Question

(4) A spherical balloon is being inflated so that its diameter is increasing at a constant rate of 6 cm/min. How quickly is the volume of the balloon increasing when the diameter is 50 cm

Answer 1

In terms of its radius $r$, the volume of the balloon is

$V(r)=\dfrac{4\pi}3r^3$

The diameter $d$ is twice the radius, so that in terms of its diameters, the balloon's volume is given by

$V(d)=\dfrac{4\pi}3\left(\dfrac d2\right)^3=\dfrac\pi6d^3$

Differentiate both sides with respect to time $t$:

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi2d^2\dfrac{\mathrm dd}{\mathrm dt}$

The diameter increases at a rate of $\frac{\mathrm dd}{\mathrm dt}=6\frac{\rm cm}{\rm min}$. When the diameter is $d=50\,\mathrm{cm}$, we have

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi2(50\,\mathrm{cm})^2\left(6\frac{\rm cm}{\rm min}\right)=7500\pi\dfrac{\mathrm{cm}^3}{\rm min}$

or about 23,562 cc/min (where cc = cubic centimeters)