Lesson 8: Related Rates of Change & Chain Rule Applications
Two quantities changing at once, tied together by a piece of geometry. Differentiate the relationship with respect to time, and substitute only afterwards — never the other way round.
🎯 Ngā Whāinga Akoranga | Learning Intentions
Related rates equations connecting multiple changing variables with respect to time (t) using the chain rule (e.g., dV/dt = (dV/dr) * (dr/dt)).
Formulate geometric relationships (spheres, cones, triangles), differentiate implicitly with respect to time t, and solve instantaneous rate of change problems.
🎥 Media Anchor & Pedagogical Scaffold
Related Rates — Khan Academy
Video (7 min 43 sec, Khan Academy): A comprehensive introduction to related rates problems, showing how to connect rates of change of different variables through geometric and implicit differentiation. Covers the classic water-tank and ladder examples.
🧠 1. Before Viewing (Activate & Predict)
If water is poured into a conical funnel at a constant rate, why does the water height rise more slowly as the cone fills up?
👁️ 2. During Viewing (Watch With a Job)
Watch the full 7 min 43 sec clip. In your calculus logbook, record:
- The related rates strategy: Write the key steps: (1) Draw and label the geometry; (2) identify given rates and target rate; (3) write a relationship equation (e.g., from Pythagorean theorem or V = πr²h); (4) differentiate both sides with respect to t; (5) substitute known values and solve for the unknown rate.
- A worked example: A 5-metre ladder leans against a wall. The base slides away from the wall at 2 m/s. When the base is 3 m from the wall, how fast is the top sliding down? (Pythagorean: x² + y² = 25; differentiate: 2x(dx/dt) + 2y(dy/dt) = 0; at x = 3, y = 4; so 6(2) + 8(dy/dt) = 0 → dy/dt = -1.5 m/s.)
- Chain rule link: Note how the chain rule appears naturally when differentiating by time.
🗣️ 3. After Viewing & Kaiako Move (Process & Apply)
Kaiako Move: Emphasise that given rates MUST be substituted ONLY AFTER differentiation, never before (common NCEA exam trap).
Immediate Task: Add to your Level 3 Calculus revision logbook (section 8): Related Rates Geometric Problems & Time Derivative Log.
⚡ Whakaoho | Do Now: Rates Already Linked (5 mins)
Write the volume of a sphere in terms of its radius. Differentiate it with respect to r. Now say in words what dV/dr means for a balloon, and how it differs from dV/dt. Today's whole method is the chain rule connecting those two.
📖 Activity 1: Chaining Two Rates Together (20 mins)
Two problems, and deliberately not the same chain. Before calculating either, write down which rate you are given and which you want. Most errors in this topic are solving for the wrong one, not differentiating incorrectly.
(a) Balloon. A spherical balloon is inflated at 100 cm³/s. How fast is the radius growing when r = 5 cm? Here the volume varies with the radius, so the chain is dV/dt = dV/dr · dr/dt.
(b) Tank. Water drains from a cylindrical tank of radius 2 m at 0.5 m³/min. How fast is the depth falling? Watch the trap: the tank's radius is fixed, so dr/dt = 0 and the balloon's chain gives you nothing. The variable that is actually changing is the depth, so the chain runs dV/dt = dV/dh · dh/dt, with V = πr²h and r = 2.
The lesson to take from the pair: the chain must run through whichever variable is actually changing. Identify that variable before you write any derivative.
📝 Activity 2: Exam-Style Practice — Set Up, Then Solve (15 mins)
A ladder 5 m long slides down a wall with its foot moving out at 0.2 m/s. Find how fast the top descends when the foot is 3 m from the wall. For Merit, state the relationship between the variables before differentiating. For Excellence, explain why the top's speed increases as the foot moves further out, using your expression rather than intuition.
For Excellence — related rates without solving for either variable first. As geothermal steam rises from a bore toward the surface it expands and cools, so its pressure falls as its volume grows. This is a scenario built to be worked by hand — a simplified, idealised expansion, not a measured field profile; real wells involve flashing, temperature change and friction losses this scenario ignores. In this scenario a rising parcel of steam obeys Boyle's Law, P·V = 90, where P is pressure in bar and V is volume in m³. At the instant V = 6 m³ the parcel is expanding at dV/dt = 2 m³/s.
- Differentiate P·V = 90 implicitly with respect to t using the product rule. (You should get P·dV/dt + V·dP/dt = 0 — note neither P nor V had to be solved for first.)
- At V = 6, P = 90 ÷ 6 = 15 bar. Find dP/dt. (You should get dP/dt = −5 bar/s.) In one sentence, say why that sign makes sense for a parcel that is expanding.
🎫 Exit Ticket: Differentiate First (5 mins)
A spherical balloon’s radius grows at 0.2 cm/s. (a) Write the chain connecting dV/dt to dr/dt. (b) Find dV/dt when r = 10 cm. (c) In one sentence: why must r = 10 be substituted only after differentiating?
🏫 Kaiako Planning & Pedagogy Notes
NCEA Level 3 alignment: Direct preparation for NCEA Level 3 Achievement Standard AS 91578, Apply differentiation methods in solving problems (external, 6 credits), against The New Zealand Curriculum (2007) Mathematics and Statistics Level 8 — Calculus. Emphasise complete algebraic working and a conclusion written in the context of the question, which is what separates Merit from Achieved.
Materials: Mini-whiteboards for the setup step, graphics calculator, calculus logbook. A physical prop helps here: a balloon being inflated, or a jug draining into a measuring cylinder, makes “which variable is actually changing” concrete before it is algebraic.
Pacing (58 mins): Do Now 5 · media anchor 13 (the clip plus the before- and after-viewing prompts) · Activity 1 20 · Activity 2 15 · exit ticket 5.
Formative assessment — what to look for: Part (c) is the named NCEA trap and the same one Activity 1 is built around: a student who substitutes r = 10 before differentiating gets a constant, whose derivative is zero. If that appears in more than a couple of scripts, put both Activity 1 problems back on the board side by side — the contrast between them is the teaching.
Differentiation. Entry: Provide the relationship equation (V = ⅔πr³, V = πr²h) and ask only for the differentiation and substitution. On level: Full problems including the ladder, with the relationship derived from a labelled diagram. Extension: Water pours into a cone at 3 m³/min. The cone’s shape gives r = h/2 at every depth. Find how fast the depth is rising when h = 4 m — and note how the r = h/2 substitution has to happen before you differentiate, not after.