0} \frac{(1+h)^2-1}{h} \end{gather*}
If we try to use the arithmetic of limits, then we see that the limit of the numerator and the limit of the denominator are both zero. Hence we should try to factor them and cancel any common factor. This gives
\begin{align*} \lim_{h \to 0} \frac{(1+h)^2-1}{h} &= \lim_{h \to 0} \frac{1+2h+h^2-1}{h}\\ &= \lim_{h \to 0} 2+h\\ &= 2 \end{align*}
Notice that even though we did this example carefully above, we have still written some text in our working explaining what we have done. You should always think about the reader and if in doubt, put in more explanation rather than less. We could make the above example even more terse
Example 1.4.16 Redoing previous example with fewer words.
Compute the following limit:
\begin{gather*} \lim_{h \to 0} \frac{(1+h)^2-1}{h} \end{gather*}
Numerator and denominator both go to zero as $$h\to 0\text{.}$$ So factor and simplify:
\begin{align*} \lim_{h \to 0} \frac{(1+h)^2-1}{h} &= \lim_{h \to 0} \frac{1+2h+h^2-1}{h}\\ &= \lim_{h \to 0} 2+h = 2 \end{align*}
A slightly harder one now
Example 1.4.17 A harder limit with cancellations.
Compute the limit
\begin{gather*} \lim_{x \to 0} \frac{x}{\sqrt{1+x}-1} \end{gather*}
If we try to use the arithmetic of limits we get
\begin{align*} \lim_{x \to 0} x &= 0\\ \lim_{x \to 0} \sqrt{1+x}-1 &= \sqrt{ \lim_{x \to 0} 1+x}-1 = 1-1 =0 \end{align*}
So doing the naive thing we'd get $$0/0\text{.}$$ This suggests a common factor that can be cancelled. Since the numerator and denominator are not polynomials we have to try other tricks 5. We can simplify the denominator $$\sqrt{1+x}-1$$ a lot, and in particular eliminate the square root, by multiplying it by its conjugate $$\sqrt{1+x}+1\text{.}$$
\begin{align*} \frac{x}{\sqrt{1+x}-1} &=\frac{x}{\sqrt{1+x}-1} \times \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1} & \text{multiply by $\frac{\text{conjugate}}{\text{conjugate}}=1$}\\ &=\frac{x \left( \sqrt{1+x}+1\right)} {\left(\sqrt{1+x}-1\right)\left(\sqrt{1+x}+1\right)} & \text{bring things together }\\ &=\frac{x \left( \sqrt{1+x}+1\right)} {\left(\sqrt{1+x}\right)^2 - 1\cdot 1} &\!\!\!\!\!\!\! \text{since $(a\!-\!b)(a\!+\!b)=a^2\!-\!b^2$}\\ &=\frac{x \left( \sqrt{1+x}+1\right)} {1+x - 1} & \text{clean up a little}\\ &=\frac{x \left( \sqrt{1+x}+1\right)}{x}\\ &= \sqrt{1+x}+1 & \text{cancel the $x$} \end{align*}
So now we have
\begin{align*} \lim_{x \to 0} \frac{x}{\sqrt{1+x}-1} &= \lim_{x \to 0} \sqrt{1+x}+1\\ &= \sqrt{1+0}+1 = 2 \end{align*}
How did we know what to multiply by? Our function was of the form
\begin{gather*} \frac{a}{\sqrt{b} - c} \end{gather*}
so, to eliminate the square root from the denominator, we employ a trick — we multiply by 1. Of course, multiplying by 1 doesn't do anything. But if you multiply by 1 carefully you can leave the value the same, but change the form of the expression. More precisely
\begin{align*} \frac{a}{\sqrt{b} - c} &= \frac{a}{\sqrt{b} - c} \cdot 1\\ &= \frac{a}{\sqrt{b} - c} \cdot \underbrace{\frac{\sqrt{b}+c}{\sqrt{b}+c}}_{=1}\\ &= \frac{a \left(\sqrt{b}+c \right)}{\left(\sqrt{b} - c\right)\left(\sqrt{b}+c \right)} & \text{expand denominator carefully}\\ &= \frac{a \left(\sqrt{b}+c \right)} {\sqrt{b} \cdot \sqrt{b} - c\sqrt{b} + c\sqrt{b} - c\cdot c} & \text{do some cancellation}\\ &= \frac{a \left(\sqrt{b}+c \right)} {b - c^2} \end{align*}
Now the numerator contains roots, but the denominator is just a polynomial.
Before we move on to limits at infinity, there is one more theorem to see. While the scope of its application is quite limited, it can be extremely useful. It is called a sandwich theorem or a squeeze theorem for reasons that will become apparent.
Sometimes one is presented with an unpleasant ugly function such as
\begin{align*} f(x) &= x^2 \sin(\pi/x) \end{align*}
It is a fact of life, that not all the functions that are encountered in mathematics will be elegant and simple; this is especially true when the mathematics gets applied to real world problems. One just has to work with what one gets. So how can we compute
\begin{gather*} \lim_{x \to 0} x^2 \sin(\pi/x) ? \end{gather*}
Since it is the product of two functions, we might try
\begin{align*} \lim_{x \to 0} x^2 \sin(\pi/x) &= \left( \lim_{x \to 0} x^2 \right) \cdot \left( \lim_{x \to 0} \sin(\pi/x) \right)\\ &= 0 \cdot \left( \lim_{x \to 0} \sin(\pi/x) \right)\\ & = 0 \end{align*}
But we just cheated — we cannot use the arithmetic of limits theorem here, because the limit
\begin{align*} \lim_{x \to 0} \sin(\pi/x) &= DNE \end{align*}
does not exist. Now we did see the function $$\sin(\pi/x)$$ before (in Example 1.3.5), so you should go back and look at it again. Unfortunately the theorem “the limit of a product is the product of the limits” only holds when the limits you are trying to multiply together actually exist. So we cannot use it.
However, we do see that the function naturally decomposes into the product of two pieces — the functions $$x^2$$ and $$\sin(\pi/x)\text{.}$$ We have sketched the two functions in the figure on the left below.
While $$x^2$$ is a very well behaved function and we know quite a lot about it, the function $$\sin(\pi/x)$$ is quite ugly. One of the few things we can say about it is the following
\begin{align*} -1 \leq \sin(\pi/x) \leq 1 && \text{provided $x \neq 0$} \end{align*}
But if we multiply this expression by $$x^2$$ we get (because $$x^2 \geq 0$$)
\begin{align*} -x^2 \leq x^2\sin(\pi/x) \leq x^2 && \text{provided $x \neq 0$} \end{align*}
and we have sketched the result in the figure above (on the right). So the function we are interested in is squeezed or sandwiched between the functions $$x^2$$ and $$-x^2\text{.}$$
If we focus in on the picture close to $$x=0$$ we see that $$x$$ approaches $$0\text{,}$$ the functions $$x^2$$ and $$-x^2$$ both approach $$0\text{.}$$ Further, because $$x^2\sin(\pi/x)$$ is sandwiched between them, it seems that it also approaches $$0\text{.}$$
The following theorem tells us that this is indeed the case:
Theorem 1.4.18 Squeeze theorem (or sandwich theorem or pinch theorem).
Let $$a \in \mathbb{R}$$ and let $$f,g,h$$ be three functions so that
\begin{gather*} f(x) \leq g(x) \leq h(x) \end{gather*}
for all $$x$$ in an interval around $$a\text{,}$$ except possibly exactly at $$x=a\text{.}$$ Then if
\begin{align*} \lim_{x \to a} f(x) &= \lim_{x \to a} h(x) = L \end{align*}
then it is also the case that
\begin{align*} \lim_{x \to a} g(x) &= L \end{align*}
Using the above theorem we can compute the limit we want and write it up nicely.
Example 1.4.19 $$\lim_{x \to 0} x^2 \sin(\pi/x)$$.
Compute the limit
$\lim_{x \to 0} x^2 \sin(\pi/x) \nonumber$
Since $$-1 \leq \sin(\theta) \leq 1$$ for all real numbers $$\theta\text{,}$$ we have
\begin{align*} -1 \leq \sin(\pi /x ) \leq 1 && \text{for all } x \neq 0 \end{align*}
Multiplying the above by $$x^2$$ we see that
\begin{align*} -x^2 \leq x^2 \sin(\pi /x ) \leq x^2 && \text{for all } x\neq0 \end{align*}
Since $$\lim_{x \to 0} x^2 = \lim_{x \to 0} (-x^2) = 0$$ by the sandwich (or squeeze or pinch) theorem we have
\begin{align*} \lim_{x \to 0} x^2 \sin(\pi/x) &= 0 \end{align*}
Notice how we have used “words”. We have remarked on this several times already in the text, but we will keep mentioning it. It is okay to use words in your answers to maths problems — and you should do so! These let the reader know what you are doing and help you understand what you are doing.
Example 1.4.20 Another sandwich theorem example.
Let $$f(x)$$ be a function such that $$1 \leq f(x) \leq x^2-2x+2\text{.}$$ What is $$\lim_{x \to 1} f(x)\text{?}$$
We are already supplied with an inequality, so it is likely that it is going to help us. We should examine the limits of each side to see if they are the same:
\begin{align*} \lim_{x \to 1} 1 &= 1\\ \lim_{x \to 1} x^2-2x+2 &= 1-2+2 = 1 \end{align*}
So we see that the function $$f(x)$$ is trapped between two functions that both approach $$1$$ as $$x \to 1\text{.}$$ Hence by the sandwich / pinch / squeeze theorem, we know that
\begin{align*} \lim_{x \to 1} f(x) & =1 \end{align*}
To get some intuition as to why the squeeze theorem is true, consider when $$x$$ is very very close to $$a\text{.}$$ In particular, consider when $$x$$ is sufficiently close to $$a$$ that we know $$h(x)$$ is within $$10^{-6}$$ of $$L$$ and that $$f(x)$$ is also within $$10^{-6}$$ of $$L\text{.}$$ That is
\begin{align*} |h(x)-L| & \lt 10^{-6} & \text{and}&& |f(x)-L| & \lt 10^{-6}. \end{align*}
This means that
\begin{align*} L-10^{-6} & \lt f(x) \leq h(x) \lt L+10^{-6} \end{align*}
since we know that $$f(x) \leq h(x)\text{.}$$
But now by the hypothesis of the squeeze theorem we know that $$f(x) \leq g(x) \leq h(x)$$ and so we have
\begin{align*} L-10^{-6} & \lt f(x) \leq g(x) \leq h(x) \lt L+10^{-6} \end{align*}
And thus we know that
$L-10^{-6} \leq g(x) \leq L+10^{-6} \nonumber$
That is $$g(x)$$ is also within $$10^{-6}$$ of $$L\text{.}$$
In this argument our choice of $$10^{-6}$$ was arbitrary, so we can really replace $$10^{-6}$$ with any small number we like. Hence we know that we can force $$g(x)$$ as close to $$L$$ as we like, by bringing $$x$$ sufficiently close to $$a\text{.}$$ We give a more formal and rigorous version of this argument at the end of Section 1.9.
Exercises
Exercise $$\PageIndex{1}$$
Suppose $$\displaystyle\lim_{x \rightarrow a} f(x)=0$$ and $$\displaystyle\lim_{x \rightarrow a} g(x)=0\text{.}$$ Which of the following limits can you compute, given this information?
1. $$\displaystyle\lim_{x \rightarrow a} \frac{f(x)}{2}$$
2. $$\displaystyle\lim_{x \rightarrow a} \frac{2}{f(x)}$$
3. $$\displaystyle\lim_{x \rightarrow a} \frac{f(x)}{g(x)}$$
4. $$\displaystyle\lim_{x \rightarrow a} f(x)g(x)$$
Exercise $$\PageIndex{2}$$
Give two functions $$f(x)$$ and $$g(x)$$ that satisfy $$\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0$$ and $$\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=10\text{.}$$
Exercise $$\PageIndex{3}$$
Give two functions $$f(x)$$ and $$g(x)$$ that satisfy $$\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0$$ and $$\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=0\text{.}$$
Exercise $$\PageIndex{4}$$
Give two functions $$f(x)$$ and $$g(x)$$ that satisfy $$\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0$$ and $$\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=\infty\text{.}$$
Exercise $$\PageIndex{5}$$
Suppose $$\displaystyle\lim_{x \rightarrow a}f(x)=\displaystyle\lim_{x \rightarrow a}g(x)=0\text{.}$$ What are the possible values of $$\displaystyle\lim_{x \rightarrow a}\dfrac{f(x)}{g(x)}\text{?}$$
Stage 2
For Questions 1.4.2.6 through 1.4.2.41, evaluate the given limits.
Exercise $$\PageIndex{6}$$
$$\displaystyle\lim_{t \rightarrow 10} \dfrac{2(t-10)^2}{t}$$
Exercise $$\PageIndex{7}$$
$$\displaystyle\lim_{y \rightarrow 0} \dfrac{(y+1)(y+2)(y+3)}{\cos y}$$
Exercise $$\PageIndex{8}$$
$$\displaystyle\lim_{x \rightarrow 3} \left(\dfrac{4x-2}{x+2}\right)^4$$
Exercise $$\PageIndex{9}$$(✳)
$$\lim_{t\to -3} \left(\frac{1-t}{\cos(t)}\right)$$
Exercise $$\PageIndex{10}$$(✳)
$$\lim_{h \to 0} \frac{(2+h)^2-4}{2h}$$
Exercise $$\PageIndex{11}$$(✳)
$$\lim_{t\to -2} \left(\frac{t-5}{t+4}\right)$$
Exercise $$\PageIndex{12}$$(✳)
$$\lim_{x\to 1} \sqrt{5x^3 + 4}$$
Exercise $$\PageIndex{13}$$(✳)
$$\displaystyle\lim_{t\rightarrow -1} \left(\frac{t-2}{t+3}\right)$$
Exercise $$\PageIndex{14}$$(✳)
$$\lim\limits_{x\rightarrow 1}\dfrac{\log(1+x)-x}{x^2}$$
Exercise $$\PageIndex{15}$$(✳)
$$\displaystyle\lim_{x\rightarrow 2} \left(\frac{x-2}{x^2-4}\right)$$
Exercise $$\PageIndex{16}$$(✳)
$$\lim\limits_{x\rightarrow 4}\dfrac{x^2-4x}{x^2-16}$$
Exercise $$\PageIndex{17}$$(✳)
$$\lim\limits_{x\rightarrow 2}\dfrac{x^2+x-6}{x-2}$$
Exercise $$\PageIndex{18}$$(✳)
$$\lim_{x \to -3} \frac{x^2-9}{x+3}$$
Exercise $$\PageIndex{19}$$
$$\displaystyle\lim_{t \rightarrow 2} \frac{1}{2}t^4-3t^3+t$$
Exercise $$\PageIndex{20}$$(✳)
$$\lim_{x\to -1} \frac{\sqrt{x^2+8}-3}{x+1}\text{.}$$
Exercise $$\PageIndex{21}$$(✳)
$$\lim_{x\to 2} \frac{\sqrt{x+7}-\sqrt{11-x}}{2x-4}\text{.}$$
Exercise $$\PageIndex{22}$$(✳)
$$\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x+2}-\sqrt{4-x}}{x-1}$$
Exercise $$\PageIndex{23}$$(✳)
$$\lim_{x\to 3} \frac{\sqrt{x-2}-\sqrt{4-x}}{x-3}\text{.}$$
Exercise $$\PageIndex{24}$$(✳)
$$\lim_{t\to 1} \frac{3t-3}{2 - \sqrt{5-t}}\text{.}$$
Exercise $$\PageIndex{25}$$
$$\displaystyle\lim_{x \rightarrow 0}-x^2\cos\left(\frac{3}{x}\right)$$
Exercise $$\PageIndex{26}$$
$$\displaystyle\lim_{x \rightarrow 0}\dfrac{x^4\sin\left(\frac{1}{x}\right)+5x^2\cos\left(\frac{1}{x}\right)+2}{(x-2)^2}$$
Exercise $$\PageIndex{27}$$(✳)
$$\lim\limits_{x\rightarrow 0}x\sin^2\left(\dfrac{1}{x}\right)$$
Exercise $$\PageIndex{28}$$
$$\displaystyle\lim_{w \rightarrow 5} \dfrac{2w^2-50}{(w-5)(w-1)}$$
Exercise $$\PageIndex{29}$$
$$\displaystyle\lim_{r \rightarrow -5} \dfrac{r}{r^2+10r+25}$$
Exercise $$\PageIndex{30}$$
$$\displaystyle\lim_{x \rightarrow -1}\sqrt{\dfrac{x^3+x^2+x+1}{3x+3}}$$
Exercise $$\PageIndex{31}$$
$$\displaystyle\lim_{x \rightarrow 0} \dfrac{x^2+2x+1}{3x^5-5x^3}$$
Exercise $$\PageIndex{32}$$
$$\displaystyle\lim_{t \rightarrow 7} \dfrac{t^2x^2+2tx+1}{t^2-14t+49}\text{,}$$ where $$x$$ is a positive constant
Exercise $$\PageIndex{33}$$
$$\displaystyle\lim_{d \rightarrow 0} x^5-32x+15\text{,}$$ where $$x$$ is a constant
Exercise $$\PageIndex{34}$$
$$\displaystyle\lim_{x \rightarrow 1} (x-1)^2\sin\left[\left(\dfrac{x^2-3x+2}{x^2-2x+1}\right)^2+15\right]$$
Exercise $$\PageIndex{35}$$(✳)
Evaluate $$\lim_{x\rightarrow 0} x^{1/101} \sin\big(x^{-100}\big)$$ or explain why this limit does not exist.
