Answer to Question #121202 in Classical Mechanics for Shanil

Question

A spherical balloon is deflated so that its volume is decreasing at a rate of 5 ftଷ/min. How
fast is the diameter of the balloon decreasing when the radius is 3 ft?

Answer 1

As per the given question,

Rate of the change in the volume of the balloons$(\frac{dV}{dt})=5 ft^3/min$

Final diameter of the volume $= 3ft$

We know that volume of the sphere $(V)=\frac{4 \pi r^3}{3}$

$\Rightarrow \frac{dV}{dt}=4\pi3r^2 \frac{dr}{3dt}=4\pi r^2\frac{dr}{dt}$

$\Rightarrow 5ft^3/min =4\pi \times 3^2 \frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{5}{4 \pi \times 9}=\frac{5}{36 \pi}ft/sec$