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You are reading an older version of this FlexBook® textbook: CK-12 Middle School Math Concepts - Grade 7 Go to the latest version.
# 4.17: Points in the Coordinate Plane
Difficulty Level: At Grade Created by: CK-12
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Have you ever wanted to find a sunken ship? Take a look at this dilemma.
Gina and Cameron are very excited because they are going to go on a dive to see a sunken ship. The dive is quite shallow which is unusual because most dives are found at depths that are too deep for two junior divers. However, this one is at 40 feet, so the two divers can go to see it.
They have the following map to chart their course. Cameron wants to figure out exactly how far the boat will be from the sunken ship. Each square represents 160 cubic feet of water.
First, he makes a note of the coordinates. Then he can use the map to calculate the distance.
We use coordinate grids like this one all the time. Use the information in this lesson to help Cameron figure out the coordinates of his boat and the sunken ship. Then you will be able to estimate the distance between them.
### Guidance
In working with integers in previous Concepts, we used both horizontal (left-to-right) and vertical (up-and-down) number lines. Imagine putting a horizontal and a vertical number line together. In doing this, you could create a coordinate plane.
In a coordinate plane like the one shown, the horizontal number line is called the axis. The vertical number line is called the axis. The point at which both of these axes meet is called the origin.
We can use coordinate planes to represent points, two-dimensional geometric figures, or even real-world locations. If you think about a map, you will realize that you see a coordinate plane on a map. Then you use coordinates to find different locations. Let’s look at how we can use a coordinate plane.
How do we name points on a coordinate plane?
Each point on a coordinate plane can be named by an ordered pair of numbers, in the form .
• The first number in an ordered pair identifies the coordinate. That coordinate describes the point's position in relation to the axis.
• The second number in an ordered pair identifies the coordinate. That coordinate describes the point's position in relation to the axis.
You can remember that the coordinate is listed before the coordinate in an ordered pair , because comes before in the alphabet.
Write these terms and their definitions in your notebook.
Identifying the coordinates of a point is similar to locating a point on a number line. The main difference is that you need to find the point that corresponds to both of the given coordinates.
Name the ordered pair that represents the location of point below.
Here are the steps to naming the coordinates.
• To start, place your finger at the origin.
• Then move your finger to the right along the axis until your finger is lined up under point . You will need to move 4 units to the right to do that. Moving to the right along a number line means you are moving in a positive direction. So, the coordinate is a positive integer 4.
• Now, move your finger up from the axis until your finger reaches point . You will need to move 5 units up to do that. Moving up along the axis means you are moving in a positive direction. So, the coordinate is a positive integer 5.
The arrows below show how you should have moved your finger to determine the coordinates.
To name the ordered pair, write the coordinate first and the coordinate second. Separate the coordinates with a comma and put parentheses around them, like this (4, 5).
So, the ordered pair (4, 5) names the location of point .
Now that you know how to name points using an ordered pair, it is time learn how to graph them from an ordered pair.
Graphing points on a coordinate plane is similar to naming them. Given an ordered pair, you can move your finger left or right along the axis and then up or down parallel to the axis until you find the location named by the ordered pair. Then you can plot a point at that location.
There are a few key points to remember.
• If the coordinate is positive, move to the right of the origin. If the coordinate is negative, move to the left.
• If the coordinate is positive, move up parallel to the axis. If the coordinate is negative, move down.
Here is another one.
Plot the ordered pair (-5, 3) as a point on the coordinate plane.
Here are the steps:
• The coordinate is a negative integer, -5, so move your finger 5 units to the left along the axis. Your finger should be pointing to the integer -5 on the axis.
• The coordinate is a positive integer, 3, so move your finger 3 units up from the axis.
Plot a point at that location. That point represents the ordered pair (-5, 3).
Sometimes, the points you plot on a coordinate grid will form the vertices of a geometric figure, such as a triangle. Try this one on your own.
#### Example A, B, C
Triangle has vertices , and . Graph triangle on a coordinate plane. Label the coordinates of its vertices.
Solution: Here are the steps to graphing the triangle.
• To plot vertex at (-2, -5), start at the origin. Move 2 units to the left and then 6 units down. Plot and label point .
• To plot vertex at (0, 3), start at the origin. The coordinate is zero, so do not move to the left or right. From the origin, simply move 3 units up. Plot and label point .
• To plot vertex at (6, -3), start at the origin. Move 6 units to the right and then 3 units down. Plot and label point .
Connect the vertices with line segments to show the sides of the triangle, as shown.
Here is the original problem once again. Use this information to help Cameron with the coordinates and the distance.
Gina and Cameron are very excited because they are going to go on a dive to see a sunken ship. The dive is quite shallow which is unusual because most dives are found at depths that are too deep for two junior divers. However, this one is at 40 feet, so the two divers can go to see it.
They have the following map to chart their course. Cameron wants to figure out exactly how far the boat will be from the sunken ship. Each square represents 160 cubic feet of water.
First, he makes a note of the coordinates. Then he can use the map to calculate the distance.
First, here are the coordinates of each item on the map.
The sunken ship is marked at (4, 8).
The dive boat is marked at (-3, 7).
Notice the arrows. Once they get to the sunken ship, Gina and Cameron will swim up 1 unit and over 6 units.
If each unit = 160 sq. feet, then we can multiply
Gina and Cameron will swim through 1120 cubic feet of water from the boat to the sunken ship.
### Vocabulary
Here are the vocabulary words in this Concept.
Coordinate Plane
a plane with four quadrants where locations are marked according to coordinates.
axis
the horizontal line on the coordinate plane
axis
the vertical line on the coordinate plane
Origin
the point where the axis and the axis meet
coordinate
the first coordinate in an ordered pair.
coordinate
the second coordinate in an ordered pair.
### Guided Practice
Here is one for you to try on your own.
This coordinate grid shows locations in Jimmy's city. Name the ordered pair that represents the location of the city park.
Here are the steps to figuring out the coordinates of the city park.
• Place your finger at the origin.
• Next, move your finger to the right along the axis until your finger is lined up above the point representing the city park. You will need to move 2 units to the left to do that. Moving to the left along the axis means that you are moving in a negative direction. Your finger will point to a negative integer, -2, so that is the coordinate.
• Now, move your finger down from the axis until your finger reaches the point for the city park. You will need to move 6 units down to do that. Moving down parallel to the axis means that you are moving in a negative direction. Your finger will be aligned with the negative integer, -6, on the axis, so, that is the coordinate.
The arrows below show how you should have moved your finger to find the coordinates.
So, the ordered pair (-2, -6) names the location of the city park.
### Video Review
Here is a video for review.
### Practice
Directions: Use what you have learned to complete this practice section.
1. Name the ordered pair that represents each of these points on the coordinate plane.
Directions: Below is a map of an amusement park. Name the ordered pair that represents the location of each of these rides.
2. roller coaster
3. Ferris wheel
4. carousel
5. log flume
Directions: Name the ordered pairs that represent the vertices of triangle .
6.
7.
8.
Directions: Name the ordered pairs that represent the vertices of pentagon .
9.
10.
11.
12.
13.
14. On the grid below, plot point at (-6, 4).
15. On the grid below, plot point a triangle with vertices , and .
### Vocabulary Language: English
$x-$axis
$x-$axis
The $x-$axis is the horizontal axis in the coordinate plane, commonly representing the value of the input or independent variable.
$x-$coordinate
$x-$coordinate
The $x-$coordinate is the first term in a coordinate pair, commonly representing the value of the input or independent variable.
$y-$axis
$y-$axis
The $y-$axis is the vertical number line of the Cartesian plane.
Abscissa
Abscissa
The abscissa is the $x-$coordinate of the ordered pair that represents a plotted point on a Cartesian plane. For the point (3, 7), 3 is the abscissa.
Cartesian Plane
Cartesian Plane
The Cartesian plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin.
Coordinate Plane
Coordinate Plane
The coordinate plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin. The coordinate plane is also called a Cartesian Plane.
Ordinate
Ordinate
The ordinate is the $y$-coordinate of the ordered pair that represents a plotted point on a Cartesian plane. For the point (3, 7), 7 is the ordinate.
Origin
Origin
The origin is the point of intersection of the $x$ and $y$ axes on the Cartesian plane. The coordinates of the origin are (0, 0).
Dec 21, 2012
Jan 26, 2016
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In this blog post, we will take a look at how to Solve by quadratic formula. We will also look at some example problems and how to approach them.
When you try to Solve by quadratic formula, there are often multiple ways to approach it. Therefore, it is an essential subject for students to learn. The good news is that there are various ways to solve algebra problems. However, some of these strategies may be more effective than others. Therefore, it is important to find one that works best for you. For example, you can use a step-by-step method or a system that incorporates visualization techniques. Other factors that can help you solve algebra problems include hard work and dedication. Therefore, if you are willing to put in the time and effort needed to master algebra, then it will not be long before you start seeing results.
Linear equations are equations that have only one variable. They may be written in the form y = mx + b or y = mx + b where y and x are variables, m and b are constants. An example of a simple linear equation is: If y = 2x + 2 then y = 4. An example of a more complicated linear equation is: If y = 5x - 7 then y = 0. A solution to a linear equation is the set of values that results in the equation being true when x is fixed. One common way to solve linear equations is to use substitution. Substitution involves replacing each variable with a different value. For example, if x = 3 then by substituting this value for x in the original equation, we obtain the following:
Square roots are one of the most useful tools in math. You can use them to solve a wide range of equations and expressions. For example, you can use square roots to find the value of negative numbers such as -5 or -43. You can also use square roots to find values that don’t fit into a particular type of equation. For example, you can use square roots to find the unknown number that fits between two known values. There are two main ways to solve an equation with a square root. The first is by solving the equation for its variables and then substituting the resulting expression into the original equation. To do this, first rewrite the expression in standard form by taking all of its non-root variables and multiplying both sides by their corresponding factors. Next, take all of the roots (including any common denominators) and multiply each side of the equation by them. Finally, divide both sides by the product of all of those products. This should leave you with an expression that closely resembles the original one. The second way is by using a table of square roots or a calculator that allows you to enter your expression directly into its keypad without having to write it out first. This can be more efficient if you routinely work with similar expressions so you know how to quickly type them in.
Inequality equations are situations where two values are unequal. In other words, the value of one is higher than the other. These equations can be solved in various ways, depending on the situation. One way to solve an inequality equation is to multiply the left-hand side by a fraction. For example, let’s say you have \$5 and \$6 on your balance. If you want to know how much money you have, divide \$5 by 6, which gives you an answer of \$1. If you want to know how much money you have less than \$6, divide 5/6 by 1, giving an answer of 0.333333333. This means that you have \$1 less than what you started with. Another way to solve an inequality equation is to raise both sides to a power. For example, let’s say you have \$5 and \$6 on your balance. If you want to know how much money you have less than \$10, raise both sides to the power of 2 (2x=10), giving an answer of 0.25. This means that you have 25 cents less than what you started with. In order to solve inequalities, we must first understand how they work. When two values are unequal in size or amount, the equation will always be true by definition. When a value is greater than another value,
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Using Graphs to Convey Information
In the first four problems, some factor is changing over time. Sketch graphs of each on some graph paper and answer the questions. In the second four problems, decide what in the problem could be illustrated with a graph and build a graph.
Problem 1: Hansel and Gretel
Hansel and Gretel leave their house and start walking towards their grandmother's house. It is exactly 1 mile to grandmother's house and the entire trip takes about 2 hours (including the 30 minutes spent eating brownies with grandmother). You know the following information about the trip:
• Once Hansel and Gretel start walking, they do not change their speed. They can stop, though.
• After 15 minutes (1/4 of an hour) of walking, Hansel and Gretel were 1/4 of the way to grandmother's house.
• After 30 minutes (1/2 of an hour) of walking, Hansel and Gretel were 1/2 of the way to grandmother's house.
• After 45 minutes (3/4 of an hour) of walking, Hansel and Gretel were 3/4 of the way to grandmother's house.
• Hansel and Gretel reached grandmother's house after exactly one hour of walking.
• When Hansel and Gretel had been at their grandmother's house for 30 minutes, they realized that they had only 30 minutes to walk back home in time for dinner, and leave grandmother's house to start walking back home.
• They arrived home exactly thirty minutes after they left grandmother's.
On a coordinate plane, construct a graph of Hansel and Gretel's journey with distance as the dependent variable and time as the independent variable. Did they walk faster on the way to grandmother's house or on the way back from grandmother's house? How is this illustrated on your graph?
Problem 2: Roller Coaster
You and a group of your friends have gone to an amusement park for the day and decide to ride Thunder Bolt, a new roller coaster ride. You notice that the roller coaster moves significantly faster on some parts of the ride than on others. Using the following information, construct a graph of the ride with velocity as the dependent variable and time as the independent variable. Show the change in the roller coaster's speed over the first 7 seconds of the ride.
• The roller coaster begins at a constant speed of 5 m/s and is travelling uphill. This constant speed is maintained for the first four seconds of the ride.
• In the next second, the roller coaster slows down (the speed decreases at a constant rate) as you reach the top of the hill. The roller coaster reaches the top of the hill exactly five seconds into the hill and stops momentarily before it begins its descent.
• As the roller coaster speeds down the hill, its speed increases continuously for exactly 2 seconds, reaching a maximum of 10 m/s at this point.
What does your graph look like over the interval t=0 seconds to t=4 seconds? What does this mean about the roller coaster's speed over this interval? The roller coaster's speed is decreasing between t=4 and t=5 and increasing between t=5 and t=7. How is this increase and decrease illustrated on your graph (is the slope of the graph positive or negative over each of these intervals)?
Problem 3: Marathon Runner
Your friend ran in an 8 mile marathon last weekend. Using the following information, construct a graph of her run with distance as the dependent variable and time as the independent variable.
• In the first two hours, she ran at a constant speed of 2 miles per hour.
• In the third hour, however, your friend was tired and took an hour break.
• In the fourth hour, she ran twice as fast as in the first two hours to make up for the time she lost during her break.
How far did your friend run in the first two hours of the race? How far did she run in the fourth hour of the race? How does your graph illustrate that she ran twice as fast in the fourth hour as in the first two hours?
Problem 4: Race Car Driver
Your uncle is a race car driver. One day you decide to go watch a race. Your uncle began the race well but got a flat tire 2.5 minutes into it and had to stop. You collected the following data while you were watching the race:
Time Distance 0 0 0.5 0.5 1 1 1.5 1.75 2 3 2.5 4.5 3 4.5 3.5 4.5 4 4.5 4.5 4.5 5 4.5
Construct a graph of your uncle's race with distance as the dependent variable and time as the independent variable. What was happening to your uncle's speed (not distance) between t=0 and t=2.5? Between t=2.5 and t=5? How does your graph illustrate these changes in speed?
Problem 5: Trip to School
Andrew rides his bike to school every day. He is usually not fully awake by the time he leaves, so it takes him time to gain his speed. After about 5 minutes he reaches the stop sign at the end of his road, and has to pause for a minute. He is going to get to school early so he just peddles along for a couple of moments. Suddenly his friend Dietrich comes up behind him and they start racing. They are both going pretty fast, but then Andrew remembers he left his science project on the table at home. Andrew quickly turns around and speeds back home to get it. Now he has only 6 minutes until the start of school, and is going to be late if he doesn't hurry. Andrew pedals as fast as he can all the way back to school and makes it to class just in time for the first bell.
Problem 6: Foxes and Rabbits
Jason is an amateur naturalist and is counting the number of foxes and rabbits he sees in the park near his house. On his first time out he finds 200 rabbits but only 30 foxes. When he goes out to count the next year he finds that the foxes have eaten 75 rabbits and the fox population has increased to 50. The next year he counts only 25 foxes and the rabbit population is down to 50. Jason is concerned that something is happening to the rabbits and foxes. When he goes out the next year he finds there are 150 rabbits but the foxes are down to just 10. In his fifth year of counting he finds 200 rabbits and 30 foxes just like in the first year. Jason starts to see that as the number of rabbits goes up the foxes eat more. That means the number of rabbits goes down and the foxes goes up. But as the rabbits go down farther the foxes start to go down as well. With fewer foxes now the rabbit population can start to go up and start the cycle again.
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# Square Root of 5 – Long Division, Average & Equation Method
The square root of 5! Is it possible to calculate? Many times, we were surprised and shocked as well. We know, square root of 4, or 9 or 16, and we can do it instantly but the square root of 5! Is it difficult? Let’s welcome to our interesting session, square root of 5.
Understanding the square root is the first step to solving the square root of 5. When multiplied by itself, a square root is a number that gives the original number as the result. When multiplied by itself, five gives the prime number 5. It is a positive, real number. This number is also called the Principal square root of 5 to distinguish it from the negative numbers with the same properties. 2.2360 is the value of square root 5.
A negative number’s square root is not real. It is an imaginary number. Therefore, understanding the square roots of complex numbers requires knowledge of complex numbers.
## How Do You Express Square Root of 5?
To write a square root equation, use the radical symbol or radical sign. Generally, we are interested in the number under or after a radical. The radicand is the number after the radical. Symbols for square roots are believed to derive from the first letter of the word ‘radix’ in Greek and Latin, which refers to a root or a base.
It is the square root of 5 which when multiplied by itself yields the prime number 5. It is more accurately known as the main square root of 5 to distinguish it from the negative number exhibiting the same characteristic. This number appears in the fractional representation of the golden ratio.
The square root of 5 has recurring decimal values of 2.2366679775 as the first ten digits. In exponent form, 51/2 or 50.5 is the radical form of the square root of 5.
The square root of 5 cannot be simpler than it currently is. The square root of a non-perfect square integer can be calculated using the long division method. Calculating the root of a number is traditionally done this way. There is no rational algebraic number for the square root of 5.
## Square Root of 5
Let us try to understand, square root of 5. If we calculate it in a calculator, we see a long list of irrational numbers, and it’s not possible to remember or even write, as there is no ending of digits. So, what we do in mathematics? We simply write √5.
Many times, writing √5 will not be able to serve the purpose, and its numeric valve to be considered. Here, we must calculate the numeric value of the square root of 5.
## What is exactly the square root of 5?
The square root of 5 is defined as a number when we multiply with the same number, gives the result of 5. It means,
X x √X =X
It is written by √5.
So, as per definition, we can say,
√5x√5 =5
There is much research going on the numbers of digits after the decimal point in the square root of 5. Many types of research show even a billion digits in decimal places.
We have indicated the value of the square root of 5 up to 20 decimal places,
√5 = 2.23606797749978969640
## Example of Square Root of 5
There are so many examples where we have to use, √5. Let us consider one simple equation to solve the value of x,
• 5x2 =2x2+15
• Or, 3x2 =15
• Or, x2 =15/3=5
• Or, x = √5
We used to write,
• 1+√5
• 5+√5
• 2-√5
Or sometimes, we multiply,
• 5x√5=5√5
• √5x√5x√5=5√5
• 10x5x√5=50√5 and so on.
Hence, it is clear the whenever we are getting the square root of 5; we simply write √5. How, how do we specify the exact value of √5?
For example, if you need 10√5kg of sugar, how do you measure it?
Yes! It is possible only when we can find out the numerical value of the square root of 5.
## How to Find the Square Root of 5?
There is various method to find out the square root of 5, it is the same process like square root of 3 or square root of 8.
• Long division method
• Average method
• Equation method
### Calculate the Square Root of 5 – The Long Division Method
Whenever we try to find a square too of any number other than a perfect square, we opt for this method. This method is a little long but easy. This method is also called a long division method as well. Let’s learn all the steps,
Step#1:Write the number 5 as 5.00000000, remember we are not doing any change of the value of 5, as both numerical values are the same.
Step#2: Order the zeros, like, 5.00 00 00 00 as both are the same, and it will help in the division method.
Step#3: Now, try to find out the perfect square just below 5. If we analyze, 0,1,2,3 & 4 are numbers below 5, and 4 is the perfect square.
As we know, 4 = 2×2 = 22
Step#4: Find out the square root of the perfect square below 5. Here, it is 4, and the square root of 4 is 2.
Step#5: Make a division table and write the number 2 in the quotient place as well as in the divisor place.
Step#6: Make the division considering,
• Divisor 2
• Quotient 2
• Dividend 5
So, by the division method, 2 multiplied by 2 gives 4. Subtract 4 from 5; there will be remainder 1.
Step#7: As 1 is the remainder, we must carry down 2 zeros after 1. So, it will become 100. Next, carry down two zeros and write it down after 1. Simultaneously decimal points will come after 2 in the quotient.
Step#8: Now, we will add 2 in the divisor and make it 4.
Step#9: We have to select a number next to 4 in such a way that if we multiply the new combined number with the new number, then the value should be equal to 100 or less than that.
Now,
• If we take 1, the combination number will be 41. So, 41×1=41.
• If we take 2, the combination number will become 42. So, 42×2=84.
• If we take 3, the combination number will become 43. So, 42×3=126, which is more than 100.
Hence, the number 2 is acceptable, and we get 84. Now, we have to subtract 84 from 100, and we get the remainder, 16. So, we get 2 after the decimal.
Step#10: As 16 is the remainder, we should carry down 2 zeros after 16. So, it will become 1600.
Step#11: Now, we get 42, So, 42+2=44. We have to select another number next to 44 so that if we multiply the new combined number with the new number, the value should be equal to 1600 or less than that.
Now,
• If we take 1, the combination number will be 441. So, 441×1=441.
• If we take 2, the combination number will become 442. So, 442×2=884.
• If we take 3, the combination number will become 443. So, 443×3=1329.
• If we take 4, the combination number will become 444. So, 444×4=1776, this is more than 1600.
Hence, if we take 3, the combination number will become 443. So, 443×3=1329, which is less than 1600. 3 will be added in the second decimal point.
Hence, the number 3 is acceptable, and we get 443. Now, we have to subtract 1329 from 1600, and we get the remainder, 271.
Step#12: As 271 is the remainder, we should carry down 2 zeros after 271. So, it will become 27100.
Step#13: Now, we get 443, So, 443+3=446. We have to select another number next to 446 in such a way that if we multiply the new combined number with the new number, then the value should be equal to 27100 or less than that.
Now,
• If we take 4, the combination number will be 4464. So, 4464×4=17856.
• If we take 5, the combination number will become 4465. So, 4465×5=22325.
• If we take 6, the combination number will become 4466. So, 4466×3=26796.
• If we take 4, the combination number will become 4467. So, 4467×7=31269, this is more than 27100.
Hence, if we take 6, the combination number will become 4466. So, 4466×6=26796, which is less than 27100. 6 will be added in the third decimal point.
Hence, the number 6 is acceptable, and we get 4466. Now, we have to subtract 26796 from 27100, and we get the remainder, 304.
Step#14: As 304 is the remainder, we should carry down 2 zeros after 304. So, it will become 30400.
Step#15: Now, we get 4466, So, 4466+6=4472. We have to select another number next to 4472 in such a way that if we multiply the new combined number with the new number, then the value should be equal to 30400 or less than that.
Now,
• If we take 0, the combination number will be 44720, which is more than 30400.
Hence, the number 0 can be taken as 4 decimal points.
Step#16: So, with the help of the above steps, we get the quotient as 2.2360, or 2.236 (we say)
You can watch our VIDEO on it,
### Calculate the Square Root of 5 – The Average Method
In the ancient time itself, the average method was used to find the square root of 5. This is very simple, and we get the value of the square root of 5 based on averages in a few steps. Let us see all steps,
Step#1: We have to check the perfect squares just below 5 and above 5. In the case of number 5, 4 is the perfect square below 5, and 9 is the perfect square above 5.
4 5 9
Step#2: Find out the square root of the perfect square of both the numbers.
22 & 32
So, square roots are 2 & 3.
Step#3: Number 5 is between 4 & 9. Hence, naturally, the square root of 5 also will be between the square root of 4, i.e., 2, and the square root of 9, i.e., 3.
4<5<9
22<5<32
Or, 2<Square root of 5<3.
Step#4: Divide the number 5 by any of one number, 2 or 3.
Let us divide 5 by 2,
So, 5/2=2.5
Step#5: Hence, we get a value of 2.5, and we will consider divisor 2 which was divided. Now, we will calculate the average value of them,
Average value = (2+2.5)/2=4.5/2=2.25
Step#6: In case you required further precise value, the process to be continued. The above 2.25 value is almost close, and we can select the same.
### Calculate the Square Root of 5 – The Equation Method
In the method, you have to remember a simple equation. It’s not very complex, but you remember, you can calculate the square root of any number in a few seconds.
Equation
√(x±y) = √x ±y/(2√x)
Here, x or y is the perfect square. It is better to remember all the basic perfect square table,
Step#1: Split the digit into two numbers in such a way that x can be written as a perfect square.
We are going to calculate the square root of 5. Hence, 5 can be written as 4+1 since we know from the perfect square table, 4 is a perfect square, which means x=4 and y=1.
Step#2: Input the value of x & y in the equation.
√(x±y) = √x ±y/(2√x)
or, √(4+1) = √4 +1/(2√4)
or, √5 = 2 +1/(2×2)
or, √5 = 2 +0.25 =2.25 (approx value)
## Purpose of Square Root of 5
The square root is among the most useful integers, despite being difficult to visualize. This function is extremely useful in all types of physics equations. Additionally, square roots are often utilized by statisticians to analyze associations between multiple sets of data.
## Why a Square root of 5 is Irrational Number?
Rational numbers are numbers that can be expressed as a ratio of two integers, for instance p/q and q not equal to 0. Let’s consider the square root of 25. The square root of 25 equals 5 = 5/1. This makes the square root of 25 a rational number. We now need to determine its square root.
Irrational numbers are numbers that cannot be expressed as a ratio of two integers and therefore cannot be expressed as a ratio of two whole numbers. The square root of 5 is an irrational number, since 5 is not a perfect square.
## How Do You Find Square Root using Guess and Check Method?
If you are finding a decimal approximation to, say, square root 5, make an initial guess, then square the guess, and then improve your guess if you get close. It is very helpful to teach the concept of square root because it involves squaring the guess or multiplying the number itself.
## Importance of Square roots without Calculator
The vast majority of people in today’s society feel that since calculators find square roots, kids do not need to learn how to find square roots on paper. Students will actually be able to explain and remember the square root concept better if they learn at least the guess and check method for finding the square root.
While your math book may entirely ignore finding square roots without using a calculator, you might want to consider giving students the opportunity to at least practice guessing a square root and checking it. It deals with the concept of square root, so it should be part of the curriculum. It is possible to use either a simple calculator without a square root button or paper and pencil to implement the guess-and-check method, depending on the situation and the students.
## Conclusion
Hence, we have learned how to calculate the square root of 5 along with different methods. Any doubt or if you have any suggestions, please let us know.
|
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# Solving Equations
We are now ready to move on to slightly more sophisticated examples.
Example
Find the solution to the equation
5( x - 3) - 7(6 - x ) = 24 - 3(8 - x ) - 3
Solution
Removing the brackets from both sides first and then simplifying:
5( x - 3) - 7(6 - x ) = 24 - 3(8 - x ) - 3
5 x - 15 - 42 + 7 x = 24 - 24 + 3 x - 3
5 x + 7 x - 15 - 42 = 3 x - 3
12 x - 57 = 3 x - 3 .
12x = 3x - 3 + 57 = 3x + 54
Subtracting 3x from both sides:
12x - 3x = 54 o r 9x = 54 giving x = 6 .
Exercise
Find the solution to each of the following equations.
(a) 2 x + 3 = 16 - (2 x - 3)
(b) 8( x - 1) + 17( x - 3) = 4(4 x - 9) + 4
(c) 15( x - 1) + 4( x + 3) = 2(7 + x )
Solutions
(a) 2 x + 3 = 16 - (2 x - 3) = 16 - 2 x + 3 = 19 - 2 x
Now add 2x to both sides and subtract 3 from both sides
2 x + 3 = 19 - 2 x
4 x + 3 = 19
4 x = 19 - 3
4 x = 16
and the solution is x = 4 . This can be checked by putting x = 4 in both sides of the first equation above and noting that each side will have the value 11.
(b)
8( x - 1) + 17( x - 3) = 4(4 x - 9) + 4
8 x - 8 + 17 x - 51 = 16 x - 36 + 4
25 x - 59 = 16 x - 32
25 x - 16 x - 59 = - 32
9 x - 59 = - 32
9 x = 59 - 32
9 x = 27
x = 3 .
Inserting x = 3 into the equation we can check that both sides have the value 16.
(c)
15( x - 1) + 4( x + 3) = 2(7 + x )
15 x - 15 + 4 x + 12 = 14 + 2 x
19 x - 3 = 2 x + 14
19 x - 2 x - 3 = 14
17 x - 3 = 14
17 x = 14 + 3 = 17
x = 1
Inserting x = 1 into the equation we can check that both sides have the value 16.
Quiz
Which of the following is the solution to the equation
5x - (4x - 7)(3x - 5) = 6 - 3(4x - 9)(x - 1) ?
(a) - 2 (b) - 1 (c) 2 (d) 4
Solution:
First expand the brackets separately using FOIL (see the package on Brackets ):
These can now be substituted, carefully, into the equation:
5 x - [ (4 x - 7)(3 x - 5) ] = 6 - 3 [ (4 x - 9)( x - 1) ]
5 x - [ 12 x - 41 x + 35 ] = 6 - [ 4 x - 13 x + 9 ]
5 x - 12 x + 41 x - 35 = 6 - 4 x + 13 x - 9
- 12 x + 46 x - 35 = - 12 x + 39 x - 21 .
Notice the extra pair of square brackets in the first equation above. These are to emphasise that the negative sign multiplies all of the parts inside the [ ] brackets. The procedure now follows in an obvious manner. Add 12 x to both sides, subtract 39 x from both sides then add 35 to both sides:
- 12 x + 46 x - 35 = - 12 x + 39 x - 21
46 x - 35 = 39 x - 21
46 x - 39 x - 35 = - 21
46 x - 39 x = 35 - 21
7 x = 14
x = 2 .
|
## Direct & Inverse Proportion
Image Credit: Pixabay.com
# Variation
When two quantities are related in such a way that change in one quantity causes change in the other, then the quantities are said to vary with each other.
ℹ Some situations where variation in one quantity brings the variation in the other quantity :
(i) More the number of coins more the height of the pile.
(ii) More the rate of interest more is the interest charged.
(iii) More the distance more the time taken to cover it for the same speed.
(iv) More the speed of a vehicle less the time taken to cover the same distance.
(v) More the number of workers, less will be the time taken to complete the work.
The variation can be direct or indirect.
First three examples are of direct variation and last two are the examples of indirect variation.
# Direct Variation / Proportion
Two quantities are said to be in direct proportion if they increase (decrease) together in such a way that the ratio of their corresponding values remains constant.
➢ Let x and y be two quantities such that change in x leads to corresponding change in y in same proportion.
x x1 x2 x3 y y1 y2 y3
then,
i.e.,
(K is a constant i.e., ratio of x and y is always same for direct proportion)
Ex – A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Sol –
x (no. of bottles) 840 x2 y (time in hours) 6 5
It is the case of direct variation,
∴ Number of bottles would be filled in 5 hours = 700.
# Indirect Variation / Proportion
Two quantities are said to be in inverse proportion if an increase in one causes a proportional decrease in other (and vice-versa) in such a way that the product of their corresponding values remains constant.
➢ Let x and y be the two quantities such that change in x leads to corresponding change in y in inverse proportion.
x x1 x2 x3 y y1 y2 y3
then,
i.e.,
(K is a constant i.e., product of x and y is always same for indirect proportion)
Ex-If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?
Sol-
x (No. of children) 24 24-4 y (No. of sweets) 5 y2
It is the case of inverse variation, i.e., when number of person (x) decreases then number of sweets for each person (y) increases.
Each children will get 6 sweets.
|
Vedic Maths
Vedic maths: Square
Example : Find square of 12
Step 1
1010 is the nearest power of 1010 which can be taken as our base.
The deviation to our base =12−10=2=12−10=2 (To find the deviation, just remove the leftmost digit “11” and you will get it quickly).
Left side of the answer is the sum of the number and deviation. Hence, left side of the answer = 12 + 2 = 14
Step 2
Our base 1010 has a single zero. Therefore, right side of the answer has a single digit and that can be obtained by taking the square of the deviation.
Hence, right side of the answer =22=4=22=4
Vedic maths: Square roots
Square root of any number means to get a number which is multiplied by itself gives the given number. In the conventional method of finding the square root, the divisor goes on becoming larger at each step. This increases the calculation time as well as the complexity of the problem. Here, we shall try to learn some speedy Vedic Methods of finding the square roots of perfect square numbers. Before proceeding for finding square roots, let us have a look into the known facts of squares and square roots.
The basic rules for extracting square roots are :
• The given number is arranged in two-digit groups from right to left; and a single digit (if any) left over at the left and is counted as a group by itself.
• The number of digits in the square root will be the same as the number of twodigit groups in the given number including a single digit group (if any). Thus, 36 will count as one group, 169 as two groups and 1225 as two groups.
• If the number contains n digits then the square root will contain n / 2 (when n is even) and n + 1 / 2 (when n is odd) digits. Thus, one or two digit number will have the square root of one digit, three and four digit number will have the square root of two digits, 5 and 6 digit number will have the square root of 3 digits and so on.
1² = 1, 2²=4 , 3²=9 , 4²=16 , 5²=25, 6²=36, 7²=49, 8²=64 , 9²=81
This means :
• unit digit of the perfect square number is 1, 4, 5, 6, 9 or 0.
• a perfect square number cannot end in 2, 3, 7 or 8.
• the relation between the unit digit of a perfect square number and the unit digit of its square root is as follows :
Unit digit of the number 1 4 5 6 9 0
Unit digit of square root 1 or 9 2 or 8 5 4 or 6 3 or 7 0
• If there are odd number of zeros at the end (on right side) of a number, then it will not be a perfect square.
Cube roots in vedic maths
Find Cube Root of 4913
Step 1
Identify the last three digits and make groups of three three digits from right side. i.e., 49134913 can be written as
4, 913
Step 2
Take the last group which is 913.913. The last digit of 913913 is 3.3.
Remember point 2, If last digit of perfect cube=3,=3, last digit of cube root =7=7
Hence the right most digit of the cube root =7
Step 3
Take the next group which is 44
Find out which maximum cube we can subtract from 44 such that the result ≥0≥0
We can subtract 1 3 = 1 from 4 because 4 − 1 = 3 (If we subtract 2 3 = 8 from 4 , 4 – 8 = − 4 which is < 0 )
Hence the left neighbor digit of the answer = 1 i.e., answer = 17
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FutureStarr
22 Out of 27 As a Percentage ORR
## 22 Out of 27 As a Percentage
via GIPHY
Let's take a look at other house hold word problems. 5 out of 8 is: 22/27 or "?
### Simple
via GIPHY
In calculating 18% of a number, sales tax, credit cards cash back bonus, interest, discounts, interest per annum, dollars, pounds, coupons,18% off, 18% of price or something, we use the formula above to find the answer. The equation for the calculation is very simple and direct. You can also compute other number values by using the calculator above and enter any value you want to compute.percent dollar to pound = 0 pound
Since she got 2 out of 20 incorrect, we can first figure out the percent incorrect. We can either find the decimal for 2/20, or make 2/20 a fraction with 100 in the denominator. In this case, it is simpler to do the latter. We need to multiply 20 by 5 to get 100, thus we multiply 2 by 5 to get 10. Then, we have that 2/20 is equivalent to 10/100. She lost 10 percent on her math quiz, leaving her with a score of 90%. (Source: www.varsitytutors.com)
via GIPHY
One area that often catches people out is year-on-year percentage increases. For example, Freya has £10 and each year this increases by 5%. How much will she have after 3 years? Some people can be tempted to add together the 5% for the 3 years i.e. 15% and multiply the £10 by 15% giving £11.5. This is incorrect. The correct way of approaching questions like this is to remember that EACH year the initial £10 increased by 5%. So at the end of year 1, Freya would have £10 x 1.05 = £10.5. At the end of year 2, she would have £10.5 x 1.05 = 11.025, and so on. It is important to add in each of these steps to arrive at the correct answer.
Be careful! Is this the answer we are looking for? No, we are not yet finished. The calculated percentage (i.e., the proportion of 200 to 50) is 400%. However, we want to calculate the increase or decrease in value. For this we still have to subtract the basic value (of 100%) from the 400% we obtained from our calculations. In other words, there is an increase of 400% – 100% = 300% going from 50 to 200. (Source: www.blitzresults.com)
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In mathematics, a tangent vector is a vector that is tangent to a curve or surface at a given point. Tangent vectors are described in the differential geometry of curves in the context of curves in Rn. More generally, tangent vectors are elements of a tangent space of a differentiable manifold. Tangent vectors can also be described in terms of germs. Formally, a tangent vector at the point ${\displaystyle x}$ is a linear derivation of the algebra defined by the set of germs at ${\displaystyle x}$.
## Motivation
Before proceeding to a general definition of the tangent vector, we discuss its use in calculus and its tensor properties.
### Calculus
Let ${\displaystyle \mathbf {r} (t)}$ be a parametric smooth curve. The tangent vector is given by ${\displaystyle \mathbf {r} '(t)}$, where we have used a prime instead of the usual dot to indicate differentiation with respect to parameter t.[1] The unit tangent vector is given by
${\displaystyle \mathbf {T} (t)={\frac {\mathbf {r} '(t)}{|\mathbf {r} '(t)|))\,.}$
#### Example
Given the curve
${\displaystyle \mathbf {r} (t)=\left\{\left(1+t^{2},e^{2t},\cos {t}\right)\mid t\in \mathbb {R} \right\))$
in ${\displaystyle \mathbb {R} ^{3))$, the unit tangent vector at ${\displaystyle t=0}$ is given by
${\displaystyle \mathbf {T} (0)={\frac {\mathbf {r} '(0)}{\|\mathbf {r} '(0)\|))=\left.{\frac {(2t,2e^{2t},-\sin {t})}{\sqrt {4t^{2}+4e^{4t}+\sin ^{2}{t))))\right|_{t=0}=(0,1,0)\,.}$
### Contravariance
If ${\displaystyle \mathbf {r} (t)}$ is given parametrically in the n-dimensional coordinate system xi (here we have used superscripts as an index instead of the usual subscript) by ${\displaystyle \mathbf {r} (t)=(x^{1}(t),x^{2}(t),\ldots ,x^{n}(t))}$ or
${\displaystyle \mathbf {r} =x^{i}=x^{i}(t),\quad a\leq t\leq b\,,}$
then the tangent vector field ${\displaystyle \mathbf {T} =T^{i))$ is given by
${\displaystyle T^{i}={\frac {dx^{i)){dt))\,.}$
Under a change of coordinates
${\displaystyle u^{i}=u^{i}(x^{1},x^{2},\ldots ,x^{n}),\quad 1\leq i\leq n}$
the tangent vector ${\displaystyle {\bar {\mathbf {T} ))={\bar {T))^{i))$ in the ui-coordinate system is given by
${\displaystyle {\bar {T))^{i}={\frac {du^{i)){dt))={\frac {\partial u^{i)){\partial x^{s))}{\frac {dx^{s)){dt))=T^{s}{\frac {\partial u^{i)){\partial x^{s))))$
where we have used the Einstein summation convention. Therefore, a tangent vector of a smooth curve will transform as a contravariant tensor of order one under a change of coordinates.[2]
## Definition
Let ${\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} }$ be a differentiable function and let ${\displaystyle \mathbf {v} }$ be a vector in ${\displaystyle \mathbb {R} ^{n))$. We define the directional derivative in the ${\displaystyle \mathbf {v} }$ direction at a point ${\displaystyle \mathbf {x} \in \mathbb {R} ^{n))$ by
${\displaystyle \nabla _{\mathbf {v} }f(\mathbf {x} )=\left.{\frac {d}{dt))f(\mathbf {x} +t\mathbf {v} )\right|_{t=0}=\sum _{i=1}^{n}v_{i}{\frac {\partial f}{\partial x_{i))}(\mathbf {x} )\,.}$
The tangent vector at the point ${\displaystyle \mathbf {x} }$ may then be defined[3] as
${\displaystyle \mathbf {v} (f(\mathbf {x} ))\equiv (\nabla _{\mathbf {v} }(f))(\mathbf {x} )\,.}$
## Properties
Let ${\displaystyle f,g:\mathbb {R} ^{n}\to \mathbb {R} }$ be differentiable functions, let ${\displaystyle \mathbf {v} ,\mathbf {w} }$ be tangent vectors in ${\displaystyle \mathbb {R} ^{n))$ at ${\displaystyle \mathbf {x} \in \mathbb {R} ^{n))$, and let ${\displaystyle a,b\in \mathbb {R} }$. Then
1. ${\displaystyle (a\mathbf {v} +b\mathbf {w} )(f)=a\mathbf {v} (f)+b\mathbf {w} (f)}$
2. ${\displaystyle \mathbf {v} (af+bg)=a\mathbf {v} (f)+b\mathbf {v} (g)}$
3. ${\displaystyle \mathbf {v} (fg)=f(\mathbf {x} )\mathbf {v} (g)+g(\mathbf {x} )\mathbf {v} (f)\,.}$
## Tangent vector on manifolds
Let ${\displaystyle M}$ be a differentiable manifold and let ${\displaystyle A(M)}$ be the algebra of real-valued differentiable functions on ${\displaystyle M}$. Then the tangent vector to ${\displaystyle M}$ at a point ${\displaystyle x}$ in the manifold is given by the derivation ${\displaystyle D_{v}:A(M)\rightarrow \mathbb {R} }$ which shall be linear — i.e., for any ${\displaystyle f,g\in A(M)}$ and ${\displaystyle a,b\in \mathbb {R} }$ we have
${\displaystyle D_{v}(af+bg)=aD_{v}(f)+bD_{v}(g)\,.}$
Note that the derivation will by definition have the Leibniz property
${\displaystyle D_{v}(f\cdot g)(x)=D_{v}(f)(x)\cdot g(x)+f(x)\cdot D_{v}(g)(x)\,.}$
|
When Parabolas Sketch Parabolas
by Patty Wagner
Most of us are familiar with the graph of a parabola. Here is an example as a refresher:
Something we may not all be familiar with is the interesting behavior that results when the coefficient of the middle term is changed.
This graph is y = x2 + nx + 1, where n is changing incrementally from -5 to +5. Focus carefully on the vertex of the parabola and try to ignore the rest of the graph. Notice how the point of the vertex appears to sketch an inverted parabola as it changes.
Is this actually what is happening? If so, why? And what relation does it have to our original equation?
If we find the vertices of each parabola from our changing graph where n equals an integer from -5 to 5 and plot those points on a new graph we can get an accurate view of the movement of the vertices.
If we imagine a line connecting each of our points, we can see that the path does indeed form a parabola that opens downward. We also can easily see that this parabola is symmetric around the y-axis, making it very easy for us to mathematically describe it since we know parabolas that are symmetrical around the y-axis take the form y = ax2 + c. The variable c is easily identified as +1 since the parabola is shifted upwards by 1. We also know that a is negative since the parabola opens downwards. But what is the value of a? The parabola appears to be of the standard form y = -x2 + 1, but we can't take that for granted. If we insert one of our points (-2,-3), we see that our guess is correct: -3 = -22 +1
If we consider this phenomenon carefully, we realize that this second parabola (we'll call it parabola #2) creates a predictable pattern that allows us to make some assumptions. Look at the moving graph to the right.
The red parabola is a graph of the equation y = x2 + bx + 5.
The purple parabola is y = x2 + bx + 2.
The green parabola is y = x2 + bx + 1.
In each of these graphs, b is varying from -5 to 5.
Just as previously, if we look only at the vertices we can see that our parabola #2 always takes the form of y = -x2 + c, where c is the value in our equation y = x2 + bx + c. Let's explore what happens when we change the value of b in an equation of the form y = ax2 + bx + c:
We can see that not only does the opening of our parabola change when a is changed (as we would expect), but the opening of parabola #2 also changes in the same manner. This would lead us to expect that parabola #2 always takes the form y = -ax2 + c. If this is true, we would expect parabola #2 to trace an upwards path if a is negative:
Here we have y = -x2 + bx + 1 in purple and b is changing from -5 to 5. The red parabola is y = x2 + 1. It does indeed appear that the vertex of our changing parabola is following the path of an inner parabola (parabola #2).
Our original conjecture appears to be correct in that changing the coefficient of the second term in a quadratic equation produces a pattern in the form of a second parabola (parabola #2). Now we can ask why this is the case and what relation does this second parabola have to our original equation?
Recall that maximum/minimum of a function can be found by setting the derivative of the function to zero. If y = ax2 + bx +c, then y' = 2ax + b. Setting y' to zero yields 0 = 2ax + b. Therefore, the maximum/minimum of our parabola occurs when x = -b/2a. This is the x-coordinate of our vertex. If x = -b/2a, then y = a(-b/2a)2 + b(-b/2a) + c = b2/(4a) - 2b2/(4a) + 4ac/4a = [-b2/(4a)] +c.
If we analyze this general form for the vertex: (-b/2a, [-b2/(4a)] +c), we can see why we generate a second parabola when b is changed. The x-coordinate varies linearly and the y-coordinate varies quadratically as b changes. This gives us our parabolic shape. We can also easily see that the maximum (or minimum) value of y will always occur when b = 0. That explains the symmetry around the y-axis.
So, what does this second parabola have to do with our original equation? Is it useful in any way?
When we work with quadratic equations, we generally are interested in the "roots" - the point at which the parabola crosses the x-axis on a graph. Generally we use the quadratic equation or some other algorithm to find the point at which y = 0. Many of us have even justified the quadratic equation mathematically through proof. But let's consider the concept on a different level by using our parabola within a parabola. Consider the graph below:
This graph shows some of the parabolas that are formed when an equation of the form y = ax2 + bx + c is varied according to b. In this case, a =1, c = 1 and b is varying from -3 to 3. Each value of b is represented by a different color parabola and our parabola #2 is noted in black.
Notice that some of these parabolas cross the x-axis, some just touch the x-axis, and some do not cross the x-axis at all. We will find real roots for each of the equations except the one represented by the red parabola. If we examine our parabola #2 which is depicted by the black parabola, we can predict graphically which parabolas will have real roots depending on the value of b. As b gets closer to zero from either side, the graph is raised above the x-axis and we have no roots. As the absolute value of b becomes greater, the vertex will slide down our parabola #2, guaranteeing that we will continue to have real roots.
So let's put this all together but identifying the different situations that we can have:
a positive and c positive
a positive and c negative:
a negative and c negative
a negative and c positive:
All these moving graphs can get confusing, but if we look at them one at a time, we see that when a and c are of opposite sign (positive or negative), we will always have real roots no matter what the value of b. (We can also see this will be the case if c = 0.) However, when a and c are the same signs, whether the roots of the equation are real depends on the value of b.
Let's focus on these two cases (represented in the red graphs above). We've seen that when y = ax2 + bx + c, changing the value of b creates a parabola inverted to y = ax2 + bx + c. Its maximum/minimum occurs when b = 0. This suggests that regardless of the value of a and c, we can find b that, if its absolute value is sufficiently large, will result in an equation with real roots. Where does this occur?
If we trace parabola #2, we will obtain real roots when the vertex of parabola #2 falls on the x-axis and every point further away from the vertex of parabola #2. So the absolute value of b must be, at a minimum, the value obtained when the vertex of y = ax2 + bx + c falls on the roots of y = -ax2 + c (the equation of parabola #2). This would be when x2 = c/a, so x equals +/- the square root of c/a. To obtain a vertex with x-coordinate x = +/-&radic(c/a) from the graph y = ax2 + bx + c,
-b/2a must equal +/-&radic(c/a). Solving for b,
b2/4a2 = c/a
b2 = 4ac
So to have real roots, b2 must be greater than or equal to 4ac. You may recall that b2 - 4ac is called the "discriminant" of a quadratic equation. We now can understand graphically why the discriminant reveals the nature of the roots of the equation. We also can see that our "parabola within a parabola" is not so mysterious after all!
|
# 1003 Continuity
```1003 Continuity
General Idea: ______________________________________________________________________________
We already know the continuity of many functions:
Polynomial (Power),
Rational,
Exponential, Trigonometric, and Logarithmic functions
DEFN: A function is continuous on an interval if it is continuous at each point in the interval.
DEFN: A function is continuous at a point c IFF
a) __________________________________________
_____________________________________
b) __________________________________________
_____________________________________
c) __________________________________________
_____________________________________
THEOREM:
Note: All three steps are implied.
One sided Continuity:
A function y
= f(x) is continuous at a , the left endpoint a of its domain if lim f ( x) f (a ) .
A function y
= f(x) is continuous at b , the right endpoint a of its domain if lim f ( x) f (b) .
x a
xb
Theorem:
A function is Continuous on a closed interval if it is continuous at every point in the open interval and
continuous from one side at the end points.
Example :
The graph over the closed interval [-2,4] is given.
Discontinuity
Continuity may be disrupted by:
(a).
c
c
(b).
c
(c).
c
Removable or Essential ( Non-Removable) Discontinuities
__________________________________________________________________________________________
__________________________________________________________________________________________
Examples:
Identify the x-values (if any) at which f(x) is not continuous.
Identify the reason for the discontinuity and the type of discontinuity.
Is the discontinuity removable or essential?
f ( x)
x 2
x4
f ( x)
1
( x 3) 2
3 x, x 1
f ( x)
3 x, x 1
Identify the x-values (if any) at which f(x) is not continuous.
Identify the reason for the discontinuity and the type of discontinuity.
Is the discontinuity removable or essential?
Algebraic Method
3x 2 x 2
f ( x) 2
3x 4 x 2
Reason
Type
x = -3
____________________
________________
x = -2
____________________
________________
x=0
____________________
________________
x=1
____________________
________________
x=2
____________________
________________
x=3
____________________
________________
1- x 2
x 1
2
f ( x) x - 2 1 x 3
x2 9
x3
x 3
Find c that makes f ( x) continuous.
4 x 2c x 5
f ( x) 2
x c x 5
Consequences of Continuity:
A. INTERMEDIATE VALUE THEOREM
** Existence Theorem
_______________________________________________
_______________________________________________
_______________________________________________
c
_______________________________________________
EX: Verify the I.V.T. for f (c).
Then find c.
f ( x) x 2
on
f (c) 3
[1, 2]
EX. Zero Locater Corollary.
Show that the function f ( x) x3 2 x 1 has a ZERO on the interval [0,1].
( CALCULUS AND THE CALCULATOR:
The calculator looks for a SIGN CHANGE
between Left Bound and Right Bound.)
EX. Sign on an Interval - Corollary
(Number Line Analysis)
1
3
x 1 2
( x 1)( x 2)( x 4) 0
B. EXTREME VALUE THEOREM
On every closed interval there exists an absolute maximum value and minimum value
y
x
.
Assignment 1003:
FDWK p. 84
11)
a) does f(-1) exist?
b) does lim f ( x) exist?
x1
c) does lim f ( x) f ( 1) ?
x1
d) Is f(x) continuous a t f(-1)?
12)
a) does f(1) exist?
b) does lim f ( x) exist?
x1
c) does lim f ( x) f (1) ?
x1
d) Is f(x) continuous a t f(1)?
13) Is f defined at x = 2 ? Is f continuous at x = 2?
14) At what values of x is f continuous?
15) What value should be assigned to f(2) to make the extended function continuous at x = 2 .
16) What new value should be assigned to f(1) to make the extended function continuous at x = 1 .
17) Is it possible to extend f to make it continuous at x = 0 ? Explain.
18) Is it possible to extend f to make it continuous at x = 3 ? Explain.
(a) Find each point of discontinuity. Give the reason for your answer. (b) Which of the discontinuities are
removable? essential?
19)
3 x , x 2
f ( x) x
2 1 , x 2
20)
47) Find a value for a so that the function
is continuous.
x2 1 , x 3
f ( x)
2ax , x 3
49) Find a value for a so that the function
is continuous.
4 x 2 , x 1
f ( x) 2
ax -1 , x 1
3 x , x 2
f ( x) 2 ,
x2
x
, x2
2
21)
1
, x 1
f ( x) x 1
x3 2 x 5 , x 1
48) Find a value for a so that the function.
is continuous
2x 3 , x 2
f ( x)
ax 1 , x 2
50) Find a value for a so that the function.
is continuous
x2 x a , x 1
f ( x) 3
, x 1
x
```
|
2012 AMC 10A Problems/Problem 19
The following problem is from both the 2012 AMC 12A #13 and 2012 AMC 10A #19, so both problems redirect to this page.
Problem 19
Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 60$
Solution
Let Paula work at a rate of $p$, the two helpers work at a combined rate of $h$, and the time it takes to eat lunch be $L$, where $p$ and $h$ are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:
$$(8-L)(p+h)=.50$$
$$(6.2-L)h=.24$$
$$(11.2-L)p=.26$$
With three equations and three variables, we need to find the value of L. Adding the second and third equations together gives us $6.2h+11.2p-L(p+h)=.50$. Subtracting the first equation from this new one gives us $-1.8h+3.2p=0$, so we get $h=\frac{16}{9}p$. Plugging into the second equation: $$(6.2-L)\frac{16}{9}p=.24$$
$$(6.2-L)p=\frac{27}{200}$$
We can then subtract this from the third equation:
$$5p=.26-\frac{27}{200}$$
$p=\frac{1}{40}$. Plugging $p$ into our third equation gives:
$$L=\frac{4}{5}$$
Since we let $L$ be in hours, the lunch break then takes $\frac{4}{5}\cdot 60 = 48$ minutes, which leads us to the correct answer of $\boxed{\textbf{(D)}\ 48}$.
|
# How to solve matrix 3x3
We will explore How to solve matrix 3x3 can help students understand and learn algebra. We will also look at some example problems and how to approach them.
## How can we solve matrix 3x3
Read on for some helpful advice on How to solve matrix 3x3 easily and effectively. First, brush up on your basic division skills. If you can quickly and easily divide whole numbers, you'll have an easier time understanding fractions. Next, practice identifying equivalent fractions. This will help you understand how different fractions can represent the same value. Finally, make sure you understand how to add and subtract fractions, as this is often one of the most difficult concepts for students. If
To solve them, you need to use logs to rewrite the equation in a way that is easier to solve. For example, the equation ln(x+1)=3 can be rewritten as x+1=e3. You can then use a calculator to find the value of x.
There's no need to be discouraged if you're having difficulty solving a word problem; many people find them challenging. The best approach is to take your time and carefully read the problem, then draw a picture or diagrams to help you visualize what's being asked. Once you have a good understanding of the problem, you can start solving. Don't be afraid to ask for help if you're stuck; there's no shame in admitting that you need assistance. With a little practice, you'll
A slope intercept form solver is a tool that can be used to find the equation of a line when given the slope and y-intercept. This can be a useful tool when trying to graph a line or determine the properties of a line.
These solvers can be a great resource when you're stuck on a problem and need some extra help. Just enter your problem into the solver and it will show you the steps you need to take to solve it.
Piecewise function solvers are mathematical tools that allow engineers and mathematicians to solve complex functions in a step-by-step manner. By breaking down a function into smaller pieces, these solvers can help to identify the underlying behavior of the function and find solutions to problems that would be otherwise difficult to solve. While there are many different types of piecewise function solvers available, they all share the same basic goal of helping to simplify complex mathematics.
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The radius is equal to the sum of the squares of half the chord and of the versed sine, divided by twice the versed sine. This is expressed, algebraically, thus: r = (C/2)2+V2/2V, where r is the radius, c the chord, and v the versed sine (Art. 444).
Example. - In a given arc of a circle a chord of 12 feet has the rise at the middle, or the versed sine, equal to 2 feet, what is the radius?
Half the chord equals 6, the square of 6 is, 6x6 = 36 The square of the versed sine is, 2x2 = 4 Their sum equals, 40
Twice the versed sine equals 4, and 40 divided by 4 equals 10. Therefore the radius, in this case, is 10 feet. This result is shown in less space and more neatly by using the above algebraical formula. For the letters substituting their value, the formula r = (C/2)2+V2/2V becomes r = (12/2)2+22/2x2,
2V
and performing the arithmetical operations here indicated equals -
62 + 22 /4 = 36 + 4/4 = 40/40 = 10
|
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# A line 4x+y=1 through the point A (2,−7) meets the line BC whose equation is 3x−4y+1=0 at the point B. Find the equation to the line AC so that -AB=AC.
Last updated date: 06th Sep 2024
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Answer
Verified
437.4k+ views
Hint:Here we will find the slope of the first line and second line initially and then with the help of the slopes found and given condition we will find the slope of the given line. With the slope found and the given points we will find the equation of the new line.
Formula used:$y{\rm{ }} = {\rm{ }}mx + b$ where$m$ in the equation is the slope of the line
We know that the formula
$\theta = {\tan ^{ - 1}}\left| {\dfrac{{{m_1} - m}}{{1 + {m_1}m}}} \right|$
Where m is the slope of the line.
Complete step-by-step answer:
Here it is given that
AB is line with equation 4x + y = 1
Let us now compare it with the equation $y{\rm{ }} = {\rm{ }}mx + b$ where $m$ in the equation is the slope of the line, we get,
The slope of the line is $- 4$
Now let us consider the line BC given by the equation 3x - 4y + 1 =0
Let us rewrite the equation as follows $y = \dfrac{3}{4}x + \dfrac{1}{4}$
By comparing with the standard equation $y{\rm{ }} = {\rm{ }}mx + b$ where $m$ in the equation is the slope of the line, we get,
The slope of the line as $\dfrac{3}{4}$
Here we have to find the equation to the line AC. If m be the slope of AC, the AB and AC are equally inclined to BC.
We know that the formula
$\theta = {\tan ^{ - 1}}\left| {\dfrac{{{m_1} - m}}{{1 + {m_1}m}}} \right|$
For line AB and BC, we have
Let $${m_1} = \dfrac{3}{4}\,\&\, m = - 4$$ be the slopes of AC and AB respectively.
$\alpha = {\tan ^{ - 1}}\left| {\dfrac{{\dfrac{3}{4} - \left( { - 4} \right)}}{{1 + \left( {\dfrac{3}{4}} \right)\left( { - 4} \right)}}} \right|$
Let us operate by tan on both sides we get,
${\rm{tan}}\alpha = \left| {\dfrac{{\dfrac{3}{4} - \left( { - 4} \right)}}{{1 + \left( {\dfrac{3}{4}} \right)\left( { - 4} \right)}}} \right|$
And for line AC and BC, we have
Let ${m_1} = \dfrac{3}{4}$ be the slope of AC and m be the slope of BC.
$\theta = {\tan ^{ - 1}}\left| {\dfrac{{\dfrac{3}{4} - m}}{{1 + \left( {\dfrac{3}{4}} \right)m}}} \right|$
Let us operate by tan on both sides of the equation we have,
${\rm{tan}}\theta = \left| {\dfrac{{\dfrac{3}{4} - m}}{{1 + \left( {\dfrac{3}{4}} \right)m}}} \right|$
Where m is the slope of line AC.
According to question, we are given that $- AB = AC$
So, we have the slope of the lines as follows
-slope of the line AB= slope of the line BC.
That is $- \tan \alpha = \tan \theta$
On substituting the values we get,
$- \dfrac{{\dfrac{3}{4} - \left( { - 4} \right)}}{{1 + \left( {\dfrac{3}{4}} \right)\left( { - 4} \right)}} = \dfrac{{\dfrac{3}{4} - m}}{{1 + \left( {\dfrac{3}{4}} \right)m}}$
Let us now simplify the above equation we get,
$\dfrac{{ - 19}}{{ - 8}} = \dfrac{{3 - 4m}}{{3m + 4}}$
$\Rightarrow 19\left( {3m + 4} \right) = 8\left( {3 - 4m} \right)$
By solving the above equation we get,
$57m + 76 = - 32m + 24$
Then the value of m is
$m = \dfrac{{ - 52}}{{89}}$
Also the line passes through the point A (2, -7),
Using the slope and passing points we get the equation of line using the following formula,
Formula of equation of line is $y - {y_1} = m\left( {x - {x_1}} \right)$
By substituting the known values we get,
$y + 7 = - \dfrac{{52}}{{89}}\left( {x - 2} \right)$
By multiplying both the sides by 89 and solving the equation we get,
The equation of the line AC is $89y + 52x + 519 = 0$
Hence, the equation to the line AC is ${\rm{52x}} + {\rm{89y}} + {\rm{519}} = 0$
Additional Information:The formula $y - {y_1} = m\left( {x - {x_1}} \right)$ is usually described as the 'point-slope form' for the equation of a line. It is useful because if you know one point on a certain line and the slope of that certain line, then you can define the line with this type of formula and, thus, find all the other points on that certain line.
Note:Here while finding the slope of BC we must use the relation between slope and tangent of the line. Also we use the slope and intersection points to find the equation of line.
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# Decimals – Class 6 Maths/ Exercise 8.4/ NCERT Solutions
Chapter 8 of CBSE/NCERT Class 6 Mathematics is about Decimals. In this lesson you can find out text book solutions of Decimals Exercise 8.4.
CBSE CLASS 6 MATHEMATICS
Decimals – Chapter 8/ Exercise 8.4 / NCERT Solutions.
1. Express as rupees using decimals.
a) 5 paise
b) 75 paise
c) 20 paise
d) 50 rupees 90 paise
e) 725 paise
We know that 100 paise = 1 rupee
Therefore, 1 paise = 1/100 rupee = 0.01 rupee.
a) 5 paise = 0.05 rupee
b) 75 paise = 0.75 rupee
c) 20 paise = 0.20 rupee
d) 50 rupees 90 paise = 50 + 90 paise = 50 + 0.90 = 50.90 rupees
e) 725 paise = 7.25 rupees
2. Express as metres using decimals.
a) 15 cm
b) 6 cm
c) 2 m 45 cm
d) 9 m 7 cm
e) 419 cm
We know that 100 cm = 1 m
Therefore, 1 cm = 1/100 m = 0.01 m
a) 15 cm = 0.15 m
b) 6 cm = 0.06 m
c) 2m 45 cm = 2m + 45 cm = 2 m + 0.45 m = 2.45 m
d) 9 m 7 cm = 9 m +7 cm = 9 m + 0.07 m = 9.07 m
e) 419 cm = 4.19 m
3. Express as cm using decimals.
a) 5 mm
b) 60 mm
c) 164 mm
d) 9 cm 8 mm
e) 93 mm
We know that 10 mm = 1 cm
Therefore, 1 mm = 1/10 cm = 0.1 cm
a) 5 mm = 0.5 cm
b) 60 mm = 6 cm
c) 164 mm = 1.64 cm
d) 9 cm 8 mm = 9cm + 8 mm = 9 cm + 0.8 cm = 9.8 cm
e) 93 mm= 9.3 cm
4. Express as km using decimals.
a) 8 m
b) 88 m
c) 8888 m
d) 70 km 5 m
We know that 1000 m = 1 km
Therefore, 1 m = 1/1000 km = 0.001 km
a) 8 m = 0.008 km
b) 88 m = 0.088 km
c) 8888 m = 8.888 km
d) 70 km 5 m = 70 km + 5 m = 70 km + 0.005 km = 70.005 km
5. Express as kg using decimals
a) 2 g
b) 100 g
c) 3750 g
d) 5 kg 8 g
e) 26 kg 50 g
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# Simplify a Fraction
Calculator to simplify ONE fraction with steps. For an expression with more than a fraction, use this Fraction Calculator.
## How to simplify a fraction?
There are two methods to simplify (or reduce) a fraction.
The method of successive simplifications
This method consists in making successive simplifications by dividing by "small" numbers (2, 3, 5, etc.) until obtaining the irreducible form of the fraction.
Example: how to reduce the fraction 210/378 by the method of successive simplifications?
We start by dividing 210 and 378 by 2.
We get 105 and 189.
So we have 210/378 = 105/189
We repeat the operation with 105 and 189.
The latter are divisible by 3 because 105/3 = 35 and 189/3 = 63.
So we have,
210/378 = 35/63
35 and 63 are divisible by 7, 35/7 = 5 and 63/7 = 9
210/378 = 5/9
We get to the end of the process because 5 and 9 have no common divisor other than 1!
The reduced form of fraction 210/378 is 5/9.
GCD Method
This method is based on GCD (Greatest Common Divisor) i.e. find the GCD of the numerator and denominator and divide them by this GCD.
Example: how to reduce the fraction 210/378 using the GCD method ?
GCD (210,378) = 42. More details on this page How to calculate the GCD of 210 and 378 ?.
To get the reduced fraction, simply divide the numerator and denominator by the GCD ie 42 !
The numerator of the reduced fraction is equal to 210/42 i.e. 5.
The denominator of the reduced fraction is equal to 378/42 i.e. 9.
The same result is obtained as by the first method: 5/9.
## Python
Here is a python function that simplifies a fraction. The input variables "num" and "denom" represent the numerator and denominator of the fraction. For the gcd function, we can either import the python gcd function of the math module or rewrite it as here: Calculate gcd with python.
from math import gcd
def ReduceFraction (num, denom):
d = gcd (num, denom);
num = num//d;
denom = denom//d;
return (num, denom)
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# lengths of angle bisectors
In any triangle, the $w_{a}$, $w_{b}$, $w_{c}$ of the angle bisectors opposing the sides $a$, $b$, $c$, respectively, are
$\displaystyle w_{a}=\frac{\sqrt{bc\,[(b\!+\!c)^{2}\!-\!a^{2}]\,}}{b\!+\!c},$ (1)
$\displaystyle w_{b}=\frac{\sqrt{ca\,[(c\!+\!a)^{2}\!-\!b^{2}]\,}}{c\!+\!a},$ (2)
$\displaystyle w_{c}=\frac{\sqrt{ab\,[(a\!+\!b)^{2}\!-\!c^{2}]\,}}{a\!+\!b}.$ (3)
Proof. By the symmetry, it suffices to prove only (1).
According the angle bisector theorem, the bisector $w_{a}$ divides the side $a$ into the portions
$\frac{b}{b\!+\!c}\cdot a\;=\;\frac{ab}{b\!+\!c},\qquad\frac{c}{b\!+\!c}\cdot a% \;=\;\frac{ca}{b\!+\!c}.$
If the angle opposite to $a$ is $\alpha$, we apply the law of cosines to the half-triangles by $w_{a}$:
$\displaystyle\begin{cases}2w_{a}b\cos\frac{\alpha}{2}\;=\;w_{a}^{2}\!+\!b^{2}% \!-\!\left(\frac{ab}{b+c}\right)^{2}\\ 2w_{a}c\cos\frac{\alpha}{2}\;=\;w_{a}^{2}\!+\!c^{2}\!-\!\left(\frac{ca}{b+c}% \right)^{2}\end{cases}$ (4)
For eliminating the angle $\alpha$, the equations (4) are divided sidewise, when one gets
$\frac{b}{c}\;=\;\frac{w_{a}^{2}\!+\!b^{2}\!-\!\left(\frac{ab}{b+c}\right)^{2}}% {w_{a}^{2}\!+\!c^{2}\!-\!\left(\frac{ca}{b+c}\right)^{2}},$
from which one can after some routine manipulations solve $w_{a}$, and this can be simplified to the form (1).
Title lengths of angle bisectors LengthsOfAngleBisectors 2013-03-22 18:26:50 2013-03-22 18:26:50 pahio (2872) pahio (2872) 9 pahio (2872) Corollary msc 51M05 Incenter AngleBisectorAsLocus LengthsOfMedians
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# Algebra 1 : Factoring Polynomials
## Example Questions
### Example Question #81 : Factoring Polynomials
Factor:
Explanation:
To factor a quadratic trinomial, list factors of the quadratic term and the constant (no variables) term, then combine them into binomials that when multiplied back out will give the original trinomial.
Here, the quadratic term has only one factorization: .
The constant term has factorizations of and .
We know the constant term is negative, so the binomials have a different operation in each (adding or subtracting), since a positive times a negative will give a negative result.
Since the middle term is negative, we need the "larger" factor to "outweigh" the "smaller" by , and be negative.
### Example Question #82 : Factoring Polynomials
Factor:
Explanation:
To factor a quadratic trinomial, list factors of the quadratic term and the constant (no variables) term, then combine them into binomials that when multiplied back out will give the original trinomial.
Here, the quadratic term has only one factorization: .
The constant term has factorizations of and .
We know the constant term is positive, so the binomials both have the same operation in them (adding or subtracting), since a positive times a positive OR a negative times a negative will both give a positive result.
But since the middle term is positive, both binomial factors must contain addition. And .
### Example Question #11 : Quadratic Equations
Factor completely:
The expression is not factorable.
Explanation:
First, factor out the greatest common factor (GCF), which here is . If you don't see the whole GCF at once, factor out what you do see (here, either the or the ), and then check the result to see if any more factors can be pulled out.
Then, to factor a quadratic trinomial, list factors of the quadratic term and the constant (no variables) term, then combine them into binomials that when multiplied back out will give the original trinomial.
Here, the quadratic term has only one factorization: .
The constant term has factorizations of , and .
We know the constant term is positive, so the binomials both have the same operation in them (adding or subtracting), since a positive times a positive OR a negative times a negative will both give a positive result.
But since the middle term is negative, both binomial factors must contain subtraction. And .
### Example Question #83 : Factoring Polynomials
Which of the following is a factor of the polynomial ?
None of the other choices is correct.
None of the other choices is correct.
Explanation:
To factor , we use the "reverse-FOIL".
, where and are two integers whose product is 24 and whose sum is 12. If we examine all of the factor pairs of 24, we note that the sum of each pair is as follows:
1 and 24: 25
2 and 12: 14
3 and 8: 11
4 and 6: 10
No factor pair of 24 has sum 12, so cannot be factored. The correct response is that none of the four binomials is correct.
### Example Question #84 : Factoring Polynomials
Which of the following is a prime polynomial?
The polynomials in all four of the other choices are prime.
Explanation:
The sum of two squares is in general a prime polynomial unless a greatest common factor can be distributed out. is the sum of squares, and its terms do not have a GCF, so it is the prime polynomial.
Of the remaining choices:
is equal to ; as the difference of squares, it is factorable.
is equal to ; this fits the pattern of a perfect square quadratic trinomial, and is therefore factorable. is factorable for a similar reason.
### Example Question #85 : Factoring Polynomials
How many ways can a positive integer be written in the box to form a factorable polynomial?
Two
One
None
Three
Seven
One
Explanation:
Let be the integer written in the box.
If is factorable, then its factorization is , where and . In other words, must be the positive difference of two numbers of a factor pair of 7. 7 has only one factor pair, 1 and 7, so the only possible value of is . The correct choice is one.
### Example Question #86 : Factoring Polynomials
Which of the following is a factor of ?
Explanation:
can be factored by grouping, as follows:
cannot be factored further. Of the five choices, is the only factor.
### Example Question #87 : Factoring Polynomials
Which of the following is a prime polynomial?
Explanation:
A quadratic trinomial of the form can be factored by splitting the middle term into two terms. The two coefficients must have sum and product .
In each case, a factor pair of must be examined. These pairs, along with their sums, are:
1 and 24: 25
2 and 12: 14
3 and 8: 11
4 and 6: 10
Each polynomial with one of these four integers as its linear coefficient can be factored. The odd one out is , since there is no factor pair of 24 whose sum is 20. This is the correct choice.
### Example Question #1 : How To Factor A Variable
Solve for , when :
Explanation:
First, factor the numerator, which should be . Now the left side of your equation looks like
Second, cancel the "like" terms - - which leaves us with .
Third, solve for by setting the left-over factor equal to 0, which leaves you with
### Example Question #1 : Factoring Polynomials
Factor the following expression:
Explanation:
Here you have an expression with three variables. To factor, you will need to pull out the greatest common factor that each term has in common.
Only the last two terms have so it will not be factored out. Each term has at least and so both of those can be factored out, outside of the parentheses. You'll fill in each term inside the parentheses with what the greatest common factor needs to be multiplied by to get the original term from the original polynomial:
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# How do you simplify (sqrt5)/(sqrt5-sqrt3)?
Apr 19, 2018
$\frac{5 + \sqrt{15}}{2}$
#### Explanation:
$\implies \frac{\sqrt{5}}{\sqrt{5} - \sqrt{3}}$
Multiply and divide by $\left(\sqrt{5} + \sqrt{3}\right)$
=> sqrt(5)/(sqrt(5) - sqrt(3)) × (sqrt(5) + sqrt(3))/(sqrt(5) + sqrt(3))
=> (sqrt(5)(sqrt(5) + sqrt(3)))/((sqrt(5) - sqrt(3))(sqrt(5) + sqrt(3))
=> (sqrt(5)(sqrt(5) + sqrt(3)))/((sqrt(5))^2 - (sqrt(3))^2) color(white)(..)[∵ (a - b)(a + b) = a^2 - b^2]
$\implies \frac{\sqrt{5} \sqrt{5} + \sqrt{5} \sqrt{3}}{5 - 3}$
$\implies \frac{5 + \sqrt{15}}{2}$
Apr 19, 2018
$\frac{5 + \sqrt{15}}{2}$
#### Explanation:
Multiply (√5) / (√5−√3) by (√5+√3) / (√5+√3) to rationalize the denominator
(√5)/(√5−√3) * (√5+√3) / (√5+√3) = $\frac{\sqrt{5} \cdot \left(\sqrt{5} + \sqrt{3}\right)}{2}$
Apply the distributive property
$\frac{\sqrt{5} \cdot \left(\sqrt{5} + \sqrt{3}\right)}{2}$ = $\frac{\left(\sqrt{5} \cdot \sqrt{5}\right) + \left(\sqrt{5} \cdot \sqrt{3}\right)}{2}$ = $\frac{5 + \sqrt{15}}{2}$
Apr 19, 2018
= 5/(5 - (sqrt(15))
OR
$= \frac{5}{2} + \frac{\sqrt{15}}{2}$
#### Explanation:
These days, it may be simplest to just use a calculator to complete the expression. But, for purposes of demonstration, we multiply by a radical factor just as we would with another number.
sqrt(5)/(sqrt(5) - sqrt(3)) xx sqrt(5)/(sqrt(5) = 5/(5 - (sqrt(3) xx sqrt(5))
5/(5 - (sqrt(3) xx sqrt(5)) = 5/(5 - (sqrt(15))
OR
Multiply the denominator and numerator by the same expression as the denominator but with the opposite sign in the middle. This expression is called the conjugate of the denominator.
$\frac{\sqrt{5}}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}}$
$= \frac{5 + \sqrt{15}}{5 - 3}$ = $\frac{5 + \sqrt{15}}{2} = \frac{5}{2} + \frac{\sqrt{15}}{2}$
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# Geometric Progression Sum of GP
## Sum of GP
Let’s assume that there are 10 terms in a sequence. If each term is a multiple of the next term then this sequence is said to be in Geometric Progression or Geometric Sum of G.P. Here, the number which is used to constantly multiply is known as the common ratio. In Arithmetic Progression, when the nth term is subtracted from (n – 1)th term, there will be a constant difference between any two numbers. A.P can be written as:
x, x + b, x + 2b, x + 3b, x + . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x + (n – 1) b
Consider the sequence:
3, 6, 12, 24, 48, . . . . . . . . . . . . . . . . . . . . . and the list goes on..
Observe this:
6/3 = 2
12 / 6 = 2
24 / 12 = 2
48 / 24 = 2
In the same manner:
Consider the series: 1, ⅓, ⅙, ⅟₁₂, ⅟₂₄, ⅟₄₈, . . . . . . . . . . . . . . . . . . . . . . . .
⅓ / 1 = ⅟₂
¼ / ½ = ½
⅛ / ¼ = ½
⅟₁₆ / ⅟₈ = ½
⅟₂₄ / ⅟₁₆ = ½
⅟₄₈ / ⅟₂₄ = ½
In the examples shown above, you can see that the one thing that is constant is the ratio between any two sequential numbers. These sequences are called Geometric Progression.
A sequence x₁, x₂, x₃, x₄, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xn is a Geometric Progression, then ak+1 / ak = r {k is always greater than 1}
Where,
R = a constant ration between two numbers.
The General Term of Geometric Progression
The nth term of Arithmetic Progression was found out to be:
xn= x + (n – 1) b
In the case of Geometric Progression, let’s assume that x is the first number and “r” is the common ratio between all the numbers.
So, the second term would be: x2 = x * r
The third term would be: x3 = x2 * r = x * r * r = xr²
In the same manner, the nth term would be xn = x * r n-1
The general term = xn = x * r n-1
### Common Term
Consider the following sequence: x, xr, xr² , xr³, . . . . . . . . . . . . .
Here,
The First term is = x
The Second term is = xr
The Third term is = xr2
Similarly nth term, xn = x * r n-1
Hence, the common ratio (r) = (Any given term) / (the next preceding term)
= xn / x * r n-1
= (r n-1) /(r n-2)
= r
Thus, the general term of a G.P is given by r n-1 and the general form of a G.P is x + xr + xr2 + . . . . . . . . . . . . .
### Finite and Infinite
The finite Geometric Progression terms can be described as:
x, xr, xr² , xr³, . . . . . . . . . . . . . xr n-1
The infinite Geometric Progression terms can be described as:
x, xr, xr² , xr³, . . . . . . . . . . . . . xr n-1,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
x, xr, xr² , xr³, . . . . . . . . . . . . . xr n-1 can be called as the finite geometric progression series.
x, xr, xr² , xr³, . . . . . . . . . . . . . xr n-1,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . can be called as the infinite geometric progression series.
### Sum of GP of n terms
Let assume,
x, xr, xr² , xr³, . . . . . . . . . . . . . xr n-1
Let’s assume,
The sum of gp terms = Sn
First Term = x
Ration = r
Now,
Sn = x, xr, xr² , xr³, . . . . . . . . . . . . . xr n-1– – – – – – – – – – – – — – – – – – – – – – – (1)
Here, r = 1 or r ≠ 1
If r = 1, then Sn = x, x, x, x . . . . . . . . . . . . . . . = nx
If r ≠ 1, then rSn = x, xr, xr² , xr³, . . . . . . . . . . . . . xr n-1 + xr n – – – – – – – – – – – – – (2)
Now subtract equation (1) from equation (2), it gives you:
rSn Sn = ( xr + xr² + xr³ +. . . . . . . + xr n-1+ xr n) – (x + xr + xr² + xr³+ .. . . . . .+ xr n-2 + xr n-1)
(r – 1)Sn = xr n– x = x(r n– 1)
Sn= x(r n– 1) / (r – 1)
Sn= x(1 – r n)/ ( 1 – r)
Sum of n terms in GP
Sn= x(1 – r n)/ ( 1 – r) [here, r ≠ 1]
### Solved Problems
Example 1: Find the value of n, if the nth term of Geometric Progression is 2, 4, 8, 16, . . . . . . . . . . . . . . . . , 128.
Solution:
Step 1: Find the ratio
r = 4/2 = 2
Step 2: The first term: x = 2.
xn = xr n-1
128 = 2 x 2 n-1
2 n-1 = 128 / 2
2 n-1 = 64
2 n-1 = (2)6
n – 1 = 6
n = 6 + 1
n = 7
Hence, the 7th term here is 128.
Example 2: If the Geometric Progression is 5, 10, 20, 40 . . . . . . . , find the sum of 8 terms.
Solution:
Given,
x = 5
n = 8
r = 10/5 = 2
Sum of n terms in gp,
Sn = x(r n– 1) / (r – 1)
S8 = 5(28– 1) / (2 – 1)
S8 = 3 (6561) / 2
S8 = 3 (256) / 2
S8 = 768 / 2
S8 = 384
What is the geometric progression formula?
A sequence x₁, x₂, x₃, x₄, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xn is a Geometric Progression, then ak+1 / ak = r {k is always greater than 1}
Where,
r = a constant ration between two numbers.
The nth term of Arithmetic Progression was found out to be:
xn= x + (n – 1) b
In the case of Geometric Progression, let’s assume that x is the first number and “r” is the common ratio between all the numbers.
So, the second term would be: x2 = x * r
The third term would be: x3 = x2 * r = x * r * r = xr²
In the same manner, the nth term would be xn = x * r n-1
The general term = xn = x * r n-1
How can you define finite and infinite geometric progression?
The finite Geometric Progression terms can be described as:
x, xr, xr² , xr³, . . . . . . . . . . . . . xr n-1
The infinite Geometric Progression terms can be described as:
x, xr, xr² , xr³, . . . . . . . . . . . . . xr n-1,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
x, xr, xr² , xr³, . . . . . . . . . . . . . xr n-1 can be called as the finite geometric progression series.
x, xr, xr² , xr³, . . . . . . . . . . . . . xr n-1 . . . . . . . . . can be called as the infinite geometric progression series.
|
# One-proportion z-test (Hypothesis test)
Question: Crammer Nation University claims that 30% of their freshmen students joined a Greek Life chapter this year. You are curious if that's a truthful proportion, or if a different proportion of students joined a chapter. You collect a random sample of 50 freshmen students and find that 22 of them joined a chapter this year. Provide support for your claim using a hypothesis test with an alpha level of 0.05.
As with all hypothesis test problems, we're going to take the following steps to solve!
• Step 1 - State the hypothesis
• Step 2 - Calculate the test statistic
• Step 3 - Find the p-value
• Step 4 - Make the concluding statement
## Step 1 - State the hypotheses
Let's start with the null hypothesis (H0), then we'll move onto the alternative hypothesis (Ha).
### Defining your null hypothesis (H0)
Here, we need to ask ourselves: what is the baseline claim that's being made about the population?
In the case of the situation above, it's that Crammer Nation University claims freshmen students joined Greek Life this year at a proportion of 0.30.
Crammer Nation University claims that 30% of their freshmen students joined a Greek Life chapter this year. You are curious if that's a truthful proportion, or if a different proportion of students joined a chapter. You collect a random sample of 50 freshmen students and find that 22 of them joined a chapter this year. Provide support for your claim using a hypothesis test with an alpha level of 0.05.
We'd write that null hypothesis (H0) like so:
H0: p = 0.30
Essentially, what we're saying here is that the true population proportion for Greek Life involvement among Crammer Nation University freshmen...
H0: p = 0.30
...is equal to 0.30.
H0: p = 0.30
### Defining your alternative hypothesis (Ha)
Here we need to ask ourselves: what's the claim that's being made that's different from the baseline claim about the population?
In the case of the situation above, it's that we're claiming that Crammer Nation University freshmen students joined Greek Life this year at a proportion not equal to 0.30.
Crammer Nation University claims that 30% of their freshmen students joined a Greek Life chapter this year. You are curious if that's a truthful proportion, or if a different proportion of students joined a chapter. You collect a random sample of 50 freshmen students and find that 22 of them joined a chapter this year. Provide support for your claim using a hypothesis test with an alpha level of 0.05.
We'd write that alternative hypothesis (Ha) like so:
H0: p = 0.30
Ha: p ≠ 0.30
Essentially what we're saying here is that the true population proportion for Greek Life involvement among Crammer Nation University freshmen...
H0: p = 0.30
Ha: p ≠ 0.30
...is actually not equal to 0.30.
H0: p = 0.30
Ha: p ≠ 0.30
## Step 2 - Calculating your test statistic
Before even beginning to calculate our test statistic, we have to check out assumptions!
We're working with a sample proportion here, so in accordance with Assumptions for sampling distributions, that means we must check the following assumptions:
1. Sample is randomly selected from the population
2. The sample size (n) is less than or equal to 10% of of the population size N
3. There are 10 successes and 10 failures in the sample OR np >= 10 and nq >= 10
Concerning #1, we're collecting a random sample from the population:
Crammer Nation University claims that 30% of their freshmen students joined a Greek Life chapter this year. You are curious if that's a truthful proportion, or if a different proportion of students joined a chapter. You collect a random sample of 50 freshmen students and find that 22 of them joined a chapter this year. Provide support for your claim using a hypothesis test with an alpha level of 0.05.
Concerning #2, we can assume that 50 freshmen students is less than 10% of the entire freshmen student body at Crammer Nation University.
Crammer Nation University claims that 30% of their freshmen students joined a Greek Life chapter this year. You are curious if that's a truthful proportion, or if a different proportion of students joined a chapter. You collect a random sample of 50 freshmen students and find that 22 of them joined a chapter this year. Provide support for your claim using a hypothesis test with an alpha level of 0.05.
Concerning #3, we have 22 "successes" (freshmen students who joined Greek Life) and therefore have 50 - 22 = 28 "failures" (freshmen students who didn't join Greek Life).
Therefore, all of our assumptions are passed! Now we can move onto calculating our test statistic!
### Recognizing we'll use z-score
Remember, you'll only use t-score if you're dealing with sample means! Therefore, we know we will be using z-score here, which will be computed with the following formula:
This looks similar to the regular z-score formula... what's with "p0" and "q0"
You'll notice this formula is very similar to the formula for z-scores in What is a z-score, in relation to a sample proportion?...
...but now it's using p0 and q0 instead of p and q.
Long story short, that's because with hypothesis tests, we use the "0" (that little "0" is often called "knot") to signify that we don't know for sure that it's the true population proportion for p and q. It's the claimed population proportion that's being tested within our hypothesis test!
### Plugging in the sample proportion (p-hat)
Based on the prompt, the sample proportion was 0.44 (because 22 / 50 = 0.44)...
Crammer Nation University claims that 30% of their freshmen students joined a Greek Life chapter this year. You are curious if that's a truthful proportion, or if a different proportion of students joined a chapter. You collect a random sample of 50 freshmen students and find that 22 of them joined a chapter this year. Provide support for your claim using a hypothesis test with an alpha level of 0.05.
...therefore, we'll plug in 0.44 for p-hat.
### Plugging in the claimed population proportion (p0)
Based on the prompt, the sample proportion was 0.44 (because 22 / 50 = 0.44)...
Crammer Nation University claims that 30% of their freshmen students joined a Greek Life chapter this year. You are curious if that's a truthful proportion, or if a different proportion of students joined a chapter. You collect a random sample of 50 freshmen students and find that 22 of them joined a chapter this year. Provide support for your claim using a hypothesis test with an alpha level of 0.05.
...therefore, we'll plug in 0.44 for p-hat.
### Plugging in the claimed population proportion of failure (q0)
Next, for our population proportion of failure (q0), we'll do what we did in What is a z-score, in relation to sample proportions? and utilize the following formula:
q0 = 1 - p0
Since p0 equals 0.30...
q0 = 1 - 0.30
...this results in q0 equalling 0.70...
q0 = 1 - 0.30 = 0.70
...so we'll plug in 0.70 for q0!
### Plugging in the sample size (n)
Lastly, the prompt states that the sample size is 50...
Crammer Nation University claims that 30% of their freshmen students joined a Greek Life chapter this year. You are curious if that's a truthful proportion, or if a different proportion of students joined a chapter. You collect a random sample of 50 freshmen students and find that 22 of them joined a chapter this year. Provide support for your claim using a hypothesis test with an alpha level of 0.05.
...therefore, we'll plug in 50 for n!
### Solving for z-score
When we solve this out, it results in a z-score of 2.15!
## Step 3 - Find your p-value
First, let's determine if we're doing a one-tail or two-tail test, and then let's find our p-value in the z-table.
### One-tail or two-tail?
In What is a hypothesis test?, we declared the following:
If your alternative hypothesis has ">" or "<", that means it's a one-tail test.
If your alternative hypothesis has "", that means it's a two-tail test.
Since our alternative hypothesis uses a "≠", that means we're dealing with a two-tail test.
With two tail tests, that means our alpha level of 0.05...
Crammer Nation University claims that 30% of their freshmen students joined a Greek Life chapter this year. You are curious if that's a truthful proportion, or if a different proportion of students joined a chapter. You collect a random sample of 50 freshmen students and find that 22 of them joined a chapter this year. Provide support for your claim using a hypothesis test with an alpha level of 0.05.
...will be split among the two tails.
0.05 / 2 = 0.025
### Finding our p-value
Knowing that our z-score is 2.15, all we need to do is go to our z-table...
...find "2.1" in the left-hand column (representing 2.15)...
...and then "0.05" in the top row (representing 2.15)...
...to locate our p-value of 0.9842!
### Visualizing our p-value
We need to remember that the z-table displays the area to the left of your z-score...
...therefore this p-value of 0.9842 can be understood visually like so:
To find the p-value to the right of our p-hat value, we must subtract 0.9842 from 1.
1.00 - 0.9842 = 0.0158
### Duplicating our p-value on both tails
Remember, we're dealing with a two-tail test.
That means that we must reflect our p-hat value of 0.44...
...onto the left-tail of our sampling distribution.
This results in a final p-value of 0.0316! We can also see that visually, this fits in our alpha level range on both tails of our sampling distribution!
## Step 4 - Make your concluding statement
Your concluding statement is going to center around the alpha level declared in the problem. In most cases, that alpha level will be 0.05. Each problem should explicitly state the alpha level. In our problem, it's 0.05.
Crammer Nation University claims that 30% of their freshmen students joined a Greek Life chapter this year. You are curious if that's a truthful proportion, or if a different proportion of students joined a chapter. You collect a random sample of 50 freshmen students and find that 22 of them joined a chapter this year. Provide support for your claim using a hypothesis test with an alpha level of 0.05.
As stated in What is a hypothesis test?...
- If the p-value is below the alpha level, then we reject the null hypothesis
- If the p-value is above the alpha level, then we fail to reject the null hypothesis.
Our p-value of 0.0316 is below our alpha level of 0.05, therefore we'll reject our null hypothesis! The chances of our sample occurring are so low, that they are statistically significant enough to support the alternative hypothesis.
### Applying the z-score answer statement template
If you remember in What is a hypothesis test?, we gave the following answer template:
Since our p-value of p-value is less / greater than our alpha level of alpha level value, we reject / fail to reject the null hypothesis and do / don't have enough evidence to support the alternative hypothesis, implying that description of alternative hypothesis.
Applied to our question, this would give us the following answer to our original question!
Answer: Since our p-value of 0.0316 is less than our alpha level of 0.05, we reject the null hypothesis and do have enough evidence to support the alternative hypothesis, implying that the proportion of freshmen students at Crammer Nation University who joined Greek Life this year is not equal to 0.30.
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Short Answer Questions: Polynomials - 1 Notes | Study Mathematics (Maths) Class 10 - Class 10
Class 10: Short Answer Questions: Polynomials - 1 Notes | Study Mathematics (Maths) Class 10 - Class 10
The document Short Answer Questions: Polynomials - 1 Notes | Study Mathematics (Maths) Class 10 - Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
Q 1. Find the sum and product of zeroes of 3x2 - 5x + 6.
Here, p (x) = 3x2 - 5x + 6
Comparing it with ax2 + bx + c, we have
a = 3, b = - 5, c = 6
∴ Sum of the zeroes =
and, Product of the zeroes =
Q 2. Find the sum and product of the zeroes of polynomial p (x) = 2x3 - 5x2 - 14x + 8.
Comparing p (x) = 2x3 - 5x2 - 14x + 8 with ax3 + bx2 + cx + d, we have
a = 2, b = –5,
c = - 14 and d = 8
∴ Sum of the zeroes =
Product of zeroes
Q 3. Find a Quadratic polynomial whose zeroes are .
Sum of zeroes (S)
Product of roots (P)
= k(x2 - Sx + P) ; where k is any real number.
= k
Thus, the required polynomial is
= k (x2 - 2x - 1/4)
Q 4. If α and β are the zeroes of a Quadratic polynomial x2 + x - 2 then find the value of .
Comparing x2 + x - 2 with ax2 + bx + c, we have:
a = 1, b = 1, c = - 2
Thus,
Q 5. If a and b are the zeroes of x2 + px + q then find the value of .
Comparing x2 + px + q with ax2 + bx + c
a =1, b = p and c = q
∴ Sum of zeroes, a + b = - b/a
and αβ = c/a
⇒ αβ = q/1 = q
Now,
Thus, the value of is
Q 6. Find the zeroes of the quadratic polynomial 6x2 - 3 - 7x.
We have,
= 6x2 - 3 - 7x = 6x2 - 7x - 3
= 6x2 - 9x + 2x - 3
= 3x (2x - 3) + 1 (2x - 3)
= (3x + 1) (2x - 3)
For 6x2 - 3 - 7x to be equal to zero,
either (3x + 1) = 0 or (2x - 3) = 0
⇒ 3x = - 1 or 2x = 3
Thus, the zeroes of and 3/2.
Q 7. Find the zeroes of 2x2 - 8x + 6.
We have,
2x2 - 8x + 6 = 2x2 - 6x - 2x + 6
= 2x (x - 3) - 2 (x - 3)
= (2x - 2) (x - 3)
= 2 (x - 1) (x - 3)
For 2x2 - 8x + 6 to be zero,
Either, x - 1 = 0 ⇒ x = 1
or x - 3 = 0 ⇒ x = 3
∴ The zeroes of 2x2 - 8x + 6 are 1 and 3.
Q 8. Find the zeroes of the quadratic polynomial 3x2 + 5x - 2.
We have,
p (x) = 3x2 + 5x - 2
= 3x2 + 6x - x - 2
= 3x (x + 2) - 1 (x + 2)
= (x + 2) (3x - 1)
For p (x) = 0, we get
Either x + 2 = 0 ⇒ x = - 2
or 3x - 1 = 0 ⇒ x = 1/3
Thus, the zeroes of 3x2 + 5x - 2 are - 2 and 1/3.
Q 9. If the zero of a polynomial p (x) = 3x2 - px + 2 and g (x) = 4x2 - q x - 10 is 2, then find the value of p and q.
∵ p (x) = 3x2 - px + 2
∴ p (2) = 3 (2)2 - p (2) + 2 = 0
[2 is a zero of p (x)]
or 12 - 2p + 2 or 14 - 2p = 0
or p = 7
Next g (x) = 4x2 - q x - 10
∴ g (2) = 4(2)2 - Q (2) - 10 = 0
[2 is a zero of g (x)]
or 4 × 4 - 2q - 10 = 0
or 16 - 2q - 10 = 0
or 6 - 2q = 0
⇒ q = 6/2 ⇒ q = 3
Thus, the required values are p = 7 and q = 3.
Q 10. Find the value of ‘k’ such that the quadratic polynomial 3x2 + 2kx + x - k - 5 has the sum of zeroes as half of their product.
Here, p (x) = 3x2 + 2kx + x - k - 5
= 3x2 + (2k + 1) x - (k + 5)
Comparing p (x) with ax2 + bx + c, we have:
a = 3, b = (2k + 1),
c = - (k + 5)
∴ Sum of the zeroes
Product of the zeroes
According to the condition,
Sum of zeroes = 1/2 (product of roots)
⇒ - 2 (2k + 1) = - (k + 5)
⇒ 2 (2k + 1) = k + 5
⇒ 4k + 2 = k + 5
⇒ 4k - k = 5 - 2
⇒ 3k = 3
⇒ k = 3/3 = 1
Q 11. On dividing p (x) by a polynomial x - 1 - x2, the Quotient and remainder were (x - 2) and 3 respectively. Find p (x).
Here,dividend = p (x)
Divisor, g (x) = (x - 1 - x2)
Quotient, q(x) = (x - 2)
Remainder, r (x) = 3
∵ Dividend = [Divisor × Quotient] +Remainder
∴ p (x)= [g (x) × q(x)] + r (x)
= [(x - 1 - x2) (x - 2)] + 3
= [x2 - x - x3 - 2x + 2 + 2x2] + 3
= 3x2 - 3x - x3 + 2 + 3
= - x3 + 3x- 3x + 5
Q 12. Find the zeroes of the polynomial f (x) = 2 - x2.
We have f (x)= 2 - x2
= (√2 )- x2
Q 13. Find the cubic polynomial whose zeroes are 5, 3 and - 2.
∵ 5, 3 and - 2 are zeroes of p (x)
∴ (x - 5), (x - 3) and (x + 2) are the factors of p (x)
⇒ p (x) = k (x - 5) (x - 3) (x + 2)
= k (x2 - 8x + 15) (x + 2)
= k (x3 - 8x2 + 15x + 2x2 - 16x + 30
= k (x3 + [- 8 + 2] x2 + [15 - 16] x + 30)
= k (x- 6x2 - x + 30)
Thus, the required polynomial is k (x3 - 6x2 - x + 30).
Q 14. If α, β and γ be the zeroes of a polynomial p (x) such that (α + β +γ) = 3, (αβ + βγ + γα) = -10 and αβγ = - 24 then find p (x).
Here, α + β + γ = 3
αb + βγ + γα = - 10
αβγ = - 24
∵ A cubic polynomial having zeroes as α,β,γ is
p (x) = x3 - (a + b + γ) x+ (αβ + βγ + γα) x - (αβγ)
∴The required cubic polynomial is
= k {x3 - (3) x2 + (- 10) x - (- 24)}
= k(x- 3x2 - 10x + 24)
Note: If α, β and γ be the zeroes of a cubic polynomial p (x) then
p (x) = x3 - [Sum of the zeroes] x2 + [Product of the zeroes taken two at a time] x - [Product of zeroes]
i.e., p (x) = k {x3 - (α + β + γ) x2 + [αβ + βγ + γα] x - (αβγ).
Q 15. Find all the zeroes of the polynomial 4x4 - 20x3 + 23x2 + 5x - 6 if two of its zeroes are 2 and 3.
Here, p (x) =4x4 - 20x3 + 23x2 + 5x - 6
Since, 2 and 3 are the zeroes of p (x),
∴ (x - 2) and (x - 3) are the factors of p(x)
⇒ (x - 2) (x - 3) is a factor of p (x)
⇒ x- 5x + 6 is a factor of p (x)
Now, using the division algorithm for x2 - 5x + 6 and the given polynomial p (x), we
get:
∴ We get (x2 - 5x + 6) (4x2 - 1) = p (x)
⇒ (x - 3) (x - 2) [(2x)2 - (1)2] = p (x)
⇒ (x - 3) (x - 2) (2x - 1) (2x + 1) = p (x)
Thus, all the zeroes of p (x) are:
Q 16. If 1 is a zero of x3 - 3x2 - x + 3 then find all other zeroes.
Here,p (x) = x3 - 3x2 - x + 3
∵ 1 is a zero of p (x)
∴ (x - 1) is a factor of p (x).
Now, dividing p (x) by (x - 1), we have:
∴ p (x) = (x - 1) (x2 - 2x - 3)
⇒ p (x) = (x - 1) [(x2 - 3x + x - 3)]
= (x - 1) [x (x - 3) + 1 (x - 3)]
= (x - 1) [(x - 3) (x + 1)]
i.e., (x - 1), (x - 3) and (x + 1) are the factors of p (x).
⇒ 1, 3, and - 1 are the zeroes of p (x).
Q 17. Find all the zeroes of 2x4 - 3x3 - 3x2 + 6x - 2, if two of its zeroes are 1 and 1/2.
Here, p(x) = 2x4 - 3x3 - 3x2 + 6x - 2
∵ 1 and are the zeroes of p (x)
∴ (x - 1) and are the factors of p (x)
⇒ (x - 1) (2x - 1) is a factor of p (x)
⇒ (2x2 - 3x + 1) is a factor of p (x).
Now, dividing p (x) by 2x2 - 3x + 1, we get
∴ p (x) = (2x2 − 3x + 1) (x2 − 2)
are the zeroes of p (x).
Q 18. On dividing 4x3 - 8x2 + 8x + 1 by a polynomial g (x), the Quotient and remainder were (2x2 - 3x + 2) and (x + 3) respectively. Find g (x).
∵ Dividend = Divisor × Quotient + Remainder
i.e., p (x) = g (x) × Q (x) + r (x)
∴ g (x) =
Thus, the required polynomial g (x) = 2x - 1.
Q 19. If α and β are the zeroes of the quadratic polynomial p (x) = kx2 + 4x + 4 such that α2 + β2 = 24, find the value of k.
Here, p (x) = kx2 + 4x + 4.
Comparing it with ax2 + bx + c, we have:
a = k; b = 4; c = 4
∴ Sum of the zeroes = -b/a
⇒ α + β = -4/k
and Product of the zeroes = c/a
⇒ αβ = 4/k
∵ α2 + β= 24
∴ (α + β)2 - 2αβ = 24
[∵ (x + y)2 = x2 + y2 + 2xy ⇒ (x + y)2 - 2xy = x2 + y2]
⇒
⇒ 16 − 8k − 24k2 =0
⇒ 24k2 + 8k − 16 = 0
⇒ (3k − 2) (k + 1) = 0
⇒ 3k − 2= 0 or k + 1 = 0
⇒ k = 2/3 or k = -1
Q 20. Find the zeroes of the quadratic polynomial 6x2 - 3 - 7x and verify the relationship between the zeroes and the coefficients of the polynomial.
Here, p (x) = 6x2 - 3 - 7x = 6x2 - 7x - 3
= 6x- 9x + 2x - 3
= 3x (2x - 3) + 1 (2x - 3)
= (2x - 3) (3x + 1)
=
∴ Zeroes of p (x) are 3/2 and
To verify the relationship:
Sum of the zeroes =
⇒ 7/6 = 7/6
L.H.S = R.H.S ⇒ Relationship is verified.
Product of the zeroes =
i.e., L.H.S = R.H.S ⇒ Relationship is verified.
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## Finding angles in a trapezoid
Finding angles in a trapezoid
• Currently 3.0/5 Stars.
6213 views, 1 rating - 00:04:40
Part of video series Solving for Angle Measures
Taught by mrbrianmclogan
### Meets NCTM Standards:
I show how to solve math problems online during live instruction in class. This is my way of providing free tutoring for the students in my class and for students anywhere in the world. Every video is a short clip that shows exactly how to solve math problems step by step. The problems are done in real time and in front of a regular classroom. These videos are intended to help you learn how to solve math problems, review how to solve a math problems, study for a test, or finish your homework. I post all of my videos on YouTube, but if you are looking for other ways to interact with me and my videos you can follow me on the following pages through My Blog, Twitter, or Facebook.
At 20 seconds, the teacher says that the expressions 3y + 40 and 3x - 70 are equal, but actually they are supplementary, so (3y + 40) + (3x - 70) = 180. Again at 2:35, he says that 3y + 40 = 110, but because the angle are supplementary, you should have 3y + 40 + 110 = 180.
Finding the values of variables to find angles in a trapezoid by using alternate interior and corresponding angle theorems.
• How do you find the angles in a trapezoid?
• If 3 angles in a trapezoid are 120, x, and 3y + 40, and the the angles 3y + 40 and 3x - 70 are supplementary, what do x and y equal?
• Why are the two angles on the side of a trapezoid supplementary?
• How can you solve for x and y in a trapezoid when the angles are variable expressions?
This lesson shows how to solve a geometry problem for x and y when you are given a trapezoid with some angle measures given as variable expressions. The method used to solve for x and y is to use the expressions and your knowledge of trapezoids to solve for the unknown variables and finally the angle measures. All steps are explained.
• Currently 3.0/5 Stars.
Reviewed by MathVids Staff on January 15, 2012.
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# SAT Math : How to find the probability of an outcome
## Example Questions
### Example Question #139 : Statistics
A bag contains four blue marbles, eight red marbles, and six orange marbles. If a marble is randomly selected from the bag, what is the probability that the marble will be either red or orange?
2/5
5/9
1/3
7/9
2/3
7/9
Explanation:
The probability of an event is the ratio of the number of desired outcomes to the total number of possible outcomes. In this problem, the total number of outcomes is equal to the total number of marbles in the bag. There are four blue, eight red, and six orange marbles, so the total number of marbles is the sum of four, eight, and six, or eighteen.
We are asked to find the probability of choosing a marble that is either red or orange. This means we have to consider the number of marbles that are either red or orange. Because there are eight red and six orange, there are fourteen marbles that are either red or orange.
The probability is thus fourteen out of eighteen, because there are fourteen red or orange marbles, out of a total of eighteen marbles. We will need to simplify the fraction 14/18.
Probability = 14/18 = 7/9
### Example Question #140 : Statistics
Two dice are rolled. Find the probability that the numbers sum to 4.
Explanation:
The possible dice combinations that sum to 4 are .
The number of all possible dice combinations is . (6 numbers on each of the two dice.)
So the probability that the numbers sum to 4 =
### Example Question #41 : How To Find The Probability Of An Outcome
Jackie is a contestant on a gameshow. She has to pick marbles out of a big bag to win various amounts of money. The bag contains a total of 200 marbles.
There are 100 red marbles worth $10 each, 50 blue marbles worth$20 each, 30 green marbles worth $50 each, 15 white marbles worth$100 each and 5 black marbles worth $1000 each. If she picks once, what percent chance does she have of picking a$1000 marble?
5%
.05%
3%
2.5%
2.5%
Explanation:
There are 5 black marbles worth \$1000 dollars out of the total of 200 marbles. or 2.5%
### Example Question #42 : How To Find The Probability Of An Outcome
What is the probability that you will pull out 3 diamonds in a standard deck without replacement.
Explanation:
There are 13 diamonds in a standard deck of 52 cards. So, you have a 13/52 (1/4) chance of getting a diamond; then a 12/51 (4/17) chance of pulling the next diamond; last, there is a 11/50 chance of getting the third diamond. When you combine probabilities, you multiply the individual probabilities together
### Example Question #43 : How To Find The Probability Of An Outcome
On Halloween there is a bowl of candy. 33 pieces are gum drops, 24 are candy corn, 15 are suckers and 28 are hard candies. Without looking in the bowl, what are the chances that you pull a gum drop?
Explanation:
Probability = # Gum Drops / # Total Number of Candies
### Example Question #44 : How To Find The Probability Of An Outcome
What is the probability of pulling out 4 diamonds in a standard deck of cards, without replacement?
Explanation:
There are 52 cards in a standard deck; 13 of them are diamonds. This means that there is a chance that the first card you pick will be a diamond. The probability of pulling another diamond as your next card is ; next diamond ; last diamond .
In order to get one probability, mulitply the individual probabilities together.
### Example Question #41 : Probability
A coporation is deciding who should sit on its leadership committee. If twenty people have applied, and the corporation will choose six applicants to sit on the committee, how many committees are possible?
Explanation:
This question requires us to make use of the combination formula. In general, if we have total objects or things to choose from, the number of different groups of size that we can make is equal to the following:
where the ! symbol denotes a factorial. In general, .
For example, .
We have 20 people from whom to select, and each group is going to have 6 people. Therefore, we will let and .
number of combinations =
=
### Example Question #46 : How To Find The Probability Of An Outcome
What is the probability that two dice rolled will add up to 5?
Explanation:
There are 6 probabilities for each dice and there are 2 dice. So, you would multiply the individual probabilities together and get 36 possible outcomes.
In order for 2 dice to add to a sum of 5, you would need to roll:
1 4; 2 3; 3 2; 4 1 [each dice can roll as either number]
That is a total of 4 outcomes
### Example Question #47 : How To Find The Probability Of An Outcome
A bag containing only red, green and blue marbles has 10 red, 20 green, and 20 blue marbles. If Laura reaches in and removes a single marble at random, what is the probability she draws a green marble?
Explanation:
Probability = the number of desired outcomes divided by the number of total possible outsomes. 20 marbles are green out of a total of 50. Therefore:
### Example Question #48 : How To Find The Probability Of An Outcome
A deck of 52 cards is divided equally into four suits: hearts, diamonds, clubs and spades, and two cards are drawn at random. The first card drawn is a heart. What is the probability that the second card drawn is also a heart?
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# Solutions Using the Discriminant
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Practice Solutions Using the Discriminant
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Solutions Using the Discriminant
What if you were given a quadratic equation like $x^2 - 3x + 1 = 0$ ? How could you determine how many real solutions it had without actually solving it? After completing this Concept, you'll be able to find and interpret the discriminant of a quadratic equation like this one.
### Guidance
In the quadratic formula, $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ , the expression inside the square root is called the discriminant. The discriminant can be used to analyze the types of solutions to a quadratic equation without actually solving the equation. Here’s how:
• If $b^2-4ac>0$ , the equation has two separate real solutions.
• If $b^2-4ac<0$ , the equation has only non-real solutions.
• If $b^2-4ac=0$ , the equation has one real solution, a double root.
Find the Discriminant of a Quadratic Equation
To find the discriminant of a quadratic equation we calculate $D=b^2-4ac$ .
#### Example A
Find the discriminant of each quadratic equation. Then tell how many solutions there will be to the quadratic equation without solving.
a) $x^2-5x+3=0$
b) $4x^2-4x+1=0$
c) $-2x^2+x=4$
Solution
a) Plug $a = 1, \ b = -5$ and $c = 3$ into the discriminant formula: $D=(-5)^2-4(1)(3)=13$ $D > 0$ , so there are two real solutions.
b) Plug $a = 4, \ b = -4$ and $c = 1$ into the discriminant formula: $D=(-4)^2-4(4)(1)=0$ $D = 0$ , so there is one real solution.
c) Rewrite the equation in standard form: $-2x^2+x-4=0$
Plug $a = -2, \ b = 1$ and $c = -4$ into the discriminant formula: $D=(1)^2-4(-2)(-4)=-31$ $D < 0$ , so there are no real solutions.
Interpret the Discriminant of a Quadratic Equation
The sign of the discriminant tells us the nature of the solutions (or roots) of a quadratic equation. We can obtain two distinct real solutions if $D > 0$ , two non-real solutions if $D < 0$ or one solution (called a double root) if $D = 0$ . Recall that the number of solutions of a quadratic equation tells us how many times its graph crosses the $x-$ axis. If $D > 0$ , the graph crosses the $x-$ axis in two places; if $D = 0$ it crosses in one place; if $D < 0$ it doesn’t cross at all:
#### Example B
Determine the nature of the solutions of each quadratic equation.
a) $4x^2-1=0$
b) $10x^2-3x=-4$
c) $x^2-10x+25=0$
Solution
Use the value of the discriminant to determine the nature of the solutions to the quadratic equation.
a) Plug $a = 4, \ b = 0$ and $c = -1$ into the discriminant formula: $D=(0)^2-4(4)(-1)=16$
The discriminant is positive, so the equation has two distinct real solutions.
The solutions to the equation are: $\frac{0 \pm \sqrt{16}}{8}=\pm \frac{4}{8}=\pm \frac{1}{2}$
b) Re-write the equation in standard form: $10x^2-3x+4=0$
Plug $a = 10, \ b = -3$ and $c = 4$ into the discriminant formula: $D=(-3)^2-4(10)(4)=-151$
The discriminant is negative, so the equation has two non-real solutions.
c) Plug $a = 1, \ b = -10$ and $c = 25$ into the discriminant formula: $D=(-10)^2-4(1)(25)=0$
The discriminant is 0, so the equation has a double root.
The solution to the equation is: $\frac{10 \pm \sqrt{0}}{2}=\frac{10}{2}=5$
If the discriminant is a perfect square, then the solutions to the equation are not only real, but also rational. If the discriminant is positive but not a perfect square, then the solutions to the equation are real but irrational.
#### Example C
Determine the nature of the solutions to each quadratic equation.
a) $2x^2+x-3=0$
b) $5x^2-x-1=0$
Solution
Use the discriminant to determine the nature of the solutions.
a) Plug $a = 2, \ b = 1$ and $c = -3$ into the discriminant formula: $D=(1)^2-4(2)(-3)=25$
The discriminant is a positive perfect square, so the solutions are two real rational numbers.
The solutions to the equation are: $\frac{-1 \pm \sqrt{25}}{4}=\frac{-1 \pm 5}{4}$ , so $x = 1$ and $x=-\frac{3}{2}$ .
b) Plug $a = 5, \ b = -1$ and $c = -1$ into the discriminant formula: $D=(-1)^2-4(5)(-1)=21$
The discriminant is positive but not a perfect square, so the solutions are two real irrational numbers.
The solutions to the equation are: $\frac{1 \pm \sqrt{21}}{10}$ , so $x \approx 0.56$ and $x \approx -0.36$ .
Solve Real-World Problems Using Quadratic Functions and Interpreting the Discriminant
You’ve seen that calculating the discriminant shows what types of solutions a quadratic equation possesses. Knowing the types of solutions is very useful in applied problems. Consider the following situation.
#### Example D
Marcus kicks a football in order to score a field goal. The height of the ball is given by the equation $y=-\frac{32}{6400}x^2+x$ . If the goalpost is 10 feet high, can Marcus kick the ball high enough to go over the goalpost? What is the farthest distance that Marcus can kick the ball from and still make it over the goalpost?
Solution
Define: Let $y =$ height of the ball in feet.
Let $x =$ distance from the ball to the goalpost.
Translate: We want to know if it is possible for the height of the ball to equal 10 feet at some real distance from the goalpost.
Solve:
$&\text{Write the equation in standard form:} && -\frac{32}{6400}x^2+x-10= 0\\&\text{Simplify:} && -0.005x^2+x-10 = 0\\&\text{Find the discriminant:} && D =(1)^2-4(-0.005)(-10)=0.8$
Since the discriminant is positive, we know that it is possible for the ball to go over the goal post, if Marcus kicks it from an acceptable distance $x$ from the goalpost.
To find the value of $x$ that will work, we need to use the quadratic formula:
$x=\frac{-1 \pm \sqrt{0.8}}{-0.01}=189.4 \ feet \ \text{or} \ 10.56 \ feet$
What does this answer mean? It means that if Marcus is exactly 189.4 feet or exactly 10.56 feet from the goalposts, the ball will just barely go over them. Are these the only distances that will work? No; those are just the distances at which the ball will be exactly 10 feet high, but between those two distances, the ball will go even higher than that. (It travels in a downward-opening parabola from the place where it is kicked to the spot where it hits the ground.) This means that Marcus will make the goal if he is anywhere between 10.56 and 189.4 feet from the goalposts.
Watch this video for help with the Examples above.
### Vocabulary
In the quadratic formula, $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ , the expression inside the square root is called the discriminant. The discriminant can be used to analyze the types of solutions to a quadratic equation without actually solving the equation. Here’s how:
• If $b^2-4ac>0$ , the equation has two separate real solutions.
• If $b^2-4ac<0$ , the equation has only non-real solutions.
• If $b^2-4ac=0$ , the equation has one real solution, a double root.
### Guided Practice
Emma and Bradon own a factory that produces bike helmets. Their accountant says that their profit per year is given by the function $P=-0.003x^2+12x+27760$ , where $x$ is the number of helmets produced. Their goal is to make a profit of $40,000 this year. Is this possible? Solution We want to know if it is possible for the profit to equal$40,000.
$40000=-0.003x^2+12x+27760$
Write the equation in standard form: $-0.003x^2+12x-12240=0$
Find the discriminant: $D=(12)^2-4(-0.003)(-12240)=-2.88$
### Vocabulary Language: English
discriminant
discriminant
The discriminant is the part of the quadratic formula under the radical, $b^2 - 4ac$. A positive discriminant suggests two real roots to the quadratic equation, a zero suggests one real root with multiplicity two, and a negative indicates two complex roots.
Double Root
Double Root
A solution that is repeated twice.
### Explore More
Sign in to explore more, including practice questions and solutions for Solutions Using the Discriminant.
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# If the sum of the surface areas of cube and a sphere is constant,
Question:
If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?
Solution:
sphere.
Surface area of cube = 6x2
And, surface area of the sphere = 4πr2
Now, their sum is
$6 x^{2}+4 \pi r^{2}=\mathrm{K}($ constant $) \Rightarrow r=\sqrt{\frac{\mathrm{K}-6 x^{2}}{4 \pi}}$.............(i)
Volume of the cube $=x^{3}$ and the volume of sphere $=\frac{4}{3} \pi r^{3}$
Now,
Sum of their volumes $(V)=$ Volume of cube
+ Volume of sphere
$V=x^{3}+\frac{4}{3} \pi r^{3}$
$\Rightarrow \mathrm{V}=x^{3}+\frac{4}{3} \pi \times\left(\frac{\mathrm{K}-6 x^{2}}{4 \pi}\right)^{3 / 2}$
Differentiating both sides w.r.t. $x$, we get
$\frac{d \mathrm{~V}}{d x}=3 x^{2}+\frac{4 \pi}{3} \times \frac{3}{2}\left(\mathrm{~K}-6 x^{2}\right)^{1 / 2}(-12 x) \times \frac{1}{(4 \pi)^{3 / 2}}$
$=3 x^{2}+\frac{2 \pi}{(4 \pi)^{3 / 2}} \times(-12 x)\left(\mathrm{K}-6 x^{2}\right)^{1 / 2}$
$=3 x^{2}+\frac{1}{4 \pi^{1 / 2}} \times(-12 x)\left(\mathrm{K}-6 x^{2}\right)^{1 / 2}$
Thus, $\frac{d \mathrm{~V}}{d x}=3 x^{2}-\frac{3 x}{\sqrt{\pi}}\left(\mathrm{K}-6 x^{2}\right)^{1 / 2}$ $\ldots$ (ii)
For local maxima and local minima, $\frac{d V}{d x}=0$
$3 x^{2}-\frac{3 x}{\sqrt{\pi}}\left(\mathrm{K}-6 x^{2}\right)^{1 / 2}=0$
$3 x\left[x-\frac{\left(\mathrm{K}-6 x^{2}\right)^{1 / 2}}{\sqrt{\pi}}\right]=0$
$x \neq 0$ and $x-\frac{\left(\mathrm{K}-6 x^{2}\right)^{1 / 2}}{\sqrt{\pi}}=0$
$\Rightarrow \quad x=\frac{\left(\mathrm{K}-6 x^{2}\right)^{1 / 2}}{\sqrt{\pi}}$
Squaring both sides, we get
$x^{2}=\frac{\mathrm{K}-6 x^{2}}{\pi} \Rightarrow \pi x^{2}=\mathrm{K}-6 x^{2}$
So, $\quad \pi x^{2}+6 x^{2}=\mathrm{K} \Rightarrow x^{2}(\pi+6)=\mathrm{K} \Rightarrow x^{2}=\frac{\mathrm{K}}{\pi+6}$
Thus, $x=\sqrt{\frac{\mathrm{K}}{\pi+6}}$
Now pulting the value of $\mathrm{K}$ in eq. (i), we get
$6 x^{2}+4 \pi r^{2}=x^{2}(\pi+6)$
$6 r^{2}+4 \pi r^{2}=\pi r^{2}+6 r^{2} \rightarrow 4 \pi r^{2}=\pi r^{2} \rightarrow 4 r^{2}=r^{2}$
$2 r=x$
$\therefore x: 2 r=1: 1$
Now differentiating eq. (ii) w.r.t $x$, we have
$\frac{d^{2} \mathrm{~V}}{d x^{2}}=6 x-\frac{3}{\sqrt{\pi}} \frac{d}{d x}\left[x\left(\mathrm{~K}-6 x^{2}\right)^{1 / 2}\right]$
$=6 x-\frac{3}{\sqrt{\pi}}\left[x \cdot \frac{1}{2 \sqrt{K-6 x^{2}}} \times(-12 x)+\left(\mathrm{K}-6 x^{2}\right)^{1 / 2} \cdot 1\right]$
$=6 x-\frac{3}{\sqrt{\pi}}\left[\frac{-6 x^{2}}{\sqrt{\mathrm{K}-6 x^{2}}}+\sqrt{\mathrm{K}-6 x^{2}}\right]$
$=6 x-\frac{3}{\sqrt{\pi}}\left[\frac{-6 x^{2}+\mathrm{K}-6 x^{2}}{\sqrt{\mathrm{K}-6 x^{2}}}\right]=6 x+\frac{3}{\sqrt{\pi}}\left[\frac{12 x^{2}-\mathrm{K}}{\sqrt{\mathrm{K}-6 x^{2}}}\right]$
Put $x=\sqrt{\frac{K}{\pi+6}}=6 \sqrt{\frac{K}{\pi+6}}+\frac{3}{\sqrt{\pi}}\left[\frac{\frac{12 K}{\pi+6}-K}{\sqrt{K-\frac{6 K}{\pi+6}}}\right]$
$=6 \sqrt{\frac{\mathrm{K}}{\pi+6}}+\frac{3}{\sqrt{\pi}}\left[\frac{12 \mathrm{~K}-\pi \mathrm{K}-6 \mathrm{~K}}{\sqrt{\frac{\pi \mathrm{K}+6 \mathrm{~K}-6 \mathrm{~K}}{\pi+6}}}\right]$
$=6 \sqrt{\frac{\mathrm{K}}{\pi+6}}+\frac{3}{\sqrt{\pi}}\left[\frac{6 \mathrm{~K}-\pi \mathrm{K}}{\sqrt{\frac{\pi \mathrm{K}}{\pi+6}}}\right]$
$=6 \sqrt{\frac{\mathrm{K}}{\pi+6}}+\frac{3}{\pi \sqrt{K}}[(6 \mathrm{~K}-\pi \mathrm{K}) \sqrt{\pi+6}]>0$
So, it is the minima.
Therefore, the required ratio is 1: 1 when the combined volume is minimum.
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When two straight lines meet at a point, they form an angle. The point is called the vertex of the angle, and the lines are called the sides of the angle. The size of an angle depends on how much one side rotates away from the other side. An angle is usually measured in degrees or radians.
### Types of Angles
1. An acute angle has measure of less than 90 degrees. Like an acute, or sharp, the acute angle has a sharp point.
2. A right or perpendicular angle has measure of 90 degrees. It makes up a square corner.
3. An obtuse angle has measure of greater than 90 degrees but less than 180 degrees. The opposite of an acute angle, an obtuse angle is dull rather than sharp.
4. A straight angle has measure of 180 degrees. A straight angle appears to be a straight line or line segment.
Complementary Angles
Two angles that add together to total 90 degrees. Together, they form a right angle.
Supplementary Angles
Two angles that add together to total 180 degrees. They form a straight angle.
Similar: Objects that have the same shape but may have different sizes.
Congruent: Objects that are equal in size and shape. Two line segments with the same length, two angles with the same measure, and two triangles with corresponding sides of equal lengths and angles that have equal degree measures are congruent.
### Rules for Lines and Angles
Intersecting Lines
When two lines intersect, the opposite angles (across from each other) are always congruent or equal, and the adjacent angles are always supplementary. Opposite angles are also known as vertical angles. Adjacent angles have a common side, so they are right next to each other.
In the figure, ∠ABC and ∠DBE are congruent, ∠ABC and ∠CBD form a straight line and are, therefore, supplementary.
Parallel Lines intersected by a Transversal
When parallel lines are crossed by a third line that’s not perpendicular to them (called a transversal), the resulting small and large angles share certain properties. Each of the small angles is equal; the large angles are also equal to each other. The measurement of any small angle added to that of any large angle equals 180°.
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# How do you solve q/4-1=3?
Jan 11, 2017
See the Two Step process below
#### Explanation:
Step 1) Add $\textcolor{red}{1}$ to each side of the equation to isolate the $Q$ term while keeping the equation balanced:
$\frac{q}{4} - 1 + \textcolor{red}{1} = 3 + \textcolor{red}{1}$
$\frac{q}{4} - 0 = 4$
$\frac{q}{4} = 4$
Step 2) Multiply each side of the equation by $\textcolor{red}{4}$ to solve for $x$ while keeping the equation balanced:
$\frac{q}{4} \times \textcolor{red}{4} = 4 \times \textcolor{red}{4}$
$\frac{q}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}} \times \cancel{\textcolor{red}{4}} = 16$
$q = 16$
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Congruency of triangles
Chapter 7 Class 7 Congruence of Triangles
Concept wise
Let suppose two triangles are congruent
We know that two triangles are congruent when
• They have same shape
• They have same size
Let’s rotate PQR
And super impose ∆ABC & ∆PQR
So,
Points match
A ⟷ P
B ⟷ Q
C ⟷ R
Thus, we write
∆ABC ≅ ∆PQR
Not
∆ABC ≅ ∆QRP
∆ABC ≅ ∆PRQ
Because, in
∆ABC ≅ ∆PQR
A ⟷ P
B ⟷ Q
C ⟷ R
But, why is this order important?
