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.

9514 1404 393

  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

__

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

_____

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.

ok thats very interesting

none of these

step-by-step explanation:

23 + 34 add up to 57 so just inform the teacher of this really stupid mistake