A right circular cone is inscribed in a hemisphere.

  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. HallsofIvy

    HallsofIvy 40,939
    Staff Emeritus
    Science Advisor

    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. do they need to be (and are they) followed by semicolons?
     
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?