Answered: A cone circumscribes a sphere of radius…

Algebra for College Students

10th Edition

ISBN:9781285195780

Author:Jerome E. Kaufmann, Karen L. Schwitters

Publisher:Jerome E. Kaufmann, Karen L. Schwitters

Chapter9: Polynomial And Rational Functions

Section9.4: Graphing Polynomial Functions

Problem 43PS

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Question

A cone circumscribes a sphere of radius 5 inches. If another sphere circumscribes this cone, what is the minimum surface area (in^2) of this sphere circumscribing the cone that has a smaller sphere inside of it?

Expert Solution

Step 1

A cone circumscribing a sphere of radius 5 inches.

Let the radius of the cone be "r" inches and height of the cone be "h" inches.

Then, the length of the slant side of the cone is: $\sqrt{r^{2} + h^{2}}$

As shown in the figure above;

Consider sine of the angle $θ$ .

From small right triangle with hypotenuse as $h - 5$ :

$\sin θ = \frac{5}{h - 5}$ ........(1)

From the large right triangle with hypotenuse as $\sqrt{r^{2} + h^{2}}$ :

$\sin θ = \frac{r}{\sqrt{r^{2} + h^{2}}}$ ...........(2)

Equate equations (1) and (2)

$\frac{r}{\sqrt{r^{2} + h^{2}}} = \frac{5}{h - 5} r (h - 5) = 5 \sqrt{r^{2} + h^{2}}$

Square on both sides.

$25 (r^{2} + h^{2}) = r^{2} (h^{2} - 10 h + 25) 25 r^{2} + 25 h^{2} = r^{2} h^{2} - 10 r^{2} h + 25 r^{2} (25 - r^{2}) h^{2} + 10 r^{2} h = 0 (25 - r^{2}) h + 10 r^{2} = 0 \Rightarrow h = \frac{- 10 r^{2}}{(25 - r^{2})}$

Step 2

A sphere circumscribing the cone above of radius r.

Let the radius of the sphere circumscribing be R inches.

As shown in the figure, the height of the cone is the sum of the segments of lengths $R a n d \sqrt{R^{2} - r^{2}}$

$\Rightarrow h = R + \sqrt{R^{2} - r^{2}}$

From the above obtained expression for height of the cone in terms of r:

$\frac{- 10 r^{2}}{(25 - r^{2})} = R + \sqrt{R^{2} - r^{2}} - \sqrt{R^{2} - r^{2}} = R + \frac{10 r^{2}}{(25 - r^{2})}$

Square on both sides:

$R^{2} - r^{2} = R^{2} + \frac{100 r^{4}}{{(25 - r^{2})}^{2}} + \frac{20 r^{2}}{(25 - r^{2})} R \Rightarrow \frac{20 r^{2}}{(25 - r^{2})} R = - r^{2} - \frac{100 r^{4}}{{(25 - r^{2})}^{2}} \Rightarrow R = \frac{- (25 - r^{2})}{20 r^{2}} [\frac{100 r^{2} + {(25 - r^{2})}^{2}}{{(25 - r^{2})}^{2}}] r^{2} \Rightarrow R = \frac{100 r^{2} + {(25 - r^{2})}^{2}}{20 (r^{2} - 25)} \Rightarrow R = \frac{r^{4} + 50 r^{2} + 625}{20 r^{2} - 500}$

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