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?
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 $\displaystyle N(t)$ represent the mass of nutrient present in the estuary at time $\displaystyle t$.
Now, we know the time rate of change of $\displaystyle N(t)$ 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 $\displaystyle N(t)$. 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 $\displaystyle t$ is just $\displaystyle N(t)$ divided by the volume of fluid in the estuary. Since the flow in is equal to the flow out, this volume remains constant at $\displaystyle 1000\text{ m}^3$.
Hence, the output rate of nutrient is:
$\displaystyle \left(100\frac{\text{m}^3}{\text{day}} \right)\left(\frac{N(t)}{1000}\,\frac{\text{kg}}{ \text{m}^3} \right)=\frac{N(t)}{10}\,\frac{\text{kg}}{ \text{day}}$
So, we now have enough information to model $\displaystyle N(t)$ with the IVP:
$\displaystyle \frac{dN}{dt}=9-\frac{N(t)}{10}$ where $\displaystyle N(0)=20$
Can you proceed from here?
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!
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?
No, the ODE is linear, as we may write it in the form:
$\displaystyle \frac{dN}{dt}+\frac{1}{10}N(t)=9$
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?