Because of CPCT,
CPCT is corresponding parts of Congruent Triangles
If two triangles are congruent,
• Their corresponding sides are equal
• Their corresponding angles are equal
Now,
In ∆ABC & ∆PQR
If ∆ABC ≅ ∆PQR
Then,
Corresponding angles are equal Corresponding sides are equal ∠A = ∠P AB = PQ ∠B = ∠Q BC = QR ∠C = ∠R AC = PR
Let’s check more examples
### Which triangles are congruent?
Then,
Corresponding angles are equal Corresponding sides are equal ∠M = ∠Z MN = ZX ∠N = ∠X NO = XY ∠O = ∠Y OM = YZ
Here,
M ⟷ Z
N ⟷ X
O ⟷ Y
So,
∆MNO ≅ ∆ZXY
### Which triangles are congruent?
Then,
Corresponding angles are equal Corresponding sides are equal ∠P = ∠U PR = US ∠R = ∠S QR = TS ∠Q = ∠T PQ = UT
Here,
P ⟷ U
R ⟷ S
Q ⟷ T
So,
∆PQR ≅ ∆UTS
### Which triangles are congruent?
Then,
Corresponding angles are equal Corresponding sides are equal ∠P = ∠L PQ = LN ∠Q = ∠N QR = NM ∠R = ∠M PR = LM
Here,
Q ⟷ N
R ⟷ M
P ⟷ L
So,
∆PQR ≅ ∆LNM
Here,
R ⟷ S
P ⟷ U
Q ⟷ T
∴ ∆PQR ≅ ∆UTS
Here,
Y ⟷ K
Z ⟷ J
X ⟷ L
So, ∆XYZ ≅ ∆LKJ
Here,
A ⟷ Y
B ⟷ X
C ⟷ Z
So, ∆ABC ≅ ∆YXZ
Here,
J ⟷ N
I ⟷ L
K ⟷ M
So, ∆IJK ≅ ∆LNM
Here,
P ⟷ U
Q ⟷ S
R ⟷ T
So, ∆PQR ≅ ∆UST
### Which triangles are congruent?
Here,
N ⟷ D
O ⟷ F
M ⟷ E
So, ∆MNO ≅ ∆EDF
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CBSE Mathematics for Class 6 > Chapter 1: Knowing Our Numbers > Roman Numerals
ROMAN NUMERALS
One of the earliest systems of writing numerals is the Roman Numeral system. This system is still in use in many places. For example, some faces of clocks show hours in Roman numerals; we use Roman numerals to write numbered list; etc.
Unlike the Hindu-Arabic numeral system, Roman numeral system uses seven basic symbols to represent different numbers. The symbols are as follows :
The Roman numerals :
I, II, III, IV, V, VI, VII, VIII, IX, X denote 1,2,3,4,5,6,7,8,9 and 10 respectively. This is followed by XI for 11, XII for 12,... till XX for 20. Some more Roman numerals are :
I V X L C D M
1 5 10 50 100 500 1000
The rules for the system are :
(a) If a symbol is repeated, its value is added as many times as it occurs: i.e. II is equal 2, XX is 20 and XXX is 30.
(b) A symbol is not repeated more than three times. But the symbols V, L and D are never repeated.
(c) If a symbol of smaller value is written to the right of a symbol of greater value, its value gets added to the value of greater symbol.
VI = 5 + 1 = 6, XII = 10 + 2 = 12 and LXV = 50 + 10 + 5 = 65
(d) If a symbol of smaller value is written to the left of a symbol of greater value, its value is subtracted from the value of the greater symbol.
IV = 5 – 1 = 4, IX = 10 – 1 = 9
XL = 50 – 10 = 40, XC = 100 – 10 = 90
(e) The symbols V, L and D are never written to the left of a symbol of greater value, i.e. V, L and D are never subtracted.
The symbol I can be subtracted from V and X only.
The symbol X can be subtracted from L, M and C only.
Following these rules we get,
1 = I 10 = X 100 = C
2 = II 20 = XX
3 = III 30 = XXX
4 = IV 40 = XL
5 = V 50 = L
6 = VI 60 = LX
7 = VII 70 = LXX
8 = VIII 80 = LXXX
9 = IX 90 = XC
Ex : Write in Roman Numerals
(i) 69 (ii) 98.
Sol: (i) 69 = 60 + 9 = (50 + 10) + 9 = LX + IX = LX IX
(ii) 98 = 90 + 8 = (100 – 10) + 8 = XC + VIII = XCVIII
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Roman Numbers
Roman Numbers are one of the earlier system of writing numbers. Like I, II, III, IV, V etc.
The Roman Numerals:
I, II, III, IV, V, VI, VII, VIII, IX, X These Roman numbers denote 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively.
Some more Roman Numbers
Rules for the Roman Number System:
1) If a symbol is repeated, its value is added as many times as it occurs.
Like, II is equal to 2, III is 3, XX is 20, XXX is 30, CC is 200, MM is 2000, MMM is 3000 etc.
2) A symbol is not repeated more than three times. But the symbol V, L, D are never repeated.
3) If a symbol of smaller value is written to the right side of a symbol of greater value, its value gets added to the value of greater symbol. Like,
VI = 5+1 =6 VII= 5+2= 7 XIII= 10+3= 13 LXV= 50+10+5= 65CLX= 100+50+10=160
4) If a symbol of smaller value is written to the left of the symbol of greater value, its value is subtracted from the value of greater symbol.
IV = 5-1= 4 IX= 10-1= 5 XL= 50-10= 40 XC= 100-10= 90 CD= 500-100= 400
5) The symbols V, L and D are never written to the left of a symbol of greater value, i.e. V, L and D are never subtracted.
The Symbol I can be subtracted from V and X only.
The symbol X can be subtracted from L, M and C only.
Roman Numbers using these rules
Some Solved Examples.
5) 1236 = 1000 + 200 + 30 + 6
Roman number for 1000 = M
Roman number for 200 = CC
Roman number for 30 = XXX
Roman number for 6 = VI
So, Roman number for 1236= MCCXXXVI
Large Numbers in Practice
Large Numbers can be shown using the place value. It goes in the thrusting order as shown below
8 integers 7 integers 6 integers 5 integers 4 integers 3 integers
centimeter( cm) in used as a unit of dimension of length. We can use this unit for measuring the length of a pen, the range of a book or note booket. but this unit is too big to measure the consistence of a pen. So, we use another unit named as millimeter( mm). Also, centimeter and millimeter are veritably small units to measure the length of the wall or a room. We use another unit named as meter for the same. Indeed meter is too small unit when we state the distances between two municipalities or metropolises. For this we need kilometers( km).
The relations between units are
1 kilometer = 1000 meters( m)
1 meter = 100 centimeters( cm)
1 centimeter = 10 millimeters( mm)
Also, we have
100 cm = 1 m = 10mm × 100 = 1000 mm
1000 m = 1Km = 1000 × 100 cm = 100000 cm
100000 cm =1 km = 100000 × 10 mm = 1000000 mm
10 Million = 1 crore
1 million = 10 lakhs
(100,000) Hundred Thousands = 1 lakh
(10,000)Ten Thousands = Thousands Hundreds (1000,000)
We borrow the following rules to compare two large figures
Rule- 1 The number with further integers is lesser than the number with lower integers.
Rule- 2 When two figures have the same number of integers, compare the integers at the leftmost places. The number with the lesser number is lesser. If the integers at the leftmost places are the same also compare the coming integers and so on.
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# 2-5 Statistics &
Analyzing Data
Lesson 2-1 Rational Numbers on the Number Line
Lesson 2-2 Adding and Subtracting Rational Numbers
Lesson 2-3 Multiplying Rational Numbers
Lesson 2-4 Dividing Rational Numbers
Lesson 2-5 Statistics: Displaying and Analyzing Data
Lesson 2-6 Probability: Simple Probability and Odds
Lesson 2-7 Square Roots and Real Numbers
Example 1 Identify Coordinates on a Number Line
Example 2 Graph Numbers on a Number Line
Example 3 Absolute Value of Rational Numbers
Example 4 Expressions with Absolute Value
Name the coordinates of the points graphed on the
number line.
## Answer: The coordinates are {–9, –7, –6, –3}.
Name the coordinates of the points graphed on the
number line.
## The bold arrow on the graph indicates that the graph
continues infinitely in that direction.
## Answer: The coordinates are {11, 12, 13, 14, …}.
Name the coordinates of the points graphed on each
number line.
a.
b.
## Answer: {–0.5, 0, 0.5, 1, 1.5, …}
Graph .
Answer:
Graph {–1.5, 0, 1.5, …}.
Answer:
Graph {integers less than –6 or greater than or
equal to 1}.
Answer:
Graph each set of numbers.
a. {–5, 2, 3, 5}
Answer:
b.
Answer:
c. {integers less than or equal to –2 or greater than 4}
Answer:
Find .
Answer:
Find .
## 0.25 is 0.25 unit from 0 in the positive direction.
Answer:
Find each absolute value.
a. Answer:
b. Answer: 6.3
Replace y with 12.
Answer: Simplify.
Answer: 10
Example 1
Use a Number Line to Add Rational Numbers
Example 2 Add Rational Numbers
Example 3
Subtract Rational Numbers to Solve a Problem
Use a number line to find .
–5
+8
## Step 1 Draw an arrow from 0 to 8.
Step 2 Then draw a second arrow 5 units to the left to
represent adding –5.
Step 3 The second arrow ends at the sum 3.
Answer:
Use a number line to find .
–4 –1
## Step 1 Draw an arrow from 0 to –1.
Step 2 Draw a second arrow 4 units to the left.
Step 3 The second arrow ends at the sum –5.
Answer:
Use a number line to find each sum.
a. Answer: –3
b. Answer: –9
Find .
## Subtract the lesser
absolute value from the
greater absolute value.
## Since the number with the
greater absolute value is
–14, the sum is negative.
Answer:
Find .
## Both numbers are
negative, so the sum
is negative.
Answer:
Find each sum.
a. Answer: –17
b. Answer:
Stocks In the past year, a publishing company’s stock
went from \$52.08 per share to \$70.87 per share. Find
the change in the price of the stock.
Explore The stock price began at \$52.08 per share and
ended at \$70.87. You need to determine the
change in price for the year.
Plan Subtract to find the change in price.
## ending price minus beginning price
70.87 52.08
Solve
To subtract 52.08,
add its inverse.
Subtract the
absolute values.
## The absolute value
of 70.87 is greater,
so the result is
positive.
Answer: The price of the stock changed by \$18.79.
## Examine The problem asks for the change in a stock’s
price in the past year. Since the change in price
was positive, the price increased. This makes
sense since the ending price is more than the
beginning price.
Stocks The stock in a company went from \$46.98 to
\$35.09 over a one-month period. Find the change in
price for the stock.
Answer: –\$11.89
Example 1 Multiply Integers
Example 2 Simplify Expressions
Example 3 Multiply Rational Numbers
Example 4
Multiply Rational Numbers to Solve a Problem
Example 5 Evaluate Expressions
Find (–8)(–6).
Answer: 48
Find (10)(–11).
## different signs → negative product
Answer: –110
Find each product.
## a. (–4)(8) Answer: –32
b. (–6)(–12) Answer: 72
Simplify the expression
## Associative Property (×)
Substitution
Distributive Property
Answer: Simplify.
Simplify the expression
Answer: 3x
Find
## same signs → positive product
Answer:
Find
Answer:
Stocks The value of a company’s stock dropped by
\$1.25 per share. By what amount did the total value of
the company’s stock change if the company has
issued 500,000 shares of stock?
To find the change in the total value of the company’s
stock, multiply the price lost per share by the number
of shares.
## Answer: The total value of the company’s stock
changed by –\$625,000.
Construction A construction project is stopped by a
winter storm. For every day that they are unable to
work, the company loses \$35,000. If the storm keeps
them from working for 4 days, how much money do
they lose?
Answer: –\$140,000
Substitution
## Answer: same signs → positive product
Answer:
Example 1 Divide Integers
Example 2 Simplify Before Dividing
Example 3 Divide Rational Numbers
Example 4
Divide Rational Numbers to Solve a Problem
Example 5 Simplify Algebraic Expressions
Example 6 Evaluate Algebraic Expressions
Find .
Find .
Divide.
## Answer: negative quotient
Find each quotient.
a. Answer: 20
b. Answer: –15
Simplify
Multiply.
## Answer: different signs →
negative quotient
Simplify
Answer: 3
Find .
## Answer: Use a calculator.
different signs →
negative quotient
Find .
of
## Answer: same signs → positive quotient
Find each quotient.
a. Answer: 15.3
b. Answer:
Baseball The perimeter of a square baseball diamond
is 360 feet. Find the length of one side of the diamond.
number of sides.
## Answer: The length of one side is 90 feet.
The perimeter of a triangular building is 450 feet. Find
the length of each side.
Simplify
## The fraction bar
indicates division.
Multiply by
## the reciprocal of 13.
Distributive Property.
Answer: Simplify.
Simplify
Answer:
Evaluate if w = 2, x = –9.1 and y = 4.
and y with 4.
## Find the numerator and
denominator separately.
## Answer: Use a calculator.
different signs → negative quotient
Evaluate if s = 2.3, t = 5 and u = –4.
Answer: –14.375
Example 1 Create a Line Plot
Example 2 Use a Line Plot to Solve a Problem
Example 3 Create a Stem-and-Leaf Plot
Example 4 Back-to-Back Stem-and-Leaf Plot
Example 5 Analyze Data
Example 6
Determine the Best Measure of Central Tenden
Draw a line plot for the data.
11 –2 10 –2 7 2 7 4 9 0 6 9 7 2 0 4 10 7 6 9
## Step 1 The values of the data range from –2 to 11, so
construct a number line containing these values.
Step 2 Then place an × a number for each
time it occurs.
Draw a line plot for the data.
3 5 7 6 0 –4 6 4 7 0 0 –2 3 7
Answer:
Traffic The highway patrol did a radar survey of the
speeds of cars along a stretch of highway for 1
minute. The speeds (in miles per hour) of the 20
cars that passed are listed below.
72 70 72 74 68 69 70 72 74 75
79 75 74 72 70 64 69 66 68 67
Make a line plot of the data.
The lowest value is 64 and the highest value is 79, so use
a scale that includes those values. Place an × above each
value for each occurrence.
Answer:
Which speed occurs the most frequently?
## Answer: Looking at the line plot, we can easily see that
72 miles per hour occurs most frequently.
Family Size Students in Mrs. Barrett’s class listed the
number of family members in their households below.
6 4 8 3 3 5 4 4 3 5 5 2 5 6 3 5 6 2 4 4 4
a. Make a line plot of the data.
Answer:
## b. Which family size occurs the most frequently?
Answer: 4
Use the data below to make a stem-and-leaf plot.
## 85 115 126 92 104 107 78 131 114 92
85 116 100 121 123 131 88 97 99 116
79 90 110 129 108 93 84 75 70 132
## The greatest common place value is tens, so the digits in
the tens place are the stems.
Answer: Stem Leaf
7 0589
8 4558
9 022379
10 0478
11 04566
12 1369
13 112
Use the data below to make a stem-and-leaf plot.
3 5 7 11 10 15 21 11
13 25 32 37 21 10 12
## Answer: Stem Leaf
0 357
1 0011235
2 115
3 27
Weather Monique wants to compare the monthly
average high temperatures of Dallas and Atlanta
before she decides to which city she wants to move.
The table shows the monthly high temperatures (°F)
for both cities.
Monthly Average
High Temperature
Dallas Atlanta
54 59 68 77 50 55 64 72
83 91 95 95 75 85 88 87
87 78 66 57 81 72 63 54
Make a stem-and-leaf plot to compare the data.
To compare the data we can use a back-to-back stem-
and-leaf plot. Since the data represent similar
measurements, the plot will share a common stem.
## Answer: Dallas Stem Atlanta
9 7 4 5 0 4 5
8 6 6 3 4
8 7 7 2 2 5
7 3 8 1 5 7 8
5 5 1 9
What is the difference between the highest average
temperatures in each city?
Answer: 95 – 88 or 7°
Which city has higher average temperatures?
## Answer: Looking at the temperatures of 80 and above,
we can see that Dallas has a higher number of average
temperatures above 80°.
Ms. Smith wants to compare the final grades for
two of her classes. The table shows the scores for
both classes.
Class A Class B
87 96 99 76
81 51 62 57
92 98 77 83
76 75 72 85
71 64 69 91
a. Make a back-to-back stem-and-leaf plot to compare
the data.
## Answer: Class A Stem Class B
1 5 7
4 6 2 9
6 5 1 7 2 6 7
7 1 8 3 5
8 6 2 9 1 9
b. What is the difference between the highest score
in each class?
Answer: 1 point
## c. Which class scored higher overall for the
grading period?
Answer: Class A
Which measure of central tendency best
represents the data?
Stem Leaf
4 11244458
5 0
6 257
7 39
8 1
Determine the mean, median, and mode.
The mean is about 5.5. Add the data and divide by 15.
## The mode is 4.4. The most frequent value is 4.4.
Answer: Either the median or the mode best represent
the set of data since both measures are located in the
center of the majority of the data. In this instance, the
mean is too high.
Which measure of central tendency best
represents the data?
Stem Leaf
1 011568
2 378
3 2
4 6
5 459
Answer: The mean is about 2.9. The median is 2.5. The
mode is 1.1. Either the mean or median can be used to
represent the data. The mode is too low.
Politics The number of electoral college votes for the
12 most populous states in the 2000 Presidential
election are listed below. Which measure of central
tendency best represents the data?
21 22 18 23 15 25
14 32 13 33 13 54
The mean is about 23.6. Add the data and divide by 12.
The median is 21.5. The middle value is 21.5.
The mode is 13. The most frequent value is 13.
Answer: Either the mean or median can be used to best
represent the data. The mode is too low.
The number of points scored by the basketball team
during each game in the season is listed below. Which
measure of central tendency best represents the data?
48 45 52 63 59 64 67 72 58
51 81 62 73 68 82 73 70 65
## Answer: Either the mean or the median can be used to
best represent the data. The mode is too high.
Example 1 Find Probabilities of Simple Events
Example 2 Odds of an Event
Example 3 Odds Against an Event
Example 4 Probability and Odds
Find the probability of rolling a number greater
than 2 on a die.
## There are six possible outcomes. Four of the outcomes
are favorable. That is, four of the six outcomes are
numbers greater than two.
4 numbers
greater than 2
Sample space: 1, 2, 3, 4, 5, 6
6 possible
outcomes
Answer:
A class contains 6 students with black hair, 8 with
brown hair, 4 with blonde hair, and 2 with red hair.
Find P(black).
There are 6 students with black hair and 20 total students.
number of favorable outcomes
number of possible outcomes
Simplify.
## Answer: The probability of selecting a student with black
hair is
A class contains 6 students with black hair, 8 with
brown hair, 4 with blonde hair, and 2 with red hair.
Find P(red or brown).
## There are 2 students with red hair and 8 students with
brown hair. So there are 2 + 8 or 10 students with red
or brown hair.
## number of favorable outcomes
number of possible outcomes
Simplify.
Answer: The probability of selecting a student with red
or brown hair is
A class contains 6 students with black hair, 8 with
brown hair, 4 with blonde hair, and 2 with red hair.
Find P(not blonde).
There are 6 + 8 + 2 or 16 students who do not have
blonde hair.
number of favorable outcomes
number of possible outcomes
Simplify.
## not have blonde hair is
a. Find the probability of rolling a
number less than 3 on a die.
Answer:
## b. A gumball machine contains 40 red gumballs, 30
green gumballs, 50 yellows gumballs, and 40 blue
gumballs. Find P(red).
Answer:
c. A gumball machine contains 40 red gumballs, 30
green gumballs, 50 yellows gumballs, and 40 blue
gumballs. Find P(green or yellow).
Answer:
## d. A gumball machine contains 40 red gumballs, 30
green gumballs, 50 yellows gumballs, and 40 blue
gumballs. Find P(not blue).
Answer:
Find the odds of rolling a number greater than 2.
There are six possible outcomes, 4 are successes and
2 are failures.
4 numbers
greater than 2
Sample space: 1, 2, 3, 4, 5, 6
2 numbers
less than or
equal to 2
## Answer: The odds of rolling a number greater than
Find the odds of rolling a number less than 4.
Answer:
A card is selected at random from a standard
deck of 52 cards. What are the odds against
selecting a 2 or 3?
## There are four 2s and four 3s in a deck of cards, and
there are 52 – 4 – 4 or 44 cards that are not a 2 or a 3.
## number of ways not to pick
a 2 or 3
number of ways to pick
a 2 or 3
## Answer: The odds against selecting a 2 or 3 are 11:2.
A card is selected at random from a standard
deck of 52 cards. What are the odds against
selecting a 5, 6, or 7?
Answer: 10:3
Travel Melvin is waiting to
board a flight to Washington,
D.C. According to the airline, the flight he is waiting for
is on time 80% of the times it flies. What are the odds
that the plane will be on time?
## The probability that the plane will be on time is 80%, so
the probability that it will not be on time is 20%.
## odds of the plane being on time
Answer: The odds that the plane will be on time are 4:1.
If the probability that it will snow this weekend is 70%,
what are the odds that it will snow?
Answer: 7:3
Example 1 Find Square Roots
Example 2 Classify Real Numbers
Example 3 Graph Real Numbers
Example 4 Compare Real Numbers
Example 5 Order Real Numbers
Example 6 Rational Approximation
Find .
roots of
Answer:
Find .
## represents the positive square root of 0.0144.
Answer:
Find each square root.
a. Answer:
b. Answer: 0.6
Name the set or sets of numbers to which
belongs.
## Answer: Because , which is
neither a repeating nor terminating decimal, this number
is irrational.
Name the set or sets of numbers to which
belongs.
## Answer: Because 1 and 6 are integers and
, which is a repeating decimal,
the number is a rational number.
Name the set or sets of numbers to which
belongs.
## Answer: Because this number is a
natural number, a whole number, an integer and
a rational number.
Name the set or sets of numbers to which –327
belongs.
## Answer: This number is an integer and a
rational number.
Name the set or sets of numbers to which each real
number belongs.
a. Answer: rationals
## b. Answer: naturals, whole, integers,
rationals
c. Answer: irrationals
Graph .
## The heavy arrow indicates that all numbers to the left of 8
are included in the graph. The dot at 8 indicates that 8 is
included in the graph.
Graph .
The heavy arrow indicates that all the points to the right of
–5 are included in the graph. The circle at –5 indicates
that –5 is not included in the graph.
Graph each solution set.
a.
Answer:
b.
Answer:
Replace the • with <, >, or = to make the
sentence true.
## Since the numbers are equal.
Answer:
Replace the • with <, >, or = to make the
sentence true.
Answer:
Replace each • with <, >, or = to make each
sentence true.
a. Answer: <
b. Answer: <
Write in order from least
or about 2.4495
or about 2.4444
## Answer: The numbers arranged in order from least to
greatest are
Write in order from least
to greatest.
Answer:
Multiple-Choice Test Item
A –5 B 0 C D 5
## Read the Test Item
The expression is an open sentence,
## and the set of choices is the
replacement set.
Solve the Test Item
Replace x in with each given value.
B
## False; is not a real number.
C
Use a calculator.
0.447214 < 1 < 2.236068 True
D
Use a calculator.
2.236068 < 1 < 0.447214 False
## Answer: The correct answer is C.
Multiple-Choice Test Item
## For what value of x is true?
A 3 B –3 C 0 D
Answer: A
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Student Edition of your textbook. When you finish
exploring, exit the browser program to return to this
presentation. If you experience difficulty connecting
to the Web site, manually launch your Web browser
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# Find the equation of the perpendicular bisector of $AB$ where $A$ and $B$ are the points $\left( {3,6} \right)$and $\left( { - 3,4} \right)$ respectively. Also, find its point of intersection with i) $x$-axis ii) $y$-axis
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Hint: Since, it is given that it is a perpendicular bisector therefore, it will divide the line into two equal parts. Take a point P such that the distance PA=PB and then solve this equation using a line formula.
It is given that $A = \left( {3,6} \right)$ and$B = \left( { - 3,4} \right)$.
Now, we know that the equation of the perpendicular bisector of $AB$ will be the locus of the points which will be equidistant from $A$ and $B$.
Let us assume a point P(x,y) on the perpendicular bisector.
Therefore,
$PA = PB$
To simplify the formula of line formula we are going to square both sides,
${\left( {PA} \right)^2} = {\left( {PB} \right)^2}$
Note: Make sure you take square of the terms only once.
On applying the simplified line formula, we get,
${\left( {x - 3} \right)^2} + {\left( {y - 6} \right)^2} = {\left( {x + 3} \right)^2} + {\left( {y - 4} \right)^2}$
On opening the brackets, we get,
${x^2} + {3^2} - 2\left( 3 \right)\left( x \right) + {y^2} + {6^2} - 2\left( 6 \right)\left( y \right) = {x^2} + {3^2} + 2\left( 3 \right)\left( x \right) + {y^2} + {4^2} - 2\left( y \right)\left( 4 \right)$
On cancelling the common terms,
$9 - 6x + 36 - 12y = 9 + 6x + 16 - 8y$
Again cancelling the common terms on the above steps, we get,
$12x + 4y - 20 = 0$
Taking 4 out common from all the terms,
$4\left( {3x + y - 5} \right) = 0$
Therefore, we can say that,
$3x + y - 5 = 0$ is the required equation.
Therefore, the equation of the perpendicular bisector of $AB$ where $A$ and $B$ are the points $\left( {3,6} \right)$and $\left( { - 3,4} \right)$ respectively.
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Question
# A 20-year old hopes to retire by age 65. To help with future expenses, they invest $6 500 today at an interest rate of 6.4% compounded annually. At age 65, what is the difference between the exact accumulated value and the approximate accumulated value $using the Rule of 72$? 232 likes 1159 views ## Answer to a math question A 20-year old hopes to retire by age 65. To help with future expenses, they invest$6 500 today at an interest rate of 6.4% compounded annually. At age 65, what is the difference between the exact accumulated value and the approximate accumulated value $using the Rule of 72$?
Jon
4.6
To find the exact accumulated value, we can use the formula for compound interest:
A = P$1 + r$^n
Where:
- A is the accumulated value
- P is the principal amount $initial investment$
- r is the interest rate per compounding period
- n is the number of compounding periods
In this case, the principal amount is $6,500, the interest rate is 6.4% or 0.064 as a decimal, and the number of compounding periods is 65 - 20 = 45 $since the investment is made for 45 years$. Let's calculate the exact accumulated value: A = 6500$1 + 0.064$^{45} Now, to find the approximate accumulated value using the Rule of 72, we can use the following formula: A \approx P \times \frac{72}{r} Where: - A is the approximate accumulated value - P is the principal amount - r is the interest rate per compounding period In this case, the principal amount is still$6,500 and the interest rate is 6.4%.
Let's calculate the approximate accumulated value:
A \approx 6500 \times \frac{72}{6.4}
Now, let's calculate both the exact accumulated value and the approximate accumulated value.
Exact accumulated value: \ 105995.3
Approximate accumulated value: \ 73125
The difference between the exact accumulated value and the approximate accumulated value is:
\text{Difference} = \text{Exact accumulated value} - \text{Approximate accumulated value}
\text{Difference}=105995.3-73125
\text{Difference}=32870.3
Therefore, the difference between the exact accumulated value and the approximate accumulated value is approximately \$32870.3.
Frequently asked questions $FAQs$
How can we find the scalar multiplication of vectors in 3D?
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Math question: In triangle ABC, angle CAB measures 35° and angle ABC measures 70°. Find the measure of angle BCA using circle theorems.
+
What is the value of f$x$ when x = 10 for the logarithmic functions f$x$ = log x and f$x$ = ln x?
+
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# What to look for in 72 divided by 9
Welcome 72 divided by 9 to our blog post on the intriguing world of division! Today, we’ll be exploring the fascinating concept of dividing numbers and unlocking the secrets hidden within. In particular, we’ll delve into the intricacies of 72 divided by 9. So whether you’re a math enthusiast looking for a deeper understanding or simply curious about how division works in everyday life, this article is for you! Get ready to embark on a journey where numbers come alive and new insights await around every corner. Let’s dive right in!
## Understanding Division
Understanding Division
Division is a fundamental mathematical operation that involves splitting a quantity or number into equal parts. It is essentially the inverse of multiplication and allows us to distribute resources, solve problems, and make sense of numerical relationships.
At its core, division involves two key components: the dividend and the divisor. The dividend represents the total amount or value being divided, while the divisor signifies how many equal parts we want to divide it into. When we perform division, we aim to determine how many times the divisor can be subtracted from the dividend without going below zero.
For example, let’s consider 72 divided by 9. In this case, 72 acts as our dividend and 9 serves as our divisor. By dividing 72 by 9, we are essentially asking ourselves how many groups of size 9 can be formed from a total of 72.
By understanding division, we gain valuable insights into fractions and ratios. It helps us comprehend proportions in real-life scenarios such as sharing equally among friends or allocating resources based on specific criteria.
So next time you encounter a division problem like 72 divided by 9, remember that it’s more than just numbers on paper – it’s a gateway to unraveling mathematical mysteries!
## The Concept of
The concept of division is a fundamental mathematical operation that allows us to split a quantity into equal parts. When we divide one number by another, we are essentially finding out how many times the second number can be subtracted from the first without resulting in a negative value.
This concept becomes particularly interesting when we look at 72 divided by 9. In this case, we want to determine how many groups of 9 can be formed from the original quantity of 72. By dividing these numbers, we find that there are indeed 8 groups of 9 within 72.
Understanding division also involves exploring factors and multiples. Factors are numbers that divide evenly into another number without leaving a remainder. For example, in the case of 72 divided by 9, both numbers share common factors such as 1, 2, and even their own values (9 and itself).
Multiples on the other hand are obtained by multiplying a given number with any whole number greater than zero. In our example above, if we multiply each factor found for both numbers individually with any whole number greater than zero they will give us multiple pairs like (18,36), (27,54) or even(63).
By grasping divisibility rules such as those for dividing by nine which states that if all digits in a given number add up to a multiple of nine then it is divisible by nine; you would understand why every time you try to divide them you get no remainders except zeros after your decimal point.
Division plays an important role not only in mathematics but also in everyday life situations. Whether it’s splitting equally among friends or family members or sharing resources among different groups or calculating average scores for grading purposes – division helps us distribute and allocate things fairly.
To sum up
Understanding the concept of division is crucial for solving various problems involving quantities and distribution. It involves knowing about factors and multiples as well as applying divisibility rules. By exploring real-life applications of division, we can see how this mathematical operation
## Factors and Multiples
Factors and multiples are fundamental concepts in mathematics that play a crucial role in division. Understanding these concepts can help us better comprehend the process of dividing numbers.
A factor is a number that divides evenly into another number without leaving a remainder. For example, 3 is a factor of 9 because when we divide 9 by 3, there is no remainder. On the other hand, multiples are numbers that result from multiplying a given number by any whole number. For instance, the multiples of 4 include 8, 12, and so on.
Knowing the factors and multiples of a number can assist us in finding common divisors between two or more numbers. This knowledge comes in handy when solving problems involving fractions or simplifying expressions.
Divisibility rules also relate to factors and multiples as they provide shortcuts for determining if one number is divisible by another without actually performing long division calculations. These rules make it easier to identify whether certain digits or patterns appear within a given number.
By understanding factors and multiples, we gain insight into how numbers interact with each other through division operations. This knowledge has practical applications beyond mathematics – such as calculating proportions for cooking recipes or planning schedules based on time intervals.
In conclusion (even though I’m not supposed to say this), having an understanding of factors and multiples enhances our mathematical abilities while also enabling us to solve real-life problems more efficiently
## Divisibility Rules
Divisibility rules are a handy tool that can help us determine if one number is evenly divisible by another. These rules provide a quick and efficient way to check divisibility without having to perform the actual division calculation.
One commonly known divisibility rule is for the number 9. If the sum of all the digits in a number is divisible by 9, then the original number itself is also divisible by 9. For example, let’s look at 72 divided by 9.
To apply this rule, we add up the digits of 72: 7 +2 = 9. Since this sum (which happens to be exactly equal to our divisor) is divisible by 9, we can conclude that 72 is indeed evenly divisible by 9.
Divisibility rules like these can save us time and effort when performing long divisions or trying to find factors of numbers. By quickly checking if certain conditions are met, we can determine divisibility without going through multiple steps.
Learning and applying these rules not only helps with mathematical calculations but also builds mental math skills. It allows for more efficient problem-solving and helps develop critical thinking abilities.
So next time you come across a division problem involving numbers like 72 divided by 9, remember to consider using divisibility rules as a helpful tool in your mathematical toolbox!
## Common Divisors in 72 divided by 9
Common Divisors in 72 divided by 9
When it comes to division, finding common divisors is a crucial step in solving the problem. In the case of 72 divided by 9, we need to identify the numbers that can evenly divide both 72 and 9. By doing so, we can determine if there are any shared factors between these two numbers.
Let’s start by listing out the factors of both 72 and 9. For 72, some of its factors include:
-1
-2
-3
-4
-6
-8
-12
-18
-and of course, itself – which is 72
Now let’s look at the factors of number nine:
1
3
-and finally itself – which is nine
By comparing these two sets of factors, we can see that three is a common divisor for both numbers. This means that three evenly divides into both seventy-two and nine without leaving any remainder.
However, it’s important to note that this doesn’t mean three is the only common divisor between these two numbers. There may be other divisors as well.
Understanding common divisors helps us grasp how different numbers interact with each other mathematically. It allows us to simplify fractions or solve complex problems efficiently.
As you continue your mathematical journey, remember that exploring concepts like common divisors will enhance your problem-solving skills and deepen your understanding of numbers’ relationships.
## Real-Life Applications of Dividing by 9
Real-Life Applications of Dividing by 9
Dividing numbers by 9 may seem like a simple mathematical concept, but its applications can extend beyond the classroom. Let’s explore some real-life situations where dividing by 9 can be useful.
1. Cooking and Baking: Have you ever needed to adjust a recipe? Dividing ingredients by 9 allows you to scale down or up your measurements accurately. This is especially handy when baking delicate desserts that require precise ratios.
2. Time Management: Imagine you have a long list of tasks to complete in one day. By dividing your available hours by 9, you can allocate equal time slots for each task. This strategy helps ensure productivity while maintaining balance throughout the day.
3. Budgeting and Finance: When managing finances, dividing expenses into equal parts over several months becomes easier using division principles such as dividing by 9. It helps distribute costs evenly and plan for future expenses effectively.
4. Sports Strategy: In team sports like soccer or basketball, strategists often analyze player positions on the field and divide them into zones based on certain factors like distance covered or defensive coverage area.
5. Construction and Design: Architects and engineers frequently use proportional scaling techniques that involve division to create accurate blueprints, models, or scaled representations of structures before construction begins.
The applications mentioned here are just a glimpse into how division by 9 can be applied in everyday life scenarios outside of mathematics classrooms! Understanding this basic math principle empowers us with practical skills applicable across various domains – from cooking to finance management! So next time you’re faced with a situation that requires proportionate allocation or adjustment, remember the power of dividing by 9!
## Conclusion
Conclusion
In this article, we have explored the concept of division and specifically focused on dividing 72 by 9. We started by understanding the basics of division and how it relates to factors and multiples. Then, we delved into divisibility rules that can help us determine if a number is divisible by another.
Applying these concepts to our specific example of 72 divided by 9, we discovered that both numbers share common divisors such as 1, 2, 3, and 6. By examining these divisors, we gained a better understanding of how the two numbers are related.
Furthermore, we also discussed real-life applications where dividing by 9 can be useful. From splitting items evenly among a group to finding ratios or proportions in everyday situations, division plays an essential role in solving various problems.
To summarize, exploring what to look for in dividing 72 by 9 has given us insights into important mathematical concepts like factors and multiples. It has also demonstrated the practicality of division in real-world scenarios.
So next time you encounter a problem involving division or need to divide something fairly among friends or colleagues, remember what you’ve learned here about dividing by nine!
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# How do you evaluate (2y ^ { 2} \cdot y ^ { 3} \cdot 3y ) ^ { 3}?
Nov 14, 2017
See a solution process below:
#### Explanation:
First, rewrite the expression as:
${\left(\left(2 \cdot 3\right) \left({y}^{2} \cdot {y}^{3} \cdot y\right)\right)}^{3} \implies$
${\left(6 \left({y}^{2} \cdot {y}^{3} \cdot y\right)\right)}^{3}$
Next, use these rules of exponents to evaluate the $y$ terms within the parenthesis:
$a = {a}^{\textcolor{red}{1}}$ and ${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$
${\left(6 \left({y}^{2} \cdot {y}^{3} \cdot y\right)\right)}^{3} \implies$
${\left(6 \left({y}^{\textcolor{red}{2}} \cdot {y}^{\textcolor{b l u e}{3}} \cdot {y}^{\textcolor{p u r p \le}{1}}\right)\right)}^{3} \implies$
${\left(6 {y}^{\textcolor{red}{2} + \textcolor{b l u e}{3} + \textcolor{p u r p \le}{1}}\right)}^{3} \implies$
${\left(6 {y}^{6}\right)}^{3}$
Now, use these rules of exponents to complete the evaluation:
$a = {a}^{\textcolor{red}{1}}$ and ${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$
${\left(6 {y}^{6}\right)}^{3} \implies$
${\left({6}^{\textcolor{red}{1}} {y}^{\textcolor{red}{6}}\right)}^{\textcolor{b l u e}{3}} \implies$
${6}^{\textcolor{red}{1} \times \textcolor{b l u e}{3}} {y}^{\textcolor{red}{6} \times \textcolor{b l u e}{3}} \implies$
${6}^{3} {y}^{18} \implies$
$216 {y}^{18}$
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# How do you write 75 as a fraction in simplest form?
## How do you write 75 as a fraction in simplest form?
Answer: 75% is written as 3/4 as a fraction in its simplest form.
### What is the equivalent of 30 to 75?
Since the simplest form of the fraction 30/75 is 2/5, the simplest form of the ratio 30:75 is also 2:5.
What is 30% expressed as a fraction in simplest form?
3/ 10
Answer: The value of 30% as a fraction in simplest form is 3/ 10.
What fraction is closest to representing 75?
Percent to Fraction Table
Percent Fraction
60 % 3 5
70 % 7 10
71.43 % 5 7
75 % 3 4
## What is a 40 out of 75?
Step 4: In the Same way, x = 40%. 75 = 100% (1). x = 40% (2)….Nearby Results.
40% of Result
75 30
75.01 30.004
75.02 30.008
75.03 30.012
### How do you simplify 45 75?
Explanation:
1. 45÷1575÷15=35.
2. 45÷575÷5=915.
3. 9÷315÷3=35.
What is fraction simplest form?
A fraction is said to be in its simplest form if 1 is the only common factor of its numerator and denominator. For example,89,because 1 is the only common factor of 8 and 9 in this fraction. We simplify fractions because it is always to work or calculate when the fractions are in the simplest form.
What is a 53 out of 75?
70.67%
The percentage score for 53 out of 75 is 70.67%. This is an C- grade.
## What is the 33% of 75 marks?
33 percent of 75 is 24.75.
### What is 45 as a fraction of 75?
Answer. Therefore, 45/75 simplified to lowest terms is 3/5.
What is the simplest form of 15 by 75?
so, the simplest form is 1:5.
What is fractional form?
Fractional notation simply means that a number is written in fraction form. It is written as a/b where neither a nor b is equal to 0. A fraction has two parts, the numerator and denominator. The numerator is the number on top of the line and it represents how many pieces of the whole we are referring to.
## What is simplified form?
1. The simplest form is the smallest possible equivalent fraction of the number.
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UPSKILL MATH PLUS
Learn Mathematics through our AI based learning portal with the support of our Academic Experts!
Working rule to construct a parallelogram:
Let us discuss the working rule to construct a parallelogram when the measure of two diagonals and one included angle of a parallelogram are given.
Example:
Construct a parallelogram $$PQRS$$ with $$PR$$ $$=$$ $$10$$ $$cm$$, $$QS$$ $$=$$ $$8$$ $$cm$$ and $$\angle POQ$$ $$=$$ $$120^{\circ}$$. Also, find its area.
Construction:
Step 1: Draw a line segment $$PR$$ $$=$$ $$10$$ $$cm$$.
Step 2: Mark the midpoint of the line segment $$PR$$ as $$O$$.
Step 3: Draw a line $$XY$$ through $$O$$ such that $$\angle POQ$$ $$=$$ $$120^{\circ}$$.
Step 4: With $$O$$ as centre, draw two arcs each of radii $$4$$ $$cm$$ on either side of $$PR$$ intersecting $$XY$$ at $$Q$$ and $$S$$.
Step 5: Join $$PQ$$, $$QR$$, $$RS$$ and $$SP$$.
Step 6: $$PQRS$$ is the required parallelogram. The measure of $$SA$$ gives the height of the parallelogram $$PQRS$$.
Area calculation:
Area of the parallelogram $$=$$ $$base \times height$$ square units
$$=$$ $$8 \times 4.3$$
$$=$$ $$34.40$$ $$cm^2$$
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# Matrix Multiplication in Excel — Markov Chains
This post is part of the “Matrix Multiplication in Excel” series. It’s composed of a math introduction, a silly interlude and an interactive tutorial (you are here). By the end of the series, you’ll learn how to perform Markov Chain calculations, which are used in some damage calculations.
Now that we know some of the basics, I’m going to introduce a toy problem where one would use matrix multiplication, step through how to calculate it in Excel and then give an example used forensic economics.
## Scenario 1: Ping Pong Problem
Suppose two ping pong players (labelled left and right) are tied near the end of a game. In order to win, one of them has to have a two point lead. Suppose that, on any given volley, the left player has a 55% chance of winning.
The game can be thought of as having five states:
1. (2,0) – The left player wins.
2. (1,0) – The left player is winning by one.
3. (0,0) – The game is tied.
4. (0,1) – The right player is winning by one.
5. (0,2) – The right player wins.
Graphically, the transitions between these states can be seen as follows.
The left player increases his score with the probability $p(LW)$, in this case 55%. How do we calculate the probability that the left player wins?
This problem can be formalized as a Markov Chain, a mathematical system in discrete-time which transitions through various states.
To simulate a Markov Chain, start by creating a matrix $P1$ of size $1 x s$ where $s$ is equal to the number of states. Element $p1_{1,j}$ gives the probability the system starts in state $j$. Next create a transition matrix $T$ of size $s x s$. Element $t_{i,j}$ gives the probability the system moves from state $i$ to state $j$.
In this example, we’ve stated the game starts at tied in state (0,0). The starting probability vector and transition matrix are displayed below.
Now we get to the value of matrix multiplication. As discussed previously, multiplying a $1 x s$ matrix by an $s x s$ matrix produces a $1 x s$ matrix. If you step through the math of matrix multiplication, you’ll see that multiplying $P1$ by matrix $T$ produces the probability the system will be in a given state in at the start of the next serve. This process can be iterated indefinitely to generate probability distributions at any given interval of time. The first ten steps are shown below.
Starting with a tie game, after two serves there’s a $30.25\%$ chance the left player will win, a $20.25\%$ chance the right player will win and a $49.50\%$ chance the game will still be tied. With enough iterations, we can see that there’s approximately a $59.90\%$ chance the left player will win.
## Scenario 2: Racquetball Problem
In the Excel book, I also included a second scenario for a racquetball-like scoring system where you only score points if you’ve served. The slightly more complex state diagram and iterations are displayed below.
By changing the inputs, one could use this model to answer questions like “what is the value of serving first?” or “how often does the ‘better’ player win?”
## Creating Matrix Multiplication Problems in Excel
Matrix multiplication in Excel is slightly different from standard formulas because the returned result is an array of numbers. Usually, excel formulas return a single result.
Instead of using a standard formula, you will have to use an array formula. To create a normal formula, simply type it into the cell and type “enter.” To create an array formula, instead press “control-shift-enter.” This will add curly braces around the formula to let you know it’s an array formula, as seen in the following screenshot. In some circles, these are called CSE formulas because of the key stroke which creates them.
To see how to create matrix multiplication CSE formula, follow these steps:
1. Clear the iterated matrix multiplication in scenario 1.
• Use Ctrl-G to select “‘S1. Matrices’!B14:F62.” Press delete to clear the contents.
2. Select range B14:F14. Leave B14 as the “active cell.”
3. Enter the following formula into B14:
• =MMULT(B4:F4,B7:F11)
• Press “Control-Shift-Enter” to have this formula applied across the entire selected range.
• I’ve named the ranges “S1.MA” and “S1.MB”, respectively. Feel free to use these “dynamic range names” in your formula instead.
4. Repeat the process for the remaining iterations, except replace “S1.MA” with the previous iteration.
• Make sure you use proper absolute/relative references so the process continues downwards.
• For row 15: =MMULT(\$B14:\$F14,\$B\$7:\$F\$11)
• For rows 16 and on, you can Copy and Paste-Special Formulas.
If you receive an error while trying to implement matrix multiplication in Excel, you’ve likely specified an incorrect number of dimensions. As we discussed in a previous post: When you multiply an $m x p$ matrix by a $p x n$ matrix, you produce an $m x n$ matrix. The number of columns in the first matrix has to match the number of rows in the second matrix. The resulting matrix will have the number of rows from the first matrix and the number of columns from the second matrix.
## Markov Chains in Economic Damage Cases
Markov Chains are commonly used to measure lost earning capacity. In any given year, a subject can be in one of several states (i.e. employed, unemployed, dead). Hopefully the analog to the ping pong problem is clear. Statistical information can give us “transition probabilities” which state the probability of a worker moving from one state to another based on the the current state. A currently employed worker is more likely to be employed the next year than a currently unemployed worker. A dead worker will stay dead in the same way a finished game of ping pong will stay finished.
In every year, we are left with a probability of the subject being in the “employed” state, necessary information for valuing his or her potential income stream.
The ping pong example is memoryless. The game of ping pong doesn’t care if this is the first or thousandth time the right player has been one away from winning. In some more sophisticated Markov Chain calculations, the transition probabilities depend on not only the current state but the state of prior years. For instance, a subject who has been employed for the past two years is more likely to be employed next year than a subject who was gainfully employed for only one year.
## Conclusion
We’ve covered a lot of ground in this series. We learned the basics of linear algebra and Markov Chains. We learned how to implement these concepts in Excel and even considered the thermodynamic consequences of living in the Matrix.
If you were intimidated by any of these concepts before, hopefully I’ve unpacked them enough to take the magic out. Like most apparently complex things, matrix multiplication is merely a few simple things done many times over.
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# Math Cracks – What is a Derivative, Really?
It seemed important to me to go over the concept of derivative of a function. The process of differentiation (this is, calculating derivatives) is one of the most fundamental operations in Calculus and even in math. In this Math Crack tutorial I will try to shed some light into the meaning and interpretation of what a derivative is and does.
First of all, for the purpose of clarifying what is the scope of this tutorial, I would like to say that we won't be practicing with solving specific practice problems involving derivatives but we will rather make an attempt to understand what we are doing when operating with derivatives. Once we understand what we are doing, we have a WAYYY better chance to solve problems.
#### DEFINITION OF A DERIVATIVE (NOT THE BORING ONE)
To get started, it is mandatory to write at least the definition of a derivative. Assume that $$f$$ is a function and $${{x}_{0}}\in dom\left( f \right)$$. Ok, we started with technicalities already? All we are saying is that $$f$$ is a function. Think of a function $$f$$ by its graphical representation shown below:
Also, when we say that "$${{x}_{0}}\in dom\left( f \right)$$", all we are saying is that $${{x}_{0}}$$ is a point where the function is well defined (so it belongs to its domain). But hold it, is it possible for a point $${{x}_{0}}$$ to make a function NOT well defined….? Certainly! Consider the following function:
$f\left( x \right)=\frac{1}{x-1}$
Such function is NOT well defined at $${{x}_{0}}=1$$. What is it not well defined at $${{x}_{0}}=1$$? Because if we plug the value of $${{x}_{0}}=1$$ in the function we get
$f\left( 1 \right)=\frac{1}{1-1}=\frac{1}{0}$
which is an INVALID operation (as you know from primary school, you cannot divide by zero, at least with the traditional arithmetic rules), so then the function is not well defined at $${{x}_{0}}=1$$. For a function to be well defined at a point means simply that the function can be evaluated at that point, without the existence of any invalid operations.
So now we can say it again, because now you know what we mean: Assume $$f$$ is a function and $${{x}_{0}}\in dom\left( f \right)$$. The derivative at the point $${{x}_{0}}$$ is defined as
$f'\left( {{x}_{0}} \right)=\lim_{x \to {x_0}}\,\frac{f\left( x \right)-f\left( {{x}_{0}} \right)}{x-{{x}_{0}}}$
when such limit exists.
Ok, that is the meat of the problem, and we will discuss it in a second. I would like to you to have some things EXTREMELY clear here:
• When the above limit exists, we call if $$f'\left( {{x}_{0}} \right)$$, and it is referred as the “derivative of the function $$f\left( x \right)$$ at the point $${{x}_{0}}$$”. So then, $$f'\left( {{x}_{0}} \right)$$ is simply a symbol that we use to refer to the derivative of the function $$f\left( x \right)$$ at the point $${{x}_{0}}$$ (when it exists). We could have used any other symbol, such as “$$deriv{{\left( f \right)}_{{{x}_{0}}}}$$” or “$$derivative\_f\_{{x}_{0}}$$”. But some aesthetic sense makes us prefer “$$f'\left( {{x}_{0}} \right)$$”.
The point is that is a MADE UP symbol to REFER to the derivative of the function $$f\left( x \right)$$ at the point $${{x}_{0}}$$. The funny thing in Math is that notation matters. Even though a concept exists regardless of the notation used to express it, a logical, flexible, compact notation can make things catch on fire as opposed to what can happen with a cumbersome, uninspired notation
#### The Role Played by Notation
(Historically, the two simultaneous developers of a usable version of the concept of derivative, Leibniz and Newton used radically different notations. Newton used $$\dot{y}$$, whereas Leibniz used $$\frac{dy}{dx}$$. Leibniz notation caught on fire and facilitated the full development of Calculus, whereas Newton’s notation caused more than one headache. Really, it was that important).
• The derivative is a POINTWISE operation. This means that it is an operation done to a function at a given point, and it needs to be verified point by point. Of course a typical domain like the real line $$\mathbb{R}$$ there is an infinite number of points, so it may take a while to check by hand if a derivative is defined at each point. BUT, there are some rules that allow to greatly simplify the work by computing the derivative at one generic point $${{x}_{0}}$$ and then analyzing for which values of $${{x}_{0}}$$ the limit the defines the derivative exists. So you can relax, because the gritty handwork won’t be to taxing, if you know what you’re doing of course.
• When the derivative of a function $$f$$ exists at a point $${{x}_{0}}$$, we say that the function is differentiable at $${{x}_{0}}$$. Also, we can say that a function is differentiable at a REGION (a region is a set of points) if the function is differentiable at EACH point of that region. So then, even though the concept of derivative is a pointwise concept (defined at a specific point), it can be understood as a global concept when it is defined for each point in a region.
• If we define $$D$$ the set of all points in the real line where the derivative of a function is defined, we can define the derivative function $$f'$$ as follows:
\begin{aligned} & f':D\subseteq \mathbb{R}\to \mathbb{R} \\ & x\mapsto f'\left( x \right) \\ \end{aligned}
This is a function because we uniquely associate each $$x$$ on $$D$$ with the value $$f'\left( x \right)$$. This means that each value of $$x$$ on $$D$$ is associated with the value $$f'\left( x \right)$$. The set of all pairs $$\left( x,f'\left( x \right) \right)$$, for $$x\in D$$ form a function, and you can do all the things you can do with functions, such as graphing them.
That should settle the question that many students have about derivatives, as they wonder how we have a derivative “function”, when the derivative is something that is computed at a certain specific point. Well, the answer is that we compute the derivative at many points, which provides the foundation to define the derivative as a function.
#### Final Words: Notation Hell
When the concept of derivative was put into the modern form we know by Newton and Leibniz (I make the emphasis on the term “modern form”, since Calculus was almost fully developed by the Greeks and others in a more intuitive and less formal way a LONG time ago), they chose radically different notations. Newton chose $$\overset{\bullet }{\mathop{y}}\,$$, whereas Leibniz chose $$\frac{dy}{dx}$$. So far so good. But the concept of derivative means much less if we don’t have powerful derivative theorems.
Using their respective notations, they both had little trouble to prove basic differentiation theorems, such as linearity and the product rule, but Newton didn’t see the need for formally stating the Chain Rule, possibly because his notation didn’t lend itself for that, whereas for Leibniz notation, the Chain Rule shows itself almost like a “Duh” rule. To be more precise, assume that $$y=y\left( x \right)$$ is a function and $$u=u\left( x \right)$$ is another function.
It is a natural question to ask if I can compute the derivative of the composition $$y\left( u\left( x \right) \right)$$ in an easy way, based on the derivatives of $$y$$ and $$u$$. The answer to this question is the Chain Rule. Using Leibniz notation the rule is
$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$
It is almost as if you could cancel the $$du$$’s like:
$\require{cancel}\frac{dy}{dx}=\frac{dy}{\cancel{du}}\frac{\cancel{du}}{dx}$
but it is not exactly like that. But that is the beauty of Leibniz notation. It has a strongly intuitive appeal (and the “canceling” $$du$$’s are almost at a reality, it is only that that it is done at the $$\Delta u$$ level and there are limits involved), but yet you need to understand what Leibniz was saying with the rule. He says:
“The derivative of the compound function $$y\left( u\left( x \right) \right)$$ is the same as the derivative of $$y$$ at the point $$u\left( x \right)$$ multiplied by the derivative of $$u$$ at the point $$x$$”
The Chain Rule using Newton’s notation gets the following form:
$\overset{\bullet }{\mathop{\left( f\circ u \right)}}\,=\overset{\bullet }{\mathop{f}}\,\left( u\left( x \right) \right)\overset{\bullet }{\mathop{u}}\,\left( x \right)$
Quite a bit less pretty, isn’t it? But guess what, Newton’s Chain Rule says EXACTLY THE SAME as
$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$
Though, this latter notation caught on fire and helped enormously to the fast development of modern Calculus, whereas Newton’s form was way less beloved. Even though the theorems were saying exactly the same, one was golden and the other one not so much. Why? NOTATION my friend.
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# Integration by substitution: Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
# Encyclopedia
In calculus, integration by substitution is a method for finding antiderivatives and integrals. Using the fundamental theorem of calculus often requires finding an antiderivative. For this and other reasons, integration by substitution is an important tool for mathematicians. It is the counterpart to the chain rule of differentiation.
Let $I \subseteq {\mathbb{R}}$ be an interval and $g : [a,b] \to I$ a continuously differentiable function. Suppose that $f : I \to \mathbb{R}$ is a continuous function. Then
$\int_a^b f(g(t))g'(t)\, dt = \int_{g(a)}^{g(b)} f(x)\,dx.$
Using Leibniz notation: the substitution x = g(t) yields dx / dt = g'(t) and thus formally $dx = g'(t)\,dt$, which is the required substitution for dx. (One may view the method of integration by substitution as a major justification of Leibniz's notation for integrals and derivatives.)
The formula is used to transform one integral into another integral that is easier to compute. Thus, the formula can be used from left to right or from right to left in order to simplify a given integral. When used in the former manner, it is sometimes known as u-substitution.
## Relation to the fundamental theorem of calculus
Integration by substitution can be derived from the fundamental theorem of calculus as follows. Let ƒ and g be two functions satisfying the above hypothesis that ƒ is continuous on I and $g'\,$ is continuous on the closed interval [a,b]. Then the function f(g(t))g'(t) is also continuous on [a,b]. Hence the integrals
$\int_{g(a)}^{g(b)} f(x)\,dx$
and
$\int_a^b f(g(t))g'(t)\,dt$
in fact exist, and it remains to show that they are equal.
Since ƒ is continuous, it possesses an antiderivative F. The composite function $F\circ g$ is then defined. Since F and g are differentiable, the chain rule gives
$(F \circ g)'(t) = F'(g(t))g'(t) = f(g(t))g'(t).$
Applying the fundamental theorem of calculus twice gives
\begin{align} \int_a^b f(g(t))g'(t)\,dt & {} = (F \circ g)(b) - (F \circ g)(a) \ & {} = F(g(b)) - F(g(a)) \ & {} = \int_{g(a)}^{g(b)} f(x)\,dx, \end{align}
which is the substitution rule.
## Examples
Consider the integral
$\int_{0}^2 x \cos(x^2+1) \,dx$
By using the substitution u = x2 + 1, we obtain du = 2x dx and
\begin{align} \int_{x=0}^{x=2} x \cos(x^2+1) \,dx & {} = \frac{1}{2} \int_{u=1}^{u=5}\cos(u)\,du \ & {} = \frac{1}{2}(\sin(5)-\sin(1)). \end{align}
Here we substituted from right to left. It is important to note that since the lower limit x = 0 was replaced with u = 02 + 1 = 1, and the upper limit x = 2 replaced with u = 22 + 1 = 5, a transformation back into terms of x was unnecessary.
For the integral
$\int_0^1 \sqrt{1-x^2}\; dx$
the formula needs to be used from left to right: the substitution x = sin(u), dx = cos(udu is useful, because √(1-sin2(u)) = cos(u):
$\int_0^1 \sqrt{1-x^2}\; dx = \int_0^\frac{\pi}{2} \sqrt{1-\sin^2(u)} \cos(u)\;du = \int_0^\frac{\pi}{2} \cos^2(u)\;du=\frac{\pi}{4}$
The resulting integral can be computed using integration by parts or a double angle formula followed by one more substitution. One can also note that the function being integrated is the upper right quarter of a circle with a radius of one, and hence integrating the upper right quarter from zero to one is the geometric equivalent to the area of one quarter of the unit circle, or pi over 4.
## Antiderivatives
Substitution can be used to determine antiderivatives. One chooses a relation between x and u, determines the corresponding relation between dx and du by differentiating, and performs the substitutions. An antiderivative for the substituted function can hopefully be determined; the original substitution between u and x is then undone.
Similar to our first example above, we can determine the following antiderivative with this method:
\begin{align} & {} \quad \int x \cos(x^2+1) \,dx = \frac{1}{2} \int 2x \cos(x^2+1) \,dx \ & {} = \frac{1}{2} \int\cos u\,du = \frac{1}{2}\sin u + C = \frac{1}{2}\sin(x^2+1) + C \end{align}
where C is an arbitrary constant of integration.
Note that there were no integral boundaries to transform, but in the last step we had to revert the original substitution u = x2 + 1.
## Substitution for multiple variables
One may also use substitution when integrating functions of several variables. Here the substitution function (v1,...,vn) = φ(u1, ..., u n ) needs to be one-to-one and continuously differentiable, and the differentials transform as
$dv_1\cdots dv_n = |\det(\operatorname{D}\varphi)(u_1, \ldots, u_n)| \, du_1\cdots du_n$
where det(Dφ)(u1, ..., u n ) denotes the determinant of the Jacobian matrix containing the partial derivatives of φ . This formula expresses the fact that the absolute value of the determinant of given vectors equals the volume of the spanned parallelepiped.
More precisely, the change of variables formula is stated in the following theorem:
Theorem. Let U, V be open sets in Rn and φ : UV an injective differentiable function with continuous partial derivatives, the Jacobian of which is nonzero for every x in U. Then for any real-valued, compactly supported, continuous function f, with support contained in φ(U),
$\int_{\varphi(U)} f(\mathbf{v})\, d \mathbf{v} = \int_U f(\varphi(\mathbf{u})) \left|\det(\operatorname{D}\varphi)(\mathbf{u})\right| \,d \mathbf{u}.$
The conditions on the theorem can be weakened in various ways. First, the requirement that φ be continuously differentiable can be replaced by the weaker assumption that φ be merely differentiable and have a continuous inverse (Rudin 1970, Theorem 7.26). This is guaranteed to hold if φ is continuously differentiable by the inverse function theorem. Alternatively, the requirement that Det(Dφ)≠0 can be eliminated by applying Sard's theorem (Spivak 1965). More general versions of this result hold. One very general version in measure theory is the following (Hewitt & Stromberg 1965, Theorem 20.3):
Theorem. Let X be a locally compact Hausdorff space equipped with a finite Radon measure μ, and let Y be a σ-compact Hausdorff space with a σ-finite Radon measure ρ. Let φ : X → Y be a continuous and absolutely continuous function (where the latter means that ρ(φ(E)) = 0 whenever μ(E) = 0). Then there exists a real-valued Borel measurable function w on X such that for every Lebesgue integrable function f : Y → R, the function (f °φ)w is Lebesgue integrable on X, and
$\int_Y f(y)\,d\rho(y) = \int_X f\circ\varphi(x)w(x)\,d\mu(x).$
Furthermore, it is possible to write
$w(x) = g\circ\varphi(x)$
for some Borel measurable function g on Y.
In geometric measure theory, integration by substitution is used with Lipschitz functions. A bi-Lipschitz function is a Lipschitz function T : URn which is one-to-one, and such that its inverse function T-1 T(U) → U is also Lipschitz. By Rademacher's theorem a bi-Lipschitz mapping is differentiable almost everywhere. In particular, the Jacobian determinant of a bi-Lipschitz mapping det DT is well-defined almost everywhere. The following result then holds:
Theorem. Let U be an open subset of Rn and T : URn be a bi-Lipschitz mapping. Let f : T(U) → R be measurable. Then
$\int_U (f\circ T)|\det DT| = \int_{T(U)}f$
in the sense that if either integral exists (or is properly infinite), then so does the other one, and they have the same value.
The above theorem was first proposed by Euler when he developed the notion of double integrals in 1769. Although generalized to triple integrals by Lagrange in 1773, and used by Legendre, Laplace, Gauss, and first generalized to n variables by Mikhail Ostrogradski in 1836, it resisted a fully rigorous formal proof for a surprisingly long time, and was first satisfactorily resolved 125 years later, by Elie Cartan in a series of papers beginning in the mid-1890s.[1][2]
## Application in probability
Substitution can be used to answer the following important question in probability: given a random variable X with probability density px and another random variable Y related to X by the equation y = Φ(x), what is the probability density for Y?
It is easiest to answer this question by first answering a slightly different question: what is the probability that Y takes a value in some particular subset S? Denote this probability $P(Y \in S)$. Of course, if Y has probability density py then the answer is
$P(Y \in S) = \int_S p_y(y)\,dy,$
but this isn't really useful because we don't know py; it's what we're trying to find in the first place. We can make progress by considering the problem in the variable X. Y takes a value in S whenever X takes a value in Φ − 1(S), so
$P(Y \in S) = \int_{\Phi^{-1}(S)} p_x(x)\,dx.$
Changing from variable x to y gives
$P(Y \in S) = \int_{\Phi^{-1}(S)} p_x(x)~dx = \int_S p_x(\Phi^{-1}(y)) ~ \left|\frac{d\Phi^{-1}}{dy}\right|~dy.$
Combining this with our first equation gives
$\int_S p_y(y)~dy = \int_S p_x(\Phi^{-1}(y)) ~ \left|\frac{d\Phi^{-1}}{dy}\right|~dy$
so
$p_y(y) = p_x(\Phi^{-1}(y)) ~ \left|\frac{d\Phi^{-1}}{dy}\right|.$
In the case where X and Y depend on several uncorrelated variables, ie. $p_x=p_x(x_1\ldots x_n)$, and y = Φ(x), py can be found by substitution in several variables discussed above. The result is
$p_y(y) = p_x(\Phi^{-1}(y)) ~ \left|\det \left[ D\Phi ^{-1}(y) \right] \right|.$
## References
1. ^ V. Katz, Change of variables in multiple integrals: Euler to Cartan, Mathematics Magazine 55 (1982) 3-11
2. ^ Anthony P. Ferzola, Euler and differentials, The College Mathematics Journal, Vol. 25, No. 2, (1994), pp. 102-111
• Hewitt, Edwin; Stromberg, Karl (1965), Real and abstract analysis, Springer-Verlag, ISBN 978-0387045597 .
• Rudin, Walter (1970), Real and complex analysis, McGraw-Hill, ISBN 978-0070542341 .
• Spivak, Michael (1965), Calculus on manifolds, Westview Press, ISBN 978-0805390216 .
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# How do you evaluate 15- 33+ 29?
Apr 8, 2017
$15 - 33 + 29 = 11$
#### Explanation:
Do the addition ($15 + 29$) first,
$15 + 29 = 44$
Now take the sum you just calculated $\left(44\right)$ and subtract $33$
$44 - 33 = 11$
Therefore,
$15 - 33 + 29 = 11$
Apr 8, 2017
11
#### Explanation:
You can perform the operations in any order, because they are all either addition or subtraction. Use of a number line may help for the 'negative' numbers or subtraction if you're now sure how they combine with multiple additions.
15 - 33 = -18 ; -18 + 29 = 11
Or,
15 + 29 = 44 ; 44 - 33 = 11
Or,
-33 + 29 = -4 ; 15 - 4 = 11
Apr 8, 2017
Just another way of thinking about it
11
#### Explanation:
We have negative 33
but positive 33 can be split into 15+18
so $- 33$ is the same as $- 15 - 18$
So write$\text{ } \textcolor{g r e e n}{15 \textcolor{red}{- 33} + 29}$ as $\textcolor{g r e e n}{15 \textcolor{red}{- 15 - 18} + 29}$
$0 - 18 + 29$
$29 - 18 = + 11$
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# What Is 2/63 as a Decimal + Solution With Free Steps
The fraction 2/63 as a decimal is equal to 0.031.
A mathematical expression consisting of numbers and basic operators is known as an algebraic expression. The basic operators are addition, subtraction, multiplication, and division. The division operator is also used to express fractions.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 2/63.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 2
Divisor = 63
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 2 $\div$ 63
This is when we go through the Long Division solution to our problem.
Figure 1
## 2/63 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 2 and 63, we can see how 2 is Smaller than 63, and to solve this division, we require that 2 be Bigger than 63.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 2, which after getting multiplied by 10 becomes 20.
Since if 2 is multiplied by 10 it becomes 20, which is still a smaller value than 63, we multiply 20 by 10 again to make it 200. For this, we add a zero in the quotient just after the decimal point. It makes 200 bigger than 63 and divisions are possible now.
Now we begin solving our dividend 200
We take this 200 and divide it by 63; this can be done as follows:
 200 $\div$ 63 $\approx$ 3
Where:
63 x 3 = 189
This will lead to the generation of a Remainder equal to 200 – 189 = 11. Now this means we have to repeat the process by Converting the 11 into 110 and solving for that:
110 $\div$ 63 $\approx$ 1Â
Where:
63 x 1 = 63
This, therefore, produces another Remainder which is equal to 110 – 63 = 47.
Finally, we have a Quotient generated after combining the three pieces of it as 0.031, with a Remainder equal to 47.
Images/mathematical drawings are created with GeoGebra.
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1 y = a x for a > 1 (exponential) y = a x for 0 < a < 1 . Graph of Quadratic Parent Function. We can do all transformation in one go using this: So it takes the square root function, and then. How to graph the reciprocal parent function and transformations of the reciprocal function. Nov 10, 2017 - This section covers: Basic Parent Functions Generic Transformations of Functions Vertical Transformations Horizontal Transformations Mixed Transformations Transformations in Function Notation Writing Transformed Equations from Graphs Rotational Transformations Transformations of Inverse Functions Applications of Parent Function Transformations More … Mathematics. You’ll probably study some “popular” parent functions and work with these to learn how to transform functions – how to move them around. ... Domain and Range of Square Root Parent Function. In the example, $$y=\sqrt{x}+3$$, the graph and key points will be moved up three units. Linear Parent Function. a For example, $$y=-\sqrt{x+2}-3$$, will move the parent down 3, left 2, and reflect it over the x-axis while $$y=\sqrt{-(x+2)}-3$$ will move the graph down 3, left 2, and reflect it over the y-axis. You could graph this by looking at how it transforms the parent function of y = sqrt (x). Next. In Figure 2(a), the parabola opens outward indefinitely, both left and right. Which transformation describes the equation from its parent function? 2 ( ) f x x B. Let’s begin by reviewing the rational and square root parent functions. Then, using the property that sqrt (ab) = sqrt (a)*sqrt (b), you can rewrite this again as: y = sqrt (2) * sqrt (x+3/2). Unit 8 Day 1 Notes Objective: Be able to identify and graph transformations of square root functions. The graph of any square root function is a transformation of the graph of the square root parent function, f (x) = 1x. slecky. STUDY. This is it. Square root functions can also be written in h,k form. Graph of Linear Parent Function. Sample Problem 1: Identify the parent function and describe the transformations. 9th - 12th grade. BUT we must add C wherever x appears in the function (we are substituting x+C for x). This transformation also may be appropriate for percentage data where the range is between 0 and 20% or between 80 and 100%.Each data point is replaced by its square root. Note: values added under the square root are subtracted from the x-values and numbers subtracted under the square root are added to the x-values. Describe the Transformation f (x) = square root of x. f (x) = √x f ( x) = x. a (x – h) + k and the square root function f (x) = a √ (x – h) + k can be transformed using methods similar to those used to transform other types of functions. A negative sign in front of x under the square root will reflect the graph over the y-axis. 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# PROMOÇÕES
## square root parent function transformations
Asymptotes for rational function. Describe the Transformations using the correct terminology. A square root function is a function with the variable under the square root. $$y=-\sqrt{x}$$ would reflect the parent graph over the x-axis and new key points would be: (0, 0), (1, -1), (4, -2), and (9, -3). Describe the Transformation y = square root of x. y = √x y = x. 9th - 12th grade . Students will be able to use algebraic parent graph function families, linear, quadratic, absolute value, square root, exponential and rational that will provide the basis for a graphic design for fabrics. These transformations can be combined. Sample Problem 3: Use the graph of parent function to graph each function. Let us start with a function, in this case it is f(x) = x2, but it could be anything: Here are some simple things we can do to move or scale it on the graph: Note: to move the line down, we use a negative value for C. Adding C moves the function to the left (the negative direction). $$y=\sqrt{x+3}$$ would move the parent function left by three units. Functions transformations-square root, quadratic, abs value. Adding or subtracting a constant on the “outside” of the square root moves the graph of the parent function up or down respectively. But you can't see it, because x2 is symmetrical about the y-axis. Similarly, a cube root function is a function with the variable under the cube root. You Show Video Lesson A negative sign in front of the square root will reflect the graph over the x-axis. Mathematics. PLAY. Students match each function card to its graph card and transformation(s) card. Subtracting 3 on the “outside” would have had the opposite effect and moved the graph down 3 units. AII.6 For absolute value, square root, cube root, rational, polynomial, exponential, and logarithmic functions, the student will a) recognize the general shape of function families; and b) use knowledge of transformations to convert between equations and the corresponding graphs of functions. To move C spaces to the left, add C to x wherever x appears: An easy way to remember what happens to the graph when we add a constant: add to y to go high Be careful when adjusting the key points for left and right translations. The sequence of graphs in Figure 2 also help us identify the domain and range of the square root function. 3 years ago. The graph of the square root parent function begins at point (0, 0) and is drawn only in quadrant I since the domain and range of the square root parent function are both greater than or equal to zero. Some of the key points on the graph of the parent function that are good to know as the graph is moved around are: (0, 0), (1, 1), (4, 2), and (9, 3). We can combine a negative value with a scaling: Example: multiplying by −2 will flip it upside down AND stretch it in the y-direction. m04 One over x y=1/x. This depends on the direction you want to transoform. Putting the negative “inside” the square root would cause the graph to reflect over the y-axis and make the x-value of the coordinates negative. The transformation from the first equation to the second one can be found by finding a a, h h, and k … Why? The graph of the square root parent function begins at point (0, 0) and is drawn only in quadrant I since the domain and range of the square root parent function are both greater than or equal to zero. Name the parent function. ACTIVITY to solidify the learning of transformations of parent functions. This video explains how to do translations and reflections to the square root parent function. How to move a function in y-direction? Go to resource library. Well imagine you will inherit a fortune when your age=25. 0. Asymptotes of Rational Functions. 0. DRAFT. Subtracting 3 would have moved the graph right by a corresponding number of units. Absolute Value Transformations. Constants added or subtracted “inside” or under the square root sign move the graph left or right respectively and subtract or add the constant from the x-value of the coordinates of the key points. y = sqrt (2 (x+3/2)). Adding 4 made it happen earlier. Cubic Parent Function. 5. It really does flip it left and right! Introduce the graph of a logarithmic function, and, if … Tags: Question 13 . Algebra. Lead students to surmise the same for other polynomial functions. This is also called reflection about the x-axis (the axis where y=0). Students create a picture on the provided Cartesian plane using transformations of parent functions with … When a number is subtracted under the square root sign, it must be ADDED to the x-values and when a number is added under the square root, it must be SUBTRACTED from the x-values to move the graph in the correct direction. $$y=-\sqrt{x+7}+3$$ will move the parent graph left 7, up 3, and reflect it over the x-axis. Choose from 500 different sets of parent functions transformations flashcards on Quizlet. In the same way that we share similar characteristics, genes, and behaviors with our own family, families of functions share similar algebraic properties, … Transformations of Square Root Functions Matching is an interactive and hands on way for students to practice matching square root functions to their graphs and transformation(s). Values would be subtracted or added to the x-values of the coordinates of the key points. Numbers added or subtracted under the square root or “inside” would have the effect of moving the graph right (-) or left (+). y = √x y = x. Please Note: It will be beneficial to watch the video to see the actual movements on the graph as a result of changes to the parent function. ESSENTIAL UNDERSTANDING Parent Function f(x) = 1x, x Ú 0 Vertical Translation Horizontal Translation y = 1x + d y = 1x-c d 7 0: shifts up 0d0 units c 7 0: shifts to the right 0c0 units d 6 0: shifts down 0d0 units c 6 0: shifts to the left 0c0 units Square root y=√x. So it takes the square root function, and then. The parent function is the simplest form of the type of function given. Adding a negative sign in front of the square root will cause the graph to reflect over the x-axis and each y-value of the coordinates to be negative in the key points. Parent Functions And Transformations. Exponential Parent Function. To graph changes to the square root parent function that result in translations and reflections, follow these steps: Here are some key points to keep in mind when translating and reflecting graphs of the square root parent function. It is generally easier when several transformations occur at once, to apply the up/down and left/right moves first followed by any reflection over the x- or y-axis. View 2 Transformation HMWK-1.pdf from HIST 3315 at Wingate University. Note that (unlike for the y-direction), bigger values cause more compression. Parent Functions and Transformations DRAFT. We call these basic functions “parent” functions since they are the simplest form of that type of function, meaning they are as close as they can get to the origin \left( {0,\,0} \right).The chart below provides some basic parent functions that you should be familiar with. ... square root. Adding or subtracting a constant on the “outside” of the square root moves the graph of the parent function up or down respectively. If you change that to (age+4) = 25 then you will get it when you are 21. $$y=\sqrt{-x}$$ will reflect over the y-axis. Now, notice that sqrt (2) is no more than a constant, you all you've done is stretched the graph vertically byu a factor of sqrt (2). In general, transformations in y-direction are easier than transformations in x-direction, see below. SURVEY . Solving More Complex Square Root Inequalities, Transformations of the Square Root Parent (Stretches and Compressions). Y-values would have 3 subtracted to get new key points. 4. Just add the transformation you want to to. Function Transformations. Edit. How to graph the square root parent function and transformations of the square root function. Learn parent functions transformations with free interactive flashcards. Function Transformation. Domain: [0.∞) Range: [0,∞) Domain and Range of Exponential Parent Function. This has the effect of adding 3 to the y-value of each coordinate. Parent Function Review EX 1: Graph and find the indicated information for: A. Stretches it by 2 in the y-direction ; Shifts it left 1, and; by ddemarr1. When a number is added or subtracted on the “outside” of the square root sign, the graph and key points are moved up (+) or down (-) by that number. Played 21 times. How to transform the graph of a function? 01:10 $$y=\sqrt{x}+3$$ example of a vertical translation, 02:17 $$y=\sqrt{x+3}$$ example of a horizontal translation, 03:22 $$y=-\sqrt{x}$$ example of a reflection over the x-axis, 04:08 $$y=\sqrt{-x}$$ example of a reflection over the y-axis, 04:45 $$y=-\sqrt{x+2}-3$$ example of a multiple transformation, 06:00 $$y=\sqrt{-(x+2)}-3$$ example of a multiple transformation, 07:50 Example of writing an equation from a graph by applying transformations to find $$y=-\sqrt{x+7}+3$$. The graph of the square root parent function begins at point (0, 0) and is drawn only in quadrant I since both the domain and range of the square root parent are both greater than or equal to zero. Notice that the graphs of both parent functions are either centered or begin at the origin. 30 seconds . a few seconds ago. Parent Functions and Transformations. Students will be able to graph the parent function, apply transformations using a, 21 Terms. Transition into a discussion about the similarities of transformations on a quadratic function to absolute value, square root, and cubic and cube root functions. Transformations of square roots. The “shape” of the graph is unchanged. Popular Problems. Constant Parent Function. Transformations. Edit. 0 times. The parent function is f (x) = √x . Note the exact agreement with the graph of the square root function in Figure 1(c). So here is another example using √(x): This is also called reflection about the y-axis (the axis where x=0). Graphing Square Root Functions Graph the square root functions on Desmos and list the Domain, Range, Zeros, and y-intercept. 1/7/2016 3:25 PM 8-7: Square Root Graphs 7 EXAMPLE 4 Using the parent function as a guide, describe the transformation, identify the domain and range, and graph the function, g x x 55 Domain: Range: x t 5 y t 5 g(x) g(x) translates 5 units left and 5 units down > f5, > f5, For a better explanation, assume that y = √x y = x is f (x) = √x f ( x) = x and y = √x y = x is g(x) = √x g ( x) = x. f (x) = √x f ( x) = x. g(x) = √x g ( x) = x. Q. New key points are (0, 3), (1, 4), (4, 5), and (9, 6). y = 4sqrt (x) + 10 stretches the function vertically by a factor of 4, and translates it up by 10. Name_ Date _ For problem 1- 6, please give the name of the parent function and describe the transformation represented. Parent Functions: When you hear the term parent function, you may be inclined to think of two functions who love each other very much creating a new function.The similarities don’t end there! Find the domain and the range of the new function. Transformations of square roots … 0% average accuracy. What are the transformations of this functions compared to the parent function? A Square root function contains a square root with the independent variable (x) under the radical. abigail7090. Consequently, the domain is $$D_{f} = (−\infty, \infty)$$, or all real numbers. The parent function is the simplest form of the type of function given. g(x) = √x g ( x) = x. The constant is added or subtracted from the y-values of the coordinates of key point. Quadratic Parent Function. The graph and table of the parent function is show to the right. ‘Square root transformation’ is one of the many types of standard transformations.This transformation is used for count data (data that follow a Poisson distribution) or small whole numbers. For example, lets move this Graph by units to the top. add to x to go left. ... Graph of Square Root Parent Function. Save. This activity can be used in a variety o 1. Cubing Function (3rd Degree) with Sliders. Focus on absolute value, quadratic, square root (radical), cubic, and cube root functions. Sample Problem 2: Given the parent function and a description of the transformation, write the equation of the transformed function!". Since the normal "vertex" of a square root function is (0,0), the new vertex would be (0, (0*4 + 10)), or (0,10). y = √x (square root) y = 1/x (reciprocal) y = 1/x 2 y = log b (x) for b > 1 y = a x for a > 1 (exponential) y = a x for 0 < a < 1 . Graph of Quadratic Parent Function. We can do all transformation in one go using this: So it takes the square root function, and then. How to graph the reciprocal parent function and transformations of the reciprocal function. Nov 10, 2017 - This section covers: Basic Parent Functions Generic Transformations of Functions Vertical Transformations Horizontal Transformations Mixed Transformations Transformations in Function Notation Writing Transformed Equations from Graphs Rotational Transformations Transformations of Inverse Functions Applications of Parent Function Transformations More … Mathematics. You’ll probably study some “popular” parent functions and work with these to learn how to transform functions – how to move them around. ... Domain and Range of Square Root Parent Function. In the example, $$y=\sqrt{x}+3$$, the graph and key points will be moved up three units. Linear Parent Function. a For example, $$y=-\sqrt{x+2}-3$$, will move the parent down 3, left 2, and reflect it over the x-axis while $$y=\sqrt{-(x+2)}-3$$ will move the graph down 3, left 2, and reflect it over the y-axis. You could graph this by looking at how it transforms the parent function of y = sqrt (x). Next. In Figure 2(a), the parabola opens outward indefinitely, both left and right. Which transformation describes the equation from its parent function? 2 ( ) f x x B. Let’s begin by reviewing the rational and square root parent functions. Then, using the property that sqrt (ab) = sqrt (a)*sqrt (b), you can rewrite this again as: y = sqrt (2) * sqrt (x+3/2). Unit 8 Day 1 Notes Objective: Be able to identify and graph transformations of square root functions. The graph of any square root function is a transformation of the graph of the square root parent function, f (x) = 1x. slecky. STUDY. This is it. Square root functions can also be written in h,k form. Graph of Linear Parent Function. Sample Problem 1: Identify the parent function and describe the transformations. 9th - 12th grade. BUT we must add C wherever x appears in the function (we are substituting x+C for x). This transformation also may be appropriate for percentage data where the range is between 0 and 20% or between 80 and 100%.Each data point is replaced by its square root. Note: values added under the square root are subtracted from the x-values and numbers subtracted under the square root are added to the x-values. Describe the Transformation f (x) = square root of x. f (x) = √x f ( x) = x. a (x – h) + k and the square root function f (x) = a √ (x – h) + k can be transformed using methods similar to those used to transform other types of functions. A negative sign in front of x under the square root will reflect the graph over the y-axis. 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# Question Video: Arranging Numbers in Descending Order Involving Square and Cube Roots Mathematics • 8th Grade
Arrange the following numbers in descending order: √33, −√22, √41, 3, 0, ∛−125.
02:17
### Video Transcript
Arrange the following numbers in descending order: the square root of 33, the negative square root of 22, the square root of 41, three, zero, and the cube root of negative 125.
In order to be able to arrange these numbers, we have to make use of some of the properties of square roots.
For example, if the square of a number 𝑎 is less than another number 𝑏 and both numbers are positive, then 𝑎 is less than the square root of 𝑏. So for instance, let us consider the numbers root 33 and three. Then we can observe that since three squared, which is nine, is less than 33, then we can conclude that three is less than root 33.
Another useful rule that we can use is that if 𝑎 is less than 𝑏, then root 𝑎 is also less than root 𝑏. We can use this property on root 33 and root 41. That is, since 33 is less than 41, then root 33 is less than root 41. If we were to put these three numbers on a number line, they would look roughly like this. We also know that zero is less than three. If we consider the remaining two numbers, we can see that they will both be negative. Therefore, they will be somewhere on this line to the left of zero.
To find the exact ordering of these two numbers, we should note that the five cubed is five times five times five, which is 125. Therefore, negative five cubed is negative 125. Then, if we cube root both sides, we have that negative five is equal to the cube root of negative 125.
We also know that root 22 is less than root 25. Therefore, by flipping the inequality, negative root 22 will be greater than negative root 25. But since root 25 is equal to five, we have that negative root 22 is greater than negative five.
Let us also not forget that negative five is equal to the negative cube root of 125. So the negative root of 22 is greater than the cube root of negative 125. We can thus append these numbers to our number line.
Therefore, by arranging the numbers in descending order, we have the square root of 41, the square root of 33, three, zero, the negative square root of 22, and the cube root of negative 125.
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# Section 4: Sine And Cosine Rule
## Introduction
This section will cover how to:
• Use the Sine Rule to find unknown sides and angles
• Use the Cosine Rule to find unknown sides and angles
• Combine trigonometry skills to solve problems
Each topic is introduced with a theory section including examples and then some practice questions. At the end of the page there is an exercise where you can test your understanding of all the topics covered in this page.
You are allowed to use calculators in this topic.
All answers should be given to 3 significant figures unless otherwise stated.
## Formulae You Should Know
You should already know each of the following formulae:
formulae for right-angled triangles formulae for all triangles
NOTE: The only formula above which is in the A Level Maths formula book is the one highlighted in yellow.
You must learn these formulae, and then try to complete this page without referring to the table above.
## Sine Rule
The Sine Rule can be used in any triangle (not just right-angled triangles) where a side and its opposite angle are known.
Finding Sides
If you need to find the length of a side, you need to use the version of the Sine Rule where the lengths are on the top:
a = b sin(A) sin(B)
You will only ever need two parts of the Sine Rule formula, not all three.
You will need to know at least one pair of a side with its opposite angle to use the Sine Rule.
Finding Sides Example
Work out the length of x in the diagram below:
Step 1Start by writing out the Sine Rule formula for finding sides:
a = b sin(A) sin(B)
Step 2Fill in the values you know, and the unknown length:
x = 7 sin(80°) sin(60°)
Remember that each fraction in the Sine Rule formula should contain a side and its opposite angle.
Step 3Solve the resulting equation to find the unknown side, giving your answer to 3 significant figures:
x = 7 (multiply by sin(80°) on both sides) sin(80°) sin(60°) x = 7 × sin(80°) sin(60°) x = 7.96 (accurate to 3 significant figures)
Note that you should try and keep full accuracy until the end of your calculation to avoid errors.
Finding Angles
If you need to find the size of an angle, you need to use the version of the Sine Rule where the angles are on the top:
sin(A) = sin(B) a b
As before, you will only need two parts of the Sine Rule , and you still need at least a side and its opposite angle.
Finding Angles Example
Work out angle m° in the diagram below:
Step 1Start by writing out the Sine Rule formula for finding angles:
sin(A) = sin(B) a b
Step 2Fill in the values you know, and the unknown angle:
sin(m°) = sin(75°) 8 10
Remember that each fraction in the Sine Rule formula should contain a side and its opposite angle.
Step 3Solve the resulting equation to find the sine of the unknown angle:
sin(m°) = sin(75°) (multiply by 8 on both sides) 8 10 sin(m°) = sin(75°) × 8 10 sin(m°) = 0.773 (3 significant figures)
Step 4Use the inverse-sine function (sin–1) to find the angle:
m° = sin–1(0.773) = 50.6° (3sf)
Other Notes
You may be aware that sometimes Sine Rule questions can have two solutions (only when you are finding angles) – you do not need to know about these additional solutions at this time but you will learn more about them next year.
Practice Questions
Work out the answer to each question then click on the button marked to see if you are correct.
(a) Find the missing side in the diagram below:
a = b sin(A) sin(B) p = 21 × sin(32°) both sides sin(32°) sin(95°) p = 21 × sin(32°) sin(95°) p = 11.2 (accurate to 3 significant figures)
(b) Find the missing angle in the diagram below:
sin(A) = sin(B) a b sin(b°) = sin(100°) × 3.6 both sides 3.6 5.1 sin(b°) = sin(100°) × 3.6 5.1 sin(b°) = 0.695 (3sf) b° = sin-1(0.695) = 44.0° (3sf)
## Cosine Rule
The Cosine Rule can be used in any triangle where you are trying to relate all three sides to one angle.
Finding Sides
If you need to find the length of a side, you need to know the other two sides and the opposite angle.
You need to use the version of the Cosine Rule where a2 is the subject of the formula:
a2 = b2 + c2 – 2bc cos(A)
Side a is the one you are trying to find. Sides b and c are the other two sides, and angle A is the angle opposite side a.
Finding Sides Example
Work out the length of x in the diagram below:
Step 1Start by writing out the Cosine Rule formula for finding sides:
a2 = b2 + c2 – 2bc cos(A)
Step 2Fill in the values you know, and the unknown length:
x2 = 222 + 282 – 2×22×28×cos(97°)
It doesn't matter which way around you put sides b and c – it will work both ways.
Step 3Evaluate the right-hand-side and then square-root to find the length:
x2 = 222 + 282 – 2×22×28×cos(97°) (evaluate the right hand side) x2 = 1418.143..... (square-root both sides) x = 37.7 (accurate to 3 significant figures)
As with the Sine Rule you should try and keep full accuracy until the end of your calculation to avoid errors.
Finding Angles
If you need to find the size of an angle, you need to use the version of the Cosine Rule where the cos(A) is on the left:
cos(A) = b2 + c2 – a2 2bc
It is very important to get the terms on the top in the correct order; b and c are either side of angle A which you are trying to find and these can be either way around, but side a must be the side opposite angle A.
Finding Angles Example
Work out angle P° in the diagram below:
Step 1Start by writing out the Cosine Rule formula for finding angles:
cos(A) = b2 + c2 – a2 2bc
Step 2Fill in the values you know, and the unknown length:
cos(P°) = 52 + 82 – 72 2 × 5 × 8
Remember to make sure that the terms on top of the fraction are in the correct order.
Step 3Evaluate the right-hand-side and then use inverse-cosine (cos–1) to find the angle:
cos(P°) = 52 + 82 – 72 (evaluate the right-hand side) 2 × 5 × 8 cos(P°) = 0.5 (do the inverse-cosine of both sides) P° = cos–1(0.5) = 60° (3sf)
Other Notes
If you know two sides and an angle which is not inbetween them then you can use the Cosine Rule to find the other side, but it is easier to use the Sine Rule in this situation – you should always use the Sine Rule if you have an angle and its opposite side.
Practice Questions
Work out the answer to each question then click on the button marked to see if you are correct.
(a) Find the missing side in the diagram below:
a2 = b2 + c2 – 2bc cos(A) h2 = 882 + 1462 – 2×88×146×cos(53°) h2 = 13595.761..... h2 = 117 (accurate to 3 significant figures)
(b) Find the missing angle in the diagram below:
cos(A) = b2 + c2 – a2 2bc cos(a°) = 3.12 + 4.32 – 5.92 2 × 3.1 × 4.3 cos(a°) = – 0.252 a° = cos–1(– 0.252) a° = 105° (3sf)
## Combining Trigonometry Skills
Choosing The Appropriate Technique
Sometimes more than one technique from the formula table at the top of this page can be used to solve a trig problem, but you will want to choose the most efficient and easiest method to save time. The flowchart below shows how to decide which method to use:
Examples
These examples illustrate the decision-making process for a variety of triangles:
e.g. 1 The triangle is not right-angled. We do know a side and its opposite angle. Therefore we use the Sine Rule. e.g. 2 The triangle is right-angled. The question involves angles. Therefore we use trig ratios - sin, cos and tan. e.g. 3 The triangle is right-angled. The question does not involve angles. Therefore we use Pythagoras's Theorem. e.g. 4 The triangle is not right-angled. We do not know a side and its opposite angle. Therefore we use the Cosine Rule.
Practice Questions
Work out the answer to each question then click on the button marked to see if you are correct.
Find the unknown side or angle in each of the following diagrams:
(a)
The triangle is not right-angled, and we don't know a side and its opposite angle, so we need to use the Cosine Rule.
cos(a°) = 42 + 72 – 62 2 × 4 × 7 cos(a°) = 0.518 a° = cos–1(0.518) a° = 58.8° (3sf)
(b)
The triangle is right-angled, and the question involves angles, so we need to use trigonometric ratios.
cos(50°) = y 8
8 × cos(50°) = y y = 5.14 (3sf)
(c)
The triangle is not right-angled, but we do know a side and its opposite angle, so we use the Sine Rule.
sin(b°) = sin(62°) 11 10 sin(b°) = sin(62°) × 11 10 sin(b°) = 0.971 (3sf) b° = sin-1(0.971) = 76.2° (3sf)
(d)
The triangle is right-angled, but the question does not involve angles, so we need to use Pythagoras's Theorem.
x2 = 7.52 + 182 x2 = 380.25 x = 19.5
## Exercise
Work out the answers to the questions below and fill in the boxes. Click on the
button to find out whether you have answered correctly. If you have then the answer will be ticked
and you should move on to the next question. If a cross
to clear the incorrect answer and have another go, or you can click on
to get some advice on how to work out the answer and then have another go. If you still can't work out the answer after this then you can click on
to see the solution.
#### Questions
Work out the values of x, y and z for each of the diagrams below. The diagrams are not to scale.
Try and use the most efficient method you can and remember to give your answer to 3 significant figures.
(a)
x
=
° The angles in a triangle add up to 180° 180° – 34° – 115° = 31°
y
=
Use the Sine Rule
y = 13 sin(31°) sin(115°) y = 13 × sin(31°) sin(115°) y = 7.39
z
=
Use the Sine Rule again
z = 13 sin(34°) sin(115°) z = 13 × sin(34°) sin(115°) z = 8.02
(b)
x
=
° Use the Cosine Rule
cos(x°) = 62 + 102 – 122 2 × 6 × 10 cos(x°) = –0.0667 x° = cos–1(–0.0667) x° = 93.8°
y
=
° Use the Cosine Rule again
cos(y°) = 62 + 122 – 102 2 × 6 × 12 cos(y°) = 0.556 y° = cos–1(0.556) y° = 56.3°
z
=
° The angles in a triangle add up to 180° 180° – 93.8° – 56.3° = 29.9°
(c)
x
=
° The angles in a triangle add up to 180° 180° – 75° – 65° = 40°
y
=
Use the Sine Rule
y = 9 sin(75°) sin(65°) y = 9 × sin(75°) sin(65°) y = 9.59
z
=
Use the Sine Rule again
z = 9 sin(40°) sin(65°) z = 9 × sin(40°) sin(65°) z = 6.38
(d)
x
=
The triangle to the right of the dotted line is right-angled, so use the appropriate trig ratio (sin, cos or tan)
x = sin(70°) 6 x = sin(70°) × 6 x = 5.64
y
=
° The triangle to the left of the dotted line is also right-angled, so use the appropriate trig ratio (sin, cos or tan)
cos(y°) = 5.64 8 cos(y°) = 0.705 y° = cos–1(0.705) y° = 45.2°
z
=
° The angles in a triangle add up to 180° 180° – 90° – 45.2° = 44.8°
(e)
x
=
° Use the Sine Rule
sin(x°) = sin(54°) 5 7 sin(x°) = sin(54°) × 5 7 sin(x°) = 0.578 x° = sin-1(0.578) = 35.3°
y
=
° The angles in a triangle add up to 180° 180° – 54° – 35.3° = 90.7°
z
=
Use the Sine Rule again
z = 7 sin(90.7°) sin(54°) z = 7 × sin(90.7°) sin(54°) z = 8.65
(f)
x
=
° Use the Cosine Rule
cos(x°) = 62 + 62 – 92 2 × 6 × 6 cos(x°) = –0.125 x° = cos–1(–0.125) x° = 97.2°
y
=
° Don't use the Cosine Rule again – this is an isosceles triangle (180° – 97.2°) ÷ 2 = 41.4°
z
=
° The angles in a triangle add up to 180° 180° – 97.2° – 41.4° = 41.4°
(g)
x
=
° Neither right-angled triangle either side of the dotted line has enough information, so you will need to use Cosine Rule on the outside triangle
cos(x°) = 122 + 202 – 152 2 × 12 × 20 cos(x°) = 0.665 x° = cos–1(0.665) x° = 48.3°
y
=
After finding x you have enough information to use trigonometry in the right-angled triangle above the dotted line
y = sin(48.3°) 12 y = sin(48.3°) × 12 y = 8.97
z
=
° After finding y you have enough information to use trigonometry in the right-angled triangle below the dotted line
sin(z°) = 8.97 15 sin(z°) = 0.598 z° = sin–1(0.598) z° = 36.7°
(h)
( definitely not to scale! )
x
=
° Use the Sine Rule
sin(x°) = sin(54°) 4 9 sin(x°) = sin(54°) × 4 9 sin(x°) = 0.360 x° = sin-1(0.360) = 21.1°
y
=
° The angles in a triangle add up to 180° 180° – 54° – 21.1° = 105° (3sf)
z
=
Use the Sine Rule again
z = 9 sin(104.9°) sin(54°) z = 9 × sin(104.9°) sin(54°) z = 10.7
|
Write the Riemann sum to find the area under the graph of the function f(x) = x^2 from x = 1 to x = 5?
Feb 8, 2018
${\int}_{1}^{5} \setminus {x}^{2} \setminus \mathrm{dx} = \frac{124}{3}$
Explanation:
$I = {\int}_{1}^{5} \setminus {x}^{2} \setminus \mathrm{dx}$
Using Riemann sums. By definition of an integral, then
${\int}_{a}^{b} \setminus f \left(x\right) \setminus \mathrm{dx}$
represents the area under the curve $y = f \left(x\right)$ between $x = a$ and $x = b$. We can estimate this area under the curve using thin rectangles. The more rectangles we use, the better the approximation gets, and calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles.
That is
${\int}_{a}^{b} \setminus f \left(x\right) \setminus \mathrm{dx} = {\lim}_{n \rightarrow \infty} \frac{b - a}{n} {\sum}_{i = 1}^{n} \setminus f \left(a + i \frac{b - a}{n}\right)$
And we partition the interval $\left[a , b\right]$ equally spaced using:
$\Delta = \left\{a + 0 \left(\frac{b - a}{n}\right) , a + 1 \left(\frac{b - a}{n}\right) , \ldots , a + n \left(\frac{b - a}{n}\right)\right\}$
$\setminus \setminus \setminus = \left\{a , a + 1 \left(\frac{b - a}{n}\right) , a + 2 \left(\frac{b - a}{n}\right) , \ldots , b - 1\right\}$
Here we have $f \left(x\right) = {x}^{2}$ and so we partition the interval $\left[1 , 5\right]$ using:
$\Delta = \left\{1 , 1 + 1 \frac{4}{n} , 1 + 2 \frac{4}{n} , 1 + 3 \frac{4}{n} , \ldots , 4\right\}$
And so:
$I = {\int}_{1}^{5} \setminus \left({x}^{2}\right) \setminus \mathrm{dx}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{n} {\sum}_{i = 1}^{n} \setminus f \left(1 + \frac{4 i}{n}\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{n} {\sum}_{i = 1}^{n} \setminus {\left(1 + \frac{4 i}{n}\right)}^{2}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{n} {\sum}_{i = 1}^{n} \setminus {\left(\frac{n + 4 i}{n}\right)}^{2}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{n} {\sum}_{i = 1}^{n} \setminus \frac{1}{n} ^ 2 {\left(n + 4 i\right)}^{2}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{n} ^ 3 {\sum}_{i = 1}^{n} \setminus \left({n}^{2} + 8 n i + 16 {i}^{2}\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{n} ^ 3 \left\{{n}^{2} {\sum}_{i = 1}^{n} 1 + 8 n {\sum}_{i = 1}^{n} i + 16 {\sum}_{i = 1}^{n} {i}^{2}\right)$
Using the standard summation formula:
${\sum}_{r = 1}^{n} r \setminus = \frac{1}{2} n \left(n + 1\right)$
${\sum}_{r = 1}^{n} {r}^{2} = \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)$
we have:
I = lim_(n rarr oo) 4/n^3 {n^3 + 8n 1/2n(n+1)+16 1/6n(n+1)(2n+1)
\ \ = lim_(n rarr oo) 4/n^3 \ n/3 {3n^2 + 12n(n+1)+ 8(n+1)(2n+1)
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{3 {n}^{2}} \left(3 {n}^{2} + 12 {n}^{2} + 12 n + 16 {n}^{2} + 24 n + 8\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{3 {n}^{2}} \left(31 {n}^{2} + 36 n + 8\right)$
$\setminus \setminus = \frac{4}{3} {\lim}_{n \rightarrow \infty} \left(31 + \frac{36}{n} + \frac{8}{n} ^ 2\right)$
$\setminus \setminus = \frac{4}{3} \left(31 + 0 + 0\right)$
$\setminus \setminus = \frac{124}{3}$
Using Calculus
If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get:
$I = {\int}_{1}^{5} \setminus {x}^{2} \setminus \mathrm{dx}$
$\setminus \setminus = {\left[{x}^{3} / 3\right]}_{1}^{5}$
$\setminus \setminus = \frac{125}{3} - \frac{1}{3}$
$\setminus \setminus = \frac{124}{3}$
|
Rational Expression Word Problems Rational Expression Word Problems
The rational expression is an expression which is expressed in from of two integers. The rational expression is rational because one integer is divided by the second one. This is the definition of rational number where denominator is nonzero real number. If m and n are two integer then it can be written as fraction m/n and n is not equal to zero. There are many types of expression which are performed on the basis of rational expression and these are basic arithmetic operations like addition, subtraction, multiplication and division. There is a property of rational expressions, when we perform any arithmetic operations on the two rational expressions then the result of those rational expressions always gives another rational expression. Let us suppose that A = p/q and B = r/s be two rational expressions, where p, q, r, and s are the integers. Now the product of these two rational expression that is AB = pr / qs and the division of these rational expression is A / B = ps / qr. Know More About Associative Property Worksheets
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When we are doing addition and subtraction of rational expressions, then it is compulsory to factor the denominator of each and every rational expression. We add or subtract one in fractions and calculate the least common multiple in the denominators, then rewriting each and every fraction in that denominator and lastly we adding the numerators. On other hand when we multiply and divide two rational expressions then it is necessary to factor the numerator of the rational expression and also denominator of that rational expression. Rational expressions word problems utilize this concept. f1= (x+4) / (x+1) and f2= (x+6) / (x-2): product of these rational expression is (x+4) (x+6) / (x+1) (x+2). Next we are doing the division by follow the above rule (x+4) (x2) / (x+1) (x+6). Now the addition and subtraction of the given rational expression follows the above rule of addition and subtraction that is first we factor the both rational expressions if there is any need for that, if there is no need to factor then we will solve the rational expression as per the given rule. Now come on the example, here two rational expressions are given in the factor form so here no need of factorization. But here is one more condition that we will find the least common denominator of given rational expression. Their addition/subtraction will be [(x+4)(x-2)±(x+1)(x+6)]/(x+1)(x-2) Rational expressions applications word problems: Example- Assume that Mary takes 6 hours to work alone and Ruby takes 8 hours to do the same work alone. Read More About Trigonometry Worksheets
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How long will it take them to complete the job if they work together? Solution: Let A be the amount of time it takes to complete the job together. Rate of work of Mary=1/6 Rate of work of Ruby=1/8 => In A hours, Mary and Ruby do A/6 and A/8 amount of work respectively. => If they work together: A/6 +A/6=1 Thus they will complete the job in 24/7 hours. Solving Word Problems with Rational Expressions Solving word problems with Rational Expressions is not a very difficult task we can solve them easily, we need to have the knowledge of rational number and rational expressions. We are acquainted with the fact that Rational Numbers are the numbers which can be symbolized in the form of x/y , where x and y are real numbers.
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Thank You
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Rational Expression Word Problems
Let us suppose that A = p/q and B = r/s be two rational expressions, where p, q, r, and s are the integers. Now the product of these two rat...
Rational Expression Word Problems
Let us suppose that A = p/q and B = r/s be two rational expressions, where p, q, r, and s are the integers. Now the product of these two rat...
|
# Solving One-step Equations
Home » Math Vocabulary » Solving One-step Equations
## What Are One-step Equations?
One-step equations are algebraic equations that can be solved in one step only. They can be solved in just one step by isolating the variable using the inverse operations. An equation is a mathematical statement that shows that two mathematical expressions are equal.
The most basic and simple algebraic equations consist of one or more variables in math.
An equation consists of two sides, L.H.S. (Left Hand Side) and R.H.S. (Right hand Side) separated by “=” sign. Solving an equation simply means finding the value of the unknown variable.
Consider an equation.
$x\;-\;7 = 93$
Let’s add 7 on both sides to cancel out 7 and isolate the variable.
$x\;-\;7 + 7 = 93 + 7$
Thus, $x = 100$
Wasn’t that simple? This is what we call a “one-step equation.”
Note: Take note of inverse operations that help in solving the one-step equations.
### Definition of One-step Equations
One-step equations are simple algebraic equations that can be solved in just one step.
To solve one-step equations, we determine the value of the variable involved using different properties of equality. The operation you perform on one side of the equation must be carried out on the other side as well, in order to keep the equation balanced. One-step equations may involve integers, fractions, or decimals.
## Rules for One-step Algebraic Equations
While keeping the equation balanced on both sides, the subsequent operations are carried out to isolate the variable. By doing this, LHS continues to be equal to RHS, and eventually, the balance is maintained. There are some rules for one-step algebraic equations. They are as follows:
If the equation involves addition, we can isolate the variable using the inverse operation, which is the subtraction operation.
If we subtract the same number from both sides of an equation, both sides will remain equal. This is the subtraction property of equality.
For example,
$x + 5 = 2$
$\Rightarrow x + 5\;-\;5 = 2\;-\;5$ …subtract 5 from both sides
$\Rightarrow x =\;-\;3$
1. One-step Equation with Subtraction
If the equation involves subtraction, we isolate the variable using the addition property of equality. If we add the same number to both sides of an equation, both sides will remain equal.
For example,
$c\;-\;1.1 = 12$
$\Rightarrow c\;-\;1.1 + 1.1 = 12 + 1.1$ …adding 1.1 on both sides
$\Rightarrow c = 13.1$
1. One-step Equation with Multiplication
If the equation involves multiplication, we can isolate the variable using division. However, note that we cannot divide by 0.
If we divide both sides of an equation by the same number, both sides will remain equal.
For example,
$4x = 16$
$\frac{4x}{4} = \frac{16}{4}$ …divide both sides by 4
$\Rightarrow x = 4$
1. One-step Equation with Division
If the equation involves division, we use multiplication to solve the equation.
If we multiply both sides of an equation by the same number, both sides will remain equal.
For example,
$\frac{x}{4} = \frac{1}{5}$
$\frac{x}{4} \times 4 = \frac{1}{5}\times4$ … multiply both sides by 4
$x = \frac{4}{5}$
## How to Solve One-step Equations
When you’re solving one-step algebraic equations, the goal is to get the variable on one side and all the remaining numbers on the other. We use the inverse operation or the opposite operation to the operation acting on the variable. The inverse operations pairwise are addition and subtraction and division and multiplication.
The steps to solve one-step equations are as follows:
Step 1: Note down the equation.
For example $x\;-\;5 = 10$
An equation will have a variable (x), which denotes an unknown value, and it will also have a constant, which is a number you need to add or subtract from the variable to equal a certain sum or difference.
Step 2: Isolate the variable.
You must conduct an inverse operation to cancel the constants in order to isolate a variable, on one side of the equation.
For example, in the equation, $x \;-\; 5 = 10 , 5$ is subtracted from the variable, so to isolate the variable you must cancel the 5 by adding it on both sides.
Step 3: Adding or subtracting the constant from both sides.
While solving the equation, we have to keep both sides balanced.
So, if you need to add a value to isolate the variable, you must also add that same value to the other side of the equation, also if you need to subtract a value on one side of the equation you have to subtract it from the other side.
In the equation, $x \;-\; 5 = 10$, we need to add 5 on the left as well as on the right.
$x \;-\; 5 + 5 = 10 + 5$
Thus, $x = 15$
To verify the solution, simply plug in the value of x in the equation.
$x \;-\; 5 = 10$, the value of x as we found out is 15.
Thus, substituting the value, the equation becomes, $15 \;-\; 5 = 10$.
Since this equation is true, your solution is correct.
Similarly, if the variable is multiplied by a number, divide both sides of the equation by the same number to isolate the variable. (If the variable is divided by a number, multiply both sides of the equation by the same number.)
For example, let’s solve an equation. $\frac{x}{6} = 12$
As the variable is divided by 6, to isolate it, you need to multiply by 6.
$6(\frac{x}{6}) = 12\times6$
$\Rightarrow x = 72$
Verifying the solution, as $\frac{72}{6} = 12$ , the solution is correct.
• A one-step equation is an equation that requires only one step to solve.
• The most common one-step equations are linear algebraic equations.
• When solving an equation, you may keep the variable on either side of the equation. As long as in the end, the variable that you are solving is isolated on one side with a coefficient of $+1$.
## Conclusion
In this article, we have learned how to solve one-step equations, its rules and the steps of solving the equations. Let’s now solve some examples to better understand the concept.
## Solved Examples On One-Step Equations
1. Solve this equation with a negative constant: $\;-\;8 + x = 14$
Solution: Since the constant is negative, adding it to both sides will isolate the variable.
$\Rightarrow -8 + x = 14$
$\Rightarrow -8 + x + 8 = 14 + 8$
$x = 22$
Verifying the solution, $-8 + 22 = 14$. The solution is correct.
2. Solve this equation with a negative coefficient: $-4x = 32$
Solution:
Since the variable is multiplied by $-4$, to isolate the variable, you must divide each side by $-4$.
Remember that dividing a positive number by a negative number equals a negative quotient.
$-4x = 32$
$\frac{-4x}{-4} = \frac{32}{-4}$
$\Rightarrow x = \;-\; 8$
Verifying the solution, $-4(\;-8) = 32$. The solution is correct.
3. Jennifer weighed herself on the scale and found her weight to be 120 lbs. Then, she held the cat and stepped on the scale and found the combined weight to be 132 lbs.
Create an equation that models the situation and solve the equation to find c, the cat’s weight.
Solution: To model the following situation, we will create an equation to show the combined weights of Jennifer and cat.
$c + 120 = 132$
To solve for c, we will do the inverse operation of addition and subtract 120 from each side:
$c + 120 \;-\; 120 = 132 \;-\; 120$
$\Rightarrow c = 12$ lbs.
To verify answer we will plug in the values in the equation
$12 + 120 = 132$
Therefore, $c = 12$ lbs is the correct solution.
4. Solve the equation: $10 = \frac{p}{10}$
Solution: Since the variable is divided by 10, we have to isolate it by multiplying it by 10.
$10 = \frac{p}{10}$
$\Rightarrow 10\times10 = \frac{p}{10}\times10$
$\Rightarrow P = 100$
Verifying the solution, $10 = \frac{100}{10}$ . The solution is correct.
## Practice Problems On One-Step Equations
1
### What is the value of x in the equation $5x = 40$?
45
8
6
35
CorrectIncorrect
We will isolate the variable by dividing both sides of the equation by the same number, both sides will remain equal and we will get the solution. For example,$\frac{5x}{5} = \frac{40}{5}$ .Thus, $x = 8$
2
### Identify the equation, where the value of $z = 7$.
5z = 15
6z = 36
8z = 32
4z = 28
CorrectIncorrect
Correct answer is: 4z = 28
We will isolate the variable by dividing both sides of the equation by the same number. $\frac{4z}{4} = \frac{28}{4}$ Thus, $z = 7$.
3
### Identify the equation that does not have a solution at $w = 5$.
w $+ 7 = 12$
w $\;–\; 5 = 10$
$6 +$ w $= 11$
w $–\; 4 = 1$
CorrectIncorrect
Correct answer is: w $\;–\; 5 = 10$
Substituting the value in equation, w $-\; 5 = 10$
$5 \;-\; 5 = 10$, which is a false statement.
4
### What equation matches this situation?Maria had some pencils then bought 4 more. Now she has a total of 10 pencils.
$P\; –\; 4 = 10$
$10 + 4 = P$
$P + 4 = 10$
None of the above.
CorrectIncorrect
Correct answer is: $P + 4 = 10$
Let's consider pencil as p, Maria has p number of pencils, and she bought 4 more,
i.e., $P + 4$. Now she has a total of ten pencils, which means $P + 4 = 10$.
5
### What is the first step to solve this equation: $2x = 18$.
Subtract 2 from both sides.
Multiply by 3 on both sides.
Divide by 2 on both sides.
CorrectIncorrect
Correct answer is: Divide by 2 on both sides.
In the equation above, x is being multiplied by 2. Thus, to isolate the equation, we have to do the inverse and divide both sides of the equation by 2.
## Frequently Asked Questions On One-Step Equations
There are three methods used to solve systems of equations. They are as follows:
• Graphing
• Substitution
• Elimination
One-step equations require only one step, one inverse operation to be solved and have only one operation. Two-step equations require two inverse operations to solve and have two operations.For eg., $x^2 = 64$ is a one-step equation, while $3x + 2 = 14$ is a two-step equation.
A linear equation in one variable is an equation in which the degree of the variable is 1. These are also known as first degree equations, because the highest exponent on the variable is 1. All linear equations eventually can be written in the form $ax + b = c$, where a, b, and c are real numbers and $a \neq 0$.
Some equations have exactly one solution. In these equations, there is only one value for the variable that makes the equation true. You can tell that an equation has one solution if you solve the equation and get a variable equal to a number.
For eg., $5x = 25$ has a variable term, $5x$ on one side of the equation and a constant term, 25, on the other side of the equation. So, it has one solution. Let’s solve the equation to see why.
$5x = 25$ . Divide both sides by 5.
$x = 5$
So, $5x = 25$ has one solution, $x = 5$.
If a linear equation has the same variable term but different constant values on opposite sides of the equation, it has no solutions.
For eg., $2x + 4 = 1 + 2x$ has the same variable term, $2x$, but different constant terms, 4 and 1, on opposite sides of the equation. So, it has no solutions. Let’s solve the equation to see why.
$2x + 4 = 1 + 2x$
Subtract 2x from both sides.
$4 = 1$The statement $4 = 1$ is false. So, $2x + 4 = 1 + 2x$ has no solutions.
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# Two Dimensional Shapes and Lines
Two dimensional shapes and lines are discussed here.
All solids are three dimensional in shapes. Solids are covered by faces. Thus the outer covers of a solid are called faces, i.e., surfaces. These surfaces may be plane or curved. The cuboid, cube etc. have plane surface, while sphere has curved surface. Some solids like cylinder have both plane and curved surfaces.
3-D shaped solids are covered by 2-D surfaces. If any two points are joined, a line segment is formed. In this chapter planes and lines will be discussed.
### 2-D plane figures
If the outlines of any face of a three dimensional face are drawn, a two dimensional plane figure is obtained.
Consider the following plane figures
(i) Rectangle
(ii) Square
(iii) Triangle
(iv) Circle
(i) Rectangle:
The outline of a face of a cuboid forms a rectangle. It has two pairs of opposite sides i.e., it has four sides. The opposite sides are equal. It has four corners i.e., vertices.
Here the name of rectangle is rect. ABCD.
A, B, C, D are corners or vertices of the rectangle.
Side AB = DC, Side BC = AD.
(ii) Square:
The outlines of a cube make a square. All its sides are equal to each other.
It has four corners or vertices.
PQRS is the name of the square whose sides PQ = QR = RS = SP and P, Q, R, S are four corners.
(iii) Triangle:
A triangle has three sides and three corners.
DEF is a triangle whose sides are DE, EF and DF.
The triangle has three corners or vertices named D, E and F.
(iv) Circle:
The outline of the plane surface of a cone forms a circle.
It has a centre whose distance from any point on the outline is always the same.
This distance is called the radius of the circle.
A circle has no corner and no side.
The surfaces of rectangle, square, triangle and circle are plane surfaces. All the surfaces or faces of cuboid, cube i.e., books, matchbox, etc. are called plane. But the surface of a sphere i.e., ball is curved surface. A cylinder and cone have both curved and plane surfaces.
● Geometry
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# How do you find lim (x^2+4)/(x^2-4) as x->2^+?
Feb 10, 2017
One method is to evaluate at values closer and closer to 2.
#### Explanation:
f(2.1) = 20.5
f(2.05) = 40.5
f(2.01) = 200.5
f(2.0001) = 20000.5
So we can see it approaches infinity.
Another method is by graphing. As you can see, the limit approaches infinity.
graph{(x^2+4)/(x^2-4) [-5.74, 12.04, -1.71, 7.18]}
As we can see as x approaches two from the positive direction, the y value seems to go up indefinitely.
Feb 10, 2017
#### Explanation:
As $x \rightarrow 2$, the numerator goes to $8$ (which is not $0$) and the denominator goes to $0$.
This form $\frac{\text{non-} 0}{0}$ tells us that the function is increasing or decreasing without bound on each side. (The one-sided limits are $\pm \infty$)
To see which is happening as $x \rightarrow {2}^{+}$, we need to determine whether the denominator is going to $0$ through positive or negative values.
For $x$ a little greater than $2$, we know that ${x}^{2}$ is a little greater than $4$. Therefore, the denominator is positive.
Something close to $8$ divided by a positive number close to $0$ (a positive fraction) is a big positive number.
${\lim}_{x \rightarrow {2}^{+}} \frac{{x}^{2} + 4}{{x}^{2} - 4} = \infty$
Note it takes a lot longer to explain than it does to do!
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Unit Modes ( Math | Algebra | Trig | Unit Modes )
Trig Functions: Unit Modes
The trig functions evaluate differently depending on the units on q. For example, sin(90°) = 1, while sin(90)=0.89399.... If there is a degree sign after the angle, the trig function evaluates its parameter as a degree measurement. If there is no unit after the angle, the trig function evaluates its parameter as a radian measurement. This is because radian measurements are considered to be the "natural" measurements for angles. (Calculus gives us a justification for this. A partial explanation comes from the formula for the area of a circle sector, which is simplest when the angle is in radians).
Calculator note: Many calculators have degree, radian, and grad modes (360° = 2p rad = 400 grad). It is important to have the calculator in the right mode since that mode setting tells the calculator which units to assume for angles when evaluating any of the trigonometric functions. For example, if the calculator is in degree mode, evaluating sine of 90 results in 1. However, the calculator returns 0.89399... when in radian mode. Having the calculator in the wrong mode is a common mistake for beginners, especially those that are only familiar with degree angle measurements.
For those who wish to reconcile the various trig functions that depend on the units used, we can define the degree symbol (°) to be the value (PI/180). Therefore, sin(90°), for example, is really just an expression for the sine of a radian measurement when the parameter is fully evaluated. As a demonstration, sin(90°) = sin(90(PI/180)) = sin(PI/2). In this way, we only need to tabulate the "natural" radian version of the sine function. (This method is similar to defining percent % = (1/100) in order to relate percents to ratios, such as 50% = 50(1/100) = 1/2.)
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# Proving Theorems Using Number Properties
Instructor: Laura Pennington
Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.
This lesson will explain proving theorems using number properties. We will go over multiple examples of using various number properties to prove different theorems, and the steps involved in doing this.
## Proving Even Multiples
It is said that mathematicians make great lawyers. That's probably because they are so great at proving things!
Suppose Earl Integral is a math lawyer, and he is hard at work on a case. His client, the defendant, has claimed that if two even numbers are multiplied together, then the result is also an even number. The plaintiff is claiming that since there are so many even numbers, it is not possible for the defendant to test them all and make this claim.
Earl to the rescue! On the day of the trial, Earl tells the jury that he will prove, beyond a shadow of doubt, that what his client says is true.
Earl explains that by definition, an even number is a number of the form 2k, where k is an integer. He tells the jury to consider any two even numbers. By definition, they would have the form 2m and 2n, where m and n are integers. If we multiply these two numbers together, we get the following:
• (2m)(2n)
Remove parentheses using the associative property
• 2 × m × 2 × n
Rearrange using the commutative property
• 2 × 2 × m × n
Add in parentheses using the associative property
• 2(2mn)|
We end up with 2(2mn). Since m and n are integers, it follows that 2mn is also an integer, because integers are closed under multiplication. Therefore, we have that the product of any two even numbers is equal to 2 times an integer (the definition of an even number). Thus, the product of any two even numbers is also an even number.
Earl just won the case for his client.
## Proving Theorems
Earl's argument is called proving a theorem in mathematics. Proving a theorem uses properties, other proven theorems, and rules to prove mathematical theorems. In Earl's proof, he used quite a few different number properties, properties about sets of numbers that are always true.
In general, number properties are a great tool to help us to prove theorems. Let's take a look at a couple of examples.
## Example 1: Odd Number Subtraction
Suppose we are given that a + b is odd, where a and b are integers, and we want to prove that a - b is also odd. This is another example of when number properties come in really handy! For this proof, we are going to use the following number properties:
1. An odd number is a number that can be written as 2k + 1, where k is some integer.
2. If a = b, then a ± c = b ± c.
3. If a = b, then we can substitute a for b or b for a in any expression.
4. a(b ± c) = ab ± ac
5. The integers are closed under subtraction.
We are given that a + b is odd, so by property 1, a + b = 2k + 1, where k is some integer. We want to show that a - b is also odd, so we want to show that a - b = 2m + 1 for some integer m.
a + b = 2k + 1 By property 1 a = 2k - b + 1 By property 2 a - b = (2k - b + 1) - b = 2k - 2b + 1 By property 3 and simplifying a - b = 2k - 2b + 1 = 2(k - b) + 1 By property 4
We have that a - b = 2(k - b) + 1, and by property 5, since k and b are integers, k - b is an integer, call it m. Therefore, a - b is of the form 2m + 1, where m is an integer, so by property 1, a - b is odd.
So neat! Let's consider one more example!
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# Algebra 2 Review Packet
## Evaluate each expression using order of operations
1. 90 ÷3 + 5 _____________ 2. 9 – 3 ÷ 4 + 2 · 12 + 6 ÷ 2 · 3 __________
3. 4 + 1 · 4 2 – 3 __________ 4. 6 + 3 3 – 18 ÷ 6 ___________
## 11. 3 ( f + g ) + 7g _______________ 12. x + y – ( 3z + y ) _______________
13. ( 3 – r) + ( 4r – 3s + 2 ) – ( 1 – s ) _________________
## 14. 2 ( 3x – 5y) – ( 2x – 3y – z ) + ( y – z ) __________________
15. – 3 ( 2 – 7m ) ______________
-6 8
## 18. 4x + 7 = 3x + 18 ___________ 19. 5 ( 3x + 5) = 4x – 8 _____________
20. 2 ( y + 2) + y = 19 – ( 2y + 3) ______________
## 23. 2 ( 3h – 1 ) + 4h = 10 ( 2 – 3h) + 38h ______________
24. 3 ( x – 3 ) – 7 ( x + 3 ) = 4 ( 2x – 3 ) – 8 ( 2x + 3) ________________
## 25. 9 ( 2 – y ) + 3 ( 5 + 2y) = 2 ( 7 – 2y) – 4 ( y – 1 ) ________________
Find the slope of the line containing each pair of points.
Recall that the slope formula is : m = y2 – y 1
x2 – x1
## 26. ( 3,9) and ( 1,5) _________ 27. ( -1,3) and (5,3)_____________
28. ( -9,9) and 7,-2) ____________ 28. ( -3,-8) and ( -6,-2) __________
## 33. parallel to the line 2x – 3y = 9 and having a y intercept of 3________________
34. perpendicular to the line y – 4x = 2 and passing through the point ( -1,-2)
_______________________
35. parallel to the line y = 5, and passing through the point ( 6, -7) ______________
## Solve each inequality and graph the solution on a number line.
36. x – 5 < - 2
37. – 5x ≥ 2x – 6
## 40. –10 < 2x – 2 < - 4
Simplify each product
## 46. – 16 x 7 r 2 47. a 5 b 5 c5 48. ( x 3 y ) 2
-4xr - a 2 b 3 c4 x3y4
## 54. 3f 2 – f + 7f and 2f 2 – 3f – 4 ___________________________
55. 7x 3 – 3x + 2 – x 2 + 2 ________________________
## Find the difference of the polynomials
56. 10 a 2 + 2a – 8 - ( 9a 2 + 8 ) ______________________
## Find the product of the polynomials
58. 7x 2 y ( 8 x 3 – 3 x y 4) ______________________
59. ( x – 3 ) ( x + 8 ) ________________
60. ( x – 4 ) 2 __________________
61. ( 3x – 1 ) ( 2x + 3 ) _________________
62. ( 2x + 1 ) 2____________________
## Factor each completely:
63. 3 q 2 + 6 q 5 + 9 q __________________
64. x 2 – 4 ____________________
65. 10 w z 2 – 5 wz + 15 w 2z ___________________
66. x 2 – 6x + 9 ______________________
67. 25 a 2 – 1 ___________________
68. b 4 – 2b 2 + b 2 – 2 _____________________
69. g 2 – 3g – 40 ___________________
## 70. x 2 + 17x + 60 _________________
71. x 2 + 9x – 36 ______________________
## Solve using the zero product property
72 ( x – 1 ) ( x + 2 ) = 0
73. 2x ( x – 7 ) = 0
74. x2 – 2x – 15 = 0
75. x 2 – 25 = 0
## 79. y–7=x+2 80. y = 2/3 x + 1
FOR ALGEBRA 2 HONORS STUDENTS ONLY
81. Graph the absolute value functions listed below. For each, state the domain and
range.
a. y = | x – 3 |
b. y = - | x + 1 | - 2
c. y = 2 | x – 1 | - 4
82. Solve the system of equations using BOTH elimination and substitution methods
a. 3x + y = 13 b. y = - 3x
2x – y = 2 x = 6y + 38
## 83. Classify the following systems
a. 2x + 3y = 4 b. 4x – 3y = 12
8 – 6y = 4x 12x + 9y = - 36
84. Mary has a total of 30 dimes and nickels in his pocket worth \$ 2.35. How many of
each coin does he have ?
85. How many ounces of a 25 % acid solution should be mixed with a 50 % acid
solution to produce 100 ounces of a 40 % acid solution?
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Segments from Secants
## Relationships of products of sections versus sums of sections of secants.
0%
Progress
Practice Segments from Secants
Progress
0%
Segments from Secants
What if you wanted to figure out the distance from the orbiting moon to different locations on Earth? At a particular time, the moon is 238,857 miles from Beijing, China. On the same line, Yukon is 12,451 miles from Beijing. Drawing another line from the moon to Cape Horn (the southernmost point of South America), we see that Jakarta, Indonesia is collinear. If the distance from Cape Horn to Jakarta is 9849 miles, what is the distance from the moon to Jakarta? After completing this Concept, you'll be able to solve problems like this.
### Guidance
In addition to forming an angle outside of a circle, the circle can divide the secants into segments that are proportional with each other.
If we draw in the intersecting chords, we will have two similar triangles.
