Chapter 11.1, Problem 15E

Solution

a)

To find: The expression for the function $f^{\text{'}}\left(t\right)$ .

The required expression is $f^{\text{'}}\left(t\right)=48-32t$ .

Given information:

The distance the rock is above at any time $t$ is given by $f\left(t\right)$ .

  $f\left(t\right)=3+48t-16t^2$

Formula used:

The derivative of the function $f$ at $x=a$ is donated by $f\text{'}\left(a\right)$ .

  $f\text{'}\left(a\right)=\underset{h\to a}{{\displaystyle \lim }}\frac{f\left(a+h\right)-f\left(a\right)}{h}$

Calculation:

Calculate the derivative of the function $f$ at $x=t$ is donated by $f\text{'}\left(t\right)$ .

  $\begin{array}{c}f\text{'}\left(t\right)=\underset{h\to 0}{{\displaystyle \lim }}\frac{f\left(t+h\right)-f\left(t\right)}{h}\\ =\underset{h\to 0}{{\displaystyle \lim }}\frac{3+48\left(t+h\right)-16{\left(t+h\right)}^2-\left(3+48t-16t^2\right)}{h}\\ =\underset{h\to 0}{{\displaystyle \lim }}\frac{48h-16h^2-32th}{h}\\ =\underset{h\to 0}{{\displaystyle \lim }}48-16h-32t\\ =48-32t\end{array}$

Therefore, the required expression is $f^{\text{'}}\left(t\right)=48-32t$ .

b)

To find: The initial velocity of the rock.

The initial velocity of the rock is $48\text{ft}/\sec$ .

Given information:

The distance the rock is above at any time $t$ is given by $f\left(t\right)$ .

  $f\left(t\right)=3+48t-16t^2$

Formula used:

The derivative of the function $f$ at $x=a$ is donated by $f\text{'}\left(a\right)$ .

  $f\text{'}\left(a\right)=\underset{h\to a}{{\displaystyle \lim }}\frac{f\left(a+h\right)-f\left(a\right)}{h}$

Calculation:

As calculated above, $f^{\text{'}}\left(t\right)=48-32t$ . The velocity is the $f^{\text{'}}\left(t\right)$ .

Calculate $f^{\text{'}}\left(0\right)$ for initial velocity at $t=0$ .

  $\begin{array}{c}{f}^{\text{'}}\left(0\right)=48-32\times 0\\ =48\end{array}$

Therefore, the initial velocity of the rock is $48\text{ft}/\sec$ .