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Homework Help: Another related rates problem

  1. Jan 22, 2007 #1
    The volume V of a cone (V = 1/3*pi*r^2*h) is increasing at the rate of 28pi cubic units per second. At the instant when the radius r of the cone is 3 units, its volume is 12pi cubic units and the radius is increasing at 1/2 unit per second. At the instant when the radius of the cone is 3 units, what is the rate of change of the area of its height h?

    Know: dV/dt = 28pi cubic units per second, dr/dt = 1/2 unit per second, and dA/dt = 3pi units square per second. Need: dh/dt
    12pi = 1/3pi*3^2*h

    V = 1/3*pi*r^2*h
    dV/dt = 1/3*pi*2r*dr/dt*h + r^2*dh/dt (use the product rule. do you multiply 1/3 pi times everything like this 1/3*pi(2r*dr/dt*h + r^2*dh/dt) or how I did it?)
    28pi = 1/3pi*2(3)*1/2*4 + 6*dh/dt
    24pi = 6*dh/dt
    dh/dt = 4pi units per second

    Is this correct?
  2. jcsd
  3. Jan 23, 2007 #2


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    What you have is wrong. You can either think of this as V= (1/3)(pi r^2h) so that dV/dt= (1/3)d(pi r^2h)/dt or do it as V= ((1/3)pir^2)(h) so that dV/dt= {d((1/3)pir^2)/dt}h+ ((1/3)pi r^2)(dh/dt). Either way, the (1/3)pi multiplies the entire derivative:
    dV/dt= (2/3)pi rh dr/dt+ (1/3)pi r^2 dh/dt.

    28pi= (1/3)pi(2)(3)(1/2)(4)+ (2/3)pi (9) dh/dt
    (Where did you get that "6"? if r= 3, r^2= 9, not 18!
    No, it isn't!
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