UNIVERSITY OF SANTO TOMAS
Faculty of Engineering
ENG 203: Calculus 1
APPLICATIONS OF THE DERIVATIVE: Tangent and Normal Lines
The derivative has many applications in many fields such as, geometry and mechanics.
In this module, we will discuss one of the four of the derivative’s many applications:
tangent and normal lines
In Unit 2.1. the derivative was mentioned as the expression of the slope 𝑚 𝑇𝐿 of line
tangent on a function 𝑦 = 𝑓(𝑥) at a specific point 𝑃(𝑥0 , 𝑦0 ). In symbols, the slope 𝑚 𝑇𝐿 at
𝑃(𝑥0 , 𝑦0 ) is
𝒎𝑻𝑳 =
𝒅𝒚
|
𝒅𝒙 (𝒙=𝒙𝟎 ,𝒚=𝒚𝟎 )
Given the said tangent line intersecting the function 𝑦 = 𝑓(𝑥) at 𝑃, there is a line
perpendicular to the tangent line that is also passing through 𝑃. This is called the normal
line.
This figure shows a function 𝑦 = 𝑓(𝑥), in black, with tangent line, in green, and normal line, in
blue, at point 𝑃(𝑥0 , 𝑦0 ).
(Photo taken from https://en.wikipedia.org/wiki/Subtangent)
Unit 3.1: Applications of the Derivative – Tangent and Normal Lines ∥ Page 1
UNIVERSITY OF SANTO TOMAS
Faculty of Engineering
ENG 203: Calculus 1
Consider two lines 𝐿1 and 𝐿2 with slopes 𝑚1 and 𝑚2 respectively, recall that their slopes
are negative reciprocals of each other. In symbols,
𝑚1 = −
1
𝑚2
or 𝑚2 = −
1
𝑚1
Applying this concept to normal lines, then one can say that 𝑚𝑁𝐿 and 𝑚 𝑇𝐿 are related as:
𝑚𝑁𝐿 = −
1
𝑚 𝑇𝐿
𝑚 𝑇𝐿 = −
1
𝑚𝑁𝐿
Moreover, since the slope of the tangent line is the derivative of the function at 𝑃, then
𝑚𝑁𝐿 = −
1
𝑚 𝑇𝐿
⇒
𝒎𝑵𝑳 = −
𝟏
𝒅𝒚
|
𝒅𝒙 (𝒙=𝒙𝟎 ,𝒚=𝒚𝟎 )
Since the point 𝑃(𝑥0 , 𝑦0 ) is a point on both the tangent and normal lines, their equations
can also be solved using the point-slope form of the line. For the tangent line, the equation
is
𝑦 − 𝑦0 = 𝑚 𝑇𝐿 (𝑥 − 𝑥0 ) ⇒
𝒚 − 𝒚𝟎 = 𝒚′ (𝒙 − 𝒙𝟎 )
On the other hand, for the normal line, the equation is
𝑦 − 𝑦0 = 𝑚𝑁𝐿 (𝑥 − 𝑥0 )
⇒
𝒚 − 𝒚𝟎 = −
𝟏
(𝒙 − 𝒙𝟎 )
𝒚′
Remember that since the point-slope form of the line is going to be used, the slope of the
line and a point on the line must be known.
Problem 1: For the function 𝑦 = 𝑥 3 − 2𝑥 + 5, determine the equations of the tangent line
at (2,9).
Solution: So far, the only known information is the point of tangency (2,9). The lacking
information is the slope of the tangent line. Recall that the slope of the tangent is the
derivative of the function at the given point.
Solving for the derivative 𝑑𝑦⁄𝑑𝑥,
𝑦 = 𝑥 3 − 2𝑥 + 5 ⇒
𝑦 ′ = 3𝑥 2 − 2
To solve for the slope of the tangent line at (2,9), we determine the value of 𝑑𝑦⁄𝑑𝑥 at the
point,
𝑦 ′ = 3𝑥 2 − 2
⇒ 𝑦 ′ (𝑥 = 2, 𝑦 = 9) = 3(2)2 − 2 = 10 ∴ 𝑦 ′ = 𝑚 𝑇𝐿 = 10
Unit 3.1: Applications of the Derivative – Tangent and Normal Lines ∥ Page 2
UNIVERSITY OF SANTO TOMAS
Faculty of Engineering
ENG 203: Calculus 1
Now the slope and a point on the line are known, the equation of the tangent can be
determined using the point-slope form.
𝑦 − 𝑦0 = 𝑚 𝑇𝐿 (𝑥 − 𝑥0 )
⇒
𝒚 − 𝟗 = 𝟏𝟎(𝒙 − 𝟐)
Problem 2: Given the function 𝑦 = 𝑥 sin 𝑥, determine the equations of the normal line at
𝑥 = 𝜋⁄2.
Solution: The only information known is the 𝑥-coordinate 𝑥 = 𝜋⁄2 of the point. Solving
for the 𝑦-coordinate,
𝑦 = 𝑥 sin 𝑥
⇒ 𝑦=
𝜋
𝜋 𝜋
sin =
2
2 2
Hence, the point being considered is (𝜋⁄2 , 𝜋⁄2). So far, the slope of the normal line is still
unknown. Solving for 𝑦 ′ ,
𝑦 = 𝑥 sin 𝑥
⇒
𝑦 ′ = 𝑥 cos 𝑥 + sin 𝑥
At the point (𝜋⁄2 , 𝜋⁄2),
𝑦 ′ = 𝑥 cos 𝑥 + sin 𝑥
𝜋 𝜋
𝜋
𝜋
𝜋
𝑦 ′ ( , ) = cos + sin = 1
2 2
2
2
2
⇒
Note that the slope of the normal line is the negative reciprocal of the derivative at the
point.
𝑚𝑁𝐿 = −
1
= −1
𝑦′
Using the point-slope form to solve of the normal line,
𝑦 − 𝑦0 = 𝑚𝑁𝐿 (𝑥 − 𝑥0 ) ⇒
𝑦−
𝜋
𝜋
= −1 ∙ (𝑥 − )
2
2
⇒
𝒚 = −𝒙 + 𝝅
Problem 3: Given the inverse function 𝑥 = (𝑦 − 1)2 (2𝑦 + 3), find the tangent line at 𝑦 = 2
Solution: Only the 𝑦-coordinate of the point is known. Solving for the 𝑥-coordinate,
𝑥 = (𝑦 − 1)2 (2𝑦 + 3)
⇒ 𝑥 = (2 − 1)2 (2(2) + 3) = 7
Therefore, the point is (7,2). Proceeding to the next portion of the solution, the derivative
that can be directly obtained is 𝑑𝑥⁄𝑑𝑦 for an inverse function.
𝑥 = (𝑦 − 1)2 (2𝑦 + 3)
Unit 3.1: Applications of the Derivative – Tangent and Normal Lines ∥ Page 3
UNIVERSITY OF SANTO TOMAS
Faculty of Engineering
ENG 203: Calculus 1
Applying the product rule,
𝑑𝑥
𝑑
𝑑
(2𝑦 + 3) + (2𝑦 + 3) ∙
(𝑦 − 1)2
= (𝑦 − 1)2 ∙
𝑑𝑦
𝑑𝑦
𝑑𝑦
𝑑𝑥
= 2(𝑦 − 1)2 + 2(2𝑦 + 3)(𝑦 − 1)
𝑑𝑦
𝑑𝑥
= 2(𝑦 − 1)(3𝑦 + 2)
𝑑𝑦
The derivative 𝑑𝑦⁄𝑑𝑥 can be easily solved by taking the reciprocal of 𝑑𝑥 ⁄𝑑𝑦.
