Chapter 2.4, Problem 25E

To determine

To find: The instantaneous rate of change of position function $y=\frac{t+1}{t}$ in feet at $t=2$ seconds.

Answer to Problem 25E

The instantaneous rate of change of position function $y=\frac{t+1}{t}$ in feet at $t=2\sec$ is equal to $-\frac{1}{4}\text{ft}/\sec$.

Explanation of Solution

Given information:

The position function is $y=\frac{t+1}{t}$ 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 $y=\frac{t+1}{t}$ 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.

  $\begin{array}{c}f\left(2+h\right)=\frac{2+h+1}{2+h}\\ f\left(2+h\right)=\frac{3+h}{2+h}\end{array}$

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

  $\begin{array}{c}f\left(t\right)=\frac{t+1}{t}\\ f\left(2\right)=\frac{2+1}{2}\\ f\left(2\right)=\frac{3}{2}\end{array}$

Substitute $\frac{3+h}{2+h}$ for $f\left(2+h\right)$ and $\frac{3}{2}$ for $f\left(2\right)$ in the formula for slope.

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

Further simplify.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}=\underset{h\to 0}{\lim }\left(\frac{-1}{2\left(2+h\right)}\right)\\ =\frac{-1}{2\left(2+0\right)}\\ =-\frac{1}{4}\end{array}$

Therefore, the instantaneous rate of change of position function $y=\frac{t+1}{t}$ in feet at $t=2\sec$ is equal to $-\frac{1}{4}\text{ft}/\sec$.