LADG Lecture Notes

Linear Algebra, Analytic and Di¤erentialGeometry

November 14, 2014

Contents

1 Geometric (Spatial) Vectors 31.1 De�nition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Addition and scalar multiplication . . . . . . . . . . . . . . . . . 41.3 Cartesian coordinates . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Dot (scalar) product of two vectors . . . . . . . . . . . . . . . . . 71.5 Cross (vector) product of two vectors . . . . . . . . . . . . . . . . 91.6 Box product of three vectors . . . . . . . . . . . . . . . . . . . . 10

2 Equations of Lines and Planes in Space 122.1 Equations of planes . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2 Equations of a straight line in space . . . . . . . . . . . . . . . . 142.3 Angles and distances in space . . . . . . . . . . . . . . . . . . . . 17

2.3.1 Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3.2 Distances . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Coordinate Transformations 203.1 Coordinate transformations in plane . . . . . . . . . . . . . . . . 20

3.1.1 Translations . . . . . . . . . . . . . . . . . . . . . . . . . . 203.1.2 Changes of bases . . . . . . . . . . . . . . . . . . . . . . . 21

3.2 Changes of bases in space . . . . . . . . . . . . . . . . . . . . . . 243.2.1 Translations . . . . . . . . . . . . . . . . . . . . . . . . . . 243.2.2 Changes of bases . . . . . . . . . . . . . . . . . . . . . . . 25

4 Conic sections 274.1 De�nition and examples . . . . . . . . . . . . . . . . . . . . . . . 274.2 Reduced equation of a conic - an elementary approach . . . . . . 31

5 Quadrics 355.1 Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355.2 Reduced canonical equations of other quadrics . . . . . . . . . . 37

5.2.1 The ellipsoidx2

a2+y2

b2+z2

c2= 1: . . . . . . . . . . . . . . 37

1

5.2.2 The hyperboloid of one sheet (H1) :x2

a2+y2

b2� z

2

c2= 1 . . 37

5.2.3 The hyperboloid of two sheetsx2

a2+y2

b2� z

2

c2= �1 . . . . 39

5.2.4 The elliptic paraboloid z =x2

a2+y2

b2. . . . . . . . . . . . 39

5.2.5 The hyperbolic paraboloid (the "saddle") (PH) : z =x2

a2� y

2

b2. . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

5.2.6 The elliptic conex2

a2+y2

b2� z

2

c2= 0 . . . . . . . . . . . . 41

5.2.7 Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

6 Generated surfaces 436.1 Cylindrical surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . 436.2 Conic surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466.3 Surfaces of revolution . . . . . . . . . . . . . . . . . . . . . . . . 48

7 Plane curves 507.1 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507.2 Arc length of a plane curve . . . . . . . . . . . . . . . . . . . . . 517.3 Contact between two intersecting curves . . . . . . . . . . . . . . 537.4 Tangent and normal line at a regular point . . . . . . . . . . . . 557.5 Osculating circle; curvature and radius of curvature . . . . . . . . 56

8 Spatial curves 618.1 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 618.2 Arc length of a space curve . . . . . . . . . . . . . . . . . . . . . 618.3 The TNB Frame (The Frenet-Serret Frame) . . . . . . . . . . . . 628.4 Curvature and torsion . . . . . . . . . . . . . . . . . . . . . . . . 66

9 Di¤erential Geometry of Surfaces 699.1 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 699.2 Curves on a surface . . . . . . . . . . . . . . . . . . . . . . . . . . 709.3 Tangent plane and normal line at a regular point . . . . . . . . . 719.4 First fundamental form. Applications . . . . . . . . . . . . . . . 74

10 Vector Spaces 7610.1 De�nition and examples . . . . . . . . . . . . . . . . . . . . . . . 7610.2 Subspaces in a vector space . . . . . . . . . . . . . . . . . . . . . 7810.3 Subspace spanned by a subset of a vector space . . . . . . . . . . 8010.4 Linear dependence and independence . . . . . . . . . . . . . . . . 8010.5 Basis and dimension of a vector space . . . . . . . . . . . . . . . 8210.6 Changes of bases . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

2

1 Geometric (Spatial) Vectors

1.1 De�nition

The notion of vector is one of the most important notions in Physics. Indeed,force, velocity, acceleration, work, momentum are all related to this notion.From now on, by "space", we will mean the physical space.Consider two points A and B in space.A bound vector

��!AB is a segment AB for which we distinguish two orienta-

tions: from A to B and from B to A: Usually, we graphically indicate boundvectors with an arrow. We denote

��!BA = ��!AB;

the vector��!BA is called the opposite of

��!AB:

For a bound vector��!AB; the point A is called the origin or the tail of the

vector, while B is called the head (tip, destination) of the vector.The zero vector is, by de�nition, the vector

�!0 =

�!AA:

The length (or the norm) ��!AB of the vector ��!AB is de�ned as the length

of the segment AB : ��!AB = AB:The direction of the bound vector

��!AB is formed by all the straight lines

which are parallel to the straight line AB:

De�nition 1 The free vector (or simply, the vector) AB is the set of allbound vectors which have the same direction, the same orientation and the samelength as the bound vector

��!AB:

We usually denote free vectors by a bar: �a:

That is, if��!AB and

��!CD are two bound vectors with the same direction,

orientation and length, as bound vectors, they are di¤erent, but as free vectors,they are equal to each other:

��!AB 6= ��!CD; AB = CD:

In the following, we will denote by E1 the set of free vectors on the line, byE2; the set of free vectors in plane and by E3; the set of free vectors in space.

3

1.2 Addition and scalar multiplication

On the set E3; we will de�ne the following operations:1. Addition: by the parallelogram rule, or by the triangle rule:

2. Scalar multiplication: the vector ��a is de�ned by:- the same directionas �a;- the length j�j �a;- the orientation of �a if � is positive, and opposite to �a; if � is negative.

Properties of the addition of free vectors:0) Addition is an internal operation on the set E3 of free vectors in space:

8�a;�b 2 E3 : �a+�b 2 E3:

1) Associativity:

8�a;�b; �c 2 E3 : (�a+�b) + �c = �a+ (�b+ �c):

2) Commutativity:8�a;�b 2 E3 : �a+�b = �b+ �a:

3) The zero vector �0 is a neutral (or identity) element:

�0 + �a = �a; 8�a 2 E3:

4) Every element �a 2 E3 admits an inverse with respect to addition; thisinverse is the opposite vector ��a :

8�a 2 E3 : �a+ (��a) = �0:

Consequently, (E3;+) is an Abelian group. The addition of free vectors inplane obeys similar properties.

Properties of scalar multiplication:0) 8� 2 R; 8�a 2 E3 : ��a 2 E3:1) 8�; � 2 R; 8�a 2 E3 : (�+ �)�a = ��a+ ��a:2) 8� 2 R; 8�a;�b 2 E3 : �(�a+�b) = ��a+ ��b:3) 8�; � 2 R; 8�a 2 E3 : �(��a) = (��)�a:4) 1�a = �a:

4

Exercise 1 Prove the above stated properties of the addition of vectors in E3:

De�nition 2 1) A set of vectors �v1; :::; �vn (n > 1) is called linearly dependentif one of the vectors �v1; :::; �vn can be expressed as a linear combination of theothers; for instance,

�vn = �1�v1 + �2�v2 + :::+ �n�1�vn�1:

In the contrary case, the vectors �v1; :::; �vn are called linearly independent.2) A set consisting of a single vector �v is called linearly independent if �v 6= �0

and linearly dependent if �v = �0:

Exercise 2 Prove that �v1; :::; �vn are linearly independent if and only if the vec-tor equation

�1�v1 + �2�v2 + :::+ �n�v = �0

(in the unknowns �1; �2; :::; �n 2 R) only admits the trivial solution �1 = �2 =::: = �n = 0:

In the following, we will examine in more detail the notions of linear depen-dence and independence in the cases n = 2; 3; 4:

A) For two vectors �a;�b; linear dependence means

�a = ��b;

where � 2 R; therefore, �a and �b have the same direction (they are collinear).Conversely, if �a and �b are collinear, we can always �nd some scalar � such

that �a = ��b: We can easily see this as follows:Case 1) If at least one of the vectors �a;�b is nonzero (let�s say, �b 6= 0), then

� :=

�b k�ak :

Case 2) If �a;�b = 0; then � can be any real numbers.We have thus proved that

�a;�b-linearly dependent, �a;�b� collinear.

B) For three vectors �a;�b; �c; linear dependence means:

�c = ��a+ ��b;

where �; � 2 R:

5

Exercise 3 Prove that:

�a;�b; �c - linearly dependent, �a;�b; �c� coplanar.

C) In the case of four vectors �a;�b; �c; �d, linear dependence is characterizedby:

�d = ��a+ ��b+ �c:

Exercise 4 Prove that, in E3; any four vectors �a;�b; �c; �d are linearly dependent.

1.3 Cartesian coordinates

Let us make the following remarks.1) On the line (E1), we can always �nd one linearly independent vector (take

any �v 6= �0), but any two vectors are linearly dependent.2) In the plane (E2); we can always �nd two linearly independent (non-

collinear) vectors, but any three vectors are linearly dependent.3) In the geometric space (E3); we can always �nd three linearly independent

(non-coplanar) vectors, but any four vectors are linearly dependent.

Taking into account the above remarks, it makes sense to introduce thefollowing notion.

De�nition 3 We call a basis of Ei (i = 1; 2; 3), a maximal set of linearly in-dependent vectors in Ei:

In other words, on the line, any nonzero vector forms a basis; in plane, abasis consists of (any) two non-collinear vectors; in space, a basis means a setconsisting of (any) three non-coplanar vectors.

In the following, we will work in the geometric space E3:

De�nition 4 A Cartesian (orthonormal) frame in E3 consists of a pointO, called the origin, together with an orthonormal basis f�{; �j; �kg; that is, abasis whose vectors are mutually perpendicular and have the length equal to 1.

Traditionally, the vectors f�{; �j; �kg are chosen in such a way that they form aright oriented frame, meaning that the orientation of �k is obtained from theorientations of �{ and �j by the right hand rule: if the index �nger of your righthand points as �{ and your middle �nger points as �j; your thumb will indicatethe orientation of �k:

In the following, unless elsewhere speci�ed, by R =�O;�{; �j; �k

; we will

always mean a right oriented, orthonormal frame.

6

According to the third remark above, for any geometric vector �a 2 E3; the set��{; �j; �k; �a

is linearly dependent, that is, one can �nd some scalars a1; a2; a3 2 R

such that:

�a = a1�{+ a2�j + a3�k:

We will denote�a(a1; a2; a3);

and we will call the triple (a1; a2; a3) 2 R3; the Cartesian coordinates of the(free) vector �a 2 E3:If �a represents the bound vector

�!OA,

�a(a1; a2; a3) = OA;

then we say that the point A has the Cartesian coordinates (a1; a2; a3) :

A(a1; a2; a3)$ OA(a1; a2; a3):

The vector�!OA is called the position vector of the point A:

Remark. Any other bound vector ~a which has the same direction, ori-entation, and length as

�!OA (that is, as free vectors, �a = OA), has the same

coordinates (a1; a2; a3):

Proposition 5 If A(xA; yA; zA); B(xB ; yB ; zB); then

AB(xB � xA; yB � yA; zB � zA):

Example: If A(2; 3; 4); B(0;�1; 2); then:

OA(2; 3; 4), OA = 2�{+ 3�j + 4�k;

OB(0;�1; 2), OB = ��j + 2�k;AB(xB � xA; yB � yA; zB � zA)) AB(�2;�4; ;�2):

1.4 Dot (scalar) product of two vectors

De�nition 6 Consider �a;�b 2 E3: The dot (inner, scalar) product �a � �b is thenumber

�a � �b = k�ak � �b � cos�;

where � is the smallest angle between the directions of �a and �b () � 2 (0; 180o)):

Note: The scalar (dot) product of two vectors is not to be confused withscalar multiplication (which refers to a number and a vector)!

Properties of the dot product:

7

1. the dot product is commutative: �a � �b = �b � �a:

2. it is distributive with respect to addition: �a � (�b+ �c) = �a � �b+ �a � �c:

3. �a � (��b) = (��a) � b = ��a � �b:

4. if �a and �b are non-zero, then �a � �b = 0 if and only is �a is perpendicular to�b : �a ? �b:

The expression in coordinates of the scalar product: If �a(a1; a2; a3)and �b(b1; b2; b3); then

�a � �b = a1b1 + a2b2 + a3b3:

Applications:

� 1. The length (norm) of the vector �a is

k�ak =p�a � �a =

qa21 + a

22 + a

23:

2. The angle between �a and �b is computed as

cos� =�a � �b

k�ak �b :

3. The unit vector having the same direction and orientation as �a is

vers(�a) =�a

k�ak :

4. the length of the projection of �a onto �b is given by pr�b�a =k�ak cos�; that is,

proj�b�a =�a � �b �b :

Exercise 5 Take A(1; 2; 3); B(2; 4; 5); C(3; 0; 4): Determine:a) the coordinates of vectors AB;AC;BC:b) the scalar products AB �AC; AB �BC; AC �BC;c) the lengths of the sides of the ABC:d) the angles of triangle ABC:

Exercise 6 Prove that the vectors �a(�2; 2; 1) and �b(2; 1; 2) are perpendicular.

Exercise 7 Consider the points A(1; 2; 0); B(1; 2; 1); C(3; 0; 1): Calculate AB �AC;

AB ; AC cos( \AB �AC); projACAB and the unit vector of AC:

8

1.5 Cross (vector) product of two vectors

De�nition 7 The cross product of �a and �b is a vector, denoted �a � �b, char-acterized by:

� direction: perpendicular to both �a and �b;

� orientation: given by the right-hand rule;

� length: �a� �b = k�ak �b sin�; where � is the smallest angle between the

directions of �a and �b:

the vector product

Properties:

1. the dot product is anticommutative: �a� �b = ��b� �a:

2. it is distributive w.r.t. addition: �a� (�b+ �c) = �a� b+ �a� �c:

3. �a� (��b) = (��a)� b = ��a� �b:

4. if �a and �b are non-zero, then �a � �b = 0, �a is collinear to �b:

Expression in coordinates:

�a� �b =

��{ �j �ka1 a2 a3b1 b2 b3

�� :Application: The length of �a � �b can be interpreted as the area of the

parallelogram having �a and �b as sides.Consequently, the area of a triangle having �a and �b as sides is half of the

area of the parallelogram, that is:

A� =1

2

�a� �b :9

Example: Take A(1; 2; 0); B(1; 2; 1); C(3; 0; 1): Calculate the area of thetriangle ABC:Solution: AB(0; 0; 1); AC(2;�2; 1) and

AB �AC =

��{ �j �k0 0 12 �2 1

�� = 2�{+ 2�j:The norm

AB �AC is then AB �AC = p22 + 22 = 2p2; which leads toA�ABC =

1

2

AB �AC = p2:Exercise 8 Consider �a(2; 1; 2); �b(3; 1; 4); �c(2; 2; 5): a) Calculate the cross prod-ucts �a��b; �a� �c; �b� �c: b) Determine the areas of the parallelograms built on �a;�band �a; �c respectively. c) Find a perpendicular vector to both �a and �b; which hasthe length equal to 2 units.

