# Tangent and Normal Lines

## Equation of a Tangent Line in Cartesian Coordinates

Suppose that a function y = f (x) is defined on the interval (a, b) and is continuous at x0 (a, b). At this point (the point M in Figure 1), the function has the value y0 = f (x0).

Let the independent variable at $${x_0}$$ has the increment $$\Delta x.$$ The corresponding increment of the function $$\Delta y$$ is expressed as

$\Delta y = f\left( {{x_0} + \Delta x} \right) - f\left( {{x_0}} \right).$

In Figure $$1,$$ the point $${M_1}$$ has the coordinates $$\left( {{x_0} + \Delta x,{y_0} + \Delta y} \right).$$ We draw the secant $$M{M_1}.$$ Its equation has the form

$y - {y_0} = k\left( {x - {x_0}} \right),$

where $$k$$ is the slope coefficient depending on the increment $$\Delta x$$ and equal

$k = k\left( {\Delta x} \right) = \frac{{\Delta y}}{{\Delta x}}.$

When $$\Delta x$$ decreases, the point $${M_1}$$ moves to the point $$M:$$ $${M_1} \to M.$$ In the limit $$\Delta x \to 0$$ the distance between the points $$M$$ and $${M_1}$$ approaches zero. This follows from the continuity of the function $$f\left( x \right)$$ at $${x_0}:$$

$\lim\limits_{\Delta x \to 0} \Delta y = 0,\;\; \Rightarrow \lim\limits_{\Delta x \to 0} \left| {M{M_1}} \right| = \lim\limits_{\Delta x \to 0} \sqrt {{{\left( {\Delta x} \right)}^2} + {{\left( {\Delta y} \right)}^2}} = 0.$

The limiting position of the secant $$M{M_1}$$ is just the tangent line to the graph of the function $$y = f\left( x \right)$$ at point $$M.$$

There are two kinds of tangent lines - oblique (slant) tangents and vertical tangents.

### Definition 1.

If there is a finite limit $$\lim\limits_{\Delta x \to 0} k\left( {\Delta x} \right) = {k_0},$$ then the straight line given by the equation

$y - {y_0} = k\left( {x - {x_0}} \right),$

is called the oblique (slant) tangent to the graph of the function $$y = f\left( x \right)$$ at the point $$\left( {{x_0},{y_0}} \right).$$

### Definition 2.

If the limit value of $$k$$ as $$\Delta x \to 0$$ is infinite: $$\lim\limits_{\Delta x \to 0} k\left( {\Delta x} \right) = \pm \infty,$$ then the straight line given by the equation

$x = {x_0},$

is called the vertical tangent to the graph of the function $$y = f\left( x \right)$$ at the point $$\left( {{x_0},{y_0}} \right).$$

It is important that

${k_0} = \lim\limits_{\Delta x \to 0} k\left( {\Delta x} \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = f'\left( {{x_0}} \right),$

that is the slope of the tangent line is equal to the derivative of the function $$f\left( {{x_0}} \right)$$ at the tangency point $${x_0}.$$ Therefore, the equation of the oblique tangent can be written in the form

$y - {y_0} = f'\left( {{x_0}} \right)\left( {x - {x_0}} \right)\;\;\text{or}\;\;y = f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) + f\left( {{x_0}} \right).$

Since the slope of a straight line is equal to the tangent of the slope angle $$\alpha,$$ which the line forms with the positive direction of the $$x$$-axis, then the following triple identity is valid:

$k = \tan \alpha = f'\left( {{x_0}} \right).$

## Equation of a Normal Line in Cartesian Coordinates

A straight line perpendicular to the tangent and passing through the point of tangency $$\left( {{x_0},{y_0}} \right)$$ is called the normal to the graph of the function $$y = f\left( x \right)$$ at this point $$\left({\text{Figure }2}\right).$$

From geometry it is known that the product of the slopes of perpendicular lines is equal to $$-1.$$ Therefore, knowing the equation of a tangent at the point $$\left( {{x_0},{y_0}} \right):$$

$y - {y_0} = f'\left( {{x_0}} \right)\left( {x - {x_0}} \right),$

we can immediately write the equation of the normal in the form

$y - {y_0} = - \frac{1}{{f'\left( {{x_0}} \right)}}\left( {x - {x_0}} \right).$

### Special Cases

1. If the derivative $$f^\prime\left( {{x_0}} \right)$$ is zero, then we have a horizontal tangent line. This means that the normal line at this point is a vertical line. It is defined by the equation
${x = {x_0}.}$
2. If the derivative $$f^\prime\left( {{x_0}} \right)$$ approaches (plus or minus) infinity, we have a vertical tangent. In this case, the normal line is a horizontal line defined by the equation
${y = {y_0}.}$

## Equations of Tangent and Normal Lines in Parametric Form

Let a plane curve be given parametrically:

