Curves given through a parameter, and the derivative recovered by the chain rule

0:00Here are three ways of being told about a curve. The first is the one you grew up with. The height is written out in terms of the across, and you differentiate it on sight. The second is the tangled one. Neither letter is on its own; they sit in one equation together, and you differentiate the whole line and then collect. There is a third, and it is the one that runs a wheel, a planet and a robot arm.

0:29Neither coordinate is given in terms of the other. Both are given in terms of a THIRD quantity, a parameter, which you can think of as time. The across is one rule in the parameter. The height is another rule in the same parameter. Feed the parameter a value and you get a point. Let the parameter run and the point traces the curve. The question of this video is the obvious one. If nobody ever wrote the height in terms of the across, what is the slope, and how would you get at it?

1:08Take a circle of radius four. The across is four times the cosine of an angle. The height is four times the sine of the same angle. There is no equation here giving the height in terms of the across. There are two separate rules, each in the angle. And yet a curve is plainly there. Turn the angle from nought and the point walks round the circle. So the curve exists. Its slope at each point exists. But the letter you would differentiate with respect to does not appear in either rule.

1:45That is the whole difficulty, and it is a difficulty about notation rather than about mathematics. The curve does not know it was given through a parameter. The parameter is scaffolding. The slope belongs to the curve. Which means the answer must not depend on which parameter you chose. Hold on to that; it comes back with teeth later. Nothing new is needed. The chain rule already covers this. Suppose, for a moment, that the height IS some function of the across, whatever it is.

2:23Then differentiating the height with respect to the parameter goes through the across in two stages. The rate of the height against the parameter equals the rate of the height against the across, times the rate of the across against the parameter. That is the chain rule, written for this situation and nothing more. Now divide both sides by the rate of the across against the parameter. The slope of the curve is the height's parameter derivative over the across's parameter derivative.

2:58Two derivatives you can do easily, divided one by the other. And division has a condition. You may divide only when the across's parameter derivative is not nought. That condition is not decoration. It is the entire content of the second half of this video. Thirty seconds on the shape of that argument, because it is worth seeing. The chain rule step differentiated the height through the across. To do that, we had to assume the height IS a differentiable function of the across.

3:36But that is exactly the thing we were trying to establish. Its derivative is what we were computing. So the derivation assumes what it delivers. That is not a scandal, and it does not make the formula false. It means the formula is a rule for FINDING the slope when there is a slope to find. It is not a proof that there is one. The proof that there is one is a separate matter, and it is settled by the same condition that lets you divide.

4:08Keep the two apart in your head. Does a slope exist here, and if it does, what is it? The formula answers only the second. Everything from here on is measured rather than asserted, and it is worth saying exactly how, because the obvious way would be worthless. A checker that computed the slope by the formula and then compared it with the formula would be agreeing with itself. So the checker behind this video never uses the formula to find the truth. It uses the curve.

4:45At a chosen parameter value it takes the across's value there. Then, for a nearby across, it HUNTS the parameter that produces it, by halving, on a window where the across has been checked to be one-to-one. Having found the parameter, it reads the height off. That gives the height as a function of the across, with no parameter left in it anywhere. And then it does the only honest thing: it takes the difference quotient of that function and shrinks the neighbourhood on both sides until the answer stops moving.

5:23That is the slope of the curve, obtained from the curve, with no formula in sight. Every candidate answer in this video is scored against that. Six candidates, then, scored by identical machinery. The height's parameter derivative over the across's. The two divided the other way round. The two multiplied. The two subtracted. The height's parameter derivative left on its own. And the height over the across. Ten pairs of rules, twenty-nine points on them, every one of which the curve gives a slope at.

6:03Out of twenty-nine, the scores are twenty-nine, one, four, nought, four and two. One candidate is right everywhere. That is the formula. And the test is sharp, not generous. Take the surviving answer and move it by a relative thousandth, up or down, and it survives none of the twenty-nine. The same formula stated as a multiplication is checked too. The slope measured on the curve, times the across's parameter derivative, gives back the height's parameter derivative at all twenty-nine points.

6:41That is the chain rule, undivided, coming back out of a measurement that never used it. Now look at those five losing scores again, because they are not all nought. At six of the twenty-nine points, at least one wrong rule is right as well. Only twenty-three of them rule out all five at once. And four of those six come from a single pair, chosen deliberately: the laziest parametrisation there is, where the parameter and the across are the same thing.

7:17Set the across equal to the parameter and its parameter derivative is one. Divide by one, multiply by one, or leave it out entirely, and you have done the same thing three times. So on that pair, three of the six candidates agree, and the example cannot tell them apart. This is the practical warning. If you test your understanding on the easiest possible example, the example will agree with almost any rule you bring it.

7:49Choose a pair where both parameter derivatives are interesting. Otherwise you are not checking; you are being flattered. The circle first. The across is four times the cosine of the angle. Its parameter derivative is minus four times the sine. The height is four times the sine. Its parameter derivative is four times the cosine. Divide. The fours cancel, and the slope is minus the cosine over the sine. Minus the cotangent of the angle.

