Question

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

• A. $54\pi c{m}^2/\min$
• B. $7\pi c{m}^2/\min$
• C. $27\pi c{m}^2/\min$
• D. None of these

Answer 1

Answer: (A)

Let $r,l$ and $h$ denote respectively the radius, slant height and height of the cone at any time $t$.
Then, $l^2=r^2+h^2$
$\Rightarrow 2l\frac{dl}{dt}=2r\frac{dr}{dt}+2h\frac{dh}{dt}$
$\Rightarrow l\frac{dl}{dt}=r\frac{dr}{dt}+h\frac{dh}{dt}$
$\Rightarrow l\frac{dl}{dt}=7\times 3+24\times (-4)$
$[\because \frac{dh}{dt}=-4$ and $\frac{dr}{dt}=3]$
$\Rightarrow l\frac{dl}{dt}=-75$
Where $r=7$ and $h=24$, we have
$l^2=7^2+24^2$
$\Rightarrow l=25$
$\therefore l\frac{dl}{dt}=-75$
$\Rightarrow \frac{dl}{dt}=-3$
Let $S$ denote the lateral surface area.
Then, $\frac{dS}{dt}=\pi \frac{d(rl)}{dt}$
$=\pi \left\{\frac{dr}{dt}l+r\frac{dl}{dt}\right\}$
$=\pi \{3\times 25+7\times (-3)\}$
$=54\pi c{m}^2/min$