Posted by Jordan on Monday, January 9, 2012 at 4:40pm.
A tank has the shape of an inverted circular cone with a base radius of 5 meters
and a height of 20 meters. If water is being pumped into the tank at 2 cubic meters
per minute, find the rate at which the water level is rising when the water is 7
meters deep

Calculus (Help please)  Jordan, Monday, January 9, 2012 at 4:42pm
is the answer 8/49pi= 0.052 meters per minute

Calculus (Help please)  Steve, Monday, January 9, 2012 at 5:06pm
I get dV/dt = pi/16 h^2 dh/dt
giving
2 = 49pi/16 dh/dt
or
dh/dt= 32/49pi
One of us has a factor of 2 out of place. Could be me ...

Calculus (Help please)  Reiny, Monday, January 9, 2012 at 5:12pm
I did not get that
Let the height of the water be h m and the radius of the water level be r m
by ratios:
r/h = 5/20 = 1/4
r = h/4
V = (1/3)πr^2 h = (1/3)π(h^2/16)(h)
= (1/48)π h^3
dV/dt = (1/16)π h^2 dh/dt
so when dV/dt = 2, h = 7
2 = (1/16)π(49) dh/dt
2(16)/(49π) = dh/dt
= 32/(49π)
= appr. .208 m/min

Calculus (Help please)  Jordan, Monday, January 9, 2012 at 5:20pm
WELL I HAVE THIS
v=1/3pi*r^2h
v=1/3pi*r2(4r)
v=4/3pi*r^3
dv/dt=4pi*r^2
2=4pi(7/4)^2
2=49pi/4 * dr/dt
dr/dt= 8/49pi = 0.052 meters per minute

Calculus (Help please)  Reiny, Monday, January 9, 2012 at 5:32pm
You have found the rate at which the rate is changing.
Unfortunately, your question asked for how fast the height is changing.

typo  Calculus (Help please)  Reiny, Monday, January 9, 2012 at 5:34pm
should say:
You have found the rate at which the radius is changing.
Unfortunately, your question asked for how fast the height is changing.
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