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Related Rates and wood nymphs

  1. Mar 1, 2009 #1
    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. jcsd
  3. Mar 1, 2009 #2
    What have you done to try and solve it?
     
  4. Mar 1, 2009 #3
    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?
     
  5. Mar 2, 2009 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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.
     
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