Q 128. - Precalculus Enhanced with Graphing Utilities

Question

Designing Fine Decorative Pieces A designer of decorative art plans to market solid gold spheres encased in clear crystal

cones. Each sphere is of fixed radius R and will be enclosed in a cone of height h and radius r. See the illustration.Many cones can be used to enclose the sphere, each having

a different slant angle u. The volume V of the cone can be expressed as a function of the slant angle $\theta$ of the cone as,

$V\left(\theta \right)=\frac{1}{3}\pi R^3\frac{{\left(1+sec\theta \right)}^3}{{\left(\tan \theta \right)}^3}$

where $\theta$ lies between $0°$ and $90°$.

What volume V is required to enclose a sphere of radius 2 cm in a cone whose slant angle $\theta$ is 30°? 45°? 60°?

Step-by-Step Solution

Answer

Volume for angle $30°$ is $\frac{8\pi }{3\sqrt{3}}\left(35+18\sqrt{3}\right)$ .Volume for angle $45°$ is $\frac{8\pi }{3}\left(11+5\sqrt{2}\right)$ .

Volume for angle $60°$ is $24\mathrm{\pi }$.

Step 1. Given information .

Consider the given given radius and angles.

Step 2. Finding volume for θ = 30 ° .

Substitute the radius 2 and angle $\theta =30°$in the given formula .

$$\begin{gathered} V\left(\theta \right)=\frac{1}{3}\pi R^3\frac{{\left(1+sec \theta \right)}^3}{{\left(\tan \theta \right)}^2} \\ V\left(30°\right)=\frac{1}{3}\pi {\left(2\right)}^3\frac{{\left(1+sec 30°\right)}^3}{{\left(\tan 30°\right)}^2} \\ =\frac{8\pi }{3\sqrt{3}}\left(35+18\sqrt{3}\right) \end{gathered}$$

Step 3. Finding the volume for θ = 45 ° .

Substitute the radius 2 and angle $45°$ in the given formula .

$$\begin{gathered} V\left(\theta \right)=\frac{1}{3}\pi R^3\frac{{\left(1+sec \theta \right)}^3}{{\left(\tan \theta \right)}^2} \\ V\left(45°\right)°=\frac{1}{3}\pi {\left(2\right)}^3\frac{{\left(1+sec 45°\right)}^3}{{\left(\tan 45°\right)}^2} \\ =\frac{8\pi }{3}\left(11+5\sqrt{2}\right) \end{gathered}$$

Step 4. Finding the volume for θ = 60 ° .

Substitute the radius 2 and angle $60°$ in the given formula .

$$\begin{gathered} V\left(\theta \right)=\frac{1}{3}\pi R^3\frac{{\left(1+ sec \theta °\right)}^3}{{\left(\tan \theta \right)}^2} \\ V\left(60°\right)=\frac{1}{3}\pi {\left(2\right)}^3\frac{{\left(1+ sec 60°\right)}^3}{{\left(\tan 60°\right)}^2} \\ =24\pi \end{gathered}$$