Question

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

• A. $80\, \pi m ^{3} / s$
• B. $144\, \pi m ^{3} / s$
• C. $80 \,m ^{3} / s$
• D. $64 \,m ^{3} / s$

Answer 1

Answer: (A)

Let the radius and height of cylinder are $r$ and $h$
Given that, $\frac{ dr }{ dt }=3 \,m / s , \frac{ d }{ dt }=-4 \,m / s$
Also, $V =\pi r ^{2} h$
On differentiating w.r.t t, we get
$\frac{ dV }{ dt }=\pi\left[ r ^{2} \frac{ dh }{ dt }+ h \cdot 2 r \frac{ dr }{ dt }\right]$
At $r =4\, m$ and $h =6 \,m$
$\frac{ dV }{ dt }=\pi[-64+144]$
$=80\, \pi m ^{2} / s$