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

What to assume they know

• The derivative as a rate, and the two notations for it, from the previous topic
• The chain rule and the product rule from Chapter 5
• Implicit differentiation of a relation with respect to a third variable
• Volume and surface area of a cube and a sphere; area and circumference of a circle; volume of a cone; area of a rectangle and of a triangle
• Similar triangles, and the ratio of corresponding sides
• Pythagoras' theorem
• Signed quantities: what a negative rate records

What they should be able to do

• State the chain rule in the form §6.2 prints it, including the condition attached to it
• Recognise the standard shape of these problems — a geometric relation, a clock, one known rate, one wanted rate
• Differentiate a geometric relation with respect to time and solve for the wanted rate
• Eliminate the intermediate rate rather than reporting it
• Carry the sign of a falling quantity correctly through a calculation with two changing lengths
• Apply the product rule where the quantity being tracked is a product of two changing lengths
• Reduce a two-variable geometric setting to one variable using a similarity ratio or a stated proportion before differentiating
• Recover a geometric relation from a described situation when no worked example supplies one
• Attach correct units to a time rate and state which quantity each unit belongs to

Where it usually goes wrong

• "The intermediate rate is part of the answer." It is scaffolding. Example 2 computes the edge rate, numbers it, uses it once and never mentions it again. A student who reports it has answered a question nobody asked.
• "I can differentiate the formula for the wanted quantity and stop." That gives a rate against a length, not against time. The chain rule exists precisely to convert one into the other, and forgetting it is the single most common failure across Exercise 6.1.
• "A decreasing quantity just means I subtract at the end." It means the rate is negative from the first line. Example 4 puts a minus sign into the data and carries it to two answers of different signs; a student who inserts the minus at the end gets the perimeter right and the area wrong.
• "Every one of these has a formula I can look up." Exercise 6.1 Q10, Q11 and Q14 do not. Each needs a relation built from the description — Pythagoras for the ladder, the printed curve for the particle, the stated proportion for the sand cone.
• "The radius of the disc is three, so I use three." Miscellaneous Example 35 states a radius of three and then evaluates at three point two. The stated value is not used. Say so, or a careful student will assume they have misread.
• "Similar triangles are a geometry topic, not a calculus one." They are how two of these problems are reduced to one variable at all. The chapter writes the similarity with a tilde and moves on in one line.
• "Units can be attached at the end." Cubic metres per hour and metres per hour are different quantities with different meanings, and the chain rule is what converts between them. Carry units through every line.
• "If the shape keeps its proportions, nothing needs eliminating." Keeping proportions is exactly what lets one variable be eliminated — the cone problem and the sand-pile problem both turn on it. Without that step there are two unknowns and one equation.

Questions to check understanding

• State the chain rule linking two time rates, with its condition
• Given one time rate and a geometric relation, find the other time rate at a stated instant
• Track two lengths changing in opposite directions and report the rates of their sum and their product
• Build the geometric relation yourself from a described situation, then differentiate it — the form of Exercise 6.1 Q10 and Q14
• Find the points on a printed curve at which one coordinate changes a stated multiple as fast as the other
• Reduce a cone or a triangle to a single variable using a fixed ratio before differentiating
• Choose the correct depth rate from four options — the form of Miscellaneous Exercise Q16

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its own exercises.

