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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.

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  • $\begingroup$ Relevant/possible dupe:economics.stackexchange.com/questions/14162/… $\endgroup$
    – EconJohn ♦
    Nov 30, 2017 at 3:52
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    $\begingroup$ 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. $\endgroup$
    – Sasaki
    Nov 30, 2017 at 4:19
  • $\begingroup$ 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. $\endgroup$
    – Pburg
    Dec 3, 2017 at 4:43

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Giskard
    Dec 3, 2017 at 10:50
  • $\begingroup$ 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. $\endgroup$
    – Sasaki
    Dec 5, 2017 at 22:21
  • $\begingroup$ 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. $\endgroup$
    – Pburg
    Dec 6, 2017 at 2:27
  • $\begingroup$ @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. $\endgroup$
    – Sasaki
    Dec 27, 2017 at 12:31
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    $\begingroup$ @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. $\endgroup$
    – Pburg
    Jan 2, 2018 at 22:07
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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 $(Q,p)$.

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  • $\begingroup$ 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. $\endgroup$
    – Sasaki
    Dec 5, 2017 at 3:07
  • $\begingroup$ @Sasaki Did you see there is also another answer to your question? It answers the linear case. $\endgroup$
    – Giskard
    Dec 5, 2017 at 9:08

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