Question

The Volume of a is sphere is increasing at the rate of 3 cubic per second. Find the rate of increase of its surface area, When the redius is 2 cm​

Answer 1

For a sphere

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

we are given

$\frac{dv}{dt} = 3 {m}^{3} per \: sec$

$\frac{dv}{dt} = \frac{4}{3} .3\pi {r}^{2} . \frac{dr}{dt} = 4\pi {r}^{2} . \frac{dr}{dt}$

so,

$\frac{dr}{dt} = \frac{dv}{dt} . \frac{1}{4\pi {r}^{2} }$

$\frac{dr}{dt} = 3. \frac{1}{4\pi {2}^{2} }$

Now,

$s.a = 4\pi {r}^{2}$

$\frac{d(s.a)}{dt} = 8\pi r. \frac{dr}{dt}$

substituting the value of dr/dt we get

rate of change of surface area at r=2 as

$3 {cm}^{2} persec$