- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,817

A solution is draining through a conical filter into an identical conical container

(both are $h=12$ and $r=4$ at top of cone)

The solution drips from the upper filter into the lower container at a rate of

$\displaystyle\frac{\pi\ cm^3}{\text{ sec}}$ and $\displaystyle V_{cone}=\frac{\pi}{3}\cdot r^2\cdot h$

since $\displaystyle r=\frac{h}{3}$ then $\displaystyle V_{cone}=\frac{\pi}{27}\cdot h^3$

When $h=6$ cm then $\displaystyle\frac{dV}{dt}=\frac{\pi}{3} 6^3 \frac{dh}{dt}=72\pi \frac{dh}{dt}$

So $\displaystyle\frac{dh}{dt}=\frac{1}{72\pi}\cdot \frac{\pi\ cm^3}{sec}=.014 \frac{cm^3}{sec}$

what is the level of the solution in the lower level when the solution level in the upper filter is a 6cm

and now fast is the level in the lower filter rising?

when the volume in the upper filter equals the volume in the lower container?

not sure that

(both are $h=12$ and $r=4$ at top of cone)

The solution drips from the upper filter into the lower container at a rate of

$\displaystyle\frac{\pi\ cm^3}{\text{ sec}}$ and $\displaystyle V_{cone}=\frac{\pi}{3}\cdot r^2\cdot h$

since $\displaystyle r=\frac{h}{3}$ then $\displaystyle V_{cone}=\frac{\pi}{27}\cdot h^3$

**a.**How fast is the level in the upper filter dropping when the solution level in the upper filter is at $6 cm$?When $h=6$ cm then $\displaystyle\frac{dV}{dt}=\frac{\pi}{3} 6^3 \frac{dh}{dt}=72\pi \frac{dh}{dt}$

So $\displaystyle\frac{dh}{dt}=\frac{1}{72\pi}\cdot \frac{\pi\ cm^3}{sec}=.014 \frac{cm^3}{sec}$

**b.**If the conical filter is initially full,what is the level of the solution in the lower level when the solution level in the upper filter is a 6cm

and now fast is the level in the lower filter rising?

**c.**How fast is the surface area of the solution in the lower filter increasingwhen the volume in the upper filter equals the volume in the lower container?

not sure that

**a.**is correct, but if it is, need some hints for**b.**and**c.**
Last edited: