Mathematics Advanced • Year 11 • Module 3 • Lesson 5
The Chain Rule
Apply the chain rule to context problems — gear ratios, thermostat dynamics, ladder slide, tangent searches, and a multi-step optimisation.
Problem 1 — Two stages of a gear chain
A cyclist's pedal turns at angular speed ωp rad/s. The chainring rotates a sprocket at speed ωs = 3 ωp (chainring is 3× larger). The sprocket turns the rear wheel at ωw = 2 ωs. The rear wheel has radius 0.35 m, so its linear speed (m/s) of the bike is v = 0.35 ωw .
Set up: What are we solving for?
(i) Express v entirely in terms of ωp. 2 marks
(ii) Hence find dv/dωp and explain what this number represents physically. 2 marks
(iii) Verify the chain rule by computing dv/dωp as the product (dv/dωw) · (dωw/dωs) · (dωs/dωp). State each factor with units, then multiply. 3 marks
Stuck? Revisit lesson § Concept — "Think of it like a gear system".Problem 2 — Thermostat and temperature
The room temperature (in °C) at time t hours after midnight is T(t) = 18 + 4 sin(π t / 12). A thermostat's heating output W (in watts) depends on the temperature deficit u = 22 − T by W(u) = 200 u² .
Set up: What are we solving for?
(i) Compute du/dT and dW/du explicitly. 2 marks
(ii) Hence find dW/dT and interpret the sign: when room temperature rises, does the heating output rise or fall? 2 marks
(iii) Evaluate dW/dT at the moment T = 20 °C and state its units. 2 marks
Problem 3 — Rocket altitude and air pressure
A rocket's altitude (in km) at time t seconds after launch is h(t) = 5 t² + 2 t . The atmospheric pressure (in kPa) at altitude h is approximately p(h) = 101 · (1 − h/40)5 (a simplified model, valid for low altitudes).
Set up: What are we solving for?
(i) Find dh/dt and dp/dh in exact form. 3 marks
(ii) Use the chain rule to find dp/dt at t = 1 s. State its units and interpret. 3 marks
(iii) Without recomputing dh/dt, explain in one sentence why the pressure falls more rapidly later in the flight (assuming the rocket stays in the model's valid range). 2 marks
Problem 4 — Tangent to a composite curve
Consider the curve y = (x² − 3)4.
Set up: What are we solving for?
(i) Use the chain rule to find dy/dx. 2 marks
(ii) Hence find the equation of the tangent to the curve at the point where x = 2. 3 marks
(iii) Find the x-coordinates of all points at which the tangent to the curve is horizontal. 2 marks
Stuck on (iii)? dy/dx = 0 when 8x (x² − 3)³ = 0. Set each factor to zero.Problem 5 — Minimum distance from a curve
The squared distance from the origin to a point (x, y) on the parabola y = x² is
D(x) = x² + (x²)² = x² + x⁴.
A student writes D(x) = u + u² where u = x², and proposes to use the chain rule to differentiate.
Set up: What are we solving for?
(i) Using the student's composition, find dD/du and du/dx, then dD/dx by the chain rule. 3 marks
(ii) Verify the same answer by differentiating D(x) = x² + x⁴ directly using the power rule (no chain rule needed). 2 marks
(iii) Find the x-value(s) at which D(x) is a minimum (i.e. dD/dx = 0). State the corresponding minimum value of D, and hence the minimum actual distance from the origin to the parabola. 3 marks
Stuck on (iii)? Actual distance = √D. Set dD/dx = 0 then take the square root of the minimum squared distance.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Gear chain
Set up. Cascade the relationships from pedal → sprocket → wheel → bike, expressing each layer as a function of the layer before, then use the chain rule.
(i) ωw = 2 ωs = 2 (3 ωp) = 6 ωp. v = 0.35 · 6 ωp = 2.1 ωp m/s.
(ii) dv/dωp = 2.1 m/s per rad/s. This means: for every extra 1 rad/s the cyclist pedals, the bike gains 2.1 m/s of forward speed.
(iii) dv/dωw = 0.35 m per rad. dωw/dωs = 2 (dimensionless). dωs/dωp = 3 (dimensionless). Chain rule: dv/dωp = 0.35 · 2 · 3 = 2.1 m per rad, matching (ii). The chain rule simply multiplies the per-stage gear ratios.
Problem 2 — Thermostat W(u) = 200 u², u = 22 − T
Set up. u depends on T, W depends on u, so W depends on T through two layers. Apply the chain rule dW/dT = dW/du · du/dT.
(i) du/dT = −1 (increasing T by 1 °C decreases the deficit u by 1). dW/du = 400 u (power rule with constant multiple).
(ii) dW/dT = 400 u · (−1) = −400 u = −400 (22 − T). Negative for T < 22 (typical room): rising room temperature reduces heating output — physically correct (thermostat backs off).
(iii) At T = 20: u = 22 − 20 = 2, so dW/dT = −400 · 2 = −800 W per °C. At this moment, each extra 1 °C of room temperature cuts heater output by about 800 W.
Problem 3 — Rocket altitude / pressure
Set up. h is a function of t, p is a function of h, so p is a function of t via the chain rule dp/dt = dp/dh · dh/dt.
(i) dh/dt = 10t + 2 km/s. dp/dh = 101 · 5 (1 − h/40)4 · (−1/40) = −(101/8) (1 − h/40)4 kPa/km.
(ii) At t = 1: h(1) = 5 + 2 = 7 km. dh/dt at t = 1 = 12 km/s. 1 − h/40 = 1 − 7/40 = 33/40. dp/dh at h = 7: −(101/8) (33/40)4. (33/40)2 = 1089/1600; (33/40)4 = (1089/1600)² ≈ 0.46347. dp/dh ≈ −(101/8) · 0.46347 ≈ −5.85 kPa/km. dp/dt = −5.85 · 12 ≈ −70.2 kPa/s. The pressure is falling at about 70 kPa per second one second after launch.
(iii) As t increases, dh/dt = 10t + 2 grows — the rocket climbs faster and faster — so the chain product dp/dh · dh/dt has an ever-larger time-derivative factor, making pressure fall ever more quickly. (The dp/dh factor also tends to be more negative at higher altitude until we exit the model's range.)
Problem 4 — Tangent to y = (x² − 3)4
Set up. Apply the chain rule with u = x² − 3, find gradient at x = 2, then write tangent and finally locate horizontal tangents.
(i) u = x² − 3; dy/du = 4 u³; du/dx = 2x. dy/dx = 4 u³ · 2x = 8x (x² − 3)³.
(ii) At x = 2: u = 1, dy/dx = 8 · 2 · 1³ = 16. Point: y = (1)4 = 1, so (2, 1). Tangent: y − 1 = 16(x − 2) ⇒ y = 16x − 31.
(iii) 8x (x² − 3)³ = 0 ⇔ x = 0 OR x² − 3 = 0. Hence x = 0, x = √3, x = −√3 (three horizontal tangents).
Problem 5 — D(x) = x² + x⁴ (squared distance to origin from y = x²)
Set up. Verify the chain rule gives the same answer as direct power rule, then minimise D to find the closest point on the parabola.
(i) u = x², so D = u + u². dD/du = 1 + 2u; du/dx = 2x. dD/dx = (1 + 2u)(2x) = 2x + 4ux = 2x + 4 x · x² = 2x + 4x³.
(ii) Direct: D(x) = x² + x⁴ ⇒ dD/dx = 2x + 4x³. ✓ Matches.
(iii) 2x + 4x³ = 0 ⇒ 2x (1 + 2x²) = 0. Since 1 + 2x² > 0 for all real x, only solution is x = 0. D(0) = 0; minimum actual distance from origin = √D = √0 = 0 m (the parabola passes through the origin, which is on the curve itself). Geometrically expected: y = x² passes through (0, 0).