Lesson 1: Related Rates and Optimization, Step by StepAP Deep Dive

The main course gave you the related-rates routine and the optimisation checklist. This deep dive adds the problems where those routines are stressed: rates hidden inside similar triangles, angular rates, shadow problems where the answer is secretly constant, and optimisation questions designed so that the careless student finds a beautiful critical point that is not the answer. Throughout, the discipline is the same: variables first, differentiate before substituting, justify in the language the graders score.

The conical tank, done to full-credit standard

Worked example 1. Water is pumped into an inverted conical tank (radius 3 m at the top, height 9 m) at 2 m³/min. How fast is the water level rising when the depth is 4 m?

Name variables and relate them *before* differentiating. By similar triangles, r/h = 3/9, so r = h/3 at every instant. Volume: V = (1/3)πr²h = (1/3)π(h/3)²h = πh³/27 — one variable only, which is the whole point of the similar-triangles step. Differentiate with respect to t: dV/dt = (πh²/9)·dh/dt. Now substitute h = 4 and dV/dt = 2: 2 = (16π/9)·dh/dt, so dh/dt = 18/(16π) = 9/(8π) ≈ 0.358 m/min. Read the answer back: the level rises more slowly as the tank fills (the same inflow spreads over a widening cross-section), which is why dh/dt decreases as h grows — a one-sentence sanity check like this catches inverted ratios. The three classic errors are all visible here: substituting h = 4 into V before differentiating (freezing the geometry), forgetting that r also changes with t, and losing the chain-rule factor dh/dt.

The shadow problem: a rate that refuses to change

Worked example 2. A streetlight stands 6 m tall. A man 1.8 m tall walks directly away from the pole at 1.5 m/s. How fast is the tip of his shadow moving?

Let x be his distance from the pole and s the length of his shadow. The tip of the shadow is at distance x + s from the pole. Similar triangles (large triangle from the lamp, small triangle from the man): s/1.8 = (x + s)/6, so 6s = 1.8x + 1.8s, giving s = (1.8/4.2)x = (3/7)x. Differentiate: ds/dt = (3/7)(1.5) = 0.643 m/s. The question asks for the tip, whose position is x + s: rate = dx/dt + ds/dt = 1.5 + 0.643 = 2.14 m/s. Notice it is constant — because s is a *linear* function of x, all the rates are proportional, unlike the tank where geometry made the relationship cubic. Recognising which problems hide a linear relation is a pattern worth marks: shadow tips and ladder angles behave linearly in the rates; tanks, spheres and sliding ladders do not. And note the scored distinction between "the shadow lengthens at 0.643 m/s" and "the tip moves at 2.14 m/s" — the FRQ asks for one of them specifically.

Extension — the angular rate. For the sliding ladder (13 ft ladder, base 5 ft from the wall sliding at 2 ft/s), let θ be the angle at the ground: cos θ = x/13. Differentiating, −sin θ·dθ/dt = (1/13)(dx/dt) = 2/13; with sin θ = 12/13 at this instant, dθ/dt = −(2/13)/(12/13) = −1/6 rad/s. Same event, three different rates (dx/dt, dy/dt = −5/6 ft/s, dθ/dt) — the exam rotates through all three, and each requires its own linking equation written *before* differentiation.

Optimisation: justification is the question

Worked example 3. An open-top box with a square base must hold 32 ft³. Minimise the surface area (material).

Let the base side be x and height h. Constraint: x²h = 32, so h = 32/x². Objective in one variable: S(x) = x² + 4xh = x² + 128/x, domain x > 0. S′(x) = 2x − 128/x² = 0 gives x³ = 64, x = 4, hence h = 2. Now the part that carries the marks — justification. Either argue S″(4) = 2 + 256/64 = 6 > 0 (local minimum, and it is the *only* critical point on the domain, with S → ∞ at both ends, so it is the absolute minimum), or use the first derivative sign change. Conclusion with numbers: minimum surface area S = 16 + 4(4)(2) = 48 ft² at a 4 × 4 × 2 box. Observe the shape — the optimal open box has height equal to *half* its base side, a memorable check.

Where students bleed points: on closed-interval problems they find the critical point and stop. If f(x) = x³ − 3x on [0, 2], the critical point x = 1 gives f(1) = −2, a *minimum*; the maximum is at the endpoint, f(2) = 2. Graders plant this deliberately. The Candidates Test — evaluate f at all critical points *and* endpoints, then compare — is not pedantry; it is the method the question is testing.

