vector calculus: are you ready? vectors in 2d and 3d...

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1

Vector Calculus: Are you ready?

Purpose: Make certain that you can define, and use in context, vector terms,

concepts and formulas listed below:

Section 7.1-7.2

find the vector defined by two points and determine the norm of the vector.add two vectors multiply a non-zero vector by a non-zero scalar.represent a non-zero vector in the xy-plane in terms of its magnitude and the angle it makes with the positive x-axis.

Vectors in 2D and 3D Space: Review

2

Vectors in 2D space

There are many quantities that are vector functions: Some Daily Use of Vectors

A wind of 80 km/h from the Southeast.A car going 80 km/h East.A vertical velocity of 20 m/s.A plane traveling 1000 km/h on a 180 heading.

These issues are described by a magnitude and a direction.

Vector algebra and vector calculus have resulted from practical engineering applications: Mechanics, Fluid flows, Wireless CommunicationsScalar: is described by a single quantity such as work, energy, potential, speed, temperature, blood pressure ..Vector: is described by a magnitude and direction such as velocity, electric force, position of a robot …

3

Some Applications of VectorsMechanics: Force, Torque, position, speed, acceleration, …Electromagnetism: Electric and magnetic fields, current density, pointing vector,…Example Walking and Different Forces

Example Mechanical System in Equilibrium

Other Examples of vector quantities

Notation ur,v⎯→⎯

AB

Acknowledgment: Most figures included in class notes are copied from the textbook by Zill and Cullen.

4

Notation and TerminologyA vector with starting point A and end point B is written asMagnitude of is written as:

⎯→⎯

AB

⎯→⎯

AB||||

⎯→⎯

AB

Two vectors with the same magnitude and direction are equal

Parallel vectors: nonzero scalar multiples of each other

Example: In 2D Cartesian Coord.:

22

21

212121

:Magnitude

],[ , ˆˆ

aaa

aaaajaiaa

+=

=><=+=r

r

5

A note about notation

The textbook uses boldface to represent vectors,I may place an arrow above general vectors and a hat over unit vectors. I would like you all to clearly identify vectors in your work.

Fr

F u u ii ˆ=

6

Addition of Vectors

The sum of two vectors is the main diagonal of the parallelogram with the vectors as sides

⎯→⎯⎯→⎯⎯→⎯

+= ACABAD

Consider two vectors and with common initial point A

⎯→⎯

AB⎯→⎯

AC

Example:

ijijijijijiji

100106664210)66()44(

=−=−++

−=−++

7

Subtraction

Subtraction: The difference of and is defined by

)(⎯→⎯⎯→⎯⎯→⎯⎯→⎯

−+=− ACABACAB

⎯→⎯

AB⎯→⎯

AC

⎯→⎯

AB⎯→⎯

− AC

⎯→⎯⎯→⎯

− ACAB is the main diagonal of the parallelogram with sides and

⎯→⎯⎯→⎯⎯→⎯

−= ACABCBOr is a vector from the end of the second vector toward the end of the first vector

8

At equilibrium: F1 + F2 + w = 0

Sphere weight=50 lb

2 supporting planes

w = -50 j lb

∴ F1 = 25.9 lb, F2 = 36.6 lb

Review Exercise (page 346): Prob. 48Find the magnitude of F1 and F2.

9

Properties of VectorsMagnitude, length, or norm of a vector a: ||a||

If then:>=< 21 aa ,a22

21 aa +=|||| a

1||||1ˆ =⎟

⎠⎞⎜

⎝⎛= u with aau

ji 212121 ,00, , aaaaaa +=><+><=><The i, j unit vectors: i=<1,0>, j=<0,1>

Example: Given a=<3,-4>, form a unit vector • in the same direction as a. Answer: <0.6,-0.8> • In the opposite direction of a. Answer: <-0.6,0.8>