Exercise 9 Calculate by means of the cross product the area of the triangleABC; where A(1; 0; 0); B(1; 0; 1); C(2; 3; 1) and the altitude from A:

Exercise 10 Find the area of the parallelogram built on the vectors �m = 2�a+3�b;

�n = �a� �b; where k�ak = 1; �b = 2; [(�a;�b) = �

6: (Hint: calculate �m� �n and take

into account that �a� �a = 0; �b� �b = 0; �b� �a = ��a� �b).

1.6 Box product of three vectors

The mixed triple product (also called the box product) is de�ned as:

(�a;�b; �c) = �a � (�b� �c):

Applications:

1. The absolute value of the box product is the volume of the paral-lelepiped built on the three vectors:

Vparall =��(�a;�b; �c)�� :

The volume of the tetrahedron built on �a;�b; �c (as non-coplanar edges)is

Vparall =1

6

��(�a;�b; �c)�� :2. The box product zero if and only if the three vectors are linearly dependent(coplanar).

3. The box product is positive if and only if the system consisting of thethree vectors �a;�b; �c is right oriented.

10

Exercise 11 Determine the mixed triple product of �a(1; 2; 3); �b(2; 3; 4) and�c(3; 3; 5):

Exercise 12 Find � 2 R such that the vectors �a(1; 2; 3); �b(2; 0; 2) and �c(4; 4; �)are coplanar.

Exercise 13 Determine the volume of the tetrahedron ABCD; where A(1; 2; 3);B(1; 2; 4); C(2; 2; 4); D(2; 3; 4); and the altitude from A of the tetrahedron.

Hint : VABCD =

��(��!AB;�!AC;��!AD)��6

=1

6:

11

2 Equations of Lines and Planes in Space

2.1 Equations of planes

Let fO;�{; �j; �kg denote a Cartesian frame in space.In the following, we will write the equation of a plane (let�s say, (�)), assum-

ing that we know:- a point M(x0; y0; z0) in the plane (�) and- a perpendicular vector �N(A;B;C) to the plane (�).

For any point P (x; y; z) in the plane, the vector �N is perpendicular to MP;which is equivalent to the fact that the dot product �N �MP is 0. In coordinates,that is written as:

A(x� x0) +B(y � y0) + C(z � z0) = 0: (1)

The above equation is called the equation of the plane throughM(x0; y0; z0); with normal direction �N(A;B;C):Computing the brackets in equation (1), we always get a relation of the form:

Ax+By + Cz +D = 0; (2)

where A;B;C;D 2 R (more precisely, D = �Ax0 � By0 � Cz0). This relationis called the general equation of the plane.

Remark 8 Relation (2) is, actually, the most general equation of degree 1 inx; y; z: Conversely, any equation of degree 1 in x; y; z geometrically means aplane in three dimensional Euclidean space.

Remark 9 The coe¢ cients of x; y and z in the general equation of a planerepresent the coordinates of the normal vector of the plane. For instance, theplane x+ 3y � z + 5 = 0 has as normal vector �N(1; 3;�1).

Example: the coordinate planes xOy; yOz; xOz:

12

The plane xOy contains the point O(0; 0; 0) and has as normal vector�k(0; 0; 1): Hence, its equation is: (xOy) : 0(x � 0) + 0(y � 0) + 1(z � 0) = 0;which is,

(xOy) : z = 0:

In a similar manner, we get

(yOz) : x = 0; (xOz) : y = 0:

Remark 10 Since two parallel planes have the same normal direction, the equa-tion of any plane which is parallel to the plane (�) : Ax + By + Cz + D = 0di¤ers from that of (�) only by its free term: Ax+By+Cz +D = �; for some� 2 R:

Exercise 14 Write the equation of the plane:

1. through A(1; 2;�1) and having as normal vector �N(2; 3; 4):

2. through A(2; 3; 5) and parallel to the plane (M1M2M3); where M1(1; 0; 0);M2(0; 1; 0); M3(0; 0; 1).

3. through A(x0; y0; z0) and parallel to the plane: a) xOy; b) yOz; c)xOz:

Another very common situation is when we know, for a plane (�):- a point M(x0; y0; z0) in the plane;- two vectors �v1(l1;m1; n1) and �v2(l2;m2; n2) parallel to (�) (or contained in

(�)).In this case, a perpendicular vector to the plane is the cross product:

�N := �v1 � �v2:

Remark 11 Suppose that M; �v1 and �v2 are known. If P (x; y; z) is an arbitrary(moving) point in this plane, then the vectors

MP (x� x0; y � y0; z � z0); �v1(l1;m1; n1); �v2(l2;m2; n2)

are coplanar. This is equivalent to the fact that

MP = ��v1 + ��v2; �; � 2 R:

In Cartesian coordinates, this is equivalent to the relations:8<: x = x0 + �l1 + �l2y = y0 + �m1 + �m2

z = z0 + �n1 + �n2

(called the parametric equations of the plane (�)).

13

Exercise 15 Write the equation of the plane:

1. containing M(1;�1; 0) and the vectors �v1(1; 2; 3); �v2(0; 1; 2);

2. through the points M1(1; 0; 1); M2(2; 0; 1); M3(2; 1; 2):

3. containing the z-axis and the point M(1; 3; 2):

2.2 Equations of a straight line in space

Basically, a straight line (also called simply, a line) is uniquely de�ned by theintersection of two non-parallel and distinct planes. This is, it can be describedby a linear system as follows:

(d) :

�A1x+B1y + C1z +D1 = 0A2x+B2y + C2z +D2 = 0

: (3)

This points out that, actually, in space, a line is described by two (!) equa-tions. Still, the system (3) does not o¤er (at least, not immediately) too muchinformation about the line, so we will generally prefer a di¤erent form of theequations of a line. We will present this form in the following.

Assume that, for the line (d) whose equation is to be determined, we know:- a point M(x0; y0; z0) on the line;- a vector �v(l;m; n) which is parallel to the line (or contained in it), called a

directing vector.

In this case, for any point P (x; y; z); the vector MP (x � x0; y � y0; z � z0)is collinear to �v(l;m; n); which is equivalent to the fact that the components ofthe two vectors are proportional:

x� x0l

=y � y0m

=z � z0n

(4)

(the canonical equations of a line in space).

Remark 12 The canonical equations provide useful information about the line:namely, the denominators l;m; n are exactly the components of the directing vec-tor �v of the line, while the quantities subtracted from x; y and z in the numeratorsgive the coordinates of some point on the line. For instance, the line

x� 11

=y � 32

=z + 1

4

has the directing vector �v(1; 2; 4) and a point on the line is M(1; 3;�1)

14

If, in (4), we denote the common value of the ratios by t we get8<: x = x0 + lty = y0 +mtz = z0 + nt

(5)

(the parametric equations of a line).Parametric equations are very useful in mechanics (in this case, the parame-

ter t usually denotes time).

Examples:

1. The equations of the coordinate axes Ox; Oy; Oz : For instance, on (Ox)we know the point O(0; 0; 0) and the vector �{(1; 0; 0); hence, the canonicalequations are:

(Ox) :x� 01

=y � 00

=z � 00

= t

It is admitted to formally write 0 in the denominators(!). This does notmean that we perform any division by 0, it is just a simpler way of writingthe fact that the numerators and the denominators are proportional. Themeaning of this can be more clearly seen from the parametric equations:

x� 0 = 1 � t = ty � 0 = 0 � t = 0z � 0 = 0 � t = 0:

Actually, 0 in a denominator means that also the corresponding numeratoris 0.

(Ox) can be also given as the intersection of the planes (xOy) and (xOz) :

(Ox) :

�y = 0z = 0

:

The equations of the axes Oy and Oz can be obtained in a similar manner.

2. Consider the intersection line of two planes:

(d) :

�x� y = 0x+ 2z = 1

:

Let us determine the canonical equations of this line.

First of all, we notice that the two planes (�1) : x � y = 0 and (�2) :x+2z = 1 indeed determine a line: since their normal vectors �N1(1;�1; 0)and �N2(1; 0; 2) are non-collinear, the planes can neither be parallel, norcoincide, hence their intersection is a line.

The vector of this line is contained both in (�1) and (�2); consequently,it is perpendicular to �N1 and �N2: This means, it is collinear to the cross

15

product �N1� �N2: Since we are interested only in its direction, we can takeas directing vector of the line

�v = �N1 � �N2 =

��{ �j �k1 �1 01 0 2

�� = �2�{� 2�j + �k:A point on the line can be obtained from any particular solution of thelinear system which describes (d): This can be done by giving particularvalues to one of the unknowns. Let us take, for instance, x = 1:We obtainy = 1; z = 0; that is, M(1; 1; 0):

The canonical equations of the line are, consequently:

x� 1�2 =

y � 1�2 =

z

1:

Another useful notion is that of sheaf of planes. If a line (d) is describedas the intersection of two planes (P ) and (Q) :

(d) :

�P � A1x+B1y + C1z +D1 = 0Q � A2x+B2y + C2z +D2 = 0

;

then the sheaf of planes through (d) (or determined by (P ) and (Q)) isthe set of all planes which contain (d): Its equation is

�P + �Q = 0; �; � 2 R:

Equivalently, with � =�

�; this can be written simply as

P + �Q = 0; � 2 R:

(the last form is simpler, still, it requires some attention: there we have elimi-nated the case � = 0; that is, the plane Q: This is, in any discussion involvingthe planes of the sheaf, we should take separately the case when the plane isQ).

Exercise 16 Write the canonical and the parametric equations of:- the line through A(1; 3; 4); having the directing vector �v(2; 0; 1);- the line AB; where A(2; 1; 2); B(0;�3; 1);- the parallel line to (d) :

x� 11

=y

�1 =z + 1

4, through A(4; 3; 2):

Exercise 17 Prove that the intersection of the planes (P ) : x + y = 3; (Q) :x � y + z = 0 is a line and �nd the canonical and the parametric equations ofthis line.

16

2.3 Angles and distances in space

2.3.1 Angles

1. the angle between two lines (d1) and (d2) is the angle between theirdirecting vectors �v1 and �v2; and its cosine can be obtained by means ofthe dot product �v1 � �v2 :

cos� =�v1 � �v2

k�v1k � k�v2k:

2. the angle between a line and a plane is equal to 90o minus the anglebetween the directing vector �v of the line and the normal vector �N of theplane:

cos(90o � �) = sin� = �v � �Nk�vk �

�N :3. the angle between two planes is the same as the angle between theirnormal vectors �N1 and �N2 :

cos� =�N1 � �N2 �N1 � �N2 :

2.3.2 Distances

1. the distance between two points A and B is, by de�nition, the normof the vector

��!AB :

dist(A;B) = ��!AB =p(xB � xA)2 + (yB � yA)2 + (zB � zA)2:

2. the distance between a point M(x0; y0; z0) and a line (d) :x� x1l

=

y � y1m

=z � z1n

is expressed as the altitude in the parallelogram built on

the vectors MM1 and �v(l;m; n) (where M1(x1; y1; z1) denotes a point onthe line):

dist(M;d) =

MM1 � �v

k�vk :

17

3. the distance between a point M(x0; y0; z0) and a plane (�) : Ax +By + Cz +D = 0 is given by

dist(M; (�)) =jAx0 +By0 + Cz0 +D0jp

A2 +B2 = C2:

4. the minimum distance between two skew (=non-coplanar) linesin space is expressed as the altitude from M1 in the parallelepiped builton the vectors M1M2; �v1 and �v2; where Mi and vi denote a point and thedirecting vector of the straight line (di); i = 1; 2 :

dist((d1); (d2)) =

��(M1M2; �v1; �v2)��

kv1 � v2k:

minimum distance between skew lines

Exercise 18 Take A(2; 1; 3); B(1; 2; 3); (d) :x

1=y

2=

z

�2 and (�) : 4x+ 3z �1 = 0: Find:

1. the distances dist(A;B); dist(A; (d)); dist(A; (�));

2. the angle between the line AB and, (d), respectively, (�):

3. the angle between (d) and (�);

4. the minimum distance between the lines AB and (d):

5. the intersection point between (d) and (�):

Exercise 19 Write the equation of a plane:

1. passing through the point O(0; 0; 0) and perpendicular to both (�1) : x +y + z = 3 and (�2) : 3x� z = 2;

18

2. passing through the point O(0; 0; 0); perpendicular to (�1) : x+ y + z = 3

and parallel to the line (d) :x� 12

=y

3=z + 2

4:

3. perpendicular to (�1) : x+ y+ z = 3 and containing the line (d) :x� 12

=

y

3=z + 2

4:

Exercise 20 Find the parameters � and � such that the linex� �1

=y

�=z

1is contained in the plane x+ y + z = 3:

Exercise 21 Consider M(2; 3; 5) and (d) :x

1=y

1=z

1: Find the projection of

M onto (d) and the re�ection of M in the line (d):

Exercise 22 Determine the re�ection of the Ox axis in the line (d) :x

1=y

1=

z

1(Hint: �nd the re�ections of two conveniently chosen points on Ox in the line

(d)).

Exercise 23 (*) Find the line which is perpendicular to both the lines Oz and

(d) :x� 21

=y + 1

3=z

4; and intersects them (the common perpendicular of the

lines Oz and (d)).

19

3 Coordinate Transformations

3.1 Coordinate transformations in plane

Let fO;�{; �jg and fO0;�{0; �j0g denote two Cartesian frames in space. Any transitionfrom the �rst frame to the latter can be regarded as the composition of twomotions:

1. Change the origin O ! O0, and leave the basis unchanged:

fO;�{; �jg ! fO0;�{; �jg:

This is called a translation.

2. Leave the origin unchanged, and perform a change of bases:

fO0;�{; �jg ! fO0;�{0; �j0g:

Let us study separately these two types of transformations.

3.1.1 Translations

Consider the transformation R =fO;�{; �jg ! R0 = fO0;�{; �jg; where O0 has, withrespect to the "old" frame R; the coordinates x0; y0: That is, its position vectorcan be written as:

OO0= x0�{+ y0�j:

Now, take an arbitrary point A in plane. Then, A has some coordinates,let�s say (x; y) in the "old" frame R and some other coordinates (which we willdenote by (x0; y0)) in the "new" frame R0; or, in other words:

�!OA = x�{+ y�j;

��!O0A = x0�{+ y0�j:

The vectors�!OA;

��!O0A and OO

0are related by:

�!OA =

��!OO0 +

��!O0A;

see the picture below:

20

In coordinates, this leads to:�x = x0 + x0y = y0 + y0

: (6)

Relations (6) are the equations of the translation of vector��!OO0(x0; y0).

Example: Take the point A(2; 3) (coordinates are considered with respectto the frame fO;�{; �jg); as the result of a translation of the frame to O0(1; 0); wewill obtain for A the coordinates

x0 = x� x0 = 2� 1 = 1; y0 = y � y0 = 3� 0 = 3

in the new frame fO0;�{; �jg:

Exercise 24 Find the equation of the line: x + 2y + 2 = 0 in the xOy planeafter a translation of the frame to O0(�2; 0):

Exercise 25 Consider, in the xOy plane, the line: ax + by + c = 0: Findthe equation of this line after a translation of the frame to an arbitrary pointO0(x0; y0) on the line.

