![\"Graph](\"https:\/\/www.mathros.net.ua\/en\/wp-content\/uploads\/2026\/07\/equation-of-a-tangent-line-and-equation-of-a-normal-line-to-a-curve1.jpg\")
<\/p>\n
Why do we need this? The normal line is used in problems where it is important to construct a perpendicular to a curve, find a perpendicular direction related to the shortest distance, describe the direction of a force, or study the geometric position of lines near a graph. So, the normal line does not repeat the tangent line. It complements it.<\/p>\n
Suppose the slope of the tangent line is<\/p>\n
\\[
\nk_t=f'(x_0).
\n\\]<\/p>\n
Since the normal line is perpendicular to the tangent line, the slopes of these two lines are related by the formula:<\/p>\n
\\[
\nk_t\\cdot k_n=-1,
\n\\]<\/p>\n
where \\( k_n \\) is the slope of the normal line. From this, if \\( f'(x_0)\\neq 0 \\), we get:<\/p>\n
\\[
\nk_n=-\\frac{1}{f'(x_0)}.
\n\\]<\/p>\n
Therefore, the equation of a normal line to the curve \\( y=f(x) \\) at the point \\( M_0(x_0,f(x_0)) \\) has the form:<\/p>\n
\\[
\ny-f(x_0)=-\\frac{1}{f'(x_0)}\\cdot(x-x_0).
\n\\]<\/p>\n
This formula works under the condition that \\( f'(x_0)\\neq 0 \\). The reason is simple: the formula contains division by \\( f'(x_0) \\), and division by zero is not allowed.<\/p>\n
If \\( f'(x_0)=0 \\), then, as we saw in the previous section, the tangent line is horizontal. So the normal line at this point will be a vertical line:<\/p>\n
\\[
\nx=x_0.
\n\\]<\/p>\n
If the tangent line at the point \\( M_0(x_0,f(x_0)) \\) is vertical and has the equation \\( x=x_0 \\), then the normal line will be horizontal:<\/p>\n
\\[
\ny=f(x_0).
\n\\]<\/p>\n
So, the normal line complements the tangent line: it gives the second important direction at the same point of the curve.<\/p>\n
Angle Between Curves: Comparing Directions Through Tangent Lines<\/h2>\n
Now let\u2019s take the next step. If a tangent line describes the direction of one curve at a point, then two tangent lines help us compare the directions of two curves at their common point.<\/p>\n
Suppose we are given two curves \\( y=f_1(x) \\) and \\( y=f_2(x) \\). Let them intersect at the point \\( M_0(x_0,f_1(x_0)) \\). Since this point belongs to both curves, the following equality holds:<\/p>\n
\\[
\nf_1(x_0)=f_2(x_0).
\n\\]<\/p>\n
The angle between two curves at their point of intersection is the angle between the tangent lines to these curves drawn at that point. Usually, the angle between curves means the smaller angle between their tangent lines.<\/p>\n
![\"Graphs](\"https:\/\/www.mathros.net.ua\/en\/wp-content\/uploads\/2026\/07\/equation-of-a-tangent-line-and-equation-of-a-normal-line-to-a-curve2.jpg\")
<\/p>\n
So, the problem of finding the angle between curves is reduced to the problem of finding the angle between two straight lines \u2014 their tangent lines.<\/p>\n
If both tangent lines are not vertical, then their slopes are:<\/p>\n
\\[
\nk_1=f’_1(x_0), \\qquad k_2=f’_2(x_0).
\n\\]<\/p>\n
Then the tangent of the angle between the tangent lines, and therefore between the curves, is calculated by the formula:<\/p>\n
\\[
\n\\tan(\\varphi)=\\left|\\frac{k_2-k_1}{1+k_1\\cdot k_2}\\right|.
\n\\]<\/p>\n
After substituting the derivatives, we get:<\/p>\n
\\[
\n\\tan(\\varphi)=\\left|\\frac{f’_2(x_0)-f’_1(x_0)}{1+f’_1(x_0)\\cdot f’_2(x_0)}\\right|.
