Water pours at a constant rate into a conical tank of height 10 feet and radius 4 feet. Let V, h and r be the volume, height, and radius if the water in the tank at a particular time t. Express the volume of the water in the tank at any time t as a function of h only.
v = 1/3 pi r^2 h
But the dimension of the tank indicate that at any time,
r/h = 4/10
So, use that to eliminate r.
To determine the volume of the water in the tank as a function of the height only, we'll use similar triangles to find a relationship between the height and the radius of the water.
Let's start by considering the similar triangles formed by the water inside the tank. As the water rises, the triangle formed by the water's surface, the tank's base, and a vertical line from the water's surface to the center of the base remains similar to the triangle formed by the original conical tank.
The small triangle formed by the water's surface, radius of the water, and height of the water is similar to the whole cone. Therefore, the ratio of the height of the water to the height of the tank is equal to the ratio of the radius of the water to the radius of the tank.
Hence, we have:
h / 10 = r / 4
Rearranging the equation gives:
r = (4h) / 10
r = 2h / 5
Now that we have the relationship between the radius and the height, we can calculate the volume using the formula for the volume of a cone:
V = (1/3)πr²h
Substituting the expression for r we derived earlier:
V = (1/3)π[(2h/5)²]h
V = (1/3)π(4h²/25)h
V = (4πh³) / 75
So, the volume of the water in the tank, expressed as a function of the height only, is given by:
V(h) = (4πh³) / 75