Question

The radius of a cylinder is increasing at the rate of $5$ cm/min so that its volume is constant. When its radius is $5$ cm and height is $3$ cm, then the rate of decreasing of its height is

• A. $6$ cm/min
• B. $3$ cm/min
• C. $4$cm/min
• D. $5$ cm/min
• E. $2$ cm/min

Answer 1

The correct option is A $6$ cm/min

Let $V$ be volume of the cylinder.

Given that:

$\frac{dr}{dt}=5$cm/min , $r=5$ cm and $h=3$ cm

We know that,

$V=\pi r^{2}h$

Differentiating it wrt $t$, we get

$\frac{dV}{dt}=\pi \times 2r\times \frac{dr}{dt}\times h+\pi r^2\times \frac{dh}{dt}$

Since, $V$ is constant $\therefore \frac{dV}{dt}=0$

$0=\pi \times 2r\times \frac{dr}{dt}\times h+\pi r^2\times \frac{dh}{dt}$

$\Rightarrow 2h\frac{dr}{dt}=-r\frac{dh}{dt}$

$\Rightarrow -5\frac{dh}{dt}=2\times 3\times 5$ cm/min

$\Rightarrow -\frac{dh}{dt}=6$ cm/min

Rate of decreasing of height,$(-\frac{dh}{dt})=6$ cm/min

Hence, A is the correct option.