$x = x\left( t \right),\;\;\;y = y\left( t \right).$

Then the slope of the tangent drawn at the point $$\left( {{x_0},{y_0}} \right)$$ can be found using the differentiation rule for parametric functions:

$k = \tan \alpha = \frac{{{y'_t}}}{{{x'_t}}}.$

The equation of the tangent is given by

$y - {y_0} = \frac{{{y'_t}}}{{{x'_t}}}\left( {x - {x_0}} \right)\;\;\text{or}\;\;\frac{{x - {x_0}}}{{{x'_t}}} = \frac{{y - {y_0}}}{{{y'_t}}}.$

Accordingly, the equation of the normal is written as

$y - {y_0} = - \frac{{{x'_t}}}{{{y'_t}}}\left( {x - {x_0}} \right)\;\;\text{or}\;\;\frac{{x - {x_0}}}{{{y'_t}}} = - \frac{{y - {y_0}}}{{{x'_t}}}.$

## Equations of Tangent and Normal Lines in Polar Coordinates

Suppose that a curve is defined by a polar equation $$r = f\left( \theta \right),$$ which expresses the dependence of the length of the radius vector $$r$$ on the polar angle $$\theta.$$ In Cartesian coordinates, this curve will be described by the system of equations

$\left\{ \begin{array}{l} x = r\cos \theta = f\left( \theta \right)\cos \theta \\ y = r\sin \theta = f\left( \theta \right)\sin\theta \end{array} \right..$

Thus, we have written the parametric equation of the curve, where the angle $$\theta$$ plays the role of a parameter. Next, it is easy to obtain an expression for the slope of the tangent to the curve at the point $$\left( {{x_0},{y_0}} \right):$$

$k = \tan \theta = \frac{{{y'_\theta }}}{{{x'_\theta }}} = \frac{{{{\left( {r\sin \theta } \right)}^\prime }}}{{{{\left( {r\cos \theta } \right)}^\prime }}} = \frac{{{r'_\theta }\sin \theta + r\cos \theta }}{{{r'_\theta }\cos\theta - r\sin \theta }}.$

As a result, the equations of the tangent and normal lines are written as follows:

$y - {y_0} = \frac{{{y'_\theta }}}{{{x'_\theta }}}\left( {x - {x_0}} \right)\;\;(\text{tangent}),$
$y - {y_0} = -\frac{{{x'_\theta }}}{{{y'_\theta }}}\left( {x - {x_0}} \right)\;\;(\text{normal}).$

The study of curves can be performed directly in polar coordinates without transition to the Cartesian system. In this case, instead of the angle $$\theta$$ with the polar axis (i.e. with the positive direction of the $$x$$-axis), it is more convenient to use the angle $$\beta$$ with the line containing the radius vector $$r$$ (Figure $$3$$).

The tangent of the angle $$\beta$$ is calculated by the formula

$\tan \beta = \frac{r}{{{r'_\theta }}}.$

The angle formed by the normal and the extended radius vector is $$\beta + \frac{\pi }{2}.$$ Using the reduction identity, we get:

$\tan \left( {\beta + \frac{\pi }{2}} \right) = - \cot \beta = - \frac{1}{{\tan \beta }} = - \frac{{{r'_\theta }}}{r}.$

## Solved Problems

### Example 1.

Find the equation of the tangent to the curve ${y = \sqrt x}$ at the point $${\left({1,1}\right)}$$ (Figure $$4$$).

Solution.

$y' = f'\left( x \right) = \left( {\sqrt x } \right)' = \frac{1}{{2\sqrt x }},$
$f'\left( {{x_0}} \right) = f'\left( 1 \right) = \frac{1}{{2\sqrt 1 }} = \frac{1}{2},$
${x_0} = 1,\;{y_0} = 1,\;\;f'\left( {{x_0}} \right) = \frac{1}{2}.$

Substitute the $$3$$ values into the equation of the tangent line:

${y - {y_0} } = f'\left( {{x_0}} \right)\left( {x - {x_0}} \right).$

This yields:

$y - 1 = \frac{1}{2}\left( {x - 1} \right)\;\; \Rightarrow y - 1 = \frac{x}{2} - \frac{1}{2}\;\; \Rightarrow y = \frac{x}{2} - \frac{1}{2} + 1\;\; \Rightarrow {y = \frac{x}{2} + \frac{1}{2}} .$

$y = \frac{x}{2} + \frac{1}{2}.$

### Example 2.

Find a point on the curve $y = {x^2} - 2x - 3$ at which the tangent is parallel to the $$x-$$axis.

Solution.

Since the tangent is parallel to the $$x-$$axis, the derivative is equal to zero at this point. Hence,

$y^\prime = \left( {{x^2} - 2x - 3} \right)^\prime = 2x - 2 = 0.$

Then we find that

${x_0} = 1.$

### Example 3.

Find the equation of the tangent line to the curve $y = {x^4}$ at the point $$\left( { - 1,1} \right).$$

Solution.