𝑑𝑦
1
=
𝑑𝑥 2(𝑦 − 1)(3𝑦 + 2)
Solving for the slope of the tangent line at (7,2),
𝑚 𝑇𝐿 =
𝑑𝑦
1
1
|
=
=
𝑑𝑥 (7,2) 2(2 − 1)(3(2) + 2) 16
The equation of the tangent line is
𝑦 − 𝑦0 = 𝑚 𝑇𝐿 (𝑥 − 𝑥0 ) ⇒
𝒚−𝟐=
𝟏
(𝒙 − 𝟕)
𝟏𝟔
Problem 3: Find the tangent and normal lines at 𝑡 = 1, given the parametric equations
𝑥=
1
𝑡+4
𝑦=
1
2𝑡 + 3
Solution: At 𝑡 = 1, the 𝑥- and 𝑦-coordinates of the point are
𝑥=
1
1
1
=
=
𝑡+4 1+4 5
𝑦=
1 1
1
1
1
∴( , )
=
=
5 5
2𝑡 + 3 2(1) + 3 5
Solving for 𝑑𝑥⁄𝑑𝑡 and 𝑑𝑦⁄𝑑𝑡,
𝑥=
1
𝑑𝑥
1
⇒
=−
,
(𝑡 + 4)2
𝑡 + 4 𝑑𝑡
𝑦=
1
𝑑𝑦
2
⇒
=−
(2𝑡 + 3)2
2𝑡 + 3
𝑑𝑡
From 𝑑𝑥⁄𝑑𝑡 and 𝑑𝑦⁄𝑑𝑡, the derivative 𝑑𝑦⁄𝑑𝑥 can be obtained.
𝑑𝑦
𝑑𝑦
= 𝑑𝑡
𝑑𝑥 𝑑𝑥
𝑑𝑡
2
𝑑𝑦 − (2𝑡 + 3)2
⇒
=
1
𝑑𝑥
−
(𝑡 + 4)2
⇒
𝑑𝑦 2(𝑡 + 4)2
=
𝑑𝑥 (2𝑡 + 3)2
Unit 3.1: Applications of the Derivative – Tangent and Normal Lines ∥ Page 4
UNIVERSITY OF SANTO TOMAS
Faculty of Engineering
ENG 203: Calculus 1
Solving for the slopes at 𝑡 = 1,
𝑚 𝑇𝐿 =
𝑑𝑦
2(1 + 4)2
|
=
=2
𝑑𝑥 𝑡=1 (2(1) + 3)2
𝑚𝑁𝐿 = −
1
1
=−
𝑚 𝑇𝐿
2
Solving for the equations of the lines,
Tangent Line: 𝑦 − 𝑦0 = 𝑚 𝑇𝐿 (𝑥 − 𝑥0 ) ⇒
𝒚−
Normal Line: 𝑦 − 𝑦0 = 𝑚𝑁𝐿 (𝑥 − 𝑥0 )
𝒚−
⇒
𝟏
𝟏
= 𝟐 (𝒙 − )
𝟓
𝟓
𝟏
𝟏
𝟏
= − (𝒙 − )
𝟓
𝟐
𝟓
Problem 4: At what point on the 𝑦-axis will the normal line at 𝑥 = 1 of the function 𝑓(𝑥) =
√2𝑥 + 7 intersect?
Solution: The problem asks at what point will the normal line intersect on the 𝑦-axis. In
other words, the 𝑦-intercept of the normal line is being asked. First, the equation of the
normal line has to be solved.
Solving for the 𝑦-coordinate of the point,
𝑦 = 𝑓(𝑥) = √2𝑥 + 7 ⇒
𝑦 = 𝑓(1) = √2(1) + 7 = 3
Solving for the slope of the normal line at the point being considered,
𝑦 = √2𝑥 + 7 ⇒ 𝑦 ′ =
1
√2𝑥 + 7
⇒ 𝑦 ′ (1,3) =
1
√2(1) + 7
=
1
∴ 𝑚𝑁𝐿 = −3
3
The equation of the normal line is
𝑦 − 𝑦0 = 𝑚𝑁𝐿 (𝑥 − 𝑥0 ) ⇒
𝑦 − 3 = −3(𝑥 − 1)
Now that the equation of the normal line is known, the 𝑦-intercept can now be solved.
Recall that the 𝑦-intercept can be determined by setting 𝑥 to be 0.
𝑦 − 3 = −3(𝑥 − 1)
⇒ 𝑦 − 3 = −3(0 − 1)
⇒
𝑦=6
Therefore, the 𝑦-intercept of the normal line is 6.
Unit 3.1: Applications of the Derivative – Tangent and Normal Lines ∥ Page 5
UNIVERSITY OF SANTO TOMAS
Faculty of Engineering
ENG 203: Calculus 1
Problem 5: Determine at which point/s on the given 𝑦 = 𝑓(𝑥) below is its slope equal to
−1⁄7?
