Question

The surface area of a ball is increasing at the rate of $2 \pi \, s cm/sec$. The rate at which the radius is increasing when the surface area is $16 \pi \, s cm$ is

• A. $0.125\, cm/sec$
• B. $0.25\, cm/sec$
• C. $0.5\, cm/sec$
• D. $1\, cm/sec$

Answer 1

Answer: (A)

$0.125\, cm/sec$

Answer 2

We have, $\frac{dS}{dt} = 2 \pi$ sq . cm /sec ? ?
? ?
....(i)
Now, $S = 4 \pi r^2$? ?
? ?
....(i)
Differentiating (ii) w.r.t, t, we get
? ? $\frac{dS}{dt} = 8\pi r \frac{dr}{dy}$ ? ? $\Rightarrow 2\pi = 8\pi r \frac{dr}{dt}$ [From (i)]
$\Rightarrow \frac{dr}{dt} =\frac{1}{4r}$ ? ?? ? ...(iii)
Now, when $S = 16 \pi \Rightarrow 4\pi r^{2} = 16\pi$
$\Rightarrow r^{2} =4 \Rightarrow r =$2 cm
Hence, $\left[\frac{dr}{dt}\right]_{r=2} = \frac{1}{4\times2} = \frac{1}{8} 0.125$ cm /sec