Question

Volume of a sphere is increasing constantly at a rate dV/dt =k. At the moment when volume is Vo, the rate of change of radius is

Answer 1

Answer:$\frac{k}{4\pi}*(4\pi/3V_o)^{2/3}$  

                                                                      Step-by-step explanation:

When Volume of sphere is $V_o$.

We have,

$V_o=\frac{4 \pi r^3}{3} \\ \\r^3=\frac{3V_o}{4\pi}\\ \\r=(\frac{3V_o}{4\pi})^{1/3}$

Also, Volume is constantly increasing at a rate of "k".

That is,

$\frac{dV}{dt}=k\\ \\=>\frac{4 \pi * 3r^2 * dr}{3dt}=k\\ \\=>4\pi r^{2}\frac{dr}{dt}=k\\ \\=>\frac{dr}{dt}=\frac{k}{4\pi r^2}\\ \\ =>\frac{k}{4\pi}*(4\pi/3V_o)^{2/3}$  

   Hence, we can say radius is increasing at a rate of dr/dt which is given as above.