Exercise $$\PageIndex{36}$$(✳)
$$\lim\limits_{x\rightarrow2}\dfrac{x^2-4}{x^2-2x}$$
Exercise $$\PageIndex{37}$$
$$\displaystyle\lim_{x \rightarrow 5} \dfrac{(x-5)^2}{x+5}$$
Exercise $$\PageIndex{38}$$
Evaluate $$\lim_{t \to \frac{1}{2}}\dfrac{\frac{1}{3t^2}+\frac{1}{t^2-1}}{2t-1}$$.
Exercise $$\PageIndex{39}$$
Evaluate $$\lim_{x \to 0}\left( 3+\dfrac{|x|}{x}\right) \text{.}$$
Exercise $$\PageIndex{40}$$
Evaluate $$\lim_{d \to -4}\dfrac{|3d+12|}{d+4}$$
Exercise $$\PageIndex{41}$$
Evaluate $$\lim_{x \to 0}\dfrac{5x-9}{|x|+2}\text{.}$$
Exercise $$\PageIndex{42}$$
Suppose $$\displaystyle\lim_{x \rightarrow -1} f(x)=-1\text{.}$$ Evaluate $$\displaystyle\lim_{x \rightarrow -1} \dfrac{xf(x)+3}{2f(x)+1}\text{.}$$
Exercise $$\PageIndex{43}$$(✳)
Find the value of the constant $$a$$ for which $$\lim\limits_{x\rightarrow -2}\dfrac{x^2+ax+3}{x^2+x-2}$$ exists.
Exercise $$\PageIndex{44}$$
Suppose $$f(x)=2x$$ and $$g(x)=\frac{1}{x}\text{.}$$ Evaluate the following limits.
1. $$\displaystyle\lim_{x \rightarrow 0} f(x)$$
2. $$\displaystyle\lim_{x \rightarrow 0} g(x)$$
3. $$\displaystyle\lim_{x \rightarrow 0} f(x)g(x)$$
4. $$\displaystyle\lim_{x \rightarrow 0} \dfrac{f(x)}{g(x)}$$
5. $$\displaystyle\lim_{x \rightarrow 2} [f(x)+g(x)]$$
6. $$\displaystyle\lim_{x \rightarrow 0} \dfrac{f(x)+1}{g(x+1)}$$
Exercise $$\PageIndex{45}$$
The curve $$y=f(x)$$ is shown in the graph below. Sketch the graph of $$y=\dfrac{1}{f(x)}\text{.}$$
Exercise $$\PageIndex{46}$$
The graphs of functions $$f(x)$$ and $$g(x)$$ are shown in the graphs below. Use these to sketch the graph of $$\dfrac{f(x)}{g(x)}\text{.}$$
Exercise $$\PageIndex{47}$$
Suppose the position of a white ball, at time $$t\text{,}$$ is given by $$s(t)\text{,}$$ and the position of a red ball is given by $$2s(t)\text{.}$$ Using the definition from Section 1.2 of the velocity of a particle, and the limit laws from this section, answer the following question: if the white ball has velocity 5 at time $$t=1\text{,}$$ what is the velocity of the red ball?
Exercise $$\PageIndex{48}$$
Let $$f(x) = \frac{1}{x}$$ and $$g(x) = \frac{-1}{x}\text{.}$$
1. Evaluate $$\displaystyle\lim_{x \rightarrow 0} f(x)$$ and $$\displaystyle\lim_{x \rightarrow 0} g(x)\text{.}$$
2. Evaluate $$\displaystyle\lim_{x \rightarrow 0} [f(x)+g(x)]$$
3. Is it always true that $$\displaystyle\lim_{x \rightarrow a} [f(x)+g(x)]= \displaystyle\lim_{x \rightarrow a} f(x)+\displaystyle\lim_{x \rightarrow a} g(x)\text{?}$$
Exercise $$\PageIndex{49}$$
Suppose
\begin{align*} f(x) &=\begin{cases} x^2+3 & \text{ if } x\gt 0\\ 0 & \text{ if } x = 0\\ x^2-3 & \text{ if } x \lt 0 \end{cases} \end{align*}
1. Evaluate $$\displaystyle\lim_{x \rightarrow 0^-} f(x)\text{.}$$
2. Evaluate $$\displaystyle\lim_{x \rightarrow 0^+} f(x)\text{.}$$
3. Evaluate $$\displaystyle\lim_{x \rightarrow 0} f(x)\text{.}$$
Exercise $$\PageIndex{50}$$
Suppose
\begin{align*} f(x) &=\begin{cases} {\displaystyle\frac{x^2+8x+16}{x^2+30x-4}} & \text{ if } x\gt -4\\ x^3+8x^2+16x & \text{ if } x \le -4 \end{cases} \end{align*}
1. Evaluate $$\displaystyle\lim_{x \rightarrow -4^-} f(x)\text{.}$$
2. Evaluate $$\displaystyle\lim_{x \rightarrow -4^+} f(x)\text{.}$$
3. Evaluate $$\displaystyle\lim_{x \rightarrow -4} f(x)\text{.}$$
This page titled 1.4: Calculating Limits with Limit Laws is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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# What Is The Median Of These Numbers?
## How do you find the mean median and mode?
– Mode-The most repetitive number.
– Median:The number in the MIDDLE when they are IN ORDER.
– Mean- The AVERAGE OF ALL NUMBERS: You add up all the numbers then you divide it by the TOTAL NUMBER of NUMBERS.
– Range – THE BIGGEST minus the Smallest!.
## What is the median of 30?
=> (30 -14) i.e 16 or more of the numbers are equal to three so the median is 3.
## What is the median of 77 and 84?
80.580.5 is an average of two given numbers 77 & 84.
## How do you find the median example?
For example, the median of 3, 3, 5, 9, 11 is 5. If there is an even number of observations, then there is no single middle value; the median is then usually defined to be the mean of the two middle values: so the median of 3, 5, 7, 9 is (5+7)/2 = 6.
## What is the median of 5 and 6?
For the data set 1, 1, 2, 5, 6, 6, 9 the median is 5. For the data set 1, 1, 2, 6, 6, 9 the median is 4.
## What is the median of 52?
Mode is 52. The median is the middle number. In this case with 8 numbers it is a bit tricky. However, if you get rid of the 3 highest numbers (61, 58, and 52) and get rid of the 3 lowest numbers ( 41, 49 and 49) you are left with 52 and 52.
## What is the median of a set of numbers?
The median of a set of numbers is the middle number in the set (after the numbers have been arranged from least to greatest) — or, if there are an even number of data, the median is the average of the middle two numbers.
## What is the median of 7?
Median: The middle number; found by ordering all data points and picking out the one in the middle (or if there are two middle numbers, taking the mean of those two numbers). Example: The median of 4, 1, and 7 is 4 because when the numbers are put in order (1 , 4, 7) , the number 4 is in the middle.
## How do I calculate the median?
MedianArrange your numbers in numerical order.Count how many numbers you have.If you have an odd number, divide by 2 and round up to get the position of the median number.If you have an even number, divide by 2. Go to the number in that position and average it with the number in the next higher position to get the median.
## What is the median of 6 numbers?
To find the median of a group of numbers: Arrange the numbers in order by size. If there is an odd number of terms, the median is the center term. If there is an even number of terms, add the two middle terms and divide by 2.
## How do u find the median of two numbers?
If there is an even number of numbers add the two middles and divide by 2. The result will be the median.
## How do you find the median quickly?
To find the median, put all numbers into ascending order and work into the middle by crossing off numbers at each end. If there are a lot of items of data, add 1 to the number of items of data and then divide by 2 to find which item of data will be the median.
## How do you find the median of a big set of numbers?
To find the median, first order the numbers from smallest to largest. Then find the middle number. For example, the middle for this set of numbers is 5, because 5 is right in the middle: 1, 2, 3, 5, 6, 7, 9.
## What is the median of first 10 whole numbers?
We get the average of 4 and 5 as 4+52=92=4.5. So, we have found the median of the first 10 whole numbers as 4.5. ∴ The median of the first 10 whole numbers is 4.5. Note: We should make sure that the terms have to be written in ascending order before finding the median of any number of terms.
## What is the median of 74?
Using the formula for median when there is an even number of values, we need to take the mean value of the n/2 ‘th and (n+2)/2 ‘th values. So that’s the 8th and the 9th values, which are 71 and 74 respectively. Then we need to take the mean of these values: (71 + 74) / 2 = 145 / 2 = 72.5 . So the median is 72.5.
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# Tips for Teaching Ratios and Rates in 6th, 7th and 8th Grade Math
Are your students hitting a bit of a roadblock when it comes to wrapping their heads around ratios and rates? Need a fun and colorful way to dive into these math essentials? Well, you’re in for a treat! In this blog post, I’m about to unveil how I teach Ratios and Rates using guided notes on a Math Wheel. It’s a lively, interactive graphic organizer that swoops in to make ratios and rates a walk in the park for your students.
So, if you’re on the lookout for a fresh and exciting way to tackle these fundamental math concepts, you’ve come to the right place. Plus, stick around to the end for some bonus resources ready to be used in your classroom today!
## How to Use the Guided Notes Math Wheel for Ratios and Rates
Liven up your lessons with these guided notes using the Ratios and Rates Math Wheel! This math wheel is perfect to use as you introduce and teach this topic or save and use as a review of the topic later in the unit. However you choose, this ratios and rates math wheel is sure to help your students understand this concept. It’s a graphic organizer and a handy study tool that students can stash in their notebooks for easy access. This specific math wheel is split into four parts!
### 1. What is a…
Let’s break down how I kick off the Ratios and Rates Math Wheel. First things first, I introduce students to our concept and we identify the topic found in the middle of the math wheel. Students love coloring in “Ratios and Rates” as they start to bring their math notes to life.
Next, we dive into the “What is a…” section, breaking down the definitions of ratios and rates.
We discuss how ratios compare two quantities, whether it’s part-to-part or part-to-whole, jotting down the definition in that section.
Then, we move on to add the definition of a rate. A rate, I explain, is essentially a ratio with a twist. It compares quantities with units of measure like inches, feet, or miles.
These definitions serve as the foundation, especially for students stepping into this concept for the first time.
### 2. Writing Ratios and Rates
Let’s dive into the next leg of our guided notes journey, which is the section titled Writing Ratios and Rates. In this section, I break down the process of expressing ratios and rates step by step.
First things first, we highlight the significance of the word “to” when jotting down these ratios. I then draw an arrow next to the line, and we work through an example. For example, 12 to 5. Next, we answer the question asking us to identify the ratio of stars to squares. On the math wheel, there are 3 stars and 2 squares, so we would write 3 to 2. I make sure to remind students that the item named first is always the first number in the ratio.
Building on that foundation, we step up our game by practicing how to express these ratios as fractions. Using our initial example of 12 to 5, we transform it into the fraction 12/5. Then we answer the next question of the ratio of squares to shapes. Again, I remind them that whichever item is named first in the sentence or question is first in the ratio. Since we have 2 squares and a total of 5 shapes, the fraction is 2/5.
We finish this section of the math wheel by exploring the third way to express ratios and rates by using a colon. Revisiting our first example, 12 to 5 becomes 12:5. If we’re pulling in units of measure, it might look something like 1 yard: 3 feet.
As we work through the math wheel, I maintain an ongoing check-in with my students to ensure we’re all on the same page and adjusting pacing as needed. We can always complete extra examples, revisit definitions or I can explain in a different way if needed.
### 3. Equivalent Fractions
In the third section of the Ratios and Rates Math Wheel, we dive into equivalent ratios. I break it down for my students, explaining that these ratios represent the same comparison of quantities. Since my students are already familiar with the concept of equivalent fractions I use that as a comparison.
I explain how we can find equivalent ratios by either multiplying or dividing both numbers in the ratio by the same number. We make sure to add this important information to the math wheel before working on the examples.
To practice, we return to familiar territory with our stars and squares. I remind my students that whatever is first in the list must be the first in the ratio. In this case, it’s the stars. With a total of 6 stars and 2 squares, we can write the ratio as 6:2. Once we have our starting ratio we set out to make some equivalent ratios. I generally do 2 or 3 examples with at least 1 using multiplication and 1 using division if that is possible. Here are some examples and how we do the math to find the equivalent ratios.
### Examples
1. Start with the ratio 6:2 and multiply each number by 2. It looks like this (6 x 2 = 12 and 2 x 2 = 4). That means 12:4 or 12 to 4 is an equivalent ratio of 6:2.
2. Start with the ratio 6:2 and multiply each number by 3. It looks like this (6 x 3 = 18 and 2 x 3 = 6). That means 18:6 or 18 to 6 is an equivalent ratio of 6:2.
3. Start with the ratio 6:2 and divide each number by 2. It looks like this (6 ➗ 2 = 3 and 2 ➗ 2 = 1). That means 3:1 or 3 to 1 is an equivalent ratio of 6:2.
Once we practice this step a few times, we move on to writing equivalent ratios for squares to shapes. We start with the ratio 2:8 because there are 2 squares and 8 total shapes.
Then, just as we did in the first example, we will find some equivalent ratios by multiplying and dividing both numbers by the same number. After we do these examples together, I have students practice what they have learned. There are ten practice problems around the wheel for us to practice the steps further.
### 4. Ratio Tables
In the final section of our guided notes math wheel, I dive into ratio tables to help organize equivalent ratios. We use the examples from the previous section to practice using tables in this section.
In the first example, we see that the first line is filled in and talks about where that information came from. The students are quick to share that there are 12 inches in 1 foot. We then use that to help us fill in the remaining sections of the table.
I make sure to model how I find the number we are multiplying by or dividing by in order to complete the table. In this example I ask the students this question: “What do I need to multiply 1 by in order to get the answer of 2?” When they tell me 2, then I remind them that they must do the same thing to the other number in the ratio. We then multiply 12 x 2 and add 24 to the table.
I continue this process for each row because I want them to begin asking themselves that question to work through ratio tables.
Next, we take on the challenge of creating a table for squares, stars, and the total number of shapes. The first column helps start because it is pre-filled with 2, 6, and 8. Multiply 2, 6, and 8 each by 2, and boom – the third column is completed, showcasing products: 4, 12, 16. Next, we multiply 2, 6, and 8 each by 3, and the fourth column becomes filled with 6, 18, and 24.
To wrap it up, we sprinkle in some division. Divide 2, 6, and 8 each by 2, and voila, the fifth column is completed with 1, 3, and 4.
## Teaching Ratios and Rates Has Never Been Easier
By working through the four sections of the Ratios and Rates Math Wheel your students will have a solid understanding of this concept, how to find equivalent ratios, and how to display ratios in a table.
When the math wheel is complete, students complete the practice problems found in the pattern around the circles. Depending on how students seem to be feeling about the concepts, we may do this as guided practice or independent practice. Then, students can color the background pattern if they’d like. After the wheels are finished, I have students keep them in their math notebooks. That way they can refer to them as needed or use them later in the year as a review. Whether deep into the unit, exploring equivalent ratios, or cruising through ratio tables, the math wheel provides structure and clarity. It becomes a reliable reference point for students, offering a visual aid and quick reminders whenever needed. Its condensed yet detailed format transforms complex concepts into an easily digestible visual aid!
## More Resources for Teaching Ratios and Rates
Looking for more easy to use resources to make teaching and learning ratios and rates a breeze? Make sure to check out the following resources to help as you teach these important math concepts.
## Ready for More Math Wheels?
Want to try out a math wheel for free? Grab this free Fraction Operations Math Wheel and give it a spin in your classroom.
## Save for Later
Remember to save this post to your favorite math Pinterest board for when you need guided notes or resources for ratios and rates!
## Ellie
### How to Use Math Small Groups in Middle School
Welcome to Cognitive Cardio Math! I’m Ellie, a wife, mom, grandma, and dog ‘mom,’ and I’ve spent just about my whole life in school! With nearly 30 years in education, I’ve taught:
• All subject areas in 4th and 5th grades
• Math, ELA, and science in 6th grade (middle school)
I’ve been creating resources for teachers since 2012 and have worked in the elearning industry for about five years as well!
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# SOLUTION: How do I graph 1/2x-4y=6?
