# Calculus

## Applications of the Derivative # Tangent and Normal Lines

### Equation of a Tangent Line in Cartesian Coordinates

Suppose that a function $$y = f\left( x \right)$$ is defined on the interval $$\left( {a,b} \right)$$ and is continuous at $${x_0} \in \left( {a,b} \right).$$ At this point (the point $$M$$ in Figure $$1$$), the function has the value $${y_0} = f\left( {{x_0}} \right).$$

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

${y – {y_0} = f’\left( {{x_0}} \right)\left( {x – {x_0}} \right)\;\;\text{or}}\;\;\;\kern-0.3pt {y = f’\left( {{x_0}} \right)\left( {x – {x_0}} \right) + f\left( {{x_0}} \right).}$

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:

${x = x\left( t \right),}\;\;\;\kern-0.3pt{y = y\left( t \right).}$

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

${y – {y_0} = \frac{{{y’_t}}}{{{x’_t}}}\left( {x – {x_0}} \right)}\;\;\;\kern-0.3pt {\text{or}\;\;\;\kern-0.3pt\frac{{x – {x_0}}}{{{x’_t}}} = \frac{{y – {y_0}}}{{{y’_t}}}.}$

Accordingly, the equation of the normal is written as

${y – {y_0} = – \frac{{{x’_t}}}{{{y’_t}}}\left( {x – {x_0}} \right)}\;\;\;\kern-0.3pt {\text{or}\;\;\;\kern-0.3pt\frac{{x – {x_0}}}{{{y’_t}}} = – \frac{{y – {y_0}}}{{{x’_t}}}.}$

### 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:

${y – {y_0} = \frac{{{y’_\theta }}}{{{x’_\theta }}}\left( {x – {x_0}} \right)}\;\;\;\kern-0.3pt {(\text{tangent}),}$

${y – {y_0} = -\frac{{{x’_\theta }}}{{{y’_\theta }}}\left( {x – {x_0}} \right)}\;\;\;\kern-0.3pt {(\text{normal}).}$

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 $$\beta + \large\frac{\pi }{2}\normalsize.$$ 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

Click or tap a problem to see the solution.

### Example 1

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

### Example 2

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

### Example 3

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

### Example 4

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

### Example 5

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

### 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.

### Example 7

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

### 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.

### Example 9

Find the equation of the normal to the graph of the function $$y = \large{\frac{{x + 1}}{{x – 1}}}\normalsize$$ at the point where $$x = 2.$$

### Example 10

Find the equations of the tangent line and normal line to the parabola $$y = 2{x^2}$$ at the point $$\left( {2,8} \right).$$

### Example 11

Write an equation of the normal to the ellipse $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1} = 1$ at the point $$\left( {1,\large\frac{{\sqrt 3 }}{2}\normalsize} \right)$$ (Figure $$5$$).

### Example 12

Find the angle at which the parabola $$y = {x^2} – \frac{1}{4}$$ intersects the positive $$x-$$axis.

### Example 13

Find the angles at which the curve $$y = {x^3} – x$$ intersects the $$x$$-axis.

### Example 14

The normal drawn to the hyperbola $$y = \large{\frac{1}{x}}\normalsize$$ at a point $${x_0}$$ makes an angle of $$45^\circ$$ with the positive $$x-$$axis. Determine the coordinate $${x_0}$$ of the point of tangency (Figure $$6$$).

### Example 15

Write equations of the tangent and normal to the graph of the function $$y = x\sqrt {x – 1}$$ at $$x = 2.$$

### Example 16

Find the equations of the tangent and normal lines to the graph of the natural logarithm function at $${x_0} = 1.$$

### Example 17

Find the equation of the normal line to the curve $$y = {\mathop{\rm arccot}\nolimits} \large{\frac{1}{x}}\normalsize$$ at $$x = 1.$$

### Example 18

Given the parabola $$y = 2{x^2}.$$ A secant is drawn through the $$2$$ points of the parabola, which have the coordinates $$x = -1$$ and $$x = 2$$ (Figure $$8$$). Find the tangent to the parabola parallel to the secant.

### Example 19

A tangent line is drawn to the graph of the function $$y = \large{\frac{1}{x}}\normalsize$$ at the point $$\left( {1,1} \right)$$ (Figure $$9$$). Find the length of the tangent segment $$AB$$ in the first quadrant.

### Example 20

The tangent and normal lines are drawn to the parabola $$y = {x^2}$$ at the point $${x_0} = 2$$ (Figure $$10$$). Find the length of the line segment $$AB$$ between the points of intersection of the lines with the $$x-$$axis.

### Example 21

Determine the area of the triangle formed by the tangent to the graph of the function $$y = 3 – {x^2}$$ drawn at the point $$\left( {1,2} \right)$$ and the coordinate axes (Figure $$11$$).