3.1.2 Changes of bases

A. General remarks.We will analyze in the following changes of frames which keep the origin

�xed (hence, only the directions of the axes are changed): R = fO;�{; �jg !R0 = fO;�{0; �j0g:

Take, again, some point A in the plane, given by:

� A(x; y) in the "old" frame fO;�{; �jg; that is, OA = x�{+ y�j;

� A(x0; y0) in the "new" frame fO;�{0; �j0g; that is, OA = x0�{0 + y0�j0:

In order to �nd the relations between the two pairs of coordinates of A; wedecompose the "new" basis vectors �{0; �j0 in terms of �{ and �j; thus �nding:�

�{0 = c11�{+ c21�j�j0 = c12�{+ c22�j

: (7)

Multiplying the �rst line by x0 and the second line by y0 and adding theobtained equalities, we get:

OA = x0�{0 + y0�j0 = (c11x0 + c12y

0)�{+ (c21x0 + c22y

0)�j:

But, on the other hand, we have: OA = x�{+ y�j: Identifying the expressions of �{and �j, we are led to the general form of the equations of a change of basesin plane:

x = c11x0 + c12y

0; y = c21x0 + c22y

0:

21

These can be written in matrix form as:

X = CX 0;

where:

X =

�xy

�; X 0 =

�x0

y0

�are column matrices containing the "old" and the "new" coordinates of A and

C =

�c11 c12c21 c22

�is called the matrix of change of bases.

Remarks.1) The matrix of change of bases C contains on its �rst column the coordinatesof �{0 in the old basis and on its second line, the coordinates of �j0:2) The fact that the bases f�{; �jg and f�{0; �j0g are orthonormal played no role

in the above considerations. Actually, all the above said also holds true forgeneral, oblique bases (also, the lengths of the basis vectors is unessential).In the following, we will particularize these results for several special changes

of bases.

B. Rotations in planeLet fO0;�{0; �j0g be a frame obtained from fO;�{; �jg after a rotation of angle

� 2 [0o; 180o]: Then, we have

�{0 = cos��{+ sin��j�j0 = � sin��{+ cos��j:

We obtain this way the matrix of change of bases:

C =

�cos� � sin�sin� cos�

�; (8)

22

The relation between the "old" coordinates (x; y) and the "new" ones (x0; y0) ofsome point A is given by X = CX 0:�

xy

�=


��x0

y0

�: (9)

Example: Find the equation of the line y = x + 5 in the plane xOy; aftera rotation of angle � = 45o:We have �

xy

�=

0B@1p2

� 1p2

1p2

1p2

1CA� x0

y0

�;

which is equivalent to 8><>:x =

1p2(x0 � y0)

y =1p2(x0 + y0)

:

Substituting these values of x and y into the equation of the line y = x+ 5; weget

x0 + y0 = x0 � y0 + 5p2 ) y0 =

5p2

2:

C. Re�ectionsAlso, an important class of changes of bases in plane are re�ections:

� re�ection in the Ox-axis ("vertical �ip"): f�{; �jg ! f�{;��jg; that is,

C =

�1 00 �1

�!�

x = x0

y = �y0 :

� re�ection in the Oy-axis ("horizontal �ip"): f�{; �jg ! f��{; �jg; thematrix of change of bases is:

C =

��1 00 1

�!�x = �x0y = y0

:

� re�ection across O : f�{; �jg ! f��{;��jg; which gives the matrix

C =

��1 00 �1

�!�x = �x0y = �y0 :

Exercise 26 Find the equation of the parabola y = x2 + 1 after: a re�ectionin the Ox-axis; a re�ection in the Oy-axis; a re�ection across O; a rotation ofangle � = 90o:

23

Exercise 27 Determine the equation of the line (d) : y = xp3 after rotating

with � = 60o the coordinate axes.

Exercise 28 Find the equation of an arbitrary curve y = f(x) after a rotationof 90o of the coordinate axes.

3.2 Changes of bases in space

Similarly, the transition from a Cartesian frame R = fO;�{; �j; �kg to anotherCartesian frame R0 = fO0;�{0; �j0; �k0g can be regarded as a composition of twotypes of transformations:- a translation fO;�{; �j; �kg ! fO0;�{; �j; �kg (shift the origin, keep the free vec-

tors f�{; �j; �kg unchanged);- a change of bases fO0;�{; �j; �kg ! fO0;�{0; �j0; �k0g (keep the origin �xed, change

the directions of the axes).Let us examine in detail these two types of transformations.

3.2.1 Translations

Assume that we have to pass from the frame R =fO;�{; �j; �kg to the frame R0 =fO0;�{; �j; �k0g. Denote by (x0; y0; z0) the coordinates of the new origin O0 in the"old" frame R:

OO0 = x0�{+ y0�j + z0�k:

Now, take an arbitrary point A in space, having the coordinates (x; y; z)and, accordingly, (x0; y0; z0), in the two frames. This is written as:

OA = x�{+ y�j + z�k; O0A = x0�{+ y0�j + z0�k:

We have, again,OA = OO0 +O0A;

which leads to the equations of the translation in space:8<: x = x0 + x0y = y0 + y0z = z0 + z0

:

Exercise 29 Find the equation of the plane (�) : x � y + 2z � 3 = 0 after atranslation of the frame to O0(2;�1; 0):

Exercise 30 Prove that, as a result of a translation of vector��!OO0(a; b; c) (where

a; b; c 2 R are arbitrary), the equation of a plane is transformed into the equationof a plane and the equations of a line are transformed into the equations of aline.

24

3.2.2 Changes of bases

In the following, we will study changes of frames which keep the origin O �xed,but shift the basis:

R =fO;�{; �j; �kg ! R0 = fO0;�{0; �j0; �k0g:

Consider a point A given by:

� A(x; y; z) with respect to the old frame R; that is: OA = x�{+ y�j + z�k;

� A(x0; y0; z0) with respect to the new frame R; that is: OA = x0�{0 + y0�j0 +z0�k0:

We decompose the vectors of R0 in terms of the old basis:8<: �{0 = s11�{+ s21�j + s31�k�j0 = s12�{+ s22�j + s32�k�k0 = s13�{+ s23�j + s33�k

: (10)

After a similar calculation to the one we did for the plane, we �nd theequations of a change of bases in space:8<: x = c11x

0 + c12y0 + c13z

0

y = c21x0 + c22y

0 + c23z0

z = c31x0 + c32y

0 + c33z0;

these can be written in the matrix form as:

X = CX 0; (11)

where

X =

0@ xyz

1A ; X 0 =

0@ x0

y0

z0

1A ; C =

0@ c11 c12 c13c21 c22 c23c31 c32 c33

1A :Remark 13 In the above, it is not compulsory that the bases f�{; �j; �kg; f�{0; �j0; �k0gbe orthonormal. Relation (11) correctly describes changes of arbitrary bases(orthonormal or not, right oriented or not). In the particular case when the basesare orthonormal, the matrix of change of bases obeys the property: C �Ct = I3;which leads to:

detC = �1:

A change of bases with detC = 1 is called a rotation. Just as a remark,in order to describe an arbitrary rotation in space, one needs to know 3 angles(Euler�s angles or Tait-Bryan angles).

25

Exercise 31 a) Prove that �{0(2

3;2

3;1

3); �j0(�2

3;1

3;2

3); �k0(

1

3;�23;2

3) form an or-

thonormal basis in space. b) Find the equation of the plane (�) : x+ y + z = 0in the basis f�{0; �j0; �k0g:

Exercise 32 Determine the equations of the following rotations in space:a) a rotation of angle � in the xOy plane (the vectors �{; �j are rotated by the

angle �; �k stays �xed);b) a rotation of angle � in the xOz plane (the vectors �k;�{ are rotated by the

angle �; �j stays �xed).

26

4 Conic sections

4.1 De�nition and examples

Conic sections (or, simply, conics) are plane curves described by equationsof degree 2 in x and y :

ax2 + bxy + cy2 + dx+ ey + f = 0; (*)

where a; b; c; d; e; f 2 R and at least one of the coe¢ cients a; b; c of the terms ofdegree 2 is nonzero. Let us present in the following several examples.

I. Ellipse x2

a2+y2

b2� 1 = 0 :

5 4 3 2 1 1 2 3 4 5

3

2

1

1

2

3

x

y

the ellipsex2

9� y

2

4= 0

The numbers a and b are called the semi-axes of the ellipse. The greaterof the two numbers is called the major semi-axis and the other, the minorsemi-axis.

The ellipsex2

a2+y2

b2� 1 = 0 has two special points F and F 0, called the

foci (singular: focus), located on the major axis, at the distance c =pja2 � b2j

from the origin:- if a > b; then: F (c; 0); F 0(�c; 0);- if a < b; then: F (0; c); F 0(0;�c):

The ellipse possesses several interesting properties, used in practical appli-cations. We will mention here just two of them:a) The sum of the distances from any point M on the ellipse to the two foci

is constant: MF + MF 0 = 2a:27

b) (Optical property of the ellipse): Any ray starting from one of the foci isre�ected by the ellipse through the other focus.

Particular case: For a = b = R; we obtain the circle with center O(0; 0)and radius R :

x2 + y2 = R2:

II. Hyperbola

x2

a2� y

2

b2� 1 = 0 (or :

y2

b2� x

2

a2� 1 = 0)

A hyperbola consists of two branches, situated along the axis correspondingto the variable appearing with a plus sign in the equation:

6 4 2 2 4 6

4

2

2

4

x

y

the hyperbolax2

9� y

2

4� 1 = 0

6 4 2 2 4 6

4

2

2

4

x

y

the hyperbola �x2

9+y2

4� 1 = 0

The numbers a and b are called the semi-axes of the hyperbola.

The hyperbolax2

a2� y

2

b2�1 = 0 described above has two asymptotes, namely,

the lines y =b

ax and y = � b

ax (the green lines in the above pictures). Also,

it admits two foci F and F 0, situated on the axis corresponding to the plussign in the equation (one focus "inside" each of the branches), at a distance ofc =

pa2 + b2 from the origin:

- For the "east-west" hyperbolax2

a2� y

2

b2� 1 = 0: F (c; 0); F 0(�c; 0);

- For the "north-south" hyperbola �x2

a2+y2

b2� 1 = 0: F (0; c); F 0(0;�c):

A remarkable property of the hyperbola, used, for instance, in the GPSnavigation system, is the following:The modulus of the di¤erence of the distances from any point M on the

hyperbola to the two foci is a constant:�� MF � MF 0 �� = 2a:28

III. Parabola

y2 = 2px (or : x2 = 2py); p 6= 0 :

x

y

parabola y2 = 2px ( p := 1=2)

x

y

parabola x2 = 2py ( p := 1=2)

Parabola admits only one focus F , situated on its symmetry axis (that is,to the axis corresponding to the non-squared variable in the equation), at a

distance ofp

2units from the vertex:

- for the parabola y2 = 2px; the focus is F (p

2; 0);

- for the parabola x2 = 2py; the focus has the coordinates F (0;p

2):

A remarkable property of the parabola, which is frequently used in practice(e.g., dish antennas, vehicle headlamps) is the so-called optical property of theparabola:Any ray coming parallel to the symmetry axis of the parabola is re�ected

through its focus.

parabola - optical property

29

Other examples:IV.

x2

a2+y2

b2+ 1 = 0

This conic section is often called an imaginary ellipse. Actually, as a set ofpoints, it is the empty set ?: We can convince ourselves of this fact by noticingthat equation (IV) has no real solutions (x; y):

V.x2

a2+y2

b2= 0 �a point (O(0; 0)).

VI.x2

a2� y

2

b2= 0 - the union of two intersecting lines, more precisely:

(d1) :x

a+y

b= 0; (d2) :

x

a� yb= 0:

Justi�cation: The equationx2

a2� y2

b2= 0 is equivalent to:�x

a+y

b

��xa� yb

�= 0: In its turn, the latter equality means

x

a+y

b= 0 or

x

a� yb= 0: (12)

Therefore, any point P (x; y) on the conic (VI) belongs to the line (d1) :x

a+y

b=

0 or to the line (d2) :x

a� yb= 0; that is, P belongs to the union of the two lines.

Conversely, the coordinates of any point P belonging to the union (d1) [ (d2)obey the equation of (d1) or the equation of (d2); that is, they obey (12). But(12) is actually equivalent to the equation of the conic section (VI).

VII. y2 = 1 - union of two parallel lines (y = 1 and y = �1)or x2 = 1 (union of the lines x = 1 and x = �1);

VIII. y2 = 0 - the line y = 0 orx2 = 0 ) the line x = 0;

IX. y2 = �1 or x2 = �1 - the empty set.

30

4.2 Reduced equation of a conic - an elementary approach

One can prove the following result:

Theorem 14 Given any conic section (*), there exists a transformation ofCartesian frame, consisting of a rotation and a translation, which brings theequation of the conic section into one of the equations (I)-(IX).

Equations (I)-(IX) are called the reduced equations of conics.The proof of the above theorem is a constructive one, i.e., it provides the

concrete way of �nding the required coordinate transformation.Let us notice, for the beginning, the distinctive features of equations (I) -

(IX):1) None of these equations contains any xy term.2) With the exception of equation (III) (representing the parabola), none of

the reduced equations contains any term of degree 1.Consequently, in order to determine the reduced equation of the general

conic:ax2 + bxy + cy2 + dx+ ey + f = 0; (*)

we have to do two things:1) Get rid of the xy term (if any).2) When no more xy terms are left, get rid (as far as possible) of the terms

of degree 1.

We will divide the discussion into two cases.Case 1. Let us �rst investigate the case when b = 0; that is, the equation

of the conic does not have any xy term:

(�) : ax2 + cy2 + dx+ ey + f = 0; (13)

where at least one of the numbers a; c is nonzero. In this case, the equation canbe brought to its reduced form just by a translation:�

x = x0 + �y = y0 + �

()�x0 = x� �y0 = y � � :

The technique to be followed is thus to try to emphasize in the equation of(�) binomials of the form (x��); (y� �) - and then denote them by x0 and y0;respectively.We have two possibilities:Case 1a) If a; c 6= 0; we rewrite the equation of (�) as:

(�) : a(x2 +d

ax) + c(y2 +

e

cy) + f = 0;

i.e., we group the terms containing x on one hand and the terms containing yon the other hand and take the coe¢ cients of the squares as common factors.Then, we complete the brackets up to some squares of binomials:

(�) : a(x2 + 2d

2ax+

d2

4a2)� d2

4a+ c(y2 + 2

e

2cy +

e2

4c2)� e2

4c2+ f = 0;

31

that is,

(�) : a(x+d

2a)| {z }

x0

2

+ c(y +e

2c)| {z }

y0

2+ f 0 = 0;

where f 0 := f � d2

4a� e2

4c2:

We denote by x0 and y0 the obtained binomials:8<: x0 := x+d

2ay0 := y +

e

2c

)

8<: x = x0 � d

2ay = y0 � e

2c

: (14)

Equations (14) represent a translation of the frame to the point

O0(� d

2a;� e

2c):

In the new coordinates, the equation of the conic will have the form:

(�) : a(x0)2 + c (y0)2+ f 0 = 0: (15)

Depending on the signs of a; c and f 0; equation (15) coincides with one ofthe equations (I), (II), (IV), (V) or (VI). In other words, the conic sectionis an ellipse, a hyperbola, an imaginary ellipse (?), a point or the union of twointersecting lines.

