Question

A tank of water in the shape of a cone is leaking water at a constant rate of 2ft3/ hour. The base radius of the tank is 7ft and the height of the tank is 11ft. a. At what rate is the depth of the water in the tank changing when the depth of the water is 6ft ? b. At what rate is the radius of the top of the water in the tank changing when the depth of the water is 6ft ?

Answer 1

a. When the water depth is 6ft, the water leaks at a rate of -0.0083 ft/hour. This means the water level is decreasing.

b. When the water depth is 6ft, the radius of the water surface changes at a rate of -0.0139 ft/hour. This means the water surface is shrinking.

Step-by-step explanation:

Part a: Water Depth Change:

Volume and Height Relationship:

The volume (V) of a cone depends on its height (h) and radius (r) by the formula: V = (1/3)πr^2h. We can express the radius in terms of height based on the cone's dimensions: r = (7/11)h.

Rate of Change Equation:

We're given the leak rate (-2ft³/hour) as the negative rate of change of volume (-dV/dt). Substituting V and differentiating V with respect to time (t) using the chain rule:

-dV/dt = -(1/3)π[(7/11)h]^2 * dh/dt

-2 = -(1/3)π * (49/121) * h^2 * dh/dt

Solving for dh/dt:

Solving for the rate of change of height (dh/dt) when the depth is 6ft (h = 6):

dh/dt = (2 * 3 * 121) / (π * 49 * 36) = -0.0083 ft/hour

Therefore, the water depth decreases at a rate of 0.0083 ft/hour when the depth is 6ft.

Part b: Water Surface Radius Change:

Similar Triangles:

Triangles formed by the cone's base radius (r), height (h), and water depth (h') are similar.

Radius Relationship:

Using the similar triangle proportions: r/h = r'/h', we can express the water surface radius (r') in terms of h and h': r' = (r/h) * h' = (7/11) * h'

Rate of Change Equation:

Differentiating both sides for both radii with respect to time:

dr'/dt = (7/11) * dh'/dt

Solving for dh'/dt:

Similar to part a, solve for the rate of change of water surface radius (dh'/dt) when the depth is 6ft (h' = 6):

dh'/dt = (11/7) * (-0.0083) = -0.0139 ft/hour

Therefore, the water surface radius shrinks at a rate of 0.0139 ft/hour when the depth is 6ft.

This approach utilizes cone geometry, volume equations, and differentiation to analyze the changing water depth and surface radius under a constant leak rate.

Answer 2

a. The rate at which the depth of the water is changing when the depth is 6 ft is -98π ft/hour. b. The rate at which the radius of the top of the water is changing when the depth is 6 ft is (3/π)(11 - 6) ft/hour.

Step-by-step explanation:

a. To find the rate at which the depth of the water is changing, we need to use the formula for the volume of a cone: V = (1/3)πr^2h, where V is the volume, π is pi, r is the radius of the base, and h is the height. Let's denote the depth as x, so the height of the water is 11 - x. The volume of the water is V = (1/3)π(7^2)(11 - x). We are given that the rate of change of volume is 2 ft^3/hour. So, we can differentiate the volume equation with respect to time to find the rate of change of depth:

dV/dt = (1/3)π(7^2)d(11-x)/dt

2 = (1/3)π(7^2)(-dx/dt)

-2(3/π)(7^2) = dx/dt

-98π = dx/dt

Therefore, at a depth of 6 ft, the rate at which the depth of the water is changing is -98π ft/hour.

b. To find the rate at which the radius of the top of the water is changing, we can use similar reasoning. Let's denote the radius as r and the rate of change of radius as dr/dt. The volume of the water is given by V = (1/3)πr^2(11 - x). Since the rate at which the volume is changing is 2 ft^3/hour, we can differentiate the volume equation with respect to time to find the rate of change of radius:

dV/dt = (1/3)π(2r)(11 - x)dr/dt

2 = (2/3)π(r)(11 - x)dr/dt

(3/π)(11 - x) = dr/dt

Therefore, at a depth of 6 ft, the rate at which the radius of the top of the water is changing is (3/π)(11 - 6) ft/hour.