Algebra -> Algebra -> Linear-equations -> SOLUTION: How do I graph 1/2x-4y=6? Log On
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Algebra: Linear Equations, Graphs, Slope Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Linear-equations Question 476141: How do I graph 1/2x-4y=6?Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!I find it easy to work with the slope intercept form of an equation to make the graph. The slope intercept form is: . . where m is the slope (the rise, a positive or negative value, divided by the horizontal distance associated with the rise). And b is the value on the y-axis where the graph crosses. . So let's begin by starting with the given equation, and working it around to get it into the slope intercept form. . You were given: . . Let's get the term with the y in it to be by itself on the left side of the equation. We can do that by subtracting from both sides of the equation to get: . . Next solve for y by dividing both sides of this equation by -4 and you have: . . This simplifies to: . . By comparing this with the equation for the point slope form, you can see that the point (b) is -3/2 which means that the graph crosses the y-axis where y is equal to minus 3/2. And you can also see that the slope (m) is the multiplier of x and it equals 1/8. This means that the graph rises 1 unit for every 8 units you go horizontally to the right. Since the slope is positive, the graph rises as you go to the right. If the slope were negative the graph would go down as you went to the right. . Next you can start at -3/2 on the y-axis. From this point move 8 units horizontally to the right. You are now at the point (8, -3/2). But since you moved to the right 8 units you know that the graph rises 1 unit (or 2/2 which is equal to 1). Therefore, from -3/2 you go up +2/2 which puts you at -1/2. You are now at the point (8, -1/2) and this point is on the graph. So you now have two points on the graph ... (0, -3/2) and (8, -1/2). If you plot those two points you can then complete the graph by using a straight edge to draw the line connecting the two points and extending beyond them in both directions. . When you get done, your graph should look like this: . . Hope this helps you to understand graphing of straight lines a little better.
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```Chapter 6
Continuous Random Variables
McGraw-Hill/Irwin
Continuous Random
Variables
6.1
6.2
6.3
6.4
Continuous Probability Distributions
The Uniform Distribution
The Normal Probability Distribution
Approximating the Binomial Distribution by
Using the Normal Distribution (Optional)
6.5 The Exponential Distribution (Optional)
6.6 The Normal Probability Plot (Optional)
6-2
LO 1: Explain what a
continuous probability
distribution is and how it
is used.
6.1 Continuous
Probability Distributions
A continuous random variable may assume any
numerical value in one or more intervals
Use a continuous probability distribution to assign
probabilities to intervals of values
The curve f(x) is the continuous probability
distribution of the random variable x if the probability
that x will be in a specified interval of numbers is the
area under the curve f(x) corresponding to the
interval
Other names for a continuous probability distribution
are probability curve and probability density function
6-3
LO1
Properties of Continuous
Probability Distributions
Properties of f(x): f(x) is a continuous
function such that
1.
2.
f(x) > 0 for all x
The total area under the f(x) curve is equal to 1
Essential point: An area under a continuous
probability distribution is a probability
6-4
LO 2: Use the uniform
distribution to compute
probabilities.
6.2 The Uniform
Distribution
If c and d are numbers on the real line (c < d), the
probability curve describing the uniform distribution is
1
for c x d
f x = d c
0
otherwise
The probability that x is any value between the values a
and b (a < b) is
ba
P a x b
d c
Note: The number ordering is c < a < b < d
6-5
LO2
The Uniform Distribution
Continued
The mean mX and standard deviation sX of a
uniform random variable x are
cd
mX
2
d c
sX
12
These are the parameters of the uniform
distribution with endpoints c and d (c < d)
6-6
LO 3: Describe the
properties of the normal
distribution and use a
cumulative normal table.
6.3 The Normal
Probability Distribution
The normal probability distribution is defined by the
equation
f( x) =
1
σ 2π
1 x m
2 s
e
2
for all values x on the real number line
m is the mean and s is the standard deviation
= 3.14159… and e = 2.71828 is the base of natural
logarithms
6-7
LO3
The Normal Probability
Distribution
Continued
The normal curve is symmetrical
The mean is in the middle under the
curve
So m is also the median
It is tallest over its mean m
The area under the entire normal
curve is 1
The area under either half of the curve
is 0.5
6-8
LO 4: Use the normal
distribution to compute
probabilities.
Find P(0 ≤ z ≤ 1)
The Standard Normal
Table Example
Find the area listed in the table corresponding to a z value
of 1.00
Starting from the top of the far left column, go down to “1.0”
Read across the row z = 1.0 until under the column headed
by “.00”
The area is in the cell that is the intersection of this row
with this column
As listed in the table, the area is 0.3413, so
P(0 ≤ z ≤ 1) = 0.3413
6-9
LO4
Calculating P(-2.53 ≤ z ≤ 2.53)
First, find P(0 ≤ z ≤ 2.53)
Go to the table of areas under the standard normal curve
Go down left-most column for z = 2.5
Go across the row 2.5 to the column headed by .03
The area to the right of the mean up to a value of z = 2.53
is value in cell that is intersection of 2.5 row and.03 column
The table value for the area is 0.4943
By symmetry, this is also the area to the left of the
mean down to a value of z = –2.53
Then P(-2.53 ≤ z ≤ 2.53) = 0.4943 + 0.4943 =
0.9886
6-10
LO 5: Find population
values that correspond
to specified normal
distribution probabilities.
1.
2.
3.
Finding Normal
Probabilities
Formulate the problem in terms of x values
Calculate the corresponding z values, and restate
the problem in terms of these z values
Find the required areas under the standard normal
curve by using the table
Note: It is always useful to draw a picture showing
the required areas before using the normal table
6-11
LO 6: Use the normal
distribution to
approximate binomial
probabilities (optional).
6.4 Approximating the Binomial
Distribution by Using the
Normal Distribution (Optional)
The figure below shows several binomial
distributions
Can see that as n gets larger and as p gets closer to
0.5, the graph of the binomial distribution tends to
have the symmetrical, bell-shaped, form of the
normal curve
6-12
LO 7: Use the
exponential distribution
to compute probabilities
(optional).
Suppose that some event occurs as a Poisson
process
That is, the number of times an event occurs is a Poisson
random variable
Let x be the random variable of the interval between
successive occurrences of the event
6.5 The Exponential
Distribution (Optional)
The interval can be some unit of time or space
Then x is described by the exponential distribution
With parameter l, which is the mean number of events that
can occur per given interval
6-13
LO7
The Exponential Distribution
Continued
If l is the mean number of events per given
interval, then the equation of the exponential
distribution is
for x 0
le lx
f x =
0
otherwise
The probability that x is any value between
given values a and b (a<b) is
Pa x b e la e lb
and
Px c 1 e lc and Px c e lc
6-14
LO 8: Use a normal
probability plot to help
decide whether data
come from a normal
distribution (optional).
6.6 The Normal
Probability Plot
A graphic used to visually check to see if
sample data comes from a normal distribution
A straight line indicates a normal distribution
The more curved the line, the less normal the
data is
6-15
```
Related documents
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# Lesson 3Reasoning about Contexts with Tape Diagrams (Part 2)
Let’s see how equations can describe tape diagrams.
### Learning Targets:
• I can match equations and tape diagrams that represent the same situation.
• If I have an equation, I can draw a tape diagram that shows the same relationship.
## 3.1Find Equivalent Expressions
Select all the expressions that are equivalent to . Explain how you know each expression you select is equivalent.
## 3.2Matching Equations to Tape Diagrams
1. Match each equation to one of the tape diagrams. Be prepared to explain how the equation matches the diagram.
2. Sort the equations into categories of your choosing. Explain the criteria for each category.
## 3.3Drawing Tape Diagrams to Represent Equations
1. Draw a tape diagram to match each equation.
1. Use any method to find values for and that make the equations true.
### Are you ready for more?
To make a Koch snowflake:
• Start with an equilateral triangle that has side lengths of 1. This is step 1.
• Replace the middle third of each line segment with a small equilateral triangle with the middle third of the segment forming the base. This is step 2.
• Do the same to each of the line segments. This is step 3.
• Keep repeating this process.
1. What is the perimeter after step 2? Step 3?
2. What happens to the perimeter, or the length of line traced along the outside of the figure, as the process continues?
## Lesson 3 Summary
We have seen how tape diagrams represent relationships between quantities. Because of the meaning and properties of addition and multiplication, more than one equation can often be used to represent a single tape diagram.
Let’s take a look at two tape diagrams.
We can describe this diagram with several different equations. Here are some of them:
• , because the parts add up to the whole.
• , because addition is commutative.
• , because if two quantities are equal, it doesn’t matter how we arrange them around the equal sign.
• , because one part (the part made up of 4 ’s) is the difference between the whole and the other part.
For this diagram:
• , because multiplication means having multiple groups of the same size.
• , because multiplication is commutative.
• , because division tells us the size of each equal part.
## Lesson 3 Practice Problems
1. Solve each equation mentally.
2. Complete the magic squares so that the sum of each row, each column, and each diagonal in a grid are all equal.
3. Draw a tape diagram to match each equation.
4. Select all the equations that match the tape diagram.
5. Each car is traveling at a constant speed. Find the number of miles each car travels in 1 hour at the given rate.
1. 135 miles in 3 hours
2. 22 miles in hour
3. 7.5 miles in hour
4. miles in hour
5. miles in hour
|
## Fractions
2.1 Fractions
We will learn to solve basic operations involving fractions. An improper fraction is a fraction with the numerator equal to or greater than the denominator. As example, $$\dfrac{24}{5}$$ is an improper fraction because the value of the numerator ($$\color{red}{24}$$) is greater than the value of the denominator ($$\color{red}{5}$$). A mixed number consists of a whole number and a proper fraction. As example,$$\dfrac{24}{5}$$ can be converted to mixed numbers as follows: Based on the above picture, $$\dfrac{24}{5}=\color{magenta}{4\dfrac{4}{5}}$$. Or, the simple steps to convert improper fractions to mixed numbers are as follows: \begin{aligned}\space \space 4 \\5\space\overline{)2\space4} \\\underline{-\space2\space0}\\4 \end{aligned}$$=\color{magenta}{4\dfrac{4}{5}}$$ The simple steps to convert mixed numbers to improper fractions are as follows: \begin {aligned} \color{black}{4\dfrac{4}{5}} &= \dfrac {4 \times 5 +4}{5} \\ &= \color{magenta}{\dfrac {24}{5}} \end {aligned} To add fractions with the same denominators, add the numerators without changing the denominators. As example, \begin{aligned} \dfrac{1}{5}+\dfrac{2}{5}&=\dfrac{1+2}{5} \\&=\color{red}{\dfrac{3}{5}} \end{aligned} To add fractions with different denominators, find the common multiple value of the denominator. As example, \begin{aligned} \dfrac{1}{6}+\dfrac{3}{4}&=\dfrac{1{\color{green}{\times2}}}{6{\color{green}{\times2}}}+\dfrac{3{\color{green}{\times3}}}{4{\color{green}{\times3}} }\\&=\dfrac{2}{12}+\dfrac{9}{12} \\&={\color{magenta}{\dfrac{11}{12}}} \end{aligned} To subtract fractions with the same denominator, the subtraction operation is performed as usual. The denominator is retained. As example, $$1-\dfrac{2}{5}=\underline{\hspace{1cm}}$$ Thus, $$1-\dfrac{2}{5}=\color{magenta}{\dfrac{3}{5}}$$. To subtract the fractions with different denominators, change to equivalent fractions with the same denominators before subtracting the numerators. Retain the denominator. As example, \begin{aligned} \dfrac{3}{5}-\dfrac{1}{2}&=\dfrac{3\color{green}{\times2}}{5\color{green}{\times2}}-\dfrac{1\color{green}{\times5}}{2\color{green}{\times5}} \\&=\dfrac{6}{10}-\dfrac{5}{10} \\&={\color{magenta}{\dfrac{1}{10}}} \end{aligned} For mixed operations involving the addition and subtraction of fractions, the calculation is done from the left to the right. As example, \begin{aligned} \dfrac{1}{7}+\dfrac{5}{7}-\dfrac{2}{7}&=\dfrac{1+5-2}{7} \\&={\color{red}{\dfrac{4}{7}}} \end{aligned} 'from' means multiply As example, find the value of $$\dfrac{5}{7}$$ from $$56$$. \begin{aligned} \dfrac{5}{7}\text{ from }56&=\dfrac{5}{7}\times56 \\&=\dfrac{5\times56}{7} \\&=\dfrac{280}{7} \\&={\color{red}{40}} \end{aligned}
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# Linear Programming in Excel
This tutorial demonstrates how to use Excel to tackle integer linear programming problems. With the Solver add-in in Microsoft Excel, users may find the solution for an integer linear program rapidly. We will walk through this process in simple, straightforward steps today. The final portion of this tutorial will also see an example of a mixed-integer linear programming problem. So, let's begin without delay.
## Integer Linear Programming
Integer linear programming uses integer variables, linear objective functions, and equations. The minimal or maximum result of a particular problem under certain conditions can be found using linear programming. It is an instrument that can be utilized to find the most effective strategy to use the scarce resources. There are several key components to all forms of linear programming. Below are their details:
1. Decision Variables: We ascertain which choices will reduce or maximize the goal function.
2. Objective Function: The objective function aids our ability to identify the decision variables. It conveys how the variables and the outcome are related.
3. Constraints: Constraints are additional functions that indicate various requirements for potential resolutions.
In addition to integer linear programming, a mixed-integer linear programming example will be shown. There are continuous and integer variables in mixed-integer linear programming.
## Excel Step-by-Step Instructions for Solving Integer Linear Programming
We'll use an example to demonstrate step-by-step methods. To understand the essential concepts, you must attentively read them. You must carefully read the question and determine the Objective function and Constraints.
Let's say that two goods that can be swapped out are produced by a machine. The machine can produce ten units of item 2 and a maximum of twenty units of item 1 each day. Another option is to set the machine to create a maximum of 22 units of item 1 and 18 units of item 2 daily. According to market analysis, the overall daily demand for the two goods is 40 units. When choosing between the two machine settings, which should be chosen given that the unit earnings for the two items are \$15 and \$18?
### Step 1: Examine the Question and Establish a Dataset
• Before anything else, we must fully comprehend and thoroughly examine the provided integer linear programming problem.
• The results of the analysis of the query above are shown below.
Decision Variables:
• X1: The product's production amount
• X2: The amount of product produced 2.
• Y: 1 in the case of the first setting picked, 0 in the case of the second setting selected.
Objective Function:
In this case, the objective function is:
Constraints:
The question above primarily leads us to three limits. These are:
• X1+X2<=40
According to the market analysis, 40 combined units of the two items are the maximum daily demand.
• 4X1-5Y<= 22
For product 1, only this constraint applies.
• X2 + 12 Y<= 18
For the second product, it serves as a limitation.
• Y = {0,1}
Y will either be zero or one.
• X1, X2> = 0
There cannot be a negative quantity of products.
• Considering the constraints, functions, and variables, we generated a dataset in the subsequent step that looks like the image below. You are free to modify it to suit your needs.
• Second, we must open Excel and load the Solver add-in. You can proceed to step 3 when the data has already been in Excel.
• Click the File tab to accomplish this.
• Next, choose Options by clicking on it in the left-bottom corner of the screen.
• Excel Options will open as a result.
• Select Add-ins from the Excel Options windows.
• Next, in the Manage box, choose Excel Add-ins, then click Go.
• Choose OK from the notification box after checking the Solver Add-in.
• Finally, you can see the Solver feature under the Analysis section of the Data tab.
### Step 3: Fill in the Objective Function and Constraints Coefficients
• Thirdly, the dataset's objective function and constraints must be filled in.
• This is where we will mostly insert the objective function and constraint coefficients.
• 2X1+3X2<=40 is our initial constraint. This indicates that the products' total should equal 40 or fewer if the initial setting is used.
• For X1 and X2, respectively, the coefficients are 2 and 3.
• Furthermore, the coefficient of Y is 1 because the equation shows the initial configuration.
• It has a <= sign.
• Furthermore, 40 is the upper limit.
• Follow the previous instructions again, entering the coefficients for each constraint.
• Next, choose Cell D10 and enter the following formula:
The SUMPRODUCT function was utilized in this formula to compute the product of the decision variables and the relevant constraints variables, which were then added. Cell A6 will multiply by Cell A10, Cell B6 by Cell B10, & Cell C6 by Cell C10. Subsequently, every product will be combined.
• Press Enter and drag the Fill Handle.
• Now, complete Cells A16 through B16 with the objective function's coefficients.
• Z = 6X1+8X2 is the objective function in our scenario.
• Choose Cell D16 once more, then enter the following formula:
• After inputting the coefficients and formulas, hit Enter to view a dataset like the one below.
### Step 4: Enter the parameters for the Solver
• Step 4 involves selecting Solver from the AnalysIis section under the Data The window for the solver parameters will open.
• You must enter the Cell holding the objective function's value within the Set Objective box.
• Thus, we have \$D\$16 typed here.
• In this case, maximizing the outcome is our goal.
• On the type \$A\$6:\$C\$6 in the "By Changing Variable Cells," the variables for making decisions are included.
### Step 5: Add in Subject to Constraints
• Subjects must be added to the constraints in the fifth stage.
• The relationship between the restrictions and the variable type-binary or integer-must be indicated.
