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Rate of change question

  1. Mar 17, 2013 #1
    1. The problem statement, all variables and given/known data
    Water runs into a cylindrical tank at the rate of 0.6m^3/s. The tank has a heigh of 2.5m and a base radius of 1.2m. How fast is the water level rising when the water level is 1.6m deep.


    2. Relevant equations
    V=∏ x r^2 x h


    3. The attempt at a solution
    I know that 0.6m^3/s is dV/dt which should mean that the equation looks something like this: dV/dt=(dV/dh)(dh/dt). When differentiating volume with respect to height though, the h is eliminated and is makes no sense if the question is giving the h variable.
     
  2. jcsd
  3. Mar 17, 2013 #2

    Doc Al

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    Staff: Mentor

    Yes, the rate of water level rise is independent of height. (Trying to trick you, no doubt.)
     
  4. Mar 17, 2013 #3
    So then that means 0.6=(∏ x 1.2^2)(dh/dt)
    0.6/(∏ x 1.2^2)=dh/dt
    0.132m^3/s=dh/dt
     
  5. Mar 17, 2013 #4

    Doc Al

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    Staff: Mentor

    Looks good. Except for the units!
     
  6. Mar 17, 2013 #5
    oh yes, sorry haha. Height isn't a volume :S
     
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