Water is leaking out of an inverted conical tank at a rate of 12200 cubic centimeters per minute 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 6.5 meters. If the water level is rising at a rate of 21 centimeters per minute when the height of the water is 1.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
You know, when you posted this earlier, with a typo, my solution should have enabled you to get the corrected answer. I assume you did that, and are now just confirming your subsequent work.
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Right?
when the water is at depth y, the radius r of the surface can be found using similar triangles:
r/y = (6.5/2)/15
r = 13/60 y
at depth y, the volume of water is
v = 1/3 pi r^2 y
= 1/3 pi (13/60)^2 y^3
= 169pi/10800 y^3
dv/dt = -12200+C = 169pi/3600 y^2 dy/dt
at y=1m=100cm, dy/dt = 21cm/min
-12200 + C = 169pi/3600 * 10000 * 21 =
C = 30971
To find the rate at which water is being pumped into the tank, we can use related rates and the known information about the leaking rate and the rising rate.
Let's start by drawing a diagram of the conical tank:
```
_
/ \
/ \
h1 / \ H
/ \
/ _______ \
Base
```
Where:
- h1 represents the height of the water in the tank in meters,
- H represents the overall height of the tank in meters,
- and the base represents the circular opening at the top of the tank.
Given data:
- The leaking rate is 12200 cubic centimeters per minute.
- The rising rate is 21 centimeters per minute when h1 = 1.0 meters.
- H (overall height) is 15 meters.
We need to find the rate at which water is being pumped into the tank (in cubic centimeters per minute).
Let's denote:
- r as the radius at height h1,
- V as the volume of water in the tank at any given time.
Using the formula for the volume of a cone, V = (1/3) * π * r^2 * h, we can find the volume in terms of h1 and r:
V = (1/3) * π * r^2 * h1
To relate the change in volume V to the change in height h1, we can differentiate both sides of the equation with respect to time t (since the rates are given in terms of minutes):
dV/dt = (1/3) * π * (2r * dr/dt * h1 + r^2 * dh1/dt)
Now, we need to express dr/dt (the change in radius with respect to time) and dh1/dt (the change in height with respect to time) in terms of the known rates:
Given that dr/dt = dh1/dt (the radius is changing at the same rate as the height is rising), we can substitute them into the equation:
dV/dt = (1/3) * π * (2r * dr/dt * h1 + r^2 * dh1/dt)
= (1/3) * π * (2r * dr/dt + r^2 * dh1/dt) * h1
Now, let's substitute the known values into the equation:
- The tank height H is 15 meters.
- The height of the water h1 is 1.0 meters.
- The rising rate dh1/dt is 21 centimeters per minute (-0.21 meters per minute).
Since the overall height H = h1 + r, we can solve for r using the given values:
r = H - h1
= 15 - 1.0
= 14.0 meters
Now, we can substitute the known values into the equation:
dV/dt = (1/3) * π * (2 * 14.0 * dr/dt + 14.0^2 * (-0.21)) * 1.0
We are left with one unknown value, dr/dt (the rate at which the radius changes with time).
To find dr/dt, we use the fact that the rate at which the water level is rising is 21 centimeters per minute:
dh1/dt = 21 centimeters per minute
= 0.21 meters per minute
Since dr/dt = dh1/dt, dr/dt also equals 0.21 meters per minute.
Substituting dr/dt = 0.21 into the equation yields:
dV/dt = (1/3) * π * (2 * 14.0 * 0.21 + 14.0^2 * (-0.21)) * 1.0
Simplifying further:
dV/dt = (1/3) * π * (5.88 - 41.16) * 1.0
dV/dt = (1/3) * π * (-35.28) * 1.0
≈ -36.91π cubic meters per minute
Since the leaking rate is also given as -12200 cubic centimeters per minute, we can convert dV/dt to cubic centimeters per minute:
-36.91π m^3/minute ≈ -116017.39 cm^3/minute
For the rising rate, a positive value is given, so we drop the negative sign and write our final answer:
116017.39 cubic centimeters per minute.
Therefore, the rate at which water is being pumped into the tank is approximately 116017.39 cubic centimeters per minute.