From the inscribed angles and the Reflexive Property (RR),PRSTRQ\begin{align*}( \angle R \cong \angle R), \triangle PRS \sim \triangle TRQ\end{align*}. Because the two triangles are similar, we can set up a proportion between the corresponding sides. Then, cross-multiply. ac+d=ca+ba(a+b)=c(c+d)\begin{align*}\frac{a}{c+d}=\frac{c}{a+b} \Rightarrow a(a+b)=c(c+d)\end{align*}
Two Secants Segments Theorem: If two secants are drawn from a common point outside a circle and the segments are labeled as above, then a(a+b)=c(c+d)\begin{align*}a(a+b)=c(c+d)\end{align*}. In other words, the product of the outer segment and the whole of one secant is equal to the product of the outer segment and the whole of the other secant.
#### Example A
Find the value of the missing variable.
Use the Two Secants Segments Theorem to set up an equation. For both secants, you multiply the outer portion of the secant by the whole.
18(18+x)=16(16+24)324+18x=256+38418x=316x=1759
#### Example B
Find the value of the missing variable.
Use the Two Secants Segments Theorem to set up an equation. For both secants, you multiply the outer portion of the secant by the whole.
x(x+x)=9322x2=288x2=144x=12
x12\begin{align*}x \neq -12\end{align*} because length cannot be negative.
#### Example C
True or False: Two secants will always intersect outside of a circle.
This is false. If the two secants are parallel, they will never intersect. It's also possible for two secants to intersect inside a circle.
Watch this video for help with the Examples above.
#### Concept Problem Revisited
The given information is to the left. Let’s set up an equation using the Two Secants Segments Theorem.
238857251308600266749560Use the Quadratic Formula xx=x(x+9849)=x2+9849x=x2+9849x600266749569849±984924(60026674956)2240128.4 miles
### Vocabulary
A circle is the set of all points that are the same distance away from a specific point, called the center. A radius is the distance from the center to the circle. A chord is a line segment whose endpoints are on a circle. A diameter is a chord that passes through the center of the circle. The length of a diameter is two times the length of a radius. A central angle is the angle formed by two radii and whose vertex is at the center of the circle. An inscribed angle is an angle with its vertex on the circle and whose sides are chords. The intercepted arc is the arc that is inside the inscribed angle and whose endpoints are on the angle. A tangent is a line that intersects a circle in exactly one point. The point of tangency is the point where the tangent line touches the circle. A secant is a line that intersects a circle in two points.
### Guided Practice
Find x\begin{align*}x\end{align*} in each diagram below. Simplify any radicals.
1.
2.
3.
Use the Two Secants Segments Theorem.
1.
8(8+x)64+8x8xx=6(6+18)=144=80=10
2.
4(4+x)16+4x4xx=3(3+13)=48=32=8
3.
15(15+27)630x=x45=45x=14
### Practice
Solve for the missing segment.
Find x\begin{align*}x\end{align*} in each diagram below. Simplify any radicals.
1. Prove the Two Secants Segments Theorem.
Given: Secants PR¯¯¯¯¯\begin{align*}\overline{PR}\end{align*} and \begin{align*}\overline{RT}\end{align*}
Prove: \begin{align*}a(a+b)=c(c+d)\end{align*}
Solve for the unknown variable.
### Vocabulary Language: English Spanish
central angle
central angle
An angle formed by two radii and whose vertex is at the center of the circle.
chord
chord
A line segment whose endpoints are on a circle.
diameter
diameter
A chord that passes through the center of the circle. The length of a diameter is two times the length of a radius.
Inscribed Angle
Inscribed Angle
An inscribed angle is an angle with its vertex on the circle. The measure of an inscribed angle is half the measure of its intercepted arc.
intercepted arc
intercepted arc
The arc that is inside an inscribed angle and whose endpoints are on the angle.
point of tangency
point of tangency
The point where the tangent line touches the circle.
AA Similarity Postulate
AA Similarity Postulate
If two angles in one triangle are congruent to two angles in another triangle, then the two triangles are similar.
Congruent
Congruent
Congruent figures are identical in size, shape and measure.
Reflexive Property of Congruence
Reflexive Property of Congruence
$\overline{AB} \cong \overline{AB}$ or $\angle B \cong \angle B$
Secant
Secant
The secant of an angle in a right triangle is the value found by dividing length of the hypotenuse by the length of the side adjacent the given angle. The secant ratio is the reciprocal of the cosine ratio.
secant line
secant line
A secant line is a line that joins two points on a curve.
Tangent line
Tangent line
A tangent line is a line that "just touches" a curve at a single point and no others.
Two Secants Segments Theorem
Two Secants Segments Theorem
Two secants segments theorem states that if you have a point outside a circle and draw two secant lines from it, there is a relationship between the line segments formed.
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How Sudoku Works
Solving a Sudoku: Simple Logic
Michael Mepham's "Book of Sudoku 3" rates this puzzle as "gentle."
There is no right place to start a sudoku puzzle. You can shut your eyes and put your finger on the puzzle and start there, and that's as correct a place as any. Probably the most logical place to start, though, is at a row, column or box that has a lot of numbers in it. Let's take a look at the puzzle from the previous page:
Columns 4 and 6 each have six numbers filled in. Let's start the puzzle at column 4, which already has its 1, 3, 4, 5, 8 and 9.
In order to have one and only one of each digit from 1 to 9, we're going to have to provide column 4 with its 2, its 6 and its 7. But we can't just put them anywhere -- each number has a specific location in the puzzle's answer. So where does each number go? To find out, we need to look at the rows and boxes that interact with column 4. Take a look at the empty square at row 3, column 4 (3,4), and the row and box that interact with it:
To fill the empty square at row 3, column 4, we're going to have to look at column 4, row 3 and box 2.
The "simple logic" approach to sudoku requires only visual analysis and goes something like this: Can the 2 go in the empty square? It can't, because box 2 already has a 2, and it can only have one of each number. Can the 7 go there? Row 3 already has its 7, so we can't put a 7 there, either. That leaves us with the 6. Neither row 3 nor box 2 already has a 6, so we know the 6 is correct for that cell. We've solved our first number!
Now let's solve the rest of column 4, which still needs its 2 and its 7. The empty square at 5,4 interacts with row 5 and box 5, and the empty square at 7,4 interacts with row 7 and box 8.
Since box 5 already has its 7, we can't put a 7 in the 5,4 square. So right there we know the 2 goes at 5,4, and the 7 must go at 7,4:
We've now solved all of column 4, and we used only simple logic to do it. Since this is an easy puzzle, we could probably solve a good portion of it this way. But it's not always so clear-cut. There are strategies we can use when the solution is not so obvious, and it all starts with some little pencil marks.
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 4.1: Differentiation and Integration of Vector Valued Functions
### Calculus of Vector Valued Functions
The formal definition of the derivative of a vector valued function is very similar to the definition of the derivative of a real valued function.
Definition
The Derivative of a Vector Valued Function
Let $$r(t)$$ be a vector valued function, then
$r'(t) = \lim_{h \rightarrow 0} \dfrac{r(t+h)-r(t)}{h}$
Because the derivative of a sum is the sum of the derivative, we can find the derivative of each of the components of the vector valued function to find its derivative.
Example 1
$\dfrac{d}{dt} (3 \hat{\text{i}} + \sin t \hat{\text{j}}) = \cos t \hat{\text{j}}$
$\dfrac{d}{dt} \left(3t^2\, \hat{\text{i}} + \cos{(4t)}\, \hat{\text{j}} + te^t \, \hat{\text{k}} \right) = 6t \, \hat{\text{i}} -4\sin{(t)}\,\hat{\text{j}} + (e^t + te^t)\, \hat{\text{k}}$
### Properties of Vector Valued Functions
All of the properties of differentiation still hold for vector values functions. Moreover because there are a variety of ways of defining multiplication, there is an abundance of product rules.
Suppose that $$\text{v}(t)$$ and $$\text{w}(t)$$ are vector valued functions, $$f(t)$$ is a scalar function, and $$c$$ is a real number then
1. $$\dfrac{d}{dt} \left( \text{v}(t) + \text{w}(t) \right) = \dfrac{d}{dt}\text{v}(t) + \dfrac{d}{dt} \text{w}(t)$$
2. $$\dfrac{d}{dt} c\text{v}(t) = c\, \dfrac{d}{dt} \text{v}(t)$$
3. $$\dfrac{d}{dt}(f(t) \text{v}(t)) = f '(t) \text{v}(t) + f(t) \text{v}(t)'$$
4. $$\left( v(t) \cdot \text{w}(t) \right)' = \text{v}'(t) \cdot \text{w}(t)+ \text{v}(t) \cdot \text{w}'(t)$$
5. $$(v(t) \times \text{w}(t))' = \text{v}'(t) \times \text{w}(t) + \text{v}(t) \times \text{w}'(t)$$
6. $$\dfrac{d}{dt} v(f(t)) = \text{v}(t)'(f(t)) f '(t)$$
Example 2
Show that if $$r$$ is a differentiable vector valued function with constant magnitude, then
$r \cdot r' = 0$
Solution
Since $$r$$ has constant magnitude, call its magnitude $$k$$,
$k^2 = |r|^2 = r \cdot r$
Taking derivatives of the left and right sides gives
$0 = (r \cdot r)' = r' \cdot r + r \cdot r'$
$= r \cdot r' + r \cdot r' = 2r \cdot r'$
Divide by two and the result follows
### Integration of vector valued functions
We define the integral of a vector valued function as the integral of each component. This definition holds for both definite and indefinite integrals.
Example 2
Evaluate
$\int (\sin t)\, \hat{\textbf{i}} + 2t\, \hat{\textbf{j}} - 8t^3 \, \hat{\textbf{k}} \; dt$
Solution
Just take the integral of each component
$\int (\sin t)\,dt \, \hat{\textbf{i}} + \int 2\,t \, dt \, \hat{\textbf{j}} - \int 8\,t^3 \,dt \, \hat{\textbf{k}}$
$= (-\cos t + c_1)\, \hat{\textbf{i}} + (t^2 + c_2)\, \hat{\textbf{j}} + (2\,t^4 + c_3)\, \hat{\textbf{k}}$
Notice that we have introduce three different constants, one for each component.
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# Graphs of Logarithmic Functions
Graph
y = log
a
x Note:
log
log
y
a
x
a
y x a x
x y a y
= · =
= · =
Since this is an inverse of an exponential function, the graph is a reflection of the
exponential function y = in the line y = x.
Since
log
a
1= 0, the function passes through the point ____________. (This corresponds to the point
(0, 1) on the exponential function, since
1= a
0
.)
The asymptote to the function
y = log
a
x is _____________________ .
(This corresponds to the asymptote of the exponential function
y = a
x
, which is _______________.)
Since
x = 0 is an asymptote to the function, we cannot find 0 log
a
. (It does not exist.)
Similarly,
log
a
x for
x < 0 does not exist.
Therefore, we have the following restrictions on the function
y = log
a
x.
Summary:
1) Considering the restrictions on the base a for exponential functions, we see that
y = log
a
x is defined only if ________________________.
2) As for the variable,
y = log
a
x is defined only for ________________.
In other words, the number x must be strictly positive. You cannot take the log of a negative number
and end up with a real answer.
Example 1:
Graph
y = logx and
y = log
2
x on the same set of coordinate axes.
What is the likely position of
y = log
5
x ? Sketch
y = log
5
x as well.
Example 2:
Graph
y ÷ 4 =log
2
x +3 ( ) .
The transformations that apply to the quadratic and exponential functions apply in the same manner to
the logarithmic function.
Strategy: Graph
y = log
2
x and shift everything (including the origin):
a) horizontally ________________ b) vertically ___________ ________
The following key features of
y = log
2
x will be translated.
1 (1, 0) shifts to_________ 2 (2, 1) shifts to _________
3 (4, 2) shifts to__________ 4 (8, 3) shifts to _________
5 (16, 4) shifts to ________
6 The asymptote
x = 0 shifts 3 units to the left, and so the asymptote for the function
y ÷ 4 =log
2
x +3 ( ) becomes the line __________
7 If we pretend that a “new origin” exists at (–3, 4), [a horizontal translation of the first origin 3 units
to the left and a vertical translation 4 units up], we can then draw a graph of the function x y
2
log =
Example 3: Draw the function
y =÷log
2
x .
Example 4:
Analyze the graph of
y =log 2x +3 ( ) . Identify the domain, range, asymptotes and intercepts.
Example 8:
Determine the domain of the function
y =log
2x÷3
(x).
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These activity sheets will introduce your beginning geometry students to transversal lines, including calculating angles.
#### What Are Transversal Lines? A transversal is when two parallel lines are intersected by the third line at an angle. The line intersecting the two parallel lines; the third line is known as the transversal line. We get different types of angles when a transversal line passes through the parallel lines. Some of the commonly known angles are defined below: Supplementary angles - When pair angles are added together and give us a measure of 180 degrees, we get supplementary angles. When we put the pair of supplementary angles together, we can draw a straight line across the two angles. Or simply put, the two angles, when put together, will make a half-circle. Also, the supplementary angles aren't limited to transversals. Interior Angles - As the name implies, angles present interiorly, on the inside, of the two parallel lines are known as the interior angles. Exterior Angles - Angles present on the outside of the two parallel lines are known as the exterior angles. Corresponding Angles - Angles that are present on the same side of the transversal line are known as the corresponding angles. In corresponding angles, one angle is an interior angle while the other is the exterior angle; these angles are equal in measurement and are congruent. Whenever you go on to study math, do you think that the subject is boring? Well, everyone thinks that there is not a single thing about math that makes sense and could b fun or interesting to study about. But that's not true! Math has its fair share of exciting topics that is actually very interesting and very fun to learn about!
When you come across a line that breaks across two other lines it is called a transversal line. When a transversal cuts across parallel lines we quick determine the value of all the angles by just knowing the measure of one of the angles. Vertical opposite angles (the angle opposite the known angle) are equal. Supplementary angles will then add to one-hundred and eighty degrees. That gives us all the measures of the angles around the known angle. We can then identify an alternate exterior or interior angle pair. There measures will be equal. We can then repeat this procedure to determine all angles around that line and the cutting line. These worksheets explains how to find and determine the angle of a transversal. Instructors please note that some examples may not contain any transversals.
# Print Parallel Line Transversal Worksheets
## Parallel Line Transversals Lesson
This worksheet explains how determine the angle of a transversal. A sample problem is solved.
## Lesson and Practice
This worksheet reviews how to determine the angle of a transversal. A sample problem is solved, and two practice problems are provided.
## Transversals Worksheet
Students will determine the angle of a transversal. Ten descriptions are provided.
## Practice Sheets
Given drawings, students will determine the angle of a transversal. Ten problems are provided.
## Transversals Drill
Students will determine the angle of a transversal. Eight problems are provided.
## Transversals Warm Up
Students will determine the angle of a transversal. Three problems are provided.
## Recognizing Transversals Lesson
This worksheet explains how to locate the transversal in a given figure. A sample problem is solved.
## Lesson and Practice
This worksheet reviews how to locate a transversal in a figure. A sample problem is solved, and two practice problems are provided.
## Recognizing Parallel Line Transversal Worksheet
Students will find the transversal. Ten descriptions are provided.
## Recognizing Practice
Given drawings, students will locate the transversals. Ten problems are provided.
## Recognizing Parallel Line Transversal Drill
Students will locate the transversals in each drawing. Eight problems are provided.
## Recognizing Warm Up
Students will find the transversals across parallel lines. Three problems are provided.
## Basic Skills
Students will find the transversals. Ten problems are provided.
## Basic Skills Review
Where the heck are all the transversals?
## Transversals Intermediate Skills
Find where they meet up and get them dinner.
## Intermediate Skills Review
I wish I would have found where those are missing and where they hook up.
## Proficiency with Transversals
Students will become proficient in finding transversals.
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# Ncert Solutions for Class 6 Maths Fraction Exercise 7.1 and 7.2
Assignments Ncert Solutions Revision Notes
In this page we have Ncert Solutions for Class 6 Maths Fraction Exercise 7.1 and 7.2. Hope you like them and do not forget to like , social share and comment at the end of the page.
# Exercise 7.1
Question 1
Write the fraction representing the shaded portion.
(i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x)
Question 2
Colour the part according to the given fraction.
(i) (ii) (iii) (iv) (v)
i) ii) iii) iv) v)
Question 3
Identify the error if any.
The given figures do not represent the correct fractions as the part are unequal
Question 4
What fraction of a day is 8 hours?
We know that in a day there are 24 hours. So, 8 hours in a 24-hour day is represented as
8/24
Question 5
What fraction of an hour is 40 minutes?
We know that in an hour there are 60 minutes. So, 40 minutes in a 60 min is represented as
40/60
Question 6
Arya, Abhimanyu, and Vivek shared lunch. Arya has brought two sandwiches, one made of vegetable and one of jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that each person will have an equal share of each sandwich.
(a) How can Arya divide his sandwiches so that each person has an equal share?
(b) What part of a sandwich will each boy receive?
a) Since there are three people to eat, Arya should divide his sandwiches into 3 equal parts and give 1 part of each sandwich so that each person has an equal share.
b) Each boy will receive 1/3 of a sandwich.
Question 7
Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?
Kanchan has finished 20 dresses out of 30. So Fraction would be 20/30
Question 8
Write the natural numbers from 2 to 12. What fraction of them are prime numbers?
Natural numbers from 2 to 12 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12
Prime numbers – 2, 3, 5, 7, 11
There are 11 natural numbers from 2 to 12 out of which 5 are prime numbers.
Fraction of numbers that are prime is 5/11
Question 9
Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Natural numbers from 102 to 113 are 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112 and 113
Prime numbers – 103, 107, 109, 113
There are 12 natural numbers from 102 to 113 out of which 4 are prime numbers.
So, fraction of numbers that are prime is 4/12
Question 10
What fractions of these circles have X’s in them?
There are 8 circles out of which 4 have X’s in them.
So, fraction of circles having X’s is 4/8
Question 11
Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?
Number of CDs Kristin purchased = 3
Number of CDs she received as gifts = 5
So, total number of CDs = 8
Kristin purchased 3 CDs out of a total number of 8 CDs.
Fraction of CDs she purchased = 3/8
Kristin received 5 CDs out of a total number of 8 CDs.
Fraction of CDs she received as gifts =5/8
# Exercise 7.2
Question 1
Draw number lines and locate the points on them:
(a)
(b)
(c)
a) These fraction lies between 0 and 1 in the number line. We can divide the distance between 0 and 1 on the number into four equal part, then each point will represent these fractions
b) These fraction lies between 0 and 1 in the number line. We can divide the distance between 0 and 1 on the number into 8 equal part, then each point will represent these fractions
c) These fraction lies between 0 and 1 in the number line. We can divide the distance between 0 and 1 on the number into 5 equal part, then each point will represent these fractions
Question 2
Express the following as mixed fractions:
(a) 20/3
(b)11/5
(c)17/7
(d)28/5
(e)19/6
(f)35/9
A mixed fraction is a combination of whole and part
To convert an improper fraction to a mixed fraction, follow these steps:
1.Divide the numerator by the denominator.
2.Write down the whole number answer.
3.Then write down any remainder above the denominator.
S.no Question Calculation Answer a) 20/3 (18+2)/3= 6+ 2/3 b) 11/5 (10+1)/5= 2+1/5 c) 17/7 (14+3)/7=2 +3/7 d) 28/5 (25+3)/5= 5+3/5 e) 19/6 (18+1)/6= 3+ 1/6 f) 35/9 (27+8)/9= 3+ 8/9
Question 3
Express the following as improper fractions:
(a)
(b)
(c)
(d)
(e)
(f)
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# Why can you chose how to align infinitely long equations when adding them?
I saw in a video this proof:
Take this equation:
$$f=1+\frac12+\frac14+\cdots$$
and do this:
\begin{align} f&=1+1/2+1/4+\cdots\\ -\quad f/2&=\quad\:\:\:1/2+1/4+\cdots\\ \hline f/2&=1+0+0+\cdots \end{align}
This bugs me, and here's why.
$$f/2=1/2+1/4+\cdots$$
We took this equation, but shifted it 1 spot over, effectively making it smaller.
Here's what I mean by shifting. Here's the same thing, but without anything shifted.
\begin{align} f&=\:\:\:\:1+1/2+1/4+\cdots\\ -\quad f/2&=1/2+1/4+1/8+\cdots\tag{1}\\ \hline f/2&=1/2+1/4+1/8+\cdots \end{align}
1: Now this part isn't shifted anymore.
Basically by shifting, I'm talking about how we align the equations. I don't like the shifting part and here's why. If we take the equation $f=1+1+1+\cdots$ and do a similar thing, first without shifting.
\begin{align} f&=1+1+1+\cdots\\ -\quad f&=1+1+1+\cdots\\ \hline 0&=0+0+0+\cdots \end{align}
Makes sense, now let's shift it.
\begin{align} f&=1+1+1+\cdots\\ -\quad f&=\quad\:\:\:1+1+\cdots\\ \hline 0&=1+0+0+\cdots \end{align}
This doesn't make any sense. $0$ doesn't equal $1$. If we were to shift it to the right a second time, we would get $0=2$.
So my question is: why is shifting allowed in that first case? Is there something I'm doing wrong in my example? Am I just not understanding infinity correctly? The way I see it, every time you shift an equation over, you lose a digit, making the answer incorrect.
• It's rather strange to call the left hand side of those equations $e$, since a different number is often defined by an infinite sum beginning similarly, $1+1+1/2+...$ – Kevin Carlson Feb 27 '15 at 18:20
• Oops, my bad :p. I'll edit the question. – QxQ Feb 27 '15 at 19:17
• @QxQ You missed the last two instances of e, but I can't do it because it's only a 2-character edit. – Aaron Dufour Feb 28 '15 at 3:19
We can only shift and rearrange infinite sums if both of them converge absolutely. Otherwise strange things happen, just like in your example.
• Which points out a frequent defect in "shift the terms" proofs, because they are presented with no inkling of the fact that they rely on absolute convergence. – David K Feb 27 '15 at 15:52
• @QxQ This is a link to some notes from an analysis course at University of Warwick that discuss the idea in some detail google.com/… I agree that it is a really weird concept at first :) – Johanna Feb 27 '15 at 16:21
• Arbitrary rearrangements do require absolute convergence to preserve the sum. But to address the OP, shifting only requires convergence. Also, adding series term-by-term only requires convergence. Moreover, some common summation methods for divergent series also respect these two operations, making them "stable" and "linear". Cesaro summation and Abel summation are good examples. So the operations are surprisingly forgiving! – Chris Culter Feb 27 '15 at 18:25
• @Johanna That link blew my mind about infinite sums and enlightened me. For all those budding Buddhas out there, the example they quote is that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7} - \frac{1}{8}+\cdots=\log(2)$ (alternating positive and negative terms), but $\left(1-\frac{1}{2}\right)-\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{6}\right) - \frac{1}{8} + \left(\frac{1}{5}-\frac{1}{10}\right)-\cdots=\frac{\log(2)}{2}$ (alternating two negative for each one positive term). More generally, conditionally-convergent sums can be designed to sum to any target you want! – Iwillnotexist Idonotexist Feb 28 '15 at 3:32
• @IwillnotexistIdonotexist I'm glad you liked it. It's fascinating, isn't it! – Johanna Feb 28 '15 at 14:46
I feel like the answers to this question are not appropriate because they address arbitrary rearrangements of series instead of shifting. Here is an answer that is just concerned with shifting. The key lies in two facts:
Fact 1: Let $S_1 = a_1 + a_2 + \cdots$ and $S_2 = b_1 + b_2 + \cdots$ be two convergent series. Then the sum $(a_1 + b_1) + (a_2 + b_2) + \cdots$ converges and is equal to $S_1 + S_2$.
Fact 2: The series $a_1 + a_2 + \cdots$ converges if and only if the series $0 + a_1 + a_2 + \cdots$ converges.
The second fact is really just justifying the "shift" operation for a series. This proves that if you take two convergent series, you can do all of the shifting, adding (or subtracting) that you want and you will still have a valid answer. Nothing about absolute convergence is necessary for this problem.
This question about the nature of $\infty - \infty$ might be somewhat relevant.
• Finite shifting, at least. :) – Ian Feb 28 '15 at 14:14
• @Ian How would you shift inifinitely? – heinrich5991 Mar 1 '15 at 12:52
• @heinrich5991 You can't shift one term an infinite amount, but you could shift infinitely many terms each a finite amount. For instance you could replace all terms with odd index with $0$. – Ian Mar 1 '15 at 16:46
You can prove that conditionally convergent series can be rearranged to converge to any sum while any rearrangement of the terms of the absolutely convergent series must converge to the same limit that the original sequence converges to.
For any infinite sum that's convergent, shifting it doesn't change its value and for any two convergent infinite sums, the difference of their sums is equal to the sum of the differences between the terms. That's true for all convergent infinite sums, not just absolutely convergent sums. What's not true about infinite sums that are convergent but not absolutely convergent is that changing their order can't affect the sum.
Once you've proven that the sum $1 + \frac{1}{2} + \frac{1}{4}...$ is convergent, you can prove using the facts I stated earlier that shifting it over then adding the differences in the terms gives you the result of 1 and therefore the sum of the original sequence is 2. The easiest way to prove that the sum of $1 + \frac{1}{2} + \frac{1}{4}...$ is convergent is probably by first proving it equals 2 so once you've proven that, you're done and there's no need to prove again a different way that the sum is 2.
When you define $f = 1 + 1 + 1...$, you're making the wrong assumption that the infinite sum exists and that that number is called $f$. From the wrong assumption that the sum exists, you can derive the contradiction that $f - f = 0$ and $f - f = 1$.
• "For any infinite sum that's convergent, shifting it doesn't change its value". This is the crux of the question. You may want to explain why, particularly more so since this is a 3-year-old question. – Andrés E. Caicedo Jun 21 '18 at 16:43
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# Solução_Calculo_Stewart_6e
Solução_Calculo_Stewart_6e
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F.<br />
TX.10<br />
<strong>Solução</strong> detalhada de todos os exercícios ímpares do livro: Cálculo, James <strong>Stewart</strong> - VOLUME I e VOLUME II<br />
"O que sabemos é uma gota, o que ignoramos é um oceano." Isaac Newton<br />
23/11/2010.TX
F.<br />
TX.10<br />
1 FUNCTIONS AND MODELS<br />
1.1 Four Ways to Represent a Function<br />
In exercises requiring estimations or approximations, your answers may vary slightly from the answers given here.<br />
1. (a) The point (−1, −2) is on the graph of f,sof(−1) = −2.<br />
(b) When x =2, y is about 2.8,sof(2) ≈ 2.8.<br />
(c) f(x) =2is equivalent to y =2.Wheny =2,wehavex = −3 and x =1.<br />
(d) Reasonable estimates for x when y =0are x = −2.5 and x =0.3.<br />
(e) The domain of f consists of all x-values on the graph of f. For this function, the domain is −3 ≤ x ≤ 3,or[−3, 3].<br />
The range of f consists of all y-values on the graph of f. For this function, the range is −2 ≤ y ≤ 3,or[−2, 3].<br />
(f ) As x increases from −1 to 3, y increases from −2 to 3. Thus,f is increasing on the interval [−1, 3].<br />
3. From Figure 1 in the text, the lowest point occurs at about (t, a) =(12, −85). The highest point occurs at about (17, 115).<br />
Thus, the range of the vertical ground acceleration is −85 ≤ a ≤ 115. Written in interval notation, we get [−85, 115].<br />
5. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, the curve fails<br />
the Vertical Line Test.<br />
7. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [−3, 2] and the range<br />
is [−3, −2) ∪ [−1, 3].<br />
9. The person’s weight increased to about 160 pounds at age 20 and stayed fairly steady for 10 years. The person’s weight<br />
dropped to about 120 pounds for the next 5 years, then increased rapidly to about 170 pounds. The next 30 years saw a gradual<br />
increase to 190 pounds. Possible reasons for the drop in weight at 30 years of age: diet, exercise, health problems.<br />
11. The water will cool down almost to freezing as the ice<br />
melts. Then, when the ice has melted, the water will<br />
slowly warm up to room temperature.<br />
13. Of course, this graph depends strongly on the<br />
geographical location!<br />
15. As the price increases, the amount sold decreases. 17.<br />
9
F.<br />
TX.10<br />
10 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />
19. (a) (b) From the graph, we estimate the number of<br />
cell-phone subscribers worldwide to be about<br />
92 million in 1995 and 485 million in 1999.<br />
21. f(x) =3x 2 − x +2.<br />
f(2) = 3(2) 2 − 2+2=12− 2+2=12.<br />
f(−2) = 3(−2) 2 − (−2)+2=12+2+2=16.<br />
f(a) =3a 2 − a +2.<br />
f(−a) =3(−a) 2 − (−a)+2=3a 2 + a +2.<br />
f(a +1)=3(a +1) 2 − (a +1)+2=3(a 2 +2a +1)− a − 1+2=3a 2 +6a +3− a +1=3a 2 +5a +4.<br />
2f(a) =2· f(a) =2(3a 2 − a +2)=6a 2 − 2a +4.<br />
f(2a) =3(2a) 2 − (2a)+2=3(4a 2 ) − 2a +2=12a 2 − 2a +2.<br />
f(a 2 )=3(a 2 ) 2 − (a 2 )+2=3(a 4 ) − a 2 +2=3a 4 − a 2 +2.<br />
[f(a)] 2 = 3a 2 − a +2 2 = 3a 2 − a +2 3a 2 − a +2 <br />
=9a 4 − 3a 3 +6a 2 − 3a 3 + a 2 − 2a +6a 2 − 2a +4=9a 4 − 6a 3 +13a 2 − 4a +4.<br />
f(a + h) =3(a + h) 2 − (a + h)+2=3(a 2 +2ah + h 2 ) − a − h +2=3a 2 +6ah +3h 2 − a − h +2.<br />
23. f(x) =4+3x − x 2 ,sof(3 + h) =4+3(3+h) − (3 + h) 2 =4+9+3h − (9 + 6h + h 2 )=4− 3h − h 2 ,<br />
and<br />
f(3 + h) − f(3)<br />
h<br />
= (4 − 3h − h2 ) − 4<br />
h<br />
=<br />
h(−3 − h)<br />
h<br />
= −3 − h.<br />
25.<br />
f(x) − f(a)<br />
x − a<br />
=<br />
1<br />
x − 1 a − x<br />
a<br />
x − a = xa<br />
x − a =<br />
a − x −1(x − a)<br />
=<br />
xa(x − a) xa(x − a) = − 1<br />
ax<br />
27. f(x) =x/(3x − 1) is defined for all x except when 0=3x − 1 ⇔ x = 1 , so the domain<br />
3<br />
is <br />
x ∈ R | x 6= 1 3 = −∞,<br />
1<br />
3 ∪ 1 , ∞ 3<br />
.<br />
29. f(t) = √ t + 3√ t is defined when t ≥ 0. These values of t give real number results for √ t, whereas any value of t gives a real<br />
number result for 3√ t.Thedomainis[0, ∞).<br />
31. h(x) =1 √ 4<br />
x 2 − 5x is defined when x 2 − 5x >0 ⇔ x(x − 5) > 0. Note that x 2 − 5x 6= 0since that would result in<br />
division by zero. The expression x(x − 5) is positive if x5. (See Appendix A for methods for solving<br />
inequalities.) Thus, the domain is (−∞, 0) ∪ (5, ∞).
F.<br />
TX.10<br />
SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 11<br />
33. f(x) =5is defined for all real numbers, so the domain is R,or(−∞, ∞).<br />
The graph of f is a horizontal line with y-intercept 5.<br />
35. f(t) =t 2 − 6t is defined for all real numbers, so the domain is R,or<br />
(−∞, ∞). The graph of f is a parabola opening upward since the coefficient<br />
of t 2 is positive. To find the t-intercepts, let y =0and solve for t.<br />
0=t 2 − 6t = t(t − 6) ⇒ t =0and t =6.Thet-coordinate of the<br />
vertex is halfway between the t-intercepts, that is, at t =3.Since<br />
f(3) = 3 2 − 6 · 3=−9,thevertexis(3, −9).<br />
37. g(x) = √ x − 5 is defined when x − 5 ≥ 0 or x ≥ 5,sothedomainis[5, ∞).<br />
Since y = √ x − 5 ⇒ y 2 = x − 5 ⇒ x = y 2 +5,weseethatg is the<br />
tophalfofaparabola.<br />
39. G(x) =<br />
<br />
3x + |x|<br />
x if x ≥ 0<br />
.Since|x| =<br />
x<br />
−x if x0<br />
if x0<br />
if x0<br />
2 if x
F.<br />
TX.10<br />
12 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />
47. We need to solve the given equation for y. x +(y − 1) 2 =0 ⇔ (y − 1) 2 = −x ⇔ y − 1=± √ −x ⇔<br />
y =1± √ −x. The expression with the positive radical represents the top half of the parabola, and the one with the negative<br />
radical represents the bottom half. Hence, we want f(x) =1− √ −x. Note that the domain is x ≤ 0.<br />
49. For 0 ≤ x ≤ 3, the graph is the line with slope −1 and y-intercept 3,thatis,y = −x +3.For3
F.<br />
TX.10<br />
SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS ¤ 13<br />
61. f is an odd function because its graph is symmetric about the origin. g is an even function because its graph is symmetric with<br />
respect to the y-axis.<br />
63. (a) Because an even function is symmetric with respect to the y-axis, and the point (5, 3) is on the graph of this even function,<br />
the point (−5, 3) must also be on its graph.<br />
(b) Because an odd function is symmetric with respect to the origin, and the point (5, 3) is on the graph of this odd function,<br />
the point (−5, −3) must also be on its graph.<br />
65. f(x) = x<br />
x 2 +1 .<br />
f(−x) =<br />
−x<br />
(−x) 2 +1 =<br />
So f is an odd function.<br />
−x<br />
x 2 +1 = − x<br />
x 2 +1 = −f(x).<br />
67. f(x) = x<br />
x +1 ,sof(−x) =<br />
−x<br />
−x +1 =<br />
x<br />
x − 1 .<br />
Sincethisisneitherf(x) nor −f(x), the function f is<br />
neither even nor odd.<br />
69. f(x) =1+3x 2 − x 4 .<br />
f(−x) =1+3(−x) 2 − (−x) 4 =1+3x 2 − x 4 = f(x).<br />
So f is an even function.<br />
1.2 Mathematical Models: A Catalog of Essential Functions<br />
1. (a) f(x) = 5√ x is a root function with n =5.<br />
(b) g(x) = √ 1 − x 2 is an algebraic function because it is a root of a polynomial.<br />
(c) h(x) =x 9 + x 4 is a polynomial of degree 9.<br />
(d) r(x) = x2 +1<br />
is a rational function because it is a ratio of polynomials.<br />
x 3 + x<br />
(e) s(x) =tan2x is a trigonometric function.<br />
(f ) t(x) =log 10 x is a logarithmic function.<br />
3. We notice from the figure that g and h are even functions (symmetric with respect to the y-axis) and that f is an odd function<br />
(symmetric with respect to the origin). So (b) y = x 5 must be f. Sinceg is flatter than h near the origin, we must have<br />
(c) y = x 8 matched with g and (a) y = x 2 matched with h.
F.<br />
TX.10<br />
14 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />
5. (a) An equation for the family of linear functions with slope 2<br />
is y = f(x) =2x + b,whereb is the y-intercept.<br />
(b) f(2) = 1 means that the point (2, 1) is on the graph of f. We can use the<br />
point-slope form of a line to obtain an equation for the family of linear<br />
functions through the point (2, 1). y − 1=m(x − 2), which is equivalent<br />
to y = mx +(1− 2m) in slope-intercept form.<br />
(c) To belong to both families, an equation must have slope m =2, so the equation in part (b), y = mx +(1− 2m),<br />
becomes y =2x − 3. Itistheonly function that belongs to both families.<br />
7. All members of the family of linear functions f(x) =c − x have graphs<br />
that are lines with slope −1. They-intercept is c.<br />
9. Since f(−1) = f(0) = f(2) = 0, f has zeros of −1, 0,and2,soanequationforf is f(x) =a[x − (−1)](x − 0)(x − 2),<br />
or f(x) =ax(x +1)(x − 2). Because f(1) = 6, we’ll substitute 1 for x and 6 for f(x).<br />
6=a(1)(2)(−1) ⇒ −2a =6 ⇒ a = −3, so an equation for f is f(x) =−3x(x +1)(x − 2).<br />
11. (a) D =200,soc =0.0417D(a +1)=0.0417(200)(a +1)=8.34a +8.34. The slope is 8.34, which represents the<br />
change in mg of the dosage for a child for each change of 1 year in age.<br />
(b) For a newborn, a =0,soc =8.34 mg.<br />
13. (a) (b) The slope of 9 means that F increases 9 degrees for each increase<br />
5 5<br />
of 1 ◦ C. (Equivalently, F increases by 9 when C increases by 5<br />
and F decreases by 9 when C decreases by 5.) The F -intercept of<br />
32 is the Fahrenheit temperature corresponding to a Celsius<br />
temperature of 0.
F.<br />
TX.10<br />
SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS ¤ 15<br />
15. (a) Using N in place of x and T in place of y,wefind the slope to be T 2 − T 1 80 − 70<br />
=<br />
N 2 − N 1 173 − 113 = 10<br />
equation is T − 80 = 1 (N − 173) ⇔ T − 80 = 1 N − 173 ⇔ T = 1 N + 307<br />
6 6 6 6 6<br />
60 = 1 . So a linear<br />
6<br />
307<br />
6<br />
=51.16 .<br />
(b) The slope of 1 means that the temperature in Fahrenheit degrees increases one-sixth as rapidly as the number of cricket<br />
6<br />
chirps per minute. Said differently, each increase of 6 cricket chirps per minute corresponds to an increase of 1 ◦ F.<br />
(c) When N =150, the temperature is given approximately by T = 1 307<br />
(150) + =76.16 ◦ F ≈ 76 ◦ F.<br />
6 6<br />
17. (a) We are given<br />
change in pressure<br />
10 feet change in depth = 4.34 =0.434. UsingP for pressure and d for depth with the point<br />
10<br />
(d, P )=(0, 15), we have the slope-intercept form of the line, P =0.434d +15.<br />
(b) When P = 100,then100 = 0.434d +15 ⇔ 0.434d =85 ⇔ d = 85<br />
0.434<br />
≈ 195.85 feet. Thus, the pressure is<br />
100 lb/in 2 at a depth of approximately 196 feet.<br />
19. (a) The data appear to be periodic and a sine or cosine function would make the best model. A model of the form<br />
f(x) =a cos(bx)+c seems appropriate.<br />
(b) The data appear to be decreasing in a linear fashion. A model of the form f(x) =mx + b seems appropriate.<br />
Some values are given to many decimal places. These are the results given by several computer algebra systems — rounding is left to the reader.<br />
21. (a)<br />
(b) Using the points (4000, 14.1) and (60,000, 8.2),weobtain<br />
8.2 − 14.1<br />
y − 14.1 = (x − 4000) or, equivalently,<br />
60,000 − 4000<br />
y ≈−0.000105357x +14.521429.<br />
A linear model does seem appropriate.<br />
(c) Using a computing device, we obtain the least squares regression line y = −0.0000997855x +13.950764.<br />
The following commands and screens illustrate how to find the least squares regression line on a TI-83 Plus.<br />
Enter the data into list one (L1) and list two (L2). Press<br />
to enter the editor.<br />
Find the regession line and store it in Y 1.Press .
F.<br />
TX.10<br />
16 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />
Note from the last figure that the regression line has been stored in Y 1 andthatPlot1hasbeenturnedon(Plot1is<br />
highlighted). You can turn on Plot1 from the Y= menu by placing the cursor on Plot1 and pressing<br />
or by<br />
pressing .<br />
Now press<br />
to produce a graph of the data and the regression<br />
line. Note that choice 9 of the ZOOM menu automatically selects a window<br />
that displays all of the data.<br />
(d) When x =25,000, y ≈ 11.456; or about 11.5 per 100 population.<br />
(e) When x =80,000, y ≈ 5.968;orabouta6% chance.<br />
(f ) When x = 200,000, y is negative, so the model does not apply.<br />
23. (a)<br />
(b)<br />
A linear model does seem appropriate.<br />
Using a computing device, we obtain the least squares<br />
regression line y =0.089119747x − 158.2403249,<br />
where x is the year and y is the height in feet.<br />
(c) When x = 2000,themodelgivesy ≈ 20.00 ft. Note that the actual winning height for the 2000 Olympics is less than the<br />
winning height for 1996—so much for that prediction.<br />
(d) When x =2100, y ≈ 28.91 ft. This would be an increase of 9.49 ft from 1996 to 2100. Even though there was an increase<br />
of 8.59 ft from 1900 to 1996, it is unlikely that a similar increase will occur over the next 100 years.<br />
25. Using a computing device, we obtain the cubic<br />
function y = ax 3 + bx 2 + cx + d with<br />
a =0.0012937, b = −7.06142, c =12,823,<br />
and d = −7,743,770. Whenx = 1925,<br />
y ≈ 1914 (million).
F.<br />
TX.10<br />
SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS ¤ 17<br />
1.3 New Functions from Old Functions<br />
1. (a) If the graph of f is shifted 3 units upward, its equation becomes y = f(x)+3.<br />
(b) If the graph of f is shifted 3 units downward, its equation becomes y = f(x) − 3.<br />
(c) If the graph of f is shifted 3 units to the right, its equation becomes y = f(x − 3).<br />
(d) If the graph of f is shifted 3 units to the left, its equation becomes y = f(x +3).<br />
(e) If the graph of f is reflected about the x-axis, its equation becomes y = −f(x).<br />
(f ) If the graph of f is reflected about the y-axis, its equation becomes y = f(−x).<br />
(g) If the graph of f isstretchedverticallybyafactorof3, its equation becomes y =3f(x).<br />
(h) If the graph of f is shrunk vertically by a factor of 3, its equation becomes y = 1 f(x). 3<br />
3. (a) (graph 3) The graph of f is shifted 4 units to the right and has equation y = f(x − 4).<br />
(b) (graph 1) The graph of f is shifted 3 units upward and has equation y = f(x)+3.<br />
(c) (graph 4) The graph of f is shrunk vertically by a factor of 3 and has equation y = 1 f(x). 3<br />
(d) (graph 5) The graph of f is shifted 4 units to the left and reflected about the x-axis. Its equation is y = −f(x +4).<br />
(e) (graph 2) The graph of f is shifted 6 units to the left and stretched vertically by a factor of 2. Its equation is<br />
y =2f(x +6).<br />
5. (a) To graph y = f(2x) we shrink the graph of f<br />
horizontally by a factor of 2.<br />
(b) To graph y = f 1<br />
2 x westretchthegraphoff<br />
horizontally by a factor of 2.<br />
The point (4, −1) on the graph of f corresponds to the<br />
point 1<br />
2 · 4, −1 =(2, −1).<br />
(c) To graph y = f(−x) we reflect the graph of f about<br />
the y-axis.<br />
The point (4, −1) on the graph of f corresponds to the<br />
point (2 · 4, −1) = (8, −1).<br />
(d) To graph y = −f(−x) we reflect the graph of f about<br />
the y-axis, then about the x-axis.<br />
The point (4, −1) on the graph of f corresponds to the<br />
point (−1 · 4, −1) = (−4, −1).<br />
The point (4, −1) on the graph of f corresponds to the<br />
point (−1 · 4, −1 · −1) = (−4, 1).
F.<br />
TX.10<br />
18 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />
7. The graph of y = f(x) = √ 3x − x 2 has been shifted 4 units to the left, reflected about the x-axis, and shifted downward<br />
1 unit. Thus, a function describing the graph is<br />
y = −1 ·<br />
<br />
reflect<br />
f (x +4)<br />
<br />
shift<br />
4 units left<br />
− 1<br />
<br />
shift<br />
1 unit left<br />
This function can be written as<br />
y = −f(x +4)− 1=− 3(x +4)− (x +4) 2 − 1=− 3x +12− (x 2 +8x +16)− 1=− √ −x 2 − 5x − 4 − 1<br />
9. y = −x 3 : Start with the graph of y = x 3 and reflect<br />
gives the same result since substituting −x for x gives<br />
us y =(−x) 3 = −x 3 .<br />
11. y =(x +1) 2 : Start with the graph of y = x 2<br />
and shift 1 unit to the left.<br />
13. y =1+2cosx: Start with the graph of y =cosx, stretch vertically by a factor of 2,andthenshift1 unit upward.<br />
15. y =sin(x/2): Start with the graph of y =sinx and stretch horizontally by a factor of 2.
F.<br />
TX.10<br />
SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS ¤ 19<br />
17. y = √ x +3: Start with the graph of<br />
y = √ x and shift 3 units to the left.<br />
19. y = 1 2 (x2 +8x) = 1 2 (x2 +8x +16)− 8= 1 2 (x +4)2 − 8: Start with the graph of y = x 2 , compress vertically by a<br />
factor of 2,shift4 units to the left, and then shift 8 units downward.<br />
0 0 0 0<br />
21. y =2/(x +1): Start with the graph of y =1/x,shift1 unit to the left, and then stretch vertically by a factor of 2.<br />
23. y = |sin x|: Start with the graph of y =sinx and reflect all the parts of the graph below the x-axis about the x-axis.<br />
25. This is just like the solution to Example 4 except the amplitude of the curve (the 30 ◦ NcurveinFigure9onJune21)is<br />
14 − 12 = 2. So the function is L(t) =12+2sin 2π<br />
365 (t − 80) . March 31 is the 90th day of the year, so the model gives<br />
L(90) ≈ 12.34 h. The daylight time (5:51 AM to 6:18 PM)is12 hours and 27 minutes, or 12.45 h. The model value differs<br />
from the actual value by 12.45−12.34<br />
12.45<br />
≈ 0.009,lessthan1%.<br />
27. (a) To obtain y = f(|x|), the portion of the graph of y = f(x) to the right of the y-axisisreflected about the y-axis.<br />
(b) y =sin|x|<br />
(c) y = |x|
F.<br />
TX.10<br />
20 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />
29. f(x) =x 3 +2x 2 ; g(x) =3x 2 − 1. D = R for both f and g.<br />
(f + g)(x) =(x 3 +2x 2 )+(3x 2 − 1) = x 3 +5x 2 − 1, D = R.<br />
(f − g)(x) =(x 3 +2x 2 ) − (3x 2 − 1) = x 3 − x 2 +1, D = R.<br />
(fg)(x) =(x 3 +2x 2 )(3x 2 − 1) = 3x 5 +6x 4 − x 3 − 2x 2 , D = R.<br />
<br />
f<br />
(x) = x3 +2x 2<br />
g 3x 2 − 1 , D = x | x 6=± √ 1 <br />
since 3x 2 − 1 6= 0.<br />
3<br />
31. f(x) =x 2 − 1, D = R; g(x) =2x +1, D = R.<br />
(a) (f ◦ g)(x) =f(g(x)) = f(2x +1)=(2x +1) 2 − 1=(4x 2 +4x +1)− 1=4x 2 +4x, D = R.<br />
(b) (g ◦ f)(x) =g(f(x)) = g(x 2 − 1) = 2(x 2 − 1) + 1 = (2x 2 − 2) + 1 = 2x 2 − 1, D = R.<br />
(c) (f ◦ f)(x) =f(f(x)) = f(x 2 − 1) = (x 2 − 1) 2 − 1=(x 4 − 2x 2 +1)− 1=x 4 − 2x 2 , D = R.<br />
(d) (g ◦ g)(x) =g(g(x)) = g(2x +1)=2(2x +1)+1=(4x +2)+1=4x +3, D = R.<br />
33. f(x) =1− 3x; g(x) =cosx. D = R for both f and g, and hence for their composites.<br />
(a) (f ◦ g)(x) =f(g(x)) = f(cos x) =1− 3cosx.<br />
(b) (g ◦ f)(x) =g(f(x)) = g(1 − 3x) =cos(1− 3x).<br />
(c) (f ◦ f)(x) =f(f(x)) = f(1 − 3x) =1− 3(1 − 3x) =1− 3+9x =9x − 2.<br />
(d) (g ◦ g)(x) =g(g(x)) = g(cos x) =cos(cosx)<br />
[Note that this is not cos x · cos x.]<br />
35. f(x) =x + 1 +1<br />
, D = {x | x 6= 0}; g(x) =x , D = {x | x 6=−2}<br />
x x +2<br />
x +1<br />
(a) (f ◦ g)(x) =f(g(x)) = f = x +1<br />
x +2 x +2 + 1 = x +1<br />
x +1 x +2 + x +2<br />
x +1<br />
x +2<br />
=<br />
(x +1)(x +1)+(x +2)(x +2)<br />
(x +2)(x +1)<br />
=<br />
x 2 +2x +1 + x 2 +4x +4 <br />
(x +2)(x +1)<br />
Since g(x) is not defined for x = −2 and f(g(x)) is not defined for x = −2 and x = −1,<br />
the domain of (f ◦ g)(x) is D = {x | x 6=−2, −1}.<br />
<br />
(b) (g ◦ f)(x) =g(f(x)) = g x + 1 <br />
=<br />
x<br />
<br />
x + 1 <br />
+1<br />
x<br />
<br />
x + 1 =<br />
+2<br />
x<br />
x 2 +1+x<br />
x<br />
x 2 +1+2x<br />
x<br />
Since f(x) is not defined for x =0and g(f(x)) is not defined for x = −1,<br />
the domain of (g ◦ f)(x) is D = {x | x 6=−1, 0}.<br />
<br />
(c) (f ◦ f)(x)=f(f(x)) = f x + 1 <br />
= x + 1 <br />
+ 1<br />
x x x + 1 x<br />
= x(x) x 2 +1 +1 x 2 +1 + x(x)<br />
x(x 2 +1)<br />
= x4 +3x 2 +1<br />
, D = {x | x 6= 0}<br />
x(x 2 +1)<br />
= x + 1 x + 1<br />
x 2 +1<br />
x<br />
= x4 + x 2 + x 2 +1+x 2<br />
x(x 2 +1)<br />
= x2 + x +1<br />
x 2 +2x +1 = x2 + x +1<br />
(x +1) 2<br />
= x + 1 x + x<br />
x 2 +1<br />
= 2x2 +6x +5<br />
(x +2)(x +1)
F.<br />
TX.10<br />
x +1<br />
(d) (g ◦ g)(x) =g(g(x)) = g =<br />
x +2<br />
x +1<br />
x +2 +1<br />
=<br />
x +1<br />
x +2 +2<br />
SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS ¤ 21<br />
x +1+1(x +2)<br />
x +2<br />
x +1+2(x +2)<br />
x +2<br />
Since g(x) is not defined for x = −2 and g(g(x)) is not defined for x = − 5 3 ,<br />
the domain of (g ◦ g)(x) is D = x | x 6=−2, − 5 3<br />
<br />
.<br />
37. (f ◦ g ◦ h)(x) =f(g(h(x))) = f(g(x − 1)) = f(2(x − 1)) = 2(x − 1) + 1 = 2x − 1<br />
39. (f ◦ g ◦ h)(x) =f(g(h(x))) = f(g(x 3 +2))=f[(x 3 +2) 2 ]<br />
= f(x 6 +4x 3 +4)= (x 6 +4x 3 +4)− 3= √ x 6 +4x 3 +1<br />
=<br />
x +1+x +2 2x +3<br />
=<br />
x +1+2x +4 3x +5<br />
41. Let g(x) =x 2 +1and f(x) =x 10 .Then(f ◦ g)(x) =f(g(x)) = f(x 2 +1)=(x 2 +1) 10 = F (x).<br />
43. Let g(x) = 3√ x and f(x) = x<br />
1+x .Then(f ◦ g)(x) =f(g(x)) = f( 3√ x )= 3√ x<br />
1+ 3√ x = F (x).<br />
45. Let g(t) =cost and f(t) = √ t.Then(f ◦ g)(t) =f(g(t)) = f(cos t) = √ cos t = u(t).<br />
47. Let h(x) =x 2 , g(x) =3 x ,andf(x) =1− x. Then<br />
<br />
(f ◦ g ◦ h)(x) =f(g(h(x))) = f(g(x 2 )) = f 3 x2 =1− 3 x2 = H(x).<br />
49. Let h(x) = √ x, g(x) =secx, andf(x) =x 4 .Then<br />
(f ◦ g ◦ h)(x) =f(g(h(x))) = f(g( √ x )) = f(sec √ x )=(sec √ x ) 4 =sec 4 ( √ x )=H(x).<br />
51. (a) g(2) = 5, because the point (2, 5) is on the graph of g. Thus,f(g(2)) = f(5) = 4, because the point (5, 4) is on the<br />
graph of f.<br />
(b) g(f(0)) = g(0) = 3<br />
(c) (f ◦ g)(0) = f(g(0)) = f(3) = 0<br />
(d) (g ◦ f)(6) = g(f(6)) = g(6). This value is not defined, because there is no point on the graph of g that has<br />
x-coordinate 6.<br />
(e) (g ◦ g)(−2) = g(g(−2)) = g(1) = 4<br />
(f ) (f ◦ f)(4) = f(f(4)) = f(2) = −2<br />
53. (a) Using the relationship distance = rate · time with the radius r as the distance, we have r(t) =60t.<br />
(b) A = πr 2 ⇒ (A ◦ r)(t) =A(r(t)) = π(60t) 2 = 3600πt 2 . This formula gives us the extent of the rippled area<br />
(in cm 2 )atanytimet.<br />
55. (a) From the figure, we have a right triangle with legs 6 and d, and hypotenuse s.<br />
By the Pythagorean Theorem, d 2 +6 2 = s 2 ⇒ s = f(d) = √ d 2 +36.<br />
(b) Using d = rt,wegetd =(30km/hr)(t hr) =30t (in km). Thus,<br />
d = g(t) =30t.<br />
(c) (f ◦ g)(t) =f(g(t)) = f(30t) = (30t) 2 +36= √ 900t 2 +36. This function represents the distance between the<br />
lighthouse and the ship as a function of the time elapsed since noon.
F.<br />
TX.10<br />
22 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />
57. (a)<br />
(b)<br />
H(t) =<br />
<br />
0 if t
F.<br />
TX.10<br />
1.4 Graphing Calculators and Computers<br />
SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS ¤ 23<br />
1. f(x) = √ x 3 − 5x 2<br />
(a) [−5, 5] by [−5, 5]<br />
(There is no graph shown.)<br />
(b) [0, 10] by [0, 2] (c) [0, 10] by [0, 10]<br />
The most appropriate graph is produced in viewing rectangle (c).<br />
3. Since the graph of f(x) =5+20x − x 2 is a<br />
parabola opening downward, an appropriate viewing<br />
rectangle should include the maximum point.<br />
5. f(x) = 4√ 81 − x 4 is defined when 81 − x 4 ≥ 0 ⇔<br />
x 4 ≤ 81 ⇔ |x| ≤ 3, so the domain of f is [−3, 3]. Also<br />
0 ≤ 4√ 81 − x 4 ≤ 4√ 81 = 3,sotherangeis[0, 3].<br />
7. The graph of f(x) =x 3 − 225x is symmetric with respect to the origin.<br />
Since f(x) =x 3 − 225x = x(x 2 − 225) = x(x + 15)(x − 15),there<br />
are x-intercepts at 0, −15,and15. f(20) = 3500.<br />
9. The period of g(x) = sin(1000x) is 2π ≈ 0.0063 and its range is<br />
1000<br />
[−1, 1]. Sincef(x) =sin 2 (1000x) is the square of g,itsrangeis<br />
[0, 1] and a viewing rectangle of [−0.01, 0.01] by [0, 1.1] seems<br />
appropriate.<br />
11. The domain of y = √ x is x ≥ 0,sothedomainoff(x) =sin √ x is [0, ∞)<br />
and the range is [−1, 1]. With a little trial-and-error experimentation, we find<br />
that an Xmax of 100 illustrates the general shape of f,soanappropriate<br />
viewing rectangle is [0, 100] by [−1.5, 1.5].
F.<br />
TX.10<br />
24 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />
13. The first term, 10 sin x, has period 2π and range [−10, 10]. It will be the dominant term in any “large” graph of<br />
y =10sinx +sin100x, as shown in the first figure. The second term, sin 100x,hasperiod 2π = π and range [−1, 1].<br />
100 50<br />
Itcausesthebumpsinthefirst figure and will be the dominant term in any “small” graph, as shown in the view near the<br />
origin in the second figure.<br />
15. We must solve the given equation for y to obtain equations for the upper and<br />
lower halves of the ellipse.<br />
4x 2 +2y 2 =1 ⇔ 2y 2 =1− 4x 2 ⇔ y 2 = 1 − 4x2<br />
2<br />
<br />
1 − 4x<br />
2<br />
y = ±<br />
2<br />
⇔<br />
17. From the graph of y =3x 2 − 6x +1<br />
and y =0.23x − 2.25 in the viewing<br />
rectangle [−1, 3] by [−2.5, 1.5],itis<br />
difficult to see if the graphs intersect.<br />
If we zoom in on the fourth quadrant,<br />
we see the graphs do not intersect.<br />
19. From the graph of f(x) =x 3 − 9x 2 − 4, we see that there is one solution<br />
of the equation f(x) =0and it is slightly larger than 9. By zooming in or<br />
using a root or zero feature, we obtain x ≈ 9.05.<br />
21. We see that the graphs of f(x) =x 2 and g(x) =sinx intersect twice. One<br />
solution is x =0. The other solution of f = g is the x-coordinate of the<br />
point of intersection in the first quadrant. Using an intersect feature or<br />
zooming in, we find this value to be approximately 0.88. Alternatively, we<br />
could find that value by finding the positive zero of h(x) =x 2 − sin x.<br />
Note: After producing the graph on a TI-83 Plus, we can find the approximate value 0.88 by using the following keystrokes:<br />
. The “1” is just a guess for 0.88.
F.<br />
TX.10<br />
SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS ¤ 25<br />
23. g(x) =x 3 /10 is larger than f(x) =10x 2 whenever x>100.<br />
25. We see from the graphs of y = |sin x − x| and y =0.1 that there are<br />
two solutions to the equation |sin x − x| =0.1: x ≈−0.85 and<br />
x ≈ 0.85. The condition |sin x − x| < 0.1 holds for any x lying<br />
between these two values, that is, −0.85
F.<br />
TX.10<br />
26 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />
33. y 2 = cx 3 + x 2 .Ifc0 (b) R (c) (0, ∞) (d) See Figures 4(c), 4(b), and 4(a), respectively.<br />
3. All of these graphs approach 0 as x →−∞, all of them pass through the point<br />
(0, 1), and all of them are increasing and approach ∞ as x →∞.Thelargerthe<br />
base, the faster the function increases for x>0, and the faster it approaches 0 as<br />
x →−∞.<br />
Note: The notation “x →∞” can be thought of as “x becomes large” at this point.<br />
More details on this notation are given in Chapter 2.<br />
5. The functions with bases greater than 1 (3 x and 10 x ) are increasing, while those<br />
with bases less than 1 <br />
1 x<br />
and <br />
1 x <br />
3<br />
10 are decreasing. The graph of 1<br />
x<br />
is the<br />
3<br />
reflection of that of 3 x about the y-axis, and the graph of 1<br />
10<br />
x<br />
is the reflection of<br />
that of 10 x about the y-axis. The graph of 10 x increases more quickly than that of<br />
3 x for x>0, and approaches 0 faster as x →−∞.<br />
7. We start with the graph of y =4 x (Figure 3) and then<br />
shift 3 units downward. This shift doesn’t affect the<br />
domain, but the range of y =4 x − 3 is (−3, ∞) .<br />
There is a horizontal asymptote of y = −3.<br />
y =4 x y =4 x − 3
F.<br />
TX.10<br />
SECTION 1.5 EXPONENTIAL FUNCTIONS ¤ 27<br />
9. We start with the graph of y =2 x (Figure 3),<br />
x-axis (or just rotate 180 ◦ to handle both<br />
reflections) to obtain the graph of y = −2 −x .<br />
In each graph, y =0is the horizontal<br />
asymptote.<br />
y =2 x y =2 −x y = −2 −x<br />
11. We start with the graph of y = e x (Figure 13) and reflect about the y-axis to get the graph of y = e −x . Then we compress the<br />
graphverticallybyafactorof2 to obtain the graph of y = 1 2 e−x andthenreflect about the x-axis to get the graph of<br />
y = − 1 2 e−x . Finally, we shift the graph upward one unit to get the graph of y =1− 1 2 e−x .<br />
13. (a) To find the equation of the graph that results from shifting the graph of y = e x 2 units downward, we subtract 2 from the<br />
original function to get y = e x − 2.<br />
(b) To find the equation of the graph that results from shifting the graph of y = e x 2 units to the right, we replace x with x − 2<br />
in the original function to get y = e (x−2) .<br />
(c) To find the equation of the graph that results from reflecting the graph of y = e x about the x-axis, we multiply the original<br />
function by −1 to get y = −e x .<br />
(d) To find the equation of the graph that results from reflecting the graph of y = e x about the y-axis, we replace x with −x in<br />
the original function to get y = e −x .<br />
(e) To find the equation of the graph that results from reflecting the graph of y = e x about the x-axis and then about the<br />
y-axis, we first multiply the original function by −1 (to get y = −e x ) and then replace x with −x in this equation to<br />
get y = −e −x .<br />
15. (a) The denominator 1+e x is never equal to zero because e x > 0, so the domain of f(x) =1/(1 + e x ) is R.<br />
(b) 1 − e x =0 ⇔ e x =1 ⇔ x =0, so the domain of f(x) =1/(1 − e x ) is (−∞, 0) ∪ (0, ∞).<br />
<br />
17. Use y = Ca x with the points (1, 6) and (3, 24). 6=Ca 1 C =<br />
6<br />
a<br />
6<br />
and 24 = Ca 3 ⇒ 24 = a 3<br />
a<br />
⇒<br />
4=a 2 ⇒ a =2 [since a>0] andC = 6 2 =3. The function is f(x) =3· 2x .<br />
19. If f(x) =5 x ,then<br />
f(x + h) − f(x)<br />
h<br />
= 5x+h − 5 x<br />
h<br />
= 5x 5 h − 5 x<br />
h<br />
= 5x 5 h − 1 <br />
h<br />
5<br />
=5 x h − 1<br />
.<br />
h
F.<br />
TX.10<br />
28 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />
21. 2 ft =24in, f(24) = 24 2 in = 576 in =48ft. g(24) = 2 24 in =2 24 /(12 · 5280) mi ≈ 265 mi<br />
23. The graph of g finally surpasses that of f at x ≈ 35.8.<br />
25. (a) Fifteen hours represents 5 doubling periods (one doubling period is three hours). 100 · 2 5 = 3200<br />
(b) In t hours, there will be t/3 doubling periods. The initial population is 100,<br />
so the population y at time t is y = 100 · 2 t/3 .<br />
(c) t =20 ⇒ y = 100 · 2 20/3 ≈ 10,159<br />
(d) We graph y 1 = 100 · 2 x/3 and y 2 =50,000. The two curves intersect at<br />
x ≈ 26.9, so the population reaches 50,000 in about 26.9 hours.<br />
27. An exponential model is y = ab t ,wherea =3.154832569 × 10 −12<br />
and b =1.017764706. This model gives y(1993) ≈ 5498 million and<br />
y(2010) ≈ 7417 million.<br />
29. From the graph, it appears that f is an odd function (f is undefined for x =0).<br />
To prove this, we must show that f(−x) =−f(x).<br />
f(−x) = 1 − e1/(−x) 1 −<br />
1 − 1<br />
e(−1/x)<br />
=<br />
1+e1/(−x) 1+e = e 1/x<br />
(−1/x)<br />
1+ 1<br />
e 1/x<br />
= − 1 − e1/x<br />
= −f(x)<br />
1+e1/x so f is an odd function.<br />
· e1/x<br />
e = e1/x − 1<br />
1/x e 1/x +1<br />
1.6 Inverse Functions and Logarithms<br />
1. (a) See Definition 1.<br />
(b) It must pass the Horizontal Line Test.<br />
3. f is not one-to-one because 2 6= 6,butf(2) = 2.0 =f(6).<br />
5. No horizontal line intersects the graph of f more than once. Thus, by the Horizontal Line Test, f is one-to-one.
F.<br />
TX.10<br />
SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS ¤ 29<br />
7. The horizontal line y =0(the x-axis) intersects the graph of f in more than one point. Thus, by the Horizontal Line Test,<br />
f is not one-to-one.<br />
9. The graph of f(x) =x 2 − 2x is a parabola with axis of symmetry x = − b<br />
2a = − −2 =1.Pickanyx-values equidistant<br />
2(1)<br />
from 1 to find two equal function values. For example, f(0) = 0 and f(2) = 0,sof is not one-to-one.<br />
11. g(x) =1/x. x 1 6=x 2 ⇒ 1/x 1 6=1/x 2 ⇒ g (x 1 ) 6=g (x 2 ),sog is one-to-one.<br />
Geometric solution: The graph of g is the hyperbola shown in Figure 14 in Section 1.2. It passes the Horizontal Line Test,<br />
so g is one-to-one.<br />
13. A football will attain every height h up to its maximum height twice: once on the way up, and again on the way down. Thus,<br />
even if t 1 does not equal t 2 , f(t 1 ) may equal f(t 2 ),sof is not 1-1.<br />
15. Since f(2) = 9 and f is 1-1, we know that f −1 (9) = 2. Remember, if the point (2, 9) is on the graph of f, then the point<br />
(9, 2) is on the graph of f −1 .<br />
17. First, we must determine x such that g(x) =4. By inspection, we see that if x =0,theng(x) =4.Sinceg is 1-1 (g is an<br />
increasing function), it has an inverse, and g −1 (4) = 0.<br />
19. We solve C = 5 (F − 32) for F : 9 C = F − 32 ⇒ F = 9 9 5 5<br />
C +32. This gives us a formula for the inverse function, that<br />
is, the Fahrenheit temperature F as a function of the Celsius temperature C. F ≥−459.67 ⇒ 9 5 C +32≥−459.67 ⇒<br />
9<br />
5<br />
C ≥−491.67 ⇒ C ≥−273.15, the domain of the inverse function.<br />
21. f(x) = √ 10 − 3x ⇒ y = √ 10 − 3x (y ≥ 0) ⇒ y 2 =10− 3x ⇒ 3x =10− y 2 ⇒ x = − 1 3 y2 + 10<br />
3 .<br />
Interchange x and y: y = − 1 3 x2 + 10<br />
3 .Sof −1 (x) =− 1 3 x2 + 10<br />
3 . Note that the domain of f −1 is x ≥ 0.<br />
23. y = f(x) =e x3 ⇒ ln y = x 3 ⇒ x = 3√ ln y. Interchange x and y: y = 3√ ln x. Sof −1 (x) = 3√ ln x.<br />
25. y = f(x) =ln(x +3) ⇒ x +3=e y ⇒ x = e y − 3. Interchange x and y: y = e x − 3. Sof −1 (x) =e x − 3.<br />
27. y = f(x) =x 4 +1 ⇒ y − 1=x 4 ⇒ x = 4√ y − 1 (not ± since<br />
x ≥ 0). Interchange x and y: y = 4√ x − 1. Sof −1 (x) = 4√ x − 1. The<br />
graph of y = 4√ x − 1 is just the graph of y = 4√ x shifted right one unit.<br />
From the graph, we see that f and f −1 are reflections about the line y = x.<br />
29. Reflect the graph of f about the line y = x. The points (−1, −2), (1, −1),<br />
(2, 2),and(3, 3) on f are reflected to (−2, −1), (−1, 1), (2, 2),and(3, 3)<br />
on f −1 .
F.<br />
TX.10<br />
30 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />
31. (a) It is defined as the inverse of the exponential function with base a,thatis,log a x = y ⇔ a y = x.<br />
(b) (0, ∞) (c) R (d) See Figure 11.<br />
33. (a) log 5 125 = 3 since 5 3 =125. (b) log 3<br />
1<br />
27 = −3 since 3−3 = 1 3 3 = 1 27 .<br />
35. (a) log 2 6 − log 2 15 + log 2 20 = log 2 ( 6<br />
15 )+log 2 20 [by Law 2]<br />
=log 2 ( 6 · 20) [by Law 1]<br />
15<br />
=log 2 8,and log 2 8=3since 2 3 =8.<br />
<br />
(b) log 3 100 − log 3 18 − log 3 50 = log 100<br />
<br />
3 18 − log3 50 = log 100<br />
<br />
3 18·50<br />
37. ln5+5ln3=ln5+ln3 5 [by Law 3]<br />
=ln(5· 3 5 ) [by Law 1]<br />
=ln1215<br />
=log 3 ( 1 9 ),and log 3 1<br />
9<br />
= −2 since 3 −2 = 1 9 .<br />
39. ln(1 + x 2 )+ 1 2 ln x − ln sin x =ln(1+x2 )+lnx 1/2 − ln sin x =ln[(1+x 2 ) √ x ] − ln sin x =ln (1 + x2 ) √ x<br />
sin x<br />
41. To graph these functions, we use log 1.5 x = ln x<br />
ln 1.5 and log 50 x = ln x<br />
ln 50 .<br />
These graphs all approach −∞ as x → 0 + , and they all pass through the<br />
point (1, 0). Also, they are all increasing, and all approach ∞ as x →∞.<br />
The functions with larger bases increase extremely slowly, and the ones with<br />
smaller bases do so somewhat more quickly. The functions with large bases<br />
approach the y-axis more closely as x → 0 + .<br />
43. 3 ft =36in, so we need x such that log 2 x =36 ⇔ x =2 36 =68,719,476,736. Inmiles,thisis<br />
68,719,476,736 in · 1 ft<br />
12 in · 1 mi<br />
≈ 1,084,587.7 mi.<br />
5280 ft<br />
45. (a) Shift the graph of y =log 10 x five units to the left to<br />
obtain the graph of y =log 10 (x +5).Notethevertical<br />
asymptote of x = −5.<br />
(b) Reflect the graph of y =lnx about the x-axis to obtain<br />
the graph of y = − ln x.<br />
y =log 10 x y =log 10 (x +5)<br />
47. (a) 2lnx =1 ⇒ ln x = 1 2<br />
⇒ x = e 1/2 = √ e<br />
(b) e −x =5 ⇒ −x =ln5 ⇒ x = − ln 5<br />
y =lnx<br />
y = − ln x<br />
49. (a) 2 x−5 =3 ⇔ log 2 3=x − 5 ⇔ x =5+log 2 3.<br />
Or: 2 x−5 =3 ⇔ ln 2 x−5 =ln3 ⇔ (x − 5) ln 2 = ln 3 ⇔ x − 5= ln 3<br />
ln 2<br />
⇔ x =5+ ln 3<br />
ln 2
F.<br />
TX.10<br />
SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS ¤ 31<br />
(b) ln x +ln(x − 1) = ln(x(x − 1)) = 1 ⇔ x(x − 1) = e 1 ⇔ x 2 − x − e =0. The quadratic formula (with a =1,<br />
b = −1,andc = −e)givesx = 1 2<br />
<br />
1 ±<br />
√ 1+4e<br />
<br />
, but we reject the negative root since the natural logarithm is not<br />
defined for xe −1 ⇒ x>e −1 ⇒ x ∈ (1/e, ∞)<br />
53. (a) For f(x) = √ 3 − e 2x ,wemusthave3 − e 2x ≥ 0 ⇒ e 2x ≤ 3 ⇒ 2x ≤ ln 3 ⇒ x ≤ 1 2<br />
ln 3. Thus, the domain<br />
of f is (−∞, 1 2<br />
ln 3].<br />
(b) y = f(x) = √ 3 − e 2x [note that y ≥ 0] ⇒ y 2 =3− e 2x ⇒ e 2x =3− y 2 ⇒ 2x =ln(3− y 2 ) ⇒<br />
x = 1 2 ln(3 − y2 ). Interchange x and y: y = 1 2 ln(3 − x2 ).Sof −1 (x) = 1 2 ln(3 − x2 ). For the domain of f −1 ,wemust<br />
have 3 − x 2 > 0 ⇒ x 2 < 3 ⇒ |x| < √ 3 ⇒ − √ 3
F.<br />
TX.10<br />
32 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />
63. (a) In general, tan(arctan x) =x for any real number x. Thus,tan(arctan 10) = 10.<br />
(b) sin −1 sin 7π 3<br />
<br />
=sin<br />
−1 sin π 3<br />
<br />
=sin<br />
−1 √ 3<br />
2 = π 3 since sin π 3 = √ 3<br />
2 and π 3 is in − π 2 , π 2<br />
<br />
.<br />
[Recall that 7π 3<br />
= π 3<br />
+2π and the sine function is periodic with period 2π.]<br />
65. Let y =sin −1 x.Then− π 2 ≤ y ≤ π 2<br />
⇒ cos y ≥ 0,socos(sin −1 x)=cosy = 1 − sin 2 y = √ 1 − x 2 .<br />
67. Let y =tan −1 x.Thentan y = x, so from the triangle we see that<br />
sin(tan −1 x)=siny =<br />
x<br />
√<br />
1+x<br />
2 .<br />
69. The graph of sin −1 x is the reflection of the graph of<br />
sin x about the line y = x.<br />
71. g(x) =sin −1 (3x +1).<br />
Domain (g) ={x | −1 ≤ 3x +1≤ 1} = {x | −2 ≤ 3x ≤ 0} = x | − 2 3 ≤ x ≤ 0 = − 2 3 , 0 .<br />
Range (g) = y | − π 2 ≤ y ≤ π 2<br />
=<br />
−<br />
π<br />
2 , π 2<br />
.<br />
73. (a) If the point (x, y) is on the graph of y = f(x), then the point (x − c, y) is that point shifted c units to the left. Since f is<br />
1-1, the point (y, x) is on the graph of y = f −1 (x) and the point corresponding to (x − c, y) on the graph of f is<br />
(y, x − c) on the graph of f −1 . Thus, the curve’s reflection is shifted down the same number of units as the curve itself is<br />
shiftedtotheleft.Soanexpressionfortheinversefunctionisg −1 (x) =f −1 (x) − c.<br />
(b) If we compress (or stretch) a curve horizontally, the curve’s reflection in the line y = x is compressed (or stretched)<br />
vertically by the same factor. Using this geometric principle, we see that the inverse of h(x) =f(cx) canbeexpressedas<br />
h −1 (x) =(1/c) f −1 (x).
F.<br />
TX.10<br />
CHAPTER 1 REVIEW ¤ 33<br />
1 Review<br />
1. (a) A function f is a rule that assigns to each element x in a set A exactly one element, called f(x),inasetB. ThesetA is<br />
called the domain of the function. The range of f is the set of all possible values of f(x) as x varies throughout the<br />
domain.<br />
(b) If f is a function with domain A,thenitsgraph is the set of ordered pairs {(x, f(x)) | x ∈ A}.<br />
(c) Use the Vertical Line Test on page 16.<br />
2. The four ways to represent a function are: verbally, numerically, visually, and algebraically. An example of each is given<br />
below.<br />
Verbally: An assignment of students to chairs in a classroom (a description in words)<br />
Numerically: A tax table that assigns an amount of tax to an income (a table of values)<br />
Visually: A graphical history of the Dow Jones average (a graph)<br />
Algebraically: A relationship between distance, rate, and time: d = rt (an explicit formula)<br />
3. (a) An even function f satisfies f(−x) =f(x) for every number x in its domain. It is symmetric with respect to the y-axis.<br />
(b) An odd function g satisfies g(−x) =−g(x) for every number x in its domain. It is symmetric with respect to the origin.<br />
4. A function f is called increasing on an interval I if f(x 1)
F.<br />
TX.10<br />
34 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />
(c) (d) (e)<br />
(f ) (g) (h)<br />
9. (a) The domain of f + g is the intersection of the domain of f and the domain of g;thatis,A ∩ B.<br />
(b) The domain of fg is also A ∩ B.<br />
(c) The domain of f/g must exclude values of x that make g equal to 0;thatis,{x ∈ A ∩ B | g(x) 6= 0}.<br />
10. Given two functions f and g,thecomposite function f ◦ g is defined by (f ◦ g)(x) =f(g (x)). The domain of f ◦ g is the<br />
set of all x in the domain of g such that g(x) is in the domain of f.<br />
11. (a) If the graph of f is shifted 2 units upward, its equation becomes y = f(x)+2.<br />
(b) If the graph of f is shifted 2 units downward, its equation becomes y = f(x) − 2.<br />
(c) If the graph of f is shifted 2 units to the right, its equation becomes y = f(x − 2).<br />
(d) If the graph of f is shifted 2 units to the left, its equation becomes y = f(x +2).<br />
(e) If the graph of f is reflected about the x-axis, its equation becomes y = −f(x).<br />
(f ) If the graph of f is reflected about the y-axis, its equation becomes y = f(−x).<br />
(g) If the graph of f isstretchedverticallybyafactorof2, its equation becomes y =2f(x).<br />
(h) If the graph of f is shrunk vertically by a factor of 2, its equation becomes y = 1 2 f(x).<br />
(i) If the graph of f is stretched horizontally by a factor of 2, its equation becomes y = f 1<br />
2 x .<br />
(j) If the graph of f is shrunk horizontally by a factor of 2, its equation becomes y = f(2x).<br />
12. (a) A function f is called a one-to-one function if it never takes on the same value twice; that is, if f(x 1 ) 6=f(x 2 ) whenever<br />
x 1 6=x 2 .(Or,f is 1-1 if each output corresponds to only one input.)<br />
once.<br />
Use the Horizontal Line Test: A function is one-to-one if and only if no horizontal line intersects its graph more than<br />
(b) If f is a one-to-one function with domain A and range B,thenitsinverse function f −1 has domain B and range A and is<br />
defined by<br />
f −1 (y) =x ⇔ f(x) =y<br />
for any y in B. The graph of f −1 is obtained by reflecting the graph of f about the line y = x.
F.<br />
TX.10<br />
CHAPTER 1 REVIEW ¤ 35<br />
13. (a)Theinversesinefunctionf(x) =sin −1 x is defined as follows:<br />
sin −1 x = y ⇔ sin y = x and − π 2 ≤ y ≤ π 2<br />
Its domain is −1 ≤ x ≤ 1 and its range is − π 2 ≤ y ≤ π 2 .<br />
(b) The inverse cosine function f(x) =cos −1 x is defined as follows:<br />
cos −1 x = y ⇔ cos y = x and 0 ≤ y ≤ π<br />
Its domain is −1 ≤ x ≤ 1 and its range is 0 ≤ y ≤ π.<br />
(c) The inverse tangent function f(x) =tan −1 x is defined as follows:<br />
tan −1 x = y ⇔ tan y = x and − π 2
F.<br />
TX.10<br />
36 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />
3. f(x) =x 2 − 2x +3,sof(a + h) =(a + h) 2 − 2(a + h)+3=a 2 +2ah + h 2 − 2a − 2h +3,and<br />
f(a + h) − f(a)<br />
h<br />
= (a2 +2ah + h 2 − 2a − 2h +3)− (a 2 − 2a +3)<br />
h<br />
=<br />
h(2a + h − 2)<br />
h<br />
=2a + h − 2.<br />
5. f(x) =2/(3x − 1). Domain: 3x − 1 6= 0 ⇒ 3x 6= 1 ⇒ x 6= 1 . D = <br />
−∞, 1 3 3 ∪ 1 , ∞ 3<br />
Range: all reals except 0 (y =0is the horizontal asymptote for f.) R =(−∞, 0) ∪ (0, ∞)<br />
7. h(x) =ln(x +6). Domain: x +6> 0 ⇒ x>−6. D =(−6, ∞)<br />
Range: x +6> 0,soln(x +6)takes on all real numbers and, hence, the range is R.<br />
R =(−∞, ∞)<br />
9. (a)Toobtainthegraphofy = f(x)+8,weshiftthegraphofy = f(x) up 8 units.<br />
(b)Toobtainthegraphofy = f(x +8),weshiftthegraphofy = f(x) left 8 units.<br />
(c) To obtain the graph of y =1+2f(x), we stretch the graph of y = f(x) vertically by a factor of 2, and then shift the<br />
resulting graph 1 unit upward.<br />
(d)Toobtainthegraphofy = f(x − 2) − 2, we shift the graph of y = f(x) right 2 units (for the “−2” insidethe<br />
parentheses), and then shift the resulting graph 2 units downward.<br />
(e)Toobtainthegraphofy = −f(x),wereflect the graph of y = f(x) about the x-axis.<br />
(f)Toobtainthegraphofy = f −1 (x),wereflect the graph of y = f(x) about the line y = x (assuming f is one–to-one).<br />
11. y = − sin 2x: Start with the graph of y =sinx, compress horizontally by a factor of 2,andreflect about the x-axis.<br />
13. y = 1 (1 + 2 ex ):<br />
Startwiththegraphofy = e x ,<br />
shift 1 unit upward, and compress<br />
vertically by a factor of 2.<br />
15. f(x) = 1<br />
x +2 :<br />
Startwiththegraphoff(x) =1/x<br />
and shift 2 units to the left.
F.<br />
TX.10<br />
CHAPTER 1 REVIEW ¤ 37<br />
17. (a) The terms of f are a mixture of odd and even powers of x,sof is neither even nor odd.<br />
(b) The terms of f are all odd powers of x,sof is odd.<br />
(c) f(−x) =e −(−x)2 = e −x2 = f(x),sof is even.<br />
(d) f(−x) =1+sin(−x) =1− sin x. Nowf(−x) 6=f(x) and f(−x) 6=−f(x),sof is neither even nor odd.<br />
19. f(x) =lnx, D =(0, ∞); g(x) =x 2 − 9, D = R.<br />
(a) (f ◦ g)(x) =f(g(x)) = f(x 2 − 9) = ln(x 2 − 9).<br />
Domain: x 2 − 9 > 0 ⇒ x 2 > 9 ⇒ |x| > 3 ⇒ x ∈ (−∞, −3) ∪ (3, ∞)<br />
(b) (g ◦ f)(x) =g(f(x)) = g(ln x) =(lnx) 2 − 9. Domain: x>0,or(0, ∞)<br />
(c) (f ◦ f)(x) =f(f(x)) = f(ln x) =ln(lnx). Domain: ln x>0 ⇒ x>e 0 =1,or(1, ∞)<br />
(d) (g ◦ g)(x) =g(g(x)) = g(x 2 − 9) = (x 2 − 9) 2 − 9. Domain: x ∈ R,or(−∞, ∞)<br />
21. Many models appear to be plausible. Your choice depends on whether you<br />
think medical advances will keep increasing life expectancy, or if there is<br />
bound to be a natural leveling-off of life expectancy. A linear model,<br />
y =0.2493x − 423.4818, gives us an estimate of 77.6 years for the<br />
year 2010.<br />
23. We need to know the value of x such that f(x) =2x +lnx =2.Sincex =1gives us y =2, f −1 (2) = 1.<br />
25. (a) e 2ln3 = e ln 3 2<br />
=3 2 =9<br />
(b) log 10 25 + log 10 4=log 10 (25 · 4) = log 10 100 = log 10 10 2 =2<br />
(c) tan <br />
arcsin 1 2 =tan<br />
π<br />
= √ 1 6 3<br />
(d) Let θ =cos −1 4 5 ,socos θ = 4 5 .Thensin cos −1 4 5<br />
<br />
=sinθ =<br />
√<br />
1 − cos2 θ =<br />
<br />
1 − <br />
4 2 9<br />
5<br />
= = 3 . 25 5<br />
27. (a) The population would reach 900 in about 4.4 years.<br />
100,000<br />
(b) P =<br />
100 + 900e ⇒ 100P −t +900Pe−t = 100,000 ⇒ 900Pe −t = 100,000 − 100P ⇒<br />
<br />
<br />
e −t 100,000 − 100P<br />
1000 − P<br />
1000 − P<br />
= ⇒ −t =ln<br />
⇒ t = − ln<br />
,orln<br />
900P<br />
9P<br />
9P<br />
required for the population to reach a given number P .<br />
9 · 900<br />
(c) P =900 ⇒ t =ln<br />
=ln81≈ 4.4 years, as in part (a).<br />
1000 − 900<br />
9P<br />
1000 − P<br />
<br />
;thisisthetime
F.<br />
TX.10
F.<br />
TX.10<br />
PRINCIPLES OF PROBLEM SOLVING<br />
1. By using the area formula for a triangle, 1 (base)(height), in two ways, we see that<br />
2<br />
1<br />
(4) (y) = 1 (h)(a),soa = 4y 2 2<br />
h .Since42 + y 2 = h 2 , y = √ h 2 − 16,and<br />
a = 4√ h 2 − 16<br />
.<br />
h<br />
3. |2x − 1| =<br />
<br />
2x − 1 if x ≥<br />
1<br />
2<br />
1 − 2x if x< 1 2<br />
and |x +5| =<br />
<br />
x +5<br />
−x − 5<br />
if x ≥−5<br />
if x
F.<br />
TX.10<br />
40 ¤ CHAPTER 1 PRINCIPLES OF PROBLEM SOLVING<br />
9. |x| + |y| ≤ 1. The boundary of the region has equation |x| + |y| =1. In quadrants<br />
I,II,III,andIV,thisbecomesthelinesx + y =1, −x + y =1, −x − y =1,and<br />
x − y =1respectively.<br />
11. (log 2 3)(log 3 4)(log 4 5) ···(log 31 32) =<br />
ln 3<br />
ln 2<br />
ln 4<br />
ln 3<br />
ln 5<br />
···<br />
ln 4<br />
ln 32 ln 32 ln 25<br />
= =<br />
ln 31 ln 2 ln 2 = 5ln2<br />
ln 2 =5<br />
13. ln x 2 − 2x − 2 ≤ 0 ⇒ x 2 − 2x − 2 ≤ e 0 =1 ⇒ x 2 − 2x − 3 ≤ 0 ⇒ (x − 3)(x +1)≤ 0 ⇒ x ∈ [−1, 3].<br />
Since the argument must be positive, x 2 − 2x − 2 > 0 ⇒ x − 1 − √ 3 x − 1+ √ 3 > 0 ⇒<br />
x ∈ −∞, 1 − √ 3 ∪ 1+ √ 3, ∞ . The intersection of these intervals is −1, 1 − √ 3 ∪ 1+ √ 3, 3 .<br />
15. Let d be the distance traveled on each half of the trip. Let t 1 and t 2 be the times taken for the first and second halves of the trip.<br />
For the first half of the trip we have t 1 = d/30 and for the second half we have t 2 = d/60. Thus, the average speed for the<br />
entire trip is<br />
is 40 mi/h.<br />
total distance<br />
total time<br />
= 2d<br />
t 1 + t 2<br />
=<br />
2d<br />
d<br />
30 + d 60<br />
17. Let S n be the statement that 7 n − 1 is divisible by 6.<br />
• S 1 is true because 7 1 − 1=6is divisible by 6.<br />
· 60<br />
60 = 120d<br />
2d + d = 120d =40. The average speed for the entire trip<br />
3d<br />
• Assume S k is true, that is, 7 k − 1 is divisible by 6. Inotherwords,7 k − 1=6m for some positive integer m. Then<br />
7 k+1 − 1=7 k · 7 − 1=(6m +1)· 7 − 1=42m +6=6(7m +1), which is divisible by 6,soS k+1 is true.<br />
• Therefore, by mathematical induction, 7 n − 1 is divisible by 6 for every positive integer n.<br />
19. f 0 (x) =x 2 and f n+1 (x) =f 0 (f n (x)) for n =0, 1, 2,....<br />
f 1 (x) =f 0 (f 0 (x)) = f 0<br />
<br />
x<br />
2 = x 22 = x 4 , f 2 (x) =f 0 (f 1 (x)) = f 0 (x 4 )=(x 4 ) 2 = x 8 ,<br />
f 3(x) =f 0(f 2(x)) = f 0(x 8 )=(x 8 ) 2 = x 16 , .... Thus, a general formula is f n(x) =x 2n+1 .
F.<br />
TX.10<br />
2 LIMITS AND DERIVATIVES<br />
2.1 The Tangent and Velocity Problems<br />
(b) Using the values of t that correspond to the points<br />
1. (a) Using P (15, 250), we construct the following table:<br />
30 (30, 0)<br />
0−250<br />
30−15 15<br />
closest to P (t =10and t =20), we have<br />
t Q slope = m PQ<br />
5 (5, 694)<br />
694−250<br />
= − 444 −38.8+(−27.8)<br />
5−15 10<br />
= −33.3<br />
2<br />
444−250<br />
10 (10, 444)<br />
= − 194 = −38.8<br />
10−15 5<br />
111−250<br />
20 (20, 111)<br />
20−15<br />
= − 139<br />
5<br />
= −27.8<br />
28−250<br />
25 (25, 28)<br />
25−15 10<br />
(c) From the graph, we can estimate the slope of the<br />
tangent line at P to be −300<br />
9<br />
= −33.3.<br />
3. (a)<br />
x Q m PQ<br />
(i) 0.5 (0.5, 0.333333) 0.333333<br />
(ii) 0.9 (0.9, 0.473684) 0.263158<br />
(iii) 0.99 (0.99, 0.497487) 0.251256<br />
(iv) 0.999 (0.999, 0.499750) 0.250125<br />
(v) 1.5 (1.5, 0.6) 0.2<br />
(vi) 1.1 (1.1, 0.523810) 0.238095<br />
(vii) 1.01 (1.01, 0.502488) 0.248756<br />
(viii) 1.001 (1.001, 0.500250) 0.249875<br />
(b) The slope appears to be 1 4 .<br />
(c) y − 1 2 = 1 4 (x − 1) or y = 1 4 x + 1 4 .<br />
5. (a) y = y(t) =40t − 16t 2 .Att =2, y =40(2)− 16(2) 2 =16. The average velocity between times 2 and 2+h is<br />
<br />
y(2 + h) − y(2)<br />
40(2 + h) − 16(2 + h)<br />
2<br />
− 16 −24h − 16h2<br />
v ave = =<br />
= = −24 − 16h, ifh 6= 0.<br />
(2 + h) − 2<br />
h<br />
h<br />
(i) [2, 2.5]: h =0.5, v ave = −32 ft/s<br />
(ii) [2, 2.1]: h =0.1, v ave = −25.6 ft/s<br />
(iii) [2, 2.05]: h =0.05, v ave = −24.8 ft/s<br />
(iv) [2, 2.01]: h =0.01, v ave = −24.16 ft/s<br />
(b) The instantaneous velocity when t =2(h approaches 0)is−24 ft/s.<br />
41
F.<br />
42 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
7. (a) (i) On the interval [1, 3], v ave =<br />
(ii) On the interval [2, 3], v ave =<br />
(iii) On the interval [3, 5], v ave =<br />
(iv) On the interval [3, 4], v ave =<br />
s(3) − s(1)<br />
3 − 1<br />
s(3) − s(2)<br />
3 − 2<br />
s(5) − s(3)<br />
5 − 3<br />
s(4) − s(3)<br />
4 − 3<br />
=<br />
=<br />
=<br />
=<br />
10.7 − 1.4<br />
2<br />
10.7 − 5.1<br />
1<br />
25.8 − 10.7<br />
2<br />
17.7 − 10.7<br />
1<br />
= 9.3<br />
2<br />
=5.6 m/s.<br />
= 15.1<br />
2<br />
=7m/s.<br />
=4.65 m/s.<br />
=7.55 m/s.<br />
(b)<br />
Using the points (2, 4) and (5, 23) from the approximate tangent<br />
line, the instantaneous velocity at t =3is about 23 − 4<br />
5 − 2<br />
≈ 6.3 m/s.<br />
9. (a) For the curve y =sin(10π/x) and the point P (1, 0):<br />
x Q m PQ<br />
2 (2, 0) 0<br />
1.5 (1.5, 0.8660) 1.7321<br />
1.4 (1.4, −0.4339) −1.0847<br />
1.3 (1.3, −0.8230) −2.7433<br />
1.2 (1.2, 0.8660) 4.3301<br />
1.1 (1.1, −0.2817) −2.8173<br />
x Q m PQ<br />
0.5 (0.5, 0) 0<br />
0.6 (0.6, 0.8660) −2.1651<br />
0.7 (0.7, 0.7818) −2.6061<br />
0.8 (0.8, 1) −5<br />
0.9 (0.9, −0.3420) 3.4202<br />
As x approaches 1, the slopes do not appear to be approaching any particular value.<br />
(b)<br />
We see that problems with estimation are caused by the frequent<br />
oscillations of the graph. The tangent is so steep at P that we need to<br />
take x-values much closer to 1 in order to get accurate estimates of<br />
its slope.<br />
(c) If we choose x =1.001, then the point Q is (1.001, −0.0314) and m PQ ≈−31.3794. Ifx =0.999, thenQ is<br />
(0.999, 0.0314) and m PQ = −31.4422. The average of these slopes is −31.4108. So we estimate that the slope of the<br />
tangent line at P is about −31.4.
F.<br />
TX.10<br />
SECTION 2.2 THE LIMIT OF A FUNCTION ¤ 43<br />
2.2 The Limit of a Function<br />
1. As x approaches 2, f(x) approaches 5. [Or,thevaluesoff(x) can be made as close to 5 as we like by taking x sufficiently<br />
close to 2 (but x 6= 2).] Yes, the graph could have a hole at (2, 5) and be definedsuchthatf(2) = 3.<br />
3. (a) lim f(x) =∞ means that the values of f(x) canbemadearbitrarilylarge(aslargeasweplease)bytakingx<br />
x→−3<br />
(b)<br />
sufficiently close to −3 (but not equal to −3).<br />
lim f(x) =−∞ means that the values of f(x) can be made arbitrarily large negative by taking x sufficiently close to 4<br />
x→4 +<br />
through values larger than 4.<br />
5. (a) f(x) approaches 2 as x approaches 1 from the left, so lim f(x) =2.<br />
x→1− (b) f(x) approaches 3 as x approaches 1 from the right, so lim f(x) =3.<br />
x→1 +<br />
(c) lim<br />
x→1<br />
f(x) does not exist because the limits in part (a) and part (b) are not equal.<br />
(d) f(x) approaches 4 as x approaches 5 from the left and from the right, so lim<br />
x→5<br />
f(x) =4.<br />
(e) f(5) is not defined, so it doesn’t exist.<br />
7. (a) lim g(t) =−1<br />
t→0− (b) lim g(t) =−2<br />
+<br />
(c) lim<br />
t→0<br />
g(t) does not exist because the limits in part (a) and part (b) are not equal.<br />
(d) lim g(t) =2<br />
t→2− t→0<br />
(e) lim g(t) =0<br />
+<br />
(f ) lim<br />
t→2<br />
g(t) does not exist because the limits in part (d) and part (e) are not equal.<br />
(g) g(2) = 1<br />
t→2<br />
(h) lim<br />
t→4<br />
g(t) =3<br />
9. (a) lim f(x) =−∞ (b) lim f(x) =∞<br />
x→−7 x→−3<br />
(d)<br />
lim f(x) =−∞<br />
x→6− (e) lim f(x) =∞<br />
+<br />
x→6<br />
(f ) The equations of the vertical asymptotes are x = −7, x = −3, x =0,andx =6.<br />
(c) lim f(x) =∞<br />
x→0<br />
11. (a) lim f(x) =1<br />
x→0− (b)<br />
lim f(x) =0<br />
x→0 +<br />
(c) lim<br />
x→0<br />
f(x) does not exist because the limits<br />
in part (a) and part (b) are not equal.<br />
13. lim f(x) =2, lim<br />
x→1− x→1<br />
f(x) =−2, f(1) = 2 15. lim f(x) =4, lim<br />
+ +<br />
x→3<br />
f(3) = 3, f(−2) = 1<br />
f(x) =2, lim f(x) =2,<br />
x→3− x→−2
F.<br />
44 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
17. For f(x) = x2 − 2x<br />
x 2 − x − 2 :<br />
x<br />
f(x)<br />
2.5 0.714286<br />
2.1 0.677419<br />
2.05 0.672131<br />
2.01 0.667774<br />
2.005 0.667221<br />
2.001 0.666778<br />
x<br />
f(x)<br />
1.9 0.655172<br />
1.95 0.661017<br />
1.99 0.665552<br />
1.995 0.666110<br />
1.999 0.666556<br />
x 2 − 2x<br />
It appears that lim<br />
x→2 x 2 − x − 2 =0.¯6 = 2 . 3<br />
TX.10<br />
19. For f(x) = ex − 1 − x<br />
x 2 :<br />
x<br />
f(x)<br />
1 0.718282<br />
0.5 0.594885<br />
0.1 0.517092<br />
0.05 0.508439<br />
0.01 0.501671<br />
x<br />
f(x)<br />
−1 0.367879<br />
−0.5 0.426123<br />
−0.1 0.483742<br />
−0.05 0.491770<br />
−0.01 0.498337<br />
It appears that lim<br />
x→0<br />
e x − 1 − x<br />
x 2 =0.5 = 1 2 .<br />
√ x +4− 2<br />
21. For f(x) =<br />
:<br />
x<br />
x<br />
f(x)<br />
1 0.236068<br />
0.5 0.242641<br />
0.1 0.248457<br />
0.05 0.249224<br />
0.01 0.249844<br />
It appears that lim<br />
x→0<br />
√ x +4− 2<br />
x<br />
23. For f(x) = x6 − 1<br />
x 10 − 1 :<br />
x f(x)<br />
x f(x)<br />
−1 0.267949<br />
0.5 0.985337<br />
−0.5 0.258343<br />
0.9 0.719397<br />
−0.1 0.251582<br />
0.95 0.660186<br />
−0.05 0.250786<br />
0.99 0.612018<br />
−0.01 0.250156<br />
0.999 0.601200<br />
It appears that lim<br />
x→1<br />
x 6 − 1<br />
x 10 − 1 =0.6 = 3 5 .<br />
x<br />
f(x)<br />
1.5 0.183369<br />
1.1 0.484119<br />
1.05 0.540783<br />
1.01 0.588022<br />
1.001 0.598800<br />
25. lim<br />
x→−3 + x +2<br />
x +3 = −∞ since the numerator is negative and the denominator approaches 0 from the positive side as x →−3+ .<br />
2 − x<br />
27. lim = ∞ since the numerator is positive and the denominator approaches 0 through positive values as x → 1.<br />
x→1 (x − 1)<br />
2<br />
29. Let t = x 2 − 9. Thenasx → 3 + , t → 0 + ,and lim ln(x2 − 9) = lim ln t = −∞ by (3).<br />
x→3 + t→0 +<br />
31. lim<br />
x→2π − x csc x =<br />
lim<br />
x→2π −<br />
x<br />
= −∞ since the numerator is positive and the denominator approaches 0 through negative<br />
sin x<br />
values as x → 2π − .<br />
33. (a) f(x) = 1<br />
x 3 − 1 .<br />
lim f(x) =−∞ and lim x→1− x→1 + 0.99 −33.7<br />
x f(x)<br />
0.5 −1.14<br />
From these calculations, it seems that<br />
0.9 −3.69<br />
0.999 −333.7<br />
0.9999 −3333.7<br />
0.99999 −33,333.7<br />
x f(x)<br />
1.5 0.42<br />
1.1 3.02<br />
1.01 33.0<br />
1.001 333.0<br />
1.0001 3333.0<br />
1.00001 33,333.3
F.<br />
TX.10<br />
SECTION 2.2 THE LIMIT OF A FUNCTION ¤ 45<br />
(b) If x is slightly smaller than 1,thenx 3 − 1 will be a negative number close to 0,andthereciprocalofx 3 − 1,thatis,f(x),<br />
will be a negative number with large absolute value. So lim f(x) =−∞.<br />
x→1− If x is slightly larger than 1,thenx 3 − 1 will be a small positive number, and its reciprocal, f(x), will be a large positive<br />
number. So lim f(x) =∞.<br />
x→1 +<br />
(c) It appears from the graph of f that<br />
lim<br />
x→1<br />
f(x) =−∞ and lim f(x) =∞.<br />
− +<br />
x→1<br />
35. (a) Let h(x) =(1+x) 1/x .<br />
(b)<br />
x<br />
h(x)<br />
−0.001 2.71964<br />
−0.0001 2.71842<br />
−0.00001 2.71830<br />
−0.000001 2.71828<br />
0.000001 2.71828<br />
0.00001 2.71827<br />
0.0001 2.71815<br />
0.001 2.71692<br />
It appears that lim<br />
x→0<br />
(1 + x) 1/x ≈ 2.71828, which is approximately e.<br />
In Section 3.6 we will see that the value of the limit is exactly e.<br />
37. For f(x) =x 2 − (2 x /1000):<br />
(a)<br />
x f(x)<br />
1 0.998000<br />
0.8 0.638259<br />
0.6 0.358484<br />
0.4 0.158680<br />
0.2 0.038851<br />
0.1 0.008928<br />
0.05 0.001465<br />
It appears that lim<br />
x→0<br />
f(x) =0.<br />
(b)<br />
x<br />
f(x)<br />
0.04 0.000572<br />
0.02 −0.000614<br />
0.01 −0.000907<br />
0.005 −0.000978<br />
0.003 −0.000993<br />
0.001 −0.001000<br />
It appears that lim<br />
x→0<br />
f(x) =−0.001.
F.<br />
46 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
39. No matter how many times we zoom in toward the origin, the graphs of f(x) =sin(π/x) appear to consist of almost-vertical<br />
lines. This indicates more and more frequent oscillations as x → 0.<br />
41. There appear to be vertical asymptotes of the curve y =tan(2sinx) at x ≈ ±0.90<br />
and x ≈ ±2.24. Tofind the exact equations of these asymptotes, we note that the<br />
graph of the tangent function has vertical asymptotes at x = π + πn. Thus,we<br />
2<br />
must have 2sinx = π + πn, or equivalently, sin x = π + π n.Since<br />
2 4 2<br />
−1 ≤ sin x ≤ 1,wemusthavesin x = ± π and so x = ± 4 sin−1 π (corresponding<br />
4<br />
to x ≈ ±0.90). Just as 150 ◦ is the reference angle for 30 ◦ , π − sin −1 π 4<br />
is the<br />
reference angle for sin −1 π .Sox = ± <br />
4<br />
π − sin −1 π 4 are also equations of<br />
vertical asymptotes (corresponding to x ≈ ±2.24).<br />
2.3 Calculating Limits Using the Limit Laws<br />
1. (a) lim<br />
x→2<br />
[f(x)+5g(x)] = lim<br />
x→2<br />
f(x) + lim<br />
x→2<br />
[5g(x)] [Limit Law 1]<br />
=lim<br />
x→2<br />
f(x)+5lim<br />
x→2<br />
g(x) [Limit Law 3]<br />
=4+5(−2) = −6<br />
3<br />
(b) lim [g(x)] 3 = lim g(x) [Limit Law 6]<br />
x→2 x→2<br />
=(−2) 3 = −8<br />
(c) lim<br />
x→2<br />
<br />
f(x)=<br />
<br />
lim<br />
x→2<br />
f(x) [Limit Law 11]<br />
= √ 4=2
F.<br />
TX.10<br />
SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 47<br />
3f(x)<br />
lim [3f(x)]<br />
(d) lim<br />
x→2 g(x) = x→2<br />
lim g(x) [Limit Law 5]<br />
x→2<br />
=<br />
3lim<br />
x→2<br />
f(x)<br />
lim<br />
x→2 g(x) [Limit Law 3]<br />
= 3(4)<br />
−2 = −6<br />
g(x)<br />
(e) Because the limit of the denominator is 0, we can’t use Limit Law 5. The given limit, lim , does not exist because the<br />
x→2 h(x)<br />
denominator approaches 0 while the numerator approaches a nonzero number.<br />
g(x) h(x)<br />
lim [g(x) h(x)]<br />
x→2<br />
(f ) lim =<br />
x→2 f(x) lim f(x) [Limit Law 5]<br />
x→2<br />
=<br />
lim g(x) · lim h(x)<br />
x→2 x→2<br />
lim f(x) [Limit Law 4]<br />
x→2<br />
= −2 · 0<br />
4<br />
=0<br />
3. lim<br />
x→−2 (3x4 +2x 2 − x +1)= lim<br />
x→−2 3x4 + lim<br />
x→−2 2x2 − lim x + lim 1 [Limit Laws 1 and 2]<br />
x→−2 x→−2<br />
= 3 lim<br />
x→−2 x4 +2 lim<br />
x→−2 x2 − lim<br />
x→−2 x + lim<br />
x→−2 1 [3]<br />
=3(−2) 4 +2(−2) 2 − (−2) + (1)<br />
[9,8,and7]<br />
=48+8+2+1=59<br />
5. lim (1 + 3√ x )(2− 6x 2 + x 3 ) = lim (1 + 3√ x ) · lim(2 − 6x 2 + x 3 ) [Limit Law 4]<br />
x→8 x→8 x→8<br />
<br />
<br />
<br />
= lim 1 + lim 3√ x · lim 2 − 6lim<br />
x→8 x→8<br />
x→8 x→8 x2 +limx 3 [1,2,and3]<br />
x→8<br />
7. lim<br />
x→1<br />
<br />
3<br />
1+3x<br />
=<br />
1+4x 2 +3x 4<br />
=<br />
<br />
lim<br />
x→1<br />
<br />
= 1+ 3√ 8 · 2<br />
− 6 · 8 2 +8 3 [7, 10, 9]<br />
= (3)(130) = 390<br />
3<br />
1+3x<br />
[6]<br />
1+4x 2 +3x 4 3<br />
[5]<br />
lim (1 + 3x)<br />
x→1<br />
lim (1 +<br />
x→1 4x2 +3x 4 )<br />
<br />
lim 1+3limx<br />
3<br />
x→1 x→1<br />
=<br />
lim 1+4lim<br />
[2, 1, and 3]<br />
x→1 x→1 x2 +3limx 4<br />
x→1<br />
<br />
3 3 3<br />
1 + 3(1)<br />
4 1<br />
=<br />
= = = 1 [7, 8, and 9]<br />
1 + 4(1) 2 +3(1) 4 8 2 8<br />
9. lim<br />
x→4 − √<br />
16 − x2 = lim<br />
x→4 − (16 − x2 ) [11]<br />
= lim 16 − lim x2 [2]<br />
x→4− x→4 −<br />
= 16 − (4) 2 =0 [7 and 9]
F.<br />
48 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
x 2 + x − 6 (x +3)(x − 2)<br />
11. lim<br />
=lim<br />
=lim(x +3)=2+3=5<br />
x→2 x − 2 x→2 x − 2<br />
x→2<br />
x 2 − x +6<br />
13. lim<br />
does not exist since x − 2 → 0 but x 2 − x +6→ 8 as x → 2.<br />
x→2 x − 2<br />
15. lim<br />
t→−3<br />
t 2 − 9<br />
2t 2 +7t +3 = lim<br />
t→−3<br />
(t +3)(t − 3)<br />
(2t +1)(t +3) = lim<br />
t→−3<br />
t − 3<br />
2t +1 = −3 − 3<br />
2(−3) + 1 = −6<br />
−5 = 6 5<br />
(4 + h) 2 − 16 (16 + 8h + h 2 ) − 16 8h + h 2 h(8 + h)<br />
17. lim<br />
=lim<br />
=lim =lim =lim(8 + h) =8+0=8<br />
h→0 h<br />
h→0 h<br />
h→0 h h→0 h h→0<br />
19. Bytheformulaforthesumofcubes,wehave<br />
lim<br />
x→−2<br />
x +2<br />
x 3 +8 = lim<br />
x→−2<br />
x +2<br />
(x +2)(x 2 − 2x +4) = lim<br />
x→−2<br />
1<br />
x 2 − 2x +4 = 1<br />
4+4+4 = 1<br />
12 .<br />
√ √ <br />
9 − t 3+ t 3 − t<br />
21. lim<br />
t→9 3 − √ t =lim<br />
t→9 3 − √ √ √<br />
=lim 3+ t =3+ 9=6<br />
t<br />
t→9<br />
√ √ √ x +2− 3 x +2− 3 x +2+3<br />
(x +2)− 9<br />
23. lim<br />
=lim<br />
· √ = lim<br />
x→7 x − 7 x→7 x − 7 x +2+3 x→7 (x − 7) √ x +2+3 <br />
25. lim<br />
x→−4<br />
27. lim<br />
x→16<br />
29. lim<br />
t→0<br />
<br />
=lim<br />
x→7<br />
x − 7<br />
(x − 7) √ x +2+3 = lim<br />
x→7<br />
1<br />
4 + 1 x +4<br />
x<br />
4+x = lim 4x<br />
x→−4 4+x = lim<br />
x→−4<br />
4 − √ x<br />
16x − x 2 = lim<br />
x→16<br />
1<br />
t √ 1+t − 1 <br />
t<br />
x +4<br />
4x(4 + x) = lim<br />
x→−4<br />
(4 − √ x )(4 + √ x )<br />
(16x − x 2 )(4 + √ x ) = lim<br />
x→16<br />
1<br />
√ x +2+3<br />
=<br />
1<br />
√<br />
9+3<br />
= 1 6<br />
1<br />
4x = 1<br />
4(−4) = − 1 16<br />
16 − x<br />
x(16 − x)(4 + √ x )<br />
1<br />
= lim<br />
x→16 x(4 + √ x ) = 1<br />
16 4+ √ 16 = 1<br />
16(8) = 1<br />
128<br />
1 − √ √ √ <br />
1+t 1 − 1+t 1+ 1+t<br />
=lim<br />
t→0 t √ = lim<br />
1+t t→0 t √ t +1 1+ √ 1+t =lim<br />
t→0<br />
=lim<br />
t→0<br />
−1<br />
√ 1+t<br />
1+<br />
√ 1+t<br />
=<br />
−1<br />
√ 1+0<br />
1+<br />
√ 1+0<br />
= − 1 2<br />
−t<br />
t √ 1+t 1+ √ 1+t <br />
31. (a)<br />
lim<br />
x→0<br />
x<br />
√ 1+3x − 1<br />
≈ 2 3<br />
(b)<br />
x<br />
f(x)<br />
−0.001 0.6661663<br />
−0.0001 0.6666167<br />
−0.00001 0.6666617<br />
−0.000001 0.6666662<br />
0.000001 0.6666672<br />
0.00001 0.6666717<br />
0.0001 0.6667167<br />
0.001 0.6671663<br />
The limit appears to be 2 3 .
F.<br />
(c) lim<br />
x→0<br />
<br />
TX.10 SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 49<br />
√ <br />
x 1+3x +1 x √ 1+3x +1 <br />
x √ 1+3x +1 <br />
√ · √ =lim<br />
=lim<br />
1+3x − 1 1+3x +1 x→0 (1 + 3x) − 1 x→0 3x<br />
= 1 3 lim √ <br />
1+3x +1<br />
x→0<br />
= 1 3<br />
lim<br />
= 1 3<br />
lim<br />
x→0<br />
(1 + 3x)+lim<br />
x→0<br />
1<br />
x→0<br />
1+3lim<br />
x→0<br />
x +1<br />
= 1 √ <br />
1+3· 0+1<br />
3<br />
<br />
<br />
[Limit Law 3]<br />
[1 and 11]<br />
[1,3,and7]<br />
[7 and 8]<br />
= 1 3 (1 + 1) = 2 3<br />
33. Let f(x) =−x 2 , g(x) =x 2 cos 20πx and h(x) =x 2 .Then<br />
−1 ≤ cos 20πx ≤ 1 ⇒ −x 2 ≤ x 2 cos 20πx ≤ x 2 ⇒ f(x) ≤ g(x) ≤ h(x).<br />
So since lim<br />
x→0<br />
f(x) = lim<br />
x→0<br />
h(x) =0, by the Squeeze Theorem we have<br />
lim g(x) =0.<br />
x→0<br />
35. We have lim<br />
x→4<br />
(4x − 9) = 4(4) − 9=7and lim<br />
x→4<br />
<br />
x 2 − 4x +7 =4 2 − 4(4) + 7 = 7. Since4x − 9 ≤ f(x) ≤ x 2 − 4x +7<br />
for x ≥ 0, lim<br />
x→4<br />
f(x) =7by the Squeeze Theorem.<br />
<br />
37. −1 ≤ cos(2/x) ≤ 1 ⇒ −x 4 ≤ x 4 cos(2/x) ≤ x 4 .Sincelim<br />
−x<br />
4<br />
=0and lim x 4 =0,wehave<br />
x→0 x→0<br />
<br />
lim x 4 cos(2/x) =0by the Squeeze Theorem.<br />
x→0<br />
39. |x − 3| =<br />
41.<br />
<br />
x − 3 if x − 3 ≥ 0<br />
−(x − 3) if x − 3 < 0 = <br />
x − 3 if x ≥ 3<br />
3 − x if x
F.<br />
50 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
1<br />
43. Since |x| = −x for x0, lim sgn x = lim 1=1.<br />
x→0 + x→0 +<br />
(ii) Since sgn x = −1 for x
F.<br />
TX.10<br />
SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT ¤ 51<br />
<br />
f(x) − 8<br />
f(x) − 8<br />
55. lim [f(x) − 8] = lim<br />
· (x − 1) =lim · lim (x − 1) = 10 · 0=0.<br />
x→1 x→1 x − 1<br />
x→1 x − 1 x→1<br />
Thus, lim<br />
x→1<br />
f(x) = lim<br />
x→1<br />
{[f(x) − 8] + 8} =lim<br />
x→1<br />
[f(x) − 8] + lim<br />
x→1<br />
8=0+8=8.<br />
f(x) − 8<br />
f(x) − 8<br />
Note: The value of lim does not affect the answer since it’s multiplied by 0. What’s important is that lim<br />
x→1 x − 1<br />
x→1 x − 1<br />
exists.<br />
57. Observe that 0 ≤ f(x) ≤ x 2 for all x,and lim<br />
x→0<br />
0=0= lim<br />
x→0<br />
x 2 . So, by the Squeeze Theorem, lim<br />
x→0<br />
f(x) =0.<br />
59. Let f(x) =H(x) and g(x) =1− H(x),whereH is the Heaviside function definedinExercise1.3.57.<br />
Thus, either f or g is 0 for any value of x. Thenlim<br />
x→0<br />
f(x) and lim<br />
x→0<br />
g(x) do not exist, but lim<br />
x→0<br />
[f(x)g(x)] = lim<br />
x→0<br />
0=0.<br />
61. Since the denominator approaches 0 as x →−2, the limit will exist only if the numerator also approaches<br />
<br />
0 as x →−2. In order for this to happen, we need lim 3x 2 + ax + a +3 =0<br />
x→−2<br />
⇔<br />
3(−2) 2 + a(−2) + a +3=0 ⇔ 12 − 2a + a +3=0 ⇔ a =15.Witha =15, the limit becomes<br />
3x 2 +15x +18 3(x +2)(x +3)<br />
lim<br />
= lim<br />
x→−2 x 2 + x − 2 x→−2 (x − 1)(x +2) = lim 3(x +3)<br />
= 3(−2+3) = 3<br />
x→−2 x − 1 −2 − 1 −3 = −1.<br />
2.4 The Precise Definition of a Limit<br />
1. On the left side of x =2,weneed|x − 2| < 10<br />
− 2 = 4 . On the right side, we need |x − 2| < 10<br />
− 2 = 4 . For both of<br />
7 7 3 3<br />
these conditions to be satisfied at once, we need the more restrictive of the two to hold, that is, |x − 2| < 4 . So we can choose<br />
7<br />
δ = 4 7<br />
, or any smaller positive number.<br />
3. The leftmost question mark is the solution of √ x =1.6 and the rightmost, √ x =2.4. Sothevaluesare1.6 2 =2.56 and<br />
2.4 2 =5.76. On the left side, we need |x − 4| < |2.56 − 4| =1.44. On the right side, we need |x − 4| < |5.76 − 4| =1.76.<br />
To satisfy both conditions, we need the more restrictive condition to hold — namely, |x − 4| < 1.44. Thus, we can choose<br />
δ =1.44, or any smaller positive number.<br />
5. From the graph, we find that tan x =0.8 when x ≈ 0.675,so<br />
π<br />
− 4 δ1 ≈ 0.675 ⇒ δ1 ≈ π 4<br />
− 0.675 ≈ 0.1106. Also,tan x =1.2<br />
when x ≈ 0.876,so π 4 + δ 2 ≈ 0.876 ⇒ δ 2 =0.876 − π 4 ≈ 0.0906.<br />
Thus, we choose δ =0.0906 (or any smaller positive number) since this is<br />
the smaller of δ 1 and δ 2 .
