Question

A spherical balloon is being inflated and the radius of the balloon is increasing at a rate of \(2\;cm/s\).

(a) Express the radius \(r\)of the balloon as a function of the time (in seconds).

(b) If \(V\)is the volume of the balloon as a function of the radius, find \(V \circ r\)and interpret it.

Short answer

(a) The radius \(r\)of the balloon as a function of thetime\(2t\).

(b) For volume of the spherical balloon \(V\) and its radius\(r\), \(V \circ r\) is \(\frac{{32}}{3}\pi {t^3}\) and represents the volume as a function of time.

Step-by-step solution

Given data

The radius of a spherical balloon is increasing at a rate of\(2\;{\rm{cm/s}}\).

Composition of two functions

For two functions \(V\left( r \right)\) and\(r\left( t \right)\)their composition is defined as

\(V \circ r = V\left( {r\left( t \right)} \right)\;\;\;\;\;.....\left( 1 \right)\)

Radius as a function of time

The rate of change of radius is

\(\frac{{dr}}{{dt}} = 2\)

Integrate to get radius as a function of time

\(r = 2t\)

The radius as a function of time is \(r = 2t\).

Volume as a function of radius (b)

Volume of a sphere as a function of radius is

\(V\left( r \right) = \frac{4}{3}\pi {r^3}\)

Thus, from equation (1),

\(\begin{array}{c}V \circ r = V\left( {2t} \right)\\ = \frac{4}{3}\pi {\left( {2t} \right)^3}\\ = \frac{{32}}{3}\pi {t^3}\end{array}\)

Thus, the composition of volume and radius is \(\frac{{32}}{3}\pi {t^3}\) which represents the volume as a function of time.