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 $4m/s$. The rate of change of volume when radius is $4m$ and altitude is $6m$, is

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

Answer 1

Answer: (A)

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