Chapter 0.2, Problem 55E

a.

To determine

To explain : circumference of the base of the cone $8\pi -x$ .Here circular piece of paper with $4$ in. radius, $a.$ cut out a paper with an arc length of $x$ .Join the two edges of the remaining portion to form a cone with radius $r$ and height $h$ .

Explanation of Solution

Given information:

The circumference of a base of a cone $8\pi -x=2\pi r$ is equal to diameter .

And the radius is $4$ inches and arc length is $x$ .

  

Concept used:

The formula for circumference of a base of a cone $C=2\pi r.$

Calculation:

Here radius is $r=4$ in.

Here cone is made from a sector with an arc length $x$ .

Thus the circumference of a base of a cone is the difference between circumference of a circle and arc length of that cut out sector.

  $\begin{array}{c}C=2\pi r-x\\ =8\pi -x\end{array}$

Thus the circumference of a circle $C=8\pi -x\text{inches}\text{.}$

b.

To determine

To write: Expression for radius $r$ in terms of $x$ .

Explanation of Solution

Given information:

The circumference is $C=8\pi -x$

Concept used:

The formula for circumference of a base of a cone $C=2\pi r.$

Calculation:

The circumference is $C=8\pi -x$

  $\begin{array}{l}2\pi r=8\pi -x\\ r=4-\frac{x}{2\pi }\end{array}$

Thus the expression $r=4-\frac{x}{2\pi }\text{cm}\text{.}$

c.

To determine

To write: The expression for height $h$ in terms of $x,\text{where}r=4-\frac{x}{2\pi }.$

Explanation of Solution

Given information:

  $r=4-\frac{x}{2\pi }$

Concept used:

Pythagorean theorem:

The sum of the square of the legs of a right triangle is equal to the square of the hypotenuse.

Calculation:

Applying Pythagorean theorem,

Then will get $h^2=16-r^2$

Substitute $r=4-\frac{x}{2\pi }$ into the equation $h^2=16-r^2$

  $\begin{array}{c}{h}^2=16-r^2\\ =16-{\left(4-\frac{x}{2\pi }\right)}^2\\ =16-16+\frac{4x}{\pi }+\frac{x^2}{4{\pi }^2}\\ =\frac{16\pi x+x^2}{4{\pi }^2}\\ h=\pm \frac{\sqrt{16\pi x+x^2}}{2\pi }\\ h=\frac{\sqrt{16\pi x+x^2}}{2\pi }\end{array}$

Thus the height is $h=\frac{\sqrt{16\pi x+x^2}}{2\pi }$ cm.

c.

To determine

To write: The expression for volume as a function of $x,\text{where}r=4-\frac{x}{2\pi },h=\frac{\sqrt{16\pi x+x^2}}{2\pi }.$ .

Explanation of Solution

Given information:

The radius $x,\text{where}r=4-\frac{x}{2\pi },h=\frac{\sqrt{16\pi x+x^2}}{2\pi }.$ ,

Concept used: . The volume of a cone $V=\frac{1}{3}\pi r^{2}h.$

Calculation: Here, $r=4-\frac{x}{2\pi },h=\frac{\sqrt{16\pi x+x^2}}{2\pi }.$

Substitute $r=4-\frac{x}{2\pi },h=\frac{\sqrt{16\pi x+x^2}}{2\pi }.$ into the equation $V=\frac{1}{3}\pi r^{2}h.$

  $\begin{array}{c}V=\frac{1}{3}\pi r^{2}h.\\ =\frac{1}{3}\times \pi \times {\left(4-\frac{x}{2\pi }\right)}^2\times \frac{\sqrt{16\pi x+x^2}}{2\pi }\\ =\frac{1}{24{\pi }^2}{\left(8\pi -x\right)}^2\left(\sqrt{16\pi x+x^2}\right)\end{array}$

Thus volume as a function of $x$ is $V\left(x\right)=\frac{1}{24{\pi }^2}{\left(8\pi -x\right)}^2\left(\sqrt{16\pi x+x^2}\right)$