PrepShorts · Study sheet · Class 11 Mathematics · Chapter 12, Limits and Derivatives
Chapter 12 · Limits and Derivatives
Averaging over shorter and shorter intervals to get a speed at an instant
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A stone falls for two seconds, and its speed at that exact instant asks for nothing divided by nothing. Every average velocity computed either side of it misses the answer.
The idea
A speed at a single instant looks undefined, because every speed you can actually compute is a distance divided by an elapsed time, and at an instant the elapsed time is zero. §12.2 never divides by zero. It computes only honest average velocities, over intervals that close on t = 2 seconds from the left and from the right, and shows the two families of averages closing in on each other from opposite sides. What makes the argument work is that the trap has two jaws: one family rises toward the answer, the other falls toward it, and the answer is squeezed into the gap between the last entry of each. That gap is where the chapter first applies the word derivative to a computation. The word itself is older than that by two pages — it is already in the opening section, which sets out the plan of the chapter — and either way it is doing work long before a definition exists: twenty-three printed pages separate its first appearance from Definition 1.
What you should be able to do
- State why an average velocity over an interval is computable and a velocity at an instant is not, in terms of what the quotient needs
- Compute the distance fallen from s = 4.9t² at any given time, and reproduce every entry of the chapter's distance table
- Compute the average velocity over an interval with one endpoint fixed at t = 2, from both the left and the right
- Read a shrinking family of average velocities as a sequence and say what it appears to approach
- State the bracket the chapter concludes with, and identify precisely which assumption lets it be stated
- Show algebraically that the average velocity over the interval from t₁ to 2 simplifies to 4.9(2 + t₁), and use that to explain why the bracket is symmetric about a single value
- Reinterpret each average velocity as the slope of a chord on the distance–time graph, and the limiting value as the slope of a tangent
- Identify what the chapter has and has not established at the end of §12.2
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| average velocity | distance covered over an interval divided by the length of that interval | printed in this chapter, §12.2, p. 217 |
| instantaneous velocity | the velocity ascribed to a single moment rather than to a stretch of time | printed in this chapter, §12.2, p. 219 |
| derivative | the rate at which a quantity changes at one instant, obtained as a limiting value of average rates | printed in this chapter from §12.1, p. 217 onward, and used throughout before §12.5 defines it on p. 240 |
| rate of change | how much one quantity moves per unit movement of another | printed in this chapter, §12.2, p. 219 |
| displacement | position measured from a fixed starting point, of which velocity is the rate of change | printed in this chapter, §12.2, p. 219 |
| tangent | the line touching a curve at a point, whose slope the chord slopes approach | printed in this chapter, §12.2, p. 220 |
| slope | the rise of a line per unit run | printed in this chapter, §12.2, p. 220 |
| chord | the straight segment joining two points on a curve | printed in this chapter, §12.5, p. 242, for the same construction drawn in Fig 12.1 |
| shrinking interval | an interval with one endpoint fixed and the other moved toward it | an added phrasing; the chapter performs the move without naming it |
| two-sided bracket | a pair of computed bounds, one from each side, between which the answer must lie | an added label; not printed in this chapter |
Where people slip up
- **"Average velocity over the last stretch just is the velocity at the end."** Table 12.2 shows every one of those averages sitting below 19.6 and none of them equal to it. The averages are correct answers to a different question.
- "The tables prove the velocity is 19.6." They do not, and the chapter is careful about this: it claims only that the value lies inside a bracket. The step from bracket to a single number needs the assumption stated on p. 218 and p. 219 — that nothing abrupt happens in the last hundredth of a second on either side. That assumption is doing real work.
- "You could get the exact answer by taking the interval to be zero." Then both the numerator and the denominator are 0, and 0/0 is not a number. The entire chapter exists because that shortcut is closed.
- "The left-hand list and the right-hand list are two different phenomena." Section 7 shows both are 4.9(2 + t); one list feeds t from below, the other from above. They were never two computations.
- "Slope of a chord and average velocity are separate ideas that happen to agree." They are the same quotient with different labels. The rise is distance covered, the run is time elapsed.
- "The chapter defines the derivative here." It uses the word on p. 219 and defines it twenty-one pages later, in §12.5, p. 240. Anything a student writes in an exam must come from the definition, not from this section's tables.
