Question

if the volume V=4/3 pi r^3 of a sphere is changing at a rate of 40 cubic cm per min, find the rate at which the surface are A=4pi r^2 is changing, when V=400/3 pi cubic cm

Answer 1

\[r=\sqrt[3]{3V/4 \pi}=\sqrt[3]{3(400 \pi/3)/4 \pi}=\sqrt[3]{100}\]
A=3V/r
dA/dt=d(3V/r)/dt
\[dA/dt=d(3V/\sqrt[3]{100})/dt=(3/\sqrt[3]{100})(dV/dt)=3/\sqrt[3]{100}(40cm^3/\min)\]

Answer 2

thanks you

Answer 3

how did you get A=3V/r?