1

Chapter 11 Vectors and Vector-Values Functions Section11.1 Vectors in the Plane Vectors Used by to indicate a quantity that has both direction and magnitude

o For example: force, displacement, velocity etc. A directed line segment, represented by an arrow.

o The length of the arrow indicates magnitude. o Arrow points in the direction. o For instance a vector that represents the wind velocity at a location. The arrow

points in the direction the wind is blowing and the length of the arrow is the wind speed.

Denoted by bold face lower case letter in text, for instance v when writing use v .

A Tail

B Head

2

Two vectors are equal if they have the same magnitude and the same direction, written u v .

Vector vs Scalar o A scalar is a quantity that has magnitude but no direction. o The zero vector, 0

, is an exception. It has length zero and no direction.

Scalar Multiplication: cv , is called a scalar multiple of v The magnitude of cv , is c times the magnitude of v . If 0c , then cv has the same direction as v . If 0c , then cv has the opposite of v

. If 0c then 0 0v .

3

Parallel Vectors: Two vectors are parallel if they are scalar multiples of each other, i.e. u is parallel to

v if u kv where k is a scalar. Addition of Vectors Geometrically

4

Definition Position Vectors and Vector Components A vector v with its tail at the origin and head at 1 2,v v is called a position vector and is written 1 2,v v . The real numbers 1v and 2v are called the x- and y-components of v . The position vectors 1 2,u u u and 1 2,v v v are equal if and only if 1 1u v and 2 2u v

5

Definition Magnitude of a Vector Given the points 1 1,P x y and 2 2,Q x y , the magnitude or length of

2 1 2 1,PQ x x y y

, denoted PQ

is the distance between P and Q

2 22 1 2 1PQ x x y y

.

The magnitude of the position vector v is 2 21 2v v v .

6

Vector Operation Suppose c is a scalar, 1 2,u u u and 1 2,v v v Vector Addition: 1 1 2 2,u v u v u v Vector Subtraction: 1 1 2 2,u v u v u v Scalar Multiplication: 1 2,cv cv cv Example: Let 7,1u , 2,5v and 6, 3w , find 5 3u v w

7

Definition Unit Vectors and Vectors of Specified Length

A unit vector is any vector of length 1. Given a nonzero vector, v , vv

are unit vectors

parallel to v , For a scalar c, c > 0, the vectors cvv

are vectors of length c parallel of v .

8

Summary: Properties of Vector Operations Let , ,u v w be vectors and , a bbe scalars. 1. Commutative property of addition: u v v u 2. Associative property of addition: u v w u v w 3. Additive identity: 0u u

4. Additive inverse: 0u u

5. Distributive property 1: a u v au av 6. Distributive property 2: a b u au bu 7. Multiplication by zero scalar: 0 0u =

8. Multiplication by zero vector: 0 0a

9. Multiplicative identity: 1u u 10. Associative property of scalar multiplication: a bu ab u

9

Example: (#44 page 691) In still air, a parachute with a payload would fall vertically at a terminal speed of 40 m/s. Find the direction and magnitude of its terminal velocity relative to the ground if it falls in a steady wind blowing horizontally from the west to the east at 10 m/s.

10

Example: (#49, page 691) Suppose that you pull a suitcase with a strap that makes a 60 angle with the horizontal . The magnitude of the force you exert is 40 lb.

a. Find the horizontal and vertical components of the force. b. Is the horizontal component of the force greater if the angle of the strap is 45

instead of 60? c. Is the vertical component of the force greater if the angle of the strap is 45 instead

of 60?

11

Section 11.2 Vectors in Three Dimensions The xyz-Coordinate System Three-dimensional rectangular coordinate system. 3 Right handed coordinate system

o If fingers of right hand are curled to palm, positive x-axis to positive y-axis, the thumb points in direction of positive z-axis.