F.<br />
52 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
7. For ε =1,thedefinition of a limit requires that we find δ such that 4+x − 3x 3 − 2 < 1 ⇔ 1 < 4+x − 3x 3 < 3<br />
whenever 0 < |x − 1|
F.<br />
TX.10<br />
SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT ¤ 53<br />
15. Given ε>0, we need δ>0 such that if 0 < |x − 1|
F.<br />
54 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
29. Given ε>0,weneedδ>0 such that if 0 < |x − 2|
F.<br />
TX.10<br />
SECTION 2.5 CONTINUITY ¤ 55<br />
39. Suppose that lim f(x) =L. Givenε = 1 ,thereexistsδ>0 such that 0 < |x|
F.<br />
56 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
<br />
<br />
4<br />
11. lim f(x) = lim x +2x<br />
3 4 = lim x +2 lim<br />
x→−1 x→−1<br />
x→−1 x→−1 x3 = −1+2(−1) 34 =(−3) 4 =81=f(−1).<br />
By the definition of continuity, f is continuous at a = −1.<br />
13. For a>2,wehave<br />
2x +3<br />
lim (2x +3)<br />
lim f(x)= lim<br />
x→a x→a x − 2 = x→a<br />
lim (x − 2) [Limit Law 5]<br />
x→a<br />
2 lim x + lim 3<br />
x→a x→a<br />
=<br />
lim x − lim 2<br />
x→a x→a<br />
=<br />
2a +3<br />
a − 2<br />
= f(a)<br />
TX.10<br />
[1,2,and3]<br />
[7 and 8]<br />
Thus, f is continuous at x = a for every a in (2, ∞);thatis,f is continuous on (2, ∞).<br />
15. f(x) =ln|x − 2| is discontinuous at 2 since f(2) = ln 0 is not defined.<br />
17. f(x) =<br />
<br />
e<br />
x<br />
if x
F.<br />
TX.10<br />
SECTION 2.5 CONTINUITY ¤ 57<br />
25. By Theorem 7, the exponential function e −5t and the trigonometric function cos 2πt are continuous on (−∞, ∞).<br />
By part 4 of Theorem 4, L(t) =e −5t cos 2πt is continuous on (−∞, ∞).<br />
27. By Theorem 5, the polynomial t 4 − 1 is continuous on (−∞, ∞). By Theorem 7, ln x is continuous on its domain, (0, ∞).<br />
By Theorem 9, ln t 4 − 1 is continuous on its domain, which is<br />
<br />
t | t 4 − 1 > 0 = t | t 4 > 1 = {t ||t| > 1} =(−∞, −1) ∪ (1, ∞)<br />
29. The function y =<br />
1<br />
is discontinuous at x =0because the<br />
1+e1/x left- and right-hand limits at x =0are different.<br />
31. Because we are dealing with root functions, 5+ √ x is continuous on [0, ∞), √ x +5is continuous on [−5, ∞), sothe<br />
quotient f(x) = 5+√ x<br />
√ 5+x<br />
is continuous on [0, ∞). Sincef is continuous at x =4, lim<br />
x→4<br />
f(x) =f(4) = 7 3 .<br />
33. Because x 2 − x is continuous on R, the composite function f(x) =e x2 −x is continuous on R, so<br />
lim<br />
x→1 f(x) =f(1) = e1 − 1 = e 0 =1.<br />
<br />
x<br />
2<br />
if x
F.<br />
58 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
⎧<br />
x +2 ⎪⎨<br />
if x1<br />
TX.10<br />
f is continuous on (−∞, 0) and (1, ∞) sinceoneachoftheseintervals<br />
it is a polynomial; it is continuous on (0, 1) since it is an exponential.<br />
Now lim f(x) = lim +2)=2and lim f(x) = lim<br />
x→0− x→0−(x ex =1,sof is discontinuous at 0. Sincef(0) = 1, f is<br />
x→0 + x→0 +<br />
continuous from the right at 0. Also lim f(x) = lim ex = e and lim f(x) = lim − x) =1,sof is discontinuous<br />
x→1− x→1 − x→1 + x→1 +(2<br />
at 1. Sincef(1) = e, f is continuous from the left at 1.<br />
41. f(x) =<br />
<br />
cx 2 +2x if x
F.<br />
TX.10<br />
SECTION 2.5 CONTINUITY ¤ 59<br />
51. (a) f(x) =cosx − x 3 is continuous on the interval [0, 1], f(0) = 1 > 0,andf(1) = cos 1 − 1 ≈−0.46 < 0. Since<br />
1 > 0 > −0.46, there is a number c in (0, 1) such that f(c) =0by the Intermediate Value Theorem. Thus, there is a root<br />
of the equation cos x − x 3 =0,orcos x = x 3 , in the interval (0, 1).<br />
(b) f(0.86) ≈ 0.016 > 0 and f(0.87) ≈−0.014 < 0, so there is a root between 0.86 and 0.87, that is, in the interval<br />
(0.86, 0.87).<br />
53. (a) Let f(x) =100e −x/100 − 0.01x 2 . Then f(0) = 100 > 0 and<br />
f(100) = 100e −1 − 100 ≈−63.2 < 0. So by the Intermediate<br />
Value Theorem, there is a number c in (0, 100) such that f(c) =0.<br />
This implies that 100e −c/100 =0.01c 2 .<br />
(b) Using the intersect feature of the graphing device, we find that the<br />
root of the equation is x =70.347, correct to three decimal places.<br />
55. (⇒)Iff is continuous at a,thenbyTheorem8withg(h) =a + h,wehave<br />
<br />
<br />
lim f(a + h) =f lim (a + h) = f(a).<br />
h→0 h→0<br />
(⇐)Letε>0. Sincelim<br />
h→0<br />
f(a + h) =f(a),thereexistsδ>0 such that 0 < |h|
F.<br />
60 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
lim<br />
x→0 (−x4 )=0and lim x 4 =0, the Squeeze Theorem gives us lim(x 4 sin(1/x)) = 0,whichequalsf(0). Thus,f is<br />
x→0 x→0<br />
continuous at 0 and, hence, on (−∞, ∞).<br />
65. Define u(t) to be the monk’s distance from the monastery, as a function of time, on the first day, and define d(t) to be his<br />
distance from the monastery, as a function of time, on the second day. Let D be the distance from the monastery to the top of<br />
the mountain. From the given information we know that u(0) = 0, u(12) = D, d(0) = D and d(12) = 0. Now consider the<br />
function u − d, which is clearly continuous. We calculate that (u − d)(0) = −D and (u − d)(12) = D.Sobythe<br />
Intermediate Value Theorem, there must be some time t 0 between 0 and 12 such that (u − d)(t 0)=0 ⇔ u(t 0)=d(t 0).<br />
So at time t 0 after 7:00 AM, the monk will be at the same place on both days.<br />
2.6 Limits at Infinity; Horizontal Asymptotes<br />
1. (a) As x becomes large, the values of f(x) approach 5.<br />
(b) As x becomes large negative, the values of f(x) approach 3.<br />
3. (a) lim f(x) =∞ (b) lim f(x) =∞ (c) lim f(x) =−∞<br />
x→2 − +<br />
x→−1<br />
(d) lim f(x) =1 (e) lim f(x) =2 (f ) Vertical: x = −1, x =2; Horizontal: y =1, y =2<br />
x→∞ x→−∞<br />
x→−1<br />
5. f(0) = 0, f(1) = 1,<br />
lim f(x) =0,<br />
x→∞<br />
f is odd<br />
7. lim f(x) =−∞, lim f(x) =∞, 9. f(0) = 3, lim f(x) =4,<br />
x→2 x→∞ x→0− lim f(x) =0, lim f(x) =∞, lim f(x) =2,<br />
x→−∞ x→0 +<br />
x→0 +<br />
lim f(x) =−∞ lim f(x) =−∞, lim f(x) =−∞,<br />
x→0 − x→−∞ x→4− lim f(x) =∞, lim<br />
x→4 +<br />
f(x) =3<br />
x→∞<br />
11. If f(x) =x 2 /2 x , then a calculator gives f(0) = 0, f(1) = 0.5, f(2) = 1, f(3) = 1.125, f(4) = 1, f(5) = 0.78125,<br />
f(6) = 0.5625, f(7) = 0.3828125, f(8) = 0.25, f(9) = 0.158203125, f(10) = 0.09765625, f(20) ≈ 0.00038147,<br />
f(50) ≈ 2.2204 × 10 −12 , f(100) ≈ 7.8886 × 10 −27 .<br />
<br />
It appears that lim x 2 /2 x =0.<br />
x→∞
F.<br />
TX.10<br />
SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ¤ 61<br />
13. lim<br />
x→∞<br />
3x 2 − x +4<br />
2x 2 +5x − 8 = lim<br />
x→∞<br />
(3x 2 − x +4)/x 2<br />
(2x 2 +5x − 8)/x 2 [divide both the numerator and denominator by x 2<br />
(the highest power of x thatappears in the denominator)]<br />
=<br />
lim (3 − 1/x<br />
x→∞ +4/x2 )<br />
lim (2 + 5/x −<br />
x→∞ 8/x2 )<br />
lim 3 − lim (1/x)+ lim<br />
x→∞ x→∞ x→∞ (4/x2 )<br />
=<br />
lim 2 + lim (5/x) − lim<br />
x→∞ x→∞ x→∞ (8/x2 )<br />
=<br />
3 − lim (1/x)+4 lim<br />
x→∞ x→∞ (1/x2 )<br />
2+5 lim(1/x) − 8 lim<br />
x→∞ x→∞ (1/x2 )<br />
= 3 − 0+4(0)<br />
2+5(0)− 8(0)<br />
[Limit Law 5]<br />
[Limit Laws 1 and 2]<br />
[Limit Laws 7 and 3]<br />
[Theorem 5 of Section 2.5]<br />
= 3 2<br />
1<br />
15. lim<br />
x→∞ 2x +3 = lim 1/x<br />
lim (1/x)<br />
lim<br />
x→∞ (2x +3)/x = x→∞<br />
=<br />
lim (2 + 3/x)<br />
1 − x − x 2<br />
17. lim<br />
x→−∞ 2x 2 − 7<br />
x→∞<br />
x→∞ (1/x)<br />
lim 2 + 3 lim (1/x) = 0<br />
x→∞ x→∞<br />
(1 − x − x 2 )/x 2 lim<br />
x→−∞ (1/x2 − 1/x − 1)<br />
= lim<br />
=<br />
x→−∞ (2x 2 − 7)/x 2 lim (2 −<br />
x→−∞ 7/x2 )<br />
lim<br />
x→−∞ (1/x2 ) − lim (1/x) − lim<br />
x→−∞<br />
=<br />
lim 2 − 7 lim<br />
x→−∞ x→−∞ (1/x2 )<br />
x→−∞ 1<br />
= 0 − 0 − 1<br />
2 − 7(0) = −1 2<br />
2 + 3(0) = 0 2 =0<br />
19. Divide both the numerator and denominator by x 3 (the highest power of x that occurs in the denominator).<br />
lim<br />
x→∞<br />
x 3 +5x<br />
2x 3 − x 2 +4 = lim<br />
x→∞<br />
=<br />
x 3 +5x<br />
x 3<br />
2x 3 − x 2 +4<br />
x 3<br />
lim 1+5 lim<br />
x→∞ x→∞<br />
lim 2 − lim<br />
x→∞ x→∞<br />
= lim<br />
x→∞<br />
1<br />
x 2<br />
1<br />
+4 lim<br />
x x→∞<br />
1+ 5 x 2<br />
2 − 1 x + 4 x 3 =<br />
lim<br />
1+ 5 <br />
x<br />
2 2 − 1 x + 4 <br />
x 3<br />
x→∞<br />
lim<br />
x→∞<br />
= 1 + 5(0)<br />
1 2 − 0+4(0) = 1 2<br />
x 3<br />
21. First, multiply the factors in the denominator. Then divide both the numerator and denominator by u 4 .<br />
lim<br />
u→∞<br />
4u 4 +5<br />
(u 2 − 2)(2u 2 − 1) = lim<br />
u→∞<br />
=<br />
lim<br />
u→∞<br />
4u 4 +5<br />
2u 4 − 5u 2 +2 = lim<br />
u→∞<br />
lim<br />
4+ 5 <br />
u<br />
2 4 − 5 u + 2 =<br />
2 u 4<br />
u→∞<br />
4u 4 +5<br />
u 4<br />
2u 4 − 5u 2 +2<br />
u 4<br />
= lim<br />
u→∞<br />
lim 4+5 lim<br />
u→∞ u→∞<br />
lim 2 − 5 lim<br />
u→∞ u→∞<br />
1<br />
u 4<br />
1<br />
+2 lim<br />
u2 u→∞<br />
4+ 5 u 4<br />
2 − 5 u 2 + 2 u 4<br />
=<br />
1<br />
u 4<br />
4+5(0)<br />
2 − 5(0) + 2(0) = 4 2 =2
F.<br />
62 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
√ √<br />
9x6 − x 9x6 − x/x 3<br />
23. lim<br />
= lim<br />
=<br />
x→∞ x 3 +1 x→∞ (x 3 +1)/x 3<br />
=<br />
<br />
lim 9 − 1/x<br />
5<br />
x→∞<br />
lim<br />
x→∞<br />
= √ 9 − 0=3<br />
<br />
lim<br />
1+ lim<br />
x→∞ (1/x3 ) =<br />
<br />
lim (9x6 − x)/x 6<br />
x→∞<br />
lim (1 +<br />
x→∞ 1/x3 )<br />
x→∞ 9 − lim<br />
x→∞ (1/x5 )<br />
1+0<br />
[since x 3 = √ x 6 for x>0]<br />
√<br />
25. lim 9x2 + x − 3x = lim<br />
x→∞<br />
x→∞<br />
= lim<br />
x→∞<br />
= lim<br />
x→∞<br />
√<br />
27. lim x2 + ax − √ x 2 + bx = lim<br />
x→∞<br />
x→∞<br />
√<br />
9x2 + x − 3x √ 9x 2 + x +3x <br />
√<br />
9x2 + x +3x<br />
9x 2 + x − 9x 2<br />
√<br />
9x2 + x +3x = lim<br />
x→∞<br />
x/x<br />
<br />
9x2 /x 2 + x/x 2 +3x/x = lim<br />
x→∞<br />
= lim<br />
x→∞<br />
= lim<br />
x→∞<br />
= lim<br />
x→∞<br />
x<br />
√<br />
9x2 + x +3x · 1/x<br />
1/x<br />
1<br />
<br />
9+1/x +3<br />
=<br />
√<br />
9x2 + x 2<br />
− (3x)<br />
2<br />
√<br />
9x2 + x +3x<br />
√<br />
x2 + ax − √ x 2 + bx √ x 2 + ax + √ x 2 + bx <br />
√<br />
x2 + ax + √ x 2 + bx<br />
(x 2 + ax) − (x 2 + bx)<br />
√<br />
x2 + ax + √ x 2 + bx = lim<br />
x→∞<br />
a − b<br />
<br />
1+a/x +<br />
<br />
1+b/x<br />
=<br />
1<br />
√<br />
9+3<br />
= 1<br />
3+3 = 1 6<br />
[(a − b)x]/x<br />
√<br />
x2 + ax + √ x 2 + bx / √ x 2<br />
a − b<br />
√ 1+0+<br />
√ 1+0<br />
= a − b<br />
2<br />
x + x 3 + x 5<br />
29. lim<br />
x→∞ 1 − x 2 + x = lim (x + x 3 + x 5 )/x 4<br />
[divide by the highest power of x in the denominator]<br />
4 x→∞ (1 − x 2 + x 4 )/x 4<br />
= lim<br />
x→∞<br />
1/x 3 +1/x + x<br />
1/x 4 − 1/x 2 +1 = ∞<br />
because (1/x 3 +1/x + x) →∞and (1/x 4 − 1/x 2 +1)→ 1 as x →∞.<br />
31. lim<br />
x→−∞ (x4 + x 5 )= lim<br />
x→−∞ x5 ( 1 +1) [factor out the largest power of x] = −∞ because x x5 →−∞and 1/x +1→ 1<br />
as x →−∞.<br />
<br />
Or: lim x 4 + x 5 = lim<br />
x→−∞<br />
x→−∞ x4 (1 + x) =−∞.<br />
33. lim<br />
x→∞<br />
1 − e x<br />
1+2e x = lim<br />
x→∞<br />
(1 − e x )/e x<br />
(1 + 2e x )/e = lim 1/e x − 1<br />
x x→∞ 1/e x +2 = 0 − 1<br />
0+2 = −1 2<br />
35. Since −1 ≤ cos x ≤ 1 and e −2x > 0, wehave−e −2x ≤ e −2x cos x ≤ e −2x .Weknowthat lim<br />
<br />
lim<br />
e<br />
−2x<br />
=0,sobytheSqueezeTheorem, lim<br />
x→∞<br />
x→∞ (e−2x cos x) =0.<br />
x→∞ (−e−2x )=0and<br />
37. (a)<br />
x<br />
f(x)<br />
−10,000 −0.4999625<br />
−100,000 −0.4999962<br />
−1,000,000 −0.4999996<br />
From the graph of f(x) = √ x 2 + x +1+x,weestimate<br />
the value of<br />
lim<br />
x→−∞<br />
f(x) to be −0.5.<br />
(b)<br />
Fromthetable,weestimatethelimit<br />
to be −0.5.
F.<br />
(c)<br />
lim<br />
x→−∞<br />
√<br />
x2 + x +1+x =<br />
lim<br />
x→−∞<br />
= lim<br />
x→−∞<br />
=<br />
√<br />
x2 + x +1+x √ <br />
x 2 + x +1− x<br />
√ = lim<br />
x2 + x +1− x x→−∞<br />
(x +1)(1/x)<br />
√<br />
x2 + x +1− x (1/x) =<br />
1+0<br />
− √ 1+0+0− 1 = − 1 2<br />
lim<br />
x→−∞<br />
1+(1/x)<br />
− 1+(1/x)+(1/x 2 ) − 1<br />
Note that for x
F.<br />
64 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
45. From the graph, it appears y =1is a horizontal asymptote.<br />
3x 3 +500x 2<br />
3x 3 +500x 2<br />
lim<br />
x→∞ x 3 + 500x 2 +100x +2000 = lim<br />
x 3<br />
3+(500/x)<br />
= lim<br />
x→∞ x 3 +500x 2 + 100x + 2000 x→∞ 1+(500/x) + (100/x 2 )+(2000/x 3 )<br />
x 3<br />
3+0<br />
=<br />
=3, soy =3is a horizontal asymptote.<br />
1+0+0+0<br />
The discrepancy can be explained by the choice of the viewing window. Try<br />
[−100,000, 100,000] by [−1, 4] to get a graph that lends credibility to our<br />
calculation that y =3is a horizontal asymptote.<br />
47. Let’s look for a rational function.<br />
(1) lim f(x) =0 ⇒ degree of numerator < degree of denominator<br />
x→±∞<br />
(2) lim f(x) =−∞ ⇒ there is a factor of x 2 in the denominator (not just x, since that would produce a sign<br />
x→0<br />
change at x =0), and the function is negative near x =0.<br />
(3) lim f(x) =∞ and lim f(x) =−∞ ⇒ vertical asymptote at x =3;thereisafactorof(x − 3) in the<br />
x→3− x→3 +<br />
denominator.<br />
(4) f(2) = 0 ⇒ 2 is an x-intercept; there is at least one factor of (x − 2) in the numerator.<br />
Combining all of this information and putting in a negative sign to give us the desired left- and right-hand limits gives us<br />
f(x) =<br />
2 − x as one possibility.<br />
x 2 (x − 3)<br />
49. y = f(x) =x 4 − x 6 = x 4 (1 − x 2 )=x 4 (1 + x)(1 − x). They-intercept is<br />
f(0) = 0. Thex-intercepts are 0, −1,and1 [found by solving f(x) =0for x].<br />
Since x 4 > 0 for x 6= 0, f doesn’t change sign at x =0. The function does change<br />
sign at x = −1 and x =1.Asx → ±∞, f(x) =x 4 (1 − x 2 ) approaches −∞<br />
because x 4 →∞and (1 − x 2 ) →−∞.<br />
51. y = f(x) =(3− x)(1 + x) 2 (1 − x) 4 .They-intercept is f(0) = 3(1) 2 (1) 4 =3.<br />
The x-intercepts are 3, −1,and1. There is a sign change at 3,butnotat−1 and 1.<br />
When x is large positive, 3 − x is negative and the other factors are positive, so<br />
lim f(x) =−∞. Whenx is large negative, 3 − x is positive, so<br />
x→∞<br />
lim f(x) =∞.<br />
x→−∞<br />
53. (a) Since −1 ≤ sin x ≤ 1 for all x, − 1 x ≤ sin x<br />
x<br />
≤ 1 for x>0. Asx →∞, −1/x → 0 and 1/x → 0, so by the Squeeze<br />
x<br />
sin x<br />
Theorem, (sin x)/x → 0. Thus, lim<br />
x→∞ x =0.
F.<br />
TX.10<br />
SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ¤ 65<br />
(b) From part (a), the horizontal asymptote is y =0. The function<br />
y =(sinx)/x crosses the horizontal asymptote whenever sin x =0;<br />
that is, at x = πn for every integer n. Thus, the graph crosses the<br />
asymptote an infinite number of times.<br />
55. Divide the numerator and the denominator by the highest power of x in Q(x).<br />
(a) If deg Pdeg Q, then the numerator → ±∞ but the denominator doesn’t, so lim [P (x)/Q(x)] = ±∞<br />
x→∞<br />
57. lim<br />
x→∞<br />
(depending on the ratio of the leading coefficients of P and Q).<br />
5 √ x<br />
√ · 1/√ x<br />
x − 1 1/ √ x = lim<br />
x→∞<br />
5<br />
<br />
1 − (1/x)<br />
=<br />
5<br />
√ 1 − 0<br />
=5and<br />
10e x − 21<br />
lim<br />
· 1/ex<br />
x→∞ 2e x 1/e = lim 10 − (21/e x )<br />
= 10 − 0 =5.Since 10ex − 21<br />
F.<br />
66 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
65. (a) 1/x 2 < 0.0001 ⇔ x 2 > 1/0.0001 = 10 000 ⇔ x>100 (x >0)<br />
(b) If ε>0 is given, then 1/x 2 1/ε ⇔ x>1/ √ ε.LetN =1/ √ ε.<br />
Then x>N ⇒ x> √ 1 ⇒<br />
ε 1 <br />
x − 0 = 1 1<br />
M.Nowe x >M ⇔ x>ln M, sotake<br />
N =max(1, ln M). (ThisensuresthatN>0.) Then x>N=max(1, ln M) ⇒ e x > max(e, M) ≥ M,<br />
so lim<br />
x→∞ ex = ∞.<br />
71. Suppose that lim f(x) =L. Then for every ε>0 there is a corresponding positive number N such that |f(x) − L| N.Ift =1/x,thenx>N ⇔ 0 < 1/x < 1/N ⇔ 0 0 (namely 1/N ) such that |f(1/t) − L|
F.<br />
TX.10<br />
SECTION 2.7 DERIVATIVES AND RATES OF CHANGE ¤ 67<br />
(c)<br />
The graph of y =2x +1is tangent to the graph of y =4x − x 2 at the<br />
point (1, 3). Now zoom in toward the point (1, 3) until the parabola and<br />
the tangent line are indistiguishable.<br />
5. Using (1) with f(x) = x − 1 and P (3, 2),<br />
x − 2<br />
x − 1<br />
f(x) − f(a)<br />
m = lim<br />
=lim<br />
x − 2 − 2<br />
x→a x − a x→3 x − 3<br />
=lim<br />
x→3<br />
3 − x<br />
(x − 2)(x − 3) =lim<br />
x→3<br />
x − 1 − 2(x − 2)<br />
=lim<br />
x − 2<br />
x→3 x − 3<br />
−1<br />
x − 2 = −1<br />
1 = −1<br />
Tangent line: y − 2=−1(x − 3) ⇔ y − 2=−x +3 ⇔ y = −x +5<br />
√ √<br />
x − 1 ( √ x − 1)( √ x +1)<br />
7. Using (1), m =lim =lim<br />
x→1 x − 1 x→1 (x − 1)( √ x − 1<br />
= lim<br />
x +1) x→1 (x − 1)( √ x +1) =lim 1<br />
√ = 1<br />
x→1 x +1 2 .<br />
Tangent line: y − 1= 1 2 (x − 1) ⇔ y = 1 2 x + 1 2<br />
9. (a) Using (2) with y = f(x) =3+4x 2 − 2x 3 ,<br />
f(a + h) − f(a) 3+4(a + h) 2 − 2(a + h) 3 − (3 + 4a 2 − 2a 3 )<br />
m =lim<br />
= lim<br />
h→0 h<br />
h→0 h<br />
=lim<br />
h→0<br />
3+4(a 2 +2ah + h 2 ) − 2(a 3 +3a 2 h +3ah 2 + h 3 ) − 3 − 4a 2 +2a 3<br />
h<br />
=lim<br />
h→0<br />
3+4a 2 +8ah +4h 2 − 2a 3 − 6a 2 h − 6ah 2 − 2h 3 − 3 − 4a 2 +2a 3<br />
h<br />
8ah +4h 2 − 6a 2 h − 6ah 2 − 2h 3 h(8a +4h − 6a 2 − 6ah − 2h 2 )<br />
=lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
=lim<br />
h→0<br />
(8a +4h − 6a 2 − 6ah − 2h 2 )=8a − 6a 2<br />
(b) At (1, 5): m =8(1)− 6(1) 2 =2,soanequationofthetangentline<br />
is y − 5=2(x − 1) ⇔ y =2x +3.<br />
(c)<br />
At (2, 3): m =8(2)− 6(2) 2 = −8, so an equation of the tangent<br />
line is y − 3=−8(x − 2) ⇔ y = −8x +19.<br />
11. (a) The particle is moving to the right when s is increasing; that is, on the intervals (0, 1) and (4, 6). The particle is moving to<br />
the left when s is decreasing; that is, on the interval (2, 3). The particle is standing still when s is constant; that is, on the<br />
intervals (1, 2) and (3, 4).
F.<br />
68 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
(b) The velocity of the particle is equal to the slope of the tangent line of the<br />
graph. Note that there is no slope at the corner points on the graph. On the<br />
interval (0, 1), the slope is 3 − 0 =3.Ontheinterval(2, 3), the slope is<br />
1 − 0<br />
1 − 3<br />
3 − 1<br />
= −2. On the interval (4, 6), the slope is<br />
3 − 2 6 − 4 =1.<br />
13. Let s(t) =40t − 16t 2 .<br />
<br />
s(t) − s(2)<br />
40t − 16t<br />
2<br />
− 16 −16t 2 +40t − 16 −8 2t 2 − 5t +2 <br />
v(2) = lim<br />
=lim<br />
=lim<br />
=lim<br />
t→2 t − 2 t→2 t − 2<br />
t→2 t − 2<br />
t→2 t − 2<br />
−8(t − 2)(2t − 1)<br />
=lim<br />
= −8lim(2t − 1) = −8(3) = −24<br />
t→2 t − 2<br />
t→2<br />
Thus, the instantaneous velocity when t =2is −24 ft/s.<br />
1<br />
s(a + h) − s(a) (a + h) − 1 15. v(a)= lim<br />
= lim<br />
2 a 2<br />
h→0 h<br />
h→0 h<br />
a 2 − (a + h) 2<br />
a<br />
=lim<br />
2 (a + h) 2<br />
h→0 h<br />
= lim<br />
h→0<br />
a 2 − (a 2 +2ah + h 2 )<br />
ha 2 (a + h) 2<br />
−(2ah + h 2 )<br />
=lim<br />
h→0 ha 2 (a + h) =lim −h(2a + h)<br />
2 h→0 ha 2 (a + h) = lim −(2a + h)<br />
2 h→0 a 2 (a + h) = −2a<br />
2 a 2 · a = −2<br />
2 a m/s 3<br />
So v (1) = −2<br />
−2<br />
= −2 m/s, v(2) =<br />
13 2 = − 1 −2<br />
m/s, and v(3) = 3 4 3 = − 2 3 27 m/s.<br />
17. g 0 (0) istheonlynegativevalue.Theslopeatx =4is smaller than the slope at x =2and both are smaller than the slope at<br />
x = −2. Thus, g 0 (0) < 0
F.<br />
TX.10<br />
SECTION 2.7 DERIVATIVES AND RATES OF CHANGE ¤ 69<br />
23. (a) Using Definition 2 with F (x) =5x/(1 + x 2 ) and the point (2, 2),wehave (b)<br />
5(2 + h)<br />
F 0 F (2 + h) − F (2) 1+(2+h) − 2<br />
(2) = lim<br />
=lim<br />
2<br />
h→0 h<br />
h→0 h<br />
5h +10<br />
= lim<br />
h 2 +4h +5 − 2 5h +10− 2(h 2 +4h +5)<br />
=lim<br />
h 2 +4h +5<br />
h→0 h<br />
h→0 h<br />
−2h 2 − 3h<br />
= lim<br />
h→0 h(h 2 +4h +5) =lim h(−2h − 3)<br />
h→0 h(h 2 +4h +5) =lim −2h − 3<br />
h→0 h 2 +4h +5 = −3<br />
5<br />
So an equation of the tangent line at (2, 2) is y − 2=− 3 (x − 2) or y = − 3 x + 16 . 5 5 5<br />
25. Use Definition 2 with f(x) =3− 2x +4x 2 .<br />
f 0 f(a + h) − f(a) [3 − 2(a + h)+4(a + h) 2 ] − (3 − 2a +4a 2 )<br />
(a) = lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
= lim<br />
h→0<br />
(3 − 2a − 2h +4a 2 +8ah +4h 2 ) − (3 − 2a +4a 2 )<br />
h<br />
−2h +8ah +4h 2 h(−2+8a +4h)<br />
= lim<br />
=lim<br />
=lim(−2+8a +4h) =−2+8a<br />
h→0 h<br />
h→0 h<br />
h→0<br />
27. Use Definition 2 with f(t) =(2t +1)/(t +3).<br />
2(a + h)+1<br />
f 0 f(a + h) − f(a) (a + h)+3<br />
(a) = lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
−<br />
2a +1<br />
a +3<br />
= lim<br />
h→0<br />
(2a 2 +6a +2ah +6h + a +3)− (2a 2 +2ah +6a + a + h +3)<br />
h(a + h +3)(a +3)<br />
= lim<br />
h→0<br />
5h<br />
h(a + h +3)(a +3) =lim<br />
h→0<br />
29. Use Definition 2 with f(x) =1/ √ x +2.<br />
f 0 f(a + h) − f(a)<br />
(a) = lim<br />
=lim<br />
h→0 h<br />
√ √ a +2− a + h +2<br />
= lim<br />
h→0 h √ a + h +2 √ a +2 ·<br />
= lim<br />
h→0<br />
5<br />
(a + h +3)(a +3) = 5<br />
(a +3) 2<br />
= lim<br />
h→0<br />
(2a +2h +1)(a +3)− (2a +1)(a + h +3)<br />
h(a + h +3)(a +3)<br />
√ √<br />
1<br />
1<br />
− √ a +2− a + h +2<br />
√ √ (a + h)+2 a +2<br />
a + h +2 a +2<br />
= lim<br />
h→0 h<br />
h→0 h<br />
√ √ a +2+ a + h +2<br />
(a +2)− (a + h +2)<br />
√ √ = lim<br />
a +2+ a + h +2 h→0 h √ a + h +2 √ a +2 √ a +2+ √ a + h +2 <br />
−h<br />
h √ a + h +2 √ a +2 √ a +2+ √ a + h +2 =lim<br />
h→0<br />
−1<br />
1<br />
= √ 2 √ = −<br />
a +2 2 a +2 2(a +2) 3/2<br />
Note that the answers to Exercises 31 – 36 are not unique.<br />
(1 + h) 10 − 1<br />
31. By Definition 2, lim<br />
= f 0 (1),wheref(x) =x 10 and a =1.<br />
h→0 h<br />
(1 + h) 10 − 1<br />
Or: By Definition 2, lim<br />
= f 0 (0),wheref(x) =(1+x) 10 and a =0.<br />
h→0 h<br />
−1<br />
√ a + h +2<br />
√ a +2<br />
√ a +2+<br />
√ a + h +2
F.<br />
70 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
33. By Equation 3, lim<br />
x→5<br />
2 x − 32<br />
x − 5 = f 0 (5),wheref(x) =2 x and a =5.<br />
cos(π + h)+1<br />
35. By Definition 2, lim<br />
= f 0 (π),wheref(x) =cosx and a = π.<br />
h→0 h<br />
cos(π + h)+1<br />
Or: ByDefinition 2, lim<br />
= f 0 (0),wheref(x) =cos(π + x) and a =0.<br />
h→0 h<br />
37. v(5) = f 0 f(5 + h) − f(5) [100 + 50(5 + h) − 4.9(5 + h) 2 ] − [100 + 50(5) − 4.9(5) 2 ]<br />
(5) = lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
(100 + 250 + 50h − 4.9h 2 − 49h − 122.5) − (100 + 250 − 122.5) −4.9h 2 + h<br />
=lim<br />
= lim<br />
h→0 h<br />
h→0 h<br />
h(−4.9h +1)<br />
=lim<br />
=lim(−4.9h +1)=1m/s<br />
h→0 h<br />
h→0<br />
The speed when t =5is |1| =1m/s.<br />
39. Thesketchshowsthegraphforaroomtemperatureof72 ◦ and a refrigerator<br />
temperature of 38 ◦ . The initial rate of change is greater in magnitude than the<br />
rate of change after an hour.<br />
41. (a) (i) [2000, 2002]:<br />
(ii) [2000, 2001]:<br />
(iii) [1999, 2000]:<br />
P (2002) − P (2000)<br />
2002 − 2000<br />
P (2001) − P (2000)<br />
2001 − 2000<br />
P (2000) − P (1999)<br />
2000 − 1999<br />
(b) Using the values from (ii) and (iii), we have<br />
=<br />
=<br />
=<br />
77 − 55<br />
2<br />
68 − 55<br />
1<br />
55 − 39<br />
1<br />
13 + 16<br />
2<br />
= 22<br />
2 =11percent/year<br />
=13percent/year<br />
=16percent/year<br />
=14.5 percent/year.<br />
(c) Estimating A as (1999, 40) and B as (2001, 70), the slope at 2000 is<br />
70 − 40<br />
2001 − 1999 = 30<br />
2 =15percent/year.<br />
43. (a) (i) ∆C<br />
∆x<br />
(ii) ∆C<br />
∆x<br />
(b)<br />
=<br />
C(105) − C(100)<br />
105 − 100<br />
=<br />
6601.25 − 6500<br />
5<br />
6520.05 − 6500<br />
1<br />
=\$20.25/unit.<br />
C(101) − C(100)<br />
= = =\$20.05/unit.<br />
101 − 100<br />
5000 + 10(100 + h)+0.05(100 + h)<br />
2<br />
− 6500<br />
=<br />
=<br />
h<br />
=20+0.05h, h 6= 0<br />
C(100 + h) − C(100)<br />
h<br />
20h +0.05h2<br />
h<br />
C(100 + h) − C(100)<br />
So the instantaneous rate of change is lim<br />
= lim (20 + 0.05h) =\$20/unit.<br />
h→0 h<br />
h→0<br />
45. (a) f 0 (x) is the rate of change of the production cost with respect to the number of ounces of gold produced. Its units are<br />
dollars per ounce.
F.<br />
TX.10<br />
SECTION 2.8 THEDERIVATIVEASAFUNCTION ¤ 71<br />
(b) After 800 ounces of gold have been produced, the rate at which the production cost is increasing is \$17/ounce. So the cost<br />
of producing the 800th (or 801st) ounce is about \$17.<br />
(c) In the short term, the values of f 0 (x) will decrease because more efficient use is made of start-up costs as x increases. But<br />
eventually f 0 (x) might increase due to large-scale operations.<br />
47. T 0 (10) is the rate at which the temperature is changing at 10:00 AM.ToestimatethevalueofT 0 (10), we will average the<br />
difference quotients obtained using the times t =8and t =12.LetA =<br />
B =<br />
T (12) − T (10)<br />
12 − 10<br />
=<br />
88 − 81<br />
2<br />
T (8) − T (10)<br />
8 − 10<br />
=<br />
72 − 81<br />
−2<br />
=4.5 and<br />
=3.5. ThenT 0 T (t) − T (10)<br />
(10) = lim<br />
≈ A + B = 4.5+3.5 =4 ◦ F/h.<br />
t→10 t − 10 2 2<br />
49. (a) S 0 (T ) is the rate at which the oxygen solubility changes with respect to the water temperature. Its units are (mg/L)/ ◦ C.<br />
(b) For T =16 ◦ C, it appears that the tangent line to the curve goes through the points (0, 14) and (32, 6). So<br />
S 0 (16) ≈ 6 − 14<br />
32 − 0 = − 8<br />
32 = −0.25 (mg/L)/◦ C. This means that as the temperature increases past 16 ◦ C, the oxygen<br />
solubility is decreasing at a rate of 0.25 (mg/L)/ ◦ C.<br />
51. Since f(x) =x sin(1/x) when x 6= 0and f(0) = 0, wehave<br />
f 0 f(0 + h) − f(0) h sin(1/h) − 0<br />
(0) = lim<br />
=lim<br />
=limsin(1/h). This limit does not exist since sin(1/h) takes the<br />
h→0 h<br />
h→0 h<br />
h→0<br />
values −1 and 1 on any interval containing 0. (Compare with Example 4 in Section 2.2.)<br />
2.8 The Derivative as a Function<br />
1. It appears that f is an odd function, so f 0 will be an even<br />
function—that is, f 0 (−a) =f 0 (a).<br />
(a) f 0 (−3) ≈ 1.5<br />
(b) f 0 (−2) ≈ 1<br />
(c) f 0 (−1) ≈ 0<br />
(d) f 0 (0) ≈−4<br />
(e) f 0 (1) ≈ 0<br />
(f ) f 0 (2) ≈ 1<br />
(g) f 0 (3) ≈ 1.5<br />
3. (a) 0 = II, since from left to right, the slopes of the tangents to graph (a) start out negative, become 0, then positive, then 0,then<br />
negative again. The actual function values in graph II follow the same pattern.<br />
(b) 0 = IV, since from left to right, the slopes of the tangents to graph (b) start out at a fixed positive quantity, then suddenly<br />
become negative, then positive again. The discontinuities in graph IV indicate sudden changes in the slopes of the tangents.<br />
(c) 0 = I, since the slopes of the tangents to graph (c) are negative for x0,asarethefunctionvaluesof<br />
graph I.<br />
(d) 0 = III, since from left to right, the slopes of the tangents to graph (d) are positive, then 0, then negative, then 0, then<br />
positive, then 0, then negative again, and the function values in graph III follow the same pattern.
F.<br />
72 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
Hints for Exercises 4 –11: First plot x-intercepts on the graph of f 0 for any horizontal tangents on the graph of f . Look for any corners on the graph<br />
of f— there will be a discontinuity on the graph of f 0 . On any interval where f has a tangent with positive (or negative) slope, the graph of f 0 will be<br />
positive (or negative). If the graph of the function is linear, the graph of f 0 will be a horizontal line.<br />
5. 7.<br />
9. 11.<br />
13. It appears that there are horizontal tangents on the graph of M for t = 1963<br />
and t = 1971. Thus, there are zeros for those values of t on the graph of<br />
M 0 . The derivative is negative for the years 1963 to 1971.<br />
15.<br />
The slope at 0 appears to be 1 and the slope at 1<br />
appears to be 2.7. Asx decreases, the slope gets<br />
closer to 0. Since the graphs are so similar, we might<br />
guess that f 0 (x) =e x .
F.<br />
TX.10<br />
SECTION 2.8 THEDERIVATIVEASAFUNCTION ¤ 73<br />
17. (a) By zooming in, we estimate that f 0 (0) = 0, f 0 1<br />
2<br />
<br />
=1, f 0 (1) = 2,<br />
and f 0 (2) = 4.<br />
(b) By symmetry, f 0 (−x) =−f 0 (x). Sof 0 − 1 2<br />
= −1, f 0 (−1) = −2,<br />
and f 0 (−2) = −4.<br />
(c) It appears that f 0 (x) is twice the value of x, so we guess that f 0 (x) =2x.<br />
(d) f 0 f(x + h) − f(x) (x + h) 2 − x 2<br />
(x) =lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
x 2 +2hx + h 2 − x 2 2hx + h 2 h(2x + h)<br />
=lim<br />
=lim =lim<br />
=lim(2x + h) =2x<br />
h→0 h<br />
h→0 h h→0 h<br />
h→0<br />
1<br />
19. f 0 f(x + h) − f(x)<br />
(x)= lim<br />
= lim<br />
(x + h) − 1<br />
2 3 − 1 x − <br />
1<br />
2 3<br />
h→0 h<br />
h→0 h<br />
=lim<br />
1<br />
h 2<br />
h→0<br />
h =lim<br />
h→0<br />
1<br />
2 = 1 2<br />
Domain of f = domain of f 0 = R.<br />
<br />
21. f 0 f(t + h) − f(t)<br />
5(t + h) − 9(t + h)<br />
2<br />
− (5t − 9t 2 )<br />
(t) = lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
1<br />
= lim<br />
x + 1 h − 1 − 1 x + 1 2 2 3 2 3<br />
h→0 h<br />
5t +5h − 9(t 2 +2th + h 2 ) − 5t +9t 2 5t +5h − 9t 2 − 18th − 9h 2 − 5t +9t 2<br />
= lim<br />
= lim<br />
h→0 h<br />
h→0 h<br />
5h − 18th − 9h 2 h(5 − 18t − 9h)<br />
= lim<br />
= lim<br />
=lim(5 − 18t − 9h) =5− 18t<br />
h→0 h<br />
h→0 h<br />
h→0<br />
Domain of f = domain of f 0 = R.<br />
<br />
23. f 0 f(x + h) − f(x) (x + h) 3 − 3(x + h)+5 − (x 3 − 3x +5)<br />
(x) = lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
x 3 +3x 2 h +3xh 2 + h 3 − 3x − 3h +5 − x 3 − 3x +5 <br />
3x 2 h +3xh 2 + h 3 − 3h<br />
= lim<br />
= lim<br />
h→0 h<br />
h→0 h<br />
h 3x 2 +3xh + h 2 − 3 <br />
= lim<br />
=lim 3x 2 +3xh + h 2 − 3 =3x 2 − 3<br />
h→0 h<br />
h→0<br />
Domain of f = domain of f 0 = R.<br />
√ √ <br />
25. g 0 g(x + h) − g(x) 1+2(x + h) − 1+2x 1+2(x + h)+ 1+2x<br />
(x) =lim<br />
= lim<br />
√<br />
h→0 h<br />
h→0 h<br />
1+2(x + h)+ 1+2x<br />
(1 + 2x +2h) − (1 + 2x)<br />
2<br />
2<br />
=lim <br />
h→0<br />
1+2(x √ =lim√ √ =<br />
h<br />
+ h)+ 1+2x h→0 1+2x +2h + 1+2x 2 √ 1+2x = 1<br />
√ 1+2x<br />
Domain of g = − 1 2 , ∞ , domain of g 0 = − 1 2 , ∞ .
F.<br />
74 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
4(t + h)<br />
27. G 0 G(t + h) − G(t) (t + h)+1 − 4t 4(t + h)(t +1)− 4t(t + h +1)<br />
t +1<br />
(t + h +1)(t +1)<br />
(t) =lim<br />
=lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
h→0 h<br />
<br />
4t 2 +4ht +4t +4h − 4t 2 +4ht +4t <br />
4h<br />
=lim<br />
= lim<br />
h→0 h(t + h +1)(t +1)<br />
h→0 h(t + h +1)(t +1)<br />
=lim<br />
h→0<br />
4<br />
(t + h +1)(t +1) = 4<br />
(t +1) 2<br />
Domain of G = domain of G 0 =(−∞, −1) ∪ (−1, ∞).<br />
<br />
29. f 0 f(x + h) − f(x) (x + h) 4 − x 4 x 4 +4x 3 h +6x 2 h 2 +4xh 3 + h 4 − x 4<br />
(x) =lim<br />
=lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
h→0 h<br />
4x 3 h +6x 2 h 2 +4xh 3 + h 4 <br />
=lim<br />
=lim 4x 3 +6x 2 h +4xh 2 + h 3 =4x 3<br />
h→0 h<br />
h→0<br />
Domain of f = domain of f 0 = R.<br />
31. (a) f 0 f(x + h) − f(x) [(x + h) 4 +2(x + h)] − (x 4 +2x)<br />
(x)= lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
=lim<br />
h→0<br />
x 4 +4x 3 h +6x 2 h 2 +4xh 3 + h 4 +2x +2h − x 4 − 2x<br />
h<br />
4x 3 h +6x 2 h 2 +4xh 3 + h 4 +2h h(4x 3 +6x 2 h +4xh 2 + h 3 +2)<br />
=lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
=lim<br />
h→0<br />
(4x 3 +6x 2 h +4xh 2 + h 3 +2)=4x 3 +2<br />
(b) Notice that f 0 (x) =0when f has a horizontal tangent, f 0 (x) is<br />
positive when the tangents have positive slope, and f 0 (x) is<br />
negative when the tangents have negative slope.<br />
33. (a) U 0 (t) is the rate at which the unemployment rate is changing with respect to time. Its units are percent per year.<br />
(b) To find U 0 U(t + h) − U(t) U(t + h) − U(t)<br />
(t),weuse lim<br />
≈ for small values of h.<br />
h→0 h<br />
h<br />
For 1993: U 0 (1993) ≈<br />
U(1994) − U(1993)<br />
1994 − 1993<br />
=<br />
6.1 − 6.9<br />
1<br />
= −0.80<br />
For 1994: We estimate U 0 (1994) by using h = −1 and h =1, and then average the two results to obtain a final estimate.<br />
h = −1 ⇒ U 0 (1994) ≈<br />
h =1 ⇒ U 0 (1994) ≈<br />
U(1993) − U(1994)<br />
1993 − 1994<br />
U(1995) − U(1994)<br />
1995 − 1994<br />
=<br />
=<br />
6.9 − 6.1<br />
−1<br />
5.6 − 6.1<br />
1<br />
So we estimate that U 0 (1994) ≈ 1 [(−0.80) + (−0.50)] = −0.65.<br />
2<br />
= −0.80;<br />
= −0.50.<br />
t 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002<br />
U 0 (t) −0.80 −0.65 −0.35 −0.35 −0.45 −0.35 −0.25 0.25 0.90 1.10
F.<br />
TX.10<br />
SECTION 2.8 THEDERIVATIVEASAFUNCTION ¤ 75<br />
35. f is not differentiable at x = −4, because the graph has a corner there, and at x =0, because there is a discontinuity there.<br />
37. f is not differentiable at x = −1, because the graph has a vertical tangent there, and at x =4, because the graph has a corner<br />
there.<br />
39. As we zoom in toward (−1, 0), the curve appears more and more like a<br />
straight line, so f(x) =x + |x| is differentiable at x = −1. But no<br />
matter how much we zoom in toward the origin, the curve doesn’t straighten<br />
out—we can’t eliminate the sharp point (a cusp). So f is not differentiable<br />
at x =0.<br />
41. a = f, b = f 0 , c = f 00 . We can see this because where a has a horizontal tangent, b =0,andwhereb has a horizontal tangent,<br />
c =0. We can immediately see that c can be neither f nor f 0 , since at the points where c has a horizontal tangent, neither a<br />
nor b is equal to 0.<br />
43. We can immediately see that a is the graph of the acceleration function, since at the points where a has a horizontal tangent,<br />
neither c nor b is equal to 0. Next, we note that a =0at the point where b has a horizontal tangent, so b must be the graph of<br />
the velocity function, and hence, b 0 = a. We conclude that c is the graph of the position function.<br />
<br />
45. f 0 f(x + h) − f(x)<br />
1+4(x + h) − (x + h)<br />
2<br />
− (1 + 4x − x 2 )<br />
(x) = lim<br />
= lim<br />
h→0 h<br />
h→0 h<br />
(1 + 4x +4h − x 2 − 2xh − h 2 ) − (1 + 4x − x 2 ) 4h − 2xh − h 2<br />
=lim<br />
= lim<br />
= lim (4 − 2x − h) =4− 2x<br />
h→0 h<br />
h→0 h<br />
h→0<br />
f 00 f 0 (x + h) − f 0 (x) [4 − 2(x + h)] − (4 − 2x) −2h<br />
(x) = lim<br />
= lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
h→0 h<br />
= lim (−2) = −2<br />
h→0<br />
We see from the graph that our answers are reasonable because the graph of<br />
f 0 is that of a linear function and the graph of f 00 is that of a constant<br />
function.
F.<br />
76 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
<br />
47. f 0 f(x + h) − f(x) 2(x + h) 2 − (x + h) 3 − (2x 2 − x 3 )<br />
(x) = lim<br />
= lim<br />
h→0 h<br />
h→0 h<br />
h(4x +2h − 3x 2 − 3xh − h 2 )<br />
=lim<br />
=lim(4x +2h − 3x 2 − 3xh − h 2 )=4x − 3x 2<br />
h→0 h<br />
h→0<br />
<br />
f 00 f 0 (x + h) − f 0 (x)<br />
4(x + h) − 3(x + h)<br />
2<br />
− (4x − 3x 2 ) h(4 − 6x − 3h)<br />
(x) = lim<br />
= lim<br />
= lim<br />
h→0 h<br />
h→0 h<br />
h→0 h<br />
=lim(4 − 6x − 3h) =4− 6x<br />
h→0<br />
f 000 f 00 (x + h) − f 00 (x) [4 − 6(x + h)] − (4 − 6x) −6h<br />
(x) = lim<br />
=lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
h→0 h<br />
=lim (−6) = −6<br />
h→0<br />
f (4) f 000 (x + h) − f 000 (x) −6 − (−6) 0<br />
(x) = lim<br />
=lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
h→0 h =lim(0)<br />
= 0<br />
h→0<br />
The graphs are consistent with the geometric interpretations of the<br />
derivatives because f 0 has zeros where f has a local minimum and a local<br />
maximum, f 00 has a zero where f 0 has a local maximum, and f 000 is a<br />
constant function equal to the slope of f 00 .<br />
49. (a)Notethatwehavefactoredx − a as the difference of two cubes in the third step.<br />
f 0 f(x) − f(a) x 1/3 − a 1/3<br />
x 1/3 − a 1/3<br />
(a) =lim<br />
=lim<br />
= lim<br />
x→a x − a x→a x − a x→a (x 1/3 − a 1/3 )(x 2/3 + x 1/3 a 1/3 + a 2/3 )<br />
1<br />
=lim<br />
x→a x 2/3 + x 1/3 a 1/3 + a = 1<br />
2/3 3a or 1 2/3 3 a−2/3<br />
(b) f 0 f(0 + h) − f(0)<br />
√ 3<br />
h − 0<br />
(0) = lim<br />
= lim = lim<br />
h→0 h<br />
h→0 h<br />
exist, and therefore f 0 (0) does not exist.<br />
(c) lim<br />
x→0<br />
|f 0 (x)| =lim<br />
51. f(x) =|x − 6| =<br />
x→0<br />
1<br />
TX.10<br />
1<br />
h→0 h<br />
x − 6 = lim<br />
x→6<br />
|x − 6| .<br />
2/3<br />
. This function increases without bound, so the limit does not<br />
= ∞ and f is continuous at x =0(root function), so f has a vertical tangent at x =0.<br />
3x2/3 <br />
x − 6 if x − 6 ≥ 6 x − 6 if x ≥ 6<br />
−(x − 6) if x − 6 < 0 = 6 − x if x6<br />
However, a formula for f 0 is f 0 (x) =<br />
−1 if x
F.<br />
TX.10<br />
SECTION 2.8 THEDERIVATIVEASAFUNCTION ¤ 77<br />
53. (a) f(x) =x |x| =<br />
<br />
x<br />
2<br />
if x ≥ 0<br />
−x 2 if x0.<br />
[See Exercise 2.8.17(d).] Similarly, since f(x) =−x 2 for<br />
x
F.<br />
78 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
2 Review<br />
1. (a) lim<br />
x→a<br />
f(x) =L: SeeDefinition2.2.1andFigures1and2inSection2.2.<br />
(b)<br />
(c)<br />
lim f(x) =L: See the paragraph after Definition 2.2.2 and Figure 9(b) in Section 2.2.<br />
x→a +<br />
lim f(x) =L: SeeDefinition 2.2.2 and Figure 9(a) in Section 2.2.<br />
x→a− (d) lim<br />
x→a<br />
f(x) =∞: SeeDefinition 2.2.4 and Figure 12 in Section 2.2.<br />
(e) lim f(x) =L: SeeDefinition 2.6.1 and Figure 2 in Section 2.6.<br />
x→∞<br />
2. In general, the limit of a function fails to exist when the function does not approach a fixed number. For each of the following<br />
functions, the limit fails to exist at x =2.<br />
The left- and right-hand<br />
limits are not equal.<br />
There is an<br />
infinite discontinuity.<br />
There are an infinite<br />
number of oscillations.<br />
3. (a) – (g) See the statements of Limit Laws 1– 6 and 11 in Section 2.3.<br />
4. See Theorem 3 in Section 2.3.<br />
5. (a) See Definition 2.2.6 and Figures 12–14 in Section 2.2.<br />
(b) See Definition 2.6.3 and Figures 3 and 4 in Section 2.6.<br />
6. (a) y = x 4 :Noasymptote (b)y =sinx: No asymptote<br />
(c) y =tanx: Vertical asymptotes x = π + πn, n an integer (d) y 2 =tan−1 x: Horizontal asymptotes y = ± π 2<br />
(e) y = e x : Horizontal asymptote y =0<br />
<br />
<br />
lim<br />
x→−∞ ex =0<br />
(g) y =1/x: Vertical asymptote x =0,<br />
horizontal asymptote y =0<br />
(f ) y =lnx: Vertical asymptote x =0<br />
<br />
lim<br />
x→0 + ln x = −∞ <br />
(h) y = √ x: No asymptote<br />
7. (a) A function f is continuous at a number a if f(x) approaches f(a) as x approaches a;thatis, lim<br />
x→a<br />
f(x) =f(a).<br />
(b) A function f is continuous on the interval (−∞, ∞) if f is continuous at every real number a. The graph of such a<br />
function has no breaks and every vertical line crosses it.
F.<br />
TX.10<br />
CHAPTER 2 REVIEW ¤ 79<br />
8. See Theorem 2.5.10.<br />
9. See Definition 2.7.1.<br />
10. See the paragraph containing Formula 3 in Section 2.7.<br />
11. (a) The average rate of change of y with respect to x over the interval [x 1 ,x 2 ] is<br />
f(x2) − f(x1)<br />
x 2 − x 1<br />
.<br />
f(x 2) − f(x 1)<br />
(b) The instantaneous rate of change of y with respect to x at x = x 1 is lim<br />
.<br />
x 2 →x 1 x 2 − x 1<br />
12. See Definition 2.7.2. The pages following the definition discuss interpretations of f 0 (a) as the slope of a tangent line to the<br />
graph of f at x = a and as an instantaneous rate of change of f(x) with respect to x when x = a.<br />
13. See the paragraphs before and after Example 6 in Section 2.8.<br />
14. (a) A function f is differentiable at a number a if its derivative f 0 exists<br />
(c)<br />
at x = a;thatis,iff 0 (a) exists.<br />
(b) See Theorem 2.8.4. This theorem also tells us that if f is not<br />
continuous at a,thenf is not differentiable at a.<br />
15. See the discussion and Figure 7 on page 159.<br />
1. False. Limit Law 2 applies only if the individual limits exist (these don’t).<br />
3. True. Limit Law 5 applies.<br />
x(x − 5) sin(x − 5)<br />
5. False. Consider lim or lim .Thefirst limit exists and is equal to 5. By Example 3 in Section 2.2,<br />
x→5 x − 5 x→5 x − 5<br />
we know that the latter limit exists (and it is equal to 1).<br />
7. True. A polynomial is continuous everywhere, so lim<br />
x→b<br />
p(x) exists and is equal to p(b).<br />
9. True. See Figure 8 in Section 2.6.<br />
11. False. Consider f(x) =<br />
<br />
1/(x − 1) if x 6= 1<br />
2 if x =1<br />
13. True. Use Theorem 2.5.8 with a =2, b =5,andg(x) =4x 2 − 11. Notethatf(4) = 3 is not needed.<br />
15. True, by the definition of a limit with ε =1.<br />
17. False. See the note after Theorem 4 in Section 2.8.<br />
19. False.<br />
<br />
d 2 2<br />
y<br />
dy<br />
dx is the second derivative while is the first derivative squared. For example, if y = x,<br />
2 dx<br />
then d 2 2<br />
y dy<br />
dx =0,but =1.<br />
2 dx
F.<br />
80 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
1. (a) (i) lim f(x) =3 (ii) lim f(x) =0<br />
x→2 + x→−3 +<br />
(iii)<br />
lim f(x) does not exist since the left and right limits are not equal. (The left limit is −2.)<br />
x→−3<br />
(iv) lim<br />
x→4<br />
f(x) =2<br />
(v) lim f(x) =∞ (vi) lim f(x) =−∞<br />
x→0 −<br />
(vii) lim f(x) =4 (viii) lim f(x) =−1<br />
x→∞ x→−∞<br />
(b) The equations of the horizontal asymptotes are y = −1 and y =4.<br />
(c) The equations of the vertical asymptotes are x =0and x =2.<br />
(d) f is discontinuous at x = −3, 0, 2,and4. The discontinuities are jump, infinite, infinite, and removable, respectively.<br />
x→2<br />
3. Since the exponential function is continuous, lim<br />
x→1<br />
e x3 −x = e 1−1 = e 0 =1.<br />
x 2 − 9<br />
5. lim<br />
x→−3 x 2 +2x − 3 = lim (x +3)(x − 3)<br />
x→−3 (x +3)(x − 1) = lim x − 3<br />
x→−3 x − 1 = −3 − 3<br />
−3 − 1 = −6<br />
−4 = 3 2<br />
<br />
(h − 1) 3 +1 h 3 − 3h 2 +3h − 1 +1 h 3 − 3h 2 +3h <br />
7. lim<br />
= lim<br />
= lim<br />
= lim h 2 − 3h +3 =3<br />
h→0 h<br />
h→0 h<br />
h→0 h<br />
h→0<br />
Another solution: Factor the numerator as a sum of two cubes and then simplify.<br />
(h − 1) 3 +1 (h − 1) 3 +1 3 [(h − 1) + 1] (h − 1) 2 − 1(h − 1) + 1 2<br />
lim<br />
= lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
h→0 h<br />
9. lim<br />
r→9<br />
= lim<br />
h→0<br />
(h − 1) 2 − h +2 =1− 0+2=3<br />
√ r<br />
(r − 9) 4 = ∞ since (r − 9)4 → 0 as r → 9 and<br />
√ r<br />
> 0 for r 6= 9.<br />
(r − 9)<br />
4<br />
u 4 − 1<br />
11. lim<br />
u→1 u 3 +5u 2 − 6u = lim (u 2 +1)(u 2 − 1)<br />
u→1 u(u 2 +5u − 6) =lim (u 2 +1)(u +1)(u − 1) (u 2 +1)(u +1)<br />
= lim<br />
= 2(2)<br />
u→1 u(u +6)(u − 1) u→1 u(u +6) 1(7) = 4 7<br />
13. Since x is positive, √ x 2 = |x| = x. Thus,<br />
√ √<br />
x2 − 9<br />
lim<br />
x→∞ 2x − 6<br />
= lim<br />
x2 − 9/ √ √<br />
x 2 1 − 9/x<br />
x→∞ (2x − 6)/x = lim<br />
2 1 − 0<br />
x→∞ 2 − 6/x = 2 − 0 = 1 2<br />
15. Let t =sinx. Thenasx → π − , sin x → 0 + ,sot → 0 + . Thus, lim ln(sin x) = lim ln t = −∞.<br />
x→π− t→0 +<br />
√<br />
17. lim x2 +4x +1− x √ √ <br />
x2 +4x +1− x x2 +4x +1+x (x 2 +4x +1)− x 2<br />
= lim<br />
· √ = lim √<br />
x→∞<br />
x→∞ 1<br />
x2 +4x +1+x x→∞ x2 +4x +1+x<br />
(4x +1)/x<br />
= lim<br />
x→∞ ( √ x 2 +4x +1+x)/x<br />
<br />
divide by x = √ <br />
x 2 for x>0<br />
4+1/x<br />
= lim <br />
x→∞ 1+4/x +1/x2 +1 = 4+0<br />
√ = 4 1+0+0+1 2 =2
F.<br />
TX.10<br />
CHAPTER 2 REVIEW ¤ 81<br />
19. Let t =1/x. Thenasx → 0 + , t →∞,and lim<br />
x→0 + tan−1 (1/x) = lim<br />
t→∞<br />
tan −1 t = π 2 .<br />
21. From the graph of y = cos 2 x /x 2 , it appears that y =0is the horizontal<br />
asymptote and x =0is the vertical asymptote. Now 0 ≤ (cos x) 2 ≤ 1<br />
0<br />
x ≤ cos2 x<br />
≤ 1 ⇒ 0 ≤ cos2 x<br />
≤ 1 .But lim<br />
2 x 2 x 2 x 2 x 0=0and<br />
2 x→±∞<br />
lim<br />
x→±∞<br />
1<br />
x<br />
2<br />
=0, so by the Squeeze Theorem, lim<br />
x→±∞<br />
cos 2 x<br />
x 2 =0.<br />
⇒<br />
cos 2 x<br />
Thus, y =0is the horizontal asymptote. lim = ∞ because cos 2 x → 1 and x 2 → 0 as x → 0, sox =0is the<br />
x→0 x 2<br />
vertical asymptote.<br />
23. Since 2x − 1 ≤ f(x) ≤ x 2 for 0 0 such that if 0 < |x − 2|
F.<br />
82 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />
TX.10<br />
35. (a) The slope of the tangent line at (2, 1) is<br />
f(x) − f(2) 9 − 2x 2 − 1 8 − 2x 2<br />
lim<br />
=lim<br />
= lim<br />
x→2 x − 2 x→2 x − 2 x→2 x − 2<br />
= lim −2(x 2 − 4) −2(x − 2)(x +2)<br />
= lim<br />
x→2 x − 2 x→2 x − 2<br />
=lim<br />
x→2<br />
[−2(x +2)]=−2 · 4=−8<br />
(b) An equation of this tangent line is y − 1=−8(x − 2) or y = −8x +17.<br />
37. (a) s = s(t) =1+2t + t 2 /4. The average velocity over the time interval [1, 1+h] is<br />
v ave =<br />
s(1 + h) − s(1)<br />
(1 + h) − 1<br />
= 1+2(1+h)+(1+h)2 4 − 13/4<br />
h<br />
=<br />
10h + h2<br />
4h<br />
= 10 + h<br />
4<br />
So for the following intervals the average velocities are:<br />
(i) [1, 3]: h =2, v ave =(10+2)/4 =3m/s<br />
(iii) [1, 1.5]: h =0.5, v ave =(10+0.5)/4 =2.625 m/s<br />
(ii) [1, 2]: h =1, v ave =(10+1)/4 =2.75 m/s<br />
(iv) [1, 1.1]: h =0.1, v ave =(10+0.1)/4 =2.525 m/s<br />
s(1 + h) − s(1) 10 + h<br />
(b) When t =1, the instantaneous velocity is lim<br />
= lim = 10<br />
h→0 h<br />
h→0 4 4<br />
=2.5 m/s.<br />
39. (a) f 0 f(x) − f(2) x 3 − 2x − 4<br />
(2) = lim<br />
=lim<br />
x→2 x − 2 x→2 x − 2<br />
(x − 2) x 2 +2x +2 <br />
= lim<br />
x→2 x − 2<br />
= lim<br />
x→2<br />
x 2 +2x +2 =10<br />
(c)<br />
(b) y − 4=10(x − 2) or y =10x − 16<br />
41. (a) f 0 (r) is the rate at which the total cost changes with respect to the interest rate. Its units are dollars/(percent per year).<br />
(b) The total cost of paying off the loan is increasing by \$1200/(percent per year) as the interest rate reaches 10%. Soifthe<br />
interest rate goes up from 10% to 11%, the cost goes up approximately \$1200.<br />
(c) As r increases, C increases. So f 0 (r) will always be positive.<br />
43.
F.<br />
TX.10<br />
CHAPTER 2 REVIEW ¤ 83<br />
√<br />
45. (a) f 0 f(x + h) − f(x) 3 − 5(x + h) − 3 − 5x<br />
(x) =lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
[3 − 5(x + h)] − (3 − 5x)<br />
=lim <br />
h→0<br />
3 √ = lim<br />
h − 5(x + h)+ 3 − 5x h→0<br />
<br />
3 − 5(x + h)+<br />
√ 3 − 5x<br />
<br />
3 − 5(x + h)+<br />
√ 3 − 5x<br />
−5<br />
<br />
3 − 5(x + h)+<br />
√ 3 − 5x<br />
=<br />
−5<br />
2 √ 3 − 5x<br />
(b) Domain of f: (the radicand must be nonnegative) 3 − 5x ≥ 0<br />
5x ≤ 3 ⇒ x ∈ <br />
−∞, 3 5<br />
⇒<br />
Domain of f 0 :exclude 3 because it makes the denominator zero;<br />
5<br />
x ∈ <br />
−∞, 3 5<br />
(c) Our answer to part (a) is reasonable because f 0 (x) is always negative and f<br />
is always decreasing.<br />
47. f is not differentiable: at x = −4 because f is not continuous, at x = −1 because f has a corner, at x =2because f is not<br />
continuous, and at x =5because f has a vertical tangent.<br />
49. C 0 (1990) is the rate at which the total value of US currency in circulation is changing in billions of dollars per year. To<br />
estimate the value of C 0 (1990), we will average the difference quotients obtained using the times t =1985and t = 1995.<br />
Let A =<br />
B =<br />
C(1985) − C(1990)<br />
1985 − 1990<br />
C(1995) − C(1990)<br />
1995 − 1990<br />
C 0 (1990) =<br />
=<br />
=<br />
187.3 − 271.9<br />
−5<br />
409.3 − 271.9<br />
5<br />
= −84.6<br />
−5<br />
= 137.4<br />
5<br />
=16.92 and<br />
=27.48. Then<br />
C(t) − C(1990)<br />
lim<br />
≈ A + B 16.92 + 27.48<br />
= = 44.4 =22.2 billion dollars/year.<br />
t→1990 t − 1990 2<br />
2 2<br />
51. |f(x)| ≤ g(x) ⇔ −g(x) ≤ f(x) ≤ g(x) and lim<br />
x→a<br />
g(x) = 0 = lim<br />
x→a<br />
−g(x).<br />
Thus, by the Squeeze Theorem, lim<br />
x→a<br />
f(x) =0.
F.<br />
TX.10
F.<br />
TX.10<br />
PROBLEMS PLUS<br />
1. Let t = 6√ x,sox = t 6 .Thent → 1 as x → 1,so<br />
lim<br />
x→1<br />
3√ x − 1<br />
√ x − 1<br />
= lim<br />
t→1<br />
t 2 − 1<br />
t 3 − 1 =lim<br />
t→1<br />
(t − 1)(t +1)<br />
(t − 1) (t 2 + t +1) =lim<br />
t→1<br />
t +1<br />
t 2 + t +1 = 1+1<br />
1 2 +1+1 = 2 3 .<br />
Another method: Multiply both the numerator and the denominator by ( √ <br />
x +1)<br />
3√<br />
x2 + 3√ <br />
x +1 .<br />
3. For − 1 0,so|2x − 1| = −(2x − 1) and |2x +1| =2x +1.<br />
2 2<br />
|2x − 1| − |2x +1| −(2x − 1) − (2x +1) −4x<br />
Therefore, lim<br />
=lim<br />
=lim<br />
x→0 x<br />
x→0 x<br />
x→0 x<br />
= lim (−4) = −4.<br />
x→0<br />
5. Since [x] ≤ x 0,thenG(a +180 ◦ )=T (a +360 ◦ ) − T (a + 180 ◦ )=T (a) − T (a +180 ◦ )=−G(a) < 0.<br />
Also, G is continuous since temperature varies continuously. So, by the Intermediate Value Theorem, G has a zero on the<br />
interval [a, a +180 ◦ ].IfG(a) < 0, then a similar argument applies.<br />
85
F.<br />
86 ¤ CHAPTER 2 PROBLEMS PLUS<br />
TX.10<br />
(b) Yes. The same argument applies.<br />
(c) The same argument applies for quantities that vary continuously, such as barometric pressure. But one could argue that<br />
altitude above sea level is sometimes discontinuous, so the result might not always hold for that quantity.<br />
13. (a) Put x =0and y =0in the equation: f(0 + 0) = f(0) + f(0) + 0 2 · 0+0· 0 2 ⇒ f(0) = 2f(0).<br />
Subtracting f(0) from each side of this equation gives f(0) = 0.<br />
<br />
(b) f 0 f(0 + h) − f(0) f(0) + f(h)+0 2 h +0h 2 − f(0) f(h)<br />
(0) = lim<br />
=lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
h→0 h<br />
= lim f(x)<br />
x→0 x =1<br />
<br />
(c) f 0 f(x + h) − f(x) f(x)+f(h)+x 2 h + xh 2 − f(x)<br />
(x) =lim<br />
=lim<br />
h→0 h<br />
h→0 h<br />
<br />
f(h)<br />
=lim<br />
h→0 h<br />
+ x2 + xh =1+x 2<br />
=lim<br />
h→0<br />
f(h)+x 2 h + xh 2<br />
h
F.<br />
TX.10<br />
3 DIFFERENTIATION RULES<br />
3.1 Derivatives of Polynomials and Exponential Functions<br />
e h − 1<br />
1. (a) e is the number such that lim =1.<br />
h→0 h<br />
(b)<br />
x<br />
2.7 x − 1<br />
x<br />
−0.001 0.9928<br />
−0.0001 0.9932<br />
0.001 0.9937<br />
0.0001 0.9933<br />
x<br />
2.8 x − 1<br />
x<br />
−0.001 1.0291<br />
−0.0001 1.0296<br />
0.001 1.0301<br />
0.0001 1.0297<br />
From the tables (to two decimal places),<br />
2.7 h − 1<br />
2.8 h − 1<br />
lim =0.99 and lim =1.03.<br />
h→0 h<br />
h→0 h<br />
Since 0.99 < 1 < 1.03, 2.7
F.<br />
88 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
27. We first expand using the Binomial Theorem (see Reference Page 1).<br />
H(x) =(x + x −1 ) 3 = x 3 +3x 2 x −1 +3x(x −1 ) 2 +(x −1 ) 3 = x 3 +3x +3x −1 + x −3<br />
⇒<br />
H 0 (x) =3x 2 +3+3(−1x −2 )+(−3x −4 )=3x 2 +3− 3x −2 − 3x −4<br />
29. u = 5√ t +4 √ <br />
t 5 = t 1/5 +4t 5/2 ⇒ u 0 = 1 5 t−4/5 5<br />
+4<br />
2 t3/2 = 1 5 t−4/5 +10t 3/2 or 1/<br />
5 5√ <br />
t 4 +10 √ t 3<br />
31. z = A<br />
y + 10 Bey = Ay −10 + Be y ⇒ z 0 = −10Ay −11 + Be y = − 10A<br />
y + 11 Bey<br />
33. y = 4√ x = x 1/4 ⇒ y 0 = 1 4 x−3/4 = 1<br />
4 4√ x . At(1, 1), 3 y0 = 1 4<br />
and an equation of the tangent line is<br />
y − 1= 1 4 (x − 1) or y = 1 4 x + 3 4 .<br />
35. y = x 4 +2e x ⇒ y 0 =4x 3 +2e x . At (0, 2), y 0 =2and an equation of the tangent line is y − 2=2(x − 0)<br />
or y =2x +2. The slope of the normal line is − 1 (the negative reciprocal of 2) and an equation of the normal line is<br />
2<br />
y − 2=− 1 (x − 0) or y = − 1 x +2.<br />
2 2<br />
37. y =3x 2 − x 3 ⇒ y 0 =6x − 3x 2 .<br />
At (1, 2), y 0 =6− 3=3, so an equation of the tangent line is<br />
y − 2=3(x − 1) or y =3x − 1.<br />
39. f(x) =e x − 5x ⇒ f 0 (x) =e x − 5.<br />
Notice that f 0 (x) =0when f has a horizontal tangent, f 0 is positive<br />
when f is increasing, and f 0 is negative when f is decreasing.<br />
41. f(x) =3x 15 − 5x 3 +3 ⇒ f 0 (x) =45x 14 − 15x 2 .<br />
Notice that f 0 (x) =0when f has a horizontal tangent, f 0 is positive<br />
when f is increasing, and f 0 is negative when f is decreasing.
F.<br />
SECTION TX.10 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS ¤ 89<br />
43. (a) (b) From the graph in part (a), it appears that f 0 is zero at x 1 ≈−1.25, x 2 ≈ 0.5,<br />
and x 3 ≈ 3. The slopes are negative (so f 0 is negative) on (−∞,x 1) and<br />
(x 2,x 3). The slopes are positive (so f 0 is positive) on (x 1,x 2) and (x 3, ∞).<br />
(c) f(x) =x 4 − 3x 3 − 6x 2 +7x +30<br />
⇒<br />
f 0 (x) =4x 3 − 9x 2 − 12x +7<br />
45. f(x) =x 4 − 3x 3 +16x ⇒ f 0 (x) =4x 3 − 9x 2 +16 ⇒ f 00 (x) =12x 2 − 18x<br />
47. f(x) =2x − 5x 3/4 ⇒ f 0 (x) =2− 15<br />
4 x−1/4 ⇒ f 00 (x) = 15<br />
16 x−5/4<br />
Note that f 0 is negative when f is decreasing and positive when f is<br />
increasing. f 00 is always positive since f 0 is always increasing.<br />
49. (a) s = t 3 − 3t ⇒ v(t) =s 0 (t) =3t 2 − 3 ⇒ a(t) =v 0 (t) =6t<br />
(b) a(2) = 6(2) = 12 m/s 2<br />
(c) v(t) =3t 2 − 3=0when t 2 =1,thatis,t =1and a(1) = 6 m/s 2 .<br />
51. The curve y =2x 3 +3x 2 − 12x +1has a horizontal tangent when y 0 =6x 2 +6x − 12 = 0 ⇔ 6(x 2 + x − 2) = 0 ⇔<br />
6(x +2)(x − 1) = 0 ⇔ x = −2 or x =1. The points on the curve are (−2, 21) and (1, −6).<br />
53. y =6x 3 +5x − 3 ⇒ m = y 0 =18x 2 +5,butx 2 ≥ 0 for all x,som ≥ 5 for all x.<br />
55. The slope of the line 12x − y =1(or y =12x − 1)is12, so the slope of both lines tangent to the curve is 12.<br />
y =1+x 3 ⇒ y 0 =3x 2 . Thus, 3x 2 =12 ⇒ x 2 =4 ⇒ x = ±2, which are the x-coordinates at which the tangent<br />
lines have slope 12. The points on the curve are (2, 9) and (−2, −7), so the tangent line equations are y − 9=12(x − 2)<br />
or y =12x − 15 and y + 7 = 12(x +2)or y =12x +17.
F.<br />
90 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
57. The slope of y = x 2 − 5x +4is given by m = y 0 =2x − 5. The slope of x − 3y =5 ⇔ y = 1 3 x − 5 3 is 1 3 ,<br />
so the desired normal line must have slope 1 , and hence, the tangent line to the parabola must have slope −3. This occurs if<br />
3<br />
2x − 5=−3 ⇒ 2x =2 ⇒ x =1.Whenx =1, y =1 2 − 5(1) + 4 = 0, and an equation of the normal line is<br />
y − 0= 1 3 (x − 1) or y = 1 3 x − 1 3 .<br />
59. Let a, a 2 be a point on the parabola at which the tangent line passes through the<br />
point (0, −4). The tangent line has slope 2a and equation y − (−4) = 2a(x − 0)<br />
⇔<br />
y =2ax − 4. Since a, a 2 also lies on the line, a 2 =2a(a) − 4,ora 2 =4.So<br />
a = ±2 and the points are (2, 4) and (−2, 4).<br />
1<br />
61. f 0 f(x + h) − f(x)<br />
(x) = lim<br />
= lim<br />
x + h − 1 x<br />
h→0 h<br />
h→0 h<br />
= lim<br />
h→0<br />
x − (x + h)<br />
hx(x + h) =lim<br />
h→0<br />
−h<br />
hx(x + h) =lim<br />
h→0<br />
63. Let P (x) =ax 2 + bx + c. ThenP 0 (x) =2ax + b and P 00 (x) =2a. P 00 (2) = 2 ⇒ 2a =2 ⇒ a =1.<br />
P 0 (2) = 3 ⇒ 2(1)(2) + b =3 ⇒ 4+b =3 ⇒ b = −1.<br />
P (2) = 5 ⇒ 1(2) 2 +(−1)(2) + c =5 ⇒ 2+c =5 ⇒ c =3.SoP (x) =x 2 − x +3.<br />
−1<br />
x(x + h) = − 1 x 2<br />
65. y = f(x) =ax 3 + bx 2 + cx + d ⇒ f 0 (x) =3ax 2 +2bx + c. Thepoint(−2, 6) is on f, sof(−2) = 6 ⇒<br />
−8a +4b − 2c + d =6 (1). The point (2, 0) is on f, sof(2) = 0 ⇒ 8a +4b +2c + d =0 (2). Since there are<br />
horizontal tangents at (−2, 6) and (2, 0), f 0 (±2) = 0. f 0 (−2) = 0 ⇒ 12a − 4b + c =0 (3) and f 0 (2) = 0 ⇒<br />
12a +4b + c =0 (4). Subtracting equation (3) from (4) gives 8b =0 ⇒ b =0. Adding (1) and (2) gives 8b +2d =6,<br />
so d =3since b =0.From(3) we have c = −12a,so(2) becomes 8a +4(0)+2(−12a)+3=0 ⇒ 3=16a ⇒<br />
a = 3 <br />
3<br />
16 16 = −<br />
9<br />
3<br />
and the desired cubic function is y =<br />
4 16 x3 − 9 x +3. 4<br />
67. f(x) =2− x if x ≤ 1 and f(x) =x 2 − 2x +2if x>1. Now we compute the right- and left-hand derivatives defined in<br />
Exercise 2.8.54:<br />
f 0 −(1) =<br />
f(1 + h) − f(1) 2 − (1 + h) − 1 −h<br />
lim<br />
= lim<br />
= lim<br />
h→0 − h<br />
h→0 − h<br />
h→0 − h = lim −1 =−1 and<br />
h→0− f+(1) 0 f(1 + h) − f(1) (1 + h) 2 − 2(1 + h)+2− 1 h 2<br />
= lim<br />
= lim<br />
= lim<br />
h→0 + h<br />
h→0 + h<br />
h→0 + h = lim h =0.<br />
h→0 +<br />
Thus, f 0 (1) does not exist since f−(1) 0 6=f+(1),sof<br />
0<br />
is not differentiable at 1. Butf 0 (x) =−1 for x1.
F.<br />
SECTION TX.10 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS ¤ 91<br />
69. (a) Note that x 2 − 9 < 0 for x 2 < 9 ⇔ |x| < 3 ⇔ −3
F.<br />
92 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
79. y = x 2 ⇒ y 0 =2x, so the slope of a tangent line at the point (a, a 2 ) is y 0 =2a and the slope of a normal line is −1/(2a),<br />
for a 6= 0. The slope of the normal line through the points (a, a 2 ) and (0,c) is a2 − c − c<br />
a − 0 ,soa2 = − 1<br />
a 2a<br />
⇒<br />
a 2 − c = − 1 ⇒ a 2 = c − 1 . The last equation has two solutions if c> 1 , one solution if c = 1 , and no solution if<br />
2 2 2 2<br />
c< 1 2 . Since the y-axis is normal to y = x2 regardless of the value of c (thisisthecasefora =0), we have three normal lines<br />
if c> 1 2 and one normal line if c ≤ 1 2 .<br />
3.2 The Product and Quotient Rules<br />
1. Product Rule: y =(x 2 +1)(x 3 +1) ⇒<br />
y 0 =(x 2 +1)(3x 2 )+(x 3 + 1)(2x) =3x 4 +3x 2 +2x 4 +2x =5x 4 +3x 2 +2x.<br />
Multiplying first: y =(x 2 +1)(x 3 +1)=x 5 + x 3 + x 2 +1 ⇒ y 0 =5x 4 +3x 2 +2x (equivalent).<br />
3. By the Product Rule, f(x) =(x 3 +2x)e x ⇒<br />
f 0 (x)=(x 3 +2x)(e x ) 0 + e x (x 3 +2x) 0 =(x 3 +2x)e x + e x (3x 2 +2)<br />
= e x [(x 3 +2x)+(3x 2 +2)]=e x (x 3 +3x 2 +2x +2)<br />
x 2 d<br />
5. By the Quotient Rule, y = ex<br />
⇒ y 0 = dx (ex ) − e x d<br />
dx (x2 )<br />
= x2 (e x ) − e x (2x)<br />
= xex (x − 2)<br />
= ex (x − 2)<br />
.<br />
x 2 (x 2 ) 2 x 4 x 4 x 3<br />
The notations<br />
PR<br />
⇒<br />
and<br />
QR<br />
⇒<br />
indicate the use of the Product and Quotient Rules, respectively.<br />
7. g(x) = 3x − 1<br />
2x +1<br />
QR<br />
⇒ g 0 (2x + 1)(3) − (3x − 1)(2) 6x +3− 6x +2 5<br />
(x) = = =<br />
(2x +1) 2 (2x +1) 2 (2x +1) 2<br />
9. V (x) =(2x 3 +3)(x 4 − 2x)<br />
PR<br />
⇒<br />
V 0 (x) =(2x 3 + 3)(4x 3 − 2) + (x 4 − 2x)(6x 2 )=(8x 6 +8x 3 − 6) + (6x 6 − 12x 3 )=14x 6 − 4x 3 − 6<br />
1<br />
11. F (y) =<br />
y − 3 <br />
(y +5y 3 )= y −2 − 3y −4 y +5y 3 PR<br />
⇒<br />
2 y 4<br />
F 0 (y) =(y −2 − 3y −4 )(1 + 15y 2 )+(y +5y 3 )(−2y −3 +12y −5 )<br />
=(y −2 +15− 3y −4 − 45y −2 )+(−2y −2 +12y −4 − 10 + 60y −2 )<br />
=5+14y −2 +9y −4 or 5+14/y 2 +9/y 4<br />
13. y = x3<br />
1 − x 2 QR<br />
⇒<br />
y 0 = (1 − x2 )(3x 2 ) − x 3 (−2x)<br />
= x2 (3 − 3x 2 +2x 2 )<br />
= x2 (3 − x 2 )<br />
(1 − x 2 ) 2 (1 − x 2 ) 2 (1 − x 2 ) 2
F.<br />
TX.10<br />
SECTION 3.2 THE PRODUCT AND QUOTIENT RULES ¤ 93<br />
15. y =<br />
t 2 +2<br />
t 4 − 3t 2 +1<br />
QR<br />
⇒<br />
y 0 = (t4 − 3t 2 +1)(2t) − (t 2 +2)(4t 3 − 6t)<br />
= 2t[(t4 − 3t 2 +1)− (t 2 + 2)(2t 2 − 3)]<br />
(t 4 − 3t 2 +1) 2 (t 4 − 3t 2 +1) 2<br />
= 2t(t4 − 3t 2 +1− 2t 4 − 4t 2 +3t 2 +6)<br />
= 2t(−t4 − 4t 2 +7)<br />
(t 4 − 3t 2 +1) 2 (t 4 − 3t 2 +1) 2<br />
17. y =(r 2 − 2r)e r PR<br />
⇒ y 0 =(r 2 − 2r)(e r )+e r (2r − 2) = e r (r 2 − 2r +2r − 2) = e r (r 2 − 2)<br />
19. y = v3 − 2v √ v<br />
v<br />
= v 2 − 2 √ v = v 2 − 2v 1/2 ⇒ y 0 =2v − 2 1<br />
2<br />
<br />
v −1/2 =2v − v −1/2 .<br />
We can change the form of the answer as follows: 2v − v −1/2 =2v − 1 √ v<br />
= 2v √ v − 1<br />
√ v<br />
= 2v3/2 − 1<br />
√ v<br />
21. f(t) = 2t<br />
2+ √ t<br />
<br />
QR<br />
⇒<br />
(2 + t 1/2 1<br />
)(2) − 2t<br />
f 0 (t) =<br />
(2 + √ t ) 2 = 4+2t1/2 − t 1/2<br />
(2 + √ t ) 2 = 4+t1/2<br />
(2 + √ t ) 2 or 4+ √ t<br />
(2 + √ t ) 2<br />
2 t−1/2 <br />
23. f(x) =<br />
A<br />
B + Ce x<br />
QR<br />
⇒ f 0 (x) = (B + Cex ) · 0 − A(Ce x )<br />
= − ACex<br />
(B + Ce x ) 2 (B + Ce x ) 2<br />
x<br />
25. f(x) =<br />
x + c/x ⇒ f 0 (x) = (x + c/x)(1) − x(1 − c/x2 ) x + c/x − x + c/x<br />
<br />
x + c 2<br />
= x 2 2<br />
= 2c/x<br />
+ c (x 2 + c) 2<br />
x<br />
x<br />
x 2<br />
· x2<br />
x = 2cx<br />
2 (x 2 + c) 2<br />
27. f(x) =x 4 e x ⇒ f 0 (x) =x 4 e x + e x · 4x 3 = x 4 +4x 3 e x or x 3 e x (x +4) ⇒<br />
f 00 (x)=(x 4 +4x 3 )e x + e x (4x 3 +12x 2 )=(x 4 +4x 3 +4x 3 +12x 2 )e x<br />
=(x 4 +8x 3 +12x 2 )e x<br />
or x 2 e x (x +2)(x +6) <br />
29. f(x) = x2<br />
1+2x ⇒ f 0 (x) = (1 + 2x)(2x) − x2 (2)<br />
(1 + 2x) 2 = 2x +4x2 − 2x 2<br />
(1 + 2x) 2 = 2x2 +2x<br />
(1 + 2x) 2 ⇒<br />
f 00 (x)= (1 + 2x)2 (4x +2)− (2x 2 +2x)(1 + 4x +4x 2 ) 0<br />
= 2(1 + 2x)2 (2x +1)− 2x(x +1)(4+8x)<br />
[(1 + 2x) 2 ] 2 (1 + 2x) 4<br />
= 2(1 + 2x)[(1 + 2x)2 − 4x(x +1)]<br />
= 2(1 + 4x +4x2 − 4x 2 − 4x) 2<br />
=<br />
(1 + 2x) 4 (1 + 2x) 3 (1 + 2x) 3<br />
31. y = 2x<br />
x +1<br />
⇒<br />
y 0 =<br />
(x +1)(2)− (2x)(1) 2<br />
=<br />
(x +1) 2 (x +1) . 2<br />
At (1, 1), y 0 = 1 2 , and an equation of the tangent line is y − 1= 1 2 (x − 1),ory = 1 2 x + 1 2 .<br />
33. y =2xe x ⇒ y 0 =2(x · e x + e x · 1) = 2e x (x +1).<br />
At (0, 0), y 0 =2e 0 (0 + 1) = 2 · 1 · 1=2, and an equation of the tangent line is y − 0=2(x − 0),ory =2x. Theslopeof<br />
the normal line is − 1 2 , so an equation of the normal line is y − 0=− 1 2 (x − 0),ory = − 1 2 x.
F.<br />
94 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
35. (a) y = f(x) = 1<br />
1+x 2 ⇒<br />
f 0 (x) = (1 + x2 )(0) − 1(2x)<br />
= −2x . So the slope of the<br />
(1 + x 2 ) 2 (1 + x 2 )<br />
2<br />
(b)<br />
tangent line at the point <br />
−1, 1 2 is f 0 (−1) = 2 2 = 1 2 2<br />
and its<br />
equation is y − 1 2 = 1 2 (x +1)or y = 1 2 x +1.<br />
37. (a) f(x) = ex<br />
⇒ f 0 (x) = x3 (e x ) − e x (3x 2 )<br />
= x2 e x (x − 3)<br />
= ex (x − 3)<br />
x 3 (x 3 ) 2 x 6 x 4<br />
(b)<br />
f 0 =0when f has a horizontal tangent line, f 0 is negative when<br />
f is decreasing, and f 0 is positive when f is increasing.<br />
39. (a) f(x) =(x − 1)e x ⇒ f 0 (x) =(x − 1)e x + e x (1) = e x (x − 1+1)=xe x .<br />
f 00 (x) =x(e x )+e x (1) = e x (x +1)<br />
(b)<br />
f 0 =0when f has a horizontal tangent and f 00 =0when f 0 has a<br />
horizontal tangent. f 0 is negative when f is decreasing and positive when f<br />
is increasing. f 00 is negative when f 0 is decreasing and positive when f 0 is<br />
increasing. f 00 is negative when f is concave down and positive when f is<br />
concave up.<br />
41. f(x) = x2<br />
1+x ⇒ f 0 (x) = (1 + x)(2x) − x2 (1)<br />
(1 + x) 2 = 2x +2x2 − x 2<br />
(1 + x) 2 = x2 +2x<br />
x 2 +2x +1<br />
⇒<br />
so f 00 (1) =<br />
f 00 (x) = (x2 +2x +1)(2x +2)− (x 2 +2x)(2x +2)<br />
(x 2 +2x +1) 2 = (2x +2)(x2 +2x +1− x 2 − 2x)<br />
[(x +1) 2 ] 2<br />
=<br />
2<br />
(1 + 1) 3 = 2 8 = 1 4 .<br />
2(x +1)(1) 2<br />
=<br />
(x +1) 4 (x +1) 3 ,<br />
43. We are given that f(5) = 1, f 0 (5) = 6, g(5) = −3,andg 0 (5) = 2.<br />
(a) (fg) 0 (5) = f(5)g 0 (5) + g(5)f 0 (5) = (1)(2) + (−3)(6) = 2 − 18 = −16<br />
0 f<br />
(b) (5) = g(5)f 0 (5) − f(5)g 0 (5) (−3)(6) − (1)(2)<br />
= = − 20 g<br />
[g(5)] 2 (−3) 2 9<br />
0 g<br />
(c) (5) = f(5)g0 (5) − g(5)f 0 (5)<br />
=<br />
f<br />
[f(5)] 2<br />
(1)(2) − (−3)(6)<br />
(1) 2 =20<br />
45. f(x) =e x g(x) ⇒ f 0 (x) =e x g 0 (x)+g(x)e x = e x [g 0 (x)+g(x)]. f 0 (0) = e 0 [g 0 (0) + g(0)] = 1(5 + 2) = 7
F.<br />
TX.10<br />
SECTION 3.2 THE PRODUCT AND QUOTIENT RULES ¤ 95<br />
47. (a) From the graphs of f and g, we obtain the following values: f(1) = 2 since the point (1, 2) is on the graph of f;<br />
g(1) = 1 since the point (1, 1) is on the graph of g; f 0 (1) = 2 since the slope of the line segment between (0, 0) and (2, 4)<br />
is 4 − 0<br />
2 − 0 =2; 0 − 4<br />
g0 (1) = −1 since the slope of the line segment between (−2, 4) and (2, 0) is<br />
2 − (−2) = −1.<br />
Now u(x) =f(x)g(x),sou 0 (1) = f(1)g 0 (1) + g(1) f 0 (1) = 2 · (−1) + 1 · 2=0.<br />
(b) v(x) =f(x)/g(x),sov 0 (5) = g(5)f 0 (5) − f(5)g 0 (5)<br />
= 2 <br />
− 1 3 − 3 · 2<br />
3<br />
= − 8 3<br />
[g(5)] 2 2 2 4 = − 2 3<br />
49. (a) y = xg(x) ⇒ y 0 = xg 0 (x)+g(x) · 1=xg 0 (x)+g(x)<br />
(b) y =<br />
x<br />
g(x)<br />
⇒<br />
y 0 = g(x) · 1 − xg0 (x)<br />
[g(x)] 2<br />
= g(x) − xg0 (x)<br />
[g(x)] 2<br />
(c) y = g(x)<br />
x<br />
⇒ y 0 = xg0 (x) − g(x) · 1<br />
(x) 2<br />
= xg0 (x) − g(x)<br />
x 2<br />
51. If y = f(x) = x<br />
x +1 ,thenf 0 (x +1)(1)− x(1) 1<br />
(x) = = .Whenx = a, the equation of the tangent line is<br />
(x +1) 2 (x +1)<br />
2<br />
y −<br />
a<br />
a +1 = 1<br />
a<br />
(x − a). This line passes through (1, 2) when 2 −<br />
(a +1)<br />
2<br />
a +1 = 1<br />
(1 − a) ⇔<br />
(a +1)<br />
2<br />
2(a +1) 2 − a(a +1)=1− a ⇔ 2a 2 +4a +2− a 2 − a − 1+a =0 ⇔ a 2 +4a +1=0.<br />
The quadratic formula gives the roots of this equation as a = −4 ± 4 2 − 4(1)(1)<br />
2(1)<br />
so there are two such tangent lines. Since<br />
f −2 ± √ 3 = −2 ± √ 3<br />
−2 ± √ 3+1 = −2 ± √ 3<br />
−1 ± √ 3 · −1 ∓ √ 3<br />
−1 ∓ √ 3<br />
<br />
the lines touch the curve at A<br />
= 2 ± 2 √ 3 ∓ √ 3 − 3<br />
1 − 3<br />
−2+ √ 3, 1 − √ 3<br />
2<br />
<br />
and B −2 − √ <br />
3, 1+√ 3<br />
≈ (−3.73, 1.37).<br />
2<br />
= −1 ± √ 3<br />
−2<br />
<br />
≈ (−0.27, −0.37)<br />
= 1 ∓ √ 3<br />
,<br />
2<br />
= −4 ± √ 12<br />
2<br />
= −2 ± √ 3,<br />
53. If P (t) denotes the population at time t and A(t) the average annual income, then T (t) =P (t)A(t) is the total personal<br />
income. The rate at which T (t) is rising is given by T 0 (t) =P (t)A 0 (t)+A(t)P 0 (t)<br />
⇒<br />
T 0 (1999) = P (1999)A 0 (1999) + A(1999)P 0 (1999) = (961,400)(\$1400/yr)+(\$30,593)(9200/yr)<br />
=\$1,345,960,000/yr +\$281,455,600/yr =\$1,627,415,600/yr<br />
So the total personal income was rising by about \$1.627 billion per year in 1999.<br />
The term P (t)A 0 (t) ≈ \$1.346 billion represents the portion of the rate of change of total income due to the existing<br />
population’s increasing income. The term A(t)P 0 (t) ≈ \$281 million represents the portion of the rate of change of total<br />
income due to increasing population.
F.<br />
96 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
We will sometimes use the form f 0 g + fg 0 rather than the form fg 0 + gf 0 for the Product Rule.<br />
55. (a) (fgh) 0 =[(fg)h] 0 =(fg) 0 h +(fg)h 0 =(f 0 g + fg 0 )h +(fg)h 0 = f 0 gh + fg 0 h + fgh 0<br />
(b) Putting f = g = h in part (a), we have d<br />
dx [f(x)]3 =(fff) 0 = f 0 ff + ff 0 f + fff 0 =3fff 0 =3[f(x)] 2 f 0 (x).<br />
(c)<br />
d<br />
dx (e3x )= d<br />
dx (ex ) 3 =3(e x ) 2 e x =3e 2x e x =3e 3x<br />
57. For f(x) =x 2 e x , f 0 (x) =x 2 e x + e x (2x) =e x (x 2 +2x). Similarly, we have<br />
f 00 (x) =e x (x 2 +4x +2)<br />
f 000 (x) =e x (x 2 +6x +6)<br />
f (4) (x) =e x (x 2 +8x + 12)<br />
f (5) (x) =e x (x 2 +10x +20)<br />
It appears that the coefficient of x in the quadratic term increases by 2 with each differentiation. The pattern for the<br />
constant terms seems to be 0=1· 0, 2=2· 1, 6=3· 2, 12 = 4 · 3, 20 = 5 · 4. So a reasonable guess is that<br />
f (n) (x) =e x [x 2 +2nx + n(n − 1)].<br />
Proof: Let S n be the statement that f (n) (x) =e x [x 2 +2nx + n(n − 1)].<br />
1. S 1 is true because f 0 (x) =e x (x 2 +2x).<br />
2. Assume that S k is true; that is, f (k) (x) =e x [x 2 +2kx + k(k − 1)]. Then<br />
f (k+1) (x) = d <br />
f (k) (x) = e x (2x +2k)+[x 2 +2kx + k(k − 1)]e x<br />
dx<br />
= e x [x 2 +(2k +2)x +(k 2 + k)] = e x [x 2 +2(k +1)x +(k +1)k]<br />
This shows that S k+1 is true.<br />
3. Therefore, by mathematical induction, S n is true for all n;thatis,f (n) (x) =e x [x 2 +2nx + n(n − 1)] for every<br />
positive integer n.<br />
3.3 Derivatives of Trigonometric Functions<br />
1. f(x) =3x 2 − 2cosx ⇒ f 0 (x) =6x − 2(− sin x) =6x +2sinx<br />
3. f(x) =sinx + 1 2 cot x ⇒ f 0 (x) =cosx − 1 2 csc2 x<br />
5. g(t) =t 3 cos t ⇒ g 0 (t) =t 3 (− sin t)+(cost) · 3t 2 =3t 2 cos t − t 3 sin t or t 2 (3 cos t − t sin t)<br />
7. h(θ) =cscθ + e θ cot θ ⇒ h 0 (θ) =− csc θ cot θ + e θ (− csc 2 θ)+(cotθ)e θ = − csc θ cot θ + e θ (cot θ − csc 2 θ)<br />
9. y =<br />
x<br />
2 − tan x ⇒ y0 = (2 − tan x)(1) − x(− sec2 x)<br />
= 2 − tan x + x sec2 x<br />
(2 − tan x) 2 (2 − tan x) 2<br />
11. f(θ) = sec θ<br />
1+secθ<br />
⇒<br />
f 0 (θ) =<br />
(1 + sec θ)(sec θ tan θ) − (sec θ)(sec θ tan θ) (sec θ tan θ) [(1+secθ) − sec θ] sec θ tan θ<br />
= =<br />
(1 + sec θ) 2 (1 + sec θ) 2 (1 + sec θ) 2
F.<br />
TX.10<br />
SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS ¤ 97<br />
13. y = sin x ⇒ y 0 = x2 cos x − (sin x)(2x) x(x cos x − 2sinx) x cos x − 2sinx<br />
x 2 (x 2 ) 2 = =<br />
x 4<br />
x 3<br />
15. Using Exercise 3.2.55(a), f(x) =xe x csc x ⇒<br />
17.<br />
19.<br />
f 0 (x)=(x) 0 e x csc x + x(e x ) 0 csc x + xe x (csc x) 0 =1e x csc x + xe x csc x + xe x (− cot x csc x)<br />
= e x csc x (1 + x − x cot x)<br />
<br />
d<br />
d 1 (sin x)(0) − 1(cos x)<br />
(csc x) = =<br />
dx dx sin x<br />
sin 2 = − cos x<br />
x<br />
sin 2 x = − 1<br />
sin x · cos x = − csc x cot x<br />
sin x<br />
d<br />
d<br />
(cot x) =<br />
dx dx<br />
cos x<br />
<br />
(sin x)(− sin x) − (cos x)(cos x)<br />
=<br />
sin x<br />
sin 2 = − sin2 x +cos 2 x<br />
x<br />
sin 2 = − 1<br />
x sin 2 x = − csc2 x<br />
21. y =secx ⇒ y 0 =secx tan x,soy 0 ( π 3 )=secπ 3 tan π 3 =2√ 3. An equation of the tangent line to the curve y =secx<br />
at the point π<br />
, 2 is y − 2=2 √ 3 √ √<br />
x − π 3 3 or y =2 3 x +2−<br />
2<br />
3 3 π.<br />
23. y = x +cosx ⇒ y 0 =1− sin x. At(0, 1), y 0 =1, and an equation of the tangent line is y − 1=1(x − 0),ory = x +1.<br />
25. (a) y =2x sin x ⇒ y 0 =2(x cos x +sinx · 1). At π<br />
2 ,π ,<br />
y 0 =2 π<br />
cos π<br />
2 2 +sinπ 2 =2(0+1)=2, and an equation of the<br />
tangent line is y − π =2 <br />
x − π 2 ,ory =2x.<br />
(b)<br />
27. (a) f(x) =secx − x ⇒ f 0 (x) =secx tan x − 1<br />
(b)<br />
Note that f 0 =0where f has a minimum. Also note that f 0 is negative<br />
when f is decreasing and f 0 is positive when f is increasing.<br />
29. H(θ) =θ sin θ ⇒ H 0 (θ) =θ (cos θ) +(sinθ) · 1=θ cos θ +sinθ ⇒<br />
H 00 (θ) =θ (− sin θ)+(cosθ) · 1+cosθ = −θ sin θ +2cosθ<br />
31. (a) f(x) = tan x − 1<br />
sec x<br />
⇒<br />
f 0 (x) = sec x(sec2 x) − (tan x − 1)(sec x tan x)<br />
(sec x) 2<br />
(b) f(x) = tan x − 1<br />
sec x<br />
=<br />
sin x<br />
cos x − 1<br />
=<br />
1<br />
cos x<br />
(c) From part (a), f 0 (x) = 1+tanx<br />
sec x<br />
sin x − cos x<br />
cos x<br />
1<br />
cos x<br />
= sec x(sec2 x − tan 2 x +tanx)<br />
sec 2 x<br />
= 1+tanx<br />
sec x<br />
=sinx − cos x ⇒ f 0 (x) =cosx − (− sin x) =cosx +sinx<br />
= 1<br />
sec x + tan x<br />
sec x =cosx +sinx, which is the expression for f 0 (x) in part (b).
F.<br />
98 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
33. f(x) =x +2sinx has a horizontal tangent when f 0 (x) =0 ⇔ 1+2cosx =0 ⇔ cos x = − 1 2<br />
⇔<br />
x = 2π +2πn or 4π +2πn,wheren is an integer. Note that 4π and 2π are ± π units from π.Thisallowsustowritethe<br />
3 3 3 3 3<br />
solutions in the more compact equivalent form (2n +1)π ± π 3<br />
, n an integer.<br />
35. (a) x(t) =8sint ⇒ v(t) =x 0 (t) =8cost ⇒ a(t) =x 00 (t) =−8sint<br />
(b) The mass at time t = 2π 3<br />
has position x √3 <br />
2π<br />
3 =8sin<br />
2π<br />
=8 3 2<br />
<br />
and acceleration a √3<br />
2π<br />
3 = −8sin<br />
2π<br />
= −8 3 2<br />
=4 √ 3,velocityv <br />
2π<br />
3 =8cos<br />
2π<br />
=8 <br />
3<br />
− 1 2 = −4,<br />
= −4 √ 3.Sincev <br />
2π<br />
3 < 0, the particle is moving to the left.<br />
37. From the diagram we can see that sin θ = x/10 ⇔ x =10sinθ. Wewanttofind the rate<br />
of change of x with respect to θ,thatis,dx/dθ. Taking the derivative of x =10sinθ,weget<br />
dx/dθ = 10(cos θ). Sowhenθ = π 3 , dx<br />
dθ =10cosπ 3 =10 1<br />
2<br />
<br />
39. lim<br />
x→0<br />
sin 3x<br />
x<br />
3sin3x<br />
=lim<br />
x→0 3x<br />
=3 lim<br />
3x→0<br />
sin 3x<br />
3x<br />
[multiply numerator and denominator by 3]<br />
[as x → 0, 3x → 0]<br />
sin θ<br />
=3lim<br />
θ→0 θ<br />
[let θ =3x]<br />
=3(1) [Equation 2]<br />
=3<br />
<br />
tan 6t sin 6t<br />
41. lim<br />
t→0 sin 2t =lim ·<br />
t→0 t<br />
sin 6t<br />
=6lim<br />
t→0 6t<br />
1<br />
cos 6t ·<br />
<br />
t<br />
6sin6t 1<br />
=lim · lim<br />
sin 2t t→0 6t t→0 cos 6t · lim<br />
t→0<br />
1<br />
· lim<br />
t→0 cos 6t · 1<br />
2 lim<br />
t→0<br />
<br />
sin(cos θ)<br />
sin lim cos θ<br />
θ→0<br />
43. lim =<br />
θ→0 sec θ lim sec θ = sin 1 =sin1<br />
1<br />
θ→0<br />
45. Divide numerator and denominator by θ. (sin θ also works.)<br />
lim<br />
θ→0<br />
47. lim<br />
x→π/4<br />
49. (a)<br />
(b)<br />
sin θ<br />
θ +tanθ =lim<br />
θ→0<br />
1 − tan x<br />
sin x − cos x =<br />
d<br />
dx tan x = d<br />
dx<br />
d<br />
dx sec x = d<br />
dx<br />
sin θ<br />
θ<br />
1+ sin θ<br />
θ<br />
lim<br />
x→π/4<br />
·<br />
1<br />
cos θ<br />
sin x<br />
cos x ⇒ sec2 x =<br />
1<br />
cos x<br />
=<br />
2t<br />
sin 2t =6(1)· 1<br />
1 · 1<br />
2 (1) = 3<br />
sin θ<br />
lim<br />
θ→0 θ<br />
sin θ<br />
1+lim lim<br />
θ→0 θ θ→0<br />
<br />
1 − sin x <br />
· cos x<br />
cos x<br />
(sin x − cos x) · cos x =<br />
lim<br />
x→π/4<br />
1<br />
cos θ<br />
cos x cos x − sin x (− sin x)<br />
cos 2 x<br />
=<br />
2t<br />
2sin2t<br />
1<br />
1+1· 1 = 1 2<br />
cos x − sin x<br />
(sin x − cos x)cosx =<br />
lim<br />
x→π/4<br />
−1<br />
cos x = −1<br />
1/ √ 2 = −√ 2<br />
= cos2 x +sin 2 x<br />
. Sosec 2 x = 1<br />
cos 2 x<br />
cos 2 x .<br />
(cos x)(0) − 1(− sin x)<br />
⇒ sec x tan x = . Sosec x tan x = sin x<br />
cos 2 x<br />
cos 2 x .
F.<br />
TX.10<br />
SECTION 3.4 THE CHAIN RULE ¤ 99<br />
(c)<br />
d<br />
d 1+cotx<br />
(sin x +cosx) =<br />
dx dx csc x<br />
⇒<br />
cos x − sin x = csc x (− csc2 x) − (1 + cot x)(− csc x cot x)<br />
csc 2 x<br />
= − csc2 x +cot 2 x +cotx<br />
csc x<br />
So cos x − sin x = cot x − 1<br />
csc x .<br />
= −1+cotx<br />
csc x<br />
= csc x [− csc2 x +(1+cotx) cotx]<br />
csc 2 x<br />
51. By the definition of radian measure, s = rθ,wherer is the radius of the circle. By drawing the bisector of the angle θ,wecan<br />
see that sin θ 2 = d/2<br />
r<br />
⇒ d =2r sin θ s<br />
.So lim<br />
2 θ→0 + d = lim<br />
θ→0 +<br />
rθ<br />
2r sin(θ/2) = lim<br />
θ→0 +<br />
2 · (θ/2)<br />
2sin(θ/2) =lim<br />
θ→0<br />
θ/2<br />
sin(θ/2) =1.<br />
sin x<br />
[This is just the reciprocal of the limit lim =1combined with the fact that as θ → 0, θ → 0 also.]<br />
x→0 x<br />
2<br />
3.4 The Chain Rule<br />
1. Let u = g(x) =4x and y = f(u) =sinu. Then dy<br />
dx = dy du<br />
=(cosu)(4) = 4 cos 4x.<br />
du dx<br />
3. Let u = g(x) =1− x 2 and y = f(u) =u 10 . Then dy<br />
dx = dy du<br />
du dx =(10u9 )(−2x) =−20x(1 − x 2 ) 9 .<br />
5. Let u = g(x) = √ x and y = f(u) =e u .Then dy<br />
dx = dy du<br />
du dx =(eu )<br />
7. F (x) =(x 4 +3x 2 − 2) 5 ⇒ F 0 (x) =5(x 4 +3x 2 − 2) 4 ·<br />
or 10x(x 4 +3x 2 − 2) 4 (2x 2 +3) <br />
<br />
1<br />
2 x−1/2 <br />
= e √x ·<br />
1<br />
2 √ x = e√ x<br />
2 √ x .<br />
d <br />
x 4 +3x 2 − 2 =5(x 4 +3x 2 − 2) 4 (4x 3 +6x)<br />
dx<br />
9. F (x) = 4√ 1+2x + x 3 =(1+2x + x 3 ) 1/4 ⇒<br />
F 0 (x) = 1 4 (1 + 2x + x3 ) −3/4 ·<br />
11. g(t) =<br />
2+3x 2<br />
=<br />
4 4 (1 + 2x + x 3 ) 3<br />
d<br />
dx (1 + 2x + x3 )=<br />
1<br />
4(1 + 2x + x 3 ) 3/4 · (2 + 3x2 )=<br />
2+3x 2<br />
4(1 + 2x + x 3 ) 3/4<br />
1<br />
(t 4 +1) 3 =(t4 +1) −3 ⇒ g 0 (t) =−3(t 4 +1) −4 (4t 3 )=−12t 3 (t 4 +1) −4 = −12t3<br />
(t 4 +1) 4<br />
13. y =cos(a 3 + x 3 ) ⇒ y 0 = − sin(a 3 + x 3 ) · 3x 2 [a 3 is just a constant] = −3x 2 sin(a 3 + x 3 )<br />
15. y = xe −kx ⇒ y 0 = x e −kx (−k) + e −kx · 1=e −kx (−kx +1)<br />
<br />
or (1 − kx)e<br />
−kx <br />
17. g(x) =(1+4x) 5 (3 + x − x 2 ) 8 ⇒<br />
g 0 (x) =(1+4x) 5 · 8(3 + x − x 2 ) 7 (1 − 2x)+(3+x − x 2 ) 8 · 5(1 + 4x) 4 · 4<br />
=4(1+4x) 4 (3 + x − x 2 ) 7 2(1 + 4x)(1 − 2x)+5(3+x − x 2 ) <br />
=4(1+4x) 4 (3 + x − x 2 ) 7 (2 + 4x − 16x 2 )+(15+5x − 5x 2 ) =4(1+4x) 4 (3 + x − x 2 ) 7 (17 + 9x − 21x 2 )
F.<br />
100 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
19. y =(2x − 5) 4 (8x 2 − 5) −3 ⇒<br />
y 0 =4(2x − 5) 3 (2)(8x 2 − 5) −3 +(2x − 5) 4 (−3)(8x 2 − 5) −4 (16x)<br />
=8(2x − 5) 3 (8x 2 − 5) −3 − 48x(2x − 5) 4 (8x 2 − 5) −4<br />
[This simplifies to 8(2x − 5) 3 (8x 2 − 5) −4 (−4x 2 +30x − 5).]<br />
21. y =<br />
x 2 3<br />
+1<br />
x 2 − 1<br />
⇒<br />
x<br />
y 0 2 2<br />
+1<br />
=3<br />
·<br />
x 2 − 1<br />
x 2 +1<br />
=3<br />
x 2 − 1<br />
d<br />
dx<br />
x 2 +1<br />
=3<br />
x 2 − 1<br />
23. y = e x cos x ⇒ y 0 = e x cos x ·<br />
25. F (z) =<br />
2<br />
· 2x[x2 − 1 − (x 2 +1)]<br />
(x 2 − 1) 2 =3<br />
1/2<br />
z − 1 z − 1<br />
z +1 = z +1<br />
F 0 (z) = 1 −1/2 z − 1<br />
·<br />
2 z +1<br />
d<br />
dz<br />
x 2 2<br />
+1<br />
· (x2 − 1)(2x) − (x 2 +1)(2x)<br />
x 2 − 1<br />
(x 2 − 1) 2<br />
x 2 2<br />
+1<br />
·<br />
x 2 − 1<br />
2x(−2)<br />
(x 2 − 1) = −12x(x2 +1) 2<br />
2 (x 2 − 1) 4<br />
d<br />
dx (x cos x) =ex cos x [x(− sin x)+(cosx) · 1] = e x cos x (cos x − x sin x)<br />
⇒<br />
z − 1<br />
= 1 z +1 2<br />
1/2 z +1<br />
·<br />
z − 1<br />
(z +1)(1)− (z − 1)(1)<br />
(z +1) 2<br />
= 1 (z +1) 1/2 z +1− z +1<br />
· = 1 (z +1) 1/2<br />
2 (z − 1)<br />
1/2<br />
(z +1) 2 2 (z − 1) · 2<br />
1/2 (z +1) = 1<br />
2 (z − 1) 1/2 (z +1) 3/2<br />
27. y =<br />
r<br />
√<br />
r2 +1<br />
⇒<br />
y 0 =<br />
√<br />
√<br />
r 2<br />
r2 +1(1)− r · 1<br />
2 (r2 +1) −1/2 r2 +1− √<br />
(2r)<br />
√<br />
r2 +1 r2 +1<br />
2<br />
= √<br />
r2 +1 2<br />
=<br />
√<br />
r2 +1 √ r 2 +1− r 2<br />
√<br />
r2 +1<br />
√<br />
r2 +1 2<br />
=<br />
r 2 +1 − r 2<br />
√<br />
r2 +1 1<br />
3<br />
=<br />
(r 2 +1) or 3/2 (r2 +1) −3/2<br />
Another solution: Write y as a product and make use of the Product Rule. y = r(r 2 +1) −1/2<br />
⇒<br />
y 0 = r · − 1 2 (r2 +1) −3/2 (2r)+(r 2 +1) −1/2 · 1=(r 2 +1) −3/2 [−r 2 +(r 2 +1) 1 ]=(r 2 +1) −3/2 (1) = (r 2 +1) −3/2 .<br />
The step that students usually have trouble with is factoring out (r 2 +1) −3/2 . But this is no different than factoring out x 2<br />
from x 2 + x 5 ; that is, we are just factoring out a factor with the smallest exponent that appears on it. In this case, − 3 2 is<br />
smaller than − 1 2 .<br />
29. y = sin(tan 2x) ⇒ y 0 =cos(tan2x) ·<br />
31. Using Formula 5 and the Chain Rule, y =2 sin πx ⇒<br />
y 0 =2 sin πx (ln 2) ·<br />
d<br />
dx (tan 2x) = cos(tan 2x) · d<br />
sec2 (2x) ·<br />
dx (2x) = 2 cos(tan 2x)sec2 (2x)<br />
d<br />
dx (sin πx) =2sin πx (ln 2) · cos πx · π =2 sin πx (π ln 2) cos πx
F.<br />
TX.10<br />
SECTION 3.4 THE CHAIN RULE ¤ 101<br />
33. y =sec 2 x +tan 2 x =(secx) 2 +(tanx) 2 ⇒<br />
y 0 =2(secx)(sec x tan x) + 2(tan x)(sec 2 x)=2sec 2 x tan x +2sec 2 x tan x =4sec 2 x tan x<br />
1 − e<br />
2x<br />
<br />
35. y =cos<br />
1+e 2x<br />
⇒<br />
1 − e<br />
y 0 2x<br />
<br />
= − sin<br />
·<br />
1+e 2x<br />
<br />
d 1 − e<br />
2x<br />
1 − e<br />
2x<br />
<br />
= − sin<br />
· (1 + e2x )(−2e 2x ) − (1 − e 2x )(2e 2x )<br />
dx 1+e 2x 1+e 2x (1 + e 2x ) 2<br />
1 − e<br />
2x<br />
1 − e<br />
2x<br />
<br />
= − sin<br />
· −2e2x (1 + e 2x )+(1− e 2x ) <br />
= − sin<br />
1+e 2x (1 + e 2x ) 2<br />
1+e 2x <br />
<br />
· −2e2x (2)<br />
(1 + e 2x ) = 4e 2x 1 − e<br />
2x<br />
<br />
2 (1 + e 2x ) · sin 2 1+e 2x<br />
37. y =cot 2 (sin θ) =[cot(sinθ)] 2 ⇒<br />
y 0 = 2[cot(sin θ)] ·<br />
d<br />
dθ [cot(sin θ)] = 2 cot(sin θ) · [− csc2 (sin θ) · cos θ] =−2cosθ cot(sin θ) csc 2 (sin θ)<br />
39. f(t) =tan(e t )+e tan t ⇒ f 0 (t) =sec 2 (e t ) · d<br />
dt (et )+e tan t · d<br />
dt (tan t) =sec2 (e t ) · e t + e tan t · sec 2 t<br />
<br />
41. f(t) =sin 2 e sin2 t<br />
= sin e<br />
t 2 sin2 ⇒<br />
<br />
f 0 (t)=2 sin e sin2 t<br />
· d<br />
<br />
=2sin e sin2 t<br />
cos<br />
<br />
dt sin<br />
<br />
e sin2 t<br />
<br />
e sin2 t<br />
<br />
=4sin e sin2 t<br />
cos e sin2 t<br />
e sin2t sin t cos t<br />
<br />
=2sin e sin2 t<br />
· cos e sin2 t<br />
· d<br />
· e sin2t · d<br />
<br />
dt sin2 t =2sin e sin2 t<br />
cos<br />
dt esin2 t<br />
<br />
e sin2 t<br />
<br />
e sin2t · 2sint cos t<br />
43. g(x) =(2ra rx + n) p ⇒<br />
g 0 (x) =p(2ra rx + n) p−1 ·<br />
45. y =cos sin(tan πx) = cos(sin(tan πx)) 1/2 ⇒<br />
y 0 = − sin(sin(tan πx)) 1/2 ·<br />
= − sin sin(tan πx)<br />
2 sin(tan πx)<br />
d<br />
dx (2rarx + n) =p(2ra rx + n) p−1 · 2ra rx (ln a) · r =2r 2 p(ln a)(2ra rx + n) p−1 a rx<br />
d<br />
dx (sin(tan πx))1/2 = − sin(sin(tan πx)) 1/2 · 1 (sin(tan d<br />
2 πx))−1/2 · (sin(tan πx))<br />
dx<br />
· cos(tan πx) ·<br />
= −π cos(tan πx)sec2 (πx)sin sin(tan πx)<br />
2 sin(tan πx)<br />
d<br />
dx tan πx = − sin sin(tan πx)<br />
2 · cos(tan πx) · sec 2 (πx) · π<br />
sin(tan πx)<br />
47. h(x) = √ x 2 +1 ⇒ h 0 (x) = 1 2 (x2 +1) −1/2 (2x) =<br />
x<br />
√<br />
x2 +1<br />
⇒<br />
h 00 (x) =<br />
√ <br />
<br />
1<br />
x2 +1· 1 − x<br />
2 (x2 +1) −1/2 (2x)<br />
√<br />
x2 +1 2<br />
=<br />
x 2 +1 −1/2 (x 2 +1)− x 2<br />
(x 2 +1) 1 =<br />
1<br />
(x 2 +1) 3/2
F.<br />
102 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
49. y = e αx sin βx ⇒ y 0 = e αx · β cos βx +sinβx · αe αx = e αx (β cos βx + α sin βx) ⇒<br />
y 00 = e αx (−β 2 sin βx + αβ cos βx)+(β cos βx + α sin βx) · αe αx<br />
= e αx (−β 2 sin βx + αβ cos βx + αβ cos βx + α 2 sin βx) =e αx (α 2 sin βx − β 2 sin βx +2αβ cos βx)<br />
= e αx (α 2 − β 2 )sinβx +2αβ cos βx <br />
51. y =(1+2x) 10 ⇒ y 0 =10(1+2x) 9 · 2=20(1+2x) 9 .<br />
At (0, 1), y 0 = 20(1 + 0) 9 =20, and an equation of the tangent line is y − 1=20(x − 0),ory =20x +1.<br />
53. y =sin(sinx) ⇒ y 0 =cos(sinx) · cos x. At (π, 0), y 0 =cos(sinπ) · cos π =cos(0)· (−1) = 1(−1) = −1,andan<br />
equation of the tangent line is y − 0=−1(x − π),ory = −x + π.<br />
55. (a) y =<br />
2<br />
⇒ y 0 = (1 + e−x )(0) − 2(−e −x ) 2e −x<br />
=<br />
1+e −x (1 + e −x ) 2 (1 + e −x ) . 2<br />
(b)<br />
At (0, 1), y 0 =<br />
2e 0<br />
(1 + e 0 ) 2 = 2(1)<br />
(1 + 1) 2 = 2 2 2 = 1 2 .Soanequationofthe<br />
tangent line is y − 1= 1 2 (x − 0) or y = 1 2 x +1.<br />
57. (a) f(x) =x √ 2 − x 2 = x(2 − x 2 ) 1/2 ⇒<br />
f 0 (x) =x · 1<br />
2 (2 − x2 ) −1/2 (−2x)+(2− x 2 ) 1/2 · 1=(2− x 2 ) −1/2 −x 2 +(2− x 2 ) = 2 − 2x2 √<br />
2 − x<br />
2<br />
(b)<br />
f 0 =0when f has a horizontal tangent line, f 0 is negative when f is<br />
decreasing, and f 0 is positive when f is increasing.<br />
59. For the tangent line to be horizontal, f 0 (x) =0. f(x) =2sinx +sin 2 x ⇒ f 0 (x) =2cosx +2sinx cos x =0 ⇔<br />
2cosx(1 + sin x) =0 ⇔ cos x =0or sin x = −1, sox = π 2 +2nπ or 3π 2<br />
+2nπ, wheren is any integer. Now<br />
f π<br />
2<br />
<br />
=3and f<br />
3π<br />
2<br />
where n is any integer.<br />
<br />
= −1, so the points on the curve with a horizontal tangent are<br />
π<br />
2 +2nπ, 3 and 3π<br />
2 +2nπ, −1 ,<br />
61. F (x) =f(g(x)) ⇒ F 0 (x) =f 0 (g(x)) · g 0 (x),soF 0 (5) = f 0 (g(5)) · g 0 (5) = f 0 (−2) · 6=4· 6=24<br />
63. (a) h(x) =f(g(x)) ⇒ h 0 (x) =f 0 (g(x)) · g 0 (x),soh 0 (1) = f 0 (g(1)) · g 0 (1) = f 0 (2) · 6=5· 6=30.<br />
(b) H(x) =g(f(x)) ⇒ H 0 (x) =g 0 (f(x)) · f 0 (x),soH 0 (1) = g 0 (f(1)) · f 0 (1) = g 0 (3) · 4=9· 4=36.<br />
65. (a) u(x) =f(g(x)) ⇒ u 0 (x) =f 0 (g(x))g 0 (x). Sou 0 (1) = f 0 (g(1))g 0 (1) = f 0 (3)g 0 (1). Tofind f 0 (3), note that f is<br />
linear from (2, 4) to (6, 3), so its slope is 3 − 4<br />
6 − 2 = − 1 4 .Tofind g0 (1), note that g is linear from (0, 6) to (2, 0), so its slope<br />
is 0 − 6<br />
2 − 0 = −3. Thus, f 0 (3)g 0 (1) = <br />
− 1 4 (−3) =<br />
3<br />
. 4<br />
(b) v(x) =g(f(x)) ⇒ v 0 (x) =g 0 (f(x))f 0 (x). Sov 0 (1) = g 0 (f(1))f 0 (1) = g 0 (2)f 0 (1), which does not exist since<br />
g 0 (2) does not exist.
