IB Applications and Interpretation SL · Differential and integral calculus · Section 2 of 7

Tangent and normal lines

3 worked examples · By Álvaro López

Key idea

The tangent line to a curve at a point is the line that matches the curve's direction at that point, and its gradient is $f^{'} (a)$ . The normal line is perpendicular to the tangent at the same point. Many tangent problems in the exam are solved with both lines.

The tangent line

A line is fixed by a point and a gradient: $y - y_{0} = m (x - x_{0})$ . For the tangent at the point with $x$ -coordinate $a$ we know both pieces of information:

the point is $(a, f (a))$ , with the $y$ -coordinate found from the function;
the gradient is $f^{'} (a)$ , the value of the derivative at $a$ .

If the question asks for the tangent parallel or perpendicular to another line, you do not need to guess the point from the graph. Just equate $f^{'} (a)$ to the gradient you are given and solve for $a$ .

The normal line

Curve y = x² + 3x − 1 for −1 ≤ x ≤ 4, with point P(1; 3), tangent line y = 5x − 2 (dashed) and normal line y = −x/5 + 16/5

How to write the two lines

Worked examples

Example 1 Basic

Tangent and normal at x = 1

Let $f (x) = x^{2} + 3 x - 1$ . Find the equations of the tangent line and the normal line to the graph of $f$ at the point with $x$ -coordinate $x = 1$ . Give both in the form $y = m x + n$ .

a) Tangent line.

b) Normal line.

Solution

First the point: $f (1) = 1 + 3 - 1 = 3$ , so the point is $(1, 3)$ .

The derivative is $f^{'} (x) = 2 x + 3$ , and the gradient at $x = 1$ is $f^{'} (1) = 5$ .

a) Tangent: $y - 3 = 5 (x - 1) \Rightarrow y = 5 x - 2$

b) The gradient of the normal is $- \frac{1}{5}$ : $y - 3 = - \frac{1}{5} (x - 1) \Rightarrow y = - \frac{1}{5} x + \frac{16}{5}$ Check: for $x = 1$ the normal gives $y = - \frac{1}{5} + \frac{16}{5} = 3$ , which is the point.

Example 2 Medium

Tangent parallel to a line

Let $f (x) = x^{3} - 6 x + 1$ .

a) Find the $x$ -coordinates of the points on the graph where the tangent is parallel to the line $y = 6 x$ .

b) Find the equation of the tangent line at $x = - 1$ .

Solution

a) Parallel lines have the same gradient. The gradient of $y = 6 x$ is $6$ , so we look for the values of $x$ with $f^{'} (x) = 6$ . The derivative is $f^{'} (x) = 3 x^{2} - 6$ : $3 x^{2} - 6 = 6 \Rightarrow 3 x^{2} = 12 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$ We check the points: $f (2) = 8 - 12 + 1 = - 3$ and $f (- 2) = - 8 + 12 + 1 = 5$ . The points are $(2, - 3)$ and $(- 2, 5)$ , and at both the gradient of the tangent is $6$ .

b) At $x = - 1$ : $f (- 1) = - 1 + 6 + 1 = 6$ and $f^{'} (- 1) = 3 - 6 = - 3$ . The tangent is $y - 6 = - 3 (x + 1) \Rightarrow y = - 3 x + 3$

Example 3 Hard

Tangent, normal and a parameter

Let $f (x) = k x^{2} + x$ , where $k$ is a real number.

a) Find the gradient of the tangent at $x = 2$ in terms of $k$ .

b) The tangent at $x = 2$ is parallel to the line $3 x - y + 1 = 0$ . Find $k$ .

c) Using the value of $k$ from the previous part, write down the equation of the normal line at $x = 2$ in the form $y = m x + n$ .

Solution

a) The derivative is $f^{'} (x) = 2 k x + 1$ , so $f^{'} (2) = 4 k + 1$ This is the gradient of the tangent at $x = 2$ .

b) The line $3 x - y + 1 = 0$ can be written as $y = 3 x + 1$ , and its gradient is $3$ . For the tangent to be parallel, $4 k + 1 = 3$ must hold, that is, $k = \frac{1}{2}$ .

c) With $k = \frac{1}{2}$ , the point is $(2, f (2))$ with $f (2) = \frac{1}{2} \cdot 4 + 2 = 4$ . The gradient of the tangent is $f^{'} (2) = 3$ , so the gradient of the normal is $- \frac{1}{3}$ : $y - 4 = - \frac{1}{3} (x - 2) \Rightarrow y = - \frac{1}{3} x + \frac{14}{3}$

In the exam: the parameter $k$ is found from a condition on the gradient, and that condition must be written down before you solve for $k$ . For the normal, do not forget the minus sign or the reciprocal of the gradient: a normal with gradient $3$ instead of $- \frac{1}{3}$ is not perpendicular.

Common mistakes

Summary

The tangent at $x = a$ is $y - f (a) = f^{'} (a) (x - a)$ : it passes through $(a, f (a))$ and has gradient $f^{'} (a)$ .
The normal is perpendicular to the tangent: $y - f (a) = - \frac{1}{f ^{'} ( a )} (x - a)$ if $f^{'} (a) \neq = 0$ .
If the tangent is parallel or perpendicular to a given line, equate $f^{'} (a)$ to the corresponding gradient and solve.
Always write the point with its $y$ -coordinate $f (a)$ before giving the equation.

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