Case 1b) If a = 0; c 6= 0; i.e., the initial equation of (�) is:

(�) : cy2 + dx+ ey + f = 0; (16)

we distinguish two subcases:Case 1b.(i) d 6= 0: In this case, we group the y-terms, take c as a common

factor and complete them up to a squared binomial of the form (y + �)2; withthe remaining terms, we will form a binomial (x+ �):

(�) : c(y2 +e

cy +

e2

4c2) + dx+ (f � e2

4c)| {z }

f 0

= 0

) (�) : c(y +e

2c| {z }y0

)2 + d(x+f 0

d| {z }x0

) = 0;

where f 0 := f� e2

4c:We have obtained this way a translation to O0(x = �f

0

d; y =

� e

2c): 8<: x0 := x+

f 0

dy0 := y +

e

2c

)

8<: x = x0 � f0

dy = y0 � e

2c

; (17)

32

as a result, the equation of the conic section becomes:

(�) : (y0)2= �d

cx0;

which is the equation of a parabola.

Case 1b.(ii) d = 0:In this case, x is completely missing from the equationof (�):

(�) : cy2 + ey + f = 0:

In order to �nd the reduced equation of (�); we group the existing terms in y;take c as a common factor and complete the bracket up to a squared binomialof the form (y + �)2:

(�) : c(y2 +e

cy +

e2

4c2) + (f � e2

4c)| {z }

f 0

= 0

) (�) : (y +e

2c| {z }y0

)2 = �f0

c:

Depending on the sign off 0

c; we can have: two parallel lines (

f 0

c< 0 ) �f

0

c>

0), a line (f 0

c= 0) or the empty set (

f 0

c> 0):

The case a 6= 0; c = 0 is treated similarly.

Case 2: b 6= 0. In this case, the equation of (�) contains a nonzero term inxy :

(�) : ax2 + bxy + cy2 + dx+ ey + f = 0: (18)

In this case, we will eliminate the xy term by a rotation of angle � 2 [0o; 90o]:�xy

�=


��x0

y0

�; (19)

in other words, we will choose � such that, in the new coordinates (x0; y0); theequation of (�) is of the form:

(�) : a0 (x0)2+ c0 (y0)

2+ d0x0 + e0y0 + f 0 = 0: (20)

The rotation angle � is obtained from the relation:

tg(2�) =b

a� c :

Once the equation of (�) is brought to the form (20), the problem is reducedto Case 1 - hence, further on, we can �nd the reduced equation of (�) by atranslation.

33

Exercise 33 Consider the conic section (�) : ax2+bxy+cy2+dx+ey+f = 0:Prove that, rotating the axes of the frame fO;�{; �jg with the angle � given bytg(2�) =

b

a� c ; the equation of (�) is transformed into an equation of the form(20) (with no xy term).

Exercise 34 Find the reduced equations, recognize and draw the following conicsections:a) x2 + y2 + 2x� 6y + 9 = 0;b) x2 + 4y2 + 4x� 4y + 1 = 0;c) 4x2 � y2 + 8x+ 2y � 1 = 0:

Exercise 35 Find the reduced equations, recognize and draw the following conicsections:d) 4x2 � y2 + 8x+ 2y + 3 = 0;e) y = x2 � 2x+ 2;f) y2 + x+ 4y = 0:

Exercise 36 Find the reduced equations, recognize and draw the following conicsections:g) x2 + 2xy + y2 � x

p2 = 0;

h) xy +p2(x� y) = 0;

i) x2 + 2xy + y2 � 2x� 2y � 3 = 0:Exercise 37 Prove that the equation x2+ y2�R2 = 0 of the circle of center Oand radius R > 0 is invariant to any rotation of the frame.

Exercise 38 Find the equation of the ellipse with center O0(x = 1; y = 2) andsemi-axes a = 3; b = 1:

Remark. Conic sections (except the pair of parallel lines y2 = 1) coincidewith the curves obtained by intersecting a right circular cone with planes:

conic sections (source: internet)

34

5 Quadrics

A quadric is a set of points in space whose coordinates (x; y; z) are related bya second degree equation:

a11x2+a22y

2+a33z2+2a12xy+2a13xz+2a23yz+2a14x+2a24y+2a34z+a44 = 0:

5.1 Sphere

The sphere is de�ned as the set of points in space situated at the same distanceR > 0 (called the radius) from a �xed point !(a; b; c); called the center.

Let P (x; y; z) denote an arbitrary point of the sphere.The fact that the distance

!P is always R is written, after eliminatingthe square root:

(x� a)2 + (y � b)2 + (z � c)2 = R2 (21)

(the equation of the sphere with center !(a; b; c) and radius R). Hence,the sphere is a quadric.

If we perform the computations in the above equation, we obtain a relationof type:

x2 + y2 + z2 +mx+ ny + pz + q = 0; (22)

(where m = �2a; n = �2b; p = �2c; q = a2 + b2 + c2 �R2), which is called thegeneral equation of a sphere.That is, any equation of a sphere looks like above. The converse is not true,

that is, not any equation of type (22) is the equation of a sphere. This can bechecked by bringing the equation (22) to the form (21).

Examples:

1. The equation x2 + y2 + z2 � 2x+ 4y � 4 = 0 can be written as:

(x2 � 2x+ 1)� 1 + (y2 + 4y + 4)� 4 + z2 � 4 = 0;

that is(x� 1)2 + (y + 2)2 + z2 = 9:

Hence, it represents the sphere of center !(1;�2; 0) and radius R = 3:

35

2. The equation: x2+ y2+ z2� 2x+4y+6 = 0 becomes, after grouping thesquares

(x� 1)2 + (y + 2)2 + z2 = �1;

that is, it cannot represent a sphere, since the equation R2 = �1 has noreal solutions.

The tangent plane to the sphere

(S) : x2 + y2 + z2 +mx+ ny + pz + q = 0; (23)

at one of its points P (x0; y0; z0) 2 (S); has the equation:

xx0 + yy0 + zz0 +mx+ x02

+ ny + y02

+ pz + z02

+ q = 0:

Exercise 39 Find out if the following equations represent spheres. If so, indi-cate the center and the radius:

1. x2 + y2 + z2 � 2x+ 4y + 4 = 0;

2. x2 + y2 + z2 � 2x+ 4y + 6 = 0;

3. x2 + y2 + 4x+ 2y + 1 = 0:

Exercise 40 Determine the sphere which passes through the points A(1; 0; 0);B(0; 2; 0); C(0; 0; 3) and O(0; 0; 0):

Exercise 41 Consider the sphere (S) : (x � 3)2 + (y � 1)2 + z2 = 25 and theplane (P�) : x+ y + z = �; � 2 R:

1. For � = 0; prove that the intersection between the sphere (S) and P0 is acircle, and determine the center and the radius of this circle.

2. Find � 2 R such that the plane (P�) is tangent to the sphere.

Exercise 42 Determine the equation of the sphere with center on the x-axis,

passing through A(1; 2; 3) and tangent to the line:x� 12

=y + 3

4=

z

�1 :

36

5.2 Reduced canonical equations of other quadrics

5.2.1 The ellipsoidx2

a2+y2

b2+z2

c2= 1:

I

The numbers a; b; c > 0 are called the semi-axes of the ellipsoid.Intersections with parallel planes to the planes of coordinates are ellipses,

points or the empty set. In order prove this fact, let us take, for instance, planes

which are parallel to (xOy); that is, z = � : we getx2

a2+y2

b2= 1� �

2

c2; which is:

an ellipse, if � 2 (�c; c); a point, if � = �c and the empty set if � 2 R n [�c; c]:If the ellipsoid has two equal semi-axes, then it is called a spheroid; if all

its semi-axes are equal, then it is a sphere.

5.2.2 The hyperboloid of one sheet (H1) :x2

a2+y2

b2� z

2

c2= 1

hyperboloid of one sheetrulings of H1(source:

internet)

37

Its intersections with horizontal planes z = � are ellipses, while the intersec-tions with planes y = const: and x = const: are hyperbolas.

The one-sheeted hyperboloid has an interesting property: it is a doublyruled surface, that is, such a shape can be built only from rectilinear beams,and this can be done in two ways (that is why we say it is "doubly" ruled). Thesebeams are called the rulings (rectilinear generatrices) of the hyperboloid.In order to obtain the equations of its rulings, we write the equation of the

hyperboloid as:x2

a2� z

2

c2= 1� y

2

b2;

and factor the di¤erences of squares:�xa+z

c

��xa� zc

�=�1 +

y

b

��1� y

b

�:

This can be written as a proportionality relation:

x

a+z

c

1 +y

b

=1� y

bx

a� zc

=: �:

Equating each of the ratios with �; we get:

(G�) :

8<:x

a+z

c= �

�1 +

y

b

��xa� zc

�= 1� y

b

:

For each value of �, (G�) represents a family of straight lines which is entirelycontained in the hyperboloid - hence, each (G�) is a ruling for this surface. Wecall the set f(G�)g; a 1-parameter family of rulings.Note. The case 1+

y

b= 0;

x

a� zc= 0 can be formally included in the above,

if we set � :=1: That is, we speak about � 2 R [ f1g:

We can also group the terms with opposite signs:

(G�) :

8<:x

a+z

c= �

�1� y

b

��xa� zc

�= 1 +

y

b

; � 2 R [ f1g

which leads to another 1-parameter family of rulings of the one-sheeted hyper-boloid. The case � =1 corresponds to the situation

x

a� zc= 0; 1� y

b= 0:

We should mention here several properties of the lines (G�) and (G�) :

� Through each point M(x0; y0; z0) of the hyperboloid there is exactly onegeneratrix (G�) and one generatrix (G�):

38

� Rulings (G�) obtained for di¤erent values of � do not intersect each other(they are non-coplanar lines). The same holds true about rulings (G�):

A notable use of one-sheeted hyperboloids is in cooling towers of powerstations. Such a shape is resistant and relatively easy to build.

5.2.3 The hyperboloid of two sheetsx2

a2+y2

b2� z

2

c2= �1

Sections with planes parallel to (xOy); that is, with z = � = const:; areellipses, for j�j > c; a point for � = �c and the empty set for j�j < c: Sectionswith planes y = const: and x = const: are hyperbolas.

5.2.4 The elliptic paraboloid z =x2

a2+y2

b2

Sections with planes z = const: are ellipses, while sections with y = const:and x = const: are parabolas.

39

5.2.5 The hyperbolic paraboloid (the "saddle") (PH) : z =x2

a2� y

2

b2

hyperbolic paraboloid rulings of (PH) (source: internet)

Sections with z = � (const.) are hyperbolas, with the exception of the planexOy; where the section consists of two intersecting lines (prove this!). Sectionswith x = const: and y = const: provide parabolas.The hyperbolic paraboloid is also a doubly ruled surface. Its rul-

ings are obtained by factoring the right hand side of the equation: z =�xa+y

b

��xa� yb

�and then grouping terms of degree 1:

(G�) :

8<:x

a+y

b= �z

��xa� yb

�= 1

; (G�) :

8<:x

a� yb= �z

��xa+y

b

�= 1

;

where �; � 2 R[f1g: The case � =1 corresponds to the line z = 0;x

a� yb= 0;

while � =1 gives the line z = 0;x

a+y

b= 0:

Properties:

� Through any point of the hyperbolic paraboloid there passes exactly oneruling (G�) and exactly one ruling (G�):

� Di¤erent rulings belonging to the same family do not intersect each other(they are non-coplanar lines).

40

5.2.6 The elliptic conex2

a2+y2

b2� z

2

c2= 0

Sections with horizontal planes z = � 6= 0 are ellipses. For a = b; one gets acircular cone.

5.2.7 Cylinders

A. the elliptic cylinderx2

a2+y2

b2= 1 :

B. the hyperbolic cylinderx2

a2� y

2

b2= 1 :

41

C. The parabolic cylinder y2 = 2px :

Exercise 43 Find the rulings of the hyperboloidx2

4+ y2 � z2

9= 1 through

A(2; 0; 0): Determine the angle between these rulings.

Exercise 44 Find the rulings of the hyperbolic paraboloidx2

4� y2 = z; which

are parallel to the plane x+ y + z = 3:

42

6 Generated surfaces

A generated surface is a surface that can be swept out by moving a givencurve. The di¤erent positions of the moving curve are called the generatricesof the surface.If the moving curve is a line, then the generated surface is called a ruled

surface. In this case, the (rectilinear) generatrices are called the rulings of thesurface.Another important class of generated surfaces is that of revolution sur-

faces, obtained by rotating a given curve around a straight line (in this case,we can regard the surface as generated by circles).

In the following, let us expose some generalities on curves and surfaces in3-dimensional Euclidean space.A surface in E3 is generally de�ned by a single equation in x; y; z:

(S) : F (x; y; z) = 0:

(Remember that a plane was characterized by a single equation!).A curve in 3-dimensional space can be de�ned as the intersection of two

surfaces, that is, it is described by a system of equations:

(C) :

�F (x; y; z) = 0G(x; y; z) = 0

:

(In particular, a straight line is described by two equations in x; y; z).

6.1 Cylindrical surfaces

A cylindrical surface is a surface swept out by a moving line having a �xeddirection, and which intersects a given curve (C); called the directrix of thesurface.In order to deduce the equation of a cylindrical surface, we consider as known:

� (d) :�P � A1x+B1y + C1z +D1 = 0Q � A2x+B2y + C2z +D2 = 0

- the "initial position" of the

moving lines. Any ruling (rectilinear generatrix) of the surface will be aline which is parallel to (d):

� (C) :�F (x; y; z) = 0G(x; y; z) = 0

- the directrix of the surface.

