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  4. The surface area of a ball is increasing at the rate of 2 π sq. cm/sec. The rate at which the radius is increasing when the surface area is 16 π sq. cm is

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

COMEDKCOMEDK 2011Application of Derivatives

Solution:

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