8:26Check it against something you know. At the top of the circle the angle is a quarter turn, the cotangent is nought, and the tangent line is flat. Correct. At the far right the angle is nought, the cotangent is infinite, and the tangent line is vertical. Also correct, and notice that the formula did not give you a number there. It refused. That refusal is the condition speaking, and we will come back to it.

8:59A parabola opening sideways. The across is a constant times the square of the parameter. The height is twice that constant times the parameter. Their parameter derivatives are twice the constant times the parameter, and twice the constant. Divide, and the constant goes. So does the two. The slope is simply one over the parameter. Look at what has just happened. The constant sets the width of the parabola, and it has vanished from the answer.

9:32That was checked: a parabola and another one seven times as wide have exactly the same slope at the same parameter value, and each matches what its own curve measures. Widen a parabola and you slide along it, but at the point carrying the same parameter the tangent line has not tilted at all. A constant that cancels is a constant the answer never has to mention. Stop here, because a great deal of wasted effort is about to be avoided.

10:08The answer we just got is one over the parameter. There is no across in it and no height in it. A lot of students see that and think they are half done. They then spend twenty minutes trying to eliminate the parameter to get the answer back into the two original letters. You do not have to. An answer in the parameter is a finished answer. Here is why that is not a cop-out. You were given the point through the parameter. To use the slope at a point you must say which point, and you say which point by giving the parameter.

10:45So the parameter is already the address of the point. An answer written in it is an answer you can actually evaluate. Eliminating the parameter is usually harder, sometimes impossible, and never required. Every worked example from here on leaves its answer exactly where it lands. The cycloid. This is the path traced by a point on the rim of a rolling wheel, and it is a lovely case because the answer needs tidying.

11:19The across is the parameter plus its sine. The height is one minus its cosine. Parameter derivatives: one plus the cosine, and the sine. So the slope is the sine over one plus the cosine. That is already a correct answer, but it can be made much prettier. Write the sine on top as twice the sine of half the angle times the cosine of half the angle. Write one plus the cosine below as twice the square of the cosine of half the angle.

11:53The twos cancel. One cosine of the half angle cancels. What is left is the tangent of half the parameter. But look hard at that cancellation. We divided by the cosine of the half angle. That is only legal where it is not nought. Where the cosine of the half angle IS nought, the wheel's point is at the bottom, at a cusp, and there is genuinely no tangent line there.

12:21The tidy answer and the ugly one agree everywhere the tidy one is allowed to exist. That is the honest way to state a simplification. So far the parametrisation has been handed to you. Sometimes you have to invent one. Take the four-cornered star curve: the across to the two-thirds power plus the height to the two-thirds power equals a constant to the two-thirds power. Try the across as the constant times the cube of a cosine, and the height as the constant times the cube of a sine.

13:00Now check that the guess actually lies on the curve. Raise a cube to the two-thirds power and you get a square. So the left side becomes the constant to the two-thirds, times the square of the cosine plus the square of the sine. That bracket is one. The equation holds. The guess is a genuine parametrisation. Never skip that check. A pair that does not satisfy the equation traces some other curve entirely, and everything after it is about the wrong shape.

13:36Now differentiate. Both parameter derivatives carry three times the constant, times a sine, times a cosine. Divide, that common factor goes, and the slope is minus the tangent of the parameter. And nothing about that answer says which parametrisation you chose. Any pair satisfying the equation would give the same slope at the same point of the curve. Back to the condition, and this is where the measurement earns its keep.

14:11The proviso is usually taught as small print: divide only when the across's parameter derivative is not nought. It is not small print. It is the difference between a curve that has a slope and one that does not. Here is why. If the across's parameter derivative is nought, the across has stopped moving. It has turned round, or paused. Then near that point the curve is not the graph of anything. Two different heights sit above the same across.

14:44So there is no function to differentiate, and no slope, and the refusal is the curve's own, not the formula's fussiness. Five such points were tested: the parabola at its nose, a corner of the star, the end the straight segment is traced to, and two more coming next. At every one of the five, two separate questions were asked. Does the curve, hunted directly, have a slope there? And does the formula refuse to divide?

15:16At all five, both answers came back the same. No slope, and no division. And at every one of the twenty-nine points that WERE scored, the across's parameter derivative stays clear of nought. The condition is doing real work in both directions. The case where the condition actually bites in an exercise looks like this. Both coordinates are built from the parameter plus its reciprocal. One is a constant raised to that bracket; the other is that same bracket squared.

15:53Differentiate each. Both parameter derivatives come out carrying the same factor: one minus the reciprocal of the square of the parameter. It is enormously tempting to cancel it and move on. You may cancel it, but only where it is not nought. It is nought when the square of the parameter is one. That is two values, plus one and minus one, and at both of them BOTH parameter derivatives vanish together.

