Tangents and normals to curves

A tangent is a straight line with the same instantaneous gradient as a curve at a chosen point. A normal is the straight line perpendicular to that tangent at the same point.

These lines turn a derivative into geometric information. They are used to approximate functions, describe motion and solve problems involving directions perpendicular to a curve.

Before you begin

You should be able to:

• find a derivative using standard differentiation rules
• calculate the gradient and equation of a straight line
• substitute a coordinate into the original curve

From derivative to tangent

For a curve $y=f(x)$, the tangent gradient at $x=a$ is

$m_{\mathrm{tangent}}=f'(a).$

The point on the curve is $(a,f(a))$. A line of gradient $m$ through $(x_1,y_1)$ has equation

$\boxed{y-y_1=m(x-x_1)}.$

This point-gradient form is usually the safest form to use. It makes the point and gradient visible, and avoids unnecessary algebra.

Example 1: tangent to a polynomial

Find the tangent to

$y=x^3-2x+4$

at $x=2$.

Differentiate:

$\frac{dy}{dx}=3x^2-2.$

At $x=2$, the tangent gradient is

$m=3(2)^2-2=10.$

Find the point from the original curve:

$y=2^3-2(2)+4=8.$

The tangent passes through $(2,8)$, so

$\boxed{y-8=10(x-2)}.$

Equivalently, $y=10x-12$.

Do not substitute $x=2$ before differentiating. Doing so would replace the whole function by the constant $8$, losing all gradient information.

The normal gradient

Two non-vertical perpendicular lines with gradients $m_1$ and $m_2$ satisfy

$m_1m_2=-1.$

Therefore, if the tangent gradient is a finite non-zero value $m$,

$\boxed{m_{\mathrm{normal}}=-\frac1m}.$

The negative reciprocal changes both the sign and the reciprocal. For example, the perpendicular gradient to $2/3$ is $-3/2$, not $3/2$ and not $-2/3$.

Example 2: tangent and normal to a rational function

For

$y=\frac{6}{x}+x,$

find the tangent and normal at $x=2$.

Write $6/x=6x^{-1}$ and differentiate:

$\frac{dy}{dx}=-6x^{-2}+1=1-\frac{6}{x^2}.$

At $x=2$,

$m_{\mathrm{tangent}}=1-\frac64=-\frac12.$

The point is

$\left(2,\frac62+2\right)=(2,5).$

Hence the tangent is

$\boxed{y-5=-\frac12(x-2)}.$

The normal gradient is

$m_{\mathrm{normal}}=-\frac{1}{-1/2}=2,$

so the normal is

$\boxed{y-5=2(x-2)}.$

A dependable method

For a tangent or normal at $x=a$:

1. Differentiate the curve.
2. Evaluate $f'(a)$ to find the tangent gradient.
3. Evaluate $f(a)$ using the original curve to find the point.
4. For a normal, take the negative reciprocal of the tangent gradient.
5. Substitute the point and gradient into $y-y_1=m(x-x_1)$.
6. Check that the final line passes through the stated point.

If a coordinate rather than an $x$ value is given, first confirm that the coordinate lies on the curve.

Finding an unknown point from a gradient

Sometimes the gradient is given and the point of contact must be found. Set the derivative equal to the required gradient, solve for $x$, then calculate every corresponding $y$ value.

Example 3: parallel tangents

Find the points on

$y=x^3-3x^2+5$

where the tangent is parallel to $y=3x-1$.

Parallel lines have equal gradients, so the required gradient is $3$. Differentiate:

$\frac{dy}{dx}=3x^2-6x.$

Set this equal to $3$:

$3x^2-6x=3,$

$x^2-2x-1=0.$

Thus

$x=1\pm\sqrt2.$

There are two answers because a curve can have the same gradient at several points. Substitute both values into the original function. Using $x^2=2x+1$ from the quadratic equation,

$x^3=x(2x+1)=2x^2+x=5x+2,$

so

$y=x^3-3x^2+5=(5x+2)-3(2x+1)+5=4-x.$

The points are therefore

$\boxed{(1+\sqrt2,\,3-\sqrt2)}$

and

$\boxed{(1-\sqrt2,\,3+\sqrt2)}.$

Horizontal and vertical special cases

The negative reciprocal rule needs care when a line is horizontal or vertical.

Tangent	Normal
Horizontal, gradient $0$	Vertical, equation $x=x_1$
Vertical, gradient undefined	Horizontal, equation $y=y_1$

There is no finite negative reciprocal of zero. A horizontal tangent and vertical normal must be written using their geometric equations.

Example 4: horizontal tangent and vertical normal

For

$y=x^3-3x+1,$

find the tangent and normal at $x=1$.

Since

$\frac{dy}{dx}=3x^2-3,$

the tangent gradient at $x=1$ is $0$. The point is $(1,-1)$.

The horizontal tangent is

$\boxed{y=-1},$

and the vertical normal is

$\boxed{x=1}.$

This point is stationary because its tangent is horizontal. Whether it is a maximum, minimum or stationary point of inflection requires further analysis, covered in stationary points and curve sketching.

