Modeling differential equation

  • #1
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Homework Statement


Liquid is pouring into a container at a constant rate of 30cm^3s^-1
At time t seconds liquid is leaking from the container at a rate of 2/15 V cm^3s^-1, where V cm^3 is the volume of liquid in the container at that time.

Show that -15 dV/dt = 2V - 450


Homework Equations

: [/B]None

The Attempt at a Solution


[/B]
At T second it leaking so the initial volume will be 30t. I don't know where to go from there.
 
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Answers and Replies

  • #2
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I assume you mean that at ##t## seconds, liquid begins leaking? This is an important condition.
You currently have a volume flow rate into the container and a volume flow rate out of the container. How could you combine these into something useful?
 
  • #3
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I assume you mean that at ##t## seconds, liquid begins leaking? This is an important condition.
You currently have a volume flow rate into the container and a volume flow rate out of the container. How could you combine these into something useful?
This is the part I do not understand, the question is not clear at all.
 
  • #4
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Here's the basic problem: you need to find some expression containing ##\frac{dV}{dt}## and prove that ##-15\frac{dV}{dt} = 2V - 450##. What does the quantity ##\frac{dV}{dt}## describe?
 
  • #5
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The change of Volume of liquid in the container with respect to time.
 
  • #6
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So can you get the change in volume as a function of time using the given quantities?
 
  • #7
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Thread moved.
Questions about differential equations belong in the Calculus & Beyond section, not the Precalculus section.
 
  • #8
Ray Vickson
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I assume you mean that at ##t## seconds, liquid begins leaking? This is an important condition.
You currently have a volume flow rate into the container and a volume flow rate out of the container. How could you combine these into something useful?

I think the (somewhat poorly worded) question means that at any time t, the leakage rate is (2/15)V(t), where V(t) = volume of liquid at time t.
 
  • #9
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I noticed how in your attempt at the solution you had the variable t yet the wording of the question doesn't have the variable t. Perhaps you're not showing the whole problem?

Anyways, anlon was hinting at taking the derivative of the equation in post 4. You need to find the equation first (anlon also told you how to make the equation in post 2). I personally recommend you take the integral of the given equation: -15 dV/dt = 2V - 450. Which honestly looks to have the variable t in there somewhere but it looks missing. Usually when you derive an equation from a volume per second the variable t will be involved somewhere. After you take the integral through separation of variables (which once again requires t), solve the problem backwards. It's a cheating method but it will give you the steps you need and you can rewrite the steps from end to beginning to make it appear as if you had the equation to start with.
 
  • #10
Ray Vickson
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I noticed how in your attempt at the solution you had the variable t yet the wording of the question doesn't have the variable t. Perhaps you're not showing the whole problem?

Anyways, anlon was hinting at taking the derivative of the equation in post 4. You need to find the equation first (anlon also told you how to make the equation in post 2). I personally recommend you take the integral of the given equation: -15 dV/dt = 2V - 450. Which honestly looks to have the variable t in there somewhere but it looks missing. Usually when you derive an equation from a volume per second the variable t will be involved somewhere. After you take the integral through separation of variables (which once again requires t), solve the problem backwards. It's a cheating method but it will give you the steps you need and you can rewrite the steps from end to beginning to make it appear as if you had the equation to start with.

That is not necessary. The problem is an elementary exercise in problem formulation, with the (unstated, but standard and well-understood) convention that t represents the time variable (because the problem writes dV/dt). It should not take much work at all to do what the problem asks. Using some technique (such as separation of variables) to find an actual solution was not given as part of the problem, and (as I said) it totally unnecessary.

In addition: we are not told the problem context. Perhaps it is in a textbook chapter that deals with, and contains, several "rate" problems, where time derivatives like dx/dt or dw/dt or dσ/dt, etc., occur, with clear ties to time behavior. In that case the meaning of the problem would be perfectly clear in context, even though it might be regarded as deficient when considered in isolation.
 
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  • #11
epenguin
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OMG it tells you the rate liquid goes in. It tells you the rate liquid leaks out. It's difficult to get an expression for the rate at whichvvolume increases?
 
  • #12
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OMG it tells you the rate liquid goes in. It tells you the rate liquid leaks out. It's difficult to get an expression for the rate at whichvvolume increases?
Sorry I am confused.

I thought I got it. Basically V is in function of t so if we find the difference that will be same as dV/dT
 
  • #13
epenguin
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I thought I got it. Basically V is in function of t so if we find the difference that will be same as dV/dT

If you mean the difference between the rate at which liquid flows in and the rate it flows out, yes.

Maybe you failed to recognise this simple fact because one rate is given as a number (a constant rate) and the other as a formula involving a variable (the volume, V).

So then you will get an equation, a simple differential equation, for dV/dt in terms involving V. Which you have to solve. Ah no, I see that solving it is not in the problem. But that might be the next problem. I hope you don't need to come here to solve it. Because it is among the most elementary and standard of differential equations. So elementary that you might not even find it in the chapter on differential equations, you might find all you need in the chapter on 'integration'..
 
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