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A right circular cone is inscribed in a hemisphere. 
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#1
Oct203, 02:32 PM

P: 46

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?



#2
Oct303, 06:47 AM

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P: 39,497

Are you sure this shouldn't be under "homework help"?
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 r^{2} so the surface area of a hemisphere is 2 &pi r^{2} and the "combinded surface area of the hemisphere and its base" is A= 3 &pi r^{2}. 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 r^{2}h (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 r^{3}. 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?) 


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