IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~12 min read

Related Rates of Change

When several quantities change together over time, their rates are LINKED through a geometric formula — and the chain rule connects them. A balloon’s volume and radius are tied by V = (4/3)πr³, so the rate of inflation forces a specific rate of radius growth. Same logic for water in a cone, a ladder sliding down a wall, a shadow lengthening, a rocket tracked by a camera. The technique is always: write the formula, differentiate with respect to t, plug in.

📘 What you need to know

The linking variable is usually time t. Every variable that changes is implicitly a function of t, even though we don’t write r(t) explicitly.
Chain rule connects rates: dA/dt = (dA/dr) · (dr/dt). The “easy” derivative dA/dr comes from the geometric formula; dr/dt is what the problem gives or asks for.
Product rule when two variables both change: for V = πr²h with both r and h changing, dV/dt = π(2r · dr/dt · h + r² · dh/dt).
Eliminate variables FIRST using geometric constraints: for a cone tank with fixed shape, similar triangles give r/h = constant — substitute BEFORE differentiating to avoid a product-rule mess.
Differentiate FIRST, substitute numbers LAST. Plugging in r = 5 too early treats r as constant and gives dV/dt = 0.
Sign of the rate: positive ⟹ quantity increasing; negative ⟹ decreasing. Draining water gives dV/dt < 0; water level falling gives dh/dt < 0. The SPEED of fall is |dh/dt|.
Units: rate of change of (X)-quantity per (time)-unit. Length/time = m/s; volume/time = cm³/s; angle/time = rad/s.
Draw a diagram. Always. Even for problems you “see” instantly — it catches sign errors and clarifies which variables are which.

Setting up — the chain rule between any two rates

The chain rule is what does all the work. If two quantities A and r are linked by a formula, and both change with time, then their time-rates are linked via:

The chain rule for related rates dAdt = dAdr · drdt

The first factor dA/dr comes from differentiating the formula. The second factor dr/dt is the “input” rate (either given, or what you’re solving for). For more than two linked quantities (like V, r, h together), the product rule generalises this.

One variable changes

A = πr²; dAdt = 2πr · drdt

single chain rule — the standard case (circle, sphere, cube)

Two variables change (product rule)

V = πr²h; dVdt = π(2rh · drdt + r² · dhdt)

product rule from implicit differentiation — both rates feature

🧭 Recipe — solve a related rates problem

Draw a diagram and list the variables and rates. Mark which rates are given, which is required, and which variables are constants (their rate is 0).
Write the formula connecting the variables — geometric (area, volume, Pythagoras), similar triangles, trig ratio, whatever fits.
Eliminate extra variables using any fixed geometric constraint. For cone tanks: similar triangles give a fixed ratio r/h, so substitute one in terms of the other to reduce to a single variable.
Differentiate both sides with respect to t using chain rule (and product rule for any term with multiple changing variables).
Substitute the numerical values at the instant of interest. Solve for the required rate and interpret the sign (positive = increasing, negative = decreasing).

The cone trap — eliminate before you differentiate

Cone-tank problems are the classic stumbling block. The mistake: keeping V = (1/3)πr²h as a formula in TWO variables, when only ONE is actually free. The cone’s fixed shape forces r/h = R/H (similar triangles, with R and H the tank’s max radius and full height).

Similar triangles: the small water-cone (radius r, height h) is similar to the full tank-cone (radius R = 4, height H = 10). The ratio r/h = R/H = 2/5 is constant; use it to eliminate r from the volume formula BEFORE differentiating.

The substitution trick: V = (1/3)πr²h with r = (2/5)h becomes V = (1/3)π · (4h²/25) · h = (4π/75)h³. Now ONE variable, no product rule needed. dV/dh = (4π/25)h².

Worked examples

WE 1

Expanding circle — oil slick

An oil slick on the surface of the sea is circular. Its radius is increasing at a rate of 0.5 cm per second. Find the rate at which the area of the slick is increasing at the moment when the radius is 8 cm.

