Question

The spherical balloon is inflated at the rate of 10 m³/sec. Find the rate at which the surface area is increasing when the radius of the sphere is 3m?

Answer 1

The balloon has a volume $V$ dependent on its radius $r$:

$V(r)=\dfrac43\pi r^3$

Differentiating with respect to time $t$ gives

$\dfrac{\mathrm dV}{\mathrm dt}=4\pi r^2\dfrac{\mathrm dr}{\mathrm dt}$

If the volume is increasing at a rate of 10 cubic m/s, then at the moment the radius is 3 m, it is increasing at a rate of

$10\dfrac{\mathrm m^3}{\mathrm s}=4\pi (3\,\mathrm m)^2\dfrac{\mathrm dr}{\mathrm dt}\implies\dfrac{\mathrm dr}{\mathrm dt}=\dfrac5{18\pi}\dfrac{\rm m}{\rm s}$

The surface area of the balloon is

$S(r)=4\pi r^2$

and differentiating gives

$\dfrac{\mathrm dS}{\mathrm dt}=8\pi r\dfrac{\mathrm dr}{\mathrm dt}$

so that at the moment the radius is 3 m, its area is increasing at a rate of

$\dfrac{\mathrm dS}{\mathrm dt}=8\pi(3\,\mathrm m)\left(\dfrac5{18\pi}\dfrac{\rm m}{\rm s}\right)=\dfrac{20}3\dfrac{\mathrm m^2}{\rm s}$