A cone with height h(t) increasing; if h' = 2 cm/s, H=10, R=5, find dV/dt when h=6 for V = (1/12)π h^3.

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Multiple Choice

A cone with height h(t) increasing; if h' = 2 cm/s, H=10, R=5, find dV/dt when h=6 for V = (1/12)π h^3.

Explanation:
As the cone grows, its volume changes in step with its height. Use related rates by expressing V in terms of h and then differentiating with respect to time. The radius scales with height for similar cones, so r/h = R/H. With R = 5 and H = 10, r = (R/H) h = (1/2) h. Then the volume is V = (1/3) π r^2 h = (1/3) π ( (1/2 h)^2 ) h = (π/12) h^3. Differentiate with respect to t: dV/dt = (π/12) * 3 h^2 * dh/dt = (π/4) h^2 dh/dt. Given dh/dt = h' = 2 cm/s and h = 6 cm, we get dV/dt = (π/4) * 36 * 2 = 18π cm^3/s. So the volume is changing at 18π cubic centimeters per second.

As the cone grows, its volume changes in step with its height. Use related rates by expressing V in terms of h and then differentiating with respect to time.

The radius scales with height for similar cones, so r/h = R/H. With R = 5 and H = 10, r = (R/H) h = (1/2) h.

Then the volume is V = (1/3) π r^2 h = (1/3) π ( (1/2 h)^2 ) h = (π/12) h^3.

Differentiate with respect to t: dV/dt = (π/12) * 3 h^2 * dh/dt = (π/4) h^2 dh/dt.

Given dh/dt = h' = 2 cm/s and h = 6 cm, we get dV/dt = (π/4) * 36 * 2 = 18π cm^3/s.

So the volume is changing at 18π cubic centimeters per second.

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