Water is leaking out of an inverted conical tank at a rate of 13500.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 12.0 meters and the diameter at the top is 5.5 meters. If the water level is rising at a rate of 16.0 centimeters per minute when the height of the water is 4.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
The radius of the water's surface at a depth of h is
r = 11/48 h
So, the volume v is
v = 1/3 πr^2 h
= π/3 (11/48)^2 h^3
so,
dv/dt = 121π/2304 h^2 dh/dt
So, when h=400,
dv/dt = (121π/2304)(400)^2(16) = 422370 cm^3/min
That is the rate at which v is increasing, even while leaking. SO, the real rate of water pumpage is
435870 cm^3/min
thank you very much! It has taken me almost 2 hours to do this
To solve this problem, we can use related rates. Let's assume that the height of the water in the conical tank is denoted by h (in meters), the radius of the water surface is denoted by r (in meters), and the rate at which water is being pumped into the tank is denoted by V (in cubic centimeters per minute).
Given information:
- Leak rate: 13500.0 cubic centimeters per minute
- Rising rate: 16.0 centimeters per minute
- Tank height: 12.0 meters
- Diameter at the top: 5.5 meters
- Target height: 4.0 meters
To start, we need to relate the height and the radius of the cone at any given time. The cone is similar to the full cone with height 12.0 meters, so we can use the ratios of corresponding lengths to find the equation:
h / r = 12.0 / (5.5 / 2)
Simplifying, we get:
h / r = 24 / 11
Next, we can take the derivative of this equation with respect to time (t):
(dh/dt) / r - h / (r^2) * (dr/dt) = (24 / 11) * (d/dt) [24 / 11]
Since we are interested in finding dV/dt, we need to find a relationship between V and the variables in the equation. The volume of a cone can be given by V = (1/3) * π * r^2 * h.
Differentiating this equation with respect to time (t), we get:
(dV/dt) = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
Since the pump is adding water at a constant rate, the derivative of the volume with respect to time is the constant rate at which water is being pumped into the tank.
Now, we have two equations representing the rates of change:
Equation 1: (dh/dt) / r - h / (r^2) * (dr/dt) = (24/11) * (d/ dt) [24 / 11]
Equation 2: (dV/dt) = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
Given that dh/dt = 16.0 centimeters per minute, h = 4.0 meters, r = (5.5 / 2) meters, and (dV/dt) = -13500.0 cubic centimeters per minute (negative sign implying a leak), we can substitute these values into Equation 1 to solve for (dr/dt) as follows:
16 / (5.5 / 2) - 4 / [(5.5 / 2) ^2] * (dr/dt) = (24 / 11) * (0) [since the 2nd half of Equation 1 = 0]
Simplifying:
32 / 5.5 - 4 / [(5.5 / 2) ^2] * (dr/dt) = 0
Fixing the square brackets:
32 / 5.5 - 4 / [(5.5 ^ 2) / 4] * (dr/dt) = 0
Simplifying further:
32 / 5.5 - 4 * 4 / (5.5 ^ 2) * (dr/dt) = 0
Multiplying through by (5.5^2):
32 * (5.5 ^ 2) - 4 * 4 * (dr/dt) = 0
Simplifying and solving for (dr/dt):
(32 * (5.5^2)) / (4 * 4) = (dr/dt)
(32 * 5.5^2) / 16 = (dr/dt)
198.375 = (dr/dt)
Thus, the rate at which water is being pumped into the tank is 198.375 cubic centimeters per minute.
To solve this problem, we can use related rates. Let's denote the rate at which water is being pumped into the tank as V(t), where t is time. We want to find the value of V(t) in cubic centimeters per minute.
Given:
- Leak rate: 13500.0 cubic centimeters per minute
- Water level rising rate: 16.0 centimeters per minute
- Height of the tank: 12.0 meters
- Diameter at the top: 5.5 meters
- Height of the water: 4.0 meters
First, let's determine the equation that relates the changing variables. We can start by finding the volume of the water in the tank.
The volume of a cone can be calculated using the formula:
V = (1/3) * π * r^2 * h
Where V is the volume, π is the mathematical constant pi, r is the radius of the base (half the diameter), and h is the height of the water.
Let's denote the radius as r(t) and the height as h(t), both are functions of time t.
Given that the diameter at the top is 5.5 meters, the radius at any time t is half of that:
r(t) = (5.5/2) = 2.75 meters
Since we want the values of the variables in centimeters, let's convert 2.75 meters to centimeters:
r(t) = 2.75 * 100 = 275 centimeters
Now, let's write the equation for the volume of the water in the tank in terms of the height:
V = (1/3) * π * r^2 * h
Since r(t) is a constant, we can simplify it further:
V = (1/3) * π * (275)^2 * h
V = (1/3) * π * 75625 * h
V = 25208.33 * h cubic centimeters (approximately)
The volume V is a function of time, so we can write it as V(t), where V is in cubic centimeters.
Next, we need to differentiate the volume function V(t) to get dV/dt (the rate at which the volume is changing with respect to time).
dV/dt = d(25208.33 * h)/dt
Now, we know that the rate at which the water level is rising is 16.0 centimeters per minute, which is dh(t)/dt. So, we can substitute dh(t)/dt into our equation:
dV/dt = 25208.33 * dh(t)/dt
At this point, we need to find the relation between h(t) and V(t) in terms of the applied leakage rate. We can do this by subtracting the leakage from the volume function:
V(t) - leakage = 25208.33 * h(t)
Since the leak rate is given as 13500.0 cubic centimeters per minute:
V(t) - 13500.0 = 25208.33 * h(t)
Now, let's apply the given values when the height of the water is 4.0 meters.
Substitute h(t) = 400 cm:
V(4.0) - 13500.0 = 25208.33 * 400
We can simplify this equation to solve for V(4.0):
V(4.0) = 25208.33 * 400 + 13500.0
Now, we can solve for V(4.0) to find the rate at which water is being pumped into the tank in cubic centimeters per minute.
V(4.0) ≈ 10,083,330 cubic centimeters per minute