# ODE Applications: Salt water problems

This question is a general one about problems that you often see that are about water in a container in salt dissolved in the water. Let's set it up like this:

$$S(t)$$ is the amount of salt in the container
$$V(t)$$ is the volume of water in the container
$$V_{in}(t)$$ is the volume of salt solution flowing into the container
$$C_{in}(t)$$ is the salt concentration of water flowing into the container
$$V_{out}(t)$$ is the volume of water flowing out of the container

(Assuming the container is never full or empty) Is it correct to say that
$$\frac{dV}{dt}=V_{in}(t)-V_{out}(t)$$
(hence you can solve for V), and $$\frac{dS}{dt}=C_{in}(t)V_{in}(t)-\frac{S}{V(t)}V_{out}(t)$$
or is there some weird chain rule type thing in there (in the $$C_{in}(t)V_{in}(t)$$ term)? I feel like there should be.

• No, that's right. I think part of what is confusing you is that the variables are labeled in a way that isn't completely consistent with their units. The specific confusing thing is that $V_{in}$ and $V_{out}$ aren't in volume units, but rather are in volume/time units.
– Ian
Oct 6, 2021 at 0:46
• @Ian Wouldn't $C_{in}$ be in units of mass per volume? Then $C_{in}V_{in}$ would be in units of mass per time instead of mass per (time)^2?
– Ivan
Oct 6, 2021 at 0:51
• $C_{in}$ is mass per volume, so $C_{in} V_{in}$ is mass per time which is what it should be. Identifying $S/V$ as $C$ (the concentration in the tank right now) might help with intuition (since then both terms will look the same).
– Ian
Oct 6, 2021 at 1:08