Rate of change of a cone

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• March 17th 2010, 02:20 PM
Frostking
Rate of change of a cone
I am trying to figure out a practice problem while studying for a calculus final and I do not know how to approach this problem. Any help would be much appreciated. The height of a cone is decreasing at 2 cm per second while its radius is increasing at 3cm per second. When the radius is 4 cm and the height is 6 cm, at what rate is the volume of the cone changing?

I started with the volume for a cone which is: 1/3 pi r^2 h

I thought I could take the derivative of volume with respect to radius and multiply that times the rate of change per sec of the radius and then add to that the derivative of volume with respect to height multiplied by the cnage per sec of h. When I do all this I get that the volume of the cone is decreasing by 13.8 cm^3 per second which does not seem reasonable since the radius is increasing by more than the height is decreasing and the radius is a squared term in the original volume equation???!!!! Thanks for all you folks time and effort! Frostking
• March 17th 2010, 02:26 PM
TKHunny
Quote:

Originally Posted by Frostking
V(t) = 1/3 pi r(t)^2 h(t)

With a little modification.

Now find the derivatvie with respect to 't'. It will take a product rule. No mysterious adding or other tomfoolery.

$\frac{dV}{dt}\;=\;\frac{\pi}{3}\cdot \left(r^{2}\frac{dh}{dt} + 2r\frac{dr}{dt}\cdot h\right)$
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