Gradients and Areas Under Graphs [H]
Gradients and Areas Under Non-Linear Graphs
For a straight-line graph, the gradient is constant throughout and easy to calculate. For a curved graph (quadratic, exponential, or other non-linear function), the gradient changes from point to point. The gradient at a specific point equals the gradient of the tangent to the curve at that point. The area under a curved graph is estimated using geometric methods.
Estimating the Gradient of a Curve at a Point
- Draw a tangent line — a straight line that just touches the curve at the given point without crossing it.
- Choose two widely spaced, clearly readable points on the tangent line.
- Calculate the gradient: \( m = \dfrac{\text{rise}}{\text{run}} = \dfrac{y_2 - y_1}{x_2 - x_1} \).
The wider apart the two chosen points, the less susceptible the gradient estimate is to reading error. Always use the tangent line — not the chord between two nearby curve points — for an instantaneous gradient.
Estimating the Area Under a Curve
The area under a non-linear graph is estimated by dividing the region into strips and approximating each strip as a trapezium (the trapezium rule). For \(n\) strips of equal width \(h\):
\[ \text{Area} \approx \frac{h}{2}\left(y_0 + 2y_1 + 2y_2 + \cdots + 2y_{n-1} + y_n\right) \]where \(y_0, y_1, ldots, y_n\) are the function values at the strip boundaries. More strips give a better estimate. The trapezium rule overestimates for concave-up curves and underestimates for concave-down curves.
Worked Examples
On a velocity–time graph of a curve, a tangent at \( t = 3 \) s passes through \( (1, 5) \) and \( (5, 21) \). Find the acceleration at \( t = 3 \) s.
\[ a = \text{gradient} = \frac{21-5}{5-1} = \frac{16}{4} = 4 \text{ m/s}^2 \]Estimate the area under \( y = x^2 \) from \( x=0 \) to \( x=4 \) using 4 strips.
Strip width \(h = 1\). Values: \(y_0=0, y_1=1, y_2=4, y_3=9, y_4=16\).
\[ \text{Area} \approx \frac{1}{2}(0 + 2(1) + 2(4) + 2(9) + 16) = \frac{1}{2}(0+2+8+18+16) = \frac{44}{2} = 22 \]The exact area is \( \frac{64}{3} \approx 21.3 \). The trapezium rule overestimates here (concave-up curve).
A velocity–time graph shows a curved relationship. The tangent at \(t=5\) s has gradient 2.4. The area under the curve from \(t=0\) to \(t=5\) is estimated as 35 m². Interpret both results.
Gradient of tangent = 2.4 m/s²: the instantaneous acceleration of the object at \(t = 5\) s is 2.4 m/s².
Area under graph = 35 m: the object has travelled 35 metres in the first 5 seconds.
Key Takeaways
- The gradient of a curve at a point = the gradient of the tangent at that point. Draw the tangent carefully; use two widely-spaced points to calculate its gradient.
- The area under a velocity–time graph gives the distance travelled. For curved graphs, estimate using the trapezium rule or by counting squares.
- Trapezium rule: \( \frac{h}{2}(y_0 + 2y_1 + 2y_2 + ldots + 2y_{n-1} + y_n) \). The first and last ordinates appear once; all middle ordinates appear twice.
- The trapezium rule overestimates area for concave-up curves and underestimates for concave-down curves. More strips → better approximation.
- In context: gradient of a distance–time graph = instantaneous speed; gradient of a velocity–time graph = instantaneous acceleration.