Chapter 2.4, Problem 26E

To determine

To find: The instantaneous rate of change of position function $y=t^3-1$ in feet at $t=2$ seconds.

Answer to Problem 26E

The instantaneous rate of change of position function $y=t^3-1$ in feet at $t=2\sec$ is equal to $12\text{ft}/\sec$.

Explanation of Solution

Given information:

The position function is $y=t^3-1$ and the time is $t=2$ seconds.

Calculation:

The formula for the instantaneous rate of change of position with respect to time at $t$ seconds is equal to $\underset{h\to 0}{\lim }\frac{f\left(t+h\right)-f\left(t\right)}{h}$.

The formula for the instantaneous rate of change of $y=t^3-1$ at $t=2$ is equal to $\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}$.

Substitute $2+h$ for $x$ in the function.

  $\begin{array}{c}f\left(t\right)=t^3-1\\ f\left(2+h\right)={\left(2+h\right)}^3-1\\ f\left(2+h\right)=8+h^3+3\left(2h^2\right)+3\left(4h\right)-1\\ f\left(2+h\right)=h^3+6h^2+12h+7\end{array}$

Substitute $2$ for $x$ in the function.

  $\begin{array}{l}f\left(t\right)=t^3-1\\ f\left(2\right)=2^3-1\\ f\left(2\right)=8-1\\ f\left(2\right)=7\end{array}$

Substitute $h^3+6h^2+12h+7$ for $f\left(2+h\right)$ and $7$ for $f\left(2\right)$ in the formula for slope.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}=\underset{h\to 0}{\lim }\frac{h^3+6h^2+12h+7-7}{h}\\ =\underset{h\to 0}{\lim }\frac{h^3+6h^2+12h}{h}\\ =\underset{h\to 0}{\lim }\left(\frac{h\left(h^2+6h+12\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(h^2+6h+12\right)\end{array}$

Further simplify,

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}=0+2\left(0\right)+12\\ =12\end{array}$

Therefore, the instantaneous rate of change of position function $y=t^3-1$ in feet at $t=2\sec$ is equal to $12\text{ft}/\sec$.