• Select the bin using the drop-down option after typing \$C\$6 within the Cell Reference box of the Add Constraint dialogue.
• Y is either 0 or 1. Cell C6 contains this value. It suggests binary numbers. We chose this particular bin for this reason.
• Click OK to continue.
• This time, fill out the Cell Reference box with \$D\$10:\$D\$12, select the <= symbol through the drop-down menu, & enter =\$F\$10:\$F\$12 in its Constraint box.
• Next, press OK.
• From the Solver Parameters window, click Add once more.
• Enter \$A\$6:\$B\$6 into the Cell Reference field now, and choose it in the drop-down list.
• The values of integers X1 and X2 are stored in cells A6 and B6.
• Once more, press OK.
### Step 6: Choose the Method for Solving
• When prompted to "Select a Solving Method," in step 6, choose Simplex LP and press Solve.
• A check for "Make Unconstrained Variables Non-Negative" should be made.
• The Solver Outcomes window will open after selecting Solve.
• After that, choose OK.
### Step 7: The Integer Linear Programming Solution
• In the Excel sheet, you will ultimately locate the solutions in the cells you want.
• We will get the greatest results in this instance using the second machine option.
### Step 8: Create an Answer Report
• You can also generate the response report.
• Choose Answer from the Reports area of the Solver Results box and click OK to accomplish that.
• The report is finally located on another sheet.
## Example of Mixed Integer Linear Programming in Excel
This part will discuss a straightforward Excel mixed-integer linear programming example. The easy steps for solving mixed-integer linear programming using Excel are as follows. Let us examine the objective function & its corresponding constraints in this scenario.
The Objective Function:
• Z = 2.52X1+1.55X2+2.55X3 + 250Y1 + 300Y2 + 400Y3
Constraints:
• X1-350Y1<=0
• X2-400Y2< = 0
• X3-450Y3<=0
• X1+X2+X3=1000
In this instance, the variables X1, X2, and X3 were integers. However, Y1, Y2, & Y3 are binary numbers. We also need to determine Z's lowest value.
Let's take the steps listed below to understand everything there is to know about the example.
STEPS:
• Construct a dataset to hold the Decision Variable, Constraint, and Objective Function coefficients.
• Second, look up the mixed coefficients for the Objective Function's variables.
• Thirdly, enter the coefficients for the Constraint variables, as shown in the image below. Let the Total column remain blank.
• Next, choose Cell G10 and input the following formula:
• To drop the fill handle, press Enter.
• Enter the following formula in cell G6 now:
• Press Enter.
• Go to the Data tab and choose Solver in the next step. The window for Solver Parameters will open as a result.
• Make the objective in Cell \$G\$6 to Min within the Solver Parameters window by adjusting the variable cells \$A\$6:\$F\$6.
• Add constraints one at a time, and select Simplex LP for the solution method.
• For further action, click Solve.
• It will then bring up the Solver Results window.
• Next, click OK.
• The outcome will eventually resemble the image below.
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# Aptitude Logical Reasoning Time and Distance 2019 CMAT Part 5
1. A boy goes to his school from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming. The distance between his house and school is?
A. 5 km
B. km
C. 6 km
D. km
Ans: C
Explanation:
Average speed = km/hr.
Distance travelled = = 12 km.
Distance between house and school = = 6 km.
2. A man on tour travels first 160 km at 64 km/he and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is?
A. km/hr
B. 36 km/hr
C. km/hr
D. 71 km/hr
Ans: C
Explanation:
Total time taken=
Average speed =
3. A boy rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. His average speed for the entire trip is approximately?
A. km/hr
B. km/hr
C. 11 km/hr
D. km/h
Ans: B
Explanation:
Total distance travelled = km /hr.
Total time taken = hrs.
Average speed = km/hr.
4. Robert is traveling on his cycle and has calculated to reach point A at 2 p.m. if he travels at 10 km/hr; he will reach there at 12 noon if he travels at 15 km/hr. At what speed must he travel to reach A at 1 p.m.?
A. 8 kmph
B. 11 kmph
C. 12 kmph
D. 14 kmph
Ans: C
Explanation:
Let the distance travelled be x km.
Then,
km.
Time taken to travel 60 km at 10 km/hr = = 6 hrs.
So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.
Required speed = = 12 kmph.
5. A train can travel 50% faster than a car. Both start from point A at the same time and reach point B at 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is?
A. 100 km/hr
B. 110 km/hr
C. 120 km/hr
D. 130 km/hr
Ans: C
Explanation:
Let speed of the car be x km/hr.
Then, speed of the train = .
km/hr.
6. Excluding stoppages, the speed of a bus is 54 km/hr and including stoppages, it is 45 km/hr. For how many minutes does the bus stop per hour?
A. 9
B. 10
C. 12
D. 20
Ans: B
Explanation:
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = = 10 min.
7. In a covering a certain distance, the speeds of A and B are in the ratio of 3:4. A takes 30 minutes more than B to reach the destination. The time taken by A to reach the destination is?
A. 1 hour
B. hour
C. 2 hour
D. hour
Ans: C
Explanation:
Ratio of speeds =
Ratio of times taken =
Suppose A takes hrs and B takes hrs to reach the destination.
Then,
Time taken by
8. In a covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay double his speed, then he would take 1 hour less than Sammer. Abhay's speed is?
A. 5 km/hr
B. 6 km/hr
C. km/hr
D. km/hr
Ans: A
Explanation:
Let Abhay's speed be x km//hr.
Then,
km/hr.
9. With a uniform speed a car covers the distance in 8 hours. Had the speed been increased by 4 km/hr, the same distance could have been covered in 7 hours. What is the distance covered?
A. 420 km
B. 480 km
C. 640 km
D. Cannot be determined
E. None of these
Ans: B
Explanation:
Let the distance be x km. Then,
km.
10. If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is?
A. 50 km
B. 56 km
C. 70 km
D. 80 km
Ans: A
Explanation:
Let the actual distance travelled be x km. Then,
km.
11. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is?
A. 1 hour
B. 2 hour
C. 3 hour
D. 4 hour
Ans: A
Explanation:
Let the duration of the flight be x hours. Then,
hr.
12. It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the car is?
A.
B.
C.
D.
Ans: C
Explanation:
Let the speed of the train be x km/hr and that of the car be y km/hr.
Then, --- (i)
And, --- (ii)
Solving (i) and (ii), we get x = 60 and y = 80
Ratio of speeds =
13. A walks around a circular field at the rate of one round per hour while B runs around it at the rate of six rounds per hour. They start in the same direction from the same point at 7.30 a.m. They shall first cross each other at?
A. 7.42 a.m.
B. 7.48 a.m.
C. 8.10 a.m.
D. 8.30 a.m.
Ans: A
Explanation:
Since A and B move in the same direction along the circle, so they will first meet each other when there is a difference of one round between the two.
Relative speed of A and B = rounds per hour.
Time taken to complete one round at this speed = hr = 12 min.
14. A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes?
A. 100 m
B. 150 m
C. 190 m
D. 200 m
Ans: A
Explanation:
Relative speed of the thief and policeman = km/hr.
Distance covered in 6 minutes = m.
Distance between the thief and policeman m.
15. A thief steals at a car at 2.30 p.m. and drives it at 60 km/hr. The theft is discovered at 3 p.m. and the owner sets off in another car at 75 km/hr. When will he overtake the thief?
A. 4.30 p.m.
B. 4.45 p.m.
C. 5 p.m.
D. 5.15 p.m.
Ans: C
Explanation:
Suppose the thief is overtaken x hrs after 2.30 p.m.
Then, distance covered by the owner in hrs.
hrs.
So, the thief is overtaken at 5 p.m.
16. The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet?
A. 10 a.m
B. 10.30 a.m
C. 11 a.m
D. 11.30 a.m
Ans: C
Explanation:
Suppose they meet x hrs after 8 a.m. Then,
So, they meet at (8 + 3) i.e., 11 a.m.
17. The jogging track in a sports complex is 726 m in circumference. Deepak and his wife start from the same point and walk in opposite directions at 4.5 km/hr and 3.75 km/hr respectively. They will meet for the first time in?
A. min
B. min
C. min
D. 6 min
Ans: B
Explanation:
Clearly, the two will meet when they are 726 m apart.
To be km apart, they take 1 hour.
To be 726 m apart, they take hrs = min = min.
18. A and B walk around a circular track. They start at 8 a.m. from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other before 9.30 a.m. ?
A. 5
B. 6
C. 7
D. 8
Ans: C
Explanation:
Relative speed = rounds per hour.
So, they cross each other 5 times in an hour and 2 times in half an hour.
Hence, they cross each other 7 times before 9.30 a.m.
19. Two cars P and Q start at the same time from A and B which are 120 km apart. If the two cars travel in opposite directions, they meet after one hour and if they travel in same direction (from A towards B), then P meets Q after 6 hours. What is the speed of car P?
A. 60 km/hr
B. 70 km/hr
C. 120 km/hr
E. None of these
Ans: B
Explanation:
Let their speed be x km/hr and y km/he respectively.
Then, --- (i)
Now, when they move in same direction:
--- (ii)
Solving (i) and (ii), we get ,
P's speed = 70 km/hr.
20. Two trains start from P and Q respectively and travel towards each other at a speed of 50 km/hr and 40 km/hr respectively. By the time they meet, the first train has travelled 100 km more than the second. The distance between P and Q is?
A. 500 km
B. 630 km
C. 660 km
D. 900 km
Ans: D
Explanation:
At the time of meeting, let the distance travelled by the second train be x km. Then, distance covered by the first train is (x + 100) km.
So, distance between P and Q km.
Doorsteptutor material for CMAT is prepared by worlds top subject experts- fully solved questions with step-by-step exaplanation- practice your way to success.
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# Missing Number Math Test: If You Are A Genius, What Is The Missing Number?
Missing Number Math Test: If you would like to know your intelligence level, take a short break and solve this riddle.
Today’s challenge will stimulate your thinking. In the image appearing on your screen, there is a mathematical challenge. It is difficult for the majority of Internet users to fail. If you have good analytical skills, the solution would be obvious.
Missing Number Math Test, To do this, use your logical reasoning. Are you eager to prove that you are a genius? Take a little time, because this test is for you.
## How Does It Work?
Missing Number Math Test, You have three triangles in the image displayed in front of you. Each of them contains 3 numbers, representing each vertex. These numbers are chosen consistently. Your task is to determine the connection between them to be able to deduce the value of the missing digit.
Take your time to find it. As you can see, this test requires concentration and perseverance. So, stay away from anything that can distract you. Go to the next paragraph to know the reasons why you should not ignore this IQ test.
## What Are The Benefits Of This Puzzle?
Missing Number Math Test, Math puzzles force your brain to be active. These challenges are not easy. You must then analyze, think carefully and develop new concepts to be able to solve them.
This is how you become more and more creative. In addition, this IQ test is a kind of game. Trying to solve it, you will have an entertaining and relaxing time.
This allows you to avoid any source of stress and eliminate anxiety. So, have you tried finding the missing number? Check your answer in the next paragraph.
## Answer: Here Is The Value That Replaces The Question Mark
The solution to this riddle is 28. The secret is that all the numbers in the same position are linked together. Their link verifies the following operation: the product of the numbers of the first triangle with those of the last gives that of the middle. Let’s take the case of the vertex at the top: 5 * 3 = 15.
Likewise, for the one on the left: 6 * 8 = 48. So, have you targeted this procedure to find the requested result?
If you found the correct answer, congratulations! This mathematical puzzle is difficult to solve. Thanks to your intelligence, you were able to elucidate it. This shows how persistent you are.
It’s a quality that can help you overcome the challenges you face on a daily basis. On the other hand, if you failed, it doesn’t matter. Now you have learned a new technique that can help you solve the other puzzles on this site.
If you liked this IQ test, you can share it with your loved ones so that they can also benefit from it.
You can always view and study more brain teaser, intellectual games, puzzles and personality tests in the entertainment section of Chashmak Website. Share them with your friends if you like. Especially those who are interested knowing themselves better and having fun. Follow us on Instagram and Facebook and share your comments and suggestions.
## Alzheimer, Brain Activity And Mental Games
Researchers have found that part of the brain disorders and the development of diseases such as forgetfulness and Alzheimer’s are related to the decrease in brain activity. Therefore, to prevent or prevent the development of these diseases, the mobility of the brain should be increased.
Mathematical questions similar to Gazer’s mathematical intelligence question can increase brain function. The correct solution of this question requires concentration and precision. In fact, the only answer to these questions is this point. Accordingly, questions like this are very useful for brain health in addition to creating entertainment.
Math Skill Challenge: Can You Complete This Math Challenge In Less Than 20 Seconds?
Container Logic Test: Are You An IQ Tester? Find Out Which Container Will Be Filled First!
Logic Test Practice: Raise Your IQ By Determining Which Glass Will Fill First!
Math Skill Challenge: Can You Complete This Math Challenge In Less Than 20 Seconds?
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# Dividing Fractions
Consider the two fractions with same or different denominators. These two fractions can be divided by taking the reciprocal of second fraction and multiply this fraction with the first fraction. That is,
Steps to divide fractions:
Step 1: Take the reciprocal of the divisor.
Step 2: Multiply the first fraction by the reciprocal of the other fraction obtained in Step 1. That is, multiply the numerators and the denominators of the fractions separately and express the result in fraction form.
Step 3: Simplify the fraction obtained in Step 2 and express the result in the standard form.
Example 1:
Consider the two fractions .
Let the dividend be and the divisor be . Then, the division of the given two fractions is obtained as follows:
Step 1: Take the reciprocal of the divisor.
That is, the reciprocal of
Step 2: Multiply the first fraction by the reciprocal of the other fraction obtained in Step 1. That is,
Step 3: Simplify the fraction obtained in Step 2 and express the result in the standard form.
Example 2:
Consider the fractions . Then, the division of these fractions by taking as dividend and as divisor is computed below:
See more Math topics
1:00
tutorial
1:00
tutorial
FOIL Method
1:00
tutorial
Finding an inverse
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Lesson Video: Simplifying Trigonometric Expressions Using Trigonometric Identities | Nagwa Lesson Video: Simplifying Trigonometric Expressions Using Trigonometric Identities | Nagwa
# Lesson Video: Simplifying Trigonometric Expressions Using Trigonometric Identities Mathematics
In this video, we will learn how to simplify trigonometric expressions by applying trigonometric identities.
14:28
### Video Transcript
In this video, we will learn how to simplify trigonometric expressions by applying trigonometric identities.
We begin by recalling that an identity is an equation that is true no matter what values are chosen. We will use combinations of these trigonometric identities, for example, the cofunction identities, shift identities, and Pythagorean identities. Before looking at some specific examples, letβs consider the properties of the unit circle.
We recall that the unit circle is a circle of radius one as shown. It enables us to measure the sin, cos, or tan of any angle π, where π is measured in the counterclockwise direction from the positive π₯-axis. The π₯-coordinate of any point on the unit circle is equal to cos π, and the π¦-coordinate is equal to sin π. The right triangle on our diagram together with the Pythagorean theorem leads us to the first Pythagorean identity. sin squared π plus cos squared π is equal to one.
Recalling the reciprocal trigonometric identities enables us to form two further Pythagorean identities. The three reciprocal functions are the cosecant, secant, and cotangent such that csc π is equal to one over sin π, sec π is equal to one over cos π, and cot π is equal to one over tan π. It is also worth noting that since tan π is equal to sin π over cos π, then cot π is equal to cos π over sin π. Dividing both sides of our first Pythagorean identity by cos squared π, we have sin squared π over cos squared π plus cos squared π over cos squared π is equal to one over cos squared π. Using the reciprocal identities, this simplifies to tan squared π plus one is equal to sec squared π.
In the same way, we can divide both sides of the first identity by sin squared π. This simplifies to one plus cot squared π is equal to csc squared π. We now have a set of three Pythagorean identities, which we will use together with the reciprocal identities to solve a couple of examples.
Simplify sin π multiplied by csc π minus cos squared π.
In this question, we are asked to simplify a trigonometric expression. One way of doing this is using the reciprocal and Pythagorean identities. In questions of this type, it is not always clear what to do first. However, as a general rule, it is worth replacing any reciprocal functions with the sine, cosine, or tangent function.
We know that csc π is equal to one over sin π. Substituting this into our expression, we have sin π multiplied by one over sin π minus cos squared π. The sin π on the numerator and denominator of our first term cancels, leaving us with one minus cos squared π. Next, we recall one of the Pythagorean identities: sin squared π plus cos squared π is equal to one. Subtracting cos squared π from both sides, this can be rewritten as sin squared π is equal to one minus cos squared π. This means that our expression can be rewritten as sin squared π. sin π multiplied by csc π minus cos squared π written in its simplest form is sin squared π.
We will now consider a second example of this type.
Simplify sin squared π plus cos squared π divided by csc squared π minus cot squared π.
