Select Page

# Calculus

Applications of the Derivative

# Tangent and Normal Lines

Page 1
Problem 1
Page 2
Problems 2-10

### 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

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

Figure 1.

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

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

i.e. 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 (Figure $$2$$).

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

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

Figure 2.

### 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

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

Figure 3.

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 on problem description to see solution.

### ✓Example 1

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 $$4$$).

### ✓Example 2

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

### ✓Example 3

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

### ✓Example 4

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 $$5$$). Find the tangent to the parabola parallel to the secant.

### ✓Example 5

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 $$6$$).

### ✓Example 6

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 7

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

### ✓Example 8

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.

### ✓Example 9

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 $$10$$).

### ✓Example 10

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 $$11$$). 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.

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 $$4$$).

#### Solution.

Find the derivative $$y’\left( x \right)$$ by implicit differentiation:

${{\left( {\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1}} \right)^\prime } = 1′,\;\;}\Rightarrow {\frac{{2x}}{4} + 2yy’ = 0,\;\;}\Rightarrow {4yy’ = – x,\;\;}\Rightarrow {y = – \frac{x}{{4y}}.}$

The derivative is the point of tangency is equal to

${y’\left( {{x_0},{y_0}} \right) = y’\left( {1,\frac{{\sqrt 3 }}{2}} \right) } = { – \frac{1}{{\frac{{4\sqrt 3 }}{2}}} } = { – \frac{1}{{2\sqrt 3 }}.}$

Figure 4.

Then the equation of the normal is written as

${y – {y_0} = – \frac{1}{{y’\left( {{x_0},{y_0}} \right)}}\left( {x – {x_0}} \right),\;\;}\Rightarrow { y – \frac{{\sqrt 3 }}{2} = – \frac{1}{{\left( { – \frac{1}{{2\sqrt 3 }}} \right)}}\left( {x – 1} \right),\;\;}\Rightarrow {y – \frac{{\sqrt 3 }}{2} = 2\sqrt 3 x – 2\sqrt 3 ,\;\;}\Rightarrow {y = 2\sqrt 3 x – 2\sqrt 3 + \frac{{\sqrt 3 }}{2},\;\;}\Rightarrow {y = 2\sqrt 3 x – \frac{{3\sqrt 3 }}{2}} \approx {3,46x – 2,60.}$
Page 1
Problem 1
Page 2
Problems 2-10