A vector that has magnitude 1 is called unit vector.A unit vector in the direction of a is:

ij u

10

7.2 Vectors in 3-SpaceRectangular or Cartesian Coordinate2D-Space: Two orthogonal axes

The three axes follow the Right Hand Rule

: Three mutually orthogonal axesSpace-3D

11

Coordinate Plane: Each pair of coordinate axes determines a coordinate plane (xy,xz and yz).Octant:The coordinate planes divide the 3-space into 8 parts known as Octants.First octant: x, y, z>0

3D-Space

12

P 1P 2

Given 2 points:

>=<= 1111 ,, zyxPOrrr

Position Vector:For a point P, the position vector is

>−−−=<−=

121212

1221

,, zzyyxxPOPOPPrrr

Vector between two points:

),,(P & ),,(P 22221111 zyxzyx

212

212

212 zzyyxxd )()()()P,P( 21 −+−+−=

Distance Formula between two points:

Examples: P1 = (1,2,3) & P2 = (1,-1,-1)

13

Component Definitions in 3D-SpaceLet and be vectors in

(i) Addition:

(ii) Scalar Multiplication:

(iii) Equality: if an only if

(iv) Negative of a vector:

(v) Subtraction:

(vi) Zero vector: 0 = <0,0,0>

(vii) Magnitude:

>=< 321 aaa ,,a >=< 321 bbb ,,b 3R>+++=<+ 332211 bababa ,,ba

>=< 321 kakakak ,,aba= 332211 bababa === ,,

>−−−=<− 321 bbb ,,b>−−−=<−+=− 332211 bababa ,,)( baba

23

22

21 aaa ++=|||| a

kjia 321321 aaaaaa ++>==< ,,

14

Unit Vectors in 3D space:

i = <1,0,0>,

j = <0,1,0>,

k = <0,0,1>

kji a 321321 ,, aaaaaa ++=><=

Example: If a = 3i - 4j + 8k and b = I - 4k, find 5a - 2b.

b = i- 0j - 4k 2b = 2i + 0j - 8k

5a = 15i - 20j + 40k

5a - 2b = 13i - 20j + 48k

15

2. Section 7.3define the dot (inner) product (a . b) and interpret it geometrically.use the dot product to determine: – work done by a force, – the angle between two vectors, – whether two vectors are perpendicular to one

another, – projections and components of vectors, – direction angles and direction cosines

7.3 Dot (scalar or inner) Product

16

Applications: Mechanics and Electromagnetism

πθθ ≤≤⋅=⋅ 0 & cos|||||||| baba

Definition:

The dot product of two vectors a and b

is the scalar

θ is the angle between a and b

Example Dot producti . i=1, j . j=1, k . k=1 since ||i||=||j||=||k||=1 and θ = 0i . j=0, j . k=0, k . i=0 since θ = 90o

Example Given:a=10i+2j-6k, b=-0.5i+4j-3k a . b=(10)(-0.5)+(2)(4)+(-6)(-3)=21

Dot (scalar or inner) Product

17

Physical Interpretation of the Dot ProductA constant force of magnitude F moves an object a distance d in the direction force, the work done by the force (W):

θθcos||||||||

||||)cos||(||d F

dF ==

⋅= dFWrr

|||||||| d F=⋅= dFWrr

When a constant force F applied to a body acts at an angle θ to the direction of motion, the work done by the force (W):

Note: if F and d are orthogonal, W=0.

Examples:

18

Work done by w (gravity force) = w . d = 0 (w ¦d)

Work done by F (applied force) = F . d = |F|.|d| cos θ (d // F)= 150 N.m

md

NF

>=<

=

3,4

30||||r

P7.3-47: Given

weight

19

Properties of Dot Producta . b = 0 if a=0 or b=0 a . b = b . a (commutative law)a . (b+c) = a . b+a . c (distributive law)a . (kb) = (ka) . b = k(a . b) k a scalara . a ≥ 0a . a = ||a||2

