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Calc Help! Related Rates?
I'm having trouble figuring out this problem on my calc homework.. please help!
The area of a cone is 1/3*pi*r^2*h. The height and radius of the cone increase at a constant rate of 1/2 when r=9 and h=6. How fast is volume changing?
1 Answer
- ?Lv 71 decade agoFavorite Answer
I think it's the volume of the cone that is V = πR²h/3, not the area. Perhaps that's why you're having trouble with the problem.
V = πR²h/3
dV = 2πRhdR/3 + πR²dh/3
dV = (πR/3)(2hdR + Rdh)
dV/dt = (πR/3)(2hdR/dt + Rdh/dt)
Assuming that the given parameters are expressed in compatible units,
dR/dt = dh/dt = 1/2
dV/dt = (πR/3)(h + R/2)
With R = 9 and h = 6,
dV/dt = 3π(6 + 4.5) = 31.5π = 98.96