Common mistakes at the top band

• Substituting the instant's values before differentiating. The equation must be true at all times; numbers belong to one instant only.
• Differentiating the constraint instead of the objective in optimisation, or optimising a two-variable function without eliminating a variable.
• Missing chain-rule factors — d/dt of x² is 2x·dx/dt, never 2x.
• Answering a rate without units or sign interpretation: "the height is decreasing at 5/6 ft/s" earns the point that "−0.833" does not.
• Optimisation without a domain. Stating x > 0 (or the closed interval) is part of the setup credit.

AP Exam Experience

1. (No calculator.) A sphere's radius is increasing at 2 cm/s. When the radius is 5 cm, the volume is increasing at

• (A) 40π cm³/s
• (B) 100π cm³/s
• (C) 200π cm³/s
• (D) 500π cm³/s

Answer: (C). V = (4/3)πr³, so dV/dt = 4πr²·dr/dt = 4π(25)(2) = 200π ≈ 628 cm³/s. Choice (B) comes from differentiating as if V were π r³·... or from using surface area incorrectly halved; (A) drops the square.

2. The maximum value of f(x) = x³ − 3x on the closed interval [0, 2] is

• (A) −2, at x = 1
• (B) 0, at x = 0
• (C) 2, at x = 2
• (D) there is no maximum, because f′(1) = 0 gives a minimum

Answer: (C). f′(x) = 3x² − 3 = 0 at x = 1 (in the interval), where f = −2 — a local minimum. The Candidates Test checks endpoints: f(0) = 0, f(2) = 8 − 6 = 2. The absolute maximum is 2 at the endpoint x = 2. Extreme Value Theorem guarantees a maximum exists, eliminating (D).

3. A point moves along the curve y = x². When x = 2, the x-coordinate is increasing at 3 units/s. At that instant, dy/dt is

• (A) 6 units/s
• (B) 12 units/s
• (C) 4 units/s
• (D) 3 units/s

Answer: (B). Differentiate implicitly with respect to t: dy/dt = 2x·dx/dt = 2(2)(3) = 12 units/s. Choice (A) forgets the factor x = 2 — the chain rule is the question.

Free-response question. (a) A 13 ft ladder leans against a vertical wall. The base slides away from the wall at 2 ft/s. When the base is 5 ft from the wall, how fast is the top of the ladder sliding down, and how fast is the angle between the ladder and the ground changing?

(b) A farmer has 200 ft of fencing to enclose a rectangular pen against an existing straight wall (the wall forms one side, so fencing is needed for only three sides). Find the dimensions that maximise the area, and justify that your answer gives the maximum.

Scoring guideline.

• (a) 3 points. 1 point for the linking equation x² + y² = 169 and its derivative 2x·dx/dt + 2y·dy/dt = 0 (the chain-rule factors dx/dt and dy/dt must appear — this is the setup point). 1 point for dy/dt: at x = 5, y = 12, so 2(5)(2) + 2(12)dy/dt = 0 gives dy/dt = −5/6 ft/s, interpreted as the top sliding down at 5/6 ft/s. 1 point for the angle: from cos θ = x/13, −sin θ·θ′ = (2/13) with sin θ = 12/13, so dθ/dt = −1/6 rad/s. A value of dy/dt = −5/6 with no equation earns only the answer's partial credit per the rubric's setup requirement — a bare number earns 0 for the part.
• (b) 5 points. 1 point for setup: with x the side perpendicular to the wall, 2x + y = 200 and A = xy. 1 point for the one-variable objective A(x) = x(200 − 2x) = 200x − 2x² with domain 0 < x < 100 stated or implied. 1 point for A′(x) = 200 − 4x = 0 → x = 50, y = 100. 1 point for the maximum area, 50 × 100 = 5000 ft². 1 point for justification by the Candidates Test (A(0) = A(100) = 0 < 5000, and x = 50 is the only critical point) or a first/second derivative argument (A″ = −4 < 0). Numerical answer without justification earns 4 of 5 — the justification point is deliberately separate, as on the real exam.

A point moving on a curve: rates in disguise

A whole family of related-rates problems hides the time derivative inside a moving point. A particle moves along the circle x² + y² = 25. At the instant it passes (3, 4), its x-coordinate is increasing at 2 units/s. How fast is y changing? Differentiate the constraint with respect to t: 2x·dx/dt + 2y·dy/dt = 0 — identical in form to the ladder equation, because it *is* the ladder equation with the wall and floor replaced by the circle. Substituting: 2(3)(2) + 2(4)dy/dt = 0, so dy/dt = −1.5 units/s: at (3, 4), moving right, the particle must move down. Push one step further, as FRQs do: the particle's speed is |v| = √((dx/dt)² + (dy/dt)²) = √(4 + 2.25) = 2.5 units/s — the rates you computed are the components of a velocity vector, and the speed combines them by Pythagoras. Recognising the shared structure (constraint → differentiate → substitute) across ladders, circles and curves is worth more than memorising any one problem.