Two quantities both changing with time, and the chain rule that links their rates

0:00A circle is spreading across the surface of a lake. Two things about it are changing. Its radius is getting longer, and the water it covers is getting larger. Both of them are changing because time is passing. One clock, two quantities. And here is the situation that a whole family of problems is built on. You are told how fast one of the two is changing. You are asked how fast the other one is.

0:30That is it. That is the entire shape. A geometric relation ties the two quantities together, a clock ties each of them to time, one of the two time rates is handed to you, and the other one is what you have to find. The new piece of machinery is a single rule, and it does exactly one job. It lets you get rid of the quantity in the middle.

0:58Suppose two quantities both depend on a third one. Call the third one the shared quantity, because it is what the other two have in common. Then the rate of the first against the second is the rate of the first against the shared quantity, divided by the rate of the second against it. Written the other way round, which is how you will actually use it, it says something simpler.

1:25The time rate of a quantity is its rate against the shared one, multiplied by the shared one's own time rate. Two rates multiplied together. There is a condition attached, and it is the only hypothesis in the whole business, and it is the one everybody skips. The rate in the denominator must not be zero. If the shared quantity is not moving at that instant, the division you are about to do is not allowed.

1:56Say it out loud once, because you will want it later. Rate against the shared quantity, times the shared quantity's own rate against time. Before any arithmetic, here is a claim, and I want to be honest that it is my framing and not something handed down. Every problem of this kind has one four-step shape. First, write down the geometric relation between the two quantities. Second, differentiate that relation with respect to time.

2:30Third, put in the rate you were given. Fourth, solve for the rate you want. I have taken twenty-two situations of this kind and worked every one of them, and the shape held at all twenty-two. So it is a good framing. But it is a framing, and by the end of this video you will see two places where step one is very much harder than it looks. Start with the longest chain of the lot.

2:59A cube is growing, and its volume is going up at nine cubic units every second. How fast is its surface area growing at the moment its edge is ten? Notice that the volume and the surface area are not directly tied to each other. They are both tied to the edge. The edge is the shared quantity, and it is the thing you have to go through. Volume is edge cubed, so the rate of volume against edge is three times edge squared.

3:31The rule says the volume's time rate is that, times the edge's own time rate. Nine equals three times a hundred, times the edge rate. So the edge is growing at three hundredths of a unit a second. Now go the other way. Surface area is six times edge squared, so its rate against the edge is twelve times the edge. Twelve times ten is a hundred and twenty, times three hundredths, and the surface area is growing at three point six square units a second.

4:08Look at what just happened to that edge rate. You computed it, you used it once, and then it vanished. Three hundredths of a unit a second is not the answer, and it is not the nine you started with either. It is scaffolding. It exists so that the two ends of the chain can talk to each other, and then it comes down. This is worth measuring rather than asserting.

4:34Across the twenty-two situations, the shared quantity's own rate is the answer at four of them, and it is scaffolding at the other eighteen. The four are exactly the problems where the thing you were asked for was the shared quantity itself. Everywhere else, a student who writes down the middle rate and stops has answered a question nobody asked. Draw the chain as three boxes with two arrows. The middle box is real, it does work, and then it goes grey.

5:09Now the lake again, with one link instead of two. The waves spread outward at four units a second, and you want the rate at which the covered area is growing when the radius is ten. Here the radius is both the shared quantity and the one whose rate you were given, so the chain has only one link in it. Area is the circle constant times radius squared. Its rate against the radius is twice the constant times the radius, which is twenty times the constant.

5:43Multiply by the radius rate of four. Eighty times the circle constant, square units a second. Same shape as the cube, one step shorter. And notice the units changed on the way through. Units of length per second went in. Square units per second came out. That conversion is not decoration. It is the rule doing its job. Here is the part of this topic that is actually worth arguing about.

6:15There are four natural things a student does instead of the rule, and rather than scold them, I scored them. Every one of the twenty-two situations was written down as an actual motion, with the quantity you are given moving in time, and the quantity you want computed from the geometry alone at every instant. Its time rate was then measured straight off that motion, with no rule anywhere in the arithmetic.

6:45Then the rule and its four rivals were all put to that measurement by the same machinery. The rule is right at twenty-two of the twenty-two. Differentiating the formula for what you want and stopping there is right at two. Reporting the shared quantity's own rate is right at four. Dividing the two rates instead of multiplying them is right at two. And taking the whole thing as a size, with the signs thrown away, is right at eighteen.

