Hard GCSE Instantaneous rate of change Questions

Challenging, exam-style GCSE Instantaneous rate of change questions with worked solutions. Stretch yourself on the hardest average vs instantaneous rate, chord vs tangent, limiting process, chords approaching a tangent problems.

average vs instantaneous ratechord vs tangentlimiting processchords approaching a tangentinstantaneous speedpercentage error
GCSE Higher34 questionsStep-by-step solutions
Question 1
5 markschallenging
The chord joining the points on the curve y=x2y = x^2 where x=2x = 2 and x=2+hx = 2 + h has gradient 4.54.5. Work out the value of hh.
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Worked solution

  1. Plan: build the chord gradient in terms of h

    gradient=y2−y1x2−x1\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1}

    The chord runs from x=2x = 2 to x=2+hx = 2 + h, so the run is hh.

  2. Find the y-value at the fixed end

    x=2:y=22=4x = 2: \quad y = 2^2 = 4

    One end of the chord is (2, 4)(2,\ 4).

  3. Write the y-value at the moving end

    x=2+h:y=(2+h)2x = 2 + h: \quad y = (2 + h)^2

    Substitute x=2+hx = 2 + h into y=x2y = x^2.

  4. Expand the bracket

    (2+h)2=4+4h+h2(2 + h)^2 = 4 + 4h + h^2

    Expanding lets the 44 cancel in the next step.

  5. Work out the rise

    rise=(4+4h+h2)−4=4h+h2\text{rise} = (4 + 4h + h^2) - 4 = 4h + h^2

    The constant terms cancel.

  6. Work out the run

    run=(2+h)−2=h\text{run} = (2 + h) - 2 = h

    The run is exactly hh.

  7. Write the chord gradient

    gradient=4h+h2h=4+h\text{gradient} = \frac{4h + h^2}{h} = 4 + h

    Every hh in the numerator cancels with the hh in the denominator (allowed because h≠0h \ne 0).

  8. Form the equation

    4+h=4.54 + h = 4.5

    The chord gradient is given as 4.54.5.

  9. Solve for h

    h=0.5h = 0.5

    Subtract 44 from both sides.

  10. Check the answer

    6.25−40.5=4.5\frac{6.25 - 4}{0.5} = 4.5

    The chord from x=2x = 2 to x=2.5x = 2.5 does have gradient 4.54.5.

  11. Look at the formula again

    gradient=4+h\text{gradient} = 4 + h

    Every chord starting at x=2x = 2 has gradient 4+h4 + h, where hh is the width of the interval.

  12. Shrink the interval

    h=0.1⇒4.1,h=0.01⇒4.01h = 0.1 \Rightarrow 4.1, \qquad h = 0.01 \Rightarrow 4.01

    As hh gets smaller the chord gradient gets closer and closer to 44.

  13. Take the limit

    h→0:4+h→4h \to 0: \quad 4 + h \to 4

    The chord gradients close in on 44, so the instantaneous rate of change at x=2x = 2 is 44.

  14. Link this to the tangent

    tangent gradient at x=2=4\text{tangent gradient at } x = 2 = 4

    A correctly drawn tangent at x=2x = 2 would have gradient 44 — the chords rotate onto the tangent as h→0h \to 0. The chord gradient 4+h4 + h is always bigger than 44, so every such chord over-estimates the instantaneous rate.

  15. State the answer

    h=0.5h = 0.5

    So h=0.5h = 0.5.

Answer
h=0.5h = 0.5
Question 2
6 markschallenging
A tank drains so that the volume of water in it, VV litres, after tt seconds is modelled by the curve V=200−2t2V = 200 - 2t^2. Work out the average rate of change of the volume from t=5t = 5 to t=5+ht = 5 + h for h=1h = 1, h=0.5h = 0.5 and h=0.1h = 0.1. Use your answers to write down the instantaneous rate of change of the volume at t=5t = 5. Give your answer in litres per second.
Show worked solution

Worked solution

  1. Plan: shrink the interval

    gradient=riserun=V(t2)−V(t1)t2−t1\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{V(t_2) - V(t_1)}{t_2 - t_1}

    Each chord gives an AVERAGE rate over its interval. As the interval shrinks onto t=5t = 5, the chord turns towards the tangent there, so the chord gradients close in on the INSTANTANEOUS rate.

  2. Find the fixed end of every chord (t=5)(t = 5)

    t=5:V=150=150t = 5: \quad V = 150 = 150

    Every chord starts at the point (5, 150)(5,\ 150); only the other end moves.

