Calculus

Applications of the Derivative

Applications of Derivative Logo

Tangent and Normal Lines

Solved Problems

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 \(11\)).

Solution.

Find the equation of the tangent. Given that

\[f'\left( x \right) = \left( {3 - {x^2}} \right) = - 2x;\;\; \Rightarrow f'\left( 1 \right) = - 2,\]

we get the equation of the tangent in the following form:

\[y - {y_M} = f'\left( {{x_M}} \right)\left( {x - {x_M}} \right),\;\; \Rightarrow y - 2 = - 2\left( {x - 1} \right),\;\; \Rightarrow y = - 2x + 4.\]

We transform it to the intercept form:

\[y = - 2x + 4,\;\; \Rightarrow y + 2x = 4,\;\; \Rightarrow \frac{y}{4} + \frac{{2x}}{4} = 1,\;\; \Rightarrow \frac{y}{4} + \frac{x}{2} = 1.\]
Area of a triangle formed by the tangent and coordinate axes
Figure 11.

It follows that the length of the segment \(OA\) is \(4,\) and the length of the segment \(OB\) is \(2.\) Then the area of the triangle \(OAB\) is given by

\[S = \frac{{\left| {OA} \right| \cdot \left| {OB} \right|}}{2} = \frac{{4 \cdot 2}}{2} = 4.\]

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 \(AOM.\)

Solution.

Area of the triangle
Figure 12.

The derivative of the natural logarithm is

\[y^\prime = \left( {\ln x} \right)^\prime = \frac{1}{x}.\]

At \({x_0} = 1,\) we have

\[y^\prime(1) = 1.\]

Then the normal line at this point is defined by the equation

\[y - {y_0} = - \frac{1}{{y^\prime\left( {{x_0}} \right)}}\left( {x - {x_0}} \right),\]
\[y - 0 = - \frac{1}{1}\left( {x - 1} \right),\]

or

\[y = - x + 1.\]

We see that the slope of the tangent line is equal to \(-1,\) so the triangle \(AOM\) is both right and isosceles. In this case, its area is

\[S = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}.\]

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 \(A\left( { - 1,1} \right).\)

Solution.

We convert the equation of the parabola to the form

\[y = {x^2} + 2x + 3 = {x^2} + 2x + 1 + 2 = {\left( {x + 1} \right)^2} + 2.\]

It can be seen that the graph of the parabola is obtained from the graph of the function \(y = {x^2}\) by parallel shifting by 1 unit to the left and 2 units up (Figure \(13\)).

Two tangents to a parabola
Figure 13.

Let us find the equations of two tangents to the parabola passing through the point \(A\left( { - 1,1} \right).\) Each of these tangents is defined by the equation

\[y - {y_A} = k\left( {x - {x_A}} \right),\;\; \Rightarrow y - 1 = k\left( {x - \left( { - 1} \right)} \right),\;\; \Rightarrow y - 1 = kx + k,\;\; \Rightarrow y = kx + k + 1,\]

where \(k\) is the slope (\({k_1}\) for the first tangent and \({k_2}\) for the second). Thus, the problem reduces to finding of the slopes of the tangents \({k_1}\) and \({k_2}.\) Take into account that at the points of tangency \(B\) and \(C\) the following condition holds:

\[\left\{ \begin{array}{l} y = kx + k + 1\\ y = {x^2} + 2x + 3 \end{array} \right.,\;\; \Rightarrow kx + k + 1 = {x^2} + 2x + 3. \]

Also at the points of tangency \(B\) and \(C,\) the slope is equal to the derivative of the function \(y = {x^2} + 2x + 3.\) Since

\[y' = \left( {{x^2} + 2x + 3} \right)^\prime = 2x + 2,\]

then we obtain another equation in the form

\[k = 2x + 2.\]

