Richard W. answered • 03/02/23
Guru Tutor with vast Knowledge in Business and Related Field
We can solve this problem using related rates. Let's denote the radius of the water level in the conical tank by r(t), where t is time, and let H(t) be the height of the water level at time t. We want to find the rate at which water is being pumped into the tank, which is the same as the rate of change of the volume of water in the tank.
We can start by finding an equation that relates r and H using similar triangles. The cone is similar to a smaller cone that is formed by a cross-section of the water at height H, so we have:
r/H = 5/15
Simplifying this equation, we get:
r = (1/3)H
Differentiating both sides with respect to time, we get:
dr/dt = (1/3)dH/dt
Next, we can use the formula for the volume of a cone to relate the volume V of the water in the tank to the radius r and height H:
V = (1/3)πr^2H
Differentiating both sides with respect to time, we get:
dV/dt = (1/3)π(2rdr/dtH + r^2dH/dt)
Substituting the expression we derived earlier for dr/dt, we get:
dV/dt = (1/9)πH^2dH/dt + (2/9)πHr^2dH/dt
Now we can substitute the given values into this equation. We know that water is leaking out of the tank at a rate of 11300 cubic centimeters per minute, so dV/dt = -11300. We also know that the height of the water is rising at a rate of 23 centimeters per minute when H = 4.0 meters, so dH/dt = 23 and H = 4.0.
Substituting these values and using the expression we derived earlier for r, we get:
-11300 = (1/9)π(4.0)^2(23) + (2/9)π(4.0/3)^2(23)dr/dt
Simplifying this equation and solving for dr/dt, we get:
dr/dt = 200π/3 ≈ 209.44
Therefore, the rate at which water is being pumped into the tank is approximately 209.44 cubic centimeters per minute.
