# Rate water has to be added to Leaking cone

1. Jul 12, 2010

### angelcase

1. The problem statement, all variables and given/known data

Water is leaking out of an inverted conical tank at a rate of 6500 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.

2. Relevant equations

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

relationships:

4/2r=6/h
therefore,
r=1/3h

Substituting h for r:
V=1/3 pi (1/3h)^2(h)
V=pi/27(h)^3
V'=dv/dt= pi/27 (3h)^2dh/dt
dv/dt=pi/27 (3*200)^2(20)

since the cone is leaking we have to find:

dv/dt-6500=pi/27(3*200)^2(20)

2. Jul 13, 2010

### Tedjn

Others may want to confirm, but this looks good to me.

3. Jul 14, 2010

### angelcase

That answer is incorrect..that is why I posted it...I can't find the correct answer..I was just showing my work so someone could tell me where I went wrong...

Does anyone know??

.....

I got it..I shouldn't have the parenthesis around the 3*200, since only the 200 is squared...

Last edited: Jul 14, 2010
4. Jul 14, 2010

### Tedjn

Yes, that's right. Sorry I missed the error before. A tip for next time: if you know your answer is wrong, let us know and we (or at least I) will look more carefully for an error.

5. Jul 14, 2010

### angelcase

Thank you..and I will do that.