Tangents and Normals

If you differentiate the equation of a curve, you will get a formula for the gradient of the curve. Before you learnt differentiation, you would have found the gradient of a curve by drawing a tangent and measuring the gradient of this. This is because the gradient of a curve at a point is equal to the gradient of the tangent at that point.

The equation of the tangent to a point on a curve can therefore be found by differentiation.

Example

Find the equation of the tangent to the curve y = x³ at the point (2, 8).
dy = 3x²
dx

Gradient of tangent when x = 2 is 3 × 2² = 12.
From the coordinate geometry section, the equation of the tangent is therefore:
y - 8 = 12(x - 2) since the gradient of the tangent is 12 and we know that it passes through (2, 8)
so y = 12x - 16

You may also be asked to find the gradient of the normal to the curve. The normal to the curve is the line perpendicular (at right angles) to the tangent to the curve at that point.

Remember, if two lines are perpendicular, the product of their gradients is -1.

So if the gradient of the tangent at the point (2, 8) of the curve y = x³ is 12, the gradient of the normal is -1/12, since -1/12 × 12 = -1 .

The equation of the normal at the point (2, 8) is therefore:

y - 8 = -1/12 (x - 2)

hence the equation of the normal at (2,8) is 12y + x = 98 .

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