Mathematics Advanced • Year 11 • Module 3 • Lesson 4
Differentiation Rules
Apply the power, constant-multiple and sum/difference rules to context problems — kinematics, marginal cost, surface area, and a tangent search.
Problem 1 — Cyclist accelerating from rest
A cyclist's position from the start line (in metres) at time t seconds is modelled by
s(t) = t³ − 4t + 1, 0 ≤ t ≤ 4.
Set up: What are we solving for?
(i) Differentiate to find the velocity function v(t) = s ′(t) and the acceleration function a(t) = v ′(t). 2 marks
(ii) Find the velocity and acceleration at t = 2. State the units. 2 marks
(iii) Find the time(s) at which the cyclist is momentarily at rest, and the position at that moment. 3 marks
Stuck on (iii)? Cyclist at rest ⇔ v(t) = 0 ⇔ s ′(t) = 0. Solve the resulting equation, accept only t in the given domain.Problem 2 — Marginal cost of producing widgets
The total cost (in dollars) of producing x widgets per day is
C(x) = 0.01 x³ − 0.6 x² + 30 x + 200, 0 ≤ x ≤ 60.
The marginal cost M(x) = C ′(x) is the rate of change of cost with respect to output — approximately the cost of producing one extra widget.
Set up: What are we solving for?
(i) Find M(x) = C ′(x). 2 marks
(ii) Calculate M(10) and M(40), and explain what each means in the production context (in one sentence each). 2 marks
(iii) Find the value of x ∈ [0, 60] at which marginal cost is minimised. Explain in one sentence what this output level represents for the firm. 3 marks
Problem 3 — Surface area of a growing sphere
A spherical balloon is being inflated. The volume V (in cm³) is related to the radius r (in cm) by V = (4/3) π r³ . The surface area is S = 4 π r² .
Set up: What are we solving for?
(i) Find dV/dr and dS/dr in exact form (leave π in your answers). 2 marks
(ii) Notice that dV/dr equals the formula for S itself. Comment in one sentence on what this physically tells you about how volume grows with radius. 2 marks
(iii) When r = 5 cm, find dS/dr and explain what this number represents (with units). 2 marks
Stuck on (ii)? "Adding a thin spherical shell of thickness Δr increases V by approximately S · Δr."Problem 4 — Tangent of prescribed gradient
Consider the curve y = x³ − 3x² + 2.
Set up: What are we solving for?
(i) Find dy/dx. 1 mark
(ii) Find the x-coordinates of all points where the tangent to the curve has gradient 9. 3 marks
(iii) Hence find the equations of the two tangent lines. 3 marks
Problem 5 — A wattage function with roots and reciprocals
The power dissipated by a particular component (in watts) as a function of current x (in amps), x > 0, is
P(x) = 4 √x + 9 / x.
Set up: What are we solving for?
(i) Rewrite P(x) in index form, then differentiate to obtain P ′(x). 2 marks
(ii) Evaluate P ′(4) in exact form, then as a decimal to 2 d.p. State its physical units. 2 marks
(iii) By setting P ′(x) = 0, find the current x > 0 at which the rate of change of power with respect to current is zero. (Hint: bring both terms to the same denominator, set the numerator equal to zero, then solve for x.) 3 marks
Stuck? After rewriting, P ′(x) = 2/√x − 9/x². Set this to zero and solve.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Cyclist s(t) = t³ − 4t + 1
Set up. Differentiate s for velocity; differentiate again for acceleration; then solve v = 0 within the time domain.
(i) v(t) = s ′(t) = 3t² − 4 m/s. a(t) = v ′(t) = 6t m/s².
(ii) v(2) = 3(4) − 4 = 8 m/s. a(2) = 6(2) = 12 m/s².
