<p>1) Yes, the steeper curve is more inelastic at all prices, if they are linear.</p>

<p>2) For linear demand curves, we have $\epsilon(P) = \frac{1}{m}\frac{P}{Q(P)}$ for a demand curve with slope $\frac{\Delta P}{\Delta Q}=m$. Let the demand curve be represented $P=b+mQ$. This will reduce to $\epsilon = \frac{P}{P-b}$ where $b$ is the $P$-intercept.</p>

<p>By hypothesis, they both share a point $(Q,P)$, so the steeper slope corresponds to a greater value of $b$, and so the curve is more inelastic. </p>

CC BY-SA 3.0
2017-12-03T10:50:22.797Z
This is a nice answer, but I think you could consider defining what you mean by steeper as this is the exact opposite in the $(Q,P)$ and the $(P,Q)$ coordinate systems. Also the last line was somewhat difficult to grasp (for me) as $\epsilon$ is negative and there $P/(P-b)$ is monotonically increasing in $b$. Your statement is true, but it may be worth spelling out.
Giskard
CC BY-SA 3.0
2017-12-05T22:21:36.277Z
Could you please elaborate a bit more on why a steeper slope implies a greater value of $b$? If we have two curves like $P=10-5Q$ and $P=10-7Q$, for instance, the latter is steeper than the former even though they have the same $P$-interpect. What am I getting wrong? Thanks very much.
Sasaki
CC BY-SA 3.0
2017-12-06T02:27:42.700Z
Ah yes, but it must be that they share a point at Q>0. So, just pick a point in the interior of the first quadrant, let P be on the vertical axis, and rotate a few linear demand curves from a single point. Steeper will imply a P-intercept at a higher point and hence be more inelastic. Think of it this way: a higher intercept corresponds to a higher choke price (highest price at which someone is willing to buy), so it must be that this indicates less price-sensitive demand.
Pburg
CC BY-SA 3.0
2017-12-27T12:31:35.940Z
@Pburg Sorry for the late response. I had kind of given up on this because I still didn't understand your explanation. But since it doesn't hurt to ask, is there a way of showing that algebraically? That would be really helpful as I honestly can't quite understand what you mean by "rotate a few linear demand curves from a single point" or how that shows what is claimed. Thanks very much for your patience.
Sasaki
CC BY-SA 3.0
2018-01-02T22:07:30.783Z
@Sasaki Here's an illustration. Take two demand curves that pass through a particular point $(Q,P)$ with $Q>0$. For concreteness, let that be $(4,4)$. Suppose one has a slope of $\frac{\Delta P}{\Delta Q}= -1$ and the other has a slope of -2. Then, the equations are $P=8-Q$ and $P=12-2Q$. They both shared a point $(4,4)$ and the steeper slope (-2) had to correspond to a higher $P$-intercept (12). With some algebra you could prove this is always the case, but it's best demonstrated with some doodling.
Pburg
CC BY-SA 3.0
2017-12-03T04:53:19.550Z
2017-12-03T04:53:19.550Z
Pburg
<p>The two demand functions $D_1(p),D_2(p)$ cross at the point $(Q,p)$. Their respective elasticities at price $p$ are
\begin{align*}
\epsilon_1(p) &amp; = \frac{\text{d}D_1(p)}{\text{d}p}\frac{p}{D_1(p)} \\
\\
\epsilon_2(p) &amp; = \frac{\text{d}D_2(p)}{\text{d}p}\frac{p}{D_2(p)}.
\end{align*}
However since both function cross at the point $(Q,p)$ we know that
$$
D_1(p) = D_2(p) = Q.
$$
But then
$$
\frac{p}{D_1(p)} = \frac{p}{Q} = \frac{p}{D_2(p)}.
$$
Meaning the only difference between their elasticities is $\text{d}D_i(p)/\text{d}p$, which is their slopes.</p>

<p>As for your 1. question, the conditions are not clear. Is the 'flatter' curve only 'flatter' locally, or for every price $p$? If you only mean locally, then no, the statement is only valid in the intersection point $(Q,p)$.</p>

CC BY-SA 3.0
2017-12-05T03:07:37.000Z
Thanks, that was helpful. As for my 1st question, I was thinking of two linear curves with different slopes. How would you go about showing that the flatter one has a greater elasticity for every $p$? Thanks.
Sasaki
CC BY-SA 3.0
2017-12-05T09:08:12.613Z
@Sasaki Did you see there is also [another answer](https://economics.stackexchange.com/a/19567/1601) to your question? It answers the linear case.
Giskard
CC BY-SA 3.0
2017-11-30T11:12:01.280Z
2017-11-30T12:49:44.473Z
2017-11-30T12:49:44.473Z
Giskard
<p>My book (Goodwin's Microeconomics in Context, pg. 117) states the following about price-elasticity of demand:</p>

<blockquote>
  <p>Given two demand curves that go through a specific point on graphs with the same scale, the flatter demand curve will represent the relatively more elastic demand and the stepper one the relatively less elastic demand.</p>
</blockquote>

<p>I have actually two questions about this:</p>

<p>1) Will the flatter demand curve be more elastic at any given point (for any given value of $p$) or just at the point that both curves pass through? </p>

<p>2) How can we show this mathematically using the definition of elasticity as  $$\epsilon=\frac{dQ}{dp}\frac{p}{Q}$$?</p>

<p>Thanks very much in advance. </p>

CC BY-SA 3.0
2017-11-30T03:52:47.130Z
Relevant/possible dupe:https://economics.stackexchange.com/questions/14162/why-is-elasticity-not-defined-simply-as-the-slope-of-the-graph/14163#14163
EconJohn
CC BY-SA 3.0
2017-11-30T04:19:27.297Z
I honestly can't see how that answers my question. I know the definitions of slope and elasticity. What I would like to know is how to show the quoted result mathematically from the definition of price-elasticity of demand.
Sasaki
CC BY-SA 3.0
2017-12-03T04:43:03.457Z
The quote requires that "specific point" to be at a strictly positive quantity. Recall that two linear demand curves with the same P-intercept will have the same elasticity at any price.
Pburg
<microeconomics><elasticity>
How exactly does elasticity relate to slope?
Sasaki
# How exactly does elasticity relate to slope?

