Chapter 1

Chapter 1: Application of differential calculus in
geometry
Lecturer: Assoc. Prof. Nguyễn Duy Tân

email: tan.nguyenduy@hust.edu.vn
Falculty of Mathematics and Informatics, HUST
March 2024
Application of differential calculus 1 / 34

Contents
Contents
1 1.1. Applications in plane geometry

1.1.1. Normal lines and tangent lines
1.1.2. Curvature
1.1.3.Envelope of a family of curves
2 1.2. Applications in the geometry of space

1.2.1. Vector functions
1.2.2. Lines
1.2.3. Surfaces

1.1. Applications in plane geometry 1.1.1. Normal lines and tangent lines
Normal lines and tangent lines

1.1.1. Normal lines and tangent lines
Problem:
In the coordinate plane Oxy , given a curve L and a point M ∈ L. Find equations
of the tangent line and the normal line of L at M.
Recall: Suppose that the curve L is defined by y = f (x). Let M(x0 , y0 ) ∈ L.

• The equation of the tangent line to L at M is
y − y0 − f ′ (x0 )(x − x0 ) = 0.
• The equation of the normal line to L at M is
f ′ (x0 )(y − y0 ) + x − x0 = 0.

Curve given by F (x, y ) = 0 (curve given in implcit form)
Suppose the curve L is given by F (x, y ) = 0.
Non-singular point
A point M(x0 , y0 ) ∈ L is called a regular point (or a non-singular point) if at least
one of Fx′ (M), Fy′ (M) is not 0, or equivalently (Fx′ (M))2 + (Fy′ (M))2 ̸= 0. A point
that is not non-singular is called singular.

Suppose M(x0 , y0 ) is a regular point of L. By using the implict function theorem,

one can obtain following.
The equation of the tangent line to L at M(x0 , y0 ) is
Fx′ (x0 , y0 )(x − x0 ) + Fy′ (x0 , y0 )(y − y0 ) = 0.
A normal vector of L at M is n⃗ = (Fx′ (x0 , y0 ), Fy′ (x0 , y0 )).

The equation of the normal line to L at M is
x − x0 y − y0
′
= ′ .
Fx (x0 , y0 ) Fy (x0 , y0 )

(
x = x(t)
Curve given by parametric equations (curve
y = y (t)
given in parametric form)
Non-singular point
A point M(x(t0 ), y (t0 )) ∈ L is said to be regular (or non-singular) if at least one
of x ′ (t0 ), y ′ (t0 ) is not 0.
The equation of the tangent line at M(x(t0 ), y (t0 )) is
x − x(t0 ) y − y (t0 )
= .
x ′ (t0 ) y ′ (t0 )
A tangent vector of L at M is (x ′ (t0 ), y ′ (t0 )).

The equation of the normal line at M(x(t0 ), y (t0 )) is
x ′ (t0 )(x − x(t0 )) + y ′ (t0 )(y − y (t0 )) = 0.

Example (GK20192)
Find the equations of the tangent line and normal line to the curve x 3 + y 3 = 9xy
at (2, 4).
Solution:
x 3 + y 3 = 9xy ⇔ F (x, y ) := x 3 + y 3 − 9xy .
Fx′ (x, y ) = 3x 2 − 9y và Fy′ (x, y ) = 3y 2 − 9x.
At (2, 4):
Fx′ (2, 4) = 3 · 22 − 9 · 4 = −24, Fy′ (2, 4) = 3 · 42 − 9 · 4 = 30.
The equation of the tangent line at (2,4):
−24(x − 2) + 30(y − 4) = 0 or equivalently 4x − 5y + 12 = 0.
The equation of the normal line at (2,4):

x −2 y −4
= or equivalently 5x + 4y − 26 = 0.
−24 30

Example (GK20181)
Find the equations of the tangent line and the normal line of x = (t 2 − 1)e 2t ,
y = (t 2 + 1)e 3t at t = 0.
Solution:
At t = 0, we have a point M(−1, 1).
x ′ = 2te 2t + 2(t 2 − 1)e 2t , y ′ = 2te 3t + 3(t 2 + 1)e 3t .
At t = 0: x ′ = −2, y ′ = 3.
The equation of the tangent line:
x +1 y −1
= or equivalently 3x + 2y + 1 = 0.
−2 3
The equation of the normal line:
−2(x + 1) + 3(y − 1) = 0 or equivalently 2x − 3y + 5 = 0

