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

  • $\begingroup$ Does it really say $\frac {dx}{dt}=Sx(a-x)?$ That would make more sense. $\endgroup$ – Ross Millikan Feb 19 '13 at 21:35
  • $\begingroup$ Yes, sorry. So that means the derivative of some function is Sx(a−x)? Or does it mean I still need to calculate the derivative and that Sx(a−x) is the original function? $\endgroup$ – koloa Feb 19 '13 at 21:40
  • $\begingroup$ I added the DEs tag, as this relates to differential equations (whether or not the asker realizes it does). (in case anyone wonders why the tag just appeared) $\endgroup$ – apnorton Feb 19 '13 at 22:22

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 $x(0)$:

enter image description here

You can see the main characteristics of the system here, namely, that $x$ either decreases or increases towards $a$ depending on where it starts.

  • $\begingroup$ Thanks so much for this. I have a few more questions...when graphing Sx(a-x) where S=1, a=1...is this picture showing dx/dt? What is the x axis and y axis? x axis is time or is it substance. y axis must be the reaction rate? $\endgroup$ – koloa Feb 19 '13 at 23:19
  • $\begingroup$ It is showing $x(t)$, the vertical axis is $x$ and the horizontal axis is $t$. The reaction rate is the slope of the curve. $\endgroup$ – user7530 Feb 19 '13 at 23:24
  • $\begingroup$ Note that this plot is a visualization aid only -- to draw it you need to know how to solve the differential equation. To find the value of $x$ at which the reaction rate is highest, you can work with the formula for $dx/dt$ directly, and use what you know from Calculus to find its critical points (take the derivative wrt $x$ and set it equal to 0; solve for $x$.) (Don't forget to also check the boundary $x\to\infty$ and $x\to -\infty$.) $\endgroup$ – user7530 Feb 19 '13 at 23:33
  • $\begingroup$ Careful -- $dx/dt$ itself is the rate of change of $x$ with respect to time. You want the values of $x$ that maximize this rate of change -- so you want to look at the rate of change with respect to x of $dx/dt$, which is $d/dx\ dx/dt$, and set it equal to 0. $\endgroup$ – user7530 Feb 19 '13 at 23:54
  • $\begingroup$ I think I understand now. To obtain the rate of change, one would need to use the derivative d/dx on Sx(a-x) since rate of change is measure by slope. Its difficult to obtain the rate of change by looking at the picture. The picture shows the rate of change of x as time moves forward. Hence the dx/dt. $\endgroup$ – koloa Feb 20 '13 at 0:04

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

$$x(t) = \frac{a x(0)}{x(0) + [a-x(0)] e^{-a S t}}$$

  • $\begingroup$ hi,I graphed the derivative, the starting derivative, and when x is maximized. How can I interpret this picture? i.imgur.com/mQy3g5H.jpg when S = 1, i.imgur.com/ApNvMjn.jpg when S = 2. $\endgroup$ – koloa Feb 19 '13 at 22:51
  • $\begingroup$ It's hard to see what you are looking for from that picture, but what you should see is that the derivative is zero at $x=0$ and $x=a$. That the derivative is zero at $x=a$ means that $x$ is not changing when it reaches $x=a$ (it is never zero). BTW the plot of the derivative against the function value is an example of a phase space plot. In this, the point $x=a$ is known as an attractor, because the final state ends up there, no matter the initial condition. $\endgroup$ – Ron Gordon Feb 19 '13 at 23:01

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