If water was leaking from an inverted conical tank at a rate of 10 000 cm³/min and at the same time water form another source is entering the tank at a constant rate Q. The tank is 6m high and the diameter at the top is 4m (r=2m=200cm). If the water level is rising at a rate of 20cm/min when the hight of water is 2m (200cm), find Q (rate going in the tank).

That is the question, I have converted all the numerics to cm, volume of cone as (1/3)pi.r²h and differentiated this with respect to time using the product rule to get:

dV/dt= (2/3)pi.r.(dr/dt)h+(1/3)pi.r².(dh/dt)

with r=(1/3)h=200/3 (this is from drawing diagram and letting (r/200)=(h/600))
h=200
(dr/dt)=(dr/dh)x(dh/dt)=(1/3)x20=20/3
(dh/dt)=20

and ended with an answer of (dV/dt), then reasoning that Q=(dV/dt)-10 000(rate going out of tank) to get 269252.6803cm³/min but the answer should have been (10000+800000.pi)/9=289252.6803 cm³/min, where have I gone wrong or am I totally off track?

If I follow your work correctly, it looks find. You probably just have an arithmetic error since your solution is so similar.

The volume of a cone is given by ... [1]

By similar triangles, we have



Sub into [1]:



Differentiate implicitly:



But,



dh/dt=20 and h=200 cm:



Solving for Q gives

You error comes in here:

Quote:

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and ended with an answer of (dV/dt), then reasoning that Q=(dV/dt)-10 000(rate going out of tank) to get 269252.6803cm³/min but the answer should have been (10000+800000.pi)/9=289252.6803 cm³/min, where have I gone wrong or am I totally off track?

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It is Q=(dV/dt)+10 000

Hey your really good! I figured this out after sitting down and playing around a bit and writing the DE again. :)

Quote:

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Originally Posted by Unknown008

You error comes in here:



It is Q=(dV/dt)+10 000

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Quote:

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Originally Posted by Vi Nguyen

Hey ur really good! I figured this out after sitting down and playing around a bit and writing the DE again. :)

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You're welcomed! :)