CHAPTER 2 Vector Valued Function

8/11/2019 CHAPTER 2 Vector Valued Function

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VECTOR-VALUED FUNCTION

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ectorposition v

aasexpressedbecanD-3inequationcurveorlineA

kjir zyx

:equationparametricin the

)(tfx )(tgy )(thz

kjir )()()((t) thtgtf

)(),(),((t) thtgtfr


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DOMAINExample 1

Determine the domain of the fo11owing function

cos ,1n 4 , 1

So1ution:

The first component is defined for a11 's.

The second component is on1y defined for 4.

The third component is on1y defined for

t t t t

t

t

r

1.

Putting a11 of these together gives the fo11owing domain.

1, 4

This is the 1argest possib1e interva1 for which a11 three

components are defined.

t


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kjir )4(3)(ofgraphSketch the(b)

line.the

sketchThen(2,3,-1)?and(1,2,2)pointsthepasses

thatlinestraightaofequationlinetheis What(a)

2ttt

Example 2


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kjir )23()2()1()(

232)21(

22)23(11)12(Hence,

Then.(2,3,-1))z,(,1whenand

)2,2,1(),(,0nthat wheSuppose(a)

111

0,00

tttt

ttz

ttyttx

,yxt

zyxt

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Solution :

001 )( PtPPP


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Thus, the line:


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3plane

on the4parabolatheisgraphtheis,chwhi

4z,3

thatfindweThus,

4,3,

arecurvetheofequationsParametric(b)

2

2

2

y

xz

xy

tzytx

Solution :


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functionfollowingtheofeachofgraphSketch the

1,)( tt r(a) (b) ttt sin3,cos6)( r

Solution :

The first thing that we need to do is plug in a few values

of tand get some position vectors. Here are a few,

(a)

Example 3


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The sketch of the curve is given as follows (red line).


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Solution :

(b)

As in Question (a), we plug in some values of t.


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The sketch of the curve is given as follows (red line).


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functionfollowingtheofeachofgraphSketch the

kjir cttatat sincos)(

Example 4

CIRCULAR HELIX


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functionsscalaraisResult)()())((

functionsvectorareResults

)()())(()()())((

)()())((

)()())((

Then

.offunctionscalaraisandoffucntionsareGandFSuppose

theorem.followingthehaveweThus

vectors.ofpropertiesloperationaeinherit thfunctionsVector

ttt

tttttt

ttt

ttt

tt

GFGF

GFGFFF

GFGF

GFGF

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THEOREM 2.1


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kji

kjikji

GFGF

GFGF

FGF

kjiG

kjiFGF

)sin5()1

()(

51)sin(

)()())(((i)

))(((iv)))(((iii)

))(((ii)))(((i)

find,51

)(and

sin)(bydefinedandfunctionsvectorFor the

2

2

2

tt

ttt

ttttt

ttt

tt

tet

ttt

tttt

t

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Example 4

Solution :


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)()sin5()sin

5(

51

sin

)51

()sin(

)()())(((iii)

)sin()()(

)())(()ii(

22

2

2

2

kji

kji

kjikji

GFGF

kji

FF

ttttt

t

tt

tt

ttt

ttttt

ttt

teteet

tete

ttt

tt

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Solution :


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Solution :

tt

ttttt

ttt

sin51

)51

()sin(

)()())(((iv)

3

2

kjikji

GFGF


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Example 5

246)(1226)(

4)52(4)(3)(

if),(and),(Find

2

32

kiFkjiF

kjiF

FF

ttttt

tt-tt

tt

Solution :


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GFG

FGF

GFG

FGF

GFGF

FF

GF

dt

d

dt

d

dt

ddt

d

dt

d

dt

d

dt

d

dt

d

dt

ddt

dcc

dt

d

)()(iv

)(iii)(

)()ii(

)((i)

then,scalaraisc

andfunctionvectorabledifferentiareandIf:4.3Theorem

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THEOREM 2.2


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tttt

tttt

ttttt

dt

d

dt

d

dt

d

sin11cos)1(5

)cossin()310(

)sin(cos)5(

)()i(

2

2

32

jikji

jikji

GFG

FGF

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Example 6

)((iii)),((ii)),((i)

find,cossin)(,5)(If

32

FFGFGF

jiGkjiF

dt

d

dt

d

dt

dttttttt

Solution :


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53

2

2323

232

62100

2)()iii(

)cos11sinsin(5

t)sin3cos(-t)cos3sin(

0cossin

3110

0sincos

5

)()ii(

ttt

dt

d

dt

d

ttttt

tttttt

tt

tt

tt

ttt

dt

d

dt

d

dt

d

FFFF

k

ji

kjikji

GFG

FGF

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Solution :


