Question

the volume of a sphere of radius ( r ) is ( v=\frac{4}{3} pi r^{3} ).
a spherical snowball is melting in such a way that its radius is decreasing at a rate of ( \frac{d r}{d t}=0.1 mathrm{~cm} / mathrm{min} ).
at what rate is the volume of the snowball decreasing when the radius is ( 11 mathrm{~cm} ) ?
(note: give the answer as a positive number).
( \frac{d v}{d t}=quad \frac{mathrm{cm}^{3}}{mathrm{~min}} )

Explanation:

Step1: Differentiate the volume formula with respect to \(t\)

Given \(V=\frac{4}{3}\pi r^{3}\). Using the chain - rule \(\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}\).
Differentiate \(V\) with respect to \(r\): \(\frac{dV}{dr}=\frac{4}{3}\pi\cdot3r^{2}=4\pi r^{2}\).

Step2: Substitute the values of \(r\) and \(\frac{dr}{dt}\)

We know that \(\frac{dr}{dt}=- 0.1\) (negative because \(r\) is decreasing) and \(r = 11\).
Substitute into \(\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}\):
\(\frac{dV}{dt}=4\pi r^{2}\cdot\frac{dr}{dt}\).
\(\frac{dV}{dt}=4\pi(11)^{2}\cdot(-0.1)\).
\(\frac{dV}{dt}=4\pi\times121\times(-0.1)=-48.4\pi\).

Since the problem asks for the rate at which the volume is decreasing (as a positive number), we take the absolute value.

Answer:

\(48.4\pi\approx152.05\) \(\frac{cm^{3}}{min}\)