𝑦=
2𝑥 − 3
𝑥+4
Solution: Recall that the slope of the function is its derivative.
𝑦=
2𝑥 − 3
𝑥+4
Applying the quotient rule,
(2𝑥 − 3)(𝑥 + 4)′ − (𝑥 + 4)(2𝑥 − 3)′
(𝑥 + 4)2
(2𝑥 − 3)(1) − (𝑥 + 4)(2)
𝑦′ =
(𝑥 + 4)2
2𝑥 − 3 − (2𝑥 + 4)
𝑦′ =
(𝑥 + 4)2
−7
𝑦′ =
(𝑥 + 4)2
𝑦′ =
The problem asks at which point/s on the function is its slope −1⁄7, so to solve for the
point, the derivative has to be equated to −1⁄7.
1
7
7
1
−
=−
2
(𝑥 + 4)
7
𝑦′ = −
(𝑥 + 4)2 = 49
𝑥 + 4 = ±7
𝑥+4=7
𝑥=3
𝑥 + 4 = −7
𝑥 = −11
For the point where 𝑥 = 3, the 𝑦-coordinate is
𝑦=
2𝑥 − 3
𝑥+4
⇒ 𝑦(3) =
2(3) − 3 3
=
3+4
7
As for the point where 𝑥 = −11,
𝑦=
2𝑥 − 3
2(−11) − 3 25
⇒ 𝑦(−11) =
=
𝑥+4
−11 + 4
7
Unit 3.1: Applications of the Derivative – Tangent and Normal Lines ∥ Page 6
UNIVERSITY OF SANTO TOMAS
Faculty of Engineering
ENG 203: Calculus 1
Based on the calculations, the points at which the slope of the function is −1⁄7 are (3, 3⁄7)
and (−11, 25⁄7).
Problem 6: Find the equation of the line passing through (1, −2) and is tangent to 𝑦 =
𝑥 2 + 1. At what point does this tangent line intersect 𝑦 = 𝑥 2 + 1?
Solution: Solving for 𝑦 ′ ,
𝑦 = 𝑥2 + 1
⇒ 𝑦 ′ = 2𝑥
The slope of the tangent line at any point on 𝑦 = 𝑥 2 + 1 is 𝑦 ′ = 2𝑥.
Unlike the previous problems, this problem asks for the equation of the tangent line,
however, the given point is not on the function. At this point, one cannot just simply
substitute the values of the point (1, −2) to the expression for 𝑦 ′ , since, again, the given
point (1, −2) is not on the graph of the given function.
The equation of all tangent lines of 𝑦 = 𝑥 2 + 1 at any point (𝑥0 , 𝑦0 ) looks like
𝑦 − 𝑦0 = 𝑚 𝑇𝐿 ∙ (𝑥 − 𝑥0 ) ⇒
𝑦 − 𝑦0 = 𝑦 ′ ∙ (𝑥 − 𝑥0 ) ⇒
𝑦 − 𝑦0 = 2𝑥 ∙ (𝑥 − 𝑥0 )
The point (1, −2) is a point on the tangent line, so the equation becomes
𝑦 − 𝑦0 = 2𝑥 ∙ (𝑥 − 𝑥0 )
⇒
𝑦 − (−2) = 2𝑥 ∙ (𝑥 − 1)
⇒
𝑦 + 2 = 2𝑥(𝑥 − 1)
The equation 𝑦 + 2 = 2𝑥(𝑥 − 1) is the form of all tangent lines on 𝑦 = 𝑥 2 + 1 passing
through (1, −2).
To continue with the solution, the relationship between 𝑥 and 𝑦 is given to be 𝑦 = 𝑥 2 + 1.
Substituting this relationship to the equation of the tangent line,
𝑦 + 2 = 2𝑥(𝑥 − 1)
⇒
𝑥 2 + 1 + 2 = 2𝑥(𝑥 − 1)
Solving for 𝑥,
𝑥 2 + 1 + 2 = 2𝑥(𝑥 − 1)
𝑥 2 + 3 = 2𝑥 2 − 2𝑥
𝑥 2 − 2𝑥 − 3 = 0
(𝑥 + 1)(𝑥 − 3) = 0
𝑥+1=0
𝑥 = −1
𝑥−3=0
𝑥=3
The meaning of the calculations so far is that at the points where 𝑥 = −1 and 𝑥 = 3, the
tangent line passes through (1, −2).
