Gradient uncertainty practice

Use error bars and limiting lines to calculate absolute and percentage uncertainty in a gradient.

Practise

Absolute gradient uncertainty

Line A is the supplied best fit. Use the larger absolute departure from its gradient to calculate the absolute gradient uncertainty. Give 2 significant figures.

Acceptable lines must cross every vertical error bar; horizontal uncertainty is neglected. Use the displayed readings and line coordinates for calculation practice. These bounds are not confidence intervals.

Read the axes and units. The values are also available in the table below.

Larger graph

Graph values

Synthetic readings; Δy is the quoted vertical uncertainty in axis units
Point	Time / s	Position / cm	± Δy
1	2	8.9	±1.2
2	4	12.8	±0.9
3	6	14.9	±0.6
4	8	18.5	±0.6
5	10	20.6	±0.9
6	12	24.5	±1.2

Line coordinates

Line A: (2, 9.2) to (12, 24.2).
Line B: (4, 11.9) to (10, 21.5).
Line C: (2, 10.1) to (12, 23.3).

Coordinates use the axis units. These are two positions on each line, not necessarily measured points.

Hint

Calculate the three gradients from line coordinates before rounding. Compare the two absolute departures from the best gradient.

Worked solution

The supplied best line is A.
Line A: gradient = (24.2 − 9.2) / (12 − 2) ≈ 1.5000 × 10⁰ cm s⁻¹. Retain the original fraction for the uncertainty calculation.
Line B: gradient = (21.5 − 11.9) / (10 − 4) ≈ 1.6000 × 10⁰ cm s⁻¹. Retain the original fraction for the uncertainty calculation.
Line C: gradient = (23.3 − 10.1) / (12 − 2) ≈ 1.3200 × 10⁰ cm s⁻¹. Retain the original fraction for the uncertainty calculation.
Use the larger of |m_max − m_best| and |m_best − m_min|. Line C is the worst acceptable line by this convention.
The two departures are approximately 1.0000 × 10^-1 and 1.8000 × 10^-1 cm s⁻¹; use the larger magnitude.
Answer: 0.18 cm s⁻¹ (2 significant figures).

Another question

Question type: Mixed practice

Report a problem

Calculate gradient uncertainty from limiting lines

Suppose the best-fit gradient is 2.40 and the accepted shallowest and steepest lines give 2.10 and 2.80, all in the same units. For the convention used here, take the larger deviation from the best fit: max(0.30, 0.40) = 0.40. Percentage uncertainty = (0.40/2.40) × 100 = 16.7%.

A limiting line must meet the stated error-bar constraints; choosing two convenient bar ends is insufficient if the line misses another bar. Follow the question’s convention. Half the range of limiting gradients is a different prescription and need not give the same result.

Practise gradients and axis units

Check every error bar

An acceptable line crosses each vertical error bar. The horizontal coordinates are treated as exact here. A line through the extremes of the first and last bars can still miss an interior bar.

The supplied best line describes the central trend. The two limiting lines have the largest and smallest signed gradients that can cross all six intervals. They need not pass through the same data points as the best line.

Calculate the gradients first

Use two well-separated coordinates on each line: m = Δy / Δx. The line-coordinate table supplies exact positions for the calculations. Keep the original fractions when finding the uncertainty; round the final answer.

Best and worst gradients

In the largest-departure exercises, Δm is the larger of |m_max − m_best| and |m_best − m_min|. Compare both limiting lines. The worst acceptable line can be either one.

Percentage uncertainty = Δm / |m_best| × 100%. For a negative gradient, use its magnitude in the denominator. Uncertainty magnitudes stay positive.

When a question specifies half-range

The half-range task asks for (m_max − m_min) / 2. This can differ from the largest departure when the best gradient is not at the centre of the range. Use the convention stated in the question; these exercises do not impose one rule on every exam question.

Uncertainty in a derived quantity

For a spring in its proportional region, ΔF = kΔx. The force–displacement gradient gives k even when displacement is measured from a loaded reference. A gradient uncertainty in N cm⁻¹ must be multiplied by 100 to express it in N m⁻¹.

What the bounds mean

The six readings are synthetic practice data with quoted vertical uncertainties. The limiting lines describe compatibility with those intervals. They are not statistical confidence intervals and do not account for a shared calibration error, uncertain horizontal coordinates or an incorrect physical model.

Give calculated gradients to 3 significant figures and uncertainty magnitudes to 2 in these exercises. This is a practice marking policy. Follow the precision and method required in your exam question.

A-level relevance

AQA 7408: 3.1.2 (partial); AQA 7408: 3.4.2.1 (selected); OCR A H556: 1.1.3(d), 1.1.4(d), 2.2.1(c), 3.4.1(c) (partial). Tasks cover the selected skills listed above.

AQA specification · AQA mechanics and materials specification · OCR A specification · AQA practical handbook · OCR mathematical skills handbook