Step 1 — list variables, rates, and the formula variables: r (radius), A (area), t (time) given: dr/dt = 0.5 cm/s; find: dA/dt at r = 8 formula: A = πr² Step 2 — differentiate with respect to t (chain rule) dA/dt = (dA/dr) · (dr/dt) = 2πr · (dr/dt) Step 3 — substitute r = 8 and dr/dt = 1/2 dA/dt = 2π(8)(1/2) = 8π cm²/s dA/dt = 8π cm²/s ≈ 25.1 cm²/s positive rate ⟹ area is increasing. Notice we never needed to find A itself; only the formula linking A and r mattered.

WE 2

Inflating sphere — balloon

A spherical balloon is being inflated at a rate of 40 cm³ per second. Find the rate at which the radius is increasing at the moment when the radius is 5 cm.

Step 1 — list variables, rates, and the formula variables: r (radius), V (volume), t (time) given: dV/dt = 40 cm³/s; find: dr/dt at r = 5 formula: V = (4/3)πr³ Step 2 — differentiate with respect to t (chain rule) dV/dt = (dV/dr) · (dr/dt) = 4πr² · (dr/dt) Step 3 — substitute and solve 40 = 4π(5)² · (dr/dt) = 100π · (dr/dt) dr/dt = 40 / (100π) = 2/(5π) cm/s dr/dt = 2/(5π) cm/s ≈ 0.127 cm/s the radius grows slowly even with rapid inflation — the volume scales with r³, so most new air goes into “thickening” the existing surface rather than pushing the boundary outward.

WE 3

Water draining from an inverted cone (similar triangles)

An inverted conical tank has top radius 4 cm and full height 10 cm. Water drains from the tank at a rate of 3 cm³ per second. Find the rate at which the depth of the water is falling at the moment when the depth is 6 cm.

Step 1 — list variables, rates, formula water variables: r (water surface radius), h (water depth), V (water volume) tank constants: R = 4, H = 10 given: dV/dt = -3 cm³/s (negative because draining); find: dh/dt at h = 6 formula: V = (1/3)π r² h Step 2 — eliminate r using similar triangles r/h = R/H = 4/10 = 2/5 ⟹ r = (2/5)h V = (1/3)π (2h/5)² h = (1/3)π (4h²/25) h = (4π/75) h³ Step 3 — differentiate with respect to t dV/dt = (dV/dh) · (dh/dt) = (4π/25) h² · (dh/dt) Step 4 — substitute h = 6 and dV/dt = -3 -3 = (4π/25)(36) · (dh/dt) = (144π/25) · (dh/dt) dh/dt = -3 · 25 / (144π) = -75/(144π) = -25/(48π) cm/s dh/dt = -25/(48π) cm/s ≈ -0.166 cm/s negative sign ⟹ water level is FALLING. The depth is falling at 25/(48π) cm per second. The classic trap: if you’d kept r as a separate variable, you’d need extra info on dr/dt — the similar-triangles substitution makes that unnecessary.

WE 4

Ladder sliding down a wall (Pythagoras with constant total)

A 5-metre ladder leans against a vertical wall. The bottom of the ladder slides away from the wall at a rate of 0.4 m per second. Find the rate at which the top of the ladder is sliding down the wall at the moment when the bottom is 3 metres from the wall.

Step 1 — diagram + variables x = horizontal distance of foot from wall y = height of top on wall ladder length = 5 m (constant); Pythagoras: x² + y² = 25 given: dx/dt = 0.4 m/s; find: dy/dt at x = 3 Step 2 — find y at x = 3 y² = 25 – 9 = 16 ⟹ y = 4 Step 3 — differentiate x² + y² = 25 with respect to t 2x · (dx/dt) + 2y · (dy/dt) = 0 ⟹ dy/dt = -(x/y) · (dx/dt) Step 4 — substitute dy/dt = -(3/4)(0.4) = -3/4 · 2/5 = -3/10 m/s top slides DOWN at 3/10 m/s = 0.3 m/s the ladder length (5 m) appears nowhere in the differentiated equation — only as a constant in deriving y from x. Constants vanish under differentiation, which is the whole point.

WE 5

Walking man’s shadow under a lamppost (similar triangles)

A man 2 metres tall walks directly away from a lamppost 6 metres tall at a constant speed of 1.5 m per second. Find the rate at which the length of his shadow is increasing.