1.1. Applications in plane geometry 1.1.2. Curvature
1.1.2. Curvature: Definition

The curvature measures how fast a curve is changing direction at a given
point.
The curvature of the curve L at M, is denoted by C (M), which is always
non-negative.
Suppose L is a smooth curve such that every point near M is regular. On L
we choose a direction, called positive direction. At every point M ′ near M
−−−→ −−→
(and at M) we choose a tangent vector M ′ T ′ ( and MT ) with direction
compatible with the positive direction.
˘′ , denoted by Ctb (MM
The mean curve of MM ˘′ ), is the ratio of α, the angle
′ ′
between MT and M T , to the arc lengh MM ′ :
˘
˘′ ) = α
Ctb (MM .
MM ′
˘
The curvature of L at M, denoted by C (M), is the limit (if exists)
C (M) = lim ˘′ ).
Ctb (MM
′ M →M

Example
The curvature of any line is 0.
Example
The curvature of a circle with radius R at any point is 1/R.

Curvature
The curve L is given by y = f (x). Let M(x, y ) in L.

The curvature at M is given by
|y ′′ |
C (M) = .
(1 + y ′ 2 )3/2
Let L be a curve given by x = x(t), y = y (t). Let M(x(t), y (t)) be a point in

L. Then
|x ′ y ′′ − y ′ (x ′′ |
C (M) = .
(x ′2 + y ′2 )3/2

Curvature
Let L be a curve defined by a polar equation r = f (φ). Then x = f (φ) cos φ và

y = f (φ) sin φ.
x ′ = r ′ cos φ − r sin φ, x ′′ = r ′′ cos φ − 2r ′ sin φ − r cos φ

y ′ = r ′ sin φ + r cos φ, y ′′ = r ′′ sin φ + 2r ′ cos φ − r sin φ.
Hence, at M(r , φ ∈ L,
|r 2 + 2r ′2 − rr ′′ |
C (M) = .
(r 2 + r ′2 )3/2

Example (GK20201)
Find the curvature of y = x 3 + x at M(1.2).
Solution:
y ′ = 3x 2 + 1, y ′′ = 6x.
At M(1, 2): y ′ (1) = 3 · 12 + 1 = 4, y ′′ (1) = 6.
The curvature at M(1, 2) is
|y ′′ (1)| |6| 6
C (M) = ′ 2 3/2
= 2 3/2
= √ .
(1 + y (1) ) (1 + 4 ) 17 17

Example (GK20192)
Find the curvature of x = 2(t − sin t), y = 2(1 − cos t) at the point corresponding
to t = −π/2.
Solution:
x ′ (t) = 2(1 − cos t), x ′′ (t) = 2 sin t, y ′ (t) = 2 sin t, y ′′ (t) = 2 cos t.
t = −π/2: x ′ (−π/2) = 2(1 − cos(−π/2)) = 2, x ′′ (t) = 2 sin(−π/2) = −2,
y ′ (−π/2) = 2 sin(−π/2) = −2, y ′′ (−π/2) = 2 cos(−π/2) = 0.
Then
|x ′ (t)y ′′ (t) − x ′′ (t)y ′ (t)| |2 · 0 − (−2)(−2)| 4 1
C (M) = ′ 2 ′ 2 3/2
= 2 2 3/2
= √ = √ .
(x (t) + y (t) ) (2 + (−2) ) 8 8 4 2

Some exercises
(GK20213) Find the curvature of (x − 2)2 + (y − 1)2 = 5 at M(4, 2).

(GK20212) Find the curvature of the curve defined by z = x 2 + y 2 , z = 2x
at A(1, 1, 2).
x2 y2
(CK20212) Let (E ) be the curve defined by + = 1. Find the curvature
16 36
of (E ) at A(4, 0).
(GK20192) Find the curvature of y = e 2x at A(0; 1).
(GK20182) Find the curvature of x = t 2 , y = t ln t, t > 0, at the point
corresponding to t = e.

1.1. Applications in plane geometry 1.1.3.Envelope of a family of curves

1.1.3. Envelope of a family of curves
Given a family L = (Lc ) of curves F (x, y , c) = 0, where c is a parameter. The

envelope of this family of curves is a curve E such that
every curve Lc in the family L is tangent to E ,
and at each point M of E , there exists a curve Lc in the family L such that
E and Lc are tangent at M.
Example
Consider the circles C : (x − c)2 + y 2 = R 2 , where c is a parameter. The envelope
of this family consists of two lines y = ±R.

