Differentiation. Rate of increase

Question

The radius of a sphere is increasing at a rate of $4 cms^{-1}.$ Find the rate of increase of the surface area when the radius is $5cm.$

Surface area $= 4\pi r^2$

Answer 1

It is given that $\frac{dr}{dt}=4$ cm/s.

As you said, $A=4\pi r^2$. Now differentiate this w.r.t. time to get

$$\frac{dA}{dt}=4 \pi \cdot 2r \frac{dr}{dt}$$ When $r=5$ cm, this will be equal to

$$8 \pi \cdot 5 \cdot 4 = 160\pi\ {cm}^2/s.$$

Answer 2

If $r$ is the radius of the sphere, then given that $dr/dt=4.$ Now if $S= 4\pi r^2$ then rate of increase of the surface area when the radius $r=5$ is given by $dS/dt=4\pi 2.r.dr/dt=4.\pi.2.5.4=160\pi$ sq.cm/sec.