Class Content I >Modeling with mathematics > Math recap > Values, change, and rates of change
2.2.5.2
Prerequisites
The basic idea of calculus is about how changes are linked. When we looked at derivatives, we paid attention to how the changes in two variables were connected when the changes were small. But we want to be able to go further than that. Suppose the changes are big? If we know what the relation of the changes are as a function of one variable (the derivative), we can build up big changes by adding up the small change. This is done with the concept of the integral.
We start with the abstract math version: We have a function, f, that depends on an independent variable (one whose value we can choose freely) x. Recall that the basic idea of a derivative is to define a function that represents the ratio of how much f changes when x changes (by a little). So
The notation contains a bit of an internal contradiction. The expression on the left, g(x), says that we are considering this derivative as a function of x so we know it at any individual point x. The expression on the right says the derivative is the change in f divided by the change in x. To get a change you have to consider two different x's. The way we get away with this is that we consider "the change" as being from "a value of x just a tiny bit smaller than x to a value of x just a tiny bit larger." So it's like looking at a movie film an saying "the velocity at frame 96 is the change in position from frame 95 to frame 97 divided by the time interval from frame 95 to frame 97." This is kind of cumbersome, but critical in thinking about what velocity -- or any derivative -- really means. People in math spend a lot of effort to get rid of this, but we don't have to worry about it.
In math the integral is conceived of as the inverse of differentiation. If you know f, you differentiate it to get g. But what do you do if you know the function g and want to find f ? This might seem strange at first. How would you know a function's derivative if you didn't know the function? But actually, this happens all the time. One of the key relations this term will be to show that an object's acceleration is determined by the totality of the forces that it feels. So we might know an object's acceleration by figuring out the forces. But the acceleration is the derivative of the velocity! So a = dv/dt and we know a, but want to find v.
Let's try it using g and f so we can focus on the general structure of the math without being distracted by the particular "what it is" for now.
If we take seriously the derivative as a ratio, we can multiply both sides of our equation by "dx" to get the following:
Read this as: "the change in f is equal to g (the derivative of f ) times the change in x." It might be clearer what this means if we used deltas instead of d's and pay attention to the ends of the interval dx. This makes it look like the following:
Since we are looking at a change, we have two values of x but g is only a function of x once. We pick the starting value, though it would probably be more aesthetic to choose to evaluate g at the midpoint of the interval (x0 + x1)/2. But everything would then look a lot messier and it turns out not to make a difference.
Now let's take many steps of size dx:
The ... means "imagine that this process continues". We'll write "N" for the top value of our last step.
If we now write how f changes in each step, we get
where again the three dots means "keep going".
Now the trick is to add up all of the equations in the line above. This gives us
Although this looks messy, something interesting happens. All the terms on the left cancel except the first and the last! (The ending point of one interval and the starting point of the next are the same but once it comes in with a + and once with a -.) But the terms on the right (with the g's) just add -- there is no cancellation. The result is that the change of f from first to last is given by the sum of lots of gdx terms. We write this sum of terms using the summation notation with a big Greek "S" (= sigma = Σ) for "sum". The right hand side of the second line below just means exactly what is written above it.
Now we imagine making the dx very small. The number of terms in the sum gets very large -- but since each term is getting smaller, these compensate and the result (usually) settles down. When we don't specify the steps in our adding up we replace our Greek Sigma by a stretched "S" or integral sign.
The final result shows that our sum (or integral) of derivatives can tell us how f changes as x moves away from the starting value. This means that our integral generates the entire function f.
Follow-ons
Joe Redish 9/4/11
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