Related rates

IB Mathematics Applications & Interpretation HL· 45 min read

1. Setting up related rates problems★★☆☆☆⏱ 10 min

The core idea of related rates is that if two quantities are related by an equation, their rates of change with respect to time are also related. We use a consistent step-by-step process to solve these problems.

📘 Definition

Related Rates

Problems that require finding the rate of change of one quantity, given the rate of change of one or more other related quantities, all changing with time $t$ .

Example:

Finding how fast the water level drops in a tank when you know how fast water drains out

List all given quantities and the quantity whose rate you need to find, labeling rates as derivatives with respect to time.
Write an equation that relates the quantities of interest.
Differentiate both sides of the equation implicitly with respect to time $t$ .
Substitute the known values and solve for the unknown rate.
Interpret your result in the context of the problem.

Exam tip:

Always label what each variable represents at the start of your working. Examiners award method marks for clear set up even if your final answer is wrong.

2. Common geometric related rates problems★★★☆☆⏱ 15 min

Most IB exam related rates questions use geometric contexts, where the relationship between quantities comes from standard geometry formulas. The most common are right triangles, circles, spheres, and cones.

📐 Worked Example

A 5m long ladder is sliding down a vertical wall, with the base sliding away from the wall at 0.8 m/s. How fast is the top of the ladder sliding down when the base is 3m from the wall?

1
Step 1: Define variables: $x =$ distance of base from wall, $y =$ height of top of ladder up wall. Given $\frac{d x}{d t} = 0.8$ m/s, find $\frac{d y}{d t}$ when $x = 3$ .
2
Step 2: Relate $x$ and $y$ via Pythagoras' theorem:
3
$x^2 + y^2 = 5^2 = 25$
4
Step 3: Differentiate both sides implicitly with respect to $t$ :
5
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \implies x \frac{dx}{dt} + y \frac{dy}{dt} = 0$
6
Step 4: Find $y$ when $x = 3$ : $y = 25 - 3^{2} = 4$ m. Substitute known values:
7
$(3)(0.8) + 4 \frac{dy}{dt} = 0$
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Step 5: Solve for $\frac{d y}{d t}$ and interpret:
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$\frac{dy}{dt} = -0.6 \text{ m/s}$
10
The negative sign indicates $y$ is decreasing, so the top slides down at 0.6 m/s.

3. Related rates for moving objects★★★★☆⏱ 15 min

Another common IB context is two objects moving perpendicular to each other, where we find the rate at which the distance between them changes. Sign conventions are especially important here.

📐 Worked Example

Car A drives south away from an intersection at 60 km/h, Car B drives east towards the intersection at 80 km/h. When Car A is 0.5 km north of the intersection and Car B is 1.2 km west, how fast is the distance between the cars changing?

1
Step 1: Define variables: $a =$ distance of A from intersection, $b =$ distance of B from intersection, $s =$ distance between cars. Given $\frac{d a}{d t} = 60$ km/h (a increasing), $\frac{d b}{d t} = - 80$ km/h (b decreasing).
2
Step 2: Relate variables:
3
$s^2 = a^2 + b^2$
4
Step 3: Differentiate with respect to $t$ :
5
$s \frac{ds}{dt} = a \frac{da}{dt} + b \frac{db}{dt}$
6
Step 4: Calculate $s = 0. 5^{2} + 1. 2^{2} = 1.3$ km, substitute values:
7
$1.3 \frac{ds}{dt} = (0.5)(60) + (1.2)(-80) = -66$
8
Step 5: Solve and interpret:
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$\frac{ds}{dt} \approx -50.8 \text{ km/h}$
10
The distance between the cars is decreasing at approximately 50.8 km/h.

✓ Quick check

Test your understanding of sign conventions:

A spherical balloon is being inflated, and the radius increases at 2 cm/s. What is the sign of $\frac{d V}{d t}$ ?
- Positive
- Negative
- Zero
Reveal answer
Positive —
Volume increases as radius increases, so the rate of change of volume with respect to time is positive.

4. Exam expectations and interpretation★★★☆☆⏱ 10 min

In IB exams, you are almost always required to include units for your final answer, and interpret the meaning of the sign of your result when asked.

5. Common Pitfalls

Wrong move:

Forgetting to apply the chain rule, and differentiating with respect to variables other than time $t$

Why:

You will not get the correct relationship between the time rates of change if you differentiate with respect to the wrong variable

Correct move:

Always differentiate both sides of the equation implicitly with respect to $t$ , using the chain rule for all variable terms

Wrong move:

Assigning the wrong sign to given rates, especially for decreasing quantities

Why:

Incorrect signs lead to wrong final values and wrong interpretations of whether the quantity is increasing or decreasing

Correct move:

Explicitly define your variables, and assign a negative rate to any quantity that decreases over time

Wrong move:

Substituting given values for the specific point in time before differentiating

Why:

A constant value will have a derivative of zero, which incorrectly removes the term from your differentiated equation

Correct move:

Differentiate the general equation first, only substitute the given values after you finish differentiating

Wrong move:

Keeping both radius and height as variables for conical problems

Why:

This requires the product rule and leads to overly complex working with more opportunities for error

Correct move:

Use the fixed proportionality of radius and height for a given cone to substitute one variable out before differentiating

Wrong move:

Forgetting to add units to the final answer

Why:

Examiners regularly deduct marks for missing units even if the numerical value is correct

Correct move:

Always add appropriate units (e.g. m/s, km/h) to your final rate answer

6. Quick Reference Cheatsheet

Step	Action
1	List given/required rates, label all variables
2	Write equation relating all quantities
3	Differentiate implicitly with respect to $t$
4	Substitute known values after differentiation
5	Solve for the unknown rate
6	Add units, interpret the sign of the result

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Conical tank draining problem
2022 · 2
Ladder sliding down wall problem
2019 · 1
Distance between moving objects

What's Next

Related rates are a core applied differentiation topic that aligns with the problem-solving focus of IB Mathematics AI HL. Mastering the process of setting up contextual calculus problems here will prepare you for extended response questions that make up a large share of your exam grade. The implicit differentiation skills you used here are also foundational for upcoming topics including parametric equations and differential equations. Next, you will move on to integration, the inverse of differentiation, which allows you to solve a wide range of additional applied problems from area calculation to kinematics.