# Calculus

Applications of the Derivative# Tangent and Normal Lines

Problem 1

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

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

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

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

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

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:

### 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 \(\left( {{x_0},{y_0}} \right):\)

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

Figure 2.

### Equations of Tangent and Normal Lines in Parametric Form

Let a plane curve be given parametrically:

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:

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

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

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

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

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.}

\]

Problem 1

Problems 2-10