# Using the Mean Value Theorem

The (classical) Mean Value Theorem states that if $f$ is a continuous function on $[a,b]$ and is differentiable on $(a,b)$, then there exists a point $c \in (a,b)$ where $f'(c) = \dfrac{f(b)-f(a)}{b-a}$.

Here is a "real-world" application of the theorem. Suppose two racers, Barry and Harry, start a race at time $t=0$ and end the race in a tie, both crossing the finish line at, say, time $t = 10$. Let $s(0)$ be the position of the starting line, and let $s(10)$ be the position of the finish line.

The position of each racer is continuous on [0,10] and (presumably) differentiable on (0,10). So the MVT tells us that at some time $t_0 \in (0,10)$, Barry's velocity equaled $s'(t_0)=v(t_0)=\dfrac{s(10)-s(0)}{10}$, and similarly there is a time $t_1 \in (0,10)$ where Harry's velocity equaled $s'(t_1)=v(t_1)=\dfrac{s(10)-s(0)}{10}$.

Question: Does the MVT also guarantee that $t_0=t_1$? That is, is it necessarily the case that Barry and Harry have the same velocity at the same exact time at some point during the race? If so, why?

• I think the MVT doesn't say that the point $c$ is unique, hence you could find other times say $t_2$ ro $t_3$ where that holds. But I don't know if this helps – Euler_Salter Nov 6 '16 at 1:49
• Let $x(t),y(t)$ be $x(t)=t$, $y(t)=\sqrt{t}$. Clearly $x(0)=y(0)=0$ and $x(1)=y(1)=1$, but $x'(t)=1=\frac{x(1)-x(0)}{1}=1$ for all $t$, while $y'(t)=1$ only when $t=\sqrt{1/2}$, so they need not be equal – user160738 Nov 6 '16 at 1:53

You're being ambiguous. If you mean to ask if their velocities will be the same at some time in the interval $[0,10]$, yes. The result follows by applying the mean value theorem to the function $x(t) = s_{b}(t) - s_{h}(t)$ on the interval $[0, 10]$. $s_b$ and $s_h$ are the position functions of Barry and Harry respectively.

However, if you apply the mean value theorem to the functions independently, the "mean value" times may not be identical. Consider the following example, where we suppose the velocities are continuous. If Barry runs very fast right at the start, almost up to the finish line, over the course of $1$ second and then slows down like a caterpillar for the other $9$ seconds, the mean value time for Barry will be somewhere around $t = 1$ seconds while he is slowing down.

If Harry does the opposite, that is, he's extremely slow for the first $9$ seconds and then lightning fast to catch up to Barry at the final second, then his "mean value time" will be somewhere around $t=9$ as he speeds up.

They will share the same velocity at some time ($g(t) = v_{b}(t) - v_{h}(t)$ is continuous, initially positive and negative at the end), but this shared velocity will not be the "mean value" velocity $\frac{s(10) - s(0)}{10}$.

• Thank you for answering my question. I don't see how I was being ambiguous (my question was very clearly stated, was it not?), but thank you nonetheless. – Mr Toad Nov 6 '16 at 5:43
• Your question was ambiguous because you ask two different things in the last paragraph of the question, and in fact they happen to have different answers. As MathematicsStudent1122 wrote, the answer to "does the MVT also guarantee that $t_0 = t_1$?" is no, but the answer to "is it necessarily the case that Barry and Harry have the same velocity at the same exact time at some point during the race?" is yes. – David Z Nov 6 '16 at 7:40
• @David Z, thanks for the clarification. – Mr Toad Nov 6 '16 at 14:53

You can use Cauchy's mean value theorem. It says that if $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$ then there exists $c\in (a,b)$ such that $$(f(b)-f(a))g'(c)=(g(b)-g(a))f'(c).$$ If we use $f$ to denote the position of Barry and $g$ to denote the position of Harry then we have $f(0)=g(0)=0$ and $f(T)=g(T)=L=$length. ($T$ denotes the total time.) So, there exists a point $c\in (0,T)$ such that $$Lg'(c)=(f(T)-f(0))g'(c)=(g(T)-g(0))f'(c)=Lf'(c).$$ That is, there exists $c$ such that $f'(c)=g'(c).$ So they must have had the same velocity at some point.

• +1 for the perfect answer. It does not appear at first that a single value of $c$ will work for $f$ and $g$ but Cauchy's MVT ensures that this is possible. – Paramanand Singh Nov 6 '16 at 7:09

Assume runner 1 runs at speed $2t$ feet per second, so that in ten seconds, runner 1 travels $\int_0^{10}2tdt=100$ feet. Assume runner $2$ runs at speed $\frac{3t^2}{1000}$ so that in $10$ seconds he also travels $100$ feet. However, $2t=\frac{3t^2}{1000}$ has roots $0$ and $\frac{2000}{3}>10$, so the runners never simultaneously travel the same speed during the race (besides the start, speed $0$). (There is nothing special about this example, I just started with a simple speed function $2t$ and tried to fudge another function to make it work. Im sure there are many more counterexamples)

• I think the example doesn't hold for runner 2. His initial speed isn't zero. – Cehhΐro Nov 6 '16 at 1:55
• @O.VonSeckendorff thanks for the catch. It is fixed. – TomGrubb Nov 6 '16 at 2:00