Question

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

(a) Follow the steps in Exercise 93 to show that the volume of the cup as a function of$\theta$isrole="math" localid="1648730117007" $\mathit{V}\mathbf{\left(}\mathit{\theta }\mathbf{\right)}\mathbf{=}\frac{\mathbf{9}}{{\mathbf{\pi }}^{\mathbf{2}}}{\mathit{\theta }}^{\mathbf{2}}\sqrt{\mathbf{4}{\mathbf{\pi }}^{\mathbf{2}}\mathbf{-}{\mathbf{\theta }}^{\mathbf{2}}}\mathbf{,}\mathbf{0}\mathbf{<}\mathit{\theta }\mathbf{<}\mathbf{2}\mathit{\pi }$.

(b) Graph the function $\mathit{V}$.

(c) For$\mathit{c}$what angle is the volume of the cup a maximum?

Short answer

(a). The final function:$\mathit{V}\mathbf{=}\frac{\mathbf{9}{\mathbf{\theta }}^{\mathbf{2}}}{{\mathbf{\pi }}^{\mathbf{2}}}\mathbf{\cdot }\sqrt{\mathbf{4}{\mathbf{\pi }}^{\mathbf{2}}\mathbf{-}{\mathbf{\theta }}^{\mathbf{2}}}$

(b). Angle is 5.13,87.062.

(c). The conical cup's volume is 5.13.

Step-by-step solution

a. Step 1. Given information.

Find the volume function$V$in terms of$\theta$ where$V$ is the volume of the cubical cup and theta is the major sector angle.

The given values:

The radius of the circular paper is 6 cm and the major sector angle$\theta =\frac{5\pi }{3}$.

To find:

The volume function in terms of$\theta$ is that $v\left(\theta \right)$.

a. Step 2. Formula Used

Formula used$C=r\theta ,V=\frac{1}{3}*\pi *r^2*h$

a. Step 3. Find the final function

$\begin{array}{l}\Rightarrow C=R\theta \\ \Rightarrow R=\frac{C}{\theta }\\ \Rightarrow r=\frac{6\theta }{2\pi }\\ \Rightarrow h=\sqrt{6^2-{\left(\frac{3\theta }{\pi }\right)}^2}\end{array}$

Put the above values in this formula of volume.

$\Rightarrow V=\frac{9{\theta }^2}{{\pi }^2}\cdot \sqrt{4{\pi }^2-{\theta }^2}$

b. Step 1. Given information.

Plot the graph for the given function$V$ by finding the different values of$V\left(\theta \right)$for different values of$\theta$.

Use the function

$V\left(\theta \right)=\frac{9{\theta }^2}{{\pi }^2}\sqrt{4{\pi }^2-{\theta }^2}\text{atcondition}\{0\le \theta \le 2\pi \}$

Given:
The function of volume$v\left(\theta \right)$

To find:

The graph between the function$v\left(\theta \right)$ and $\theta$.

b. Step 2. Finding the maximum value of Volume

Solve $V\left(\theta \right)$for different values of$\theta$such as for$\theta =5.13$ radian, 0radian, and$2\pi$.

$\begin{array}{l}\Rightarrow V\left(\theta \right)=\frac{9{\theta }^2}{{\pi }^2}\sqrt{4{\pi }^2-{\theta }^2}\\ \Rightarrow V\left(\theta \right)=\frac{9*{\left(5.13\right)}^2}{{\pi }^2}*\sqrt{4{\pi }^2-{\left(5.13\right)}^2}\end{array}$

$V\left(5.13\right)=87.062isthemaximumvalueofvolume.$

$V\left(0\right)=0andV\left(2\pi \right)=0$.

b. Step 3. Plotting the graph

Plot the values of $V\left(\theta \right)$for values of$\theta$.

The graph is:

c. Step 1. Given information

To find:
The value of angle for the maximum value of the conical cup's volume

To get the maximum value use the graph between $V\left(\theta \right)$

role="math" localid="1648730803468" $V\left(\theta \right)=\frac{9{\theta }^2}{{\pi }^2}\sqrt{4{\pi }^2-{\theta }^2}$and its graph.

c. Step 2. Draw the graph.

The graph is:

Hence, we can see that value of angle for the maximum value of the cup's volume in the graph is 5.13