Water is leaking out of an inverted conical tank at a rate of 1100.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 15.0 meters and the diameter at the top is 5.0 meters. If the water level is rising at a rate of 300 centimeters per minute when the height of the water is 2.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
To solve this problem, we can use the related rates method.
Let's define some variables:
- V(t) represents the volume of water in the tank at time t (in cubic centimeters).
- r(t) represents the radius of the water level in the tank at time t (in centimeters).
- h(t) represents the height of the water level in the tank at time t (in centimeters).
- R(t) represents the rate at which water is being pumped into the tank at time t (in cubic centimeters per minute).
Based on the given information, we have:
- dV/dt = R(t) - 1100.0 (the rate of change of volume is the difference between the rate of pumping in and the rate of leaking out)
- dh/dt = 300 (the rate at which the height of the water level is rising)
- h(t) = 250 (height of the water is given as 2.5 meters = 250 centimeters)
We need to find dV/dt when h(t) = 250.
To relate the variables h(t) and r(t), we use similar triangles:
The larger triangle formed by the tank has height H = 1500 centimeters (given as 15.0 meters = 1500 centimeters) and base diameter D = 500 centimeters (given as 5.0 meters = 500 centimeters).
The smaller triangle formed by the water level has height h(t) and base diameter d(t) (we need to find this).
Using similar triangles, we have:
r(t) / h(t) = D / H
Substituting the given values, we can solve for r(t):
r(t) / 250 = 500 / 1500
r(t) = 250 * (500 / 1500)
r(t) = 250 / 3
Now, we differentiate both sides of the equation r(t) = 250 / 3 with respect to t:
dr/dt = (1 / 3) * (dh/dt)
Since dh/dt = 300, we have:
dr/dt = (1 / 3) * 300
dr/dt = 100
So, the rate at which the radius of the water level is changing is 100 centimeters per minute.
Next, we need to find the expression for V(t) in terms of r(t). The volume of a cone can be expressed as:
V(t) = (1/3) * π * r(t)^2 * h(t)
Substituting the known values, we have:
V(t) = (1/3) * π * ((250 / 3)^2) * 250
Now, let's differentiate V(t) with respect to t:
dV/dt = (1/3) * π * [(250 / 3)^2] * (dh/dt) + (1/3) * π * [(dh/dt) * (250 / 3)^2]
Using the given information, we have:
dV/dt = (1/3) * π * (250 / 3)^2 * 300 + (1/3) * π * (300^2) * (250 / 3)^2
Simplifying the expression above, we can find the value of dV/dt when h(t) = 250.
I'll calculate the numerical value for you.
To solve this problem, we can use related rates, which involves calculating the rates at which different quantities are changing with respect to time.
Let's start by identifying the known quantities and the rates at which they change:
Known quantities:
- The height of the tank, h = 15.0 meters
- The diameter at the top of the tank, D = 5.0 meters
- The rate at which water is leaking out, dV/dt = -1100.0 cubic centimeters per minute (negative because water is being lost from the system)
- The rate at which water is being pumped into the tank, dV/dt (what we need to find)
Rates of change:
- The rate at which the height is changing, dh/dt = 300 centimeters per minute
- The rate at which the radius of the water surface is changing, dr/dt (also what we need to find)
To relate the different variables, we can use similar triangles. The ratio of the radius of the water surface, r, to the height of the water, h, remains constant throughout. This means we have the following relationship:
r / h = (D / 2) / h
Now, let's differentiate both sides of this equation with respect to time, t, using the chain rule:
(d/dr)(r/h) = (d/dr)((D / 2) / h)
dr/dt * (1 / h) - r * dh/dt * (1 / h^2) = (0 - (D / 2) * dh/dt) / h^2
Simplifying, we get:
dr/dt * (1 / h) - r * (dh/dt / h^2) = - (D / 2) * (dh/dt / h^2)
Next, we substitute the known values and solve for dr/dt:
dr/dt * (1 / 15) - r * (300 / 15^2) = - (5 / 2) * (300 / 15^2)
dr/dt * (1 / 15) - 300r / 15^2 = -5 / 2 * 300 / 15^2
dr/dt * (1 / 15) - 20r / 15^2 = -5 / 2 * 300 / 15^2
dr/dt * (1 / 15) - 20r / 225 = -5 / 2 * 300 / 225
dr/dt * (1 / 15) - 20r / 225 = -1
dr/dt * (1 / 15) = -1 + 20r / 225
dr/dt * (1 / 15) = (20r - 225) / 225
dr/dt = (20r - 225) / (15 * 225)
Now, we need to find the value of r to substitute into this equation. We know that the height of the water is 2.5 meters when the rate of change of the height is 300 centimeters per minute.
Using similar triangles, we can find the corresponding radius:
2.5 / 15 = r / (D / 2)
Substituting the given values, we have:
2.5 / 15 = r / (5 / 2)
Cross-multiplying, we get:
2.5 * (5 / 2) = 15 * r
r = 2.5 * 5 / 15
r = 0.833333 meters
Now, we can substitute this value of r into the equation for dr/dt:
dr/dt = (20 * 0.833333 - 225) / (15 * 225)
dr/dt = (16.66666 - 225) / (15 * 225)
dr/dt = -208.33334 / 3375
dr/dt ≈ -0.06176 meters per minute
Lastly, we need to convert this rate from meters per minute to centimeters per minute:
-0.06176 * 100 = -6.176 centimeters per minute
Since water is pumped into the tank, the rate should be positive, so we take the absolute value:
|-6.176| = 6.176 centimeters per minute
Therefore, the rate at which water is being pumped into the tank is approximately 6.176 cubic centimeters per minute.