Chapter 4.4, Problem 64E

a.

To determine

Find a formula for the volume $V$ of the cone in terms

Answer to Problem 64E

The volume of cone is $V(x)=\frac{\pi }{3}{\left(\frac{2\pi a-x}{2\pi }\right)}^2\sqrt{a^2-{\left(\frac{2\pi a-x}{2\pi }\right)}^2}$ .

Explanation of Solution

Given:

  $\text{Height of cone }=h,$

  $\text{Radius of cone }=r$ .

A circular disk of radius $a$ inch.

Also, given that the cone is constructed by removing a sector $AOC$ of arc length $x$ in.

Concept used:

Circumference of circle of radius $a$ = $2\pi a$ units.

Calculation:

Circumference of circular disk of radius $a$ = $2\pi a$ inch.

Circumference of circular disk by removing a sector $AOC$ of arc length $x$ in $=\left(2\pi a-x\right)\text{in}$

The circumference of base of cone with height $h$ and radius $r=2\pi r$

As given that the cone is constructed by removing a sector $AOC$ of arc length $x$ in then

  $\begin{array}{c}\text{The circumference of base of cone }=\text{ Circumference of circular disk}\\ 2\pi r=\left(2\pi a-x\right)\\ r=\frac{2\pi a-x}{2\pi }\end{array}$

Height of cone in the form of $x$and $a$:

  

In triangle $PQR,$

  $\begin{array}{c}{h}^2+r^2=a^2\\ h^2=a^2-r^2\\ h=\sqrt{a^2-{\left(\frac{2\pi a-x}{2\pi }\right)}^2}in\end{array}$

Volume of cone:

  $\begin{array}{c}V(x)=\frac{1}{3}\pi r^{2}h\\ V(x)=\frac{\pi }{3}{\left(\frac{2\pi a-x}{2\pi }\right)}^2\sqrt{a^2-{\left(\frac{2\pi a-x}{2\pi }\right)}^2}i{n}^3.\end{array}$

Conclusion:

The volume of cone is $V(x)=\frac{\pi }{3}{\left(\frac{2\pi a-x}{2\pi }\right)}^2\sqrt{a^2-{\left(\frac{2\pi a-x}{2\pi }\right)}^2}$ cubic in.

b.

To determine

Find the $r\text{ and }h$ in the cone of maximum volume for $a=4,5,6,8$ .

Answer to Problem 64E

The values of radius and height :

$a$	$r(in.)$	$h(in.)$
4	3.265	2.309
5	4.082	2.887
6	4.899	3.464
8	6.532	4.619

Explanation of Solution

Given:

Radius of cone: $r=\frac{2\pi a-x}{2\pi }$

Height of cone: $h=\sqrt{a^2-{\left(\frac{2\pi a-x}{2\pi }\right)}^2}$

Volume of cone: $V(x)=\frac{\pi }{3}{\left(\frac{2\pi a-x}{2\pi }\right)}^2\sqrt{a^2-{\left(\frac{2\pi a-x}{2\pi }\right)}^2}$

Values of $a=4,5,6,8$ respectively.

Calculation:

Case 1:- when $a=4,$

Volume of cone:

Put $a=4,$

  $\begin{array}{c}V(x)=\frac{\pi }{3}{\left(\frac{2\pi a-x}{2\pi }\right)}^2\sqrt{a^2-{\left(\frac{2\pi a-x}{2\pi }\right)}^2}\\ V(x)=\frac{\pi }{3}{\left(\frac{2\pi \left(4\right)-x}{2\pi }\right)}^2\sqrt{{\left(4\right)}^2-{\left(\frac{2\pi \left(4\right)-x}{2\pi }\right)}^2}\\ V(x)=\frac{\pi }{3}{\left(\frac{8\pi -x}{2\pi }\right)}^2\sqrt{16-{\left(\frac{8\pi -x}{2\pi }\right)}^2}\end{array}$

Draw the graph of $V(x)$ ,

  

Here, $X-axis$ denotes values of $x$ .

  $Y-axis$ denotes values of $V\left(x\right)$ .

Now, according to graph $V(x)$ is maximum at $x=4.612,45.654$ .

Case 1:- when $a=4,x=4.612$

Put $a=4,x=4.612$

Radius of cone:

  $\begin{array}{c}r=\frac{2\pi a-x}{2\pi }\\ r=\frac{2\pi \left(4\right)-4.612}{2\pi }\\ r=3.265in\end{array}$

Height of cone:

  $\begin{array}{l}h=\sqrt{a^2-{\left(\frac{2\pi a-x}{2\pi }\right)}^2}\\ h=\sqrt{4^2-{\left(\frac{2\pi \left(4\right)-4.612}{2\pi }\right)}^2}\\ h=2.309in\end{array}$

Case 2:- when $a=4,x=45.654$

Put $a=4,x=45.654$

Radius of cone:

  $\begin{array}{c}r=\frac{2\pi a-x}{2\pi }\\ r=\frac{2\pi \left(4\right)-45.654}{2\pi }\\ r=3.265in\end{array}$

Height of cone:

  $\begin{array}{l}h=\sqrt{a^2-{\left(\frac{2\pi a-x}{2\pi }\right)}^2}\\ h=\sqrt{4^2-{\left(\frac{2\pi \left(4\right)-\text{45}\text{.654}}{2\pi }\right)}^2}\\ h=2.309\end{array}$

So, $\text{Radius }=3.265 \text{in and height}=2.309 \text{in, } \text{if }a=4$ .

Similarly, solve for the remaining values of $a$ .

$a$	$r(in.)$	$h(in.)$
4	3.265	2.309
5	4.082	2.887
6	4.899	3.464
8	6.532	4.619

c.

To determine

Find a relation between $r\text{ and }h$ .

Answer to Problem 64E

The relation between radius $r$ and height $h$ is $r^2=2h^2$ .

Explanation of Solution

Calculation:

Given that,

  

In triangle $PQR,$

  $\begin{array}{c}{h}^2+r^2=a^2\\ r^2=a^2-h^2\end{array}$

Also, volume of cone:

  $V=\frac{1}{3}\pi r^{2}h$

Simplify it,

  $\begin{array}{c}V=\frac{1}{3}\pi r^{2}h\\ =\frac{1}{3}\pi h\left(a^2-h^2\right)\\ =\frac{1}{3}\pi \left(a^{2}h-h^3\right)\end{array}$

For maxima minima differentiate the $V$ with respect to $h$ ,

  $\begin{array}{c}V=\frac{1}{3}\pi \left(a^{2}h-h^3\right)\\ \frac{d}{dh}\left(V\right)=\frac{d}{dh}\left(\frac{1}{3}\pi \left(a^{2}h-h^3\right)\right)\\ \frac{dV}{dh}=\frac{1}{3}\pi \left(a^2-3h^2\right)\end{array}$

Put $\frac{dV}{dh}=0$ ,

  $\begin{array}{c}\frac{dV}{dh}=0\\ \frac{1}{3}\pi \left(a^2-3h^2\right)=0\\ \left(a^2-3h^2\right)=0\\ a^2=3h^2\end{array}$

Now,

  $\begin{array}{l}{r}^2=a^2-h^2\\ r^2=3h^2-h^2\\ r^2=2h^2\end{array}$

This is the relation between radius $r$ and height $h$ .

To get this relation first write the radius $r$ in the form of height $h$ then put the values in the formula of volume of cone.

For maximum value of volume differentiate the equation. Now simplify it for relation.

Conclusion:

The relation between radius $r$ and height $h$ is $r^2=2h^2$ .