The rates of change of some other functions

In an earlier homework assignment we saw that it will be possible to compute the rate of change of the exponential function, $\exp:(-\infty,\infty)\rightarrow(-\infty,\infty)$, $\exp(x) = e^x$, at any value a if we could compute the rate of change at a=0, and that the rate of change at appears to be 1. We will demonstrate by analytical means later that this correct, and that, remarkably, the rate of change of the exponential function at a is $\exp(a)$!

This is an extremely important observation for the following reason. There are many physical systems which satisfy the following law: The rate of change of the system is proportional to the current state of the system. Some examples are:

• The rate at which bacteria grow is proportional to the number of bacteria present;
• The rate at which a radioactive substance decays is proportional to the amount of the substance present;
• The rate at which the temperature difference between an object and it's surrounding changes is proportional to the temperature difference.

In mathematical language, if we take the proportionality constants to be 1, and denote the state of the system at time x to be S(x), we want S'(x) = S(x), that is we want a function whose rate of change at x is the value of the function at x. In other words, we want the exponential function.

Two more important functions to study in this regard are $\sin$ and $\cos$. We shall demonstrate below that their rates of change can be computed from the rate of change of $\sin$ at x=0 by using trigonometric identities, just as the rate of change of $\exp$ could be computed from its rate of change at and an identity for exponents.

First, to get an idea of what is happening, consider the plot of the difference quotient for $\sin$ at x=0 as a function of h:

$$\frac{\sin(0+h)-\sin(0)}{0+h-0} = \frac{\sin(h)}{h}$$

It would appear that the rate of change of $\sin$ is 1 at x=0.

Now consider the plot of

$$\frac{\sin(a+h)-\sin(a)}{a+h-a} = \frac{\sin(a+h)-\sin(a)}{h}$$

for h=0.01 as we vary a from $-2\pi$ to $2\pi$.

Sure looks like the graph of $y = \cos(x)$ on the same interval. Let's try to see why:

$$\begin{array} {rcl} {\displaystyle\frac{\sin(a+h)-\sin(a)}{a... ...- \sin(a)\frac{\sin(h)}{h}\frac{\sin(h)}{1+\cos(h)}}\end{array}$$

In this last expression, the first term should approach $\cos(a)\times 1$while the second term should approach $-\sin(a)\times 1 \times 0$ as h approaches , confirming what we saw in the graph. Indeed, by analytical methods we will be able to confirm what we see here, that the rate of change of $\sin$ at x=a is $\cos(a)$.

The function $\cos$ is handled in the same way. Here is a plot of the difference quotient for cosine, with h=0.01, for values from $-2\pi$ to $2\pi$. Observe that the graph looks like that of $y = -\sin(x)$ over the same range.

Here the algebra to support this observation. It is nearly identical to that for $\sin$. Can you explain why?

$$\begin{array} {rcl}{\displaystyle \frac{\cos(a+h)-\cos(a)}{a... ...- \cos(a)\frac{\sin(h)}{h}\frac{\sin(h)}{1+\cos(h)}}\end{array}$$

This time the first term approaches $-\sin(a)$ while the second term approaches , confirming what we saw in the graph.