R15: Instantaneous Rate of Change
Calculate the instantaneous rate of change by drawing or estimating a tangent
📋 Key Concepts
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.
Key Terms
- Tangent: A straight line that touches a curve at exactly one point
- Instantaneous rate of change: The rate of change at a specific instant, not averaged over time
- Estimating gradients: Draw a tangent and calculate its gradient
📝 Drawing a Tangent
Method:
- Locate the point on the curve
- Place a ruler so it touches only that point
- Adjust until the ruler follows the curve direction at that point
- Draw the tangent line extending both directions
Example 1
A curve has equation y = x². Draw the tangent at x = 2.
Solution:
Point: (2, 4) lies on the curve
Draw a straight line that just touches the curve at (2, 4)
Extend the line to cross grid lines for easy reading
📝 Finding the Gradient of a Tangent
Gradient = (change in y) ÷ (change in x)
Pick two points on the tangent line (not on the curve)
Pick two points on the tangent line (not on the curve)
Example 2
A tangent at point (3, 10) on a curve passes through (1, 2) and (5, 18). Find the gradient.
Solution:
Gradient = (18 - 2) / (5 - 1)
Gradient = 16 / 4 = 4
Instantaneous rate of change = 4
📝 Real-World Applications
Distance-Time Graphs: The tangent at any point gives the speed at that instant.
Velocity-Time Graphs: The tangent gives the acceleration at that instant.
Example 3
A distance-time graph shows a curve. At time t = 4 seconds, the tangent has gradient 15. Interpret this.
Solution:
The gradient represents speed.
At t = 4 seconds, the instantaneous speed is 15 m/s
Example 4
The depth of water in a tank is plotted against time. At t = 2 minutes, the tangent gradient is -5 cm/min. What is happening?
Solution:
Negative gradient means decreasing depth.
At t = 2 minutes, water level is falling at 5 cm per minute
📝 Estimating Without Drawing
Method: Use two points very close to the point of interest to estimate the gradient.
Example 5
Estimate the instantaneous rate of change of y = x² at x = 3 by using values at x = 2.9 and x = 3.1.
Solution:
At x = 2.9: y = 2.9² = 8.41
At x = 3.1: y = 3.1² = 9.61
Gradient ≈ (9.61 - 8.41) / (3.1 - 2.9) = 1.20 / 0.20 = 6
Estimated rate of change = 6
(Actual value is 6 - the derivative is 2x = 2(3) = 6)
🧮 Worked Examples — Exam Style
Instantaneous rate = gradient of the TANGENT at that point. Draw the tangent as long as the grid allows, then rise ÷ run on two clean grid points.
A tangent to a distance-time graph at t = 3 s passes through (1, 10) and (5, 50). Find the speed at t = 3.
Gradient = (50 − 10)/(5 − 1) = 40/4 = 10 m/s
M1 for the gradient, A1 for 10. Use points ON THE TANGENT (not the curve) — any two of them give the same gradient.
On a cooling curve, the tangent at t = 2 min has gradient −8 °C/min. Interpret.
The liquid is cooling at 8 degrees per minute at that moment — the instantaneous (not average) rate.
M1 for the magnitude with units, A1 for 'at that moment'. Negative gradient = decreasing temperature.
Exam wording to expect: "draw a tangent at t = … and find its gradient", "estimate the speed at …".
❓ Practice Questions
Q1: A tangent at (4, 12) on a curve passes through (2, 4) and (6, 20). Find the gradient.
Q2: On a distance-time graph, the tangent at t = 5s has gradient 8. What is the instantaneous speed?
Q3: A tangent is drawn at (2, 9). If the tangent passes through (0, 1), what is the gradient?
Q4: Estimate the rate of change of y = x² at x = 4 using x = 3.9 and x = 4.1.
Q5: A velocity-time curve has a tangent at t = 3 with gradient -2. Interpret this.
Q6: A distance-time curve has a tangent at t = 4 s passing through (2, 6) and (7, 26). Find the instantaneous speed
Q7: Why does the tangent method give a different answer from the chord between two distant points?
✅ Answers
- 4
- 8 m/s
- 4
- 8 (approximately)
- Deceleration of 2 m/s² at t = 3 seconds
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. (20 − 4)/(6 − 2) = 16/4 = 4
Q2. 8 m/s — the tangent gradient IS the instantaneous speed
Q3. (9 − 1)/(2 − 0) = 4
Q4. y(3.9) = 15.21, y(4.1) = 16.81 → (16.81 − 15.21)/0.2 = 8
Q5. Deceleration of 2 m/s² at t = 3 s
Q6. (26 − 6)/(7 − 2) = 20/5 = 4 m/s
Q7. A chord measures the AVERAGE rate over the interval; the tangent measures the rate AT one instant — they differ whenever the rate is changing
📌 Key Facts to Memorise
- Instantaneous rate = tangent gradient at the point
- Average rate = chord gradient (rise ÷ run between two curve points)
- Draw tangents LONG — long tangents reduce reading error
- Small-interval trick: f(x + h) − f(x − h) over 2h approximates the tangent
- Units: distance-time tangent → speed at that instant
🗺️ In the Exam
Instantaneous rates: Higher-boundary content in this folder — 3–4 marks on Higher Papers 2/3. The tangent-drawing is marked for touching the curve at the point only.
Links to: gradients/areas (A15), rates from graphs (R14), real-life graphs (A14).
🧭 Board Focus — AQA
Spec 8300. Three papers, each 80 marks and 1h30: Paper 1 non-calculator, Papers 2 and 3 calculator. Foundation/Higher tiers. Roughly a third of the marks sit under AO2 (reasoning) and AO3 (problem-solving), so show every line of working — method marks survive arithmetic slips.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 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
🧠 Problem-Solving Strategies
Problem-Solving
Draw a tangent at the point of interest to find the instantaneous rate of change. Calculate the gradient of the tangent using rise/run. The steeper the tangent, the greater the rate of change. On a distance-time graph, the tangent gradient gives instantaneous speed. On a speed-time graph, it gives instantaneous acceleration.
Multi-Step Problem
The distance (d metres) of a particle from its starting point after t seconds is given by d = t² + 3t. Find the instantaneous speed at t = 2 by drawing a tangent.
Solution: At t = 2, d = 4 + 6 = 10. Draw a tangent at (2, 10). Using the gradient: pick two points on the tangent, e.g. (1, 4) and (3, 16). Gradient = (16−4)/(3−1) = 12/2 = 6 m/s. (Check: using calculus, dd/dt = 2t + 3 = 7 at t = 2 — tangent drawing gives an approximation.)
⚠️ 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 specific point.
✍️ 6-Mark Exam 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.
(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
📊 AO3: Reason & Interpret
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 (exponential decay). It gets arbitrarily close but never exactly reaches 20°C in finite time, though practically it would reach 20°C within measurement precision.
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