Question

The radius of the base of a cone is increasing at the rate of $3cm/min$ and the altitude is decreasing at the rate of $4cm/min.$ The rate of change of lateral surface area when the radius is $7cm$ and the altitude is $24cm$ is

• A. $54\pi c{m}^2/min$
• B. $7\pi c{m}^2/min$
• C. $27\pi c{m}^2/min$
• D. $108\pi c{m}^2/min$

Answer 1

The correct option is A

$54\pi c{m}^2/min$

Let $r,l,$ and $h$ denote, respectively, the radius, slant height, and altitude of the cone at any time $t.$

Then, $l^2=r^2+h^2$

Differentiating, we get

$2l\frac{dl}{dt}=2r\frac{dr}{dt}+2h\frac{dh}{dt}$

$=l\frac{dl}{dt}=r\frac{dr}{dt}+h\frac{dh}{dt}$

$\Rightarrow l\frac{dl}{dt}=7\times 3+24\times (-4)$ $[\because \frac{dh}{dt}=-4$ and $\frac{dr}{dt}=3]$

$\Rightarrow l\frac{dl}{dt}=-75$

Also $l^2=7^2+{24}^2$ $[\because r=7$ and $h=24]$

$\Rightarrow l=\sqrt{625}=25cm$

$\therefore \frac{dl}{dt}=\frac{-75}{25}=-3cm/min$

Let $S$ denote the lateral surface area. Then

$S=\pi rl$

$\Rightarrow \frac{dS}{dt}=\frac{d}{dt}\left(\pi rl\right)=\pi (\frac{dr}{dt}l+r\frac{dl}{dt})$

$=\pi (3\times 25+7\times (-3\left)\right)$

$=54\pi c{m}^2/min$