How Fast, Exactly Here
Last timeA Rule, and the Picture of It
A rate over an interval is a division anyone can do. The rate at a single instant is what that division settles on as the interval shrinks, and it usually exists.
A car covers sixty miles in an hour. Its average speed was sixty miles an hour, and
that statement is complete, honest and says nothing about whether it was ever doing
sixty. The speedometer says something different and much stranger: it claims a speed
at an instant, during which no distance was covered and no time passed.
This lesson is about making that second claim mean something.
- the rise: how much the answer changed between the two inputs. For a rule that is falling this is negative, and everything below works unchanged
- the width of the interval. It is never zero anywhere in this expression, which is what keeps the division legal at every stage
- the rate at the single input x, pronounced f prime of x. It is a new rule, built from the old one, which answers at every input where the settling happens
- the quotients settle on a value as h shrinks. Settling means they can be made as close to it as anybody demands by taking h small enough, and stay that close
Watching it settle
Take the squaring rule and ask about its rate at an input of two. Measure across
intervals that get smaller, and read the last column downwards.
| step | width of the interval | rise across it | rise divided by width | what happened |
|---|---|---|---|---|
| 1 | 1 | 5 | 5 | A crude measurement: from two to three the answer rose by five. |
| 2 | 0.5 | 2.25 | 4.5 | Half the interval. The rise more than halved, so the quotient has dropped. |
| 3 | 0.25 | 1.0625 | 4.25 | Halving again cuts the quotient by half of what it dropped before. |
| 4 | 0.1 | 0.41 | 4.1 | The pattern is now obvious: the quotient is four plus the width. |
| 5 | 0.01 | 0.0401 | 4.01 | And it stays obvious. Nothing is approaching zero over zero here. |
Doing it with algebra instead
The table suggests four. Here is why it is exactly four, with no appeal to how the
numbers looked.
Write the rise across an interval of width h starting at x. The answers are x plus h,
all squared, and x squared. Multiply out the first: it is x squared plus twice x times
h plus h squared. Subtract x squared and the leading term is gone, leaving twice x
times h plus h squared. Every term now contains h, so dividing by h leaves twice x
plus h, with nothing underneath.
That last expression is the whole point. Before the cancelling, setting h to zero gave
nothing over nothing and was meaningless. After the cancelling, h appears only as
something being added, and letting it go to zero is obviously allowed. The answer is
twice x, which at an input of two is four.
The lesson stops here
4 more paragraphs to go
You have read the opening. The rest of the argument, the problems that check whether it landed, and the lines worth keeping at the end all come with a plan.
The first lesson of every course in the library reads the whole way through, free, so you can see exactly what the rest of them are.
See the planThe contentsThis is the reading half
Starting the course gives you your own copy of it. Every idea on every page has problems standing under it, marked with a reason rather than a tick, and any sentence you do not believe can be opened and argued with. None of that can happen on a page nobody owns.
The contents