Lesson 11: Applications: Tangents and Optimisation

The derivative earns its keep when it answers questions about the world: where is a function heading, where does it peak, what is the best design? This lesson (AA Topic 5) turns Lesson 10's rules into analysis: tangents and normals, the sign of f′ as a direction detector, stationary points and their classification, and optimisation — modelling a quantity, differentiating, and interpreting the answer in context. It also introduces kinematics: if s(t) is displacement, then v(t) = s′(t) is velocity and a(t) = v′(t) = s″(t) is acceleration.

Tangents and normals

At the point (a, f(a)), the tangent has gradient f′(a) and equation y − f(a) = f′(a)(x − a). The normal is perpendicular to the tangent, so its gradient is −1/f′(a) (when f′(a) ≠ 0 — a horizontal tangent has a vertical normal x = a, and an examiner will absolutely test that edge case). Geometry questions chain these: find the tangent, find where it meets an axis, find the area of the triangle it makes — each step feeding the next, with follow-through marks protecting you after an early slip.

Worked Example 1 — tangent and normal (past-paper style). Let f(x) = x² − 3x + 1. Find the equations of the tangent and the normal at x = 2.

Solution. f(2) = 4 − 6 + 1 = −1, so the point is (2, −1). f′(x) = 2x − 3, so f′(2) = 1. Tangent: y + 1 = 1(x − 2), i.e. y = x − 3. Normal gradient −1/1 = −1: y + 1 = −1(x − 2), i.e. y = −x + 1. Two accuracy habits: use the *point* in the point–gradient form (not the origin), and simplify only at the end so follow-through stays clean.

Increasing, decreasing and stationary points

f is increasing where f′ > 0 and decreasing where f′ < 0. Stationary points occur where f′(x) = 0. Classify them either by the first derivative test (sign of f′ either side: + to − is a maximum, − to + a minimum, no change is a horizontal inflection) or the second derivative test (f″(a) < 0 → maximum; f″(a) > 0 → minimum; f″(a) = 0 → inconclusive, fall back to the first test). State which test you used; a naked "it's a maximum" after f′ = 0 earns nothing.

Worked Example 2 — a full curve analysis. Analyse f(x) = x³ − 3x² − 9x + 2 (from Lesson 10): find its stationary points, classify them, and find the range of x where f is increasing.

Solution. f′(x) = 3(x − 3)(x + 1): stationary at x = −1 and x = 3. f″(x) = 6x − 6: f″(−1) = −12 < 0 → maximum at (−1, f(−1)) = (−1, 7). f″(3) = 12 > 0 → minimum at (3, f(3)) = (3, −25). Sign chart for f′: positive outside the roots, negative between, so f increases on (−∞, −1) and (3, ∞), decreases on (−1, 3). On Paper 2, the GDC's graph confirms the shape — quote both: analytic classification first, calculator check second.

Optimisation: the five-step script

Every optimisation problem in the course runs the same script: (1) define variables and draw a diagram; (2) write the objective as a function of ONE variable, using the constraint to eliminate the others; (3) state the feasible domain from the context; (4) differentiate, solve f′ = 0, verify it's the optimum you want (test + domain); (5) answer in context, with units. Marks cluster at steps 2 and 5 — the algebra of elimination and the interpretation are the examined skills; differentiation is just the engine.

Worked Example 3 — the box problem. A closed box with a square base is made from 600 cm² of card. Find the dimensions that maximise its volume.

Solution. Base side x, height h. Constraint (surface area): x² + 4xh = 600, so h = (600 − x²)/(4x). Objective: V(x) = x²h = x(600 − x²)/4 = 150x − x³/4, domain 0 < x < √600. Differentiate: V′(x) = 150 − 3x²/4 = 0 → x² = 200 → x = 10√2 ≈ 14.1 cm (positive root only, by domain). V″ = −3x/2 < 0, confirming a maximum. Then h = (600 − 200)/(4 × 10√2) = 400/(40√2) = 5√2 ≈ 7.07 cm, and V_max = x²h = 200 × 5√2 = 1000√2 ≈ 1414 cm³. Notice the optimum has h = x/2 — a pattern worth remembering for square-based boxes, and a check on the arithmetic.

Kinematics

If s(t) = t³ − 6t² + 9t (metres), then v(t) = 3t² − 12t + 9 and a(t) = 6t − 12. "When is the particle at rest?" is v(t) = 0 (t = 1 and t = 3). "When is it moving backwards?" is v < 0 (1 < t < 3). Total distance travelled needs the turning points — displacement alone (s(final) − s(start)) hides the backtracking, the classic Section B trap of this topic.