(Asked by: username_0 on 2017-11-30)

My book (Goodwin's Microeconomics in Context, pg. 117) states the following about price-elasticity of demand:

> Given two demand curves that go through a specific point on graphs with the same scale, the flatter demand curve will represent the relatively more elastic demand and the stepper one the relatively less elastic demand.

I have actually two questions about this:

1) Will the flatter demand curve be more elastic at any given point (for any given value of $p$) or just at the point that both curves pass through?

2) How can we show this mathematically using the definition of elasticity as $$\epsilon=\frac{dQ}{dp}\frac{p}{Q}$$?

Thanks very much in advance.

**Tags:** microeconomics, elasticity
> **Comments:**
> **username_1**: Relevant/possible dupe:https://economics.stackexchange.com/questions/14162/why-is-elasticity-not-defined-simply-as-the-slope-of-the-graph/14163#14163  
> **username_0**: I honestly can't see how that answers my question. I know the definitions of slope and elasticity. What I would like to know is how to show the quoted result mathematically from the definition of price-elasticity of demand.  
> **username_2**: The quote requires that "specific point" to be at a strictly positive quantity. Recall that two linear demand curves with the same P-intercept will have the same elasticity at any price.  

---

## 2 answers

### Answer by username_2 on 2017-12-03. Score: 2 | Accepted answer

1) Yes, the steeper curve is more inelastic at all prices, if they are linear.

2) For linear demand curves, we have $\epsilon(P) = \frac{1}{m}\frac{P}{Q(P)}$ for a demand curve with slope $\frac{\Delta P}{\Delta Q}=m$. Let the demand curve be represented $P=b+mQ$. This will reduce to $\epsilon = \frac{P}{P-b}$ where $b$ is the $P$-intercept.

By hypothesis, they both share a point $(Q,P)$, so the steeper slope corresponds to a greater value of $b$, and so the curve is more inelastic.

> **Comments:**
> **username_3**: This is a nice answer, but I think you could consider defining what you mean by steeper as this is the exact opposite in the $(Q,P)$ and the $(P,Q)$ coordinate systems. Also the last line was somewhat difficult to grasp (for me) as $\epsilon$ is negative and there $P/(P-b)$ is monotonically increasing in $b$. Your statement is true, but it may be worth spelling out.  
> **username_0**: Could you please elaborate a bit more on why a steeper slope implies a greater value of $b$? If we have two curves like $P=10-5Q$ and $P=10-7Q$, for instance, the latter is steeper than the former even though they have the same $P$-interpect. What am I getting wrong? Thanks very much.  
> **username_2**: Ah yes, but it must be that they share a point at Q>0. So, just pick a point in the interior of the first quadrant, let P be on the vertical axis, and rotate a few linear demand curves from a single point. Steeper will imply a P-intercept at a higher point and hence be more inelastic. Think of it this way: a higher intercept corresponds to a higher choke price (highest price at which someone is willing to buy), so it must be that this indicates less price-sensitive demand.  
> **username_0**: @Pburg Sorry for the late response. I had kind of given up on this because I still didn't understand your explanation. But since it doesn't hurt to ask, is there a way of showing that algebraically? That would be really helpful as I honestly can't quite understand what you mean by "rotate a few linear demand curves from a single point" or how that shows what is claimed. Thanks very much for your patience.  
> **username_2**: @Sasaki Here's an illustration. Take two demand curves that pass through a particular point $(Q,P)$ with $Q>0$. For concreteness, let that be $(4,4)$. Suppose one has a slope of $\frac{\Delta P}{\Delta Q}= -1$ and the other has a slope of -2. Then, the equations are $P=8-Q$ and $P=12-2Q$. They both shared a point $(4,4)$ and the steeper slope (-2) had to correspond to a higher $P$-intercept (12). With some algebra you could prove this is always the case, but it's best demonstrated with some doodling.  

---

### Answer by username_3 on 2017-11-30. Score: 7

The two demand functions $D\_1(p),D\_2(p)$ cross at the point $(Q,p)$. Their respective elasticities at price $p$ are
\begin{align\*}
\epsilon\_1(p) & = \frac{\text{d}D\_1(p)}{\text{d}p}\frac{p}{D\_1(p)} \\
\\
\epsilon\_2(p) & = \frac{\text{d}D\_2(p)}{\text{d}p}\frac{p}{D\_2(p)}.
\end{align\*}
However since both function cross at the point $(Q,p)$ we know that
$$
D\_1(p) = D\_2(p) = Q.
$$
But then
$$
\frac{p}{D\_1(p)} = \frac{p}{Q} = \frac{p}{D\_2(p)}.
$$
Meaning the only difference between their elasticities is $\text{d}D\_i(p)/\text{d}p$, which is their slopes.

As for your 1. question, the conditions are not clear. Is the 'flatter' curve only 'flatter' locally, or for every price $p$? If you only mean locally, then no, the statement is only valid in the intersection point $(Q,p)$.

> **Comments:**
> **username_0**: Thanks, that was helpful. As for my 1st question, I was thinking of two linear curves with different slopes. How would you go about showing that the flatter one has a greater elasticity for every $p$? Thanks.  
> **username_3**: @Sasaki Did you see there is also [another answer](https://economics.stackexchange.com/a/19567/1601) to your question? It answers the linear case.  

---