16:24The point stops dead. The curve has a corner there, and no slope. So the answer is correct everywhere except at those two values, and saying so is part of the answer, not an afterthought. This is the one place in the whole topic where the proviso is exercised rather than quoted, and it is worth doing slowly. A short one, and it is a good test of whether you are reading or reciting.

16:55Suppose both coordinates are given as constants times the cosine of the same angle. Three times the cosine across, five times the cosine up. The two constants differ. The trigonometric function does not. Differentiate: minus three sine, and minus five sine. Divide, the sines cancel, and the slope is five thirds. A pure number. No parameter in the answer at all. That is because this pair does not trace a curve. The height is always five thirds of the across. It traces a straight segment through the origin.

17:36And here is the pleasing part. This is the one place in the whole population where the crude slip of reading the slope off as the height over the across is actually correct. It is correct because a straight line through the origin is the one curve for which it is true. A right answer for the wrong reason is still a wrong method. It scored two out of twenty-nine.

18:04Now the promise made at the start gets paid. The slope belongs to the curve, not to the parameter. Different parameters, same curve, same slope. Take the circle of radius four again, and reach it two completely different ways. The first is the angle. The second is a parameter that is not an angle at all: the across is four times one minus its square over one plus its square, and the height is eight times it over one plus its square.

18:38That second pair has no trigonometry in it whatsoever, and it traces the same circle. Meet them at the point three fifths of the way along. The angle's two parameter derivatives there are minus sixteen fifths and twelve fifths. The other parameter's two are minus a hundred and twenty-eight twenty-fifths and ninety-six twenty-fifths. Four completely different numbers. Both ratios come to minus three quarters. And minus three quarters is what the circle itself measures there.

19:13The scaffolding cancelled. That is what it means for the answer to be about the curve. Sometimes the tidy answer is not in the parameter at all, but in the two coordinates. Take a pair built from the two angles that add to a quarter turn: the across is a constant raised to half the angle whose sine is the parameter, and the height is that same constant raised to half the angle whose cosine it is.

19:44Take logarithms of each and you see the structure at once. Each coordinate's logarithm is a constant multiple of an inverse trigonometric function. And the two inverse derivatives are exact negatives of each other, because the two angles add to a constant. So the across's parameter derivative is the across times something, and the height's is MINUS the height times the same something. Divide and the something cancels. The slope is minus the height over the across.

20:21That was checked at two parameter values, and at both it agrees with what the curve measures. An answer in the coordinates is just as finished as an answer in the parameter. What matters is that you can evaluate it at the point you care about. There is a question type that looks like a separate technique and is not. Differentiate the square of the sine with respect to the exponential of the cosine.

20:52The words are strange, but read them literally. You are being asked for the slope of a curve whose across is the exponential of the cosine and whose height is the square of the sine. The input is playing the parameter's part. That is the whole trick. So use the same formula. Differentiate the top with respect to the input: twice the sine times the cosine. Differentiate the bottom: minus the sine times the exponential of the cosine.

21:25Divide. The sine cancels, and the answer is minus twice the cosine, over the exponential of the cosine. Checked at three inputs, against the curve, and right at all three. Whenever you are asked to differentiate one thing with respect to another thing, you have been handed a parametric pair with the parameter renamed. One last move, and it is the one that goes wrong most often. You have a first derivative. You want the second.

21:59The reflex is to do again what worked before: divide the height's SECOND parameter derivative by the across's SECOND parameter derivative. That is wrong, and it is not slightly wrong. The second derivative means the derivative of the slope with respect to the ACROSS. The slope is currently a rule in the parameter. So you must go through the parameter again. Differentiate the slope with respect to the parameter, and then divide, once more, by the across's parameter derivative.

22:35Two divisions, not one. The second division is the step people drop. Both were scored against the bend measured on the curve, at five points on three pairs. The correct route matches at all five. The reflex matches at none. And the correct route moved a relative thousandth matches at none either, so the agreement is real. Take the sideways parabola. Its bend at a parameter of two is minus a forty-eighth, and at a half it is minus four thirds.

23:10The reflex answers nought at both, because the height's second parameter derivative is nought. Nought is a very confident wrong answer. Three things. The formula is the chain rule divided through. The slope is the height's parameter derivative over the across's, and no new rule was invented to get it. The condition is the mathematics. Where the across stops moving, the curve is not the graph of anything, and there is no slope to find. Look for those parameter values before you cancel anything.

23:50And an answer left in the parameter is a finished answer. So is one written in the coordinates. Do not spend your time eliminating scaffolding that was never in the way. One habit to take with you: whenever you cancel a common factor out of a ratio of two parameter derivatives, write down where that factor is nought. That single line is the difference between an answer and a correct answer.

24:20All of it was measured. Forty-two claims, none of them wrong, fourteen liveness controls, and four deliberately planted faults, all four caught. Ninety deliberate defects were then planted in the checker one at a time and every one of them was caught, alongside twenty-two rewrites that change nothing, none of which was.