For nonzero vectors a and b(i) a . b > 0 if and only if θ is acute(ii) a . b < 0 if and only if θ is obtuse, and(iii) a . b = 0 if and only of if cos θ =0 (Orthogonal vectors)

Theorem 7.1 Criterion for Orthogonal VectorsTwo nonzero vectors a and b are orthogonal if and only if a . b=0

20

Angle Between Two Vectors:

||||||||||||||||cos 332211

b ab ababababa ++

=•

=rr

θ

27214cos =θ o9.4477.0

942cos 1 ≈≈⎟⎟

⎞⎜⎜⎝

⎛=∴ −θ

ExampleFind the angle between a = 2i+3j+k & b = -i+5j+k.

27||||,14|||| == baa . b=14,

Solution:

21

For a nonzero vector in 3D-Space the angles α, β and γ with i, j, and k are called direction angles of a.

Direction cosines for

ooo 4.53,8.41,7.72 ≈≈≈⇒ γβα

kjia 321 aaa ++=

||||||ˆ||||||ˆ

cos 1

a aa

iia

=•

=r

α

||||cos 2

aa

=β||||

γcosa

3a=

1coscoscos 222 =++ γβα

Example Find the direction cosines and direction angles of the vector a = 2i+5j+4k. ||a||=6.71

Direction Angles

22

Component of a on b:

( ) vector

ˆ ˆ||||

)comp(proj

=

⋅=⎟⎟⎠

⎞⎜⎝

⎛= bba

bbaa bb

r

scalar

ˆ||||

cos||||comp

=

⋅=⋅

== bab

baaab θ

Projection of a in the direction of b:

Example: a = < -1,-2,7 > & b = < 6,-3,-2 >

23

3. Section 7.4

define the vector (cross) product (a x b) and interpret it geometrically.determine the cross product of vectors and combinations of vectors, use to determine torquefind unit vectors that are perpendicular to two given vectors.

7.4 Cross (Vector) Product

24

Cross (Vector) ProductThe vector product of 2 vectors A and B is given by

amparalleogr of area ||BA|| BA

B B BA A Ak j i

BA

zyx

zyx

∗∗=

−−−==×

n

BABAjBABAi xzzxyzzy

ˆ)sin(

)(ˆ)(ˆˆˆˆ

θ

)(ˆ xyyx BABAk −+

where n is a unit vector perpendicular to A and B, pointing in the direction given by the right hand screw rule (i.e. the direction in which a screw would advance if it were turned from A to B.

A

AxB

n

Example: a = < -1,-2,7 >

& b = < 6,-3,-2 >

25

Typical Applications

b) AREA OF A Triangle with edges a and b:

Area = 1/2⎪⎢a × b ⎪⎢= 1/2 ⎪⎢a⎪⎢⎪⎢b⎪⎢ sin θ

a) AREA OF A PARALLELOGRAM with edges a and b:

Area = ⎪⎢a × b ⎪⎢= ⎪⎢a⎪⎢⎪⎢b⎪⎢ sin θ

Example (p7.4, # 48): Find area of the triangle through:

p1 = (0,0,0), p2 = (0,1,2), P3 = (2,2,0)

26

Typical Applications

Volume of a parallelepiped (with edges: a, b & c)

Volume = (area of base) . (height)

= ⎪⎢b × c ⎪⎢ .⎮comp b × c a⎮

= ⎪⎢b × c ⎪⎢. ⎮a • (b × c) ⎮ / ⎪⎢b × c ⎪⎢

∴Volume =⎮a • (b × c)⎮

Example: a = < 3,1,1 >,

b = < 1,4,1 >

& c = < 1,1,5 >

27

Typical applicationsMOMENT OF A FORCE

In mechanics the moment m of a force F about a point Q is defined as the product

m =⎪⎢F ⎪⎢ d

d = ⎪⎢r ⎢⎢ sin θ

rFdrFm rrr×=∗=∴ ˆ sin θ

m is called the moment vector or vector moment of F about Q

where d is the (perpendicular) distance between Q and the line of action L on F.