Let us now deduce the equation of the cylindrical surface.Step 1: Write the equations of all straight lines which have the same direc-

tion as (d) (that is, all lines which are parallel to (d)). Since (d) is given as theintersection of two planes (P ) and (Q); this can be done by intersecting a planewhich is parallel to (P ) with a plane which is parallel to (Q) :

(G��) :

�P = �Q = �

; �; � 2 R:

43

Step 2: The rulings of the surface are exactly those lines (G��) whichintersect the curve (C): In other words, the system8>><>>:

(G��) :

�P = �Q = �

(�)

(C) :

�F (x; y; z) = 0G(x; y; z) = 0

(24)

has to be consistent. The system (24) is a (generally, nonlinear) system with3 unknowns x; y; z and 4 equations. Basically, in order to solve the system,one needs 3 of the 4 equations (24). The solution (x; y; z) will depend on theparameters � and � :

x = x(�; �); y = y(�; �); z = z(�; �):

Then, substituting the obtained expressions of x; y and z into the fourth (un-used) equation, we �nd a condition of consistency of the form:

�(�; �) = 0; (25)

which relates the parameters � and �:The equation of the surface can now be obtained by simply substituting into

the condition of consistency the expressions (�) of � and � in terms of x; y; z :

� = P; � = Q;

that is, the equation of the cylindrical surface is

�(P;Q) = 0:

The reason for the above said is the following: writing the equation of thesurface means to �nd a relation between the coordinates x; y; z of an arbitrarypoint of the surface. Or, points of the surface are precisely the points of thoselines (G��) which intersect (C); that is, of those (G��) for which � and � arerelated by �(�; �) = 0: But, once � and � obey this relation, the coordinatesx; y; z are necessarily related by �(P;Q) = 0:

Example: Write the equation of the cylindrical surface generated by linesparallel to (d) :

x

1=y

2=z

1; and which intersect the curve (C) : y = sinx;

z = 0:First, let us write the equations of (d) in the form P = 0; Q = 0: We get

(d) :

�P � 2x� y = 0Q � x� z = 0 :

Then, any line which is parallel to (d) has the equations

(G��) :

�2x� y = �x� z = � : (�)

44

The intersection between (G��) and the directrix (C) is given by8>><>>:(G��) :

�2x� y = �x� z = � :

(C) :

�y � sinx = 0z = 0

:

We eliminate x; y; z; in order to obtain the condition of consistency. Using thesecond and the last equation, we get x = �: From the �rst equation, we havenow y = 2�� : The third equation then leads to

�(�; �) � 2�� sin� = 0:

The coordinates of the points of the surface are related to � and � by (�):Substituting into � the expressions (�); we obtain the equation of the surface:

2x� 2z � 2x+ y � sin(x� z) = 0;

that is,(S) : y � 2z � sin(x� z) = 0:

Exercise 45 Find the equation of the cylindrical surface generated by lines

which are parallel to (d) :�

x+ y = 0x� z � 1 = 0 and intersect the ellipse (C) :8<: x2 +

y2

4= 1

z = 0::

Exercise 46 Determine the equation of the cylindrical surface with rulings par-

allel to �v(1; 2; 2); and having as directrix the hyperbola (C) :

8<: x2 � y2

9= 1

z = 0::

45


allel to Oz; having as directrix the curve (C) :

8<: x2 +y2

9= 1

z = 0:: Recognize and

draw the curve.


allel to Oz; having as directrix the curve (C) :�x2 � y2 = 1z = 0:

: Recognize and

draw the curve.


allel to Oz; having as directrix the curve (C) :�y2 = 2xz = 0:

: Recognize and draw

the curve.

Exercise 50 Determine the equation of the tangent cylinder to the sphere (S) :

x2 + y2 + z2 = 4; having the rulings parallel to (d) :x

1=y + 4

2=

z

�1 :

6.2 Conic surfaces

A conic surface is swept out by straight lines which pass through a �xed point(called vertex or apex) and intersect a given curve, called the directrix.Assume that the vertex V has the coordinates (x0; y0; z0) and the directrix

(C) is again described as the intersection of two surfaces:

(C) :

�F (x; y; z) = 0G(x; y; z) = 0:

Step 1: Write the equations of all lines passing through V :

(G��) :x� x0�

=y � y0�

=z � z01

)

8><>:� =

x� x0z � z0

� =y � y0z � z0

:

(Actually, in this writing, lines parallel to the plane xOy are missing; if needed,we can consider them separately).Step 2: The lines (G��) have to intersect (C); that is, the corresponding

system of equations has to be consistent:8>><>>:(G��) :

�x� x0 = �(z � z0)y � y0 = �(z � z0)

(C) :

�F (x; y; z) = 0G(x; y; z) = 0:

:

We get the condition of consistency: �(�; �) = 0; and, by replacing into this

condition the expressions � =x� x0z � z0

; � =y � y0z � z0

; we get the equation of the

46

surface:�(x� x0z � z0

;y � y0z � z0

) = 0:

Example: write the equation of the cone having as vertex V (0; 0; 0) and asdirectrix curve, the parabola: (C) : y2 = 2x; z = 1:Lines passing through V are given by

(G��) :x

�=y

�=z

1)

�x = �yx = �z:

(�)

Intersection with (C) leads to the system8>><>>:(G��) :

�x = �yx = �z:

(C) :

�y2 = 2xz = 1

:

By the second and the last equation, we get x = �; from the �rst, y =�

�: Now,

from the third equation, we get

�(�; �) ��

�2� 2� = 0:

But, from (�); we get � = x

y; � =

x

z: Hence, the equation of the surface is

�yz

�2� 2x

z= 0) y2 = 2xz:

Exercise 51 Find the equation of the surface generated by lines passing throughO(0; 0; 0) and which intersect the circle (C) : x2 + y2 = 1; z = 5:

47

Exercise 52 Determine the equation of the conic surface with vertex A(1; 4; 5)

and having as directrix the hyperbola (C) :

8<: x2 � y2

9= 1

z = 0::

Exercise 53 Determine the equation of the cone of vertex V (3; 4; 5); tangentto the sphere (S) : x2 + y2 + z2 = 4:

6.3 Surfaces of revolution

A surface of revolution is obtained by rotating a plane curve (C) (called thedirectrix) around a straight line, called the axis of rotation.Let (C) : F (x; y; z) = 0; G(x; y; z) = 0 denote the directrix, and

(d) :x� x0l

=y � y0m

=z � z0n

be the axis of rotation. Then, the surface can be regarded as swept out bycircles having their centers oh the axis of rotation, and situated in planes whichare perpendicular to the axis. These can be obtained by intersecting of spheresof arbitrary radius � 2 R+; and centers at an arbitrary �xed point on (d) (forinstance, at (x0; y0; z0)), with planes perpendicular to (d): This is, the generatingcircles can be described as:

(G��) :

�(x� x0)2 + (y � y0)2 + (z � z0)2 = �2

lx+my + nz = �: (�)

Intersecting them with (C); the resulting system8>><>>:(G��) :

�(x� x0)2 + (y � y0)2 + (z � z0)2 = �2lx+my + nz = �

:

(C) :

�F (x; y; z) = 0G(x; y; z) = 0

:

(3 unknowns x; y; z, 4 equations) requires a condition of consistency �(�2; �) =0: Substituting the values of �2 and � from (�); we get the equation of thesurface:

��(x� x0)2 + (y � y0)2 + (z � z0)2; lx+my + nz

�= 0:

Example: Find the equation of the surface obtained by rotating the circle(C) : (x� 4)2 + y2 = 1; z = 0 in the plane xOy; around Oy.We have (d) = (Oy) :

x

0=y

1=z

0; that is, the generating circles are

(G��) :

�x2 + y2 + z2 = �2

y = �:

48

Intersection with (C) is given by the system:8>><>>:x2 + y2 + z2 = �2

y = �(x� 4)2 + y2 = 1

z = 0

:

We get x2 = �2 � �2 ) x = �p�2 � �2. The condition of consistency of the

above system is:

�(�; �) ��q�2 � �2 � 4

�2+ �2 � 1 = 0:

Since �2 = x2 + y2 + z2; � = y; we have:��px2 + z2 � 4

�2+ y2 � 1 = 0:

The "donut" shape we obtained is called a torus:

Exercise 54 Find the equations of the surfaces obtained by rotating the parabola(C) : y = x2; z = 0 around Oy: Recognize and draw the obtained surface.

Exercise 55 Find the equation of the surface obtained by rotating the curve(C) : x2 � 4y2 = 1; z = 0 around: a) Ox and b) Oy. Recognize and draw theobtained surfaces.

Exercise 56 Find the equation of the surface obtained by rotating the curve(C) : y = sinx; z = 0 around Ox.

49

7 Plane curves

7.1 Representations

A plane curve is described by:

1. y = f(x) (explicit representation) or

2. F (x; y) = 0 (implicit representation) or

3.�x = x(t)y = y(t)

; t 2 [a; b] 2 R (parametric representation).

The parametric representation 3) is also used in its vector variant

OM � �r(t) = x(t)�{+ y(t)�j:

A point M(x0; y0) of a curve (C) is called regular if the involved functionsf; F; x(t); y(t) are di¤erentiable (and continuous) in M and:

1. in the implicit representation, the partial derivatives of F do not simulta-neously vanish at M(x0; y0) :

(F 0x(x0; y0))2+�F 0y(x0; y0)

�2> 0;

2. in the parametric representation 3), the derivativesdx

dt(t0) and

dy

dt(t0)

do not simultaneously vanish, where t0 is the value of the parameter tcorresponding to M(x0; y0):

An arc of a curve which consists only of regular points is called a regulararc.In the neighborhood of a regular point one can always pass between the

representations 1),2) and 3).If at some point M; the functions which describe the curve are di¤erentiable

of a certain class p � 1; we say that the curve is of class p at that point.A point of the curve which is not regular is called a singular point. In-

tuitively, singular points M can be points where the functions which appearare not continuous (the curve "breaks" at M), or points where the curve hasan "angle" (which usually corresponds to the situation when the involved func-tions are continuous, but not di¤erentiable, or x0(t) = y0(t) = 0), or "multiplepoints", through which the curve passes more than once).

50

A curve with singular points

7.2 Arc length of a plane curve

Let (C) be given in explicit representation (C) : y = f(x) and_

AB a regular arcof (C):

The length of the arc_

AB is de�ned as the limiting case of the length of a

polygonal line with n sides inscribed in_

AB; joining A and B.

�g. 1.2

This polygonal line is chosen such that the length of its greatest side tends to0, when n tends to in�nity. The length Ln of such a polygonal line is computed

by Pythagoras�theorem. If_

AB is a regular arc, then it can be shown that thislimit always exists.

Theorem 15 1. If (C) is given in explicit representation

y = f (x) ; x 2 [xA; xB ] � (a; b) ; (26)

51

then the length of the regular arc_

AB is

L _AB

=

xBZxA

q1 + (f 0(x))

2dx: (27)

Assume now that (C) is given in the parametric representation x = x(t); y =y(t): Then, the dependence of y on t is given by

y(t) = f(x(t))) dy

dt=df

dx

dx

dt= f 0(x)

dx

dt:

Usually, derivatives with respect to the parameter t are denoted by dots:

_y =dy

dt; _x =

dx

dt:

With these notations, we get

f 0(x) =_y

_x:

Substituting the value of the derivative f 0(x) into the formula of the arc length(27), we get

Theorem 16 If the arc_

AB of (C) is parametrically represented by�x = x (t)y = y (t)

t 2 [tA; tB ] � (t1; t2) ; (28)

then its length is

L _AB

=

tBZtA

q�x2(t) +

�y2(t)dt; (29)

where�x =

dx

dt; _y =

dy

dt:

An alternative expression is obtained in terms of di¤erentials. If we takeinto account that

dx = _xdt; dy = _ydt;

then the formula of the arc length can be also written as

L _AB

=

tBZtA

pdx2 + dy2:

The integrand

ds =

q�x2(t) +

�y2(t)dt =

pdx2 + dy2

52

or, in explicit representation,

ds =p1 + (f 0(x))2dx

is called the element of arc length of the curve (C): With this notation, the

length of the arc_

AB is expressed as:

L _AB

=

Z_AB

ds:

The function s =tRt0

ds (which measures the length of the curve between some

starting point t0 and the current point t) can be used as a parameter on thecurve, called the natural parameter.Intuitively, the element of arc length ds =

pdx2 + dy2 measures the length

of a small (in�nitesimal) arc of the curve, which can be approximated by thelength of a line segment.

the element of arc length

Exercise 57 Compute the element of arc length ds =p_x2 + _y2dt for the cy-

cloid:x = a(t� sin t); y = a(1� cos t); t 2 R

and the length of its arc between the points corresponding to t = 0 and t = 2�:

7.3 Contact between two intersecting curves

Given two curves that intersect at some point P; the notion of contact givesinformation about how close to each other the two curves pass in a neighborhoodof the intersection point.We assume that the functions which de�ne the curves are of class n+1; n �

0:

53

De�nition 17 Let (C) and (C 0) be two curves which intersect at some pointP: We say that (C) and (C 0) have an (n+ 1)-point contact (or, equivalently,a contact of order n), if they have at P (n+ 1) coinciding common points.

In order to establish the order of the contact, the algorithm is the following:we write the system of equations which gives the intersection of (C) and (C 0):Eliminating all the unknowns except one, we will �nd some relation �(x) = 0;�(y) = 0 or �(t) = 0. If the value of the unknown corresponding to the commonpoint P is a multiple root of order (n+1) of �; then at the point P we have an(n+ 1)-point contact.Consider, for instance, the case when the two curves are given in explicit

representation:(C) : y = f1(x); (C

0) : y = f2(x)

and let P (x0; y0) denote a common point.The intersection of (C) and (C 0) is given by:�

y = f1(x)y = f2(x)

:

Subtracting the two equations, y is eliminated and we get:

�(x) � f1(x)� f2(x) = 0:

At the point P we have an (n + 1)-point contact, or an n-th order contact, ifand only if x0 is a multiple root of order (n+ 1) of �; that is, if and only if:

�(x0) = �0(x0) = �00(x0):::: = �

(n)(x0) = 0;

�(n+1)(x0) 6= 0:

Actually, when we say "n-th order contact", n refers to the last derivativeof � which vanishes at x0; while the expression "n+ 1-point contact" refers tothe number of common coinciding points at P:

Also, if we have

(C) : F (x; y) = 0; (C 0) : x = x(t); y = y(t);

and the common point P (x0; y0) corresponds to the value t0 of the parameteron (C 0); then, the system which characterizes the intersection between the twocurves is �

F (x; y) = 0x = x(t); y = y(t)

:

We get�(t) � F (x(t); y(t)) = 0:

Then, an (n+ 1)-point contact at P means

�(t0) = �0(t0) = �00(t0):::: = �

(n)(t0) = 0;

�(n+1)(t0) 6= 0:

54

Exercise 58 Find the order of the contact at the point P (0; 1); between thecurves (C) : y = ex and:

� (C 0) : y = 1 + x;

� (C 0) : y = 1 + x+ x2

2;

� (C 0) : y = 1 + x+ x2

2+x3

6+ :::+

xn

n!; n � 3:

7.4 Tangent and normal line at a regular point

The tangent line to a curve (C) at a point P 2 (C) is de�ned as the limitingposition of a secant line PP 0; when P 0 tends to P: In other words, the tangentline is the straight line which has at least a 2-point contact with the curve at agiven point.In order to write down the equation of a line in plane, one needs a point

M(x0; y0) on the line and its slope m: Then, the equation of the line is

y � y0 = m(x� x0):

This is, once we know a regular point M(x0; y0) of a curve (C); we only needthe slopes of the tangent and of the normal line.