First we find the derivative of the function:

$f'\left( x \right) = \left( {{x^4}} \right)' = 4{x^3}.$

Calculate the value of the derivative at $${x_0} = - 1:$$

$f'\left( {{x_0}} \right) = f'\left( { - 1} \right) = 4 \cdot {\left( { - 1} \right)^3} = - 4.$

Substitute the $$3$$ known numbers and find the equation of the tangent line:

$y - {y_0} = f'\left( {{x_0}} \right)\left( {x - {x_0}} \right),$
$y - 1 = - 4\left( {x - \left( { - 1} \right)} \right),$
$y - 1 = - 4\left( {x + 1} \right),$
$y - 1 = - 4x - 4,$
$y = - 4x - 3.$

### Example 4.

Find the equation of the tangent line to the curve $y = {x^3}$ at $${x_0} = 1.$$

Solution.

First we find the derivative:

$y^\prime = f^\prime\left( x \right) = \left( {{x^3}} \right)^\prime = 3{x^2}.$

The value of the derivative at the point of tangency is

$f^\prime\left( {{x_0}} \right) = 3 \cdot {1^2} = 3.$

Calculate $${y_0}:$$

${y_0} = {\left( {{x_0}} \right)^3} = {1^3} = 1.$

Substitute this in the equation of tangent:

$y - 1 = 3\left( {x - 1} \right),$
$y - 1 = 3x - 3,$
$y = 3x - 2.$

### Example 5.

Find the equation of the tangent line to the curve $y = \ln {x^2}$ that is parallel to the straight line $$y = x.$$

Solution.

The derivative of the function is given by

$y' = \left( {\ln {x^2}} \right)' = \frac{1}{{{x^2}}} \cdot 2x = \frac{2}{x}.$

The slope of the tangent line must be equal to $$1$$ as it follows from the equation of the straight line. This allows to find the tangency point:

$\frac{2}{x} = 1, \Rightarrow {x_0} = 2.$

Calculate the value of the function at this point:

${y_0} = y\left( 2 \right) = \ln {2^2} = \ln 4.$

Now we can write the equation of the tangent line:

$y - {y_0} = f'\left( {{x_0}} \right)\left( {x - {x_0}} \right),$
$y - \ln 4 = 1 \cdot \left( {x - 2} \right),$
$y - \ln 4 = x - 2,$
$y = x + \ln 4 - 2.$

### Example 6.

Find a point on the curve $y = \sqrt x,$ where the tangent makes an angle of 45 degrees with the positive $$x-$$axis.

Solution.

We use the triple identity

$k = \tan \alpha = f'\left( {{x_0}} \right).$

This yields

$k = \tan 45^\circ = 1,$

so the derivative is equal to

$f'\left( {{x_0}} \right) = 1.$

From the other side,

$\left( {\sqrt x } \right)' = \frac{1}{{2\sqrt x }},\;\; \Rightarrow \frac{1}{{2\sqrt {{x_0}} }} = 1.$

Therefore

$2\sqrt {{x_0}} = 1,\;\;\Rightarrow \sqrt {{x_0}} = \frac{1}{2},\;\;\Rightarrow {x_0} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}.$

### Example 7.

Find the equation of the normal line to the curve $y = {x^3} + {e^x}$ at $${x_0} = 0.$$

Solution.

Determine the value of the function at $${x_0} = 0.$$

${y_0} = y\left( 0 \right) = {0^3} + {e^0} = 1.$

The derivative is given by

$y^\prime\left( x \right) = \left( {{x^3} + {e^x}} \right)^\prime = 3{x^2} + {e^x}.$

At the point $${x_0} = 0,$$ it equals

$y^\prime\left( 0 \right) = 3 \cdot {0^2} + {e^0} = 1.$

Thus, the equation of the normal is written as follows:

$y - {y_0} = - \frac{1}{{y^\prime\left( {{x_0}} \right)}}\left( {x - {x_0}} \right),$
$y - 1 = - \frac{1}{1}\left( {x - 0} \right),$
$y = - x + 1.$

### Example 8.

The equation of the tangent line to the graph of a function at $${x_0} = 1$$ is defined by the equation $2x + y - 4 = 0.$ Find the equation of the normal line passing through this point.

Solution.

We rewrite the equation of the tangent as

$y = - 2x + 4$

and find the $$y-$$coordinate of the tangency point:

${y_0} = - 2 \cdot 1 + 4 = 2.$

The slope of the tangent line is $$-2.$$ Since the slope of the normal line is the negative reciprocal of the slope of the tangent line, we get that the slope of the normal is equal to $$\frac{1}{2}.$$ So the equation of the normal can be written as

$y - {y_0} = k\left( {x - {x_0}} \right),$
$y - 2 = \frac{1}{2}\left( {x - 1} \right),$
$y - 2 = \frac{x}{2} - \frac{1}{2},$
$2y - 4 = x - 1,$

or

$x - 2y + 3 = 0.$

See more problems on Pages 2,3.