In order to answer this question, we need to recall the Pythagorean identities. Firstly, we have sin squared π plus cos squared π is equal to one. Dividing each term by sin squared π and using our knowledge of the reciprocal trigonometric functions, we have one plus cot squared π is equal to csc squared π. We notice that the left-hand side of the first identity is identical to the numerator of our expression. Subtracting cot squared π from both sides of the second identity, we have one is equal to csc squared π minus cot squared π. The right-hand side of this is the same as the denominator of our expression.
As sin squared π plus cos squared π equals one and csc squared π minus cot squared π also equals one, our expression simplifies to one divided by one. And this is equal to one. The expression sin squared π plus cos squared π over csc squared π minus cot squared π is equal to one.
Before looking at one final example, we will recall the cofunction and shift identities. Once again, we begin by considering the unit circle. Since the angles in a triangle sum to 180 degrees, the third angle in our right triangle is equal to 90 degrees minus π. Letβs consider what happens if we redraw this triangle such that the angle between the positive π₯-axis and the hypotenuse is 90 degrees minus π. The coordinates of the point marked on the unit circle will be cos of 90 degrees minus π, sin of 90 degrees minus π. We notice that the distance in the π₯-direction here is the same as the distance in the π¦-direction in our first triangle. This means that the cos of 90 degrees minus π must be equal to sin π. Likewise, the sin of 90 degrees minus π is equal to cos π.
Since sin π over cos π is equal to tan π, the tan of 90 degrees minus π is equal to cos π over sin π. And using our knowledge of the reciprocal functions, this is equal to cot π. It follows that the sec of 90 degrees minus π is equal to csc π. The csc of 90 degrees minus π is equal to sec π. And the cot of 90 degrees minus π is equal to tan π. These six identities are known as the cofunction identities.
We can also use the unit circle to find identities involving angles such as 180 degrees minus π, 180 degrees plus π, and 360 degrees minus π. In our final example, we will use these identities together with the Pythagorean identities to simplify an expression.
Simplify one plus cot squared three π over two minus π over one plus tan squared π over two minus π.
In order to answer this question, we will need to use a variety of trigonometric identities. There are many ways to start here. However, we will begin by trying to rewrite the expression simply in terms of π. By firstly sketching the unit circle, we recall that π radians is equal to 180 degrees. This means that π over two radians is equal to 90 degrees. The denominator of our expression can therefore be rewritten as one plus tan squared of 90 degrees minus π. One of our cofunction identities states that tan of 90 degrees minus π is equal to cot π. This means that tan squared of 90 degrees minus π is equal to cot squared π. And the denominator of our expression is therefore equal to one plus cot squared π.
Letβs now consider the angle three π over two minus π. Once again, we can see from the unit circle that three π over two radians is equal to 270 degrees. This means that the numerator of our expression is equal to one plus cot of 270 degrees minus π. If π lies in the first quadrant, as shown in our right triangle, then three π over two minus π, or 270 degrees minus π, lies in the third quadrant. It is clear from the diagram that cos of three π over two minus π is equal to negative sin π and sin of three π over two minus π is equal to negative cos π. Since sin π over cos π is tan π and cos π over sin π is cot π, then cot of 270 degrees minus π is equal to tan π. Squaring both sides of this identity, we can rewrite the numerator of our expression as one plus tan squared π.
Our next step is to recall two of the Pythagorean identities. Firstly, tan squared π plus one is equal to sec squared π. And secondly, one plus cot squared π is equal to csc squared π. Our expression simplifies to sec squared π over csc squared π. And this can be rewritten as sec squared π multiplied by one over csc squared π. Recalling the reciprocal identities sec π is equal to one over cos π and csc π is equal to one over sin π, we have one over cos squared π multiplied by sin squared π, which can be rewritten as sin squared π over cos squared π and, in turn, is equal to tan squared π. The expression one plus cot squared three π over two minus π over one plus tan squared π over two minus π written in its simplest form is tan squared π.
We will now summarize the key points from this video. In this video, we simplified trigonometric expressions using a variety of trigonometric identities. We used the three Pythagorean identities sin squared π plus cos squared π is equal to one, tan squared π plus one is equal to sec squared π, and one plus cot squared π is equal to csc squared π. We also used the reciprocal identities csc π is equal to one over sin π, sec π is equal to one over cos π, and cot π is equal to one over tan π. Since tan π is equal to sin π over cos π, we also saw that cot π is equal to cos π over sin π.
We also recalled the cofunction identities sin of 90 degrees minus π is equal to cos π and cos of 90 degrees minus π is equal to sin π. And using the reciprocal identities above, these led us to four further cofunction identities. Finally, we used the unit circle to determine other related angle identities as demonstrated in our final example. In many examples, we need to apply more than one identity or type of identity to simplify a trigonometric expression.
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# Difference between revisions of "2004 AMC 10B Problems/Problem 21"
## Problem
Let $1$; $4$; $\ldots$ and $9$; $16$; $\ldots$ be two arithmetic progressions. The set $S$ is the union of the first $2004$ terms of each sequence. How many distinct numbers are in $S$?
$\mathrm{(A) \ } 3722 \qquad \mathrm{(B) \ } 3732 \qquad \mathrm{(C) \ } 3914 \qquad \mathrm{(D) \ } 3924 \qquad \mathrm{(E) \ } 4007$
## Solution 1
The two sets of terms are $A=\{ 3k+1 : 0\leq k < 2004 \}$ and $B=\{ 7l+9 : 0\leq l<2004\}$.
Now $S=A\cup B$. We can compute $|S|=|A\cup B|=|A|+|B|-|A\cap B|=4008-|A\cap B|$. We will now find $|A\cap B|$.
Consider the numbers in $B$. We want to find out how many of them lie in $A$. In other words, we need to find out the number of valid values of $l$ for which $7l+9\in A$.
The fact "$7l+9\in A$" can be rewritten as "$1\leq 7l+9 \leq 3\cdot 2003 + 1$, and $7l+9\equiv 1\pmod 3$".
The first condition gives $0\leq l\leq 857$, the second one gives $l\equiv 1\pmod 3$.
Thus the good values of $l$ are $\{1,4,7,\dots,856\}$, and their count is $858/3 = 286$.
Therefore $|A\cap B|=286$, and thus $|S|=4008-|A\cap B|=\boxed{(A) 3722}$.
## Solution 2
We can start by finding the first non-distinct term from both sequences. We find that that number is $16$. Now, to find every
other non-distinct terms, we can just keep adding $21$. We know that the last terms of both sequences are $1+3\cdot 2003$ and
$9+7\cdot 2003$. Clearly, $1+3\cdot 2003$ is smaller and that is the last possible common term of both sequences. Now, we can
create the inequality $16+21k \leq 1+3\cdot 2003$. Using the inequality, we find that there are $286$ common terms. There are 4008
terms in total. $4008-286=\boxed{(A) 3722}$
~Hithere22702
2004 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
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Integration Using Euler's Formula
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Integration Using Euler's Formula
In integral calculus, Euler's formula for complex numbers may be used to evaluate integrals involving trigonometric functions. Using Euler's formula, any trigonometric function may be written in terms of complex exponential functions, namely ${\displaystyle e^{ix}}$ and ${\displaystyle e^{-ix}}$ and then integrated. This technique is often simpler and faster than using trigonometric identities or integration by parts, and is sufficiently powerful to integrate any rational expression involving trigonometric functions.[1]
## Euler's formula
Euler's formula states that [2]
${\displaystyle e^{ix}=\cos x+i\,\sin x.}$
Substituting ${\displaystyle -x}$ for ${\displaystyle x}$ gives the equation
${\displaystyle e^{-ix}=\cos x-i\,\sin x}$
because cosine is an even function and sine is odd. These two equations can be solved for the sine and cosine to give
${\displaystyle \cos x={\frac {e^{ix}+e^{-ix}}{2}}\quad {\text{and}}\quad \sin x={\frac {e^{ix}-e^{-ix}}{2i}}.}$
## Examples
### First example
Consider the integral
${\displaystyle \int \cos ^{2}x\,dx.}$
The standard approach to this integral is to use a half-angle formula to simplify the integrand. We can use Euler's identity instead:
{\displaystyle {\begin{aligned}\int \cos ^{2}x\,dx\,&=\,\int \left({\frac {e^{ix}+e^{-ix}}{2}}\right)^{2}dx\\[6pt]&=\,{\frac {1}{4}}\int \left(e^{2ix}+2+e^{-2ix}\right)dx\end{aligned}}}
At this point, it would be possible to change back to real numbers using the formula e2ix + e-2ix = 2 cos 2x. Alternatively, we can integrate the complex exponentials and not change back to trigonometric functions until the end:
{\displaystyle {\begin{aligned}{\frac {1}{4}}\int \left(e^{2ix}+2+e^{-2ix}\right)dx&={\frac {1}{4}}\left({\frac {e^{2ix}}{2i}}+2x-{\frac {e^{-2ix}}{2i}}\right)+C\\[6pt]&={\frac {1}{4}}\left(2x+\sin 2x\right)+C.\end{aligned}}}
### Second example
Consider the integral
${\displaystyle \int \sin ^{2}x\cos 4x\,dx.}$
This integral would be extremely tedious to solve using trigonometric identities, but using Euler's identity makes it relatively painless:
{\displaystyle {\begin{aligned}\int \sin ^{2}x\cos 4x\,dx&=\int \left({\frac {e^{ix}-e^{-ix}}{2i}}\right)^{2}\left({\frac {e^{4ix}+e^{-4ix}}{2}}\right)dx\\[6pt]&=-{\frac {1}{8}}\int \left(e^{2ix}-2+e^{-2ix}\right)\left(e^{4ix}+e^{-4ix}\right)dx\\[6pt]&=-{\frac {1}{8}}\int \left(e^{6ix}-2e^{4ix}+e^{2ix}+e^{-2ix}-2e^{-4ix}+e^{-6ix}\right)dx.\end{aligned}}}
At this point we can either integrate directly, or we can first change the integrand to 2 cos 6x - 4 cos 4x + 2 cos 2x and continue from there. Either method gives
${\displaystyle \int \sin ^{2}x\cos 4x\,dx=-{\frac {1}{24}}\sin 6x+{\frac {1}{8}}\sin 4x-{\frac {1}{8}}\sin 2x+C.}$
## Using real parts
In addition to Euler's identity, it can be helpful to make judicious use of the real parts of complex expressions. For example, consider the integral
${\displaystyle \int e^{x}\cos x\,dx.}$
Since cos x is the real part of eix, we know that
${\displaystyle \int e^{x}\cos x\,dx=\operatorname {Re} \int e^{x}e^{ix}\,dx.}$
The integral on the right is easy to evaluate:
${\displaystyle \int e^{x}e^{ix}\,dx=\int e^{(1+i)x}\,dx={\frac {e^{(1+i)x}}{1+i}}+C.}$
Thus:
{\displaystyle {\begin{aligned}\int e^{x}\cos x\,dx&=\operatorname {Re} \left({\frac {e^{(1+i)x}}{1+i}}\right)+C\\[6pt]&=e^{x}\operatorname {Re} \left({\frac {e^{ix}}{1+i}}\right)+C\\[6pt]&=e^{x}\operatorname {Re} \left({\frac {e^{ix}(1-i)}{2}}\right)+C\\[6pt]&=e^{x}{\frac {\cos x+\sin x}{2}}+C.\end{aligned}}}
## Fractions
In general, this technique may be used to evaluate any fractions involving trigonometric functions. For example, consider the integral
${\displaystyle \int {\frac {1+\cos ^{2}x}{\cos x+\cos 3x}}\,dx.}$
Using Euler's identity, this integral becomes
${\displaystyle {\frac {1}{2}}\int {\frac {6+e^{2ix}+e^{-2ix}}{e^{ix}+e^{-ix}+e^{3ix}+e^{-3ix}}}\,dx.}$
If we now make the substitution ${\displaystyle u=e^{ix}}$, the result is the integral of a rational function:
${\displaystyle -{\frac {i}{2}}\int {\frac {1+6u^{2}+u^{4}}{1+u^{2}+u^{4}+u^{6}}}\,du.}$
One may proceed using partial fraction decomposition.
## References
1. ^ Kilburn, Korey. "Applying Euler's Formula to Integrate". American Review of Mathematics and Statistics. American Research Institute for Policy Development. 7: 1-2. doi:10.15640/arms.v7n2a1. eISSN 2374-2356. ISSN 2374-2348 – via http://armsnet.info/.
2. ^ Weisstein, Eric W. "Euler Formula". mathworld.wolfram.com. Retrieved .
This article uses material from the Wikipedia page available here. It is released under the Creative Commons Attribution-Share-Alike License 3.0.
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Algebra 1 Online! Henrico County Public Schools, Virginia
Module - Polynomials
Introduction - Animal Populations
Lessons 1- Multiplying Monomials 2- Dividing Monomials 3- Scientific Notation 4- Degree, Ascending, and Descending Order 5- Adding and Subtracting Polynomials 6- Multiplying Polynomials by Monomials 7- Multiplying Polynomials 8- Special Products 9- Module Review
Anyone who hunts knows that they can only do this during the designated season. This is to make sure that the animal population is not depleted. However, if left uncontrolled, an animal population develops polynomially. This means that if one animal has and average of 3 offspring, then each of the 3 babies will have an average of 3 offspring (a total of 9), and those 9 babies will average 3 offspring each (a total of 27), and so on.
Generation Descendants 1 3 2 3 + 9 = 12 3 3 + 9 + 27 = 39 4 3 + 9 + 27 + 81 = 120
The number of descendants of one animal, written as a polynomial, can be expressed as x + x^2 + x^3 + x^4 where x is the number of offspring an animal has. So, if an animal has an average of 8 offspring in its lifetime, it will probably have 4680 (8 + 8^2 + 8^3 + 8^4) descendants in four generations. Luckily, animal populations are controlled naturally by predators, disease, starvation, and natural competition.
Upon completion of the activities in this unit, you should be able to:
• Use the following terms in a written paragraph to describe the key concepts of this unit.
• polynomial
• monomial
• binomial
• trinomial
• simplify
• decimal notation
• scientific notation
• degree
• constants
• FOIL method
• Multiply monomials and simplify expressions with powers of monomials.
• Divide monomials and simplify expressions with quotients of monomials and negative exponents.
• Express numbers in scientific and decimal notation.
• Find products and quotients of numbers expressed in scientific notation.
• Find the degree of a polynomial.
• Arrange the terms of a polynomial in ascending and descending order.
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# Fractal Geometry
## Fractal Dimension
Ordinary dimension is always given as an integer. For instance, a line is one-dimensional, a square is two-dimensional, and a cube is three-dimensional. However, fractal dimension can be a non-integer, i.e. fractal dimension can be a fraction!
To understand why fractal dimension can be a non-integer, let's first determine why the dimension of a square is two. If we join the midpoints of opposite sides of a square, we will decompose the square into 4 self-similar copies of itself. Each self-similar copy has a magnification factor of 2. That is to say that if we magnify any of the copies by a factor of 2, we get the original square.
If we trisect the sides of the square, we decompose the square into 9 self-similar copies of itself, each with a magnification factor of 3. By continuing the process, a pattern emerges. See the table below.
Decomposition of a Square
Self-Similar Copies (S) Magnification Factor (N)
4 2
9 3
16 4
25 5
Now suppose that on each face of a cube, we join the midpoints of opposite edges. The cube is decomposed into 8 self-similar copies of the itself, each with a magnification factor of 2. If we trisect opposite edges of each face, we decompose the cube into 27 self-similar copies of itself, each with a magnification factor of 3.
Continue the process and the pattern becomes clear. The number of self-similar copies equals the cube of the magnification factor, i.e. $S={N}^{3}.$
For a square, a 2-dimensional figure, the exponent of the magnification factor is 2. For a cube, a 3-dimensional figure, the exponent of the magnification factor is 3. It appears that the exponent of the magnification factor is the dimension. In general, if a figure is decomposed into S self-similar parts, each with magnification factor N, then the exponent of the magnification factor is the dimension. Let D = dimension and we have
${N}^{D}=S$
Solving for D we find
$\begin{array}{l}\mathrm{log}{N}^{D}=\mathrm{log}S\\ D\mathrm{log}N=\mathrm{log}S\\ \therefore D=\frac{\mathrm{log}S}{\mathrm{log}N}\end{array}$
Thus, we can define fractal dimension as follows.
### Dimension of the Sierpinski Triangle
One of the simplest geometric fractals is the Sierpinski triangle. It is constructed by joining the midpoints of the sides of an equilateral triangle and removing the triangle formed. Repeat the process to get successive stages of the Sierpinski triangle.
The second figure consists of 3 self-similar copies of the original triangle, each with a magnification factor of 2. The third stage of the Sierpinski triangle consists of 9 self-similar copies, each with magnification factor 4. This kind of self-similarity is a defining characteristic of fractals. So what is the dimension of the Sierpinski triangle?
