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Related rates and sand

  1. Oct 8, 2005 #1
    Sand is falling onto a cone-shaped pile at the rate of 9pi cubic feet per minute. the diameter of the base is always 3 times the height of the cone. At what rate is the height of the pile changing when the pile is 12 feet high.


    work:

    dV/dt=+9 ft^3/min

    d=3h then r=3h/2 dr/dt=(3/2) dH/dt

    Find dH/dt when h=12 ft. which means r=18 ft

    V=pi(r)^2h/3

    dV/dt=(pi/3)(r^2*dh/dt+h*2r*(3/2)dH/dt)

    9pi ft^3/min=(pi/3)(r^2*dh/dt+h*2r*(3/2)dH/dt)

    would this be the correct way to find the answer and get dH/dt i get 27/792 ft/min is this correct?
     
  2. jcsd
  3. Oct 8, 2005 #2

    mezarashi

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    Homework Helper

    You nearly have it right. However if you have two variables r and h, when differentiating you would've needed to have a dr/dt as well, since r is also a function of t. However, to make things easier, you can sub in 2r = d = 3h.

    V = 1/3(base)(height)
    V = pi/12(diameter^2)(height)
    V = pi/12(9height^2)(height)
    V = 9pi/12(height^3)

    Now you can easily differentiate both V and h with respect to time.
     
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