The Mean Value Theorem

A ball is dropped from a height of 100 ft. Its position in seconds after it's dropped is modeled by the function $s\left(t\right)=-16t^2+100$.

1. How long does the ball take to hit the ground after it's dropped?
2. Find the ball's average velocity between when it is released and when it hits the ground.
3. Then, find the time guaranteed by the Mean Value Theorem when the instantaneous velocity of the ball is equal to the ball's average velocity.

Step 1: Ensure that s(t) is continuous and differentiable

Before using the Mean value Theorem, we must ensure our function meets the theorem's requirements.

Since $s\left(t\right)$ is a polynomial, we know it is continuous and differentiable on the entire interval!

Step 2: Find the number of seconds it takes for the ball to hit the ground

The ball is dropped from a height of 100 ft. Let the ground be at 0 ft. So, to find when the ball hits the ground, we can set $s\left(t\right)=0$ and solve for $t$!

$s\left(t\right)=0=-16t^2+100-100=-16t^2\pm \frac{5}{2}=t$

As our time in seconds cannot be negative, the ball must hit the ground at $\frac{5}{2}$, or 2.5, seconds.

Step 3: Find the average velocity of the ball

We will use the time the ball is released, $t=0$, and the time the ball hits the ground $t=\frac{5}{2}$ (which is the total time that the ball is in free fall) to find the average velocity of the ball over the course of its fall. Averaging over both values...

$\begin{array}{rcl}{v}_{avg}& =& \frac{s\left(\frac{5}{2}\right)-s\left(0\right)}{\frac{5}{2}-0}\\ & =& \frac{\left(-16{\left(\frac{5}{2}\right)}^2+100\right)-\left(-16{\left(0\right)}^2+100\right)}{\frac{5}{2}}\\ & =& \frac{-100}{\frac{5}{2}}\\ & =& -40ft/sec\end{array}$

Therefore, the average velocity of the ball is $40ft/sec$ (down) during the time it is in the air. The sign of the velocity is negative here because the ball travels in the negative (in this case, down) direction.

Step 4: Apply the Mean Value Theorem

The Mean Value Theorem states that there is at least one point on$\left[0,\frac{5}{2}\right]$ seconds where the ball has an instantaneous velocity of $40ft/sec$ (down).

We'll start by taking the derivative of the position function s(t).

$s\text{'}\left(t\right)=-32t$

$$\begin{gathered} -32t=-40 \\ t=\frac{5}{4} \end{gathered}$$

So, the ball reaches a velocity of -40 ft per second at time $\frac{5}{4}$, or 1.25, seconds.

Assume that a function f(x) is continuous and differentiable on the interval [5, 15]. Given f(5) = 4 and f'(x)$\le$10. Find the largest possible value that f(15) can take on.

The problem specifies that f(x) is in fact continuous and differentiable. Thus, we can apply the Mean Value Theorem.

Step 1: Rearrange the Mean Value Theorem equation

We are looking for the value of f(15), or f(b). So, let's rearrange the equation to solve for f(b).

$\begin{array}{rcl}f\text{'}\left(c\right)& =& \frac{f\left(b\right)-f\left(a\right)}{b-a}\\ f\text{'}\left(c\right)(b-a)& =& f\left(b\right)-f\left(a\right)\\ f\left(b\right)& =& f\text{'}\left(c\right)(b-a)+f\left(a\right)\end{array}$

Step 2: Plug in known values

If f'(x) has a maximum value of 10, then f'(c) also has a maximum value of 10. So, we can substitute 10 in for f'(c) into the Mean Value Theorem.

$\begin{array}{rcl}f\left(b\right)& =& 10(15-5)+4\\ & =& 104\end{array}$

So, the maximum value that f(15) can be is 104.