Since the Sierpinski triangle consists of 3 self-similar copies, each with a magnification factor of 2, we compute the dimension as
$D=\frac{\mathrm{log}3}{\mathrm{log}2}\approx 1.5850$
However, the third stage of the Sierpinski triangle consists of 9 self-similar copies, each with magnification factor 4. So we can compute the dimension as
$D=\frac{\mathrm{log}9}{\mathrm{log}4}=\frac{\mathrm{log}{3}^{2}}{\mathrm{log}{2}^{2}}=\frac{2\mathrm{log}3}{2\mathrm{log}2}=\frac{\mathrm{log}3}{\mathrm{log}2}\approx 1.5850$
As before, the dimension is roughly 1.59.
### Dimension of the Koch Snowflake
The Koch snowflake is constructed by first trisecting each side of an equilateral triangle, forming 3 self-similar parts. The self-similar parts each have a magnification factor of 3. Each side of the triangle is then reconstructed using 4 self-similar parts. Thus, the dimension of the Koch snowflake is
$D=\frac{\mathrm{log}4}{\mathrm{log}3}\approx 1.2619$
## Conclusion
This brief introduction to fractal geometry barely scratches the surface of this intriguing branch of modern mathematics. One often gets the impression that all mathematics originated in the distant past. Names such as Isaac Newton (1642 - 1727) and Euclid (circa 300 BC) come to mind as the Who's Who of mathematicians. However, fractal geometry would not have been possible without the advent of modern computer technology. Fractal geometry is a new and still emerging branch of mathematics. Diverse, groundbreaking, and completely unexpected practical applications of fractal geometry are constantly surfacing.
Hopefully you are now inspired to dig deeper and learn more about this relatively new and fascinating area of mathematics. If not, then watch this PBS video about fractals. That should do the trick.
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# What is the antiderivative of x^3/(1+x^2)?
Jul 7, 2016
$\frac{1}{2} \left\{{x}^{2} + 1 - \ln \left({x}^{2} + 1\right)\right\} + C .$
OR
$\frac{1}{2} \left\{{x}^{2} - \ln \left({x}^{2} + 1\right)\right\} + {C}_{1} ,$ where ${C}_{1} = C + \frac{1}{2.}$
#### Explanation:
Let $I = \int {x}^{3} / \left(1 + {x}^{2}\right) \mathrm{dx}$
We take substn. ${x}^{2} + 1 = t$, so that, $2 x \mathrm{dx} = \mathrm{dt} .$
Also, ${x}^{2} + 1 = t \Rightarrow {x}^{2} = t - 1$
Now, $I = \int {x}^{3} / \left(1 + {x}^{2}\right) \mathrm{dx} = \frac{1}{2} \int \frac{{x}^{2} \cdot 2 x}{1 + {x}^{2}} \mathrm{dx} = \frac{1}{2} \int \frac{t - 1}{t} \mathrm{dt}$
$= \frac{1}{2} \int \left\{\frac{t}{t} - \frac{1}{t}\right\} \mathrm{dt} = \frac{1}{2} \int \left\{1 - \frac{1}{t}\right\} \mathrm{dt} = \frac{1}{2} \left\{t - \ln t\right\} = \frac{1}{2} \left\{{x}^{2} + 1 - \ln \left({x}^{2} + 1\right)\right\} + C .$
$I$ is also$= \frac{1}{2} {x}^{2} + \frac{1}{2} - \frac{1}{2} \ln \left({x}^{2} + 1\right) + C$
$= \frac{1}{2} \left\{{x}^{2} - \ln \left({x}^{2} + 1\right)\right\} + {C}_{1} ,$ where ${C}_{1} = C + \frac{1}{2}$
Is this not enjoyable?! Enjoy maths.!
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# Series - Parallel Resistances the Combination
Whenever we work with circuits in the real world, they are seldom as straightforward as a simple series or parallel circuit. Normally, they are a combination of the two, called a SERIES-PARALLEL circuit. While they look forbidding at first, you must keep in mind that ALL circuits can be broken down into smaller parts. They can be made simpler to work with. Such is the case with the Series-Parallel circuit.
If you look at the example on the right, it has 3 resistors and 1 battery. R 1 and R 2 are both 10 Ω in parallel.
1
------------
1 1
---- + ----
R 1 R 2
If we do the math (reciprocating the reciprocals), we come up with a total of 5Ω for these two.
We can say that:
R 1&2 = 5Ω.
We also have R 3 in the circuit, which is 20Ω. Once we have combined the 2 parallel resistors, we have a simpler circuit.... 2 series resistors. R 1&2 and R 3 . If we add the value of these two resistors, we come up with
R Total =R 1&2 +R 3 .
So R Total =5Ω+20Ω=25Ω.
Then if we know the voltage, we can find the current through the entire circuit, and through each individual resistor. Go ahead and try plugging in a voltage (like 25V) and finding the currents. You'll be surprised at how simple it is.
Let's try another example. In the circuit on the right, we have 3 resistors again. But this time, they are configured differently. Do you see how you would combine them for the total resistance? First, you must add the 10 Ω resistors by adding them. This is simple because they are in Series.
10 Ω + 10 Ω = 20 Ω.
Using our parallel circuit formula then:
1
------------
1 1
---- + ----
R 1 R 2
We find that our total resistance for the circuit = 10Ω. Note that if we have the same battery (25V), our current turns out much different through the entire circuit. So the same components, configured differently, will create a vast difference in the way the circuit works.
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## What is the slope of 4x 3y 12?
Therefore the slope is 4/3.
## What type of border line should the inequality have 4x 3y 12?
The boundary line will be solid because the inequality operator contains an “or equal to” clause.
## Is 4x 3y 12 a linear equation in two variables?
The equation 4x – 3y = 12 is a linear equation in standard form.
## What is 4x 3y in slope intercept form?
Norma, I’m going to illustrate with a different equation, 4x – 3y = 7. The slope intercept form of a line is y = mx + b.
## What is the solution of 4x 3y 12?
Linear Equations in Two Variables
We plot the points (3, 0) and (0, 4) on a graph paper and join the same by a ruler to get the line which is the graph of the equation 4x + 3y = 12. Three different solutions of the equation 4x + 3y = 12 from the graph are (0, 4), (3, 0) and (6, –4).
## What is the slope of a line that is parallel to the graph of 4x 3y 1?
Using the slope-intercept form, the slope is −43 . All lines that are parallel to y=−4×3+13 y = – 4 x 3 + 1 3 have the same slope of −43 .
## At what point is the graph of the linear equation 4x 3y 12 cuts the y-axis?
Hence (0,-4) is a solution of equation 4x-3y=12 which cuts at y axis.
## What is a parallel slope?
Note that two lines are parallel if their slopes are equal and they have different y-intercepts. Perpendicular Lines and Their Slopes. In other words, perpendicular slopes are negative reciprocals of each other. Here is a quick review of the slope/intercept form of a line.
## How do you find the slope of a line parallel?
Explanation: First, you should put the equation in slope intercept form (y = mx + b), where m is the slope. The slope of the line is -3/8. A parallel line will have the same slope, thus -3/8 is the correct answer.
## What is a parallel equation?
Parallel lines have the same slope but different y-intercepts. When the equations of two lines are the same they have infinitely many points in common, whereas parallel lines have no points in common. Our equation is given in slope-intercept form, \displaystyle y=mx+b.
## What is the graph of parallel?
Parallel lines have slopes that are the same. All of the lines shown in the graph are parallel because they have the same slope and different y-intercepts. Lines that are perpendicular intersect to form a 90∘ angle. The slope of one line is the negative reciprocal of the other.
## How are lines KL and MN related?
How are lines KL and MN related? … The lines are perpendicular. The lines do not have slopes.
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# What is the slope of any line perpendicular to the line passing through (16,6) and (-2,-13)?
Apr 3, 2018
$- \frac{18}{19}$
#### Explanation:
Let's first find the slope of the line passing through the aforementioned points
$\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} \rightarrow$ Finding a slope via two points formula
$\frac{- 13 - 6}{- 2 - 16} \rightarrow$ Plug in the points
$\frac{- 19}{-} 18$
$\frac{19}{18} \rightarrow$ This is the slope of the line
Perpendicular slopes are opposite reciprocals of each other
To make something the opposite of another number, add a negative sign in front of it (a positive number's opposite will be negative, a negative number's opposite will be positive)
To find the reciprocal of a number, switch the numerator and denominator
$\frac{19}{18}$
$- \frac{19}{18} \rightarrow$ The opposite
$- \frac{18}{19} \rightarrow$ The (opposite) reciprocal
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# Transpose of a matrix - properties and formulas -
Definition
- Definition
- Examples
Properties
- Product
- Inverse
- Determinant
- Trace
- Linearity
Definition
A matrix in which the row and column elements of a matrix are exchanged is called the transposed matrix of that matrix. The transpose of a matrix $A$ is represented as $A^ {T}$. The $i$-th row, j-th column element of $A^T$ is the $j$-th row, $i$-th column element of $A:$
Explanation
The $i$-th row, $j$-th column element of $A^T$ is the $j$-th row, $i$-th column element of $A$. For example, if $i=1$ and $j=2$,
And if $i=2$ and $j=1$,
By transposition, each element is exchanged corssing diagonal elements.
Since $(A^{T})_{ii} = A_{ii}$, each diagonal element is unchanged.
In general, the size of a traspose matrix is different from its original one. If $A$ is an $m \times n$ matrix, then $A^{T}$ is an $n \times m$ matrix.
If $A$ is a square matrix, the size of $A^{T}$ is the same as that of $A$.
A column matrix has only one column but any number of rows. A row matrix has only one row but any number of columns. Transpose transforms a column matrix to a row matrix, and vice versa.
Various matrices are defined by transpose. For example,
• If $A^{T}=A$, $A$ is called a symmetric matrix.
• If $A^{T}A=AA^{T}=I$, $A$ is called an orthogonal matrix.
Examples
The transposed matrix of matrix
is
. The transposed matrix of matrix
is
.
Product
The transposed matrix of the product of two matrices is equal to the product of the transposed matrices in which the order of the products is reversed:
.
Proof
Let $A$ be a $l \times m$ matrix, and $B$ be a $m \times n$ matrix, written as
Using the definition of transposed matrix and the definition of product of matrix, for the $i$-th row, $j$-th column element of $(AB)^T$, we see that
. Since this holds for any element, we obtain
.
Inverse
If a matrix $A$ is invertible, the transpose of $A$ is also invertible, and the inverse matrix of $A^{T}$ is $(A^{-1})^{T}$, that is,
.
Proof
Let $A$ be an invertible matrix. There exists $A^{-1}$ for which the following holds
$$\tag{4.1}$$ , where $I$ is the identity matrix. By the propery of transpose of product of two matrices, and the first equation of $(4.1)$, we see that
$$\tag{4.2}$$ Simimary, by the second equation of $(4.2)$, we see that
$$\tag{4.3}$$ Eq. $(4.2)$ and $(4.3)$ give
. This expression shows that $(A ^ {-1}) ^ {T}$ is the inverse matrix of $A ^ {T}$, that is,
.
Determinant
Let $A$ be a square matrix, and $|A^{T}|$ be the determinant of transpose of $A$. $|A^{T}|$ is equal to the determinant of $A$, that is,
Proof
By definition, the determinant of transpose of an arbitray $n \times n$ matrix $A$ is written as
, where $\sigma$ denotes a function, called permutation, that reorders the set of integers, $\{ 1,2,\cdots,n \}$ $S_{n}$ denotes the set of all such permutations, $\mathrm{sgn}(\sigma)$ is a sign depending on the permutation, and a sum $\sum_{\sigma \in S_{n}}$ involves all permutations. (For details, see "definition of determinant".)
Since $A_ {ij} ^ {T} = A_ {ji}$ from the definition of transpose,
. The permutation is bijective. Therefore, for an arbitray permutation $\sigma$, there exists an inverse map $\sigma^{-1}$ such that
, where $i=1,2,\cdots,n$. Using this, we have
$$\tag{5.1}$$ . Since $\sigma$ is a map that maps $\{1,2, \cdots, n \}$ to the same set $\{1,2, \cdots, n \}$, there exists $j$ $(j=1,\cdots,n)$ such that $\sigma(j) = 1$. Therefore $(5.1)$ can be written as
, where ,in the last line, the order of multiplication is just changed. Similary, there exists $k$ $(k=1,\cdots,n, k\neq j)$ such that $\sigma(k) = 2$. We see that
By repeating the same procedure to the end, we obtain
$$\tag{5.2}$$ Since $\sigma^{-1}$ of an even permutation $\sigma$ is also an even permutation, and $\sigma^{-1}$ of an odd permutation $\sigma$ is also an odd permutation, $\mathrm{sgn}(\sigma^{-1}) = \mathrm{sgn}(\sigma)$ holds. Therefore, we can write
$\sigma$ is a map that maps $\{1,2, \cdots, n \}$ to the same set $\{1,2, \cdots, n \}$, and similary, $\sigma^{-1}$ is a map that maps $\{1,2, \cdots, n \}$ to the same set $\{1,2, \cdots, n \}$. The set of all $\sigma^{-1}$ is the same as the set of all $\sigma$, Therefore, in Eq.$(5.2)$, $\sum_{\sigma \in S_{n}}$ can be replaced with $\sum_{\sigma^{-1} \in S_{n}}$
Since $\sigma^{-1}$ is one of permutation, we can rewrite $\sigma^{-1}$ as $\xi \hspace{1mm} (\in S_{n})$,
In this way, the determinant of transpose of a square matrix is equal to the determinant of the square matrix.
Trace
Let $A$ a squared matrix and $A ^ {T}$ be the transpose of $A$. The trace of $A^{T}$ is equal to the trace of $A$:
Proof
Let $A$ an $n \times n$ matrix, and $A_{ij}$ be the $i$-th row, $j$-th column element of $A$. By the definition of transpose, we have
, whrere ($i,j=1,2 \cdots , n$). This and the defintion of the trace give
Linearity
Let $A$ and $B$ be $m \times n$ matrices and alpha be a scalar. The transpse is a linear transformation:
Proof
Suppose $A$ and $B$ are represented as
Then
We see that
and
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Set (mathematics)
Informally, a set is thought of as any collection of distinct elements. The relevant formal definitions are given in a branch of mathematics known as set theory.
Introduction
The basic property of sets is that they are solely determined by the elements they contain (this is called extensionality). Thus, we can identify sets by listing their elements. For instance, we can talk about the set that has as its elements the numbers 1, 2 and 3. This set is denoted {1, 2, 3}.
A consequence of this basic property is that a set cannot contain an element twice. The set {1, 2, 2, 3} contains the elements 1, 2 and 3 and is thus the same as the set {1, 2, 3}. This is the difference between sets and multisets; considered as multisets, {1, 2, 2, 3} and {1, 2, 3} are different.
For the same reason, the order in which the elements are listed does not matter. The sets {1, 2, 3} and {3, 2, 1} have the same elements and thus these two sets are equal. However, there are many contexts in which we want to consider structures that have elements in a certain order and these elements may be the same. Such a structure is called a tuple or a sequence. The tuple containing the elements 1, 2 and 3 (in that order) is different from the tuple containing the elements 3, 2 and 1. These tuples are denotes (1, 2, 3) and (3, 2, 1) respectively, with round brackets (or angle brackets) instead of curly brackets to emphasize the difference between tuples and sets.
Despite the intuitive definition, a set is usually not defined formally in terms of other mathematical objects; rather it is defined by the laws (called axioms) that is satisfies. For instance, one commonly requires that no set may be an element of itself. Because sets are defined by themselves, they are fundamental structures in mathematics and logic. Mathematicians have found ways to define many mathematical objects, such as the real numbers, in terms of sets.
The number of elements that a set contains does not have to be finite. Sets that contain a finite number of elements are called finite sets. Sets that contain an infinite number of elements are called infinite sets. The number of elements that a finite set contains is called that set's cardinality. The concept of cardinality can also be applied to infinite sets, though the concept is less intuitive, and relies upon bijections between sets.
Notation
Some sets can be denoted by a list of objects separated with commas, enclosed with curly brackets. As mentioned before, {1, 2, 3} is the set of the numbers 1, 2, and 3. We say that 1, 2, and 3 are its members (or elements).
There are many other ways to write out sets. For example,
A = {x | 1 < x < 10, x is a natural number}
can be read as follows: A is the set of all x, where x is between 1 and 10, and x is a natural number. A could also be written as:
A = {2, 3, 4, 5, 6, 7, 8, 9}
Membership in a set is expressed with the ∈ symbol. To say that the set A contains the number 2 as an element (or that 2 is an element of A), we write
2 ∈ A
The cardinality of a set is expressed by placing bars around the name of the set. For example, one would express the cardinality of the above set as such:
|A| = 8
Some special sets
Some sets that are ubiquitous in the mathematical literature have special symbols:
• $\emptyset$, the empty set, sometimes written {};
• $\mathbb{N}$, the set of natural numbers;
• $\mathbb{Z}$, the set of integers;
• $\mathbb{Q}$, the set of rational numbers;
• $\mathbb{R}$, the set of real numbers;
• $\mathbb{C}$, the set of complex numbers.
Sometimes also
• $\mathbb{F}$, the set of irrational numbers;
• $\mathbb{P}$, the set of prime numbers.
