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 $ and altitude is $ 6\,m $ , is:

• A. $ 80\text{ }\pi \text{ }cu\text{ }m/s $
• B. $ \text{144 }\pi \text{ }cu\text{ }m/s $
• C. 80 cu m/s
• D. 64 cu m/s
• E. $ -\text{ }80\text{ }\pi \text{ }cu\text{ }m/s $

Answer 1

Answer: (A)

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