Common mistakes

• Normal gradient used for the tangent or the negative reciprocal forgotten; also using the gradient at the wrong point.
• Stationary points classified by f″ alone when f″ = 0 — the second derivative test can be inconclusive; the first derivative test always works.
• Optimisation without a domain: accepting a negative length root, or never checking endpoints when the domain is closed (on a closed interval the optimum can sit at an endpoint where f′ ≠ 0).
• Two-variable objective differentiated anyway — eliminate first, using the constraint.
• Units and interpretation dropped at the final step: "x = 14.1" instead of "the base should be 14.1 cm square".
• Velocity confused with speed: speed is |v|; "at rest" and "direction" questions hinge on the sign.

IB Exam Experience

Question 1 [2 marks]. Find the equation of the tangent to y = ln x at x = e. *Marking notes:* M1: y′(x) = 1/x so gradient at e is 1/e, point (e, 1); A1: y − 1 = (1/e)(x − e), i.e. y = x/e. (An exact, elegant answer — decimals earn nothing here.)

Question 2 [3 marks]. Let f(x) = 2x³ − 3x² − 12x. Find the x-coordinates of the stationary points and justify which is a maximum. *Marking notes:* M1: f′(x) = 6x² − 6x − 12 = 6(x − 2)(x + 1); A1: x = 2, x = −1; A1: justification — f″(−1) = −18 < 0 so x = −1 is the maximum (or a correct sign-chart argument).

Question 3 [4 marks]. A particle moves with displacement s(t) = t³ − 9t² + 24t metres, t ≥ 0. Find when the particle is at rest, and its acceleration at those times. *Marking notes:* M1: v(t) = 3t² − 18t + 24; M1: setting v = 0 and dividing by 3: t² − 6t + 8 = 0; A1: t = 2 s and t = 4 s; A1: a(t) = 6t − 18 gives a(2) = −6 m/s² and a(4) = 6 m/s². Units required for full credit.

Extended problem (Paper 2 style) [10 marks]. A farmer uses 240 m of fencing to enclose a rectangular field against a long straight wall. The fence forms the two short sides and the side parallel to the wall, plus one internal divider parallel to the short sides. Let each short side be x m. (a) Show that the area of the field is A(x) = 240x − 3x². [3] (b) Find the maximum area and the dimensions that achieve it. [5] (c) State the domain of A in this context and explain why the endpoints are excluded. [2] *Marking notes:* (a) M1 for the fencing count: two short sides + one divider + the long side L gives 3x + L = 240 (the wall replaces the fourth side — forgetting either the divider or the wall's free side is the classic modelling slip); M1 for L = 240 − 3x; A1 for A(x) = Lx = 240x − 3x². (b) M1: A′(x) = 240 − 6x; M1: setting A′ = 0 gives x = 40; A1: L = 240 − 120 = 120 m; A1: A(40) = 4800 m²; A1: verification that it is a maximum (A″ = −6 < 0). (c) A1: 0 < x < 80; A1: as x → 0 the field vanishes in width, and at x = 80 the long side L falls to 0 — both endpoints are degenerate, so the interior stationary point is the only candidate. Examiner's note: this question's first mark is reading comprehension — the divider is stated in the first sentence and missed by a third of candidates.

Exploration (IA) idea: optimise a real container — a drinks can's dimensions versus the theoretical optimum for its volume — and investigate why real cans deviate (manufacturing, grip, stacking). Superb personal engagement, honest modelling, clean calculus.

Going deeper: optimisation with two constraints and endpoints

Harder optimisation problems add a second constraint or a closed domain — and on a closed domain the maximum may sit at an endpoint where f′ ≠ 0, which is why the script ends with "check the endpoints" whenever the domain has them.

Extra worked example. A rectangular enclosure of area 200 m² is built against a wall (no fence needed on the wall side). Fencing costs $12 per metre for the two short sides and $20 per metre for the long side (parallel to the wall). Minimise the cost.

Solution. Let the short sides be x and the long side y, with xy = 200 → y = 200/x. Cost C(x) = 2(12x) + 20y = 24x + 4000/x, x > 0. C′(x) = 24 − 4000/x² = 0 → x² = 4000/24 = 500/3 → x ≈ 12.9 m, y ≈ 15.5 m. C″ = 8000/x³ > 0: minimum. Minimum cost C ≈ 24(12.9) + 4000/12.9 ≈ $620. Note the shape: the optimum balances the two cost contributions (24x ≈ 310 ≈ 4000/x) — an echo of AM–GM, and a useful estimate check.

Extra exam practice [3 marks]. A particle's velocity is v(t) = 12t − 3t² m/s for t ≥ 0. Find the greatest distance from the starting point that the particle reaches while it is moving in the positive direction. *Marking notes:* M1: the particle advances while v > 0, i.e. until t = 4 (v = 3t(4 − t)); M1: displacement = ∫₀⁴ v dt = [6t² − t³]₀⁴ = 96 − 64 = 32; A1: 32 m, with units. (Velocity changing sign is what "returns" means — the integral of v, not of |v|, to the turning time.)