Question

A spherical balloon is being inflated. Find the rate of increase of the surface area \(\left( {S = 4\pi {r^2}} \right)\) with respect to the radius \(r\) when \(r\) is (a) 1 ft, (b) 2 ft, and (c) 3 ft. What conclusion can you make?

Short answer

(a).\(8\pi {\rm{ f}}{{\rm{t}}^2}{\rm{/ft}}\)

(b).\(16\pi {\rm{ f}}{{\rm{t}}^2}{\rm{/ft}}\)

(c).\(24\pi {\rm{ f}}{{\rm{t}}^2}{\rm{/ft}}\)

Step-by-step solution

Surface Area of the Sphere

The Surface Area of a Spherecan be defined as the total amount of surface occupied by any spherical objectin any three dimensional space where the projection of the surface is in two dimensional plane

Deriving function for the surface area of the sphere:

(a)

It isgiven that the spherical balloon is being inflated.

Then, the rate of increase of surface area with respect to radius is given by:

\(\begin{array}{c}S = 4\pi {r^2}\\\frac{{dS}}{{dr}} = 8\pi r\end{array}\)

Now, the rate of increase of surface area when radius is 1 feet is given as:

\(S'\left( r \right) = 8\pi r = 8\pi {\rm{ f}}{{\rm{t}}^2}{\rm{/ft}}\)

Calculating this rate of change at 2 feet:

(b)

The rate of increase of surface area when radius is 2 feet is given as:

\(\begin{array}{l}S'\left( r \right) = 8\pi r\\S'\left( 2 \right) = 16\pi {\rm{ f}}{{\rm{t}}^2}{\rm{/ft}}\end{array}\)

Calculating this rate of change at 3 feet:

(c)

Now, the rate of increase of surface area when radius is 3 feet is given as:

\(\begin{array}{l}S'\left( r \right) = 8\pi r\\S'\left( 3 \right) = 24\pi {\rm{ f}}{{\rm{t}}^2}{\rm{/ft}}\end{array}\)

Hence, it can be concluded that the surface area of the balloon is increasing when the radius of the sphere is increasing.