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Related Rates in a water tank

  1. Nov 15, 2011 #1
    1. The problem statement, all variables and given/known data
    A water tank has the shape of an inverted right-circular cone, with radius at the top 15 meters and depth 12 meters. Water is flowing into the tank at rate of 2 cublic meters per minute. How fast is the depth of water in the tank increasing at the instant when the depth is 8 meters


    2. Relevant equations
    V=(1/3)(pi)(r^2)(h)

    3. The attempt at a solution
    dv/dt=2 meter^3/min
    dv/dt=(1/3)(pi)(2rh(dr/dt)+(r^2)(dh/dt))
    but that has 2 unknow varibles in it.
     
  2. jcsd
  3. Nov 15, 2011 #2
    You need to relate the radius in terms of the height of the cone since both are changing with respect to time and because you only know the rate of change of the height. Once you do this, you will be able to differentiate it since you know the rate of change of the height. Try to find a way to relate the radius in terms of height. With this information can you figure out how to get rid of r and put it in terms of h?
     
  4. Nov 15, 2011 #3
    Yeah thanks that helps a lot
     
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