# Rates of change and connected rates of change

> Edexcel International GCSE Further Pure Mathematics · 4PM1
> Source: https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s9-rates-of-change-and-connected/

This guide teaches core rates of change skills for Edexcel IGCSE Further Pure Maths 4PM1, including connected rates using the chain rule and small change approximation, with exam-aligned worked problems.

**Prerequisites:** [Differentiation using power, chain, product and quotient rules](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s9-basic-differentiation/); [Mensuration formulae for 3D shapes](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s9-kinematics-volumes/)

## Learning objectives

- Calculate rates of change of quantities using differentiation
- Apply the chain rule to solve connected rates of change problems
- Use the small change approximation δy ≈ (dy/dx)δx for small δx
- Interpret units and context correctly for all rates problems

## 1. Basic Rates of Change

The rate of change of a quantity \(y\) with respect to a second quantity \(x\) is given by the first derivative \(\frac{dy}{dx}\). When the independent variable is time \(t\), this represents how fast \(y\) increases or decreases over time, with units matching \(y\) per unit time.

**Rate of Change** — The derivative of a dependent variable with respect to an independent variable, measuring the speed of change of the dependent variable. A positive value indicates an increasing quantity, negative indicates decreasing.

*Example:* If \(V\) is the volume of a balloon in cm³ and \(t\) is time in seconds, \(\frac{dV}{dt}\) has units cm³/s, measuring how fast the balloon inflates or deflates.

**Worked example:** The surface area of a cube is given by \(A = 6x^2\), where \(x\) is the side length in cm. Find the rate of change of surface area with respect to side length when \(x = 3\) cm.

1. Step 1: Differentiate the area function with respect to \(x\)
2. $$\frac{dA}{dx} = \frac{d}{dx}(6x^2) = 12x$$
3. Step 2: Substitute \(x = 3\) cm into the derivative
4. $$\frac{dA}{dx} = 12 \times 3 = 36 \text{ cm}^2/\text{cm}$$
5. Final answer: The rate of change of surface area is 36 cm² per cm of side length increase.

> **Exam tip:** Always include units in your final answer for rates questions; marks are explicitly awarded for correct units as well as correct numerical values.

## 2. Connected Rates of Change

When two quantities are related by a known formula, their rates of change can be linked using the chain rule. This is the core of connected rates problems, where you will be given one rate and asked to find a second related rate.

**Connected Rates of Change** — Rates of change of two or more related quantities, linked via the chain rule to allow calculation of an unknown rate from a known rate.

*Example:* For an expanding sphere, the rate of change of volume \(\frac{dV}{dt}\) can be linked to the rate of change of radius \(\frac{dr}{dt}\) via the chain rule: \(\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}\).

**Worked example:** A spherical balloon is inflated so that its radius increases at a constant rate of 2 cm/s. Find the rate of increase of the volume of the balloon when the radius is 5 cm. (Volume of a sphere: \(V = \frac{4}{3}\pi r^3\), given on the formula sheet.)

1. Step 1: List known and unknown values
2. $$\text{Known: } \frac{dr}{dt} = 2 \text{ cm/s}, r = 5 \text{ cm} \quad \text{Unknown: } \frac{dV}{dt}$$
3. Step 2: Write the chain rule relationship linking the two rates
4. $$\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$$
5. Step 3: Differentiate the volume formula with respect to \(r\)
6. $$\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2$$
7. Step 4: Substitute all known values into the chain rule
8. $$\frac{dV}{dt} = 4\pi (5)^2 \times 2 = 200\pi \text{ cm}^3/\text{s}$$
9. Final answer: The rate of increase of volume is \(200\pi\) cm³/s, or 628 cm³/s to 3 significant figures if a numerical value is requested.

> **tip**
>
> If you are given a decreasing quantity (e.g. water leaking from a tank), the rate will be negative. Carry the negative sign through your calculation to avoid sign errors in your final answer.

*Calculator:* allowed

## 3. Small Change Approximation

For very small changes in an independent variable \(x\), we can approximate the corresponding change in the dependent variable \(y\) using the gradient of the function at the point of interest. This approximation is widely used in estimation problems on the 4PM1 exam.

**Small Change Approximation** — An approximate formula for the change in \(y\) (\(\delta y\)) resulting from a small change in \(x\) (\(\delta x\)). The approximation becomes more accurate as \(\delta x\) gets smaller.

*Notation:* \(\delta y \approx \frac{dy}{dx} \delta x\)

*Example:* If \(y = x^2\), a small increase of \(\delta x = 0.01\) at \(x = 10\) gives \(\delta y \approx 2(10)(0.01) = 0.2\), very close to the exact change of \(10.01^2 - 10^2 = 0.2001\).