  3. Write down the chord gradient formula

    gradient=riserun=V(5+h)−V(5)h\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{V(5 + h) - V(5)}{h}

    The run is always hh, so only the rise changes as hh shrinks.

  4. Chord with h=1h = 1

    t=6:V=128=128⇒gradient=riserun=128−1501=−22t = 6: \quad V = 128 = 128 \quad \Rightarrow \quad \text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{128 - 150}{1} = -22

    The chord from t=5t = 5 to t=6t = 6 has gradient −22-22 — the average rate of change of the volume over that short interval.

  5. Chord with h=0.5h = 0.5

    t=5.5:V=2792=139.5⇒gradient=riserun=139.5−1500.5=−21t = 5.5: \quad V = \frac{279}{2} = 139.5 \quad \Rightarrow \quad \text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{139.5 - 150}{0.5} = -21

    The chord from t=5t = 5 to t=5.5t = 5.5 has gradient −21-21 — the average rate of change of the volume over that short interval.

  6. Chord with h=0.1h = 0.1

    t=5.1:V=23448366977721674502603172520919158456325028528675187087900672=147.98⇒gradient=riserun=147.98−1500.1=−20.2t = 5.1: \quad V = \frac{23448366977721674502603172520919}{158456325028528675187087900672} = 147.98 \quad \Rightarrow \quad \text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{147.98 - 150}{0.1} = -20.2

    The chord from t=5t = 5 to t=5.1t = 5.1 has gradient −20.2-20.2 — the average rate of change of the volume over that short interval.

  7. List the chord gradients in order

    −22, −21, −20.2-22,\ -21,\ -20.2

    Writing them in order (as hh shrinks) makes the pattern easy to see.

  8. Describe the trend

    −22, −21, −20.2 ⟶ −20-22,\ -21,\ -20.2 \ \longrightarrow \ -20

    The gradients are increasing and are closing in on −20-20.

  9. Check how close the last one is

    ∣−20.2−−20∣=0.2|-20.2 - -20| = 0.2

    The smallest interval already gives a gradient within 0.20.2 of −20-20, and the gap keeps shrinking as hh gets smaller.

  10. State the limiting value

    −20 litres per second-20\text{ litres per second}

    As hh gets closer and closer to 00 the chord gradients get closer and closer to −20-20, so the instantaneous rate of change of the volume at t=5t = 5 is −20-20 litres per second.

  11. Link this to the tangent

    tangent gradient at t=5=−20\text{tangent gradient at } t = 5 = -20

    The limiting value is exactly what a correctly drawn tangent at t=5t = 5 would give: the shrinking chords rotate onto the tangent.

  12. Note whether the chords are above or below

    −22 vs −20-22 \ \text{vs} \ -20

    Every one of these chord gradients is less than −20-20, so each is an under-estimate of the instantaneous rate — the shape of the curve decides this, so it must be checked, not guessed.

  13. Give the units

    −20 litres per second-20\text{ litres per second}

    The rate is a change in VV per unit change in tt, so the units are litres per second.

  14. Ask what a still smaller step would give

    h=0.01 ⇒ gradient=−20.02h = 0.01 \ \Rightarrow \ \text{gradient} = -20.02

    Taking hh ten times smaller again brings the chord gradient closer still to −20-20 — the chords never quite reach the tangent, they close in on it.

  15. Write down the answer

    −20 litres per second-20\text{ litres per second}

    The instantaneous rate of change of the volume at t=5t = 5 is −20-20 litres per second.

Answer
−20 litres per second-20\text{ litres per second}
Question 3
6 markschallenging
For the curve y=x2y = x^2, Jon estimates the instantaneous rate of change of yy at x=2x = 2 by drawing a tangent at that point; his drawn tangent passes through (1, 0)(1,\ 0) and (3, 8)(3,\ 8). Priya instead works out the gradients of the chords from x=2x = 2 to x=2+hx = 2 + h for h=1h = 1, h=0.5h = 0.5, h=0.1h = 0.1 and h=0.01h = 0.01. Which statement about their two methods is correct?
Show worked solution

Worked solution

  1. Jon: use the two points on his tangent

    (1, 0)and(3, 8)(1,\ 0) \quad \text{and} \quad (3,\ 8)

    Both points lie on the straight tangent Jon has drawn.

  2. Jon: work out his tangent gradient

    gradient=riserun=8−03−1=4\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{8 - 0}{3 - 1} = 4

    Jon’s estimate of the instantaneous rate at x=2x = 2 is 44.

