Worked example 1
Find a tangent
Find the tangent to y = x² + 3x at x = 2.
- Differentiate: dy/dx = 2x + 3
- At x = 2, m = 7
- The point is (2, 10).
- Use y − 10 = 7(x − 2).
Answer: y = 7x − 4
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Higher Mathematics
Tangents, normals and rates of change
Find gradient functions, tangent equations, normal equations and rates of change.
Before you start
- Find a straight-line equation from a point and gradient.
- Understand perpendicular gradients.
- Differentiate powers accurately.
Explanation
A tangent gradient comes from dy/dx at the point. A normal is perpendicular to the tangent, so its gradient is the negative reciprocal.
Rates of change use the same derivative idea but are interpreted in context.
Visual support
Method and rules
- Tangent: m = dy/dx at the point
- Normal gradient = −1m when m ≠ 0
- Line through (a, b): y − b = m(x − a)
Worked examples
Worked example 2
Find a normal
Find the normal to y = x² at x = 3.
- dy/dx = 2x, so tangent gradient is 6
- Normal gradient is −16
- Point is (3, 9).
So: y − 9 = −16(x − 3)
Worked example 3
Interpret a rate
If s = t³ − 4t, find the rate of change at t = 2.
- Differentiate s with respect to t.
- ds/dt = 3t² − 4
- Substitute t = 2
Answer: Rate of change = 8
Watch out
- Using the y-coordinate instead of the derivative as the gradient.
- Forgetting to find the point on the curve.
- Using the same gradient for the normal.
Exam reminder
Tangent and normal questions need both the gradient and the point. If the point is not given directly, substitute into the curve to find it.