Question

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

• A. $54 \,\pi \,cm ^{2} / \min$
• B. $7 \,\pi \,cm ^{2} / \min$
• C. $27 \,\pi \,cm ^{2} / \min$
• 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, $l^{2}=r^{2}+h^{2}$
$\Rightarrow 2 l \frac{dl}{dt}=2r \frac{dr}{dt}+2h \frac{dh}{dt}$
$\Rightarrow l \frac{dl}{dt}=r \frac{dr}{dt}+h \frac{dh}{dt}$
$\Rightarrow l\frac{dl}{dt}=7 \times 3+24 \times(-4) $
$[\because \frac{d h}{d t}=-4$ and $\left.\frac{d r}{d t}=3\right]$
$\Rightarrow l \frac{dl}{d t}=-75$
Where $r=7$ and $h=24$, we have
$ l^{2}=7^{2}+24^{2}$
$\Rightarrow l=25$
$\therefore l \frac{d l}{d t}=-75$
$\Rightarrow \frac{dl}{d t}=-3$
Let $S$ denote the lateral surface area.
Then, $\frac{dS}{dt}=\pi \frac{d(rl)}{dt}$
$=\pi\left\{\frac{dr}{dt} l+r \frac{d l}{d t}\right\}$
$=\pi\{3 \times 25+7 \times(-3)\}$
$=54 \,\pi \,cm ^{2} / min$