Question

A spherical balloon is inflated so that its volume is increasing at the rate of 2.7 ft3/min. How rapidly is the diameter of the balloon increasing when the diameter is 1.3 feet

Answer 1

The formula for volume of a sphere is
$V=\frac{4}{3}\pi r^{3}$
$\rightarrow\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$

If the diameter is 1.3, then $r=\frac{1.3}{2}=0.65$

We are given
$\frac{dV}{dt}=2.7$ and $r=0.65$
$\rightarrow2.7=4\pi(0.65)^{2}\frac{dr}{dt}$
$\rightarrow\frac{dr}{dt}=\frac{2.7}{4\pi(0.65)^{2}}\approx0.5085\ ft/min$

Change in diameter is twice change in radius, so we get 2*0.5085=1.017 ft/min