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Worked answers: Exercise 12.1 · Exercise 12.2 · Miscellaneous Exercise
Transcript1,873 words
A stone is dropped from a cliff. Two seconds later, how fast is it going? That sounds like an ordinary question, and it is not, because of what a speed actually is. Every speed you can compute is a distance divided by a time. So pick your moment, ask how far the stone went and how long it took, and divide. But at a single instant no time passes at all.
The distance is nothing and the elapsed time is nothing, and nothing divided by nothing is not a number. The honest thing to do is refuse. Everything that follows is an attempt to get an answer anyway, without ever once dividing by zero. Start from the one fact we are handed. A dropped body covers four point nine times the square of the elapsed seconds, in metres. That is a rule for distance, not for speed, and the difference is the whole of the difficulty.
Feed it thirteen moments and it gives thirteen distances. At one and a half seconds, four point nine times two point two five is eleven point zero two five metres. At one point nine five, four point nine times three point eight zero two five is eighteen point six three two two five. At two and a half, four point nine times six point two five is thirty point six two five.
Every one of the thirteen comes out exactly, with nothing rounded anywhere. And at the moment we care about, two seconds, the stone has fallen nineteen point six metres. Here is the trick, and it is completely honest. We cannot ask about an instant, but we can ask about a stretch of time ending at that instant. Take the stretch from zero to two seconds: nineteen point six metres in two seconds, so nine point eight metres per second.
That is a real average, correctly computed, and it is nowhere near what we want. So shorten the stretch. From one second to two: fourteen point seven. From one point five: seventeen point one five. From one point eight, one point nine, one point nine five: eighteen point six two, nineteen point one one, nineteen point three five five. Seven averages, and every one of them is a genuine distance divided by a genuine time.
Now push closer still, to a stretch lasting one hundredth of a second. From one point nine nine to two. And here something happens that is worth stopping for. You cannot look this one up, because our distance table has no entry at one point nine nine. You have to compute it: four point nine times three point nine six zero one is nineteen point four zero four four nine metres.
Then nineteen point six minus that, divided by one hundredth, gives nineteen point five five one. Of the fourteen averages this video uses, the number whose moment the distance table does not carry is two. That one, and its mirror on the other side. Every other entry is a lookup; these two are work. Now come at the same instant from the other side. Stretches that start at two seconds and end later.
From two to four seconds: seventy-eight point four minus nineteen point six, over two, which is twenty-nine point four. From two to three: twenty-four point five. Then twenty-two point zero five, twenty point five eight, twenty point zero nine, nineteen point eight four five. And the last one needs work again: four point nine times four point zero four zero one is nineteen point seven nine six four nine. That gives nineteen point six four nine.
Seven averages on this side too, and not one of them was obtained by any rule other than distance over time. Two families of perfectly ordinary arithmetic, aimed at the same unreachable moment. Put the two lists side by side, and look at what they are doing. Going down the left list, of the six steps from one entry to the next, the number that fails to rise is zero.
Going down the right list, the number that fails to fall is zero. One family is climbing, the other is dropping, and they are climbing and dropping toward each other. The last entry on the left is nineteen point five five one. The last on the right is nineteen point six four nine. The gap between them is exactly nought point nought nine eight, and whatever the answer is, it is caught in there.
That is what makes this an argument rather than a hopeful pattern: the trap has two jaws. Of all fourteen averages, the number that actually lies inside that final gap is two, and those two are the jaws themselves. Before going further, kill the most tempting misreading. It is very natural to think the average over the last little stretch just is the speed at the end of it. It is not, and the numbers say so plainly.
Of the fourteen averages, the number equal to the value they are closing on is zero. Of the seven from the left, the number sitting below it is seven. Of the seven from the right, the number sitting above it is seven. Every single one of them misses. They are not bad answers to our question; they are correct answers to a different one. The question they answer is about a stretch of time, and the one we asked is about a moment.
So how do we get the value at all, if no computation ever lands on it? By elimination, and nothing else. Keep building brackets from shorter and shorter stretches: a tenth of a second, a hundredth, a thousandth, and on down to a hundred millionth. That is eight brackets, and of the seven steps inward, the number that fails to be strictly narrower than the one before is zero. None of them has zero width, because none of them was allowed to.
