# Related Rates problem

• April 19th 2010, 08:12 PM
Related Rates problem
Hey everybody. I need help understanding this problem

A solution is passing through a conical filter 24 cm deep and 16 cm across the top into a cylindrical vessel of diameter 12 cm.
At what rate is the level of the solution in the cylinder rising if when the depth of the solution in the conical filter is 12 cm

Its level is falling at the rate of 1 cm/min in the filter
• April 19th 2010, 11:20 PM
ione
Quote:

Hey everybody. I need help understanding this problem

A solution is passing through a conical filter 24 cm deep and 16 cm across the top into a cylindrical vessel of diameter 12 cm.
At what rate is the level of the solution in the cylinder rising if when the depth of the solution in the conical filter is 12 cm

Its level is falling at the rate of 1 cm/min in the filter

The volume(v) of solution in the conical filter is

$v=\frac{1}{3}\pi r^2h$

$\frac{r}{h}=\frac{8}{24}$

$r=\frac{h}{3}$

$v=\frac{1}{3}\pi (\frac{h}{3})^2h=\frac{1}{27}\pi h^3$

$\frac{dv}{dt}=\frac{1}{9}\pi h^2 \frac{dh}{dt}$

The level(h) of the solution in the conical filter is falling at the rate of 1 cm/min so

$\frac{dh}{dt}=-1$

When $h=12$

$\frac{dv}{dt}=\frac{1}{9}\pi (12)^2(-1)=-16\pi$

The volume(V) of solution in the cylinder is

$V=\pi(6)^2H=36\pi H$

$\frac{dV}{dt}=36\pi \frac{dH}{dt}$

When $h=12$, $\frac{dv}{dt}=-16\pi$

So $\frac{dV}{dt}=16\pi$

$16\pi=36\pi \frac{dH}{dt}$

Solve for $\frac{dH}{dt}$
• April 20th 2010, 04:14 AM