The surface area of a cube is changing at the rate of 8 cm^2/s. How fast is the volume changing when the surface area is 60cm?
Hi scubasteve94,
if it is a regular cube, then
$\displaystyle surface\ area=6s^2$
$\displaystyle volume=s^3$
where s=length of all edges
$\displaystyle SA=60\ cm^2=6s^2\ \Rightarrow\ s^2=10\ cm^2\ \Rightarrow\ s=\sqrt{10}\ cm$
$\displaystyle \frac{d}{dt}(SA)=8\ cm^2/sec\ =\ \frac{ds}{dt}\ \frac{d}{ds}6s^2=\frac{ds}{dt}12s=\frac{ds}{dt}12\ sqrt{10}\ cm$
hence
$\displaystyle \frac{ds}{dt}=\frac{8}{12\sqrt{10}}=\frac{2}{3\sqr t{10}}\ cm/sec$..... which is the rate of change of the length of an edge of the cube.
$\displaystyle \frac{dV}{dt}=\frac{dV}{ds}\ \frac{ds}{dt}=3s^2\ \frac{2}{3\sqrt{10}}=30\ \frac{2}{3\sqrt{10}}=\frac{20}{\sqrt{10}}\ cm^3/sec$