1. If (C) : y = f(x) is given in explicit representation, then, the slope ofthe tangent line at M is

mtg = f0(x0):

The normal line is de�ned as the perpendicular atM to the tangent line,therefore, its slope is

mN = �1

mtg) mN = �

1

f 0(x0):

2. If (C) : F (x; y) = 0 is given in implicit representation, then (accordingto Math Analysis), in a neighborhood of the regular point M; y = f(x) is

de�ned and its derivative at x0 is f 0(x0) = �F 0xF 0yj(x0;y0): Hence,

mtg = �F 0xF 0yj(x0;y0); mN =

F 0yF 0xj(x0;y0):

Since at a regular point, F 0x and F0y do not simultaneously vanish, these two

lines are always well determined (the situation with vanishing denominatorcorresponds to a vertical line, formally, m = �1).

55

3. In parametric representation, (C) : x = x(t); y = y(t) : as shown

above, passing to the explicit representation y(t) = f(x(t)); we getdy

dt=

f 0(x)dx

dt; that is, f 0(x) =

_y(t)

_x(t): Then, at the point P corresponding to

the value t0 of the parameter, we have

mtg =_y(t0)

_x(t0)) mN = �

_x(t0)

_y(t0):

Proposition 18 Two curves (C) and (C 0) have at least a 2-point contact (acontact of order at least 1) at a common point P if and only if they have thesame tangent line at P:

Exercise 59 Write the equations of the tangent and of the normal lines of thefollowing curves:

1. (C) : y = ex at P (x0 = 0);

2. (C) : x2=3 + y2=3 � 1 = 0 (the astroid), at P ( 1

2p2;1

2p2):

the astroid

3. the ellipse (C) : x = 2 cos t; y = sin t; at A(t0 = 0) and at B(0; 1):

Exercise 60 Prove Proposition (18) for the curves (C) : y = f1(x) and (C 0) :y = f2(x); at the common point P (x0; y0):

7.5 Osculating circle; curvature and radius of curvature

Let(C) : x = x(t); y = y(t)

denote a parametrized curve of class at least 2, and P (x0; y0) 2 (C) a regularpoint, corresponding to some value t0 of the parameter.

56

We are looking for a circle that has a maximal order contact with (C) atthe point P . In order to uniquely de�ne a circle, on needs three non-collinearpoints. So, we might expect that (C) should have at P (at least) a 3-pointcontact with a given circle (�):

De�nition 19 The osculating circle (or the circle of curvature) of thecurve (C) at the point P 2 (C) (if it exists) is the circle which has at least a3-point contact with (C) at the point P:

osculating circles of a plane curve

In order to deduce the equation of a circle, one needs the coordinates of itscenter !(a; b) and its radius Rc; once we know these, its equation is:

(�) : (x� a)2 + (y � b)2 �R2c = 0:

At the point P (x0; y0) $ t0, (C) and (�) have at least a 3-point contact ifand only if the function

�(t) � (x� a)2 + (y � b)2 �R2c

has the property:�(t0) = �

0(t0) = �00(t0) = 0:

Performing all the computations (we will skip them here), one can prove thatthe coordinates a; b of the center and the radius Rc are given by

a = x0 �_y( _x2 + _y2)

_x��y � ��

x _yjt0 ; b = y0 +

_x( _x2 + _y2)

_x��y � ��

x _yjt0

Rc =( _x2 + _y2)3=2�� _x��y � ��

x _y�� jt0 :

where the notation jt0 means that all the involved derivatives are calculated att = t0:

57

Remark 20 At the points P where the denominator _x��y� ��x _y vanishes, the curve

has no osculating circle. Such points are called in�ection points of the curve.

Intuitively, the bigger the radius Rc; the less the curve is "bent" at P andconversely: the smaller Rc; the more is (C) bent at P: This "bending" of a curveis seized in the mathematical notion of curvature.

De�nition 21 The number

K =_x��y � ��

x _y

( _x2 + _y2)3=2jt0

is called the curvature of (C) at the point P (t = t0); and the number

R =1

K=( _x2 + _y2)3=2

_x��y � ��

x _yjt0

is called the radius of curvature of (C) at P:

Remark 22 The radius of curvature R is actually the radius of the osculatingcircle, taken with a plus or a minus sign:

jRj = Rc:

This sign tells "on which side" of the curve the osculating circle is located.A change of the sign of R (equivalently, of the curvature K) means that theosculating circle passes on the other side of the curve. The points at which thecurvature vanishes are precisely the in�ection points of the curve.

Remark 23 The centers of all osculating circles of the curve (C) describe acurve called the evolute of (C):

Remark 24 If two curves have at a common point at least a 3-point contact,then at that point, they have the same osculating circle and the same curvature.

Example: Find the osculating circle and the curvature of the parabola(C) : y = x2 at the point A(0; 0):The osculating circle (�) : (x� a)2+(y� b)2�R2c = 0 has at least a 3-point

contact with (C) at A:Intersecting (C) with (�), we get

�(x) � (x� a)2 + (x2 � b)2 �R2c ;

58

the conditions of 3-point contact at A(x = 0) are �(0) = �0(0) = �00(0) = 0; indetail:

�(0) � a2 + b2 �R2c = 0�0(x)jx=0 � 2(x� a) + 2(x2 � b) � 2x jx=0 = 0�00(x)jx=0 � 2 + 12x2 � 4bjx=0 = 0;

that is, 8<: a2 + b2 �R2c = 0�2a = 02� 4b = 0;

which leads to a = 0; b =1

2; Rc =

1

2: Obviously, then, the curvature must be 2

or -2.In order to compute the curvature, we need a parametric representation of

the curve. The simplest choice is by setting t as one of the coordinates x or y;for instance:

x = t

y = t2:

The point A(x = 0; y = 0) is obtained for t0 = 0: Then, _x = 1; �x = 0;_y = 2tjt0=0 = 0; �y = 2; and we get, at the point A

_x��y � ��

x _y = 2; _x2 + _y2 = 1:

Then, the curvature is

K =_x��y � ��

x _y

( _x2 + _y2)3=2=2

1= 2;

and the radius of curvature is R =1

K=1

2:

Exercise 61 Prove that the radius of curvature R =1

Kof the circle (C) : x =

r cos t; y = r sin t; t 2 [0; 2�] at an arbitrary point A(t = t0) coincides withits radius r: Equivalently: the curvature of a circle is a constant, namely, the

inverse of its radius, K =1

r:

Exercise 62 Prove that the curvature of the straight line (d) : y = mx + n atany of its points is 0 (all points of a line are in�ection points).

Exercise 63 Determine the osculating circle, the curvature and the radius ofcurvature of the ellipse

x = 2 cos t; y = sin t

at the point A(0; 1):

59

Exercise 64 For a curve (C) given in explicit representation y = f(x); showthat:

1. in�ection points A(x = x0) are characterized by f 00(x0) = 0:

2. the curvature K at an arbitrary point P (x; y) is given by

K =f 00(x)

[1 + (f 0(x))2)]3=2:

3. There holds the equivalence: f is convex at P (that is, f 00(x) � 0) if andonly if the curvature of (C) at P is positive; f is concave at P if and onlyif its graph (C) has negative curvature at P:

(Hint: choose t = x as a parameter).

60

8 Spatial curves

8.1 Representations

A spatial curve is described by:

1.�F (x; y; z) = 0G(x; y; z) = 0

(as the intersection of two surfaces) or

2.

8<: x = x(t)y = y(t)z = z(t)

; t 2 [a; b] 2 R (parametric representation), which is also

used in its vector variant:

OM � �r(t) = x(t)�{+ y(t)�j + z(t)�k:;

where OM is the position vector of the arbitrary point M of the curve.

A point M(x0; y0; z0) of a curve (C) is called regular if the involved func-tions F;G; x(t); y(t); z(t) are di¤erentiable (and continuous) in M and

1. in implicit representation, the Jacobian of F and G does not vanish atM(x0; y0; z0) : �� F 0x F 0y F 0z

G0x G0y G0z

�� 6= 0 at (x0; y0; z0);

2. in the parametric representation 3), the derivativesdx

dt(t0),

dy

dt(t0) and

dz

dt(t0) do not simultaneously vanish, where t0 is the value of the parameter

t corresponding to M(x0; y0; z0):

An arc of a curve which consists only of regular points is called a regulararc. If at a point M; the functions which describe the curve are di¤erentiableof a certain class p � 1; we say that the curve is of class p at that point.In the neighborhood of a regular point one can always pass between the

representations 1) and 2).

8.2 Arc length of a space curve

Suppose that (C) is given in parametric representation:

�r(t) = x(t)�{+ y(t)�j + z(t)�k

and_

AB denotes a regular arc of (C); corresponding to t 2 [tA; tB ]:The length of the arc

_

AB is de�ned in the same way as for plane curves, thatis, as the limiting case of the length of a polygonal line with n sides inscribed

in_

AB, chosen such that the length of its greatest side tends to 0, when n tends

to in�nity. If_

AB is a regular arc, then it can be proven that its arc length iswell de�ned (=it uniquely exists).

61

Theorem 25 if the arc_

AB of (C) is parametrically represented by8<: x = x (t)y = y (t)z = z(t)

t 2 [tA; tB ] ; (30)

then its length is

L _AB

=

tBZtA

p_x2(t) + _y2(t) + _z2(t)dt; (31)

where the dots denote derivatives w.r.t. t :

_x =dx

dt; _y =

dy

dt; _z =

dz

dt:

In terms of di¤erentials, we have dx = _xdt; dy = _ydt; dz = _zdt; hence, theformula of the arc length can be also written as

L _AB

=

tBZtA

pdx2 + dy2 + dz2:

The integrand

ds =p_x2(t) + _y2(t) + _z2(t)dt =

pdx2 + dy2 + dz2

is called the element of arc length of the curve (C): With this notation, the

length of the arc_

AB is expressed as:

L _AB

=

Z_AB

ds:

The function s =tRt0

ds (which measures the length of the curve between some

starting point t0 and the current point t) can be used as a parameter on thecurve, called the natural parameter.The element of arc length ds =

pdx2 + dy2 + dz2 measures the length of a

small (in�nitesimal) arc of the curve.

8.3 The TNB Frame (The Frenet-Serret Frame)

The TNB-frame (also known as the Frenet-Serret frame) is an orthonormalframe, moving along the curve. The initials "TNB" come from the vectors whichconstitute the frame, namely: the unit tangent vector �� ; the unit normalvector (or, more rigorously, the unit principal normal vector) ��; and the unitbinormal vector ��:

62

Let us describe in the following, the faces and the edges of this frame. Tothis aim, let us recall some basic things about lines and planes in space:- a line can be uniquely de�ned by a point M(x0; y0; z0) and a direction

�v(l;m; n); this way:

(line) :x� x0l

=y � y0m

=z � z0n

:

- the plane which passes through the point M(x0; y0; z0) and has as normaldirection �N = �v(l;m; n); has the equation:

(plane) : l(x� x0) +m(y � y0) + n(z � z0) = 0:

Consider a curve (C) of class at least 2, given in parametric representation:

�r(t) = x(t)�{+ y(t)�j + z(t)�k;

and let M(x0; y0; z0) denote a regular point of the curve, corresponding to thevalue t0 of the parameter.

� The tangent line (tg) and the normal plane (�N ) :The tangent line atM(t = t0) has as directing vector, the tangent vector(velocity vector, derivative vector), having as Cartesian coordinates,the derivatives at t0 of the "trajectory" coordinates x; y; z :

��r(t0) = _x(t0)�{+ _y(t0)�j + _z(t0)�k: (32)

Shortly, if we omit the argument t0, we can write:��r( _x; _y; _z): Since M is

a regular point, the tangent vector at M exists and does not vanish. Theequations of the tangent line are:

(tg) :x� x0_x(t0)

=y � y0_y(t0)

=z � z0_z(t0)

The normal plane (�N ) is de�ned as the plane through M; which isperpendicular to the tangent line. Hence, its equation is:

(�N ) : _x(t0)(x� x0) + _y(t0) (y � y0) + _z(t0) (z � z0) = 0:

� The binormal line (b) and the osculating plane (�Osc):Suppose that at the point M; there exists the second derivative

��r(t0) = �x(t0)�{+ �y(t0)�j + �z(t0)�k;

(the acceleration vector) and the cross product��r(t0)�

��r(t0) does not

vanish, that is, the velocity vector��r(t0) and the acceleration vector

��r(t0)

63

are nonvanishing and non-collinear. Then, the two vectors, together withthe point M; de�ne a plane, called the osculating plane of (C) at M:The vector

�b(t0) =��r(t0)�

��r(t0); (33)

which is actually, the normal vector of the osculating plane, is called thebinormal vector of (C) at M: Let us suppose that, after computingthe cross product, we get for �b(t0) the coordinates �b(l;m; n): Then, thebinormal line has the equations:

(b) :x� x0l

=y � y0m

=z � z0n

:

The osculating plane is de�ned by:

(�Osc) : l(x� x0) +m(y � y0) + n(z � z0) = 0:

� The (principal) normal line (np) and the rectifying plane (�R) :The normal line (or the principal normal line) of the curve is de�nedas the intersection between the normal plane (�N ) and the osculatingplane (�Osc). The normal line is perpendicular to both the tangent vectorand to the binormal one. Consequently, its directing vector �np can beobtained as the cross product of the two vectors:

�np = �b(t0)��r(t0); (34)

that is,

�np =� ��r(t0)�

��r(t0)

��

��r(t0): (35)

This choice of the order �b(t0) ��r(t0) (instead of

��r(t0) � �b(t0)) insures

that the vectors��r, �np and �b constitute a right-oriented system at M (the

orientation of �b can be obtained by the right hand rule from those of��r

and �np).

Consequently, if, by performing computations, we are led to

�np(A;B;C);

then, the equations of the principal normal line are

(np) :x� x0A

=y � y0B

=z � z0C

;

while the rectifying plane (which is de�ned as the plane perpendicularto np at M) is given by:

(�R) : A(x� x0) +B(y � y0) + C(z � z0) = 0:

64

TNB frame

By (32), (33) and (34)-(35), we get

Theorem 26 The unit vectors of the TNB frame at the point M(t = t0) are:

� the unit tangent vector �� =��r(t0) ��r(t0) ;

� the unit (principal) normal vector �� =

� ��r(t0)�

��r(t0)

��

��r(t0) � ��r(t0)� ��

�r(t0)��

��r(t0)

;� the unit binormal vector �� =

��r(t0)�

��r(t0) ��r(t0)� ��r(t0)

Example: Find the unit vectors of the TNB (Frenet) frame of the curve

(C) : �r(t) = et�{+ e�t�j + t2�k at the point M(1; 1; 0):The point M has the Cartesian coordinates x = 1; y = 1; z = 0; hence

x0 = 1 = et0

y0 = 1 = e�t0

z0 = 0 = t20:

From all the three relations above, we obtain for M the value t0 = 0:

65

The derivatives of the position vector �r at M are:

��r(t0) = (et�{� e�t�j + 2t�k)jt0=0 = �{� �j��r(t0) = (et�{+ e�t�j + 2�k)jt0=0 = �{+ �j + 2�k;

that is,��r(1;�1; 0);

��r(1; 1; 2): The unit tangent vector is �� =

��r(t0) ��r(t0) )

��(1p2;� 1p

2; 0):

In order to calculate the unit binormal vector, we need

�b =��r(t0)�

��r(t0) =

��{ �j �k1 �1 01 1 2

�� = �2�{� 2�j + 2�k:Then, �� =

�b �b ) ��(� 1p3;� 1p

3;1p3):

The principal normal vector is �np = �b��r; in coordinates:

�np =

��{ �j �k�2 �2 21 �1 0

�� = 2�{+ 2�j + 4�k:Its unit vector is �� =

�npk�npk

) ��(1p6;1p6;2p6):

Exercise 65 Determine the equations of the axes and of the faces of the Frenetframe for the following curves:

1. �r(t) = cos t �{+ sin t �j + t�k at A(t0 =�

2):

2. �r(t) = t�{+ t2�j + ln t�k at M(1; 1; 0):

8.4 Curvature and torsion

Consider a curve (C) : �r(t) = x(t)�{+ y(t)�j + z(t)�k of class at least 3, and take

a regular point M(t = t0) on (C); with��r �

��r 6= 0; the latter condition insures

that the TNB frame exists at M .The curvature of (C) at the point M(t = t0) is a number which measures

the "deviation" of the curve from its tangent line in a neighborhood of M: It isgiven by:

K(t0) =

��r � ��r ��r 3 jt0 :

66

A point at which the curvature vanishes (that is,��r �

��r = 0) is called an

in�ection point of the curve. At in�ection points, there is no TNB frame.(!) Do not mistake this formula for the one of the curvature of a plane

curve! The curvature of a space curve, as de�ned above, is always nonnegative.On the contrary, for plane curves, the curvature can have a minus sign.

