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Average and instantaneous rates of change

3 exam-style questions. Answers and the mark for each step are on the last page.

  1. Question 1Calculator · 2 marks

    (a) A curved graph shows the volume of water V litres in a tank against time t seconds. At t = 6, a drawn tangent passes through (2, 7) and (10, 31). Use this tangent to estimate the rate at which the volume is increasing at t = 6. (2)

  2. Question 2Non-calculator · 3 marks

    (a) Volume V litres is modelled by V = 4t24t^{2} + 3, where t is time in seconds. The tangent at t = 3.75 has gradient 30 litres per second. By what percentage is this instantaneous rate greater than the average rate from t = 1 to t = 5? You must show your working. (3)

  3. Question 3Non-calculator · 5 marks

    A ball is thrown upwards. Its height, hh metres, after tt seconds is h=20t−5t2h = 20t - 5t^2.

    (a) Work out the average rate of change of height between t=1t = 1 and t=3t = 3. (2)

    (b) Interpret your answer to part (a). (1)

    (c) Use a chord from t=2t = 2 to t=2.1t = 2.1 to estimate the rate of change of height at t=2t = 2. (2)

Answers and marks

Question 1

(a) 33 litres/second

  • P1 Establishing 31−710−2\frac{31-7}{10-2} or an equivalent valid method.
  • A1 Correct answer: 33 litres/second

Question 2

(a) 2525%

  • P1 Establishing 103−75−1\frac{103-7}{5-1} or an equivalent valid method.
  • P1 Establishing (30−24)/24×100(30-24)/24\times 100 or an equivalent valid method.
  • A1 Correct answer: 2525%

Question 3

(a) 00 m/s

  • M1 Finding h(1)=15h(1) = 15 and h(3)=15h(3) = 15.
  • A1 The correct answer, 00.

(b) The ball is at the same height at t = 1 and t = 3: it rises and then falls back.

  • C1 Saying the ball is at the same height at both times (up then down), so the average rate is zero although it moves.

(c) −0.5-0.5 m/s

  • P1 Finding h(2.1)=19.95h(2.1) = 19.95.
  • A1 The correct answer, −0.5-0.5.

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Average and instantaneous rates of change

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Independent practice for Pearson Edexcel GCSE Mathematics (1MA1), not endorsed by Pearson.

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