Plane containing the x-axis and y-axis is called the xy-plane ( 0z ). The plane containing the x-axis and z-axis, is the xz-plane ( 0y ). The plane containing the y-axis and z-axis is the yz-plane ( 0x ).

xyz-axis creates 8 octants.

12

Equations of Simple Planes 0x (yz-plane), 0y (xz-plane), 0z (xy-plane) What about 2x , 3x , 15y , or 2z ?

13

Distance Formula in xyz-space The distance between the points 1 1 1, ,P x y z and 2 2 2, ,Q x y z is

2 2 22 1 2 1 2 1x x y y z z .

14

Spheres and Balls A sphere centered at , ,a b c with radius r is the set of points satisfying the equation

2 2 2 2x a y b z c r A ball centered at , ,a b c with radius r is the set of points satisfying the equation

2 2 2 2x a y b z c r Example: Give a geometric description of 2 2 2 6 6 8 2 0x y z x y z

15

Vectors in 3 Definition: If vis a three-dimensional vector in the plane equal to the vector with initial point at the origin and terminal point 1 2 3, ,v v v then the component form of v is

1 2 3, ,v v v v v To determine the component form of a vector that has initial point 1 1 1, ,A x y z and terminal point 2 2 2, ,B x y z then 2 1 2 1 2 1, ,AB v x x y y z z

Example: Suppose 5,1,3A , 7,9, 1B and 1, 15,11C then find AB

, AC

16

The magnitude or length of v , 2 2 21 2 3v v v v . Recall in two-dimensions

2 21 2v v v

Example: Find the magnitude of AB

from the previous example.

17

Definition: Vector Operations in 3 Let 1 2 3, ,u u u u and 1 2 3, ,v v v v be vectors and c be a scalar then

Addition: 1 1 2 2 3 3, ,u v u v u v u v + Subtraction: 1 1 2 2 3 3, ,u v u v u v u v Scalar Multiplication: 1 2 3, ,cu cu cu cu

Example: Suppose 1,0,2u and 2,1, 4v , find

1. u v 2. 2 3u v

18

Example: Are the points P, Q, and R collinear (lie on a line)? 1. 1,6, 5 , 2,5, 3 , 4,3,1P Q R 2. 1,2,3 , 2, 3,6 , 3, 1,9P Q R

19

Unit Vectors A vector of length one is called a unit vector. Standard basis vectors:

o 1,0,0i o 0,1,0j o 0,0,1k

Any vector can be written as a linear combination of the standard basis/unit vectors. o 1 2 3 1 2 3 1 2 3, , 1,0,0 0,1,0 0,0,1v v v v v v v v v v i j k

If v 0 , then

o vv

is a unit vector in the direction of v

o The equation cvv

is a vector of length c in the direction of v .

20

Example: Find a vector of magnitude 7 in the direction of 12 5 v i k . (If we wanted the opposite direction?)

21

11.3 Dot Product Definition: Given two nonzero vectors u and v in two or three dimensions, their dot product is

cosu v u v

Where is the angle between u and v with 0 . If 0u or 0v

, then 0u v and is undefined. Note: If and v u are parallel in the same direction, then 0 and v tu where 0t .

cos u vu v

If and v u are parallel in opposite directions, then and v tu where 0t .

If and v u are orthogonal (perpendicular) if 2 and 0u v .

22

Definition: Orthogonal Vectors Two vectors u and v are orthogonal if and only if 0u v . The zero vector is orthogonal to all vectors. In two or three dimensions, two nonzero vectors orthogonal vectors are perpendicular to each other.

23

Theorem 11.1: Dot Product If 1 2 3, ,v v v v and 1 2 3, ,u u u u then the dot product of and v u is

1 1 2 2 3 3u v u v u v u v Example: 1,7,4 6,2, 1/ 2

24

Theorem 11.2 Properties of Dot Products Let , , u v w be vectors and c be a scalar. 1. Commutative property: v u u v 2. Associative property: cv u c v u v cu 3. Distributive property? u v w u v u w 0 0v

25

Projections: How much of u points in the direction of v ?

