1. differential equations

Hi,

I need a bit of help with this question:

'A polluted river with a nutrient concentration of 90g.m^3 is flowing at a rate of 100m^3/day into an estuary of volume 1000m^3. At the same time, water from the estuary is flowing into the ocean at 100m^3/day. The initial nutrient concentration in the estuary is 20g/min^3.
(i) Let N(t) be the amount of nutrient (in grams) in the estuary at time t. Write down and slove an appropriate differential equation for N(t) along with the appropriate initial condition.'

My workings have got me this far:

N'(t)=(rate of nutrient going in)-(rate of nutrient going out)=((N(t)g nutrient)/1000m^3)-(100m^3/days)=(N(t)g of nutrient)/((10+t)days)

I think the differential equation should be along the lines of:

N'(t)=9000-(N(t)/(10+t)) => N'(t)+N(t)/(10+t)=9000

but I'm really not too sure. Should that be the end of the equation, or have I missed a few steps?

2. Re: differential equations

Take a look at the following topic:

exponential growth

3. Re: differential equations

Thanks markFL2, so does my equation is correct for the diffrential form, what about if I am caculating what the concentration would be after a long time? I have changes it into the form N(t)=((20t+t^2)/(2(10)+t))+(D/(10+t)), and found that D=10, so the for would now be N(t)=((20t+t^2)/(2(10)+t))+(10/(10+t)). Now I tried to change t=100 so 100 days (which I consider a long time) to make 5.5454g/m^3

4. Re: differential equations

Originally Posted by NettieL
Hi,

I need a bit of help with this question:

'A polluted river with a nutrient concentration of 90g.m^3 is flowing at a rate of 100m^3/day into an estuary of volume 1000m^3. At the same time, water from the estuary is flowing into the ocean at 100m^3/day. The initial nutrient concentration in the estuary is 20g/min^3.
(i) Let N(t) be the amount of nutrient (in grams) in the estuary at time t. Write down and slove an appropriate differential equation for N(t) along with the appropriate initial condition.'

My workings have got me this far:

N'(t)=(rate of nutrient going in)-(rate of nutrient going out)=((N(t)g nutrient)/1000m^3)-(100m^3/days)=(N(t)g of nutrient)/((10+t)days)
"(rate of nutrient going in)- (rate of nutrient going out)" is good but what you have is NOT "rate of nutrient going in" or "rate of nutrient going out". There is water coming in at $100 m^3/day$ and each cubic meter contains 90 g of nutrient. The amount of nutrient going into the 90(100)= 9000 g per day. That does not depend on the amount of nutrient already in the estuary. Using N(t) as the amount of nutrient (in g) the amount of nutrient per cubic meter is N(t)/1000. Since the water outflow is also 100 cubic meters per day, the outflow of nutrient is (N/1000)(100)= N/100 g per day.

I think the differential equation should be along the lines of:

N'(t)=9000-(N(t)/(10+t)) => N'(t)+N(t)/(10+t)=9000

but I'm really not too sure. Should that be the end of the equation, or have I missed a few steps?

5. Re: differential equations

Originally Posted by HallsofIvy
" Since the water outflow is also 100 cubic meters per day, the outflow of nutrient is (N/1000)(100)= N/100 g per day.
apologies for sounding very thick here, but I thought (N/1000)(100)= 100/1000=1/10 ??

6. Re: differential equations

Let's let $N(t)$ represent the mass in kg of nutrient present in the estuary at time $t$.

Now, we know the time rate of change of $N(t)$ is equal to the rate in minus the rate out. We have already determined the rate in. The rate out will be a function of $N(t)$. We will assume the concentration of nutrient is uniform in the estuary. That is, the concentration of nutrient in any part of the estuary at time $t$ is just $N(t)$ divided by the volume of fluid in the estuary. Since the flow in is equal to the flow out, this volume remains constant at $1000\text{ m}^3$.

Hence, the output rate of nutrient is:

$\left(100\frac{\text{m}^3}{\text{day}} \right)\left(\frac{N(t)}{1000}\,\frac{\text{kg}}{ \text{m}^3} \right)=\frac{N(t)}{10}\,\frac{\text{kg}}{ \text{day}}$

So, we now have enough information to model $N(t)$ with the IVP:

$\frac{dN}{dt}=9-\frac{N(t)}{10}$ where $N(0)=20$

The ODE is linear, and written in standard linear form is:

$\frac{dN}{dt}+\frac{1}{10}N(t)=9$

Next, you want to calculate the integrating factor $\mu(t)$...

7. Re: differential equations

what is the integrating factor used for?

8. Re: differential equations

I tried to find out the integral of what was used in that example given by MarkFL2, but I just came out with: de^(t/10)(10t-100)=9(sqrt^10(e)) t or

9. Re: differential equations

The integrating factor allows the left-hand side of a linear ODE to be expressed as the derivative of the product of two functions. Then integration is possible.

You should have already studied the theory behind solving linear ODEs prior to the applications of first order ODEs. Now, I'm not saying you should instantly recall every detail in the proof of why this works or how to apply it, but you should at least be familiar with the concept of using an integrating factor. Does it sound at all familiar?

10. Re: differential equations

Originally Posted by NettieL
I tried to find out the integral of what was used in that example given by MarkFL2, but I just came out with: de^(t/10)(10t-100)=9(sqrt^10(e)) t or
It appears you correctly computed $\mu(t)=e^{\frac{t}{10}}$ and so you should have:

$\int\,d\left(e^{\frac{t}{10}}N(t) \right)=9\int e^{\frac{t}{10}}\,dt$

Now, can you integrate both sides?

12. Re: differential equations

and yes, it does sound familiar

13. Re: differential equations

$\int\,d\left(e^{\frac{t}{10}}N(t) \right)=9\int e^{\frac{t}{10}}\,dt$

Integrating, you should get:

$e^{\frac{t}{10}}N(t)=90e^{\frac{t}{10}}+C$

Do you see why?

14. Re: differential equations

I think I have the formula now, thank you so much for your help