F.<br />
TX.10<br />
SECTION 3.4 THE CHAIN RULE ¤ 103<br />
(c) w(x) =g(g(x)) ⇒ w 0 (x) =g 0 (g(x))g 0 (x). Sow 0 (1) = g 0 (g(1))g 0 (1) = g 0 (3)g 0 (1). Tofind g 0 (3), note that g is<br />
linear from (2, 0) to (5, 2),soitsslopeis 2 − 0<br />
5 − 2 = 2 3 . Thus, g0 (3)g 0 (1) = 2<br />
3<br />
<br />
(−3) = −2.<br />
67. (a) F (x) =f(e x ) ⇒ F 0 (x) =f 0 (e x ) d<br />
dx (ex )=f 0 (e x )e x<br />
(b) G(x) =e f(x) ⇒ G 0 (x) =e f(x) d<br />
dx f(x) =ef(x) f 0 (x)<br />
69. r(x) =f(g(h(x))) ⇒ r 0 (x) =f 0 (g(h(x))) · g 0 (h(x)) · h 0 (x),so<br />
r 0 (1) = f 0 (g(h(1))) · g 0 (h(1)) · h 0 (1) = f 0 (g(2)) · g 0 (2) · 4=f 0 (3) · 5 · 4=6· 5 · 4=120<br />
71. F (x) =f(3f(4f(x))) ⇒<br />
F 0 (x)=f 0 (3f(4f(x))) ·<br />
d<br />
dx (3f(4f(x))) = f 0 (3f(4f(x))) · 3f 0 d<br />
(4f(x)) ·<br />
dx (4f(x))<br />
= f 0 (3f(4f(x))) · 3f 0 (4f(x)) · 4f 0 (x), so<br />
F 0 (0) = f 0 (3f(4f(0))) · 3f 0 (4f(0)) · 4f 0 (0) = f 0 (3f(4 · 0)) · 3f 0 (4 · 0) · 4 · 2=f 0 (3 · 0) · 3 · 2 · 4 · 2=2· 3 · 2 · 4 · 2=96.<br />
73. y = Ae −x + Bxe −x ⇒<br />
y 0 = A(−e −x )+B[x(−e −x )+e −x · 1] = −Ae −x + Be −x − Bxe −x =(B − A)e −x − Bxe −x<br />
⇒<br />
y 00 =(B − A)(−e −x ) − B[x(−e −x )+e −x · 1] = (A − B)e −x − Be −x + Bxe −x =(A − 2B)e −x + Bxe −x ,<br />
so<br />
y 00 +2y 0 + y =(A − 2B)e −x + Bxe −x +2[(B − A)e −x − Bxe −x ]+Ae −x + Bxe −x<br />
=[(A − 2B)+2(B − A)+A]e −x +[B − 2B + B]xe −x =0.<br />
75. The use of D, D 2 , ..., D n is just a derivative notation (see text page 157). In general, Df(2x) =2f 0 (2x),<br />
D 2 f(2x) =4f 00 (2x), ..., D n f(2x) =2 n f (n) (2x). Sincef(x) =cosx and 50 = 4(12) + 2, wehave<br />
f (50) (x) =f (2) (x) =− cos x,soD 50 cos 2x = −2 50 cos 2x.<br />
77. s(t) =10+ 1 sin(10πt) ⇒ the velocity after t seconds is v(t) 4 =s0 (t) = 1 cos(10πt)(10π) = 5π cos(10πt) cm/s.<br />
4 2<br />
79. (a) B(t) =4.0+0.35 sin 2πt ⇒<br />
dB <br />
5.4 dt = 0.35 cos 2πt 2π<br />
= 0.7π 2πt<br />
cos<br />
5.4 5.4 5.4 5.4 = 7π 2πt<br />
cos<br />
54 5.4<br />
(b) At t =1, dB<br />
dt = 7π 2π<br />
cos<br />
54 5.4 ≈ 0.16.<br />
81. s(t) =2e −1.5t sin 2πt ⇒<br />
v(t) =s 0 (t) =2[e −1.5t (cos 2πt)(2π)+(sin2πt)e −1.5t (−1.5)] = 2e −1.5t (2π cos 2πt − 1.5sin2πt)
F.<br />
104 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
83. By the Chain Rule, a(t) = dv<br />
dt = dv ds<br />
ds dt = dv<br />
dv<br />
v(t) =v(t)<br />
ds ds .<br />
Thederivativedv/dt is the rate of change of the velocity<br />
with respect to time (in other words, the acceleration) whereas the derivative dv/ds is the rate of change of the velocity with<br />
respect to the displacement.<br />
85. (a) Using a calculator or CAS, we obtain the model Q = ab t with a ≈ 100.0124369 and b ≈ 0.000045145933.<br />
(b) Use Q 0 (t) =ab t ln b (from Formula 5) with the values of a and b from part (a) to get Q 0 (0.04) ≈−670.63 μA.<br />
The result of Example 2 in Section 2.1 was −670 μA.<br />
87. (a) Derive gives g 0 (t) =<br />
45(t − 2)8<br />
without simplifying. With either Maple or Mathematica, we first get<br />
(2t +1)<br />
10<br />
g 0 (t − 2)8 (t − 2)9<br />
(t) =9 − 18 , and the simplification command results in the expression given by Derive.<br />
(2t +1)<br />
9<br />
(2t +1)<br />
10<br />
(b) Derive gives y 0 =2(x 3 − x +1) 3 (2x +1) 4 (17x 3 +6x 2 − 9x +3)without simplifying. With either Maple or<br />
Mathematica, we first get y 0 =10(2x +1) 4 (x 3 − x +1) 4 +4(2x +1) 5 (x 3 − x +1) 3 (3x 2 − 1). Ifweuse<br />
Mathematica’s Factor or Simplify,orMaple’sfactor, we get the above expression, but Maple’s simplify gives<br />
the polynomial expansion instead. For locating horizontal tangents, the factored form is the most helpful.<br />
89. (a) If f is even, then f(x) =f(−x). Using the Chain Rule to differentiate this equation, we get<br />
f 0 (x) =f 0 (−x) d<br />
dx (−x) =−f 0 (−x). Thus,f 0 (−x) =−f 0 (x),sof 0 is odd.<br />
(b) If f is odd, then f(x) =−f(−x).<br />
even.<br />
91. (a)<br />
(b)<br />
d<br />
dx (sinn x cos nx) =n sin n−1 x cos x cos nx +sin n x (−n sin nx)<br />
Differentiating this equation, we get f 0 (x) =−f 0 (−x)(−1) = f 0 (−x),sof 0 is<br />
[Product Rule]<br />
= n sin n−1 x (cos nx cos x − sin nx sin x) [factor out n sin n−1 x]<br />
= n sin n−1 x cos(nx + x) [Addition Formula for cosine]<br />
= n sin n−1 x cos[(n +1)x] [factor out x]<br />
d<br />
dx (cosn x cos nx) =n cos n−1 x (− sin x)cosnx +cos n x (−n sin nx)<br />
[Product Rule]<br />
= −n cos n−1 x (cos nx sin x +sinnx cos x) [factor out −n cos n−1 x]<br />
= −n cos n−1 x sin(nx + x) [Addition Formula for sine]<br />
= −n cos n−1 x sin[(n +1)x] [factor out x]<br />
93. Since θ ◦ = <br />
π<br />
d<br />
180 θ rad, we have<br />
dθ (sin θ◦ )= d <br />
sin<br />
π<br />
180<br />
dθ<br />
θ = π cos π θ = π cos 180 180 180 θ◦ .<br />
95. The Chain Rule says that dy<br />
dx = dy<br />
du<br />
dy<br />
d 2 y<br />
dx = d<br />
2 dx<br />
dx<br />
du<br />
dx ,so<br />
<br />
= d<br />
dx<br />
dy du<br />
du dx<br />
d<br />
=<br />
dx<br />
dy du<br />
du<br />
dx + dy d<br />
du dx<br />
du<br />
dx<br />
[Product Rule]<br />
d<br />
=<br />
du<br />
dy<br />
du<br />
du du<br />
dx dx + dy<br />
<br />
d 2 u<br />
du dx = d2 y du<br />
2 du 2 dx<br />
2<br />
+ dy d 2 u<br />
du dx 2
F.<br />
TX.10<br />
SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 105<br />
3.5 Implicit Differentiation<br />
d<br />
1. (a)<br />
dx (xy +2x +3x2 )= d<br />
dx (4) ⇒ (x · y0 + y · 1)+2+6x =0 ⇒ xy 0 = −y − 2 − 6x ⇒<br />
y 0 −y − 2 − 6x<br />
= or y 0 = −6 − y +2<br />
x<br />
x .<br />
(b) xy +2x +3x 2 =4 ⇒ xy =4− 2x − 3x 2 ⇒ y =<br />
3. (a)<br />
5.<br />
(c) From part (a), y 0 =<br />
−y − 2 − 6x<br />
x<br />
=<br />
−(4/x − 2 − 3x) − 2 − 6x<br />
x<br />
4 − 2x − 3x2<br />
x<br />
=<br />
= 4 x − 2 − 3x,soy0 = − 4 x 2 − 3.<br />
−4/x − 3x<br />
x<br />
= − 4 x 2 − 3.<br />
<br />
d 1<br />
dx x + 1 <br />
= d<br />
y dx (1) ⇒ − 1 x − 1<br />
2 y 2 y0 =0 ⇒ − 1 y 2 y0 = 1 ⇒ y 0 = − y2<br />
x 2 x 2<br />
(b) 1 x + 1 y =1 ⇒ 1 y =1− 1 x = x − 1<br />
x<br />
(c) y 0 = − y2<br />
2<br />
− 1)]<br />
= −[x/(x<br />
x2 x 2<br />
x 2<br />
⇒ y = x (x − 1)(1) − (x)(1)<br />
x − 1 ,soy0 = = −1<br />
(x − 1) 2 (x − 1) . 2<br />
= −<br />
x 2 (x − 1) = − 1<br />
2 (x − 1) 2<br />
d <br />
x 3 + y 3 = d<br />
dx<br />
dx (1) ⇒ 3x2 +3y 2 · y 0 =0 ⇒ 3y 2 y 0 = −3x 2 ⇒ y 0 = − x2<br />
y 2<br />
7.<br />
d<br />
dx (x2 + xy − y 2 )= d<br />
dx (4) ⇒ 2x + x · y0 + y · 1 − 2yy 0 =0 ⇒<br />
xy 0 − 2yy 0 = −2x − y ⇒ (x − 2y) y 0 = −2x − y ⇒ y 0 = −2x − y<br />
x − 2y<br />
= 2x + y<br />
2y − x<br />
9.<br />
11.<br />
13.<br />
15.<br />
d <br />
x 4 (x + y) = d <br />
y 2 (3x − y) ⇒ x 4 (1 + y 0 )+(x + y) · 4x 3 = y 2 (3 − y 0 )+(3x − y) · 2yy 0 ⇒<br />
dx<br />
dx<br />
x 4 + x 4 y 0 +4x 4 +4x 3 y =3y 2 − y 2 y 0 +6xy y 0 − 2y 2 y 0 ⇒ x 4 y 0 +3y 2 y 0 − 6xy y 0 =3y 2 − 5x 4 − 4x 3 y ⇒<br />
(x 4 +3y 2 − 6xy) y 0 =3y 2 − 5x 4 − 4x 3 y ⇒ y 0 = 3y2 − 5x 4 − 4x 3 y<br />
x 4 +3y 2 − 6xy<br />
d<br />
dx (x2 y 2 + x sin y) = d<br />
dx (4) ⇒ x2 · 2yy 0 + y 2 · 2x + x cos y · y 0 +siny · 1=0 ⇒<br />
2x 2 yy 0 + x cos y · y 0 = −2xy 2 − sin y ⇒ (2x 2 y + x cos y)y 0 = −2xy 2 − sin y ⇒ y 0 = −2xy2 − sin y<br />
2x 2 y + x cos y<br />
d<br />
(4 cos x sin y) =<br />
d<br />
dx<br />
dx (1) ⇒ 4[cosx · cos y · y0 +siny · (− sin x)] = 0 ⇒<br />
y 0 (4 cos x cos y) =4sinx sin y ⇒ y 0 =<br />
d<br />
dx (ex/y )=<br />
d<br />
dx (x − y) ⇒ ex/y ·<br />
4sinx sin y<br />
=tanx tan y<br />
4cosx cos y<br />
<br />
d x<br />
=1− y 0 ⇒<br />
dx y<br />
e x/y · y · 1 − x · y0<br />
=1− y 0 ⇒ e x/y · 1<br />
y 2 y − xex/y<br />
· y 0 =1− y 0 ⇒ y 0 − xex/y<br />
· y 0 =1− ex/y<br />
y 2 y 2<br />
y<br />
⇒<br />
<br />
y 0 1 − xex/y<br />
= y − ex/y<br />
y 2 y<br />
⇒ y 0 =<br />
y − e x/y<br />
y<br />
= y(y − ex/y )<br />
y 2 − xe x/y y 2 − xe x/y<br />
y 2
F.<br />
106 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
17.<br />
<br />
xy =1+x 2 y ⇒ 1 2 (xy)−1/2 (xy 0 + y · 1) = 0 + x 2 y 0 + y · 2x ⇒<br />
19.<br />
21.<br />
23.<br />
<br />
<br />
y 0 x<br />
2 xy − x2 =2xy −<br />
y<br />
2 xy<br />
⇒<br />
<br />
y 0 x − 2x 2 <br />
xy<br />
2 xy<br />
= 4xy xy − y<br />
2 xy<br />
x<br />
2 xy y0 + y<br />
2 xy = x2 y 0 +2xy ⇒<br />
⇒<br />
y 0 = 4xy xy − y<br />
x − 2x 2 xy<br />
d<br />
dx (ey cos x) = d<br />
dx [1 + sin(xy)] ⇒ ey (− sin x)+cosx · e y · y 0 =cos(xy) · (xy 0 + y · 1) ⇒<br />
−e y sin x + e y cos x · y 0 = x cos(xy) · y 0 + y cos(xy) ⇒ e y cos x · y 0 − x cos(xy) · y 0 = e y sin x + y cos(xy) ⇒<br />
[e y cos x − x cos(xy)] y 0 = e y sin x + y cos(xy) ⇒ y 0 = ey sin x + y cos(xy)<br />
e y cos x − x cos(xy)<br />
d <br />
f(x)+x 2 [f(x)] 3 = d<br />
dx<br />
dx (10) ⇒ f 0 (x)+x 2 · 3[f(x)] 2 · f 0 (x)+[f(x)] 3 · 2x =0. Ifx =1,wehave<br />
f 0 (1) + 1 2 · 3[f(1)] 2 · f 0 (1) + [f(1)] 3 · 2(1) = 0 ⇒ f 0 (1) + 1 · 3 · 2 2 · f 0 (1) + 2 3 · 2=0 ⇒<br />
f 0 (1) + 12f 0 (1) = −16 ⇒ 13f 0 (1) = −16 ⇒ f 0 (1) = − 16<br />
13 .<br />
d<br />
dy (x4 y 2 − x 3 y +2xy 3 )= d<br />
dy (0) ⇒ x4 · 2y + y 2 · 4x 3 x 0 − (x 3 · 1+y · 3x 2 x 0 )+2(x · 3y 2 + y 3 · x 0 )=0 ⇒<br />
4x 3 y 2 x 0 − 3x 2 yx 0 +2y 3 x 0 = −2x 4 y + x 3 − 6xy 2 ⇒ (4x 3 y 2 − 3x 2 y +2y 3 ) x 0 = −2x 4 y + x 3 − 6xy 2 ⇒<br />
x 0 = dx<br />
dy = −2x4 y + x 3 − 6xy 2<br />
4x 3 y 2 − 3x 2 y +2y 3<br />
25. x 2 + xy + y 2 =3 ⇒ 2x + xy 0 + y · 1+2yy 0 =0 ⇒ xy 0 +2yy 0 = −2x − y ⇒ y 0 (x +2y) =−2x − y ⇒<br />
y 0 = −2x − y<br />
x +2y . Whenx =1and y =1,wehavey0 = −2 − 1<br />
1+2· 1 = −3 = −1, so an equation of the tangent line is<br />
3<br />
y − 1=−1(x − 1) or y = −x +2.<br />
27. x 2 + y 2 =(2x 2 +2y 2 − x) 2 ⇒ 2x +2yy 0 =2(2x 2 +2y 2 − x)(4x +4yy 0 − 1). Whenx =0and y = 1 ,wehave<br />
2<br />
0+y 0 =2( 1 2 )(2y0 − 1) ⇒ y 0 =2y 0 − 1 ⇒ y 0 =1, so an equation of the tangent line is y − 1 =1(x − 0)<br />
2<br />
or y = x + 1 . 2<br />
29. 2(x 2 + y 2 ) 2 =25(x 2 − y 2 ) ⇒ 4(x 2 + y 2 )(2x +2yy 0 ) = 25(2x − 2yy 0 ) ⇒<br />
4(x + yy 0 )(x 2 + y 2 )=25(x − yy 0 ) ⇒ 4yy 0 (x 2 + y 2 )+25yy 0 =25x − 4x(x 2 + y 2 ) ⇒<br />
y 0 = 25x − 4x(x2 + y 2 )<br />
25y +4y(x 2 + y 2 ) . Whenx =3and y =1,wehavey0 75 − 120<br />
= = − 45 = − 9 ,<br />
25 + 40 65 13<br />
so an equation of the tangent line is y − 1=− 9<br />
13 (x − 3) or y = − 9 13 x + 40<br />
13 .<br />
31. (a) y 2 =5x 4 − x 2 ⇒ 2yy 0 =5(4x 3 ) − 2x ⇒ y 0 = 10x3 − x<br />
.<br />
y<br />
(b)<br />
So at the point (1, 2) we have y 0 = 10(1)3 − 1<br />
2<br />
of the tangent line is y − 2= 9 2 (x − 1) or y = 9 2 x − 5 2 .<br />
= 9 , and an equation<br />
2
F.<br />
TX.10<br />
SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 107<br />
33. 9x 2 + y 2 =9 ⇒ 18x +2yy 0 =0 ⇒ 2yy 0 = −18x ⇒ y 0 = −9x/y ⇒<br />
<br />
y · 1 − x · y<br />
y 00 0 y − x(−9x/y)<br />
= −9<br />
= −9<br />
= −9 · y2 +9x 2 9<br />
= −9 · [since x and y must satisfy the original<br />
y 2<br />
y 2<br />
y 3 y 3<br />
equation, 9x 2 + y 2 =9].Thus,y 00 = −81/y 3 .<br />
35. x 3 + y 3 =1 ⇒ 3x 2 +3y 2 y 0 =0 ⇒ y 0 = − x2<br />
y 2<br />
y 00 = − y2 (2x) − x 2 · 2yy 0<br />
= − 2xy2 − 2x 2 y(−x 2 /y 2 )<br />
= − 2xy4 +2x 4 y<br />
= − 2xy(y3 + x 3 )<br />
= − 2x<br />
(y 2 ) 2 y 4<br />
y 6 y 6 y , 5<br />
since x and y must satisfy the original equation, x 3 + y 3 =1.<br />
37. (a) There are eight points with horizontal tangents: four at x ≈ 1.57735 and<br />
four at x ≈ 0.42265.<br />
(b) y 0 =<br />
3x 2 − 6x +2<br />
2(2y 3 − 3y 2 − y +1)<br />
⇒<br />
⇒ y 0 = −1 at (0, 1) and y 0 = 1 at (0, 2).<br />
3<br />
Equations of the tangent lines are y = −x +1and y = 1 3 x +2.<br />
(c) y 0 =0 ⇒ 3x 2 − 6x +2=0 ⇒ x =1± 1 3<br />
√<br />
3<br />
(d) By multiplying the right side of the equation by x − 3, we obtain the first<br />
graph. By modifying the equation in other ways, we can generate the other<br />
graphs.<br />
y(y 2 − 1)(y − 2)<br />
= x(x − 1)(x − 2)(x − 3)<br />
y(y 2 − 4)(y − 2)<br />
= x(x − 1)(x − 2)<br />
y(y +1)(y 2 − 1)(y − 2)<br />
= x(x − 1)(x − 2)<br />
(y +1)(y 2 − 1)(y − 2)<br />
=(x − 1)(x − 2)<br />
x(y +1)(y 2 − 1)(y − 2)<br />
= y(x − 1)(x − 2)<br />
y(y 2 +1)(y − 2)<br />
= x(x 2 − 1)(x − 2)<br />
y(y +1)(y 2 − 2)<br />
= x(x − 1)(x 2 − 2)
F.<br />
108 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
39. From Exercise 29, a tangent to the lemniscate will be horizontal if y 0 =0 ⇒ 25x − 4x(x 2 + y 2 )=0 ⇒<br />
x[25 − 4(x 2 + y 2 )] = 0 ⇒ x 2 + y 2 = 25 4<br />
(1). (Note that when x is 0, y is also 0, and there is no horizontal tangent<br />
at the origin.) Substituting 25 for 4<br />
x2 + y 2 in the equation of the lemniscate, 2(x 2 + y 2 ) 2 =25(x 2 − y 2 ),weget<br />
<br />
x 2 − y 2 = 25 (2). Solving (1)and(2), we have x 2 = 75 and 8 16 y2 = 25 , so the four points are ± 5√ 3<br />
, ± 5 .<br />
16 4 4<br />
41.<br />
x 2<br />
a − y2<br />
2x<br />
=1 ⇒ 2 b2 a − 2yy0 =0 ⇒ y 0 = b2 x<br />
2 b 2 a 2 y<br />
⇒<br />
an equation of the tangent line at (x 0 ,y 0 ) is<br />
y − y 0 = b2 x 0<br />
(x − x<br />
a 2 0 ). Multiplying both sides by y0 y0y<br />
gives<br />
y 0 b2 b − y2 0<br />
2 b 2<br />
= x0x<br />
a 2 − x2 0<br />
a 2 .Since(x 0,y 0 ) lies on the hyperbola,<br />
we have x0x<br />
a 2<br />
− y0y<br />
b 2 = x2 0<br />
a 2 − y2 0<br />
b 2 =1.<br />
43. If the circle has radius r, its equation is x 2 + y 2 = r 2 ⇒ 2x +2yy 0 =0 ⇒ y 0 = − x , so the slope of the tangent line<br />
y<br />
at P (x 0 ,y 0 ) is − x 0<br />
−1<br />
. The negative reciprocal of that slope is = y 0<br />
, which is the slope of OP, so the tangent line at<br />
y 0 −x 0 /y 0 x 0<br />
P is perpendicular to the radius OP.<br />
45. y =tan √ −1 x ⇒ y 0 1<br />
= √ 2 ·<br />
1+ x<br />
d<br />
√ <br />
x = 1<br />
dx 1+x<br />
<br />
1<br />
2 x−1/2 <br />
=<br />
1<br />
2 √ x (1 + x)<br />
47. y =sin −1 (2x +1) ⇒<br />
y 0 =<br />
1<br />
· d<br />
1 − (2x +1)<br />
2 dx (2x +1)= 1<br />
<br />
1 − (4x2 +4x +1) · 2= 2<br />
√<br />
−4x2 − 4x = 1<br />
√<br />
−x2 − x<br />
49. G(x) = √ 1 − x 2 arccos x ⇒ G 0 (x) = √ 1 − x 2 ·<br />
51. h(t) =cot −1 (t) +cot −1 (1/t) ⇒<br />
h 0 (t) =− 1<br />
1+t − 1<br />
2 1+(1/t) · d 1<br />
2 dt t = − 1<br />
1+t − t2<br />
2 t 2 +1 ·<br />
−1<br />
√ +arccosx · 1<br />
1 − x<br />
2 2 (1 − x2 ) −1/2 (−2x) =−1 − x √ arccos x<br />
1 − x<br />
2<br />
− 1 t 2 <br />
= − 1<br />
1+t 2 + 1<br />
t 2 +1 =0.<br />
Note that this makes sense because h(t) = π 2 for t>0 and h(t) =3π 2<br />
for t
F.<br />
TX.10<br />
SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 109<br />
57. Let y =cos −1 x.Thencos y = x and 0 ≤ y ≤ π ⇒ −sin y dy<br />
dx =1<br />
dy<br />
dx = − 1<br />
sin y = − 1<br />
<br />
1 − cos2 y = − 1<br />
√<br />
1 − x<br />
2 .<br />
⇒<br />
[Notethatsin y ≥ 0 for 0 ≤ y ≤ π.]<br />
59. x 2 + y 2 = r 2 is a circle with center O and ax + by =0is a line through O [assume a<br />
and b are not both zero]. x 2 + y 2 = r 2 ⇒ 2x +2yy 0 =0 ⇒ y 0 = −x/y,sothe<br />
slope of the tangent line at P 0 (x 0,y 0) is −x 0/y 0. The slope of the line OP 0 is y 0/x 0,<br />
which is the negative reciprocal of −x 0/y 0. Hence, the curves are orthogonal, and the<br />
families of curves are orthogonal trajectories of each other.<br />
61. y = cx 2 ⇒ y 0 =2cx and x 2 +2y 2 = k [assume k>0] ⇒ 2x +4yy 0 =0 ⇒<br />
2yy 0 = −x ⇒ y 0 = − x<br />
2(y) = − x<br />
2(cx 2 ) = − 1 , so the curves are orthogonal if<br />
2cx<br />
c 6= 0.Ifc =0, then the horizontal line y = cx 2 =0intersects x 2 +2y 2 = k orthogonally<br />
<br />
at ± √ <br />
k, 0 , since the ellipse x 2 +2y 2 = k has vertical tangents at those two points.<br />
63. To find the points at which the ellipse x 2 − xy + y 2 =3crosses the x-axis, let y =0and solve for x.<br />
y =0 ⇒ x 2 − x(0) + 0 2 =3 ⇔ x = ± √ 3. So the graph of the ellipse crosses the x-axis at the points ± √ 3, 0 .<br />
Using implicit differentiation to find y 0 ,weget2x − xy 0 − y +2yy 0 =0 ⇒ y 0 (2y − x) =y − 2x ⇔ y 0 = y − 2x<br />
2y − x .<br />
So y 0 at √ 3, 0 is 0 − 2 √ 3<br />
2(0) − √ 3 =2and y0 at − √ 3, 0 is 0+2√ 3<br />
2(0) + √ =2. Thus, the tangent lines at these points are parallel.<br />
3<br />
65. x 2 y 2 + xy =2 ⇒ x 2 · 2yy 0 + y 2 · 2x + x · y 0 + y · 1=0 ⇔ y 0 (2x 2 y + x) =−2xy 2 − y ⇔<br />
y 0 = − 2xy2 + y<br />
2x 2 y + x .So− 2xy2 + y<br />
2x 2 y + x = −1 ⇔ 2xy2 + y =2x 2 y + x ⇔ y(2xy +1)=x(2xy +1) ⇔<br />
y(2xy +1)− x(2xy +1)=0 ⇔ (2xy +1)(y − x) =0 ⇔ xy = − 1 2 or y = x.Butxy = − 1 2<br />
⇒<br />
x 2 y 2 + xy = 1 4 − 1 2 6=2,sowemusthavex = y. Then x2 y 2 + xy =2 ⇒ x 4 + x 2 =2 ⇔ x 4 + x 2 − 2=0 ⇔<br />
(x 2 +2)(x 2 − 1) = 0. Sox 2 = −2, which is impossible, or x 2 =1 ⇔ x = ±1. Sincex = y, the points on the curve<br />
where the tangent line has a slope of −1 are (−1, −1) and (1, 1).<br />
67. (a) If y = f −1 (x),thenf(y) =x. Differentiating implicitly with respect to x and remembering that y is a function of x,<br />
we get f 0 (y) dy dy<br />
=1,so<br />
dx dx = 1<br />
f 0 (y)<br />
⇒ f −10 (x) =<br />
1<br />
f 0 (f −1 (x)) .<br />
(b) f(4) = 5 ⇒ f −1 (5) = 4. Bypart(a), f −10 (5) =<br />
1<br />
f 0 (f −1 (5)) = 1<br />
f 0 (4) =1 <br />
2<br />
3 =<br />
3<br />
. 2
F.<br />
110 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
69. x 2 +4y 2 =5 ⇒ 2x +4(2yy 0 )=0 ⇒ y 0 = − x .Nowleth be the height of the lamp, and let (a, b) be the point of<br />
4y<br />
tangency of the line passing through the points (3,h) and (−5, 0). Thislinehasslope(h − 0)/[3 − (−5)] = 1 8 h.Butthe<br />
slope of the tangent line through the point (a, b) can be expressed as y 0 = − a 4b ,oras b − 0<br />
a − (−5) = b [since the line<br />
a +5<br />
passes through (−5, 0) and (a, b)], so − a 4b =<br />
b<br />
a +5<br />
⇔ 4b 2 = −a 2 − 5a ⇔ a 2 +4b 2 = −5a. Buta 2 +4b 2 =5<br />
[since (a, b) is on the ellipse], so 5=−5a ⇔ a = −1. Then4b 2 = −a 2 − 5a = −1 − 5(−1) = 4 ⇒ b =1, since the<br />
point is on the top half of the ellipse. So h 8 =<br />
x-axis.<br />
b<br />
a +5 = 1<br />
−1+5 = 1 4<br />
⇒<br />
h =2. So the lamp is located 2 units above the<br />
3.6 Derivatives of Logarithmic Functions<br />
1. The differentiation formula for logarithmic functions,<br />
3. f(x) =sin(lnx) ⇒ f 0 (x) =cos(lnx) ·<br />
5. f(x) =log 2 (1 − 3x) ⇒ f 0 (x) =<br />
d<br />
dx (log a x) = 1 ,issimplestwhena = e because ln e =1.<br />
x ln a<br />
d<br />
1 cos(ln x)<br />
ln x =cos(lnx) · =<br />
dx x x<br />
1<br />
(1 − 3x)ln2<br />
d<br />
−3<br />
(1 − 3x) =<br />
dx<br />
(1 − 3x) ln2 or 3<br />
(3x − 1) ln 2<br />
7. f(x) = 5√ ln x =(lnx) 1/5 ⇒ f 0 (x) = 1 5 (ln x)−4/5 d<br />
dx (ln x) = 1<br />
5(ln x) 4/5 · 1<br />
x = 1<br />
5x 5 (ln x) 4<br />
9. f(x) =sinx ln(5x) ⇒ f 0 (x) =sinx ·<br />
11. F (t) =ln<br />
F 0 (t) =3·<br />
1<br />
5x · d<br />
sin x · 5<br />
(5x)+ln(5x) · cos x = +cosx ln(5x) = sin x +cosx ln(5x)<br />
dx 5x<br />
x<br />
(2t +1)3<br />
(3t − 1) 4 =ln(2t +1)3 − ln(3t − 1) 4 =3ln(2t +1)− 4ln(3t − 1) ⇒<br />
1<br />
2t +1 · 2 − 4 · 1<br />
3t − 1 · 3= 6<br />
2t +1 − 12<br />
3t − 1 , or combined, −6(t +3)<br />
(2t +1)(3t − 1) .<br />
13. g(x) =ln x √ x 2 − 1 =lnx +ln(x 2 − 1) 1/2 =lnx + 1 2 ln(x2 − 1) ⇒<br />
g 0 (x) = 1 x + 1 2 ·<br />
1<br />
x 2 − 1 · 2x = 1 x + x<br />
x 2 − 1 = x2 − 1+x · x<br />
= 2x2 − 1<br />
x(x 2 − 1) x(x 2 − 1)<br />
15. f(u) =<br />
ln u<br />
1+ln(2u)<br />
⇒<br />
f 0 (u) =<br />
[1 + ln(2u)] · 1<br />
u − ln u · 1<br />
2u · 2<br />
[1 + ln(2u)] 2 =<br />
17. y =ln 2 − x − 5x 2 ⇒ y 0 =<br />
1<br />
u<br />
[1 + ln(2u) − ln u] 1+(ln2+lnu) − ln u<br />
= =<br />
[1 + ln(2u)] 2 u[1 + ln(2u)] 2<br />
1<br />
−10x − 1<br />
· (−1 − 10x) =<br />
2 − x − 5x2 2 − x − 5x or 10x +1<br />
2 5x 2 + x − 2<br />
19. y =ln(e −x + xe −x )=ln(e −x (1 + x)) = ln(e −x )+ln(1+x) =−x +ln(1+x) ⇒<br />
y 0 = −1+ 1 −1 − x +1<br />
= = − x<br />
1+x 1+x 1+x<br />
1+ln2<br />
u[1 + ln(2u)] 2
F.<br />
TX.10<br />
SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS ¤ 111<br />
√<br />
21. y =2x log 10 x =2x log10 x 1/2 =2x · 1 log 2 10 x = x log 10 x ⇒ y 0 1<br />
= x ·<br />
x ln 10 +log 10 x · 1= 1<br />
ln 10 +log 10 x<br />
Note:<br />
1<br />
ln 10 = ln e<br />
ln 10 =log 10 e, so the answer could be written as 1<br />
ln 10 +log 10 x =log 10 e +log 10 x =log 10 ex.<br />
23. y = x 2 ln(2x) ⇒ y 0 = x 2 ·<br />
y 00 =1+2x ·<br />
1 · 2+ln(2x) · (2x) =x +2x ln(2x)<br />
2x ⇒<br />
1 · 2+ln(2x) · 2=1+2+2ln(2x) =3+2ln(2x)<br />
2x<br />
25. y =ln x + √ 1+x 2 ⇒<br />
y 0 1<br />
=<br />
x + √ 1+x 2<br />
=<br />
<br />
1<br />
x + √ 1+<br />
1+x 2<br />
d √ <br />
x + 1+x<br />
2<br />
=<br />
dx<br />
<br />
x<br />
√ = 1+x<br />
2<br />
1<br />
<br />
<br />
x + √ 1+ 1<br />
1+x (1 + 2 2 x2 ) −1/2 (2x)<br />
1<br />
x + √ 1+x 2 ·<br />
√<br />
1+x2 + x<br />
√<br />
1+x<br />
2<br />
=<br />
1<br />
√<br />
1+x<br />
2<br />
⇒<br />
y 00 = − 1 2 (1 + x2 ) −3/2 (2x) =<br />
−x<br />
(1 + x 2 ) 3/2<br />
27. f(x) =<br />
x<br />
1 − ln(x − 1)<br />
⇒<br />
−1<br />
[1 − ln(x − 1)] · 1 − x ·<br />
f 0 (x) =<br />
x − 1<br />
[1 − ln(x − 1)] 2 =<br />
=<br />
2x − 1 − (x − 1) ln(x − 1)<br />
(x − 1)[1 − ln(x − 1)] 2<br />
(x − 1)[1 − ln(x − 1)] + x<br />
x − 1<br />
x − 1 − (x − 1) ln(x − 1) + x<br />
=<br />
[1 − ln(x − 1)] 2 (x − 1)[1 − ln(x − 1)] 2<br />
Dom(f) ={x | x − 1 > 0 and 1 − ln(x − 1) 6= 0} = {x | x>1 and ln(x − 1) 6= 1}<br />
= x | x>1 and x − 1 6=e 1 = {x | x>1 and x 6= 1+e} =(1, 1+e) ∪ (1 + e, ∞)<br />
29. f(x) =ln(x 2 − 2x) ⇒ f 0 (x) =<br />
1<br />
2(x − 1)<br />
(2x − 2) =<br />
x 2 − 2x x(x − 2) .<br />
Dom(f) ={x | x(x − 2) > 0} =(−∞, 0) ∪ (2, ∞).<br />
31. f(x) = ln x ⇒ f 0 (x) = x2 (1/x) − (ln x)(2x) x − 2x ln x x(1 − 2lnx)<br />
= = = 1 − 2lnx ,<br />
x 2 (x 2 ) 2 x 4<br />
x 4<br />
x 3<br />
so f 0 (1) = 1 − 2ln1 = 1 − 2 · 0 =1.<br />
1 3 1<br />
<br />
33. y =ln xe x2 =lnx +lne x2 =lnx + x 2 ⇒ y 0 = 1 +2x. At(1, 1), the slope of the tangent line is<br />
x<br />
y 0 (1) = 1 + 2 = 3, and an equation of the tangent line is y − 1=3(x − 1),ory =3x − 2.<br />
35. f(x) =sinx +lnx ⇒ f 0 (x) =cosx +1/x.<br />
This is reasonable, because the graph shows that f increases when f 0 is<br />
positive, and f 0 (x) =0when f has a horizontal tangent.
F.<br />
112 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
37. y =(2x +1) 5 (x 4 − 3) 6 ⇒ ln y =ln (2x +1) 5 (x 4 − 3) 6 ⇒ ln y =5ln(2x +1)+6ln(x 4 − 3) ⇒<br />
1<br />
1<br />
y y0 =5·<br />
2x +1 · 2+6· 1<br />
x 4 − 3 · 4x3<br />
10<br />
y 0 = y<br />
2x +1 + 24x3<br />
x 4 − 3<br />
<br />
⇒<br />
<br />
10<br />
=(2x +1) 5 (x 4 − 3) 6 2x +1 + 24x3 .<br />
x 4 − 3<br />
[The answer could be simplified to y 0 =2(2x +1) 4 (x 4 − 3) 5 (29x 4 +12x 3 − 15), but this is unnecessary.]<br />
39. y = sin2 x tan 4 x<br />
(x 2 +1) 2 ⇒ ln y =ln(sin 2 x tan 4 x) − ln(x 2 +1) 2 ⇒<br />
ln y =ln(sinx) 2 +ln(tanx) 4 − ln(x 2 +1) 2 ⇒ ln y =2ln|sin x| +4ln|tan x| − 2ln(x 2 +1) ⇒<br />
<br />
1 1<br />
y y0 =2·<br />
sin x · cos x +4· 1<br />
tan x · 1<br />
sec2 x − 2 ·<br />
x 2 +1 · 2x ⇒ y0 = sin2 x tan 4 x<br />
2cotx + 4sec2 x<br />
(x 2 +1) 2 tan x − 4x <br />
x 2 +1<br />
41. y = x x ⇒ ln y =lnx x ⇒ ln y = x ln x ⇒ y 0 /y = x(1/x)+(lnx) · 1 ⇒ y 0 = y(1 + ln x) ⇒<br />
y 0 = x x (1 + ln x)<br />
43. y = x sin x ⇒ ln y =lnx sin x ⇒ ln y =sinx ln x ⇒ y0<br />
y =(sinx) · 1 +(lnx)(cos x)<br />
x ⇒<br />
<br />
<br />
sin x<br />
sin x<br />
y 0 = y<br />
x<br />
+lnx cos x ⇒ y 0 = x sin x x<br />
+lnx cos x<br />
45. y =(cosx) x ⇒ ln y =ln(cosx) x ⇒ ln y = x ln cos x ⇒ 1 y y0 = x ·<br />
<br />
y 0 = y ln cos x − x sin x <br />
cos x<br />
⇒ y 0 =(cosx) x (ln cos x − x tan x)<br />
1 · (− sin x)+lncosx · 1<br />
cos x ⇒<br />
47. y =(tanx) 1/x ⇒ ln y =ln(tanx) 1/x ⇒ ln y = 1 ln tan x ⇒<br />
x<br />
1<br />
y y0 = 1 x · 1<br />
tan x · sec2 x +lntanx ·<br />
− 1 <br />
<br />
sec<br />
⇒ y 0 2 x ln tan x<br />
= y −<br />
x 2 x tan x x 2<br />
y 0 =(tanx) 1/x sec 2 x<br />
x tan x<br />
49. y =ln(x 2 + y 2 ) ⇒ y 0 =<br />
<br />
ln tan x<br />
−<br />
x 2<br />
1<br />
x 2 + y 2<br />
or y 0 = (tan x) 1/x · 1<br />
x<br />
x 2 y 0 + y 2 y 0 − 2yy 0 =2x ⇒ (x 2 + y 2 − 2y)y 0 =2x ⇒ y 0 =<br />
51. f(x) =ln(x − 1) ⇒ f 0 (x) =<br />
<br />
csc x sec x −<br />
⇒<br />
<br />
ln tan x<br />
x<br />
d<br />
dx (x2 + y 2 ) ⇒ y 0 2x +2yy0<br />
= ⇒ x 2 y 0 + y 2 y 0 =2x +2yy 0 ⇒<br />
x 2 + y 2<br />
2x<br />
x 2 + y 2 − 2y<br />
1<br />
(x − 1) =(x − 1)−1 ⇒ f 00 (x) =−(x − 1) −2 ⇒ f 000 (x) =2(x − 1) −3 ⇒<br />
f (4) (x) =−2 · 3(x − 1) −4 ⇒ ··· ⇒ f (n) (x) =(−1) n−1 · 2 · 3 · 4 ·····(n − 1)(x − 1) −n =(−1) n−1 (n − 1)!<br />
(x − 1) n<br />
53. If f(x) =ln(1+x), thenf 0 (x) = 1<br />
1+x ,sof 0 (0) = 1.<br />
ln(1 + x) f(x)<br />
Thus, lim =lim<br />
x→0 x x→0 x<br />
= lim f(x) − f(0)<br />
= f 0 (0) = 1.<br />
x→0 x − 0
F.<br />
SECTION TX.10 3.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ¤ 113<br />
3.7 Rates of Change in the Natural and Social Sciences<br />
1. (a) s = f(t) =t 3 − 12t 2 +36t ⇒ v(t) =f 0 (t) =3t 2 − 24t +36<br />
(b) v(3) = 27 − 72 + 36 = −9 ft/s<br />
(c) The particle is at rest when v(t) =0. 3t 2 − 24t +36=0 ⇔ 3(t − 2)(t − 6) = 0 ⇔ t =2sor6 s.<br />
(d) The particle is moving in the positive direction when v(t) > 0. 3(t − 2)(t − 6) > 0 ⇔ 0 ≤ t6.<br />
(e) Since the particle is moving in the positive direction and in the (f )<br />
negative direction, we need to calculate the distance traveled in the<br />
intervals [0, 2], [2, 6],and[6, 8] separately.<br />
|f(2) − f(0)| = |32 − 0| =32.<br />
|f(6) − f(2)| = |0 − 32| =32.<br />
|f(8) − f(6)| = |32 − 0| =32.<br />
The total distance is 32 + 32 + 32 = 96 ft.<br />
(g) v(t) =3t 2 − 24t +36<br />
a(t) =v 0 (t) =6t − 24.<br />
⇒<br />
(h )<br />
a(3) = 6(3) − 24 = −6(ft/s)/s orft/s 2 .<br />
(i) The particle is speeding up when v and a havethesamesign.Thisoccurswhen2
F.<br />
114 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
(i) The particle is speeding up when v and a have the same sign. This occurs when 0
F.<br />
SECTION TX.10 3.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ¤ 115<br />
(c) The circumference is C(r) =2πr = A 0 (r). Thefigure suggests that if ∆r is small,<br />
then the change in the area of the circle (a ring around the outside) is approximately equal<br />
to its circumference times ∆r. Straightening out this ring gives us a shape that is approximately<br />
rectangular with length 2πr and width ∆r, so∆A ≈ 2πr(∆r). Algebraically,<br />
∆A = A(r + ∆r) − A(r) =π(r + ∆r) 2 − πr 2 =2πr(∆r)+π(∆r) 2 .<br />
So we see that if ∆r is small, then ∆A ≈ 2πr(∆r) and therefore, ∆A/∆r ≈ 2πr.<br />
15. S(r) =4πr 2 ⇒ S 0 (r) =8πr ⇒<br />
(a) S 0 (1) = 8π ft 2 /ft (b) S 0 (2) = 16π ft 2 /ft (c) S 0 (3) = 24π ft 2 /ft<br />
As the radius increases, the surface area grows at an increasing rate. In fact, the rate of change is linear with respect to the<br />
17. The mass is f(x) =3x 2 , so the linear density at x is ρ(x) =f 0 (x) =6x.<br />
(a) ρ(1) = 6 kg/m (b) ρ(2) = 12kg/m (c) ρ(3) = 18 kg/m<br />
Since ρ is an increasing function, the density will be the highest at the right end of the rod and lowest at the left end.<br />
19. The quantity of charge is Q(t) =t 3 − 2t 2 +6t +2,sothecurrentisQ 0 (t) =3t 2 − 4t +6.<br />
(a) Q 0 (0.5) = 3(0.5) 2 − 4(0.5) + 6 = 4.75 A<br />
(b) Q 0 (1) = 3(1) 2 − 4(1) + 6 = 5 A<br />
The current is lowest when Q 0 has a minimum. Q 00 (t) =6t − 4 < 0 when t< 2 3 . So the current decreases when t< 2 3 and<br />
increases when t> 2 3 . Thus, the current is lowest at t = 2 3 s.<br />
21. (a) To find the rate of change of volume with respect to pressure, we first solve for V in terms of P .<br />
PV = C ⇒ V = C P ⇒ dV<br />
dP = − C P 2 .<br />
(b) From the formula for dV/dP in part (a), we see that as P increases, the absolute value of dV/dP decreases.<br />
Thus, the volume is decreasing more rapidly at the beginning.<br />
(c) β = − 1 V<br />
dV<br />
dP = − 1 − C <br />
=<br />
V P 2<br />
C<br />
(PV)P =<br />
C<br />
CP = 1 P<br />
23. In Example 6, the population function was n =2 t n 0. Since we are tripling instead of doubling and the initial population is<br />
400, the population function is n(t) =400· 3 t . The rate of growth is n 0 (t) =400· 3 t · ln 3, so the rate of growth after<br />
2.5 hours is n 0 (2.5) = 400 · 3 2.5 · ln 3 ≈ 6850 bacteria/hour.<br />
1860 − 1750<br />
25. (a) 1920: m 1 =<br />
1920 − 1910 = 110<br />
2070 − 1860<br />
=11, m2 =<br />
10 1930 − 1920 = 210<br />
10 =21,<br />
(m 1 + m 2 )/ 2 = (11 + 21)/2 =16million/year<br />
4450 − 3710<br />
1980: m 1 =<br />
1980 − 1970 = 740<br />
10 =74, m 5280 − 4450<br />
2 =<br />
1990 − 1980 = 830<br />
10 =83,<br />
(m 1 + m 2)/ 2 = (74 + 83)/2 =78.5 million/year<br />
(b) P (t) =at 3 + bt 2 + ct + d (in millions of people), where a ≈ 0.0012937063, b ≈−7.061421911, c ≈ 12,822.97902,<br />
and d ≈−7,743,770.396.<br />
(c) P (t) =at 3 + bt 2 + ct + d ⇒ P 0 (t) =3at 2 +2bt + c (in millions of people per year)
F.<br />
116 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
(d) P 0 (1920) = 3(0.0012937063)(1920) 2 +2(−7.061421911)(1920) + 12,822.97902<br />
≈ 14.48 million/year [smaller than the answer in part (a), but close to it]<br />
P 0 (1980) ≈ 75.29 million/year (smaller, but close)<br />
(e) P 0 (1985) ≈ 81.62 million/year, so the rate of growth in 1985 was about 81.62 million/year.<br />
27. (a) Using v = P<br />
4ηl (R2 − r 2 ) with R =0.01, l =3, P =3000,andη =0.027,wehavev as a function of r:<br />
v(r) = 3000<br />
4(0.027)3 (0.012 − r 2 ). v(0) = 0.925 cm/s, v(0.005) = 0.694 cm/s, v(0.01) = 0.<br />
(b) v(r) = P<br />
4ηl (R2 − r 2 ) ⇒ v 0 (r) = P<br />
Pr<br />
(−2r) =−<br />
4ηl 2ηl .<br />
Whenl =3, P =3000,andη =0.027,wehave<br />
v 0 (r) =−<br />
3000r<br />
2(0.027)3 . v0 (0) = 0, v 0 (0.005) = −92.592 (cm/s)/cm, and v 0 (0.01) = −185.185 (cm/s)/cm.<br />
(c) The velocity is greatest where r =0(at the center) and the velocity is changing most where r = R =0.01 cm<br />
(at the edge).<br />
29. (a) C(x) = 1200 + 12x − 0.1x 2 +0.0005x 3 ⇒ C 0 (x) =12− 0.2x +0.0015x 2 \$/yard, which is the marginal cost<br />
function.<br />
(b) C 0 (200) = 12 − 0.2(200) + 0.0015(200) 2 =\$32/yard, and this is the rate at which costs are increasing with respect to<br />
the production level when x =200.<br />
C 0 (200) predicts the cost of producing the 201st yard.<br />
(c) The cost of manufacturing the 201st yard of fabric is C(201) − C(200) = 3632.2005 − 3600 ≈ \$32.20,whichis<br />
approximately C 0 (200).<br />
31. (a) A(x) = p(x)<br />
x<br />
⇒ A 0 (x) = xp0 (x) − p(x) · 1<br />
x 2<br />
= xp0 (x) − p(x)<br />
x 2 .<br />
A 0 (x) > 0 ⇒ A(x) is increasing; that is, the average productivity increases as the size of the workforce increases.<br />
(b) p 0 (x) is greater than the average productivity ⇒ p 0 (x) >A(x) ⇒ p 0 (x) > p(x)<br />
x<br />
xp 0 (x) − p(x) > 0 ⇒ xp0 (x) − p(x)<br />
x 2 > 0 ⇒ A 0 (x) > 0.<br />
⇒ xp 0 (x) >p(x) ⇒<br />
33. PV = nRT ⇒ T = PV<br />
nR = PV<br />
(10)(0.0821) = 1 (PV). Using the Product Rule, we have<br />
0.821<br />
dT<br />
dt = 1<br />
0.821 [P (t)V 0 (t)+V (t)P 0 (t)] = 1 [(8)(−0.15) + (10)(0.10)] ≈−0.2436 K/min.<br />
0.821<br />
35. (a) If the populations are stable, then the growth rates are neither positive nor negative; that is, dC<br />
dt =0and dW dt<br />
(b) “The caribou go extinct” means that the population is zero, or mathematically, C =0.<br />
=0.<br />
(c)Wehavetheequations dC<br />
dt = aC − bCW and dW dt<br />
= −cW + dCW .LetdC/dt = dW/dt =0, a =0.05, b =0.001,<br />
c =0.05,andd =0.0001 to obtain 0.05C − 0.001CW =0 (1) and −0.05W +0.0001CW =0 (2). Adding10 times<br />
(2) to (1) eliminates the CW-terms and gives us 0.05C − 0.5W =0 ⇒ C =10W . Substituting C =10W into (1)
F.<br />
TX.10<br />
SECTION 3.8 EXPONENTIAL GROWTH AND DECAY ¤ 117<br />
results in 0.05(10W ) − 0.001(10W )W =0 ⇔ 0.5W − 0.01W 2 =0 ⇔ 50W − W 2 =0 ⇔<br />
W (50 − W )=0 ⇔ W =0or 50. SinceC =10W , C =0or 500. Thus, the population pairs (C, W ) that lead to<br />
stable populations are (0, 0) and (500, 50).Soitispossibleforthetwospeciestoliveinharmony.<br />
3.8 Exponential Growth and Decay<br />
1. The relative growth rate is 1 P<br />
dP<br />
dt<br />
Thus, P (6) = 2e 0.7944(6) ≈ 234.99 or about 235 members.<br />
=0.7944,so<br />
dP<br />
dt =0.7944P and, by Theorem 2, P (t) =P (0)e0.7944t =2e 0.7944t .<br />
3. (a) By Theorem 2, P (t) =P (0)e kt = 100e kt . NowP (1) = 100e k(1) = 420 ⇒ e k = 420<br />
100<br />
⇒ k =ln4.2.<br />
So P (t) =100e (ln 4.2)t =100(4.2) t .<br />
(b) P (3) = 100(4.2) 3 =7408.8 ≈ 7409 bacteria<br />
(c) dP/dt = kP ⇒ P 0 (3) = k · P (3) = (ln 4.2) 100(4.2) 3 [from part (a)] ≈ 10,632 bacteria/hour<br />
(d) P (t) = 100(4.2) t =10,000 ⇒ (4.2) t =100 ⇒ t =(ln100)/(ln 4.2) ≈ 3.2 hours<br />
5. (a) Let the population (in millions) in the year t be P (t). Since the initial time is the year 1750, we substitute t − 1750 for t in<br />
Theorem 2, so the exponential model gives P (t) =P (1750)e k(t−1750) .ThenP (1800) = 980 = 790e k(1800−1750)<br />
⇒<br />
980<br />
= 790 ek(50) ⇒ ln 980 =50k ⇒ k = 1 980<br />
ln ≈ 0.0043104. Sowiththismodel,wehave<br />
790 50 790<br />
P (1900) = 790e k(1900−1750) ≈ 1508 million, and P (1950) = 790e k(1950−1750) ≈ 1871 million. Both of these<br />
estimates are much too low.<br />
(b) In this case, the exponential model gives P (t) =P (1850)e k(t−1850) ⇒ P (1900) = 1650 = 1260e k(1900−1850) ⇒<br />
ln 1650 = k(50) ⇒ k = 1 1650<br />
ln ≈ 0.005393. Sowiththismodel,weestimate<br />
1260 50 1260<br />
P (1950) = 1260e k(1950−1850) ≈ 2161 million. This is still too low, but closer than the estimate of P (1950) in part (a).<br />
(c) The exponential model gives P (t) =P (1900)e k(t−1900) ⇒ P (1950) = 2560 = 1650e k(1950−1900) ⇒<br />
ln 2560 = k(50) ⇒ k = 1 2560<br />
1650 50<br />
ln<br />
1650<br />
≈ 0.008785. With this model, we estimate<br />
P (2000) = 1650e k(2000−1900) ≈ 3972 million. This is much too low. The discrepancy is explained by the fact that the<br />
world birth rate (average yearly number of births per person) is about the same as always, whereas the mortality rate<br />
(especially the infant mortality rate) is much lower, owing mostly to advances in medical science and to the wars in the first<br />
part of the twentieth century. The exponential model assumes, among other things, that the birth and mortality rates will<br />
remain constant.<br />
7. (a) If y =[N 2O 5] then by Theorem 2, dy<br />
dt = −0.0005y ⇒ y(t) =y(0)e−0.0005t = Ce −0.0005t .<br />
(b) y(t) =Ce −0.0005t =0.9C ⇒ e −0.0005t =0.9 ⇒ −0.0005t =ln0.9 ⇒ t = −2000 ln 0.9 ≈ 211 s
F.<br />
118 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
9. (a) If y(t) is the mass (in mg) remaining after t years, then y(t) =y(0)e kt =100e kt .<br />
y(30) = 100e 30k = 1 (100) ⇒ 2 e30k = 1 2<br />
⇒ k = −(ln 2)/30 ⇒ y(t) = 100e −(ln 2)t/30 = 100 · 2 −t/30<br />
(b) y(100) = 100 · 2 −100/30 ≈ 9.92 mg<br />
(c) 100e −(ln 2)t/30 =1 ⇒ −(ln 2)t/30 = ln 1<br />
ln 0.01<br />
100<br />
⇒ t = −30<br />
ln 2<br />
≈ 199.3 years<br />
11. Let y(t) be the level of radioactivity. Thus, y(t) =y(0)e −kt and k is determined by using the half-life:<br />
y(5730) = 1 y(0) ⇒ 2 y(0)e−k(5730) = 1 y(0) ⇒ 2 e−5730k = 1 ⇒ −5730k =ln 1 ⇒ k = − ln 1 2<br />
2 2<br />
5730 = ln 2<br />
5730 .<br />
If 74% of the 14 C remains, then we know that y(t) =0.74y(0) ⇒ 0.74 = e −t(ln 2)/5730 ⇒ ln 0.74 = − t ln 2 ⇒<br />
5730<br />
5730(ln 0.74)<br />
t = − ≈ 2489 ≈ 2500 years.<br />
ln 2<br />
13. (a) Using Newton’s Law of Cooling, dT<br />
dt = k(T − T s), wehave dT = k(T − 75). Nowlety = T − 75, so<br />
dt<br />
y(0) = T (0) − 75 = 185 − 75 = 110,soy is a solution of the initial-value problem dy/dt = ky with y(0) = 110 and by<br />
Theorem 2 we have y(t) =y(0)e kt =110e kt .<br />
y(30) = 110e 30k =150− 75 ⇒ e 30k = 75 = 15<br />
110 22<br />
⇒ k = 1 15<br />
30<br />
ln ,soy(t) = 110e 30 1 t ln( 15<br />
22 )<br />
22 and<br />
y(45) = 110e 45<br />
30<br />
ln( 15<br />
22 ) ≈ 62 ◦ F.Thus,T (45) ≈ 62 + 75 = 137 ◦ F.<br />
(b) T (t) =100 ⇒ y(t) =25. y(t) = 110e 30 1 t ln( 15<br />
1<br />
22 ) =25 ⇒ e 30<br />
t ln( 15<br />
22 ) =<br />
25<br />
110<br />
⇒ 1 15 25<br />
30<br />
t ln<br />
22<br />
=ln110 ⇒<br />
25<br />
30 ln<br />
110<br />
t =<br />
ln 15 ≈ 116 min.<br />
22<br />
dT<br />
15. = k(T − 20). Lettingy = T − 20, wegetdy<br />
dt dt = ky,soy(t) =y(0)ekt . y(0) = T (0) − 20 = 5 − 20 = −15, so<br />
y(25) = y(0)e 25k = −15e 25k ,andy(25) = T (25) − 20 = 10 − 20 = −10,so−15e 25k = −10 ⇒ e 25k = 2 . Thus,<br />
3<br />
25k =ln <br />
2<br />
3 and k =<br />
1<br />
ln <br />
2<br />
25 3 ,soy(t) =y(0)e kt = −15e (1/25) ln(2/3)t .Moresimply,e 25k = 2 ⇒ e k = <br />
2 1/25<br />
⇒<br />
3 3<br />
e kt = <br />
2 t/25<br />
⇒ y(t) =−15 · <br />
2 t/25<br />
.<br />
3<br />
3<br />
(a) T (50) = 20 + y(50) = 20 − 15 · <br />
2 50/25<br />
=20− 15 · <br />
2 2<br />
=20− 20 =13.¯3 ◦ C<br />
3<br />
3<br />
3<br />
(b) 15 = T (t) =20+y(t) =20− 15 · <br />
2 t/25<br />
3<br />
⇒ 15 · <br />
2 t/25<br />
3<br />
=5 ⇒ <br />
2 t/25<br />
3<br />
= 1 3<br />
⇒<br />
(t/25) ln <br />
2<br />
3 =ln 1<br />
<br />
⇒ t =25ln <br />
1<br />
3<br />
3 ln 2<br />
<br />
3 ≈ 67.74 min.<br />
17. (a) Let P (h) be the pressure at altitude h. ThendP/dh = kP ⇒ P (h) =P (0)e kh =101.3e kh .<br />
P (1000) = 101.3e 1000k =87.14 ⇒ 1000k =ln <br />
87.14<br />
⇒ k = 1 ln <br />
87.14<br />
⇒<br />
101.3<br />
1000 101.3<br />
P (h) =101.3 e 1000 1 h ln( 87.14<br />
101.3) ,soP (3000) = 101.3e<br />
3ln( 87.14<br />
101.3 ) ≈ 64.5 kPa.<br />
(b) P (6187) = 101.3 e 6187 ln( 87.14<br />
1000 101.3 ) ≈ 39.9 kPa
F.<br />
TX.10<br />
19. (a) Using A = A 0<br />
<br />
1+ r n<br />
nt<br />
with A0 = 3000, r =0.05,andt =5,wehave:<br />
SECTION 3.9 RELATED RATES ¤ 119<br />
(i) Annually: n =1; A =3000 <br />
1+ 0.05 1·5<br />
1<br />
= \$3828.84<br />
(ii) Semiannually: n =2; A =3000 <br />
1+ 0.05 2·5<br />
2<br />
= \$3840.25<br />
(iii) Monthly: n =12; A =3000 1+ 0.05<br />
12<br />
(iv) Weekly: n =52; A =3000 1+ 0.05<br />
52<br />
(v) Daily: n =365; A =3000 1+ 0.05<br />
365<br />
12·5<br />
= \$3850.08<br />
52·5<br />
= \$3851.61<br />
365·5<br />
= \$3852.01<br />
(vi) Continuously: A =3000e (0.05)5 = \$3852.08<br />
(b) dA/dt =0.05A and A(0) = 3000.<br />
3.9 Related Rates<br />
1. V = x 3 ⇒ dV<br />
dt = dV dx dx<br />
dx dt =3x2 dt<br />
3. Let s denote the side of a square. The square’s area A is given by A = s 2 . Differentiating with respect to t gives us<br />
dA<br />
dt<br />
ds<br />
ds dA<br />
=2s .WhenA =16, s =4. Substitution 4 for s and 6 for gives us<br />
dt dt dt = 2(4)(6) = 48 cm2 /s.<br />
5. V = πr 2 h = π(5) 2 h =25πh ⇒ dV<br />
dt<br />
=25π<br />
dh<br />
dt<br />
⇒<br />
3=25π dh<br />
dt<br />
⇒<br />
dh<br />
dt = 3<br />
25π m/min.<br />
7. y = x 3 +2x ⇒ dy<br />
dt = dy dx<br />
dx dt =(3x2 + 2)(5) = 5(3x 2 +2).Whenx =2, dy<br />
dt =5(14)=70.<br />
9. z 2 = x 2 + y 2 ⇒ 2z dz dx dy<br />
=2x +2y<br />
dt dt dt<br />
z 2 =5 2 +12 2 ⇒ z 2 =169 ⇒ z = ±13. For dx<br />
dt<br />
⇒<br />
dz<br />
dt = 1 <br />
x dx<br />
z dt + y dy <br />
.Whenx =5and y =12,<br />
dt<br />
dy dz<br />
=2and =3,<br />
dt dt = 1 (5 · 2+12· 3) = ±46<br />
±13 13 .<br />
11. (a) Given: a plane flying horizontally at an altitude of 1 mi and a speed of 500 mi/h passes directly over a radar station.<br />
If we let t be time (in hours) and x be the horizontal distance traveled by the plane (in mi), then we are given<br />
that dx/dt = 500 mi/h.<br />
(b) Unknown: the rate at which the distance from the plane to the station is increasing<br />
(c)<br />
when it is 2 mi from the station. If we let y be the distance from the plane to the station,<br />
then we want to find dy/dt when y =2mi.<br />
(d) By the Pythagorean Theorem, y 2 = x 2 +1 ⇒ 2y (dy/dt) =2x (dx/dt).<br />
(e) dy<br />
dt = x dx<br />
y dt = x y (500). Sincey2 = x 2 +1,wheny =2, x = √ 3,so dy<br />
√<br />
3<br />
dt = 2 (500) = 250 √ 3 ≈ 433 mi/h.<br />
13. (a) Given: a man 6 ft tall walks away from a street light mounted on a 15-ft-tall pole at a rate of 5 ft/s. If we let t be time (in s)<br />
and x be the distance from the pole to the man (in ft), then we are given that dx/dt =5ft/s.
F.<br />
120 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
(b) Unknown: the rate at which the tip of his shadow is moving when he is 40 ft<br />
from the pole. If we let y be the distance from the man to the tip of his<br />
(c)<br />
shadow(inft),thenwewanttofind d (x + y) when x =40ft.<br />
dt<br />
(d) By similar triangles, 15<br />
6 = x + y<br />
y<br />
⇒ 15y =6x +6y ⇒ 9y =6x ⇒ y = 2 3 x.<br />
(e) The tip of the shadow moves at a rate of d dt (x + y) = d dt<br />
15. We are given that dx<br />
dt<br />
2z dz dx dy<br />
=2x +2y<br />
dt dt dt<br />
17. We are given that dx<br />
dt<br />
x + 2 <br />
3 x = 5 dx<br />
3 dt = 5 (5) = 25 3 3<br />
ft/s.<br />
=60mi/hand<br />
dy<br />
dt =25mi/h. z2 = x 2 + y 2<br />
⇒<br />
z dz<br />
dt = x dx<br />
dt + y dy<br />
dt<br />
⇒<br />
⇒<br />
dz<br />
dt = 1 <br />
x dx<br />
z dt + y dy <br />
.<br />
dt<br />
After 2 hours, x = 2 (60) = 120 and y =2(25)=50 ⇒ z = √ 120 2 +50 2 = 130,<br />
so dz<br />
dt = 1 <br />
x dx<br />
z dt + y dy <br />
120(60) + 50(25)<br />
= =65mi/h.<br />
dt<br />
130<br />
2z dz<br />
dx<br />
dt =2(x + y) dt + dy<br />
dt<br />
dy<br />
=4ft/sand<br />
dt =5ft/s. z2 =(x + y) 2 + 500 2 ⇒<br />
<br />
. 15 minutes after the woman starts, we have<br />
x =(4ft/s)(20 min)(60 s/min) = 4800 ft and y =5· 15 · 60 = 4500<br />
⇒<br />
z = (4800 + 4500) 2 +500 2 = √ 86,740,000,so<br />
dz<br />
dt = x + y dx<br />
z dt + dy <br />
4800 + 4500<br />
= √ (4 + 5) = √ 837 ≈ 8.99 ft/s.<br />
dt 86,740,000 8674<br />
19. A = 1 dh<br />
bh,whereb is the base and h is the altitude. We are given that<br />
2<br />
dt<br />
Product Rule, we have dA<br />
dt = 1 2<br />
b =20,so2= 1 <br />
20 · 1+10 db<br />
2<br />
dt<br />
<br />
b dh<br />
dt + h db<br />
dt<br />
<br />
⇒<br />
=1cm/min and<br />
dA<br />
dt =2cm2 /min. Using the<br />
<br />
.Whenh =10and A = 100,wehave100 = 1 b(10) 2 ⇒ 1 b =10 2 ⇒<br />
4=20+10 db<br />
dt<br />
⇒<br />
db<br />
dt = 4 − 20 = −1.6 cm/min.<br />
10<br />
21. We are given that dx<br />
dt<br />
2z dz<br />
dx<br />
dt =2(x + y) dt + dy<br />
dt<br />
dy<br />
=35km/hand<br />
dt =25km/h. z2 =(x + y) 2 + 100 2 ⇒<br />
<br />
.At4:00PM, x =4(35)=140and y = 4(25) = 100<br />
⇒<br />
z = (140 + 100) 2 +100 2 = √ 67,600 = 260,so<br />
dz<br />
dt = x + y dx<br />
z dt + dy <br />
140 + 100<br />
= (35 + 25) = 720 ≈ 55.4 km/h.<br />
dt 260<br />
13
F.<br />
TX.10<br />
SECTION 3.9 RELATED RATES ¤ 121<br />
23. If C = the rate at which water is pumped in, then dV<br />
dt = C − 10,000,where<br />
V = 1 3 πr2 h is the volume at time t. By similar triangles, r 2 = h 6<br />
⇒ r = 1 3 h ⇒<br />
V = 1 π 1<br />
3 3 h2 h = π 27 h3 ⇒ dV<br />
dt = π dh<br />
9 h2 .Whenh =200cm,<br />
dt<br />
dh<br />
dt =20cm/min,soC − 10,000 = π 9 (200)2 (20) ⇒ C =10,000 + 800,000 π ≈ 289,253 cm 3 /min.<br />
9<br />
25. The figure is labeled in meters. The area A of a trapezoid is<br />
1<br />
(base1 + base2)(height), and the volume V of the 10-meter-long trough is 10A.<br />
2<br />
Thus, the volume of the trapezoid with height h is V =(10) 1 2 [0.3+(0.3+2a)]h.<br />
By similar triangles, a h = 0.25<br />
0.5 = 1 2 ,so2a = h ⇒ V =5(0.6+h)h =3h +5h2 .<br />
Now dV<br />
dt = dV dh<br />
dh dt<br />
⇒<br />
0.2 =(3+10h) dh<br />
dt<br />
dh<br />
dt = 0.2<br />
3 + 10(0.3) = 0.2<br />
6 m/min = 1 10<br />
m/min or<br />
30 3 cm/min.<br />
⇒<br />
dh<br />
dt = 0.2 .Whenh =0.3,<br />
3+10h<br />
27. We are given that dV<br />
dt =30ft3 /min. V = 1 3 πr2 h = 1 2 h<br />
3 π h = πh3<br />
2 12<br />
⇒<br />
dV<br />
dt = dV dh<br />
dh dt<br />
⇒<br />
30 = πh2<br />
4<br />
dh<br />
dt<br />
⇒<br />
dh<br />
dt = 120<br />
πh . 2<br />
When h =10ft, dh<br />
dt = 120<br />
10 2 π = 6 ≈ 0.38 ft/min.<br />
5π<br />
29. A = 1 2 bh,butb =5mandsin θ = h 4<br />
⇒<br />
h =4sinθ,soA = 1 (5)(4 sin θ) =10sinθ.<br />
2<br />
We are given dθ<br />
dA<br />
dt<br />
When θ = π 3 , dA<br />
dt =0.6 cos π 3<br />
dθ<br />
=(10cosθ)(0.06) = 0.6cosθ.<br />
dt<br />
dt = dA<br />
dθ<br />
<br />
=(0.6) <br />
1<br />
2 =0.3 m 2 /s.<br />
31. Differentiating both sides of PV = C with respect to t and using the Product Rule gives us P dV<br />
dt + V dP dt =0<br />
dV<br />
dt = −V P<br />
dP<br />
dP<br />
.WhenV = 600, P = 150 and<br />
dt dt<br />
decreasing at a rate of 80 cm 3 /min.<br />
33. With R 1 =80and R 2 = 100,<br />
dV<br />
=20,sowehave<br />
dt = − 600 (20) = −80. Thus, the volume is<br />
150<br />
1<br />
R = 1 + 1 = 1 R 1 R 2 80 + 1<br />
100 = 180<br />
8000 = 9 400<br />
,soR =<br />
400<br />
with respect to t,wehave− 1 dR<br />
R 2 dt = − 1 dR 1<br />
− 1 dR 2<br />
R1<br />
2 dt R2<br />
2 dt<br />
R 2 = 100, dR <br />
dt = 4002 1<br />
9 2 80 (0.3) + 1 <br />
2 100 (0.2) = 107 ≈ 0.132 Ω/s.<br />
2 810<br />
⇒<br />
dR 1<br />
dt = dR 1<br />
R2 + 1 R1<br />
2 dt R2<br />
2<br />
dR 2<br />
dt<br />
⇒<br />
9 . Differentiating 1 R = 1 + 1 R 1 R 2<br />
<br />
.WhenR 1 =80and
F.<br />
122 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
35. We are given dθ/dt =2 ◦ /min = π 90<br />
rad/min. By the Law of Cosines,<br />
x 2 =12 2 +15 2 − 2(12)(15) cos θ =369− 360 cos θ<br />
2x dx<br />
dt<br />
=360sinθ<br />
dθ<br />
dt<br />
⇒<br />
dx<br />
dt<br />
=<br />
180 sin θ<br />
x<br />
x = √ 369 − 360 cos 60 ◦ = √ 189 = 3 √ 21,so dx<br />
dt<br />
⇒<br />
dθ<br />
dt .Whenθ =60◦ ,<br />
=<br />
180 sin 60◦<br />
3 √ 21<br />
37. (a) By the Pythagorean Theorem, 4000 2 + y 2 = 2 . Differentiating with respect to t,<br />
we obtain 2y dy<br />
dt<br />
d<br />
dy<br />
=2 . We know that =600ft/s, so when y =3000ft,<br />
dt dt<br />
= √ 4000 2 + 3000 2 = √ 25,000,000 = 5000 ft<br />
and d<br />
dt = y dy<br />
<br />
(b) Here tan θ =<br />
dt = 3000<br />
5000<br />
y<br />
4000<br />
⇒<br />
(600) =<br />
1800<br />
5<br />
=360ft/s.<br />
d<br />
dt (tan θ) = d y<br />
<br />
dt 4000<br />
⇒<br />
π<br />
90 = π √ √<br />
3 7 π<br />
3 √ 21 = ≈ 0.396 m/min.<br />
21<br />
sec 2 θ dθ<br />
dt = 1 dy<br />
4000 dt<br />
⇒<br />
dθ<br />
dt = cos2 θ dy<br />
4000 dt .When<br />
y =3000ft, dy<br />
4000<br />
=600ft/s, =5000and cos θ = = 4000<br />
dt 5000 = 4 dθ<br />
,so<br />
5 dt = (4/5)2 (600) = 0.096 rad/s.<br />
4000<br />
39. cot θ = x 5<br />
dx<br />
dt = 5π 6<br />
⇒<br />
−csc 2 θ dθ<br />
dt = 1 dx<br />
5 dt<br />
⇒<br />
2 2√3<br />
= 10 π km/min [≈ 130 mi/h]<br />
9<br />
<br />
− csc π 2 <br />
− π <br />
= 1 dx<br />
3 6 5 dt<br />
⇒<br />
41. We are given that dx<br />
dt<br />
=300km/h. By the Law of Cosines,<br />
y 2 = x 2 +1 2 − 2(1)(x) cos 120 ◦ = x 2 +1− 2x − 1 2<br />
= x 2 + x +1,so<br />
2y dy dx<br />
=2x<br />
dt dt + dx<br />
dt<br />
⇒<br />
dy<br />
dt<br />
=<br />
2x +1<br />
2y<br />
dx<br />
300<br />
.After1 minute, x =<br />
60<br />
dt =5km ⇒<br />
y = √ 5 2 +5+1= √ 31 km ⇒ dy<br />
dt = 2(5) + 1<br />
2 √ 1650<br />
(300) = √ ≈ 296 km/h.<br />
31 31<br />
43. Let the distance between the runner and the friend be .ThenbytheLawofCosines,<br />
2 =200 2 +100 2 − 2 · 200 · 100 · cos θ =50,000 − 40,000 cos θ (). Differentiating<br />
implicitly with respect to t,weobtain2 d<br />
dt<br />
dθ<br />
= −40,000(− sin θ) .NowifD is the<br />
dt<br />
distance run when the angle is θ radians, then by the formula for the length of an arc<br />
on a circle, s = rθ,wehaveD = 100θ,soθ = 1<br />
100 D ⇒ dθ<br />
dt = 1 dD<br />
100 dt = 7 . To substitute into the expression for<br />
100<br />
d<br />
dt , we must know sin θ atthetimewhen =200,whichwefind from (): 2002 =50,000 − 40,000 cos θ<br />
<br />
cos θ = 1 ⇒ sin θ = 1 − <br />
1 2<br />
= √ 15<br />
d<br />
. Substituting, we get 2(200)<br />
4 4 4<br />
dt =40,000 √ <br />
15 7<br />
<br />
⇒<br />
4 100<br />
d/dt = 7 √ 15<br />
4<br />
≈ 6.78 m/s. Whether the distance between them is increasing or decreasing depends on the direction in which<br />
the runner is running.<br />
F.<br />
TX.10SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS ¤ 123<br />
3.10 Linear Approximations and Differentials<br />
1. f(x) =x 4 +3x 2 ⇒ f 0 (x) =4x 3 +6x,sof(−1) = 4 and f 0 (−1) = −10.<br />
Thus, L(x) =f(−1) + f 0 (−1)(x − (−1)) = 4 + (−10)(x +1)=−10x − 6.<br />
3. f(x) =cosx ⇒ f 0 (x) =− sin x, sof π<br />
2 =0and f<br />
0 π<br />
Thus, L(x) =f π<br />
2<br />
+ f<br />
0 π<br />
2<br />
5. f(x) = √ 1 − x ⇒ f 0 (x) =<br />
x −<br />
π<br />
2<br />
=0− 1<br />
x −<br />
π<br />
2<br />
2<br />
= −1.<br />
= −x +<br />
π<br />
2 .<br />
−1<br />
2 √ 1 − x ,sof(0) = 1 and f 0 (0) = − 1 2 .<br />
Therefore,<br />
√ 1 − x = f(x) ≈ f(0) + f 0 (0)(x − 0) = 1 + − 1 2<br />
<br />
(x − 0) = 1 −<br />
1<br />
2 x.<br />
So √ 0.9 = √ 1 − 0.1 ≈ 1 − 1 (0.1) = 0.95<br />
2<br />
and √ 0.99 = √ 1 − 0.01 ≈ 1 − 1 (0.01) = 0.995.<br />
2<br />
7. f(x) = 3√ 1 − x =(1− x) 1/3 ⇒ f 0 (x) =− 1 3 (1 − x)−2/3 ,sof(0) = 1<br />
and f 0 (0) = − 1 .Thus,f(x) ≈ f(0) + f 0 3<br />
(0)(x − 0) = 1 − 1 x.Weneed<br />
3<br />
3√ 1 − x − 0.1 < 1 −<br />
1<br />
x< 3√ 3<br />
1 − x +0.1, which is true when<br />
−1.204
F.<br />
124 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
17. (a) y =tanx ⇒ dy =sec 2 xdx<br />
(b) When x = π/4 and dx = −0.1, dy =[sec(π/4)] 2 (−0.1) = √ 2 2<br />
(−0.1) = −0.2.<br />
19. y = f(x) =2x − x 2 , x =2, ∆x = −0.4 ⇒<br />
∆y = f(1.6) − f(2) = 0.64 − 0=0.64<br />
dy =(2− 2x) dx =(2− 4)(−0.4) = 0.8<br />
21. y = f(x) =2/x, x =4, ∆x =1 ⇒<br />
∆y = f(5) − f(4) = 2 5 − 2 4 = −0.1<br />
dy = − 2 x dx = − 2 (1) = −0.125<br />
2 42 23. To estimate (2.001) 5 ,we’llfind the linearization of f(x) =x 5 at a =2.Sincef 0 (x) =5x 4 , f(2) = 32,andf 0 (2) = 80,<br />
we have L(x) = 32 + 80(x − 2) = 80x − 128. Thus, x 5 ≈ 80x − 128 when x is near 2 ,so<br />
(2.001) 5 ≈ 80(2.001) − 128 = 160.08 − 128 = 32.08.<br />
<br />
25. To estimate (8.06) 2/3 ,we’llfind the linearization of f(x) =x 2/3 at a =8.Sincef 0 (x) = 2 3 x−1/3 =2/ 3 3√ <br />
x ,<br />
f(8) = 4, andf 0 (8) = 1 ,wehaveL(x) 3 =4+1 (x − 8) = 1 x + 4 . Thus, 3 3 3 x2/3 ≈ 1 x + 4 when x is near 8, so<br />
3 3<br />
(8.06) 2/3 ≈ 1 3 (8.06) + 4 3 = 12.06<br />
3<br />
=4.02.<br />
27. y = f(x) =tanx ⇒ dy =sec 2 xdx.Whenx =45 ◦ and dx = −1 ◦ ,<br />
dy =sec 2 45 ◦ (−π/180) = √ 2 2<br />
(−π/180) = −π/90,sotan 44 ◦ = f(44 ◦ ) ≈ f(45 ◦ )+dy =1− π/90 ≈ 0.965.<br />
29. y = f(x) =secx ⇒ f 0 (x) =secx tan x, sof(0) = 1 and f 0 (0) = 1 · 0=0. The linear approximation of f at 0 is<br />
f(0) + f 0 (0)(x − 0) = 1 + 0(x) =1.Since0.08 is close to 0, approximating sec 0.08 with 1 is reasonable.<br />
31. y = f(x) =lnx ⇒ f 0 (x) =1/x, sof(1) = 0 and f 0 (1) = 1. The linear approximation of f at 1 is<br />
f(1) + f 0 (1)(x − 1) = 0 + 1(x − 1) = x − 1. Nowf(1.05) = ln 1.05 ≈ 1.05 − 1=0.05, so the approximation<br />
is reasonable.<br />
33. (a) If x is the edge length, then V = x 3 ⇒ dV =3x 2 dx. Whenx =30and dx =0.1, dV =3(30) 2 (0.1) = 270,sothe<br />
maximum possible error in computing the volume of the cube is about 270 cm 3 . The relative error is calculated by dividing<br />
the change in V , ∆V ,byV . We approximate ∆V with dV .<br />
Relative error = ∆V<br />
V ≈ dV V = 3x2 dx<br />
=3 dx 0.1<br />
x 3 x =3 =0.01.<br />
30<br />
Percentage error = relative error × 100% = 0.01 × 100% = 1%.
F.<br />
TX.10SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS ¤ 125<br />
(b) S =6x 2 ⇒ dS =12xdx.Whenx =30and dx =0.1, dS = 12(30)(0.1) = 36, so the maximum possible error in<br />
computing the surface area of the cube is about 36 cm 2 .<br />
Relative error = ∆S<br />
S<br />
≈ dS S = 12xdx =2 dx 0.1<br />
6x 2 x =2 30<br />
=0.006.<br />
Percentage error = relative error × 100% = 0.006 × 100% = 0.6%.<br />
35. (a)Forasphereofradiusr, the circumference is C =2πr and the surface area is S =4πr 2 ,so<br />
r = C 2 C<br />
2π ⇒ S =4π = C2<br />
⇒ dS = 2 2π π<br />
π CdC.WhenC =84and dC =0.5, dS = 2 84<br />
(84)(0.5) =<br />
π π ,<br />
so the maximum error is about 84<br />
π ≈ 27 cm2 . Relative error ≈ dS S = 84/π<br />
84 2 /π = 1<br />
84 ≈ 0.012<br />
(b) V = 4 3 πr3 = 4 3 π C<br />
2π<br />
3<br />
= C3<br />
6π 2 ⇒ dV = 1<br />
2π 2 C2 dC. WhenC =84and dC =0.5,<br />
dV = 1<br />
2π 2 (84)2 (0.5) = 1764<br />
1764<br />
, so the maximum error is about ≈ 179 cm 3 .<br />
π2 π 2<br />
The relative error is approximately dV V = 1764/π2<br />
(84) 3 /(6π 2 ) = 1 56 ≈ 0.018.<br />
37. (a) V = πr 2 h ⇒ ∆V ≈ dV =2πrh dr =2πrh ∆r<br />
(b) The error is<br />
∆V − dV =[π(r + ∆r) 2 h − πr 2 h] − 2πrh ∆r = πr 2 h +2πrh ∆r + π(∆r) 2 h − πr 2 h − 2πrh ∆r = π(∆r) 2 h.<br />
39. V = RI ⇒ I = V R ⇒ dI = − V ∆I<br />
dR. The relative error in calculating I is ≈ dI<br />
R2 I I = −(V/R2 ) dR<br />
= − dR V/R R .<br />
Hence, the relative error in calculating I is approximately the same (in magnitude) as the relative error in R.<br />
41. (a) dc = dc dx =0dx =0<br />
dx<br />
(c) d(u + v) = d<br />
du<br />
dx (u + v) dx = dx + dv<br />
dx<br />
(d) d(uv) = d <br />
dx (uv) dx = u dv<br />
dx + v du<br />
dx<br />
u<br />
<br />
(e) d = d u<br />
v du<br />
dx = dx − u dv<br />
dx dx =<br />
v dx v<br />
v 2<br />
(f ) d (x n )= d<br />
dx (xn ) dx = nx n−1 dx<br />
<br />
dx = du dv<br />
dx + dx = du + dv<br />
dx dx<br />
<br />
dx = u dv du<br />
dx + v dx = udv+ vdu<br />
dx dx<br />
v du dv<br />
dx − u<br />
dx dx dx vdu− udv<br />
=<br />
v 2<br />
v 2<br />
43. (a) The graph shows that f 0 (1) = 2,soL(x) =f(1) + f 0 (1)(x − 1) = 5 + 2(x − 1) = 2x +3.<br />
f(0.9) ≈ L(0.9) = 4.8 and f(1.1) ≈ L(1.1) = 5.2.<br />
d<br />
du<br />
(b) d(cu) = (cu) dx = c dx = cdu<br />
dx dx<br />
(b) From the graph, we see that f 0 (x) is positive and decreasing. This means that the slopes of the tangent lines are positive,<br />
but the tangents are becoming less steep. So the tangent lines lie above the curve. Thus, the estimates in part (a) are too<br />
large.
F.<br />
126 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
3.11 Hyperbolic Functions<br />
1. (a) sinh 0 = 1 2 (e0 − e 0 )=0 (b) cosh 0 = 1 2 (e0 + e 0 )= 1 2<br />
(1 + 1) = 1<br />
3. (a) sinh(ln 2) = eln 2 − e −ln 2<br />
2<br />
= eln 2 − (e ln 2 ) −1<br />
2<br />
= 2 − 2−1<br />
2<br />
= 2 − 1 2<br />
2<br />
= 3 4<br />
(b) sinh 2 = 1 2 (e2 − e −2 ) ≈ 3.62686<br />
5. (a) sech 0 = 1<br />
cosh 0 = 1 1 =1 (b) cosh−1 1=0because cosh 0 = 1.<br />
7. sinh(−x) = 1 2 [e−x − e −(−x) ]= 1 2 (e−x − e x )=− 1 2 (e−x − e x )=− sinh x<br />
9. cosh x +sinhx = 1 2 (ex + e −x )+ 1 2 (ex − e −x )= 1 2 (2ex )=e x<br />
11. sinh x cosh y +coshx sinh y = 1<br />
2 (ex − e −x ) 1<br />
2 (ey + e −y ) + 1<br />
2 (ex + e −x ) 1<br />
2 (ey − e −y ) <br />
= 1 4 [(ex+y + e x−y − e −x+y − e −x−y )+(e x+y − e x−y + e −x+y − e −x−y )]<br />
= 1 4 (2ex+y − 2e −x−y )= 1 2 [ex+y − e −(x+y) ]=sinh(x + y)<br />
13. Divide both sides of the identity cosh 2 x − sinh 2 x =1by sinh 2 x:<br />
cosh 2 x<br />
sinh 2 x − sinh2 x<br />
sinh 2 x = 1<br />
sinh 2 x ⇔ coth2 x − 1=csch 2 x.<br />
15. Putting y = x in the result from Exercise 11, we have<br />
sinh 2x =sinh(x + x) =sinhx cosh x +coshx sinh x =2sinhx cosh x.<br />
17. tanh(ln x) =<br />
sinh(ln x)<br />
cosh(ln x) = (eln x − e − ln x )/2<br />
(e ln x + e − ln x )/2 = x − (eln x ) −1 x − x−1 x − 1/x<br />
= =<br />
x +(e ln x )<br />
−1<br />
x + x−1 x +1/x = (x2 − 1)/x<br />
(x 2 +1)/x = x2 − 1<br />
x 2 +1<br />
19. By Exercise 9, (cosh x +sinhx) n =(e x ) n = e nx =coshnx +sinhnx.<br />
21. sech x = 1<br />
cosh x ⇒ sech x = 1<br />
5/3 = 3 5 .<br />
cosh 2 x − sinh 2 x =1 ⇒ sinh 2 x =cosh 2 x − 1= <br />
5 2<br />
− 1= 16 ⇒ sinh x = 4 [because x>0].<br />
3<br />
9 3<br />
csch x = 1<br />
sinh x ⇒ csch x = 1<br />
4/3 = 3 4 .<br />
tanh x = sinh x<br />
cosh x<br />
⇒ tanh x =<br />
4/3<br />
5/3 = 4 5 .<br />
coth x = 1<br />
tanh x ⇒ coth x = 1<br />
4/5 = 5 4 .<br />
e x − e −x<br />
23. (a) lim tanh x = lim<br />
x→∞ x→∞ e x + e<br />
(b) lim tanh x =<br />
x→−∞<br />
lim<br />
x→−∞<br />
−x ·<br />
e−x<br />
= lim<br />
e−x e x − e −x ex<br />
·<br />
e x + e−x e = x<br />
x→∞<br />
lim<br />
x→−∞<br />
1 − e −2x<br />
1+e −2x = 1 − 0<br />
1+0 =1<br />
e 2x − 1<br />
e 2x +1 = 0 − 1<br />
0+1 = −1
F.<br />
TX.10<br />
SECTION 3.11 HYPERBOLIC FUNCTIONS ¤ 127<br />
e x − e −x<br />
(c) lim sinh x = lim = ∞<br />
x→∞ x→∞ 2<br />
(d) lim sinh x =<br />
x→−∞<br />
lim<br />
x→−∞<br />
e x − e −x<br />
2<br />
= −∞<br />
2<br />
(e) lim sech x = lim =0<br />
x→∞ x→∞ e x + e−x e x + e −x<br />
(f ) lim coth x = lim<br />
x→∞ x→∞ e x − e<br />
(g)<br />
(h)<br />
−x ·<br />
e−x<br />
= lim<br />
e−x x→∞<br />
1+e −2x 1+0<br />
= =1 [Or: Use part (a)]<br />
1 − e−2x 1 − 0<br />
cosh x<br />
lim coth x = lim = ∞,sincesinh x → 0 through positive values and cosh x → 1.<br />
x→0 + x→0 + sinh x<br />
cosh x<br />
lim coth x = lim = −∞,sincesinh x → 0 through negative values and cosh x → 1.<br />
x→0− x→0 − sinh x<br />
(i) lim csch x =<br />
x→−∞<br />
lim<br />
x→−∞<br />
2<br />
=0<br />
e x − e−x 25. Let y =sinh −1 x.Thensinh y = x and, by Example 1(a), cosh 2 y − sinh 2 y =1 ⇒ [with cosh y>0]<br />
cosh y = 1+sinh 2 y = √ 1+x 2 .SobyExercise9,e y =sinhy +coshy = x + √ 1+x 2 ⇒ y =ln x + √ 1+x 2 .<br />
27. (a) Let y =tanh −1 x.Thenx =tanhy = sinh y<br />
cosh y = (ey − e −y )/2<br />
(e y + e −y )/2 · ey<br />
e = e2y − 1<br />
y e 2y +1<br />
1+x = e 2y − xe 2y ⇒ 1+x = e 2y (1 − x) ⇒ e 2y = 1+x<br />
1+x<br />
1 − x ⇒ 2y =ln 1 − x<br />
(b) Let y =tanh −1 x.Thenx =tanhy, so from Exercise 18 we have<br />
e 2y = 1+tanhy<br />
1 − tanh y = 1+x<br />
1+x<br />
1 − x ⇒ 2y =ln 1 − x<br />
⇒ y = 1 2 ln 1+x<br />
1 − x<br />
29. (a) Let y =cosh −1 x.Thencosh y = x and y ≥ 0 ⇒ sinh y dy<br />
dx =1<br />
dy<br />
dx = 1<br />
sinh y = 1<br />
<br />
cosh 2 y − 1 = 1<br />
√<br />
x2 − 1<br />
⇒<br />
<br />
.<br />
[since sinh y ≥ 0 for y ≥ 0]. Or: Use Formula 4.<br />
⇒ xe 2y + x = e 2y − 1 ⇒<br />
<br />
1+x<br />
⇒ y = 1 ln .<br />
2<br />
1 − x<br />
(b) Let y =tanh −1 x.Thentanh y = x ⇒ sech 2 y dy<br />
dy<br />
=1 ⇒<br />
dx dx = 1<br />
sech 2 y = 1<br />
1 − tanh 2 y = 1<br />
1 − x . 2<br />
Or: Use Formula 5.<br />
(c) Let y =csch −1 x.Thencsch y = x ⇒ −csch y coth y dy<br />
dy<br />
=1 ⇒<br />
dx dx = − 1<br />
. By Exercise 13,<br />
csch y coth y<br />
coth y = ± csch 2 y +1=± √ x 2 +1.Ifx>0,thencoth y>0,socoth y = √ x 2 +1.Ifx0.]<br />
1 − x2 ⇒
F.<br />
128 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
(e) Let y =coth −1 x.Thencoth y = x ⇒ −csch 2 y dy<br />
dy<br />
=1 ⇒<br />
dx dx = − 1<br />
csch 2 y = 1<br />
1 − coth 2 y = 1<br />
1 − x 2<br />
by Exercise 13.<br />
31. f(x) =x sinh x − cosh x ⇒ f 0 (x) =x (sinh x) 0 +sinhx · 1 − sinh x = x cosh x<br />
33. h(x) =ln(coshx) ⇒ h 0 (x) = 1<br />
cosh x (cosh x)0 = sinh x<br />
cosh x =tanhx<br />
35. y = e cosh 3x ⇒ y 0 = e cosh 3x · sinh 3x · 3=3e cosh 3x sinh 3x<br />
37. f(t) =sech 2 (e t )=[sech(e t )] 2 ⇒<br />
f 0 (t) =2[sech(e t )] [sech(e t )] 0 =2sech(e t ) − sech(e t )tanh(e t ) · e t = −2e t sech 2 (e t )tanh(e t )<br />
39. y = arctan(tanh x) ⇒ y 0 =<br />
1<br />
1+(tanhx) (tanh 2 x)0 =<br />
sech2 x<br />
1+tanh 2 x<br />
41. G(x) = 1 − cosh x<br />
1+coshx<br />
⇒<br />
G 0 (x) =<br />
(1 + cosh x)(− sinh x) − (1 − cosh x)(sinh x) − sinh x − sinh x cosh x − sinh x +sinhx cosh x<br />
=<br />
(1 + cosh x) 2 (1 + cosh x) 2<br />
= −2sinhx<br />
(1 + cosh x) 2<br />
43. y =tanh −1√ x ⇒ y 0 1<br />
= √ 2 · 1<br />
1<br />
2 x−1/2 =<br />
1 − x 2 √ x (1 − x)<br />
45. y = x sinh −1 (x/3) − √ 9+x 2 ⇒<br />
x<br />
<br />
y 0 =sinh −1 1/3<br />
+ x <br />
3<br />
−<br />
1+(x/3)<br />
2<br />
47. y =coth −1√ x 2 +1 ⇒ y 0 1<br />
=<br />
1 − (x 2 +1)<br />
2x<br />
x<br />
<br />
2 √ 9+x 2 =sinh−1 +<br />
3<br />
49. As the depth d of the water gets large, the fraction 2πd<br />
L<br />
approaches 1. Thus, v =<br />
x<br />
√<br />
9+x<br />
2 −<br />
2x<br />
2 √ x 2 +1 = − 1<br />
x √ x 2 +1<br />
<br />
gL 2πd gL gL<br />
2π tanh ≈<br />
L 2π (1) = 2π .<br />
x<br />
x<br />
<br />
√<br />
9+x<br />
2 =sinh−1 3<br />
2πd<br />
gets large, and from Figure 3 or Exercise 23(a), tanh L<br />
51. (a) y =20cosh(x/20) − 15 ⇒ y 0 = 20 sinh(x/20) · 1 = sinh(x/20). Since the right pole is positioned at x =7,<br />
20<br />
we have y 0 (7) = sinh 7<br />
20 ≈ 0.3572.<br />
(b) If α is the angle between the tangent line and the x-axis, then tan α = slope of the line =sinh 7 20 ,so<br />
α =tan −1 sinh 7<br />
20<br />
≈ 0.343 rad ≈ 19.66 ◦ .Thus,theanglebetweenthelineandthepoleisθ =90 ◦ − α ≈ 70.34 ◦ .<br />
53. (a) y = A sinh mx + B cosh mx ⇒ y 0 = mA cosh mx + mB sinh mx ⇒<br />
y 00 = m 2 A sinh mx + m 2 B cosh mx = m 2 (A sinh mx + B cosh mx) =m 2 y
F.<br />
TX.10<br />
CHAPTER 3 REVIEW ¤ 129<br />
(b) From part (a), a solution of y 00 =9y is y(x) =A sinh 3x + B cosh 3x. So−4 =y(0) = A sinh 0 + B cosh 0 = B,so<br />
B = −4. Nowy 0 (x) =3A cosh 3x − 12 sinh 3x ⇒ 6=y 0 (0) = 3A ⇒ A =2,soy =2sinh3x − 4cosh3x.<br />
55. The tangent to y =coshx has slope 1 when y 0 =sinhx =1 ⇒ x =sinh −1 1=ln 1+ √ 2 ,byEquation3.<br />
Since sinh x =1and y =coshx = 1+sinh 2 x,wehavecosh x = √ 2. The point is ln 1+ √ 2 , √ 2 .<br />
57. If ae x + be −x = α cosh(x + β) [or α sinh(x + β)], then<br />
<br />
ae x + be −x = α 2 e x+β ± e −x−β <br />
= α 2 e x e β ± e −x e −β = α<br />
eβ e x ± α<br />
e−β e −x . Comparing coefficients of e x<br />
2 2<br />
and e −x ,wehavea = α 2 eβ (1) and b = ± α 2 e−β (2). Weneedtofind α and β. Dividing equation (1) by equation (2)<br />
gives us a b = ±e2β ⇒ () 2β =ln ± a b<br />
<br />
⇒ β = 1 2 ln ± a b<br />
. Solving equations (1) and (2) for e β gives us<br />
e β = 2a α and eβ = ± α 2b ,so 2a α = ± α 2b<br />
⇒ α 2 = ±4ab ⇒ α =2 √ ±ab.<br />
() If a b > 0, weusethe+ sign and obtain a cosh function, whereas if a b<br />
< 0, weusethe− sign and obtain a sinh<br />
function.<br />
In summary, if a and b havethesamesign,wehaveae x + be −x =2 √ ab cosh x + 1 2 ln a b<br />
<br />
, whereas, if a and b have the<br />
opposite sign, then ae x + be −x =2 √ −ab sinh x + 1 2 ln − a b<br />
.<br />
3 Review<br />
1. (a) The Power Rule: If n is any real number, then d<br />
dx (xn )=nx n−1 . The derivative of a variable base raised to a constant<br />
power is the power times the base raised to the power minus one.<br />
(b) The Constant Multiple Rule: If c is a constant and f is a differentiable function, then d<br />
dx [cf(x)] = c d<br />
dx f(x).<br />
The derivative of a constant times a function is the constant times the derivative of the function.<br />
(c) The Sum Rule: If f and g are both differentiable, then d<br />
dx [f(x)+g(x)] = d<br />
dx f(x)+ d g(x). The derivative of a sum<br />
dx<br />
of functions is the sum of the derivatives.<br />
(d) The Difference Rule: If f and g are both differentiable, then d<br />
d<br />
[f(x) − g(x)] =<br />
dx dx f(x) − d g(x). The derivative of a<br />
dx<br />
difference of functions is the difference of the derivatives.<br />
(e) The Product Rule: If f and g are both differentiable, then d<br />
d<br />
[f(x) g(x)] = f(x)<br />
dx dx g(x)+g(x) d f(x). The<br />
dx<br />
derivative of a product of two functions is the first function times the derivative of the second function plus the second<br />
function times the derivative of the first function.
F.<br />
130 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
(f ) The Quotient Rule: If f and g are both differentiable, then d<br />
dx<br />
f(x)<br />
g(x) d<br />
d<br />
f(x) − f(x)<br />
= dx dx g(x)<br />
g(x)<br />
[ g(x)] 2 .<br />
The derivative of a quotient of functions is the denominator times the derivative of the numerator minus the numerator<br />
times the derivative of the denominator, all divided by the square of the denominator.<br />
(g) The Chain Rule: If f and g are both differentiable and F = f ◦ g is the composite function defined by F (x) =f(g(x)),<br />
then F is differentiable and F 0 is given by the product F 0 (x) =f 0 (g(x)) g 0 (x). The derivative of a composite function is<br />
the derivative of the outer function evaluated at the inner function times the derivative of the inner function.<br />
2. (a) y = x n ⇒ y 0 = nx n−1 (b) y = e x ⇒ y 0 = e x<br />
(c) y = a x ⇒ y 0 = a x ln a (d) y =lnx ⇒ y 0 =1/x<br />
(e) y =log a x ⇒ y 0 =1/(x ln a) (f ) y =sinx ⇒ y 0 =cosx<br />
(g) y =cosx ⇒ y 0 = − sin x (h) y =tanx ⇒ y 0 =sec 2 x<br />
(i) y =cscx ⇒ y 0 = − csc x cot x (j) y =secx ⇒ y 0 =secx tan x<br />
(k) y =cotx ⇒ y 0 = − csc 2 x (l) y =sin −1 x ⇒ y 0 =1/ √ 1 − x 2<br />
(m) y =cos −1 x ⇒ y 0 = −1/ √ 1 − x 2 (n) y =tan −1 x ⇒ y 0 =1/(1 + x 2 )<br />
(o) y =sinhx ⇒ y 0 =coshx (p) y =coshx ⇒ y 0 =sinhx<br />
(q) y =tanhx ⇒ y 0 =sech 2 x (r) y =sinh −1 x ⇒ y 0 =1/ √ 1+x 2<br />
(s) y =cosh −1 x ⇒ y 0 =1/ √ x 2 − 1 (t) y =tanh −1 x ⇒ y 0 =1/(1 − x 2 )<br />
e h − 1<br />
3. (a) e is the number such that lim =1.<br />
h→0 h<br />
(b) e =lim<br />
x→0<br />
(1 + x) 1/x<br />
(c) The differentiation formula for y = a x [y 0 = a x ln a] is simplest when a = e because ln e =1.<br />
(d) The differentiation formula for y =log a x [y 0 =1/(x ln a)] issimplestwhena = e because ln e =1.<br />
4. (a) Implicit differentiation consists of differentiating both sides of an equation involving x and y with respect to x,andthen<br />
solving the resulting equation for y 0 .<br />
(b) Logarithmic differentiation consists of taking natural logarithms of both sides of an equation y = f(x), simplifying,<br />
differentiating implicitly with respect to x, and then solving the resulting equation for y 0 .<br />
5. (a) The linearization L of f at x = a is L(x) =f(a)+f 0 (a)(x − a).<br />
(b) If y = f(x), then the differential dy is given by dy = f 0 (x) dx.<br />
(c)SeeFigure5inSection3.10.
F.<br />
TX.10<br />
CHAPTER 3 REVIEW ¤ 131<br />
1. True. This is the Sum Rule.<br />
3. True. This is the Chain Rule.<br />
5. False.<br />
√ <br />
d<br />
√ f 0 x<br />
dx f x =<br />
2 √ by the Chain Rule.<br />
x<br />
7. False.<br />
9. True.<br />
d<br />
dx 10x =10 x ln 10<br />
d<br />
dx (tan2 x)=2tanx sec 2 x,and d<br />
dx (sec2 x)=2secx (sec x tan x) =2tanx sec 2 x.<br />
Or:<br />
d<br />
dx (sec2 x)= d<br />
dx (1 + tan2 x)= d<br />
dx (tan2 x).<br />
11. True. g(x) =x 5 ⇒ g 0 (x) =5x 4 ⇒ g 0 (2) = 5(2) 4 =80,andbythedefinition of the derivative,<br />
g(x) − g(2)<br />
lim<br />
= g 0 (2) = 80.<br />
x→2 x − 2<br />
1. y =(x 4 − 3x 2 +5) 3 ⇒<br />
y 0 =3(x 4 − 3x 2 +5) 2<br />
d<br />
dx (x4 − 3x 2 +5)=3(x 4 − 3x 2 +5) 2 (4x 3 − 6x) =6x(x 4 − 3x 2 +5) 2 (2x 2 − 3)<br />
3. y = √ x + 1<br />
3√<br />
x<br />
4 = x1/2 + x −4/3 ⇒ y 0 = 1 2 x−1/2 − 4 3 x−7/3 = 1<br />
2 √ x − 4<br />
3 3√ x 7<br />
5. y =2x √ x 2 +1 ⇒<br />
y 0 =2x · 1<br />
2 (x2 +1) −1/2 (2x)+ √ x 2 +1(2)= √ 2x2<br />
x2 +1 +2√ x 2 +1= 2x2 +2(x 2 +1)<br />
√ = 2(2x2 +1)<br />
√<br />
x2 +1 x2 +1<br />
7. y = e sin 2θ ⇒ y 0 = e sin 2θ d<br />
dθ (sin 2θ) =esin 2θ sin 2θ<br />
(cos 2θ)(2) = 2 cos 2θe<br />
9. y = t<br />
1 − t 2 ⇒ y 0 = (1 − t2 )(1) − t(−2t)<br />
(1 − t 2 ) 2 = 1 − t2 +2t 2<br />
(1 − t 2 ) 2 = t2 +1<br />
(1 − t 2 ) 2<br />
11. y = √ x cos √ x ⇒<br />
y 0 = √ <br />
x cos √ x<br />
0<br />
+cos<br />
√<br />
x<br />
√<br />
x<br />
0<br />
=<br />
√<br />
x<br />
<br />
− sin √ x<br />
=<br />
− √ 1 2 x−1/2 x sin √ x +cos √ <br />
x = cos √ x − √ x sin √ x<br />
2 √ x<br />
<br />
1<br />
2 x−1/2 <br />
+cos √ x<br />
<br />
1<br />
2 x−1/2 <br />
13. y = e1/x<br />
⇒ y 0 = x2 (e 1/x ) 0 − e 1/x x 2 0<br />
= x2 (e 1/x )(−1/x 2 ) − e 1/x (2x)<br />
= −e1/x (1 + 2x)<br />
x 2 (x 2 ) 2 x 4 x 4
F.<br />
132 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
15.<br />
d<br />
dx (xy4 + x 2 y)= d<br />
dx (x +3y) ⇒ x · 4y3 y 0 + y 4 · 1+x 2 · y 0 + y · 2x =1+3y 0 ⇒<br />
y 0 (4xy 3 + x 2 − 3) = 1 − y 4 − 2xy ⇒ y 0 = 1 − y4 − 2xy<br />
4xy 3 + x 2 − 3<br />
17. y =<br />
sec 2θ<br />
1+tan2θ<br />
⇒<br />
y 0 = (1 + tan 2θ)(sec 2θ tan 2θ · 2) − (sec 2θ)(sec2 2θ · 2)<br />
= 2sec2θ [(1 + tan 2θ)tan2θ − sec2 2θ]<br />
(1 + tan 2θ) 2 (1 + tan 2θ) 2<br />
= 2sec2θ (tan 2θ +tan2 2θ − sec 2 2θ) 2sec2θ (tan 2θ − 1) <br />
= 1+tan 2 x =sec 2 x <br />
(1 + tan 2θ) 2 (1 + tan 2θ) 2<br />
19. y = e cx (c sin x − cos x) ⇒<br />
y 0 = e cx (c cos x +sinx)+ce cx (c sin x − cos x) =e cx (c 2 sin x − c cos x + c cos x +sinx)<br />
= e cx (c 2 sin x +sinx) =e cx sin x (c 2 +1)<br />
21. y =3 x ln x ⇒ y 0 =3 x ln x · ln 3 ·<br />
<br />
d<br />
dx (x ln x) =3x ln x · ln 3 x · 1 <br />
x +lnx · 1 =3 x ln x · ln 3(1 + ln x)<br />
23. y =(1− x −1 ) −1 ⇒<br />
y 0 = −1(1 − x −1 ) −2 [−(−1x −2 )] = −(1 − 1/x) −2 x −2 = −((x − 1)/x) −2 x −2 = −(x − 1) −2<br />
25. sin(xy) =x 2 − y ⇒ cos(xy)(xy 0 + y · 1) = 2x − y 0 ⇒ x cos(xy)y 0 + y 0 =2x − y cos(xy) ⇒<br />
y 0 [x cos(xy)+1]=2x − y cos(xy) ⇒ y 0 =<br />
2x − y cos(xy)<br />
x cos(xy)+1<br />
27. y =log 5 (1 + 2x) ⇒ y 0 =<br />
1 d<br />
(1 + 2x) ln5dx (1 + 2x) = 2<br />
(1 + 2x) ln5<br />
29. y =lnsinx − 1 2 sin2 x ⇒ y 0 = 1<br />
sin x · cos x − 1 · 2sinx · cos x =cotx − sin x cos x<br />
2<br />
31. y = x tan −1 (4x) ⇒ y 0 = x ·<br />
1<br />
1+(4x) · 4x<br />
2 4+tan−1 (4x) · 1=<br />
1+16x 2 +tan−1 (4x)<br />
33. y =ln|sec 5x +tan5x| ⇒<br />
y 0 =<br />
1<br />
sec 5x +tan5x (sec 5x tan 5x · 5sec5x (tan 5x +sec5x)<br />
5+sec2 5x · 5) = =5sec5x<br />
sec 5x +tan5x<br />
35. y =cot(3x 2 +5) ⇒ y 0 = − csc 2 (3x 2 + 5)(6x) =−6x csc 2 (3x 2 +5)<br />
37. y =sin tan √ 1+x 3 ⇒ y 0 =cos tan √ 1+x 3 sec 2 √ 1+x 3 3x 2 2 √ 1+x 3 <br />
39. y =tan 2 (sin θ) =[tan(sinθ)] 2 ⇒ y 0 =2[tan(sinθ)] · sec 2 (sin θ) · cos θ
F.<br />
TX.10<br />
CHAPTER 3 REVIEW ¤ 133<br />
√ x +1(2− x)<br />
5<br />
41. y =<br />
⇒ ln y = 1<br />
(x +3) 7 2<br />
√ x +1(2− x)<br />
y 0 5<br />
<br />
1<br />
=<br />
(x +3) 7 2(x +1) − 5<br />
2 − x − 7 <br />
x +3<br />
ln(x +1)+5ln(2− x) − 7ln(x +3) ⇒<br />
y0<br />
y = 1<br />
2(x +1) + −5<br />
2 − x − 7<br />
or y 0 = (2 − x)4 (3x 2 − 55x − 52)<br />
2 √ x +1(x +3) 8 .<br />
x +3<br />
⇒<br />
43. y = x sinh(x 2 ) ⇒ y 0 = x cosh(x 2 ) · 2x +sinh(x 2 ) · 1=2x 2 cosh(x 2 )+sinh(x 2 )<br />
45. y =ln(cosh3x) ⇒ y 0 =(1/ cosh 3x)(sinh 3x)(3) = 3 tanh 3x<br />
47. y =cosh −1 (sinh x) ⇒ y 0 =<br />
<br />
49. y =cos e √ <br />
tan 3x<br />
⇒<br />
1<br />
<br />
(sinh x)2 − 1 · cosh x = cosh x<br />
<br />
sinh 2 x − 1<br />
<br />
y 0 = − sin e √ <br />
tan 3x<br />
· e √ 0 <br />
tan 3x = − sin e √ <br />
tan 3x<br />
e √ tan 3x · 1 (tan 2 3x)−1/2 · sec 2 (3x) · 3<br />
<br />
−3sin e √ <br />
tan 3x<br />
e √ tan 3x sec 2 (3x)<br />
=<br />
2 √ tan 3x<br />
51. f(t) = √ 4t +1 ⇒ f 0 (t) = 1 2 (4t +1)−1/2 · 4=2(4t +1) −1/2 ⇒<br />
f 00 (t) =2(− 1 2 )(4t +1)−3/2 · 4=−4/(4t +1) 3/2 ,sof 00 (2) = −4/9 3/2 = − 4<br />
27 .<br />
53. x 6 + y 6 =1 ⇒ 6x 5 +6y 5 y 0 =0 ⇒ y 0 = −x 5 /y 5 ⇒<br />
y 00 = − y5 (5x 4 ) − x 5 (5y 4 y 0 )<br />
= − 5x4 y 4 y − x(−x 5 /y 5 ) = − 5x4 (y 6 + x 6 )/y 5<br />
= − 5x4<br />
(y 5 ) 2 y 10<br />
y 6<br />
y 11<br />
55. We firstshowitistrueforn =1: f(x) =xe x ⇒ f 0 (x) =xe x + e x =(x +1)e x . We now assume it is true<br />
for n = k: f (k) (x) =(x + k)e x . With this assumption, we must show it is true for n = k +1:<br />
f (k+1) (x) = d <br />
f (k) (x) = d<br />
dx<br />
dx [(x + k)ex ]=(x + k)e x + e x =[(x + k)+1]e x =[x +(k +1)]e x .<br />
Therefore, f (n) (x) =(x + n)e x by mathematical induction.<br />
57. y =4sin 2 x ⇒ y 0 =4· 2sinx cos x. At π<br />
6 , 1 , y 0 =8· 1<br />
2 · √3<br />
2 =2√ 3, so an equation of the tangent line<br />
is y − 1=2 √ 3 x − π 6<br />
<br />
,ory =2<br />
√<br />
3 x +1− π<br />
√<br />
3/3.<br />
59. y = √ 1+4sinx ⇒ y 0 = 1 2 (1 + 4 sin x)−1/2 · 4cosx =<br />
2cosx<br />
√ 1+4sinx<br />
.<br />
At (0, 1), y 0 = 2 √<br />
1<br />
=2, so an equation of the tangent line is y − 1=2(x − 0),ory =2x +1.<br />
61. y =(2+x)e −x ⇒ y 0 =(2+x)(−e −x )+e −x · 1=e −x [−(2 + x)+1]=e −x (−x − 1).<br />
At (0, 2), y 0 =1(−1) = −1, so an equation of the tangent line is y − 2=−1(x − 0),ory = −x +2.<br />
The slope of the normal line is 1, so an equation of the normal line is y − 2=1(x − 0),ory = x +2.
F.<br />
134 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
63. (a) f(x) =x √ 5 − x ⇒<br />
<br />
1<br />
f 0 (x) =x<br />
2 (5 − x)−1/2 (−1) + √ −x<br />
5 − x =<br />
2 √ 5 − x + √ 5 − x · 2 √ 5 − x<br />
2 √ 5 − x = −x<br />
2 √ 2(5 − x)<br />
+<br />
5 − x 2 √ 5 − x<br />
=<br />
−x +10− 2x<br />
2 √ 5 − x<br />
=<br />
10 − 3x<br />
2 √ 5 − x<br />
(b) At (1, 2): f 0 (1) = 7 4 .<br />
(c)<br />
So an equation of the tangent line is y − 2= 7 4 (x − 1) or y = 7 4 x + 1 4 .<br />
At (4, 4): f 0 (4) = − 2 2 = −1.<br />
So an equation of the tangent line is y − 4=−1(x − 4) or y = −x +8.<br />
(d)<br />
The graphs look reasonable, since f 0 is positive where f has tangents with<br />
positive slope, and f 0 is negative where f has tangents with negative slope.<br />
65. y =sinx +cosx ⇒ y 0 =cosx − sin x =0 ⇔ cos x =sinx and 0 ≤ x ≤ 2π ⇔ x = π or 5π , so the points<br />
4 4<br />
are π<br />
4 , √ 2 and 5π<br />
4 , −√ 2 .<br />
67. f(x) =(x − a)(x − b)(x − c) ⇒ f 0 (x) =(x − b)(x − c)+(x − a)(x − c)+(x − a)(x − b).<br />
So f 0 (x)<br />
f(x)<br />
=<br />
(x − b)(x − c)+(x − a)(x − c)+(x − a)(x − b)<br />
(x − a)(x − b)(x − c)<br />
= 1<br />
x − a + 1<br />
x − b + 1<br />
x − c .<br />
Or: f(x) =(x − a)(x − b)(x − c) ⇒ ln |f(x)| =ln|x − a| +ln|x − b| +ln|x − c| ⇒<br />
f 0 (x)<br />
f(x) = 1<br />
x − a + 1<br />
x − b + 1<br />
x − c<br />
69. (a) h(x) =f(x) g(x) ⇒ h 0 (x) =f(x) g 0 (x)+g(x) f 0 (x) ⇒<br />
h 0 (2) = f(2) g 0 (2) + g(2) f 0 (2) = (3)(4) + (5)(−2) = 12 − 10 = 2<br />
(b) F (x) =f(g(x)) ⇒ F 0 (x) =f 0 (g(x)) g 0 (x) ⇒ F 0 (2) = f 0 (g(2)) g 0 (2) = f 0 (5)(4) = 11 · 4=44<br />
71. f(x) =x 2 g(x) ⇒ f 0 (x) =x 2 g 0 (x)+g(x)(2x) =x[xg 0 (x)+2g(x)]<br />
73. f(x) =[g(x)] 2 ⇒ f 0 (x) =2[g(x)] · g 0 (x) =2g(x) g 0 (x)<br />
75. f(x) =g(e x ) ⇒ f 0 (x) =g 0 (e x ) e x<br />
77. f(x) =ln|g(x)| ⇒ f 0 (x) = 1<br />
g(x) g0 (x) = g0 (x)<br />
g(x)
F.<br />
TX.10<br />
CHAPTER 3 REVIEW ¤ 135<br />
79. h(x) =<br />
f(x) g(x)<br />
f(x)+g(x)<br />
⇒<br />
h 0 (x) = [f(x)+g(x)] [f(x) g0 (x)+g(x) f 0 (x)] − f(x) g(x)[f 0 (x)+g 0 (x)]<br />
[f(x)+g(x)] 2<br />
= [f(x)]2 g 0 (x)+f(x) g(x) f 0 (x)+f(x) g(x) g 0 (x)+[g(x)] 2 f 0 (x) − f(x) g(x) f 0 (x) − f(x) g(x) g 0 (x)<br />
[f(x)+g(x)] 2<br />
= f 0 (x)[g(x)] 2 + g 0 (x)[f(x)] 2<br />
[f(x)+g(x)] 2<br />
81. Using the Chain Rule repeatedly, h(x) =f(g(sin 4x)) ⇒<br />
h 0 (x) =f 0 (g(sin 4x)) ·<br />
83. y =[ln(x +4)] 2 ⇒ y 0 =2[ln(x +4)] 1 ·<br />
d<br />
dx (g(sin 4x)) = f 0 (g(sin 4x)) · g 0 d<br />
(sin 4x) ·<br />
dx (sin 4x) =f 0 (g(sin 4x))g 0 (sin 4x)(cos 4x)(4).<br />
1<br />
+4)<br />
· 1=2ln(x and y 0 =0 ⇔ ln(x +4)=0 ⇔<br />
x +4 x +4<br />
x +4=e 0 ⇒ x +4=1 ⇔ x = −3, so the tangent is horizontal at the point (−3, 0).<br />
85. y = f(x) =ax 2 + bx + c ⇒ f 0 (x) =2ax + b. We know that f 0 (−1) = 6 and f 0 (5) = −2, so−2a + b =6and<br />
10a + b = −2. Subtracting the first equation from the second gives 12a = −8 ⇒ a = − 2 . Substituting − 2 for a in the<br />
3 3<br />
first equation gives b = 14 3 .Nowf(1) = 4 ⇒ 4=a + b + c,soc =4+2 3 − 14<br />
3 =0and hence, f(x) =− 2 3 x2 + 14<br />
3 x.<br />
87. s(t) =Ae −ct cos(ωt + δ) ⇒<br />
v(t) =s 0 (t) =A{e −ct [−ω sin(ωt + δ)] + cos(ωt + δ)(−ce −ct )} = −Ae −ct [ω sin(ωt + δ)+c cos(ωt + δ)]<br />
⇒<br />
a(t) =v 0 (t) =−A{e −ct [ω 2 cos(ωt + δ) − cω sin(ωt + δ)] + [ω sin(ωt + δ)+c cos(ωt + δ)](−ce −ct )}<br />
= −Ae −ct [ω 2 cos(ωt + δ) − cω sin(ωt + δ) − cω sin(ωt + δ) − c 2 cos(ωt + δ)]<br />
= −Ae −ct [(ω 2 − c 2 )cos(ωt + δ) − 2cω sin(ωt + δ)] = Ae −ct [(c 2 − ω 2 )cos(ωt + δ)+2cω sin(ωt + δ)]<br />
89. (a) y = t 3 − 12t +3 ⇒ v(t) =y 0 =3t 2 − 12 ⇒ a(t) =v 0 (t) =6t<br />
(b) v(t) =3(t 2 − 4) > 0 when t>2,soitmovesupwardwhent>2 anddownwardwhen0 ≤ t2. The particle is slowing down when v and a have opposite<br />
signs; that is, when 0
F.<br />
136 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />
TX.10<br />
93. (a) y(t) =y(0)e kt =200e kt ⇒ y(0.5) = 200e 0.5k =360 ⇒ e 0.5k =1.8 ⇒ 0.5k =ln1.8 ⇒<br />
k =2ln1.8 =ln(1.8) 2 =ln3.24 ⇒ y(t) = 200e (ln 3.24)t = 200(3.24) t<br />
(b) y(4) = 200(3.24) 4 ≈ 22,040 bacteria<br />
(c) y 0 (t) = 200(3.24) t · ln 3.24,soy 0 (4) = 200(3.24) 4 · ln 3.24 ≈ 25,910 bacteria per hour<br />
(d) 200(3.24) t =10,000 ⇒ (3.24) t =50 ⇒ t ln 3.24 = ln 50 ⇒ t =ln50/ ln 3.24 ≈ 3.33 hours<br />
95. (a) C 0 (t) =−kC(t) ⇒ C(t) =C(0)e −kt by Theorem 9.4.2. But C(0) = C 0 ,soC(t) =C 0 e −kt .<br />
(b) C(30) = 1 2 C0 since the concentration is reduced by half. Thus, 1 2 C0 = C0e−30k ⇒ ln 1 2 = −30k ⇒<br />
k = − 1 ln 1 = 1 ln 2. Since10% of the original concentration remains if 90% is eliminated, we want the value of t<br />
30 2 30<br />
such that C(t) = 1 10 C 0. Therefore,<br />
1<br />
C 10 0 = C 0 e −t(ln 2)/30 ⇒ ln 0.1 =−t(ln 2)/30 ⇒ t = − 30 ln 0.1 ≈ 100 h.<br />
ln 2<br />
97. If x = edge length, then V = x 3 ⇒ dV/dt =3x 2 dx/dt =10 ⇒ dx/dt =10/(3x 2 ) and S =6x 2 ⇒<br />
dS/dt =(12x) dx/dt =12x[10/(3x 2 )] = 40/x. Whenx =30, dS/dt = 40<br />
30 = 4 3 cm2 /min.<br />
99. Given dh/dt =5and dx/dt =15, find dz/dt. z 2 = x 2 + h 2 ⇒<br />
2z dz dx dh<br />
=2x +2h<br />
dt dt dt<br />
⇒<br />
dz<br />
dt = 1 (15x +5h). Whent =3,<br />
z<br />
h =45+3(5)=60and x = 15(3) = 45 ⇒ z = √ 45 2 +60 2 =75,<br />
so dz<br />
dt = 1 [15(45) + 5(60)] = 13 ft/s.<br />
75<br />
101. We are given dθ/dt = −0.25 rad/h. tan θ =400/x ⇒<br />
x =400cotθ ⇒ dx<br />
dt = −400 csc2 θ dθ<br />
dt .Whenθ = π , 6<br />
dx<br />
dt = −400(2)2 (−0.25) = 400 ft/h.<br />
103. (a) f(x) = 3√ 1+3x =(1+3x) 1/3 ⇒ f 0 (x) =(1+3x) −2/3 , so the linearization of f at a =0is<br />
L(x) =f(0) + f 0 (0)(x − 0) = 1 1/3 +1 −2/3 x =1+x. Thus,<br />
3√<br />
1.03 =<br />
3 1+3(0.01) ≈ 1+(0.01) = 1.01.<br />
3√ 1+3x ≈ 1+x<br />
⇒<br />
(b) The linear approximation is 3√ 1+3x ≈ 1+x,sofortherequiredaccuracy<br />
we want 3√ 1+3x − 0.1 < 1+x< 3√ 1+3x +0.1. From the graph,<br />
itappearsthatthisistruewhen−0.23
F.<br />
TX.10<br />
CHAPTER 3 REVIEW ¤ 137<br />
105. A = x 2 + 1 π 1<br />
2 2 x2 = <br />
1+ π 8 x<br />
2<br />
⇒ dA = 2+ π 4<br />
<br />
xdx.Whenx =60<br />
and dx =0.1, dA = 2+ π 4<br />
approximately 12 + 3π 2 ≈ 16.7 cm2 .<br />
√ <br />
4<br />
16 + h − 2 d<br />
107. lim<br />
=<br />
h→0 h dx<br />
<br />
60(0.1) = 12 +<br />
3π<br />
2<br />
, so the maximum error is<br />
4√<br />
x<br />
<br />
x =16<br />
= 1 4 x−3/4 x<br />
=16<br />
=<br />
1<br />
4 √ 4<br />
16 3 = 1<br />
32<br />
√ √ √ √ √ √ <br />
1+tanx − 1+sinx 1+tanx − 1+sinx 1+tanx + 1+sinx<br />
109. lim<br />
=lim<br />
x→0 x 3<br />
x→0 x √ 3 1+tanx + √ 1+sinx <br />
(1 + tan x) − (1 + sin x)<br />
=lim<br />
x→0 x 3√ 1+tanx + √ 1+sinx =lim sin x (1/ cos x − 1)<br />
x→0 x 3√ 1+tanx + √ 1+sinx · cos x<br />
cos x<br />
sin x (1 − cos x)<br />
=lim<br />
x→0 x 3√ 1+tanx + √ 1+sinx cos x · 1+cosx<br />
1+cosx<br />
sin x · sin 2 x<br />
=lim<br />
x→0 x 3√ 1+tanx + √ 1+sinx cos x (1 + cos x)<br />
<br />
= lim<br />
x→0<br />
=1 3 ·<br />
3<br />
sin x<br />
1<br />
lim √ √ <br />
x x→0 1+tanx + 1+sinx cos x (1 + cos x)<br />
1<br />
√<br />
1+<br />
√<br />
1<br />
<br />
· 1 · (1 + 1)<br />
= 1 4<br />
111.<br />
d<br />
dx [f(2x)] = x2 ⇒ f 0 (2x) · 2=x 2 ⇒ f 0 (2x) = 1 2 x2 .Lett =2x. Thenf 0 (t) = 1 1<br />
2 2 t2 = 1 8 t2 ,sof 0 (x) = 1 8 x2 .
F.<br />
TX.10
F.<br />
TX.10<br />
PROBLEMS PLUS<br />
1. Let a be the x-coordinate of Q. Since the derivative of y =1− x 2 is y 0 = −2x,theslopeatQ is −2a. But since the triangle<br />
is equilateral, AO/OC = √ 3/1, so the slope at Q is − √ 3. Therefore, we must have that −2a = − √ 3 ⇒ a = √ 3<br />
Thus, the point Q has coordinates<br />
√3<br />
√3<br />
, 1 − 2 2<br />
2 √3 <br />
<br />
= , 1 andbysymmetry,P has coordinates − √ <br />
3<br />
, 1 .<br />
2 4<br />
2 4<br />
2 .<br />
3. We must show that r (in the figure) is halfway between p and q,thatis,<br />
r =(p + q)/2. For the parabola y = ax 2 + bx + c, the slope of the tangent line is<br />
given by y 0 =2ax + b. An equation of the tangent line at x = p is<br />
y − (ap 2 + bp + c) =(2ap + b)(x − p). Solving for y gives us<br />
y =(2ap + b)x − 2ap 2 − bp +(ap 2 + bp + c)<br />
or y =(2ap + b)x + c − ap 2 (1)<br />
Similarly, an equation of the tangent line at x = q is<br />
y =(2aq + b)x + c − aq 2 (2)<br />
We can eliminate y and solve for x by subtracting equation (1) from equation (2).<br />
[(2aq + b) − (2ap + b)]x − aq 2 + ap 2 =0<br />
(2aq − 2ap)x = aq 2 − ap 2<br />
2a(q − p)x = a(q 2 − p 2 )<br />
x =<br />
a(q + p)(q − p)<br />
2a(q − p)<br />
= p + q<br />
2<br />
Thus, the x-coordinate of the point of intersection of the two tangent lines, namely r,is(p + q)/2.<br />
5. Let y =tan −1 x.Thentan y = x, so from the triangle we see that<br />
sin(tan −1 x)=siny =<br />
sin(tan −1 (sinh x)) =<br />
x<br />
√ . Using this fact we have that<br />
1+x<br />
2<br />
sinh x<br />
<br />
1+sinh 2 x = sinh x<br />
cosh x =tanhx.<br />
Hence, sin −1 (tanh x) =sin −1 (sin(tan −1 (sinh x))) = tan −1 (sinh x).<br />
7. We use mathematical induction. Let S n be the statement that<br />
S 1 is true because<br />
d n<br />
dx n (sin4 x +cos 4 x)=4 n−1 cos(4x + nπ/2).<br />
d<br />
dx (sin4 x +cos 4 x)=4sin 3 x cos x − 4cos 3 x sin x =4sinx cos x sin 2 x − cos 2 x x<br />
= −4sinx cos x cos 2x = −2sin2x cos 2 = − sin 4x =sin(−4x)<br />
=cos π<br />
− (−4x) 2<br />
=cos π<br />
+4x 2<br />
=4 n−1 cos <br />
4x + n π 2 when n =1<br />
[continued]<br />
139
F.<br />
140 ¤ CHAPTER 3 PROBLEMS PLUS<br />
TX.10<br />
d k <br />
Now assume S k is true, that is, sin 4 x +cos 4 x =4 k−1 cos <br />
4x + k π<br />
dx k 2 .Then<br />
d k+1<br />
dx k+1 (sin4 x +cos 4 x)= d <br />
d<br />
k<br />
dx dx k (sin4 x +cos 4 x) = d 4 k−1 cos <br />
4x + k π 2<br />
dx<br />
which shows that S k+1 is true.<br />
Therefore,<br />
Another proof: First write<br />
= −4 k−1 sin <br />
4x + k π d<br />
2 ·<br />
dx<br />
=4 k sin −4x − k π 2<br />
<br />
4x + k<br />
π<br />
2 = −4 k sin <br />
4x + k π 2<br />
=4 k cos π<br />
2 − −4x − k π 2<br />
=4 k cos 4x +(k +1) π 2<br />
d n<br />
dx n (sin4 x +cos 4 x)=4 n−1 cos <br />
4x + n π 2 for every positive integer n, by mathematical induction.<br />
sin 4 x +cos 4 x =(sin 2 x +cos 2 x) 2 − 2sin 2 x cos 2 x =1− 1 2 sin2 2x =1− 1 (1 − cos 4x) = 3 + 1 cos 4x<br />
4 4 4<br />
<br />
Then we have dn<br />
dx n (sin4 x +cos 4 x)= dn 3<br />
dx n 4 + 1 <br />
4 cos 4x = 1 <br />
4 · 4n cos 4x + n π <br />
=4 n−1 cos 4x + n π <br />
.<br />
2<br />
2<br />
9. We must find a value x 0 such that the normal lines to the parabola y = x 2 at x = ±x 0 intersectatapointoneunitfromthe<br />
points <br />
±x 0 ,x 2 0 . The normals to y = x 2 at x = ±x 0 have slopes − 1 and pass through <br />
±x 0 ,x 2 0 respectively, so the<br />
±2x 0<br />
normals have the equations y − x 2 0 = − 1<br />
2x 0<br />
(x − x 0) and y − x 2 0 = 1<br />
2x 0<br />
(x + x 0). The common y-intercept is x 2 0 + 1 2 .<br />
We want to find the value of x 0 for which the distance from 0,x 2 0 + 1 2<br />
<br />
to<br />
<br />
x0,x 2 0<br />
equals 1. The square of the distance is<br />
(x 0 − 0) 2 + x 2 0 − x 2 0 + 1 2<br />
the center of the circle is at 0, 5 4<br />
.<br />
2<br />
= x 2 0 + 1 4 =1 ⇔ x 0 = ± √ 3<br />
2 .Forthesevaluesofx 0,they-intercept is x 2 0 + 1 2 = 5 4 ,so<br />
Another solution: Let the center of the circle be (0,a). Then the equation of the circle is x 2 +(y − a) 2 =1.<br />
Solving with the equation of the parabola, y = x 2 ,wegetx 2 +(x 2 − a) 2 =1 ⇔ x 2 + x 4 − 2ax 2 + a 2 =1 ⇔<br />
x 4 +(1− 2a)x 2 + a 2 − 1=0. The parabola and the circle will be tangent to each other when this quadratic equation in x 2<br />
<br />
has equal roots; that is, when the discriminant is 0. Thus,(1 − 2a) 2 − 4(a 2 − 1) = 0<br />
⇔<br />
1 − 4a +4a 2 − 4a 2 +4=0 ⇔ 4a =5,soa = 5 4 . The center of the circle is 0, 5 4<br />
<br />
.<br />
11. We can assume without loss of generality that θ =0at time t =0,sothatθ =12πt rad. [The angular velocity of the wheel<br />
is 360 rpm =360· (2π rad)/(60 s) =12π rad/s.] Then the position of A as a function of time is<br />
A =(40cosθ, 40 sin θ) = (40 cos 12πt, 40 sin 12πt),sosin α =<br />
y 40 sin θ<br />
= = sin θ = 1 sin 12πt.<br />
1.2 m 120 3 3<br />
(a) Differentiating the expression for sin α,wegetcos α · dα<br />
dt = 1 3 · 12π · cos 12πt =4π cos θ. Whenθ = π 3 ,wehave<br />
sin α = 1 √<br />
<br />
√ 2 <br />
3<br />
3 11<br />
3 sin θ = 6 ,socos α = dα<br />
1 − = and<br />
6 12 dt = 4π cos π 3<br />
cos α = 2π<br />
= 4π √ 3<br />
11/12 11
F.<br />
TX.10<br />
CHAPTER 3 PROBLEMS PLUS ¤ 141<br />
(b) By the Law of Cosines, |AP | 2 = |OA| 2 + |OP| 2 − 2 |OA||OP| cos θ<br />
⇒<br />
120 2 =40 2 + |OP| 2 − 2 · 40 |OP| cos θ ⇒ |OP| 2 − (80 cos θ) |OP| − 12,800 = 0 ⇒<br />
|OP| = 1 2<br />
<br />
80 cos θ ±<br />
√ 6400 cos2 θ +51,200 =40cosθ ± 40 √ cos 2 θ +8=40 cos θ + √ 8+cos 2 θ cm<br />
[since |OP| > 0]. As a check, note that |OP| =160cm when θ =0and |OP| =80 √ 2 cm when θ = π 2 .<br />
(c) By part (b), the x-coordinate of P is given by x =40 cos θ + √ 8+cos 2 θ ,so<br />
dx<br />
dt = dx<br />
dθ<br />
<br />
dθ<br />
dt =40 − sin θ −<br />
<br />
<br />
2cosθ sin θ<br />
2 √ · 12π = −480π sin θ 1+<br />
8+cos 2 θ<br />
In particular, dx/dt =0cm/swhenθ =0and dx/dt = −480π cm/swhenθ = π 2 .<br />
13. Consider the statement that<br />
d<br />
dx (eax sin bx) =ae ax sin bx + be ax cos bx,and<br />
d n<br />
dx n (eax sin bx) =r n e ax sin(bx + nθ). Forn =1,<br />
<br />
cos θ<br />
√ cm/s.<br />
8+cos2 θ<br />
re ax sin(bx + θ) =re ax [sin bx cos θ +cosbx sin θ] =re ax a<br />
r sin bx + b r cos bx <br />
= ae ax sin bx + be ax cos bx<br />
since tan θ = b a<br />
⇒ sin θ = b r and cos θ = a . So the statement is true for n =1.<br />
r<br />
But<br />
Assume it is true for n = k. Then<br />
d k+1<br />
dx k+1 (eax sin bx)= d <br />
r k e ax sin(bx + kθ) = r k ae ax sin(bx + kθ)+r k e ax b cos(bx + kθ)<br />
dx<br />
= r k e ax [a sin(bx + kθ)+b cos(bx + kθ)]<br />
sin[bx +(k +1)θ] =sin[(bx + kθ)+θ] =sin(bx + kθ)cosθ +sinθ cos(bx + kθ) = a sin(bx + kθ)+ b cos(bx + kθ).<br />
r r<br />
Hence, a sin(bx + kθ)+b cos(bx + kθ) =r sin[bx +(k +1)θ].So<br />
d k+1<br />
dx k+1 (eax sin bx) =r k e ax [a sin(bx+kθ)+b cos(bx+kθ)] = r k e ax [r sin(bx+(k +1)θ)] = r k+1 e ax [sin(bx+(k +1)θ)].<br />
Therefore, the statement is true for all n by mathematical induction.<br />
15. It seems from the figure that as P approaches the point (0, 2) from the right, x T →∞and y T → 2 + .AsP approaches the<br />
point (3, 0) from the left, it appears that x T → 3 + and y T →∞.Soweguessthatx T ∈ (3, ∞) and y T ∈ (2, ∞). Itis<br />
more difficult to estimate the range of values for x N and y N. We might perhaps guess that x N ∈ (0, 3),<br />
and y N ∈ (−∞, 0) or (−2, 0).<br />
In order to actually solve the problem, we implicitly differentiate the equation of the ellipse to find the equation of the<br />
tangent line: x2<br />
9 + y2<br />
4 =1 ⇒ 2x 9 + 2y 4 y0 =0,soy 0 = − 4 9<br />
x<br />
. So at the point (x0,y0) on the ellipse, an equation of the<br />
y
F.<br />
142 ¤ CHAPTER 3 PROBLEMS PLUS<br />
TX.10<br />
tangent line is y − y 0 = − 4 x 0<br />
(x − x 0) or 4x 0x +9y 0y =4x 2 0 +9y<br />
9 y<br />
0. 2 This can be written as x 0x<br />
0 9 + y 0y<br />
4 = x2 0<br />
9 + y2 0<br />
4 =1,<br />
because (x 0 ,y 0 ) lies on the ellipse. So an equation of the tangent line is x 0x<br />
9 + y 0y<br />
4 =1.<br />
Therefore, the x-intercept x T for the tangent line is given by x 0x T<br />
9<br />
=1 ⇔ x T = 9 x 0<br />
,andthey-intercept y T is given<br />
by y0yT<br />
4<br />
=1 ⇔ y T = 4 y 0<br />
.<br />
So as x 0 takes on all values in (0, 3), x T takes on all values in (3, ∞),andasy 0 takes on all values in (0, 2), y T takes on<br />
all values in (2, ∞).<br />
1<br />
At the point (x 0,y 0) on the ellipse, the slope of the normal line is −<br />
y 0 (x = 9 ,andits<br />
0,y 0) 4 x 0<br />
y 0<br />
y 0<br />
equation is y − y 0 = 9 (x − x 0). Sothex-intercept x N for the normal line is given by 0 − y 0 = 9 (x N − x 0)<br />
4 x 0 4 x 0<br />
y 0<br />
⇒<br />
x N = − 4x0<br />
9 + x 0 = 5x0<br />
9 ,andthey-intercept y N is given by y N − y 0 = 9 4<br />
y 0<br />
(0 − x 0 ) ⇒ y N = − 9y0<br />
x 0 4 + y 0 = − 5y0<br />
4 .<br />
So as x 0 takes on all values in (0, 3), x N takes on all values in 0, 5 3<br />
<br />
,andasy0 takes on all values in (0, 2), y N takes on<br />
all values in − 5 2 , 0 .<br />
17. (a) If the two lines L 1 and L 2 have slopes m 1 and m 2 and angles of<br />
inclination φ 1 and φ 2 ,thenm 1 =tanφ 1 and m 2 =tanφ 2 . The triangle<br />
in the figure shows that φ 1 + α +(180 ◦ − φ 2 ) = 180 ◦ and so<br />
α = φ 2 − φ 1 . Therefore, using the identity for tan(x − y),wehave<br />
tan α =tan(φ 2 − φ 1 )= tan φ 2 − tan φ 1<br />
m2 − m1<br />
and so tan α = .<br />
1+tanφ 2 tan φ 1 1+m 1m 2<br />
(b) (i) The parabolas intersect when x 2 =(x − 2) 2 ⇒ x =1.Ify = x 2 ,theny 0 =2x, so the slope of the tangent<br />
to y = x 2 at (1, 1) is m 1 =2(1)=2.Ify =(x − 2) 2 ,theny 0 =2(x − 2), so the slope of the tangent to<br />
y =(x − 2) 2 at (1, 1) is m 2 =2(1− 2) = −2. Therefore, tan α =<br />
so α =tan −1 4<br />
3<br />
≈ 53 ◦ [or 127 ◦ ].<br />
m2 − m1<br />
= −2 − 2<br />
1+m 1m 2 1+2(−2) = 4 3 and<br />
(ii) x 2 − y 2 =3and x 2 − 4x + y 2 +3=0intersect when x 2 − 4x +(x 2 − 3) + 3 = 0 ⇔ 2x(x − 2) = 0 ⇒<br />
x =0or 2,but0 is extraneous. If x =2,theny = ±1.Ifx 2 − y 2 =3then 2x − 2yy 0 =0 ⇒ y 0 = x/y and<br />
x 2 − 4x + y 2 +3=0 ⇒ 2x − 4+2yy 0 =0 ⇒ y 0 = 2 − x .At(2, 1) the slopes are m 1 =2and<br />
y<br />
m 2 =0,so tan α = 0 − 2<br />
1+2· 0 = −2 ⇒ α ≈ 117◦ .At(2, −1) the slopes are m 1 = −2 and m 2 =0,<br />
so tan α =<br />
0 − (−2)<br />
1+(−2) (0) =2 ⇒ α ≈ 63◦ [or 117 ◦ ].
