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AP calculus problem

  1. Oct 4, 2007 #1
    1. The problem statement, all variables and given/known data

    Water is drained from a conical tank with height 12ft and diameter 8ft into a cylindrical tank that has a base with area 400Π square feet. the deep, h, in feet, of the water in the conical tank is changing at the rate of 12 ft/min

    a. write an expression for the volume of water int eh conical tank as a function of h
    b. at what rate is the volume of water in the conical tank changing when h=3?
    c. At the same time, h=3, at what rate is the radius changing?
    d. let y be the depth, in feet, of the water in the cylindrical tank. at what rate is y changing when h=3?

    2. Relevant equations
    V= 1/3Πr^2h


    3. The attempt at a solution

    for a I think is just rearranging the volume equation right?, for part B I had dV/dt = 324Π, can somebody check is that is ok, and for c and d I'm lost.


    Thanks in advance for all your help!!!!
     
  2. jcsd
  3. Oct 4, 2007 #2

    dynamicsolo

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    What do you mean by "rearranging"? It will help a reader if you show your calculations.
     
  4. Oct 4, 2007 #3
    I mean that V=1/3Πr^2h

    then 3V=Πr^2h
    3V/Πr^2 = h
     
  5. Oct 4, 2007 #4

    dynamicsolo

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    That isn't what it meant by "expressing the volume as a function of h". What they're looking for is the volume function for this cone written with h as the only variable. So you have to use some information to eliminate the variable r .
     
  6. Oct 4, 2007 #5
    so would it be:
    V= 3Πh^3
     
  7. Oct 4, 2007 #6

    dynamicsolo

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    I don't quite agree (though I think I know what you did) -- what is r in terms of h? (If you look at the cross-section of the conical tank with water in it, you'll see that you can use similar triangles to show that the relationship remains the same for any water level.)
     
  8. Oct 4, 2007 #7
    ok, so:
    V=1/27Πh^3
     
  9. Oct 4, 2007 #8

    dynamicsolo

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    Much better! So how would you find (dV/dt) at the moment when h = 3 ft?
     
  10. Oct 4, 2007 #9
    dV/dt = 3(1/27)Πh^2. dh/dt
    dv/dt = 12Π

    so for c do I use the same triangle relation to find dr/dt?
     
  11. Oct 4, 2007 #10

    dynamicsolo

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    Yes. (Don't forget the (ft^3)/min units on dV/dt .)
     
  12. Oct 4, 2007 #11
    and for d, how is Dv/dt from the cone the same as Dy/dt?
     
  13. Oct 4, 2007 #12

    dynamicsolo

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    The dV/dt for the cylinder will be the negative of the dV/dt for the cone, since the cylinder is being filled up.

    Oh, by the way, I forgot to mention that for your calculation

    dV/dt = 3(1/27)(pi)(h^2) · dh/dt

    that dh/dt = -12 ft/min because that tank is being drained, so the level is decreasing and we should have

    dv/dt = -12(pi) .

    So dV/dt for the cylindrical tank is +12(pi) (ft^3)/min. You'll need a formula for the volume of a cylinder, so you can turn it into a rate equation. We don't know how much water is already in the tank, but that doesn't matter, since you'll be working with rates of change...
     
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