Ok, we did it in class but I still can't get it right because I suck at related rates.

Water is leaking out of an inverted conical tank at a rate of 13000 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.

V= 1/3 * pi * r^2 *h
dv/dt = x -13000
dh/dt = 20

I have to find the radius and I am pretty sure that is what I'm doing wrong.

The way we did it in class was set up two triangles. So the tank is 6 m tall and the r = 2, the height of the water is 2 m and I need to find the radius at that height. So I did r/2=2/6. That gives me r =2/3.

So V= 1/3 * pi * (2/3)^2 * h
V= (4*pi*h)/27
dv/dt = ((4*pi)/27) * dh/dt
dv/dt = 9.3
9.3 = x - 13000
x = wrong answer

Now, I know that I didn't convert the m to cm, but I have done it since and the answer is still wrong. So somewhere I am screwing up the process. Any help is appreciated.

Thanks

10/23/2006 8:13:28 PM

You plugged in too soon!

The water is in the shape of a cone. When the volume of this water increases, not only does the height of this cone change, but its radius changes as well. So you can't just plug in r=2/3 and then take the derivative. You gotta take the derivative first.

10/23/2006 8:45:56 PM

ok, I got

dv/dt = 2/3 * pi * r * dr/dt * dh/dt

but I have no clue how to find dr/dt?!

10/23/2006 9:43:00 PM

You differentiated wrong. Remember that you have to use the product rule here.

Earlier you found r when h=2. Use the same idea to find dr/dt.

10/23/2006 10:21:15 PM