The radius of a cone is decreasing at a constant rate of 6 feet per second, and the volume is increasing at a rate of 42 cubic feet per second. At the instant when the height of the cone is 33 feet and the volume is 246 cubic feet, what is the rate of change of the height? The volume of a cone can be found with the equation V, equals, one third, pi, r, squared, h, .V= 31 πr 2 h. Round your answer to three decimal places.
Question
The radius of a cone is decreasing at a constant rate of 6 feet per second, and the volume is increasing at a rate of 42 cubic feet per second. At the instant when the height of the cone is 33 feet and the volume is 246 cubic feet, what is the rate of change of the height? The volume of a cone can be found with the equation V, equals, one third, pi, r, squared, h, .V= 31 πr 2 h. Round your answer to three decimal places.
Solution
The problem involves related rates, a concept in calculus. We are given the rates of change of the radius and volume of a cone, and we are asked to find the rate of change of the height.
The volume V of a cone is given by the formula V = 1/3 * π * r² * h, where r is the radius and h is the height.
We are given that dV/dt = 42 cubic feet per second and dr/dt = -6 feet per second. We are asked to find dh/dt when r = sqrt((3V)/(πh)) = sqrt((3246)/(π33)) = 3 feet and h = 33 feet.
We can differentiate the volume formula with respect to time t to get a relationship between these rates:
dV/dt = π * r * h * dr/dt + 1/3 * π * r² * dh/dt.
Substituting the given values and solving for dh/dt, we get:
42 = π * 3 * 33 * (-6) + 1/3 * π * 3² * dh/dt 42 = -594π + π * dh/dt dh/dt = (42 + 594π) / π dh/dt ≈ 0.133 feet per second.
So, the height of the cone is increasing at a rate of approximately 0.133 feet per second.
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