7:17That last one is the dangerous number. A reader who ignores signs entirely will get eighteen of these right, feel confident, and then be wrong about exactly the ones where the sign was the whole point. Look at the two rivals that scored two each. Differentiating and stopping, and dividing instead of multiplying. They score two, and it is the same two problems. Both of them are the particle on a curve, and there is a reason.

7:51At those two, the shared quantity moves at exactly one unit a second. And when the middle rate is one, multiplying by it, dividing by it, and ignoring it altogether all give you the same number. So a student who has only ever seen that example cannot tell the rule from either of its rivals. It is not that they have understood it. It is that the example was not able to ask.

8:20This is worth knowing when you choose what to practise on. An example where the middle rate is one is an example that tests nothing. Now let both quantities move. A rectangle is ten units long and six wide. Its length is shrinking at three units a minute while its width grows at two. The word shrinking is doing real work there, and it does that work at the very first line, not at the last one.

8:51The length rate is minus three. Not three with a minus stuck on at the end. Minus three from the start. The distance round the rectangle is twice the sum of the two sides, so its rate is twice the sum of the two rates. Twice minus three plus two, which is minus two units a minute. The rectangle's perimeter is falling. The area is where the sign earns its keep.

9:20Area is length times width, and both of them are moving, so you need the rule for the rate of a product. The area's rate is the length times the width's rate, plus the width times the length's rate. Ten times two is twenty, plus six times minus three, which is minus eighteen. Twenty minus eighteen is two. So the perimeter is falling at two units a minute while the area is rising at two square units a minute.

9:50Opposite signs, out of one single pair of inputs. A reader who worked in sizes and attached the minus at the end would have got the perimeter right and the area wrong, which is the worst possible outcome, because getting one right feels like getting it. I scored the readings here too, on two different rectangles. Only the reading that adds both terms matches the measured area rate. Taking either term on its own, multiplying the two side rates together, and adding the two terms with their signs thrown away are each wrong at both rectangles.

10:32Now the harder half of this topic, which is step one. Sand is falling onto a pile in the shape of a cone. A cone's volume is built from two lengths, its radius and its depth. So if all you know is the volume, you do not know the depth. Here are two cones with exactly the same volume, one four units deep and one a single unit deep. One equation, two unknowns, and no amount of differentiating will help you.

11:04What rescues it is a fact about the shape that does not change. This pile always keeps its proportions, so the radius is always six times the depth. Put that in, and of those two cones only the first survives. The volume becomes twelve times the circle constant, times the depth cubed. One length, one equation, and now you can differentiate. Keeping its proportions is not a detail of the setting.

11:34It is the thing that makes the problem solvable at all. Finish the sand pile. The volume is twelve times the constant times the depth cubed, so its rate against the depth is thirty-six times the constant times the depth squared. At depth four, that is five hundred and seventy-six times the constant. Sand arrives at twelve cubic units a second, so the depth is rising at one over forty-eight times the constant.

12:08A note on the picture: that proportion makes the pile six times as wide as it is tall, which is an extremely flat heap. The arithmetic does not care, and neither should you. Now the same idea upside down. Liquid is running into a cone standing on its point, and the angle at the apex is fixed. A fixed apex angle is a fixed proportion by another name: it makes the surface radius half the depth.

12:39The volume becomes the constant times depth cubed over twelve, and at depth four the level is rising at five over four times the constant. Two problems, one move. Find the proportion, cut one length out, then differentiate. Triangles of the same shape are the other way to cut a length out, and they are usually treated as a geometry topic rather than as part of this one. That is a mistake, because in these problems they are the whole of step one.

13:15A lamp post stands six units high. A person two units tall walks away from it at five units an hour. How fast is the tip of their shadow running away from them? The small triangle made by the person and their shadow has the same shape as the large one made by the post and the whole distance to the shadow's tip. Corresponding sides are in one ratio, and working that ratio out gives something clean.

13:46The shadow is always half the walker's distance from the post. Half. So the shadow lengthens at half their speed, two and a half units an hour. And here is the pleasing part, which I measured at three different positions along the path. Two units out, seven units out, thirteen units out. The answer is two and a half every single time. Where they are does not enter the answer at all.

14:16Now the item that catches almost everybody. A ladder five metres long leans against a wall, and its foot is being pulled away from the wall at two centimetres a second. How fast is the top sliding down when the foot is four metres out? Stop and notice what is missing. There is no formula here. Nobody handed you a relation between the height and the distance out. You have to build one, and the thing that builds it is the relation among the three sides of a right angle.

