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Posted by **Jordan** on Monday, January 9, 2012 at 4:40pm.

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 4:42pmis the answer 8/49pi= 0.052 meters per minute

- Calculus (Help please) -
**Steve**, Monday, January 9, 2012 at 5:06pmI 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 5:12pmI 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 5:20pmWELL 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 5:32pmYou 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 5:34pmshould 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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