12.4 Derivatives  (Page 4/18)

The velocity at $t=1$ and $t=3$ is the instantaneous rate of change of distance per time, or velocity. Notice that the initial height is 6 feet. To find the instantaneous velocity, we find the derivative    and evaluate it at $t=1$ and $t=3:$

$$\begin{array}{ll}{f}^{\prime }(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}\hfill & \hfill \\ =\underset{h\to 0}{\lim }\frac{-16{(t+h)}^2+64(t+h)+6-(-16t^2+64t+6)}{h}\hfill & \text{Substitute }s(t+h)\text{ and }s(t).\hfill \\ =\underset{h\to 0}{\lim }\frac{-16t^2-32ht-h^2+64t+64h+6+16t^2-64t-6}{h}\hfill & \text{Distribute}.\hfill \\ =\underset{h\to 0}{\lim }\frac{-32ht-h^2+64h}{h}\hfill & \text{Simplify}.\hfill \\ =\underset{h\to 0}{\lim }\frac{\cancel{h}(-32t-h+64)}{\cancel{h}}\hfill & \text{Factor the numerator}.\hfill \\ =\underset{h\to 0}{\lim }-32t-h+64\hfill & \text{Cancel out the common factor }h.\hfill \\ s^{\prime }(t)=-32t+64\hfill & \text{Evaluate the limit by letting }h=0.\hfill \end{array}$$

For any value of $t$ , $s^{\prime }\left(t\right)$ tells us the velocity at that value of $t.$

Evaluate $t=1$ and $t=3.$

The velocity of the ball after 1 second is 32 feet per second, as it is on the way up.

The velocity of the ball after 3 seconds is $\mathrm{-32}$ feet per second, as it is on the way down.

To find the functional value, $f\left(a\right),$ find the y -coordinate at $x=a.$

To find the derivative    at $x=a,$ $f^{\prime }\left(a\right),$ draw a tangent line at $x=a,$ and estimate the slope of that tangent line. See [link] .

1. $f(0)$ is the y -coordinate at $x=0.$ The point has coordinates $\left(0,1\right),$ thus $f(0)=1.$
2. $f(2)$ is the y -coordinate at $x=2.$ The point has coordinates $\left(2,1\right),$ thus $f(2)=1.$
3. $f^{\prime }(0)$ is found by estimating the slope of the tangent line to the curve at $x=0.$ The tangent line to the curve at $x=0$ appears horizontal. Horizontal lines have a slope of 0, thus $f^{\prime }(0)=0.$
4. $f^{\prime }(2)$ is found by estimating the slope of the tangent line to the curve at $x=2.$ Observe the path of the tangent line to the curve at $x=2.$ As the $x$ value moves one unit to the right, the $y$ value moves up four units to another point on the line. Thus, the slope is 4, so $f^{\prime }(2)=4.$