TopMyGrade

GCSE/Mathematics/OCR· Higher tier

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

Notes

Instantaneous rate of change

OCR J560 Higher (J560/04–06) expects students to find the gradient of a curve at a specific point and interpret it as a real-world rate. This is the precursor to A-level differentiation.

The tangent method

The gradient at a point P on a curve is defined as the gradient of the tangent to the curve at P.

To find it from a graph:

  1. Mark the point P on the curve.
  2. Draw a straight line just touching the curve at P (and not crossing it locally) — this is the tangent.
  3. Pick two clear points on the tangent (not on the curve!).
  4. Compute (Δy)/(Δx) between those two points.

The longer the tangent line you draw, the more accurate the gradient reading.

Why it matters

On a distance–time curve, the gradient at a point gives the instantaneous speed at that moment. On a speed–time curve, the gradient gives the acceleration at that instant.

Worked example

A distance–time curve passes through (3, 18) where the tangent at this point has been drawn. The tangent passes through (1, 8) and (5, 28).

Gradient = (28 − 8) / (5 − 1) = 20/4 = 5 m/s.

So at t = 3 s the object is travelling at 5 m/s.

Estimation note

Reading a gradient from a hand-drawn tangent has uncertainty. OCR mark schemes typically allow a ±10% tolerance and award A1 for any answer within range, with M1 for a sensible tangent and reading.

Average vs instantaneous rate

  • Average rate of change between t = a and t = b: chord gradient = (f(b) − f(a)) / (b − a).
  • Instantaneous rate of change at t = a: tangent gradient at t = a.

These differ unless the curve is straight.

OCR mark scheme conventions

  • M1 for drawing a sensible tangent at the correct point.
  • M1 for picking two readable points on the tangent and computing (Δy)/(Δx).
  • A1 for the numerical gradient (within tolerance).
  • B1 for stating the rate with units and meaning ("acceleration is 3 m/s² at t = 4 s").

Common mistakes

  1. Picking points on the curve instead of on the tangent.
  2. Drawing too short a tangent — small Δx amplifies error.
  3. Ignoring units when stating the rate.
  4. Confusing average with instantaneous gradient.

AI-generated · claude-opus-4-7 · v3-ocr-maths-leaves

Practice questions

Try each before peeking at the worked solution.

  1. Question 14 marks

    Gradient of a tangent

    OCR J560/05 — Higher (calculator)

    A tangent has been drawn to a distance–time curve at the point (4, 12) where time is in seconds and distance in metres. The tangent is a straight line that passes through (2, 4) and (6, 20).

    (a) Calculate the gradient of the tangent. [2]
    (b) Interpret this gradient in context. [2]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-ocr-maths-leaves

  2. Question 24 marks

    Average vs instantaneous

    OCR J560/04 — Higher (non-calculator)

    A speed–time graph is a curve where v(0) = 0, v(2) = 6, v(4) = 16. A tangent drawn at t = 2 passes through (0, −2) and (4, 14).

    (a) Find the average acceleration between t = 0 and t = 4. [2]
    (b) Find the instantaneous acceleration at t = 2. [2]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-ocr-maths-leaves

  3. Question 34 marks

    Cooling curve interpretation

    OCR J560/06 — Higher (calculator)

    A cooling curve plots temperature T (°C) against time t (minutes). At t = 5, a tangent has been drawn that passes through (0, 70) and (10, 30).

    (a) Calculate the gradient. [2]
    (b) Interpret the gradient with units. [2]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-ocr-maths-leaves

Flashcards

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

7-card SR deck for OCR GCSE Mathematics J560 (leaf top-up — batch 4) topic R15

7 cards · spaced repetition (SM-2)