A water tank is shaped like an inverted right circular cone with a base radius of 14 feet and a height of 25 feet high. If water flows into the tank at a rate of 20 ft^3/min, how fast is the depth of the water increasing when the water is 18 feet deep?
To find the rate at which the depth of the water is increasing, we need to use related rates. We can use the formula for the volume of a cone to relate the variables:
V = (1/3) * π * r^2 * h
where V is the volume of the water in the tank, π is a constant (approximately 3.14159), r is the radius of the base of the cone, and h is the depth of the water.
We are given that the water flows into the tank at a rate of 20 ft^3/min, which means that dV/dt = 20 ft^3/min. Our goal is to find dh/dt, the rate at which the depth is increasing.
First, let's find an expression for V in terms of h. We know that the radius of the base of the cone is 14 feet and the height is 25 feet. We can use similar triangles to find the radius of the cone at height h:
r = (14/25) * h
Next, substitute this expression for r into the formula for V:
V = (1/3) * π * [(14/25) * h]^2 * h
V = (1/3) * π * (196/625) * h^3
Now, differentiate both sides of the equation with respect to time t:
dV/dt = (1/3) * π * (196/625) * 3h^2 * dh/dt
Substitute the given value dV/dt = 20 ft^3/min and the known value for h = 18 feet:
20 = (1/3) * π * (196/625) * 3(18)^2 * dh/dt
Now, solve for dh/dt:
dh/dt = 20 / [(1/3) * π * (196/625) * 3(18)^2]
Simplifying this expression will give us the rate at which the depth of the water is increasing when the water is 18 feet deep.