26

The scalar component of u onto v Magnitude of the vector projection, cosu

scalvu vu

v

o Will be positive if 02

o Will be negative if 2

Definition: Projection of u onto v

2proj scalv vv u v v u vu u vv v v v

27

Example: Find the scalar projection of u onto v and the vector projection of u onto v when

1. 1,6, 2u and 2, 3,1v

28

2. 2u i j k and 2 3v i j k

29

Work done by a constant force F moving an object through distance d is W F d . This applies only when the force is directed along the line of motion of the object. However, if the force points in some other directions, PRF

, while the moving the objection from

P to Q the displacement vector is PQD

Then work is defined to be the product of the component of force along D and distance moved.

cosW F D F D Example: A constant force 2,4,1F

moves an object from 0,0,0 to 2,4,6 , find the

work done.

30

Section 11.4 Cross Product Cross product, u v Second of the two vector multiplication methods. If two vectors, u and v , are in a plane, then the cross product is a vector normal

(perpendicular) to the plane containing both u and v . Right hand rule, if fingers curl from u to v (angle between two vectors is less than

180°) then thumb points in the direction of u v× Definition: Cross Product Given two nonzero vectors u and v in 3 , the cross product u v is a vector with magnitude

sinu v u v where 0 is the angle between u and v .

31

Evaluating the Cross Product: If 1 2 3, ,v v v v and 1 2 3, ,v u u u then the cross product of u and v is

2 3 3 2 3 1 1 3 1 2 2 1, ,u v u v u v u v u v u v u v ×

Calculating Cross Products Using Determinants

If 1 2 3u u u u i j k and 1 2 3v v v v i j k then 1 2 3

1 2 3

u v u u uv v v

i j k

×

32

Example: Suppose 5,1,4u and 1,0,2v , find u v× .

33

Theorem 11.3: Geometry of the Cross Product Let u and v be two nonzero vectors in 3 . 1. The vectors u and v are parallel ( 0 or ) if and only if 0u v

. 2. If u and v are two sides of a parallelogram, then the area of the parallelogram is

sinu v u v × Note: Half the area of the parallelogram is the area of a triangle (formed by u , v , and u v- ).

34

Example: 1. Find the area of the triangle determined by the points 1,1,1P , 2,1,3Q , and

3, 1,1R .

2. Find a unit vector orthogonal to plane PQR .

35

Theorem: The vector u v is orthogonal to both u and v . Proof: To show that u v and u are orthogonal, we need to show that their dot product is zero.

2 3 1 3 1 21 2 3

2 3 1 3 1 2

1 2 3 3 2 2 3 1 1 3 3 1 2 2 1

1 2 3 1 3 2 2 3 1 2 1 3 3 1 2 3 2 1

0

u u u u u uu v u u u u

v v v v v v

u u v u v u u v u v u u v u vu u v u u v u u v u u v u u v u u v

×

Similarly, 0u v v × . Therefore, u v is orthogonal to both uand v .

36

Theorem 11.4 Properties of the Cross Product Let , , u v w be vectors in 3 and let a and b be scalars. 1. Anticommutative property: v u u v × 2. Associative property: au bv ab u v 3. Distributive property: u v w u v u w 4. Distributive property: u v w u w v w + × ×

37

Theorem: Triple Scalar or Box Product and Volume The volume, V of the parallelepiped determined by the vectors, u , v , and w is the absolute value of the scalar triple product.

V u v w × Note in the theorem the vectors v and w form the base plane of the parallelepiped. If we let u and v form the base or u and w form the base, we see that

u v w v u w w u v × × ×

Calculating the Triple Scalar Product

1 2 3

1 2 3

1 2 3

u u uu v w v v v

w w w

Example: Find the volume of the parallelepiped determined by u i j k ,

2 2 v i j k , and 2 w i j k .

