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Rate of change

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data
    A bathtub with shape semi-circle has radius 5 m is filled with water with rate 2 m^3/minute. How fast will the height of water in the bathtub change when it is 2 m from the bottom? Given that the volume of water when the radius = R and height = y is πy^2(R - y/3)


    2. Relevant equations
    differential


    3. The attempt at a solution
    The volume of bathtub (constant) = 2/3 πr^3

    The rate of change of volume of water :
    [tex]\frac{dV}{dt}=2\frac{m^3}{minute}[/tex]

    I have to find dy/dt. I think I have to find the relation between R and y, but I don't know how...

    Thanks
     
  2. jcsd
  3. Oct 10, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    R doesn't change. R=5 m. Only V and y change. And they already gave you the relation between them. V=πy^2(R - y/3). That part of the job is done. You just have to start with the differentiation.
     
  4. Oct 10, 2009 #3
    Hi Dick

    Oh I Think R is the radius of the water, not the bathtub. I just realized that's not possible.

    Thanks a lot
     
  5. Oct 10, 2009 #4

    Mark44

    Staff: Mentor

    Can you provide the exact statement of the problem? Saying that "A bathtub with shape semi-circle" doesn't clearly describe it. Is a vertical cross-section of the tub semicircular? That would probably be the most reasonable interpretation, but I suppose it could have a horizontal cross-section that is semicircular. In either case, a radius of 5 m. seems very large for a bathtub. A radius of .5 m seems more realistic.
     
  6. Oct 10, 2009 #5
    Hi Mark44

    I got this question from my friend and that's all what he told me. I interpreted it as the vertical cross-section. Will it be different if it's horizontal cross-section? I think it will be the same.
    About the radius, yeah 5 m is ridiculous for a bathtub. Maybe it's a pool :smile:
     
  7. Oct 11, 2009 #6

    Mark44

    Staff: Mentor

    You are given a formula for the volume of water in the pool when the water is y feet deep, and the radius of the pool is R feet:
    V = [itex]\pi y^2(R - y/3)[/itex]

    Differentiate both sides with respect to t, using the chain rule. That will give you
    dV/dt = (something involving y)*dy/dt

    Solve this equation for dy/dt, and substitute in all the other known quantities at the time when y = 2 feet.
     
  8. Oct 11, 2009 #7
    Hi Mark44

    Ok Thanks a lot Mark :smile:
     
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