14:51The height squared plus the distance squared equals the ladder squared, and the ladder does not change. That constant is what makes the whole thing work. Differentiate with respect to time and one rate comes out as minus the other, times the ratio of the two distances. At four hundred centimetres out the height is three hundred, and the top is falling at eight thirds of a centimetre a second.

15:20Minus eight thirds, because it is going down. I checked how many of the twenty-two have this character. At sixteen of them, the quantity you want is a plain power of the shared length, which is to say a formula you could look up. At six of them it is not, and those six are where the real work is. The particle on a curve is the third of those.

15:47A particle moves along a curve where six times its height is its distance along, cubed, plus two. Find the points at which the height changes eight times as fast as the distance along. Differentiate the relation with respect to time. Six times the height rate equals three times the square of the distance, times the distance rate. Set the height rate to eight times the distance rate, and everything cancels down to the square of the distance equalling sixteen.

16:21And here is the whole point of the item. The square of something is sixteen. That is two answers, not one. Four, and minus four. Four gives a height of eleven. Minus four gives a height of minus thirty-one thirds. Two points, and the second one is the one that gets dropped, every time, by people who take a square root and write down one number. One more, and this one asks something the worked examples never do.

16:54A triangle has two equal sides and a fixed base. The two equal sides are shrinking at three units a second. How fast is its area changing at the instant the equal sides are the same length as the base? You are not given a number for the base. Call it b and carry it. The height is the square root of four times the equal side squared, less the base squared, all over two.

17:23So the area is the base times that root, over four. Differentiate with respect to the equal side and you get the base times the side, over the root. At the moment the two are equal, the root is the base times the square root of three, and the whole thing collapses to the base over the square root of three. Multiply by the side rate of minus three. The area is falling at the square root of three, times the base, square units a second.

17:55I ran it on bases of six, ten and fourteen. Three different answers, and each one divided by its own base gives the same number: minus the square root of three. The answer carries a symbol, and that is not a failure to finish. It is the answer. Now something I did not expect to find, and it is worth thirty seconds of your attention. Two of the answers you will see quoted for this material are not quite right.

18:28The cone standing on its point: the level rises at five over four times the circle constant, and that gets quoted as thirty-five over eighty-eight. Those are not the same number. They agree only if you take the circle constant as twenty-two sevenths. The real gap is in the fourth decimal place. A cylinder of radius ten filling at three hundred and fourteen cubic units an hour: the level rises at three hundred and fourteen over a hundred times the constant, and that gets quoted as exactly one unit an hour.

19:07It is not one. It is a little under one, and it becomes exactly one only if you take the constant as three point one four. And no single schoolroom value rescues both, because they need different ones. This matters for a reason beyond pedantry. When I scored the rule and its rivals, I scored them against rates measured off the motion, not against the answers as they are usually quoted.

19:37Score against the quoted numbers instead and the rule itself comes out wrong at two of the twenty-two, and those two are exactly these two. If you check your work against an answer, you are also checking the answer. A smaller thing, but it stops careful students dead, so it is worth naming. A metal disc is heated so that its radius grows at five hundredths of a unit a second.

20:06The problem introduces the disc with a radius of three, and then asks for the area rate at radius three point two. The three is never used. Work it at three point two and the area is growing at thirty-two hundredths of the circle constant. Work it at three and you get thirty hundredths, which is a different number and is not the answer. If you found yourself re-reading that question, you had not misread it.

20:36The earlier value simply is not needed. Two honest limits before the summary. First, everything here has been a rate with respect to time, because that is what one clock buys you. The same rule works between any two quantities that share a third one, and time has no special status in it. Second, every relation used here was exact and every motion was smooth. The condition attached to the rule, the one about the denominator not being zero, has not been stressed anywhere in these examples, because in all of them the shared quantity was genuinely moving.

21:20There are settings where it stops moving for an instant, and there the rule as stated has nothing to say. That is not a gap in your understanding. It is a gap in the rule, and it is written into it. Three things to take away. One: the shape. A relation, a clock, one rate given and one wanted, and a shared quantity in the middle that you compute and then throw away.

21:49It held at all twenty-two situations I tested it on, and the middle rate was scaffolding at eighteen of them. Two: the sign is not a decoration. It goes in at the first line and it comes out at the last one, and a reader who works in sizes gets eighteen of twenty-two right and is wrong about precisely the four that mattered. Three: step one is the hard step.

22:16At sixteen of the twenty-two the relation was a formula you could look up. At six it was not, and every one of those six needed you to build the relation yourself, out of a right angle, a curve, or a proportion that does not change. The differentiating is the easy part. Getting to something worth differentiating is where the mathematics is.