Water is leaking out of an inverted conical tank at a rate of 6700.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 7.0 meters and the diameter at the top is 4.0 meters. If the water level is rising at a rate of 28.0 centimeters per minute when the height of the water is 3.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
To solve this problem, we need to apply the principle of related rates. Let's break it down into steps:
Step 1: Find the volume of the tank as a function of time.
The volume V of the inverted conical tank can be calculated using the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
where π is the mathematical constant pi, r is the radius of the base of the cone, and h is the height of the cone.
Given that the diameter at the top is 4.0 meters, the radius r can be calculated as half of the diameter:
r = 4.0 / 2 = 2.0 meters
The height h of the water in the tank is changing over time. Let's denote it as h(t), where t is the time in minutes.
Now, we can express the volume V as a function of time t:
V(t) = (1/3) * π * (2.0)^2 * h(t)
Step 2: Find the rate at which the volume of water is changing with respect to time.
This rate is given as -6700.0 cubic centimeters per minute, which means the volume is decreasing. Let's denote it as dV(t)/dt.
dV(t)/dt = -6700.0 cm^3/min
Step 3: Find the rate at which the height of the water is changing with respect to time.
This rate is given as 28.0 centimeters per minute, which means the height is increasing. Let's denote it as dh(t)/dt.
dh(t)/dt = 28.0 cm/min
Step 4: Relate the rates of change using the chain rule.
Using the chain rule, we can relate the rates of change as follows:
dV(t)/dt = dV(t)/dh(t) * dh(t)/dt
Step 5: Find dV(t)/dh(t) using the volume formula.
Differentiating the volume formula with respect to h(t), we get:
dV(t)/dh(t) = (1/3) * π * (2.0)^2
dV(t)/dh(t) = (4/3) * π
Step 6: Substitute the known values into the chain rule equation.
Substituting the known values into the chain rule equation, we get:
-6700.0 = (4/3) * π * dh(t)/dt
Step 7: Solve for dh(t)/dt.
Dividing both sides of the equation by (4/3) * π, we get:
dh(t)/dt = -6700.0 / [(4/3) * π]
Step 8: Substitute the known value of dh(t)/dt.
Substituting the known value of dh(t)/dt calculated in step 3, we get:
28.0 = -6700.0 / [(4/3) * π]
Step 9: Solve for [(4/3) * π].
Solving for [(4/3) * π], we get:
[(4/3) * π] = -6700.0 / 28.0
Step 10: Calculate the rate at which water is being pumped into the tank.
Now that we know [(4/3) * π], we can substitute it back into the equation in step 8 to find the rate at which water is being pumped into the tank:
28.0 = -6700.0 / [(4/3) * π]
Let's calculate this:
π ≈ 3.14159
[(4/3) * 3.14159] ≈ 4.18879
28.0 = -6700.0 / 4.18879
Multiplying both sides by 4.18879, we get:
28.0 * 4.18879 = -6700.0
117.2532 = -6700.0
Dividing both sides by -6700.0, we get:
117.2532 / -6700.0 ≈ -0.0175
However, the negative rate doesn't make sense in this context. We need a positive value for the rate at which water is being pumped into the tank. Therefore, we made an error somewhere in the calculations. Please review the steps to see where the mistake occurred.
To solve this problem, we can use the concept of related rates. Related rates involve finding the rate at which one quantity is changing with respect to the rate of change of another related quantity.
Let's define the variables:
- Let V represent the volume of water in the tank (in cubic centimeters).
- Let h represent the height of the water in the tank (in meters).
- Let r represent the radius of the water surface at height h (in meters).
We know the following information:
- The water is leaking out of the tank at a rate of 6700.0 cubic centimeters per minute. This means that dV/dt = -6700.0 cm^3/min.
- The water level is rising at a rate of 28.0 centimeters per minute. This means that dh/dt = 28.0 cm/min.
- The height of the tank is 7.0 meters, so at h = 7.0 meters, we can find the rate at which the volume is changing, dV/dt.
We need to find the rate at which water is being pumped into the tank, so we are looking for dV/dt. To find this, we need to find the rate at which the volume changes with respect to time, dV/dt.
First, we need to find an equation that relates the variables V and h. The volume of a cone can be given by V = (1/3)πr^2h, where r is the radius of the cone.
From the given information, we can determine the radius r of the water surface at height h. Since the diameter at the top is 4.0 meters, the radius can be found using the proportion r/h = (1/2)(4.0/7.0). Solving for r, we get r = (2/7)h.
Now we can find V in terms of h by substituting the value of r into the equation for the volume:
V = (1/3)π((2/7)h)^2h
V = (4/21)πh^3
Now, differentiate both sides of the equation with respect to time:
dV/dt = (4/21)π * 3h^2 * dh/dt
Substitute the given values into the equation:
- dV/dt = -6700.0 cm^3/min (the rate at which water is leaking out)
- dh/dt = 28.0 cm/min (the rate at which the water level is rising)
- h = 3.0 meters (the given height when the water level is rising)
Plugging in these values, we can solve for dV/dt:
-6700.0 = (4/21)π * 3(3.0)^2 * 28.0
Now we can solve the equation to find the value of dV/dt, which represents the rate at which water is being pumped into the tank:
dV/dt = (-6700.0) / [(4/21)π * 3(3.0)^2 * 28.0]