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

[Mathman2013]

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?

 

[Mark44]

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).

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.

Any idea? On how to approach this problem differently?

No need for a different approach.

 

[pasmith]

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?

$V$ is strictly increasing if and only if $V(t) < 125/8\,\mathrm{L}$ and strictly decreasing if and only if $V(t) > 125/8\,\mathrm{L}$. Which of those regimes does $V(0) = 25\,\mathrm{L}$ fall into?

 

[vela]

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.