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Rate of change - cones

  1. Mar 14, 2012 #1
  2. jcsd
  3. Mar 14, 2012 #2


    Staff: Mentor

    Please post the problem and your work here. IMO, it's a pain in the butt to have to open a web page to see the problem and the work, plus I can't insert a comment at the appropriate place where there's an error.
  4. Mar 14, 2012 #3

    dv/dt = 4 cm3 min-1

    tan(60) = r/h

    r = √3* h

    V = (1/3) π r2h

    dv/dh = (1/3) π r2

    dh/dt = (dh/dv) * (dv/dt)

    = 1/(1/3) π r2 * 4

    = 12/(pi r2)

    for h = 4

    dh/dt = 0.079577

    The answer given is 0.0265 cm/min. why?

    I don't know how do use latex - see post for a clearer solution if you get lost!
  5. Mar 14, 2012 #4
    question was:

    A hollow cone with a semi-vertical angle of 60 degrees is held vertex down with its axis vertical.
    Water drips into the cone at 4 cm3/min
    Find the rate at which the depth of water is increasing when the water is 4 cm deep
  6. Mar 14, 2012 #5


    Staff: Mentor

    You're ignoring the relationship between r and h.

    Substitute for r in your volume equation. Then you'll have V purely as a function of h.
    What you have is not correct, because V is a function of r and h.
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