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 is $7cm$ and altitude $24cm$ is

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

Answer 1

The correct option is A $54\pi c{m}^2$/min

Given $\frac{dr}{dt}=3cm/minute,\frac{dh}{dt}=-4cm/minute$

We have $l^2=r^2+h^2$

$\Rightarrow l^2=49+576=625$

$\Rightarrow l=25cm$

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

$25\frac{dl}{dt}=21-96$

$\Rightarrow \frac{dl}{dt}=-3cm/sec$

Curved surface area $S=\pi rl$

$\Rightarrow \frac{dS}{dt}=\pi [r\frac{dl}{dt}+l\frac{dr}{dt}]$

$\Rightarrow \frac{dS}{dt}=\pi [-21+75]=54\pi c{m}^2/sec$