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December 22, 2014

December 22, 2014

Posted by **Mehak** on Saturday, March 12, 2011 at 6:04pm.

- Calculus -
**Damon**, Saturday, March 12, 2011 at 6:29pmThe important thing to realize here is that the rate of water loss is the surface area times the rate of depth change.

dV/dt = pi r^2 dh/dt

where h is the depth and both dV/dt and dh/dt are negative

-1000 m^3/min = pi (200^2)dh/dt

when h = 50

then what is r when h = 20?

r is proportional to h (cone geometry

so

20/50 = r/200

r = 400/5 = 80

so then

-1000 = pi(80^2) dh/dt

- Calculus -
**Mehak**, Saturday, March 12, 2011 at 6:45pmThaanks that helped a lot <3

- Calculus -
**Mehak**, Sunday, March 13, 2011 at 1:35amShouldnt you have used the formula of volume of cone which is 1/3 pi r^2 h . And the answer should be 8 or 50 but by your procedure i m not getting it .I appreciate you help.

- Calculus -
**Damon**, Sunday, March 13, 2011 at 6:35amNo, you do not need volume of cone, just the surface area of the water.

Look more formally

V = (1/3) pi r^2 h

where r = (200/50) h = 4 h

V = (1/3) pi (16) h^3

dV/dh = 16 pi h^2

dV = 16 pi h^2 dh

dV/dt = 16 pi h^2 dh/dt

but h = r/4

so

h^2 = r/16

so

dV/dt = pi r^2 dh/dt

which you could see immediately by considering a slice of depth dh at the surface.

- Calculus -
**Mehak**, Sunday, March 13, 2011 at 12:21pmOkaay thaat makes sense but do i get 8 or 50 as your answer if u solve it fully .thanks .

- Calculus -
**Damon**, Sunday, March 13, 2011 at 3:50pm-1000 m^3/min = pi (200^2)dh/dt

when h = 50

-----------------------

so

dh/dt = -.008 meters/minute = - 0.8 cm/min

-------------------

then what is r when h = 20?

r is proportional to h (cone geometry

so

20/50 = r/200

r = 400/5 = 80

so then

-1000 = pi(80^2) dh/dt

------------------

so

dh/dt = - .05 m/min = - 5 cm/min

- Calculus -
**Mehak**, Sunday, March 13, 2011 at 7:49pmThanks

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