Chapter 3.5, Problem 40PS

Solution

a.

To show: The radius $r=\frac{1}{\sqrt{\pi }}\sqrt{S+\frac{\pi }{4}}-\frac{1}{2}$ Where surface area $S$ of a right circular cone with a slant height of $1$ is given by $S=$ $S=\pi r+\pi r^2.$

Given information:

The given surface area is $S=\pi r+\pi r^2$ and height $h=1$ unit.

Formula used:

Surface area of cone is $S=\pi r+\pi r^2.$

Quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculations:

Here surface area $S=\pi r+\pi r^2.$

Re write the equation $S=\pi r+\pi r^2\Rightarrow r^2+r-\frac{s}{\pi }=0$

Solve for $r$ by using quadratic formula:

  $\begin{array}{c}r=\frac{-1\pm \sqrt{1+4\frac{S}{\pi }}}{2}\\ =\frac{-1}{2}\pm \sqrt{\frac{\pi +4S}{4\pi }}\\ =\frac{-1}{2}+\frac{1}{\sqrt{\pi }}\sqrt{\frac{\pi }{4}+S},\sin cerisapositivevalue\end{array}$

Thus $r=\frac{1}{\sqrt{\pi }}\sqrt{S+\frac{\pi }{4}}-\frac{1}{2}$ is obtained from the surface area of the cone.

b.

To plot : The equation $r=\frac{1}{\sqrt{\pi }}\sqrt{S+\frac{\pi }{4}}-\frac{1}{2}$ Where surface area

  $S$ of a right circular cone.

Given information:

The given $r=\frac{1}{\sqrt{\pi }}\sqrt{S+\frac{\pi }{4}}-\frac{1}{2}$ Where surface area

  $S$ of a right circular cone.

Plot the graph:

Consider the equation The equation $r=\frac{1}{\sqrt{\pi }}\sqrt{S+\frac{\pi }{4}}-\frac{1}{2}$ Where surface area $S$ of a right circular cone.

The graph of the equation is given below.

  

c.

To find: The radius if surface area $S=\frac{3\pi }{4}\text{square}\text{units}$ of a right circular cone with a slant height of $1$ is given by $S=$

  $S=\pi r+\pi r^2.$

The radius $r=0.50$ units.

Given information:

The given surface area is $S=\frac{3\pi }{4}\text{square}\text{units}$ and height $h=1$ unit.

Formula used:

Surface area of cone is $S=\pi r+\pi r^2.$

Quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculations:

Here surface area $S=\frac{3\pi }{4}$

Substitute $S=\frac{3\pi }{4}$ in to the equation .

  $\begin{array}{c}S=\pi r+\pi r^2\\ \frac{3\pi }{4}=\pi \left(r+r^2\right)\\ 4\left(r^2+r\right)=3\\ 4r^2+4r-3=0\end{array}$

Solve for $r$ by using quadratic formula:

  $\begin{array}{c}4r^2+4r-3=0\\ r=\frac{-4\pm \sqrt{16+4\times 4\times 3}}{2\times 4}\\ =\frac{-4\pm 8}{8}\sin cerisapositivenumber\\ =\frac{4}{8}\\ =0.50unit.\end{array}$

  $\begin{array}{c}r=\frac{-1\pm \sqrt{1+4\frac{S}{\pi }}}{2}\\ =\frac{-1}{2}\pm \sqrt{\frac{\pi +4S}{4\pi }}\\ =\frac{-1}{2}+\frac{1}{\sqrt{\pi }}\sqrt{\frac{\pi }{4}+S},\sin cerisapositivevalue\end{array}$

Thus the radius $r=0.50$ units.