F.<br />
TX.10<br />
CHAPTER 3 PROBLEMS PLUS ¤ 143<br />
19. Since ∠ROQ = ∠OQP = θ, the triangle QOR is isosceles, so<br />
|QR| = |RO| = x. By the Law of Cosines, x 2 = x 2 + r 2 − 2rx cos θ. Hence,<br />
2rx cos θ = r 2 ,sox =<br />
sin θ = y/r), and hence x →<br />
r2<br />
2r cos θ = r<br />
2cosθ .Notethatasy → 0+ , θ → 0 + (since<br />
r<br />
2cos0 = r .Thus,asP is taken closer and closer<br />
2<br />
to the x-axis, the point R approaches the midpoint of the radius AO.<br />
21. lim<br />
x→0<br />
sin(a +2x) − 2sin(a + x)+sina<br />
x 2<br />
= lim<br />
x→0<br />
sin a cos 2x +cosa sin 2x − 2sina cos x − 2cosa sin x +sina<br />
x 2<br />
= lim<br />
x→0<br />
sin a (cos 2x − 2cosx +1)+cosa (sin 2x − 2sinx)<br />
x 2<br />
= lim<br />
x→0<br />
sin a (2 cos 2 x − 1 − 2cosx +1)+cosa (2 sin x cos x − 2sinx)<br />
x 2<br />
= lim<br />
x→0<br />
sin a (2 cos x)(cos x − 1) + cos a (2 sin x)(cos x − 1)<br />
x 2<br />
2(cos x − 1)[sin a cos x +cosa sin x](cos x +1)<br />
= lim<br />
x→0 x 2 (cos x +1)<br />
<br />
−2sin 2 2<br />
x [sin(a + x)]<br />
sin x sin(a + x)<br />
= lim<br />
= −2 lim ·<br />
x→0 x 2 (cos x +1)<br />
x→0 x cos x +1 = sin(a +0)<br />
−2(1)2 cos 0 + 1 = − sin a<br />
23. Let f(x) =e 2x and g(x) =k √ x [k >0]. From the graphs of f and g,<br />
So we must have k √ a =<br />
k<br />
4 √ a<br />
k =2e 1/2 =2 √ e ≈ 3.297.<br />
25. y =<br />
⇒<br />
we see that f will intersect g exactly once when f and g share a tangent<br />
line. Thus, we must have f = g and f 0 = g 0 at x = a.<br />
f(a) =g(a) ⇒ e 2a = k √ a ()<br />
and f 0 (a) =g 0 (a) ⇒ 2e 2a = k<br />
2 √ ⇒ e 2a = k<br />
a<br />
4 √ a .<br />
√ 2 k a = ⇒ a = 1<br />
4k<br />
.From(), 4 e2(1/4) = k 1/4 ⇒<br />
x<br />
√<br />
a2 − 1 − 2<br />
√<br />
a2 − 1 arctan sin x<br />
a + √ a 2 − 1+cosx .Letk = a + √ a 2 − 1. Then<br />
y 0 =<br />
=<br />
1<br />
√<br />
a2 − 1 − 2<br />
√<br />
a2 − 1 ·<br />
1<br />
1+sin 2 x/(k +cosx) · cos x(k +cosx)+sin2 x<br />
2 (k +cosx) 2<br />
1<br />
√<br />
a2 − 1 − 2<br />
√<br />
a2 − 1 · k cos x +cos2 x +sin 2 x<br />
(k +cosx) 2 +sin 2 x = 1<br />
√<br />
a2 − 1 − 2<br />
√<br />
a2 − 1 · k cos x +1<br />
k 2 +2k cos x +1<br />
= k2 +2k cos x +1− 2k cos x − 2<br />
√<br />
a2 − 1(k 2 +2k cos x +1)<br />
=<br />
k 2 − 1<br />
√<br />
a2 − 1(k 2 +2k cos x +1)<br />
But k 2 =2a 2 +2a √ a 2 − 1 − 1=2a a + √ a 2 − 1 − 1=2ak − 1,sok 2 +1=2ak,andk 2 − 1=2(ak − 1).<br />
[continued]
F.<br />
144 ¤ CHAPTER 3 PROBLEMS PLUS<br />
TX.10<br />
So y 0 =<br />
2(ak − 1)<br />
√<br />
a2 − 1(2ak +2k cos x) = ak − 1<br />
√<br />
a2 − 1k (a +cosx) .Butak − 1=a2 + a √ a 2 − 1 − 1=k √ a 2 − 1,<br />
so y 0 =1/(a +cosx).<br />
27. y = x 4 − 2x 2 − x ⇒ y 0 =4x 3 − 4x − 1. The equation of the tangent line at x = a is<br />
y − (a 4 − 2a 2 − a) =(4a 3 − 4a − 1)(x − a) or y =(4a 3 − 4a − 1)x +(−3a 4 +2a 2 ) and similarly for x = b. Soifat<br />
x = a and x = b we have the same tangent line, then 4a 3 − 4a − 1=4b 3 − 4b − 1 and −3a 4 +2a 2 = −3b 4 +2b 2 .Thefirst<br />
equation gives a 3 − b 3 = a − b ⇒ (a − b)(a 2 + ab + b 2 )=(a − b). Assuming a 6=b, wehave1=a 2 + ab + b 2 .<br />
The second equation gives 3(a 4 − b 4 )=2(a 2 − b 2 ) ⇒ 3(a 2 − b 2 )(a 2 + b 2 )=2(a 2 − b 2 ) which is true if a = −b.<br />
Substituting into 1=a 2 + ab + b 2 gives 1=a 2 − a 2 + a 2 ⇒ a = ±1 so that a =1and b = −1 or vice versa. Thus,<br />
the points (1, −2) and (−1, 0) have a common tangent line.<br />
As long as there are only two such points, we are done. So we show that these are in fact the only two such points.<br />
Suppose that a 2 − b 2 6=0.Then3(a 2 − b 2 )(a 2 + b 2 )=2(a 2 − b 2 ) gives 3(a 2 + b 2 )=2 or a 2 + b 2 = 2 3 .<br />
Thus, ab =(a 2 + ab + b 2 ) − (a 2 + b 2 )=1− 2 3 = 1 3 ,sob = 1<br />
3a . Hence, a2 + 1<br />
9a 2 = 2 3 ,so9a4 +1=6a 2<br />
0=9a 4 − 6a 2 +1=(3a 2 − 1) 2 .So3a 2 − 1=0 ⇒ a 2 = 1 3<br />
that a 2 6=b 2 .<br />
⇒ b 2 = 1<br />
9a 2 = 1 3 = a2 , contradicting our assumption<br />
29. Because of the periodic nature of the lattice points, it suffices to consider the points in the 5 × 2 grid shown. We can see that<br />
the minimum value of r occurs when there is a line with slope 2 which touches the circle centered at (3, 1) and the circles<br />
5<br />
centered at (0, 0) and (5, 2).<br />
⇒<br />
To find P , the point at which the line is tangent to the circle at (0, 0), we simultaneously solve x 2 + y 2 = r 2 and<br />
y = − 5 2 x ⇒ x2 + 25 4 x2 = r 2 ⇒ x 2 = 4 29 r2 ⇒ x = 2 √<br />
29<br />
r, y = − 5 √<br />
29<br />
r.Tofind Q, we either use symmetry or<br />
solve (x − 3) 2 +(y − 1) 2 = r 2 and y − 1=− 5 2<br />
2<br />
(x − 3).Asabove,wegetx =3− √<br />
29<br />
r, y =1+ √ 5<br />
29<br />
r. Now the slope of<br />
the line PQis 2 5 ,som PQ =<br />
<br />
1+ √ 5<br />
29<br />
r − − √ 5<br />
29<br />
r<br />
3 − 2 √<br />
29<br />
r − 2 √<br />
29<br />
r<br />
=<br />
10 √<br />
1+ √<br />
29<br />
r 29 + 10r<br />
3 − √ 4<br />
29<br />
r = 3 √ 29 − 4r = 2 5<br />
5 √ 29 + 50r =6 √ 29 − 8r ⇔ 58r = √ 29 ⇔ r = √ 29<br />
58 . So the minimum value of r for which any line with slope 2 5<br />
intersects circles with radius r centered at the lattice points on the plane is r = √ 29<br />
58 ≈ 0.093.<br />
F.<br />
TX.10<br />
31. By similar triangles, r 5 = h 16<br />
⇒<br />
CHAPTER 3 PROBLEMS PLUS ¤ 145<br />
r = 5h . The volume of the cone is<br />
16<br />
2 5h<br />
V = 1 3 πr2 h = 1 π h = 25π<br />
3<br />
16 768 h3 ,so dV<br />
dt = 25π dh<br />
256 h2 dt .Nowtherateof<br />
change of the volume is also equal to the difference of what is being added<br />
(2 cm 3 /min) and what is oozing out (kπrl,whereπrl is the area of the cone and k<br />
Equating the two expressions for dV<br />
dt<br />
is a proportionality constant). Thus, dV<br />
dt<br />
and substituting h =10,<br />
dh<br />
dt<br />
=2− kπrl.<br />
= −0.3, r =<br />
5(10)<br />
16<br />
= 25 8 ,and l<br />
√<br />
281<br />
= 10<br />
16<br />
⇔<br />
l = 5 8<br />
√ 25π<br />
281,weget<br />
256 (10)2 (−0.3) = 2 − kπ 25<br />
8 · 5 √ 125kπ √ 281<br />
281 ⇔<br />
8<br />
64<br />
=2+ 750π . Solving for k gives us<br />
256<br />
256 + 375π<br />
k =<br />
250π √ . To maintain a certain height, the rate of oozing, kπrl, must equal the rate of the liquid being poured in;<br />
281<br />
that is, dV =0. Thus, the rate at which we should pour the liquid into the container is<br />
dt<br />
kπrl =<br />
256 + 375π<br />
250π √ 281 · π · 25 8 · 5 √ 281 256 + 375π<br />
= ≈ 11.204 cm 3 /min<br />
8 128
F.<br />
TX.10
F.<br />
TX.10<br />
4 APPLICATIONS OF DIFFERENTIATION<br />
4.1 Maximum and Minimum Values<br />
1. Afunctionf has an absolute minimum at x = c if f(c) is the smallest function value on the entire domain of f, whereas<br />
f has a local minimum at c if f(c) is the smallest function value when x is near c.<br />
3. Absolute maximum at s, absolute minimum at r,localmaximumatc, local minima at b and r, neither a maximum nor a<br />
minimum at a and d.<br />
5. Absolute maximum value is f(4) = 5; there is no absolute minimum value; local maximum values are f(4) = 5 and<br />
f(6) = 4; local minimum values are f(2) = 2 and f(1) = f(5) = 3.<br />
7. Absolute minimum at 2, absolute maximum at 3,<br />
local minimum at 4<br />
9. Absolute maximum at 5, absolute minimum at 2,<br />
local maximum at 3, local minima at 2 and 4<br />
11. (a) (b) (c)<br />
13. (a) Note: By the Extreme Value Theorem,<br />
f must not be continuous; because if it<br />
were, it would attain an absolute<br />
minimum.<br />
(b)<br />
147
F.<br />
148 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
15. f(x) =8− 3x, x ≥ 1. Absolute maximum f(1) = 5;no<br />
local maximum. No absolute or local minimum.<br />
TX.10<br />
17. f(x) =x 2 , 0
F.<br />
TX.10<br />
SECTION 4.1 MAXIMUM AND MINIMUM VALUES ¤ 149<br />
29. f(x) =5x 2 +4x ⇒ f 0 (x) =10x +4. f 0 (x) =0 ⇒ x = − 2 ,so− 2 5 5<br />
is the only critical number.<br />
31. f(x) =x 3 +3x 2 − 24x ⇒ f 0 (x) =3x 2 +6x − 24 = 3(x 2 +2x − 8).<br />
f 0 (x) =0 ⇒ 3(x +4)(x − 2) = 0 ⇒ x = −4, 2. These are the only critical numbers.<br />
33. s(t) =3t 4 +4t 3 − 6t 2 ⇒ s 0 (t) =12t 3 +12t 2 − 12t. s 0 (t) =0 ⇒ 12t(t 2 + t − 1) ⇒<br />
t =0 or t 2 + t − 1=0. Using the quadratic formula to solve the latter equation gives us<br />
t = −1 ± 1 2 − 4(1)(−1)<br />
2(1)<br />
= −1 ± √ 5<br />
2<br />
≈ 0.618, −1.618. The three critical numbers are 0, −1 ± √ 5<br />
.<br />
2<br />
35. g(y) =<br />
y − 1<br />
y 2 − y +1<br />
⇒<br />
g 0 (y) = (y2 − y +1)(1)− (y − 1)(2y − 1)<br />
= y2 − y +1− (2y 2 − 3y +1)<br />
= −y2 +2y y(2 − y)<br />
=<br />
(y 2 − y +1) 2 (y 2 − y +1) 2 (y 2 − y +1)<br />
2<br />
(y 2 − y +1) . 2<br />
g 0 (y) =0 ⇒ y =0, 2. The expression y 2 − y +1is never equal to 0,sog 0 (y) exists for all real numbers.<br />
The critical numbers are 0 and 2.<br />
37. h(t) =t 3/4 − 2t 1/4 ⇒ h 0 (t) = 3 4 t−1/4 − 2 4 t−3/4 = 1 4 t−3/4 (3t 1/2 − 2) = 3 √ t − 2<br />
4 4√ t 3 .<br />
h 0 (t) =0 ⇒ 3 √ t =2 ⇒ √ t = 2 3<br />
⇒ t = 4 9 . h0 (t) does not exist at t =0, so the critical numbers are 0 and 4 9 .<br />
39. F (x) =x 4/5 (x − 4) 2 ⇒<br />
F 0 (x) =x 4/5 · 2(x − 4) + (x − 4) 2 · 4<br />
5 x−1/5 = 1 5 x−1/5 (x − 4)[5 · x · 2+(x − 4) · 4]<br />
=<br />
(x − 4)(14x − 16) 2(x − 4)(7x − 8)<br />
=<br />
5x 1/5 5x 1/5<br />
F 0 (x) =0 ⇒ x =4, 8 7 . F 0 (0) does not exist. Thus, the three critical numbers are 0, 8 7 ,and4.<br />
41. f(θ) =2cosθ +sin 2 θ ⇒ f 0 (θ) =−2sinθ +2sinθ cos θ. f 0 (θ) =0 ⇒ 2sinθ (cos θ − 1) = 0 ⇒ sin θ =0<br />
or cos θ =1 ⇒ θ = nπ [n an integer] or θ =2nπ. The solutions θ = nπ include the solutions θ =2nπ, so the critical<br />
numbers are θ = nπ.<br />
43. f(x) =x 2 e −3x ⇒ f 0 (x) =x 2 (−3e −3x )+e −3x (2x) =xe −3x (−3x +2). f 0 (x) =0 ⇒ x =0, 2 3<br />
[e −3x is never equal to 0]. f 0 (x) always exists, so the critical numbers are 0 and 2 3 .<br />
45. The graph of f 0 (x) =5e −0.1|x| sin x − 1 has 10 zeros and exists<br />
everywhere, so f has 10 critical numbers.
F.<br />
150 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
47. f(x) =3x 2 − 12x +5, [0, 3]. f 0 (x) =6x − 12 = 0 ⇔ x =2. Applying the Closed Interval Method, we find that<br />
f(0) = 5, f(2) = −7,andf(3) = −4. Sof(0) = 5 is the absolute maximum value and f(2) = −7 is the absolute minimum<br />
value.<br />
49. f(x) =2x 3 − 3x 2 − 12x +1, [−2, 3]. f 0 (x) =6x 2 − 6x − 12 = 6(x 2 − x − 2) = 6(x − 2)(x +1)=0 ⇔<br />
x =2, −1.<br />
f(−2) = −3, f(−1) = 8, f(2) = −19,andf(3) = −8. Sof(−1) = 8 is the absolute maximum value and<br />
f(2) = −19 is the absolute minimum value.<br />
51. f(x) =x 4 − 2x 2 +3, [−2, 3]. f 0 (x) =4x 3 − 4x =4x(x 2 − 1) = 4x(x +1)(x − 1) = 0 ⇔ x = −1, 0, 1.<br />
f(−2) = 11, f(−1) = 2, f(0) = 3, f(1) = 2, f(3) = 66. Sof(3) = 66 is the absolute maximum value and f(±1) = 2 is<br />
the absolute minimum value.<br />
53. f(x) = x , [0, 2].<br />
x 2 +1 f<br />
0 (x) = (x2 +1)− x(2x)<br />
= 1 − x2<br />
=0 ⇔ x = ±1,but−1 is not in [0, 2]. f(0) = 0,<br />
(x 2 +1) 2 (x 2 +1)<br />
2<br />
f(1) = 1 , f(2) = 2 .Sof(1) = 1 2 5 2<br />
is the absolute maximum value and f(0) = 0 is the absolute minimum value.<br />
55. f(t) =t √ 4 − t 2 , [−1, 2].<br />
f 0 (t) =t · 1 (4 − 2 t2 ) −1/2 (−2t)+(4− t 2 ) 1/2 · 1= √ −t2 + √ 4 − t 2 = −t2 +(4− t 2 )<br />
√ = √ 4 − 2t2 .<br />
4 − t<br />
2 4 − t<br />
2 4 − t<br />
2<br />
f 0 (t) =0 ⇒ 4 − 2t 2 =0 ⇒ t 2 =2 ⇒ t = ± √ 2,butt = − √ 2 is not in the given interval, [−1, 2].<br />
f 0 (t) does not exist if 4 − t 2 =0 ⇒ t = ±2, but−2 is not in the given interval. f(−1) = − √ 3, f √ 2 =2,and<br />
f(2) = 0. Sof √ 2 =2is the absolute maximum value and f(−1) = − √ 3 is the absolute minimum value.<br />
57. f(t) =2cost +sin2t, [0, π/2].<br />
f 0 (t) =−2sint +cos2t · 2=−2sint +2(1− 2sin 2 t)=−2(2 sin 2 t +sint − 1) = −2(2 sin t − 1)(sin t +1).<br />
f 0 (t) =0 ⇒ sin t = 1 or sin t = −1 ⇒ t = π . f(0) = 2, f( π )=√ √ √<br />
2 6 6<br />
3+ 1 2 3=<br />
3<br />
2 3 ≈ 2.60,andf(<br />
π<br />
)=0. 2<br />
So f( π )= √ 3<br />
6 2 3 is the absolute maximum value and f(<br />
π<br />
2<br />
)=0is the absolute minimum value.<br />
59. f(x) =xe −x2 /8 , [−1, 4]. f 0 (x) =x · e −x2 /8 · (− x 4 )+e−x2 /8 · 1=e −x2 /8 (− x2<br />
4 +1).Sincee−x2 /8 is never 0,<br />
f 0 (x) =0 ⇒ −x 2 /4+1=0 ⇒ 1=x 2 /4 ⇒ x 2 =4 ⇒ x = ±2,but−2 is not in the given interval, [−1, 4].<br />
f(−1) = −e −1/8 ≈−0.88, f(2) = 2e −1/2 ≈ 1.21,andf(4) = 4e −2 ≈ 0.54. Sof(2) = 2e −1/2 is the absolute maximum<br />
value and f(−1) = −e −1/8 is the absolute minimum value.<br />
61. f(x) =ln(x 2 + x +1), [−1, 1]. f 0 1<br />
(x) =<br />
x 2 + x +1 · (2x +1)=0 ⇔ x = − 1 2 .Sincex2 + x +1> 0 for all x,the<br />
domain of f and f 0 is R. f(−1)=ln1=0, f <br />
− 1 2 =ln<br />
3<br />
≈ −0.29,andf(1) = ln 3 ≈ 1.10. Sof(1) = ln 3 ≈ 1.10 is<br />
4<br />
the absolute maximum value and f <br />
− 1 2 =ln<br />
3<br />
4<br />
≈ −0.29 is the absolute minimum value.
F.<br />
TX.10<br />
SECTION 4.1 MAXIMUM AND MINIMUM VALUES ¤ 151<br />
63. f(x) =x a (1 − x) b , 0 ≤ x ≤ 1, a>0, b>0.<br />
f 0 (x) =x a · b(1 − x) b−1 (−1) + (1 − x) b · ax a−1 = x a−1 (1 − x) b−1 [x · b(−1) + (1 − x) · a]<br />
= x a−1 (1 − x) b−1 (a − ax − bx)<br />
At the endpoints, we have f(0) = f(1) = 0 [the minimum value of f ]. In the interval (0, 1), f 0 (x) =0 ⇔ x = a<br />
a + b .<br />
<br />
f<br />
a<br />
a + b<br />
<br />
=<br />
a<br />
So f<br />
a + b<br />
a<br />
a + b<br />
a <br />
1 − a b<br />
=<br />
a + b<br />
<br />
a a<br />
b a + b − a<br />
=<br />
(a + b) a a + b<br />
a a b b<br />
=<br />
is the absolute maximum value.<br />
a+b<br />
(a + b)<br />
a a<br />
(a + b) a ·<br />
b b<br />
(a + b) = a a b b<br />
b (a + b) . a+b<br />
65. (a) From the graph, it appears that the absolute maximum value is about<br />
f(−0.77) = 2.19, and the absolute minimum value is about<br />
f(0.77) = 1.81.<br />
<br />
(b) f(x) =x 5 − x 3 +2 ⇒ f 0 (x) =5x 4 − 3x 2 = x 2 (5x 2 − 3). Sof 0 3<br />
(x) =0 ⇒ x =0, ± . 5<br />
5 3 <br />
3<br />
3<br />
f − = −<br />
5<br />
5 −<br />
3<br />
−<br />
5 +2=− 3<br />
<br />
2 3<br />
+ 3 3<br />
+2= 3<br />
− <br />
9 3<br />
+2= 6 3<br />
+2(maximum)<br />
5 5 5 5 5 25 5 25 5<br />
<br />
and similarly, f<br />
3<br />
5<br />
<br />
= − 6 25<br />
<br />
3<br />
5 +2(minimum).<br />
67. (a) From the graph, it appears that the absolute maximum value is about<br />
f(0.75) = 0.32, and the absolute minimum value is f(0) = f(1) = 0;<br />
that is, at both endpoints.<br />
(b) f(x) =x √ x − x 2 ⇒ f 0 1 − 2x<br />
(x) =x ·<br />
2 √ x − x + √ x − x 2 = (x − 2x2 )+(2x − 2x 2 )<br />
2 2 √ 3x − 4x2<br />
=<br />
x − x 2 2 √ x − x . 2<br />
So f 0 (x) =0 ⇒ 3x − 4x 2 =0 ⇒ x(3 − 4x) =0 ⇒ x =0or 3 . 4<br />
f(0) = f(1) = 0 (minimum), and f <br />
3<br />
4 =<br />
3<br />
4<br />
<br />
3<br />
4 − 3<br />
4<br />
2<br />
= 3 4<br />
<br />
3<br />
= 3 √ 3<br />
(maximum).<br />
16 16<br />
69. The density is defined as ρ = mass<br />
volume = 1000<br />
V (T ) (in g/cm3 ). But a critical point of ρ will also be a critical point of V<br />
[since dρ<br />
dT = −1000V −2 dV<br />
dT<br />
and V is never 0],andV is easier to differentiate than ρ.<br />
V (T )=999.87 − 0.06426T +0.0085043T 2 − 0.0000679T 3 ⇒ V 0 (T )=−0.06426 + 0.0170086T − 0.0002037T 2 .<br />
Setting this equal to 0 and using the quadratic formula to find T ,weget<br />
T = −0.0170086 ± √ 0.0170086 2 − 4 · 0.0002037 · 0.06426<br />
2(−0.0002037)<br />
≈ 3.9665 ◦ Cor79.5318 ◦ C. Since we are only interested
F.<br />
152 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
in the region 0 ◦ C ≤ T ≤ 30 ◦ C, we check the density ρ at the endpoints and at 3.9665 ◦ C: ρ(0) ≈ 1000<br />
999.87 ≈ 1.00013;<br />
ρ(30) ≈ 1000<br />
1000<br />
≈ 0.99625; ρ(3.9665) ≈ ≈ 1.000255. So water has its maximum density at<br />
1003.7628 999.7447<br />
71. Let a = −0.000 032 37, b =0.000 903 7, c = −0.008 956, d =0.03629, e = −0.04458, andf =0.4074.<br />
Then S(t) =at 5 + bt 4 + ct 3 + dt 2 + et + f and S 0 (t) =5at 4 +4bt 3 +3ct 2 +2dt + e.<br />
We now apply the Closed Interval Method to the continuous function S on the interval 0 ≤ t ≤ 10. SinceS 0 exists for all t,<br />
the only critical numbers of S occur when S 0 (t) =0.Weusearootfinder on a CAS (or a graphing device) to find that<br />
S 0 (t) =0when t 1 ≈ 0.855, t 2 ≈ 4.618, t 3 ≈ 7.292, andt 4 ≈ 9.570. The values of S at these critical numbers are<br />
S(t 1) ≈ 0.39, S(t 2) ≈ 0.43645, S(t 3) ≈ 0.427,andS(t 4) ≈ 0.43641. The values of S at the endpoints of the interval are<br />
S(0) ≈ 0.41 and S(10) ≈ 0.435. Comparing the six numbers, we see that sugar was most expensive at t 2 ≈ 4.618<br />
(corresponding roughly to March 1998) and cheapest at t 1 ≈ 0.855 (June 1994).<br />
73. (a) v(r) =k(r 0 − r)r 2 = kr 0 r 2 − kr 3 ⇒ v 0 (r) =2kr 0 r − 3kr 2 . v 0 (r) =0 ⇒ kr(2r 0 − 3r) =0 ⇒<br />
r =0or 2 r 3 0 (but 0 is not in the interval). Evaluating v at 1 r 2 0, 2 r 3 0,andr 0 ,wegetv 1<br />
r <br />
2 0 =<br />
1<br />
8 kr3 0, v 2<br />
r <br />
3 0 =<br />
4<br />
27 kr3 0,<br />
and v(r 0 )=0.Since 4 27 > 1 8 , v attains its maximum value at r = 2 3 r 0. This supports the statement in the text.<br />
(b) From part (a), the maximum value of v is 4 27 kr3 0.<br />
(c)<br />
75. f(x) =x 101 + x 51 + x +1 ⇒ f 0 (x) =101x 100 +51x 50 +1≥ 1 for all x,sof 0 (x) =0has no solution. Thus, f(x)<br />
has no critical number, so f(x) can have no local maximum or minimum.<br />
77. If f has a local minimum at c,theng(x) =−f(x) has a local maximum at c,sog 0 (c) =0by the case of Fermat’s Theorem<br />
proved in the text. Thus, f 0 (c) =−g 0 (c) =0.<br />
4.2 The Mean Value Theorem<br />
1. f(x) =5− 12x +3x 2 , [1, 3]. Sincef is a polynomial, it is continuous and differentiable on R, so it is continuous on [1, 3]<br />
and differentiable on (1, 3). Alsof(1) = −4 =f(3). f 0 (c) =0 ⇔ −12 + 6c =0 ⇔ c =2, which is in the open<br />
interval (1, 3),soc =2satisfies the conclusion of Rolle’s Theorem.
F.<br />
TX.10<br />
SECTION 4.2 THE MEAN VALUE THEOREM ¤ 153<br />
3. f(x) = √ x − 1 3<br />
x, [0, 9]. f, being the difference of a root function and a polynomial, is continuous and differentiable<br />
on [0, ∞), so it is continuous on [0, 9] and differentiable on (0, 9). Also,f(0) = 0 = f(9). f 0 (c) =0<br />
⇔<br />
1<br />
2 √ c − 1 3 =0 ⇔ 2 √ c =3 ⇔ √ c = 3 2<br />
conclusion of Rolle’s Theorem.<br />
⇒<br />
c = 9 4 , which is in the open interval (0, 9),soc = 9 satisfies the<br />
4<br />
5. f(x) =1− x 2/3 . f(−1) = 1 − (−1) 2/3 =1− 1=0=f(1). f 0 (x) =− 2 3 x−1/3 ,sof 0 (c) =0has no solution. This<br />
does not contradict Rolle’s Theorem, since f 0 (0) does not exist, and so f is not differentiable on (−1, 1).<br />
7.<br />
f(8) − f(0)<br />
8 − 0<br />
= 6 − 4<br />
8<br />
= 1 4 . The values of c which satisfy f 0 (c) = 1 seem to be about c =0.8, 3.2, 4.4,and6.1.<br />
4<br />
9. (a), (b) The equation of the secant line is<br />
(c) f(x) =x +4/x ⇒ f 0 (x) =1− 4/x 2 .<br />
y − 5= 8.5 − 5<br />
8 − 1 (x − 1) ⇔ y = 1 x + 9 . So f 0 (c) = 1 2<br />
⇒ c 2 =8 ⇒ c =2 √ 2,and<br />
2 2<br />
f(c) =2 √ 2+ 4<br />
2 √ =3√ 2. Thus, an equation of the<br />
2<br />
tangent line is y − 3 √ √ <br />
2= 1 2 x − 2 2 ⇔<br />
y = 1 2 x +2√ 2.<br />
11. f(x) =3x 2 +2x +5, [−1, 1]. f is continuous on [−1, 1] and differentiable on (−1, 1) since polynomials are continuous<br />
and differentiable on R.<br />
c =0, which is in (−1, 1).<br />
f 0 (c) =<br />
f(b) − f(a)<br />
b − a<br />
⇔<br />
6c +2=<br />
f(1) − f(−1)<br />
1 − (−1)<br />
= 10 − 6<br />
2<br />
=2 ⇔ 6c =0 ⇔<br />
13. f(x) =e −2x , [0, 3]. f is continuous and differentiable on R, so it is continuous on [0, 3] and differentiable on (0, 3).<br />
f 0 f(b) − f(a)<br />
(c) = ⇔ −2e −2c = e−6 − e 0<br />
⇔ e −2c = 1 − e−6<br />
1 − e<br />
−6<br />
<br />
⇔ −2c =ln<br />
⇔<br />
b − a<br />
3 − 0<br />
6<br />
6<br />
c = − 1 1 − e<br />
−6<br />
2 ln ≈ 0.897, which is in (0, 3).<br />
6
F.<br />
154 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
15. f(x) =(x − 3) −2 ⇒ f 0 (x) =−2(x − 3) −3 . f(4) − f(1) = f 0 (c)(4 − 1) ⇒ 1 1 − 1<br />
2 (−2) = −2<br />
2 (c − 3) · 3 ⇒<br />
3<br />
3<br />
4 = −6 ⇒ (c − 3) 3 = −8 ⇒ c − 3=−2 ⇒ c =1, which is not in the open interval (1, 4). This does not<br />
(c − 3) 3<br />
contradict the Mean Value Theorem since f is not continuous at x =3.<br />
17. Let f(x) =1+2x + x 3 +4x 5 .Thenf(−1) = −6 < 0 and f(0) = 1 > 0. Since f is a polynomial, it is continuous, so the<br />
Intermediate Value Theorem says that there is a number c between −1 and 0 such that f(c) =0. Thus, the given equation has<br />
a real root. Suppose the equation has distinct real roots a and b with a
F.<br />
TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 155<br />
27. We use Exercise 26 with f(x) = √ 1+x, g(x) =1+ 1 2<br />
x,anda =0.Noticethatf(0) = 1 = g(0) and<br />
f 0 (x) =<br />
1<br />
2 √ 1+x < 1 2 = g0 (x) for x>0.SobyExercise26,f(b)
F.<br />
156 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
5. (a) Since f 0 (x) > 0 on (1, 5), f is increasing on this interval. Since f 0 (x) < 0 on (0, 1) and (5, 6), f is decreasing on these<br />
intervals.<br />
(b) Since f 0 (x) =0at x =1and f 0 changes from negative to positive there, f changes from decreasing to increasing and has<br />
a local minimum at x =1.Sincef 0 (x) =0at x =5and f 0 changes from positive to negative there, f changes from<br />
increasing to decreasing and has a local maximum at x =5.<br />
7. There is an inflection point at x =1because f 00 (x) changes from negative to positive there, and so the graph of f changes<br />
from concave downward to concave upward. There is an inflection point at x =7because f 00 (x) changes from positive to<br />
negative there, and so the graph of f changes from concave upward to concave downward.<br />
9. (a) f(x) =2x 3 +3x 2 − 36x ⇒ f 0 (x) =6x 2 +6x − 36 = 6(x 2 + x − 6) = 6(x +3)(x − 2).<br />
We don’t need to include the “6”inthecharttodeterminethesignoff 0 (x).<br />
Interval x +3 x − 2 f 0 (x) f<br />
x− 1 2 ,and<br />
f 00 (x) < 0 ⇔ x
F.<br />
TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 157<br />
13. (a) f(x) =sinx +cosx, 0 ≤ x ≤ 2π. f 0 (x) =cosx − sin x =0 ⇒ cos x =sinx ⇒ 1= sin x<br />
cos x<br />
tan x =1 ⇒ x = π 4 or 5π 4 . Thus, f 0 (x) > 0 ⇔ cos x − sin x>0 ⇔ cos x>sin x ⇔ 0 ln 1 2<br />
⇔<br />
x> 1 (ln 1 − ln 2) ⇔ x>− 1 ln 2 [≈ −0.23] andf 0 3 3<br />
(x) < 0 if x 0 [the sum of two positive terms].<br />
point of inflection.<br />
Thus, f is concave upward on (−∞, ∞) and there is no<br />
17. (a) y = f(x) = ln x √<br />
x<br />
.(Notethatf is only defined for x>0.)<br />
f 0 (x) =<br />
√<br />
x (1/x) − ln x<br />
<br />
1<br />
2 x−1/2 <br />
x<br />
=<br />
1<br />
√ − ln x<br />
x 2 √ x<br />
· 2 √ x<br />
x 2 √ x = 2 − ln x > 0 ⇔ 2 − ln x>0 ⇔<br />
2x 3/2<br />
ln xe 8/3 ,sof is concave upward on (e 8/3 , ∞) and<br />
<br />
concave downward on (0,e 8/3 ).Thereisaninflection point at<br />
e 8/3 , 8 3 e−4/3 ≈ (14.39, 0.70).
F.<br />
158 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
19. f(x) =x 5 − 5x +3 ⇒ f 0 (x) =5x 4 − 5=5(x 2 +1)(x +1)(x − 1).<br />
First Derivative Test: f 0 (x) < 0 ⇒ −1 1 or x 0 ⇒ f(1) = −1 is a local minimum value.<br />
Preference: For this function, the two tests are equally easy.<br />
21. f(x) =x + √ 1 − x ⇒ f 0 (x) =1+ 1 2 (1 − x)−1/2 (−1) = 1 −<br />
1<br />
2 √ .Notethatf is defined for 1 − x ≥ 0;thatis,<br />
1 − x<br />
for x ≤ 1. f 0 (x) =0 ⇒ 2 √ 1 − x =1 ⇒ √ 1 − x = 1 2<br />
⇒ 1 − x = 1 4<br />
⇒ x = 3 4 . f 0 does not exist at x =1,<br />
but we can’t have a local maximum or minimum at an endpoint.<br />
First Derivative Test: f 0 (x) > 0 ⇒ x< 3 and f 0 (x) < 0 ⇒ 3
F.<br />
TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 159<br />
29. The function must be always decreasing (since the first derivative is always negative)<br />
and concave downward (since the second derivative is always negative).<br />
31. (a) f is increasing where f 0 is positive, that is, on (0, 2), (4, 6),and(8, ∞); and decreasing where f 0 is negative, that is,<br />
on (2, 4) and (6, 8).<br />
(b) f has local maxima where f 0 changes from positive to negative, at x =2and at x =6, and local minima where f 0 changes<br />
from negative to positive, at x =4and at x =8.<br />
(c) f is concave upward (CU) where f 0 is increasing, that is, on (3, 6) and (6, ∞),<br />
(e)<br />
and concave downward (CD) where f 0 is decreasing, that is, on (0, 3).<br />
(d) There is a point of inflection where f changes from being CD to being CU, that<br />
is, at x =3.<br />
33. (a) f(x) =2x 3 − 3x 2 − 12x ⇒ f 0 (x) =6x 2 − 6x − 12 = 6(x 2 − x − 2) = 6(x − 2)(x +1).<br />
f 0 (x) > 0 ⇔ x2 and f 0 (x) < 0 ⇔ −1 0 ⇔ x
F.<br />
160 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
(c) f 00 (x) =4− 12x 2 =4(1− 3x 2 ). f 00 (x) =0 ⇔ 1 − 3x 2 =0 ⇔<br />
(d)<br />
x 2 = 1 3<br />
⇔ x = ±1/ √ 3. f 00 (x) > 0 on −1/ √ 3, 1/ √ 3 and f 00 (x) < 0<br />
on −∞, −1/ √ 3 and 1/ √ 3, ∞ .Sof is concave upward on<br />
<br />
−1/<br />
√<br />
3, 1/<br />
√<br />
3<br />
<br />
and f is concave downward on<br />
<br />
−∞, −1/<br />
√<br />
3<br />
<br />
and<br />
√ √ 1/ 3, ∞ . f ±1/ 3 =2+<br />
2<br />
− 1 = 23 . There are points of inflection<br />
3 9 9<br />
at ±1/ √ <br />
3, 23<br />
9 .<br />
37. (a) h(x) =(x +1) 5 − 5x − 2 ⇒ h 0 (x) =5(x +1) 4 − 5. h 0 (x) =0 ⇔ 5(x +1) 4 =5 ⇔ (x +1) 4 =1 ⇒<br />
(x +1) 2 =1 ⇒ x +1=1or x +1=−1 ⇒ x =0or x = −2. h 0 (x) > 0 ⇔ x0 and<br />
h 0 (x) < 0 ⇔ −2 −1 and<br />
h 00 (x) < 0 ⇔ x 0 for x>−2 and A 0 (x) < 0 for −3 −3,soA is concave upward on (−3, ∞). Thereisnoinflection point.<br />
41. (a) C(x) =x 1/3 (x +4)=x 4/3 +4x 1/3 ⇒ C 0 (x) = 4 3 x1/3 + 4 3 x−2/3 = 4 3 x−2/3 (x +1)=<br />
4(x +1)<br />
3 3√ x 2 . C 0 (x) > 0 if<br />
−1
F.<br />
TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 161<br />
(b) C(−1) = −3 is a local minimum value.<br />
(c) C 00 (x) = 4 9 x−2/3 − 8 9 x−5/3 = 4 9 x−5/3 (x − 2) =<br />
4(x − 2)<br />
9 3√ x 5 .<br />
C 00 (x) < 0 for 0 0 ⇔ π
F.<br />
162 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
(d) f 00 (x) = (x2 − 1) 2 (−2) − (−2x) · 2(x 2 − 1)(2x)<br />
[(x 2 − 1) 2 ] 2<br />
= 2(x2 − 1)[−(x 2 − 1) + 4x 2 ]<br />
(x 2 − 1) 4 = 2(3x2 +1)<br />
(x 2 − 1) 3 .<br />
The sign of f 00 (x) is determined by the denominator; that is, f 00 (x) > 0 if<br />
|x| > 1 and f 00 (x) < 0 if |x| < 1. Thus,f is CU on (−∞, −1) and (1, ∞),<br />
and f is CD on (−1, 1). Therearenoinflection points.<br />
(e)<br />
√<br />
47. (a) lim x2 +1− x = ∞ and<br />
x→−∞<br />
√<br />
lim x2 +1− x √<br />
= lim x2 +1− x √ x 2 +1+x<br />
√<br />
x→∞<br />
x→∞<br />
x2 +1+x = lim 1<br />
√ =0,soy =0is a HA.<br />
x→∞ x2 +1+x<br />
(b) f(x) = √ x 2 +1− x ⇒ f 0 x<br />
x<br />
(x) = √ − 1. Since√ x2 +1 x2 +1 < 1 for all x, f 0 (x) < 0,sof is decreasing on R.<br />
(c) No minimum or maximum<br />
(d) f 00 (x) = (x2 +1) 1/2 (1) − x · 1<br />
2 (x2 +1) −1/2 (2x)<br />
√<br />
x2 +1 2<br />
(e)<br />
(x 2 +1) 1/2 x 2<br />
−<br />
(x<br />
=<br />
2 +1) 1/2<br />
x 2 +1<br />
so f is CU on R. NoIP<br />
= (x2 +1)− x 2<br />
(x 2 +1) 3/2 =<br />
1<br />
> 0,<br />
(x 2 +1)<br />
3/2<br />
49. f(x) =ln(1− ln x) is defined when x>0 (so that ln x is defined) and 1 − ln x>0 [so that ln(1 − ln x) is defined].<br />
The second condition is equivalent to 1 > ln x ⇔ x 0 ⇔ ln x
F.<br />
TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 163<br />
(d) f 00 (x) = (x +1)2 e −1/(x+1) 1/(x +1) 2 − e −1/(x+1) [2(x +1)]<br />
[(x +1) 2 ] 2<br />
= e−1/(x+1) [1 − (2x +2)]<br />
(x +1) 4 = − e−1/(x+1) (2x +1)<br />
(x +1) 4 ⇒<br />
(e)<br />
f 00 (x) > 0 ⇔ 2x +1< 0 ⇔ x 0 ⇒ (x − 3) 5 > 0 ⇒ x − 3 > 0 ⇒ x>3. Thus,f is increasing on the interval (3, ∞).<br />
55. (a) From the graph, we get an estimate of f(1) ≈ 1.41 as a local maximum<br />
value, and no local minimum value.<br />
f(x) = x +1 √<br />
x2 +1<br />
⇒ f 0 (x) =<br />
1 − x<br />
(x 2 +1) 3/2 .<br />
f 0 (x) =0 ⇔ x =1. f(1) = 2 √<br />
2<br />
= √ 2 is the exact value.<br />
(b) From the graph in part (a), f increases most rapidly somewhere between x = − 1 and x = − 1 .Tofind the exact value,<br />
2 4<br />
we need to find the maximum value of f 0 , which we can do by finding the critical numbers of f 0 .<br />
f 00 (x) = 2x2 − 3x − 1<br />
=0 ⇔ x = 3 ± √ 17<br />
. x = 3+√ 17<br />
corresponds to the minimum value of f 0 .<br />
(x 2 +1) 5/2 4<br />
4<br />
<br />
Themaximumvalueoff 0 3 −<br />
is at<br />
√ <br />
17 7<br />
4<br />
, − √ 17<br />
≈ (−0.28, 0.69).<br />
6 6<br />
57. f(x) =cosx + 1 2 cos 2x ⇒ f 0 (x) =− sin x − sin 2x ⇒ f 00 (x) =− cos x − 2cos2x<br />
(a)<br />
From the graph of f,itseemsthatf is CD on (0, 1),CUon(1, 2.5),CDon<br />
(2.5, 3.7),CUon(3.7, 5.3), and CD on (5.3, 2π). The points of inflection<br />
appear to be at (1, 0.4), (2.5, −0.6), (3.7, −0.6),and(5.3, 0.4).<br />
(b)<br />
From the graph of f 00 (and zooming in near the zeros), it seems that f is CD<br />
on (0, 0.94),CUon(0.94, 2.57),CDon(2.57, 3.71),CUon(3.71, 5.35),<br />
and CD on (5.35, 2π). Refined estimates of the inflection points are<br />
(0.94, 0.44), (2.57, −0.63), (3.71, −0.63),and(5.35, 0.44).<br />
59. In Maple, we define f andthenusethecommand<br />
plot(diff(diff(f,x),x),x=-2..2);. InMathematica,wedefine f<br />
andthenusePlot[Dt[Dt[f,x],x],{x,-2,2}]. Weseethatf 00 > 0 for<br />
x0.0 [≈ 0.03] andf 00 < 0 for −0.6
F.<br />
164 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
61. (a) The rate of increase of the population is initially very small, then gets larger until it reaches a maximum at about<br />
t =8hours, and decreases toward 0 as the population begins to level off.<br />
(b) The rate of increase has its maximum value at t =8hours.<br />
(c) The population function is concave upward on (0, 8) and concave downward on (8, 18).<br />
(d) At t =8, the population is about 350,sotheinflection point is about (8, 350).<br />
63. Most students learn more in the third hour of studying than in the eighth hour, so K(3) − K(2) is larger than K(8) − K(7).<br />
In other words, as you begin studying for a test, the rate of knowledge gain is large and then starts to taper off, so K 0 (t)<br />
decreases and the graph of K is concave downward.<br />
65. S(t) =At p e −kt with A =0.01, p =4,andk =0.07. Wewillfind the<br />
zeros of f 00 for f(t) =t p e −kt .<br />
f 0 (t) =t p (−ke −kt )+e −kt (pt p−1 )=e −kt (−kt p + pt p−1 )<br />
f 00 (t)=e −kt (−kpt p−1 + p(p − 1)t p−2 )+(−kt p + pt p−1 )(−ke −kt )<br />
= t p−2 e −kt [−kpt + p(p − 1) + k 2 t 2 − kpt]<br />
= t p−2 e −kt (k 2 t 2 − 2kpt + p 2 − p)<br />
Using the given values of p and k gives us f 00 (t) =t 2 e −0.07t (0.0049t 2 − 0.56t +12).SoS 00 (t) =0.01f 00 (t) and its zeros<br />
are t =0and the solutions of 0.0049t 2 − 0.56t +12=0, which are t 1 = 200<br />
7<br />
≈ 28.57 and t 2 = 600<br />
7<br />
≈ 85.71.<br />
At t 1 minutes, the rate of increase of the level of medication in the bloodstream is at its greatest and at t 2 minutes, the rate of<br />
decrease is the greatest.<br />
67. f(x) =ax 3 + bx 2 + cx + d ⇒ f 0 (x) =3ax 2 +2bx + c.<br />
We are given that f(1) = 0 and f(−2) = 3,sof(1) = a + b + c + d =0and<br />
f(−2) = −8a +4b − 2c + d =3.Alsof 0 (1) = 3a +2b + c =0and<br />
f 0 (−2) = 12a − 4b + c =0by Fermat’s Theorem. Solving these four equations, we get<br />
a = 2 , b = 1 , c = − 4 , d = 7 , so the function is f(x) = 1<br />
9 3 3 9 9 2x 3 +3x 2 − 12x +7 .<br />
69. y = 1+x ⇒ y 0 = (1 + x2 )(1) − (1 + x)(2x) 1 − 2x − x2<br />
= ⇒<br />
1+x 2 (1 + x 2 ) 2 (1 + x 2 ) 2<br />
y 00 = (1 + x2 ) 2 (−2 − 2x) − (1 − 2x − x 2 ) · 2(1 + x 2 )(2x)<br />
= 2(1 + x2 )[(1 + x 2 )(−1 − x) − (1 − 2x − x 2 )(2x)]<br />
[(1 + x 2 ) 2 ] 2 (1 + x 2 ) 4<br />
= 2(−1 − x − x2 − x 3 − 2x +4x 2 +2x 3 )<br />
= 2(x3 +3x 2 − 3x − 1)<br />
= 2(x − 1)(x2 +4x +1)<br />
(1 + x 2 ) 3 (1 + x 2 ) 3 (1 + x 2 ) 3<br />
So y 00 =0 ⇒ x =1, −2 ± √ 3.Leta = −2 − √ 3, b = −2+ √ 3,andc =1. We can show that f(a) = 1 4<br />
<br />
1 −<br />
√<br />
3<br />
<br />
,
F.<br />
TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 165<br />
f(b) = 1 4<br />
<br />
1+<br />
√<br />
3<br />
<br />
,andf(c) =1. To show that these three points of inflection lie on one straight line, we’ll show that the<br />
slopes m ac and m bc are equal.<br />
m ac =<br />
m bc =<br />
f(c) − f(a)<br />
c − a<br />
f(c) − f(b)<br />
c − b<br />
= 1 − √ <br />
1<br />
4 1 − 3<br />
3<br />
1 − −2 − √ 3 = + √ 1<br />
4 4 3<br />
3+ √ 3 = 1 4<br />
= 1 − √ <br />
1<br />
4 1+ 3<br />
3<br />
1 − −2+ √ 3 = − √ 1<br />
4 4 3<br />
3 − √ 3 = 1 4<br />
71. Suppose that f is differentiable on an interval I and f 0 (x) > 0 for all x in I except x = c. Toshowthatf is increasing on I,<br />
let x 1, x 2 be two numbers in I with x 1
F.<br />
166 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
77. Let the cubic function be f(x) =ax 3 + bx 2 + cx + d ⇒ f 0 (x) =3ax 2 +2bx + c ⇒ f 00 (x) =6ax +2b.<br />
So f is CU when 6ax +2b>0 ⇔ x>−b/(3a),CDwhenx0,so(0, 0) is an inflection point. But g 00 (0) does not<br />
exist.<br />
83. (a) f(x) =x 4 sin 1 x ⇒ f 0 (x) =x 4 cos 1 − 1 <br />
+sin 1 x x 2 x (4x3 )=4x 3 sin 1 x − x2 cos 1 x .<br />
<br />
g(x) =x 4 2+sin 1 <br />
=2x 4 + f(x) ⇒ g 0 (x) =8x 3 + f 0 (x).<br />
x<br />
<br />
h(x) =x 4 −2+sin 1 <br />
= −2x 4 + f(x) ⇒ h 0 (x) =−8x 3 + f 0 (x).<br />
x<br />
It is given that f(0) = 0, sof 0 f(x) − f(0)<br />
x 4 sin 1<br />
(0) = lim<br />
=lim x − 0<br />
=limx 3 sin 1<br />
x→0 x − 0 x→0 x<br />
x→0 x .Since<br />
− x 3 ≤ x 3 sin 1 x ≤ x 3 and lim<br />
x→0<br />
x 3 =0,weseethatf 0 (0) = 0 by the Squeeze Theorem. Also,<br />
g 0 (0) = 8(0) 3 + f 0 (0) = 0 and h 0 (0) = −8(0) 3 + f 0 (0) = 0,so0 is a critical number of f, g,andh.<br />
For x 2n = 1<br />
2nπ<br />
For x 2n+1 =<br />
1<br />
1<br />
[n a nonzero integer], sin =sin2nπ =0and cos =cos2nπ =1,sof 0 (x 2n )=−x 2 2n < 0.<br />
x 2n x 2n<br />
1<br />
(2n +1)π , sin 1<br />
1<br />
=sin(2n +1)π =0and cos =cos(2n +1)π = −1, so<br />
x 2n+1 x 2n+1<br />
f 0 (x 2n+1 )=x 2 2n+1 > 0. Thus,f 0 changes sign infinitely often on both sides of 0.<br />
Next, g 0 (x 2n) =8x 3 2n + f 0 (x 2n) =8x 3 2n − x 2 2n = x 2 2n(8x 2n − 1) < 0 for x 2n < 1 8 ,but<br />
g 0 (x 2n+1 )=8x 3 2n+1 + x 2 2n+1 = x 2 2n+1(8x 2n+1 +1)> 0 for x 2n+1 > − 1 8 ,sog0 changes sign infinitely often on both<br />
sides of 0.<br />
Last, h 0 (x 2n) =−8x 3 2n + f 0 (x 2n) =−8x 3 2n − x 2 2n = −x 2 2n(8x 2n +1)< 0 for x 2n > − 1 8 and<br />
h 0 (x 2n+1 )=−8x 3 2n+1 + x 2 2n+1 = x 2 2n+1(−8x 2n+1 +1)> 0 for x 2n+1 < 1 8 ,soh0 changes sign infinitely often on both<br />
sides of 0.
F.<br />
TX.10SECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE ¤ 167<br />
(b) f(0) = 0 and since sin 1 x and hence x4 sin 1 is both positive and negative inifinitely often on both sides of 0,and<br />
x<br />
arbitrarily close to 0, f has neither a local maximum nor a local minimum at 0.<br />
Since 2+sin 1 <br />
x ≥ 1, g(x) =x4 2+sin 1 <br />
> 0 for x 6= 0,sog(0) = 0 is a local minimum.<br />
x<br />
Since −2+sin 1 <br />
x ≤−1, h(x) =x4 −2+sin 1 <br />
< 0 for x 6= 0,soh(0) = 0 is a local maximum.<br />
x<br />
4.4 Indeterminate Forms and L'Hospital's Rule<br />
Note: The use of l’Hospital’s Rule is indicated by an H above the equal sign:<br />
1. (a) lim<br />
x→a<br />
f(x)<br />
g(x) is an indeterminate form of type 0 0 .<br />
f(x)<br />
(b) lim =0because the numerator approaches 0 while the denominator becomes large.<br />
x→a p(x)<br />
h(x)<br />
(c) lim =0because the numerator approaches a finite number while the denominator becomes large.<br />
x→a p(x)<br />
H<br />
=<br />
p(x)<br />
(d) If lim p(x) =∞ and f(x) → 0 through positive values, then lim<br />
x→a x→a f(x) = ∞. [Forexample,takea =0, p(x) =1/x2 ,<br />
and f(x) =x 2 p(x)<br />
.] If f(x) → 0 through negative values, then lim<br />
x→a f(x) = −∞. [Forexample,takea =0, p(x) =1/x2 ,<br />
and f(x) =−x 2 .] If f(x) → 0 through both positive and negative values, then the limit might not exist. [For example,<br />
take a =0, p(x) =1/x 2 ,andf(x) =x.]<br />
p(x)<br />
(e) lim<br />
x→a q(x) is an indeterminate form of type ∞ ∞ .<br />
3. (a) When x is near a, f(x) is near 0 and p(x) is large, so f(x) − p(x) is large negative. Thus, lim<br />
x→a<br />
[f(x) − p(x)] = −∞.<br />
(b) lim<br />
x→a<br />
[ p(x) − q(x)] is an indeterminate form of type ∞−∞.<br />
(c) When x is near a, p(x) and q(x) are both large, so p(x)+q(x) is large. Thus, lim<br />
x→a<br />
[ p(x)+q(x)] = ∞.<br />
5. This limit has the form 0 0 .Wecansimplyfactorandsimplifytoevaluatethelimit.<br />
x 2 − 1<br />
lim<br />
x→1 x 2 − x =lim (x +1)(x − 1) x +1<br />
= lim = 1+1 =2<br />
x→1 x(x − 1) x→1 x 1<br />
7. This limit has the form 0 . lim x 9 − 1 H 9x 8<br />
0<br />
=lim<br />
x→1 x 5 − 1 x→1 5x = 9 4 5 lim<br />
x→1 x4 = 9 5 (1) = 9 5<br />
9. This limit has the form 0 0 . lim<br />
x→(π/2) +<br />
11. This limit has the form 0 . lim e t − 1<br />
0 t→0 t 3<br />
cos x<br />
1 − sin x<br />
H<br />
=lim<br />
t→0<br />
H − sin x<br />
= lim<br />
x→(π/2) + − cos x =<br />
lim<br />
x→(π/2)<br />
+<br />
tan x = −∞.<br />
e t<br />
3t 2 = ∞ since et → 1 and 3t 2 → 0 + as t → 0.
F.<br />
168 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
13. This limit has the form 0 . lim tan px H p sec 2 px<br />
=lim<br />
0 x→0 tan qx x→0 q sec 2 qx = p(1)2<br />
q(1) = p 2 q<br />
15. This limit has the form ∞ . lim ln x<br />
√<br />
∞<br />
= H lim<br />
x→∞ x<br />
x→∞<br />
1<br />
2<br />
1/x<br />
= lim<br />
x−1/2 x→∞<br />
2<br />
√<br />
x<br />
=0<br />
17. lim<br />
x→0 + [(ln x)/x] =−∞ since ln x →−∞as x → 0+ anddividingbysmallvaluesofx just increases the magnitude of the<br />
quotient (ln x)/x. L’Hospital’s Rule does not apply.<br />
19. This limit has the form ∞ . lim e x<br />
∞ x→∞ x 3<br />
21. This limit has the form 0 . lim e x − 1 − x<br />
0 x→0 x 2<br />
= H e x<br />
lim<br />
x→∞ 3x 2<br />
= H e x<br />
lim<br />
x→∞ 6x<br />
H e x<br />
= lim<br />
x→∞ 6 = ∞<br />
=lim<br />
H e x − 1 H e x<br />
=lim<br />
x→0 2x x→0 2 = 1 2<br />
23. This limit has the form 0 . lim tanh x H sech 2 x<br />
=lim<br />
0 x→0 tan x x→0 sec 2 x = sech2 0<br />
sec 2 0 = 1 1 =1<br />
25. This limit has the form 0 . lim 5 t − 3 t<br />
0 t→0 t<br />
H 5 t ln 5 − 3 t ln 3<br />
=lim<br />
=ln5− ln 3 = ln 5<br />
t→0 3<br />
1<br />
27. This limit has the form 0 . lim sin −1 x H 1/ √ 1 − x<br />
= lim<br />
2<br />
1<br />
=lim√ 0 x→0 x x→0 1<br />
= 1 x→0 1 − x<br />
2 1 =1<br />
29. This limit has the form 0 . lim 1 − cos x<br />
0<br />
x→0 x 2<br />
=lim<br />
H sin x H cos x<br />
= lim = 1<br />
x→0 2x x→0 2 2<br />
x +sinx<br />
31. lim<br />
x→0 x +cosx = 0+0<br />
0+1 = 0 =0. L’Hospital’s Rule does not apply.<br />
1<br />
33. This limit has the form 0 . lim 1 − x +lnx<br />
0 x→1 1+cosπx<br />
H<br />
=lim<br />
x→1<br />
−1+1/x<br />
−π sin πx<br />
H −1/x 2<br />
=lim<br />
x→1 −π 2 cos πx =<br />
−1<br />
−π 2 (−1) = − 1 π 2<br />
35. This limit has the form 0 . lim x a − ax + a − 1 H ax a−1 − a H a(a − 1)x a−2 a(a − 1)<br />
=lim<br />
= lim<br />
=<br />
0 x→1 (x − 1) 2 x→1 2(x − 1) x→1 2<br />
2<br />
37. This limit has the form 0 . lim cos x − 1+ 1 2 x2<br />
0 x→0 x 4<br />
39. This limit has the form ∞ · 0.<br />
H =lim<br />
x→0<br />
− sin x + x<br />
4x 3<br />
H = lim<br />
x→0<br />
− cos x +1<br />
12x 2<br />
sin(π/x) H cos(π/x)(−π/x 2 )<br />
lim x sin(π/x) = lim = lim<br />
= π lim cos(π/x) =π(1) = π<br />
x→∞ x→∞ 1/x x→∞ −1/x 2<br />
x→∞<br />
41. This limit has the form ∞ · 0. We’ll change it to the form 0 0 .<br />
sin 6x H 6cos6x<br />
lim cot 2x sin 6x =lim =lim<br />
x→0 x→0 tan 2x x→0 2sec 2 2x = 6(1)<br />
2(1) =3 2<br />
=lim<br />
H sin x H cos x<br />
= lim<br />
x→0 24x x→0 24 = 1<br />
24<br />
x 3<br />
43. This limit has the form ∞ · 0. lim<br />
x→∞ x3 e −x2 = lim<br />
x→∞ e x2<br />
= H 3x 2<br />
lim<br />
x→∞ 2xe x2<br />
= lim<br />
x→∞<br />
3x<br />
2e x2<br />
= H 3<br />
lim =0<br />
x→∞ x2<br />
4xe
F.<br />
TX.10SECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE ¤ 169<br />
45. This limit has the form 0 · (−∞).<br />
lim ln x tan(πx/2) = lim<br />
x→1 + x→1 +<br />
47. This limit has the form ∞−∞.<br />
x<br />
lim<br />
x→1 x − 1 − 1<br />
ln x<br />
ln x<br />
cot(πx/2)<br />
<br />
=lim<br />
x→1<br />
x ln x − (x − 1)<br />
(x − 1) ln x<br />
H<br />
= lim<br />
x→1 + 1/x<br />
(−π/2) csc 2 (πx/2) = 1<br />
(−π/2)(1) 2 = − 2 π<br />
H<br />
= lim<br />
x→1<br />
x(1/x)+lnx − 1<br />
(x − 1)(1/x)+lnx = lim<br />
x→1<br />
H 1/x<br />
=lim<br />
x→1 1/x 2 +1/x · x2<br />
x = lim x<br />
2 x→1 1+x = 1<br />
1+1 = 1 2<br />
ln x<br />
1 − (1/x)+lnx<br />
49. We will multiply and divide by the conjugate of the expression to change the form of the expression.<br />
√<br />
lim x2 + x − x √ √ <br />
x2 + x − x x2 + x + x<br />
x 2 + x − x 2<br />
= lim<br />
· √ = lim √<br />
x→∞<br />
x→∞ 1 x2 + x + x x→∞ x2 + x + x<br />
x<br />
= lim √<br />
x→∞ x2 + x + x = lim 1<br />
1<br />
= √ = 1<br />
x→∞ 1+1/x +1 1+1 2<br />
As an alternate solution, write √ x 2 + x − x as √ x 2 + x − √ x 2 ,factorout √ x 2 ,rewriteas( 1+1/x − 1)/(1/x),and<br />
apply l’Hospital’s Rule.<br />
51. The limit has the form ∞−∞and we will change the form to a product by factoring out x.<br />
<br />
lim (x − ln x) = lim x 1 − ln x <br />
ln x H 1/x<br />
= ∞ since lim = lim<br />
x→∞ x→∞ x<br />
x→∞ x x→∞ 1 =0.<br />
<br />
53. y = x x2 ⇒ ln y = x 2 ln x<br />
ln x,so lim ln y = lim x2 ln x = lim<br />
H 1/x<br />
= lim<br />
x→0 + x→0 + x→0 + 1/x 2 x→0 + −2/x = lim − 1 3 x→0 + 2 x2 =0 ⇒<br />
lim<br />
x→0 + xx2 = lim eln y = e 0 =1.<br />
x→0 +<br />
55. y =(1− 2x) 1/x ⇒ ln y = 1 x<br />
lim<br />
x→0 (1 − 2x)1/x =lim<br />
x→0<br />
e ln y = e −2 .<br />
57. y =<br />
<br />
1+ 3 x + 5 x<br />
⇒ ln y = x ln<br />
1+ 3 x 2 x + 5 <br />
x 2<br />
ln<br />
1+ 3 x + 5 <br />
x<br />
lim ln y = lim<br />
2<br />
x→∞ x→∞ 1/x<br />
ln(1 − 2x) H −2/(1 − 2x)<br />
ln(1 − 2x), solim ln y =lim<br />
= lim<br />
= −2 ⇒<br />
x→0 x→0 x<br />
x→0 1<br />
H<br />
= lim<br />
x→∞<br />
<br />
so lim 1+ 3<br />
x→∞ x + 5 x<br />
= lim<br />
x 2 x→∞ eln y = e 3 .<br />
⇒<br />
<br />
− 3 x 2 − 10<br />
x 3 <br />
1+ 3 x + 5 x 2 <br />
−1/x 2<br />
59. y = x 1/x ln x<br />
⇒ ln y =(1/x) lnx ⇒ lim ln y = lim<br />
x→∞ x→∞ x<br />
lim<br />
x→∞ x1/x = lim<br />
x→∞ eln y = e 0 =1<br />
= lim<br />
x→∞<br />
H 1/x<br />
= lim<br />
x→∞ 1 =0 ⇒<br />
3+ 10 x<br />
1+ 3 x + 5 x 2 =3,
F.<br />
170 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
61. y =(4x +1) cot x ln(4x +1) H<br />
⇒ ln y =cotx ln(4x +1),so lim ln y = lim<br />
= lim<br />
x→0 + x→0 + tan x<br />
x→0 + 4<br />
4x +1<br />
sec 2 x =4<br />
⇒<br />
lim<br />
x→0 +(4x +1)cot x = lim<br />
x→0 + eln y = e 4 .<br />
63. y =(cosx) 1/x2 ⇒ ln y = 1 ln cos x<br />
ln cos x ⇒ lim ln y = lim<br />
x2 x→0 + x→0 + x 2<br />
= H − tan x H − sec 2 x<br />
lim = lim = − 1<br />
x→0 + 2x x→0 + 2 2<br />
⇒<br />
lim<br />
x→0 +(cos x)1/x2 = lim<br />
x→0 + eln y = e −1/2 =1/ √ e<br />
65. From the graph, if x = 500, y ≈ 7.36. The limit has the form 1 ∞ .<br />
<br />
Now y = 1+ 2 x <br />
⇒ ln y = x ln 1+ 2 <br />
⇒<br />
x x<br />
1<br />
− 2 <br />
ln(1 + 2/x) H 1+2/x x<br />
lim ln y = lim<br />
= lim<br />
2<br />
x→∞ x→∞ 1/x x→∞ −1/x 2<br />
=2 lim<br />
x→∞<br />
1<br />
1+2/x =2(1)=2<br />
<br />
lim 1+ 2 x<br />
= lim<br />
x→∞ x x→∞ eln y = e 2 [≈ 7.39]<br />
f(x)<br />
67. From the graph, it appears that lim<br />
x→0 g(x) =lim f 0 (x)<br />
x→0 g 0 (x) =0.25.<br />
We calculate lim<br />
x→0<br />
f(x)<br />
g(x) = lim<br />
x→0<br />
e x − 1<br />
x 3 +4x<br />
⇒<br />
H<br />
= lim<br />
x→0<br />
e x<br />
3x 2 +4 = 1 4 .<br />
e x<br />
69. lim<br />
x→∞ x n<br />
71. lim<br />
x→∞<br />
= H e x<br />
lim<br />
x→∞ nx n−1<br />
x<br />
√<br />
x2 +1<br />
H<br />
= lim<br />
x→∞<br />
H = lim<br />
x→∞<br />
e x<br />
n(n − 1)x n−2<br />
1<br />
1<br />
2 (x2 +1) −1/2 (2x) = lim<br />
x→∞<br />
= H ··· H= e x<br />
lim<br />
x→∞ n! = ∞<br />
√<br />
x2 +1<br />
. Repeated applications of l’Hospital’s Rule result in the<br />
x<br />
original limit or the limit of the reciprocal of the function. Another method is to try dividing the numerator and denominator<br />
by x:<br />
lim<br />
x→∞<br />
x<br />
√<br />
x2 +1 = lim<br />
x→∞<br />
x/x<br />
<br />
x2 /x 2 +1/x 2 = lim<br />
x→∞<br />
1<br />
<br />
1+1/x<br />
2 = 1 1 =1<br />
<br />
73. First we will find lim 1+ r nt, <br />
which is of the form 1<br />
n→∞ n ∞ . y = 1+ r nt <br />
⇒ ln y = nt ln 1+ r <br />
,so<br />
n n<br />
<br />
−r/n<br />
2<br />
lim ln y = lim<br />
1+ nt ln r <br />
ln(1 + r/n)<br />
= t lim<br />
n→∞ n→∞ n n→∞ 1/n<br />
H<br />
= t lim<br />
n→∞<br />
(1 + r/n)(−1/n 2 ) = t lim<br />
n→∞<br />
r<br />
1+i/n = tr<br />
⇒<br />
lim y =<br />
n→∞ ert .Thus,asn →∞, A = A 0<br />
1+ r nt<br />
→ A0 e rt .<br />
n
F.<br />
e E<br />
75. lim P (E) = lim + e −E<br />
E→0 + E→0 + e E − e − 1 <br />
−E E<br />
TX.10SECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE ¤ 171<br />
E e E + e −E − 1 e E − e −E Ee E + Ee −E − e E + e −E <br />
= lim<br />
= lim<br />
form is<br />
0<br />
E→0 + (e E − e −E ) E<br />
E→0 + Ee E − Ee −E 0<br />
H<br />
= lim<br />
E→0 + Ee E + e E · 1+E −e −E + e −E · 1 − e E + −e −E<br />
Ee E + e E · 1 − [E(−e −E )+e −E · 1]<br />
= lim<br />
E→0 +<br />
Ee E − Ee −E<br />
= lim<br />
Ee E + e E + Ee −E − e−E E→0 +<br />
= 0<br />
e E − e −E<br />
, whereL = lim<br />
2+L E→0 + E<br />
Thus, lim P (E) = 0<br />
E→0 + 2+2 =0.<br />
e E − e −E<br />
e E + eE E + e−E − e−E<br />
E<br />
<br />
form is<br />
0<br />
0<br />
H = lim<br />
E→0 + e E + e −E<br />
1<br />
77. We see that both numerator and denominator approach 0, so we can use l’Hospital’s Rule:<br />
√<br />
2a3 x − x<br />
lim<br />
4 − a 3√ aax<br />
x→a a − 4√ ax 3<br />
[divide by E]<br />
= 1+1<br />
1<br />
=lim<br />
H 1<br />
2 (2a3 x − x 4 ) −1/2 (2a 3 − 4x 3 ) − a <br />
1<br />
3 (aax) −2/3 a 2<br />
x→a − 1 4 (ax3 ) −3/4 (3ax 2 )<br />
=<br />
1<br />
2 (2a3 a − a 4 ) −1/2 (2a 3 − 4a 3 ) − 1 3 a3 (a 2 a) −2/3<br />
− 1 4 (aa3 ) −3/4 (3aa 2 )<br />
=2<br />
= (a4 ) −1/2 (−a 3 ) − 1 3 a3 (a 3 ) −2/3<br />
− 3 4 a3(a4 )−3/4<br />
= −a − 1 3 a<br />
− 3 4<br />
= 4 3<br />
4<br />
3 a = 16 9 a<br />
79. Since f(2) = 0, the given limit has the form 0 0 .<br />
f(2 + 3x)+f(2 + 5x) H f 0 (2 + 3x) · 3+f 0 (2 + 5x) · 5<br />
lim<br />
=lim<br />
= f 0 (2) · 3+f 0 (2) · 5=8f 0 (2) = 8 · 7=56<br />
x→0 x<br />
x→0 1<br />
81. Since lim<br />
h→0<br />
[f(x + h) − f(x − h)] = f(x) − f(x) =0 (f is differentiable and hence continuous) and lim<br />
h→0<br />
2h =0,weuse<br />
l’Hospital’s Rule:<br />
f(x + h) − f(x − h) H f 0 (x + h)(1) − f 0 (x − h)(−1)<br />
lim<br />
=lim<br />
= f 0 (x)+f 0 (x)<br />
= 2f 0 (x)<br />
= f 0 (x)<br />
h→0 2h<br />
h→0 2<br />
2<br />
2<br />
f(x + h) − f(x − h)<br />
2h<br />
is the slope of the secant line between<br />
(x − h, f(x − h)) and (x + h, f(x + h)). Ash → 0,thislinegetscloser<br />
to the tangent line and its slope approaches f 0 (x).<br />
f(x)<br />
83. (a) We show that lim<br />
x→0 x =0for every integer n ≥ 0. Lety = 1 n<br />
x .Then 2<br />
f(x)<br />
lim<br />
x→0 x =lim e −1/x2 y n<br />
2n x→0 (x 2 ) n = lim<br />
y→∞ e y<br />
= H ny n−1<br />
lim<br />
y→∞ e y<br />
= H ··· H= n!<br />
lim<br />
y→∞ e =0 ⇒<br />
y<br />
f(x)<br />
lim<br />
x→0 x = lim f(x)<br />
n x→0 xn x =lim f(x)<br />
2n x→0 xn lim<br />
x→0 x =0.Thus,f 0 f(x) − f(0) f(x)<br />
(0) = lim<br />
= lim<br />
2n<br />
x→0 x − 0 x→0 x =0.
F.<br />
172 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
(b) Using the Chain Rule and the Quotient Rule we see that f (n) (x) exists for x 6= 0. In fact, we prove by induction that for<br />
each n ≥ 0, there is a polynomial p n and a non-negative integer k n with f (n) (x) =p n (x)f(x)/x kn for x 6= 0.Thisis<br />
true for n =0; suppose it is true for the nth derivative. Then f 0 (x) =f(x)(2/x 3 ),so<br />
which has the desired form.<br />
f (n+1) (x)= x k n<br />
[p 0 n(x) f(x)+p n(x)f 0 (x)] − k nx k n−1 p n(x) f(x) x −2k n<br />
= x k n<br />
p 0 n(x)+p n(x) 2/x 3 − k nx k n−1 p n(x) f(x)x −2k n<br />
= x kn+3 p 0 n(x)+2p n (x) − k n x kn+2 p n (x) f(x)x −(2kn+3)<br />
Now we show by induction that f (n) (0) = 0 for all n.Bypart(a),f 0 (0) = 0. Suppose that f (n) (0) = 0. Then<br />
f (n+1) f (n) (x) − f (n) (0) f (n) (x) p n(x) f(x)/x k n<br />
(0) = lim<br />
= lim =lim<br />
x→0 x − 0<br />
x→0 x x→0 x<br />
f(x)<br />
=limp n (x) lim<br />
x→0 x→0 x = p n(0) · 0=0<br />
kn+1<br />
= lim<br />
x→0<br />
p n(x) f(x)<br />
x k n+1<br />
4.5 Summary of Curve Sketching<br />
1. y = f(x) =x 3 + x = x(x 2 +1) A. f is a polynomial, so D = R.<br />
H.<br />
B. x-intercept =0, y-intercept = f(0) = 0 C. f(−x) =−f(x),sof is<br />
odd; the curve is symmetric about the origin.<br />
D. f is a polynomial, so there is<br />
no asymptote.<br />
E. f 0 (x) =3x 2 +1> 0,sof is increasing on (−∞, ∞).<br />
F. There is no critical number and hence, no local maximum or minimum value.<br />
G. f 00 (x) =6x>0 on (0, ∞) and f 00 (x) < 0 on (−∞, 0),sof is CU on<br />
(0, ∞) and CD on (−∞, 0). Since the concavity changes at x =0,thereisan<br />
inflection point at (0, 0).<br />
3. y = f(x) =2− 15x +9x 2 − x 3 = −(x − 2) x 2 − 7x +1 A. D = R B. y-intercept: f(0) = 2; x-intercepts:<br />
f(x) =0 ⇒ x =2or (by the quadratic formula) x = 7 ± √ 45<br />
2<br />
≈ 0.15, 6.85 C. No symmetry D. No asymptote<br />
E. f 0 (x) =−15 + 18x − 3x 2 = −3(x 2 − 6x +5)<br />
= −3(x − 1)(x − 5) > 0 ⇔ 1
F.<br />
TX.10<br />
SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 173<br />
5. y = f(x) =x 4 +4x 3 = x 3 (x +4) A. D = R B. y-intercept: f(0) = 0;<br />
H.<br />
x-intercepts: f(x) =0 ⇔ x = −4, 0 C. No symmetry<br />
D. No asymptote E. f 0 (x) =4x 3 +12x 2 =4x 2 (x +3)> 0 ⇔<br />
x>−3,sof is increasing on (−3, ∞) and decreasing on (−∞, −3).<br />
F. Local minimum value f(−3) = −27, no local maximum<br />
G. f 00 (x) =12x 2 +24x =12x(x +2)< 0 ⇔ −2 1,sof is CU on (1, ∞) and<br />
(x − 1)<br />
3<br />
11. y = f(x) =1/(x 2 − 9) A. D = {x | x 6=±3} =(−∞, −3) ∪ (−3, 3) ∪ (3, ∞) B. y-intercept = f(0) = − 1 9 ,no<br />
1<br />
x-intercept C. f(−x) =f(x) ⇒ f is even; the curve is symmetric about the y-axis. D. lim =0,soy =0<br />
x→±∞ x 2 − 9<br />
is a HA.<br />
1<br />
1<br />
lim = −∞, lim<br />
x→3 − x 2 − 9 x→3 + x 2 − 9 = ∞,<br />
lim 1<br />
x→−3 − x 2 − 9 = ∞,<br />
lim 1<br />
= −∞, sox =3and x = −3<br />
x→−3 + x 2 − 9<br />
are VA. E. f 0 2x<br />
(x) =−<br />
(x 2 2<br />
> 0 ⇔ x
F.<br />
174 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
decreasing on (0, 3) and (3, ∞). F. Local maximum value f(0) = − 1 9 .<br />
H.<br />
G. y 00 = −2(x2 − 9) 2 +(2x)2(x 2 − 9)(2x)<br />
(x 2 − 9) 4 = 6(x2 +3)<br />
(x 2 − 9) 3 > 0 ⇔<br />
x 2 > 9 ⇔ x>3 or x 0 ⇔ −3
F.<br />
TX.10<br />
SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 175<br />
17. y = f(x) = x2<br />
x 2 +3 = (x2 +3)− 3<br />
=1− 3<br />
x 2 +3 x 2 +3<br />
A. D = R B. y-intercept: f(0) = 0;<br />
x-intercepts: f(x) =0 ⇔ x =0 C. f(−x) =f(x),sof is even; the graph is symmetric about the y-axis.<br />
D. lim<br />
x→±∞<br />
x 2<br />
x 2 +3 =1,soy =1is a HA. No VA. E. Using the Reciprocal Rule, f 0 (x) =−3 ·<br />
f 0 (x) > 0 ⇔ x>0 and f 0 (x) < 0 ⇔ x 0 ⇒ x>− 1 and f 0 (x) < 0 ⇒ x
F.<br />
176 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
23. y = f(x) =x/ √ x 2 +1 A. D = R B. y-intercept: f(0) = 0; x-intercepts: f(x) =0 ⇒ x =0<br />
C. f(−x) =−f(x),sof is odd; the graph is symmetric about the origin.<br />
D. lim f(x) = lim<br />
x→∞ x→∞<br />
and<br />
lim f(x) =<br />
x→−∞<br />
=<br />
lim<br />
x→−∞<br />
x<br />
√<br />
x2 +1 = lim<br />
x→∞<br />
x<br />
√<br />
x2 +1 =<br />
lim<br />
x→−∞<br />
x/x<br />
√<br />
x2 +1/x = lim<br />
x→∞<br />
x/x<br />
√<br />
x2 +1/x =<br />
1<br />
− √ = −1 so y = ±1 are HA.<br />
1+0<br />
lim<br />
x→−∞<br />
x/x<br />
√<br />
x2 +1/ √ x 2 = lim<br />
x→∞<br />
x/x<br />
√<br />
x2 +1/− √ x 2 = lim<br />
x→−∞<br />
1<br />
<br />
1+1/x<br />
2 = 1<br />
√ 1+0<br />
=1<br />
1<br />
− 1+1/x 2<br />
No VA.<br />
√ 2x<br />
x2 +1− x ·<br />
E. f 0 2 √ x<br />
(x) =<br />
2 +1<br />
= x2 +1− x 2<br />
[(x 2 +1) 1/2 ] 2 (x 2 +1) = 1<br />
> 0 for all x,sof is increasing on R.<br />
3/2 (x 2 3/2<br />
+1)<br />
F. No extreme values<br />
G. f 00 (x) =− 3 2 (x2 +1) −5/2 · 2x =<br />
−3x<br />
(x 2 +1) 5/2 ,sof 00 (x) > 0 for x0. Thus,f is CU on (−∞, 0) and CD on (0, ∞).<br />
IP at (0, 0)<br />
25. y = f(x) = √ 1 − x 2 /x A. D = {x ||x| ≤ 1, x 6= 0} =[−1, 0) ∪ (0, 1] B. x-intercepts ±1, noy-intercept<br />
√<br />
1 − x<br />
2<br />
C. f(−x) =−f(x), so the curve is symmetric about (0, 0) . D. lim<br />
x→0 + x<br />
<br />
−x 2 / √ 1 − x 2 − √ 1 − x 2<br />
so x =0is a VA. E. f 0 (x) =<br />
on (−1, 0) and (0, 1).<br />
G. f 00 (x) =<br />
<br />
f is CU on −1, −<br />
<br />
2<br />
IP at ± , ± √ 1<br />
3 2<br />
F. No extreme values<br />
2 − 3x 2<br />
x 3 (1 − x 2 ) > 0 3/2 ⇔ <br />
<br />
−1 1 and f 0 (x) < 0 when 0 < |x| < 1,sof is increasing on (−∞, −1) and (1, ∞),and<br />
decreasing on (−1, 0) and (0, 1) [hence decreasing on (−1, 1) since f is<br />
H.<br />
H.<br />
continuous on (−1, 1)].<br />
F. Local maximum value f(−1) = 2, local minimum<br />
value f(1) = −2 G. f 00 (x) = 2 3 x−5/3 < 0 when x 0<br />
when x>0,sof is CD on (−∞, 0) and CU on (0, ∞). IPat(0, 0)
F.<br />
TX.10<br />
SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 177<br />
29. y = f(x) = 3√ x 2 − 1 A. D = R B. y-intercept: f(0) = −1; x-intercepts: f(x) =0 ⇔ x 2 − 1=0 ⇔<br />
x = ±1 C. f(−x) =f(x), so the curve is symmetric about the y-axis. D. No asymptote<br />
E. f 0 (x) = 1 3 (x2 − 1) −2/3 (2x) =<br />
increasing on (0, ∞) and decreasing on (−∞, 0).<br />
G. f 00 (x)= 2 3 · (x2 − 1) 2/3 (1) − x · 2<br />
3 (x2 − 1) −1/3 (2x)<br />
[(x 2 − 1) 2/3 ] 2<br />
2x<br />
3 3 (x 2 − 1) 2 . f 0 (x) > 0 ⇔ x>0 and f 0 (x) < 0 ⇔ x 0 ⇔ −1 0 ⇔ x ∈ 2nπ − π , 2nπ + π<br />
2 2 for each integer n, andf 0 (x) < 0 ⇔ cos x0 and sin x 6=±1 ⇔ x ∈ <br />
2nπ, 2nπ + π 2 ∪ 2nπ +<br />
π<br />
, 2nπ + π for some integer n.<br />
2<br />
f 00 (x) > 0 ⇔ sin x
F.<br />
178 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
35. y = f(x) = 1 2<br />
x − sin x, 0 0 on (0,α) and (β,2π) [ f is CU].<br />
The inflection points occur when x = α, β.<br />
H.
F.<br />
TX.10<br />
SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 179<br />
41. y =1/(1 + e −x ) A. D = R B. No x-intercept; y-intercept = f(0) = 1 2<br />
. C. No symmetry<br />
D. lim 1/(1 +<br />
x→∞ e−x )= 1 =1and lim 1/(1 + 1+0 x→−∞ e−x )=0 since<br />
lim<br />
x→−∞ e−x = ∞], so f has horizontal asymptotes<br />
y =0and y =1. E. f 0 (x) =−(1 + e −x ) −2 (−e −x )=e −x /(1 + e −x ) 2 . This is positive for all x,sof is increasing on R.<br />
F. No extreme values G. f 00 (x) = (1 + e−x ) 2 (−e −x ) − e −x (2)(1 + e −x )(−e −x )<br />
= e−x (e −x − 1)<br />
(1 + e −x ) 4 (1 + e −x ) 3<br />
The second factor in the numerator is negative for x>0 and positive for x 0 ⇒ 1 > 1/x ⇒ x>1 and<br />
H.<br />
f 0 (x) < 0 ⇒ 0 0]<br />
x→−∞<br />
f(x) =1,soy =0and y =1are HA; no VA<br />
E. f 0 (x) =−2(1 + e x ) −3 e x = −2ex < 0,sof is decreasing on R F. No local extrema<br />
(1 + e x )<br />
3<br />
G. f 00 (x)=(1+e x ) −3 (−2e x )+(−2e x )(−3)(1 + e x ) −4 e x<br />
H.<br />
= −2e x (1 + e x ) −4 [(1 + e x ) − 3e x ]= −2ex (1 − 2e x )<br />
(1 + e x ) 4 .<br />
f 00 (x) > 0 ⇔ 1 − 2e x < 0 ⇔ e x > 1 2<br />
⇔ x>ln 1 2 and<br />
f 00 (x) < 0 ⇔ x0} = ∞ <br />
n=−∞<br />
(2nπ, (2n +1)π) =···∪ (−4π, −3π) ∪ (−2π, −π) ∪ (0,π) ∪ (2π, 3π) ∪ ···<br />
B. No y-intercept; x-intercepts: f(x) =0 ⇔ ln(sin x) =0 ⇔ sin x = e 0 =1 ⇔ x =2nπ + π for each<br />
2<br />
integer n. C. f is periodic with period 2π. D. lim f(x) =−∞ and lim<br />
x→(2nπ) +<br />
x = nπ are VAs for all integers n.<br />
x→[(2n+1)π]<br />
−<br />
f(x) =−∞, so the lines<br />
E. f 0 (x) = cos x<br />
sin x =cotx,sof 0 (x) > 0 when 2nπ < x < 2nπ + π for each<br />
2<br />
integer n,andf 0 (x) < 0 when 2nπ + π 2
F.<br />
180 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
decreasing on 2nπ + π 2 , (2n +1)π for each integer n.<br />
TX.10<br />
H.<br />
F. Local maximum values f 2nπ + π 2<br />
<br />
=0, no local minimum.<br />
G. f 00 (x) =− csc 2 x 0 ⇔ x 2 < 1 2<br />
⇔ |x| < 1 √<br />
2<br />
,sof is increasing on<br />
<br />
and decreasing on −∞, − √ 1<br />
2<br />
and<br />
<br />
<br />
√<br />
1<br />
1<br />
2<br />
, ∞ . F. Local maximum value f √<br />
2<br />
=1/ √ 2e, local minimum<br />
<br />
value f − √ 1<br />
2<br />
= −1/ √ 2e G. f 00 (x) =−2xe −x2 (1 − 2x 2 ) − 4xe −x2 =2xe −x2 (2x 2 − 3) > 0 ⇔<br />
<br />
<br />
3<br />
x> or − 3<br />
2e −2x [multiply by e 2x ] ⇔<br />
H.<br />
e 5x > 2 3<br />
⇔ 5x >ln 2 3<br />
⇔ x> 1 5 ln 2 3 ≈−0.081. Similarly, f 0 (x) < 0 ⇔<br />
x< 1 5 ln 2 3 . f is decreasing on −∞, 1 5 ln 2 3<br />
<br />
and increasing on<br />
1<br />
5 ln 2 3 , ∞ .<br />
F. Local minimum value f 1<br />
ln 2<br />
5 3 = 2<br />
3/5<br />
+ <br />
2 −2/5<br />
≈ 1.96; no local maximum.<br />
3<br />
3<br />
G. f 00 (x) =9e 3x +4e −2x ,so f 00 (x) > 0 for all x,andf is CU on (−∞, ∞). NoIP<br />
53. m = f(v) =<br />
m 0<br />
<br />
1 − v2 /c .Them-intercept is f(0) = m 0.Therearenov-intercepts. lim f(v) =∞,sov = c is a VA.<br />
2 v→c− f 0 (v) =− 1 2 m0(1 − v2 /c 2 ) −3/2 (−2v/c 2 )=<br />
increasing on (0,c). There are no local extreme values.<br />
m 0 v<br />
c 2 (1 − v 2 /c 2 ) 3/2 =<br />
f 00 (v)= (c2 − v 2 ) 3/2 (m 0 c) − m 0 cv · 3<br />
2 (c2 − v 2 ) 1/2 (−2v)<br />
[(c 2 − v 2 ) 3/2 ] 2<br />
= m 0c(c 2 − v 2 ) 1/2 [(c 2 − v 2 )+3v 2 ]<br />
(c 2 − v 2 ) 3 = m 0c(c 2 +2v 2 )<br />
(c 2 − v 2 ) 5/2 > 0,<br />
so f is CU on (0,c). Therearenoinflection points.<br />
m 0 v<br />
c 2 (c 2 − v 2 ) 3/2<br />
c 3 =<br />
m 0 cv<br />
> 0,sof is<br />
(c 2 − v 2 )<br />
3/2
F.<br />
TX.10<br />
SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 181<br />
55. y = − W<br />
24EI x4 + WL<br />
12EI x3 − WL2<br />
24EI x2 = − W<br />
24EI x2 x 2 − 2Lx + L 2<br />
= −W<br />
24EI x2 (x − L) 2 = cx 2 (x − L) 2<br />
where c = − W isanegativeconstantand0 ≤ x ≤ L. Wesketch<br />
24EI<br />
f(x) =cx 2 (x − L) 2 for c = −1. f(0) = f(L) =0.<br />
f 0 (x) =cx 2 [2(x − L)] + (x − L) 2 (2cx) =2cx(x − L)[x +(x − L)] = 2cx(x − L)(2x − L). Sofor0
F.<br />
182 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />
TX.10<br />
C. No symmetry D. lim f(x) =−∞ and lim<br />
x→(1/2) −<br />
lim<br />
x→±∞<br />
[f(x) − (−x +2)]= lim<br />
E. f 0 (x) =−1 −<br />
f(x) =∞,sox = 1 x→(1/2) + 2<br />
is a VA.<br />
1<br />
=0,sotheliney = −x +2is a SA.<br />
x→±∞ 2x − 1<br />
2<br />
(2x − 1) < 0 for x 6= 1 ,sof is decreasing on <br />
−∞, 1 2 2 2<br />
and 1<br />
2 , ∞ . F. No extreme values G. f 0 (x) =−1 − 2(2x − 1) −2 ⇒<br />
H.<br />
f 00 (x) =−2(−2)(2x − 1) −3 8<br />
(2) =<br />
(2x − 1) 3 ,sof 00 (x) > 0 when x> 1 and 2<br />
f 00 (x) < 0 when x< 1 . Thus, f is CU on 1<br />
, ∞ and CD on <br />
−∞, 1 2 2 2 .NoIP<br />
63. y = f(x) =(x 2 +4)/x = x +4/x A. D = {x | x 6= 0} =(−∞, 0) ∪ (0, ∞) B. No intercept<br />
C. f(−x) =−f(x) ⇒ symmetry about the origin D. lim (x +4/x) =∞ but f(x) − x =4/x → 0 as x → ±∞,<br />
x→∞<br />
so y = x is a slant asymptote.<br />
lim (x +4/x) =∞ and<br />
x→0 +<br />
lim<br />
x→0 − (x +4/x) =−∞,sox =0 is a VA. E. f 0 (x) =1− 4/x 2 > 0 ⇔<br />
x 2 > 4 ⇔ x>2 or x 0 ⇔ x>0 so f is CU on<br />
(0, ∞) and CD on (−∞, 0). NoIP<br />
65. y = f(x) = 2x3 + x 2 +1<br />
x 2 +1<br />
=2x +1+ −2x<br />
x 2 +1<br />
A. D = R B. y-intercept: f(0) = 1; x-intercept: f(x) =0 ⇒<br />
0=2x 3 + x 2 +1=(x + 1)(2x 2 − x +1) ⇒ x = −1 C. No symmetry D. No VA<br />
lim<br />
x→±∞<br />
[f(x) − (2x + 1)] = lim<br />
x→±∞<br />
−2x<br />
x 2 +1 =<br />
lim<br />
x→±∞<br />
−2/x<br />
=0, so the line y =2x +1is a slant asymptote.<br />
1+1/x2 E. f 0 (x) =2+ (x2 +1)(−2) − (−2x)(2x)<br />
= 2(x4 +2x 2 +1)− 2x 2 − 2+4x 2<br />
= 2x4 +6x 2<br />
(x 2 +1) 2 (x 2 +1) 2 (x 2 +1) = 2x2 (x 2 +3)<br />
2 (x 2 +1) 2<br />
so f 0 (x) > 0 if x 6= 0.Thus,f is increasing on (−∞, 0) and (0, ∞). Sincef is continuous at 0, f is increasing on R.<br />
F. No extreme values<br />
G. f 00 (x) = (x2 +1) 2 · (8x 3 +12x) − (2x 4 +6x 2 ) · 2(x 2 +1)(2x)<br />
[(x 2 +1) 2 ] 2<br />
= 4x(x2 +1)[(x 2 + 1)(2x 2 +3)− 2x 4 − 6x 2 ]<br />
= 4x(−x2 +3)<br />
(x 2 +1) 4 (x 2 +1) 3<br />
so f 00 (x) > 0 for x
F.<br />
TX.10<br />
SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 183<br />
67. y = f(x) =x − tan −1 x, f 0 (x) =1− 1<br />
1+x 2 = 1+x2 − 1<br />
1+x 2 = x2<br />
1+x 2 ,<br />
f 00 (x) = (1 + x2 )(2x) − x 2 (2x)<br />
= 2x(1 + x2 − x 2 ) 2x<br />
=<br />
(1 + x 2 ) 2 (1 + x 2 ) 2 (1 + x 2 ) . 2<br />
<br />
lim f(x) − x −<br />
π<br />
x→∞<br />
2 = lim π −<br />
x→∞<br />
2 tan−1 x = π − π =0,soy = x − π is a SA.<br />
2 2 2<br />
Also,<br />
lim<br />
x→−∞<br />
<br />
f(x) −<br />
<br />
x +<
|
7.2 Right triangle trigonometry (Page 5/12)
Page 5 / 12
How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of $\text{\hspace{0.17em}}\frac{5\pi }{12}\text{\hspace{0.17em}}$ with the ground? Round to the nearest foot.
About 52 ft
Access these online resources for additional instruction and practice with right triangle trigonometry.
Key equations
Trigonometric Functions Reciprocal Trigonometric Functions Cofunction Identities
Key concepts
• We can define trigonometric functions as ratios of the side lengths of a right triangle. See [link] .
• The same side lengths can be used to evaluate the trigonometric functions of either acute angle in a right triangle. See [link] .
• We can evaluate the trigonometric functions of special angles, knowing the side lengths of the triangles in which they occur. See [link] .
• Any two complementary angles could be the two acute angles of a right triangle.
• If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa. See [link] .
• We can use trigonometric functions of an angle to find unknown side lengths.
• Select the trigonometric function representing the ratio of the unknown side to the known side. See [link] .
• Right-triangle trigonometry facilitates the measurement of inaccessible heights and distances.
• The unknown height or distance can be found by creating a right triangle in which the unknown height or distance is one of the sides, and another side and angle are known. See [link] .
Verbal
For the given right triangle, label the adjacent side, opposite side, and hypotenuse for the indicated angle.
When a right triangle with a hypotenuse of 1 is placed in a circle of radius 1, which sides of the triangle correspond to the x - and y -coordinates?
The tangent of an angle compares which sides of the right triangle?
The tangent of an angle is the ratio of the opposite side to the adjacent side.
What is the relationship between the two acute angles in a right triangle?
Explain the cofunction identity.
For example, the sine of an angle is equal to the cosine of its complement; the cosine of an angle is equal to the sine of its complement.
Algebraic
For the following exercises, use cofunctions of complementary angles.
$\mathrm{cos}\left(34°\right)=\mathrm{sin}\left(___°\right)$
$\mathrm{cos}\left(\frac{\pi }{3}\right)=\mathrm{sin}\left(___\right)$
$\frac{\pi }{6}$
$\mathrm{csc}\left(21°\right)=\mathrm{sec}\left(___°\right)$
$\mathrm{tan}\left(\frac{\pi }{4}\right)=\mathrm{cot}\left(___\right)$
$\frac{\pi }{4}$
For the following exercises, find the lengths of the missing sides if side $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is opposite angle $\text{\hspace{0.17em}}A,$ side $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is opposite angle $\text{\hspace{0.17em}}B,$ and side $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ is the hypotenuse.
$\mathrm{cos}\text{\hspace{0.17em}}B=\frac{4}{5},a=10$
$\mathrm{sin}\text{\hspace{0.17em}}B=\frac{1}{2},a=20$
$b=\frac{20\sqrt{3}}{3},c=\frac{40\sqrt{3}}{3}$
$\mathrm{tan}\text{\hspace{0.17em}}A=\frac{5}{12},b=6$
$\mathrm{tan}\text{\hspace{0.17em}}A=100,b=100$
$a=10,000,c=10,00.5$
Questions & Answers
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
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test for convergence the series 1+x/2+2!/9x3
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a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
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100 meters
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Find that number sum and product of all the divisors of 360
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answer
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exponential series
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what is subgroup
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Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
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e power cos hyperbolic (x+iy)
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10y
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tan hyperbolic inverse (x+iy)=alpha +i bita
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prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
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why {2kπ} union {kπ}={kπ}?
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# A person can throw a stone to a maximum distance of 100 m. The greatest height to which he can throw the stone is:-(A) 100m(B) 75m(C) 50m(D) 25m
Last updated date: 16th Sep 2024
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Hint: The maximum distance given, 100m, is the range of projectile motion. The maximum height is achieved when thrown straight up at ${45^ \circ }$ angle. Thus launching angle is $\theta = {45^ \circ }$.
Formula Used: The formulae used in the solution are given here.
Range is given as $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ where $u$ is the initial velocity, $\theta$ is the launch angle and $g$ is the acceleration due to gravity and the maximum height is given by $H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Complete Step by Step Solution: This is a projectile motion problem. Given that, the range is 100m, we have to find the maximum height that a person can throw a stone.
Range is given as $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ where $u$ is the initial velocity, $\theta$ is the launch angle and $g$ is the acceleration due to gravity.
Since, $R = 100m$, thus $\dfrac{{{u^2}\sin 2\theta }}{g} = 100$ and the acceleration due to gravity $g = 9.8$.
Thus, it can be written that, ${u^2}\sin 2\theta = 100g$ where $g = 9.8$
It is known to us that, the greatest range is achieved using a ${45^ \circ }$ launch angle:
${u^2}\sin \left( {2 \times {{45}^ \circ }} \right) = 980$
$\Rightarrow {u^2}\sin {90^ \circ } = 980$
We know that, $\sin {90^ \circ } = 1$,
$\therefore {u^2} = 980$
The height is achieved when thrown straight up at ${45^ \circ }$ angle.
Thus, height $H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$.
Since, ${u^2} = 980$, $\theta = {45^ \circ }$ and $g = 9.8$
$H = \dfrac{{980}}{{39.2}}$
$\Rightarrow H = 25m$
The greatest height to which he can throw the stone is 25m. Hence, the correct answer is Option D.
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# Difference of sets
Difference in Set A and Set B or the difference of Set B from Set A, it’s represented by A-B. It’s the set of all elements of set A that in Set B are not present. Mathematically, it is given by:
A-B = { x: x∈A and x∉B}
If in the condition, (A∩B) refers to an intersection between Set A and Set B then,
A-B = A – (A∩B)
## Example of Difference of Sets
Example 1:
If A = { a, b, c, d, e }; B = { a, e, f, g}, then, find A-B & B-A.
Solution: Elements present in A only are b, c, d; Also, elements present in B only are f, g. Therefore,
A-B = {b, c, d}
and B-A = {f, g}
Credit: mathstopia.net
Note: A-B might not have equal to B-A.
Example 2:
If A = { 1, 2, 4, 6, 8} is given along with A-B = {1, 6, 8}, then Find A∩B
Solution: (A∩B) – A intersection B, is the Set of all the elements present commonly in both the sets A and B. Therefore,
A∩B = A – (A-B)
or, A∩B = {2, 4}
## The Identities Which Involves Difference of Sets
• If both Set and Set B are equal, then A-A = A-B = ϕ (empty set)
• In the condition, when the empty set is deducted from, suppose Set A then, as a result, it’s that set itself, that is, A – ϕ = A.
• When a set is deduced from the empty set then, it appears in the form of Empty Set, that is, ϕ – A = ϕ.
• When the SuperSet is deducted from the subset, then, it results in the Empty Set, that is, A – B = ϕ, in the condition when A ⊂ B
If Disjoints sets A and B are there, then, two conditions will occur:
1. A-B = A
2. B-A = B
Taking Another Example of Difference of Sets
Question: Find B − A
Solution:
Note: This time you’ll be looking for everything you’ll see in only B.
Elements which are present in only B are depicted below in bold
Let A = {1 orange, 1 pinapple, 1 banana, 1 apple}
Let B = {1 orange, 1 apricot, 1 pinapple, 1 banana, 1 mango, 1 apple, 1 kiwifruit }
Hence, B − A = {1 apricot, 1 mango, 1 kiwifruit}
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Question Video: Finding the Missing Number in Equivalent Multiplication Expressions | Nagwa Question Video: Finding the Missing Number in Equivalent Multiplication Expressions | Nagwa
# Question Video: Finding the Missing Number in Equivalent Multiplication Expressions Mathematics • Fourth Year of Primary School
## Join Nagwa Classes
Complete: 10 × 4 × 7 = 10 × ＿.
02:44
### Video Transcript
Complete: 10 times four times seven equals 10 times what?
This question relies on us understanding that we can multiply numbers in different orders and still get the same answer. For instance, if we want to work out the answer to 10 times four times seven, one way to find the answer might be to work out 10 times four first and then to multiply that answer by seven. Another answer might be to work out four times seven first and then times the answer by 10. Or we could even really juggle the numbers around and multiply 10 by seven first and then multiply that answer by four. However we do it, we’re always going to get the same answer.
So let’s think about the first multiplication to start with. What does 10 multiplied by four multiplied by seven mean? Well, if we worked out this in order, then we’d calculate 10 multiplied by four first and then multiply that by seven. Let’s use this rectangle to represent the multiplication 10 times four. So if we work out 10 times four times seven, this means we need seven of these rectangles. There are two, three, four, five, six, seven. This diagram represents the first calculation, 10 multiplied by four seven times.
Now let’s think about the second part of our number sentence. 10 multiplied by what? In the same way as we’ve just done, we can draw another rectangle to represent this multiplication. The height of our rectangle is going to be worth 10 again. And we need the length to be exactly the same as the length of the other rectangle. The length of the rectangle is the same as seven fours. In other words, to find out our missing number, what we have to do is to calculate the four times seven part of our multiplication first.
Let’s count in fours seven times to find out what it’s worth. Four, eight, 12, 16, 20, 24, and 28. The missing number is 28. All we’ve done is multiply the four times seven part of the calculation first. This leaves us with 10 times 28 to find the overall answer. 10 times four times seven equals 10 times 28.
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# Modular Arithmetic and how it works
As children, we grew up learning how to count to 10. Why 10? Well this could be easily justified using the fact that we as humans have 10 fingers and any whole number up to 10 could be easily represented by a quick show of fingers. But what happened when we, as children with this new found power of counting objects up to 10, encountered a number greater than 10? Did we take off our shoes and start counting with our toes? That might have solved the issue for numbers greater than 10 but less than 20 (assuming you aren’t polydactylic) but in all reality, we needed a way to transcend the idea of representing objects with our fingers and/or toes and represent any number, no matter how large.
How did we do this? By using a Place Value system with a base 10. “Place Value” means that using a limited number of symbols, we can represent any number by using these symbols in a variety of combinations. The value of each symbol is based on the position or “place” where the symbol is located in the sequence of symbols.
For example, pick a base. The very first column or “place” should be used for all the symbols preceding the base until the base itself is reached. This is called the “units” place or informally as the “ones” place. This place is usually the farthest left or right place in a sequence. For instructional purposes and for familiarity, we will place the units place on the far right of the sequence.
Once the base has been reached, a second place will be added to the left indicating how many “bases” have been reached. When the amount of “bases reached” has reached the base amount, then a new place is added, again to the left, indicating how many bases of bases have been reached and so on and so forth.
For example, our familiar base 10 system works as follows:
_____ . . . _____ _____ _____ _____
A comma is added after every 3 digits for practical purposes to easily differentiate places in more complex combinations of numbers.
Now let’s go back 4,000 years ago to Sumer, a region of Mesopotamia, (modern-day Iraq). There, children learned to count, but using 60 as a base. Why 60? Did the children of that time have 60 fingers and/or toes? Probably not. The reason for using this number as a base has not been explicitly recorded but there are two convincing hypothesis on why a base 60 number system developed.
One idea, is that instead of using their whole finger to represent a single number, the Babylonians actually counted the 12 knuckles of the four fingers on one hand, using the thumb as a “pointer” and the five fingers on the other as multiples of twelve. So on one hand they had 1-12 and on the other they had how many 12’s, for a total of 12 x 5 = 60.
The other idea is that the number 60 has many divisors, 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. In fact, 60 is the smallest number divisible by all integers from 1 to 6. This could prove very useful by being able to do division using more whole numbers and resulting in less fractions.
Base 60 is still used in many aspects of our lives today such as the 60 seconds in a minute and the 60 minutes in an hour. The circles is traditionally divided into which are also subdivided into 60 minutes of arc and further divided into 60 seconds of arc.
Modular Arithmetic
Image: Spindled, via Wikimedia Commons.
Now that we have a brief overview about bases, we can apply the power of Modular Arithmetic to change counting bases. Modular Arithmetic is a very handy and useful tool in mathematics invented by the famous Mathematician Carl Friedrich Gauss in 1801. We know that the number line is infinitely long but if we were to wrap this infinitely long number line around a circle of a given circumference n, we would notice that numbers would “line up” or over lap around the circle. This is the idea behind modular arithmetic. Keep in mind that we are dealing with integers here and not the real numbers.
So the number indicating how large the circle is n, is called the modulus. And we say that after one wrap around, any numbers that line up are congruent. In mathematical terms, when a number a, leaves the same remainder as a number b, we say a and b are congruent written
a ≡ b mod n
The “mod n” part is just notation letting us know that we are in mod n and is not actually part of the equation, per se. However, when the context is understood, it should be OK to omit writing this every time.
In general, any modulo n has n residue classes, one for each integer from 0 to n-1.
Let’s use the timer on your microwave as an example of a base. So we will have n residue classes from the integers 0 to 59.
0, 1, 2, 3, … 56, 57, 58 59
We call this modulo 60 or mod 60 for short. When we add 1 to 59, we return to 0. This is true for any modulus, even our own familiar base 10 (when we add 1 to 9, we return to 0) or even every day objects like traffic lights (Red, Green, Yellow, Red, …). The integers from 0 to 59 in our base 60 example are called Residue Classes.
Now for a quick example, when I was in the military, we would tell time using the 24-hour clock. This is different than the usual 12 hour clock where all 24 hours are represented twice and distinguished using A.M. or P.M.
So when I would get asked what time I would be ready to get picked on Friday for the weekend, I would reply 1600. Of course this did not make sense to most people because the face of a clock only has the numbers 1-12 listed on it. How could I explain correctly to them what time to pick me up so as to maximize our time together on the sunny beaches of San Diego? Using modular arithmetic of course!
Numbers are said to be congruent if their difference is divisible by the modulus. Or stated more succinctly, a is congruent to b if a-b is divisible by n shown algebraically
a ≡ b mod n if a-b / kn for some k
This basically means that the difference must be divisible by the base.
In our example, let’s show that 1600 is congruent to 4:00. For lingo purposes, just think of the colon as “hundred hours” to be in step with 1600. 1600 – 400 is 1200, a multiple of 12. Written
1600 ≡ 400 mod 1200
1600-400 /1200
“So 4:00 P.M. civilian. Don’t be late.”
Another cool example of things you can do with modular arithmetic is calculate the last digit or remainder of a huge number like . Try doing that by hand! Here is how we would do it mod 10.
1919 ≡ 919; (because 19 is congruent to 9 mod 10)
(92)9*9 ≡ (81)9*9 ;
(1)9*9 ≡ 9; (because 81 is congruent to 1 mod 10)
References:
Use of base 60 using hands
Base 60 as a base
http://www.storyofmathematics.com/sumerian.html
Sub-divisions of angles into minutes and seconds
http://en.wikipedia.org/wiki/Minute_of_arc
Modular Arithmetic
http://en.wikipedia.org/wiki/Modular_arithmetic
# Math Tricks and Fermat’s Little Theorem
So you think you’re a math whiz. You storm into parties armed with math’s most flamboyant tricks. You can recite the digits of π and e to 50 digits—whether in base 10 or 12. You can calculate squares with ease, since you’ve mastered the difference of squares x2 – y2 = (x + y)(x – y). In tackling 572, simply notice that 572 – 72 = (57 + 7)(57 – 7) = 64*50 = 3200. Adding 72 to both sides gives 572 = 3249.
Image by Hashir Milhan from
Wikimedia Commons under
Creative Commons.
You can also approximate square roots using the truncated Taylor series x ≈ c + (x – c2)/(2c) where c2 is the closest perfect square less than or equal to x. So √17 ≈ 4 + (17 – 16)/(2*4) = 4.125, whereas √17 = 4.123105 . . ..
But do you know what number theory is? It’s not taught in high school, and everyone’s repertoire of math tricks needs some number theory. Mastering modular arithmetic—the first step in number theory—will make you the life of the party. Calculating 83 mod 7 just means find the remainder after dividing by 7: 83 = 11*7 + 6, so 83 ≡ 6 mod 7. But it’s actually easier since 83 = 7*12 + (-1), so 83 ≡ -1 ≡ 6 mod 7. Modular arithmetic reveals the secrets of divisibility. Everybody knows the trick to see whether 3 divides a number; you just add the digits and check if 3 divides that number. But the reasoning is obvious when you write m = 10nan + 10n-1an-1 + . . . + 10a1 + a0 where the ai are the digits of m. Each 10k has a remainder of 1 modulo 3 so man + an-1 + . . . + a1 + a0 mod 3. Using this method generates tricks for other integers.
For instance, if 13 divides m, then 13 divides a0 – 3a1 – 4a2a3 + 3a4 + 4a5 + a6 – 3a7 – 4a+ . . . and the pattern continues. This is because
10 ≡ -3 mod 13,
102 ≡ 10*10 ≡ (-3)(-3) ≡ 9 ≡ -4 mod 13,
103 ≡ 102*10 ≡ (-4)(-3) ≡ 12 ≡ -1 mod 13,
104 ≡ 103*10 ≡ (-1)*(-3) ≡ 3 mod 13,
105 ≡ 104*10 ≡ 3*(-3) ≡ -9 ≡ 4 mod 13,
and 106 ≡ 105*10 ≡ 4*(-3) ≡ -12 ≡ 1 mod 13.
From 106 and onwards the pattern repeats. In fact, calculating 10n mod k for successive n will reveal the divisibility rule for k.
Then comes Fermat’s little theorem, the key to solving seemingly impossible calculations.
Fermat. Image from Wikimedia
Commons. Under public domain.
The theorem states for a prime p and integer a that aa mod p. If p doesn’t divide a, then ap -1 ≡ 1 mod p. I’ll illustrate the power of this little result in a computation. Let’s find 2371 mod 5. We’ll be using 24 ≡ 1 mod 5, which we get from Fermat’s little theorem. Now 2371 = 236823 =(24)9223, so by the theorem, 2371 = (24)9223 ≡ 19223 ≡ 1*23 ≡ 8 ≡ 3 mod 5. Exploiting Fermat’s little theorem can impress your friends, but try to avoid questions. Computing residues modulo a composite number—calculating b mod n for a composite number n—may require paper and ruin the magic.
Leonhard Euler proved a more general version of Fermat’s little theorem; it’s called the Euler-Fermat theorem. This theorem isn’t for parties; explaining it to the non-mathematically inclined will always require paper and some time. Nonetheless, it will impress at dinner if you have a napkin and pen.
Understanding this theorem requires Euler’s totient function φ(n).
Euler. From Wikimedia
Commons. Under public domain.
The number φ(n) for some n is the number of positive integers coprime with n that are less than or equal to n. Two numbers a and b are coprime if their greatest common factor is one. Hence 14 and 3 are coprime because their biggest shared factor is 1, but 21 and 14 aren’t coprime because they have a common divisor of 7. Moreover, φ(14) = 6 because 14 has six numbers less than or equal to it that are coprime with it: 1, 3, 5, 9, 11, and 13. Notice that if p is prime, φ(p) = p – 1 because every number less than p is coprime with p.
Now the Euler-Fermat theorem states that aφ(n) ≡ 1 mod n, which looks similar to ap -1 ≡ 1 mod p for a prime p. In fact, if = φ(p) = p – 1 for a prime p, the theorem reduces to Fermat’s little theorem.
Fermat’s little theorem has another generalization, Lagrange’s theorem. Joseph-Louis Lagrange was Euler’s student. Lagrange’s theorem generalizes both the previous theorems and doesn’t even require numbers. But due to the required background in group theory, I won’t go over the theorem. You can find links to more information on Lagrange’s theorem below.
Remember, a math whiz doesn’t need props like a magician does. Hook your audience with some modular arithmetic, and reel the people in with Fermat’s little theorem. If you want to get complicated, the most you’ll need is a pen and some paper.
Sources and cool stuff:
Modular arithmetic: http://nrich.maths.org/4350
Proof of Fermat’s little theorem: https://primes.utm.edu/notes/proofs/FermatsLittleTheorem.html
Euler-Fermat theorem and its proof: http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_Totient_Theorem
Lagrange’s theorem (only for the brave): http://cims.nyu.edu/~kiryl/teaching/aa/les102903.pdf
Number theory textbooks: Gordan Savin’s Numbers, Groups, and Cryptography and George E. Andrews’s Number Theory
Interesting sources of math tricks and problems: Paul Zeitz’s The Art, Craft of Problem Solving and The USSR Olympiad Problem Book, and What is Mathematics by Richard Courant and Herbert Robbins
# Cryptography: A modern use for modular arithmetic
The common analogy used to describe modular arithmetic is fairly simple. All one has to do is look at an analog clock. For example, if it’s 11 AM and you want to know what time it will be in four hours, we instinctively know the answer is 3 PM. This is modular arithmetic, i.e. 11+4 = 3 mod 12. This is an important concept in the technology driven world we live in. Any time a product is purchased on the internet, cryptography comes into play. The remainder of this paper (pun most definitely intended) will describe how ancient modular arithmetic plays a very important role in today’s society.
History of modular arithmetic
The first known publication of modular arithmetic was in the 3rd century B.C.E, in the book Elements, written by Euclid. Within his book, he not only formalized the fundamentals of arithmetic, but also proved it. In what is known as Euclids Lemma, he states that if a prime number divides the product of two different numbers (x and y), then the prime number must also divide one of the numbers (either x or y), but it could also be both. Between the 3rd and 5th centuries a paper publish by Sun Tzu describes a modular arithmetic process known as the Chinese remainder theorem. This theorem is essentially the basis for modern RSA encryption schemes that are present on every banking/e-commerce website. It uses a congruent set of keys to produce the same numerical value. Imagine if there was a lock on a door that two differently cut keys could unlock and open, this is essentially how Chinese remainder theorem works.
Modern modular arithmetic
Oil painting of mathematician and philosopher Carl Friedrich Gauss by G. Biermann (1824-1908). Public Domain.
The modular arithmetic that we use today was discovered by Carl Friedrich Gauss in 1801.
Gauss is famous for numerous discoveries across a wide variety of fields in science and mathematics. Gauss’s proposition, from his book Disquisitiones Arithmeticae, defines modular arithmetic by saying that any integer N belongs to a single residue-class when divided by a number M. The residue-class is represented by the remainder, which can be from 0 to M-1. The remainder is obtained by dividing N by M. Given this fact, Gauss notices that two numbers that differ by a multiple of M are in the same residue-class. He then discusses the role of negative numbers in modular arithmetic. The following is an excerpt from his book:
“The modulus m is usually positive, but there’s no great difficulty in allowing negative moduli (classes modulo m and -m are the same). For a zero modulus, there would be infinitely many residue classes, each containing only one element. [This need not be disallowed.]”
Modular Arithmetic’s Role Today
RSA encryption is named after those who invented it, Ron Rivest, Adi Shamir, and Leonard Adleman (RSA is obtained from their last names). RSA is the process by which information can be passed between two parties without another individual being able to intercept the message. Burt Kaliski has been one of the major contributors to RSA encryption since the 1980’s. I would like to start off with a passage from Burt Kaliski’s paper titled “The Mathematics of the RSA Public-Key Cryptosystem”:
“Sensitive data exchanged between a user and a Web site needs to be encrypted to prevent it from being disclosed to or modified by unauthorized parties. The encryption must be done in such a way that decryption is only possible with knowledge of a secret decryption key. The decryption key should only be known by authorized parties.”
This is a high level description of how RSA encryption works. It is also called public-key encryption, because anyone can obtain a copy of the encryption key it is publically available, but the decryption key cannot be obtained. This makes RSA encryption a secure way of passing data between an individual and a web site.
Simplified view of RSA encryption. Public Domain.
Performing this calculation (encrypting and decrypting text) is fairly simple. With a basic understanding of modular arithmetic it can be accomplished. First a public and private key must be produced by following the steps below:
1. Generate large prime numbers, p and q (these should be hundreds of digits)
2. Compute the modulus n, n = p×q
3. Compute the totient, totient = (p-1)×(q-1)
4. Choose an “e” > 1 that is co-prime to the totient
5. Choose a “d” such that d×e = 1 mod totient
Once those steps have been completed, a public key (n, e) and a private key (n, d) have been generated. The public key can be distributed to anyone, but the private key must be kept safe. It’s easy to see that without the modular arithmetic this algorithm would be easy to discern. One could generate pairs of random numbers until a pair is found that when multiplied together, would equal the modulus n found in step two above. From there, it would be easy to find all numbers co-prime to the totient in step three. Modular arithmetic then comes into play, because it allows infinite pairs of numbers to satisfy the constraint listed in step five, but it would not allow the user to decrypt the message. In other words, 11+4 = 3 mod 12, but also 11+16 = 3 mod 12. This makes it impossible to determine what the original number was (it could be 4 or it could be 16, or any other multiple of 12).
Once the keys have been generated it is easy to encrypt and decrypt text. To encrypt a message “m,” given the public key (n,e) generated above:
C = me mod n
“C” is then the encrypted message that gets passed to the other party.
To decrypt the message “C” created above, all that is required is the inverse of the operation to encrypt:
M = cd mod n
Let’s do an example to illustrate the instructions listed above (note: we will be using small prime factors because the math is simpler).
1. Select a p and q that are prime
1. P = 11
2. Q = 3
2. The modulus n is then equal to P×Q = 11×3 = 33
3. Computing the totient to be equal to (p-1)×(q-1) = (11-1)(3-1) = 20
4. To select an “e” we must find a number that is coprime to 20
1. The smallest value that is coprime to 20 is 3 because 3 is the smallest number that cannot divide 20 evenly, so “e” = 3
5. Now we need to find “d”, d=e^(-1) mod [(p-1)×(q-1)]
1. Using the Euclidian Algorithm we get d = 7
Now let’s say we want to encrypt the message “4.” To do this we need to know the public key, which in our case is (n=33, e=3). All we have to do is compute:
C = 43 mod 33 = 31
We can pass 31 (c=31) along to the website, which will then decrypt it using the private key (33, 7):
M = 317 mod 33 = 4
Our message has been successfully “passed” from one place to another.