Rules to find the envelope
Theorem
Let F (x, y , c) = 0 be a family of curve, where c is a parameter (tham số). If every
curve in the family has no singular points then the® parametric equations of the
F (x, y , c) = 0
envelope are defined by the system of equations
Fc′ (x, y , c) = 0.
Eliminating the parameter c from these equations, we can get the equation of the
envelope in implicit form.
Remark:
For us, we should find the equation of the envelope in implicit form (or
explictly form).
If any curve has a singular point, we must exclude the singular points.

Example (GK20201)
Find the envelope of (Γc ): 2x cos c + y sin c = 1.
Solution:
2x cos c + y sin c = 1 ⇔ F (x, y , c) := 2x cos c + y sin c − 1 = 0.
Fx′ = 2 cos c, Fy′ = sin c. The systems Fx′ = Fy′ = 0 has no solutions. The
family (Γc ) has no singular points.

®
F (x, y , c) = 0
®
2x cos c + y sin c = 1 x = 1 cos c
⇔ ⇔ 2
Fc′ (x, y , c) = 0 −2x sin c + y cos c = 0 y = sin c
The envelope is the ellipse

4x 2 + y 2 = 1.

Some exercises
x y
(GK20213) Find the envelope of + = 1, where c is a parameter.
c4 (1 − c)4
(GK20212) Find the envelope of y = 2cx 2 + c 2 + 1, where c ≤ 0 is a
parameter.
(GK20192) Find the envelope of x 2 + y 2 − 4yc + 2c 2 = 0, where c ̸= 0 is a
parameter.
(GK20192) Find the envelope of y = 4cx 3 + c 4 , where c ̸= 0 is a parameter.
(GK20182) Find the envelope of (x + c)2 + (y − c)2 = 2, where c ̸= 0 is a
parameter.
(GK20181) Find the envelope of x = 2cy 2 + 3c 2 , where c ̸= 0 is a parameter.

1.2. Applications in the geometry of space 1.2.1. Vector functions
1.2.1. Vector functions
Let I be an interval in R.
A map ⃗r : I → Rn , t 7→ ⃗r (t) is called a vector function of t defined in I .
Let n = 3 and write ⃗r (t) = (x(t), y (t), z(t)) = x(t)⃗i + y (t)⃗j + z(t)⃗k. The set
of all points M(x(t), y (t), z(t)) with t in I is called the graph of the function
r . We also say that a space curve L has the equation
x = x(t), y = y (t), z = z(t).
Limit: The function ⃗r (t) has limit ⃗a as t approaches to t0 if
lim ||⃗r (t) − ⃗a|| = 0, denoted by lim ⃗r (t) = ⃗a.
t→t0 t→t0
Continuous: The function ⃗r (t) defined in I is continuous at t0 ∈ I if
lim ⃗r (t) = ⃗r (t0 ). (This is the same as x(t), y (t), z(t) are continuous at t0 .)
t→t0

1.2. Applications in the geometry of space 1.2.1. Vector functions
Derivative
The limit (if exists)
∆⃗r ⃗r (t0 + h) − ⃗r (t0 )

lim = lim
h→0 h h→0 h
d⃗r (t0 )
is called the derivative o ⃗r (t) at t0 , denoted by ⃗r ′ (t0 ) or .
dt
When ⃗r (t) has the derivative at t0 , we say ⃗r (t) is differentiable t t0 .
Remark: If x(t), y (t), z(t) are differentiable at t0 , then ⃗r (t) is differentiable
at t0 and
⃗r ′ (t0 ) = x ′ (t0 )⃗i + y ′ (t0 )⃗j + z ′ (t0 )⃗k.

1.2. Applications in the geometry of space 1.2.2. Lines

1.2.2. Tangent
Let L be a space curve with parametrized equations x = x(t), y = y (t),

z = z(t). The corresponding vector function is ⃗r (t) = (x(t), y (t), z(t)).
Let M(x(t0 ), y (t0 ), z(t0 )) ∈ L be a non-singular point (at least one of
x ′ (t0 ), y ′ (t0 ), z ′ (t0 ) is non-zero).
Then ⃗r ′ (t0 ) = (x ′ (t0 ), y ′ (t0 ), z ′ (t0 )) is called the tangent vector of L at M.
The equation of the tangent line at M:
x − x(t0 ) y − y (t0 ) z − z(t0 )