• The chain rule display (§6.2, Part I p. 147). The chapter supposes both quantities are functions of a third variable, then writes one rate as the quotient of the other two, with a condition on the denominator printed to its right. Section 2 should read the condition aloud: it is the only hypothesis in the whole of §6.2 and students skip it every time.
• Example 2 (Part I p. 148). A cube whose volume grows at nine cubic centimetres per second; the surface-area rate is wanted at edge ten. The chapter differentiates volume with respect to time by the chain rule, solves for the edge rate, substitutes into the differentiated surface area, and numbers the intermediate result so it can be quoted. Verified: the edge rate is three over the edge squared; the surface-area rate is thirty-six over the edge; at edge ten that is three point six square centimetres per second. This is the longest chain in §6.2 and the right one for sections 4 and 5.
• Example 3 (Part I p. 149). Circular waves spreading at four centimetres per second; the area rate is wanted at radius ten. Verified: the area rate is two pi times radius times the radius rate, which is eighty pi square centimetres per second. One link instead of two, and the same shape — use it immediately after Example 2 so the shape is visible rather than asserted.
• Example 4 (Part I p. 149). A rectangle whose length shrinks at three centimetres per minute while its width grows at two. Perimeter and area rates are wanted at length ten and width six. Verified: the perimeter rate is twice the sum of the two rates, which is minus two centimetres per minute; the area rate needs the product rule and comes to minus eighteen plus twenty, which is two square centimetres per minute. Two things earn their thirty seconds here — the minus sign entered at the top and survived to the answer, and the two answers have opposite signs from the same pair of inputs.
• Exercise 6.1 Q2 to Q6 (Part I pp. 150–151). Five drills on the shape: another cube, a circle whose radius grows uniformly, a growing cube edge, a second stone in a second lake, and a circle whose circumference rate is wanted. Verified: Q2 gives eight over three square centimetres per second at edge twelve; Q3 gives sixty pi square centimetres per second; Q4 gives nine hundred cubic centimetres per second; Q5 gives eighty pi square centimetres per second; Q6 gives one point four pi centimetres per second.
• Exercise 6.1 Q7 (Part I p. 151). Example 4 with four numbers changed. Verified: the perimeter rate is minus two centimetres per minute and the area rate is two square centimetres per minute — the same two answers as the worked example, from different inputs. Worth showing beside Example 4; the coincidence is a good hook and nothing more.
• Exercise 6.1 Q8 and Q12 (Part I p. 151). Two spheres tracked in time. Q8: gas pumped in at nine hundred cubic centimetres per second, radius rate wanted at radius fifteen. Verified: the volume rate is four pi times radius squared times the radius rate, so the radius rate is one over pi centimetres per second. Q12: radius growing at half a centimetre per second, volume rate wanted at radius one. Verified: two pi cubic centimetres per second. Put Q8 and Q12 beside Exercise 6.1 Q9 and Q13 from the previous topic: same sphere, and the second quantity is time in one pair and a length in the other.
• Exercise 6.1 Q10 (Part I p. 151). A five metre ladder against a wall, foot pulled away at two centimetres per second; the wall height is falling and its rate is wanted with the foot four metres out. Verified: the two distances satisfy Pythagoras' relation with a constant hypotenuse, so differentiating gives one rate as minus the other times the ratio of the distances; the height is three metres, and the height falls at eight thirds of a centimetre per second. This is the first item where the student must supply the geometric relation. No worked example in §6.2 does that.
• Exercise 6.1 Q11 (Part I p. 151). A particle on a printed cubic relation; find the points where one coordinate changes eight times as fast as the other. Verified: differentiating with respect to time gives six times the vertical rate equal to three times the horizontal rate times the square of the horizontal coordinate; setting the vertical rate to eight times the horizontal one leaves the square equal to sixteen, so the horizontal coordinate is four or minus four, and the two points are four and eleven, and minus four and minus thirty-one thirds. The negative root is the item's whole point and it is easy to drop.
• Exercise 6.1 Q14 (Part I p. 151). Sand forming a cone whose height is fixed at one sixth of the base radius; the height rate is wanted at height four. Verified: radius is six times the height, so the volume is twelve pi times the height cubed; the volume rate is thirty-six pi times the height squared times the height rate, giving one over forty-eight pi centimetres per second. Note: the printed relation makes the cone six times as wide as it is tall, which is a very flat pile; the arithmetic is unaffected and a student who pictures a tall cone will still get the right answer.
• Miscellaneous Example 31 and Fig 6.20 (Part I p. 178). An inverted cone filling at five cubic metres per hour, with the apex angle fixed. The chapter turns the angle into a fixed ratio of radius to depth, eliminates the radius, and differentiates. Verified: radius is half the depth, so the volume is pi times depth cubed over twelve; the volume rate is a quarter pi times depth squared times the depth rate; at depth four that gives five over four pi, which the chapter converts using the fractional value of pi to thirty-five over eighty-eight metres per hour. The figure is an inverted cone with the depth drawn as a dashed axis, the radius across the top, and the angle marked at the apex.
• Miscellaneous Example 32 and Fig 6.21 (Part I pp. 178–179). A man walking from a lamp post; the rate at which his shadow lengthens is wanted. Verified from the printed similarity: the shadow is half the man's distance from the post, so the shadow lengthens at half his speed, two and a half kilometres per hour. The figure is a right triangle with the post upright at one end, the man drawn as a short upright segment, and the two ground distances marked with dashed arrows. This is the topic's cleanest instance of section 10: the answer does not depend on where he is.
• Miscellaneous Example 35 (Part I p. 181). A disc heated so its radius grows at five hundredths of a centimetre per second; the area rate is wanted at radius three point two. Verified: two pi times three point two times five hundredths, which is thirty-two hundredths of pi. Two things need saying and both are in Notes: the disc's stated starting radius is never used, and the final line is written as an increment while its answer is quoted as a rate.
• Miscellaneous Exercise Q2 (Part I p. 183). An isosceles triangle on a fixed base whose two equal sides shrink at three centimetres per second; the area rate is wanted at the moment the equal sides match the base. Verified: with base b and equal side a, the height is the square root of four a squared less b squared, all over two, so the area is b times that root over four; the derivative of area with respect to the equal side is a b over the root, and at a equal to b the root is b times the square root of three, so the area falls at the square root of three times b square centimetres per second. Note that the answer carries the base as a symbol, which no worked example in the chapter does.
• Miscellaneous Exercise Q16 (Part I p. 185). A cylinder of radius ten filling at three hundred fourteen cubic metres per hour; the depth rate is offered as four options. Verified: the volume is one hundred pi times the depth, so the depth rate is three hundred fourteen over one hundred pi, which is one metre per hour on the chapter's own fractional value of pi — option A. This item is a multiple-choice question printed with no instruction line above it; see Notes.
• The Summary's Chain Rule bullet (Part I p. 185). Restates the display on Part I p. 147 with its condition intact. Use it as the end card.

