Question:
A cylindrical vessel of radius $0.5 \mathrm{~m}$ is filled with oil at the rate of $0.25 \pi \mathrm{m}^{3} /$ minute. The rate at which the surface of the oil is rising, is
Solution:
(a) 1 m/minute
Let $r$ be the radius, $h$ be the height and $V$ be the volume of the cylindrical vessel at any time $t$. Then,
$V=\pi r^{2} h$
$\Rightarrow \frac{d V}{d t}=\pi r^{2} \frac{d h}{d t}$
$\Rightarrow \frac{d h}{d t}=\frac{1}{\pi r^{2}} \frac{d V}{d t}$
$\Rightarrow \frac{d h}{d t}=\frac{0.25 \pi}{\pi(0.5)^{2}}$
$\Rightarrow \frac{d h}{d t}=\frac{0.25}{0.25}$
$\Rightarrow \frac{d h}{d t}=1 \mathrm{~m} / \min$
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