The altitude of a cone is 20 cm

Question:

The altitude of a cone is 20 cm and its semi-vertical angle is 30°. If the semi-vertical angle is increasing at the rate of 2° per second, then the radius of the base is increasing at the rate of

(a) $30 \mathrm{~cm} / \mathrm{sec}$

(b) $\frac{160}{3} \mathrm{~cm} / \mathrm{sec}$

(c) $10 \mathrm{~cm} / \mathrm{sec}$

(d) $160 \mathrm{~cm} / \mathrm{sec}$

Solution:

(b) $\frac{160}{3} \mathrm{~cm} / \mathrm{sec}$

Let $r$ be the radius, $h$ be the height and $\alpha$ be the semi $-$ vertical angle of the cone.

Then,

$\tan \alpha=\frac{r}{h}$

$\Rightarrow \sec ^{2} \alpha\left(\frac{d \alpha}{d t}\right)=\frac{d r}{h d t}$

$\Rightarrow \frac{d r}{d t}=h \times \sec ^{2} \alpha\left(\frac{d \alpha}{d t}\right)$

$\Rightarrow \frac{d r}{d t}=20 \times \sec ^{2} 30 \times 2$        $\left[\because h=20 \mathrm{~cm}, \alpha=30^{\circ}\right.$ and $\frac{d \alpha}{d t}=2^{\circ}$ per second $]$

$\Rightarrow \frac{d r}{d t}=40 \times\left(\frac{2}{\sqrt{3}}\right)^{2}$

$\Rightarrow \frac{d r}{d t}=\frac{160}{3} \mathrm{~cm} / \mathrm{sec}$