\n\\]<\/p>\n
This formula shows that the angle between curves depends on the slopes of their tangent lines at the point of intersection. In other words, the main role is played not by the function values themselves, but by the derivative values \\( f’_1(x_0) \\) and \\( f’_2(x_0) \\).<\/p>\n
If<\/p>\n
\\[
\n1+f’_1(x_0)\\cdot f’_2(x_0)=0,
\n\\]<\/p>\n
then the tangent lines are perpendicular to each other. In this case, the angle between the curves is \\( 90^\\circ \\).<\/p>\n
So, the tangent line, the normal line, and the angle between curves give us a convenient way to describe the local geometry of graphs through the derivative.<\/p>\n
Practical Part: Equation of a Tangent Line, Normal Line, and Angle Between Curves<\/h2>\n
Now let\u2019s move on to practice. Here it is important not just to memorize the formulas, but to see how they work in actual problems. First, we will find the tangent line and the normal line at a given point, then we will look at tangent lines that pass through an external point, and after that we will calculate the angle between two curves.<\/p>\n
Example 1. Write the equation of a tangent line and the equation of a normal line to the curve \\( y=5\\cdot x^3+3\\cdot x \\) at the point \\( M_0(1,8) \\)<\/h3>\n
According to the condition of the problem, \\( x_0=1 \\). First, let\u2019s check the value of the function at this point:<\/p>\n
\\[
\nf(x_0)=f(1)=5\\cdot 1^3+3\\cdot 1=8.
\n\\]<\/p>\n
So, the point \\( M_0(1,8) \\) really belongs to the given curve.<\/p>\n
Now let\u2019s find the derivative at the point \\( x_0=1 \\). For clarity, we will do this using the definition of the derivative<\/a>:<\/p>\n \\[ Substitute the given function:<\/p>\n \\[ Expand the brackets and simplify the expression:<\/p>\n \\[ Cancel \\( h \\):<\/p>\n \\[ So, the slope of the tangent line is \\( k=f'(1)=18 \\).<\/p>\n Now write the equation of a tangent line:<\/p>\n \\[ After substituting \\( x_0=1 \\), \\( f(x_0)=8 \\), \\( f'(x_0)=18 \\), we get:<\/p>\n \\[ Or, after simplification:<\/p>\n \\[ Now let\u2019s write the equation of the normal line. Since \\( f'(1)=18\\neq 0 \\), we can use the formula:<\/p>\n \\[ Substitute the values we found:<\/p>\n \\[ Or, after simplification:<\/p>\n \\[ So, the tangent line has the equation \\( y-8=18\\cdot(x-1) \\), and the normal line has the equation \\( y-8=-\\frac{1}{18}\\cdot(x-1) \\).<\/p>\n Here the point \\( M(-1,5) \\) is not given as the point of tangency. It only has to lie on the tangent line we are looking for. So, we will denote the point of tangency separately.<\/p>\n Let \\( x_0 \\) be the x-coordinate of the point of tangency to the graph of the function<\/p>\n \\[ Then the value of the function at the point of tangency is:<\/p>\n \\[ Find the derivative of the given function:<\/p>\n \\[ Therefore, at the point of tangency, we have:<\/p>\n \\[ Write the equation of a tangent line to the graph of the function at the point with x-coordinate \\( x_0 \\):<\/p>\n \\[ Substitute the expressions we found:<\/p>\n \\[ From this:<\/p>\n \\[ Expand the brackets:<\/p>\n \\[ Now express \\( y \\):<\/p>\n \\[ This is the equation of the tangent line, but it still contains the unknown parameter \\( x_0 \\). How do we find it? We use the condition that the tangent line passes through the point \\( M(-1,5) \\). So, the coordinates of this point must satisfy the equation of the tangent line.