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GCSE/Mathematics/AQA· Higher tier

R15Interpret gradient at a point on a curve as instantaneous rate of change

Notes

Instantaneous rate of change: gradient at a point

For a straight line, the gradient is constant. For a curve, the gradient varies — at each point it tells you the instantaneous rate of change at that moment.

What "instantaneous" means

On a distance–time graph that's a curve, the average speed over an interval is (Δd)/(Δt) — but the speed at a single instant requires the gradient at the point, not over an interval.

Drawing a tangent

The gradient at a point on a curve = gradient of the tangent line at that point. To estimate it manually:

  1. Draw a tangent line that touches the curve at the point and just kisses it (doesn't cross).
  2. Pick two points on the tangent (not the curve) far apart for accuracy.
  3. Compute (Δy)/(Δx).

Worked example: at point P on a distance–time curve, you draw a tangent passing through (3, 5) and (7, 21). Speed at P = (21 − 5)/(7 − 3) = 16/4 = 4 m/s (or whatever the units are).

Sketch tips

  • Use a ruler to draw the tangent.
  • Pick points where coordinates are easy to read.
  • Check the tangent doesn't cross the curve nearby.

Average vs instantaneous rate

Method
Average rateChord between two points: (y₂ − y₁)/(x₂ − x₁)
Instantaneous rateTangent at a single point — gradient of the tangent

Worked example: a velocity–time curve at t = 5 s has tangent through (4, 8) and (8, 20). Acceleration at t = 5 s = (20 − 8)/(8 − 4) = 3 m/s².

Common mistakes

  1. Treating the curve like a chord — using (y₂ − y₁)/(x₂ − x₁) on the curve gives the average rate, not instantaneous.
  2. Drawing a chord, not a tangent — tangent touches, chord cuts.
  3. Reading tangent points from the curve instead of from the tangent line.
  4. Choosing tangent points too close — small denominator inflates errors.
  5. Forgetting units — m/s for speed; m/s² for acceleration; £ per unit for unit cost.

Try thisQuick check

A tangent at point Q is drawn passing through (1, 10) and (5, 2). Gradient at Q = (2 − 10)/(5 − 1) = −2.

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Practice questions

Try each before peeking at the worked solution.

  1. Question 12 marks

    Tangent reading

    (H1) A tangent at point P passes through (2, 5) and (6, 17). Find the gradient at P.

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  2. Question 22 marks

    Speed from distance-time curve

    (H2) A distance-time curve has a tangent at t = 4 s passing through (3, 6) and (5, 14) where d is in metres. Find the instantaneous speed at t = 4 s.

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  3. Question 32 marks

    Acceleration from v-t curve

    (H3) A velocity-time curve has tangent at t = 6 s passing through (4, 7) and (10, 25) where v is m/s. Find the acceleration at t = 6 s.

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  4. Question 44 marks

    Average vs instantaneous

    (H4) Distance-time curve passes through (0, 0), (2, 10) and (5, 30). The tangent at t = 2 passes through (1, 4) and (3, 14). (a) Find the average speed over 0 ≤ t ≤ 5. (b) Find the instantaneous speed at t = 2.

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  5. Question 52 marks

    Negative tangent gradient

    (H5) A graph of remaining fuel against distance has a tangent at point R passing through (50, 30) and (150, 20) where x is in km and y in litres. Find the gradient and interpret it.

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  6. Question 62 marks

    Estimate instantaneous rate

    (H6) A population P (in thousands) is plotted against time t (in years). A tangent at t = 8 passes through (6, 50) and (10, 90). Find the rate of population change at t = 8.

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  7. Question 73 marks

    Spotting maxima/minima

    (H7) On a curve, a tangent is horizontal at point M. (a) State the gradient. (b) State what this implies about the rate of change at M. (c) For a velocity-time graph, what physical situation does this represent?

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Flashcards

R15 — Interpret gradient at a point on a curve as instantaneous rate of change

12-card SR deck for AQA GCSE Maths topic R15

12 cards · spaced repetition (SM-2)