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Rate of change

  1. Feb 16, 2013 #1
    1. The problem statement, all variables and given/known data
    An inverted right circular cone of vertical angle 120 is collecting water from a tap at a steady rate of 18∏ cm^3/min. Find
    a) the depth of water after 12min,
    b) rate of increase of depth at this instant



    2. Relevant equations



    3. The attempt at a solution

    All I know is that dV/dt = 18cm3/min
    so shouldn't the depth after 12 mins be 12 * dv/dt?
     
  2. jcsd
  3. Feb 16, 2013 #2

    haruspex

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    .. which would imply a depth of what, 216 cm3? Anything strike you as odd about that?
     
  4. Feb 16, 2013 #3
    Well yeah that is way too much :S
     
  5. Feb 16, 2013 #4

    haruspex

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    No, I meant depth measured in cm3 (!)
     
  6. Feb 16, 2013 #5
    Depth would be dh/dt, h being that height
     
  7. Feb 16, 2013 #6

    vela

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    That would give you units of cm/min, which isn't the units of depth. Based on your answer, I'm not sure you understand what the word depth means in the context of this problem.

    Why don't you start by looking up the formula for the volume of a cone? Sketch a picture and tell us how the variables in the formula relate to the physical quantities in this problem.
     
  8. Feb 16, 2013 #7

    haruspex

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    The depth of the water is the height of the surface measured from the point of the cone. So it's a distance, not a volume, not a rate. dh/dt would be the rate of change of the depth.
     
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