Calculate average change and estimate a tangent slope
Calculate an average rate on a stated interval · Estimate a local rate with a tangent drawn to a curve · Interpret units and distinguish secant from tangent
Section Paper 2 · Historical Extended · non-calculator
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Gradient from two points
- Function graphs and units
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Calculate an average rate on a stated interval
- Estimate a local rate with a tangent drawn to a curve
- Interpret units and distinguish secant from tangent
Average rate joins two curve points
For distinct inputs a and b, average rate is (f(b) − f(a))/(b − a). It is the gradient of the secant joining the two curve points. The denominator is the input change, not the ending input.
A distance in metres plotted against time in seconds gives a rate in metres per second. For nonlinear graphs this average need not equal the rate at either endpoint.
A tangent estimates change at one point
Draw a straight line touching the curve in its local direction at the chosen point. Pick two widely separated, readable points on that tangent, then calculate rise/run. The chosen points usually are not on the curve itself.
A graph-based tangent is an estimate sensitive to scale and drawing. This 0444 method does not require analytic differentiation. A secant over a long interval answers an average-rate question, not an instantaneous-rate question.
Sources for this lesson
Pause and explain
Which line gives the average rate between two inputs?
Worked example
For s(t) = t² metres, find the average rate from t = 0 to 4 seconds. A tangent at t = 3 is drawn through (2, 3) and (4, 15); estimate its gradient.
Show the worked solution
- Average rate = (16 − 0)/(4 − 0) = 4 m/s.
- The drawn tangent slope is (15 − 3)/(4 − 2) = 6 m/s.
- The tangent points are on the drawn line, not both on s = t².
Answer Average 4 m/s; tangent estimate at t = 3 is 6 m/s.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
f(2) = 5 and f(6) = 17. Find the average rate from 2 to 6.
Give me a hint
Use both changes.
Compare my reasoning
- Output change is 12; input change is 4.
- 12/4 = 3.
3 output units per input unit.
Look for these in your work
- Subtract corresponding values
- Include rate units
A tangent to a graph is drawn through (1, 2) and (5, 10). Estimate its slope. Must these two points lie on the curve?
Give me a hint
Calculate the slope of the drawn straight line.
Compare my reasoning
- Slope = (10 − 2)/(5 − 1) = 2.
- Tangent points used for rise/run need not be curve points, apart from its contact point.
Estimated slope 2; the two chosen line points need not lie on the curve.
Look for these in your work
- Use tangent rather than curve points
- Label a graph reading as an estimate
A tank has 10 litres at minute 2 and 34 litres at minute 8. At minute 5 a drawn tangent passes through (3, 12) and (7, 36). Compare average and estimated local filling rates.
Give me a hint
Compute a secant and a tangent separately.
Compare my reasoning
- Average on [2, 8] = (34 − 10)/6 = 4 L/min.
- Tangent at minute 5 = (36 − 12)/4 = 6 L/min.
- The local estimate can exceed the interval average if the rate varies.
Average 4 L/min; local tangent estimate 6 L/min at minute 5.
Look for these in your work
- State each interval or instant
- Do not claim a constant filling rate
Common mistakes
- Do not divide by b unless a = 0.
- Do not use arbitrary curve points as if they were tangent points.
What can you explain now?
Which line gives the average rate between two inputs?
Compare with the explanation
The secant joining their two curve points.
Its rise/run equals the output change divided by the input change.
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Make it stick
Which line gives the average rate between two inputs?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.