Question

The volume of a closed hemisphere increases at the rate of 4 cm³/sec. Find the rate of increase of its surface area, when the radius is 4 cm.

Answer 1

Let V is the volume of the hemisphere.

Given:   $\frac{dv}{dt}=4cm^{3}/s$

                    r= 4cm

We know V can be given as:

 $V=\frac{2}{3} \pi r^{3}$

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

$\frac{dr}{dt} =\frac{2}{\pi r^{2} } cm/s$

Let Surface area is S  

 $S=2\pi r^{2}$

$\frac{dS}{dt}=4\pi r\frac{dr}{dt}$

$\frac{dS}{dt}= 4\pi r*\frac{2}{\pi r^{2} } =\frac{8}{r}$

at r =4cm

$\frac{dS}{dt}=2cm^{2}/s$

Hope it helps

Thanks

With Regards

Answer 2

Dear Student,

Answer: 2 sq- cm/sec

Solution:

Given that dV/dt = 4 cm³/sec

Volume of Hemisphere =
$\frac{2}{3} \pi {r}^{3} \\ \\ \frac{dV}{dt} = 2\pi {r}^{2} \frac{dr}{dt} \\ 4 = 2\pi {r}^{2} \frac{dr}{dt} \\ \\ \frac{dr}{dt} = \frac{4}{2\pi {r}^{2} } \\ \frac{dr}{dt} at \: \: r = 4 = \frac{1}{8\pi}$
Surface area of Hemisphere A =
$2\pi {r}^{2} \\ \\ \frac{dA}{dt} = 4\pi \:r \frac{dr}{dt} \\ \\ \frac{dA}{dt} = 4\pi (4)\times \frac{1}{8\pi} \\ \\ \frac{dA}{dt} = 2$
rate of increase of surface area is equal to 2 cm square per second