38

39

Section 11.5 Lines and Curves in Space

Vector-Values Functions A function of the form , ,r t x t y t z t may be view two ways: It is a set of three parametric equations that describe a curve in space. A vector-valued function which means the three dependent variables are the

components of r and each component varies with respect to a single independent variable.

In the xy-plane, a line is determined by a point on the line and the direction, slope (or angle of inclination). In 3-dimensional space, a line L is determined by a point on the line, 0 0 0 0, ,P x y z and the direction of L, a vector , ,v a b c .

40

Vector Equation for a Line L through 0 0 0 0, ,P x y z parallel to v is 0r t r tv for t

Where r is the position vector of a point , ,P x y z and 0r is the position vector of

0 0 0 0, ,P x y z . Each value of the parameter t gives the position vector r of a point on the line L. So as t varies, the line is traced out by the tip of the vector r . Note, 0 0 0, ,r t x at y bt z ct where t , so the Parametric Equations for a Line, L through 0 0 0 0, ,P x y z parallel to v a b c i j k is

0 0 0, , x x at y y bt z z ct for t

41

Example: Determine the line through the point 1,0, 3 and parallel to the vector, 2 4 5v i j k

42

Recall that parametric equations for a curve in the plane are not unique; this is true in space as well. For instance the point 3, 4,2 is also a point on the line in the example above so another parametric representation might be

3 2 , 4 4 , 2 5x t y t z t Or suppose 10, 10,25v (this is 5 times the original v ) so we could have

1 10 , 20 , 3 25x t y t z t

43

Example: Find parametric equation of the line that passes through the points, 6,1, 3A and 2,4,5B .

44

To determine a line segment from 0r to 1r

the vector equation is given by

1

0 1 0

1 for 0 1or t t r tr t

r t r r

If we let 1 0v r r then we will have the line segment from A to B.

45

Example: Determine whether the lines 1L and 2L are parallel, skew, or intersecting. If they intersect find the point of intersection.

1

2:

: 1 2 , 3 , 2

1 4 1 3L x t y t z tL x s y s z s

46

E

Example: FFind the di

Dist

istance from

tance fromThrough

d

m the poin

m a Point Sh P paralle

PS vv

nt 2,1, 1

S to a Lineel to v

to the line

e

2 , x t y 1 2 , zt

47

2t .

48

49

Space Curves Suppose , , and f t g t h t are continuous functions on an interval I, then the set of points defined by the parametric equations , ,x f t y g t z h t as t varies throughout I is a space curve.

50

Vector Functions Let tr denote a vector function whose domain is and range is a set of vectors.

, ,r t f t g t h t f t g t h t i j k Where t and , ,f t g t h t are the component functions ofr . The domain of r t is the set of values for which r is defined (i.e. set of values for

which , , and f t g t h t are all defined). When a particle travels along a space curve, then tr is the position vector of a

particle at time, t. The components of r are scalar functions of t.

Example: Find the domain of 22 ,sin , ln 92

tr t t tt

.

51

52

Graphing vector-valued functions with Maple: With (plots): Spacecurve

, , , .. ,axes=normal,scaling=constrained,numpoints _ ;f t g t h t t a b

53

Definition: Let r t f t g t h t i j k be a vector function and L a vector. We say that r has limit L at t approaches a and we write

limt a

r t

L

So lim lim ,lim ,limt a t a t a t a

r t f t g t h t

provided the limits of the components exist.

54

Example: 1.

0lim cos ,sin , lnt

t t t t

=

2. 2 lnlim arctan , ,t

t

tt et

=

55

Definition: A vector function is continuous at a point t a in its domain, if lim

t ar t r a

. The

function is continuous, if it is continuous at every point in its domain. In other words, r is continuous at a if and only if f, g, and h are continuous at a.

