# A cone whose height is always equal

Question:

A cone whose height is always equal to its diameter is increasing in volume at the rate of $40 \mathrm{~cm}^{3} / \mathrm{sec}$. At what rate is the radius increasing when its circular base area is $1 \mathrm{~m}^{2}$ ?

(a) $1 \mathrm{~mm} / \mathrm{sec}$

(b) $0.001 \mathrm{~cm} / \mathrm{sec}$

(c) $2 \mathrm{~mm} / \mathrm{sec}$

(d) $0.002 \mathrm{~cm} / \mathrm{sec}$

Solution:

(d) $0.002 \mathrm{~cm} / \mathrm{sec}$

Let $r$ be the radius, $h$ be the height and $V$ be the volume of the cone at any time $t .$ Then,

$V=\frac{1}{3} \pi r^{2} h$

$\Rightarrow V=\frac{2}{3} \pi r^{3} \quad[\because h=2 r]$

$\Rightarrow \frac{d V}{d t}=2 \times 10^{4} \frac{d r}{d t} \quad\left[\because \pi r^{2}=1 \mathrm{~m}^{2}\right.$ or $\left.10^{4} \mathrm{~cm}^{2}\right]$

$\Rightarrow \frac{d r}{d t}=\frac{1}{2 \times 10^{4}} \frac{d V}{d t}$

$\Rightarrow \frac{d r}{d t}=\frac{40}{2 \times 10^{4}}$

$\Rightarrow \frac{d r}{d t}=0.002 \mathrm{~cm} / \mathrm{sec}$