AP Calculus AB Exam Question:

A water tank in the shape of a right circular cone is being filled with water at a constant rate of 2 cubic feet per minute. The height of the tank is 10 feet and the radius of the top of the tank is 6 feet.

a) At what rate is the depth of the water increasing when the water is 5 feet deep?

b) At what rate is the radius of the water surface increasing when the water is 5 feet deep?

Answer:

a) To find the rate at which the depth of the water is increasing, we need to determine the relationship between the height of the water and the radius of the water surface.

Let's denote the height of the water as 'h' (in feet) and the radius of the water surface as 'r' (in feet). Since the shape of the tank is a right circular cone, we can use similar triangles to set up a proportion:

h/(10) = r/(6)

Cross-multiplying and simplifying, we get:

6h = 10r

Differentiating both sides with respect to time 't', we get:

6dh/dt = 10dr/dt

We are given that dh/dt = 2 ft/min (rate at which water is being added), and we need to find dr/dt when h = 5 ft.

Plugging the given values into the equation, we have:

6(2) = 10(dr/dt)

Simplifying, we find:

12 = 10(dr/dt)

Dividing both sides by 10, we get:

dr/dt = 1.2 ft/min

Therefore, the depth of the water is increasing at a rate of 1.2 feet per minute when the water is 5 feet deep.

b) To find the rate at which the radius of the water surface is increasing, we can use the relationship between the height of the water and the radius of the water surface (which we derived earlier):

6h = 10r

Differentiating both sides with respect to time 't', we get:

6dh/dt = 10dr/dt

We are given that dh/dt = 2 ft/min (rate at which water is being added), and we need to find dh/dt when h = 5 ft.

Plugging the given values into the equation, we have:

6(2) = 10(dr/dt)

Simplifying, we find:

12 = 10(dr/dt)

Dividing both sides by 10, we get:

dr/dt = 1.2 ft/min

Therefore, the radius of the water surface is increasing at a rate of 1.2 feet per minute when the water is 5 feet deep.

Note: In this particular case, the rates of change for the depth of the water and the radius of the water surface are the same since they are related through the shape of the cone.