1. ## exponential growth

Can anyone explain more about exponential growth of the concentration on nutrients in an estuary when water is flowing into and out of the estuary at the same rate?

2. ## Re: exponential growth

Are you given initial conditions and the concentration of nutrients flowing in?

3. ## Re: exponential growth

the flow in rate is 100m^3/day with concentration of 90g/m^3.
Initial concentration in the estuary is 20g/m^3
vol 1000m^3
flow out rate is also 100m^3/day

4. ## Re: exponential growth

First, you want to determine the mass of nutrient initially present. Amount per volume times volume equals amount. What do you find?

5. ## Re: exponential growth

20g/m^3 x 1000m^3 = 20000g/m^3 initial concentration
and I have 90 x 100/day = 9000 flowing in

6. ## Re: exponential growth

You have the correct initial mass of 20,000 g, but you don't want to call this a concentration, it is simply an amount. Also, you are correct that 9000 grams is flowing in. So, let's then use smaller numbers and say the initial mass is 20 kg and 9 kg is flowing in per day.

Let's let $N(t)$ represent the mass 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 asume 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 $N(0)=20$

Can you proceed from here?

7. ## Re: exponential growth

would this mean that on day zero, my flow out amount is 9 -20/10 = 7?
we are supposed to write down and solve an appropriate differential equation for N(t) along with the appropriate initial condition. That would somehow be my initial amount plus the flow in rate, minus the flow out rate? My apologies for sounding so thick!

8. ## Re: exponential growth

The initial output rate of nutrient would be 7 kg/day, but that is really only applicable at the very start of the first day, the initial moment.

I have already stated the appropriate differential equation/initial condition. This is called an initial value problem (IVP).

Now, to solve the ordinary differential equation (ODE), do you recognize what type of equation it is, and if so, what you need to do to solve it?

9. ## Re: exponential growth

By the way, are you supposed to find the mass of nutrient at time t, or the concentration? Since the concentration is just mass/volume and the volume is constant it is simple to convert between the two, I was just curious.

10. ## Re: exponential growth

we are supposed to find the amount at time t and then the concentration after a long time

11. ## Re: exponential growth

Okay, good, then we are set to do just that.

12. ## Re: exponential growth

it's an exponential growth equation.
I need the initial amount plus the flow in amount = 20 +9^t minus the flow out amount

13. ## Re: exponential growth

No, the ODE is linear, as we may write it in the form:

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

Do you recognize this form? Do you recall how to transform the equation so that we have the product of a differentiation on the left?

14. ## Re: exponential growth

I'm just reading this from my text book. The equation is already in the required form(dN/dt + P(t)N= Q(t)

15. ## Re: exponential growth

Yes, so what do we need to do next?

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