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A balloon is pumped at the rate of $4 \mathrm{~cm}^{3}$ per second. What is the rate at which its surface area increases and radius is $4 \mathrm{~cm} ?$
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Verified Answer
The correct answer is:
$2 \mathrm{~cm}^{2} / \mathrm{s}$
Let $r$ be the radius of balloon.
Balloon is like a sphere and volume of sphere $=\frac{4}{3} \pi r^{3}$
$\therefore \quad V=\frac{4}{3} \pi r^{3}$
Differentiate both side w.r.t ' $\mathrm{t}^{\prime}$
$\Rightarrow \frac{d V}{d t}=\frac{4}{3} \pi \cdot 3 r^{2} \frac{d r}{d t}$
$\Rightarrow 4=\frac{4}{3} \pi .3(4)^{2} \frac{d r}{d t} \quad\left(\because \frac{d V}{d t}=4 \mathrm{~cm}^{3} / \mathrm{s}\right)$
$\Rightarrow \frac{d r}{d t}=\frac{1}{16 \pi}$
Now, surface area of balloon $=S=4 \pi r^{2}$
$$
\begin{aligned}
\frac{d S}{d t} &=4 \pi \cdot 2 r \frac{d r}{d t} \\
&=4 \pi \cdot 2 \times 4 \frac{1}{16 \pi}(\text { from }(1))=2 \mathrm{~cm}^{2} / \mathrm{s}
\end{aligned}
$$
Balloon is like a sphere and volume of sphere $=\frac{4}{3} \pi r^{3}$
$\therefore \quad V=\frac{4}{3} \pi r^{3}$
Differentiate both side w.r.t ' $\mathrm{t}^{\prime}$
$\Rightarrow \frac{d V}{d t}=\frac{4}{3} \pi \cdot 3 r^{2} \frac{d r}{d t}$
$\Rightarrow 4=\frac{4}{3} \pi .3(4)^{2} \frac{d r}{d t} \quad\left(\because \frac{d V}{d t}=4 \mathrm{~cm}^{3} / \mathrm{s}\right)$
$\Rightarrow \frac{d r}{d t}=\frac{1}{16 \pi}$
Now, surface area of balloon $=S=4 \pi r^{2}$
$$
\begin{aligned}
\frac{d S}{d t} &=4 \pi \cdot 2 r \frac{d r}{d t} \\
&=4 \pi \cdot 2 \times 4 \frac{1}{16 \pi}(\text { from }(1))=2 \mathrm{~cm}^{2} / \mathrm{s}
\end{aligned}
$$
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