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instantaneous rate of change

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4 detailed 50-minute lessons with teaching scripts, worked examples, parent guides, and assessment criteria.

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Lesson Overview

Total Lessons: 4
Tier: Foundation and Higher
Duration: 50 minutes per lesson (200 minutes total)
Exam Boards: AQA, Edexcel, OCR, Eduqas, CCEA

Learning Objectives

Prerequisites

Materials & Equipment

Lesson 1: Introduction: instantaneous rate of change

Duration: 50 minutes

Starter Activity (5 minutes)

Quick Recall

Write down everything you already know about instantaneous rate of change. Then check against the key terms: Higher Tier Only. Use a mini-whiteboard or paper.

Main Content (35 minutes)

Parent/Teacher Guide:
Before lesson: Read the script below. Pre-teach key vocab: Higher Tier Only.
If stuck: Re-read the revision notes (link above), then break the content into smaller steps.
Extension: See the Stretch & Challenge ideas in Lesson 4.
Teaching Script (35 mins):
Mins 0-5 - Hook: "Today: instantaneous rate of change. By the end you will be able to answer exam questions on it unaided. It connects to the rest of Mathematics because the ideas here recur across the spec."
Mins 5-20 - Direct Instruction: Work through the core ideas below one at a time; after each, ask your student to explain it back in their own words.
Mins 20-30 - Guided Practice: Model the worked example together, then let your student attempt the first practice question with guidance.
Mins 30-35 - Independent Practice: 2-3 practice questions from Lesson 3 below, with immediate feedback.
First Look

Start with the revision notes summary, then attempt: A tangent at (4, 12) on a curve passes through (2, 4) and (6, 20). Find the gradient.

Plenary (5 minutes)

Check Out

Your student states one thing they learned and one question they still have about instantaneous rate of change.

Lesson 2: Core Concepts: instantaneous rate of change

Duration: 50 minutes

Starter Activity (5 minutes)

Review Previous Lesson

Quick recap: write 3 key points from Lesson 1 on instantaneous rate of change. Check them against the notes below.

Main Content (35 minutes)

Higher Tier Only: The instantaneous rate of change at a point on a curve is the gradient of the tangent to the curve at that point.

Practice (10 minutes)

Q: A tangent at (4, 12) on a curve passes through (2, 4) and (6, 20). Find the gradient.

Answer: 4

Plenary (5 minutes)

Explain Back

Your student teaches the key points back to you without looking. Fill any gaps immediately.

Lesson 3: Application: instantaneous rate of change

Duration: 50 minutes

Starter Activity (5 minutes)

Quick Recall

Recall the key terms: Higher Tier Only. Define each in one sentence.

Main Content (35 minutes)

Parent/Teacher Guide: Let your student attempt each question alone first, then compare with the model answer. Award method marks for correct working even if the final answer is wrong.

Q1: A tangent at (4, 12) on a curve passes through (2, 4) and (6, 20). Find the gradient.

Answer: 4

Q2: On a distance-time graph, the tangent at t = 5s has gradient 8. What is the instantaneous speed?

Answer: 8 m/s

Q3: A tangent is drawn at (2, 9). If the tangent passes through (0, 1), what is the gradient?

Answer: 4

Q4: Estimate the rate of change of y = x² at x = 4 using x = 3.9 and x = 4.1.

Answer: 8 (approximately)

Q5: A velocity-time curve has a tangent at t = 3 with gradient -2. Interpret this.

Answer: Deceleration of 2 m/s² at t = 3 seconds

Plenary (5 minutes)

Error Review

Review any questions answered incorrectly. Identify whether the error was knowledge, method, or reading the question.

Lesson 4: Exam Practice: instantaneous rate of change

Duration: 50 minutes

Starter Activity (5 minutes)

Command Words

Review what these command words require: state (one point), describe (say what happens), explain (say why), compare (both sides), evaluate (judgement).

Main Content (35 minutes)

Extended Answer

Extended question: Extended Answer 6 marks: Water is poured into a cone. The volume of water V cm³ after t seconds is shown on a V-t graph that curves upwards. At t = 4, V = 30. A tangent drawn at this point passes through (2, 12) and (6, 60). (a) Find the instantaneous rate of change at t = 4. (b) The average rate of change from t = 0 to t = 4 is 7.5 cm³/s. Is the instantaneous rate at t = 4 greater or less than the average? What does this tell you about the graph? (c) Explain why the rate of change increases over time for water filling a cone. <div class="

(a) Gradient = (60−12)/(6−2) = 48/4 = 12 cm³/s. (b) Instantaneous rate (12) > average rate (7.5). This means the graph is getting steeper — the rate is increasing, so the curve is convex (bends upwards). (c) A cone is wider at the top. As water level rises, the cross-sectional area increases, so each cm of depth adds more volume. The rate of volume increase speeds up even though the water is poured at a constant rate of height increase. Mark scheme: M1 gradient calculation, A1 12 cm³/s, M1 comparison, A1 correct conclusion about shape, M1 explanation, A1 cone geometry reasoning

Exam Tips: Draw tangents carefully - they touch at one point only | Extend the tangent to make reading coordinates easier | Pick two clear points ON THE TANGENT LINE | Include units in your answer | Positive gradient = increasing, negative = decreasing
Common Errors: Watch Out! 1. Wrong: Drawing a tangent that also passes through the curve (a secant, not a tangent) Correct: A tangent touches the curve at exactly one point and has the same gradient as the curve at that point. Draw it carefully so it just grazes the curve. 2. Wrong: Using the origin and the point of interest to find the gradient Correct: Use two points ON THE TANGENT LINE, not on the curve. The line from the origin to the point gives the average rate, not the instantaneous rate. 3. Wrong: Confusing average rate of change with instantaneous rate of change Correct: Average rate = change in y / change in x between two points on the curve. Instantaneous rate = gradient of the tangent at one sp
AO3 - Reasoning & Interpretation: Reasoning and Interpretation A temperature-time graph shows coffee cooling in a room at 20°C. At t = 0, T = 90°C. At t = 5 min, T = 65°C. At t = 10 min, T = 48°C. (a) Is the rate of cooling constant? How can you tell? (b) Estimate the instantaneous rate of cooling at t = 5 minutes. (c) Will the coffee ever reach exactly 20°C? Explain. Answers: (a) No — the temperature drops 25°C in first 5 min but only 17°C in next 5 min. The rate is decreasing. The curve is concave (flattening). (b) Approximate tangent at t = 5: using (0,90) and (10,48) gives average = −4.2°C/min. A better tangent estimate might be around −4 to −5°C/min. (c) Mathematically, the temperature approaches 20°C asymptotically (ex
Stretch & Challenge (Grade 8-9):
  • Synoptic links: explain how instantaneous rate of change connects to another Mathematics topic you have studied
  • Real-world: research one real-world use or example of instantaneous rate of change
  • Critical: "What are the limitations of the models used in instantaneous rate of change?"

Plenary (5 minutes)

Assessment Criteria
  • Got it: Confident explanation + correct worked examples
  • Getting there: Main points OK, needs support with detail
  • Not yet: Confused on key concepts - re-run Lesson 2

Homework & Consolidation

Recommended Resources

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