Question

A cone whose height is always equal to its diameter is increasing in volume at the rate of 40 $\frac{c{m}^3}{sec}$. At the what rate is the radius increasing when its circular base area is 1 $m^2$?

• A. $1\frac{mm}{sec}$
• B. $20.001\frac{cm}{sec}$
• C. $2\frac{mm}{sec}$
• D. $0.002\frac{cm}{sec}$

Answer 1

The correct option is D $0.002\frac{cm}{sec}$

Given $h=2r$ ,$\frac{dV}{dt}=40c{m}^3/sec$

Also, Area of base $=\pi r^2=1m^2=10000c{m}^2$

Volume of cone $V=\frac{1}{3}\pi r^{2}h$

$\Rightarrow V=\frac{2}{3}\pi r^3$

$\Rightarrow \frac{dV}{dt}=2\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=0.002cm/sec$