As a result, we have the system of two equations

\[\left\{ \begin{array}{l} kx + k + 1 = {x^2} + 2x + 3\\ k = 2x + 2 \end{array} \right.\]

with two unknowns \(k\) and \(x.\) Solving this system, we find the values of \(k\) and \(x\) (i.e. the slopes of the tangents \({k_1},\) \({k_2}\) and \(x\)-coordinates of the points of tangency \(B\) and \(C\)):

\[\left\{ \begin{array}{l} kx + k + 1 = {x^2} + 2x + 3\\ k = 2x + 2 \end{array} \right.,\;\; \Rightarrow \left( {2x + 2} \right)x + 2x + 2 + 1 = {x^2} + 2x + 3,\;\; \Rightarrow 2{x^2} + 2x + 2x + 3 = {x^2} + 2x + 3,\;\; \Rightarrow {x^2} + 2x = 0,\;\; \Rightarrow {x_1} = - 2,\;{x_2} = 0.\]

The first solution \({x_1} = - 2\) corresponds to point \(B.\) The second solution \({x_2} = 0\) is the coordinate of the point of tangency \(C.\) The slopes have the following values:

  1. tangent \(AB:\;\) \({x_1} = -2,\) \({k_1} = -2;\)
  2. tangent \(AC:\;\) \({x_2} = 0,\) \({k_2} = 2.\)

Then the equations of the tangents to the parabola are given by

  1. tangent \(AB:\;\) \(y = -2x - 1;\)
  2. tangent \(AC:\;\) \(y = 2x + 3.\)

Example 24.

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

Solution.

This curve is given in implicit form. So we differentiate the equation implicitly to determine the derivative:

\[\left( {{y^4} - 4{x^4} - 6xy} \right)^\prime = 0^\prime,\]
\[4{y^3}y^\prime - 16{x^3} - 6\left( {y + xy^\prime} \right) = 0,\]
\[4{y^3}y^\prime - 16{x^3} - 6y - 6xy^\prime = 0,\]
\[y^\prime\left( {2{y^3} - 3x} \right) = 3y + 8{x^3},\]
\[y^\prime = \frac{{3y + 8{x^3}}}{{2{y^3} - 3x}}.\]

Calculate the value of the derivative at the point \(M\left( {1,2} \right):\)

\[y^\prime(1,2) = \frac{{3 \cdot 2 + 8 \cdot {1^3}}}{{2 \cdot {2^3} - 3 \cdot 1}} = \frac{{14}}{{13}}.\]

Now we can write the equation of the tangent line:

\[y - 2 = \frac{{14}}{{13}}\left( {x - 1} \right),\]
\[y - 2 = \frac{{14}}{{13}}x - \frac{{14}}{{13}},\]
\[13y - 26 = 14x - 14,\]
\[14x - 13y + 12 = 0.\]

Example 25.

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

Solution.

The given function is defined implicitly. We follow the same steps to get the equation of the normal as for an explicit function. First we find the derivative (using implicit differentiation):

\[\left( {{x^3} + {y^2} + 2x - 6} \right)^\prime = 0^\prime,\]
\[3{x^2} + 2yy^\prime + 2 = 0,\]
\[2yy^\prime = - 3{x^2} - 2,\]
\[y^\prime = - \frac{{3{x^2} + 2}}{{2y}}.\]

Substitute the coordinates of the tangency point:

\[y^\prime = - \frac{{3 \cdot {{\left( { - 1} \right)}^2} + 2}}{{2 \cdot 3}} = - \frac{5}{6}.\]

Then we have the equation of the normal in the following form

\[y - {y_0} = - \frac{1}{{y^\prime\left( {{x_0}} \right)}}\left( {x - {x_0}} \right),\]
\[y - 3 = - \frac{1}{{\left( { - \frac{5}{6}} \right)}}\left( {x - \left( { - 1} \right)} \right),\]
\[y - 3 = \frac{6}{5}\left( {x + 1} \right),\]
\[y = \frac{6}{5}x + \frac{6}{5} + 3,\]
\[y = \frac{6}{5}x + \frac{{21}}{5},\]
\[6x - 5y + 21 = 0.\]

Example 26.

Prove that the curves \({x^2} - {y^2} = 3\) and \(xy = 2\) intersect at the right angle.

Solution.

These curves are hyperbolas shown schematically in Figure \(14.\)

Two curves intersecting at the right angle
Figure 14.

