Water is leaking out of an inverted conical tank at a rate of 1100 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 meters and the diameter at the top is 4.5 meters. If the water level is rising at a rate of 23 centimeters per minute when the height of the water is 3 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
find the surface area of the water in the tank at 3 meters from the bottom (pointy end), pi r^2 (in CENTIMETERS squared)
then
dh (pi r^2 = dV
pi r^2 (dh/dt ) = dV/dt
so
dh/dt = (1/pir^2) dV/dt
23 = (1/pir^2)(flow rate in - 1100)
To solve this problem, we can use the concept of related rates. We are given the rate at which water is leaking out of the tank, the rate at which the water level is rising, and we need to find the rate at which water is being pumped into the tank.
Let's start by drawing a diagram to better understand the problem:
```
-----
/ \
/ h \
| \ |
| r1 \ |
\ /
\ /
\ /
v
```
In the diagram, h represents the height of the water in the tank, r1 represents the radius of the water surface, and v represents the volume of water in the tank.
From the problem statement, we know that:
- The water is leaking at a rate of 1100 cubic centimeters per minute.
- The water level is rising at a rate of 23 centimeters per minute.
- The height of the tank is 15 meters and the diameter at the top is 4.5 meters.
To find the rate at which water is being pumped into the tank, we need to find how the volume of water in the tank changes over time. We can relate the volume, height, and radius using the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
We need to find dV/dt (the rate of change of volume with respect to time). To do this, we will differentiate the formula with respect to time:
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
Now let's plug in the given values:
- r = r1 (the radius at any given time)
- dr/dt = 0 (since the radius is not changing)
- dh/dt = 23 cm/min (the rate at which the height is changing)
- h = 3 meters (the height when we want to find the rate)
Plugging these values into the formula, we get:
dV/dt = (1/3) * π * (2r1 * 0 * 3 + r1^2 * 23)
Simplifying the equation, we get:
dV/dt = (23/3) * π * r1^2
Now we need to solve for r1. We know that r1 can be found using similar triangles. Since the tank is a cone, the ratio of the height to the radius remains constant:
h / r1 = H / R, where H is the height of the tank and R is the radius at the top.
Plugging in the given values, we get:
3 / r1 = 15 / (4.5/2)
Simplifying the equation, we get:
r1 = (3/15) * (4.5/2)
Now we can substitute this value back into the equation for dV/dt:
dV/dt = (23/3) * π * [(3/15) * (4.5/2)]^2
Simplifying the equation, we get:
dV/dt = (23/3) * π * (9/30)^2
Now we can calculate the value of dV/dt:
dV/dt = (23/3) * π * (9/30)^2
Finally, let's solve this equation to find dV/dt:
dV/dt ≈ 2,005.60 cubic centimeters per minute
Therefore, the rate at which water is being pumped into the tank is approximately 2,005.60 cubic centimeters per minute.