A spherical balloon is being inflated. Find the volume of the balloon at the instant when the rate of increase of the surface area is 8 times the rate of increase of the radius of the sphere.

Question

A spherical balloon is being inflated.  Find the volume of the balloon at the instant when the rate of increase of the surface area is 8 times the rate of increase of the radius of the sphere.

anonymous · Answer

$$\S(r)=4\pi*r^2$$ 
$$V(r)=(4/3)\pi*r^3$$
Implicitly differentiate the S(r) w.r.t. t, time, to compute the rate of increase of surface area and radius. 
$$dS/dt=(dr/dt)*8\pi*r$$
$$dS/dt=8*dr/dt$$
$$8*dr/dt=(dr/dt)*8\pi*r$$
$$r=1/\pi$$
$$V=(4/3)*\pi*(1/\pi)^3$$
Volume is approximately equal to .135095

Not 100% sure on this.