6 , 7 2. vector functions lesson 6 7
TRANSCRIPT
Vector FunctionsLesson 6
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Objectives
At the end of the lesson you should be able to:
1. Define a vector function.
2. Find the limit of a vector function.
3. Differentiate a vector function.
4. Evaluate a given integral.
Vector Function
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Definition:
A vector function is a function whose domain is a set of real numbers and whose range is a set of vectors.
In notation,
t
were
kthjtgitftr
thtgtftr
,
)()()()(
)(),(),()(
a real number
Examples of Vector Functions
ktjtittr
ttttr
)sin()(cos)()2
5,1,)()1 3
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Limit
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The limit of a vector function r(t) is defined by taking
limits of its component function,
Definition:
If
)(lim,)(lim),(lim)(lim thtgtftratatatat
provided the limits of of the component functions exists.
What is a Continuous Function?
A vector function r is continuous at a if )()(lim artrat
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That is if and only if f,g, and h are continuous at a.
To find the limit of the vector function,
1. Substitute in the function.
2. Otherwise use L’Hospital’s Rule.
L’Hospital’s Rule:
atast
.0
0mindet
)(
)(lim0)('
)('
)('lim
)(
)(lim
orliketypeformateerin
becomesxg
xfandxgwere
xg
xf
xg
xf
ax
axax
Example of Limit
1. Evaluate the limit of
4/)(sin)(cos)( tasktjtittr
Solution:
4,707.0,707.0)(lim
4707.0707.0)(lim
4)
4lim(sin)
4lim(cos)(lim
4
4
4
t
t
t
tr
kjitr
kjitr
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Example 2
Evaluate the limit of
ttttt
ln,sin,coslim0
For the first two functions, limit exist, i.e. substitute t = 0. For the third function, substituting the value of t, limit does not exist. By L’Hospital’s Rule,
0lim
1
1
lim
1
lim1
lnlim)ln(lim
0
2
02000
t
tttt
tt
ttt
t
ttt
0,0,1)(lim tr
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tt
t
t
et
t 1
3,
11,
1lim
0
Use L’Hospital’s Rule for the first two components.
Your answer should be
3,2
1,1)(lim tr
Example 3:
Evaluate the
Derivatives of Vector Functions
The derivative r’(t) of a vector function r(t) is
defined as much the same way as for real valued
functions.
h
trhtr
dt
dr
trdt
dr
h
)()(lim
)('
0
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Geometric Interpretation of the Derivative
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vectoranttrhtr
vectorgenttr
sec)()(
tan)('
P)(' tr
Q
)( htr
x y
z
C
0
See Figure!
r(t)
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Theorem:
If
handgfwere
kthjtgitftr
thtgtftr
,
)()()()(
)(,)(),()(
are differentiable functions, then
kthjtgitftr
thtgtftr
)(')(')(')('
)('),('),(')('
Rules for derivatives are similar to rules for real valued
functions. In addition are the following:[ v and u are
differentiable vector functions]
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1.
RuleChaintfutftfudt
d
tvtutvtutvtudt
d
tutvtvtutvtudt
d
,))((')(')}]({[
)(')()()(')]()([
)()(')()(')]()([
2.
3.
Added Rules are the Following:
Second Derivative
)' ' ( ) ( "r t r
Rules are the same as the real valued functions
Smooth Curves
A curve given by a vector function r(t) on an interval I
is called smooth if r’(t) is continuous and r’(t) is not
equal to 0. (except possibly at any endpoints of I)
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Example 1 Derivative
a) Find the first derivative of
kjtittr 21 1sin)(
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0,1
,1
1)('
0,12
2,
1
1)('
22
22
t
t
ttr
t
t
ttr
Answer
Example 2: Derivative of a Dot Product
If and
find .
ktjtitu 32 32)( ktjtittv sincos)(
)()(')()(')()( tutvtvtutvtudt
d
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Solution:
tttttt
tttttttt
ttttttttt
tttttdt
d
cos3sin11cos41
cos3sin21sin9cos40
3,2,1cos,sin,1sin,cos,9,4,0
sin,cos,3,2,1
32
322
322
32
Derivative of a Cross Product
Example : Differentiate
22 4,,)1(2,, tttttat
Answer: ttattattt 232,2412,416 223
Rule: )(')()()(')()( tvtutvtutvtuDt
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Example 3: Second Derivative
Find the second derivative of
ttbtattr cos,sin,3cos)(
,sin,cos,3cos3sin3)('
sin,cos,3cos)3sin3()('
cos,sin,3cos)(
ttbtatattr
ttbtatattr
ttbttatr
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The second derivative is
ttbtatattr
ttb
taattattr
cos,sin,3sin63cos9)("
cos,sin
,3sin3)3(3sin3cos33)("
Use the same rules and find the second derivative for each component.
On page 897, do problems 9-16.
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Integrals
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The definite integral of a continuous vector r(t) can be defined in much the same way as for real valued functions except that of the integral is a vector.
b
a
b
a
b
a
b
a
kdtthjdttgidttfdttr ))(())(())(()(
The Fundamental Theorem of Calculus
)()()}({)( aRbRtRdttrb
a
ba
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Example 1 : Integrals
Evaluate 1
0
32 )( dtktjtti
kji
tttdtttt
4
1
3
1
2
1
4
1,
3
1,
2
1
4,
3,
2,,
1
0
432321
0
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4
0)sin2sin2(cos.2 dtkttjtit
Example 2 : Integrals
)4
1(2
2,2/1,2/1
)]0sin0()4/sin4/cos4/[(
,)0cos2/(cos2/1,)0sin2/sin(2/1
sincos,2cos2
1,2sin
2
1
sin,2sin,2cos
4
0
4
0
ttttt
dttttt
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Practice More!!!
