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Related Rate Problem Including a Cone

  1. Oct 26, 2009 #1
    A reservoir in the shape of an inverted cone has a radius of 2 meters at the top and a depth of 6 meters. Wine is poured into the reservoir at a rate of 1 m^3/sec. At what rate is the depth of the wine increasing when the depth is 4 meters?

    Help?
     
  2. jcsd
  3. Oct 29, 2009 #2

    LCKurtz

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    Have you written a formula for the volume of the wine in terms of its depth?
     
  4. Nov 1, 2009 #3
    dV/dt=1 we are asked to find dh/dt when h is 4m.
    The volume and the height are related by:
    V=(1/3)*Pi*R^2*h
    We have to eliminate R.
    Sketch the inverted cone: if h is the wine level (and r is the radius of the cone at that h) and the cone height is 9 you can use similar triangles:
    r/h=2/6 and so:
    r=h/3
    The expression is now:
    V=(1/3)*Pi*(h/3)^2 *h
    you can now differentiate each side.
     
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