The widest is forty-nine fiftieths wide; the narrowest is about a hundred-millionth. Now offer forty-one candidate values, a thousandth apart, and throw away any that some bracket excludes. The number still standing at the end is one. Nobody wrote that value down; it is simply the only thing the brackets left alive. There is a piece of algebra hiding under both lists, and it explains everything at once. The average from a moment t up to two seconds is four point nine times the difference of the squares, over the difference of the times.
Two squared minus t squared factorises into two minus t, times two plus t. The two minus t on the top cancels the one underneath, and what is left is four point nine times two plus t. Scored across eighty different moments either side, the number where that tidy expression disagrees with the honest quotient is zero. So the two lists were never two computations. But watch the one place they differ, because it is the whole topic in a single line.
At the instant itself the quotient still refuses, and the tidy expression cheerfully answers. Cancelling is only legal while there is something to cancel, and at that one moment there is not. Now draw it, because the picture makes the last step obvious. Put time across and distance up, and the falling law is a curve. Mark the point at two seconds, and mark a second point later on. The straight line joining them has a slope: how much it rises, divided by how far it runs.
The rise is distance covered and the run is time elapsed, so that slope is an average velocity. Not a similar idea, the same quotient with different words on it. Across all eighty moments, the number where the slope of the joining line disagrees with the average velocity is zero. And a joining line with no run has no slope, which is the refusal from the very first scene, drawn.
Now slide the far end down the curve toward the fixed point, and watch the line settle. As the second point slides in, the joining lines stop swinging and lean onto one particular line. That line touches the curve at our point instead of cutting across it. Its slope is the value the brackets squeezed out, and now you can see why the answer exists. A curve that is smooth has a definite direction at every point, and direction is exactly what a slope measures.
The joining lines are approximations to that direction; the touching line is the direction itself. Notice what changed and what did not. Not one number in either list moved. What moved is what we are willing to say those numbers were about. And now the hard question, the one that is easy to skate past. Do those two lists actually prove the answer is what we said? They do not, and here is a construction that settles it.
Build a second law of falling, deliberately rigged to agree with the first at every single moment either table mentions. Of the thirteen distances, the number the second law gets wrong is zero. Of the fourteen averages, the number it gets wrong is zero. Every number we have written down all video is reproduced exactly, and yet the two laws are genuinely different. Of the eight brackets, the number the two laws build identically is two — and they are exactly the two our tables reach.
They part company at a stretch of one thousandth of a second, which is finer than anything we tabulated. Follow that second law down and see where it goes. Take the four brackets narrower than anything our tables carry. Under the first law, the number of candidates surviving all four is one, and it is nineteen point six. Under the second law, the number surviving is also one, and it is eighteen point six.
The number of values the two closings have in common is zero. There is more. Under the first law the brackets nest, one strictly inside the next, with zero failures out of seven. Under the second, the number of steps where the bracket is not strictly inside the one before is two, so its averages do not even settle tidily. Both laws produce our tables. Only one of them is the one we meant.
So what did the two lists earn us, exactly? Count the candidates that survive everything the tables alone can rule out: ninety-nine of them. Ninety-nine values are still consistent with every number we tabulated, and the one we want is merely one of those ninety-nine. What closes the gap is an assumption, and it should be said out loud rather than smuggled. The assumption is that nothing abrupt happens to the stone in the last hundredth of a second on either side.
That is a physical belief about falling stones, not a fact about arithmetic, and it is doing real work. The second law is exactly what that assumption rules out. So here is the honest summary of where we stand. The averages are earned, the bracket is earned, and the single value is earned only once you say what you are assuming. That last step is the whole of what is coming next.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- Why writing a pair in order carries information a set cannotClass 11 · Ch 2, Relations and Functions
Comes up again in
- Approaching from the left and from the right, and when the two disagreeClass 11 · Ch 12, Limits and Derivatives
- The rate of change at a point, defined as a limit of average ratesClass 11 · Ch 12, Limits and Derivatives
Either side of this one
- Applying the right-triangle rule twice to get out of the planeClass 11 · Ch 11, Introduction to Three Dimensional Geometry