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 cm/sec
(b) $\frac{160}{3}\mathrm{cm}/\sec$
(c) 10 cm/sec
(d) 160 cm/sec

Answer 1

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

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



$$\begin{gathered} \mathrm{Then}, \\ \tan \alpha =\frac{r}{h} \\ \Rightarrow se{c}^2\alpha \left(\frac{d\alpha }{dt}\right)=\frac{dr}{hdt} \\ \Rightarrow \frac{dr}{dt}=h\times se{c}^2\alpha \left(\frac{d\alpha }{dt}\right) \\ \Rightarrow \frac{dr}{dt}=20\times se{c}^{2}30\times 2\left[\because h=20\mathrm{cm},\alpha =30°\mathrm{and}\frac{d\alpha }{dt}=2°\mathrm{per}\mathrm{second}\right] \\ \Rightarrow \frac{dr}{dt}=40\times {\left(\frac{2}{\sqrt{3}}\right)}^2 \\ \Rightarrow \frac{dr}{dt}=\frac{160}{3}\mathrm{cm}/\sec \end{gathered}$$