# How exactly does elasticity relate to slope?

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.

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 $$\epsilon=\frac{dQ}{dp}\frac{p}{Q}$$?

• Relevant/possible dupe:economics.stackexchange.com/questions/14162/… – EconJohn Nov 30 '17 at 3:52
• 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 Nov 30 '17 at 4:19
• 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 Dec 3 '17 at 4:43

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.

• 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 Dec 3 '17 at 10:50
• 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 Dec 5 '17 at 22:21
• 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 Dec 6 '17 at 2:27
• @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 Dec 27 '17 at 12:31
• @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 Jan 2 '18 at 22:07

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 $(Q,p)$.

• 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 Dec 5 '17 at 3:07
• @Sasaki Did you see there is also another answer to your question? It answers the linear case. – Giskard Dec 5 '17 at 9:08