Among other such well known sets are the fibonacci numbers, even numbers, odd numbers, quaternions, octonions and the Hamiltonian integers.
Some examples of sets
• The set consisting of all tuples (a,b), where a is any real number and ditto for b. This set is known as $\mathbb{R}$x$\mathbb{R}$ or $\mathbb{R}$2.
• The three element set {Red, Yellow, Green}.
• The set consisting of the two elements Brake, Accelerate.
• The set consisting of all tuples (a,b) where a is any element in the set {Red, Yellow, Green} and b is any element in the set {Brake, Accelerate}.
• The set of all functions from the set {Red, Yellow, Green} to the set {Brake, Accelerate}.
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## NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1
Free download NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.1, Ex 5.2, Ex 5.3, Ex 5.4, Ex 5.5, Ex 5.6 and Ex 5.7, 5.8 and 5.9 Understanding Elementary Shapes PDF for CBSE 2020 Exams.
### Understanding Elementary Shapes Class 6 Ex 5.1
Ex 5.1 Class 6 Maths Question 1.
What is the disadvantage in comparing line segment by metre observation?
Solution:
Comparing the lengths of two line segments simply by ‘observation’ may not be accurate. So we use divider to compare the length of the given line segments.
Ex 5.1 Class 6 Maths Question 2.
Why is it better to use a divider than a ruler, while measuring the length of a line segment?
Solution:
Measuring the length of a line segment using a ruler, we may have the following errors:
(i) Thickness of the ruler
(ii) Angular viewing
These errors can be eradicated by using the divider. So, it is better to use a divider than a ruler, while measuring the length of a line segment.
Ex 5.1 Class 6 Maths Question 3.
Draw any line segment, say $$\overline { AB }$$. Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AB = AC + CB?
[Note: If A, B, C are any three points on a line such AC + CB = AB, then we can be sure that C lies between A and B]
Solution:
Let us consider
A, B and C such that C lies between A and B and AB = 7 cm.
AC = 3 cm, CB = 4 cm.
∴ AC + CB = 3 cm + 4 cm = 7 cm.
But, AB = 7 cm.
So, AB = AC + CB.
Ex 5.1 Class 6 Maths Question 4.
If A, B, C are three points on a line such that AB = 5 cm, BC = 3 cm and AC = 8 cm, which one of them lies between the other two?
Solution:
We have, AB = 5 cm; BC = 3 cm
∴ AB + BC = 5 + 3 = 8 cm
But, AC = 8 cm
Hence, B lies between A and C.
Ex 5.1 Class 6 Maths Question 5.
Verify, whether D is the mid point of $$\overline { AG }$$ .
Solution:
From the given figure, we have
AG = 7 cm – 1 cm = 6 cm
AD = 4 cm – 1 cm = 3 cm
and DG = 7 cm – 4 cm = 3 cm
∴ AG = AD + DG.
Hence, D is the mid point of $$\overline { AG }$$.
Ex 5.1 Class 6 Maths Question 6.
If B is the mid point of $$\overline { AC }$$ and C is the mid point of $$\overline { BD}$$ , where A, B, C, D lie on a straight line, say why AB = CD?
Solution:
We have
B is the mid point of $$\overline { AC }$$ .
∴ AB = BC …(i)
C is the mid-point of $$\overline { BD }$$ .
BC = CD
From Eq.(i) and (ii), We have
AB = CD
Ex 5.1 Class 6 Maths Question 7.
Draw five triangles and measure their sides. Check in each case, if the sum of the length of any two sides is always less than the third side.
Solution:
Case I. In ∆ABC
Let AB = 2.5 cm
BC = 4.8 cm
and AC = 5.2 cm
AB + BC = 2.5 cm + 4.8 cm
= 7.3 cm
Since, 7.3 > 5.2
So, AB + BC > AC
Hence, sum of any two sides of a triangle is greater than the third side.
Case II. In ∆PQR,
Let PQ = 2 cm
QR = 2.5 cm
and PR = 3.5 cm
PQ + QR = 2 cm + 2.5 cm = 4.5 cm
Since, 4.5 > 3.5
So, PQ + QR > PR
Hence, sum of any two sides of a triangle is greater than the third side.
Case III. In ∆XYZ,
Let XY = 5 cm
YZ = 3 cm
and ZX = 6.8 cm
XY + YZ = 5 cm + 3 cm
= 8 cm
Since, 8 > 6.8
So, XY + YZ > ZX
Hence, the sum of any two sides of a triangle is greater than the third side.
Case IV. In ∆MNS,
Let MN = 2.7 cm
NS = 4 cm
MS = 4.7 cm
and MN + NS = 2.7 cm + 4 cm = 6.7 cm
Since, 6.7 >4.7
So, MN + NS > MS
Hence, the sum of any two sides of a triangle is greater than the third side.
Case V. In ∆KLM,
Let KL = 3.5 cm
LM = 3.5 cm
KM = 3.5 cm
and KL + LM = 3.5 cm + 3.5 cm = 7 cm
7 cm > 3.5 cm
Solution:
(i) For one-fourth revolution, we have
So, KL + LM > KM
Hence, the sum of any two sides of a triangle is greater than the third side.
Hence, we conclude that the sum of any two sides of a triangle is never less than the third side.
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### Understanding Elementary Shapes Class 6 Ex 5.2
Ex 5.2 Class 6 Maths Question 1.
What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from
(a) 3 to 9
(b) 4 to 7
(c) 7 to 10
(d) 12 to 9
(e) 1 to 10
(f) 6 to 3
Solution:
(a) 3 to 9
9 – 3 = 6 ÷ 12 = $$\frac { 1 }{ 2 }$$ of a revolution
(b) 4 to 7
7 – 4 = 3 ÷ 12 = $$\frac { 1 }{ 4 }$$ of a revolution
(c) 7 to 10
10 – 7 = 3 ÷ 12 = $$\frac { 1 }{ 4 }$$ of a revolution
(d) 12 to 9 i.e., 0 to 9
9 – 0 = 9 ÷ 12 = $$\frac { 3 }{ 4 }$$ of a revolution
(e) 1 to 10
10 – 1 = 9 ÷ 12 = $$\frac { 3 }{ 4 }$$ of a revolution
(f) 6 to 3 i.e., 6 to 12 and then 12 to 3
6 to 12 = 12 – 6 = 6 and 12 to 3 = 0 to 3 = 3 – 0 = 3
6 + 3 = 9 ÷ 12 = $$\frac { 3 }{ 4 }$$ of a revolution
Ex 5.2 Class 6 Maths Question 2.
Where will the hand of a clock stop if it
(a) starts at 12 and makes $$\frac { 1 }{ 2 }$$ of a revolution, clockwise?
(b) starts at 2 and makes $$\frac { 1 }{ 2 }$$ of a revolution, clockwise?
(c) starts at 5 and makes $$\frac { 1 }{ 2 }$$ of a revolution, clockwise?
(d) starts at 5 and makes $$\frac { 1 }{ 2 }$$ of a revolution, clockwise?
Solution:
(a) Starting from 12 and making $$\frac { 1 }{ 2 }$$ of a revolution, the clock hand stops at 6.
(b) Starting from 2 and making $$\frac { 1 }{ 2 }$$ of a revolution, the clock hand stops at 8.
(c) Starting from 5 and making $$\frac { 1 }{ 2 }$$ of a revolution, the clock hand stops at 8.
(d) Starting from 5 and making $$\frac { 1 }{ 2 }$$ of a revolution, the clock hand stops at 2.
Ex 5.2 Class 6 Maths Question 3.
Which direction will you face if you start facing
(a) east and make $$\frac { 1 }{ 2 }$$ of a revolution clockwise? z
(b ) east and make $$1\frac { 1 }{ 2 }$$ of a revolution clockwise? z
(c) west and make $$\frac { 3 }{ 4 }$$ of a revolution anticlockwise?
(d) south and make one full revolution? (Should we specify clockwise or anticlockwise for this last question? Why not?)
Solution:
Taking one full revolution we will reach back to the original (starting) position. Therefore, it make no difference whether we turn clockwise or anticlockwise.
Ex 5.2 Class 6 Maths Question 4.
What part of a revolution have you turned through if you stand facing
(a) east and turn clockwise to face north?
(b) south and turn clockwise to face east?
(c) west and turn clockwise to face east?
Solution:
(a) If we start from east and reach at north (turning clockwise) $$\frac { 3 }{ 4 }$$ of a revolution is required.
(b) If we start from south turning clockwise to face east, $$\frac { 3 }{ 4 }$$ of a revolution is required.
(c) If we start from west turning clockwise to face east, $$\frac { 1 }{ 2 }$$ of a revolution is required.
Ex 5.2 Class 6 Maths Question 5.
Find the number of right angles turned through by the hour hand of a clock when it goes from
(a)3 to 6
(b) 2 to 8
(c) 5 to 11
(d) 10 to 1
(e) 12 to 9
(f) 12 to 6
Solution:
(a) 3 to 6
Starting from 3 to 6, the hour hand turns through 1 right angle.
(b) 2 to 8
Starting from 2 to 8, the hour hand turns through 2 right angles.
(c) 5 to 11
Starting from 5 to 11, the hour hand turns through 2 right angles.
(d) 10 to 1
Starting from 10 to 1, the hour hand turns through 1 right angle.
(e) 12 to 9
Starting from 12 to 9, the hour hand turns through 3 right angles.
(f) 12 to 6
Starting from 12 to 6, the hour hand turns through 2 right angles.
Ex 5.2 Class 6 Maths Question 6.
How many right angles do you make if you start facing
(a) south and turn clockwise to west?
(b) north and turn anticlockwise to east?
(c) west and turn to west?
(d) south and turn to north?
Solution:
Ex 5.2 Class 6 Maths Question 7.
Where will the hour hand of a clock stop if it starts
(a) from 6 and turns through 1 right angle?
(b) from 8 and turns through 2 right angles?
(c) from 10 and turns through 3 right angles?
(d) from 7 and turns through 2 straight angles?
Solution:
(a) Starting from 6 and turning through 1 right angle, the hour hand stops at 9.
(b) Starting from 8 and turning through 2 right angles, the hour hand stops at 2.
(c) Starting from 10 and turning through 3 right angles, the hour hand stops at 7.
(b) Starting from 7 and turning through 2 right angles, the hour hand stops at 7.
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### Understanding Elementary Shapes Class 6 Ex 5.3
Ex 5.3 Class 6 Maths Question 1.
Match the following:
(i) Straight angle         (a) Less than one-fourth of a revolution.
(ii) Right angle           (b) More than half a revolution.
(iii) Acute angle          (c) Half of a revolution.
(iv) Obtuse angle         (d) One-fourth of a revolution.
(v) Reflex angle           (e) Between $$\frac { 1 }{ 4 }$$ and $$\frac { 1 }{ 2 }$$ of a revolution.
–Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (f) One complete revolution.
Solution:
(i) Straight angle     ↔    (c) Half of a revolution.
(ii) Right angle       ↔    (d) One-fourth of a revolution.
(iii) Acute angle      ↔    (a) Less than one-fourth of a revolution.
(iv) Obtuse angle     ↔    (e) Between $$\frac { 1 }{ 4 }$$ and $$\frac { 1 }{ 2 }$$ of a revolution.
(v) Reflex angle      ↔    (f) One complete revolution, right, acute, obtuse or reflex.
Ex 5.3 Class 6 Maths Question 2.
Classify each one of the following angles
Solution:
(a) Acute angle
(b) Obtuse angle
(c) Right angle
(d) Reflex angle
(e) Straight angle
(f) Acute angle
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### Understanding Elementary Shapes Class 6 Ex 5.4
Ex 5.4 Class 6 Maths Question 1.
What is the measure of (i) a right angle (ii) a straight angle?
Solution:
(i) Measure of a right angle = 90°
(ii) Measure of a straight angle = 180°
Ex 5.4 Class 6 Maths Question 2.
Say True or False:
(a) The measure of an acute angle < 90°
(b) The measure of an obtuse angle < 90°
(c) The measure of a reflex angle > 180°
(d) The measure of one complete revolution = 360°
(e) If m ∠A = 53° and ∠B = 35°, then m∠A > m∠B.
Solution:
(a) True
(b) False
(c) True
(d) True
(e) True
Ex 5.4 Class 6 Maths Question 3.
Write down the measures of
(a) some acute angles
(b) some obtuse angles
Solution:
(a) 25°, 63° and 72° are acute angles.
(b) 105°, 120° and 135° are obtuse angles.
Ex 5.4 Class 6 Maths Question 4.
Measure the angles given below using the protractor and write down the measure.
Solution:
(a) 45°
(b) 125°
(c) 90°
(d) ∠1 = 60°, ∠2 = 90°, ∠3 = 125°
Ex 5.4 Class 6 Maths Question 5.
Which angle has a large measure? First estimate and then measure.
Measure of Angle A =
Measure of Angle B =
Solution:
Measure of Angle A = 40°
Measure of Angle B = 60°.
Ex 5.4 Class 6 Maths Question 6.
From these two angles which has large measure? Estimate and then confirm by measuring them.
Solution:
The opening of angle (b) is more than angle (a).
∴ Measure of angle (a) = 45°
and the measure of angle (b) = 60°
Ex 5.4 Class 6 Maths Question 7.
Fill in the blanks with acute, obtuse, right or straight:
(a) An angle whose measure is less than that of a right angle is ……… .
(b) An angle whose measure is greater than that of a right angle is ……… .
(c) An angle whose measure is the sum of the measures of two right angles is ……… .
(d) When the sum of the measures of two angles is that of a right angle, then each one of them is ……… .
(e) When the sum of the measures of two angles is that of a straight angle and if one of them is acute then the other should be ……… .
Solution:
(a) acute
(b) obtuse
(c) straight
(d) acute
(e) obtuse
Ex 5.4 Class 6 Maths Question 8.
Find the measure of the angle shown in each figure. (First estimate with your eyes and than find the actual measure with a protractor).
Solution:
(a) Measure of the angle = 40°
(b) Measure of the angle = 130°
(c) Measure of the angle = 65°
(d) Measure of the angle = 135°.
Ex 5.4 Class 6 Maths Question 9.
Find the angle measure between the hands of the clock in each figure:
Solution:
(i) The angle between hour hand and minute hand of a clock at 9.00 a.m = 90°
(ii) The angle between the hour hand and minute hand of a clock at 1.00 p.m = 30°
(iii) The angle between the hour hand and minute hand of a clock at 6.00 p.m = 180°.
Ex 5.4 Class 6 Maths Question 10.
Investigate: In the given figure, the angle measures 30°. Look at the same figure through a magnifying glass. Does the angle becomes larger? Does the size of the angle change?
Solution:
It is an activity. So try it yourself.
Ex 5.4 Class 6 Maths Question 11.
Measure and classify each angle:
Angle Measure Type ∠AOB ∠AOC ∠BOC ∠DOC ∠DOA ∠DOB
Solution:
Angle Measure Type ∠AOB 40° Acute angle ∠AOC 125° Obtuse angle ∠BOC 85° Acute angle ∠DOC 95° Obtuse angle ∠DOA 140° Obtuse angle ∠DOB 180° Straight angle
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### Understanding Elementary Shapes Class 6 Ex 5.5
Ex 5.5 Class 6 Maths Question 1.
Which of the following are models for perpendicular lines:
(a) The adjacent edges of a table top.
(b) The lines of a railway track.
(c) The line segments forming a letter ‘L’.
(d) The letter V.
Solution:
(a) Yes, the adjacent edges of a table top are the models of perpendicular lines.
(b) No, the lines of a railway tracks are parallel to each other. So they are not a model for perpendicular lines.
(c) Yes, the two line segments of‘L’ are the model for perpendicular lines.
(d) No, the two line segments of ‘V’ are not a model for perpendicular lines.
Ex 5.5 Class 6 Maths Question 2.
Let $$\overline { PQ }$$ be the perpendicular to the line segment $$\overline { XY }$$ . Let $$\overline { PQ }$$ and $$\overline { XY }$$ intersect at in the point A. What is the measure of ∠PAY?
Solution:
Since $$\overline { PQ }$$ ⊥ XY
∴ ∠PAY = 90°
Ex 5.5 Class 6 Maths Question 3.
There are two set-squares in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?
Solution:
The figures of the two set-squares are given below:
The measure angles of triangle (a) are : 30°, 60° and 90°.
The measure angles of triangle (b) are 45°, 45° and 90°.
Yes, they have a common angle of measure 90°.
Ex 5.5 Class 6 Maths Question 4.
Study the diagram. The line l is perpendicular to line m.
(a) Is CE = EG?
(b) Does PE bisects CG?
(c) Identify any two line segments for which PE is the perpendicular bisector.
(d) Are these true?
(i) AC > FG
(ii) CD = GH
(iii) BC < EH
Solution:
(a) Yes,
Since, CE = 2 units and EG = 2 units
Hence, CE = EG.