= = .
x ′ (t0 ) y ′ (t0 ) z ′ (t0 )

Normal plane
Let L be a space curve with parametrized equations x = x(t), y = y (t),

z = z(t). The corresponding vector function is ⃗r (t) = (x(t), y (t), z(t)). Let
M(x(t0 ), y (t0 ), z(t0 )) ∈ L be a non-singular point.
The plane passing through M and is perpendicular to the tangent line of L at
M is called the normal plane of the curve L at M.
−−→
The normal plane of L at M consists of all points P such that the vector MP
is perpendicular to the vector ⃗r ′ (t0 ) = (x ′ (t0 ), y ′ (t0 ), z ′ (t0 )). The equation of
the normal plane of L at M is
x ′ (t0 )(x − x(t0 )) + y ′ (t0 )(y − y (t0 )) + z ′ (t0 )(z − z(t0 )) = 0.

Curvature
The curvature measures how fast a curve is changing direction at a given point.
Let L be a space curve defined by x = x(t), y = y (t), z = z(t). Let
M(x(t0 ), y (t0 ), z(t0 )) in L. The curvature of L at M is
s
2 2 2
x′ y′ y′ z′ z′ x′
+ +
x ′′ y ′′ y ′′ z ′′ z ′′ x ′′
C (M) =
(x ′2 + y ′2 + z ′2 )3/2
Remark: Let ⃗r (t) = (x(t), y (t), z(t)). Then
||⃗r ′ ∧ ⃗r ′′ ||
C (M) = .
||⃗r ′ ||3

Example (CK20182)
Find the equations of the tangent line and the normal plane of the curve defined
by x = t cos 2t, y = t sin 2t, z = 3t at t = π/2.
Giải:
At t = π/2, we have a point M(−π/2, 0, 3π/2).
x ′ (t) = cos 2t − 2t sin 2t, y ′ (t) = sin 2t + 2t cos 2t, z ′ (t) = 3.
At t = π/2: x ′ (π/2) = −1, y ′ (π/2) = −π, z ′ (π/2) = 3.
The equation of the tangent line is
x + π/2 y z − 3π/2
= = .
−1 −π 3
The equation of the normal plane is
(−1)(x + π/2) − πy + 3(z − 3π/2) = 0 hay − x − πy + 3z − 5π = 0.

1.2. Applications in the geometry of space 1.2.3. Surfaces

1.2.3. Tangent planes and normal line
Given the surface S and a point M ∈ S. The line MT is called a tangent line
of S at M if it is a tangent line at M to a curve lying in S.
Let S be the surface defined by the equation f (x, y , z) = 0. A point M ∈ S is
called non-singular if at least one of fx′ (M), fy′ (M), fz′ (M) is not zero.
Theorem
The set of all tangent lines of S at a non-singular point M forms a plane.
The plane containing all tangent lines of S at a non-singular point M is called

the tangent plane of S at M.
The line through M and is perpendicular to the tangent plane is called the
normal line of S at M.

Formula
Given the surface S defined by the equationf (x, y , z) = 0. Let M(x0 , y0 , z0 ) ∈ S

be a non-singular point.
The equation of the tangent plane of S at M is
fx′ (M)(x − x0 ) + fy′ (M)(y − y0 ) + fz′ (M)(z − z0 ) = 0.
The equation of the normal line of S at M is
x − x0 y − y0 z − z0
′
= ′ = ′ .
fx (M) fy (M) fz (M)

Example (GK20201)
Find the equations of the tangent plane and the normal line of z = ln(2x + y ) at
M(−1, 3, 0).
Solution:
Let F (x, y , z) = ln(2x + y ) − z.
Fx′ = 2/(2x + y ), Fy′ = 1/(2x + y ), Fz′ = −1.
At M(−1, 3, 0): Fx′ (M) = 2, Fy′ (M) = 1, Fz′ (M) = −1.
The equation of the tangent plane
2(x + 1) + (y − 3) − z = 0 or 2x + y − z − 1 = 0.
The equation of the normal line

x +1 y −3 z
= = .
2 1 −1

Some exercises
(GK20213) Find the equations of the tangent plane and the normal line
ofz = 2x 2 + y 2 at M(1, 1, 3).
(GK20212) Find the equations of the tangent plane and the normal line of
arctan(x + y 2 ) + z = 0 at M(−1, 1, 0).
(CK20193) Find the equations of the tangent plane and the normal line of
x 2 − 2y 3 + 3z 2 = 11 at A(1; 1; 2).
x = 2(t − sin t), y = 2(1 − cos t) at t = π/2.
x = sin t, y = cos t, z = e 2t at M(0; 1; 1).
x 2 + y 2 − e z − 2xyz = 0 at M(1; 0; 0).
ln(2x + y 2 ) + 3z 3 = 3 at M(0; −1; 1).
(CK20171) Find the equations of the tangent plane and the normal line of
z = ln(4 − x 2 − 2y 2 ) at A(−1; 1; 0).

