Answer to Question #242767 in Calculus for Jaison

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

$V=\frac{4}{3}\pi r^3=\frac{1}{6}\pi d^3$

$\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)=\frac{4}{3}\pi\cdot 3r^2\frac{dr}{dt}=3$

$\frac{dr}{dt}=\frac{3}{4\pi r^2}$

$\frac{dD}{dt}=\frac{3}{\pi D^2}$

$\frac{dD}{dt}(4)=\frac{3}{\pi 4^2}=\frac{3}{16\pi }=0.06$ ft/min