Unit 3.1: Applications of the Derivative – Tangent and Normal Lines ∥ Page 7
UNIVERSITY OF SANTO TOMAS
Faculty of Engineering
ENG 203: Calculus 1
At 𝑥 = −1, the slope of the tangent line is
𝑦 ′ = 2𝑥
⇒ 𝑦 ′ (−1) = 2(−1) = −2
Knowing that the tangent line is passing through (1, −2), the equation of the slope is
𝑦 − 𝑦0 = 𝑚 𝑇𝐿 (𝑥 − 𝑥0 ) ⇒
𝒚 + 𝟐 = −𝟐(𝒙 − 𝟏)
At 𝑥 = 3, the slope of the tangent line is
𝑦 ′ = 2𝑥
⇒ 𝑦 ′ (3) = 2(3) = 6
and the equation of the tangent line is
𝑦 − 𝑦0 = 𝑚 𝑇𝐿 (𝑥 − 𝑥0 ) ⇒
𝒚 + 𝟐 = 𝟔(𝒙 − 𝟏)
Problem 7: Using the concept of the derivatives, determine the vertex of the parabola 𝑦 2 −
3𝑥 − 2𝑦 − 8 = 0.
Solution: The given parabola has a quadratic 𝑦-term, suggesting it is a parabola facing
either to the left or to the right. Either way, if a tangent line is drawn at the vertices of
these parabolas, the slope would be a vertical line. If a line is vertical, then its slope is ∞,
meaning its denominator is zero. This makes sense because the slope is “rise over run,”
and a vertical line “only rises but never runs.”
𝑚 𝑇𝐿 = ∞
With this, one can say that 𝑦 ′ at the vertex of these parabola is infinite, i.e. the
denominator is zero.
Solving for 𝑦 ′ ,
𝑦 2 − 3𝑥 − 2𝑦 − 8 = 0
Applying implicit differentiation,
2𝑦𝑦 ′ − 3 − 2𝑦 ′ − 8 = 0
2(𝑦 − 1)𝑦 ′ = 11
𝑦′ =
11
2(𝑦 − 1)
Equating the denominator to be zero,
2(𝑦 − 1) = 0 ⇒
𝑦=1
Unit 3.1: Applications of the Derivative – Tangent and Normal Lines ∥ Page 8
UNIVERSITY OF SANTO TOMAS
Faculty of Engineering
ENG 203: Calculus 1
This means that at the point where 𝑦 = 1, the tangent line is vertical. Moreover, this is
also the 𝑦-coordinate of the vertex. Solving for the 𝑥-coordinate,
𝑦 2 − 3𝑥 − 2𝑦 − 8 = 0
(1)2 − 3𝑥 − 2(1) − 8 = 0
−3𝑥 = 9
𝑥 = −3
Therefore, the vertex of the parabola is at (−3,1).
To summarize, this can be applied to other conic sections with vertices. The solution is
just a matter of determining the slope of the tangent line at their vertices.
PROBLEM SET
Practice Problem: Determine the equations of the tangent and normal lines at 𝑃(1,1) given
𝑥 3 𝑦 − 2𝑥𝑦 2 + 𝑥 2 − 3𝑦 + 3 = 0.
Answer: tangent line 3𝑥 − 4𝑦 + 1 = 0; normal line 4𝑥 + 3𝑦 − 7 = 0;
Practice Problem: Determine the vertices of the hyperbola 9𝑥 2 − 16𝑦 2 − 18𝑥 − 96𝑦 + 9 =
0. Hint: The major axis of this hyperbola is parallel to the 𝑦-axis.
Answer: The vertices of the hyperbola are (1,0) and (1, −6).
Practice Problem: Determine the endpoints of the major and of the minor axes of the
ellipse described by the equation 4𝑥 2 + 𝑦 2 − 16𝑥 − 2𝑦 + 13 = 0. Hint: The major axis of this
ellipse is parallel to the 𝑦-axis.
Answer: The endpoints of the major axis are (2, −1) and (2,3), while that of the minor
axis are (1,1) and (3,1).
Unit 3.1: Applications of the Derivative – Tangent and Normal Lines ∥ Page 9