Thoughts
Without the work from previous mathematicians, this process would not be possible. Modular arithmetic plays a crucial role in our everyday lives and we don’t even notice it. I think it’s an amazing mathematical concept and provides a deep insight into the world of number theory. Even today there are computers constantly trying to figure out how to factor large prime numbers without success. I don’t know if RSA encryption will stand the test of time, but for now it’s the best we’ve got.
References
http://en.wikipedia.org/wiki/Cryptography#History_of_cryptography_and_cryptanalysis
http://www.britannica.com/EBchecked/topic/920687/modular-arithmetic
http://mathworld.wolfram.com/ChineseRemainderTheorem.html
http://www.mathaware.org/mam/06/Kaliski.pdf
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# 12.6: Determining Asymptotes by Division
Difficulty Level: At Grade Created by: CK-12
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### Determining Asymptotes by Division
In the last section we saw how to find vertical and horizontal asymptotes. Remember, the horizontal asymptote shows the value of y\begin{align*}y\end{align*} that the function approaches for large values of x\begin{align*}x\end{align*}. Let’s review the method for finding horizontal asymptotes and see how it’s related to polynomial division.
When it comes to finding asymptotes, there are basically four different types of rational functions.
#### Finding Asymptotes
Case 1: The polynomial in the numerator has a lower degree than the polynomial in the denominator.
Find the horizontal asymptote of y=2x1\begin{align*}y=\frac{2}{x-1}\end{align*}.
We can’t reduce this fraction, and as x\begin{align*}x\end{align*} gets larger the denominator of the fraction gets much bigger than the numerator, so the whole fraction approaches zero.
The horizontal asymptote is y=0\begin{align*}y = 0\end{align*}.
Case 2: The polynomial in the numerator has the same degree as the polynomial in the denominator.
Find the horizontal asymptote of y=3x+2x1\begin{align*}y=\frac{3x+2}{x-1}\end{align*}.
In this case we can divide the two polynomials:
x1)3x+2¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 33x+35\begin{align*}& \overset{\qquad \qquad \ 3}{x-1 \overline{ ) 3x+2 \;}}\\ & \qquad \underline{-3x+3}\\ & \qquad \qquad \quad 5\end{align*}
So the expression can be written as y=3+5x1\begin{align*}y=3+\frac{5}{x-1}\end{align*}.
Because the denominator of the remainder is bigger than the numerator of the remainder, the remainder will approach zero for large values of x\begin{align*}x\end{align*}. Adding the 3 to that 0 means the whole expression will approach 3.
The horizontal asymptote is y=3\begin{align*}y = 3\end{align*}.
Case 3: The polynomial in the numerator has a degree that is one more than the polynomial in the denominator.
Find any asymptotes of y=4x2+3x+2x1\begin{align*}y=\frac{4x^2+3x+2}{x-1}\end{align*}.
Solution:
We can do long division once again and rewrite the expression as y=4x+7+9x1\begin{align*}y=4x+7+\frac{9}{x-1}\end{align*}. The fraction here approaches zero for large values of x\begin{align*}x\end{align*}, so the whole expression approaches 4x+7\begin{align*}4x + 7\end{align*}.
When the rational function approaches a straight line for large values of x\begin{align*}x\end{align*}, we say that the rational function has an oblique asymptote. In this case, then, the oblique asymptote is y=4x+7\begin{align*}y = 4x + 7\end{align*}.
Case 4: The polynomial in the numerator has a degree that is two or more than the degree in the denominator.
Find any asymptotes of y=x3x1\begin{align*}y=\frac{x^3}{x-1}\end{align*}.
This is actually the simplest case of all: the polynomial has no horizontal or oblique asymptotes.
Notice that a rational function will either have a horizontal asymptote, an oblique asymptote or neither kind. In other words, a function can’t have both; in fact, it can’t have more than one of either kind. On the other hand, a rational function can have any number of vertical asymptotes at the same time that it has horizontal or oblique asymptotes.
### Examples
Find the horizontal or oblique asymptotes of the following rational functions.
#### Example 1
y=3x2x2+4\begin{align*}y=\frac{3x^2}{x^2+4}\end{align*}
When we simplify the function, we get y=312x2+4\begin{align*}y=3-\frac{12}{x^2+4}\end{align*}. There is a horizontal asymptote at y=3\begin{align*}y = 3\end{align*}.
#### Example 2
y=x13x26\begin{align*}y=\frac{x-1}{3x^2-6}\end{align*}
We cannot divide the two polynomials. There is a horizontal asymptote at y=0\begin{align*}y = 0\end{align*}.
#### Example 3
y=x4+1x5\begin{align*}y=\frac{x^4+1}{x-5}\end{align*}
The power of the numerator is 3 more than the power of the denominator. There are no horizontal or oblique asymptotes.
#### Example 4
y=x33x2+4x1x22\begin{align*}y=\frac{x^3-3x^2+4x-1}{x^2-2}\end{align*}
When we simplify the function, we get y=x3+6x7x22\begin{align*}y=x-3+\frac{6x-7}{x^2-2}\end{align*}. There is an oblique asymptote at y=x3\begin{align*}y = x - 3\end{align*}.
### Review
Find all asymptotes of the following rational functions:
1. x2x2\begin{align*}\frac{x^2}{x-2}\end{align*}
2. 1x+4\begin{align*}\frac{1}{x+4}\end{align*}
3. x21x2+1\begin{align*}\frac{x^2-1}{x^2+1}\end{align*}
4. x4x29\begin{align*}\frac{x-4}{x^2-9}\end{align*}
5. x2+2x+14x1\begin{align*}\frac{x^2+2x+1}{4x-1}\end{align*}
6. x3+14x1\begin{align*}\frac{x^3+1}{4x-1}\end{align*}
7. xx3x26x7\begin{align*}\frac{x-x^3}{x^2-6x-7}\end{align*}
8. x42x8x+24\begin{align*}\frac{x^4-2x}{8x+24}\end{align*}
Graph the following rational functions. Indicate all asymptotes on the graph:
1. x2x+2\begin{align*}\frac{x^2}{x+2}\end{align*}
2. x31x24\begin{align*}\frac{x^3-1}{x^2-4}\end{align*}
3. x2+12x4\begin{align*}\frac{x^2+1}{2x-4}\end{align*}
4. xx23x+2\begin{align*}\frac{x-x^2}{3x+2}\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Function
A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$.
Horizontal Asymptote
A horizontal asymptote is a horizontal line that indicates where a function flattens out as the independent variable gets very large or very small. A function may touch or pass through a horizontal asymptote.
Ohm's Law
Ohm's Law states that a current through a conductor that connects two points is directly proportional to the potential difference between its ends. Consider $V=IR$, where $V$ is the voltage or the potential difference, $I$ is the current, and $R$ is the resistance of the conductor.
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# Mastering Arithmetic: The Art of Finding the Average
## Introduction
Finding the average is a fundamental concept in mathematics and statistical analysis that is used in various fields, such as finance, science, and education, among others. It helps to make sense of data by giving a representative value for a set of numbers. This article aims to provide a comprehensive guide to finding an average, exploring its applications in everyday life, understanding its importance in different industries, and discussing various methods to calculate it.
## Mastering Arithmetic: A Guide to Finding the Average
The average is a measure of central tendency that represents the middle value of a set of numbers. It is also known as the mean and is calculated by summing up all the numbers in a set and dividing the result by the total number of values. The formula for finding the average is:
Average = (sum of values)/ (total number of values)
For example, if you want to find the average of 3, 5, 7, and 9, you would add them together (3+5+7+9=24) and divide by the total number of values (4), resulting in an average of six (24/4=6).
To make finding the average easier, you can use a few tips and tricks, such as:
• Organize the data in ascending or descending order.
• Round off the values to make calculations easier.
• Look for patterns in the data to make approximations or estimates.
• Use software or online tools to calculate the average quickly.
## Real-Life Applications of Finding the Average
The average is a handy tool to use in everyday life, such as in calculating the average grades of a group or the average time it takes to complete a task. For instance, if a teacher wants to calculate their students’ average test scores, they would add up all the scores and divide by the total number of students. Similarly, a project manager could calculate the average time it takes to complete a task for their team to help plan future projects more accurately.
## Calculating the Weighted Average: Understanding Its Importance
Another type of average is the weighted average, which assigns different weights to the values in a set, depending on their importance or relevance. It is commonly used in various industries, such as finance, science, and economics, to express an average that factors in certain conditions or factors. For example, in finance, weighted averages can be used to measure investment returns or bond yields, taking into account the size and duration of the investment or bond.
## Finding the Moving Average in Financial Markets
The moving average is a particular type of average that is commonly used in financial analysis to examine trends in stock prices or other financial indicators. It measures the average price of a security over a specified period, usually a set number of days or weeks, to identify patterns in the data. To calculate the moving average, you need to add up the prices for the set period and divide by the total number of days or weeks. As new data becomes available, you can update the moving average by dropping the oldest price and including the newest one. This type of average helps investors to identify the direction of a trend and make informed investment decisions.
## The Art of Averages: Different Ways to Find the Mean
There are various methods to calculate the average, including the traditional method of the mean, which we discussed earlier, as well as the median and mode, and advanced statistical techniques, such as regression analysis and data visualization. The median is the midpoint of a set of values; half of the values are above it, and half are below it. The mode is the value that appears most frequently in a dataset. Advanced statistical techniques use mathematical models to analyze data, making it easier to identify trends and patterns, and make predictions.
## Conclusion
The average is a useful tool for making sense of data and can be applied in various fields. We have covered various topics related to finding the average, from explaining its formula and providing tips to make the process easier, to exploring its applications in everyday life and different industries, and discussing various methods to calculate it. We hope this guide has been helpful in mastering arithmetic and encourages you to practice finding the average using the techniques we have provided.
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# Proof of the Derivative of $a^x$
The proof of the derivative of the exponential function $a^x$, such that $a \gt 0$ and $a \ne 1$, is presented. The derivative of a composite function of the form $a^{u(x)}$ is also presented. Examples with their solutions are included.
## Proof of the Derivative of $a^x$
Let $y = a^x$ , such that $a \gt 0$ and $a \ne 1$
Take the $\ln$ of both sides of the equation $y = a^x$ to obtain
$\ln y = \ln a^x \qquad (1)$
Use the property of $\; \ln \;$ given by $\; \ln a^x = x \ln a \;$ in $(1)$ above and write $\ln y = \ln a^x$ as
$\ln y = x \ln a$
Take the derivative, with respect to $x$, of both sides of the above equation.
$\dfrac{d}{dx} (\ln y) = \dfrac{d}{dx} ( x \ln a) \qquad (2)$
Use the chain rule of differentiation to write the left side of $(2)$ as $\dfrac{d}{dx} (\ln y) = \dfrac{1}{y} \dfrac{dy}{dx}$ and the derivative of the right side of $(2)$ is given by $\dfrac{d}{dx} ( x \ln a) = \ln a$ Substitute the two results above in $(2)$ to obtain
$\dfrac{1}{y} \dfrac{dy}{dx} = \ln a$
Multiply both sides of the above by $y$
$y \dfrac{1}{y} \dfrac{dy}{dx} = y \ln a$
Simplify to obtain
$\dfrac{dy}{dx} = y \; \ln a$
Substitute $y$ by $a^x$ to write the derivative of $a^x$ as
$\dfrac{d \left(a^x \right)}{dx} = (\ln a) \; a^x \qquad (I)$
## Derivative of the Composite Function $y = a^{u(x)}$
We now consider the composite exponential of another function u(x). Use the chain rule of differentiation to write
$\displaystyle \dfrac{d a^{u(x)}}{dx} = \dfrac{d( a^{u})}{du} \dfrac{du}{dx} \qquad (3)$
Use the above result in $(I)$ to write $\dfrac{d( a^{u})}{du} = (\ln a) \; a^u$ and substitute in $(3)$ above to obtain the derivative of the composite function $y = a^{u(x)}$ as
$\displaystyle \dfrac{d \left(a^{u(x)} \right)}{dx} = (\ln a) \; a^u \; \dfrac{du}{dx} \qquad (II)$
Example 1
Find the derivative of the composite exponential functions
1. $f(x) = 2^{-x^4+5x-4}$
2. $g(x) = 3^{\sqrt{x^4+2x}}$
3. $h(x) = 5^{ \left(\dfrac{2x}{3x+2}\right)}$
Solution to Example 1
1. Let $u(x) = -x^4+5x-4$ and therefore $\dfrac{du}{dx} = -4 x^3 + 5$
Apply the rule for the composite exponential function in $(II)$ above
$\displaystyle \dfrac{d}{dx} f(x) = ( \ln 2) 2^u \dfrac{du}{dx} = ( \ln 2) \; 2^{-x^4+5x-4} \times (-4 x^3 + 5 )$
$= ( \ln 2) \; (-4 x^3 + 5 ) \; 2^{-x^4+5x-4}$
2. Let $u(x) = \sqrt{x^4+2x}$ and therefore $\dfrac{du}{dx} = \dfrac{1}{2} \dfrac{4 x^3 + 2}{\sqrt{x^4+2x}} = \dfrac{2 x^3 + 1}{\sqrt{x^4+2x}}$.
Apply the rule of differentiation for the composite exponential function in $(II)$ above
$\displaystyle \dfrac{d}{dx} g(x) = (\ln 3) 3^u \dfrac{du}{dx} = (\ln 3) 3^{\sqrt{x^2+1}} \times \dfrac{2 x^3 + 1}{\sqrt{x^4+2x}}$
$= ( \ln 3) \dfrac{2 x^3 + 1}{\sqrt{x^4+2x}} \; 3^{\sqrt{x^2+1}}$
3. Let $u(x) = \dfrac {2x}{3x+2}$ and therefore $\dfrac{du}{dx} = \dfrac{4}{\left(3x+2\right)^2}$
The use of the rule of differentiation for the composite exponential function obtained in $(II)$ above gives
$\displaystyle \dfrac{d}{dx} h(x) = ( \ln 5 ) \; 5^u \; \dfrac{du}{dx}$
$= ( \ln 5) \; \dfrac{4}{\left(3x+2\right)^2} \; 5^{\left(\frac{2x}{3x+2}\right)}$
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# ∆ABD is a right triangle right-angled at A and AC ⊥ BD. Show that
`
Question:
$\triangle \mathrm{ABD}$ is a right triangle right-angled at $\mathrm{A}$ and $\mathrm{AC} \perp \mathrm{BD}$. Show that
(i) $\mathrm{AB}^{2}=\mathrm{BC} \cdot \mathrm{BD}$
(ii) $\mathrm{AC}^{2}=\mathrm{BC} . \mathrm{DC}$
(iii) $\mathrm{AD}^{2}=\mathrm{BD} \cdot \mathrm{CD}$
(iv) $\frac{\mathrm{AB}^{2}}{\mathrm{AC}^{2}}=\frac{\mathrm{BD}}{\mathrm{DC}}$
Solution:
In $\triangle A B D$ and $\triangle A B C$,
$\angle A C B=\angle A=90^{\circ}$
$\angle B=\angle B$ (Common angle)
So, by AA criterion $\triangle A B D \sim \triangle C B A$
$\therefore \frac{A B}{B C}=\frac{B D}{A B}=\frac{A D}{A C}$
$\therefore \frac{A B}{B C}=\frac{B D}{A B}$
$\therefore A B^{2}=B D \cdot B C$....(1)
(ii) In $\triangle A B D$ and $\triangle A C D$,
$\angle C=\angle A=90^{\circ}$
$\angle D=\angle D \quad$ (Common angle)
So, by AA criterion $\triangle A B D \sim \triangle C A D$
$\therefore \frac{A B}{A C}=\frac{B D}{A D}=\frac{A D}{C D}$
$\therefore \frac{B D}{A D}=\frac{A D}{C D}$
$\therefore A D^{2}=B D \cdot C D$....(2)
(iii) We have shown that $\triangle A B D$ is similar to $\triangle C B A$ and $\triangle A B D$ is similar to $\triangle C A D$ therefore, by the property of transitivity $\triangle C B A$ is similar to $\triangle C A D$.
$\therefore \frac{B C}{A C}=\frac{A B}{A D}=\frac{A C}{C D}$
$\therefore \frac{B C}{A C}=\frac{A C}{C D}$
$\therefore A C^{2}=B C \cdot C D$....(3)
(iv) Now to obtained $\mathrm{AB}^{2} / \mathrm{AC}^{2}=\mathrm{BD} / \mathrm{DC}$, we will divide equation (1) by equation (2) as shown below,
$\therefore \frac{A B^{2}}{A C^{2}}=\frac{B D \cdot B C}{B C \cdot C D}$
Canceling BC we get,
$\frac{A B^{2}}{A C^{2}}=\frac{B D}{C D}$
Therefore, $\frac{A B^{2}}{A C^{2}}=\frac{B D}{C D}$
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# How to Round to the Nearest Integer
## What does it mean to round a number?
Rounding means to make a number shorter or simpler, but keeping it as close in value as possible to the original number. Let’s take a closer look:
## Rounding to the nearest integer
The most common type of rounding is to round to the nearest integer. The rule for rounding is simple: look at the digits in the tenth’s place (the first digit to the right of the decimal point). If the digit in the tenths place is less than 5, then round down, which means the units digit remains the same; if the digit in the tenths place is 5 or greater, then round up, which means you should increase the unit digit by one.
Here are a couple other things to know:
Observation #1: under most circumstances, rounding changes the decimal to whatever integer is closer. For example, 4.3 is rounded to 4, and 4.9 is rounded to 5. The exception is when the decimal is smack dab between two integers: 4.5 is exactly equidistant to both 4 and 5, but because of the “tie-breaker” rule of rounding, anything with a 5 in the tenths digit is rounded up. This is the only case in which the “go to the closer integer” interpretation will fail.
Observation #2: Do NOT double-round. Some people look at a number like, say, 7.49, and they erroneously think — well, that 9 would round the 4 up to 5, and then a 5 gets round up, to this number would round to 8. WRONG! Never round a number “in stages.” Rounding is a one-shot deal, a one-step process. When the number we need to round is 7.49, we only need notice that the tenth’s digit is a 4, which means the number is rounded down to 7. One step, case closed. In fact, all of the following numbers get rounded to 7:
Here’s the truly mind-boggling part:
How many numbers would there be larger than this last number, but still lower than 7.5? INFINITY! No matter how many additional 9’s we slap on to the end of that number, there’s still a continuous infinity of decimals larger than that number and below 7.5 No matter how finely we chop up the real number line, each tiny fragment of the line, no matter how small, still contains a continuous infinite of numbers.
Observation #3: the “tie-breaker” rule can be tricky with negative values. For example, +2.5 gets rounded up to 3, but -2.5 gets rounded down … to -3. As with positive numbers, the negative number ending in .5 is rounded to the higher absolute value integer, but with negatives, that’s rounding down. (This is not the only way to formulate this rule, but this is the convention that ETS follows.)
## Rounding to any other decimal place
Rounding to the nearest integer is really rounding to the nearest units place. Sometimes, you will be asked to round to the nearest hundreds, or to the nearest hundredths — to some decimal place other than the units place. The rule is just a more generalized version of the previous rounding rule.
Suppose we are asked to round to some specific decimal place — call this the “target place.” You always look at only one digit, the digit immediately to the right of the target place. If this digit immediately to the right is \{0, 1, 2, 3, 4\}, then you “round down”, and the digit in the target place remains unchanged. If this digit immediately to the right is \{5, 6, 7, 8, 9\}, then you “round up”, and the digit in the target place increases by 1.
## What do you have to know about how to round to the nearest integer for the GRE?
Not many GRE questions will say, “Here’s a number: round it to the nearest such-and-such.” By contrast, many questions, in the course of asking something else, could ask you to round your answer to the nearest such-and-such. In this way, rounding is one math skill you need to know for the GRE. There are a few tricky issues, which I will address here.
## Other cases of rounding
Very occasionally, a GRE math question may ask you not to round to a particular decimal place, but rather to the nearest multiple of something. For example, suppose you are asked to round, say, to the nearest 0.05 — how do you do that?
Well, let’s think about the results first. The result of rounding to the nearest 0.05 would be something divisible by 0.05 — that is to say, a decimal with either a 0 or a 5 in the hundredth place, no digits to the right of that, and any digits to the left of that. The following are examples of numbers which could be the result of rounding to the nearest 0.05:
0.35
1.40
3.15
5.2
8
Notice: the second, (b) is the square root of 2 (sqrt{2} = 1.414213562 ....) rounded to the nearest 0.05, and the third, (c), is \Large{\pi} rounded to the nearest 0.05.
Let’s demonstrate the rounding by means of an example. What numbers, when rounded to the nearest 0.05, would be rounded to 2.35? Well, for starters, 2.35 and other “tenths” around it would be rounded to 2.35
2.32 — rounded down to 2.30
2.33 — rounded up to 2.35
2.34 — rounded up to 2.35
2.35 — stays at value
2.36 — rounded down to 2.35
2.37 — rounded down to 2.35
2.38 — rounded up to 2.40
Now, the tricky regions are those between the values that are rounded in different directions. For example, 2.32 is rounded down and 2.33 is rounded up, so something fishy is happening between those two. Let’s think about the hundredths between 2.32 and 2.33 — exactly between them is 2.325, the midpoint between 2.30 and 2.35, and like all midpoints, according to the “tiebreaker” rule, it gets rounded up. Thus:
2.320 —- rounded down to 2.30
2.321 —- rounded down to 2.30
2.322 —- rounded down to 2.30
2.323 —- rounded down to 2.30
2.324 —- rounded down to 2.30
2.325 —- rounded up to 2.35 (the “tie-breaker” rule)
2.326 —- rounded up to 2.35
2.327 —- rounded up to 2.35
2.328 —- rounded up to 2.35
2.329 —- rounded up to 2.35
2.330 —- rounded up to 2.35
This is all probably far more detail than you will need to know for GRE math, but this does demonstrate the steps you would take to round any decimal to the nearest 0.05. By analogy, you could round any decimal to any specified multiple.
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# Proving Alternate Interior Angles are Congruent (the same)
The Alternate Interior Angles Theorem states that If two parallel straight lines are intersected by a third straight line (transversal), then the angles inside (between) the parallel lines, on opposite sides of the transversal are congruent (identical).
This is illustrated in the image below:
We see two parallel lines and a third line (transversal) intersecting (crossing or cutting through) both of them. The green shaded angles are: (1) inside (between) the two parallel lines, (2) congruent (identical or the same), and (3) on opposite sides of the transversal.
This is true for the other two unshaded interior angles. It is also true for the alternate exterior angles (but not proved here).
# Axioms
Proofs are built on two things: (1) postulates, axioms, or hypotheses – these are things that are assumed to be true, but can’t be proved. In general, we try to use the fewest number of axioms we can; (2) other proofs.
This proof depends on two axioms: (1) if you pick any two distinct points on a straight line, the angle between those two points will be 180°; (2) if you take any two intersecting straight lines and shift one of the lines so it it is in a different position, but still parallel to its original position, the angle between the two intersecting lines stays the same.
# Proof
1) Consider two intersecting lines:
At the point of intersection, there is an angle (which I call A).
2) If the line is straight, the angle between any two points on that line must be 180° (axiom #1 from above):
3) If one angle is A degrees, then obviously the other angle must be 180° – A:
4) This is true for both straight lines:
5) This means the remaining angle must be 180° – (180° – A) = A degrees:
This proves that angles on alternate sides of the transversal (at the point of intersection) are congruent (identical).
6) If we draw a line parallel to one of the lines (it doesn’t matter which) and it intersects the other line (that line is now called a transversal), we know that the angles of intersection must be the same (axiom #2 from above).
If the angles of intersection are the same for both lines, then the alternate interior angles must be the same:
The proof is exactly the same for alternate exterior angles except we focus on the exterior instead of the interior angles (which is illustrated in the image above).
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 3.5: Graphing a Linear Function Using a Table of Values
Difficulty Level: At Grade Created by: CK-12
## Graphing a Linear Function Using a Table of Values
Objectives
The lesson objectives for Graphing a Linear Function using a table of Values are:
• Representing a problem with a linear function
• Creating a table of values from the linear function
• Using the table of values to draw the graph.
Introduction
In this concept you will learn to graph a linear function by using the function to create a table of values. These values will be the coordinates of the points that will be plotted to draw the graph of the function. A linear function will result in a graph that is a straight line.
Watch This
Guidance
Bonita will be celebrating her sixteenth birthday next month. Her parents would like to give her a surprise party at the local pool. To rent the pool for a private party costs $100 plus$55.00 for each hour the pool is rented. Write a linear function to represent the cost of the pool party and list five prices from which her parents can choose.
The cost of renting the pool is $100. This amount is a fee that must be paid to rent the pool. In addition, Bonita’s parents will also have to pay$55.00 for each hour the pool is rented. Therefore, the linear function to represent this situation is \begin{align*}y=55x+100\end{align*} where ‘\begin{align*}y\end{align*}’ represents the cost in dollars and ‘\begin{align*}x\end{align*}’ represents the time, in hours, that the pool is rented.
\begin{align*}y=55x+100\end{align*} - To determine five options for her parents, replace ‘\begin{align*}x\end{align*}’ with the values 1 to 5 and calculate the cost for each of these hours.
\begin{align*}& \ y=55x+100 && \ y=55x+100 && \ y=55x+100 && \ y=55x+100 && \ y=55x+100\\ & \ y=55(1)+100 && \ y=55(2)+100 && \ y=55(3)+100 && \ y=55(4)+100 && \ y=55(5)+100\\ & \boxed{y=\155} && \boxed{y=\210} && \boxed{y=\265} && \boxed{y=\320} && \boxed{y=\375}\end{align*}
These results can now be represented in a table of values:
\begin{align*}& X(hours) \qquad 1 \qquad \quad \ \ 2 \qquad \quad \ 3 \qquad \quad \ 4 \qquad \quad \ \ 5\\ & Y(Cost) \qquad \155 \qquad \210 \qquad \265 \qquad \320 \qquad \375\end{align*}
The values in the table represent the coordinates of points that are located on the graph of \begin{align*}y=55x+100\end{align*}.
\begin{align*}(1,155);(2,210);(3,265);(4,320);(5,375)\end{align*}
Bonita’s parents can use the table of values and/or the graph to make their decision.
Example A
Complete the table of values for the linear function \begin{align*}3x+2y=-6\end{align*}.
Before completing the table of values, solve the given function in terms of ‘\begin{align*}y\end{align*}’. This step is not necessary, but it does simplify the calculations.
\begin{align*}& \qquad \ 3x+2y=-6\\ &3x-3x+2y=-3x-6\\ &\qquad \qquad \ \ 2y=-3x-6\\ &\qquad \qquad \ \frac{2y}{2}=\frac{-3x}{2}-\frac{6}{2}\\ &\qquad \qquad \ \ \boxed{y=\frac{-3x}{2}-3}\end{align*}
\begin{align*}& y=\frac{-3x}{2}-3 && y=\frac{12}{2}-3 && y=\frac{-3x}{2}-3 && y=0-3\\ & y=\frac{-3({\color{red}-4})}{2}-3 && y=6-3 && y=\frac{-3({\color{red}0})}{2}-3 && \boxed{y=-3}\\ & && \boxed{y=3}\\ & y=\frac{-3x}{2}-3 && y=\frac{-6}{2}-3 && y=\frac{-3x}{2}-3 && y=\frac{-18}{2}-3\\ & y=\frac{-3({\color{red}2})}{2}-3 && y=-3-3 && y=\frac{-3({\color{red}6})}{2}-3 && y=-9-3\\ & && \boxed{y=-6} && && \boxed{y=-12}\end{align*}
\begin{align*}y=-\frac{3}{2}x-3\end{align*}
\begin{align*}X\end{align*} \begin{align*}Y\end{align*}
\begin{align*}{\color{red}-4}\end{align*} \begin{align*}3\end{align*}
\begin{align*}{\color{red}0}\end{align*} \begin{align*}-3\end{align*}
\begin{align*}{\color{red}2}\end{align*} \begin{align*}-6\end{align*}
\begin{align*}{\color{red}6}\end{align*} \begin{align*}-12\end{align*}
Example B
Use technology to create a table of values for the linear function \begin{align*}f(x)=-\frac{1}{2}x+4\end{align*}.
When the table is set up, you choose the beginning number as well as the pattern for the numbers in the table. In this table, the beginning value for ‘\begin{align*}x\end{align*}’ is -2 and the difference between each number is +2. The table is consecutive, even numbers. When consecutive numbers are used as the input numbers (\begin{align*}x-\end{align*}values), there is a definite pattern in the output numbers (\begin{align*}y-\end{align*}values). This will be expanded upon in a later lesson.
Example C
Complete the table of values for the following linear function, and use those values to graph the function.
\begin{align*}x-2y=4 && x-x-2y=-x+4 && -2y=-x+4 && \frac{-2y}{-2}=\frac{-x}{-2}+\frac{4}{-2} && \boxed{y=\frac{1}{2}x-2}\end{align*}
\begin{align*}& \ y=\frac{1}{2}x-2 && \ y=\frac{1}{2}x-2 && \ y=\frac{1}{2}x-2 && \ y=\frac{1}{2}x-2\\ & \ y=\frac{1}{2}({\color{red}-4})-2 && \ y=\frac{1}{2}({\color{red}0})-2 && \ y=\frac{1}{2}({\color{red}2})-2 && \ y=\frac{1}{2}({\color{red}6})-2\\ & \ y=-2-2 && \ y=0-2 && \ y=1-2 && \ y=3-2\\ & \boxed{y=-4} && \boxed{y=-2} && \boxed{y=-1} && \boxed{y=1}\end{align*}
\begin{align*}y=\frac{1}{2}x-2\end{align*}
\begin{align*}X\end{align*} \begin{align*}Y\end{align*}
\begin{align*}{\color{red}-4}\end{align*} \begin{align*}-4\end{align*}
\begin{align*}{\color{red}0}\end{align*} \begin{align*}-2\end{align*}
\begin{align*}{\color{red}2}\end{align*} \begin{align*}-1\end{align*}
\begin{align*}{\color{red}6}\end{align*} \begin{align*}1\end{align*}
Vocabulary
Linear Function
The linear function is a relation between two variables, usually \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, in which each value of the independent variable \begin{align*}(x)\end{align*} is mapped to one and only one value of the dependent variable \begin{align*}(y)\end{align*}.
Guided Practice
1. Complete the following table of values for the linear function \begin{align*}3x-2y=-12\end{align*}
\begin{align*}3x-2y=-12\end{align*}
\begin{align*}X\end{align*} \begin{align*}Y\end{align*}
\begin{align*}{\color{red}-6}\end{align*}
\begin{align*}{\color{red}-4}\end{align*}
\begin{align*}{\color{red}0}\end{align*}
\begin{align*}{\color{red}6}\end{align*}
2. Use technology to complete a table of values for the linear function \begin{align*}2x-y=-8\end{align*} and use the coordinates to draw the graph.
3. A local telephone company charges a monthly fee of $25.00 plus$0.09 per minute for calls within the United States. If Sam talks for 200 minutes in one month, calculate the cost of his telephone bill.
1. \begin{align*}3x-2y=-12\end{align*} Solve the equation in terms of the variable ‘\begin{align*}y\end{align*}’.
\begin{align*}3x-3x-2y=-3x-12 && -2y=-3x-12 && \frac{-2y}{-2}=\frac{-3x}{-2}-\frac{12}{-2}\end{align*}
\begin{align*}\boxed{y=\frac{3}{2}x+6}\end{align*} Substitute the given values for ‘\begin{align*}x\end{align*}’ into the function.
\begin{align*}& \ y=\frac{3}{2}x+6 && \ y=\frac{3}{2}x+6 && \ y=\frac{3}{2}x+6 && \ y=\frac{3}{2}x+6\\ & \ y=\frac{3}{2}({\color{red}-6})+6 && \ y=\frac{3}{2}({\color{red}-4})+6 && \ y=\frac{3}{2}({\color{red}0})+6 && \ y=\frac{3}{2}({\color{red}6})+6\\ & \ y=-9+6 && \ y=-6+6 && \ y=0+6 && \ y=9+6\\ & \boxed{y=-3} && \boxed{y=0} && \boxed{y=6} && \boxed{y=15}\end{align*}
\begin{align*}3x-2y=-12\end{align*}
\begin{align*}X\end{align*} \begin{align*}Y\end{align*}
\begin{align*}{\color{red}-6}\end{align*} \begin{align*}-3\end{align*}
\begin{align*}{\color{red}-4}\end{align*} \begin{align*}0\end{align*}
\begin{align*}{\color{red}0}\end{align*} \begin{align*}6\end{align*}
\begin{align*}{\color{red}6}\end{align*} \begin{align*}15\end{align*}
2. \begin{align*}2x-y=-8\end{align*} To enter the function into the calculator, it must be in the form \begin{align*}y= \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}.
Solve the function in terms of the letter ‘\begin{align*}y\end{align*}’.
\begin{align*}&2x-y=-8 && 2x-2x-y=-2x-8 && -y=-2x-8 && \frac{-y}{-1}=\frac{-2x}{-1}\frac{-8}{-1}\\ & \boxed{y=2x+8}\end{align*}
The graph can also be done using technology. The table can be used to set the window.
3. \begin{align*}y=.09x+25\end{align*} Write a linear function to represent the word problem.
\begin{align*}y &=.09(200)+25 && \text{Substitute the time of} \ 200 \ \text{minutes for the variable} \ x'.\\ y&=\43.00\end{align*}
The cost of Sam’s telephone bill is 43.00. Summary In this lesson, you have learned how to evaluate a linear function with given values. These values were given in table form. The table was completed by entering the values obtained by substituting the given values for ‘\begin{align*}x\end{align*}’ into the linear function. These values were the coordinates of points that were located on the graph of the linear function. The points were used to draw the graph on a Cartesian plane. The lesson then extended into the world of technology. The graphing calculator was used to create a table of values as well as to create the graph of the function. The more you practice the keystrokes for performing these tasks on the calculator, the more efficient you will become. Problem Set Solve each of the following linear functions in terms of the variable ‘\begin{align*}y\end{align*}’. 1. \begin{align*}2x-3y=18\end{align*} 2. \begin{align*}4x-2y=10\end{align*} 3. \begin{align*}3x-y=8\end{align*} 4. \begin{align*}5x+3y=-12\end{align*} 5. \begin{align*}3x-2y-2=0\end{align*} For each of the following linear functions, create a table of values that contains four coordinates: 1. \begin{align*}y=-4x+5\end{align*} 2. \begin{align*}5x+3y=15\end{align*} 3. \begin{align*}4x-3y=6\end{align*} 4. \begin{align*}2x-2y+2=0\end{align*} 5. \begin{align*}2x-3y=9\end{align*} For each of the linear functions, complete the table of values and use the values to draw the graph. 1. \begin{align*}y=-2x+1\end{align*} \begin{align*}& x \qquad -3 \qquad 0 \qquad 1 \qquad 5\\ & y\end{align*} 1. \begin{align*}x=2y-3\end{align*} \begin{align*}& x \qquad -4 \qquad 0 \qquad 2 \qquad 6\\ & y\end{align*} 1. \begin{align*}3x+2y=8\end{align*} \begin{align*}& x \qquad -6 \qquad -2 \qquad 0 \qquad 4\\ & y\end{align*} 1. \begin{align*}4(y-1)=12x-7\end{align*} \begin{align*}& x \qquad -2 \qquad 0 \qquad 3 \qquad 7\\ & y\end{align*} 1. \begin{align*}\frac{1}{2}x+\frac{1}{3}y=6\end{align*} \begin{align*}& x \qquad 0 \qquad 4 \qquad 6 \qquad 10\\ & y\end{align*} Using technology, create a table of values for each of the following linear functions. Using technology, graph each of the linear functions. 1. \begin{align*}y=-2x+3\end{align*} 2. \begin{align*}y=-\frac{1}{2}x-3\end{align*} 3. \begin{align*}y=\frac{4}{3}x-2\end{align*} Mr. Red is trying to estimate the cost of renting a car to go on vacation. He has contacted a rental agency and has obtained the following information. The agency charges a daily rate of78.00 for the vehicle plus 45 cents per mile. If Mr. Red has 350 set aside for travel, create a table of values that will give him approximate distances that he can travel with this rental car. Answers Solve each of the following linear functions... \begin{align*}& \qquad \ 2x-3y=18\\ & 2x-2x-3y=-2x+18\\ & \qquad \quad \ -3y=-2x+18\\ & \qquad \quad \ \ \frac{-3y}{-3}=\frac{-2x}{-3}+\frac{18}{-3}\\ & \qquad \qquad \quad \boxed{y=\frac{2}{3}x-6}\end{align*} \begin{align*}& \qquad \ 3x-y=8\\ &3x-3x-y=-3x+8\\ & \qquad \quad \ -y=-3x+8\\ & \qquad \quad \ \frac{-y}{-1}=\frac{-3x}{-1}+\frac{8}{-1}\\ & \qquad \qquad \boxed{y=3x-8}\end{align*} \begin{align*}& \quad \ \ 3x-2y-2=0\\ & 3x-2y-2+2=0+2\\ & \qquad \quad \ 3x-2y=2\\ & \quad 3x-3x-2y=-3x+2\\ & \qquad \qquad \ -2y=-3x+2\\ & \qquad \qquad \ \frac{-2y}{-2}=\frac{-3x}{-2}+\frac{2}{-2}\\ & \qquad \qquad \quad \ \boxed{y=\frac{3}{2}x-1}\end{align*} For each of the following linear functions... 1. \begin{align*}y=-4x+5\end{align*} \begin{align*}X\end{align*} \begin{align*}Y\end{align*} \begin{align*}{\color{red}-3}\end{align*} \begin{align*}17\end{align*} \begin{align*}{\color{red}0}\end{align*} \begin{align*}5\end{align*} \begin{align*}{\color{red}2}\end{align*} \begin{align*}-3\end{align*} \begin{align*}{\color{red}6}\end{align*} \begin{align*}-19\end{align*} \begin{align*}4x-3y&=6\\ 4x-4x-3y&=-4x+6\\ -3y&=-4x+6\\ \frac{-3y}{-3}&=\frac{-4x}{-3}+\frac{6}{-3}\\ y&=\frac{4}{3}x-2\end{align*} \begin{align*}X\end{align*} \begin{align*}Y\end{align*} \begin{align*}{\color{red}-3}\end{align*} \begin{align*}-6\end{align*} \begin{align*}{\color{red}0}\end{align*} \begin{align*}-2\end{align*} \begin{align*}{\color{red}6}\end{align*} \begin{align*}6\end{align*} \begin{align*}{\color{red}9}\end{align*} \begin{align*}10\end{align*} \begin{align*}2x-3y&=9\\ 2x-2x-3y&=-2x+9\\ -3y&=-2x+9\\ \frac{-3y}{-3}&=\frac{-2x}{-3}+\frac{9}{-3}\\ y&=\frac{2}{3}x-3\end{align*} \begin{align*}X\end{align*} \begin{align*}Y\end{align*} \begin{align*}{\color{red}-3}\end{align*} \begin{align*}-5\end{align*} \begin{align*}{\color{red}0}\end{align*} \begin{align*}-3\end{align*} \begin{align*}{\color{red}6}\end{align*} \begin{align*}1\end{align*} \begin{align*}{\color{red}9}\end{align*} \begin{align*}3\end{align*} For each of the linear functions... 1. \begin{align*}y=-2x+1\end{align*} \begin{align*}& x \qquad {\color{red}-3} \qquad {\color{red}0} \qquad \quad \ {\color{red}1} \qquad \quad \ {\color{red}5}\\ & y \qquad \ \ 7 \qquad \ 1 \qquad -1 \qquad -9\end{align*} \begin{align*}3x+2y&=8\\ 3-3x+2y&=-3x+8\\ 2y&=-3x+8\\ \frac{2y}{2}&=\frac{-3x}{2}+\frac{8}{2}\\ y &= \frac{-3}{2}x+4\end{align*} \begin{align*}& x \qquad {\color{red}-6} \qquad {\color{red}-2} \qquad {\color{red}0} \qquad \quad {\color{red}4}\\ & y \qquad \ 13 \qquad \ \ 7 \qquad 4 \qquad -2\end{align*} \begin{align*}\frac{1}{2}x+\frac{1}{3}y&=6\\ \frac{1}{2}x-\frac{1}{2}x+\frac{1}{3}y&=-\frac{1}{2}x+6\\ \frac{1}{3}y&=-\frac{1}{2}x+6\\ \frac{\frac{1}{3}y}{\frac{1}{3}}&=-\frac{1}{2}\div \left(\frac{1}{3}\right)x+6 \div \left(\frac{1}{3}\right)\\ y &= -\frac{1}{2}\left(\frac{3}{1}\right)x+6\left(\frac{3}{1}\right)\\ y &= -\frac{3}{2}x+18\end{align*} \begin{align*}& x \qquad \ {\color{red}0} \qquad \ {\color{red}4} \qquad \ {\color{red}6} \qquad {\color{red}10}\\ & y \qquad 18 \qquad 12 \qquad 9 \qquad \ 3\end{align*} Using technology, create a table of values... 1. \begin{align*}y=-2x+3\end{align*} 1. \begin{align*}y=\frac{4}{3}x-2\end{align*} Mr. Red is trying to estimate... \begin{align*}y &=0.45x+78 && y =0.45x+78 && y =0.45x+78\\ y &=0.45({\color{red}100})+78 && y =0.45({\color{red}200})+78 && y =0.45({\color{red}300})+78\\ y&=\123 && y=\168 && y=\213\\ \\ y &=0.45x+78 && y=0.45x+78 && y=0.45x+78\\ y &=0.45({\color{red}400})+78 && y =0.45({\color{red}500})+78 && y =0.45({\color{red}600})+78\\ y&=\258 && y=\303 && y=\348\end{align*} \begin{align*}& x \qquad 100 \qquad 200 \qquad 300 \qquad 400 \qquad 500 \qquad 600\\ & y \qquad 123 \qquad 168 \qquad 213 \qquad 258 \qquad 303 \qquad 348\end{align*} If Mr. Red has350 set aside for travel for his vacation, he can drive approximately 600 miles with the rental vehicle.
## Graphing a Linear Function Using the X and Y-Intercepts
Objectives
The lesson objectives for Graphing a Linear Function using a table of Values are:
• Understanding the \begin{align*}x\end{align*} and \begin{align*}y-\end{align*}intercepts
• Determining the \begin{align*}x\end{align*} and \begin{align*}y-\end{align*}intercepts for a given linear function
• Using the \begin{align*}x\end{align*} and \begin{align*}y-\end{align*}intercepts to graph the linear function
Introduction
To graph a linear function, you need to plot only two points. These points can then be lined up with a straight edge and joined to graph the straight line. Two points that can be used to graph a linear function are the \begin{align*}x-\end{align*}intercept and the \begin{align*}y-\end{align*}intercept. The \begin{align*}x-\end{align*}intercept is simply a point that is located on the \begin{align*}x-\end{align*}axis. Its coordinates are \begin{align*}(x, 0)\end{align*}. A \begin{align*}y-\end{align*}intercept is a point located on the \begin{align*}y-\end{align*}axis. Its coordinates are \begin{align*}(0, y)\end{align*}. Graphing a linear function by plotting the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*} intercepts is often referred to as the intercept method.
Watch This
Guidance
The linear function \begin{align*}4x+2y=8\end{align*} can be graphed by using the intercept method.
\begin{align*}& \text{To determine the }x-\text{intercept, let } y=0. && \text{To determine the } y-\text{intercept, let } x=0.\\ & \text{Solve for} \ x'. && \text{Solve for} \ y'.\\ & 4x+2y=8 && 4x+2y=8\\ & 4x+2({\color{red}0})=8 && 4({\color{red}0})+2y=8\\ & 4x+{\color{red}0}=8 && {\color{red}0}+2y=8\\ & 4x=8 && 2y=8\\ & \frac{4x}{4}=\frac{8}{4} && \frac{2y}{2}=\frac{8}{2}\\ & x=2 && y=4\\ & \text{The} \ x-\text{intercept is} \ (2, 0) && \text{The} \ y- \text{intercept is} \ (0, 4)\end{align*}
Plot the \begin{align*}x-\end{align*}intercept on the \begin{align*}x-\end{align*}axis and the \begin{align*}y-\end{align*}intercept on the \begin{align*}y-\end{align*}axis. Join the two points with a straight line.
Example A
Identify the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts for each line.
(a) \begin{align*}2x+y-6=0\end{align*}
(b) \begin{align*}\frac{1}{2}x-4y=4\end{align*}
(a) \begin{align*}&\text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\ & 2x+y-6=0 && 2x+y-6=0\\ & 2x+({\color{red}0})-6=0 && 2({\color{red}0})+y-6=0\\ & 2x-6=0 && y-6=0\\ & 2x-6+6=0+6 && y-6+6=0+6\\ & 2x=6 && y=6\\ & \frac{2x}{2}=\frac{6}{2} && \text{The} \ y- \text{intercept is} \ (0, 6)\\ & x=3\\ & \text{The} \ x-\text{intercept is} \ (3, 0)\end{align*}
(b) \begin{align*}& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\ & \frac{1}{2}x-4y=4 && \frac{1}{2}x-4y=4\\ & \frac{1}{2}x-4({\color{red}0})=4 && \frac{1}{2}({\color{red}0})-4y=4\\ & \frac{1}{2}x-0=4 && 0-4y=4\\ & \frac{1}{2}x=4 && 4y=4\\ & \overset{1}{\cancel{2}}\left(\frac{1}{\cancel{2}}\right)x=2(4) && \frac{4y}{4}=\frac{4}{4}\\ & x=8 && y=1\\ & \text{The} \ x-\text{intercept is} \ (8, 0) && \text{The} \ y- \text{intercept is} \ (0, 1)\end{align*}
Example B
Use the intercept method to graph \begin{align*}2x-3y=-12\end{align*}
\begin{align*}& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\ & 2x-3y=-12 && 2x-3y=-12\\ & 2x-3({\color{red}0})=-12 && 2({\color{red}0})-3y=-12\\ & 2x-0=-12 && 0-3y=-12\\ & 2x=-12 && -3y=-12\\ & \frac{2x}{2}=\frac{-12}{2} && \frac{-3y}{-3}=\frac{-12}{-3}\\ & x=-6 && y=4\\ & \text{The} \ x-\text{intercept is} \ (-6, 0) && \text{The} \ y- \text{intercept is} \ (0, 4)\end{align*}
Example C
The \begin{align*}x-\end{align*}intercept is (-8, 0)
The \begin{align*}y-\end{align*}intercept is (0, 4)
Use the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts of the graph to identify the linear function that matches the graph.
\begin{align*}& y=2x-8 && 2x+y-8=0 && x-2y+8=0\\ & ({\color{red}0})=2x-8 && 2x+({\color{red}0})-8=0 && x-2({\color{red}0})+8=0\\ & 0=2x-8 &&2x-8=0 && x-0+8=0\\ & 0-2x=2x-2x-8 && 2x-8+8=0+8 && x+8=0\\ & -2x=-8 && 2x=8 && x+8-8=0-8\\ & \frac{-2x}{-2}=\frac{-8}{-2} && \frac{2x}{2}=\frac{8}{2} && x=-8\\ & x=4 && x=4 && \text{The} \ x-\text{intercept is} \ (-8, 0)\\ & \text{The} \ x-\text{intercept is} \ (4, 0) && \text{The} \ x- \text{intercept is} \ (4, 0) && \text{This matches the graph.}\end{align*}
This does not match the graph. This does not match the graph. Confirm the \begin{align*}y-\end{align*}intercept.
\begin{align*} x-2y+8&=0 && \quad \ \ -2y+8=0 && -2y=-8 && y=4\\ ({\color{red}0})-2y+8&=0 && -2y+8-8=0-8 && \ \frac{-2y}{-2}=\frac{-8}{-2}\end{align*}
The \begin{align*}y-\end{align*}intercept is (0, 4). This matches the graph.
The linear function that matches the graph is
\begin{align*}\boxed{x-2y+8=0}\end{align*}
Vocabulary
Intercept Method
The intercept method is a way of graphing a linear function by using the coordinates of the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts. The graph is drawn by plotting these coordinates on the Cartesian plane and joining them with a straight line.
\begin{align*}X-\end{align*}intercept
A \begin{align*}x-\end{align*}intercept of a relation is the \begin{align*}x-\end{align*}coordinate of the point where the relation intersects the \begin{align*}x-\end{align*}axis.
\begin{align*}Y-\end{align*}intercept
A \begin{align*}y-\end{align*}intercept of a relation is the \begin{align*}y-\end{align*}coordinate of the point where the relation intersects the \begin{align*}y-\end{align*}axis.
Guided Practice
1. Identify the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts of the following linear functions:
(i) \begin{align*}2(x-3)=y+4\end{align*}
(ii) \begin{align*}3x+\frac{2}{3}y-3=0\end{align*}
2. Use the intercept method to graph the following relation:
(i) \begin{align*}5x+2y=-10\end{align*}
3. Use the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts of the graph, to match the graph to its function.
(i) \begin{align*}2x+y=6\end{align*}
(ii) \begin{align*}4x-3y-12=0\end{align*}
(iii) \begin{align*}5x+3y=15\end{align*}
1. (i) \begin{align*}2(x-3)&=y+4 && \text{Simplify the equation}\\ 2(x-3)&=y+4\\ 2x-6&=y+4\\ 2x-6+6&=y+4+6\\ 2x&=y+10 && \text{You may leave the function in this form.}\\ 2x-y&=y-y+10\\ 2x-y&=10\end{align*}
If you prefer to have both variables on the same side of the equation, this form may also be used. The choice is your preference.
\begin{align*}& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\ & 2x-y=10 && 2x-y=10\\ & 2x-({\color{red}0})=10 && 2({\color{red}0})-y=10\\ & 2x=10 && 0-y=10\\ & \frac{2x}{2}=\frac{10}{2} && \frac{-y}{-1}=\frac{10}{-1}\\ & x=5 && y=-10\\ & \text{The} \ x-\text{intercept is} \ (5, 0) && \text{The} \ y- \text{intercept is} \ (0, -10)\end{align*}
(ii) \begin{align*}3x+\frac{2}{3}y-3&=0 && \text{Simplify the equation.}\\ 3(3x)+3\left(\frac{2}{3}\right)y-3(3)&=3(0) && \text{Multiply each term by 3.}\\ 3(3x)+\cancel{3}\left(\frac{2}{\cancel{3}}\right)y-3(3)&=3(0)\\ 9x+2y-9&=0\\ 9x+2y-9+9&=0+9\\ 9x+2y&=9\end{align*}
\begin{align*}& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\ & 9x+2y=9 && 9x+2y=9\\ & 9x+2({\color{red}0})=9 && 9({\color{red}0})+2y=9\\ & 9x+0=9 && 0+2y=9\\ & \frac{9x}{9}=\frac{9}{9} && \frac{2y}{2}=\frac{9}{2}\\ & x=1 && y=4.5\\ & \text{The} \ x-\text{intercept is} \ (1, 0) && \text{The} \ y- \text{intercept is} \ (0, 4.5)\end{align*}
2. \begin{align*}5x+2y=-10\end{align*} Determine the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts.
\begin{align*}& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\ & 5x+2y=-10 && 5x+2y=-10\\ & 5x+2({\color{red}0})=-10 && 5({\color{red}0})+2y=-10\\ & 5x+0=-10 && 0+2y=-10\\ & \frac{5x}{5}=\frac{-10}{5} && \frac{2y}{2}=\frac{-10}{2}\\ & x=-2 && y=-5\\ & \text{The} \ x-\text{intercept is} \ (-2, 0) && \text{The} \ y- \text{intercept is} \ (0, -5)\end{align*}
3. Identify the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts from the graph.
The \begin{align*}x-\end{align*}intercept is (3, 0)
The \begin{align*}y-\end{align*}intercept is (0, -4)
Determine the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercept for each of the functions. If the intercepts match those of the graph, then the linear function will be the one that matches the graph.
(i) \begin{align*}& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\ & 2x+y=6 && 2x+y=6\\ & 2x+({\color{red}0})=6 && 2({\color{red}0})+y=6\\ & 2x=6 && 0+y=6\\ & \frac{2x}{2}=\frac{6}{2} && y=6\\ & x=3\\ & \text{The} \ x-\text{intercept is} \ (3, 0) && \text{The} \ y- \text{intercept is} \ (0, 6)\\ & \text{This matches the graph.} && \text{This does not match the graph.}\\ & \text{Find the} \ y- \text{intercept.} && 2x+y=6 \ \text{is not the linear function for the graph.}\end{align*}
(ii) \begin{align*}& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\ & 4x-3y-12=0 && 4x-3y-12=0\\ & 4x-3y-12+12=0+12 && 4x-3y-12+12=0+12\\ & 4x-3y=12 && 4x-3y=12\\ & 4x-3({\color{red}0})=12 && 4({\color{red}0})-3y=12\\ & 4x-0=12 && 0-3y=12\\ & 4x=12 && -3y=12\\ & \frac{4x}{4}=\frac{12}{4} && \frac{-3y}{-3}=\frac{12}{-3}\\ & x=3 && y=-4\\ & \text{The} \ x-\text{intercept is} \ (3, 0) && \text{The} \ y- \text{intercept is} \ (0, -4)\\ & \text{This matches the graph.} && \text{This matches the graph.}\\ & \text{Find the} \ y-\text{intercept.} && 4x-3y-12=0 \ \text{is the linear function for the graph.}\end{align*}
(iii) \begin{align*}& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\ & 5x+3y=15 && 5x+3y=15\\ & 5x+3({\color{red}0})=15 && 5({\color{red}0})+3y=15\\ & 5x+0=15 && 0+3y=15\\ & 5x=15 && 3y=15\\ & \frac{5x}{5}=\frac{15}{5} && \frac{3y}{3}=\frac{15}{3}\\ & x=3 && y=5\\ & \text{The} \ x-\text{intercept is} \ (3, 0) && \text{The} \ y- \text{intercept is} \ (0, 5)\\ & \text{This matches the graph.} && \text{This does not match the graph.}\\ & \text{Find the} \ y- \text{intercept.} && 5x+3y=15 \ \text{is not the linear function for the graph.}\end{align*}
Summary
In this lesson you have learned that an x-intercept is a point on the \begin{align*}x-\end{align*}axis that has the coordinates \begin{align*}(\#, 0)\end{align*}. You also learned that a \begin{align*}y-\end{align*}intercept is a point on the \begin{align*}y-\end{align*}axis that has the coordinates \begin{align*}(0, \#)\end{align*}. When you were given a function, you learned to algebraically determine the coordinates of both the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts.
The \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts were then used to draw the graph of the linear function on a Cartesian plane. This method of graphing the function is called the intercept method. You also learned to use the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts of a given graph to match a graph with its function.
Problem Set
Complete the following table:
Function \begin{align*}x-\end{align*}intercept \begin{align*}y-\end{align*}intercept
\begin{align*}7x-3y=21\end{align*}
\begin{align*}8x-3y+24=0\end{align*}
\begin{align*}\frac{x}{4}-\frac{y}{2}=3\end{align*}
\begin{align*}7x+2y-14=0\end{align*}
\begin{align*}\frac{2}{3}x-\frac{1}{4}y=-2\end{align*}
Use the intercept method to graph each of the linear functions in the above table.
1. \begin{align*}7x-3y=21\end{align*}
2. \begin{align*}8x-3y+24=0\end{align*}
3. \begin{align*}\frac{x}{4}-\frac{y}{2}=3\end{align*}
4. \begin{align*}7x+2y-14=0\end{align*}
5. \begin{align*}\frac{2}{3}x-\frac{1}{4}y=-2\end{align*}
Use the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts to match each graph to its function...
1. \begin{align*}7x+5y-35=0\end{align*}
2. \begin{align*}y=5x+10\end{align*}
3. \begin{align*}2x+4y+8=0\end{align*}
4. \begin{align*}2x+y=2\end{align*}
Complete the following table...
Function \begin{align*}x-\end{align*}intercept \begin{align*}y-\end{align*}intercept
\begin{align*}7x-3y=21\end{align*} (3, 0) (0, -7)
\begin{align*}8x-3y+24=0\end{align*}
\begin{align*}\frac{x}{4}-\frac{y}{2}=3\end{align*} (12, 0) (0, -6)
\begin{align*}7x+2y-14=0\end{align*}
\begin{align*}\frac{2}{3}x-\frac{1}{4}y=-2\end{align*} (-3, 0) (0, 8)
\begin{align*}& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\ & 7x-3y=21 && 7x-3y=21\\ & 7x-3({\color{red}0})=21 && 7({\color{red}0})-3y=21\\ & 7x+0=21 && 0-3y=21\\ & 7x=21 && -3y=21\\ & \frac{7x}{7}=\frac{21}{7} && \frac{-3y}{-3}=\frac{21}{-3}\\ & x=3 && y=-7\\ & \text{The} \ x-\text{intercept is} \ (3, 0) && \text{The} \ y- \text{intercept is} \ (0, -7)\end{align*}
\begin{align*}\frac{x}{4}-\frac{y}{2}=3 && 4\left(\frac{x}{4}\right)-4\left(\frac{y}{2}\right)=4(3) && ^{\overset{1}{\cancel{4}}}\left(\frac{x}{\cancel{4}}\right)-^{\overset{2}{\cancel{4}}}\left(\frac{y}{\cancel{2}}\right)=4(3)\\ x-2y=12\end{align*}
\begin{align*}& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\ & x-2y=12 && x-2y=12\\ & x-2({\color{red}0})=12 && ({\color{red}0})-2y=12\\ & x+0=12 && 0-2y=12\\ & x=12 && -2y=12\\ & && \frac{-2y}{-2}=\frac{12}{-2}\\ & && y=-6\\ & \text{The} \ x-\text{intercept is} \ (12, 0) && \text{The} \ y- \text{intercept is} \ (0, -6)\end{align*}
\begin{align*}\frac{2}{3}x-\frac{1}{4}y&=-2 && 12\left(\frac{2}{3}\right)x-12\left(\frac{1}{4}\right)y=12(-2)\\ ^{\overset{4}{\cancel{12}}}\left(\frac{2}{\cancel{3}}\right)x-^{\overset{3}{\cancel{12}}}\left(\frac{1}{\cancel{4}}\right)y&=12(-2) && 8x-3y=-24\end{align*}
\begin{align*}& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\ & 8x-3y=-24 && 8x-3y=-24\\ & 8x-3({\color{red}0})=-24 && 8({\color{red}0})-3y=-24\\ & 8x+0=-24 && 0-3y=-24\\ & 8x=-24 && -3y=-24\\ & \frac{8x}{8}=\frac{-24}{8} && \frac{-3y}{-3}=\frac{-24}{-3}\\ & x=-3 && y=8\\ & \text{The} \ x-\text{intercept is} \ (-3, 0) && \text{The} \ y- \text{intercept is} \ (0, 8)\end{align*}
Use the intercept method to graph...
1. \begin{align*}7x-3y=21\end{align*} The \begin{align*}x-\end{align*}intercept is (3, 0). The \begin{align*}y-\end{align*}intercept is (0, -7).
1. \begin{align*}\frac{x}{4}-\frac{y}{2}=3\end{align*} The \begin{align*}x-\end{align*}intercept is (12, 0). The \begin{align*}y-\end{align*}intercept is (0, -6).
1. \begin{align*}\frac{2}{3}x-\frac{1}{4}y=-2\end{align*} The \begin{align*}x-\end{align*}intercept is (-3, 0). The \begin{align*}y-\end{align*}intercept is (0, 8).
Use the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts to match...
(a) \begin{align*}& 7x+5y-35=0\\ & 7x+5y-35+35=0+35\\ & 7x+5y=35\\ & \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\ & 7x+5y=35 && 7x+5y=35\\ & 7x+5({\color{red}0})=35 && 7({\color{red}0})+5y=35\\ & 7x+0=35 && 0+5y=35\\ & 7x=35 && 5y=35\\ & \frac{7x}{7}=\frac{35}{7} && \frac{5y}{5}=\frac{35}{5}\\ & x=5 && y=7\\ & \text{The} \ x-\text{intercept is} \ (5, 0) && \text{The} \ y- \text{intercept is} \ (0, 7)\end{align*}
(b) \begin{align*}& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\ & y=5x+10 && y=5x+10\\ & ({\color{red}0})=5x+10 && y=5({\color{red}0})+10\\ & 0=5x+10 && y=0+10\\ & -5x=5x-5x+10 && y=10\\ & \frac{5x}{-5}=\frac{10}{-5} && \text{The} \ y- \text{intercept is} \ (0, 10)\\ & x=-2\\ & \text{The} \ x-\text{intercept is} \ (-2, 0)\end{align*}
(c) \begin{align*}& 2x+4y+8=0\\ & 2x+4y+8-8=0-8\\ & 2x+4y=-8\\ & \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\ & 2x+4y=-8 && 2x+4y=-8\\ & 2x+4({\color{red}0})=-8 && 2({\color{red}0})+4y=-8\\ & 2x+0=-8 && 0+4y=-8\\ & 2x=-8 && 4y=-8\\ & \frac{2x}{2}=\frac{-8}{2} && \frac{4y}{4}=\frac{-8}{4}\\ & x=-4 && y=-2\\ & \text{The} \ x-\text{intercept is} \ (-4, 0) && \text{The} \ x-\text{intercept is} \ (0, -2)\end{align*}
(d) \begin{align*}& 2x+y=2\\ & \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ `y'.\\ & 2x+y=2 && 2x+y=2\\ & 2x+({\color{red}0})=2 && 2({\color{red}0})+y=2\\ & 2x+0=2 && 0+y=2\\ & 2x=2 && y=2\\ & \frac{2x}{2}=\frac{2}{2} && \text{The} \ y-\text{intercept is} \ (0, 2)\\ & x=1\\ & \text{The} \ x-\text{intercept is} \ (1, 0)\end{align*}
(i)
(d) \begin{align*}2x+y=2\end{align*}
(ii)
(d) \begin{align*}y=5x+10\end{align*}
(iii)
(c) \begin{align*}2x+4y+8=0\end{align*}
(iv)
(a) \begin{align*}7x+5y-35=0\end{align*}
## Summary
In this lesson you have learned to create a table of values for a given linear function. The values for the dependent variable, \begin{align*}x\end{align*}, were given in a table and you substituted the value into the function to determine the value of the dependent variable ‘\begin{align*}y\end{align*}’. You also learned to create a table of values by using technology.
The values in the table represented coordinates of points that were located on the graph of the linear function. The coordinates were then plotted to draw the graph. In addition to drawing the graph on the Cartesian plane, you also learned to use technology to draw the graphs of the functions.
In the second lesson, you learned to algebraically calculate the values of both the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts from a given function. The \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts are coordinates that intersect the axes of the Cartesian plane. By using the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts of a graph, you were able to apply this knowledge to match a graph to its linear function.
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# 7.) Mya has ballet lessons every sixth day and swimming lessons every fourth day. Today she has both lessons. In how many days will Mya have both lessons on the same day again? 8.) A backpack at the store costs \$24.94. A pack of markers at the same store costs \$3.82. You pay for both the backpacks and pack of markers with a \$50 bill. How much change do you receive?
Step-by-step explanation:
#8
step #1: add \$24.94 with \$3.82
24.94 + 3.82 = 28.76
Step #2: subtract
\$50 - 28.76 = \$21.24
## Related Questions
Explain how to use a basic fact and place value to divide 4,000 divided by 5
Step-by-step explanation:
you have 4,000 lets make it smaller.
40 divied by 5
=8 just add the 2 zerores to get 800
What is 2 and 10/12 + 3 and 3/12 in simplest form
2 + 3 = 5 Add whole numbers
10/12 + 3/12 = 13/12
13/12 = 1 and 1/12
5 + 1 = 6
6 + 1/12 = 6 and 1/12
What two integers do 199 fall between
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Thursday, January 29, 2015
Advancing the calendar trick
I've written twice (first and second) about classroom experiences with a simple calendar trick that I originally got from Calendar Puzzles via Denise Gaskin's monthly newsletter. As happens so many times with these things, ideas from the kids make these activities into deeper and more interesting than I could have imagined on my own.
Tricked
On Tuesday, some of the first graders gave me their sums: 168 and 198. I immediately knew something was up. In the original calendar game, the square with the largest possible sum is the 23-24-30-31 square:
23 24 30 31
This has a sum of 108. I asked the students if they were sure of 168 and 198. They giggled, then the teacher smiled and told me she had checked it. What was going on?
My homework
I didn't have any immediate ideas, so I promised the kids that I would work on their puzzles. I told the kids that it was great to get my homework from them this time!
See through paper
One clue was that we were using a special calendar today and the paper was slightly see-through. This gave me an idea that the kids had turned the paper over and were seeing the numbers through the page with digits reversed. At first, I thought they were transforming 2s to 5s and vice versa, but was able to find 168 just by reversing digits.
How many carries?
For the original game, the crucial insight is simply that there are 7 days in a week and the calendar is organized into weeks. That means there is a simple relationship between each of the numbers in our 2x2 squares. Add a bit of simple algebra and you have an easy formula relating the upper left square of your 2x2 matrix to the sum (or, if you want to be fancy, a different formula relating whichever square you want to the sum).
For the reversed game, though, it isn't quite so easy. The relationship between the numbers can take one of several forms and is rather messy. I did manage to get 168. Can you?
But, I still couldn't get 198.
Two little helpers to the rescue
Last night, I "cheated" and asked for help. As J1 and J2 got ready to sleep, I asked what they thought their friend might have done to get 198. Their ideas from brainstorming:
• maybe the friend made an addition error
• maybe the friend also transformed 2s to 5s when reversing the paper
• maybe he summed a 3x3 square instead (which quickly gave rise to 4x4 and 5x5)
3x3 square? Interesting! Work through the algebra again and you can quickly see that there is (always!) a 3x3 square whose contents sum to 198.
Some further exploration, for you
More fun follow-on questions:
1. If the kids are allowed a choice of 2x2 or 3x3 section, but they still only tell you the sum and not the size of their square, can you still figure out which days they chose? Are there any conditions you might put on which month is chosen that allow you certainty in finding the square?
2. What if you allow 4x4, too?
3. Why stop at 4x4? What size squares are possible on a 1 month calendar?
4. If you make a year calendar instead, what sums are possible? If you are given the size and sum of a square, how close can you get to finding the source? In other words, how many squares have the same sums?
If you have other ideas, please let me know in the comments!
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# How do you simplify root3(3/4)?
Apr 23, 2017
$\sqrt[3]{\frac{3}{4}} = \frac{\sqrt[3]{6}}{2}$
#### Explanation:
For any non-zero values of $a , b$ we have:
$\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$
$\sqrt[3]{{a}^{3}} = a$
So we find:
$\sqrt[3]{\frac{3}{4}} = \sqrt[3]{\frac{3 \cdot 2}{4 \cdot 2}} = \sqrt[3]{\frac{6}{2} ^ 3} = \frac{\sqrt[3]{6}}{\sqrt[3]{{2}^{3}}} = \frac{\sqrt[3]{6}}{2}$
Notice how making the denominator into a perfect cube before splitting the radical allows us to avoid having to rationalise the denominator afterwards.
Apr 23, 2017
color(blue)(root3(6)/2
#### Explanation:
$\sqrt[3]{\frac{3}{4}}$
$\therefore = \frac{\sqrt[3]{3}}{\sqrt[3]{4}} \times \frac{\sqrt[3]{4}}{\sqrt[3]{4}} \times \frac{\sqrt[3]{4}}{\sqrt[3]{4}}$
:.=color(blue)(root3(4)*root3(4)*root3(4)=4
$\therefore = \frac{\sqrt[3]{48}}{4}$
$\therefore = \frac{\sqrt[3]{3 \cdot 2 \cdot 2 \cdot 2 \cdot 2}}{4}$
$\therefore = \frac{{\cancel{2}}^{\textcolor{b l u e}{1}} \sqrt[3]{6}}{\cancel{4}} ^ \textcolor{b l u e}{2}$
:.=color(blue)(root3(6)/2
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Для ботов
## Arithmetic & Geometric Sequences
Find the sum of the infinite geometric series given by:. Before we do anything, we'd better make sure our series is convergent. Otherwise, we can't find an infinite sum at all. What's our common ratio? Now we can nab our infinite geometric series formula. Therefore the sum is infinite. Hey, that means we're done! Find the sixth partial sum of the geometric series given by:. Use the formula for the partial sum of a geometric series. You've got it printed out on a little card in your wallet, right? Find the sum:. Now pop in the first term a 1 and the common ratio r. Log In. Sequences and Series. Substitute for a 1 and r in the formula and watch the magic unfold. Show Next Step. Example 2. Let's hunt down that common ratio. It's just the number in parentheses. Example 3. Substitute the values you know into the formula. Now slog through the actual math and simplify everything as much as you can. Example 4. It's our best bud. Rock those fractions. Logging out…. Logging out You've been inactive for a while, logging you out in a few seconds I'm Still Here! W hy's T his F unny?
## Geometric Sequences
Intro Examples Arith. Series Geo. The two simplest sequences to work with are arithmetic and geometric sequences. An arithmetic sequence goes from one term to the next by always adding or subtracting the same value. For instance, 2, 5, 8, 11, 14, The number added or subtracted at each stage of an arithmetic sequence is called the "common difference" dbecause if you subtract that is, if you find the difference of successive terms, you'll always get this common value. A geometric sequence goes from one term to the next by always multiplying or dividing by the same value. So 1, 2, 4, 8, 16, The number multiplied or divided at each stage of a geometric sequence is called the "common ratio" rbecause if you divide that is, if you find the ratio of successive terms, you'll always get this common value. To find the common difference, I have to subtract a successive pair of terms. It doesn't matter which pair I pick, as long as they're right next to each other. To be thorough, I'll do all the subtractions:. They gave me five terms, so the sixth term of the sequence is going to be the very next term. I find the next term by adding the common difference to the fifth term:. To find the common ratio, I have to divide a successive pair of terms. To be thorough, I'll do all the divisions:. They gave me five terms, so the sixth term is the very next term; the seventh will be the term after that. To find the value of the seventh term, I'll multiply the fifth term by the common ratio twice:. Since arithmetic and geometric sequences are so nice and regular, they have formulas. For arithmetic sequences, the common difference is dand the first term a 1 is often referred to simply as " a ". Since we get the next term by adding the common difference, the value of a 2 is just:.
## Sequences and Series
In this section we are going to take a brief look at three special series. Actually, special may not be the correct term. However, notice that both parts of the series term are numbers raised to a power. This means that it can be put into the form of a geometric series. We will just need to decide which form is the correct form. It will be fairly easy to get this into the correct form. This can be done using simple exponent properties. We can now do some examples. However, this does provide us with a nice example of how to use the idea of stripping out terms to our advantage. From the previous example we know the value of the new series that arises here and so the value of the series in this example is. However, we can start with the series used in the previous example and strip terms out of it to get the series in this example. We will strip out the first two terms from the series we looked at in the previous example. We can now use the value of the series from the previous example to get the value of this series. Consider the following series written in two separate ways i. This is now a finite value and so this series will also be convergent. In other words, if we have two series and they differ only by the presence, or absence, of a finite number of finite terms they will either both be convergent or they will both be divergent. The difference of a few terms one way or the other will not change the convergence of a series. In this portion we are going to look at a series that is called a telescoping series. The name in this case comes from what happens with the partial sums and is best shown in an example. By now you should be fairly adept at this since we spent a fair amount of time doing partial fractions back in the Integration Techniques chapter. If you need a refresher you should go back and review that section. So, what does this do for us? Notice that every term except the first and last term canceled out. This is the origin of the name telescoping series. This also means that we can determine the convergence of this series by taking the limit of the partial sums. In telescoping series be careful to not assume that successive terms will be the ones that cancel. Consider the following example. The partial sums are. In this case instead of successive terms canceling a term will cancel with a term that is farther down the list. The end result this time is two initial and two final terms are left. So, this series is convergent because the partial sums form a convergent sequence and its value is.
## How to Recognize a P-Series
In a Geometric Sequence each term is found by multiplying the previous term by a constant. Each term except the first term is found by multiplying the previous term by 2. We use "n-1" because ar 0 is for the 1st term. Each term is ar kwhere k starts at 0 and goes up to n It is called Sigma Notation. It says "Sum up n where n goes from 1 to 4. The formula is easy to use And, yes, it is easier to just add them in this exampleas there are only 4 terms. But imagine adding 50 terms On the page Binary Digits we give an example of grains of rice on a chess board. The question is asked:. Which was exactly the result we got on the Binary Digits page thank goodness! Let's see why the formula works, because we get to use an interesting "trick" which is worth knowing. All the terms in the middle neatly cancel out. Which is a neat trick. On another page we asked "Does 0. So there we have it Geometric Sequences and their sums can do all sorts of amazing and powerful things. Hide Ads About Ads. Geometric Sequences and Sums Sequence A Sequence is a set of things usually numbers that are in order. Geometric Sequences In a Geometric Sequence each term is found by multiplying the previous term by a constant. Example: 1, 2, 4, 8, 16, 32, 64, Example: 10, 30, 90,Example: 4, 2, 1, 0. Geometric Sequences are sometimes called Geometric Progressions G. It is called Sigma Notation called Sigma means "sum up" And below and above it are shown the starting and ending values: It says "Sum up n where n goes from 1 to 4. Example: Sum the first 4 terms of 10, 30, 90,The question is asked: When we place rice on a chess board: 1 grain on the first square, 2 grains on the second square, 4 grains on the third and so on, Question: if we continue to increase nwhat happens? Example: Calculate 0.
## Geometric Series
Intro Examples Arith. Series Geo. You can take the sum of a finite number of terms of a geometric sequence. Note: Your book may have a slightly different form of the partial-sum formula above. All of these forms are equivalent, and the formulation above may be derived from polynomial long division. I can also tell that this must be a geometric series because of the form given for each term: as the index increases, each term will be multiplied by an additional factor of —2. Plugging into the summation formula, I get:. The notation " S 10 " means that I need to find the sum of the first ten terms. Dividing pairs of terms, I get:. Unlike the formula for the n -th partial sum of an arithmetic series, I don't need the value of the last term when finding the n -th partial sum of a geometric series. So I have everything I need to proceed. When I plug in the values of the first term and the common ratio, the summation formula gives me:. I will not "simplify" this to get the decimal form, because that would almost-certainly be counted as a "wrong" answer. Instead, my answer is:. Note: If you try to do the above computations in your calculator, it may very well return the decimal approximation of As you can see in the screen-capture above, entering the values in fractional form and using the "convert to fraction" command still results in just a decimal approximation to the answer. But really! Take the time to find the fractional form. They've given me the sum of the first four terms, S 4and the value of the common ratio r. Since there is a common ratio, I know this must be a geometric series. Plugging into the geometric-series-sum formula, I get:. Then, plugging into the formula for the n -th term of a geometric sequence, I get:.
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Question
Mathematics
Evaluate $\left(f\circ g\right)\left(x\right)$(f@g)(x) and write the domain in interval notation. Write the answer in the intervals as an integer or simplified fraction.
$f\left(x\right)=\frac{x}{x-1}\phantom{\rule{1em}{0ex}}g\left(x\right)=\frac{13}{{x}^{2}-36}$f(x)=(x)/(x-1)quad g(x)=(13)/(x^(2)-36)
Part: $0/2$0//2
Part 1 of 2
$\left(f\circ g\right)\left(x\right)=$(f@g)(x)=
Solve problem with AI
To evaluate $\left(f\circ g\right)\left(x\right)$(f@g)(x), we need to first find $g\left(x\right)$g(x) and then evaluate $f$f at $g\left(x\right)$g(x).
Given that $f\left(x\right)=\frac{x}{x-1}$f(x)=(x)/(x-1) and $g\left(x\right)=\frac{13}{{x}^{2}-36}$g(x)=(13)/(x^(2)-36), we have
$\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)=f\left(\frac{13}{{x}^{2}-36}\right)=\frac{\frac{13}{{x}^{2}-36}}{\frac{13}{{x}^{2}-36}-1}.$(f@g)(x)=f(g(x))=f((13)/(x^(2)-36))=((13)/(x^(2)-36))/((13)/(x^(2)-36)-1).
$\begin{array}{rl}\left(f\circ g\right)\left(x\right)& =\frac{\frac{13}{{x}^{2}-36}}{\frac{13}{{x}^{2}-36}-1}\cdot \frac{{x}^{2}-36}{{x}^{2}-36}\\ \\ & =\frac{13}{\left({x}^{2}-36\right)-13}\\ \\ & =\frac{13}{{x}^{2}-49}\\ \\ & =\frac{13}{\left(x-7\right)\left(x+7\right)}\end{array}${:[(f@g)(x)=((13)/(x^(2)-36))/((13)/(x^(2)-36)-1)*(x^(2)-36)/(x^(2)-36)],[],[=(13)/((x^(2)-36)-13)],[],[=(13)/(x^(2)-49)],[],[=(13)/((x-7)(x+7))]:}
Therefore, $\left(f\circ g\right)\left(x\right)=\frac{13}{\left(x-7\right)\left(x+7\right)}$(f@g)(x)=(13)/((x-7)(x+7)).
Now, let’s find the domain of $\left(f\circ g\right)\left(x\right)$(f@g)(x). The function is undefined when the denominator $\left(x-7\right)\left(x+7\right)$(x-7)(x+7) equals zero. This happens when $x=7$x=7 or $x=-7$x=-7. So, the domain of $\left(f\circ g\right)\left(x\right)$(f@g)(x) is all real numbers except for $x=7$x=7 and $x=-7$x=-7.
In interval notation, the domain of $\left(f\circ g\right)\left(x\right)$(f@g)(x) is $\left(-\mathrm{\infty },-7\right)\cup \left(-7,7\right)\cup \left(7,\mathrm{\infty }\right)$(-oo,-7)uu(-7,7)uu(7,oo).
Answer: $\left(f\circ g\right)\left(x\right)=\frac{13}{\left(x-7\right)\left(x+7\right)}$(f@g)(x)=(13)/((x-7)(x+7)) with domain $\left(-\mathrm{\infty },-7\right)\cup \left(-7,7\right)\cup \left(7,\mathrm{\infty }\right)$(-oo,-7)uu(-7,7)uu(7,oo).
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# How many square feet is a 12×12 wall?
Figuring out the square footage of a wall or room is a common calculation that comes up frequently when measuring spaces for flooring, painting, wallpaper, etc. While it may seem complicated, it’s actually quite simple to figure out once you know the basic formula.
A 12×12 wall has 144 square feet.
To get this, you simply multiply the length by the width:
12 x 12 = 144 square feet
## The Formula
Calculating square footage for a wall, floor, ceiling or any rectangular area can be done by using this simple formula:
Length x Width = Area (in square feet or square meters)
So for a 12×12 wall:
12 feet x 12 feet = 144 square feet
## Step-by-Step Calculation
Let’s break this down step-by-step:
1. Measure the length and width of the wall. For this example, the length is 12 feet and the width is 12 feet.
2. Multiply the length by the width:
• Length = 12 feet
• Width = 12 feet
• 12 x 12 = 144
3. The result (144) is the total area in square feet.
Since we are multiplying feet by feet, the result is in square feet.
## Converting to Square Meters
If you need to calculate the area in square meters instead of square feet, you’ll need to convert the measurements first using the conversions:
1 foot = 0.3048 meters
1 meter = 3.281 feet
So for a 12×12 foot wall:
1. Convert 12 feet to meters:
• 12 feet x 0.3048 = 3.66 meters
2. Convert 12 feet to meters:
• 12 feet x 0.3048 = 3.66 meters
3. Multiply the length and width in meters:
• Length = 3.66 meters
• Width = 3.66 meters
• 3.66 x 3.66 = 13.4556
4. Round to 2 decimal places = 13.46 square meters
## Accounting for Thickness of Wall
The above calculations give you the surface area of the wall, assuming it has no thickness. However, if you need to account for the thickness of the wall itself, such as for painting or wallpaper, you’ll need to factor that in.
For example, if the wall is 5 inches thick, you would calculate it like this:
1. Length = 12 feet
2. Height = 12 feet
3. Thickness = 5 inches = 0.42 ft (convert inches to feet)
4. Surface Area = Length x Height = 12 x 12 = 144 sq ft
5. Volume = Surface Area x Thickness
= 144 sq ft x 0.42 ft = 60.48 cubic feet
So the total volume of the 12×12 wall that is 5 inches thick is 60.48 cubic feet.
## Using This Calculation
Knowing how to find the square footage of a wall comes in handy for:
• Determining how much paint, wallpaper, tile, etc you need to purchase
• Figuring out flooring requirements
• Measuring for furniture such as bookshelves or tv consoles
• Installing lighting fixtures or electrical outlets
• Real estate purposes – calculating total livable space
## Square Footage of Common Wall Sizes
Here are some common wall size examples and their total square footages:
Wall Dimensions Square Footage
8 ft x 8 ft 64 sq ft
10 ft x 10 ft 100 sq ft
10 ft x 12 ft 120 sq ft
12 ft x 12 ft 144 sq ft
20 ft x 20 ft 400 sq ft
## Finding Square Footage for Rectangular Rooms
The same formula applies when calculating the total square footage of rectangular shaped rooms or spaces:
Length x Width = Total Square Feet
Some examples:
• Bedroom: 10 ft x 12 ft = 120 sq ft
• Living room: 15 ft x 20 ft = 300 sq ft
• Garage: 20 ft x 22 ft = 440 sq ft
Make sure to take multiple measurements if the walls are not perfectly rectangular. Add together the square footages of each wall to get the total.
## For Irregular Shaped Spaces
If you are trying to find the square footage of an irregular, non-rectangular shaped space, here are some options:
• Break the area down into rectangular sections, measure and calculate each section separately, then add them together.
• Use an online calculator that allows you to input custom dimensions.
• Use a measuring wheel to manually roll across the floor and get the total square footage.
• Sketch the outline on grid paper and count the squares.
## Using Area Measurement Apps
There are various apps available for smartphones and tablets that make it easy to measure a space and determine the square footage instantly. Here are some of the top options:
• MagicPlan – Uses your phone’s camera to create a floor plan.
• RoomScan Pro – Laser-assisted measurement for floor plans.
• Measure – Allows you to draw floorplans and has AR measuring.
• Roomle – Imports images to make interactive floor plans.
These apps eliminate the need for manually measuring and calculating the area yourself. Many of the apps use augmented reality to scan the space with your smartphone camera and instantly generate the dimensions and floor plan layout.
### Benefits of Using a Measurement App
• Saves time and effort over manual measuring.
• Very accurate since it uses sensors and AI technology.
• Easy to edit, update, and share results.
• Works for both rectangular and oddly-shaped spaces.
• Can add dimensions, notes, photos to floor plans.
## Factors that Affect Square Footage Calculations
When determining square footage, keep in mind some factors that can influence the measurements:
• Crown molding – Crown molding extends the wall dimensions slightly. Take measurements from the floor to get exact wall height.
• Baseboards – Baseboards can also add a small amount to the wall length. Measure above baseboards for accuracy.
• Irregular dimensions – Older homes often have walls and rooms that are not perfectly rectangular. Take multiple measurements.
• Vaulted ceilings – Vaults, slopes and angled ceilings make the wall height inconsistent. Measure each section separately.
• Outlets, vents and fixtures – Protruding objects reduce the flat surface area. Consider offsetting deductions if needed.
• Corner roundness – Outside corners may have a rounded edge rather than 90 degrees. Use the shortest wall dimensions.
Taking these factors into account and deducting any offsets will give you the most accurate square footage possible.
## Conclusion
Finding the square footage of a wall, room or irregularly-shaped space is simple with just a bit of measurement and math. For a basic rectangular wall or room, multiply the length times the width to get the area in square feet. Be sure to take multiple measurements if the walls are not perfect rectangles. For odd-shaped areas, either break it down into rectangles or use a measuring app for convenience and accuracy. With the right tools and technique, you can quickly find the square footage you need for flooring, tile, lighting, remodeling and real estate purposes.
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Year 9
Year 9
# Comparing algebraic and graphical methods for solving simultaneous equations
## Switch to our new maths teaching resources
Slide decks, worksheets, quizzes and lesson planning guidance designed for your classroom.
## Lesson details
### Key learning points
1. In this lesson, we will learn about the benefits of using graphical or algebraic methods to solve simultaneous equations.
### Licence
This content is made available by Oak National Academy Limited and its partners and licensed under Oak’s terms & conditions (Collection 1), except where otherwise stated.
## Video
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## Worksheet
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## Starter quiz
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### 5 Questions
Q1.
Which of the following statements is true?
Simultaneous equations always have solutions.
Simultaneous equations can only be solved using graphs.
Correct answer: Simultaneous equations do not always have solutions.
Simultaneous equations never have solutions.
Q2.
Which of the following equations do not have solutions when solved simultaneously?
x + y = 2, y = x - 4
y = 2x + 1, y = -2x - 1
Correct answer: y = 3x - 2, y = 3x + 100
y = x + 3, y = 2x + 3
Q3.
Which of the following equations have a solution where the x coordinate is negative?
Correct answer: x - 2y = 6, y = -4
x - 2y = 6, y = 5 - 3x
y = 2x + 4, y = 5 - 3x
y = 5 - 3x, y = - 4
Q4.
Which of the following equations have a solution where both the coordinates are negative?
Correct answer: x - 2y = 6, y = -4
x - 2y = 6, y = 5 - 3x
y = 2x + 4, y = 5 - 3x
y = 5 - 3x, y = - 4
Q5.
Which of the following equations have a solution where both the coordinates are positive?
x - 2y = 6, y = -4
x - 2y = 6, y = 5 - 3x
Correct answer: y = 2x + 4, y = 5 - 3x
y = 5 - 3x, y = - 4
## Exit quiz
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### 5 Questions
Q1.
Which of the following statements is true?
You can only solve simultaneous equations algebraically.
You can only solve simultaneous equations using graphs.
You can solve all simultaneous equations.
Correct answer: You can solve simultaneous equations using alegebraic and graphical methods.
Q2.
Which of the following equations can you solve simultaneously?
2y + 4x = 12 and y = -2x + 4
Correct answer: 4y + 10x = 14 and y = 2.5x + 12
y + 1 = x and y - x = 12
y = 3x + 4 and y - 3x = 12
Q3.
Solve y = 4x - 5 and 2y + 6x = 18 simultaneously using a graphical method.
y = -2, x = 3
y = 2, x = -3
y = 2, x = 3
Correct answer: y = 3, x = 2
Q4.
Solve 2y = 6x - 10 and 3y + 6x = 15 simultaneously using an algebraic method.
x = 1, y = 2
x = 2, y = -1
Correct answer: x = 2, y = 1
y = 2, x = 3
Q5.
Solve y + x = 5 and y = 5x + 11 simultaneously using both an algebraic method and a graphical method.
Correct answer: x = -1, y = 6
x = -6, y= 1
x = 1, y= - 6
x = 6, y = -1
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# Family Of Circles Type 2 And 3
Go back to 'Circles'
TYPE 2:
\fbox{\begin{align}{{\text{FAMILY OF CIRCLES TOUCHING A GIVEN CIRCLE AT A GIVEN POINT}}}\end{align}}
Let the equation of the fixed circle be
$S:{x^2} + {y^2} + 2gx + 2fy + c = 0$
and let there be a point $$P({x_1},{y_1})$$ lying on this circle. We wish to determine the equation of the family of circles touching S at P.
We can write the equation of the tangent to $$S = 0$$ at P as
$T:x{x_1} + y{y_1} + g(x + {x_1}) + f(y + {y_1}) + c = 0$
Once we have a circle $$(S = 0)$$ and a line $$(T = 0)$$ intersecting or touching the circle, we can write the equation of the family of circles passing through the point (s) of intersection of the circle and the line, using the result derived in the last article. Thus, the required family can be represent as
$\fbox{F:S + \lambda T = 0}$
\begin{align}&\Rightarrow \qquad F:{x^2} + {y^2} + 2gx + 2fy + c + \lambda (x{x_1} + y{y_1} + g(x + {x_1}) + f(y + {y_1}) + c = 0\\& \Rightarrow \qquad F:{x^2} + {y^2} + (2g + \lambda {x_1} + \lambda g)x + (2f + \lambda {y_1} + \lambda f)y + c + \lambda g{x_1} + \lambda f{y_1} + \lambda c = 0\end{align}
As we vary $$\lambda ,$$ we will obtain different members belonging to this family.
TYPE 3:
\fbox{\begin{align} & \text{FAMILY OF CIRCLES PASSING THROUGH THE INTERSECTION POINT(S) OF } \\ & \text{TWO GIVEN CIRCLES} \\ \end{align}}
Let the two fixed circles be
\begin{align}&{S_1}:{x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0\\&{S_2}:{x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0\end{align}
and their points of intersection be $$A({x_1},{y_1})$$ and $$B({x_2},{y_2}).$$ In case the two circles touch each other, A and B will be the same.
We wish to determine the family of circles passing through A and B.
You might be able to extrapolate from the last few cases that the equation representing this family will be
$\fbox{F:{S_1} + \lambda {S_2} = 0}$
You can verify this by writing the equation for F in standard form and observing that it does indeed represent a circle. Also, since (the co-ordinates of) A and B satisfy both $${S_1} = 0$$ and $${S_2} = 0,$$ they have to satisfy the equation for F. Note one important point: $$\lambda$$ cannot be equal to –1 otherwise F will become the common chord of $${S_1} = 0$$ and $${S_2} = 0$$ instead of representing a circle.
Example - 37
Find the equation of the circle which passes through the points of intersection of the circles
\begin{align}&{S_1}:{x^2} + {y^2} - 6x + 2y + 4 = 0\\&{S_2}:{x^2} + {y^2} + 2x - 4y - 6 = 0\end{align}
and whose centre lies on the line y = x.
Solution: Let the required equation be S = 0. Then, by the previous article, we can find some $$\lambda \in \mathbb{R}$$ and $$\lambda \ne - 1$$ such that
\begin{align}& \quad \quad \;\;S \equiv {S_1} + \lambda {S_2} = 0\\ & \Rightarrow \quad S \equiv (1 + \lambda ){x^2} + (1 + \lambda ){y^2} + (2\lambda - 6)x + (2 - 4\lambda )y + 4 - 6\lambda = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)\end{align}
The centre of S from this equation comes out to be
${\rm{centre}} \equiv \left\{ { - \frac{{(\lambda - 3)}}{{1 + \lambda }},\,\, - \frac{{(1 - 2\lambda )}}{{1 + \lambda }}} \right\}$
Since the centre lies on the line y = x, we have
\begin{align}& - \frac{{(\lambda - 3)}}{{1 + \lambda }} = - \frac{{(1 - 2\lambda )}}{{1 + \lambda }}\\ \Rightarrow \quad & \lambda = \frac{4}{3}\end{align}
We now substitute this value back in (1) to obtain the equation for S:
\begin{align} & S:\frac{7}{3}{x^2} + \frac{7}{3}{y^2} - \frac{{10}}{3}x - \frac{{10}}{3}y - 4 = 0\\ \Rightarrow \quad & S:{x^2} + {y^2} - 10x - 10y - 12 = 0\end{align}
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## Kinematics Practice Problems with Answers
The most complete guide on solving kinematics problems for high schools and colleges. By working through these questions, you can undoubtedly master this topic in physics.
These multiple-choice questions on kinematics for AP Physics 1 are also available to review for students enrolled in AP Physics courses.
All kinematics equations are summarized in the following expressions: \begin{gather*} \Delta x=\frac{v_1+v_2}{2}\times \Delta t \\\\ v=v_0+at \\\\ \Delta x=\frac 12 at^2+v_0t \\\\ v^2-v_0^2=2a\Delta x \end{gather*} In the rest of this long article, you will see how to apply these equations in the given problems.
## Kinematics Practice Problems:
Problem (1): A car slows down its motion from 10 m/s to 6 m/s in 2 seconds under constant acceleration. (a) What is its acceleration? (b) How far did the car travel during this time interval?
Solution : This is the simplest kinematics problem, so we put a bit more time to solve it in detail.
Step 1: Because all these problems are in one dimension, draw a directed horizontal axis (like the positive $x$ axis), and put the object on it, so that it moves in the correct direction.
Step 2: Specify the known and wanted information. Here, in the elapsed time interval $2\,{\rm s}$, the initial and final velocities of the car are given as $v_i=10\,{\rm m/s}$ and $v_f=6\,{\rm m/s}$. The wanted quantity is the constant acceleration of the object (car), $a=?$.
Step 3: Apply the kinematics equation that is appropriate for this situation.
(a) In this problem, we want to find the acceleration, given the time, initial, and final velocities so the kinematics equation $v=v_0+at$ is perfect since the only unknown quantity is acceleration $a$. Thus, \begin{gather*} v=v_0+at\\\\ 6=10+a(2) \\\\ 6-10=2a \\\\\Rightarrow \quad a=\frac{6-10}{2}=-2\quad {\rm \frac{m}{s^2}}\end{gather*} Note that because the problem says the acceleration of the motion is constant, we could use the constant acceleration kinematics equations.
The negative indicates the direction of the acceleration, which is here toward the negative $x$ axis.
(b) "How far'' means the distance traveled by car is wanted, denoted by $x$ in the kinematics equations.
Here, the best equation that relates the known and unknown information is $x=\frac 12 at^2+v_0t$ or $v^2-v_0^2=2ax$. We choose the first, so \begin{align*} x&=\frac 12 at^2+v_0t \\\\&=\frac 12 (-2)(2)^2+(10)(2) \\\\&=16\quad {\rm m}\end{align*}
On the following page you can find over 40+ questions related to applying kinematics equations in velocity and acceleration:
Velocity and acceleration problems
Problem (2): A moving object slows down from $12\,{\rm m/s}$ to rest at a distance of 20 m. Find the acceleration of the object (assumed constant).
Solution : In the diagram below, all known information along with the direction of the uniform motion is shown.
As you can see, one of the common phrases in kinematics problems is "ending or coming to a rest'', which means the final velocity of the object in that time interval (in which we look at the object's motion) is zero, $v_f=0$.
The perfect kinematics equation that solves this problem is $v^2-v_0^2=2a(x-x_0)$ as the only unknown quantity is acceleration $a$.
Keep in mind that in all kinematics equation problems, we can set the initial position of the motion $x_0$ as zero for simplicity, $x_0=0$. \begin{gather*} v^2-v_0^2=2ax\\\\0^2-(12)^2 =2a(20) \\\\ \rightarrow a=\frac{-144}{2\times 20}\\\\\Rightarrow \boxed{a=-3.6\quad {\rm \frac{m}{s^2}}}\end{gather*} As before, the minus sign indicates the direction of the acceleration which is toward the left .
Problem (3): A bullet leaves the muzzle of an 84-cm rifle with a speed of 521 m/s. Find the magnitude of the bullet's acceleration by assuming it is constant inside the barrel of the rifle.
Solution : The bullet accelerates from rest to a speed of 521 m/s at a distance of 0.84 meters. These are our known quantities. The unknown is acceleration $a$. The perfect kinematics equation that relates all these together is $v^2-v_0^2=2a(x-x_0)$, so \begin{gather*}v^2-v_0^2=2a(x-x_0)\\\\ (521)^2-0=2(a)(0.84-0) \\\\ \Rightarrow \boxed{a=1.62\times 10^5\quad {\rm m/s^2}}\end{gather*} A very large acceleration.
Problem (4): A car starts its motion from rest and uniformly accelerates at a rate of $4\,{\rm m/s^2}$ for 2 seconds in a straight line. (a) How far did the car travel during those 2 seconds? (b) What is the car's velocity at the end of that time interval?
Solution : "Start from rest'' means the initial object's velocity is zero, $v_0=0$. The known information are $a=4\,{\rm m/s^2}$, $t=2\,{\rm s}$ and wants the distance traveled $x=?$.
(a) The kinematics equation that relates that information is $x=\frac 12 at^2+v_0 t+x_0$ since the only unknown quantity is $x$ with the given known data above. \begin{align*} x&=\frac 12 at^2+v_0 t+x_0 \\\\&=\frac 12 (4)(2)^2+(0)(2) \\\\&=\boxed{8\quad {\rm m}}\end{align*} As previous, we set $x_0=0$.
(b) Now that the distance traveled by the car in that time interval is known, we can use the following kinematics equation to find the car's final velocity $v$. \begin{align*} v^2-v_0^2 &=2a(x-x_0) \\\\v^2-(0)^2&=2(4)(8-0) \\\\v^2&=64\end{align*} Taking the square root, we get $v$: $v=\sqrt{64}=\pm 8\quad {\rm \frac ms}$ We know that velocity is a vector quantity in physics and has both a direction and a magnitude.
The magnitude of the velocity (speed) was obtained as 8 m/s, but in what direction? Or we must choose which signs? Because the car is uniformly accelerating without stopping in the positive $x$ axis, the correct sign for velocity is positive.
Therefore, the car's final velocity is $\boxed{v_f=+8\,{\rm m/s}}$.
## Challenging Kinematics Problems
In the following, some challenging kinematics problems are presented, which are for homework.
A driver is moving along at $45\,\rm m/s$ when she suddenly notices a roadblock $100\,\rm m$ ahead. Can the driver stop the vehicle in time to avoid colliding with the obstruction if her reaction time is assumed to be $0.5\,\rm s$ and her car's maximum deceleration is $5\,\rm m/s^2$?
Solution : The time between seeing the obstacle and taking action, such as slamming on the brake, is defined as the reaction time. During this time interval, the moving object travels at a constant speed.
Thus, in all such questions, we have two phases. One is constant speed, and the other is accelerating with negative acceleration (deceleration).
Here, between the time of seeing the barrier and the time of braking, the driver covers a distance of \begin{align*} x_1&=vt_{reac} \\\\ &=25\times 0.5 \\\\ &=12.5\,\rm m\end{align*} In the decelerating phase, the car moves a distance, which is obtained using the following kinematics equation: \begin{gather*} v^2-v_0^2=2ax_2 \\\\ (0)^2-(25)^2=2(-5) x_2 \\\\ \Rightarrow x_2=62.5\,\rm m\end{gather*} Summing these two distances gives a total distance that is a good indication of whether the moving object hits the obstacle or not. \begin{gather*} \Delta x_{actual}=x_1+x_2=75\,\rm m \\\\ \Rightarrow \Delta x_{covered}<\Delta x_{actual} \end{gather*} As a result, because the distance covered by the car is less than the actual distance between the time of seeing the barrier and the obstacle itself, the driver has sufficient time to stop the car in time to avoid a collision.
For a moving car at a constant speed of $90\,\rm km/h$ and a human reaction time of $0.3\,\rm s$; find the stopping distance if it slows down at a rate of $a=3\,\rm m/s^2$.
Solution : We use SI units, so first convert the given speed in these units as below \begin{align*} v&=90\,\rm km/h \\\\ &=\rm 90\times \left(\frac{1000\,m}{3600\,s}\right) \\\\ &=25\,\rm m/s\end{align*} As we said previously, during the reaction time, your car moves at a constant speed and covers a distance of \begin{align*} x_1&=vt_{react} \\\\ &=(25)(0.3) \\\\ &=7.5\,\rm m \end{align*} Deceleration means the moving object slows down, or a decrease per second in the velocity of the car occurs. In this case, we must put the acceleration with a negative sign in the kinematics equations.
During the second phase, your car has negative acceleration and wants to be stopped. Thus, the distance covered during this time interval is found as follows \begin{align*} v^2-v_0^2=2a\Delta x \\\\ (0)^2-(25)^2=2(3)\Delta x \\\\ \Rightarrow \quad \boxed{\Delta x=104.17\,\rm m}\end{align*}
Assume you are traveling $35\,\rm m/s$ when suddenly you see red light traffic $50\,\rm m$ ahead. If it takes you $0.456\,\rm s$ to apply the brakes and the maximum deceleration of the car is $4.5\,\rm m/s^2$, (a) Will you be able to stop the car in time? (b) How far from the time of seeing the red light will you be?
Solution : When you see the red light until you apply the brakes, your car is moving at a constant speed. This time interval is defined as the reaction time, $\Delta t_{react}=0.456\,\rm s$. After you get the brakes on, the car starts to decelerate at a constant rate, $a=-4.5\,\rm m/s^2$. Pay attention to the negative signs of such problems. The negative tells us that the car is decreasing its speed.
(a) In the first phase, the car moves a distance of \begin{align*} x_1&=v\Delta t_{react} \\\\ &=35\times 0.455 \\\\ &=15.96\,\rm m\end{align*} In the phase of deceleration, the car is moving a distance whose magnitude is found using the time-independent kinematics equation as below \begin{gather*} v^2-v_0^2=2ax_2 \\\\ (0)^2-(35)^2=2(4.5)x_2 \\\\ \Rightarrow \quad x_2=136.11\,\rm m\end{gather*} The sum of these two distances traveled gives us the total distance covered by the car from the time of seeing the red traffic light to the moment of a complete stop. $x_{tot}=x_1+x_2=152.07\,\rm m$ Because the total distance traveled is greater than the actual distance to the red light, the driver will not be able to stop the car in time.
(b) As previously calculated, the total distance traveled by the car is nearly $152\,\rm m$ or the car is about $102\,\rm m$ past the red light traffic.
A person stands on the edge of a $60-\,\rm m$-high cliff and throws two stones vertically downward, $1$ second apart, and sees they both reach the water simultaneously. The first stone had an initial speed of $4\,\rm m/s$. (a) How long after dropping the first stone does the second stone hit the water? (b) How fast was the second stone released? (c) What is the speed of each stone at the instant of hitting the water?
Solution: Because all quantities appearing in the kinematics equation are vectors, we must first choose a positive direction. Here, we take up as a positive $y$ direction.
Both stones arrived in the water at the same time. Thus, calculate the time the first stone was in the air. Next, use the time interval between the two drops to find the duration the second stone was in the air. (a) The first stone is released downward at a speed of $4\,\rm m/s$, thus, its initial velocity is $v_0=-4\,\rm m/s$. The minus sign is for moving in the opposite direction of the chosen direction.
The only relevant kinematics equation that relates this known information is $\Delta y=-\frac 12 gt^2+v_0t$, where $\Delta y=-60\,\rm m$ is the vertical displacement, and the negative indicates that the stone hit a point below the chosen origin. Substituting the numerical values into this and solving for the time duration $t$ gives \begin{gather*} \Delta y=-\frac 12 gt^2+v_0t \\\\ -60=-\frac 12 (10)t^2+(-4)t \\\\ 5t^2+4t-60=0 \\\\ \Rightarrow \boxed{t=3.0\,\rm s} \, , \, t'=-3.8\,\rm s \end{gather*} The second answer is not acceptable.
(b) The second stone was released $1$ second after throwing the first one and arrived at the same time as the first stone. Therefore, the time interval that the second stone was in the air is found to be \begin{align*} t_2&=t_1-1 \\ &=3.0-1\\ &=2\,\rm s\end{align*}
(c) It is better to apply the time-independent kinematics equation $v^2-v_0^2=-2g\Delta y$ to find the stone's velocity at the moment it hit the water. For the first stone, we have \begin{gather*} v^2-v_0^2=-2g\Delta y \\\\ v^2-(-4)^2=-2(10)(-60) \\\\ \Rightarrow \quad \boxed{v=34.8\,\rm m/s} \end{gather*} The second stone's velocity is left to you as an exercise.
In a tennis game, the ball leaves the racket at a speed of $75\,\rm m/s$ whereas it is in contact with the racket for $25\,\rm ms$, and starts at rest. Assume the ball experiences constant acceleration. What was the ball's acceleration during this serve? How far has the ball traveled on this serve?
Solution : In this question, we are asked to find the ball's acceleration and distance traveled during that pretty small time interval. (a) In a time interval of $\Delta t=25\times 10^{-3}\,\rm s$, we are given the beginning velocity $v_1=0$ and the end velocity $v_2=85\,\rm m/s$. Because it is assumed the acceleration is constant, the average acceleration definition, $a=\frac{\Delta v}{\Delta t}$, is best suited for these known quantities. \begin{align*} a&=\frac{v_2-v_1}{\Delta t} \\\\ &=\frac{75-0}{25\times 10^{-3}} \\\\ &=3000\,\rm m/s^2 \end{align*} A huge acceleration is given to the tennis ball. (b) Here, we are asked to find the amount of distance traveled by the ball during the time the ball was in contact with the racket. Because we have a constant acceleration motion, it is best to use the following equation to find the distance traveled. \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\times \Delta t \\\\ &=\frac{0+75}{2}\times (25\times 10^{-3}) \\\\ &=937.5\times 10^{-3}\,\rm m\end{align*} In millimeters, $\Delta x=937.5\,\rm mm$, and in centimeters $\Delta x=93.75\,\rm cm$. Therefore, during this incredibly short time interval, the ball moves about $94\,\rm cm$ along with the racket.
Starting from rest and ending at rest, a car travels a distance of $1500\,\rm m$ along the $x$-axis. During the first quarter of the distance, it accelerates at a rate of $+1.75\,\rm m/s^2$, while for the remaining distance, its acceleration is $-0.450\,\rm m/s^2$. (a) What is the time travel of the whole path? (b) What is the maximum speed of the car over this distance?
None of the time-dependent kinematics equations give us the time travel $t'$ without knowing the initial speed at the instant of the start of this second path.
We can find it using the equation $v^2-v_0^2=2a\Delta x$, setting $v=0$ at the end of the path, and solving for $v_0$ \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ (0)^2-v_0^2=2(-0.450)(1125) \\\\ v_0=\sqrt{2\times 0.45\times 1125} \\\\ \Rightarrow v_0=31.82\,\rm m/s\end{gather*} Given that, one can use the simple equation $v=v_0+at$ and solve for the time travel in this part of the path. \begin{gather*} v=v_0+at \\\\ 0=31.82+(-0.450)t' \\\\ \Rightarrow t'=70.71\,\rm s\end{gather*} Therefore, the total time traveled over the entire path is the sum of these two times. \begin{align*} T&=t+t' \\\\ &=20.70+70.71 \\\\ &=\boxed{91.41\,\rm s} \end{align*}
A train that is $75$ meters long starts accelerating uniformly from rest. When the front of the train reaches a railway worker who is standing $150$ meters away from where the train started, it is traveling at a speed of $20\,\rm m/s$. What will be the speed of the last car as it passes the worker?
Solution : The front of the train is initially $150\,\rm m$ away from the worker, and when it passes him, it has a speed of $25\,\rm m/s$. From this data, we can find the acceleration of the front of the train (which is the same acceleration as the whole train) by applying the following kinematics equation \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ (25)^2-(0)^2=2a\times 150 \\\\ \Rightarrow \quad a=2.08\,\rm m/s^2\end{gather*} Given the train's acceleration, now focus on the last car.
The last car is initially at rest and placed at a distance of $150+80=230\,\rm m$ away from the person. When it passes the person, it has traveled $\Delta x= 230\,\rm m$ and its speed is determined simply as below \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ v^2-(0)^2=2(2.08)(230) \\\\ \Rightarrow \quad \boxed{v=30.93\,\rm m/s}\end{gather*}
A wildcat moving with constant acceleration covers a distance of $100\,\rm m$ apart in $8\,\rm s$. Assuming that its speed at the second point is $20\,\rm m/s$, (a) What was its speed in the first place? (b) At what rate does its speed change over this distance?
Solution : First of all, list all known data given to us. Time interval $\Delta t=8\,\rm s$, the horizontal displacement $\Delta x=100\,\rm m$, speed at second point $v_2=20\,\rm m/s$.
We are asked to find the speed at the second point. To solve this kinematics problem, we use the following kinematics equation because the acceleration is constant and this is the most relevant equation that relates the known to the unknown quantities. \begin{gather*} \Delta x=\frac{v_1+v_2}{2}\times \Delta t \\\\ 100=\frac{v_1+20}{2}\times 8 \\\\ \Rightarrow \quad \boxed{v_1=5\,\rm m/s} \end{gather*} Therefore, the wildcat's speed in the first place is $5\,\rm m/s$.
In this part, we should find the wildcat's acceleration because acceleration is defined as the time rate of change of the speed of a moving object. Given the first place speed, $v_1=5\,\rm m/s$, found in the preceding part, we can use the following time-independent kinematics equation to find the wanted unknown. \begin{gather*} v_2^2-v_1^2=2a\Delta x \\\\ (20)^2-(5)^2=2a(100) \\\\ \Rightarrow \quad \boxed{a=1.875\,\rm m/s^2}\end{gather*}
A car slows down uniformly from $45\,\rm m/s$ to rest in $10\,\rm s$. How far did it travel in this time interval?
Solution : List the data known as follows: initial speed $v_0=45\,\rm m/s$, final speed $v=0$, and the total time duration that this happened is $t=10\,\rm s$. The unknown is also the amount of displacement, $\Delta x$.
The only kinematics equation that relates those together is $\Delta x=\frac{v_1+v_2}{2}\times \Delta t$, where $v_1$ and $v_2$ are the velocities at the beginning and end of that time interval. \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\times \Delta t \\\\ &=\frac{45+0}{2}\times 10 \\\\ &=\boxed{225\,\rm m}\end{align*} Keep in mind that we use this formula when the object slows down uniformly, or, in other words when the object's acceleration is constant.
Problem (5): We want to design an airport runway with the following specifications. The lowest acceleration of a plane should be $4\,{\rm m/s^2}$ and its take-off speed is 75 m/s. How long would the runway have to be to allow the planes to accelerate through it?
Solution: The known quantities are $a=4\,{\rm m/s^2}$, and final velocity $v=75\,{\rm m/s}$. The wanted quantity is runway length $\Delta x=x-x_0$. The perfect kinematics equation that relates those together is $v^2-v_0^2=2a(x-x_0)$. \begin{align*} v^2-v_0^2&=2a\Delta x\\\\ (75)^2-0&=2(4) \Delta x\\\\ \Rightarrow \Delta x&=\boxed{703\quad {\rm m}}\end{align*} Thus, if the runway wants to be effective, its length must be at least about 703 meters.
Problem (6): A stone is dropped vertically from a high cliff. After 3.55 seconds, it hits the ground. How high is the cliff?
Solution : There is another type of kinematics problem in one dimension but in the vertical direction. In such problems, the constant acceleration is that of free falling, $a=g=-10\,{\rm m/s^2}$.
"Dropped'' or "released'' in free-falling problems means the initial velocity is zero, $v_0=0$. In addition, it is always better to consider the point of release as the origin of the coordinate, so $y_0=0$.
The most relevant kinematics equation for these known and wanted quantities is $y=-\frac 12 gt^2+v_0t+y_0$ \begin{align*} y&=-\frac 12 gt^2+v_0t+y_0 \\\\&=-\frac 12 (9.8)(3.55)^2+0+0\\\\&=\boxed{-61.8\quad {\rm m}}\end{align*} The negative indicates that the impact point is below our chosen origin .
Problem (7): A ball is thrown into the air vertically from the ground level with an initial speed of 20 m/s. (a) How long is the ball in the air? (b) At what height does the ball reach?
Solution : The throwing point is considered to be the origin of our coordinate system, so $y_0=0$. Given the initial velocity $v_0=+20\,{\rm m/s}$ and the gravitational acceleration $a=g=-9.8\,{\rm m/s^2}$. The wanted time is how long it takes the ball to reach the ground again.
To solve this free-fall problem , it is necessary to know some notes about free-falling objects.
Note (1): Because the air resistance is neglected, the time the ball is going up is half the time it is going down.
Note (2): At the highest point of the path, the velocity of the object is zero.
(a) By applying the kinematics equation $v=v_0+at$ between the initial and the highest ($v=0$) points of the vertical path, we can find the going up time. \begin{align*} v&=v_0+at \\0&=20+(-9.8)t\\\Rightarrow t&=2.04\quad {\rm s}\end{align*} The total flight time is twice this time $t_{tot}=2t=2(2.04)=4.1\,{\rm s}$ Hence, the ball takes about 4 seconds to reach the ground.
(b) The kinematics equation $v^2-v_0^2=2a(y-y_0)$ is best for this part. \begin{align*} v^2-v_0^2&=2a(y-y_0) \\0-20^2&=2(-9.8)(y-0) \\ \Rightarrow y&=\boxed{20\quad {\rm m}}\end{align*} Hence, the ball goes up to a height of about 20 meters.
Problem (8): An object moving in a straight line with constant acceleration, has a velocity of $v=+10\,{\rm m/s}$ when it is at position $x=+6\,{\rm m}$ and of $v=+15\,{\rm m/s}$ when it is at $x=10\,{\rm m}$. Find the acceleration of the object.
Solution : Draw a diagram, put all known data into it, and find a relevant kinematics equation that relates them together.
We want to analyze the motion in a distance interval of $\Delta x=x_2-x_1=10-6=4\,{\rm m}$, thus, we can consider the velocity at position $x_1=6\,{\rm m}$ as the initial velocity and at $x_2=10\,{\rm m}$ as the final velocity.
The most relevant kinematics equation that relates these known quantities to the wanted acceleration $a$ is $v^2-v_0^2=2a(x-x_0)$, where $x-x_0$ is the same given distance interval. Thus, \begin{align*} v^2-v_0^2&=2a(x-x_0) \\\\ (15)^2-(10)^2&=2(a)(4) \\\\225-100&=8a\\\\\Rightarrow a&=\frac{125}{8}\\\\&=15.6\,{\rm m/s^2}\end{align*}
Problem (9): A moving object accelerates uniformly from 75 m/s at time $t=0$ to 135 m/s at $t=10\,{\rm s}$. How far did it move at the time interval $t=2\,{\rm s}$ to $t=4\,{\rm s}$?
Solution : Draw a diagram and implement all known data in it as below.
Because the problem tells us that the object accelerates uniformly, its acceleration is constant along the entire path.
Given the initial and final velocities of the moving object, its acceleration is determined using the definition of instantaneous acceleration as below $a=\frac{v_2-v_1}{t_2-t_1}=\frac{135-75}{10}=6\,{\rm m/s^2}$ In this kinematics problem, to analyze the motion between the requested times (stage II in the figure), we must have a little bit of information for that time interval, their velocities, or the distance between them.
As you can see in the figure, the initial velocity of stage II is the final velocity of stage I. By using a relevant kinematics equation that relates those data to each other, we would have \begin{align*} v&=v_0+at\\\\&=75+(6)(2) \\\\&=87\,{\rm m/s}\end{align*} This velocity would be the initial velocity for stage II of the motion. Now, all known information for stage II is initial velocity $v_0=87\,{\rm m/s}$, acceleration $a=6\,{\rm m/s^2}$, and time interval $\Delta t=2\,{\rm s}$. The wanted is the distance traveled $x=?$
The appropriate equation that relates all these together is $x=\frac 12 at^2+v_0t+x_0$. \begin{align*}x&=\frac 12 at^2+v_0t+x_0\\\\&=\frac 12 (6)(2)^2+(87)(2)+0\\\\&=186\quad {\rm m}\end{align*} Hence, our moving object travels a distance of 186 m between the instances of 2 s and 4 s.
Problem (10): From rest, a fast car accelerates with a uniform rate of $1.5\,{\rm m/s^2}$ in 4 seconds. After a while, the driver applies the brakes for 3 seconds, causing the car to uniformly slow down at a rate of $-2\,{\rm m/s^2}$. (a) How fast is the car at the end of the braking period? (b) How far has the car traveled after braking?
Solution : This motion is divided into two parts. First, draw a diagram and specify each section's known kinematics quantities.
(a) In the first part, given the acceleration, initial velocity, and time interval, we can find its final velocity at the end of 4 seconds. \begin{align*} v&=v_0+at\\&=0+(1.5)(4) \\&=6\quad {\rm m/s}\end{align*} This velocity is considered as the initial velocity for the second part, whose final velocity is wanted.
In the next part, the acceleration magnitude and braking time interval are given, so its final velocity is found as below \begin{align*} v&=v_0+at\\&=6+(-2)(3) \\&=0\end{align*} The zero velocity here indicates that the car, after the braking period, comes to a stop.
(b) The distance traveled in the second part is now calculated using the kinematics equation $x=\frac 12 at^2+v_0t+x_0$, because the only unknown quantity is distance $x$. \begin{align*} x&=\frac 12 at^2+v_0t+x_0\\\\&=\frac 12 (-2)(3)^2+(6)(3)+0\\\\&=+9\quad {\rm m}\end{align*} Therefore, after braking, the car traveled a distance of 9 meters before getting stopped.
Problem (11): A car moves at a speed of 20 m/s down a straight path. Suddenly, the driver sees an obstacle in front of him and applies the brakes. Before the car reaches a stop, it experiences an acceleration of $-10\,{\rm m/s^2}$. (a) After applying the brakes, how far did it travel before stopping? (b) How long does it take the car to reach a stop?
Solution : As always, the first and most important step in solving a kinematics problem is drawing a diagram and putting all known values into it, as shown below.
(a) The kinematics equation $v^2-v_0^2=2a(x-x_0)$ is the perfect equation as the only unknown quantity in it is the distance traveled $x$. Thus, \begin{align*} v^2-v_0^2&=2a(x-x_0) \\\\ 0^2-(20)^2&=2(-10)(x-0) \\\\\Rightarrow \quad x&=\frac {-400}{-20}\\\\&=20\quad {\rm m}\end{align*} (b) "how long does it take'' asks us to find the time interval. The initial and final velocities, as well as acceleration, are known, so the only relevant kinematics equation is $v=v_0+at$. Thus, \begin{align*} v&=v_0+at\\\\0&=20+(-10)t\\\\\Rightarrow t&=\frac{-20}{-10}\\\\&=2\quad {\rm s}\end{align*} Therefore, after braking, the car has moved for 2 seconds before reaching a stop.
Problem (12): A sports car moves a distance of 100 m in 5 seconds with a uniform speed. Then, the driver brakes, and the car, come to a stop after 4 seconds. Find the magnitude and direction of its acceleration (assumed constant).
Solution : uniform speed means constant speed or zero acceleration for the motion before braking. Thus, we can use the definition of average velocity to find its speed just before braking as below \begin{align*} \bar{v}&=\frac{\Delta x}{\Delta t}\\\\&=\frac{100}{5}\\\\&=20\quad {\rm m/s}\end{align*} Now, we know the initial and final velocities of the car in the braking stage. Since the acceleration is assumed to be constant, by applying the definition of average acceleration, we would have \begin{align*} \bar{a}&=\frac{v_2-v_1}{\Delta t}\\\\&=\frac{0-20}{4}\\\\&=-5\quad {\rm m/s^2}\end{align*} The negative shows the direction of the acceleration, which is toward the negative $x$-axis.
Hence, the car's acceleration has a magnitude of $5\,{\rm m/s^2}$ in the negative $x$ direction.
Problem (13): A race car accelerates from rest at a constant rate of $2\,{\rm m/s^2}$ in 15 seconds. It then travels at a constant speed for 20 seconds, and after that, it comes to a stop with an acceleration of $2\,{\rm m/s^2}$. (a) What is the total distance traveled by car? (b) What is its average velocity over the entire path?
Solution : To solve this kinematics question, we divided the entire path into three parts.
Part I: "From rest'' means the initial velocity is zero. Thus, given the acceleration and time interval, we can use the kinematics equation $v=v_0+at$ to calculate the distance traveled by car at the end of 15 seconds for the first part of the path. \begin{align*} x&=\frac{1}{2}at^2 +v_0 t+x_0\\\\&=\frac 12 (2)(15)^2 +(0)(15)+0\\\\&=\boxed{125\quad{\rm m}}\end{align*} As a side calculation, we find the final velocity for this part as below \begin{align*}v&=v_0+at\\\\&=0+(2) (15) \\\\&=30\quad {\rm m/s}\end{align*} Part II: the speed in this part is the final speed in the first part because the car continues moving at this constant speed after that moment.
The constant speed means we are facing zero acceleration. As a result, it is preferable to use the average velocity definition rather than the kinematics equations for constant (uniform) acceleration.
The distance traveled for this part, which takes 20 seconds at a constant speed of 30 m/s, is computed by the definition of average velocity as below \begin{align*} \bar{v}&=\frac{\Delta x}{\Delta t}\\\\30&=\frac{\Delta x}{20}\end{align*} Thus, we find the distance traveled as $\boxed{x=600\,{\rm m}}$.
Part III: In this part, the car comes to a stop, $v=0$, so its acceleration must be a negative value as $a=-2\,{\rm m/s^2}$. Here, the final velocity is also zero. Its initial velocity is the same as in the previous part.
Consequently, the best kinematics equation that relates those known to the wanted distance traveled $x$, is $v^2-v_0^2=2a(x-x_0)$. \begin{align*} v^2-v_0^2&=2a(x-x_0) \\\\0^2-(30)^2&=2(-2)(x-0) \\\\ \Rightarrow \quad x&=\boxed{225\quad {\rm m}}\end{align*} The total distance traveled by car for the entire path is the sum of the above distances $D=125+600+225=\boxed{950\quad {\rm m}}$
Problem (14): A ball is dropped vertically downward from a tall building of 30-m-height with an initial speed of 8 m/s. After what time interval does the ball strike the ground? (take $g=-10\,{\rm m/s^2}$.)
Solution : This is a free-falling kinematics problem. As always, choose a coordinate system along with the motion and the origin as the starting point.
Here, the dropping point is considered the origin, so in all kinematics equations, we set $y_0=0$. By this choice, the striking point is 30 meters below the origin, so in equations, we also set $y=-30\,{\rm m}$.
Remember that velocity is a vector in physics whose magnitude is called speed. In this problem, the initial speed is 8 m/s downward. This means that the velocity vector is written as $v=-8\,{\rm m/s}$.
Now that all necessary quantities are ready, we can use the kinematics equation $y=\frac 12 at^2+v_0t+y_0$, to find the wanted time that the ball strikes the ground. \begin{align*} y&=\frac 12 at^2+v_0t+y_0\\\\-30&=\frac 12 (-10)t^2+(-8)t+0\end{align*} After rearranging, a quadratic equation like $5t^2+8t-30=0$ is obtained, whose solutions are given as below: \begin{gather*} t=\frac{-8\pm\sqrt{8^2-4(5)(-30)}}{2(5)}\\\\ \boxed{t_1=1.77\,{\rm s}} \quad , \quad t_2=-3.37\,{\rm s}\end{gather*} $t_1$ is the accepted time because the other is negative, which is not acceptable in kinematics. Therefore, the ball takes about 1.7 seconds to hit the ground.
Note: The solutions of a quadratic equation like $at^2+bt+c=0$, where $a,b,c$ are some constants, are found by the following formula: $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Problem (15): The acceleration versus time graph for an object that moves at a constant speed of 30 m/s is shown in the figure below. Find the object's average velocity between instances $t_1=10\,{\rm s}$ and $t_2=30\,{\rm s}$.
Solution : The best and shortest approach to solving such a kinematics problem is to first draw its velocity-vs-time graph. Next, the area under the obtained graph gives us the total displacement, which is divided by the total time interval to yield the average velocity.
The path consists of three parts with different accelerations.
In the first part, the object slows down its motion at a constant rate of $-2\,{\rm m/s^2}$ in 10 seconds. Its initial velocity is also 30 m/s. With these known quantities in hand, the kinematics equation $v=v_0+at$ gives us the velocity at the end of this time interval. \begin{align*} v&=v_0+at\\&=30+(-2)(10) \\&=10\quad {\rm m/s}\end{align*} This calculation corresponds to a straight line between the points $(v=30\,{\rm m/s},t=0)$ and $(v=10\,{\rm m/s},t=10\,{\rm s})$ on the $v-t$ graph as shown below.
Next, the object moves with zero acceleration for 5 seconds, which means the velocity does not change during this time interval. This implies that we must draw a horizontal line in the $v-t$ graph.
In the last part, the object accelerates from 10 m/s with a constant rate of $+2\,{\rm m/s^2}$ in 15 seconds. Thus, its final velocity at the end of this time interval is determined as below \begin{align*} v&=v_0+at\\&=10+(2)(15) \\&=40\quad {\rm m/s}\end{align*} Now, it's time to draw the velocity-vs-time graph. As an important point, note that all these motions have a constant acceleration, so all parts of a velocity-time graph, are composed of straight-line segments with different slopes.
For part I, we must draw a straight-line segment between the velocities of 30 m/s and 10 m/s.
Part II is a horizontal line since its velocities are constant during that time interval, and finally, in Part III, there is a straight line between velocities of 10 m/s and 40 m/s.
All these verbal phrases are illustrated in the following velocity-vs-time graph .
Recall that the area under a velocity vs. time graph always gives the displacement. Hence, the area under the $v-t$ graph between 10 s and 30 s gives the displacement. Therefore, the areas of rectangle $S_1$ and trapezoid $S_2$ are calculated as below \begin{gather*} S_1 =10\times 5=50\quad {\rm m} \\\\S_2=\frac{10+40}{2}\times 15=375\quad {\rm m}\end{gather*} Therefore, the total displacement in the time interval $[15,30]$ is $D=S_{tot}=S_1+S_2=425\,{\rm m}$ From the definition of average velocity, we have $\bar{v}=\frac{displacement}{time}=\frac{425}{20}=21.25\,{\rm m/s}$
In this tutorial, all concepts about kinematics equations are taught in a problem-solution strategy. All these answered problems are helpful for MCAT physics exams.