<\/p>\n Substitute \\( x=-1 \\), \\( y=5 \\):<\/p>\n \\[ So,<\/p>\n \\[ Move everything to one side:<\/p>\n \\[ Factor out the common factor:<\/p>\n \\[ From this, we get two values: \\( x_0=0 \\) or \\( x_0=-2 \\). So, there are two tangent lines.<\/p>\n If \\( x_0=0 \\), then<\/p>\n \\[ Therefore, the first tangent line has the equation:<\/p>\n \\[ If \\( x_0=-2 \\), then<\/p>\n \\[ From this:<\/p>\n \\[ So,<\/p>\n \\[ Thus, the required equations of the tangent lines are \\( y=5 \\) and \\( y=12\\cdot x+17 \\).<\/p>\n First, let\u2019s find the points of intersection of the given curves. To do this, equate the right-hand sides of the equations:<\/p>\n \\[ Move everything to one side:<\/p>\n \\[ Factor out the common factor:<\/p>\n \\[ From this, \\( x=0 \\) or \\( x=1 \\). So, we have two points of intersection: \\( O(0,0) \\) and \\( M(1,1) \\).<\/p>\n Now find the derivatives of both functions:<\/p>\n \\[ First, consider the point \\( O(0,0) \\). At this point:<\/p>\n \\[ So, the tangent lines to both curves at the point \\( O(0,0) \\) have the same slope. Both tangent lines coincide with the x-axis. Therefore, the angle between the curves at this point is \\( 0^\\circ \\).<\/p>\n Now consider the point \\( M(1,1) \\). At this point, the slopes of the tangent lines are:<\/p>\n \\[ We define the angle between the curves as the angle between their tangent lines. Use the formula:<\/p>\n \\[ Substitute \\( k_1=2 \\), \\( k_2=3 \\):<\/p>\n \\[ We get:<\/p>\n \\[ Therefore,<\/p>\n \\[ If we write the angle approximately in degrees, we get \\( \\varphi\\approx 8.13^\\circ \\).<\/p>\n So, the curves \\( y=x^2 \\) and \\( y=x^3 \\) intersect at the point \\( O(0,0) \\) at an angle of \\( 0^\\circ \\), and at the point \\( M(1,1) \\) at an angle of \\( 8.13^\\circ \\).<\/p>\n Next, it is worth moving on to topics that help you work with derivatives faster and more confidently. They are needed in almost every calculus<\/a> problem. So these articles are a good continuation of this topic.<\/p>\n If you are interested in programming, this topic can easily be turned into a small computational project. The flowchart below shows an algorithm that checks the point of intersection for two curves, finds the slopes of the tangent lines, and calculates the angle between the curves in degrees.<\/p>\n
\nf'(1)=\\lim_{h\\to 0}\\frac{f(1+h)-f(1)}{h}.
\n\\]<\/p>\n
\nf'(1)=\\lim_{h\\to 0}
\n\\frac{5\\cdot(1+h)^3+3\\cdot(1+h)-8}{h}.
\n\\]<\/p>\n
\nf'(1)=\\lim_{h\\to 0}
\n\\frac{18\\cdot h+15\\cdot h^2+5\\cdot h^3}{h}.
\n\\]<\/p>\n
\nf'(1)=\\lim_{h\\to 0}(18+15\\cdot h+5\\cdot h^2)=18.
\n\\]<\/p>\n
\ny-f(x_0)=f'(x_0)\\cdot(x-x_0).
\n\\]<\/p>\n
\ny-8=18\\cdot(x-1).
\n\\]<\/p>\n
\ny=18\\cdot x-10.
\n\\]<\/p>\n
\ny-f(x_0)=-\\frac{1}{f'(x_0)}\\cdot(x-x_0).
\n\\]<\/p>\n
\ny-8=-\\frac{1}{18}\\cdot(x-1).
\n\\]<\/p>\n
\ny=-\\frac{1}{18}\\cdot x+\\frac{145}{18}.
\n\\]<\/p>\nExample 2. Write the equations of all tangent lines to the graph of the function \\( y=-3\\cdot x^2+5 \\) that pass through the point \\( M(-1,5) \\)<\/h3>\n
![\"Graph](\"https:\/\/www.mathros.net.ua\/en\/wp-content\/uploads\/2026\/07\/equation-of-a-tangent-line-and-equation-of-a-normal-line-to-a-curve3.jpg\")
<\/p>\n
\ny=-3\\cdot x^2+5.