  3. Jon: note why his answer is an estimate

    tangent drawn by eye\text{tangent drawn by eye}

    The tangent is drawn by hand, so a slightly different line would give a slightly different gradient.

  4. Priya: find the fixed end (x=2)(x = 2)

    x=2:y=4=4x = 2: \quad y = 4 = 4

    All of Priya’s chords start at (2, 4)(2,\ 4).

  5. Priya: chord with h=1h = 1

    gradient=riserun=9−41=5\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{9 - 4}{1} = 5

    The chord from x=2x = 2 to x=3x = 3 has gradient 55 — an average rate over that interval.

  6. Priya: chord with h=0.5h = 0.5

    gradient=riserun=6.25−40.5=4.5\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{6.25 - 4}{0.5} = 4.5

    The chord from x=2x = 2 to x=2.5x = 2.5 has gradient 4.54.5 — an average rate over that interval.

  7. Priya: chord with h=0.1h = 0.1

    gradient=riserun=4.41−40.1=4.1\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{4.41 - 4}{0.1} = 4.1

    The chord from x=2x = 2 to x=2.1x = 2.1 has gradient 4.14.1 — an average rate over that interval.

  8. Priya: chord with h=0.01h = 0.01

    gradient=riserun=4.0401−40.01=4.01\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{4.0401 - 4}{0.01} = 4.01

    The chord from x=2x = 2 to x=2.01x = 2.01 has gradient 4.014.01 — an average rate over that interval.

  9. Priya: list her chord gradients

    5, 4.5, 4.1, 4.015,\ 4.5,\ 4.1,\ 4.01

    Each one is an average rate over a shorter and shorter interval.

  10. Priya: describe the trend

    5, 4.5, 4.1, 4.01 ⟶ 45,\ 4.5,\ 4.1,\ 4.01 \ \longrightarrow \ 4

    The values are decreasing and getting closer and closer to 44.

  11. Priya: state her limiting value

    44

    Her chord gradients close in on 44, so the instantaneous rate at x=2x = 2 is 44.

  12. Compare the two methods

    4=44 = 4

    Jon’s drawn tangent and Priya’s shrinking chords give the same value, so the two methods agree.

  13. Say why they must agree

    chord→tangent as h→0\text{chord} \to \text{tangent as } h \to 0

    As the interval shrinks, the chord rotates onto the tangent. The tangent gradient IS the limit of the chord gradients.

  14. Rule out calling a tangent gradient an average

    tangent touches at one point\text{tangent touches at one point}

    A tangent meets the curve at a single point, so its gradient cannot be an average across an interval.

  15. Choose the correct statement

    4=44 = 4

    The instantaneous rate of change at x=2x = 2 is 44 from Jon’s tangent, and the chord gradients are 55, 4.54.5, 4.14.1 and 4.014.01, which get closer and closer to 44: the two methods agree.

Answer
4=44 = 4
Question 4
6 markschallenging
The chord joining the points on the curve y=x2+4xy = x^2 + 4x where x=1x = 1 and x=bx = b has gradient 1010, where b>1b > 1. Work out the value of bb.
Show worked solution

Worked solution

  1. Plan: build the chord gradient in terms of b

    gradient=y2−y1x2−x1\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1}

    One end of the chord is fixed at x=1x = 1; the other end is at the unknown x=bx = b.

  2. Find the y-value at the fixed end

    x=1:y=12+4(1)=5x = 1: \quad y = 1^2 + 4(1) = 5

    So one end of the chord is (1, 5)(1,\ 5).

  3. Write the y-value at the moving end

    x=b:y=b2+4bx = b: \quad y = b^2 + 4b

    Substitute x=bx = b into the equation of the curve.

  4. Work out the rise

    rise=b2+4b−5\text{rise} = b^2 + 4b - 5

    The change in yy between the two ends of the chord.

  5. Work out the run

    run=b−1\text{run} = b - 1

    The change in xx between the two ends of the chord.

  6. Write the chord gradient

    gradient=b2+4b−5b−1\text{gradient} = \frac{b^2 + 4b - 5}{b - 1}

    This must equal 1010.

  7. Factorise the numerator

    b2+4b−5=(b+5)(b−1)b^2 + 4b - 5 = (b + 5)(b - 1)

    Two numbers multiplying to −5-5 and adding to 44 are +5+5 and −1-1.

