# Linear Differential Problem(Pollutants into a lake)

1. Jun 1, 2010

### paraboloid

Hi,

This is my problem:

Consider a lake of constant volume V containing at time t an amount Q(t) of pollutant,
evenly distributed throughout the lake with a concentration c(t), where c(t) = Q(t)/V .
Assume that water containing a concentration k of pollutant enters the lake at a rate r ,
and that water leaves the lake at the same rate. Suppose that pollutants are also added
directly to the lake at a constant rate P.

If at time t = 0 the concentration of pollutant is c0 , find an expression for the concen-
tration c(t) at any time. What is the limiting concentration as t → ∞?

I have no clue how to solve this notably because pollutants are flowing in and being added. I've been at it for an hour and half with nothing to show so I thought I'd turn to the forums.

Any help would be great,
Thanks

2. Jun 1, 2010

### lanedance

you nee dto write down a differential equation then attempt to solve

so you know:
- initial concentration
- the incoming rate
- the outogoing rate

so make the assumption everything in the lake is perfectly mixed, so will be the same as the outgoing concentration & try to write down the rate of chaneg of contaminant in the lake

it may be easier to do in terms of total volume of contaimnant, though is the same

3. Jun 2, 2010

### Susanne217

So we view this this $$\frac{dc}{dt} = \frac{Q}{V}$$ with the intial condition

$$c(0) = c_0$$

and isn't this equal to

$$dc = \frac{Q}{V} dt$$ and integrating on both sides

$$c(t) = (\frac{Q}{V})\cdot t + k$$ and inserting the intial condition is that then

k = $$c_0$$ giving the result $$c(t) = (\frac{Q}{V})\cdot t + c_0$$

Last edited: Jun 2, 2010
4. Jun 3, 2010

### lanedance

think about what that solution is telling you, c(t) increases without out bound, which doesn't make a heap of sense...

its probably easier to write in terms of the contaminant in the lake, then solve for concentration later.

So at time t=0, assume we know the pollutant in the lake is
$$Q(0) = Q_0$$

the rate of change of the total pollutant can be broken into 2 contributions, one for the incoming pollutant and one for the outgoing:
$$\frac{dQ(t)}{dt} = (\frac{dQ(t)}{dt})_{in} + (\frac{dQt)}{dt})_{out}$$

see if you can write down what they are in terms of k, r and c(t) or Q(t)
$$(\frac{dQ(t)}{dt})_{in}$$
$$(\frac{dQt)}{dt})_{out}=?$$

that will give you the differential equation you need to solve for Q and c will follow

5. Jun 3, 2010

### lanedance

just to point out where this went wrong you need to start at
$$c(t) = \frac{Q(t)}{V(t)}$$

then differentiating w.r.t. t gives
$$c'(t) = \frac{dc(t)}{dt} = \frac{d}{dt}(\frac{Q(t)}{V(t)})$$

now is we assume V(t) = V is constant, which we can as the same amount of water is going in as out, it becomes:
$$c'(t) = \frac{d}{dt}(\frac{Q(t)}{V}) = \frac{dQ(t)}{dt}\frac{1}{V} = \frac{Q'(t)}{V}$$

then Q(t) isn't constant, but is a function of t, so we can't simply integrate