Calculus

Applications of the Derivative

Applications of Derivative Logo

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).

Equation of a tangent line in Cartesian coordinates
Figure 1.

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 \(\left({\text{Figure }2}\right).\)

Equation of a normal line in Cartesian coordinates
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

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

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 \(3\)).

Equations of tangent and normal lines in polar coordinates
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.

Tangent line to the graph of the square root function.
Figure 4.
\[y' = f'\left( x \right) = \left( {\sqrt x } \right)' = \frac{1}{{2\sqrt x }},\]
\[f'\left( {{x_0}} \right) = f'\left( 1 \right) = \frac{1}{{2\sqrt 1 }} = \frac{1}{2},\]
\[{x_0} = 1,\;{y_0} = 1,\;\;f'\left( {{x_0}} \right) = \frac{1}{2}.\]

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:

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

Answer:

\[y = \frac{x}{2} + \frac{1}{2}.\]

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.

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