The rate of change of surface area of a sphere of radius r when the radius is increasing at the rate of 2 cm/ sec is proportional to

A
$\frac{1}{r^{2}}$
B
$\frac{1}{r}$
C
$r^{2}$
D
$r$

Solution:

Let $S$ be the surface area and $r$ be the radius of the

sphere at any time $t$ . Then,

$\begin{array}{l} S = 4 π r^{2} \\ \Rightarrow \frac{d S}{d t} = 8 π r \frac{d r}{d t} \\ \Rightarrow \frac{d S}{d t} = 8 π r \times 2 [\frac{d r}{d t} = 2 cm / \sec] \end{array}$

$\Rightarrow \frac{d S}{d t} = 16 π r \Rightarrow \frac{d S}{d t} \propto r$

The rate of change of surface area of a sphere of radius $r$ when the radius is increasing at the rate of $2 c m / s e c$ is proportional to

Solution:

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