Chapter 3.6, Problem 7MRPS

Solution

a.

To state: The inverse function if an object is launched upward from the ground level and reaches a maximum height of h feet The initial velocity v (in feet per second) of the object is given by the function $v=8\sqrt{h}$ .

The inverse function is $h=\frac{v^2}{64}$ .

Given information:

The given function is $v=8\sqrt{h}$ .

Calculation:

Write the given function:

  $v=8\sqrt{h}$ .

Divide both sides of the equation by 8,

  $\begin{array}{c}\frac{v}{8}=\frac{8\sqrt{h}}{8}\\ \frac{v}{8}=\sqrt{h}\end{array}$

Square both sides of the equation,

  $\begin{array}{c}{\left(\frac{v}{8}\right)}^2={\left(\sqrt{h}\right)}^2\\ \frac{v^2}{64}=h\\ h=\frac{v^2}{64}\end{array}$

Therefore, the inverse function is $h=\frac{v^2}{64}$ .

b.

To state: A problem that can be solved using an inverse function also show how to solve the problem.

The problem is “The time period of a pendulum, in seconds, can be calculated using the function $T=1.11\sqrt{l}$ , where l is the length of the pendulum in inches. Find the inverse function which gives the length in terms of the time period”.

Given information:

The given statement says to write a problem that can be solved using the inverse function also show how to solve the problem.

Explanation:

The time period of a pendulum, in seconds, can be calculated using the function $T=1.11\sqrt{l}$ , where l is the length of the pendulum in inches. Find the inverse function which gives the length in terms of the time period.

The function is $T=1.11\sqrt{l}$ .

Divide both sides of the function by 1.11,

  $\begin{array}{c}\frac{T}{1.11}=\frac{1.11\sqrt{l}}{1.11}\\ \frac{T}{1.11}=\sqrt{l}\end{array}$

Square both sides of the function,

  $\begin{array}{c}{\left(\frac{T}{1.11}\right)}^2={\left(\sqrt{l}\right)}^2\\ {\left(\frac{T}{1.11}\right)}^2=l\\ l={\left(\frac{T}{1.11}\right)}^2\end{array}$

Therefore, the inverse function is $l={\left(\frac{T}{1.11}\right)}^2$ .