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A right circular cone is inscribed in a hemisphere.

  1. Oct 2, 2003 #1
    A right circular cone is inscribed in a hemisphere. The figure is expanding in such a way that the combinded surface area of the hemisphere and its base is increasing at a constant rate of 18 in^2 per second. At what rate is the volume of the cone changing when the radius of the common base is 4 in?
     
    Last edited by a moderator: Feb 6, 2013
  2. jcsd
  3. Oct 3, 2003 #2

    HallsofIvy

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    Are you sure this shouldn't be under "homework help"? :smile:

    I assume that the cone is inscribed in the hemisphere so that its base is the circular base of the hemispher.

    If I remember correctly, the surface area of a sphere is 4 &pi r2 so the surface area of a hemisphere is 2 &pi r2 and the "combinded surface area of the hemisphere and its base" is A= 3 &pi r2. Knowing that dA= 18 square inches per second, you should be able to find the rate of change of r from that.

    The volume of a right circular cone is given by V= (1/3) &pi r2h (I confess I looked that up). In this case, the height of the cone, as well as the radius of its base, is the radius of the hemisphere so V= (1/3) &pi r3. Since you have already calculated dr/dt, you can use that to find dV/dt.

    (Why are the "& codes" that Greg Barnhardt noted not working for me?)
     
  4. Oct 6, 2003 #3
    do they need to be (and are they) followed by semicolons?
     
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