Question

The volume of a sphere increases at the rate 8 cm³/sec. Find the rate of increase of its surface area, when the radius is 4 cm.

Answer 1

Dear Student:

Let V is the volume of the sphere.

Given:   $\frac{dv}{dt}=8cm^{3}/s$

                         r= 4cm

We know V can be given as:

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

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

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

Let Surface area is S

                  $S=4\pi r^{2}$

$\frac{dS}{dt}=8\pi r\frac{dr}{dt}$

$\frac{dS}{dt}= 8\pi r*\frac{2}{\pi r^{2} } =\frac{16}{r}$

at r =4cm

$\frac{dS}{dt}=4cm^{2}/s$

Answer 2

volume of sphere increase at the rate 8cm³/sec.
Let V denotes volume of sphere and r denotes radius of sphere.
then , a/c to question, $\bf{\frac{dV}{dt}=8cm^3/sec}$
and r = 4cm

we know,Volume of sphere ,$\bf{V=\frac{4}{3}\pi r^3}$
differentiate both sides with respect to time,
$\bf{\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}}$

$\bf{8=4\pi(4)^2\frac{dr}{dt}}$

$\bf{\frac{dr}{dt}=\frac{8}{4\pi(4)^2}}$

$\bf{\frac{dr}{dt}=\frac{1}{8\pi}cm/sec}$....(1)

now, surface area of sphere , A = 4πr²

differentiate both sides with respect to time,

dA/dt = 8πr. dr/dt

put r = 4cm and dr/dt = 1/8π from equation (1),

dA/dt = 8π × (4) × 1/8π = 4cm²/sec

hence,rate of increase of its surface area is 4cm²/sec