AP Calculus AB Exam Question - Related Rates (Advanced)

A cylindrical tank is being filled with water at a rate of 5 cubic feet per minute. The tank has a height of 10 feet and a radius of 3 feet. At what rate is the water level rising when the height of the water is 8 feet?

Note: Assume that the tank is initially empty.

Solution:

Let's denote:

• V(t) as the volume of water in the tank at time t.
• h(t) as the height of the water in the tank at time t.
• r(t) as the radius of the water surface in the tank at time t.

The volume V and the height h are related by the formula for the volume of a cylinder:

V(t) = π * (r(t))^2 * h(t)

Given that the water is being filled at a rate of 5 cubic feet per minute, we know that dV/dt = 5 ft^3/min. We want to find the rate at which the water level is rising, so we need to find dh/dt when h = 8 ft.

We can derive an equation relating the variables r(t) and h(t) by using similar triangles. From the given information, we know that the radius of the cylindrical tank is 3 feet and the height of the tank is 10 feet. Therefore, at any time t, we have:

r(t) / h(t) = 3 / 10

Solving for r(t), we get:

r(t) = (3/10) * h(t)

Now, we can substitute this expression for r(t) in terms of h(t) into the volume equation:

V(t) = π * [(3/10) * h(t)]^2 * h(t)

Simplifying the expression, we get:

V(t) = (9π/100) * h(t)^3

To find dh/dt, we differentiate both sides of the equation with respect to t:

dV/dt = d/dt ((9π/100) * h(t)^3)

5 = (27π/100) * h(t)^2 * dh/dt

Now, we can solve for dh/dt by plugging in the values we know:

5 = (27π/100) * (8)^2 * dh/dt

5 = (27π/100) * 64 * dh/dt

dh/dt = 5 / (27π/100 * 64)

dh/dt ≈ 0.098 ft/min (rounded to three decimal places)

Therefore, the water level is rising at a rate of approximately 0.098 feet per minute when the height of the water is 8 feet.