Proposition 27 For straight lines, the curvature is 0 at each point.(a straightline consists only of in�ection points). Conversely, if the curvature of a curveis 0 at each point, then the curve is a straight line.

This is easy to see, if we take into account that a straight line is describedparametrically by:

x = x0 + lt; y = y0 +mt; z = z0 + nt;

where x0; y0; z0; l;m; n are constants. Then, we get��r(l;m; n);

��r = �0; which

leads to K = 0:

The torsion � of the space curve (C) at a regular point M(t = t0) is anumber which indicates how much the curves deviates from a plane (namely,from the osculating plane at M) in a neighborhood of M: It is given by theformula:

�(t0) =

� ��r ;

��r ;

��r�

��r � ��r 2 jt0 :

Proposition 28 For plane curves, the torsion is identically 0. Conversely, ifthe torsion identically vanishes, then the curve is contained in a plane.

A space curve which is contained in no plane is called a skew curve.The sign of the torsion provides information about the side of the osculating

plane in which the curve "bends" around M: Namely:

� if � > 0 at M; then, in a neighborhood of M; the curve "bends" on theside indicated by the binormal vector �b;

� if � < 0 at M; then, in a neighborhood of M; the curve "bends" on theopposite side of �b;

� if � changes the sign at M; then the curve "pierces" the osculating planeat M:

Exercise 66 Calculate the curvature and the torsion of the curve (C) : �r(t) =et�{+ e�t�j + t2�k at point M(1; 1; 0):

67

Exercise 67 Find the curvature and the torsion of the curve (C) : �r(t) =a cos t�{+ a sin t�j + bt�k (where a; b 2 R are constants) at an arbitrary point.

Exercise 68 Prove that for plane curves (C) : �r(t) = x(t)�{+ y(t)�j +0�k (in the

plane xOy), the formula of the curvature of a space curve Kspace =

��r � ��r ��r 3 jt0

gives the absolute value of the curvature of (C) computed by the "plane" formula

Kplane =_x�y � �x _y

j _x2 + _y2j3=2:

68

9 Di¤erential Geometry of Surfaces

9.1 Representations

A surface (in a general sense) is a set (�) of points in Euclidean space whoseposition vectors obey an equation of type:

(�) : �r(u; v) = x(u; v)�{+ y(u; v)�j + z(u; v)�k; (u; v) 2 A�B � R2;

that is, �r = �r(u; v); or, equivalently,

(�) :

8<: x = x(u; v);y = y(u; v);z = z(u; v):

(36)

This is called the parametric representation of (�); while u and v are calledthe parameters (or the curvilinear coordinates) of the representation.

We will denote the partial derivatives of �r(u; v) by indices, as follows:

�r1 =@�r

@u; �r2 =

@�r

@u; �r11 =

@2�r

@u2; �r12 =

@2�r

@u@vetc.

A point of M the surface is called regular, if, at M; �r1 and �r2 exist andtheir cross product is nonzero:

�r1 � �r2 6= 0

(that is, the two vectors are nonzero and non-collinear). A point of the surfacewhich is not regular is called a singular point.

� Other description of a surface is

(�) : z = z(x; y); (37)

which is called the explicit Cartesian equation. At regular points, z has tobe (continuous and) di¤erentiable.

� Finally, a surface or a fragment of surface can be represented also as:

F (x; y; z) = 0; (38)

which is called the implicit Cartesian equation. At regular points, Fhas to be di¤erentiable and to obey the condition:

(F 0x)2 + (F 0x)

2 + (F 0x)2 > 0;

i.e., at least one of the partial derivatives F 0x; F0y; F

0z has to be nonzero.

69

If the functions which describe the surface are di¤erentiable of some classp � 1 at some point, we say that the surface is of class p at that point.

Example: For the elliptic paraboloid z = x2 + y2 (! explicit representa-tion), we have

� the implicit representation: x2 + y2 � z = 0:

� a parametric representation is x = u cos v; y = u sin v; z = u2; or, in vectorform,

~r(u; v) = u cos v~i+ u sin v~j + u2~k:

9.2 Curves on a surface

Let (S) be a surface of class p � 1 in the representation

�r = �r(u; v); (39)

and P 2 (S) a regular point. A curve (C) on (S) is de�ned by establishing alink between the parameters, for instance:

u = u(t); v = v(t); t 2 A � R; (40)

or, as well,v = v(u) or u = u(v) or h(u; v) = 0: (41)

A peculiar importance have the curves u = u0 (constant) and v = v0 (con-stant), called coordinate curves; they compose a grid, such that through everypoint P (u = u0; v = v0) of (S); there passes exactly one curve of each family,namely: (�u) : v = v0; (C2) : (�v) : u = u0:

coordinate curves (�u); (�v)

70

In other words, P represents the intersection point of the coordinate curvesv = v0 and u = u0; the situation is similar to the one in the plane xOy; whereM(x0; y0) lies at the intersection of the straight lines x = x0 and y = y0; there,(x0,y0) were called rectilinear (or Cartesian) coordinates. This similarity is whatgives us the right to call u and v; curvilinear coordinates on (S):Let now P denote a point of an arbitrary curve (C) � (S); represented as

in (40). That is, along (C); the position vector is �r = �r(u(t); v(t)): Then, thetangent vector at P to (C) is

��r =

d�r

dt=@�r

@u

du

dt+@�r

@v

dv

dt= �r1 _u+ �r2 _v; (42)

where we used the notations �r1 =@�r

@u; �r2 =

@�r

@v: This relation points out that

actually, the tangent vector��r is coplanar with �r1 and �r2:

Example: For the coordinate curves to a surface �r = �r(u; v), we have:(�u) : (u = t; v = const:) ) _u = 1; _v = 0 and (�v) : (u = const:; v = t) )_u = 0; _v = 1: Therefore, the tangent vectors to the coordinate curves v = const

are �r1 =@�r

@u; similarly, the tangent vectors to the coordinate curves u = const

are �r2 =@�r

@v:

If P 2 (S) is a regular point, then, at P; the two vectors �r1 and �r2 are nonzeroand non-collinear (since �r1 � �r2 6= 0). Hence, together with the point P , theyuniquely de�ne a plane P: Moreover, by (42), we conclude that all tangent linesto curves through P belong to this plane.

Exercise 69 Describe the coordinate curves of the surface: ~r(u; v) = cosu ~i+sinu ~j + v~k: Also, �nd the implicit Cartesian equation of the surface and recog-nize it.

9.3 Tangent plane and normal line at a regular point

De�nition 29 The tangent plane at a regular point P (x0; y0; z0) 2 (S) is theplane which contains the tangent lines at P of all curves on (S); passing throughP:

71

tangent plane to a surface

Suppose that P corresponds to u = u0; v = v0: As shown above, the tangentplane is actually the plane which passes through P and contains the directions�r1 and �r2: That is, its equation is:��

x� x0 y � y0 z � z0@x

@u(u0; v0)

@y

@u(u0; v0)

@z

@u(u0; v0)

@x

@v(u0; v0)

@y

@v(u0; v0)

@z

@v(u0; v0)

�� = 0: (43)

The normal line at a regular point P 2 (S) to (S) is the perpendicularthrough P to the tangent plane of (S) at P: Since the vectors �r1; �r2 belong tothe tangent plane at P to (S), it follows that the normal vector to the surfaceat P is �!

N = (�r1 � �r2) j(u0;v0): (44)

The unit surface normal at the point P is:

�!n =�!N �!N = �r1 � �r2

k�r1 � �r2k: (45)

Example: Consider the surface (S) : x = u cos v; y = u sin v; z = hv (thehelicoid):

72

Determine the tangent plane and the normal line at the point A(1; 0; 0):The point A corresponds to the values u0 = 1; v0 = 0 of the parameters.

We have: ~r1 = cos v~i + sin v~jjA = ~i; ~r2 = �u sin v~i + u cos v~j + h~kjA = ~j + h~k:Then, the tangent plane is described by the equation:��

x� 1 y z1 0 00 1 h

�� = 0;or �hy + z = 0: The surface normal vector at A is the normal vector of this

plane, hence ~N(0;�h; 1); the unit surface normal is ~n�0;

�hp1 + h2

;1p1 + h2

�;

and the equations of the normal line are:

x� 10

=y

�h =z

1:

If (S) is given by in implicit form (38):

F (x; y; z) = 0;

then the equation of the tangent plane to the surface (S) : F (x; y; z) = 0 at aregular point P (x0; y0; z0) 2 (S) is:

(x� x0)F 0x(x0; y0; z0) + (y� y0)F 0y(x0; y0; z0) + (z � z0)F 0z(x0; y0; z0) = 0: (46)

The surface normal�!N; which is also denoted by gradF (and called the gradient

of F ), is�!N � gradF (F 0x; F 0y; F 0z)j(x0;y0;z0) (47)

Exercise 70 Find the tangent planes and the corresponding normal lines to thesurface:(S) : ~r(u; v) = u cos v~i+ u sin v~j + u~k; at A(u = 1; v = 0):Find the implicit equation of the surface, recognize and draw it.

Exercise 71 Find the tangent planes and the corresponding normal lines to thesurfaces:1) (S) : x2 + y2 + z2 �R2 = 0 (the sphere with centre O and radius R) at a

point A(x0; y0; z0) 2 (S);2) (S) : z = x2 � y2 (the "saddle"), at A(1; 1; 0):

73

9.4 First fundamental form. Applications

Consider a surface(S) : ~r = ~r(u; v)

of class p � 1; (C) : u = u(t); v = v(t); a curve on (S) and a regular arcaAB

of the curve (C); corresponding to the values t 2 [a; b] : A(t = a); B(t = b):

We intend to calculate the length of the arcaAB � (S): Taking into account

thataAB is an arc of a spatial curve, its arc length is de�ned by the integral

bRa

��r(t) dt.We have

��r = ~r1 _u+ ~r2 _v, hence: ��r 2 = �

�r ��r = (�r1 � �r1) _u2 + 2(�r1 � �r2) _u _u2 + (�r2 � �r2) _v2:

Denoting

E = �r1 � �r1 = k�r1k2 ; F = �r1 � �r2; G = �r2 � �r2 = k�r2k2 ; (48)

we get: ��r 2 = E _u2 + 2F _u _v +G _v2; (49)

This leads to the formula of the arc length of a curve on (S):

L aAB

=

bZa

pE _u2 + 2F _u _v +G _v2dt: (50)

Moreover, from (49), we can see that the arc length element of the curve (C)is

ds2 = Edu2 + 2Fdudv +Gdv2: (51)

De�nition 30 The di¤erential quadratic form ds2 in (51) is called the �rstfundamental form of the surface (S):

Other applications of the �rst fundamental form are to the calculus of anglesbetween curves on the surface and areas of regions of (�):1) The angle � between two intersecting curves (C) and (C�) on (S);

is, by de�nition, is the angle between the tangent lines to (C) and (C�) at theircommon point.To this aim, let us consider the curves (C) and (C�) given on (S) by

(C) : u = u(t); v = v(t)

(C�) : u� = u�(t�); v� = v�(t�):

If P 2 (C1) \ (C2) is a common point, then, there holds

74

Proposition 31 The angle � between two curves on (S) is given by:

cos� =E _u _u� + F ( _u _v� + _u� _v) +G _v _v�p

E _u2 + 2F _u _v +G _v2pE _u�2 + 2F _u� _v� +G _v�2

: (52)

Example: If (C) are (C�) the coordinate curves u = const: and v = const:;

then _u = 0; _v� = 0; which means that�~r = ~r2 _v and

�~r�= ~r1 _u

�: It follows thatthe angle between the coordinate curves is

cos� =~r1 � ~r2k~r1k k~r2k

=FpEG

; (53)

2) The area A(�) of a regular fragment (�) of the surface (S) is

A(�) =

Z ZU

pEG� F 2dudv; (54)

where U � R2 is the domain of the values (u; v) corresponding to the fragment(�): The expression

d� =pEG� F 2dudv (55)

is called the area element of the surface (S); and represents the area of asmall (in�nitesimal) "curvilinear parallelogram" of the surface (�); determinedby two pairs of neighboring coordinate curves.

Exercise 72 For the cone ~r(u; v) = u cos v~i+u sin v~j+u~k; v 2 [0; 2�); u 2 R;determine:a) its �rst fundamental form;b) the length of the curve u = 1 between the points corresponding to v0 = 0

and v1 = �;c) the angle of the curves u = 1 and u+ v = 1;d) the area of the curvilinear parallelogram built on the curves: u = 1; u = 2;

v = 0; v =�

2:

75

10 Vector Spaces

10.1 De�nition and examples

Informally speaking, a vector space (or a linear space) is a set of objects (gener-ically called vectors) that may be scaled and added according to some rules;these objects can be numbers, (geometrical) free vectors, matrices, functionsetc. The theory of vector spaces provides a powerful tool for other branches ofmathematics, such as: geometry, analysis, di¤erential equations, coding theory.

Let V 6= ? denote a non-empty set, and (K;+; �) a commutative �eld. Letus de�ne two operations:

+ : V � V ! V; (u; v)! u+ v;

(internal operation on V ) which will be called vector addition and

�K : K � V ! V; (�; v)! �v;

(external operation) called further, scalar multiplication.

