Example 2: Conceptual

Felix’s parent called him and told him he needed to come home from his friend’s house immediately. Felix left his friend’s house and made it back to his house 13 miles away in 15 minutes. His parent told him thanks for coming home, but they would appreciate it if he didn’t break the neighborhood speed limit of 45mph.

How did his parent know that he had been speeding?

Step 1: Assure that the requirements of the Mean Value Theorem are met by your situation or equation.

In a real-world example you must first try to determine the function that you are working with. In this example you are working with a position function , s(t) .

You know the starting position and starting time.

(0,0)

You also know the ending position and the ending time.

(.25,13)

You want to be careful of units in a real-world problem. Your units from your actual function need to match the units of your rate of change. This problem does not say specifically which units you need to use to compare value.

I would say it is easier to convert minutes into hours than it is to convert miles per hour to miles per minute.

You would want to convert 15 minutes to .25 hours .

Also, most real-world problems that involve time have an implied start, initial time, and that start is usually considered time zero, so t = 0 . Time is then taken as time since that moment.

This position function is known to be:

i)                     continuous on the closed interval [0,.25]

You cannot go from one position to the next without going through every position in between. The same is true of time, you cannot get from 0 hours to .25 hours without being at every time in between them.

ii)                   differentiable on the open interval (0,.25)

You position would be a smooth and continuous graph.

Step 2: Find the averagerate of change, slope ,between the two x-value endpoints, [ a , b ] of your function.

$\mathit{average }\mathit{rate\; of\; change}=\frac{f\left(b\right)–f\left(a\right)}{b–a}=\frac{y_2–y_1}{x_2–x_1}$

Don’t overthink these. Remember that averagerate of change is just fancy math talk for slope as you have always known it, $\frac{y_2–y_1}{x_2–x_1}$.

[ a , b ]= [0,.25]

s(t) = position

$\left(x_1,y_1\right)=\left(0,0\right)$

$\left(x_2,y_2\right)=\left(0.25,13\right)$

$\frac{f\left(b\right)–f\left(a\right)}{b–a}=\frac{y_2–y_1}{x_2–x_1}=\frac{13–0}{.25–0}=\frac{13}{.25}=52\frac{\mathit{miles}}{\mathit{hour}}$

Step 3: Find the derivative of your equation, $f\prime \left(x\right)$.

You don’t need to know the specific position equation, s(t) . You only need to remember your physics relationships. If you start with position the 1 ^st Derivative of position is velocity .

s(t) = position

$s^{\prime }\left(t\right)=v\left(t\right)=\mathit{velocity}$

Step 4: Set your derivative, $f\prime \left(x\right)$, equal to your $\mathit{average }\mathit{rate\; of\; change}=\frac{f\left(b\right)–f\left(a\right)}{b–a}$, then solve that algebra problem for x .

$\frac{f\left(b\right)–f\left(a\right)}{b–a}=s^{\prime }\left(t\right)$

$52\frac{\mathit{miles}}{\mathit{hour}}=v\left(t\right)=\mathit{velocity}$

Final Results:

By the MVT your parents know that at some point in between t=0 hours and t=.25 hours your velocity had to be $52\frac{\mathit{miles}}{\mathit{hour}}$.

Notice that the problem does not ask you to find that specific point in time, it only wants you to prove that it had to exist.