  8. Cancel the common factor

    (b+5)(b−1)b−1=b+5\frac{(b + 5)(b - 1)}{b - 1} = b + 5

    Since b>1b > 1, the factor b−1b - 1 is not zero, so it cancels. The chord gradient is simply b+5b + 5.

  9. Form the equation

    b+5=10b + 5 = 10

    The chord gradient is given as 1010.

  10. Solve for b

    b=5b = 5

    Subtract 55 from both sides.

  11. Check the answer

    45−55−1=10\frac{45 - 5}{5 - 1} = 10

    The chord from x=1x = 1 to x=5x = 5 does have gradient 1010.

  12. Say what the 10 is

    average rate=10\text{average rate} = 10

    It is the AVERAGE rate of change from x=1x = 1 to x=5x = 5.

  13. Check the rate at the left-hand end

    x=1→x=1.1:gradient=6.1x = 1 \to x = 1.1: \quad \text{gradient} = 6.1

    A very short chord starting at x=1x = 1 has gradient 6.16.1, so the instantaneous rate at x=1x = 1 is about 66, not 1010.

  14. Notice where the rate really is 10

    x=3:instantaneous rate=10x = 3: \quad \text{instantaneous rate} = 10

    The instantaneous rate equals the average rate 1010 at the midpoint x=3x = 3 of the interval.

  15. State the answer

    b=5b = 5

    So b=5b = 5.

Answer
b=5b = 5
Question 5
5 markschallenging
For the curve y=x3−9xy = x^3 - 9x, a tangent is drawn at the point where x=2x = 2 and it passes through (0, −16)(0,\ -16) and (4, −4)(4,\ -4). Work out the difference between the average rate of change of yy from x=2x = 2 to x=4x = 4 and the instantaneous rate of change of yy at x=2x = 2.
Show worked solution

Worked solution

  1. Plan: one chord and one tangent

    chord: x=2→x=4,tangent at x=2\text{chord: } x = 2 \to x = 4, \qquad \text{tangent at } x = 2

    The average rate of change comes from the gradient of the CHORD across the interval. The instantaneous rate of change comes from the gradient of the TANGENT at the single point x=2x = 2.

  2. Find the value when x=2x = 2

    x=2:y=−10=−10x = 2: \quad y = -10 = -10

    Substitute x=2x = 2 into the equation of the curve.

  3. Find the value when x=4x = 4

    x=4:y=28=28x = 4: \quad y = 28 = 28

    Substitute x=4x = 4 into the equation of the curve.

  4. Write down the ends of the chord

    (2, −10)and(4, 28)(2,\ -10) \quad \text{and} \quad (4,\ 28)

    These are the two points on the curve that the chord joins.

  5. Work out the rise of the chord

    rise=28−−10=38\text{rise} = 28 - -10 = 38

    This is the total change in yy across the interval.

  6. Work out the run of the chord

    run=4−2=2\text{run} = 4 - 2 = 2

    This is the total change in xx across the interval.

  7. Work out the chord gradient

    gradient=riserun=382=19\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{38}{2} = 19

    So the average rate of change across the interval is 1919.

  8. Say what the chord gradient means

    chord gradient=19\text{chord gradient} = 19

    This is an AVERAGE: it treats the whole interval as if the rate were steady at 1919, which it is not.

  9. Write down the two points on the drawn tangent

    (0, −16)and(4, −4)(0,\ -16) \quad \text{and} \quad (4,\ -4)

    These points lie on the straight tangent drawn at x=2x = 2, not on the curve.

  10. Work out the rise of the tangent

    rise=−4−−16=12\text{rise} = -4 - -16 = 12

    Use only the two points on the tangent line.

  11. Work out the run of the tangent

    run=4−0=4\text{run} = 4 - 0 = 4

    Use the same two points on the tangent line.

  12. Work out the tangent gradient

    gradient=riserun=124=3\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{12}{4} = 3

    So the instantaneous rate of change at x=2x = 2 is about 33.

  13. Say what the tangent gradient means

    tangent gradient=3\text{tangent gradient} = 3

    This is an INSTANTANEOUS rate: it describes one instant, x=2x = 2, and nothing else. It is an estimate because the tangent was drawn by eye.

  14. Compare the two rates

    19and319 \quad \text{and} \quad 3

    The average rate (1919) is larger the instantaneous rate (33), so a chord over an interval and a tangent at a point genuinely measure different things.

  15. Work out the difference

    ∣19−3∣=16|19 - 3| = 16

    The difference between the two rates is 1616.

Answer
1616

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