Gradients and Areas of Graphs
When a graph isn’t a straight line, how can we find its steepness or the area underneath it? This guide explores the key higher-tier techniques for estimating the gradient of a curve at a point and estimating the area under a curve – essential skills for understanding changing rates like speed and acceleration.
Key Methods for Curved Graphs
Estimating the Gradient
The gradient of a curve is always changing. To estimate the gradient at a specific point, we draw a tangent – a straight line that just touches the curve at that point.
The gradient of the curve at that point is the same as the gradient of the tangent. You can then calculate the tangent’s gradient using “rise over run”.
Estimating the Area
To estimate the area under a curve, we can split it into a series of vertical strips and treat each strip as a trapezium.
By calculating the area of each trapezium and adding them together, we can find a good approximation for the total area. This is known as the trapezium rule.
What Does It Mean in Context?
Distance-Time Graphs
On a distance-time graph, the gradient represents the speed (or velocity). A steep gradient means high speed, while a flat gradient (zero) means the object is stationary.
Velocity-Time Graphs
On a velocity-time graph, the gradient represents acceleration (how quickly the velocity is changing), and the area under the graph represents the total distance travelled.
Worked Examples
Example 1: Estimating Speed (Gradient)
Estimate the speed of a car at $t=6$ seconds from the distance-time graph.
- Draw a tangent to the curve at the point where $t=6$.
- Pick two clear points on your tangent line. For example, $(4, 18)$ and $(8, 38)$.
- Calculate the gradient:
Gradient = $\frac{\text{change in distance}}{\text{change in time}} = \frac{38 – 18}{8 – 4} = \frac{20}{4} = 5$.
Answer: Approx. $5 \text{ m/s}$. (Your answer may vary slightly depending on your tangent).
Example 2: Estimating Distance (Area)
Estimate the distance travelled in 60s using 3 trapeziums.
The width of each strip ($h$) is $60 \div 3 = 20$s. Read the velocities: $v_0=0, v_1=21, v_2=33, v_3=40$.
Area = $\frac{h}{2}[v_0 + v_3 + 2(v_1+v_2)]$
$= \frac{20}{2}[0 + 40 + 2(21+33)]$
$= 10[40 + 2(54)] = 10[40+108] = 10 \times 148$.
Answer: Approx. 1480 metres.
Tutor Insights
🤔 Common Misunderstandings
- Drawing the tangent: Drawing a line that crosses the curve (a chord) instead of one that just touches it.
- Points for gradient: Picking points on the curve instead of on the tangent line you’ve drawn.
- Trapezium Rule: Forgetting to double the “middle” heights in the formula.
📝 Common Exam Mistakes
- Inaccurate tangent: If your tangent is drawn poorly, your gradient will be outside the accepted range.
- Misreading graph scales: Always check what each square on the grid is worth.
- Missing units in the final answer (e.g., m/s for speed, m for distance).
Practice Questions
- For the velocity-time graph in Example 2, is the area calculated using the trapezium rule an over- or under-estimate? Explain your answer.
- A cooling cup of tea’s temperature is recorded. What would the gradient of its temperature-time graph represent?
Show Answers
- The estimate is an over-estimate. Because the curve is bending downwards (concave down), the straight tops of the trapeziums will lie slightly above the curve, including small extra slivers of area.
- The gradient would represent the rate of cooling, likely in degrees Celsius per minute ($^\circ\text{C/min}$). It would be negative because the temperature is decreasing.