(iii) v(t) = 0 ⇔ 3t² − 4 = 0 ⇔ t² = 4/3 ⇔ t = 2/√3 ≈ 1.155 s (taking positive root within [0, 4]). Position: s(2/√3) = (2/√3)³ − 4(2/√3) + 1 = 8 / (3√3) − 8/√3 + 1 = 8/(3√3) − 24/(3√3) + 1 = −16/(3√3) + 1 ≈ −2.08 m (cyclist is 2.08 m behind the start line at the moment of rest).
Problem 2 — Marginal cost M(x) = C ′(x)
Set up. Differentiate the cost function to get marginal cost, evaluate at two output levels, then minimise M(x).
(i) M(x) = C ′(x) = 0.03 x² − 1.2 x + 30 dollars per widget.
(ii) M(10) = 0.03(100) − 1.2(10) + 30 = 3 − 12 + 30 = $21/widget (adding the 11th widget costs about $21). M(40) = 0.03(1600) − 1.2(40) + 30 = 48 − 48 + 30 = $30/widget (adding the 41st widget costs about $30 — production is becoming more expensive at the margin).
(iii) Differentiate M: M ′(x) = 0.06x − 1.2. Set = 0: x = 1.2 / 0.06 = 20 widgets/day. (At this output the marginal cost itself is at its minimum: M(20) = 0.03(400) − 1.2(20) + 30 = 12 − 24 + 30 = $18. Producing fewer or more widgets pushes the cost-per-widget at the margin higher — this is the firm's most efficient scale at the margin.)
Problem 3 — Sphere V = (4/3) π r³, S = 4 π r²
Set up. Apply power rule to find dV/dr and dS/dr; compare; then evaluate at r = 5 with units.
(i) dV/dr = (4/3) π · 3r² = 4 π r². dS/dr = 4 π · 2r = 8 π r.
(ii) dV/dr = 4 π r² = S, i.e. the rate at which the volume grows with respect to radius equals the surface area. Physically: adding a thin spherical shell of thickness Δr to a sphere increases the volume by approximately S · Δr (surface area × shell thickness).
(iii) dS/dr at r = 5: 8 π (5) = 40 π cm²/cm ≈ 125.66 cm²/cm. This is the rate of change of surface area with respect to radius: each extra centimetre of radius adds about 125.66 cm² of surface area (at the moment r = 5).
Problem 4 — Tangent of gradient 9 to y = x³ − 3x² + 2
Set up. Differentiate, set dy/dx = 9, solve the resulting quadratic, then write a tangent at each solution.
(i) dy/dx = 3x² − 6x.
(ii) 3x² − 6x = 9 ⇒ 3x² − 6x − 9 = 0 ⇒ x² − 2x − 3 = 0 ⇒ (x − 3)(x + 1) = 0 ⇒ x = 3 or x = −1.
(iii) At x = 3: y = 27 − 27 + 2 = 2; tangent y − 2 = 9(x − 3) ⇒ y = 9x − 25. At x = −1: y = −1 − 3 + 2 = −2; tangent y − (−2) = 9(x − (−1)) ⇒ y = 9x + 7.
Problem 5 — P(x) = 4 √x + 9/x
Set up. Rewrite as powers, apply the power rule, evaluate, then solve for stationary point.
(i) P(x) = 4 x1/2 + 9 x−1. P ′(x) = 4 · (1/2) x−1/2 + 9 · (−1) x−2 = 2 x−1/2 − 9 x−2 = 2 / √x − 9 / x².
(ii) P ′(4) = 2 / √4 − 9 / 16 = 2/2 − 9/16 = 1 − 9/16 = 7/16 ≈ 0.44 W/A.
(iii) Set P ′(x) = 0: 2 / √x = 9 / x². Cross-multiply: 2 x² = 9 √x. Square both sides (both sides positive for x > 0): 4 x⁴ = 81 x ⇒ 4 x³ = 81 ⇒ x³ = 81/4 ⇒ x = (81/4)1/3 ≈ 2.73 A. (At this current the instantaneous rate of change of dissipated power with respect to current is zero — corresponding to the minimum of P, where the two terms 4√x (rising) and 9/x (falling) balance their rates.)