# Related rates

> IB Mathematics Applications & Interpretation HL · Calculus
> Source: https://www.owlsprep.com/study/ib-math-ai-hl-u4-related-rates/

Related rates use the chain rule to relate the rates of change of multiple quantities changing over time. This subtopic covers common applied contexts and problem-solving methods tested in IB HL exams.

**Prerequisites:** [Basic differentiation rules](https://www.owlsprep.com/study/ib-math-ai-hl-u3-basic-differentiation/)

## Learning objectives

- Identify and set up related rates problems from contextual scenarios
- Differentiate related quantities with respect to time using the chain rule to relate rates of change
- Solve applied problems involving related rates for common IB contexts
- Interpret results in the context of the original problem

## Setting up related rates problems

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.

**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

1. List all given quantities and the quantity whose rate you need to find, labeling rates as derivatives with respect to time.
2. Write an equation that relates the quantities of interest.
3. Differentiate both sides of the equation with respect to time $t$.
4. Substitute the known values and solve for the unknown rate.
5. 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.

## Common geometric related rates problems

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 $\frac{dx}{dt} = 0.8$ m/s, find $\frac{dy}{dt}$ 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 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 $x=3$: $y = \sqrt{25 - 3^2} = 4$ m. Substitute known values:
7. $$(3)(0.8) + 4 \frac{dy}{dt} = 0$$
8. Step 5: Solve for $\frac{dy}{dt}$ and interpret:
9. $$\frac{dy}{dt} = -0.6 \text{ m/s}$$
10. The negative sign indicates $y$ is decreasing, so the top slides down at 0.6 m/s.

> **tip**
>
> For conical volumes, use the fixed ratio of radius to height (from the constant vertex angle) to substitute one variable out before differentiating, simplifying your working.

## Related rates for moving objects

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 $\frac{da}{dt} = 60$ km/h (a increasing), $\frac{db}{dt} = -80$ 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 $s = \sqrt{0.5^2 + 1.2^2} = 1.3$ km, substitute values:
7. $$1.3 \frac{ds}{dt} = (0.5)(60) + (1.2)(-80) = -66$$
8. Step 5: Solve and interpret:
9. $$\frac{ds}{dt} \approx -50.8 \text{ km/h}$$
10. The distance between the cars is decreasing at approximately 50.8 km/h.

**Check your understanding**

Test your understanding of sign conventions:

1. A spherical balloon is being inflated, and the radius increases at 2 cm/s. What is the sign of $\frac{dV}{dt}$?

   - Positive
   - Negative
   - Zero

   *Why:* Volume increases as radius increases, so the rate of change of volume with respect to time is positive.

## Exam expectations and interpretation

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.

**Exam command terms**

Common command terms for related rates questions:

- **Find the rate** — Calculate the numerical value of the unknown rate, include correct units *(No extra interpretation required beyond units)*

- **State whether the quantity is increasing/decreasing** — Use the sign of your derivative to answer this question *(Negative rate = decreasing, positive rate = increasing)*

> **info**
>
> Always round your final answer to 3 significant figures, which is the standard requirement for IB Mathematics exams unless stated otherwise.

## Common pitfalls

- **Wrong:** Forgetting to apply the chain rule, and differentiating with respect to variables other than time $t$
  - Why it fails: You will not get the correct relationship between the time rates of change if you differentiate with respect to the wrong variable
  - Correct: Always differentiate both sides of the equation with respect to $t$, using the chain rule for all variable terms
- **Wrong:** Assigning the wrong sign to given rates, especially for decreasing quantities
  - Why it fails: Incorrect signs lead to wrong final values and wrong interpretations of whether the quantity is increasing or decreasing
  - Correct: Explicitly define your variables, and assign a negative rate to any quantity that decreases over time
- **Wrong:** Substituting given values for the specific point in time before differentiating
  - Why it fails: A constant value will have a derivative of zero, which incorrectly removes the term from your differentiated equation
  - Correct: Differentiate the general equation first, only substitute the given values after you finish differentiating
- **Wrong:** Keeping both radius and height as variables for conical problems
  - Why it fails: This requires the product rule and leads to overly complex working with more opportunities for error
  - Correct: Use the fixed proportionality of radius and height for a given cone to substitute one variable out before differentiating
- **Wrong:** Forgetting to add units to the final answer
  - Why it fails: Examiners regularly deduct marks for missing units even if the numerical value is correct
  - Correct: Always add appropriate units (e.g. m/s, km/h) to your final rate answer

## Cheatsheet

| Step | Action |
| --- | --- |
| 1 | List given/required rates, label all variables |
| 2 | Write equation relating all quantities |
| 3 | Differentiate 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 |

## 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 chain-rule 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.

- [Antiderivatives and indefinite integration](https://www.owlsprep.com/study/ib-math-ai-hl-u4-antiderivatives-and-indefinite-integration/)
- [Definite integrals and area under curves](https://www.owlsprep.com/study/ib-math-ai-hl-u4-definite-integrals-and-area-under/)
- [Integration by substitution](https://www.owlsprep.com/study/ib-math-ai-hl-u4-integration-techniques-substitution-and-by/)

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