Determine their points of intersection:

\[ \left\{ \begin{array}{l} {x^2} - {y^2} = 3\\ xy = 2 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} {x^2} - {y^2} = 3\\ y = \frac{2}{x} \end{array} \right.,\;\; \Rightarrow {x^2} - {\left( {\frac{2}{x}} \right)^2} = 3,\;\; \Rightarrow \frac{{{x^2} - 4}}{{{x^2}}} = 3,\;\; \Rightarrow {x^4} - 3{x^2} - 4 = 0,\;\; \Rightarrow D = 9 + 16 = 25,\;\; \Rightarrow {x^2} = \frac{{3 \pm 5}}{2} = - 1;\;4.\]

Obviously, the points of intersection can be determined from the condition \({x^2} = 4:\)

\[{x^2} = 4,\;\; \Rightarrow {x_{1,2}} = \pm 2.\]

Compute the \(y\)-coordinates of these points:

\[ {x_1} = - 2,\;\; \Rightarrow {y_1} = \frac{2}{{\left( { - 2} \right)}} = - 1;\]
\[{x_2} = 2,\;\; \Rightarrow {y_2} = \frac{2}{2} = 1.\]

Now we find the derivatives of the given functions. We calculate the derivative of the first function using implicit differentiation:

\[{x^2} - {y^2} = 3,\;\; \Rightarrow {\left( {{x^2} - {y^2}} \right)^\prime } = 3',\;\; \Rightarrow 2x - 2yy' = 0,\;\; \Rightarrow y' = \frac{x}{y}.\]

The derivative of the second function is written as

\[y' = {\left( {\frac{2}{x}} \right)^\prime } = - \frac{2}{{{x^2}}}.\]

Determine the values of the derivatives at \(x = -2\) (thus we find the slopes of the tangents to each hyperbola at this point) and make sure that the product of the slopes of the tangents at this point is equal to \(-1:\)

\[{k_1} = {\left( {\frac{x}{y}} \right)_{\substack{ x = -2\\ y= -1}}} = \frac{{\left( { - 2} \right)}}{{\left( { - 1} \right)}} = 2;\;\;{k_2} = {\left( { - \frac{2}{{{x^2}}}} \right)_{x = - 2}} = - \frac{1}{2};\;\;\Rightarrow {k_1}{k_2} = 2 \cdot \left( { - \frac{1}{2}} \right) = - 1.\]

We apply the same test for the second point of intersection:

\[{k_1} = {\left( {\frac{x}{y}} \right)_{\substack{ x = 2\\ y= 1}}} = \frac{{2}}{{1}} = 2;\;\;{k_2} = {\left( { - \frac{2}{{{x^2}}}} \right)_{x = 2}} = - \frac{1}{2};\;\; \Rightarrow {k_1}{k_2} = 2 \cdot \left( { - \frac{1}{2}} \right) = - 1.\]

Thus, the curves intersect at the right angle at each of these points.

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.

Solution.

Intersection of two curves
Figure 15.

Firstly, we determine the points of intersection of the two curves. Solve the following equation:

\[{x^2} = \sqrt x , \;\Rightarrow {x^4} = x, \;\Rightarrow {x^4} - x = 0,\; \Rightarrow x\left( {{x^3} - 1} \right) = 0.\]

The equation has two roots:

\[{x_1} = 0,\;{x_2} = 1.\]

Find the angle between the curves at the first point \({x_1} = 0.\) Calculate the values of the derivatives of both functions at this point:

\[{f_1}^\prime\left( x \right) = {\left( {{x^2}} \right)^\prime } = 2x,\;\Rightarrow {f_1}^\prime\left( {{x_1}} \right) = 2{x_1} = 0;\]
\[{f_2}^\prime\left( x \right) = {\left( {\sqrt x } \right)^\prime } = \frac{1}{{2\sqrt x }},\; \Rightarrow {f_2}^\prime\left( {{x_1}} \right) = \frac{1}{{2\sqrt {{x_1}} }} = \frac{1}{{2 \cdot 0}} = \infty .\]

