Question

Helium is pumped into a spherical balloon at the rate of 3pi * m ^ 3 / min . At what rate is the radius of the balloon increasing when the volume is 36pi cubic meters?

Answer 1

To find:

At what rate is the radius of the balloon increasing when the volume is 36pi cubic meters.

Solution:

It is known that the volume of the sphere is given by:

$V=\frac{4}{3}\pi r^3$

Differentiate the volume with respect to time:

$\begin{gathered} \frac{dV}{dt}=\frac{4}{3}\times3\pi r^2(\frac{dr}{dt}) \\ \frac{dV}{dt}=4\pi r^2(\frac{dr}{dt}) \end{gathered}$

Here, given that dV/dt = 3pi, V = 36pi. First find the radius when V = 36pi.

$\begin{gathered} V=\frac{4}{3}\pi r^3 \\ 36\pi=\frac{4}{3}\pi r^3 \\ 27=r^3 \\ r=3 \end{gathered}$

Now, the rate of change of radius can be obtained as follows:

$\begin{gathered} 3\pi=4\pi(3)^2\times\frac{dr}{dt} \\ \frac{3\pi}{36\pi}=\frac{dr}{dt} \\ \frac{1}{12}=\frac{dr}{dt} \end{gathered}$

Thus, the radius of the sphere is increasing at the rate of 1/12 m/min when the volume is 36pi.