56

Section 11.6 Calculus of Vector-Values Functions Definition: The vector function r t f t g t h t i j k has a derivative (is differentiable) at t if f, g, and h have derivatives at t. The derivative of the vector function

0

' limt

t t td df dg dhtdt t dt dt dt

r rrr i j k

The vector 'r t points in the directions of the curve at P. For this reason, ' 0r t

is a tangent vector at P (provided it is not the zero vector). The vector 'r t is the derivative of r with respect to t; it gives the rate of change

of the function r t . r is differentiable if it is differentiable at every point of its domain. r is a smooth curve if /dr dt is continuous and never 0 , i.e. f, g, and h have

continuous first derivatives that are not simultaneously zero. o For a smooth curve, /dr dt is never zero; the particle does not stop or reverse

direction. A curve that is made up of a finite number of smooth curves pieced together in a

continuous fashion is called a piecewise smooth curve.

57

Definition: Unit Tangent Vector Let r t f t i g t j h t k

be a smooth parameterized curve for a t b . The unit tangent vector for as particular value of t is

''

r tT t

r t

Orientation of Curves: A parameterized curve described by r t has a natural

direction or orientation. o The positive direction is the direction in which the curve is generated as t

increases from t a to t b .

58

Example:

Let 4sec tan3

r t t i t j tk is the position of a particle in space at time,

6t .

Find the unit tangent vector.

59

Example: Find the parametric equations for the line that is tangent to the curve

cos sin sin 2r t t i t j t k at

2t .

60

Theorem 11.7 Derivative Rules Let u and vbe differentiable vector-valued functions and let f any differentiable scalar-valued function, all at t . Let c be a constant vector. The following rules apply:

1. Constant Function Rule: 0d cdt

2. Sum/Difference Rule: d d du t v t u t v tdt dt dt

3. Scalar Product Rule: d d df t u t f t u t f t u tdt dt dt

4. Dot Product Rule: d d du t v t u t v t u t v tdt dt dt

5. Cross Product Rule: d d du t v t u t v t u t v tdt dt dt

6. Chain Rule: ' 'd u f t f t u f tdt

61

Example: Let 4 4,2 1,2t t tr t e e e , compute ' , " , '"r t r t r t .

62

Definition: Indefinite Integral of a Vector-Valued Function Let r f i g j hk

be a vector function and let R F i G j H k

, where , ,F G Hare antiderivatives of , ,f g h , respectively. The indefinite integral of r is

r t dt R t C

Where C

is an arbitrary constant vector. Example: Evaluate 2cos ,2sin 3 ,4cos8t t t dt

63

Example: Find r if 2

2 2

2' , ,1 4

tt tr t tet t

and 30 1, , 32

r .

64

65

Definition: Definite Integral of a Vector-Valued Function Let r t f t i g t j h t k

, where , ,f g and hare integrable on the interval ,a b

b b b b

a a a a

r t dt f t dt i g t dt j h t dt k

Example: Evaluate /3

0

sec tan tan 2cos sint t t t t dt

i j k

66

67

Section 11.7 Motion in Space Definition: Position, Velocity, Speed and Acceleration Let the position of an object moving in three-dimensional space be given by , ,r t x t y t z t , for 0t . The velocity of the object is

' ' , ' , 'v t r t x t y t z t The speed of the object is the scalar function

2 2 2' ' 'v t x t y t z t The acceleration of the object is ' "a t v t r t .

68

Example:

Let 4sec tan3

t t t t r i j k is the position of a particle in space at time, 6

t . Find

the particles 1. Velocity vector

2. Acceleration vector

3. Speed at time t.

69

Theorem 11.8 Motion with Constant r Let r describe a path on which r is constant (motion on a circle in a plane or motion on a sphere in three dimensions). Then 0r v , which means the position vector and the velocity vector are orthogonal at all times for which the function is defined.

Suppose r has constant length then

2

0

0

2 0

t t cd t tdt

d dt t t tdt dt

dt tdt

r r

r r

r r r r

r r

Thus if r is a differentiable vector of constant length, then 0ddt

rr

70

Example: A particle traveling in a straight line is located at the point 1, 1,2 and has speed 2 at 0t . The particle moves toward the point 3,0,3 with constant acceleration 2 i j k . Find its position vector r t at time t.

