Use the derivative as the gradient to find the equations of the tangent and the normal to a curve at a given point

What this dot point is asking

SEAB wants you to find the equations of the tangent and the normal to a curve at a given point, using the derivative for the gradient. The tangent just touches the curve and has gradient equal to the derivative there; the normal is perpendicular to the tangent at the same point.

The answer

The tangent gradient

The gradient of the tangent at a point is the value of the derivative there. So:

1. Differentiate to get $\dfrac{dy}{dx}$.
2. Substitute the $x$-coordinate of the point to get the numerical gradient $m$.

If only the $x$-coordinate is given, find the $y$-coordinate first by substituting into the curve.

The tangent equation

With the gradient $m$ and the point $(x_1, y_1)$, use the point-gradient form:

This is the equation of the tangent line.

The normal gradient

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent gradient:

If the tangent is horizontal ($m = 0$), the normal is vertical, and vice versa.

The normal equation

Use the same point with the normal gradient in the point-gradient form to write the normal line.

Tangents parallel to a given line

A question may ask where a tangent is parallel to a given line, or horizontal. A parallel tangent has the same gradient as that line, so set the derivative equal to that gradient and solve for $x$; a horizontal tangent has gradient zero, the condition for a stationary point.

Where a tangent meets the axes

After finding a tangent or normal equation, you may be asked where it crosses the axes. Set $y = 0$ for the $x$-intercept and $x = 0$ for the $y$-intercept, then use those points (for instance to find the area of the triangle the tangent makes with the axes), a common Paper 2 extension.

Examples in context

Example 1. Direction of motion on a path. A particle following a curved track has its instantaneous direction given by the tangent to the path; the tangent line at a point shows the heading at that moment.

Example 2. Reflecting off a curved mirror. The normal to a curved mirror at the point of incidence is the line about which light reflects, so finding the normal is the first step in tracing a ray off a parabolic dish.

Try this

Q1. Find the tangent gradient to $y = x^3$ at $x = 1$. [2 marks]

• Cue. $\dfrac{dy}{dx} = 3x^2$, so at $x = 1$ the gradient is $3$.

Q2. Find the equation of the tangent to $y = x^2 + 1$ at $(1, 2)$. [3 marks]

• Cue. Gradient $2x = 2$; tangent $y - 2 = 2(x - 1)$, so $y = 2x$.

Q3. State the normal gradient to a curve where the tangent gradient is $\dfrac{1}{4}$. [1 mark]

• Cue. Negative reciprocal: $-4$.