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Exam questions · Maths · Functions, Sequences and Rates of Change

Rates of Change and Areas Under Graphs

  • 6 exam questions
  • 20 marks
  • 9 quick checks
  1. 1 Work out [3 marks]

    The diagram shows the graph of \(y = x^2\) and the tangent to the curve at the point \((2, 4)\). Work out the gradient of the curve at the point \((2, 4)\). (3 marks)

    A curve with a tangent drawn at a marked point, on a grid.
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    Model answer

    The tangent passes through \((1, 0)\) and \((3, 8)\). Gradient \(= \dfrac{8 - 0}{3 - 1} = 4\).

    Mark scheme

    • Two points read from the tangent, such as \((1, 0)\) and \((3, 8)\) — M1
    • \(\dfrac{8 - 0}{3 - 1}\) — M1
    • \(4\) — A1
  2. 2 Work out [3 marks]

    The graph shows the distance, \(s\) metres, travelled by a runner after \(t\) seconds. The line is the tangent to the curve at \(t = 4\). Work out the speed of the runner at \(t = 4\). (3 marks)

    A distance-time graph with a tangent drawn at a marked point.
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    Model answer

    The tangent passes through \((2, 0)\) and \((6, 32)\). Gradient \(= \dfrac{32}{4} = 8\), so the speed is 8 m/s.

    Mark scheme

    • Two points read from the tangent, such as \((2, 0)\) and \((6, 32)\) — M1
    • \(\dfrac{32 - 0}{6 - 2}\) — M1
    • \(8\) m/s — A1
  3. 3 Work out [2 marks]

    The distance, \(s\) metres, travelled by a car after \(t\) seconds is given by \(s = t^2\). Work out the average speed of the car between \(t = 1\) and \(t = 4\). (2 marks)

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    Model answer

    \(s = 1\) when \(t = 1\) and \(s = 16\) when \(t = 4\). Average speed \(= \dfrac{16 - 1}{4 - 1} = 5\) m/s.

    Mark scheme

    • \(\dfrac{16 - 1}{4 - 1}\) — M1
    • \(5\) m/s — A1
  4. 4 Work out [5 marks]

    The graph shows the velocity, \(v\) m/s, of a particle at time \(t\) seconds. (a) Use 4 strips of equal width to estimate the distance travelled between \(t = 0\) and \(t = 8\). (3 marks) (b) Is your answer an underestimate or an overestimate? Give a reason for your answer. (2 marks)

    A velocity-time curve with vertical lines at equal intervals, used to estimate the area under it.
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    Model answer

    (a) The heights are 0, 12, 16, 12 and 0. Area \(= \dfrac{1}{2} \times 2 \times (0 + 12) + \dfrac{1}{2} \times 2 \times (12 + 16) + \dfrac{1}{2} \times 2 \times (16 + 12) + \dfrac{1}{2} \times 2 \times (12 + 0) = 12 + 28 + 28 + 12 = 80\) m. (b) An underestimate, because the curve bends downwards and the straight tops of the trapezia are below the curve.

    Mark scheme

    • (a) Reads the heights 12, 16 and 12 — B1
    • (a) Uses \(\dfrac{1}{2}(a + b)h\) for each strip — M1
    • (a) \(80\) — A1
    • (b) Underestimate — B1
    • (b) The curve is above the straight tops of the trapezia — C1
  5. 5 Work out [3 marks]

    The graph shows the velocity, \(v\) m/s, of a car at time \(t\) seconds. The line is the tangent to the curve at \(t = 4\). Work out an estimate of the acceleration of the car at \(t = 4\). (3 marks)

    A velocity-time graph with a tangent drawn at a marked point.
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    Model answer

    The tangent passes through \((2, 0)\) and \((6, 16)\). Gradient \(= \dfrac{16}{4} = 4\), so the acceleration is 4 m/s\(^2\).

    Mark scheme

    • Two points read from the tangent, such as \((2, 0)\) and \((6, 16)\) — M1
    • \(\dfrac{16 - 0}{6 - 2}\) — M1
    • \(4\) m/s\(^2\) — A1
  6. 6 Work out [4 marks]

    The velocity of a particle was measured every second. \(t\) (s): 0, 1, 2, 3, 4 \(v\) (m/s): 0, 3, 8, 15, 24 Use trapezia to estimate the distance travelled in the first 4 seconds. State whether your answer is an underestimate or an overestimate. Give a reason. (4 marks)

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    Model answer

    Area \(= \dfrac{1}{2}(0 + 3) + \dfrac{1}{2}(3 + 8) + \dfrac{1}{2}(8 + 15) + \dfrac{1}{2}(15 + 24) = 1.5 + 5.5 + 11.5 + 19.5 = 38\) m. It is an overestimate, because the velocity curve bends upwards, so the straight tops of the trapezia are above the curve.

    Mark scheme

    • \(\dfrac{1}{2}(0 + 3) + \dfrac{1}{2}(3 + 8) + \ldots\), with strips of width 1 — M1
    • \(38\) — A1
    • Overestimate — B1
    • The curve bends upwards, so the straight tops are above the curve — C1

Quick check

  1. 1

    What do you draw to find the gradient of a curve at a point?

    1. AA tangent
    2. BA chord
    3. CA normal
    4. DA diameter
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    A: A tangent

    A tangent touches the curve at that point.

  2. 2

    What is a chord?

    1. AA line that touches a curve at one point
    2. BA line through the origin
    3. CThe highest point of a curve
    4. DA straight line joining two points on a curve
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    D: A straight line joining two points on a curve

    The chord joins two points on the curve.

  3. 3

    What does the gradient of a distance-time graph represent?

    1. AAcceleration
    2. BDistance
    3. CSpeed
    4. DTime
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    C: Speed

    Distance divided by time is speed.

  4. 4

    What does the area under a velocity-time graph represent?

    1. ASpeed
    2. BDistance travelled
    3. CAcceleration
    4. DTime
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    B: Distance travelled

    Velocity multiplied by time is distance.

  5. 5

    A tangent passes through \((1, 0)\) and \((3, 8)\). What is its gradient?

    1. A4
    2. B8
    3. C2
    4. D\(\dfrac{1}{4}\)
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    A: 4

    \(\dfrac{8 - 0}{3 - 1} = 4\).

  6. 6

    What is the average rate of change of \(y = x^2\) between \(x = 1\) and \(x = 4\)?

    1. A3
    2. B15
    3. C4
    4. D5
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    D: 5

    \(\dfrac{16 - 1}{4 - 1} = 5\).

  7. 7

    What is the area of a trapezium with parallel sides 4 and 6 and width 2?

    1. A20
    2. B5
    3. C10
    4. D24
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    C: 10

    \(\dfrac{1}{2}(4 + 6) \times 2 = 10\).

  8. 8

    A velocity-time curve bends downwards. Is a trapezium estimate of the area an over- or underestimate?

    1. AAn overestimate
    2. BAn underestimate
    3. CIt is exact
    4. DIt depends on the width
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    B: An underestimate

    The straight tops lie below the curve.

  9. 9

    A velocity-time graph has heights 0, 8, 8, 0 at times 0, 2, 4, 6. What is the trapezium estimate of the distance?

    1. A32
    2. B16
    3. C48
    4. D24
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    A: 32

    \(8 + 16 + 8 = 32\).