For an explicit differentiable function $y=f(x)$, a vertical tangent usually arises where the displayed derivative is undefined or becomes unbounded. Parametric and implicit descriptions often represent such points more naturally.

Tangents to implicit curves

For a relation involving both $x$ and $y$, obtain $dy/dx$ by implicit differentiation, then evaluate it at the stated point. The line method is unchanged.

Example 5: tangent and normal to a circle

Find the tangent and normal to

$x^2+y^2=25$

at $(3,4)$.

Differentiate both sides with respect to $x$:

$2x+2y\frac{dy}{dx}=0.$

Hence

$\frac{dy}{dx}=-\frac{x}{y}.$

At $(3,4)$, the tangent gradient is $-3/4$. Therefore the tangent is

$\boxed{y-4=-\frac34(x-3)}.$

The normal gradient is $4/3$, giving

$\boxed{y-4=\frac43(x-3)}.$

The normal passes through the centre $(0,0)$, as the geometry of a circle predicts. Indeed, rearranging its equation gives $y=4x/3$.

For a complete treatment of differentiating relations such as this, see implicit differentiation.

Tangents to parametric curves

If $x$ and $y$ are both functions of a parameter $t$, first find

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}.$

Then use the given parameter value to calculate both the gradient and the point. The detailed method, including vertical tangents and second derivatives, is covered in parametric differentiation.

Applications

Instantaneous direction of motion

If a particle follows the graph $y=f(x)$, its tangent gives its instantaneous direction of travel. A direction perpendicular to the path lies along the normal.

For example, a track has equation

$y=\frac14x^2+2.$

At $x=4$,

$\frac{dy}{dx}=\frac12x=2.$

The track makes an angle $\theta$ with the positive $x$ axis satisfying

$\tan\theta=2.$

Thus

$\theta=\tan^{-1}(2)\approx63.4^\circ.$

An angle obtained from a gradient may need interpretation. Calculator values from $\tan^{-1}$ give a principal angle, while the context determines the actual direction.

Tangent line approximation

Near $x=a$, a smooth function is approximately equal to its tangent line:

$f(x)\approx f(a)+f'(a)(x-a).$

Use this to estimate $\sqrt{4.1}$. Let $f(x)=\sqrt{x}$ and choose the nearby convenient value $a=4$.

$f(4)=2, \qquad f'(x)=\frac{1}{2\sqrt{x}}, \qquad f'(4)=\frac14.$

Therefore

$\sqrt{4.1}\approx2+\frac14(4.1-4)=2.025.$

The true value is approximately $2.02485$, so the tangent gives an accurate local estimate. It is accurate nearby because the curve and tangent share both a point and a gradient.

Common mistakes

• Finding the point from $dy/dx$ instead of from the original curve.
• Using $f(a)$ as the gradient and $f'(a)$ as the $y$ coordinate.
• Taking a reciprocal but forgetting the negative sign for a normal.
• Treating gradient $0$ as if it had a finite negative reciprocal.
• Assuming a line is tangent merely because it meets a curve at one point.
• Solving $f'(x)=m$ and keeping only one solution when several points have gradient $m$.
• Expanding point-gradient form incorrectly. Leaving a correct equation unexpanded is acceptable unless another form is requested.
• Using degrees or radians inconsistently when converting a gradient to an angle.

Check your understanding

1. Find the tangent and normal to $y=x^2+2x-3$ at $x=1$.
2. Find every point on $y=x^3-3x$ where the tangent is horizontal, and write the corresponding normal equations.
3. The curve $x^2+xy+y^2=7$ passes through $(1,2)$. Find the tangent there.
4. Use a tangent to estimate $\sqrt[3]{8.12}$.

Answers

1

$\frac{dy}{dx}=2x+2.$

At $x=1$, the gradient is $4$ and the point is $(1,0)$. Thus

$\boxed{y=4(x-1)}$

and

$\boxed{y=-\frac14(x-1)}.$

2

$\frac{dy}{dx}=3x^2-3=0$

gives $x=\pm1$. The points are $(-1,2)$ and $(1,-2)$. Both tangents are horizontal. The normal equations are

$\boxed{x=-1 \quad\text{and}\quad x=1}.$

3

Implicit differentiation gives

$2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0.$

Therefore

$\frac{dy}{dx}=-\frac{2x+y}{x+2y}.$

At $(1,2)$, the gradient is $-4/5$, so the tangent is

$\boxed{y-2=-\frac45(x-1)}.$

4

Let $f(x)=x^{1/3}$ and use $a=8$. Then

$f(8)=2, \qquad f'(x)=\frac{1}{3x^{2/3}}, \qquad f'(8)=\frac1{12}.$

Hence

$\sqrt[3]{8.12}\approx2+\frac1{12}(0.12)=\boxed{2.01}.$

What to learn next

Study stationary points and curve sketching to classify points with horizontal tangents. Then extend the same line methods to implicit differentiation and parametric differentiation.