Q. 1 - 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.

Calculate the average rate of change of the height of the bowling ball from $t=3$to $t=3+h$seconds in the cases where h is equal to $0.5, 0.25, 0.1,$and$0.01.$

Step-by-Step Solution

Answer

The average rate of change of the height of the bowling ball from $t=3$to $t=3+h$seconds at $h=0.5, 0.25, 0.1, \& 0.01$will be $-104,-100,-97.6,\&-96.16$respectively.

Step 1. Given information.

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

Given height values are $h=0.5, 0.25, 0.1, \& 0.01.$

Step 2. The average rate of change in the height

The average rate of change of the height of the bowling ball from $t=3$to $t=3+h$seconds is as follows.

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

Step 3. The average rate of change in the height for different values of h .

Substitute $h=0.5, 0.25, 0.1, \& 0.01$in the expression for the average rate of change of the height of the ball.

$$\begin{gathered} r=-16h-96 \\ {r}_{h=0.5}=-16\left(0.5\right)-96 \\ {r}_{h=0.5}=-104 \\ {r}_{h=0.25}=-16\left(0.25\right)-96 \\ {r}_{h=0.25}=-100 \\ {r}_{h=0.1}=-16\left(0.1\right)-96 \\ {r}_{h=0.1}=-97.6 \\ {r}_{h=0.01}=-16\left(0.01\right)-96 \\ {r}_{h=0.01}=-96.16 \end{gathered}$$