71

72

Section 11.8 Length of Curves Definition: Arc Length of Vector Functions Consider the parameterized curve r t x t i y t j z t k

, where 'f , 'g , and 'h are continuous, and the curve is traversed once for a t b . The arc length of the curve between , ,f a g a h a and , ,f b g b h b is

2 2 2

'

bb

aa

dx dy dzL dt r t dtdt dt dt

Example: Find the length of the curve 6sin 2 6cos2 5r t t t t i j k for 0 t .

73

74

75

Example: Find the length of the curve sin , cos ,t t tr t e t e t e for 0 ln 2t .

76

77

Arc Length of a Polar Curve Let f have a continuous derivative on the interval , . The arc length of the polar curve r f for is

2 2'L f f d

Example: Find the length of the polar curve 24r for 0 6 .

78

Section 11.9 Curvature and Normal Vectors Theorem 11.9 Arc Length as a Function of a Parameter Let r t describe a smooth curve for t a . The arc length is given by

0

t

s t v u du

Where 'v r . Equivalently, 0ds v tdt

. If 1v t for all t a the parameter t is

arc length.

79

Speed on a Smooth Curve Since ' , ' , and 'x u y u z u are continuous (the curve is smooth), then the Fundamental Theorem of Calculus tells us that s is a differentiable function of t with derivative

ds v tdt

So the rate at which the particle covers distance along its path is independent of how far away it is from its base point.

Note, 0dsdt

since by definition 0v for a smooth curve. Thus s is an increasing

function of t.

80

Example: Find the arc length parameter along the curve cos sin sin cosr t t t t t t t i j from the point where 0t .

81

82

Definition: Curvature

Let r describe a smooth parameterized curve. If s denotes arc length and ''

rTr

is the

unit tangent vector, the curvature is dTsds

.

83

Theorem 11.10 Formula for Curvature Let r t describe a smooth parameterized curve, where t is any parameter. If 'v r is the velocity and T

is the unit tangent vector, then the curvature is

'1'

TdTtv dt r

Theorem 11.11 Alternative Curvature Formula Let r be the position of an object moving on a smooth curve. The curvature at a point on the curve is

3

a vt

v

Example: What is the curvature of cos , sinr a t a t ?

84

85

Definition Principle Unit Normal Vector Let r describe a smooth parameterized curve. The principal unit normal vector at a point P on the curve at which 0 is

/ 1/

dT ds dTNdsdT ds

.

In practice, we use the equivalent formula //

dT dtNdT dt

Evaluated at the value of t corresponding to P . Example: Find the principal unit normal vector for 3sin ,5cos ,4sinr t t t t

86

87

Theorem 11.12 Properties of the Principal Unit Normal Vector Let r describe a smooth parameterized curve with unit tangent vector T

and principal

unit normal vector N

. 1. T

and N

are orthogonal at all points of the curve; that is, 0T t N t

at all points where N

is defined.

2. The principal unit normal vector points to the inside of the curve—in the direction that the curve is turning

88

Theorem 11.13 Tangential and Normal Components of the Acceleration The acceleration vector of an object moving in space along a smooth curve has the following representation in terms of its tangential component Ta (in the direction of T

)

and its normal component Na (in the direction of N

) T Na a T a N

Where 2N

a va v

v

and 2

2Td sadt

.

Example: Find the tangential and normal components of the acceleration vector fi 3 23 3r t t t t i j

89

90

Summary Formulas for Curves in Space Position function: , ,r t x t y t z t Velocity: 'v r Acceleration: 'a v

Unit Tangent Vector: vTv

Principal Unit Normal Vector: //

dT dtNdT dt

(provided / 0dT dt

)

Curvature: 31 a vdT dT

ds v dt v

Components of Acceleration: T Na a T a N where 2

N

a va v

v

and 2

2Td sadt

.

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