Question

The surface area of a snowball melting decreases at a rate of 6cm/min. how is the volume changing when the radius is 3

Answer 1

bear in mind that, ds/dt is a negative 6, because the area is "decreasing".

$\bf \textit{surface area of a sphere}\\\\ s=4\pi r^2\implies \boxed{\cfrac{s}{4\pi }=r^2} \\\\\\ \cfrac{1}{4\pi }\cdot \cfrac{ds}{dt}=\stackrel{chain~rule}{2r\cfrac{dr}{dt}}\implies \boxed{\cfrac{\frac{ds}{dt}}{8\pi r}=\cfrac{dr}{dt}}\\\\ -------------------------------$

$\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}\implies V=\cfrac{4\pi }{3}\cdot r^2\cdot r\implies V=\cfrac{\underline{4\pi} }{3}\cdot \boxed{\cfrac{s}{\underline{4\pi} }}\cdot r \\\\\\ V=\cfrac{1}{3}sr\implies \cfrac{dV}{dt}=\cfrac{1}{3}\left( \cfrac{ds}{dt}\cdot r+s\cdot \cfrac{dr}{dt} \right)\\\\ -------------------------------$

$\bf \textit{now, when the radius is 3}\qquad \begin{cases} \frac{ds}{dt}=-6\\\\ s=4\pi 3^2\\ \qquad 36\pi \\\\ \frac{dr}{dt}=\frac{-6}{8\pi 3}\\\\ \qquad -\frac{1}{4\pi } \end{cases} \\\\\\ \cfrac{dV}{dt}=\cfrac{1}{3}\left(-6\cdot 3~~+~~36\pi\cdot -\cfrac{1}{4\pi } \right)\implies \cfrac{dV}{dt}=\cfrac{1}{3}(-18-9) \\\\\\ \cfrac{dV}{dt}=-\cfrac{27}{3}\implies \cfrac{dV}{dt}=-9$