Thread: cylindrical tank - how fast water drops

1. cylindrical tank - how fast water drops

Suppose that we pump water into an inverted right-circular conical tank at the rate of 5 cubic feet per minute (i.e., the tank stands point down.) The tank has a height of 6ft and the radius on top is 3ft. What is the rate at which the water level is rising when the water is 2ft deep? (Note that the volume of a right circular cone of radius $r$ and height $h$ is $V = \frac{1}{3}\pi r^2h$.

2. Originally Posted by centenial
Suppose that we pump water into an inverted right-circular conical tank at the rate of 5 cubic feet per minute (i.e., the tank stands point down.) The tank has a height of 6ft and the radius on top is 3ft. What is the rate at which the water level is rising when the water is 2ft deep?
$\frac{dV}{dt} = 5 \, ft^3/min$

the question asks for the value of $\frac{dh}{dt}$ when $h = 2 \, ft$

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

by similar triangles ... $r = \frac{h}{2}$

substitute $\frac{h}{2}$ for $r$ in the volume equation to get $V$ as a function of $h$ only ... take the time derivative, substitute in your given values and determine the value of $\frac{dh}{dt}$

3. just curious what answer you came up with?

4. Does this look good?

$V=\frac{1}{3}\pi(\frac{h}{2})^2h$

$V=\frac{1}{12}\pi(h)^2h$

$V=\frac{1}{12}\pi h^3$

$\frac{dV}{dt}=\frac{1}{12}\pi 3h^2(\frac{dh}{dt})$

$\frac{dV}{dt}=\frac{1}{4}\pi h^2(\frac{dh}{dt})$

$\frac{dV}{dt}=\frac{1}{4}\pi (4)(\frac{dh}{dt})$

$\frac{dV}{dt}=\pi (\frac{dh}{dt})$

$\frac{\frac{dV}{dt}}{\pi}=(\frac{dh}{dt})$

$\frac{5}{\pi}=\frac{dh}{dt}$

$\frac{dh}{dt}=1.59 ft/min$