The radius of a spherical soap bubble is increasing at the rate

Question:

The radius of a spherical soap bubble is increasing at the rate of $0.2 \mathrm{~cm} / \mathrm{sec}$. Find the rate of increase of its surface area, when the radius is $7 \mathrm{~cm}$.

Solution:

Let $r$ be the radius and $S$ be the surface area of the spherical ball at any time $t .$ Then,

$S=4 \pi r^{2}$

$\Rightarrow \frac{d S}{d t}=8 \pi r \frac{d r}{d t}$

$\Rightarrow \frac{d S}{d t}=8 \pi \times 7 \times 0.2$        $\left[\because r=7 \mathrm{~cm}\right.$ and $\left.\frac{d r}{d t}=0.2 \mathrm{~cm} / \mathrm{sec}\right]$

$\Rightarrow \frac{d S}{d t}=11.2 \pi \mathrm{cm}^{2} / \mathrm{sec}$

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