Example 6 from Algebra and Trigonometry, 5.7 Inverses and Radical Functions

Solving an Application with a Cubic Function

A mound of gravel is in the shape of a cone with the height equal to twice the radius. The volume of the cone in terms of the radius is given by

V = \frac{2}{3} π r^{3}

Find the inverse of the function $V = \frac{2}{3} π r^{3}$ that determines the volume $V$ of a cone and is a function of the radius $r .$ Then use the inverse function to calculate the radius of such a mound of gravel measuring 100 cubic feet. Use $π = 3.14.$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Start with the given function for $V .$ Notice that the meaningful domain for the function is $r > 0$ since negative radii would not make sense in this context nor would a radius of 0. Also note the range of the function (hence, the domain of the inverse function) is $V > 0.$ Solve for $r$ in terms of $V,$ using the method outlined previously. Note that in real-world applications, we do not swap the variables when finding inverses. Instead, we change which variable is considered to be the independent variable.

$\begin{matrix} V & = & \frac{2}{3} π r^{3} \\ r^{3} & = & \frac{3 V}{2 π} & Solve for r^{3} . \\ r & = & \sqrt[3]{\frac{3 V}{2 π}} & Solve for r . \end{matrix}$
This is the result stated in the section opener. Now evaluate this for $V = 100$ and $π = 3.14.$

$\begin{matrix} r & = & \sqrt[3]{\frac{3 V}{2 π}} \\ = & \sqrt[3]{\frac{3 \cdot 100}{2 \cdot 3.14}} \\ \approx & \sqrt[3]{47.7707} \\ \approx & 3.63 \end{matrix}$

Therefore, the radius is about 3.63 ft.

How did it go?

Solution

Solution