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INTEGR TION OF VECTOR

FUNCTIONS

))(())(())(()(

thenb],[a,onofand,,

functionsintegrablesomefrom)()()()(If

ise.componentwdonealsoisfunctionsvectorofnIntegratio

b

akjiF

kjiF

b

a

b

a

b

adtthdttgdttfdtt

thgf

thtgtft


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Example 7

Solution :

kji

kji

kjiF

kjiF

F

802-42

])5()2[(

4)52()43()(

4t)52()43()(

if)(Find

3

1

4223

3

1

3

3

1

3

1

3

1

2

32

3

1

ttttt

dttdttdtttdtt

tttt

dtt


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b

a dt

dydtdxL

22

b

a dt

dz

dt

dy

dt

dxL

222

In 2-space

In 3-space

In general,

b

a dt

dL

r


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Notes: Smooth Curve

The graph of the vector function defined byr(t) is smooth on any interval of t where is

continuous and .

The graph is piecewise smooth on an interval

that can be subdivided into a finite number of

subintervals on which ris smooth.

r

0t r


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Find the arc length of the parametric curve

4

30;2,sin,cos)( 33

tztytxa

10;2,,)( ttzeyexb tt

Find the arc length of the graph of r(t)

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1)()( 23 ttttta kjir

20;2sin3cos3)()( tttttb kjir

4

3: LAns

1: eeLAns

58: LAns

5

132:

LAns

Example 8

Example 9


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If r(t) is a vector function that defines a smooth graph,then at each point a unit tangent vectoris

tt

t

rT

r

UNIT TANGENT VECTOR

3a) Find the derivative of 1 sin 2b) Find the unit tangent vector at the point where 0.

tt t te t

t

r i j k

Example 10


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curve.thetont vectorunit tangetheiswhere

asbydenoted),(curvethevector to

normalunitprinciplethedefinewe,0If

T

Nr

T

t

dtd

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)('

)('

T

T

t

t

dtd

dtd

T

T

N

UNIT NORMAL VECTOR


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).(curvethetoly,respectivevector

unitprincipaltheandnt vectorunit tangetheareandwhere

asdefinediscurveaofvectorbinormalThe

tr

NT

B

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BINORMAL VECTOR

NTB


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Find the unit normal and binormal

vectors for the circular helixcos sint t t t r i j k

Example 11


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curve.thetont vectorunit tangetheiswhere

asdefinedis)(curvesmoothaofcurvatureThe

T

rt

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)('

)('

t

t

dtd

dtd

r

T

r

T

CURVATURE

3

r r

r


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Curvature is the measure of how sharply a curve r(t) in

2-space or 3-space bends.


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Find the curvature of the helix traced out by

2sin ,2cos ,4t t t t r

Example 12


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Radius of Curvature

asdefinediscurvatureofradiusitsthen

),(curvesmooththeofcurvaturetheisIf

tr

1


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thentime,iswherer(t),ectorposition v

bygivencurvethealongmovesparticleaIf

t

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dt

dt

rv )(velocity

2

2

)(onacceleratidt

d

dt

dt

rva

dt

dst )(speed v


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Example 13

.2when

particletheofonacceleratiandspeedvelocity,theFind

sincos)(

bygivenisafter timeparticleaofectorposition vThe

3

t

tttt

t

jir

Solution :

kji

kjiv

kji

r

v

1242.09.0

)2(3)2(cos)2(sin

2when

)3()(cos)(sin

velocityobtain thewe,w.r.tatingDifferenti

2

2

t

tttdt

d

t


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kji

kjia

kjiv

a

a

v

129.00.42

)2(6)2sin()2cos(,2when

6)sin()cos(

bygivenisonacceleratiThe

04.12)2(91,2when

91)3()(cos)sin(

bygivenisany timeforspeedThe

4

42222

t

tttdt

d

t

tttt-

t

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Example 14

kjir

r

kjiv

v

2)0(particle

theof)(ectorposition vtheFind

2cos)(

bygivenismotioninparticleaofVelocity

2

t

ttet t


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C

Ce

cc

Ct

te

ct

ctce

tdtdttdtet

dtd

t

t

t

i

kjir

kji

kji

kji

kjir

rv

2

)0(2sin)0(

3

1)0(

cCwhere

2

2sin

3

1

)

2

2sin()

3

1()(

2cos)(

havewe,Since

30

321

3

32

3

1

2

Solution :


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kji

kjikjir

kjikjii

r

)12

2sin()131()1(

)2

2sin(

3

1)(

obtainweHence

C2

obtainwe),0(ofegiven valutheusingBy

3

3

tte

ttet

C

t

t

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Find the position vector R(t), given the

velocity V(t) and the initial position R(0) for

2 2 ; 0 4tt t e t V i j k R i j k

Example 15


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CHAPTER 2 Vector Valued Function

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