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It's time, however, to go back to the dashboard: we're not focused on finding speeds from changes in distance now, but rather we're going to focus on the odometer: how do tell how far we've travelled given a history of our speeds? Back to Day02:
The big ideas of Calc I are contained in Gil Strang's "Auto Analogy": he says that the "The central question of calculus is the relation between [speed and distance traveled]."
Imagine that the car is operated on a long, straight road. (Just for the sake of convenience, we won't allow it to travel in reverse! Otherwise we'd be talking about velocity -- speed and direction -- rather than just speed.)
minute | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
speed (mph) | 0 | 30 | 45 | 30 | 70 | 65 | 70 | 70 | 45 | 30 | 35 |
Estimate the distance travelled.
This strategy gives rise to the so-called "Trapezoidal rule", which we'll encounter in a moment.
Last time we did a "thought experiment", about traveling down a straight road in different ways:
If we were going a constant speed, then everything is easy:
\[ D(b)=p'(0)(b-0) = p(b)-p(0) \]
We also noticed that this is equal to the area under the rectangle formed by the speed function (a constant, say 5 mph):
If you walk 5 mph, then in two hours you will walk 10 miles: $5\frac{miles}{hour}*2hours=10$ miles.
If we sample the speed at $n$ equally spaced moments $t_i$ (like your black box does!), then we discovered that we could estimate the displacement, or distance travelled, as a sum:
\[ D(b)=\sum_{i=0}^{n-1}p'(t_i)\Delta t \]
where we use "summation notation" to represent a sum of terms, with $i$ running from 0 to $n-1$ (notice that gives $n$ values of $i$, and hence $n$ subintervals).
$\Delta t = \frac{b}{n}$ is the length of an interval (in your black box this might be every hundredth of a second).
Now it's true that \[ D(b)=\sum_{i=0}^{n-1}p'(t_i)\Delta t \approx p(b)-p(0) \] Now if we want to get this exactly right, and replace that $\approx$ with an equals sign, we need to continue to subdivide the interval more finely; in fact, we must subdivide so finally that we let $\Delta t \to 0$ (or, alternatively, we let $n \to \infty$):
\[ D(b)=\lim\limits_{n \to \infty}\sum_{i=0}^{n-1}p'(t_i)\Delta t = p(b)-p(0) \]
Finally we introduced notation for that horrible limiting sum on the left: we used integral notation, and wrote (more cleanly)
\[ D(b)=\int_{0}^{b}\ p'(t)\ dt = p(b)-p(0) \]
where the integral sign $\int$ is thought of as an elongated S, for "sum" (only we're adding up an infinite number of infinitesimally small rectangles (there's a word for you, a big word for a very small thing!!)).
The rectangles have a real, finite height $p'(t)$, but their width has gone to zero, but in such a way that they remain... well... like pixie dust, let's say!:) So we call the width $dt$ an infinitesimal.
You've actually seen this before: in \[ \frac{df}{dt}=\lim\limits_{dt\to 0}\frac{f(t+dt)-f(t)}{dt} \]
$dt$ has been going to zero for quite some time, and it never quite gets there -- but it's smaller than any real number. You know -- pixie dust!
So now we recast our problem as one of computing areas -- but we won't forget our amazing result, that the distance travelled (the odometer reading) can be determined from the speed (ometer): if you know your rate of change at every moment, then you can figure out your position.
We noticed last time that the choice of the speed was a little arbitrary: why do we use the speed at the left of the interval in time to make our estimates of the distance travelled?
We could just as easily have taken an estimate at the end of an interval, leading to what we call the "right rectangle rule":
\[ D(b)=\sum_{i=1}^{n}p'(t_i)\Delta t \approx p(b)-p(0) \]
It looks a lot like the other rule, only $i$ is running from 1 to $n$, and to the $t_i$ sample are right endpoints, not left endpoints.
So let's have a look at several different "integration rules", and how they relate to one another.
There are several useful rectangular rules, such as Left/Right/Midpoint rules:
(notice that $a+N\Delta{x=b}$ that is, that $\Delta{x}=\frac{b-a}{N}$)
Then, in fact,
where by $\bar{f}\mbox{}$ we mean an average value of $f$.