Question

The rate of change of the surface are of a sphere of radius r when the radius is increasing at the rate of 2 cm/sec is _________________.

Answer 1

Let r be the radius of sphere at any time t.

It is given that,

$\frac{dr}{dt}$ = 2 cm/sec

Surface area of the sphere, S = 4$\mathrm{\pi }$r²

S = 4$\mathrm{\pi }$r²

Differentiating both sides with respect to t, we get

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

$\Rightarrow \frac{dS}{dt}=4\mathrm{\pi }\times 2r\frac{dr}{dt}$

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

Putting $\frac{dr}{dt}$ = 2 cm/sec, we get

$\frac{dS}{dt}=8\mathrm{\pi }r\times 2$ = 16$\mathrm{\pi }$r cm²/sec

Thus, the rate of change of the surface are of a sphere of radius r when the radius is increasing at the rate of 2 cm/sec is 16$\mathrm{\pi }$r cm²/sec.


The rate of change of the surface are of a sphere of radius r when the radius is increasing at the rate of 2 cm/sec is ___16$\mathrm{\pi }$r cm²/sec___.