### Example 22

The normal drawn to the curve $$y = \ln x$$ at the point $$M\left( {1,0} \right)$$ intersects the $$y-$$axis at the point $$A$$ (Figure $$12$$). Find the area $$S$$ of the triangle $$AOM.$$

### Example 23

A parabola is defined by the equation $$y = {x^2} + 2x + 3.$$ Write equations of the tangents to the parabola passing through the point $$A\left( { – 1,1} \right).$$

### Example 24

Find the equation of the tangent line drawn to the curve $${y^4} – 4{x^4} – 6xy = 0$$ at the point $$M\left( {1,2} \right).$$

### Example 25

Find the equation of the normal line to the curve $${x^3} + {y^2} + 2x – 6 = 0$$ at the point $$\left( { – 1,3} \right).$$

### Example 26

Prove that the curves $${x^2} – {y^2} = 3$$ and $$xy = 2$$ intersect at the right angle.

### Example 27

At what points do the curves $$y = {x^2}$$ and $$y = \sqrt x$$ intersect? Find the angles of intersection between the curves at these points.

### Example 28

Find the distance between the origin and the normal to the curve $$y = {e^{2x}} + {x^2}$$ at the point where $$x = 0$$ (Figure $$16$$).

### Example 29

Find the angle between the tangent to the cardioid $$r = a\left( {1 + \cos \theta } \right)$$ and the radius vector of the point of tangency (Figure $$17$$).

### Example 30

Find equation of the tangent and normal to the astroid $$x = a\,{\cos ^3}t,$$ $$y = a\,{\sin ^3}t$$ at the point $$t = \large\frac{\pi }{4}\normalsize$$ (Figure $$18$$).

### Example 31

A tangent is drawn to the graph of the function $$y =\cos x$$ at the point $$M\left( {{x_0},{y_0}} \right),$$ where $$0 \lt {x_0} \lt \large\frac{\pi }{2}\normalsize$$ (Figure $$19$$). Find the value of $${x_0},$$ at which the area of the triangle formed by the tangent and the coordinate axes will be the least.

### Example 1.

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

Solution.

${y’ = f’\left( x \right) } = {{{\left( {\sqrt x } \right)’ } = {\frac{1}{{2\sqrt x }}}, }}\\ {{f’\left( {{x_0}} \right) = f’\left( 1 \right) } = {\frac{1}{{2\sqrt 1 }} = \frac{1}{2}},}\\ {{{x_0} = 1,\;{y_0} = 1,\;\;}\kern0pt {f’\left( {{x_0}} \right) = \frac{1}{2}}}$

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 $$\large{\frac{1}{2}}\normalsize .$$ 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

$x – 2y + 3 = 0.$

### Example 9.

Find the equation of the normal to the graph of the function $$y = \large{\frac{{x + 1}}{{x – 1}}}\normalsize$$ at the point where $$x = 2.$$

Solution.

Differentiate the given function using the quotient rule:

${\require{cancel}y^\prime = \left( {\frac{{x + 1}}{{x – 1}}} \right)^\prime }={ \frac{{\left( {x + 1} \right)^\prime\left( {x – 1} \right) – \left( {x + 1} \right)\left( {x – 1} \right)^\prime}}{{{{\left( {x – 1} \right)}^2}}} }={ \frac{{x – 1 – \left( {x + 1} \right)}}{{{{\left( {x – 1} \right)}^2}}} }={ \frac{{\cancel{x} – 1 – \cancel{x} + 1}}{{{{\left( {x – 1} \right)}^2}}} }={ \frac{{ – 2}}{{{{\left( {x – 1} \right)}^2}}}.}$

At the point $${x_0} = 2,$$ the function and the derivative have the following values:

$y\left( 2 \right) = \frac{{2 + 1}}{{2 – 1}} = 3,$

${y^\prime\left( 2 \right) = f^\prime\left( 2 \right) }={ – \frac{2}{{{{\left( {2 – 1} \right)}^2}}} = – 2.}$

So the equation of the normal is given by

$y – {y_0} = – \frac{1}{{f^\prime\left( {{x_0}} \right)}}\left( {x – {x_0}} \right),$

$y – 3 = – \frac{1}{{\left( { – 2} \right)}}\left( {x – 2} \right),$

$y – 3 = \frac{x}{2} – 1,$

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

### Example 10.

Find the equations of the tangent line and normal line to the parabola $$y = 2{x^2}$$ at the point $$\left( {2,8} \right).$$

Solution.

Calculate the derivative of the function:

${y^\prime = \left( {2{x^2}} \right)^\prime = 4x,}\;\; \Rightarrow {y^\prime\left( 2 \right) = 8.}$

Write the equation of the tangent line:

$y – {y_0} = y^\prime\left( {{x_0}} \right)\left( {x – {x_0}} \right),$

$y – 8 = 8\left( {x – 2} \right),$

$y – 8 = 8x – 16,$

$8x – y – 8 = 0.$

Similarly, we get the equation of the normal line:

$y – {y_0} = – \frac{1}{{y^\prime\left( {{x_0}} \right)}}\left( {x – {x_0}} \right),$

$y – 8 = – \frac{1}{8}\left( {x – 2} \right),$

$y – 8 = – \frac{x}{8} + \frac{1}{4},$

$8y – 64 = – x + 2,$

$x + 8y – 66 = 0.$

So the answer is given by

${8x – y – 8 = 0,\;\;}\kern0pt{x + 8y – 66 = 0.}$

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Problems 1-10
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Problems 11-31