De�nition 32 (V;+; �K) is called a vector space over K if:

1. (V;+) is an Abelian group, that is:

(a) V is closed under vector addition: 8 u; v 2 V : u+ v 2 V ;(b) vector addition is associative: for all u; v; w 2 V , we have u + (v +

w) = (u+ v) + w;

(c) vector addition is commutative: 8v; w 2 V : v + w = w + v;(d) vector addition has an identity (a neutral) element: 90 2 V; such that

v + 0 = v, 8v 2 V ;(e) any element v 2 V admits a symmetric (or an additive inverse)

(�v) 2 V such that v + (�v) = 0:

2. the scalar multiplication �K has the following properties:

(a) V is closed under scalar multiplication: 8� 2 K; 8v 2 V : �v 2 V ;(b) �K is distributive w.r.t. vector addition: 8� 2 K; 8u; v 2 V : �(u +

v) = �u+ �v;

(c) �K is distributive w.r.t. the addition in K: 8�; � 2 K; 8v 2 V :(�+ �)v = �v + �v;

(d) scalar multiplication is compatible with multiplication in K ("asso-ciativity"): 8�; � 2 K; 8v 2 V : �(�v) = (��)v;

(e) the neutral (identity) element 1 2 K is an identity element for scalarmultiplication: 8v 2 V : 1v = v:

76

The elements u; v; ::: 2 V are called vectors, while the elements �; �; ::: 2 Kare called scalars.

Rules of computation in vector spaces:

1. Scalar multiplication with the zero vector 0 2 V gives the zero vector:

8� 2 K : �0 = 0:

2. Scalar multiplication by the scalar 0 2 K gives the zero vector:

8v 2 V : 0v = 0:

3. No other scalar multiplication leads to the zero vector: if �v = 0; theneither � = 0; or v = 0:

4. The result of scalar multiplication (�1) is the additive inverse of the vector:

8v 2 V : (�1)v = �v:

If the �eld of scalars K is R; then the vector space is called a real vectorspace; if K = C; then V is called a complex vector space.

Examples of vector spaces:

1. Take K = R; and Rn = fx = (x1; x2; :::; xn) j xi 2 R; i = 1; ng; the setof all ordered n-uples of real numbers, and the operations:�

x = (x1; x2; :::; xn) 2 Rny = (y1; y2; :::; yn) 2 Rn

! x+ y = (x1 + y1; x2 + y2; :::; xn + yn)

andx = (x1; x2; :::; xn) 2 Rn ! �x = (�x1; �x2; :::; �xn):

Then, (Rn;+; �R) is a vector space, called the real coordinate space Rn:This is perhaps the most important example of vector space; especially,the cases n = 2 and n = 3 are important for Geometry.

2. Let K be a �eld and Mm�n(K) the space of m � n-type matrices withentries aij 2 K. Then, (Mm�n(K);+; �K) is a vector space over K: Inparticular, for K = R; (Mm�n(R);+; �R) is a real vector space.

3. The set K[X] of all polynomials with coe¢ cients in a �eld K is a vectorspace over K:

77

4. The set F(I) of all real-valued functions de�ned over an interval I � R;

F(I) = ff j f : I ! Rg;

with the usual addition and scalar multiplication of functions,

(f + g)(x) = f(x) + g(x);

(�f)(x) = � � f(x); 8x 2 I; � 2 R;

is a real vector space.

5. ("the last, but not the least"), the set E3 of all free vectors �a in space,with the usual addition (e.g., using the parallelogram rule) and scalarmultiplication is a real vector space.

Exercise 73 Consider V = (0;1) and K = R: We de�ne the operations:

u; v 2 V ! u� v = u � v� 2 R; v 2 V ! �� v = v�:

Prove that (V;�;�) is a vector space over R:

10.2 Subspaces in a vector space

Let (V;+; �K) denote a vector space.

De�nition 33 A non-empty subset S of V is called a subspace in V if S; to-gether with the operations + and �K of the space V; restricted to S, is a vectorspace.

In order to check that a subset of a vector space is a subspace, one has thefollowing criterion:

Criterion 34 A non-empty set S of a vector space (V;+; �K) is a subspace ifand only if:

1. for any u; v 2 S; the sum u+ v belongs to S and

2. for any scalar � 2 K and any vector u 2 S; we have: �u 2 S:

The above criterion is equivalent to:

Criterion 35 A non-empty set S of a vector space (V;+; �K) is a subspace ifand only if:

8�; � 2 V;8u; v 2 S : �u+ �v 2 S:

78

This is, a subspace is a subset of a vector space which is closed under additionand scalar multiplication, or, equivalently, under formation of linear combina-tions.

Exercise 74 Prove Criterion (35).

Examples:1) For any space (V;+; �K), the sets f0g and V represent subspaces of V:Let us check the above statement:

� For S = f0g; we have: 0 + 0 = 0 2 S and �0 = 0 2 S; 8� 2 K: Hence,S = f0g is a subspace.

� For S = V; there holds: 8 u; v 2 V : u+ v 2 V; that is, V is closed underaddition. Also, by the de�nition of a vector space, V is closed under scalarmultiplication: 8� 2 K; 8 u 2 V : �u 2 V; which proves the statement.

Exercise 75 Prove that the solutions of the system:�x1 + 2x2 + x3 = 0x1 � 3x3 = 0

(56)

form a vector subspace of R3:

Exercise 76 Prove that the solutions of the system�x1 + 2x2 + x3 = 1x1 � 3x3 = 2

(57)

do not constitute a vector subspace of R3: This is, if in the system (56) wechange the free terms, then solution set of the system does no longer representa vector subspace.

Exercise 77 Prove that, in the space of polynomials (C[X];+; �C); the set

Pn = ff 2 C[X] j f = anXn + an�1Xn�1 + ::::+ a1X + a0g

of all polynomials of degree at most n, is a vector subspace.

Exercise 78 Consider a vector space (V;+; �K) and two subspaces U; S � V .Prove that:

� the intersection U \ S is a subspace of V ;

� the sum U + S = fu+ s j u 2 U; s 2 Sg is a subspace of V ;

� the union U [ S is generally not a subspace (�nd a counterexample!).

79

10.3 Subspace spanned by a subset of a vector space

Assume that (V;+; �K) is a vector space and U � V a non-empty subset. Then,the set of all �nite linear combinations of elements in U; namely

[U ] = f�1u1 + ::::�nun j �i 2 K; ui 2 U; i = 1; n; n 2 N�g

is a vector subspace of V; called the subspace spanned by U; or simply, thespan of U: U is called a spanning set (a set of generators) for [U ]:Moreover, it can be proved that [U ] is the "smallest" subspace of V which

contains U � that is, it is contained in any other subspace containing U:

In particular, if U = fu1; u2; ::::; ung is a �nite set, then its span [U ] is simplythe set of all linear combinations of its elements:

[U ] = f�1u1 + ::::�nun j �i 2 K; i = 1; ng:

Examples:1) Take U = f(1; 2); (3; 4)g � R2: Then,

[U ] = f�(1; 2) + �(3; 4) j �; � 2 Rg = f(�+ 3�; 2�+ 4� j �; � 2 Rg:

2) If U contains only one vector U = fug; then the subspace spanned by Uconsists of all multiples of u :

[U ] = f�u j � 2 Kg:

Exercise 79 Prove that the following subsets of R3 are subspaces and �nd aspanning subset for each of them:

1. S1 = f(�; 2�; 3�) j� 2 Rg;

2. S2 = f(�+ �; �� ; 2�) j �; � 2 Rg;

3. S3 = f(x1; x2; x3) j x1 + x2 = 0; x1 � 2x3 = 0g:

10.4 Linear dependence and independence

Consider a vector space (V;+; �K) and a subset U � V .

De�nition 36 A set U of vectors in V is called:

� linearly dependent if one of the vectors of U can be written as a �nitelinear combination of other vectors of U ; for instance

un+1 = �1u1 + ::::+ �nun; �i 2 K; ui 2 U ;

80

� linearly independent if none of the vectors of U can be written as alinear combination of other vectors in U:

For instance, the vectors u = (1; 0); v2 = (0; 1); w = (1; 2) in R2 are linearlydependent, because w is a linear combination of u and v:

w = u+ 2v

Still, in practice it would be hard to guess whether and which of the vectorsof a set U could be written as a linear combination of the others. This is whywe will use the following criterion:

Criterion 37 The vectors u1; u2; :::un 2 V are linearly independent if and onlyif the equation

�1u1 + �2u2 + ::::+ �nun = 0;

(with the unknowns �1; ::; �n 2 K) has the unique solution �1 = �2 = ::: =�n = 0:

Remark 38 An in�nite set U � V is linearly independent if any �nite subsetof U is linearly independent.

Exercise 80 Which of the following systems of vectors in R2 are linearly inde-pendent:

1. u = (1; 2);

2. u = (1; 2); v = (3;�1);

3. u = (1; 2); v = (3;�1); w = (4; 1)?

Exercise 81 Prove that, in a vector space (V;+; �K) :

� a system consisting of one vector fvg � V is linearly independent ()v 6= 0;

� two vectors u; v 2 V are linearly dependent if and only if one of them is amultiple of the other, that is, there exists an � 2 K such that v = �u:

81

10.5 Basis and dimension of a vector space

Suppose (V;+; �K) is a vector space.

De�nition 39 A subset B 6= ? of V is called a basis of V if:

1. B is linearly independent;

2. B spans the entire space V : any v 2 V can be written as a �nite linearcombination of elements of B :

v = �1e1 + :::+ �nen;

where e1; :::; en 2 B; �1; :::; �n 2 K:

Example (the standard basis in Rn): Take, in Rn; B = fe1; :::; eng;where: 8>><>>:

e1 = (1; 0; 0; :::; 0)e2 = (0; 1; 0; :::; 0)

:::en = (0; 0; 0; :::; 1)

:

The above de�ned set B is a basis of Rn; called the standard or the canonicalbasis.Let us check that B is indeed a basis:

� linear independence: the equality �1e1 + ::::+ �nen = 0 means:

�1(1; 0; :::; 0) + ::::+ �n(0; 0; ::::; 1) = 0;

which is, (�1; ::::; �n) = (0; 0; ::::; 0):

� the "spanning property": Let v = (v1; v2; :::; vn) 2 Rn be arbitrary. Then,the equation

v = �1e1 + :::+ �nen

has the solution �1 = v1; :::; �n = vn; that is, v can indeed be written asa linear combination of e1; ::::; en 2 B:

Properties:

1. Any non-zero vector space admits at least a basis. From now on, whenreferring to vector spaces, we will only mean non-zero ones.

2. Any linearly independent system of vectors can be completed up to abasis( , can be included in a basis).

3. From any spanning set of a vector space, one can extract a basis.

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That is, we can summarize the situation in the following way:

linearly independent set � basis � spanning set. (58)

A vector space which admits a basis B consisting of a �nite number of vectorsis called a �nite-dimensional vector space.For �nite-dimensional vector spaces, there exists a "shortcut" for proving

that a certain set is a basis. Namely, it relies on the notion of dimension of avector space.

Theorem 40 If V is a �nite-dimensional vector space, then any two bases ofV have the same number of vectors. This number is called the dimension ofthe vector space V :

n = dimV:

Example: The dimension of Rn is n (since the standard basis has n vectors).

From (58) we can now infer:

Theorem 41 If V is a vector space of dimension n; then any set consisting ofprecisely n linearly independent vectors of V is a basis of V:

Example: In R2; the vectors u = (1; 2) and v = (2; 1) are linearly indepen-dent (prove it!); since the dimension of R2 is 2, we get that fu; vg is a basis ofR2:

Exercise 82 Prove that the vectors u = (1; 0; 0); v = (1; 2; 0) and w = (1; 2; 3)form a basis of R3:

Another important property of bases is the following:

Theorem 42 Consider a �nite dimensional vector space (V;+; �K) and denoteby B = fe1; e2; :::; eng a basis of V: Any vector x 2 V admits a unique repre-sentation:

x = x1e1 + x2e2 + :::+ xnen; (59)

with x1; :::; xn 2 K (scalars). The scalars x1; :::; xn are called the coordinatesof x with respect to the basis B (in the basis B).

Proof. Let x 2 V be arbitrary. The "spanning property" in the de�nitionof a basis insures the existence of the scalars x1; :::; xn 2 K such that x =x1e1 + x2e2 + :::+ xnen:Let us prove the uniqueness of this representation: if x01; ::::; x

0n 2 K is

another set of scalars with x = x01e1 + x02e2 + :::+ x

0nen:

83

Subtracting the above equalities, we are led to:

0 = (x1 � x01)e1 + :::+ (xn � x0n)en:

Since e1; :::; en are linearly independent, we get x1 � x01 = ::: = xn � x0n = 0;q.e.d.Once the basis B is �xed, the column matrix having as entries the coordinates

of some vector x 2 V in the basis B;

X =

[email protected]

1CCCAis called the coordinate vector of x 2 V:

Example: If B = fe1; :::; eng is the standard basis of Rn; we have shownabove that for any x = (x1; x2; :::; xn); we have x = x1e1 + :::: + xnen; whichmeans that, with respect to the standard basis, the coordinates of a vector x 2 Rncoincide with its components. The coordinate vector X is, in this case, preciselythe transpose matrix xt:For instance, the vector v = (1; 3) 2 R2 has, with respect to the standard

basis, the coordinates

X =

�13

�:

10.6 Changes of bases

Let (V;+; �K) denote a �nite-dimensional vector space and

B = fe1; ::::; eng; B0 = fe01; :::; e0ng

two bases of V . Then, any vector x 2 V admits a unique decomposition

x = x1e1 + ::::+ xnen

with respect to B; and a unique decomposition

x = x01e01 + ::::+ x

0ne0n

with respect to B0:The problem which arises is obviously the following: which is the relation

between the two sets of coordinates of x; namely, between

X =

[email protected]

1CCCA and X 0 =

[email protected]

1CCCA?

84

In order to answer this question, we decompose, one by one, the vectors ofthe "new" basis B0 in terms of the basis B: We get:

e01 = s11e1 + s21e2 + :::+ sn1en

e02 = s12e1 + s22e2 + :::+ sn2en

:::

e0n = s1ne1 + s2ne2 + :::+ snnen;

with sij 2 K; i; j = 1; :::; n:The matrix S = (sij)i;j=1;n; having as columns the coordinate vectors of

e01; :::; e0n respectively, in the basis B; is called the matrix of change of bases

from B to B0: We denoteB S! B0:

Then, we get

x =nXj=1

x0je0j =

nXj=1

x0j

nXi=1

sijei

!=

nXi=1

0@ nXj=1

sijx0j

1A ei:On the other side, x =

nPi=1

xiei; since this decomposition is unique, we get:

xi =nPj=1

sijx0j ; or, in terms of matrices, X = SX 0: We have thus proved:

Proposition 43 If S is the matrix of change of bases from B to B0; then thecorresponding sets of coordinates of a vector x 2 V are related as follows:

X = SX 0:

The matrix of change S is always invertible, hence we can also write

X 0 = S�1X:

Exercise 83 Take B1 = fe1 = (1; 1; 1); e2 = (1; 1; 0); e3 = (1; 2; 3)g and B2 =fe01 = (0; 1; 1); e02 = (1; 0; 0); e03 = (3; 2; 1)ga) Prove that B1 and B2 are bases of R3:b) Find the expression of the vector x = (2; 2; 1) in the bases B1 and B2:c) Find the matrix of change B1

S! B2 and check, for the vector x above, therelation X = SX 0:d) Find the matrices of change from the standard basis B to B1 and B2:What

do you notice? Why?

85

LADG Lecture Notes

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