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March 31st 2009, 03:48 PM
tradar

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Coffee is draining from a conical filter into a cylindrical coffeepot at the rate of 5 cubic centimeters per minute. The conical filter measures 10 cm tall and 10 cm across its top. The diameter of the coffeepot is 10 cm.

How fast is the level of coffee in the pot rising when the level in the filter is 8 cm deep? Remember...the diameter of the coffeepot is 15 cm.

dV/dt= +10 cm^3/min
V=TTr^2h
r is a constant
V=TT(7.5)^2h
I know I have to take the time derivative and solve for dh/dt, but I can't figure it out...
March 31st 2009, 05:19 PM
skeeter

Quote:

Originally Posted by tradar

Coffee is draining from a conical filter into a cylindrical coffeepot at the rate of 5 cubic centimeters per minute. The conical filter measures 10 cm tall and 10 cm across its top. The diameter of the coffeepot is 10 cm.

How fast is the level of coffee in the pot rising when the level in the filter is 8 cm deep? Remember...the diameter of the coffeepot is 15 cm.

so ... which is it? 10 or 15 cm for the coffeepot diameter ?

level in the cone doesn't matter ... you were given the rate of volume flow from the cone to the pot ... 5 cc/min

cylindrical coffeepot volume ... (r is a constant)

$V = \pi r^2 h$

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

$5 = \pi r^2 \frac{dh}{dt}$

$\frac{5}{\pi r^2} = \frac{dh}{dt}$

$\frac{dh}{dt}$ will be in cm/min
April 2nd 2009, 01:55 PM
tradar

Sorry...it was 10 cm.

Ok...

dV/dt =TTr^2 dh/dt

5 = TTr^2 dh/dt

5/TTr^2 = dh/dt

5/TT(5)^2 = dh/dt

0.0636619772 cm/min Is that right?

Do I have to change that into cm^3?
April 4th 2009, 02:31 PM
skeeter

Quote:

Originally Posted by tradar

Was my answer correct, and did I need to change it to cm^3?

as stated previously, $\frac{dh}{dt}$ is in cm/min ... cm^3 is a measure of volume, not the rate of change of height w/r to time.

$\frac{dh}{dt} = \frac{1}{5\pi} \approx .064$ cm/min