As the slope of the tangent line is equal to the derivative of the function at the point of tangency, then we immediately conclude that the tangent line to the first function is horizontal \(({\tan{\alpha _1} = 0,\;} \Rightarrow {{\alpha _1} = 0)},\) and the tangent line to the second function is vertical \(\left({\tan{\alpha _2} = \infty ,\;} \Rightarrow {{\alpha _2} = \frac{\pi }{2}}\right).\) Therefore the angle \({\beta _1}\) between the curves at the point \({x_1} = 0\) is equal to

\[{\beta _1} = {\alpha _2} - {\alpha _1} = \frac{\pi }{2} - 0 = \frac{\pi }{2}.\]

Now consider intersection of the curves at the second point \({x_2} = 1.\) Similarly, calculate the derivatives:

\[{f_1}^\prime\left( {{x_2}} \right) = 2{x_2} = 2;\]
\[{f_2}^\prime\left( {{x_2}} \right) = \frac{1}{{2\sqrt {{x_2}} }} = \frac{1}{2}.\]

Next, we use the remarkable tangent subtraction formula

\[\tan \left( {{\alpha _2} - {\alpha _1}} \right) = \frac{{\tan {\alpha _2} - \tan {\alpha _1}}}{{1 + \tan {\alpha _2}\tan {\alpha _1}}}.\]

Given that \(\tan {\alpha _1} = 2\) and \(\tan {\alpha _2} = \frac{1}{2},\) we have

\[\tan{\beta _2} = \tan \left( {{\alpha _2} - {\alpha _1}} \right) = \frac{{\tan {\alpha _2} - \tan {\alpha _1}}}{{1 + \tan {\alpha _2}\tan {\alpha _1}}} = \frac{{\frac{1}{2} - 2}}{{1 + \frac{1}{2} \cdot 2}} = \frac{{ - \frac{3}{2}}}{2} = - \frac{3}{4}.\]

So, the angle between the curves at the second point of intersection is equal to

\[{\beta _2} = \arctan \left( { - \frac{3}{4}} \right) = - \arctan \frac{3}{4}.\]

In the final answer, we indicate the absolute value of the angle, i.e. \({\beta _2} = \arctan \frac{3}{4}.\)

We also compute the \(y-\)values of the points of intersection:

\[y({x_1}) = x_1^2 = \sqrt {{x_1}} = 0;\]
\[y({x_2}) = x_2^2 = \sqrt {{x_2}} = 1.\]

Thus, the final answer is given by

\[\left( {0,0} \right):{\beta _1} = \frac{\pi }{2};\;\;\left( {1,1} \right):{\beta _2} = \arctan \frac{3}{4}.\]

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 \(16\)).

Solution.

Distance between the origin and normal
Figure 16.

Calculate the value of the function at the point \(x = 0:\)

\[y\left( 0 \right) = {e^0} + {0^2} = 1.\]

Find the derivative:

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

Then the derivative value at \(x = 0\) is

\[y^\prime\left( 0 \right) = 2{e^0} + 2 \cdot 0 = 2.\]

Write the equation of the normal passing through the point \(M\left( {0,1} \right):\)

\[y - {y_0} = - \frac{1}{{y^\prime\left( {{x_0}} \right)}}\left( {x - {x_0}} \right),\]
\[y - 1 = - \frac{1}{2}\left( {x - 0} \right),\]
\[y = - \frac{x}{2} + 1.\]

We find from the equation that

\[ {ON} = 2,\; {OM} = 1.\]

The distance \(L\) between the origin and the normal line is the line segment \(OP\), which is the altitude of the right triangle \(OMN.\) The altitude \(L=OP\) is given by the formula

\[L = OP = \frac{{ON \cdot OM}}{{MN}}.\]

Using the Pythagorean theorem, we have

\[MN = \sqrt {O{N^2} + O{M^2}}.\]

Thus, the distance \(L\) is equal to

\[L = OP = \frac{{ON \cdot OM}}{{MN}} = \frac{{ON \cdot OM}}{{\sqrt {O{N^2} + O{M^2}} }} = \frac{{2 \cdot 1}}{{\sqrt {{2^2} + {1^2}} }} = \frac{2}{{\sqrt 5 }}.\]

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 \(17\)).