Figures to have open

• A single changing figure for section 1 — a circle with a growing radius is enough — carrying two labels and a clock. Not in the book; §6.2 prints no figure.
• A three-box chain diagram for section 5, with the middle box dimmed after use. An added device, and the load-bearing picture of this topic.
• A signed-rate pair for section 7: one arrow shrinking, one growing, both labelled with their rates. Not in the book.
• A redraw of Fig 6.20 (Part I p. 178): an inverted cone, apex at the bottom, the depth as a dashed vertical axis, the surface radius across the top, and the half-angle marked at the apex. The chapter's own figure.
• A redraw of Fig 6.21 (Part I p. 179): a right triangle with the lamp post upright at the left, the man as a short upright segment part-way along the base, and the two ground distances marked beneath with dashed arrows. The chapter's own figure.
• A ladder-against-wall sketch for section 12 with the two distances and the fixed length marked. Exercise 6.1 Q10 prints no figure and needs one.

Where this sits in the book

• NCERT Class 12 Mathematics, Part I, Chapter 6 "Application of Derivatives", §6.2, the Chain Rule display and its condition, Part I p. 147
• Examples 2, 3 and 4, Part I pp. 148–149
• Exercise 6.1, questions 2 to 8, 10 to 12 and 14, Part I pp. 150–151
• Miscellaneous Examples 31, 32 and 35, with Fig 6.20 and Fig 6.21, Part I pp. 178–179 and 181
• Miscellaneous Exercise on Chapter 6, questions 2 and 16, Part I pp. 183 and 185
• Summary, the Chain Rule bullet, Part I p. 185