\n\\]<\/p>\n
\nf(x_0)=-3\\cdot x_0^2+5.
\n\\]<\/p>\n
\nf'(x)=-6\\cdot x.
\n\\]<\/p>\n
\nf'(x_0)=-6\\cdot x_0.
\n\\]<\/p>\n
\ny-f(x_0)=f'(x_0)\\cdot(x-x_0).
\n\\]<\/p>\n
\ny-(-3\\cdot x_0^2+5)=-6\\cdot x_0\\cdot(x-x_0).
\n\\]<\/p>\n
\ny+3\\cdot x_0^2-5=-6\\cdot x_0\\cdot(x-x_0).
\n\\]<\/p>\n
\ny+3\\cdot x_0^2-5=-6\\cdot x\\cdot x_0+6\\cdot x_0^2.
\n\\]<\/p>\n
\ny=3\\cdot x_0^2-6\\cdot x\\cdot x_0+5.
\n\\]<\/p>\n
\n5=3\\cdot x_0^2-6\\cdot(-1)\\cdot x_0+5.
\n\\]<\/p>\n
\n5=3\\cdot x_0^2+6\\cdot x_0+5.
\n\\]<\/p>\n
\n3\\cdot x_0^2+6\\cdot x_0=0.
\n\\]<\/p>\n
\n3\\cdot x_0\\cdot(x_0+2)=0.
\n\\]<\/p>\n
\ny=3\\cdot 0^2-6\\cdot x\\cdot 0+5=5.
\n\\]<\/p>\n
\ny=5.
\n\\]<\/p>\n
\ny=3\\cdot(-2)^2-6\\cdot x\\cdot(-2)+5.
\n\\]<\/p>\n
\ny=12+12\\cdot x+5.
\n\\]<\/p>\n
\ny=12\\cdot x+17.
\n\\]<\/p>\nExample 3. Find at what angles the curves ( y=f_1(x)=x^2 ) and ( y=f_2(x)=x^3 ) intersect<\/h3>\n
\nx^2=x^3.
\n\\]<\/p>\n
\nx^3-x^2=0.
\n\\]<\/p>\n
\nx^2\\cdot(x-1)=0.
\n\\]<\/p>\n
\nf’_1(x)=2\\cdot x, \\qquad f’_2(x)=3\\cdot x^2.
\n\\]<\/p>\n
\nf’_1(0)=2\\cdot 0=0, \\qquad f’_2(0)=3\\cdot 0^2=0.
\n\\]<\/p>\n![\"Graphs](\"https:\/\/www.mathros.net.ua\/en\/wp-content\/uploads\/2026\/07\/equation-of-a-tangent-line-and-equation-of-a-normal-line-to-a-curve4.jpg\")
<\/p>\n
\nk_1=f’_1(1)=2\\cdot 1=2, \\qquad k_2=f’_2(1)=3\\cdot 1^2=3.
\n\\]<\/p>\n
\n\\tan(\\varphi)=\\left|\\frac{k_2-k_1}{1+k_1\\cdot k_2}\\right|.
\n\\]<\/p>\n
\n\\tan(\\varphi)=\\left|\\frac{3-2}{1+2\\cdot 3}\\right|.
\n\\]<\/p>\n
\n\\tan(\\varphi)=\\frac{1}{7}.
\n\\]<\/p>\n
\n\\varphi=\\arctan\\left(\\frac{1}{7}\\right).
\n\\]<\/p>\nWhat to Study Next: Topics for Further Learning<\/h2>\n
\n
Equation of a Tangent Line in Code: From Formula to Program<\/h2>\n
![\"Flowchart](\"https:\/\/www.mathros.net.ua\/en\/wp-content\/uploads\/2026\/07\/equation-of-a-tangent-line-and-equation-of-a-normal-line-to-a-curve5.jpg\")
<\/p>\n","protected":false},"excerpt":{"rendered":"