**Worked example:** The volume of a cube is given by \(V = x^3\), where \(x\) is the side length. Use the small change approximation to estimate the change in volume when the side length increases from 8 cm to 8.02 cm.

1. Step 1: Identify values for \(x\) and \(\delta x\)
2. $$x = 8 \text{ cm}, \delta x = 8.02 - 8 = 0.02 \text{ cm}$$
3. Step 2: Differentiate the volume function with respect to \(x\)
4. $$\frac{dV}{dx} = 3x^2$$
5. Step 3: Substitute into the small change approximation formula
6. $$\delta V \approx \frac{dV}{dx} \delta x = 3(8)^2 \times 0.02 = 3.84 \text{ cm}^3$$
7. Final answer: The approximate increase in volume is 3.84 cm³.

**Check your understanding**

1. When is the small change approximation most accurate?

   - When \(\delta x\) is very large
   - When \(\delta x\) is very small
   - When \(\frac{dy}{dx}\) is zero
   - When \(y\) is a linear function

   *Answer:* When \(\delta x\) is very small

   *Why:* Correct! The approximation relies on the curve being almost straight over very small intervals of \(x\).

## 4. Solving Mixed Exam-Style Rates Problems

Most exam questions will combine context, unit conversion, and either connected rates or small change approximation. Follow a consistent step-by-step approach to avoid mistakes.

1. Read the question carefully to identify known values, unknown values, and the relationship between quantities.
2. Write down any relevant formulae (use the formula sheet for mensuration values where allowed).
3. Set up the chain rule (for connected rates) or small change approximation formula explicitly.
4. Substitute values, paying close attention to units and the sign of rates (negative for decreasing quantities).
5. Check your final answer has the correct units and makes sense in the context of the problem.

**Exam command terms**

Common command terms for rates questions on 4PM1 exams:

- **Find the rate of change** — Calculate the derivative of the relevant quantity with respect to the given independent variable, include units. *(Find the rate of change of volume when r=3cm.)*

- **Use the small change approximation** — You must use the \(\delta y \approx \frac{dy}{dx}\delta x\) formula, not calculate the exact change. No marks are awarded for exact calculation here.

## Common pitfalls

- **Wrong:** Forgetting to include the chain rule link in connected rates problems, e.g. writing \(\frac{dV}{dt} = 4\pi r^2\) without multiplying by \(\frac{dr}{dt}\).
  - Why it fails: You are directly differentiating with respect to time instead of linking via the intermediate variable.
  - Correct: Always write out the full chain rule relationship explicitly before substituting values.
- **Wrong:** Using a positive rate for a decreasing quantity (e.g. water leaking from a tank).
  - Why it fails: The sign of the rate indicates direction of change, so incorrect signs lead to wrong final answers.
  - Correct: Assign negative values to rates of decreasing quantities, and check your final answer sign matches the context.
- **Wrong:** Calculating the exact change instead of using the small change approximation when instructed.
  - Why it fails: The question specifically tests your knowledge of the approximation formula, so exact calculations get zero marks.
  - Correct: Use only the \(\delta y \approx \frac{dy}{dx}\delta x\) formula when the question explicitly asks for the small change approximation.
- **Wrong:** Forgetting to include units in the final answer.
  - Why it fails: Exam marks are explicitly awarded for correct units for rates questions.
  - Correct: Always check units match the quantity being measured (e.g. cm³/s for volume rate of change).
- **Wrong:** Mixing up the order of variables in the chain rule, e.g. writing \(\frac{dV}{dt} = \frac{dr}{dV} \times \frac{dr}{dt}\).
  - Why it fails: The chain rule requires the intermediate variable to cancel out (e.g. \(dr\) in numerator and denominator).
  - Correct: Write the chain rule so the intermediate variable appears once in the numerator and once in the denominator to cancel correctly.

## Cheatsheet

| Concept | Formula | Key Notes |
| --- | --- | --- |
| Basic Rate of Change | \(\frac{dy}{dx}\) | Units = (units of y) / (units of x) |
| Connected Rates (Chain Rule) | \(\frac{dy}{dt} = \frac{dy}{du} \times \frac{du}{dt}\) | Intermediate variable \(u\) cancels out |
| Small Change Approximation | \(\delta y \approx \frac{dy}{dx} \delta x\) | Only valid for very small \(\delta x\) |
| Sign Convention | Negative = decreasing, Positive = increasing | Check sign matches problem context |

## What's next

Now that you have mastered rates of change and connected rates for Edexcel IGCSE Further Pure Maths 4PM1, you are ready to move to more advanced calculus topics. These skills are foundational for all subsequent calculus work, including integration applications and kinematics problems. Make sure to practice past paper questions to get comfortable with the context and problem structures commonly tested, and always double-check your chain rule setup and units to avoid easy mark losses.

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