Question

A spherical balloon's volume increasinq at a rate of $0.001m^{3/s}$. The rate change its surface area $\left(in{m}^2/s\right)$ when the radius the balloon is $2m$ is

• A. $0.01$
• B. $0.001$
• C. $0.002$
• D. $0.05$

Answer 1

The correct option is B $0.001$

$\frac{dv}{dt}=4\pi r^2\frac{dr}{dt}=0.001$

$\Rightarrow \frac{dr}{dt}=0.0000199$

for $r=2$m

$\frac{ds}{dt}=8\pi r\frac{dr}{dt}=8\times 3.14\times 2\times 0.0000199$

for $r=2$m

$\therefore \frac{ds}{dt}=0.001m^2/s$.