Chapter 6.1, Problem 88E

Conical Cup In this exercise we find the volume of the conical cup in Exercise 93 for any angle θ.

1. (a) Follow the steps in Exercise 93 to show that the volume of the cup as a function of θ is

$V(\theta )=\frac{9}{{\pi }^2}{\theta }^2\sqrt{4{\pi }^2-{\theta }^2},0<\theta <2\pi$

1. (b) Graph the function V.
2. (c) For what angle θ is the volume of the cup a maximum?

To determine

To show: The volume of the cup as a function of $\theta$ is $V\left(\theta \right)=\frac{9}{{\pi }^2}{\theta }^2\sqrt{4{\pi }^2-{\theta }^2},0<\theta <2\pi$.

Explanation of Solution

From Exercise $93$, the radius of the circular piece of paper is $6\mathrm{cm}$ and slant height of the cone is $6\mathrm{cm}$.

The radius and height of the cone to be taken in terms of $\theta$.

Formula for arc length is $s=r\theta$.

Substitute $6$ for $r$ in $s=r\theta$ as $s=6\theta$.

The arc length is $6\theta$.

Circumference of the circle is $2\pi r$ which is the total arc length of the circle. That is,

$\begin{array}{l}2\pi r=s\\ 2\pi r=6\theta \\ r=\frac{6\theta }{2\pi }\end{array}$

The radius of the circle in terms of $\theta$ is $\frac{6\theta }{2\pi }$.

Formula for the Pythagorean Theorem is $l^2=h^2+r^2$.

Substitute $6$ for $l$ and $\frac{6\theta }{2\pi }$ for $h$ in $l^2=h^2+r^2$.

$\begin{array}{l}{\left(6\right)}^2=h^2+{\left(\frac{6\theta }{2\pi }\right)}^2\\ h=\sqrt{36-\frac{36{\theta }^2}{4{\pi }^2}}\\ h=\sqrt{\frac{36}{4{\pi }^2}\left(4{\pi }^2-{\theta }^2\right)}\\ h=\frac{3}{\pi }\sqrt{4{\pi }^2-{\theta }^2}\end{array}$

The height of the cone in terms of $\theta$ is $\frac{3}{\pi }\sqrt{4{\pi }^2-{\theta }^2}$.

Formula for the volume of cone is $\mathrm{V}=\frac{1}{3}\pi r^{2}h$.

Substitute $\frac{6\theta }{2\pi }$ for $r$ and $\frac{3}{\pi }\sqrt{4{\pi }^2-{\theta }^2}$ for $h$.

$\begin{array}{c}V\left(\theta \right)=\frac{1}{3}\times \pi \times {\left(\frac{6\theta }{2\pi }\right)}^2\times \frac{3}{\pi }\sqrt{4{\pi }^2-{\theta }^2}\\ ={\left(\frac{3\theta }{\pi }\right)}^2\sqrt{4{\pi }^2-{\theta }^2}\\ =\frac{9{\theta }^2}{{\pi }^2}\sqrt{4{\pi }^2-{\theta }^2}\end{array}$

Hence, the volume of the cup as a function of $\theta$ is $V\left(\theta \right)=\frac{9{\theta }^2}{{\pi }^2}\sqrt{4{\pi }^2-{\theta }^2}$.

To determine

To sketch: A graph of the function $V\left(\theta \right)=\frac{9}{{\pi }^2}{\theta }^2\sqrt{4{\pi }^2-{\theta }^2},0<\theta <2\pi$.

Explanation of Solution

The given function for volume of the cup is $V\left(\theta \right)=\frac{9}{{\pi }^2}{\theta }^2\sqrt{4{\pi }^2-{\theta }^2}$.

The domain for $\theta$ is between $0$ and $2\pi$, for $0<\theta <2\pi$.

Graph $V\left(\theta \right)=\frac{9}{{\pi }^2}{\theta }^2\sqrt{4{\pi }^2-{\theta }^2}$ in the domain $0$ and $2\pi$ as shown below:

From Figure 1, the graph is in shape of sharp parabolas between $0$ and $2\pi$.

To determine

The angle $\theta$ for which the volume of cup is maximum.

Answer to Problem 88E

The volume of the cup is maximum at $\underset{¯}{\theta =5.13}$.

Explanation of Solution

From Figure 1, note that the cup has maximum volume 87.062 at 5.13. That is, when $\theta =5.13$.

Thus, the volume of the cup is maximum at $\underset{¯}{\theta =5.13}$.