Related Rates of Change (AA HL)
Related rates problems connect two or more quantities that are both changing with time, even though the question only ever gives you a fixed geometric or algebraic relationship between them. The trick is to differentiate that relationship implicitly with respect to \(t\), so the rates themselves appear in the equation. It's part of the broader Kinematics & Related Rates topic.
13 questions on this sub-topic.
The method, not a formula
Covered under IB syllabus reference AHL5.14. There's no formula to memorise here - related rates is a technique built entirely on implicit differentiation, so the two steps below are what actually earns the marks.
Connect the variables
Write a single equation linking the quantities using a fixed relationship - Pythagoras' theorem, a volume or area formula, or a given trig identity.
Not in the formula booklet - comes from the geometry of the problemDifferentiate implicitly w.r.t. \(t\)
Differentiate every term of the connecting equation with respect to time before substituting any known numbers, so each rate of change survives in the equation.
Not in the formula booklet - techniqueNeed the full syllabus wording and formula-booklet reference table? See Kinematics & Related Rates, including the GDC guidance there.
Worked examples
A circular oil slick expands so its radius grows at 0.5 m/s.
Find the rate of increase of its area when \(r = 10\) m.
Worked solution
\(A = \pi r^2\), so \(\dfrac{dA}{dt}\) M1
\(\dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt}.\) A1
\(\dfrac{dA}{dt} = 2\pi(10)(0.5)\) M1 A1
\(\dfrac{dA}{dt} = 10\pi \approx 31.4\) m²/s. A1
A 5 m ladder leans against a wall. The base slides away at 0.5 m/s.
Find the rate at which the top slides down when the base is 3 m from the wall.
Worked solution
\(x^2 + y^2 = 25.\) Differentiate: \(2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt}\) M1 \(= 0.\) A1
at \(x=3,\ y=4.\) M1 \(2(3)(0.5) + 2(4)\dfrac{dy}{dt} = 0\) A1 M1 \(\Rightarrow \dfrac{dy}{dt} = -0.375\) m/s. A1
Common mistakes
- Dropping a rate when differentiating. Differentiating \(x^2+y^2=25\) with respect to \(t\) needs \(2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0\) - forgetting either \(\dfrac{dx}{dt}\) or \(\dfrac{dy}{dt}\) breaks the whole equation.
- Substituting known values before differentiating. If you plug in a fixed number for a variable first, its derivative vanishes from the equation - differentiate the full connecting equation first, then substitute.
- Losing the sign of a rate. A shrinking or falling quantity has a negative rate of change - if the ladder's top is "sliding down", the answer should come out negative before you describe it in words.
Ready to practise properly?
13 related-rates questions, marked instantly like the real exam.
Quick answers
How do I set up a related rates of change problem?
Write a fixed equation connecting the quantities involved - Pythagoras' theorem, a volume or area formula, a trig relationship. Then differentiate both sides implicitly with respect to time and substitute the known rate and value to solve for the unknown rate.
Why can't I substitute the known values before differentiating?
Substituting a fixed value too early turns a variable into a constant, so its derivative silently disappears from the equation. Differentiate the whole connecting equation first, with every quantity still a variable, and only substitute known numbers afterwards.