Q. 3 - Calculus

Question

A limit representing an instantaneous rate of change: After t seconds, a bowling ball dropped from $350$feet has height $h\left(t\right)=350-16t^2,$measured in feet.

Take the limit as $h\to 0^{+}$of the formula you found for average rate of change in the previous problem. What does this limit represent in real-world terms?

Step-by-Step Solution

Answer

Limit $h\to 0^{+}$represents the initial time of observation of the average rate of change in height that is $-96.$

Step 1. Given information.

The given function for the height is $h\left(t\right)=350-16t^2.$

Given limit is  $h\to 0^{+}$

Step 2. The average rate of change in the height

The average rate of change in the height of the bowling ball is as follows.

$$\begin{gathered} r=\frac{h(h+3)-h\left(3\right)}{(t+3)-3} \\ r=\frac{\left[350-16{(h+3)}^2\right]-\left[350-16\left(3^2\right)\right]}{h+3-3} \\ r=\frac{350-16h^2-96h-144-350+144}{h} \\ r=\frac{-16h^2-96h}{h} \\ r=-16h-96 \end{gathered}$$

Take limit $h\to 0^{+}$for The average rate of change in the height.

$$\begin{gathered} \underset{h\to 0^{+}}{\lim }r=\underset{h\to 0^{+}}{\lim }\left(-16h-96\right) \\ =-16\left(0\right)-96 \\ =-96 \end{gathered}$$

Limit $h\to 0^{+}$represents the initial time of observation of the average rate of change in height that is $-96.$