calculus
posted by joseph .
At the instant when the radius of a cone is 3 inches, the volume of the cone is increasing at the rate of 28.27433388 cubic inches per minute. If the height is always 3 times the radius, find the rate of change of the radius at that instant?

For a cone, V = (1/3)(pi)(r^2)(h).
But h=3r, so V = (pi)r^3.
Differentiate with respect to time.
dV/dt = (3pi)(r^2)(dr/dt)
dr/dt = (3pi)(r^2)/(dV/dt)
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