Its about the rate of change of pond A and pond B. The question is attached. I just don't know how to make a start. I know that in the first year that 20,000 kg of pollutant goes into A which has a volume of 4000. Also given is that 200 km3 of polluted volume goes in and the same amount goes out. I cant cant work out what the constant, a, might be as there is no pollutant to begin with so I thought it might be zero- it not. I did sketch the situation. If i can have some help with the pond A then I might guess how to work out the rates of change in pond B.
Q on Differential equation.
- Thread starter Sonal7
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It says in the question let x amd y be the pollution in kg/km^3 in pond A and pond B respectively. I sure that t is time in years. I don't understand why both the constants aren't 0. It says in the Q that there is no pollution to begin with. Okay hang on. I think the rate is 20,000 ÷ 4200 for pond A to start off with. Don't get confused with my answers as they haven't been submitted yet. However the rate changes continuously so I not sure.
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Let's look first at the amount \(A\) of pollutant in lake A:
[MATH]\d{A}{t}=\left(200\frac{\text{km}^3}{\text{yr}}\right)\left(\frac{20000\text{ kg}}{200\text{ km}^3}\right)-\left(200\frac{\text{km}^3}{\text{yr}}\right)\left(\frac{A\text{ kg}}{V_A\text{ km}^3}\right)[/MATH]
[MATH]\d{A}{t}=20000-\frac{200}{V_A}A[/MATH]
Where \(V_A\) is the volume of lake A.
And so the level of pollution in lake A is:
[MATH]x=\frac{A}{V_A}[/MATH]
, hence:
[MATH]\d{x}{t}=\frac{1}{V_A}\cdot\d{A}{t}[/MATH]
Can you proceed?
[MATH]\d{A}{t}=\left(200\frac{\text{km}^3}{\text{yr}}\right)\left(\frac{20000\text{ kg}}{200\text{ km}^3}\right)-\left(200\frac{\text{km}^3}{\text{yr}}\right)\left(\frac{A\text{ kg}}{V_A\text{ km}^3}\right)[/MATH]
[MATH]\d{A}{t}=20000-\frac{200}{V_A}A[/MATH]
Where \(V_A\) is the volume of lake A.
And so the level of pollution in lake A is:
[MATH]x=\frac{A}{V_A}[/MATH]
, hence:
[MATH]\d{x}{t}=\frac{1}{V_A}\cdot\d{A}{t}[/MATH]
Can you proceed?
HallsofIvy
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x is the amount of pollutant in lake A. dx/dt is the rate at which that changes. So how does it change? Well, some comes in and some goes out. You are told "a polluted river flows into lake A at the rate of 200 km^2 per year". I presume that were you got the "200" you entered for (a). But that is the volume of water that flows in, not pollutant. The very next sentence says "20000 kg of chemical pollutant enters the lake every year". THAT is what (a) is. How does pollution flow out of lake A? Well, water flows out at the same rate as it flows in which is that previous 200 km^2. But how much pollutant is in that water? With x as the amount of pollutant per cubic km, the amount of pollutant is x kg/km^3* 200 km= 200x kg. But that is going out so (b) is -200. Since there is no water, so no pollutant, flowing from B to A, (c) is 0. Do the same thing for Lake B.
Steven G
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If x and y were both 0, that is they were both constants then dx/dt and dy/dt would both be 0. Do you see that? The problem is that dx/dt and dy/dt are both changing causing x and y not to be constants and hence dx/dt and dy/dt can't be 0. Possible x and y may initially be 0 but again they change from there.
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Using my previous post, I would continue as follows:
[MATH]\frac{1}{V_A}\cdot\d{A}{t}=\frac{1}{V_A}\left(20000-\frac{200}{V_A}A\right)[/MATH]
[MATH]\d{x}{t}=\frac{1}{V_A}20000-\frac{200}{V_A}x[/MATH]
And then using the given \(V_A=4000\text{ km}^3\)
[MATH]\d{x}{t}=\frac{20000}{4000}-\frac{200}{4000}x[/MATH]
And finally:
[MATH]\d{x}{t}=(5)+\left(-\frac{1}{20}\right)x+(0)y[/MATH]
And would therefore conclude:
[MATH](a,b,c)=\left(5,-\frac{1}{20},0\right)[/MATH]
Can you work the second part of the problem now?
I already did. The answers were correct but as rates keep changing I am doubtful the solution is that straightforward. I can see the answer but I dont think its right answer as the 20,000 goes into 4200 km3, and not 4000. Just my feeling. I find it confusing that you used A and then used X which are both the same variables. I think if you left it as X it would be easier follow you thought.
The volume that goes in 200, so it doesn't matter that much goes out as far as I can see. the contaminant is going into 4200 total of water. Its not a workable solution as the rates will change throughout the year. It wont be constant. When there is 0 pollutant the rate be relatively higher and at higher pollution later on in the year, the rate will change slowly as its relative to the existent concentration. Its a bad Q.
