Calculating the Changing Rate of Water Height in a Cone-Shaped Funnel

Explanation:

Related Rates and Calculus:

To find the rate at which the height of the water in the funnel is changing when the water level is at a height of 2 ft, we must use calculus and the concept of related rates. The volume of water in the funnel can be described by the formula for the volume of a cone, which is V = (1/3)πr^2h, where V is volume, r is radius, and h is height.

Constant Ratio:

Since the funnel tapers down uniformly, the ratio of the height to the radius remains constant, which means that r/h = R/H where R is the known radius of 5ft at the height H of 8ft.

Given Information:

We are given that the volume of water V is decreasing at a rate of 0.4 ft³/sec, so we have dV/dt = -0.4 ft³/sec. Using the chain rule, we can differentiate both sides of the volume formula with respect to time (t) to find the relationship between the rates of change of the volume, radius, and height.

Finding the Rate of Change:

Through substitution and solving for dh/dt, we'll find the rate at which the height is changing when h is 2ft.

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