# A balloon is being filled with helium at the rate if 4"ft"^3/"min". What is the rate, in "ft"^2/"min", at which the surface area is increasing when the volume is (32pi)/3"ft"^3?

Jul 31, 2018

The answer is $= 4 \frac{f {t}^{2}}{\min}$

#### Explanation:

The volume of a balloon is

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

The volume is $V = \frac{32}{3} \pi$

Therefore,

$\frac{32}{3} \pi = \frac{4}{3} \pi {r}^{3}$

${r}^{3} = 8$

$r = 2$

So,

Differentiating wrt $t$

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

As,

$\frac{\mathrm{dV}}{\mathrm{dt}} = 4$

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

$\frac{\mathrm{dr}}{\mathrm{dt}} = \frac{1}{\pi {r}^{2}}$

The surface area is

$A = 4 \pi {r}^{2}$

Differentiating wrt $t$

$\frac{\mathrm{dA}}{\mathrm{dt}} = 8 \pi r \frac{\mathrm{dr}}{\mathrm{dt}} = 8 \pi \cdot 2 \cdot \frac{1}{\pi {2}^{2}} = 4 \frac{f {t}^{2}}{\min}$