Solution.

The given angle (Figure \(17\)) is calculated by the formula

\[\tan \omega = \frac{r}{{{r'_\theta }}}.\]
Angle between the tangent to a cardioid and the radius vector.
Figure 17.

Here

\[r'_\theta = \left[ {a\left( {1 + \cos \theta } \right)} \right]^\prime = - a\sin \theta .\]

Consequently,

\[ \tan \omega = \frac{{a\left( {1 + \cos \theta } \right)}}{{\left( { - a\sin \theta } \right)}} = - \frac{{1 + \cos \theta }}{{\sin\theta }} = - \frac{{\cancel{2}{{\cos }^{\cancel{2}}}\frac{\theta }{2}}}{{\cancel{2}\sin \frac{\theta }{2}\cancel{\cos \frac{\theta }{2}}}} = - \cot \frac{\theta }{2} = \tan \left( {\frac{\theta }{2} + \frac{\pi }{2}} \right).\]

In the last expression we have used a reduction identity. Thus, the angle between the tangent and the radius vector is

\[\omega = \frac{\theta }{2} + \frac{\pi }{2}.\]

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 \(t = \frac{\pi }{4}\) (Figure \(18\)).

Solution.

Calculate the derivatives of the parametric function:

\[{x'_t} = \left( {a\,{{\cos }^3}t} \right)^\prime = - 3a\,{\cos ^2}t\sin t;\;\;{y'_t} = {\left( {a\,{{\sin }^3}t} \right)^\prime } = 3a\,{\sin ^2}t\cos t.\]

Consequently,

\[{y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{3a\,{{\sin }^2}t\cos t}}{{\left( { - 3a\,{{\cos }^2}t\sin t} \right)}} = - \frac{{\sin t}}{{\cos t}} = - \tan t.\]

By the reduction formula, we can write

\[ - \tan t = \tan \left( {\pi - t} \right).\]

Since \(\tan \alpha = {y'_x} = \tan \left( {\pi - t} \right),\) the angle \(\alpha\) is equal to

\[\alpha = \pi - t = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4} = 135^{\circ}.\]

Then the derivative of the cardioid and, accordingly, the slope of the tangent at the point of tangency are equal

\[{y'_x}\left( {\frac{\pi }{4}} \right) = \tan \frac{{3\pi }}{4} = - 1.\]

Find the coordinates of the point of tangency:

\[{x_0} = x\left( {\frac{\pi }{4}} \right) = a\,{\cos ^3}\frac{\pi }{4} = a{\left( {\frac{{\sqrt 2 }}{2}} \right)^3} = \frac{{a\sqrt 2 }}{4},\]
\[{y_0} = y\left( {\frac{\pi }{4}} \right) = a\,{\sin ^3}\frac{\pi }{4} = a{\left( {\frac{{\sqrt 2 }}{2}} \right)^3} = \frac{{a\sqrt 2 }}{4}.\]

Now we can write the equation of the tangent:

\[y - {y_0} = {y'_x}\left( {{x_0}} \right)\left( {x - {x_0}} \right),\;\; \Rightarrow y - \frac{{a\sqrt 2 }}{4} = - 1\left( {x - \frac{{a\sqrt 2 }}{4}} \right),\;\; \Rightarrow y - \frac{{a\sqrt 2 }}{4} = - x + \frac{{a\sqrt 2 }}{4},\;\; \Rightarrow y = - x + \frac{{a\sqrt 2 }}{2}\]

and the equation of the normal:

\[y - {y_0} = - \frac{1}{{{{y'}_x}\left( {{x_0}} \right)}}\left( {x - {x_0}} \right),\;\; \Rightarrow y - \frac{{a\sqrt 2 }}{4} = - \frac{1}{{\left( { - 1} \right)}}\left( {x - \frac{{a\sqrt 2 }}{4}} \right),\;\; \Rightarrow y - \cancel{\frac{{a\sqrt 2 }}{4}} = x - \cancel{\frac{{a\sqrt 2 }}{4}},\;\; \Rightarrow y = x.\]
Tangent and normal to the astroid
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 \(0 \lt {x_0} \lt \frac{\pi }{2}\) (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.

