Question

A spherical balloon is deflated so that its volume is decreasing at a rate of 3 ftଷ/min. How fast is the diameter of the balloon decreasing when the radius is 2 ft?

Answer 1

V=4πr³/3, where r=radius, and r=D/2 where D is the diameter.

dV/dr=4πr², dV/dt=(dV/dr)(dr/dt)=-3ft³/min,

4πr²dr/dt=-3, and when r=2ft, this is 16πdr/dt=-3, making dr/dt=-3/(16π).

dr/dt=½dD/dt, so dD/dt=2dr/dt=-3/(8π)ft/min. So the diameter is decreasing at the rate of 3/(8π)=0.1194ft/min approx.