Q d=r sin θ

If r is the vector from Q to any point Aon L, then

28

• Torque T = r x F

• Force on a moving charge due to a magnetic field due to

FB

vqF = q v x B

• Velocity of a rotating body

v = w x rω= angular speed

|w| = ω and directed along axis of rotation

w

r

v

ω

o

More Applications

29

Two non-zero vectors A and B are parallel if & if : 0BA =×rr

jkiijkkij

jikikjkjiˆˆˆ ˆˆˆ ˆˆˆ

ˆˆˆ ˆˆˆ ˆˆˆ

−=×−=×−=×

=×=×=×

• Circular Mnemonic:

• More Cross Product Properties:

Cross product is not commutative:

ABBA

ABBArrrr

rrrr

×−=×

×≠×

Cross product is not associative: C)BA()CB(Arrrrrr

××≠××Example (p7.4, # 13): A = <2,7,-4>, B = <1,1,-1>

Find a vector that is perpendicular to A and B

30

Purpose: Make certain that you can define, and use in context, vector

terms, concepts and formulas listed below:

4. Section 7.5express a line as a: vector parameterization, and scalar parameterization,use vectors to determine whether two lines intersect, and if so, the point of intersection.use vectors to find the distance from a point to a line.express a plane as a scalar equation and as a vector equation.find whether two planes intersect, and if so, the angle of intersection and a vector parameterization of the line formed by the intersection.unit normal for a plane.

7.5 Lines and Planes

31

Equation of a “straight” Line

>=<>=<

22222

11111

,,:P& ,,:P

zyxrzyxr

r

r

aatrrtrrtrr

rrrr

rrrr

of direction thein is line the,parameterscalar ),(

2

122

+==−+=

12

2

12

2

12

2

zzzz

yyyy

xxxx

−−

=−−

=−−

1. Vector equation of the line through r1 & r2:

2. Parametric & symmetric equations of the line:

Given two points in 3D:

Two forms for the line through P1 & P2:

If a is a unit vector, then its components are direction cosines of the line.

Examples to follow:

32

Examples: 7.5, # 3

parameter scalar =−+= trrtrr ),( 211rrrr

Find the vector equation of a line through: (1/2, -1/2, 1) & (-3/2, 5/2, -1/2).

Examples: 7.5, # 27

Show that the two lines:

r = t <1,1,1> and r = <6,6,6> + t <-3,-3,-3>

are the same.

33

7.5 Equation of a Plane

kcjbian ˆ ˆ ˆ ++=r

0 z y x =+++ dcba

1. The equation of a plane perpendicular to a normal vector

is given by:

2. The equation of a plane contains 3 points: P1(r1), P2(r2), P3(r3) is given by:

[ ] 0)()()( 11312 =−•−×− rrrrrr rrrrrr

Examples to follow:

Two forms:

a vector form.

0( =• n)r-r 1rrr

34

Examples: 7.5, # 39

0 z y x =+++ dcba

Examples: 7.5, # 51

Find the equation of a plane contains: (5,1,3) & perpendicular to <2,-3,4>

< r-r1 > . n = 0

Two methods:

Find the equation of a plane contains: (2,3,-5) & parallel to x + y - 4z = 1

Or:

Answer:

35

Intersection of Two PlanesLeta1 x + b1 y + c1 z = d1 &a2 x + b2 y + c2 z = d2

be two non parallel planes.

We get a system of two equations and three unknowns.

Choose one variable arbitrary, say x = t, and solve the new system of two equations and two unknowns y and z.

parametric equations for the line of intersection

36

Example Find the parametric equation for the line of intersection of

2x – 3y + 4z = 1 and x – y – z = 5

SolutionLet choose z = t, sub in the 2 equatinsand solve for x and y from

2x – 3y = 1 – 4t and x – y = 5 + t

Then, x = 14 + 7t, y = 9 + 6t, z = t

END of selected materials from Chapter 7.

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