Gradients and areas under curves Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Gradients and areas under curves questions. See exactly how to solve problems on chord gradient, average rate of change, tangent gradient, instantaneous rate of change.

chord gradientaverage rate of changetangent gradientinstantaneous rate of changerise over runtrapezium rule
GCSE Higher70 questionsStep-by-step solutions
Question 1
2 markseasy
The curve y=x2y = x^2 is drawn for 0≤x≤40 \le x \le 4. Work out the average rate of change of yy between x=1x = 1 and x=3x = 3 by finding the gradient of the chord joining the two points on the curve.

Worked solution

  1. Find the two points on the curve

    x1=1, y1=1x2=3, y2=9x_1 = 1,\ y_1 = 1 \qquad x_2 = 3,\ y_2 = 9

    Substitute x=1x = 1 and x=3x = 3 into x2x^2 to get the two end points of the chord.

  2. Gradient of the chord = rise ÷ run

    gradient=9−13−1=82=4\text{gradient} = \frac{9 - 1}{3 - 1} = \frac{8}{2} = 4

    The average rate of change over an interval is the gradient of the straight chord joining the two end points.

  3. State the answer

    44

    That is the average rate of change across the interval.

Answer
44
Question 2
1 markeasy
Work out the gradient of the chord joining the points on the curve y=x2+1y = x^2 + 1 where x=0x = 0 and x=2x = 2.

Worked solution

  1. Find the two points on the curve

    x1=0, y1=1x2=2, y2=5x_1 = 0,\ y_1 = 1 \qquad x_2 = 2,\ y_2 = 5

    Substitute x=0x = 0 and x=2x = 2 into x2+1x^2 + 1 to get the two end points of the chord.

  2. Gradient of the chord = rise ÷ run

    gradient=5−12−0=42=2\text{gradient} = \frac{5 - 1}{2 - 0} = \frac{4}{2} = 2

    The average rate of change over an interval is the gradient of the straight chord joining the two end points.

  3. State the answer

    22

    That is the average rate of change across the interval.

Answer
22
Question 3
2 markseasy
A tangent is drawn to the curve y=x2y = x^2 at the point where x=3x = 3. The tangent passes through the points (2, 3)(2,\ 3) and (4, 15)(4,\ 15). Use the tangent to estimate the gradient of the curve at x=3x = 3.

Worked solution

  1. Use the two points on the tangent

    (2, 3)and(4, 15)(2,\ 3) \quad \text{and} \quad (4,\ 15)

    The tangent is the straight line that just touches the curve at the point. Its gradient is the gradient of the curve there.

  2. Gradient = rise ÷ run

    gradient=15−34−2=122=6\text{gradient} = \frac{15 - 3}{4 - 2} = \frac{12}{2} = 6

    Use the two given points on the straight tangent line, not points on the curve.

  3. State the estimate

    66

    So the gradient of the curve at that point is estimated to be 6. It is an estimate because it depends on how accurately the tangent was drawn.

Answer
66
Question 4
2 markseasy
A tangent is drawn to the curve y=x2+1y = x^2 + 1 at the point where x=2x = 2. The tangent passes through (0, −3)(0,\ -3) and (4, 13)(4,\ 13). Estimate the gradient of the curve at x=2x = 2.

Worked solution

  1. Use the two points on the tangent

    (0, −3)and(4, 13)(0,\ -3) \quad \text{and} \quad (4,\ 13)

    The tangent is the straight line that just touches the curve at the point. Its gradient is the gradient of the curve there.

  2. Gradient = rise ÷ run

    gradient=13−−34−0=164=4\text{gradient} = \frac{13 - -3}{4 - 0} = \frac{16}{4} = 4

    Use the two given points on the straight tangent line, not points on the curve.

  3. State the estimate

    44

    So the gradient of the curve at that point is estimated to be 4. It is an estimate because it depends on how accurately the tangent was drawn.

Answer
44
Question 5
2 markseasy
Use the trapezium rule with 22 strips, each of width 11, to estimate the area under the curve y=x2y = x^2 between x=0x = 0 and x=2x = 2.

Worked solution

  1. Write down the 3 ordinates

    y0=0,y1=1,y2=4y_0 = 0,\quad y_1 = 1,\quad y_2 = 4

    Substitute x=0x = 0, 1, 2 into x2x^2. The strips each have width h=1h = 1.

  2. Apply the trapezium rule

    A≈12[0+4+2(1)]=12×6A \approx \frac{1}{2}\Big[0 + 4 + 2\big(1\big)\Big] = \frac{1}{2} \times 6

    Add the two end ordinates, add twice each middle ordinate, then multiply by half the strip width.

  3. State the estimate

    A≈3A \approx 3

    The area under the curve is approximately 3. The curve is convex here (it bends upwards), so every chord lies ABOVE the curve. Each trapezium therefore contains a sliver of extra area, and the estimate is an OVER-estimate.

Answer
33

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