(b) Yes, PE bisects CG
(c) Required line segments for which PE is perpendicular bisector are: $$\overline { BG }$$ and $$\overline { DF }$$
(d) (i) True (ii) True (iii) True
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### Understanding Elementary Shapes Class 6 Ex 5.6
Ex 5.6 Class 6 Maths Question 1.
Name the types of following triangles:
(а) Triangle with lengths of sides 7 cm, 8 cm and 9 cm.
(b) ∆ABC with AB = 8.7 cm, AC = 7 cm and BC = 6 cm.
(c) ∆PQR such that PQ = QR = PR = 5 cm.
(d) ∆DEF with m∠D = 90°
(e) ∆XYZ with m∠Y = 90° and XY = YZ.
(f) ∆LMN with m∠L = 30° m∠M = 70° and m∠N = 80°.
Solution:
(a) Lengths of the sides of a triangle are given as: 7 cm, 8 cm and 9 cm.
Since, all sides of the given triangle are different.
Hence, it is a Scalene triangle.
(b) Given that: AB = 8.7 cm, AC = 7 cm and BC = 6 cm
Here AB ≠AC ≠BC Hence, ∆ABC is Scalene triangle.
(c) Given that: PQ = QR = PR = 5 cm
Since all sides are equal.
Hence, it is an equilateral triangle.
(d) Given that: In ∆DEF, m∠D = 90°
Hence it is a right angled triangle.
(e) Given that: In ∆XYZ, m∠Y = 90° and XY = YZ
Hence it is a right angled triangle.
(f) Given that: ∆LMN, m∠L = 30°, m ∠M = 70° and m∠N = 80°.
Hence it is an acute angled triangle.
Ex 5.6 Class 6 Maths Question 2.
Match the following:
Measure of triangle                           Type of triangle
(i) 3 sides of equal length                          (a) Scalene
(ii) 2 sides of equal length                         (b) Isosceles right angled
(iii) All sides are of different length                 (c) Obtuse angled
(iv) 3 acute angles                                (d) Right angled
(v) 1 right angle                                   (e) Equilateral
(vi) 1 obtuse angle                                 (f) Acute angled
(vii) 1 right angle with two sides of equal length      (g) Isosceles
Solution:
(i) ↔ (e)
(ii) ↔ (g)
(iii) ↔ (a)
(iv) ↔ (f)
(v) ↔ (d)
(vi) ↔ (c)
(vii) ↔ (b)
Ex 5.6 Class 6 Maths Question 3.
Name each of the following triangles in two different ways: (You may judge the nature of the angle by observation)
Solution:
(a) (i) Acute angled triangle
(ii) Isosceles triangle
(b) (i) Right angled triangle
(ii) Scalene triangle
(c) (i) Obtuse angled triangle
(ii) Isosceles triangle
(d) (i) Right angled triangle
(ii) Isosceles triangle
(e) (i) Acute angled triangle
(ii) Equilateral triangle
(f) (i) Obtuse angled triangle
(ii) Scalene triangle.
Ex 5.6 Class 6 Maths Question 4.
Try to construct triangles using matchsticks. Some are shown here. Can you make a triangle with
(a) 3 matchsticks?
(b) 4 matchsticks?
(c) 5 matchsticks?
(d) 6 matchsticks?
(Remember you have to use all the available matchsticks in each case)
Name the type of triangle in each case.
If you cannot make a triangle, give of reasons for it.
Solution:
(a) Yes, we can make an equilateral triangle with 3 matchsticks.
(b) No, we cannot make a triangle with 4 matchsticks.
(c) Yes, we can make an isosceles triangle with five matchsticks.
(d) Yes, we can make an equilateral triangle with 6 matchsticks.
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### Understanding Elementary Shapes Class 6 Ex 5.7
Ex 5.7 Class 6 Maths Question 1.
Say True or False:
(a) Each angle of a rectangle is a right angle.
(b) The opposite sides of a rectangle are equal in length.
(c) The diagonals of a square are perpendicular to one another.
(d) All the sides of a rhombus are of equal length.
(e) All the sides of a parallelogram are of equal length.
(f) The opposite sides of a trapezium are parallel.
Solution:
(a) True
(b) True
(c) True
(d) True
(e) False
(f) False
Ex 5.7 Class 6 Maths Question 2.
Give reasons for the following:
(a) A square can be thought of as a special rectangle.
(b) A rectangle can be thought of as a special parallelogram.
(c) A square can be thought of as a special rhombus.
(d) Square, rectangles, parallelograms are all quadrilaterals.
(e) Square is also a parallelogram.
Solution:
(a) A square has all the properties as that of rectangle. So, it is a special rectangle.
(b) A rectangle has the same properties as that of parallelogram. So, it is a special parallelogram.
(c) A square has the same properties as that of a rhombus. So, it is a special rhombus.
(d) Square, rectangles and parallelogram are all quadrilateral as they are all enclosed by four sides.
Ex 5.7 Class 6 Maths Question 3.
A figure is said to be regular if its sides are equal in length and angles are equal in measure. Can you identify the regular quadrilateral?
Solution:
Square is only the regular quadrilateral with equal sides and equal angles.
Therefore, square is a regular quadrilateral.
Move to Top → Understanding Elementary Shapes
### Understanding Elementary Shapes Class 6 Ex 5.8
Ex 5.8 Class 6 Maths Question 1.
Examine whether the following are polygons. If any one among them is not, say why?
Solution:
(a) The given figure is not closed. Therefore, it is not a polygon.
(b) The given figure is a polygon.
(c) The given figure is not a polygon because every polygon is enclosed with line segments.
(d) The given figure is not a polygon because it is enclosed by an arc and two line segments.
Ex 5.8 Class 6 Maths Question 2.
Name the polygon.
Make two more examples of each of these.
Solution:
(b) A Triangle
Examples:
(c) A Pantagon
Examples:
(d) A Octagon
Examples:
Ex 5.8 Class 6 Maths Question 3.
Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn.
Solution:
ABCDEF is a rough sketch of a regular hexagon. If we join any three vertices like D, A and B, we get a scalene triangle DAB.
But if we join the alternate vertices, we get an equilateral triangle EAC.
Ex 5.8 Class 6 Maths Question 4.
Draw a rough sketch of a regular octagon. (Using squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.
Solution:
ABCDEFGH is a rough sketch of regular octagon. GHCD is the rectangle formed by joining the four vertices of the given octagon.
Ex 5.8 Class 6 Maths Question 5.
A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals.
Solution:
A B C D E is the rough sketch of a pentagon.
By joining its any two vertices, we get, the following diagonals.
$$\overline { AD }$$ , $$\overline { AC }$$ , $$\overline { BE }$$ , $$\overline { BD }$$ and $$\overline { CE }$$
Move to Top → Understanding Elementary Shapes
### Understanding Elementary Shapes Class 6 Ex 5.9
Ex 5.9 Class 6 Maths Question 1.
Match the following:
Give two examples of each shape.
Solution:
(a) 4 ↔ (ii)
Examples:
(i) An ice-cream cone
(ii) Birthday cap
(b) ↔ (iv)
Examples:
(i) Tennis ball
(ii) Cricket ball
(c) ↔ (v)
Examples:
(ii) A lawn roller
(d) ↔ (iii)
Examples:
(i) Math book
(ii) A brick
(e) ↔ (i)
Examples:
(i) A diamond
(ii) Egypt-Pyramids
Ex 5.9 Class 6 Maths Question 2.
What shape is
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# Math
• ### Eureka Math – 3rd – Lesson 7
Compare fractions – How can we compare fractions? Join Ms. Roose as she partitions wholes into unit fractions and reasons about their size.
• ### Eureka Math – 3rd – Lesson 8
Compare fractions with different wholes – How can we compare fractions when our wholes are different? Join Ms. Roose as she explores comparing fractions with uniquely shaped wholes.
• ### Eureka Math – 3rd – Lesson 9
Find whole from unit fraction -How can we find the whole when we only know the unit fraction? In this lesson, Ms. Roose uses one part as different unit fractions, finding the different wholes as the unit fractions change.
• ### Eureka Math – 1st – Lesson 7
Use symbols to compare numbers – How can you use symbols to compare numbers? Join Mrs. Coleman for this lesson as students are introduced to comparison symbols to compare pairs of two-digit numbers. Students will need a whiteboard and marker or pencil and paper.
• ### Eureka Math – Kindergarten – Lesson 7
Take apart 8 – What are all the ways to take apart 8? Join Mr. Hammer for this lesson as we use story situations, pictures, and number bonds to find number pairs that make 8.
• ### Eureka Math – 2nd – Lesson 7
Repeated addition word problems – Let’s use what we’ve learned to solve word problems! In this lesson, Mr. Waldorf and students will apply their understanding of equal rows and columns and repeated addition. Students will need a whiteboard and marker or pencil and paper.
• ### Eureka Math – Kindergarten – Lesson 5
Sort shapes like a math detective. Do you remember how to sort shapes like a math detective? Join Mr. Hammer for this lesson on sorting shapes to make pictorial and numeric number bonds up to 5.
• ### Eureka Math – Kindergarten – Lesson 6
Take apart 7. What are all of the ways to take apart 7? Join Mr. Hammer for this lesson as we use story situations, sets, and number bonds to find number pairs that make 7.
• ### Eureka Math – 1st – Lesson 5
10 more, 10 less What is 10 more than 21? Join Ms. Lassiter for this lesson that encourages students to use materials and drawings to find 10 more, 10 less, 1 more, and 1 less than a given number.
• ### Eureka Math – 1st – Lesson 6
Compare 2-digit numbers. Is 18 greater than or less than 21? Join Ms. Lassiter for this lesson as students compare numbers by saying the comparison sentence from left to right. Students will need up to 40 objects that they can organize and count, 4 dimes and 10 pennies, as well as a whiteboard and marker or pencil and paper
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# Evaluate the following integrals:
Question:
Evaluate the following integrals:
Solution:
Let $t=\sin ^{2} x$
$d t=2 \sin x \cos x d x$
we know $\sin 2 x=2 \sin 2 x \cos 2 x$
therefore, $d t=\sin 2 x d x$
$\int \frac{\sin 2 x}{\sqrt{\sin ^{4} x+4 \sin ^{2} x-2}} d x=\int \frac{d t}{\sqrt{t^{2}+4 t-2}}$
Add and subtract $2^{2}$ in denominator
$=\int \frac{d t}{\sqrt{t^{2}+4 t-2}}=\int \frac{d t}{\sqrt{t^{2}+2 \times 2 t+2^{2}-2^{2}-2}}$
Let $\mathrm{t}+2=\mathrm{u}$
$\mathrm{dt}=\mathrm{du}$
$\left.\left.=\int \mathrm{dt} / \sqrt{(}(\mathrm{t}+2)^{2}-6\right)=\int \mathrm{dt} / \sqrt{(} \mathrm{u}^{2}-6\right)$
Since, $\int \frac{1}{\sqrt{\left(x^{2}-a^{2}\right)}} d x=\log \left[x+\sqrt{\left.\left(x^{2}-a^{2}\right)\right]+c}\right.$
$\left.=\int \mathrm{dt} / \sqrt{(} \mathrm{u}^{2}-6\right)=\log \left[\mathrm{u}+\sqrt{\mathrm{u}^{2}-6}+\mathrm{c}\right.$
$=\log \left[t+2+\sqrt{(t+2)^{2}-6}+c\right.$
$=\log \left[t+2+\sqrt{(t+2)^{2}-6}+c=\log \left[\sin ^{2} x+2+\right.\right.$
$\sqrt{\left(\sin ^{2} x+2\right)^{2}-6}+c$
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# RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions Exercise 9.5 (Updated for 2021-22)
RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5: The selection of terms in an A.P. is the main focus of the RD Sharma Class 10 Maths Solutions chapter 9 exercise 9.5. Kopykitab created answers to these exercise problems with the primary goal of dispelling doubts and enhancing knowledge in A.P. The RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.5 PDF is also available for download.
## Download RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 Free PDF
RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5
### Access answers to RD Sharma Solutions Class 10 Maths Chapter 9 Exercise 9.5- Important Question with Answers
Question 1.
Find the value of x for which (8x + 4), (6x – 2) and (2x + 7) are in A.P.
Solution:
(8x + 4), (6x – 2) and (2x + 7) are in A.P.
(6x – 2) – (8x + 4) = (2x + 7) – (6x – 2)
⇒ 6x – 2 – 8x – 4 = 2x + 7 – 6x + 2
⇒ -2x – 6 = -4x + 9
⇒ -2x + 4x = 9 + 6
⇒ 2x = 15
Hence x = 152
Question 2.
If x + 1, 3x and 4x + 2 are in A.P., find the value of x.
Solution:
x + 1, 3x and 4x + 2 are in A.P.
3x – x – 1 = 4x + 2 – 3x
⇒ 2x – 1 = x + 2
⇒ 2x – x = 2 + 1
⇒ x = 3
Hence x = 3
Question 3.
Show that (a – b)², (a² + b²) and (a + b)² are in A.P.
Solution:
(a – b)², (a² + b²) and (a + b)² are in A.P.
If 2 (a² + b²) = (a – b)² + (a + b)²
If 2 (a² + b²) = a² + b² – 2ab + a² + b² + 2ab
If 2 (a² + b²) = 2a² + 2b² = 2 (a² + b²)
Which is true
Hence proved.
Question 4.
The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceed the second term by 6, find three terms.
Solution:
Let the three terms of an A.P. be a – d, a, a + d
Sum of three terms = 21
⇒ a – d + a + a + d = 21
⇒ 3a = 21
⇒ a = 7
and product of the first and 3rd = 2nd term + 6
⇒ (a – d) (a + d) = a + 6
a² – d² = a + 6
⇒ (7 )² – d² = 7 + 6
⇒ 49 – d² = 13
⇒ d² = 49 – 13 = 36
⇒ d² = (6)²
⇒ d = 6
Terms are 7 – 6, 7, 7 + 6 ⇒ 1, 7, 13
Question 5.
Three numbers are in A.P. If the sum of these numbers is 27 and the product 648, find the numbers.
Solution:
Let the three numbers of an A.P. be a – d, a, a + d
According to the conditions,
Sum of these numbers = 27
a – d + a + a + d = 27
⇒ 3a = 27
Question 6.
Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.
Solution:
Let the four terms of an A.P. be (a – 3d), (a – d), (a + d) and (a + 3d)
Now according to the condition,
Sum of these terms = 50
⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 50
⇒ a – 3d + a – d + a + d + a – 3d= 50
⇒ 4a = 50
⇒ a = 252
and greatest number = 4 x least number
⇒ a + 3d = 4 (a – 3d)
⇒ a + 3d = 4a – 12d
⇒ 4a – a = 3d + 12d
Question 7.
The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.
Solution:
Question 8.
Divide 56 into four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5: 6. [CBSE 2016]
Solution:
Question 9.
The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the angles.
Solution:
Let the four angles of a quadrilateral which are in A.P., be
a – 3d, a – d, a + d, a + 3d
Common difference = 10°
Now sum of angles of a quadrilateral = 360°
a – 3d + a – d + a + d + a + 3d = 360°
⇒ 4a = 360°
⇒ a = 90°
and common difference = (a – d) – (a – 3d) = a – d – a + 3d = 2d
2d = 10°
⇒ d = 5°
Angles will be
a – 3d = 90° – 3 x 5° = 90° – 15° = 75°
a – d= 90° – 5° = 85°
a + d = 90° + 5° = 95°
and a + 3d = 90° + 3 x 5° = 90° + 15°= 105°
Hence the angles of the quadrilateral will be
75°, 85°, 95° and 105°
Question 10.
Split 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623. [NCERT Exemplar]
Solution:
Let the three parts of the number 207 are (a – d), a and (a + d), which are in A.P.
Now, by given condition,
⇒ Sum of these parts = 207
⇒ a – d + a + a + d = 207
⇒ 3a = 207
a = 69
Given that, product of the two smaller parts = 4623
⇒ a (a – d) = 4623
⇒ 69 (69 – d) = 4623
⇒ 69 – d = 67
⇒ d = 69 – 67 = 2
So, first part = a – d = 69 – 2 = 67,
Second part = a = 69
and third part = a + d = 69 + 2 = 71
Hence, required three parts are 67, 69, 71.
Question 11.
The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles. [NCERT Exemplar]
Solution:
Given that, the angles of a triangle are in A.P.
Question 12.
The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7: 15. Find the number. [NCERT Exemplar]
Solution:
or, d = ± 2
So, when a = 8, d = 2,
the numbers are 2, 6, 10, 14.
We have provided complete details of RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5. If you have any queries related to CBSE Class 10, feel free to ask us in the comment section below.
## FAQs on RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5
### Is the RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 available on the Kopykitab website?
Yes, the PDFs of RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 are available. These solutions are created in a unique method by Kopykitab’s expert faculty.
### What are the key benefits of learning RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.5?
The RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 answers were quite helpful. This makes it simple to clear any questions about arithmetic progression. Easily answer all of the questions that the Class 10 students have given for exercise.
### Where can I get RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 Free PDF?
You can get RD Sharma Class 10 Solutions Chapter 9 Exercise 9.5 Free PDF from the above article.
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