GCSE Instantaneous rate of change Practice Questions

Free GCSE Instantaneous rate of change practice questions with full step-by-step worked solutions. Covers chord gradient, average rate of change, negative rate of change, reciprocal curve. Practise exam-style problems and check your method.

chord gradientaverage rate of changenegative rate of changereciprocal curvetangent gradientinstantaneous rate of change
GCSE Higher70 questionsStep-by-step solutions
Question 1
2 markseasy
The points where x=1x = 1 and x=4x = 4 on the curve y=x2+3y = x^2 + 3 are joined by a chord. Work out the gradient of that chord.
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Worked solution

  1. Find the value at each end of the chord

    x=1:y=4=4x=4:y=19=19x = 1: \quad y = 4 = 4 \qquad x = 4: \quad y = 19 = 19

    Substitute each xx-value into the equation of the curve to get the two points the chord joins.

  2. Use rise over run for the chord

    gradient=riserun=19−44−1=153=5\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{19 - 4}{4 - 1} = \frac{15}{3} = 5

    The chord is a straight line joining the two points, so its gradient is the change in yy divided by the change in xx.

  3. State the average rate of change

    55

    A chord gradient is an AVERAGE rate of change over the whole interval from x=1x = 1 to x=4x = 4. It is not the rate at either end of the interval — that would need a tangent.

Answer
55
Question 2
2 markseasy
A ball's distance travelled, ss metres, after tt seconds is modelled by the curve s=t2s = t^2. A tangent is drawn to the curve at t=4t = 4 and it passes through (2, 0)(2,\ 0) and (6, 32)(6,\ 32). Which statement correctly says what the gradient of this tangent tells you, and gives its value?
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Worked solution

  1. Use the two points on the drawn tangent

    (2, 0)and(6, 32)(2,\ 0) \quad \text{and} \quad (6,\ 32)

    A tangent touches the curve at one point only; read two points off the straight tangent line.

  2. Work out the gradient of the tangent

    gradient=riserun=32−06−2=8\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{32 - 0}{6 - 2} = 8

    So the tangent has gradient 88.

  3. State what the tangent gradient means

    8 m/s8\text{ m/s}

    The tangent gradient is the instantaneous rate at that instant: the instantaneous speed at t=4t = 4 is 88 m/s.

Answer
8 m/s8\text{ m/s}
Question 3
2 marksintermediate
An oven cools so that its temperature, CC degrees Celsius, after tt minutes is modelled by the curve C=90−t2C = 90 - t^2. A tangent is drawn to the curve at t=3t = 3 and it passes through (1, 93)(1,\ 93) and (5, 69)(5,\ 69). Use the drawn tangent to estimate the instantaneous rate of change of the temperature at t=3t = 3. Give your answer in degrees Celsius per minute.
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Worked solution

  1. Know what a tangent measures

    tangent at t=3\text{tangent at } t = 3

    A chord gives an average rate over an interval. To get the rate at the single instant t=3t = 3 you draw the tangent — the straight line that just touches the curve there — and find its gradient.

  2. Write down the two points on the drawn tangent

    (1, 93)and(5, 69)(1,\ 93) \quad \text{and} \quad (5,\ 69)

    These two points lie on the straight tangent line, so they can be used for rise and run. Points on the curve itself would give a chord instead.

  3. Work out the rise on the tangent

    rise=69−93=−24\text{rise} = 69 - 93 = -24

    This is the change in CC between the two points on the tangent.

  4. Work out the run on the tangent

    run=5−1=4\text{run} = 5 - 1 = 4

    This is the change in tt between the same two points.

  5. Divide the rise by the run

    gradient=riserun=−244=−6\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{-24}{4} = -6

    This is the gradient of the tangent line.

  6. Interpret the answer

    −6 degrees Celsius per minute-6\text{ degrees Celsius per minute}

    The point of contact is (3, 81)(3,\ 81), and the tangent there has gradient −6-6, so the instantaneous rate of change of the temperature at t=3t = 3 is about −6-6 degrees Celsius per minute. The tangent has been drawn by eye, so the gradient found from it is an ESTIMATE of the instantaneous rate; a slightly different drawn tangent would give a slightly different value.

Answer
−6 degrees Celsius per minute-6\text{ degrees Celsius per minute}
Question 4
4 markshard
A boat's distance travelled, ss metres, after tt seconds is modelled by the curve s=16t−t2s = 16t - t^2. A tangent is drawn to the curve at t=6t = 6 and it passes through (4, 52)(4,\ 52) and (8, 68)(8,\ 68). By how much does the boat's average speed from t=0t = 0 to t=6t = 6 exceed its instantaneous speed at t=6t = 6? Give your answer in m/s.
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Worked solution

  1. Plan: one chord and one tangent

    chord: t=0→t=6,tangent at t=6\text{chord: } t = 0 \to t = 6, \qquad \text{tangent at } t = 6

    The average speed comes from the gradient of the CHORD across the interval. The instantaneous speed comes from the gradient of the TANGENT at the single point t=6t = 6.

  2. Find the value when t=0t = 0

    t=0:s=0=0t = 0: \quad s = 0 = 0

    Substitute t=0t = 0 into the equation of the curve.

  3. Find the value when t=6t = 6

    t=6:s=60=60t = 6: \quad s = 60 = 60

    Substitute t=6t = 6 into the equation of the curve.

  4. Work out the rise of the chord

    rise=60−0=60\text{rise} = 60 - 0 = 60

    This is the total change in ss across the interval.

  5. Work out the chord gradient

    gradient=riserun=606=10\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{60}{6} = 10

    So the average speed across the interval is 1010 m/s.

  6. Say what the chord gradient means

    chord gradient=10 m/s\text{chord gradient} = 10\text{ m/s}

    This is an AVERAGE: it treats the whole interval as if the rate were steady at 1010 m/s, which it is not.

  7. Write down the two points on the drawn tangent

    (4, 52)and(8, 68)(4,\ 52) \quad \text{and} \quad (8,\ 68)

    These points lie on the straight tangent drawn at t=6t = 6, not on the curve.

  8. Work out the tangent gradient

    gradient=riserun=164=4\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{16}{4} = 4

    So the instantaneous speed at t=6t = 6 is about 44 m/s.

  9. Compare the two rates

    10and410 \quad \text{and} \quad 4

    The average rate (1010) is larger the instantaneous rate (44), so a chord over an interval and a tangent at a point genuinely measure different things.

  10. Work out the difference

    10−4=610 - 4 = 6

    The difference between the two rates is 66 m/s.

Answer
6 m/s6\text{ m/s}
Question 5
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

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