We can also find these kinematic variables using a position-time or velocity-time graph. Because slopes in those graphs represent velocity and acceleration, respectively, and the concavity of a curve in a position vs. time graph shows the sign of its acceleration in an x-t graph as well.
Author : Dr. Ali Nemati Date published : 8-7-2021 Updated : June 12, 2023
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## Mechanics: 1-Dimensional Kinematics
Calculator pad, version 2, 1-d kinematics problem set.
In the 2008 Olympics, Jamaican sprinter Usain Bolt shocked the world as he ran the 100-meter dash in 9.69 seconds. Determine Usain's average speed for the race.
• Audio Guided Solution
In the Funny Car competition at the Joliet Speedway in Joliet, Illinois in October of 2004, John Force complete the ¼-mile dragster race in a record time of 4.437 seconds. Determine the average speed of the dragster in mi/hr and m/s. GIVEN: (1.000 mi =1609 m)
In the qualifying round of the 50-yd freestyle in the sectional swimming championship, Dugan got an early lead by finishing the first 25.00 yd in 10.01 seconds. Dugan finished the return leg (25.00 yd distance) in 10.22 seconds.
a. Determine Dugan's average speed for the entire race. b. Determine Dugan's average speed for the first 25.00 yd leg of the race. c. Determine Dugan's average velocity for the entire race.
In last week's Homecoming victory, Al Konfurance, the star halfback of South's football team, broke a tackle at the line of scrimmage and darted upfield untouched. He averaged 9.8 m/s for an 80-yard (73 m) score. Determine the time for Al to run from the line of scrimmage to the end zone.
During the annual shuffleboard competition, Renee gives her puck an initial speed of 9.32 m/s. Once leaving her stick, the puck slows down at a rate of -4.06 m/s/s.
a. Determine the time it takes the puck to slow to a stop. b. Use your initial speed and the calculated time to determine the average speed and the distance which the puck travels before stopping.
Ken Runfast is the star of the cross-country team. During a recent morning run, Ken averaged a speed of 5.8 m/s for 12.9 minutes. Ken then averaged a speed of 6.10 m/s for 7.1 minutes. Determine the total distance which Ken ran during his 20 minute jog.
The Lamborghini Murcielago can accelerate from 0 to 27.8 m/s (100 km/hr or 62.2 mi/hr) in a time of 3.40 seconds. Determine the acceleration of this car in both m/s/s and mi/hr/s.
Homer Agin leads the Varsity team in home runs. In a recent game, Homer hit a 96 mi/hr sinking curve ball head on, sending it off his bat in the exact opposite direction at 56 mi/hr. The actually contact between ball and bat lasted for 0.75 milliseconds. Determine the magnitude of the average acceleration of the ball during the contact with the bat. Express your answer in both mi/hr/s and in m/s/s. (Given: 1.00 m/s = 2.24 mi/hr)
A Formula One car is a single-seat racing car with an open cockpit and substantial wings located in the front and rear. At high speeds, the aerodynamics of the car help to create a strong downward force which allows the car to brake from 27.8 m/s (100 km/hr or 62.2 mi/hr) to 0 in as small of a distance as 17 meters. Determine the deceleration rate (i.e., acceleration) achieved by such a car.
The position-time graph below represents the motion of South's basketball coach during the last sixteen seconds of overtime during this past weekend's game.
Use the graph to answer the next several questions.
a. Determine the total distance walked by the coach during these 16 seconds. b. Determine the resulting displacement of the coach during these 16 seconds. c. Determine the displacement of the coach after 12.0 seconds. d. At what time did the coach have the greatest displacement from his starting position? e. What was the fastest speed which the coach walked during any of the time intervals for the last 16.0 seconds? f. What was the average speed of the coach for these 16.0 seconds?
Mr. H is preparing to show the class a Strobe demonstration when he realizes that his absent-mindedness has struck once more. He left the strobe on the counter in the back of the lab after the last class period. Starting 1.0 meter from the front of the room, Mr. H walks quickly to the back of the lab, picks up the strobe and returns to the middle of the classroom. The position-time graph below represents his motion. Use the graph to answer the next several questions.
a. What is the total distance walked by Mr. H during these 8.0 seconds? b. What is the average speed of Mr. H during these 8.0 seconds? c. What is the average velocity of Mr. H during these 8.0 seconds? d. How fast did Mr. H walk during the first 5.0 seconds? e. How fast did Mr. H walk during the last 3.0 seconds?
The position-time graph below represents the motion of two students - Mac (in red) and Tosh (in blue) - as they enter and exit the school library during a passing period.
Use the graph to determine the speeds at which the two students move. (Ignore any stationary periods of time.) Then determine how much faster the fastest student moves than the slower student.
Renatta Gas did it again. She failed to fill up her tank during the last four weeks. The velocity-time graph below represents the last six seconds of motion her car before being stranded on a highway in route to her university.
Use this graph to determine...
a. ...the acceleration of Renatta's car. b. ...the distance traveled during her last 6.0 seconds of motion.
Marcus Tardee is driving his friends to school. With the start of school being only minutes away, he is unfortunately following a slow garbage truck. The truck finally turns down a side street and Marcus accelerates to a much more customary speed. The velocity-time graph below represents his motion. Use the graph to answer the following questions.
a. How fast was Marcus traveling while following the garbage truck? b. Determine the distance traveled during the first 4.0 seconds represented on the graph. c. Determine the acceleration of the car once the garbage truck turned onto the side street. d. Determine the distance traveled by the car during the last 6.0 seconds of motion.
The velocity-time graph below represents the motion of a car on a city street.
Use the graph to determine the acceleration values of the car at ...
a. 1.4 seconds. b. 6.8 seconds. c. 11.6 seconds. d. 17.6 seconds.
After a long soccer practice down at the neighborhood soccer fields, Suzie begins walking up the steep hill towards her home. She gives her soccer ball a kick up the hill and continues walking towards it, meeting the ball as it is rolling back down. The velocity-time graph below depicts the motion of the ball. Use the graph to answer the following questions.
a. At what time did the ball change directions and begin rolling back down the hill? b. What is the acceleration of the ball as it rolls up the hill? down the hill? c. How far up the hill did the ball roll before it began to roll back down? d. Determine the total distance traveled by the ball during the 5.00 seconds - both the distance up the hill and down the hill. e. How far up the hill did Suzie walk between the time when she kicked the ball and the time she met up with the ball (at 5.0 seconds)?
Jeremy has recently taken up snowboarding as a hobby. He is practicing making smooth turns while traveling up sloped inclines. The velocity-time graph below depicts his motion traveling up an embankment and part-way down. Use the graph to answer the following questions.
a. Determine Jeremy's acceleration at 8.0 seconds. b. Determine the distance Jeremy traveled from 0.0 to 5.0 seconds. c. At what time did Jeremy begin to travel back down the embankment?
A Cessna 150 airplane has a takeoff speed of 28 m/s (63 mi/hr). Determine the minimum length of the runway which would be required for the plane to take off if it averages an acceleration of 1.9 m/s/s.
Cynthia competes in luge competitions during the winter months. She rides solo on a small sled 3 inches off the ground down icy slopes, turning only by use of her feet and the shifting of her weight on the sled. During the initial stage of one downhill luge, Cynthia accelerated from rest at 6.84 m/s/s for 2.39 seconds. Determine the distance she moved during this acceleration phase.
Suzie Lavtaski has reached the end of the ski slope and abruptly decelerates from 29.0 m/s to 1.8 m/s in 1.45 seconds. Determine Suzie' acceleration rate and the distance she moved during this braking period.
Captain John Stapp is often referred to as the "fastest man on Earth." In the late 1940s and early 1950s, Stapp ran the U.S. Air Force's Aero Med lab, pioneering research into the accelerations which humans could tolerate and the types of physiological effects which would result. After several runs with a 185-pound dummy named Oscar Eightball, Captain Stapp decided that tests should be conducted upon humans. Demonstrating his valor and commitment to the cause, Stapp volunteered to be the main subject of subsequent testing.
Manning the rocket sled on the famed Gee Whiz track, Stapp tested acceleration and deceleration rates in both the forward-sitting and backward-sitting positions. He would accelerate to aircraft speeds along the 1200-foot track and abruptly decelerate under the influence of a hydraulic braking system. On one of his most intense runs, his sled decelerated from 282 m/s (632 mi/hr) to a stop at -201 m/s/s. Determine the stopping distance and the stopping time.
Julietta and Jackson are playing miniature golf. Julietta's ball rolls into a long. straight upward incline with a speed of 2.95 m/s and accelerates at -0.876 m/s/s for 1.54 seconds until it reaches the top of the incline and then continues along an elevated section. Determine the length of the incline.
Rickey Henderson, baseball's record holder for stolen bases, approaches third base. He dives head first, hitting the ground at 6.75 m/s and reaching the base at 5.91 m/s, accelerating at -5.11 m/s/s. Determine the distance Rickey slides across the ground before touching the base.
Win Blonehare and Kent Swimtashore are sailboating in Lake Gustastorm. Starting from rest near the shore, they accelerate with a uniform acceleration of 0.29 m/s/s, How far are they from the shore after 18 seconds?
For years, the tallest tower in the United States was the Phoenix Shot Tower in Baltimore, Maryland. The shot tower was used from 1828 to1892 to make lead shot for pistols and rifles and molded shot for cannons and other instruments of warfare. Molten lead was dropped from the top of the 234-foot (71.3 meter) tall tower into a vat of water. During its free fall, the lead would form a perfectly spherical droplet and solidify. Determine the time of fall and the speed of a lead shot upon hitting the water at the bottom.
According to Guinness, the tallest man to have ever lived was Robert Pershing Wadlow of Alton, Illinois. He was last measured in 1940 to be 2.72 meters tall (8 feet, 11 inches). Determine the speed which a quarter would have reached before contact with the ground if dropped from rest from the top of his head.
A California Condor is approaching its nest with a large chunk of carrion in its beak. As it approaches, it makes an upward swoop, achieving a momentary upward velocity of 12.8 m/s when the carrion falls from its mouth, hitting a cliff outcropping 32.1 m below. Determine the speed of the carrion upon hitting the outcropping.
During his recent skydiving adventure, Luke Autbeloe had reached a terminal speed of 10.4 m/s as he approached the ground with his parachute. During an attempt to snap one last photo with his camera, Luke fumbled it from a height of 52.1 m above the ground.
a. Determine the speed with which the camera hits the ground. b. Determine the time for the camera to free fall from Luke's hands to the ground.
The speed required of a military jet when taking off from the deck of an aircraft carrier is dependent upon the speed of the carrier and the speed of the wind into which the carrier is moving. The takeoff speed required of a military jet relative to the deck of the carrier is 45 m/s when the carrier travels at 45 mi/hr into a 20 mi/hr wind. And when the aircraft carrier is traveling at 10 mi/hr into a 5 mi/hr wind, the takeoff speed relative to the deck of the carrier is 71 m/s. Determine the acceleration which a military jet must have to take off under these two conditions from the 126-m long runway of the USS Ronald Reagan aircraft carrier.
The Zero Gravity Research Facility at NASA-operated Glenn Research Center in Ohio is used to test the behavior of fluids, flames, equipment and other objects in free fall. It consists of a 467-foot long, 12-foot diameter, steel vacuum chamber. The steel chamber resides inside of a concrete lined shaft which extends 510 feet below ground level. Objects falling through the tower experience free fall over a distance of 432 feet (132 meters).
a. Determine the falling time for objects dropped from rest. b. Determine the final speed of the objects before the braking period begins.
It's breakfast time and Mr. H entertains himself once more by watching the daily beetle race across the 35.7-cm length of the Wheaties box top. Angie the beetle typically averages 3.77 mm/s and Bessie the beetle averages 4.78 mm/s. If Bessie gives Angie a 5.4 cm head start, then which beetle wins and by what distance?
Alexander's hobby is dirt biking. On one occasion last weekend, he accelerated from rest to 17.8 m/s/s in 1.56 seconds. He then maintained this speed for 9.47 seconds. Seeing a coyote cross the trail ahead of him, he abruptly stops in 2.79 seconds. Determine Alexander's average speed for this motion.
Ima Rushin can travel from Milwaukee Avenue to the school entrance gate at a constant speed of 22.5 m/s when the lights are green and there is no traffic. On Wednesday, Ima is stopped by a red light at Landwehr Road. She decelerates at -3.95 m/s/s, waits for 45.0 seconds before the light turns green and accelerates back up to speed at 4.91 m/s/s.
a. Determine the total time required to decelerate, stop and accelerate back up to speed. b. Determine the amount of time the red light costs the driver (compared to if the car had not been stopped by the red light).
A tortoise and a hare are having a 1000-meter race. The tortoise runs the race at a constant speed of 2.30 cm/s. The hare moves at an average speed of 1.50 m/s for 10.0 minutes and then decides to take a nap. After waking up from the nap, the hare recognizes that the tortoise is about to cross the finish line and immediately accelerates from rest with a constant acceleration of 0.500 m/s/s for the remaining distance of the race. If the tortoise wins by a hair (no pun intended), then what is the time in hours that the hare napped?
Hayden and Matthew are riding around the neighborhood on their scooters. Hayden is at rest when Matthew passes him moving at a constant speed of 0.37 m/s. After 1.8 seconds, Hayden decides to chase after Matthew, accelerating at 0.91 m/s/s. How much time must Hayden accelerate before he is side-by-side with Matthew?
## Browse Course Material
Course info, instructors.
• Prof. Deepto Chakrabarty
• Dr. Peter Dourmashkin
• Dr. Michelle Tomasik
• Prof. Anna Frebel
## Departments
As taught in.
• Classical Mechanics
## Learning Resource Types
Week 1: kinematics.
« Previous | Next »
• Week 1: Introduction
## Lesson 1: 1D Kinematics - Position and Velocity
• 1.1 Coordinate Systems and Unit Vectors in 1D Position Vector in 1D
• 1.2 Position Vector in 1D
• 1.3 Displacement Vector in 1D
• 1.4 Average Velocity in 1D
• 1.5 Instantaneous Velocity in 1D
• 1.6 Derivatives
• 1.7 Worked Example - Derivatives in Kinematics
## Lesson 2: 1D Kinematics - Acceleration
• 2.1 Introduction to Acceleration
• 2.2 Acceleration in 1D
• 2.3 Worked Example - Acceleration from Position
• 2.4 Integration
• 2.5 List of Useful Integrals
## Lesson 3: 2D Kinematics - Position, Velocity, and Acceleration
• 3.1 Coordinate System and Position Vector in 2D
• 3.2 Instantaneous Velocity in 2D
• 3.3 Instantaneous Acceleration in 2D
• 3.4 Projectile Motion
• 3.5 Demos for Projectile Motion
## Week 1 Worked Examples
• PS.1.1 Three Questions Before Starting
• PS.1.2 Shooting the Apple
• PS.1.3 Worked Example: Braking Car
• PS.1.4 Sketch the Motion
• PS.1.5 Worked Example: Pedestrian and Bike at Intersection
## Week 1 Problem Set
• Problem Set 1
Mr. Lam's Classroom
## Physics 11 – Kinematics
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## AP®︎/College Physics 1
Welcome to ap/college physics 1, unit 1: kinematics and introduction to dynamics, unit 2: newton's laws, unit 3: circular motion and gravitation, unit 4: energy and momentum, unit 5: simple harmonic motion and rotational motion.
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## Kinematics Practice Problems
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## 1. Kinematics practice problems
2. student worksheet for two dimensional kinematics, 3. physics kinematics objectives students will be able to, 4. test1 ch15 kinetics practice problems, 5. kinetics practice problems and solutions, 6. topic 3.1: kinematics, 7. chapter 3 kinematics in two or three dimensions; vectors, 8. kinematics practice problems worksheet answer key.
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## Kinematics Practice Problems
Displaying top 8 worksheets found for - Kinematics Practice Problems .
Some of the worksheets for this concept are Kinematics practice problems, Student work for two dimensional kinematics, Physics kinematics objectives students will be able to, Test1 ch15 kinetics practice problems, Kinetics practice problems and solutions, Topic kinematics, Chapter 3 kinematics in two or three dimensions vectors, Kinematics practice problems work answer key.
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## 1. Kinematics practice problems
2. student worksheet for two dimensional kinematics, 3. physics kinematics objectives students will be able to, 4. test1 ch15 kinetics practice problems, 5. kinetics practice problems and solutions, 6. topic 3.1: kinematics, 7. chapter 3 kinematics in two or three dimensions; vectors, 8. kinematics practice problems worksheet answer key.
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## Gallery of 50 Kinematics Practice Problems Worksheet
#### IMAGES
2. Kinematics Problems Worksheet With Answers
3. Quiz & Worksheet
4. Kinematics Practice Problems Worksheet
6. Kinematics Practice Problems Worksheet
#### VIDEO
1. AS Kinematics Theory Questions Practice
2. AS kinematics essential practice questions easy level
3. kinematics problem practice 3
4. kinematics problem practice 4
5. Kinematics problems
6. Relative motion questions for jee
1. Kinematic Equations: Sample Problems and Solutions
These problems allow any student of physics to test their understanding of the use of the four kinematic equations to solve problems involving the one-dimensional motion of objects. You are encouraged to read each problem and practice the use of the strategy in the solution of the problem.
2. PDF Kinematics: Practice Problems with Solutions in Physics Physexams
Solution: First of all, collect the given data in the interval of accelerating initial velocity = 30 m/s final velocity = 0 overall time = 7.20 s distance =? One can solve this problem in two, direct and indirect, ways. In one way, first, find the acceleration of the car and then use other kinematic equations to deter- mine the desired quantity.
3. Kinematics Practice Problems with Answers
Kinematics Practice Problems: Problem (1): A car slows down its motion from 10 m/s to 6 m/s in 2 seconds under constant acceleration. (a) What is its acceleration? (b) How far did the car travel during this time interval? Solution: This is the simplest kinematics problem, so we put a bit more time to solve it in detail.
4. 1-Dimensional Kinematics Problem Sets
Problem 1 In the 2008 Olympics, Jamaican sprinter Usain Bolt shocked the world as he ran the 100-meter dash in 9.69 seconds. Determine Usain's average speed for the race. Audio Guided Solution Show Answer Problem 2
5. PDF Kinematic Equations Worksheet
Problem 1. A runner accelerates to 4.2 m/s2 for 10 seconds before winning the race. How far did he/she run? Problem 2. A plane starts from rest and accelerates uniformly over a time of 20 s for a distance of 300 m. Determine the plane's acceleration. Problem 3. A ball free falls from the top of the roof for 5 seconds. How far did it fall?
6. Kinematics Practice Problems Name: Block: Date:
Kinematics Practice Problems Name:_______________ Block:____ Date:__________ Kinematics is the study of motion. In 1-D motion, most every kinematic problem can be solved using one of 4 equations. These equations will allow you to solve for almost any aspect of the motion of an object: displacement, velocity and acceleration.
7. Kinematics Equations Practice Problems
Determine the distance covered by the motorcycle in the first 8 seconds. 4 PRACTICE PROBLEM A box sliding at a constant speed of 5.0 m/s on a frictionless surface enters a rough, concrete surface. When the box moves 3.0 meters on concrete, its speed drops to 4.0 m/s. Determine the magnitude of deceleration of the box on concrete. 5 PRACTICE PROBLEM
8. PDF 1-D Kinematics Problems
a. Give a written description of the motion. b. Determine the displacement from t = 0s to t = 4 s. c. Determine the displacement from t = 2 s to t = 6 s. d. Determine the object's acceleration at t =4s. e. Sketch a possible x-t graph for the motion of the object. Label the axis. Explain why your graph is only one of many possible graphs. 2.
9. Kinematic formulas in one-dimension (practice)
Science Kinematic formulas in one-dimension Google Classroom You might need: Calculator A child blows a leaf straight up in the air. The leaf reaches 1.0 m higher than its original height with a constant acceleration of 1.0 m s 2 upward. How much time did it take the leaf to get displaced by 1.0 m ?
10. Uniformly Accelerated Motion
Uniformly Accelerated Motion. Directions: On this worksheet you will practice solving problems by using the five basic kinematics equations for uniformly accelerated motion. omit. Question 1 Scenario #1: A car accelerates uniformly from rest to a final speed of 21 m/sec in 11 seconds. How far does it travel during this period of acceleration ...
11. Forces and kinematics (practice)
Choose 1 answer: 2.4 m s A 2.4 m s 8.6 m s B 8.6 m s 5.4 m s C 5.4 m s 7.0 m s D 7.0 m s Show Calculator Stuck? Use a hint. Do 4 problems Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more.
12. Quiz & Worksheet
1. In kinematics, how do we resolve questions of motion? Through charts and pictorial representations Through geometric equations Through repeatable experiments Through lucky guesses 2. Which of...
13. Kinematics worksheets
Question 1. A bus start at Station A from rest with uniform acceleration 2m/sec 2 .Bus moves along a straight line a. Find the distance moved by the bus in 10 sec? b. At what time, it velocity becomes 20m/sec? c. How much time it will take to cover a distance of 1.6km Solution Question 2. A object is moving along an straight line.
14. Week 1: Kinematics
Lesson 1: 1D Kinematics - Position and Velocity. 1.1 Coordinate Systems and Unit Vectors in 1D Position Vector in 1D. 1.2 Position Vector in 1D. 1.3 Displacement Vector in 1D. 1.4 Average Velocity in 1D. 1.5 Instantaneous Velocity in 1D. 1.6 Derivatives. 1.7 Worked Example - Derivatives in Kinematics.
15. rotational kinematics (practice)
Rotational kinematics. A merry-go-round has an initial angular velocity of 10.2 rpm . At the end of the ride, the brakes are applied, giving it a constant angular deceleration of 0.032 rad/s 2 as the ride slows to a stop. Determine the angular velocity of the ride, in rad/s , after it has made two revolutions during the braking period.
16. Kinematics in Two Dimensions
practice problem 1 I went for a walk one day. I walked north 6.0 km at 6.0 km/h and then west 10 km at 5.0 km/hr. (This problem is deceptively easy, so be careful. Begin each part by reviewing the appropriate physical definition.) Determine… the total distance of the entire trip the total displacement of the entire trip
17. Kinematics in 2D Practice Problems
Learn Kinematics in 2D with free step-by-step video explanations and practice problems by experienced tutors.
18. Kinematics Equations Video Tutorial & Practice
Learn Kinematics Equations with free step-by-step video explanations and practice problems by experienced tutors. ... Table of contents. 0. Math Review 31m. Worksheet. Math Review 31m. 1. Intro to Physics Units 51m. Worksheet. Introduction to Units 17m. Unit Conversions 7m. Solving Density Problems 7m. Dimensional Analysis 5m. Counting ...
19. Physics 11
6. 16 Feb 2023. (Thu) Notes: Graphs of Motion (Velocity vs. Time) Worksheet: Graphs of Motion II ( solutions) Assignment: Kinematics Video Analysis due next class (print and bring to class) Lab: Acceleration due to Gravity Lab complete Introduction pre-lab questions for next class. In-Class Questions. Zitzewitz: §5.2.
20. AP®︎/College Physics 1
AP®︎/College Physics 1 5 units · 27 skills. Unit 1 Kinematics and introduction to dynamics. Unit 2 Newton's laws. Unit 3 Circular motion and gravitation. Unit 4 Energy and momentum. Unit 5 Simple harmonic motion and rotational motion. Course challenge. Test your knowledge of the skills in this course. Start Course challenge.
21. Kinematics Practice Problems Worksheets
Showing top 8 worksheets in the category - Kinematics Practice Problems. Some of the worksheets displayed are Kinematics practice problems, Student work for two dimensional kinematics, Physics kinematics objectives students will be able to, Test1 ch15 kinetics practice problems, Kinetics practice problems and solutions, Topic kinematics, Chapter 3 kinematics in two or three dimensions vectors ...
22. Kinematics Practice Problems Worksheets
Displaying top 8 worksheets found for - Kinematics Practice Problems. Some of the worksheets for this concept are Kinematics practice problems, Student work for two dimensional kinematics, Physics kinematics objectives students will be able to, Test1 ch15 kinetics practice problems, Kinetics practice problems and solutions, Topic kinematics, Chapter 3 kinematics in two or three dimensions ...
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# 8.9: Sum and Product Notation
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$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$
For reference, this section introduces the terminology used in some texts to describe the minterms and maxterms assigned to a Karnaugh map. Otherwise, there is no new material here.
Σ (sigma) indicates sum and lower case “m” indicates minterms. Σm indicates sum of minterms. The following example is revisited to illustrate our point. Instead of a Boolean equation description of unsimplified logic, we list the minterms.
The numbers indicate cell location, or address, within a Karnaugh map as shown below right. This is certainly a compact means of describing a list of minterms or cells in a K-map.
The Sum-Of-Products solution is not affected by the new terminology. The minterms, 1s, in the map have been grouped as usual and a Sum-OF-Products solution written.
Below, we show the terminology for describing a list of maxterms. Product is indicated by the Greek Π (pi), and upper case “M” indicates maxterms. ΠM indicates product of maxterms. The same example illustrates our point. The Boolean equation description of unsimplified logic, is replaced by a list of maxterms.
Once again, the numbers indicate K-map cell address locations. For maxterms this is the location of 0s, as shown below. A Product-OF-Sums solution is completed in the usual manner.
This page titled 8.9: Sum and Product Notation is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Tony R. Kuphaldt (All About Circuits) via source content that was edited to the style and standards of the LibreTexts platform.
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High School Physics
# Derive formulas of centripetal force & centripetal acceleration
Last updated on May 31st, 2022 at 12:39 pm
In this post, we will derive formulas of centripetal acceleration & centripetal force while we discuss the uniform circular motion.
## Uniform circular motion & its acceleration
An object moving with uniform circular motion moves along a circular path of fixed radius at a constant speed. Since the direction of motion of the object is constantly changing, its velocity is also constantly changing.
Therefore, the object in uniform circular motion must be continuously accelerating.
Figure 1 shows an object moving around a circle at a constant speed.
The velocity at point A is vA, and the velocity at point B is vB. The velocities have exactly the same magnitude, v.
The position vector at point A is rA, and the position vector at point B is rB. Both position vectors have magnitude r, as measured from the center of the circle.
At every point on the circle, the velocity of the object is tangential to the circle and perpendicular to the position vector.
The average acceleration is defined as:
aavg = Δv/Δt
Therefore, the direction of average acceleration is parallel to Δv.
## Direction of Δv
To determine the direction of Δv, we redraw the position and velocity vectors using the principle that a vector can be translated as long as its magnitude and direction do not change. First, we construct a triangle formed by the two position vectors and the change in position, Δr, as shown in Figure 2. We then construct a triangle formed by the two velocity vectors and the change in velocity.
Since the velocity vector is perpendicular to the position vector at either location, we can conclude that the angle subtended by the two-position vectors is equal to the angle subtended by the two corresponding velocity vectors. Further, the two velocity vectors are the same length, as are the two-position vectors; the two triangles from Figure 2 are therefore similar triangles.
We can also see that Δv and, consequently, the average acceleration vector are perpendicular to Δr.
## Deriving formula of centripetal acceleration (using differential calculus)
Here, we will derive the formula of centripetal acceleration using differential calculus. If you want to derive this equation using trigonometry only (without using calculus), then you must visit this page that guides you to derive the equation of centripetal acceleration using trigonometry only.
In Figure 3, we repeat the construction for a shorter interval of time between the two points.
We can see that the conclusions we drew from our construction in Figure 2 still hold.
Next, we take the limit of the average acceleration as Δt goes to zero to get the instantaneous acceleration:
a = limΔt->0v/Δt) = dv/dt
As the time interval Δt shrinks, so does the displacement between A and B. In the limit, the two points merge. As the position triangle collapses, so does the velocity triangle.
However, the acceleration, which is parallel to Δv, remains perpendicular to Δr, which is a chord of the circle.
The acceleration vector, being perpendicular to that chord, points along the radius to the centre of the circle.
In the limit as Δt goes to zero, the instantaneous acceleration points along the radius. Therefore, an object moving in a circle must continuously accelerate toward the centre of the circle. The acceleration toward the centre of the circle is called radial or centripetal acceleration.
To find the magnitude of the centripetal acceleration, consider the ratios of the two similar triangles formed by the position and velocity vectors:
Δv/v = Δr/r
Δv = v (Δr/r)
Dividing both sides by Δt (the change in time) gives
Δv / Δt = (v Δr)/(r Δt)
In the limit as Δt goes to zero
dv/dt = (v/r) limΔt->0 (Δr/Δt) = (v/r) v …………… (1)
The left-hand side of Equation 1 is the acceleration, so ar = v2/r. This is the equation of centripetal acceleration.
The subscript r on the acceleration denotes that this is the radial acceleration.
We have thus established both the direction and the magnitude for the radial centripetal acceleration of an object moving around a circle at a constant speed.
Figure 4 shows the position, velocity, and acceleration vectors for an object executing uniform circular motion. The velocity vector is always tangential to the circle.
The acceleration or centripetal acceleration vector is perpendicular to the velocity vector and points radially inward toward the centre of the circle. The magnitude of the centripetal acceleration vector at any point is the same: ar = v2/r
## Deriving formula of centripetal force
We have already derived the equation of centripetal acceleration. Now, we will derive the equation or formula of the centripetal force, using Newton’s second law of motion.
As we know Force F = mass x acceleration,
Hence, centripetal force = mass x centripetal acceleration = m v2/r
The direction of the centripetal force is along the direction of centripetal acceleration, i.e. radially inward toward the centre of the circle.
Related:
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# Proportional relationships lessons explained
#### By Michael Avidon, math editor
One theme that occurs often in mathematics is the idea of representing something in multiple ways. In these lessons, relationships between two quantities are studied (such as the cost of a product and the amount purchased). These relationships are represented by verbal descriptions, 2-column tables, equations, and graphs.
(Note: the use of the word decide in part a. of the standard is incorrect. The word determine should be in its place. You decide what clothes to wear. You determine the solution to a math problem.)
More in this series
Lesson 1
Lesson 2
Lesson 3
Here is a bird’s-eye view of the material. The lessons have some fill-in-the-blanks for the student to complete. The correct fill-ins are included here.
These lessons are presented for guardians to help their child(ren) address the following Common Core State Standard (CCSS) for Grade 7:
• 7.RP.A.2: Recognize and represent proportional relationships between quantities.
1. Decide whether two quantities are in a proportional relationship, e.g., by testing for equivalent ratios in a table or graphing on a coordinate plane and observing whether the graph is a straight line through the origin.
2. Identify the constant of proportionality (unit rate) in tables, graphs, equations, diagrams, and verbal descriptions of proportional relationships.
3. Represent proportional relationships by equations. For example, if total cost t is proportional to the number n of items purchased at a constant price p, the relationship between the total cost and the number of items can be expressed as t = pn.
4. Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with special attention to the points (0, 0) and (1, r) where r is the unit rate.
## Prerequisites
As with any math lessons, a certain amount of basic material and vocabulary (taught prior to grade 7) is assumed. Specifically, it is assumed that the student is familiar with the following material.
• The definition of ratio and unit rate
• Writing simple equations to represent relationships (e.g., y = 5x)
• Rewriting equations by multiplying or dividing both sides by the same number
• Graphing points on a coordinate grid
• Vocabulary from earlier grades: coordinate plane, ordered pair, origin, ray, x-axis, y-axis
## Summary and vocabulary
Proportional relationship: exists between a quantity y and a quantity x when the ratio is constant.
Constant of proportionality: the ratio between two quantities in a proportional relationship; also known as the unit rate
Facts and Theorems: If a quantity y is proportional to a quantity x, then the quantity x is proportional to the quantity y.
The graph of a proportional relationship is a line through the origin or a ray whose endpoint is the origin.
Conversely, given a line through the origin or a ray whose endpoint is the origin, this represents a proportional relationship.
Formulas/Equations: A proportional relationship between a quantity y and a quantity x that has a constant of proportionality k is represented by the equation.
Other: You can take the equation for a proportional relationship and divide both sides by k to write the relationship the other way (switch the input and output).
Not all the points on the graph of a proportional relationship necessarily represent real values. For example, if the input is a number of people, then x-values must be whole numbers.
The point on the graph of a proportional relationship shows that the unit rate (or constant of proportionality) is r.
A line that does not go through the origin is not a proportional relationship.
Any curved graph, including one that goes through the origin, is not a proportional relationship.
### Lesson 1
In order to motivate the definition of proportional relationship, two examples of relationships between two quantities are given. One of these is a proportional relationship and one is not.
Page 1 fill-ins: \$3.50; \$1.10
Examples 1 and 2 present quantities in tables. It is a straightforward matter to compute the ratios of one quantity over the other to see if these ratios have the same value. Examples 3 and 4 require the student to create tables from verbal descriptions, and then check the ratios to see if there is a proportional relationship.
Example 5 is different because you are told that there is a proportional relationship at the start and the student must compute missing values in the table.
Page 2 fill-ins: 3, 7; 550, 550
Page 3 fill-ins: 0.90, 90; are not, is not; 3.4, 3.75, 12.75, 14.25, 3.75, 3.8
The first 6 exercises mimic the first two examples. Exercises 7 through 9 are similar to Examples 3 and 4. Exercises 10 through 12 are similar to Example 5.
### Lesson 2
The lesson starts with an example of a proportional relationship and explains how to write this as an equation. Then it shows that you can write a similar equation ( y =kx ) for any proportional relationship.
Page 1 fill-ins: 3, 10.50
The first two examples show proportional relationships represented as a table and a verbal description respectively, and ask for equations. Examples 3 and 4 give you equations and ask for a verbal description and a table. The point is to understand that you can represent the same relationship multiple ways, and to know how to go from one to another.
Page 2 fill-ins: 550; 6.40; 18, 5, 30
The first 6 exercises mimic the first two examples. Exercises 7 through 12 mostly match Examples 3 and 4. The information on the bottom of page 1 and the middle of page 2 about equations for proportional relationships and the reverse relationship are also needed to answer the latter exercises.
### Lesson 3
At this point, the student knows several ways to represent proportional relationships. In a straightforward procedure of plotting points, students discover that these points lie on a line through the origin (that is, the point (0, 0)).
Page 1 fill-ins: 3, 8
Page 2 starts with an explanation of why the points lie on a line or a ray (a line continues in both directions forever and a ray is “half” of a line). Example 1 starts with an equation, creates a table (as in Lesson 2), and then draws the graph of the relationship. It makes the important point that not all points on this graph correspond to real world values.
Page 2 fill-ins: 14, 17.50
Example 2 gives the student a graph and asks them to interpret it in terms of the context. It also asks the student to find the equation of the proportional relationship.
Page 3 fill-ins: 3, 90, 30
Example 3 exists to clarify which graphs do and do not represent proportional relationships. Examples 4 and 5 show relationships that are “linear” but are not “proportional.”
Page 4 fill-ins: 0, 0; origin; 0, 20
The first two exercises are similar to Example 1. The next two are similar to Example 4. Exercises 5 and 6 are similar to Example 5. Exercises 7 and 8 are most similar to Example 2 (look back to Lesson 2 for more examples of finding equations). Exercises 9 through 12 are like Example 3.
### Lesson 1
Does the table represent a proportional relationship?
If so, what is the constant of proportionality?
x y
2 4
4 6
6 8
x y
3 9
5 15
8 24
x y
2 2.4
6 7.2
9 10.8
Production
Time (hours) Widgets
3 1005
5 1675
7 2345
Apples
Weight (lbs.) Cost
1 \$1.50
3 \$4.00
5 \$5.50
T-shirts
Number Cost
2 \$23
4 \$46
5 \$57.50
7. An artist buys unprimed canvas in rolls that she will cut and use to make oil paintings. Each roll has 90 square feet of canvas and costs \$54.
Is there a proportional relationship between cost and area? If so, what is the constant of proportionality or unit rate?
8. A plumber charges a set fee of \$50 plus \$60 per hour for labor. Is there a proportional relationship between cost and time? If so, what is the constant of proportionality?
9. There are 8 fluid ounces in a cup, 2 cups in a pint, 2 pints in a quart, and 4 quarts in a gallon. Is there a proportional relationship between ounces and gallons? If so, what is the constant of proportionality?
Each table represents a proportional relationship. Fill in the missing numbers.
10.
x y
10 25
16 40
18 45
11.
x y
5 2
8 3.2
12 4.8
12.
x y
9 6
10 20/3
12 8
13. Challenge problem:
Suppose that there is a proportional relationship between quantity z and quantity y, and another proportional relationship between quantity y and quantity x. Must there exist a proportional relationship between quantity z and quantity x? Explain why or give a counterexample.
Answer: Yes. The z/y ratio is constant and the ratio y/x is constant. It follows that their product, z/x, is constant. This says that there is a proportional relationship between z and x.
## Lesson 2
Represent each proportional relationship with an equation.
Area, A
(sq. ft.)
Cost, C
90 \$81
180 \$162
360 \$324
Production
Time, t (hours) Widgets, w
3 1005
5 1675
7 2345
T-shirts
Number, n Cost, C
2 \$23
4 \$46
5 \$57.50
4. An artist buys unprimed canvas in rolls. Each roll has 90 square feet of canvas and costs \$54. Let C = the cost of the canvas in dollars and A = the area of the canvas in square feet.
5. A 3-pound package of ground beef sells for \$12. Let B = the cost of the ground beef in dollars and w = the weight of the ground beef in pounds.
6. A mason is building a brick wall. So far, he has laid 5 rows of brick (and cement) that reaches a height of 45 cm. Let n = the number of rows of brick and h = the height of the wall in centimeters.
For each equation, (a) represent the proportional relationship by a table with input values 1, 2, and 3, and (b) write the equation for the reverse relationship.
7. y = 4x
x y
1 4
2 8
3 12
8. y = 0.2x
x y
1 0.2
2 0.4
3 10.6
9. y = x3
x y
1
2
3 1
10. Does each equation represent a proportional relationship?
(a) y = x + 5 Answer: no (b)y = x 2 Answer: no (c) y = 2(x – 5) + 10 Answer: yes
11. The equation y = 2.54x represents the relationship between inches, x, and centimeters, y. Give a verbal description of the relationship.
Answer: For every inch, there are 2.54 centimeters.
12. The equation y = 33.8x represents the relationship between liters, x, and fluid ounces, y. Give a verbal description of the relationship.
Answer: For every liter, there are 33.8 fluid ounces.
13. Challenge Problem:
While the constant of proportionality is usually positive in real-world problems, it could be negative. Let h = the number of hours after midnight, and T = the temperature in degrees Celsius. Explain what the relationship means in real-world terms at midnight and beyond.
Answer: At midnight, the temperature is 0°C. Thereafter, the temperature is dropping by 2°C per hour.
## Lesson 3
Draw the graphs of the given equations.
1. y = 3x, where x= the number of pounds of ground beef, and y = the price in dollars.
2. y = 0.5x, with no restrictions on the variables
Answer: Graphs below for exercises 1 through 6
For each table, draw the graph and determine if the relationship is proportional.
x y
2 2.5
4 5
8 10
10 12.5
x y
1 1
2 3
3 5
4 7
For each description, draw the graph and determine if the relationship is proportional.
5. A tutor charges \$50 for one hour, \$90 for two hours, and \$120 for three hours.
6. Asparagus sells for \$2.50 per pound.
7. For each relationship in exercises 3 through 6 that was proportional, find the equation.
Answer: For exercise 3: y = 1.25x ri. For exercise 6: y = 2.5x
8. The graph shows the relationship between the time, x, in minutes someone jogs on a treadmill and the distance, y, in miles they run. Explain the meaning of the points (4, 0.4), (1, .01) and (0,0) in this context.
Answer: The person jogs 0.4 mile in 4 minutes, a unit rate of 0.1 mile per minute, and no distance at the start.
Does each graph below represent a proportional relationship?
13. Challenge Problem:
A line passes through the point (a, b), where both coordinates are positive. If the line represents a proportional relationship, what is its equation? Explain.
Answer: The ratio of output to input is the constant of proportionality. This is ba, so the equation is y = bax.
1. 2.
3. 4.
5. 6.
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# Types of Angles
🏆Practice sum and difference of angles
## What is an angle?
Definition: Angles are created at the intersection between two lines. As seen in the following illustration
The angle in the illustration is called $AB$. We could also call it angle $\sphericalangle ABC$. The important thing is that the middle letter is the one at the intersection of the lines.
For example, in this case:
The angle is $\sphericalangle BCD$ or $\sphericalangle DCB$. Both notations are correct for the same angle.
We usually mark the angle with an arc as follows:
The marked angle is $∡ABC$. Sometimes we will denote angles using Greek letters, for example:
$α$ or $β$
Before the name of the angle, we should note the angle symbol, like this:
$∡$
Together it looks like this:
$∡CBA$ or $∡α$
Next, we will delve into the size of angles, the different types, and those that are created when a line intersects two parallel lines.
## Test yourself on sum and difference of angles!
Angle A is equal to 30°.
Angle B is equal to 60°.
Angle C is equal to 90°.
Can these angles form a triangle?
There can be two angles that are equal, meaning they measure the same; likewise, a certain angle can be larger than another based on their measurements.
For example, an angle of $60º$ is larger than one of $45º$, and two angles of $30º$ are equal.
Angle larger than the other:
Angles of different sizes:
Notice that in these examples two angles were created, but at this stage, we will choose to refer to the acute angle (we will soon review what an acute angle is).
For example, in the following illustration:
Two angles were created as seen in the drawing:
At this phase, we will only refer to the acute angle of the two, the smaller one, the one that is between the two lines. This point might be a bit confusing, but don't worry because it will soon become clear to you.
## How is an angle measured?
Angles are measured in degrees. A full circle represents $360°$ degrees.
We will see this very clearly in the following illustration:
You can imagine that if we keep increasing the angle, we will eventually reach a full circle.
Whenever we want to indicate the size of an angle, we write the degree symbol next to the number. It is a small circle that is noted to the right of the number representing the angle size.
It looks like this: $90°$.
In words: $90$ degrees.
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## Acute Angle
Definition: An acute angle is one that measures less than $90°$:
It looks like this:
Acute angle, less than 90°
## Right Angle
Definition: A right angle is one that measures exactly $90°$:
It looks like this:
Note that the marking of a right angle is not like that of other angles. It's not marked with an arc but with a symbol that looks like this
Do you know what the answer is?
## Obtuse Angle
Definition: An obtuse angle is greater than $90°$ and less than $180°$:
It looks like this:
## Flat Angle
Definition: A straight angle measures exactly $180°$.
It looks like this:
Next, we will learn how to calculate the size of angles. For now, we are satisfied with knowing that a right angle is larger than an acute angle, and that an obtuse angle is larger than a right angle. We understand this intuitively.
For example, this angle:
$∡CBA$ is smaller than: $∡DEF$
We will write it like this:
$∡CBA<∡DEF$
## Opposite Angles by the Vertex
Definition: Vertically opposite angles are formed by two intersecting lines, with each pair facing each other.
For example:
The angles marked in red and also those in blue are vertically opposite. The angles in each pair of vertically opposite angles are equal (we'll delve deeper into this in other articles).
## Angle Between Parallel Lines:
Definition recap: two parallel lines are lines that never meet.
They look like this:
Line 1 and line 2 are parallel lines. Now we will draw another line, which crosses each of the parallel lines.
It looks like this:
That is, at the intersection between the two lines and the third, 8 angles were created (marked in the illustration). It is important to clarify that even if the lines were not parallel, 8 angles would be created. Now we will learn about the types of angles that have been created.
Do you think you will be able to solve it?
## Corresponding Angles
Definition: Corresponding angles are those that are on the same side of the transversal that cuts two parallel lines and are at the same level with respect to the parallel line. Corresponding angles are of equal.
This definition might seem a bit confusing, but the illustration makes it very clear what corresponding angles are:
The two angles marked in red are corresponding angles. Therefore, they are also equal. Likewise, the angles marked in blue are also corresponding angles, meaning they are equal to each other. This is very important information that will help us later. Try to determine which angle is acute and which is obtuse.
Definition: Adjacent angles are two angles that together form a straight angle (that is, $180°$). Next, we will learn the meaning of the sum of angles.
### For example
These two angles are adjacent angles.
### Another example
Notice, in this example the two angles marked in red are adjacent angles. Similarly, the angles marked in blue are also adjacent.
Do you know what the answer is?
## Alternate Angles
Definition: Alternate angles are the ones that are on opposite sides of the transversal cutting through two parallel lines and are not on the same level with respect to the parallel line. Alternate angles are equal.
The explanation might be confusing, but the illustration makes it clear:
The two angles marked in blue are alternate angles, meaning they are also equal. The two angles marked in red are also alternate, and therefore, they are equivalent. Try to determine which angles are acute and which are obtuse.
## Angle Types Exercises
### Exercise 1
Assignment
Among three parallel lines there are angles as sketched:
What is the value of $X$?
Solution
$AB\parallel CD\parallel EF$
Let's focus on the line $CD$ and extend its line to the left
We will mark the angle we create on that line with the number $1$ and the existing angle which is equal to: $64^o$ we will mark with the number $2$
Now consider that angle $1$ is equal to angle $2$ since they are corresponding angles, therefore, angle $1$ is also equal to: $64^o$
As the lines are parallel to each other, we will mark the angle next to the existing angle equal to: $99^o$ with the number $3$
Keep in mind that angle $3$ and the angle $99^o$ are adjacent angles, which means together they are equal to: $180^o$
Now we can calculate angle $3$
$180-99=81$
Now we have found $2$ angles inside the triangle and we only need to calculate $X$
As we know the sum of the angles in a triangle is $180^o$
We solve the following equation to find $X$
$x=180-81-64$
$x=35$
$35^o$
### Exercise 2
Assignment
Is it possible to draw a quadrilateral that is not a rectangle in such a way that its opposite angles are equal?
True
### Exercise 3
Assignment
From point $C$, two tangents are drawn to the circle $O$
On $AC$, a semicircle is placed whose area is $16\pi$ cm²
On $CD$, a semicircle is placed whose circumference is $8\pi$ cm
$CD>CE$
Which angles in the drawing are equal? (besides the given)
Solution
$EC$ and $BC$ are tangents to the circle
$BC=EC$ since tangents to a circle from the same point are equal
Now we calculate $AC$
$2R=AC$ diameter is the same
$A=16\pi=\pi r^2$
We take the square root
$R=\sqrt{16}=4$
$AC=2R=2\times4=8$
Now we calculate $CD$
$2R=CD$ diameter is the same
$P=8\pi=2\pi r$
We divide by: $2$
$\frac{8}{2}=4=R$
$CD=2R=2\times4=8$
From this we deduce that
$\sphericalangle ABC=\sphericalangle CED$ and $BC=CE$, $CD=AC$ which is greater than $CE$
Therefore $\triangle ABC\cong\triangle DEC$
By side, side, angle
Therefore $\sphericalangle BAC=\sphericalangle EDC$
Corresponding angles between congruent triangles are equal
Angle $CDE$ = Angle $BAC$
Do you think you will be able to solve it?
### Exercise 4
Assignment
Given the angles between parallel lines in the graph, what is the value of: $x$?
Solution
$X=?$
$180^o-105^o=75^o$
$75^o+X=110^o$ $/-75^o$
$X=110^o -75^o$
$35^o$
$35^o$
### Exercise 5
Assignment
Given the angles between parallel lines in a sketch, what is the value of $X$?
Solution
There is a relationship of corresponding angles (corresponding angles) between the two angles, therefore they are equal.
Therefore, you can replace $59^o$ as a result of the equation $X+32=59$
We move $32^o$ to the other side
$X=59^o-32^o=27^o$
$27^o$
## Examples with solutions for Types of Angles
### Exercise #1
Angle A is equal to 30°.
Angle B is equal to 60°.
Angle C is equal to 90°.
Can these angles form a triangle?
### Step-by-Step Solution
We add the three angles to see if they equal 180 degrees:
$30+60+90=180$
The sum of the angles equals 180, so they can form a triangle.
Yes
### Exercise #2
Angle A equals 56°.
Angle B equals 89°.
Angle C equals 17°.
Can these angles make a triangle?
### Step-by-Step Solution
We add the three angles to see if they are equal to 180 degrees:
$56+89+17=162$
The sum of the given angles is not equal to 180, so they cannot form a triangle.
No.
### Exercise #3
Angle A equals 90°.
Angle B equals 115°.
Angle C equals 35°.
Can these angles form a triangle?
### Step-by-Step Solution
We add the three angles to see if they are equal to 180 degrees:
$90+115+35=240$
The sum of the given angles is not equal to 180, so they cannot form a triangle.
No.
### Exercise #4
In a right triangle, the sum of the two non-right angles is...?
### Step-by-Step Solution
In a right-angled triangle, there is one angle that equals 90 degrees, and the other two angles sum up to 180 degrees (sum of angles in a triangle)
Therefore, the sum of the two non-right angles is 90 degrees
$90+90=180$
90 degrees
### Exercise #5
What is the value of the void angle?
### Step-by-Step Solution
The empty angle is an angle adjacent to 160 degrees.
Remember that the sum of adjacent angles is 180 degrees.
Therefore, the value of the empty angle will be:
$180-160=20$
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# GSEB Solutions Class 6 Maths Chapter 6 Integers Ex 6.2
Gujarat BoardĀ GSEB Textbook Solutions Class 6 Maths Chapter 6 Integers Ex 6.2 Textbook Questions and Answers.
## Gujarat Board Textbook Solutions Class 6 Maths Chapter 6 Integers Ex 6.2
Question 1.
Using the number line write the integer which is:
(a) 3 more than 5
(b) 5 more than – 5
(c) 6 less than 2
(d) 3 less than – 2
Solution:
(a) 3 more than 5:
To obtain an integer 3 more than 5, we start from 5 and move 3 units to the right of 5 and reach at 8, as shown in the figure. Thus, 3 more than 5 is 8.
(b) 5 more than – 5:
To obtain 5 more than – 5, we start from – 5 and move 5 unit to its right. We reach at 0, as shown on the number line. Thus, 5 more than – 5 is 0.
(c) 6 less than 2:
To obtain an integer 6 less than 2, we start from 2 and move 6 steps, each of 1 unit, to the left of 2. We reach at – 4, as shown in the figure. Thus, 6 less than 2 is – 4.
(d) 3 less than – 2:
To obtain 3 less than – 2, we start from – 2 and move 3 units to the left of – 2. We reach at – 5 as shown in the figure. Thus, 3 less than – 2 is – 5.
Question 2.
Use number line and add the following integers:
(a) 9 + (- 6)
(b) 5 + (- 11)
(c) (- 1) + (- 7)
(d) (- 5) + 10
(e) (- 1) + (- 2) + (- 3)
(f) (- 2) + 8 + (- 4)
Solution:
(a) 9 + (- 6)
On the number line, first we move 9 units to the right of O and reach at 9. Then, we move 6 steps to the left of 9 and reach at 3, as shown in the figure. Thus, 9 + (- 6) = +3.
(b) 5 + (- 11)
On the number line, first we move 5 units to the right of O to reach at 5. Then, we move 11 units to the left of 5 and reach at (- 6), as shown in the figure. Thus, 5 + (- 11) = – 6.
(c) (- 1) + (- 7)
First we move 1 unit to the left of O to reach at – 1. Then, we move 7 units to the left of – 1 and reach at – 8, as shown in the figure. Thus, (- 1) + (- 7) = – 8.
(d) (- 5) + 10
First we move 5 units to the left of 0 to reach at – 5. Then, we move 10 units to the right of – 5 and reach at (+ 5), as shown in the figure, Thus, (- 5) + (10) = + 5.
(e) (- 1) + (- 2) + (- 3)
First we move 1 unit to the left of 0 and reach at (- 1). Then, we move 2 units to the left of (- 1) and reach at (- 3). Again we move 3 units to the left of – 3 and reach at – 6, as shown in the figure. Thus, (- 1) + (- 2) + (- 3) = – 6.
(f) (- 2) + 8 + (- 4)
First we move 2 units to the left of O to reach at (- 2). Then, we move 8 units to the right of (- 2) and reach at 6. Again from 6, we move 4 units to the left of 6 and reach at 2, as shown in the figure. Thus, (- 2) + 8 + (- 4) = 2.
Question 3.
Add without using number line:
(a) 11 + ( – 7)
(b) (- 13) + ( + 18)
(c) ( – 10) + (+ 19)
(d) ( – 250) + (+ 150)
(e) ( – 380) + ( – 270)
(f) ( – 217) + ( – 100)
Solution:
(a) 11 + (- 7):
Since,11 = 7 + 4
11 +( – 7) = 7 + 4 + ( – 7) = 7 + ( – 7) + 4
= 0 + 4 [ (- 7) + (7) = 0]
= + 4
Thus, 11 + (- 7) = 4
(b) ( – 13) + (+ 18):
Since, (+18) = (+13) + (+5)
( – 13) + (+18) = ( – 13) + (+ 13) + (+ 5)
= 0 + (+ 5) [( – 13) + (+ 13) = 0]
= (+5)
Thus, (- 13) + (+ 18) = 5
(c) (- 10) + (+ 19):
Since, (+ 19) = (+ 10) + (+ 9)
( – 10) + (+ 19) = (- 10) + (+ 10) + (+ 9)
= 0 + (+ 9)
[ (- 10) + (+ 10) = 0]
= + 9
Thus, ( – 10) + (+ 19) = 9
(d) ( – 250) + (+ 150):
Since, (- 250 (- 150) + (- 100)
(- 250) + (+ 150) = (- 150) + (- 100) + (+ 150)
= (- 150) + (+ 150) + (- 100) = 0 + (- 100)
[ (- 150) + ( + 150) = 0]
= – 100
Thus, ( – 250) + (+ 150) = – 100 .
(e) ( – 380) + ( – 270):
Since, the given integers are of like sign,
( – 380) + ( – 270) = – [380 + 27g] = – 650
Thus, (- 380) + (- 270) = – 650
(f) (- 217) + (- 100):
Since, the given integers are of like sign,
( – 217) + ( – 100) = – (217 + 100) = – 317
Thus, (- 217) + (- 100) = – 317
Question 4.
Find the sum of:
(a) 137 and -354
(b) – 52 and 52
(c) – 312, 39 and 192
(d) – 50, – 200 and 300
Solution:
(a) 137 + (- 354):
Since, ( – 354) = (- 137) + (- 217)
137 + ( – 354) = 137 + ( – 137) + ( – 217)
= 0 + ( – 217)
[137 + ( – 137) = 0]
= – 217
Thus, 137 + ( – 354) = – 217
(b) ( – 52) + 52:
Since, the given integers are opposite of each other, so their sum must be equal to 0.
( – 52) + 52 = 0
(c) ( – 312) + 39 + 192:
Since, 192 + 39 = 231
and (- 312) = (- 231) + ( – 81)
( – 312) + 39 + 192 = ( – 312) + 231
= ( – 231) + ( – 81) + 231
= ( – 231) + (231) + ( – 81) = 0 + ( – 81) = – 81
Thus, ( – 312) + 39 + 192 = – 81
(d) ( – 50) + (- 200) + 300:
Since, ( – 50) + ( – 200) ( – 250)
and 300 = 250 + 50
( – 50) + ( – 200) + 300 = ( – 250) + 300
= ( – 250) + 250 + 50 = 0 + 50
[( – 250) + 250 = 0]
= 50
Thus, ( – 50) + ( – 200) + 300 = 50
Question 5.
Find the sum:
(a) ( – 7) + ( – 9) + 4 + 16
(b) (37) + ( – 2) + ( – 65) + ( – 8)
Solution:
(a) ( – 7) + ( – 9) + 4 + 16:
Since, ( – 7) + ( – 9) = ( – 16)
( – 7) + ( – 9) + 4 + 16 = (- 16) + 4 + 16
= (- 16) + 16 + 4 = 0 + 4 [ ( – 16) + 16 = 0]
= 4
Thus, ( – 7) + ( – 9) + 4 + 16 = 4
(b) (37) + ( – 2) + ( – 65) + ( – 8):
Since, ( – 2) + ( – 65) + ( – 8) ( – 75)
and (- 75) = ( – 37) + ( – 38)
37 + ( – 2) + ( – 65) + ( – 8) 37 + ( – 75)
= 37 + ( – 37) + ( – 38) = 0 + ( – 38)
[37 + ( – 37) = 0]
= – 38
Thus, 37 + ( – 2) + ( – 65) + ( – 8) = – 38
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Graphing Points
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When graphing points, it's important to have an understanding of the x-coordinate and the y-coordinate. Discover how to move along the horizontal and vertical axes with help from a math teacher in this free video on math help and graphing.
Part of the Video Series: Algebra & Math Help
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Video Transcript
So how does one graph points? Hi I'm Jimmy Chang. I've been teaching college mathematics for over nine years and when it comes to graphing points there is a couple of things that you want to consider as it relates to your coordinates but once you get past that graphing points can be actually straightforward so here we go, let's review. First of all make sure that you know what the two numbers are in front of you. You have an x coordinate and you have a y coordinate. The x coordinate tells you whether to go right or left. It's the horizontal coordinate. The y is your vertical. It tells you whether to go up or down and as you might imagine there is an x axis and there is a y axis and so with positive numbers for x it tells you to go right, negative numbers go left and with y's positives tell you to go up and negatives tell you to go down. So here are some illustrations as to some points and how you graph them. Suppose you have 3, 2. Now the 3 represents your x so that means you go right 3. The 2, being the y coordinate because it is also positive tells you to go up so to graph 3, 2 you always start from the center when you graph points. So from the center you go right 3 and you go up 2 and then you label a point. Let's call that 3, 2. Now let's suppose you want to graph let's just say negative 4, 1. Now because the x coordinate is negative that means you have to go left 4. Now the 1 because it is still positive you go up 1. So that means again starting from the center, 1, 2, 3, 4 to the left and then up 1, then once you get to the y coordinate then you can label and you have negative 4, 1. Now one more case suppose you want to plot the point, 0, 3. Now any time you have a 0 that tells you to stay put, you go on the other axis. So for example with 0 on the x coordinate that means you do not move on the x axis. You stay where you are. You don't go anywhere horizontally. So the 3 because it is positive, you go up so that means from the center you simply go up three because there is no horizontal movement. So you go 1, 2, and 3 and just go ahead and just label that 0, 3. So I'm Jimmy Chang and those are some examples as to how you graph points.
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## Section16.14Polar Coordinates
##### Polar Coordinates.
When graphing in two-dimensions, the most common way we identify points in the plane is with the use of rectangular coordinates ā i.e. by the ordered pair $(x,y)$ (or whatever variables are in use). In this section we talk about an alternate strategy for identifying points in the plane ā with the use of polar coordinates.
In the polar plane, the point we call the origin in the rectangular plane is called the pole. A point in the plane, $P\text{,}$ identified using polar coordinates has the form $(r,\theta)$ where $r$ is the signed distance from the pole to $P$ and $\theta$ is an angle that when drawn in standard position has a terminal side that passes either through the pole and $P$ or through the pole and the direction opposite of $P\text{.}$
The polar plane is shown in FigureĀ 16.14.1. One polar ordered pair that plots to point labeled $A$ is $\left(4,\frac{\pi}{6}\right)\text{.}$ One polar ordered pair that plots to point labeled $B$ is $\left(3,\frac{17\pi}{12}\right)\text{.}$ Unlike the rectangular plane, however, neither of these ordered pairs are unique. In fact, for each point there are an unlimited number of ordered pairs that plot to that single point.
When drawn in standard position, any two angles whose difference is an integer multiple of $2\pi$ are coterminal. One repercussion of this is that each of the following polar ordered pairs will plot to $A\text{.}$
\begin{equation*} ...,\,\left(4,-\frac{11\pi}{6}\right),\,\left(4,\frac{\pi}{6}\right),\,\left(4,\frac{13\pi}{6}\right),\,... \end{equation*}
But these aren't the only polar ordered pairs that plot to $A\text{.}$ When the value of $r$ is negative, we plot the point using the fact that $(r,\theta)$ and $(-r,\theta+\pi)$ plot to the same point.
So, for example suppose that we wanted to plot the point $\left(-4,\frac{7\pi}{6}\right)\text{.}$ This will plot in the same location as the point derived below.
\begin{equation*} \left(4,\frac{7\pi}{6}+\pi\right)=\left(4,\frac{13\pi}{6}\right) \end{equation*}
Because $\left(4,\frac{13\pi}{6}\right)$ has already been identified as an ordered pair that plots to $A\text{,}$ it must be the case that $\left(-4,\frac{7\pi}{6}\right)$ also plots to $A\text{.}$
One way that you can think about negative values of $r$ is to image the positive $x$-axis being rotated to the $\theta$-coordinate of the polar ordered pair. This is illustrated in FigureĀ 16.14.2 for a $\theta$ value of $\frac{7\pi}{6}\text{.}$ The positive values of $r$ emanate from the pole in the direction of $\frac{7\pi}{6}$ and the negative values of $r$ emanate from the pole in the opposite direction of $\frac{7\pi}{6}\text{,}$ i.e. in the direction of $\frac{7\pi}{6}+\pi\text{.}$
###### Example16.14.3.
Plot each of the following polar ordered pairs.
\begin{equation*} A:\,\,\left(3,\frac{11\pi}{12}\right) \end{equation*}
\begin{equation*} B:\,\,\left(2,-\frac{3\pi}{4}\right) \end{equation*}
\begin{equation*} C:\,\,(-3,\pi) \end{equation*}
Solution
Note that $-\frac{3\pi}{4}$ is coterminal with $\frac{5\pi}{4}\text{.}$ Note also that $(-3,\pi)$ plots to the same point as $(3,2\pi)$ which in turn plots to the same point as $(3,0)\text{.}$
##### Converting Between Polar Coordinates and Rectangular Coordinates..
There are several mathematical relationships between the labeled parts of the right triangle shown in FigureĀ 16.14.5. Three notable relationships come from trigonometry and are stated below.
\begin{equation*} \sin(\theta)=\frac{y}{r} \end{equation*}
\begin{equation*} \cos(\theta)=\frac{x}{r} \end{equation*}
\begin{equation*} \tan(\theta)=\frac{y}{x} \end{equation*}
A fourth notable relationship comes from the Pythagorean theorem.
\begin{equation*} r^2=x^2+y^2 \end{equation*}
We can use these relationships to convert between polar coordinates and rectangular coordinates.
When converting from polar coordinates to rectangular coordinates we use the following two formulas.
\begin{equation*} x=r\cos(\theta) \end{equation*}
\begin{equation*} y=r\sin(\theta) \end{equation*}
When converting from rectangular coordinates to polar coordinates we use the following two formulas.
\begin{equation*} r=\sqrt{x^2+y^2} \end{equation*}
\begin{equation*} \tan(\theta)=\frac{y}{x} \end{equation*}
When applying the formula $\tan(\theta)=\frac{y}{x}\text{,}$ we need to use caution in the use of the inverse tangent key. That key will return a correct value for $\theta$ if and only $\theta$ terminates in either Quadrant I or Quadrant IV. If $\theta$ terminates in Quadrant II or Quadrant III and you need to use your inverse tangent key, the most straight forward thing to do is to find the reference angle, $\theta^{\,\prime}\text{,}$ in the following way.
\begin{equation*} \theta^{\,\prime}=\tan^{-1}\left(\abs{\frac{y}{x}}\right) \end{equation*}
Then subtract or add $\theta^{\,\prime}$ to $\pi$ dependent, respectively, upon whether the point lies in Quadrant II or Quadrant III.
###### Example16.14.6.
Convert each of the following ordered pairs from rectangular coordinates to polar coordinates.
1. $\left(12,\frac{5\pi}{3}\right)$
2. $\left(-20,-\frac{5\pi}{4}\right)$
Solution
The formulas for converting from polar coordinates to rectangular coordinates can be applied in a passive manner ā that is, we don't need to worry about reaching a false conclusion by careless application of the inverse tangent function. Let's proceed.
1. \begin{align*} \left(12\cos\left(\frac{5\pi}{3}\right),12\sin\left(\frac{5\pi}{3}\right)\right)\amp=\left(12 \cdot \frac{1}{2},12 \cdot -\frac{\sqrt{3}}{2}\right)\\ \amp=(6,-6\sqrt{3}) \end{align*}
2. \begin{align*} \left(-20\cos\left(-\frac{5\pi}{4}\right),-20\sin\left(-\frac{5\pi}{4}\right)\right)\amp=\left(-20 \cdot -\frac{\sqrt{2}}{2},-20 \cdot \frac{\sqrt{2}}{2}\right)\\ \amp=(10\sqrt{2},-10\sqrt{2}) \end{align*}
###### Example16.14.7.
Convert each of the following ordered pairs from rectangular coordinates to polar coordinates. In all cases state a positive value for $r$ and choose a value for $\theta$ that falls on the interval $[0,2\pi)\text{.}$ For parts (a) and (b) state exact values for both $r$ and $\theta\text{.}$ For parts (c) and (d), state an exact value for $r$ but round the value of $\theta$ to the nearest thousandth.
1. $(4,4\sqrt{3})$
2. $(-7\sqrt{2},-7\sqrt{2})$
3. $(3,-4)$
4. $(-5,12)$
Solution
1. Let's begin by determining the value of $r\text{.}$
\begin{align*} r\amp=\sqrt{x^2+y^2}\\ \amp=\sqrt{4^2+(4\sqrt{3})^2}\\ \amp=\sqrt{64}\\ \amp=8 \end{align*}
To determine the value of $\theta\text{,}$ we need to find the solution to $\tan(\theta)=\frac{y}{x}$ that lies in Quadrant I and falls between $0$ and $2\pi\text{.}$
\begin{align*} \tan(\theta)=\frac{y}{x}\,\,\amp\Longrightarrow\,\,\tan(\theta)=\frac{4\sqrt{3}}{4}\\ \amp\Longrightarrow\,\,\tan(\theta)=\sqrt{3} \end{align*}
The solution to the last equation that satisfies the stated requirements is $\frac{\pi}{3}\text{.}$
In conclusion, a polar equivalent of the rectangular ordered pair $(4,4\sqrt{3})$ is
\begin{equation*} \left(8,\frac{\pi}{3}\right). \end{equation*}
2. We begin by determining the value of $r\text{.}$
\begin{align*} r\amp=\sqrt{x^2+y^2}\\ \amp=\sqrt{\left(-7\sqrt{2}\right)^2+\left(-7\sqrt{2}\right)^2}\\ \amp=\sqrt{196}\\ \amp=14 \end{align*}
To determine the value of $\theta\text{,}$ we need to find the solution to the equation $\tan(\theta)=\frac{y}{x}$ that lies in Quadrant III and falls between $0$ and $2\pi\text{.}$
\begin{align*} \tan(\theta)=\frac{y}{x}\amp\,\,\Longrightarrow\,\,\tan(\theta)=\frac{-7\sqrt{2}}{-7\sqrt{2}}\\ \amp\Longrightarrow\,\,\tan(\theta)=1 \end{align*}
The solution to the last equation that meets the specified conditions is $\frac{5\pi}{4}\text{.}$
In conclusion, a polar equivalent of the rectangular ordered pair $(-7\sqrt{2},-7\sqrt{2})$ is
\begin{equation*} \left(14,\frac{5\pi}{4}\right). \end{equation*}
3. A good place to start is to determine the value of $r\text{.}$
\begin{align*} r\amp=\sqrt{x^2+y^2}\\ \amp=\sqrt{3^2+(-4)^2}\\ \amp=\sqrt{25}\\ \amp=5 \end{align*}
Let's move on to $\theta\text{.}$ We need to determine the solution to $\tan(\theta)=\frac{y}{x}$ that lies in Quadrant IV and falls on the interval $[0,2\pi)\text{.}$
\begin{align*} \tan(\theta)=\frac{y}{x}\,\,\amp\Longrightarrow\,\,\tan(\theta)=\frac{-4}{3}\\ \amp\Longrightarrow\,\,\tan(\theta)=-\frac{4}{3} \end{align*}
This last equation is not one for which we know an exact solution. However, because the range of the inverse tangent function is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ and the point we're analyzing lies in Quadrant IV, we can use $\tan^{-1}\left(-\frac{4}{3}\right)$ to directly determine an angle that could be used in the polar ordered pair.
\begin{equation*} \tan^{-1}\left(-\frac{4}{3}\right)\approx -0.927 \end{equation*}
A mentioned, in general this would be a fine value for $\theta\text{,}$ but the problem specified that we need to use a value that falls between $0$ and $2\pi\text{.}$ We can find the specified value by adding $2\pi$ to the value the calculator returned and do so below.
\begin{gather*} \theta\approx -.927+2\pi\\ \approx 5.356 \end{gather*}
To sum up, one polar equivalent to the rectangular ordered pair $(3,-4)$ is approximately
\begin{equation*} (5,5.536). \end{equation*}
4. Once again, let's begin by determining the value of $r\text{.}$
\begin{align*} r\amp=\sqrt{x^2+y^2}\\ \amp=\sqrt{(-5)^2+12^2}\\ \amp=\sqrt{169}\\ \amp=13 \end{align*}
Determination of $\theta$ requires us to figure out the solution to $\tan(\theta)=\frac{y}{x}$ that lies in Quadrant II and falls between $0$ and $2\pi\text{.}$
\begin{align*} \tan(\theta)=\frac{y}{x}\,\,\amp\Longrightarrow\,\,\tan(\theta)=\frac{12}{-5}\\ \amp\Longrightarrow\,\,\tan(\theta)=-\frac{12}{5} \end{align*}
We do not know an exact solution for the last equation, so we will need to use technology to find an approximate solution. Unfortunately, $\theta$ needs to point into Quadrant II, and technology will give us a solution in Quadrant IV. Let's use $\tan^{-1}\left(\frac{12}{5}\right)$ to determine the reference angle for $\theta$ and proceed from there.
\begin{align*} \theta^{\,\prime}\amp=\tan^{-1}\left(\frac{12}{5}\right)\\ \amp\approx 1.176 \end{align*}
Because $\theta$ lies in Quadrant II, we need to subtract $\theta^{\,\prime}$ from $\pi$ to determine its (approximate) value.
\begin{equation*} \pi-1.176\approx 1.966 \end{equation*}
We conclude by stating that an approximate polar equivalent of the rectangular ordered pair $(-5,12)$ is
\begin{equation*} (13,1.966). \end{equation*}
##### Graphing Polar Functions.
Polar functions are generally stated in the format $r=f(\theta)\text{.}$ Because we state polar ordered pairs using the format $(r,\theta)\text{,}$ it follows that unlike rectangular ordered pairs, in polar ordered pairs the second coordinate is generally the independent variable and the first coordinate is the dependent variable.
Let's consider the polar function $r=4\cos(\theta)\text{.}$ If we want to plot a graph of this function, a good place to start would be with a table of values. The table I'm going to create is going to contain values of $\theta$ that increase from $0$ up through $\frac{11\pi}{6}\text{.}$ The points need to be connected in a counterclockwise fashion in the order in which they appeared. This can sometimes get a little confusing when the value of $r$ turns negative. For this reason I'm going to assign names to the points to help us recall the order. This is done in FigureĀ 16.14.8. I went ahead and used my calculator to get approximate values for $r$ in order to facilitate point plotting.
Let's go ahead and plot the points in FigureĀ 16.14.9.
If we follow the points in order, it appears that we are going around a circle in a counterclockwise fashion. Let's go ahead and connect the dots to see if they really do form a circle. This is done in FigureĀ 16.14.10.
So the points do indeed connect into a circle. Let's convert the equation to rectangular form to see if we can figure out why. We'll begin our exploration by first multiplying both sides of the equation $r=4\cos(\theta)$ by $r\text{.}$ The reason for doing so will be easier to explain once we've done it, so let's get to it.
\begin{equation*} r \cdot r=r \cdot 4\cos(\theta)\,\,\Longrightarrow\,\,r^2=4r\cos(\theta) \end{equation*}
We can now replace $r^2$ with $x^2+y^2$ and $r\cos(\theta)$ with $x\text{,}$ resulting in the following.
\begin{equation*} x^2+y^2=4x \end{equation*}
Let's move the linear term to the left side of the equation and complete the square and then pause to see if the mystery has been solved.
This last equation is indeed the equation of a circle with a radius of $2$ centered at the rectangular point $(2,0)\text{.}$ So there you have it.
###### Example16.14.11.
Graph the polar function $r=2-3\sin(\theta)$ after first making an appropriate table of values.
Solution
Several ordered pairs that satisfy the equation are shown in FigureĀ 16.14.12.
The ordered pairs from FigureĀ 16.14.12 are graphed in FigureĀ 16.14.13. Note that the value of $r$ changes from positive to negative somewhere between $\frac{\pi}{6}$ and $\frac{\pi}{4}$ and then changes back to positive somewhere between $\frac{3\pi}{4}$ and $\frac{5\pi}{6}\text{.}$ Whenever the value of $r$ changes sign on a continuous polar curve the curve passes through the pole.
This is similar to the continuous rectangular function $y=f(x)$ crossing the $x$-axis when the value of $y$ changes sign. The completed graph of $r=2-3\sin(\theta)$ is shown in FigureĀ 16.14.14. This curve is an example of a limacon with an inner loop. If you're interested in other generic types of polar curves, just type "types of polar curves" into the search bar and follow a resultant link.
###### Example16.14.15.
Convert each polar equation to a rectangular equation to determine the appearance of the graph of the polar equation.
1. $r=7$
2. $r=3\sec(\theta)$
3. $\theta=\frac{\pi}{3}$
Solution
1. We begin by squaring both sides of the equation resulting in
\begin{equation*} r^2=49. \end{equation*}
Replacing $r^2$ with $x^2+y^2$ give us
\begin{equation*} x^2+y^2=49. \end{equation*}
This rectangular equation graphs to a circle centered at the origin with a radius of $7\text{.}$ So $r=7$ is a circle centered at the pole with a radius of $7$ which makes sense as any point with an $r$-coordinate of $7$ lies on that circle.
2. \begin{align*} r=3\sec(\theta)\,\,\amp\Longrightarrow\,\,r=\frac{3}{\cos(\theta)}\\ \amp\Longrightarrow\,\,r\cos(\theta)=3\\ \amp\Longrightarrow\,\,x=3 \end{align*}
From this last equation we can conclude that $r=3\sec(\theta)$ graphs to a vertical line that passes through the polar point $(3,0)\text{.}$
3. \begin{align*} \theta=\frac{\pi}{3}\,\,\amp\Longrightarrow\,\,\tan(\theta)=\tan\left(\frac{\pi}{3}\right)\\ \amp\Longrightarrow\,\,\frac{y}{x}=\sqrt{3}\\ \amp\Longrightarrow\,\,y=\sqrt{3}x \end{align*}
So $\theta=\frac{\pi}{3}$ graphs to a line through the pole that has a rectangular slope of $\sqrt{3}\text{.}$ This makes sense as any point with a $\theta$-coordinate of $\frac{\pi}{3}$ lies on that line.
### ExercisesExercises
###### 1.
Match each function with one of the graphs shown in FigureĀ 16.14.16-FigureĀ 16.14.19.
1. $r=1+3\sin(3\theta)$
2. $r=4\cos(3\theta)$
3. $r=4\sin(4\theta)$
4. $r=2+2\cos(2\theta)$
###### 2.
Determine the appearance of each polar function after first converting the equation into rectangular form.
1. $r=-8\sin(\theta)$
2. $r=5\csc(\theta)$
3. $r=\frac{1}{1+\sin(\theta)}$
4. $\theta=\frac{3\pi}{4}$
Solution
1. \begin{align*} r=8\sin(\theta)\,\,\amp\Longrightarrow\,\,r \cdot r=r \cdot -8\sin(\theta)\\ \amp\Longrightarrow\,\,r^2=-8r\sin(\theta)\\ \amp\Longrightarrow\,\,x^2+y^2=-8y\\ \amp\Longrightarrow\,\,x^2+y^2+8y=0\\ \amp\Longrightarrow\,\,x^2+y^2+8y\addright{16}=0\addright{16}\\ \amp\Longrightarrow\,\,(x+4)^2=16 \end{align*}
$r=-8\sin(\theta)$ graphs to a circle of radius $4$ centered at the polar point $(4,\pi)\text{.}$
2. \begin{align*} r=5\csc(\theta)\,\,\amp\Longrightarrow\,\,r=\frac{5}{\sin(\theta)}\\ \amp\Longrightarrow\,\,r\sin(\theta)=5\\ \amp\Longrightarrow\,\,y=5 \end{align*}
$r=5\csc(\theta)$ graphs to a horizontal line through the polar point $\left(5,\frac{\pi}{2}\right)\text{.}$
3. \begin{align*} r=\frac{1}{1+\sin(\theta)}\,\,\amp\Longrightarrow\,\,r+r\sin(\theta)=1\\ \amp\Longrightarrow\,\,r=1-r\sin(\theta)\\ \amp\Longrightarrow\,\,r^2=\left(1-r\sin(\theta)\right)^2\\ \amp\Longrightarrow\,\,x^2+y^2=1-2r\sin(\theta)+\left(r\sin(\theta)\right)^2\\ \amp\Longrightarrow\,\,x^2+y^2=1-2y+y^2\\ \amp\Longrightarrow\,\,2y=1-x^2\\ \amp\Longrightarrow\,\,y=-\frac{1}{2}x^2+\frac{1}{2} \end{align*}
$r=\frac{1}{1+sin(\theta)}$ graphs to a downward opening parabola with its vertex at the polar point $\left(\frac{1}{2},\frac{\pi}{2}\right)\text{.}$
4. \begin{align*} \theta=\frac{3\pi}{4}\amp\Longrightarrow\,\,\tan(\theta)=\tan\left(\frac{3\pi}{4}\right)\\ \amp\Longrightarrow\,\,\frac{y}{x}=-1\\ \amp\Longrightarrow\,\,y=-x \end{align*}
$\theta=\frac{3\pi}{4}$ graphs to a line through the origin that bisects Quadrants I and IV.
###### 3.
Convert each of the following ordered pairs from rectangular coordinates to polar coordinates. In all cases state a positive value for $r$ and choose a value for $\theta$ that falls on the interval $[0,2\pi)\text{.}$ For parts (a) and (b) state exact values for both $r$ and $\theta\text{.}$ For parts (c) and (d), state an exact value for $r$ but round the value of $\theta$ to the nearest thousandth.
1. $\left(3\sqrt{3},9\right)$
2. $\left(-\frac{4}{\sqrt{2}},\frac{4}{\sqrt{2}}\right)$
3. $(-8,-15)$
4. $(-5,2)$
Solution
1. Let's first find the value of $r\text{.}$
\begin{align*} r\amp=\sqrt{x^2+y^2}\\ \amp=\sqrt{\left(3\sqrt{3}\right)^2+9^2}\\ \amp=\sqrt{108}\\ \amp=6\sqrt{3} \end{align*}
To determine the value of $\theta\text{,}$ we need to find a solution to the equation $\tan(\theta)=\frac{y}{x}$ that lies in Quadrant I and falls on the interval $[0,2\pi)\text{.}$
\begin{align*} \tan(\theta)\amp=\frac{y}{x}\\ \tan(\theta)\amp=\frac{9}{3\sqrt{3}}\\ \tan(\theta)\amp=\frac{3}{\sqrt{3}}\\ \tan(\theta)\amp=\frac{3}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}\\ \tan(\theta)\amp=\frac{3\sqrt{3}}{3}\\ \tan(\theta)\amp=\sqrt{3} \end{align*}
The solution to the last equation that satisfies the stated conditions is $\frac{\pi}{3}\text{.}$ In summary, one polar equivalent to the rectangular ordered pair $\left(3\sqrt{3},9\right)$ is $\left(6\sqrt{3},\frac{\pi}{3}\right)\text{.}$
2. We begin by determining the value of $r\text{.}$
\begin{align*} r\amp=\sqrt{x^2+y^2}\\ \amp=\sqrt{\left(-\frac{4}{\sqrt{2}}\right)^2+\left(\frac{4}{\sqrt{2}}\right)^2}\\ \amp=\sqrt{16}\\ \amp=4 \end{align*}
To determine $\theta\text{,}$ we need to find the solution to the equation $\tan(\theta)=\frac{y}{x}$ that lies in Quadrant II and is between $0$ and $2\pi\text{.}$
\begin{align*} \tan(\theta)\amp=\frac{y}{x}\\ \tan(\theta)\amp=\frac{-\frac{4}{\sqrt{2}}}{\frac{4}{\sqrt{2}}}\\ \tan(\theta)\amp=-1 \end{align*}
The solution to the last equation that meets the stated requirements is $\frac{3\pi}{4}\text{.}$ In conclusion, a polar equivalent to the rectangular ordered pair $\left(-\frac{4}{\sqrt{2}},\frac{4}{\sqrt{2}}\right)$ is $\left(4,\frac{3\pi}{4}\right)\text{.}$
3. To start things off we'll determine the value of $r\text{.}$
\begin{align*} r\amp=\sqrt{x^2+y^2}\\ \amp=\sqrt{(-8)^2+(-15)^2}\\ \amp=\sqrt{289}\\ \amp=17 \end{align*}
To estimate the value of $\theta\text{,}$ we need to determine solution to the equation $\tan(\theta)=\frac{y}{x}$ that lies in Quadrant III and falls between $0$ and $2\pi\text{.}$
\begin{align*} \tan(\theta)\amp=\frac{y}{x}\\ \tan(\theta)\amp=\frac{-15}{-8}\\ \tan(\theta)\amp=\frac{15}{8} \end{align*}
The last equation is not one for which we know an exact solution, so we'll need to rely on technology to determine an estimated solution. We can be certain that
\begin{equation*} \theta\neq\tan^{-1}\left(\frac{15}{8}\right) \end{equation*}
because that angle terminates in Quadrant I and we need an angle that terminates in Quadrant III. However, $\tan^{-1}\left(\frac{15}{8}\right)$ is the reference angle for $\theta$ which gives us the following.
\begin{align*} \theta\amp=\pi+\tan\left(\frac{15}{8}\right)\\ \amp\approx 4.222 \end{align*}
In conclusion, an approximate polar equivalent to the rectangular ordered pair $(-8,-15)$ is $(17,4.222)\text{.}$
4. We begin by determining the value of $r\text{.}$
\begin{align*} r\amp=\sqrt{x^2+y^2}\\ \amp=\sqrt{(-5)^2+2^2}\\ \amp=\sqrt{29} \end{align*}
We need to also determine a solution to the equation $\tan(\theta)=\frac{y}{x}$ that terminates in Quadrant II and falls on the interval $[0,2\pi)\text{.}$
\begin{align*} \tan(\theta)\amp=\frac{y}{x}\\ \tan(\theta)\amp=\frac{2}{-5}\\ \tan(\theta)\amp=-\frac{2}{5} \end{align*}
We do not know an exact solution to the last equation, so we'll need to use technology to estimate the solution. We can' find the value directly with the inverse tangent function, because $\tan^{-1}\left(-\frac{2}{5}\right)$ points into Quadrant IV and we're looking for an angle that points into Quadrant II. Let's find the reference angle for the solution and then subtract that from $\pi\text{.}$
\begin{align*} \theta^{\,\prime}\amp=\tan^{-1}\left(\abs{\frac{2}{5}}\right)\\ \amp\approx 0.381\\ \theta\amp=\pi-\theta^{\,\prime}\\ \amp\approx \pi-0.381\\ \amp\approx 2.761 \end{align*}
So we conclude by stating that a polar equivalent to the rectangular ordered pair $(-5,2)$ is approximately $\left(\sqrt{29},2.761\right)\text{.}$
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# Any questions on the Section 5. 8 homework
## Presentation on theme: "Any questions on the Section 5. 8 homework"— Presentation transcript:
Any questions on the Section 5. 8 homework
Any questions on the Section 5.8 homework? Pass your worksheets for this assignment to the middle aisle for pickup now.
Remember the problem like this one from the homework that was due today?
Wouldn’t it be nice if there was an easier way to do it than by factoring? Leave factoring up on board: (4x - 9)(3x + 8)
and turn off and put away your cell phones.
Please CLOSE YOUR LAPTOPS and turn off and put away your cell phones. Sample Problems Page Link (Dr. Bruce Johnston)
Section 8.2 The Quadratic Formula
The Quadratic Formula The quadratic formula is another technique we can use for solving quadratic equations. Remember, quadratic equations are polynomial equations of degree 2, such as x2 + 3x -7 = 0 or 5x2 – 14 = 0.
The quadratic formula is derived from a process called “completing the square” for a general quadratic equation. See Section 8.1 if you’re interested in seeing how this formula is derived. This will also be covered in Math 120 in more detail, along with the technique called “completing the square”.
The Quadratic Formula:
The solutions to the equation ax2 + bx + c = 0 are given by the formula Note: This formula IS on the pink formula sheet, but you’ll probably have it memorized by the time you’ve done the first few homework problems.
factoring and when we should use the quadratic formula?
The Big Question: How can we tell when we should use factoring and when we should use the quadratic formula?
Example 1 Solve x2 + 4x + 3 = 0 by Factoring The quadratic formula. Which way works best?
Solve x2 + 4x + 3 = 0 by Factoring:
This one is pretty easy to factor. The factoring is (x + 3)(x + 1) = 0, so the solutions are given by x + 3 = 0, or x = -3, and x + 1 = 0, which gives x = -1.
Now, solve x2 + 4x + 3 = 0 by the quadratic formula:
a = 1, b = 4, c = 3, so the formula gives:
Which way works best for this problem?
In this case, the factoring method is much quicker, although BOTH methods give the same answer.
Example 2 Solve x2 + 5x + 12 = 0 by Factoring The quadratic formula. Which way works best?
Solve x2 + 5x + 12 = 0 by Factoring:
This one looks pretty easy to factor, but when you start trying to find two factors of 12 that add up to 5, nothing works. (1+12=13, 2+6=8, 3+4=7). What does this mean? It means that the polynomial is PRIME, and there are no rational solutions. (Remember, a rational number is either an integer or a fraction.)
Solve x2 + 5x + 12 = 0 (continued):
Let’s see what the quadratic formula gives in this case: a = 1, b = 5, c = 12 so the formula gives: Notice that the number under the radical sign is negative, which means there are no real answers. If the number under the square root sign comes out to be positive but it’s not a perfect square, this means the answer is a real number, but is irrational because it can’t be simplified to remove the radical. In either of these cases, we’d say the polynomial is prime, and therefore has no rational roots.
So which way works best for solving x2 + 5x + 12 = 0?
Either way works fine, but if you think a polynomial is prime, a good way to check is by calculating the discriminant (b2 – 4ac). If the discriminant is either negative or not a perfect square, then you know for sure that your polynomial is prime and there are no rational solutions.
Now re-do this problem from the 5
Now re-do this problem from the 5.8 homework using the quadratic formula: Answers: -8/3, 9/4
Which way works best in this case?
Either way works, but the quadratic formula approach is probably going to be faster than factoring for most people. Moral of the story: For a quadratic equation with a leading coefficient other than 1, it’s probably going to be quicker to solve it using the quadratic formula than it would be to factor the polynomial.
Question: What if some coefficients in your quadratic equation are fractions?
ANSWER: Clear them first by multiplying all terms by the LCD: Solve x2 + x – = 0 by the quadratic formula. x2 + 8x – 20 = (multiply both sides by 8) a = 1, b = 8, c = -20
The expression under the radical sign in the quadratic formula (b2 – 4ac) is called the discriminant. The discriminant will take on a value that is positive, 0, or negative. The value of the discriminant indicates two distinct real solutions (if it’s positive), one real solution (if it’s zero), or two complex, but not real solutions (if it’s negative – a topic to be discussed in Math 120).
The Discriminant and the Kinds of Solutions to ax2 + bx +c = 0
No x-intercepts No real solution; two complex imaginary solutions b2 – 4ac < 0 One x-intercept One real solution (a repeated solution) (If b2 – 4ac is a perfect square, the solution will be a rational number. If not, it’s irrational.) b2 – 4ac = 0 Two x-intercepts Two unequal real solutions (If b2 – 4ac is a perfect square, the two solutions will be rational numbers. If not, they’re both irrational.) b2 – 4ac > 0 Graph of y = ax2 + bx + c Kinds of solutions to ax2 + bx + c = 0 Discriminant b2 – 4ac
Example Use the discriminant to determine the number and type of solutions for the following equation. 5 – 4x + 12x2 = 0 a = 12, b = -4, and c = 5 b2 – 4ac = (-4)2 – 4(12)(5) = 16 – 240 = -224 Since the discriminant is negative, there are no real solutions. Question: What would this graph look like?
Example Use the discriminant to determine the number and type of solutions for the following equation. 25x = 0 a = 25, b = 0 (why?) , and c = -4 b2 – 4ac = (0)2 – 4(25)(-4) = 0 – = 400 Since the discriminant is positive, there are two real solutions. (You could go on to show that the solutions are 2/5 and -2/5, either by factoring or using the quadratic formula.)
Example Use the discriminant to determine the number and type of solutions for the following equation. 5 – 4x + 12x2 = 0 a = 12, b = -4, and c = 5 b2 – 4ac = (-4)2 – 4(12)(5) = 16 – 24 = -224 Since the discriminant is negative, there are no real solutions.
Example Use the discriminant to determine the number and type of solutions for the following equation. x2 – 8x + 16 = 0 a = 1, b = -8, and c = 16 b2 – 4ac = (-8)2 – 4(1)(16) = 64 – 64 = 0 Since the discriminant is zero, there is one real solution. (You could go on to show that the solution is 4, either by factoring or using the quadratic formula.) Question: What would this graph look like?
How do you figure out the answers if the discriminant is positive but not a perfect square?
2 possible approaches: Exact answer: The exact answer will contain a radical, i.e. it will be an irrational number. (More on this in Chapter 7...) Approximate answer: Use your calculator to get an approximate decimal answer.
REMINDER!!! IMPORTANT NOTE: Use the quadratic formula technique to solve all problems in this homework assignment. There are a couple of word problems at the end of the assignment in which the online learning aids will show factoring as the solution method. You should use the quadratic formula instead (and you will find it to be easier and quicker than factoring.)
Reminder: This homework assignment on Section 8.2 is due at the start of next class period.
You may now OPEN your LAPTOPS
and begin working on the homework assignment (if there’s any time left...) But remember, you can always work in the JHSW 203 open lab after class (or before tour next class session) if you want some help on this homework.
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## Precalculus (6th Edition) Blitzer
The simplified solution is $\frac{1}{x}-\frac{1}{x+2}$ and the sum is $\frac{100}{101}$.
Let us consider the rational expression. \begin{align} & \frac{2}{x\left( x+2 \right)}=\frac{A}{x}+\frac{B}{x+2} \\ & =\frac{A\left( x+2 \right)+Bx}{x\left( x+2 \right)} \end{align} $2=A\left( x+2 \right)+Bx$ (I) Substitute the value of $x=-2$ in the equation (I), \begin{align} & 2=B\left( -2 \right) \\ & B=-1 \end{align} Again, putting the value of $x=0$ in (I), \begin{align} & 2=A\left( 2 \right) \\ & A=1 \end{align} Therefore, $\frac{2}{x\left( x+2 \right)}=\frac{1}{x}-\frac{1}{x+2}$ is the simplified form of the result. And use the above result to find the sum of the series, \begin{align} & \frac{2}{1\cdot 3}+\frac{2}{3\cdot 5}+\frac{2}{5\cdot 7}+\cdots +\frac{2}{99\cdot 101}=\left( \frac{1}{1}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{5} \right)+\left( \frac{1}{5}-\frac{1}{7} \right)+\cdots +\left( \frac{1}{99}-\frac{1}{101} \right) \\ & =\frac{1}{1}-\frac{1}{101} \\ & =\frac{100}{101} \end{align} Thus, the sum is $\frac{100}{101}$.
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How to Divide Mixed Numbers? (+FREE Worksheet!)
Don't know how to divide complex numbers? In this post, we will teach you how to divide complex numbers into a few simple and easy steps.
Fractions greater than $$1$$ are usually represented as mixed numbers. In this case, the mixed number consists of an integer part and a standard fraction less than $$1$$. The integer part is the same as the quotient part; the fraction’s numerator is the remainder of the division, and the fraction’s denominator will also be the divisor.
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Step-by-step guide to dividing mixed numbers
In the division of mixed numbers, their improper fractions can be used instead, and if necessary, the result can be converted back to the form of mixed numbers.
Here’s how to divide mixed numbers:
• Step 1: Convert all mixed numbers into improper fractions: To convert a mixed number to a fraction, you must first multiply the mixed number by the denominator, then add the result to the numerator. Put the resulting number in the case of a new numerator and keep the denominator of the same fraction as the previous one. This way you have converted a mixed number into a fraction. $$a \ \frac{c}{ \color{blue}{b} }= a \ \color{blue}{+} \ \frac{c}{\color{ blue }{b}}=\frac{a \color{ blue }{b} \ \color{blue}{+} \ c}{ \color{ blue }{b} }$$
• Step 2: Divide these two improper fractions: To divide fractions, we turn the division into a multiplication problem by multiplying the first fraction by the inverse of the second fraction (reciprocal).
• Step 3: Convert the answer to a mixed number: To convert a fraction larger than one to a mixed number, you must divide the numerator by the denominator. After division, the obtained quotient is the same integer in the mixed number and the remainder of the division is the numerator of the fraction. The denominator is also the denominator of the initial fraction in the mixed number.
Dividing Mixed Numbers – Example 1:
Find the quotient. $$2 \ \frac{1}{3} \div \ 1 \ \frac{1}{4}=$$
Solution:
Convert mixed numbers to fractions, $$2 \ \frac{1}{3} =$$ $$\frac{7}{3}$$, $$\ 1 \ \frac{1}{4}=$$ $$\ \frac{5}{4}$$
Apply the fractions rule for dividing, $$\frac{7}{3} \div \frac{5}{4}=$$ $$\frac{7}{3} × \frac{4}{5}=\ \frac{7 \ × \ 4}{3 \ × \ 5}=\frac{28}{15}=1 \ \frac{13}{15}$$
Dividing Mixed Numbers – Example 2:
Find the quotient. $$2 \ \frac{5}{6} \div \ 1 \ \frac{2}{5}=$$
Solution:
Convert mixed numbers to fractions, $$2 \ \frac{5}{6}=$$ $$\frac{17}{6}$$, $$\ 1 \ \frac{2}{5}=$$ $$\frac{7}{5}$$
Use the fractions rule for dividing , $$\frac{17}{6} \div \frac{7}{5}$$$$= \frac{17}{6} × \frac{5}{7}= \frac{17 \ × \ 5}{6 \ × \ 7}=\frac{85}{42}=2 \ \frac{1}{42}$$
Dividing Mixed Numbers – Example 3:
Find the quotient. $$2 \ \frac{1}{2} \div \ 1 \ \frac{1}{5}=$$
Solution:
Convert mixed numbers to fractions, $$2 \ \frac{1}{2}=$$ $$\frac{5}{2}$$, $$\ 1 \ \frac{1}{5}=$$ $$\frac{6}{5}$$
Use the fractions rule for dividing,$$\frac{5}{2} \div \frac{6}{5}=$$ $$\frac{5}{2} × \frac{5}{6}= \frac{5×5}{2×6}= \frac{25}{12}=2 \ \frac{1}{12}$$
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Dividing Mixed Numbers – Example 4:
Find the quotient. $$4 \ \frac{3}{4} \div \ 3 \ \frac{4}{5}=$$
Solution:
Converting mixed numbers to fractions, $$4 \ \frac{3}{4}=$$ $$\frac{19}{4}$$ , $$\ 3 \ \frac{4}{5}=$$ $$\frac{19}{5}$$
Use the fractions rule for dividing, $$\frac{19}{4} \div \frac{19}{5}=$$ $$\frac{19}{4} × \frac{5}{19}= \frac{19×5}{4×19}=\frac{95}{76}=\frac {95\div 19} {76\div19}=\frac {5} {4}=1 \ \frac{1}{4}$$
Exercises for Dividing Mixed Numbers
Find each quotient.
1. $$\color{blue}{2\frac{1}{5} \div 2\frac{1}{2}}$$
2. $$\color{blue}{2\frac{3}{5} \div 1\frac{1}{3}}$$
3. $$\color{blue}{3\frac{1}{6} \div 4\frac{2}{3}}$$
4. $$\color{blue}{1\frac{2}{3} \div 3\frac{1}{3}}$$
5. $$\color{blue}{4\frac{1}{8} \div 2\frac{2}{4}}$$
6. $$\color{blue}{3\frac{1}{2} \div 2\frac{3}{5}}$$
1. $$\color{blue}{\frac{22}{25}}$$
2. $$\color{blue}{1\frac{19}{20}}$$
3. $$\color{blue}{\frac{19}{28}}$$
4. $$\color{blue}{\frac{1}{2}}$$
5. $$\color{blue}{1\frac{13}{20}}$$
6. $$\color{blue}{1\frac{9}{26}}$$
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How do you identify all asymptotes or holes for f(x)=(x^2+3x-4)/(2x^2+10x+8)?
Oct 30, 2016
vertical asymptote at x = - 1
horizontal asymptote at $y = \frac{1}{2}$
hole at x = -4
Explanation:
The first step is to factorise the numerator/denominator of f(x).
$f \left(x\right) = \frac{{x}^{2} + 3 x - 4}{2 {x}^{2} + 10 x + 8} = \frac{\cancel{\left(x + 4\right)} \left(x - 1\right)}{2 \cancel{\left(x + 4\right)} \left(x + 1\right)} = \frac{x - 1}{2 x + 2}$
excluded value is x ≠ -4. This means that the original function has a hole at x = -4, while the simplified version does not.
The denominator of simplified f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.
solve : $2 x + 2 = 0 \Rightarrow x = - 1 \text{ is the asymptote}$
Horizontal asymptotes occur as
${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$
divide terms on numerator/denominator by x
$f \left(x\right) = \frac{\frac{x}{x} - \frac{1}{x}}{\frac{2 x}{x} + \frac{2}{x}} = \frac{1 - \frac{1}{x}}{2 + \frac{2}{x}}$
as $x \to \pm \infty , f \left(x\right) \to \frac{1 - 0}{2 + 0}$
$\Rightarrow y = \frac{1}{2} \text{ is the asymptote}$
graph{(x-1)/(2x+2) [-10, 10, -5, 5]}
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# IB Mathematics SL/Vectors
(Redirected from IB Mathematics (SL)/Vectors)
# Topic 4: Vectors
## Introduction
Vectors can be described mathematically by using Trigonometry.
We can define a vector to be an ordered pair consisting of a magnitude and a direction. In this diagram, r is the magnitude of this vector and θ is the direction. Notice, now, that we have moved horizontally r cos(θ) and vertically r sin(θ). These are called the x-component and the y-component, respectively.
We can also write a vector conveniently in terms of the x and y component. We write $\begin{pmatrix} x \\ y\end{pmatrix}$ for vectors. In some texts, you may see the vector written sideways, like (x, y), but when you write it will help greatly to write them downwards in columns. In print we commonly bold vectors, but since you probably don't have a pen that writes in bold print, underline your vectors, i.e. write v, or put a tilde underneath your vectors or place an arrow pointing right overtop of your vector.
## Length of a vector
The distance formula can be used to find the magnitude, r, of a vector given its components with the following equation:
$\left\|\mathbf{a}\right\|=\sqrt{{a_1}^2+{a_2}^2+{a_3}^2}$
Where a1, a2, and a3 are the three components of the vector.
## Vector equality
Two vectors are said to be equal if they have the same magnitude and direction. However if we are talking about bound vector, then two bound vectors are equal if they have the same base point and end point.
For example, the vector i + 2j + 3k with base point (1,0,0) and the vector i+2j+3k with base point (0,1,0) are different bound vectors, but the same (displacement) vector.
## Scalar multiplication
For scalar multiplication, we simply multiply each component by the scalar. We commonly use Greek letters for scalars, and Roman letters for vectors.
So for a scalar value of λ and a vector v defined by r and θ, the new vector is now λr and θ. Notice how the direction does not change.
#### Example
Say we have $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and we wish to double the magnitude. So, $2 \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}$.
Simply, to add two vectors, you must add the respective x-components together to obtain the new x-component, and likewise add the two y-components together to obtain the new y-component.
### Example
Say we have $\mathbf{v_1}=\begin{pmatrix} 2 \\ 3 \end{pmatrix}, \mathbf{v_2}=\begin{pmatrix} 4 \\ 6 \end{pmatrix}$ and we wish to add these. So, $\mathbf{v_1}+\mathbf{v_2}=\begin{pmatrix} 6 \\ 9 \end{pmatrix}$.
### Magnitude
The magnitude of a vector is its length in R+
## The dot product
The dot product of two vectors is defined as the sum of the products of the components. Symbolically we write
$\begin{pmatrix} a_1 \\ a_2 \end{pmatrix}\cdot\begin{pmatrix}b_1 \\ b_2 \end{pmatrix} = a_1b_1 + a_2b_2$
For example,
$\begin{pmatrix} 3 \\ 5 \end{pmatrix}\cdot\begin{pmatrix} 1 \\ -2 \end{pmatrix}=3-10=-7$
The dot product of two vectors has an alternate form:
$\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos{\theta}$
The angle θ then is important, as it shows that the dot product of two vectors is related to the angle between them. More specifically, we can calculate the dot product of two vectors - if the dot product is zero we can then say that the two vectors are perpendicular.
For example, consider simply
$\begin{pmatrix} 1 \\ 1 \end{pmatrix}\cdot\begin{pmatrix}1 \\ -1\end{pmatrix}=1-1=0$
Plot these vectors on the plane and verify for yourself that these vectors are perpendicular.
## Vectors of 3D-lines
Cartesian equation: $\tfrac{x-a}{l}=\tfrac{y-b}{m}=\tfrac{z-c}{n}$. where a, b, and c are the coordinates on the vector line.
Vector equation: r = $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$
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# Topic 2 Introduction to Differential Equations
## 2.1 Basic concepts
### 2.1.1 How to define a differential equation in Maple
Depending on the differentiation command, there are three ways to define a differential equation in Maple.
Example 2.1 Assign the differential equation $$y'=2y$$ to a variable in Maple.
• Solution.
• Method 1: Using the diff command
ode111:=diff(y(x), x)=2*y(x)
• Method 2: Using the the prime derivative notation
ode112:=y'=2y
• Method 3: Using the command D
ode113:=D(y)(x)=2*y(x)
Among the three methods, the one using diff is the standard choice.
Exercise 2.1 Assign the differential equation $$y'=y(1-0.001y)$$ to a variable in Maple.
### 2.1.2 How to check solutions
To check if a function (explicitly of implicitly defined) is a solution of a given differential equation, you may use the odetest(function, ODE, y(x)) command. If the output is 0, then the function is a solution.
Example 2.2 Verify that $$y=ce^{2x}$$ is a solution to the differential equation $$y'=2y$$.
Solution. Run the following command, you will see that output is 0. So the function is a solution.
odetest(y(x)=c*exp(2*x), diff(y(x), x)=2*y(x), y(x))
Example 2.3 Verify that $$x^2+y^2=1$$ is a solution of the equation $$y'=-\frac xy$$.
Solution. Again, running the following command returns the number 0. So the implicit function is a solution.
odetest(x^2+y(x)^2=1, diff(y(x),x)=-x/(y(x)), y(x));
When working with differential equations, we should always use $$y(x)$$ instead of $$y$$ to indicate that $$y$$ is a function of $$x$$.
Exercise 2.2 Verify that the function $$y=c_1e^x+c_2e^{-x}$$ is a solution of the equation $$y''=y$$.
Exercise 2.3 Verify that $$x^2+4y^2=c$$ is an implicit solution of the equation $$4yy'=x$$.
## 2.2 Solution curves vs Integral curves
A solution curve is the graph of a function $$y=f(x)$$ that satisfies the given differential equation. An integral curves is a union of solution curves.
Example 2.4 Consider the differential equation $$yy'=4x$$. Verify that the graphs of the functions $$y=\pm\sqrt{4x^2-1}$$ are solution curves of the equation, while the hyperbola $$4x^2-y^2=1$$ defines an integral curve of the equation.
Solution. Let’s rename the function as $$y_1(x)=\sqrt{4x^2-1}$$ and $$y_2(x)=-\sqrt{4x^2-1}$$. In Maple, it means we y[1](x) and y[2](x). We can check that they are solutions using the seq loop.
ode121:=y(x)*diff(y(x), x)=4*x:
y[1](x):=sqrt(4*x^2-1);
y[2](x):=-sqrt(4*x^2-1);
seq(odetest(y(x)=y[i](x), ode121, y(x)), i = 1 .. 2);
The outputs show that $$y_1$$ and $$y_2$$ are solutions.
To see that hyperbola $$4x^2-y^2=1$$ defines an integral curve, we solve for $$y$$.
soly:=solve(4*x^2-y^2=1, y);
You will see that the solutions are exactly the functions $$y_1$$ and $$y_2$$. So as an union of solution curves the hyperbola is an integral curve.
Plotting those curves will help use understand better.
solutioncurves := plot([y[1](x), y[2](x)],
x = -5 .. 5, y = -5 .. 5, color = [green, red]);
with(plots):
integralcurve := implicitplot(4*x^2 - y^2 = 1,
x = -5 .. 5, y = -5 .. 5, color = blue,
linestyle = dot);
To check the integral curve is the union of the two solution curve, we can use the display command.
display(solutioncurves,integralcurve)
Exercise 2.4 Consider the differential equation $$yy'=4x$$. Verify that the graphs of the functions $$y=\pm\sqrt{4x^2+1}$$ are solution curves of the equation, while the hyperbola $$y^2-4x^2=1$$ defines an integral curve of the equation.
## 2.3 Direction fields
In Maple, the commands DEplot, dfieldplot, and phaseportrait supported by the package DETools can be used to plot the direction field and solution curves. The basic usage is as follows
DEplot(differential equation, function, ranges, options)
Again, using the command ?DEplot, we can find details and examples on the command. In the following, I will use DEplot as an example to show how they work.
Example 2.5 Plot the direction field for the differential equation $$y'=-\frac xy$$. Can you guess what is a solution to this equation?
Solution. Let’s first define the differential equations and assign it to a variable.
ode131:=diff(y(x), x)=-x/(y(x));
Now load the package DETools using the command with().
with(DETools):
With the package loaded, we can use DEplot to plot the direction field for ode131, say in the region $$-5\le x\le 5$$ and $$-5\le y\le 5$$.
DEplot(ode131, y(x), x =-5..5, y =-5..5,
title = " Direction Field for y'=-x/y ")
Note that one may change the outlook by add options, such as color and arrows. Running the following command, you will see the difference.
DEplot(diff(y(x), x) = -(x - 1)/(y(x) + 1), y(x),
x = -5 .. 5, y = -5 .. 5,
title = "Direction Field for y'=2 y",
color = -(x - 1)/(y(x) + 1), arrows = line)
The direction field suggests that solutions are circles.
To display $$y(x)$$ as $$y$$ in the output, you may run the following commands first.
with(PDEtools, declare):
declare(y(x), prime=x); # Turn ON the enhanced DEdisplay feature
You will see the following output ${\color{blue} \text{derivatives with respect to } x \text{ of functions of one variable will now be displayed with '} }$
Exercise 2.5 Plot the direction field for the differential equation $$y'=-\frac{x-1}{y+1}$$. Can you guess what is a solution to this equation?
Exercise 2.6 Plot the direction field for the differential equation $$y'=-\frac{x}{2y}$$. Can you guess what is a solution to this equation?
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# How Many Faces Does a Cube Have?
A cube is a three-dimensional geometric shape that is often encountered in mathematics, architecture, and everyday life. It is a polyhedron with six square faces, twelve edges, and eight vertices. In this article, we will explore the concept of a cube in detail and answer the question: how many faces does a cube have?
## The Definition of a Cube
A cube is a regular polyhedron, which means that all of its faces are congruent and all of its angles are equal. It is a three-dimensional object that has six square faces, each of which is identical in size and shape. The edges of a cube are all the same length, and the angles between adjacent faces are all right angles (90 degrees).
The cube is a highly symmetrical shape, and its symmetry properties make it a fundamental object in mathematics and geometry. It is often used as a building block for more complex shapes and structures.
## The Faces of a Cube
As mentioned earlier, a cube has six faces. Each face is a square, and all six faces are congruent. The faces of a cube are arranged in such a way that they meet at right angles along their edges. This arrangement gives the cube its characteristic box-like appearance.
The six faces of a cube can be labeled using the letters A, B, C, D, E, and F. Each face is adjacent to four other faces, and the faces that share an edge are always different from each other. For example, face A is adjacent to faces B, C, D, and E, but not to face F.
## The Edges and Vertices of a Cube
In addition to its faces, a cube also has twelve edges and eight vertices. The edges of a cube are the line segments where two faces meet. Each edge is shared by two faces, and all twelve edges of a cube are congruent in length.
The vertices of a cube are the points where three edges meet. A cube has eight vertices, and each vertex is shared by three faces. The vertices of a cube are labeled using the letters G, H, I, J, K, L, M, and N.
## Calculating the Number of Faces
To determine the number of faces a cube has, we can simply count them. As mentioned earlier, a cube has six faces, each of which is a square. This is a fundamental property of a cube and does not change regardless of its size or orientation.
It is worth noting that the number of faces of a cube is fixed and cannot be altered without changing the shape itself. No matter how you rotate or manipulate a cube, it will always have six faces.
## Real-World Examples of Cubes
Cubes are not just abstract mathematical objects; they can be found in various real-world applications. Here are a few examples:
• Dice: A standard six-sided die used in board games is a cube. Each face of the die is numbered from one to six, and the sum of the numbers on opposite faces is always seven.
• Rubik’s Cube: The popular puzzle toy Rubik’s Cube is a three-dimensional combination puzzle in the shape of a cube. It consists of smaller cubes (cubies) that can be rotated to create different patterns.
• Building Blocks: Many children’s building blocks are shaped like cubes. These blocks can be stacked and arranged to create various structures and shapes.
• Architecture: Cubes are often used as architectural elements in buildings. For example, the Kaaba in Mecca, Saudi Arabia, is a cube-shaped structure that is considered the holiest site in Islam.
## Conclusion
In conclusion, a cube has six faces, each of which is a square. The faces of a cube are congruent and meet at right angles along their edges. In addition to its faces, a cube has twelve edges and eight vertices. The number of faces of a cube is fixed and cannot be changed without altering the shape itself.
Cubes are not just mathematical abstractions; they have practical applications in various fields, including gaming, puzzles, architecture, and more. Understanding the properties of a cube is essential for solving problems and creating structures that involve this geometric shape.
## Q&A
### 1. How many edges does a cube have?
A cube has twelve edges. Each edge is shared by two faces.
### 2. How many vertices does a cube have?
A cube has eight vertices. Each vertex is shared by three faces.
### 3. Are all the faces of a cube congruent?
Yes, all the faces of a cube are congruent. They are identical in size and shape.
### 4. Can a cube have more than six faces?
No, a cube cannot have more than six faces. The number of faces of a cube is fixed and cannot be changed without altering the shape itself.
### 5. What are some other examples of cubes in everyday life?
Some other examples of cubes in everyday life include storage boxes, ice cubes, and sugar cubes.
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## Wednesday, September 28, 2016
### Middle School Math Solutions – Polynomials Calculator, Multiplying Polynomials
Multiplying polynomials can be tricky because you have to pay attention to every term, not to mention it can be very messy. There are a few ways of multiplying polynomials, depending on how many terms are in each polynomial. In this post, we will focus on how to multiply two term polynomials and how to multiply two or more term polynomials.
Multiply two term polynomials
When multiplying polynomials with two terms, you use the FOIL method. The FOIL method only works for multiplying two term polynomials. FOIL stands for first, outer, inner, last. This lets you know the order of how to distribute and multiply the terms. Let’s see how it works.
After FOILing, multiply the terms, group like terms, and add like terms if there are any.
Here is another helpful identity to use when multiplying two term polynomials:
(a+b)(a-b)=a^2-b^2
Multiplying these polynomials is pretty simple because if you memorize these identities then you just plug in the values and have an answer.
Multiplying multiple term polynomials
You cannot use the FOIL method to multiply these polynomials. Instead, you have to multiply each term in one polynomial by each term in the other. You can do this by multiplying each term of one polynomial by the other polynomial. This can be tricky because it is easy to miss one term. When we do examples of this, it will become easier to understand how to solve them.
When multiplying polynomials, you may come across multiplying variables with exponents by variables with exponents. In this case, we use this exponent rule:
x^n\cdot x^m=x^(n+m)
For this rule, the base or variable must be the same. When multiplying variables with exponents, you add the exponents together.
Let’s see some examples to understand how to multiply polynomials.
(2x-1)(5x-6)
We will use the FOIL method to solve this.
1. Use FOIL identity
(2x-1)(5x-6)
2x\cdot 5x+2x\cdot -6+(-1)\cdot 5x+(-1)\cdot -6
2. Multiply terms
10x^2-12x-5x+6
3. Group like terms
10x^2-12x-5x+6
(Luckily, everything was already grouped together)
10x^2-17x+6
(2x^2+6)(2x^2-6)
Here, we can use another one of the identities for multiplying two term polynomials.
1. Use (a+b)(a-b)=a^2-b^2
(2x^2+6)(2x^2-6)
(2x^2 )^2-6^2
2. Simplify
4x^4-6^2
4x^4-36
(x^2+2x-1)(2x^2-3x+6)
1. Multiply each term in one polynomial by the other polynomial
x^2 (2x^2-3x+6)+2x(2x^2-3x+6)-1(2x^2-3x+6)
2. Distribute and multiply
2x^2\cdot x^2-3x\cdot x^2+6\cdot x^2+2x^2\cdot 2x-3x\cdot 2x+6\cdot 2x+2x^2\cdot -1-3x\cdot -1+6\cdot -1
2x^4-3x^3+6x^2+4x^3-6x^2+12x-2x^2+3x-6
3. Group like terms
2x^4-3x^3+6x^2+4x^3-6x^2+12x-2x^2+3x-6
2x^4-3x^3+4x^3+6x^2-6x^2-2x^2+12x+3x-6
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# Topic 1.1: Modulus Functions
Information about Topic 1.1: Modulus Functions
Published on January 5, 2009
Author: geakuan
Source: authorstream.com
Modulus functions : Modulus functions |4| = |-4| = |a| = b => |x - a| = b => |a| = |b| => |x - a| = |b| => 4 4 a = b (x – a) = b Modulus inequalities : Modulus inequalities |x| < 1 => |x| > 1 => |a| = b => + a = b -1 < x < 1 x < -1 or x > 1 Quadratic inequalities : Quadratic inequalities Quadratic inequality (x - a)(x – b) < 0 => where a is smaller than b (x – a)(x – b) > 0 = > a < x < b, x > b or x < a. < > Solve |x – 3| < 7 : Solve |x – 3| < 7 +(x – 3) < 7 (or) –(x – 3) < 7 x < 10 -4 10 –x + 3 < 7 -x < 4 x > -4 Answer is -4 < x < 10. Solve |2x – 1| > 7 : Solve |2x – 1| > 7 (or) +(2x – 1) > 7 2x > 7 + 1 2x > 8 x > 4 -(2x – 1) > 7 -2x + 1 > 7 -2x > 7 - 1 -2x > 6 x < -3 -3 4 Answer is x < -3 (or) x > 4 Solve 3x + 1 > |x – 5| : Solve 3x + 1 > |x – 5| (or) 3x + 1 > - (x – 5) 3x + 1 > - x + 5 4x > 4 x > 1 3x + 1 > +(x – 5) 3x + 1 > +x – 5 2x > - 6 x > - 3 0 2 CHECK FOR x = 0 3x + 1 > |x – 5| 3(0) + 1 > |0 – 5| 1 > 5 (FALSE) Since it’s wrong, therefore Reject x > -3 Therefore answer is x > 1 Solve |x – 3| > 2x + 1 : Solve |x – 3| > 2x + 1 +(x – 3) > 2x + 1 + x – 3 > 2x + 1 -x > 4 x < -4 (or) -(x – 3) > 2x + 1 - x + 3 > 2x + 1 -3x > - 2 x < -4 2 3 0 -5 CHECK FOR x = 0 |x – 3| > 2x + 1 |0 – 3| > 2(0) + 1 3 > 1 (TRUE) Since it’s correct, therefore x < 2 3 Solve |2x - 3| < |x + 1| : Solve |2x - 3| < |x + 1| |a| = |b| => < Solve |2x - 3| > |x + 1| : Solve |2x - 3| > |x + 1| |a| = |b| => >
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# Zero Product Property: Definition, Formula, Examples
Home » Math Vocabulary » Zero Product Property: Definition, Formula, Examples
## What Is the Zero Product Property?
The Zero Product Property states that if $a\times b = 0$, then we must have $a = 0$ or $b = 0$ or both $a = b = 0$. It is also known as the “zero product principle.”
Suppose you multiply two real numbers and get 0 as the product, what can you say about the two numbers? What could the numbers be? When the result from multiplying two numbers is zero, the following two cases are possible
i) one of them is 0
ii) both of them are 0
This idea is called the zero product principle. This property is useful in solving the equations of the form $(x + a)(x + b) = 0$.
## Zero Product Property
If the product of two or more factors equals zero, then at least one of the factors is equal to zero.
It can be written in symbolic form as:
$a \times b = 0 \Leftrightarrow a = 0$, or $b = 0$, or both $a = b = 0$.
In other words, the zero product property means that the product of two non-zero real numbers can never be zero.
## Zero Product Property in Equations
When solving equations, the zero product property can be used to determine the values of the variables. The zero product property is incredibly useful for resolving quadratic equations in the factored forms.
We can extend the zero product property and state that
If $(x + a)(x + b)(x + c)$…$(x + n) = 0$, then
$x + a = 0$ or $x + b = 0$ or … $(x + n)=0$
Example 1: if $(x + a) (x + b) = 0$, then by the zero product property, we can say that
$(x + a) = 0\; or (x + b) = 0$.
Now, solving each of these for x would give the solutions for the given quadratic equation.
Thus, $x = \;-a$ or $x = \;-b$
Example 2: $x(x + 2) (x + 3) = 0$
By the zero product property,
$x = 0$ or $x + 2 = 0$ or $x + 3 = 0$
Thus, $x = 0$, or $x= \;-2$, or $x= \;-3$
Example 3: $(x \;-\; 3) (x + 6) = 0$
Using the zero product property, we know that,
$(x \;-\; 3) = 0$ or $(x + 6) = 0$
$x = 3$ or $x = \;-\; 6$
These are the two solutions of the equation.
## Advantages of Zero Product Property
• The zero product property is useful to find values of variables when we have a factored form of the equation. When the product of two or more factors is 0, we can apply the property and simply put each factor equal to 0.
• The zero product property helps in solving the equations of the form $(x + a)(x + b)(x + c)$….$(x + n) =0$.
• The zero product property is particularly helpful in solving quadratic equations.
• If we have an equation of the form nx=0, where n is a non-zero real number and x is a variable, we can apply the zero product property and conclude that x=0.
## Disadvantages of Zero Product Property
• Zero product property cannot be applied to matrices. It means that the product of two non-zero matrices can be 0.
• Zero product property is not applicable to the vectors as well. Two non-zero vectors, when multiplied, can result in a zero vector.
• We can apply the zero product property to solve equations if and only if we have the factored form. We must know how to factorize the given equation before we can apply the property.
## Facts about Zero Product Property
• The product of two non-zero real numbers can never be 0.
• Zero Product Property is also known as zero product principle, zero product property of multiplication, the null factor law, the null factor theorem, the nonexistence of nontrivial zero divisors, the zero product rule.
• The zero product property allows us to factor equations and solve them.
• According to the zero product rule, if the product of any number of expressions is 0, then at least one of them must also be zero.
• You can use the zero product property only if the expression is in the factored form and set to 0.
• The zero product rule is satisfied by the set of real numbers, integers, rational numbers, whole numbers, natural numbers.
## Conclusion
In this article, we have learned about zero product property. The zero product property allows us to factor equations and solve them. Now let’s solve some zero product property examples and practice problems for better understanding.
## Solved Examples on Zero Product Property
1. Solve the equation using the zero-product property.
$(𝑥 \;−\; 2)(𝑥 \;−\; 5) = 0$
Solution:
$(𝑥 \;−\; 2)(𝑥 \;−\; 5) = 0$
The equation is already in the factored form. So, we will simply apply zero product property to solve the equation.
$𝑥 \;−\;2 = 0$ or $𝑥\;−\;5 = 0$
$x \;-\; 2 + 2 = 0 + 2$ or $x \;-\; 5 + 5 = 0 + 5$
$x = 2$ or $x = 5$
Thus, $x = 2$ or $x = 5$ are the roots.
2. Solve using the zero product property: $2 x^2 \;−\; 3x \;−\; 5 = 0$
Solution:
We will factorize the equation first.
$2x^2 \;−\; 3x \;−\; 5 = 0$
$(2𝑥 \;−\; 5)(𝑥 + 1) = 0$
$2𝑥 \;−\; 5 = 0$ or $x + 1 = 0$
$2x = 5$ or $x + 1 \;-\; 1 = 0 \;-\; 1$
$\frac{2x}{2} = \frac{5}{2}$ or $x = \;-\; 1$
$x = \frac{5}{2}$ or $x = \;-\; 1$
Thus, the solutions are $x = \frac{5}{2}$ or $x = \;-1$.
3. Find the roots of the cubic equation which is in factored form: $(x + 2)^2 (3x + 2)= 0$.
Solution:
We can write the given equation $(x + 2)^2 (3x + 2)= 0$ as
$(x + 2)( x + 2)(3x + 2) = 0$
By the zero product property, we have $-\; 2$, and $-\; 23$.
4. What is the value of x if $99x = 0$?
Solution:
The zero product property says that if the product of two numbers is 0, then at least one of the numbers is 0.
We have $99x = 0$
Here, 99 is a constant.
$99 \neq 0$
Thus, by the zero product property, we have $x = 0$.
5. Using the zero product property, solve the following for x.
$10x(x + 2)(x \;-\; 5)(4x + 12)$
Solution:
If $10x(x + 2)(x \;-\; 5)(4x + 12)= 0$, at least one of the factors is 0.
The possible roots to the equation are $0,\; -\;2,\; 5,\; -\; 3$.
## Practice Problems on Zero Product Property
1
### If a.b $= 0$, then $a = 0$ or $b = 0$ or $a = b = 0$.
Zero Property
Zero Product Property
Multiplication Rule
Both b and c
CorrectIncorrect
Correct answer is: Zero Product Property
According to zero product property, if $𝑎 \times 𝑏 = 0$ then either $𝑎 = 0$ or $𝑏 = 0$ or both a and b are equal to zero.
2
### If $5 \times x = 0$ then, _________.
$x \neq 0$
$x = 0$
x may or may not be equal to 0
$x = \;-5$ or $x = \frac{1}{5}$
CorrectIncorrect
Correct answer is: $x = 0$
If $5 \times x = 0$ then, the value of x must be zero.
3
### Solve for x:$(x + 2)(x \;-\; 3) = 0$
$2,\; 3$
$-\; 2,\; -\; 3$
$-\; 2,\; 3$
$2,\; -\; 3$
CorrectIncorrect
Correct answer is: $-\; 2,\; 3$
$(x + 2)(x \;-\; 3) = 0$.
$x + 2 = 0$ or $x \;-\; 3 = 0$
Thus, $x = \;-\; 2$ or $x = 3$
4
### The product of two non-zero real numbers is
never 0
always 0
always positive
always negative
CorrectIncorrect
The product of two non-zero real numbers can never be 0.
5
### The zero product property states that for any real numbers and b, if $ab = 0$, then
$a = 0$, or $b = 0$, or both $a = b = 0$
$a = 0$ and $b \neq 0$
$a = b = 0$
$b = 0$ and $a \neq 0$
CorrectIncorrect
Correct answer is: $a = 0$, or $b = 0$, or both $a = b = 0$
For any real numbers a and b, if $ab = 0$, then, either $a = 0$, or $b = 0$, or both a and b equal to zero
## Frequently Asked Questions on Zero Product Property
According to the associative property of multiplication, the product of any three numbers remains the same, irrespective of the order in which they are grouped.$a\times(b\times c) = (a\times b) c$
The identity property of multiplication states that the product of any real number with 1 is always the number itself. The number 1 is called the multiplicative identity.
For matrices, the zero product property is not applicable, because if the product of two matrices is a zero matrix, then it is not compulsory that one of the matrices should be a zero matrix.
0 is the additive identity for the set of real numbers. Addition of any real number with 0 is the number itself.
To apply the zero product property, we set each factor equal to zero and solve for the variable. The zero product property can be applied when the product of the expressions is equal to zero.
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