Instantaneous rate of change Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Instantaneous rate of change questions. See exactly how to solve problems on chord gradient, average rate of change, negative rate of change, reciprocal curve.

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.

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
The points where x=1x = 1 and x=4x = 4 on the curve y=x2−2xy = x^2 - 2x are joined by a chord. Work out the gradient of that chord.

Worked solution

  1. Find the value at each end of the chord

    x=1:y=−1=−1x=4:y=8=8x = 1: \quad y = -1 = -1 \qquad x = 4: \quad y = 8 = 8

    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=8−−14−1=93=3\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{8 - -1}{4 - 1} = \frac{9}{3} = 3

    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

    33

    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
33
Question 3
2 markseasy
A chord joins the points where x=1x = 1 and x=3x = 3 on the curve y=20−x2y = 20 - x^2. Work out the gradient of the chord.

Worked solution

  1. Find the value at each end of the chord

    x=1:y=19=19x=3:y=11=11x = 1: \quad y = 19 = 19 \qquad x = 3: \quad y = 11 = 11

    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=11−193−1=−82=−4\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{11 - 19}{3 - 1} = \frac{-8}{2} = -4

    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

    −4-4

    A chord gradient is an AVERAGE rate of change over the whole interval from x=1x = 1 to x=3x = 3. It is not the rate at either end of the interval — that would need a tangent.

Answer
−4-4
Question 4
2 markseasy
For the curve y=x3y = x^3, work out the average rate of change of yy from x=0x = 0 to x=2x = 2.

Worked solution

  1. Find the value at each end of the chord

    x=0:y=0=0x=2:y=8=8x = 0: \quad y = 0 = 0 \qquad x = 2: \quad y = 8 = 8

    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=8−02−0=82=4\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{8 - 0}{2 - 0} = \frac{8}{2} = 4

    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

    44

    A chord gradient is an AVERAGE rate of change over the whole interval from x=0x = 0 to x=2x = 2. It is not the rate at either end of the interval — that would need a tangent.

Answer
44
Question 5
2 markseasy
A chord joins the points where x=2x = 2 and x=6x = 6 on the curve y=24xy = \frac{24}{x}. Work out the gradient of the chord.

Worked solution

  1. Find the value at each end of the chord

    x=2:y=12=12x=6:y=4=4x = 2: \quad y = 12 = 12 \qquad x = 6: \quad y = 4 = 4

    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=4−126−2=−84=−2\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{4 - 12}{6 - 2} = \frac{-8}{4} = -2

    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

    −2-2

    A chord gradient is an AVERAGE rate of change over the whole interval from x=2x = 2 to x=6x = 6. It is not the rate at either end of the interval — that would need a tangent.

Answer
−2-2

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