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VECTOR-VALUED FUNCTION
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Prepared by :MISS RAHIMAH JUSOH @
AWANG
ectorposition v
aasexpressedbecanD-3inequationcurveorlineA
kjir zyx 
:equationparametricin the
)(tfx  )(tgy 
)(thz 
kjir )()()((t) thtgtf 
)(),(),((t) thtgtfr
DOMAIN
Example 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  

rahimahj@ump.edu.my
kjir )4(3)(ofgraphSketch the(b)
line.the
sketchThen(2,3,-1)?and(1,2,2)pointsthepasses
thatlinestraightaofequationlinetheisWhat(a)
2
ttt 
Example 2
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
tty
ttx
,yxt
zyxt
rahimahj@ump.edu.my
Solution :
001 )( PtPPP 
rahimahj@ump.edu.my
Thus, the line:
rahimahj@ump.edu.my
3plane
on the4parabolatheisgraphtheis,chwhi
4z,3
thatfindweThus,
4,3,
arecurvetheofequationsParametric(b)
2
2
2




y
xz
xy
tzytx
Solution :
rahimahj@ump.edu.my
rahimahj@ump.edu.my
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 t and get some position vectors. Here are a few,
(a)
Example 3
rahimahj@ump.edu.my
The sketch of the curve is given as follows (red line).
rahimahj@ump.edu.my
Solution :
(b)
As in Question (a), we plug in some values of t.
rahimahj@ump.edu.my
The sketch of the curve is given as follows (red line).
functionfollowingtheofeachofgraphSketch the
kjir cttatat  sincos)(
Example 4
CIRCULAR HELIX
Applied Calculus Chapter  2 vector valued function
functionsscalaraisResult)()())((
functionsvectorareResults
)()())((
)()())((
)()())((
)()())((
Then
.offunctionscalaraisandoffucntionsareGandFSuppose
theorem.followingthehaveweThus
vectors.ofpropertiesloperationaeinherit thfunctionsVector
ttt
ttt
ttt
ttt
ttt
tt
GFGF
GFGF
FF
GFGF
GFGF














rahimahj@ump.edu.my
THEOREM 2.1
kji
kjikji
GFGF
GFGF
FGF
kjiG
kjiFGF
)sin5()
1
()(
5
1
)sin(
)()())(((i)
))(((iv)))(((iii)
))(((ii)))(((i)
find,5
1
)(and
sin)(bydefinedandfunctionsvectorFor the
2
2
2
t
t
ttt
t
tttt
ttt
tt
tet
t
tt
tttt
t







rahimahj@ump.edu.my
Example 4
Solution :
)()sin5()
sin
5(
5
1
sin
)5
1
()sin(
)()())(((iii)
)sin()()(
)())(()ii(
22
2
2
2
kji
kji
kjikji
GFGF
kji
FF
ttttt
t
t
t
t
t
ttt
t
tttt
ttt
teteet
tete
ttt
tt






rahimahj@ump.edu.my
Solution :
rahimahj@ump.edu.my
Solution :
tt
t
tttt
ttt
sin51
)5
1
()sin(
)()())(((iv)
3
2



kjikji
GFGF
rahimahj@ump.edu.my
Example 5
246)(
1226)(
4)52(4)(3)(
if),(and),(Find
2
32
kiF
kjiF
kjiF
FF
tt
ttt
tt-tt
tt




Solution :
G
FG
FGF
G
FG
FGF
GF
GF
F
F
GF




dt
d
dt
d
dt
d
dt
d
dt
d
dt
d
dt
d
dt
d
dt
d
dt
d
cc
dt
d
)()(iv
)(iii)(
)()ii(
)((i)
then,scalaraisc
andfunctionvectorabledifferentiareandIf:4.3Theorem
rahimahj@ump.edu.my
THEOREM 2.2
tttt
tttt
ttttt
dt
d
dt
d
dt
d
sin11cos)1(5
)cossin()310(
)sin(cos)5(
)()i(
2
2
32




jikji
jikji
G
FG
FGF
rahimahj@ump.edu.my
Example 6
)((iii)),((ii)),((i)
find,cossin)(,5)(If 32
FFGFGF
jiGkjiF


dt
d
dt
d
dt
d
ttttttt
Solution :
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