Solution.

The least triangle formed by the tangent and coordinates axes
Figure 19.

As the derivative of the cosine is

\[y'\left( x \right) = \left( {\cos x} \right)^\prime = - \sin x,\]

the slope of the tangent is given by

\[\tan \alpha = - \sin {x_0} = y'\left( {{x_0}} \right).\]

Then the equation of the tangent has the form

\[y - {y_0} = y'\left( {{x_0}} \right)\left( {x - {x_0}} \right),\;\; \Rightarrow y - \cos {x_0} = - \sin {x_0}\left( {x - {x_0}} \right),\;\; \Rightarrow y - \cos {x_0} = \left( { - \sin {x_0}} \right)x - \left( { - \sin {x_0}} \right){x_0},\;\; \Rightarrow y + \left( {\sin {x_0}} \right)x = \cos {x_0} + \left( {\sin {x_0}} \right){x_0}.\]

We represent it in the intercept form:

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

Hence,

\[\frac{y}{{\cos {x_0} + \left( {\sin {x_0}} \right){x_0}}} + \frac{{\left( {\sin {x_0}} \right)x}}{{\cos {x_0} + \left( {\sin {x_0}} \right){x_0}}} = 1,\]

that is, the legs of the right triangle \(OAB\) are

\[\left| {OA} \right| = q = \cos {x_0} + \left( {\sin {x_0}} \right){x_0},\;\;\left| {OB} \right| = p = \frac{{\cos {x_0} + \left( {\sin {x_0}} \right){x_0}}}{{\sin {x_0}}}.\]

For convenience, we denote \({x_0} = z\) and express the area of the triangle \(OAB\) as a function \(S\left( z \right):\)

\[S = S\left( z \right) = \frac{{pq}}{2} = \frac{{{{\left( {\cos z + z\sin z} \right)}^2}}}{{2\sin z}}.\]

Investigate the extreme values of the function \(S\left( z \right).\) Its derivative is given by

\[S'\left( z \right) = \frac{1}{2}{\left[ {\frac{{{{\left( {\cos z + z\sin z} \right)}^2}}}{{\sin z}}} \right]^\prime } = \frac{{\left( {\cos z + z\sin z} \right)}}{{2\,{{\sin }^2}z}} \cdot \left[ {{z\cos z\sin z} - {{\cos }^2}z} \right].\]

Since in the interval \(0 \lt z \lt {\frac{\pi }{2}}\)

\[\cos z + z\sin z \gt 0,\]

then the derivative has only one critical point that is determined by the condition

\[z\cos z\sin z - {\cos ^2}z = 0,\;\; \Rightarrow \cos z\left( {z\sin z - \cos z} \right) = 0,\;\; \Rightarrow z - \cot z = 0.\]

This equation can be solved numerically. However, one can see that if \(z = {\frac{\pi }{4}},\) then the left-hand side is negative:

\[z = \frac{\pi }{4}:\;\;z - \cot z = \frac{\pi }{4} - \cot \frac{\pi }{4} = \frac{\pi }{4} - 1 \approx - 0,21 \lt 0,\]

and at \(z = {\frac{\pi }{3} },\) the left-hand side is positive:

\[z = \frac{\pi }{3}:\;\;z - \cot z = \frac{\pi }{3} - \cot \frac{\pi }{3} = \frac{\pi }{3} - \frac{1}{{\sqrt 3 }} \approx 0,47 \gt 0.\]

Hence, an extreme point of the function \(S\left( z \right)\) is in the range of angles \(\left( {\frac{\pi }{4},\frac{\pi }{3}} \right)\) (Figure \(20\)) and this point is a point of minimum (judging by the change of sign of the derivative).

Solution of the equation z-cot(z)=0.
Figure 20.

The approximate coordinate of the point of minimum is about \(0.86\;\text{rad}\) or \(49,3^{\circ}.\)

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