Question

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

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

Answer 1

The correct option is A

$6cm/min$

Explanation for the correct answer

Given,

$\left(\frac{dr}{dt}\right)=5\frac{\mathrm{cm}}{\min }$

$\frac{dV}{dt}=0$

We need to find the rate of decrease in height$\frac{dh}{dt}$

We know that volume of the cylinder is

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

On differentiating the volume concerning the radius

$\frac{dV}{dt}=\pi r^2\left(\frac{dh}{dt}\right)+2\mathrm{\pi rh}\left(\frac{d\mathrm{r}}{d\mathrm{t}}\right)$

At $r=5m$and $h=3m,$

$\begin{array}{rcl}0& =& 2\pi rh\left(\frac{dr}{dt}\right)+\pi r^2\left(\frac{dh}{dt}\right)\\ \pi r^2\left(\frac{dh}{dt}\right)& =& –2\pi rh\left(\frac{dr}{dt}\right)\\ \pi {\left(5\right)}^2\left(\frac{dh}{dt}\right)& =& –2\pi \times 5\times 5\times 3\\ \left(\frac{dh}{dt}\right)& =& -6\frac{\mathrm{cm}}{\min }\end{array}$

As we need a decreasing rate of its height$\frac{dh}{dt}=6\frac{\mathrm{cm}}{\min }$, a negative sign implies decreasing.

Hence the rate of decreasing height of the cylinder is $6\frac{\mathrm{cm}}{\min }$