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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
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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8/11/2019 CHAPTER 2 Vector Valued Function
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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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8/11/2019 CHAPTER 2 Vector Valued Function
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functionsscalaraisResult)()())((
functionsvectorareResults
)()())(()()())((
)()())((
)()())((
Then
.offunctionscalaraisandoffucntionsareGandFSuppose
theorem.followingthehaveweThus
vectors.ofpropertiesloperationaeinherit thfunctionsVector
ttt
tttttt
ttt
ttt
tt
GFGF
GFGFFF
GFGF
GFGF
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
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
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
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
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
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)
42;62
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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8/11/2019 CHAPTER 2 Vector Valued Function
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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
)('
)('
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
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
)('
)('
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
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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All our dreams can come true, if we have the
courage to pursue them