Final answer:
a) The volume of the tank is approximately 2419.36 cubic feet. b) The surface area of the tank is approximately 1088.29 square feet. c) The rate at which the water level is changing when the water is 10 feet deep is approximately 0.01 feet per minute.
Step-by-step explanation:
a) Calculate the volume of the tank:
The volume of a cone can be calculated using the formula:
V = (1/3)πr^2h
Where V is the volume, π is a constant approximately equal to 3.14, r is the radius of the base, and h is the height of the cone.
For the given water tank, the radius, r, is 12 feet and the height, h, is 18 feet. Plugging these values into the formula, we have:
V = (1/3)π(12^2)(18)
V ≈ 2419.36 cubic feet
b) Determine the surface area of the tank:
The surface area of a cone can be calculated using the formula:
A = πr(r + √(r^2 + h^2))
For the given water tank, plugging in the values r = 12 feet and h = 18 feet, we have:
A = π(12)(12 + √(12^2 + 18^2))
A ≈ 1088.29 square feet
c) Find the rate at which the water level is changing when the water is 10 feet deep:
To find the rate at which the water level is changing, we can use similar triangles. Let y represent the height of the water in feet and x represent the radius of the water surface in feet. We can setup the proportion:
(18 - y) / (12 - x) = y / x
Since the question asks for the rate at which the water level is changing when y = 10 feet, we can substitute y = 10 feet into the equation and solve for dx/dt, the rate at which x is changing with respect to time:
(18 - 10) / (12 - x) = 10 / x
8 / (12 - x) = 10 / x
8x = 10(12 - x)
8x = 120 - 10x
18x = 120
x ≈ 6.67
Substituting this value back into the proportion, we have:
(18 - y) / 5.33 = 10 / 6.67
(18 - y) / 5.33 ≈ 1.5
18 - y ≈ 1.5(5.33)
18 - y ≈ 7.99
y ≈ 10.01
Therefore, when the water is 10 feet deep, the rate at which the water level is changing is approximately 0.01 feet per minute.