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The height of a right circular cylinder is decreasing while is diameter is increasing at a rate of $4 \mathrm{~cm} / \mathrm{see}$ so as to keep its volume unchanged. the rate of change in its lateral surface area (in $\mathrm{cm}^2 / \mathrm{sec}$ ) at the instant when its diameter is $8 \mathrm{~cm}$ and height is $12 \mathrm{~cm}$, is
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Verified Answer
The correct answer is:
$48 \pi$
Given, $\frac{d(2 r)}{d t}=4 \mathrm{~cm} / \mathrm{s}$
$\Rightarrow \quad \frac{d r}{d t}=2 \mathrm{~cm} / \mathrm{s}$
Volume of cylinder $=\pi r^2 h$
$$
\begin{aligned}
v & =\pi r^2 h \\
\Rightarrow \quad \frac{d v}{d t} & =\pi\left(r^2 \frac{d h}{d t}+2 r h \frac{d r}{d t}\right)
\end{aligned}
$$
Volume is unchanged and height of cylinder is decreasing
$$
\begin{aligned}
& \therefore & \pi r^2 h & =\pi\left(-r^2 \frac{d h}{d t}+2 r h(2)\right) \\
\Rightarrow & & r h & =-r \frac{d h}{d t}+4 h \\
\Rightarrow & & \frac{d h}{d t} & =\frac{4 h-r h}{r}
\end{aligned}
$$
Lateral surface area of cylinder $(S)=2 \pi r h$
$$
\frac{d S}{d t}=2 \pi\left(r \frac{d h}{d t}+h \frac{d r}{d t}\right)
$$
$$
\begin{aligned}
& \frac{d S}{d t}=2 \pi\left[\frac{r(4 h-r h)}{r}+h(2)\right] \\
& \frac{d S}{d t}=2 \pi(4 h-r h+2 h) \\
& \Rightarrow \quad \frac{d S}{d t}=2 \pi(6 h-r h) \\
& \frac{d S}{d t}=2 \pi(72-48) \quad[\because h=12, r=4] \\
& \frac{d S}{d t}=48 \pi \\
&
\end{aligned}
$$
$\Rightarrow \quad \frac{d r}{d t}=2 \mathrm{~cm} / \mathrm{s}$
Volume of cylinder $=\pi r^2 h$
$$
\begin{aligned}
v & =\pi r^2 h \\
\Rightarrow \quad \frac{d v}{d t} & =\pi\left(r^2 \frac{d h}{d t}+2 r h \frac{d r}{d t}\right)
\end{aligned}
$$
Volume is unchanged and height of cylinder is decreasing
$$
\begin{aligned}
& \therefore & \pi r^2 h & =\pi\left(-r^2 \frac{d h}{d t}+2 r h(2)\right) \\
\Rightarrow & & r h & =-r \frac{d h}{d t}+4 h \\
\Rightarrow & & \frac{d h}{d t} & =\frac{4 h-r h}{r}
\end{aligned}
$$
Lateral surface area of cylinder $(S)=2 \pi r h$
$$
\frac{d S}{d t}=2 \pi\left(r \frac{d h}{d t}+h \frac{d r}{d t}\right)
$$
$$
\begin{aligned}
& \frac{d S}{d t}=2 \pi\left[\frac{r(4 h-r h)}{r}+h(2)\right] \\
& \frac{d S}{d t}=2 \pi(4 h-r h+2 h) \\
& \Rightarrow \quad \frac{d S}{d t}=2 \pi(6 h-r h) \\
& \frac{d S}{d t}=2 \pi(72-48) \quad[\because h=12, r=4] \\
& \frac{d S}{d t}=48 \pi \\
&
\end{aligned}
$$
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