F
FFF
k
ji
kjikji
G
FG
FGF
rahimahj@ump.edu.my
Solution :
rahimahj@ump.edu.my
INTEGRATION OF VECTOR
FUNCTIONS
))(())(())(()(
thenb],[a,onofand,,
functionsintegrablesomefrom)()()()(If
ise.componentwdonealsoisfunctionsvectorofnIntegratio
b
a
kjiF
kjiF
  

b
a
b
a
b
a
dtthdttgdttfdtt
thgf
thtgtft
rahimahj@ump.edu.my
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
rahimahj@ump.edu.my
 












b
a
dt
dy
dt
dx
L
22
 


















b
a
dt
dz
dt
dy
dt
dx
L
222
In 2-space
In 3-space
In general,
 
b
a
b
a
tL
dt
d
L )('or r
r
Notes: Smooth Curve
The graph of the vector function defined by
r(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 r is smooth.
r
  0t r
rahimahj@ump.edu.my
Find the arc length of the parametric curve
4
3
0;2,sin,cos)( 33 
 tztytxa
10;2,,)(  
ttzeyexb tt
Find the arc length of the graph of r(t)
42;6
2
1
)()( 23
 ttttta kjir
20;2sin3cos3)()(  tttttb kjir
4
3
: LAns
1
: 
 eeLAns
58: LAns
132: LAns
Example 8
Example 9
Applied Calculus Chapter  2 vector valued function
If r(t) is a vector function that defines a smooth graph,
then at each point a unit tangent vector is
 
 
 
t
t
t



r
T
r
UNIT TANGENT VECTOR
   3
a) Find the derivative of 1 sin 2
b) Find the unit tangent vector at the point where 0.
t
t t te t
t

   

r i j k
Example 10
curve.thetont vectorunit tangetheiswhere
asbydenoted),(curvethevector to
normalunitprinciplethedefinewe,0If
T
Nr
T
t
dtd 
rahimahj@ump.edu.my
)('
)('
T
T
t
t
dtd
dtd
T
T
N 
UNIT NORMAL VECTOR
).(curvethetoly,respectivevector
unitprincipaltheandnt vectorunit tangetheareandwhere
asdefinediscurveaofvectorbinormalThe
tr
NT
B
rahimahj@ump.edu.my
BINORMAL VECTOR
NTB 
 
Find the unit normal and binormal
vectors for the circular helix
cos sint t t t  r i j k
Example 11
curve.thetont vectorunit tangetheiswhere
asdefinedis)(curvesmoothaofcurvatureThe
T
r t
rahimahj@ump.edu.my
)('
)('
t
t
dtd
dtd
r
T
r
T

CURVATURE
3

 


r r
r
Curvature is the measure of how “sharply” a curve r(t) in
2-space or 3-space “bends”.
Find the curvature of the helix traced out by
  2sin ,2cos ,4t t t tr
Example 12
Radius of Curvature
asdefinediscurvatureofradiusitsthen
),(curvesmooththeofcurvaturetheisIf

 tr


1

thentime,iswhere),r(ectorposition v
bygivencurvethealongmovesparticleaIf
tt
rahimahj@ump.edu.my
dt
d
t
r
v  )(velocity
2
2
)(onaccelerati
dt
d
dt
d
t
rv
a 
dt
ds
t  )(speed v
rahimahj@ump.edu.my
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
ttt
dt
d
t
kji
kjia
kji
v
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
ttt
dt
d
t
tttt-
t



rahimahj@ump.edu.my
rahimahj@ump.edu.my
Example 14
kjir
r
kjiv
v


2)0(particle
theof)(ectorposition vtheFind
2cos)(
bygivenismotioninparticleaofVelocity
2
t
ttet t
rahimahj@ump.edu.my
     
C
Ce
cc
C
t
te
c
t
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 :
kji
kjikjir
kji
kjii
r
)1
2
2sin
()1
3
1
()1(
)
2
2sin
(
3
1
)(
obtainweHence
C
2
obtainwe),0(ofegiven valutheusingBy
3
3




t
te
t
tet
C
t
t
rahimahj@ump.edu.my
Find the position vector R(t), given the
velocity V(t) and the initial position R(0) for
   2 2
; 0 4t
t t e t     V i j k R i j k
Example 15
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