Calculating Minimum Water Level in a Barrel using ODE: Analysis and Approach

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In summary: Any idea? On how to approach this problem differently?One approach would be to set up the integral for the volume function and then use optimization techniques to find the minimum value. Another approach could be to graph the function and visually determine the minimum value.
  • #1
Mathman2013
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Homework Statement
Find the lower bound of the ODE?
Relevant Equations
dV/dt = 5 - 2*V(t)^(1/3)
I have following differential equation dV/dt = 5 - 2 * V(t)^(1/3) which represents a the time its take to drain a barrel of rain water which contain 25 Liter of water, at t = 0.

I am suppose to calculate the least amount of water in barrel during this process.

If I set the rate of growth to zero, meaning dV/dt = 0, then 5 - 2 * V^(1/3) = 0, and thusly V = (5/2)^(3) = 15.65, meaning the least amount of water which is in the barrel is 15.65 L,if the rate of change is zero.

But question, is this correct why at handling this problem? Because I am not really using the info, that t = 0, the barrel contains 25 L.

Any idea? On how to approach this problem differently?
 
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  • #2
Mathman2013 said:
Homework Statement:: Find the lower bound of the ODE?
Relevant Equations:: dV/dt = 5 - 2*V(t)^(1/3)

I have following differential equation dV/dt = 5 - 2 * V(t)^(1/3) which represents a the time its take to drain a barrel of rain water which contain 25 Liter of water, at t = 0.

I am suppose to calculate the least amount of water in barrel during this process.

If I set the rate of growth to zero, meaning dV/dt = 0, then 5 - 2 * V^(1/3) = 0, and thusly V = (5/2)^(3) = 15.65, meaning the least amount of water which is in the barrel is 15.65 L,if the rate of change is zero.
You're off by a little bit -- I get the least amount as 15.625 L (125/8 == 15.625).
Mathman2013 said:
But question, is this correct why at handling this problem? Because I am not really using the info, that t = 0, the barrel contains 25 L.
Yes, this is the right approach. Since water is draining from the barrel, dV/dt will be negative, so at a time when dV/dt = 0, the volume will be at a minimum.
One interpretation for this scenario is that the barrel has a hole in it somewhere above the middle of the barrel.
Mathman2013 said:
Any idea? On how to approach this problem differently?
No need for a different approach.
 
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  • #3
Mathman2013 said:
Homework Statement:: Find the lower bound of the ODE?
Relevant Equations:: dV/dt = 5 - 2*V(t)^(1/3)

I have following differential equation dV/dt = 5 - 2 * V(t)^(1/3) which represents a the time its take to drain a barrel of rain water which contain 25 Liter of water, at t = 0.

I am suppose to calculate the least amount of water in barrel during this process.

If I set the rate of growth to zero, meaning dV/dt = 0, then 5 - 2 * V^(1/3) = 0, and thusly V = (5/2)^(3) = 15.65, meaning the least amount of water which is in the barrel is 15.65 L,if the rate of change is zero.

But question, is this correct why at handling this problem? Because I am not really using the info, that t = 0, the barrel contains 25 L.

Any idea? On how to approach this problem differently?

[itex]V[/itex] is strictly increasing if and only if [itex]V(t) < 125/8\,\mathrm{L}[/itex] and strictly decreasing if and only if [itex]V(t) > 125/8\,\mathrm{L}[/itex]. Which of those regimes does [itex]V(0) = 25\,\mathrm{L}[/itex] fall into?
 
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  • #4
Mathman2013 said:
But question, is this correct why at handling this problem? Because I am not really using the info, that t = 0, the barrel contains 25 L.
You would need that information if you wanted to determine how long it takes for the volume to reach the minimum. That would obviously depend on how much you start with.
 
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Related to Calculating Minimum Water Level in a Barrel using ODE: Analysis and Approach

1. How do you calculate the minimum water level in a barrel?

To calculate the minimum water level in a barrel, you can use the ODE (ordinary differential equations) approach. This involves setting up a differential equation that represents the rate of change of water level in the barrel, and then solving it using mathematical techniques such as separation of variables or integrating factors.

2. What factors affect the minimum water level in a barrel?

The minimum water level in a barrel is affected by several factors, including the shape and size of the barrel, the material it is made of, the temperature and humidity of the surrounding environment, and the rate at which water is added or removed from the barrel.

3. Can you provide an example of calculating the minimum water level in a barrel using ODE?

Yes, for example, let's say we have a cylindrical barrel with a radius of 1 meter and a height of 2 meters. The barrel is initially filled with 3 cubic meters of water, and water is being added to the barrel at a rate of 0.5 cubic meters per minute. Using the ODE approach, we can set up the differential equation: dV/dt = 0.5 - (πr^2)/h * dh/dt, where V is the volume of water in the barrel, r is the radius, h is the height, and t is time. Solving this equation, we can find the minimum water level in the barrel at any given time.

4. Are there any limitations to using the ODE approach for calculating minimum water level in a barrel?

Yes, there are some limitations to using the ODE approach. One limitation is that it assumes a constant rate of water addition or removal, which may not always be the case in real-world scenarios. Additionally, the ODE approach may not be suitable for more complex barrel shapes or situations where there are multiple variables affecting the water level.

5. How can the results of calculating minimum water level in a barrel using ODE be applied in practical situations?

The results of calculating minimum water level in a barrel using ODE can be applied in various practical situations, such as designing and optimizing water storage systems, determining the capacity of barrels for specific purposes, and predicting water levels in barrels over time. This information can also be useful for managing and conserving water resources in agricultural, industrial, and residential settings.

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