Gradients of curves
Objective. Estimate instantaneous rates using tangents.
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- 2. Worked example
- 3. Practice
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Learn: the key idea
Gradients of curves are estimated at a single point by drawing a tangent line that just touches the curve at that point without crossing it, then calculating the gradient of that straight tangent using rise over run between two points on it. Unlike a straight line, a curve's gradient changes continuously, so the tangent must be drawn freshly at each point of interest. The more carefully the tangent is drawn and the further apart the chosen points on it, the more accurate the gradient estimate becomes. For example, if a tangent at a point passes through (1,2) and (4,11), its gradient is (11−2)/(4−1)=3.
Worked example
A tangent to a curve at the point (2, 5) also passes through (6, 17). Find the gradient of the curve at (2, 5).
- 1Identify two points on the tangent line: (2, 5) and (6, 17)
- 2Find the change in y: 17−5=12
- 3Find the change in x: 6−2=4
- 4Divide to find the gradient: 12÷4=3
Answer: Gradient = 3
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Practice
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Start practisingCommon mistake. Pupils often draw a chord connecting two points on the curve itself instead of a true tangent touching at one point, which estimates the average rate of change over an interval rather than the instantaneous gradient at the point required.
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