Please help me understand this. $\frac{dx}{dt} = S x (a-x)$. What does it mean for some constant $S$? How to find $x$ for fastest/slowest growth? I am having some trouble understanding this problem. There is this function that calculates reaction rate of a substance for some constant positive $S$.
$a$ = original amount of the first substance 
$x$ = some amount of substance
First question, what does $\frac{dx}{dt} = S x (a-x)$ mean? Does this mean that the rate of change of $x$ in the equation $(S x (a-x))$ is affected by the change in time? So if there was no '$x$' in the equation, than change in time would not affect the equation right? *This is the first time I am encountering '$dt$' in my derivative assignments. When finding derivatives for simple equations, its mostly been of d/dx notation.
When I graphed $S x (a-x)$, I substituted random numbers for $S$ and $a$. Does this tell me anything about the function for rate of increase and decrease? Should I have solved for $a$? I notice the function increases and then decreases as $x$ moves away from $0$. I also know that when the derivative crosses the $x$ axis, the original function (which I don't know) will start to decrease.
What do I need to do to determine the fastest/slowest growth rate using the derivative? 
Thanks
 A: It means that the amount $x(t)$ is a function of time (makes sense) and tells you that the rate of change of $x$ is a function of $x$ itself.
If the right hand side were a constant, then the rate of change would be constant: every second the amount $x$ would increase by $S$.
I'm sure in Calculus you've already seen examples of functions where the right-hand side depended on $t$. If that were the case, then you could calculate $x(t)$, given the formula for $dx/dt$, by integrating both sides, e.g. finding the area under the curve of $dx/dt$.
Here the situation is a bit different, since the derivative depends on $x$ itself, and not $t$. It takes some getting used to but conceptually, the formula tells you the slope of $x(t)$, given the current value of $x(t)$.
In this particular case, if there is no material ($x=0$) or there is exactly $a$ amount of material, then $dx/dt=0$ and the system is in equilibrium: the slope is 0 and $x$ does not change over time.
If $x$ is between 0 and $a$, then $dx/dt$ is positive and $x$ will increase over time. As $x$ increases and approaches $a$, the rate of increase decreases.
If $x$ is over $a$, then $dx/dt$ is negative and $x$ decreases over time towards $a$.
The maximum increase will occur when $Sx(a-x)$ is most positive; you can find the value of $x$ as usual, by taking the derivative of $Sx(a-x)$ with respect to $x$ and setting it equal to zero.
The equation $dx/dt = Sx(t)[a-x(t)]$ is called an ordinary differential equation and given the initial value of $x$ at time 0, $x(0)$, you can solve it for the function $x(t)$ which gives $x$ at all times, although you have probably not learned the techniques yet. Here are some plots of $x(t)$ for $S=1$, $a=1$, and several different values of $x(0)$:

You can see the main characteristics of the system here, namely, that $x$ either decreases or increases towards $a$ depending on where it starts.
A: The explanation given by @user7530 is so excellent that I have little to add.  That said, in case you are curious as to the actual solution of the differential equation you posted, it is actually not at all hard to solve with a little Calc II knowledge.  I will not derive unless you ask, (and I am sure that @user7530 has done it, given he has graphed it) and the result is
$$x(t) = \frac{a x(0)}{x(0) + [a-x(0)] e^{-a S t}}$$
