Question

The radius of a cylinder is increasing at the rate of $3m/s$ and its altitude is decreasing at the rate of $4m/s$. The rate of change of volume, when radius is $4m$and altitude is $6m$, is

• A. $80\pi cum/s$
• B. $144\pi cum/s$
• C. $80cum/s$
• D. $64cum/s$

Answer 1

The correct option is A

$80\pi cum/s$

Explanation for the correct option

Let the radius of the of the cylinder is $r$ and $h$ the altitude/ height of the cylinder

Given, rate of increase of radius is $3m/s$

That is $\frac{dr}{dt}=3m/s$

The rate of decrease of altitude is $4m/s$

That is $\frac{dh}{dt}=-4m/s$

We know that volume of the cylinder is $V=\pi r^{2}h$

On differentiating the volume with respect to $t$

$\Rightarrow \frac{dV}{dt}=\pi r^2\left(\frac{dh}{dt}\right)+2\pi rh\left(\frac{dr}{dt}\right)$

At $r=4m$and $h=6m,$

$\begin{array}{rcl}\therefore \left(\frac{dV}{dt}\right)& =& \pi {\left(4\right)}^2\left(-4\right)+2\pi \left(4\right)\left(6\right)\left(3\right)\\ & =& -64\pi +144\pi \\ & =& 80\pi m^3/s\end{array}$

Hence option A is correct