chapter 2 vector valued function.pptx
TRANSCRIPT
CHAPTER 2VECTOR-VALUED FUNCTION
BUM2123 APPLIED CALCULUS
Prepared by :MISS RAHIMAH JUSOH @ AWANG
2.1 VECTOR FUNCTIONS SPACE CURVES
ectorposition v
a as expressed becan D-3in equation curveor lineA
kjir zyx
:equation parametric in the
)(tfx )(tgy )(thz
kjir )()()((t) thtgtf
)(),(),((t) thtgtfr
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
kjir )4(3)( ofgraph Sketch the (b)
line. the
sketch Then (2,3,-1)? and (1,2,2) points thepasses
that linestraight a ofequation line theis What (a)
2ttt
Example 2
kjir )23()2()1()(
232)21(
22)23(
11)12( Hence,
Then . (2,3,-1))z,(, 1 when and
)2,2,1(),( , 0n that wheSuppose (a)
111
0,00
tttt
ttz
tty
ttx
,yxt
zyxt
Solution :
001 )( PtPPP
3 plane
on the 4 parabola theisgraph theis,ch whi
4z ,3
thatfind weThus,
4,3,
are curve theof equations Parametric (b)
2
2
2
y
xz
xy
tzytx
Solution :
function following theofeach ofgraph Sketch 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
function following theofeach ofgraph Sketch the
kjir cttatat sincos)(
Example 4
CIRCULAR HELIX
functionsscalar a isResult )()())((
functions vector are Results
)()())((
)()())((
)()())((
)()())((
Then
. offunction scalar a is and of fucntions areG and F Suppose
theorem.following thehave weThus
vectors.of properties loperationa einherit th functionsVector
ttt
ttt
ttt
ttt
ttt
tt
GFGF
GFGF
FF
GFGF
GFGF
THEOREM 2.1
kji
kjikji
GFGF
GFGF
FGF
kjiG
kjiFGF
)sin5()1
()(
51
)sin(
)()( ))(( (i)
))(( (iv) ))(( (iii)
))(( (ii) ))(( (i)
find,51
)( and
sin)(by defined and functions vector For the
2
2
2
tt
ttt
ttttt
ttt
tt
tet
ttt
tttt
t
Example 4
Solution :
)()sin5()sin
5(
51
sin
)51
()sin(
)()())(( (iii)
)sin()()(
)())(( )ii(
22
2
2
2
kji
kji
kjikji
GFGF
kji
FF
tttttt
tt
tt
ttt
ttttt
ttt
teteet
tetettt
tt
Solution :
2.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS
Example 5
246)(
1226)(
4)52(4)(3 )(
if ),( and ),( Find
2
32
kiF
kjiF
kjiF
FF
tt
ttt
tt-tt
tt
Solution :
GFG
FGF
GFG
FGF
GFGF
FF
GF
dt
d
dt
d
dt
ddt
d
dt
d
dt
ddt
d
dt
d
dt
ddt
dcc
dt
d
)( )(iv
)( iii)(
)( )ii(
)( (i)
then,scalar a is c
andfunction vector abledifferenti are and If :4.3 Theorem
THEOREM 2.2
ttt t
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
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
FFFF
k
ji
kjikji
GFG
FGF
Solution :
INTEGRATION OF VECTOR FUNCTIONS
))(())(())(()(
thenb],[a,on of and ,,
functions integrable some from )()()()( If
ise.componentw done also is functions vector ofn Integratio
b
akjiF
kjiF
b
a
b
a
b
adtthdttgdttfdtt
thgf
thtgtft
Example 7
Solution :
kji
kji
kjiF
kjiF
F
802-42
])5()2[(
4)52()43( )(
4t)52()43()(
if )( Find
31
4223
3
1
33
1
3
1
3
1
2
32
3
1
ttttt
dttdttdtttdtt
tttt
dtt
2.3 ARC LENGTH AND CURVATURE
b
a dt
dy
dt
dxL
22
b
a dt
dz
dt
dy
dt
dxL
222
In 2-space
In 3-space
In general,
b
a
b
a
tLdt
dL )('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
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 t tttta kjir
20; 2sin3cos3)( )( tttttb kjir
4
3: LAns
1: eeLAns
58: LAns
132: LAns
Example 8
Example 9
UNIT TANGENT AND PRINCIPAL UNIT NORMAL VECTORS
If r(t) is a vector function that defines a smooth graph, then at each point a unit tangent vector is
t
tt
r
Tr
UNIT TANGENT VECTOR
3a) Find the derivative of 1 sin 2
b) Find the unit tangent vector at the point where 0.
tt t te t
t
r i j k
Example 10
curve. thetont vector unit tange theis where
as by denoted ),( curve thevector to
normalunit principle thedefine we,0 If
T
Nr
T
t
dtd
)('
)('
T
T
t
t
dtd
dtd
T
TN
UNIT NORMAL VECTOR
).( curve the toly,respectivevector
unit principal theandnt vector unit tange theare and where
as defined is curve a of vector binormal The
tr
NT
B
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 vector unit tange theis where
as defined is )( curvesmooth a of curvature The
T
r t
)('
)('
t
t
dtd
dtd
r
T
r
T
CURVATURE
3
r r
rOR
Curvature is the measure of how “sharply” a curve r(t) in 2-space or 3-space “bends”.
CURVATURE OF A HELIXFind the curvature of the helix traced out by
2sin ,2cos ,4t t t tr
Example 12
Radius of Curvature
as defined is curvature of radius itsthen
),( curvesmooth theof curvature theis If
tr
1
then time,is where),r(ector position v
by given curve thealong moves particle a If
tt
2.4 MOTION IN SPACE : VELOCITY AND ACCELERATION
dt
dt
rv )( velocity
2
2
)( on acceleratidt
d
dt
dt
rva
dt
dst )( speed v
Example 13
.2when
particle theofon accelerati and speed velocity, theFind
sincos)(
bygiven is after time particle a ofector position v The3
t
tttt
t
jir
Solution :
kji
kjiv
kjir
v
1242.09.0
)2(3)2(cos)2(sin
2when
)3()(cos)(sin
velocityobtain the we, w.r.t atingDifferenti
2
2
t
tttdt
d
t
kji
kjia
kjiv
a
a
v
129.00.42
)2(6)2sin()2cos( ,2 when
6)sin()cos(
bygiven is on accelerati The
04.12)2(91 ,2when
91)3()(cos)sin(
bygiven is any timefor speed The
4
42222
t
tttdt
d
t
tttt-
t
Example 14
kjir
r
kjiv
v
2)0( particle
theof )(ector position v theFind
2cos)(
bygiven ismotion in particle a of Velocity 2
t
ttet t
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(
cC where2
2sin
3
1
)2
2sin()
3
1()(
2cos)(
have we, Since
30
321
3
323
1
2
Solution :
kji
kjikjir
kji
kjii
r
)12
2sin()1
3
1()1(
)2
2sin(
3
1)(
obtain weHence
C
2
obtain we),0( of egiven valu theusingBy
3
3
tte
ttet
C
t
t
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
The end… =)
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