Question

The radius of a sphere is increasing at a rate of 4 mm/s. How fast is the volume increasing when the diameter is 80 mm?

Short answer

The required answer is \(25600\pi \,\;{\rm{m}}{{\rm{m}}^{\rm{3}}}{\rm{/s}}\).

Step-by-step solution

The volume of the sphere

It is known that the volume of the sphere whose radius is \(r\) equals to \(V = \frac{4}{3}\pi {r^3}\).

The rate of change of volume

The rate of change of volume with respect to time is given by \(\frac{{dV}}{{dt}}\).

Here, \(V = \frac{4}{3}\pi {r^3}\). Differentiate \(V\) with respect to \(t\):

\(\begin{array}{l}\frac{{dV}}{{dt}} = \frac{d}{{dt}}\left( {\frac{4}{3}\pi {r^3}} \right)\\\frac{{dV}}{{dt}} = \frac{4}{3}\pi \left( {3{r^2}} \right) \cdot \frac{{dr}}{{dt}}\\\frac{{dV}}{{dt}} = 4\pi {r^2} \cdot \frac{{dr}}{{dt}}\end{array}\)

Hence, \(\frac{{dV}}{{dt}} = 4\pi {r^2} \cdot \frac{{dr}}{{dt}}\).

Increase in area

Here, given that radius of the sphere is increasing at the rate of \(4\,\,{\rm{mm/s}}\) when the radius of the sphere is \(80\,{\rm{mm}}\), means \(\frac{{dr}}{{dt}} = 4\) and \(r = 80\).

Substitute in \(\frac{{dV}}{{dt}} = 4\pi {r^2} \cdot \frac{{dr}}{{dt}}\) :

\(\begin{array}{c}\frac{{dV}}{{dt}} = 4\pi {\left( {\frac{{80}}{2}} \right)^2} \cdot 4\\ = 25600\pi \end{array}\)

Hence, the volume of the sphere is increasing at \(25600\pi \,\;{\rm{m}}{{\rm{m}}^3}{\rm{/s}}\).