Curves defined by intersection of surfaces
(CK20181) Find the tangent vector at M(1; −1; 1) of the curve defined by
®
x + y + 2z − 2 = 0
x 2 + 2y 2 − 2z 2 − 1 = 0
(CK20142) Find the equations of the tangent line and the normal plane at
A(1; −2; 5) of the curve defined by z = x 2 + y 2 , z = 2x + 3.

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Chapter 1: Application of differential calculus in

Lecturer: Assoc. Prof. Nguyễn Duy Tân

Falculty of Mathematics and Informatics, HUST

Application of differential calculus 1 / 34

1 1.1. Applications in plane geometry

2 1.2. Applications in the geometry of space

Application of differential calculus 2 / 34

Normal lines and tangent lines

Application of differential calculus 3 / 34

1.1.1. Normal lines and tangent lines

Recall: Suppose that the curve L is defined by y = f (x). Let M(x0 , y0 ) ∈ L.

• The equation of the normal line to L at M is

Application of differential calculus 4 / 34

Curve given by F (x, y ) = 0 (curve given in implcit form)

Suppose the curve L is given by F (x, y ) = 0.

Application of differential calculus 5 / 34

Suppose M(x0 , y0 ) is a regular point of L. By using the implict function theorem,

Fx′ (x0 , y0 )(x − x0 ) + Fy′ (x0 , y0 )(y − y0 ) = 0.

A normal vector of L at M is n⃗ = (Fx′ (x0 , y0 ), Fy′ (x0 , y0 )).

Application of differential calculus 6 / 34

The equation of the tangent line at M(x(t0 ), y (t0 )) is

A tangent vector of L at M is (x ′ (t0 ), y ′ (t0 )).

x ′ (t0 )(x − x(t0 )) + y ′ (t0 )(y − y (t0 )) = 0.

Application of differential calculus 7 / 34

−24(x − 2) + 30(y − 4) = 0 or equivalently 4x − 5y + 12 = 0.

The equation of the normal line at (2,4):

Application of differential calculus 8 / 34

The equation of the normal line:

−2(x + 1) + 3(y − 1) = 0 or equivalently 2x − 3y + 5 = 0

Application of differential calculus 9 / 34

1.1.2. Curvature: Definition

Application of differential calculus 10 / 34

Application of differential calculus 11 / 34

The curve L is given by y = f (x). Let M(x, y ) in L.

Let L be a curve given by x = x(t), y = y (t). Let M(x(t), y (t)) be a point in

Application of differential calculus 12 / 34

Let L be a curve defined by a polar equation r = f (φ). Then x = f (φ) cos φ và

x ′ = r ′ cos φ − r sin φ, x ′′ = r ′′ cos φ − 2r ′ sin φ − r cos φ

Application of differential calculus 13 / 34

Application of differential calculus 14 / 34

Application of differential calculus 15 / 34

(GK20213) Find the curvature of (x − 2)2 + (y − 1)2 = 5 at M(4, 2).

Application of differential calculus 16 / 34

Application of differential calculus 17 / 34

1.1.3. Envelope of a family of curves

Given a family L = (Lc ) of curves F (x, y , c) = 0, where c is a parameter. The

Application of differential calculus 18 / 34

Rules to find the envelope

Application of differential calculus 19 / 34

The envelope is the ellipse

Application of differential calculus 20 / 34

Application of differential calculus 21 / 34

1.2.1. Vector functions

Application of differential calculus 22 / 34

The limit (if exists)

∆⃗r ⃗r (t0 + h) − ⃗r (t0 )

⃗r ′ (t0 ) = x ′ (t0 )⃗i + y ′ (t0 )⃗j + z ′ (t0 )⃗k.

Application of differential calculus 23 / 34

Application of differential calculus 24 / 34

Let L be a space curve with parametrized equations x = x(t), y = y (t),