Step 1 — set up similar triangles x = distance of man from base of lamppost s = length of his shadow similar triangles: 6 / (x + s) = 2 / s ← lamppost tip, man’s top, shadow tip Step 2 — solve for s in terms of x 6s = 2(x + s) = 2x + 2s 4s = 2x ⟹ s = x/2 Step 3 — differentiate with respect to t ds/dt = (1/2) · (dx/dt) Step 4 — substitute dx/dt = 1.5 ds/dt = (1/2)(1.5) = 0.75 m/s shadow lengthens at 0.75 m/s the relationship s = x/2 is independent of x — no matter where the man stands, his shadow grows at exactly half his walking speed. The tip of the shadow itself moves at dx/dt + ds/dt = 1.5 + 0.75 = 2.25 m/s.

WE 6

Camera tracking a rocket (angle of elevation, trig)

A camera positioned 200 m from a launch pad tracks a rocket as it rises vertically. The rocket rises at a constant speed of 50 m per second. Find the rate at which the camera’s angle of elevation is changing at the moment when the rocket is 150 m above the ground.

Step 1 — set up trig relationship θ = angle of elevation of camera from horizontal h = height of rocket above ground camera distance = 200 m (constant) tan θ = h / 200 given: dh/dt = 50 m/s; find: dθ/dt at h = 150 Step 2 — differentiate with respect to t (chain rule on tan θ) sec²θ · (dθ/dt) = (1/200) · (dh/dt) Step 3 — find sec²θ at h = 150 at h = 150: hypotenuse from camera to rocket = √(200² + 150²) = √62500 = 250 cos θ = adj/hyp = 200/250 = 4/5 sec θ = 5/4 ⟹ sec²θ = 25/16 Step 4 — substitute (25/16) · (dθ/dt) = (1/200)(50) = 1/4 dθ/dt = (1/4) · (16/25) = 4/25 rad/s dθ/dt = 4/25 rad/s ≈ 0.16 rad/s use of sec²θ = 1 + tan²θ would also work: at h = 150, tan θ = 150/200 = 3/4, so sec²θ = 1 + 9/16 = 25/16 ✓. Both ways land on 4/25 rad/s — the chain rule for tan θ is the key step.

💡 Top tips

Always draw a diagram — even when you can “see” the geometry. The labels and arrows on the diagram catch sign errors and missing constraints.
Use similar triangles to eliminate a variable BEFORE differentiating when working with cones — avoids product-rule complications.
Differentiate first, substitute last. Numerical values are time-specific; the differentiation must be done with everything still symbolic.
Track the sign of every rate: draining ⟹ negative dV/dt; falling level ⟹ negative dh/dt; sliding away ⟹ positive dx/dt. Sign discipline matters.
Check units at the end — they should match the question’s expected dimension (cm/s, m/s, rad/s, cm³/s, etc.).

⚠ Common mistakes

Substituting numerical values before differentiating — this treats variables as constants. If you write V = (1/3)π(2)²·(6) = 8π and try to differentiate, you get 0.
Keeping r and h independent in cone problems — they’re linked by the cone’s shape. Forgetting this gives a product-rule expression with an unknown dr/dt.
Sign confusion when interpreting — “falling at” vs “dy/dt = “. If the question asks the SPEED at which the top falls, give the magnitude (positive number); if it asks dy/dt, keep the sign.
For Pythagorean problems, forgetting that L is constant — the rate equation 2x · dx/dt + 2y · dy/dt = 0 (with zero on the right) only works because L² is constant.
Trig problems: not converting between sec²θ and other trig identities — at the substitution step, you need sec²θ from the right triangle (or from 1 + tan²θ).

Up next: Differentiating Reciprocal Trigonometric Functions. With chain rule and product rule firmly in place, you’ll learn the derivatives of sec x, cosec x, and cot x — three “new” trig functions defined as reciprocals of cos, sin, and tan. Each derivative comes out as a product of TWO reciprocal-trig functions (e.g., d/dx[sec x] = sec x · tan x), proved using the quotient rule.

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