The volume of a sphere is increasing at the rate of 4$\pi$ cm3/sec. The rate of increase of the radius when the volume is 288 $\pi$ cm3, is

Question

KollegeApply · Accepted Answer

$\frac{1}{36}$ \text { Let r be the radius and V be the volume of the sphere at any time t. Then }, V=$\frac{4}{3}$$\pi$ r^3 \Rightarrow $\frac{4}{3}$$\pi$ r^3 =288$\pi$ \Rightarrow r^3 = 288 $\times$ $\frac{3}{4}$ \Rightarrow r^3 = 216 \Rightarrow r = 6 \Rightarrow dV/dt = 4$\pi$ r^2 dr/dt \Rightarrow dV/dt = 4$\pi$ \left( 6 \right)^2 dr/dt \Rightarrow 4$\pi$ = 144$\pi$dr/dt \Rightarrow dr/dt = $\frac{1}{36}$

AChoose Your Answer