Calculus
posted by Mehak .
a pond that lies in a crater is being drained at the rate of 1000m3/min.If the crater has the shape of an inverted cone of radius 200(at he original water level) and original depth is 50 m (at the center) find that rate at which the water level begins to fall.How fast is it falling when the water at the center is only 20 m deep?

The 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 
Thaanks that helped a lot <3

Shouldnt 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.

No, 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. 
Okaay thaat makes sense but do i get 8 or 50 as your answer if u solve it fully .thanks .

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 
Thanks
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