Related Rates Help

1. Mar 1, 2009

1.Suppose the wood nymphs and satyrs are having a hot party in honor of Bacchus and the wine is flowing freely from the bottom of a giant cone-shaped barrel which is 12 feet deep and 6 feet in radius at the top. if the wine is disappearing at a rate of 6 cubic feet per hour, at what rate is the depth of wine in the tank going down when the depth is 4 feet?

2. Relevant equations
volume of a cone: 1/3(pi)(r^2)(h)

2. Mar 1, 2009

Bacat

What have you done to try and solve it?

3. Mar 1, 2009

well,
i figured that dv/dt= -6
and that im looking for dh/dt
so:
v=(1/3)(pi)(r^2)h
v=(1/3)(pi)(4)h
v=(4/3)(pi)h
dv/dt=(4/3)(pi)(dh/dt)
-6=(4/3)(pi)(dh/dt)
-24/3(pi)= dh/dt

is that right?

4. Mar 2, 2009

HallsofIvy

Staff Emeritus
You need to have V as a function of h only, not both h and r. You seem to have done that by replacing r by 2 and I can see no reason to do that.

You are told that the entire cone has h= 12 and r= 6 and, as the level reduces it must retain that shape and that ratio: r/h= 6/12= 1/2 so r= (1/2)h.

Oh, I see where you got r= 2: when h= 4, r must be half that, 2. But when you differentiate, you need r as a function of h, not just the value at that particular time.