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The radius of a cone is decreasing at a constant rate of 6 meters per minute. The volume remains a constant 154 cubic meters. At the instant when the height of the cone is 33 meters, what is the rate of change of the height? The volume of a cone can be found with the equation V=\frac{1}{3}\pi r^2 h. Round your answer to three decimal places.

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sqdancefan

9514 1404 393

Answer:

  h' ≈ 187.588 . . . . m/min

Step-by-step explanation:

We need to know the radius at the given height.

  V = (1/3)πr^2h

  r^2 = 3V/(πh)

  r = √(3V/(πh)) = √(3·154/(π·33)) = √(14/π) . . . . meters

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The rate of change of height can be found by solving for height, then differentiating with respect to radius.

  V = (1/3)πr^2h

  h = 3V/(πr^2)

  h' = -6V/(πr^3)r' = -6(154)/(π(14/π)^(3/2))(-6) = 198√(2π/7)

  h' ≈ 187.588 . . . . m/min

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Using a graphing calculator, we can put an expression for r(t) in the equation for h(t), and use the calculator's differentiating function to find the rate of change of height.

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