On page 898 Solve Nos. 33-38
Arc Length and CurvatureLesson 7
At the end of the lesson you should be able to :
1. Define arc length.
2. Find arc length.
3. Reparametrize the curve with respect to arc length.
4. Define curvature.
5. Find curvature.
6. Define TNB
7. Find TNB.
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Arc Length
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Definition:
The length of a smooth curve r(t) = x(t)I + y(t)j + z(t)k,
that is traced exactly once a t increases
from a to b is (arc length)
1.
b
a
b
a
dtdt
dz
dt
dy
dt
dxL
dtthtgtfL
222
222 )]([)]('[)]('[
bta
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Example 1: Arc Length
Find the length of a given curve
Use
1010,cos2,5,sin2)( tttttr
dt
tt
thtgtftrwheredttrb
a
29
29
)sin2(5)cos2(
)]('[)]('[)]('[)(',)('
10
10
222
222
Answer: How? 2920
Then,
Example 2: Arc Length
Find the length of the curve given
.80,)3/2()( 2/3 tktittr
3
52)127(
3
2
)1(3
21
)(01
80
38
0
22/1228
0
L
tdttL
dttL
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Solution:
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2. For a plane curve with vector equation
dtdt
dy
dt
dxL
dttgtfL
continuousaregandf
tgyandtfx
equationsparametricwithbta
tgtftr
b
a
b
a
22
22 )]('[)]('[
.,','
)(,)(
)(),()(
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Our formulas therefore are the following:
.
)]('[)]('[)]('[)('
)(')(')(')(')
.
)]([)]('[)(')
)(')(')('
)('
222
22
curvespacefor
thtgtftr
kthjtgitftrb
curveplanefor
tgtftra
jtgitftrwere
dttrLb
a
Reparametrize the Curve with Respect to Arc Length
How to do this?
The parameter t must be expressed in terms of another parameter s.
dudu
dz
du
dy
du
dxduurtss
t
a
t
a
222
)(')(
Differentiating both sides you get
)(' trdt
ds
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Arc Length Parametrization
Example 2:
Find an arc length parametrization of the circle of radius 4 centered at the origin.
Solution:
The arc length from u=0 to u=t is given by
20,sin4)(;cos4)(: tttgyttfxC
4,
4)cos4()sin4()(
))('())('()(
22
0
22
0
sttherefore
tduuuts
dtuguftsL
t
t
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Substitute this to C.
Example 3: Reparametrize a Curve
Reparametrize the curve r(t) = 2t i +(1-3t)j + (5+4t) k with respect to arc length from the point t = 0 in the direction of increasing t.
29
294)3(2)(
)(')(
222
0
0
st
tdutss
duurts
t
t
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Then using the given curve, replace t by
29
)(
29
tss
ks
js
is
tsr
tstststsr
)29
45()
29
31(
29
2))((
29
)(45,
29
)(31,
29
)(2))((
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Curvature
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At any given point the curvature is a measure of how
quickly the curve changes directly at that point. Or, it is
the magnitude of the rate of change of the unit tangent
vector T with respect to arc length.
.)('
)(")('
,sin
)('
)('
3tr
trxtr
Theoremtheguor
tr
tT
ds
dT
where the curve is traced out by the vector function r(t).
JBJ/UTP/2004 12
For a special case , like the plane curve, where y= f(x)
the curvature is
2/32)]("[1
)(")(
xf
xfx
Curvature is a scalar quantity!!!
The TNB
y
z
x
TB
r
N
T - the unit Tangent vector
represents the forward
direction.
N- the unit Normal vector
represents the direction in
which your turning
B- is the Binormal vector a
tendency of your motion to
“twist”out of the plane
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Unit Tangent Vector
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vectorgenttrweretr
trtT tan)('
/)('/
)(')( It is
xy
z
C
PQ
)(' tr
)( tr)( htr
)()( hrhtr
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Example 4: Unit Tangent Vector
Find the unit Tangent vector T(t) of
.1,2,4,6)( 35 tttttr
Use .)('
)(')(
tr
trtT
262
1,6,15
2622
1,6,152)1(
1048
2,12,30)1(
1,2)12()30(
2,12,30)(
2222
24
T
T
twhentt
tttT
Answer is
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Example 5. Curvature
Find the curvature of
at the point (1,0,0) .
ttetetr tt ,sin,cos)(
To find the curvature you can use
3)('
)(")(')(
)('
)(')(
tr
trtrtor
tr
tTt
The given point (1,0,0) is satisfied when t = 0. ( substitute t = 0 in r(t)).
Solution
1,sincos,cossin)('
,sin,cos)(
tetetetetr
ttetetrtttt
tt
a)
When t = 0,
3)1('1,1,1)0(' randr
b)
0,2,0)0("
0,cos2,sin2)("
0,sincoscossin
,sincossincos)("
r
tetetr
tetetete
tetetetetr
tt
tttt
tttt
c)
2,0,2
0,2,01,1,1)0(")0(' rr
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69
2
3
2
3
2
)3(
8
1,1,1
2,0,2
)0('
)0(")0(')0(
3
3
r
rr
Your answer!
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Using the other formula will give you the same answer. Decide which formula would be easier to use…
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Normal and Binormal Vectors
The Unit Normal of the Principal Unit Vector is defined as:
/)('/
)(')(
tT
tTtN
The Binormal Vector is defined as a vector
perpendicular to both T and N and also a unit vector.
)()()( tNtTtB
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See the Figure!!!
C
)( tT)( tB
)( tN
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Example 6: Normal Vector and Binormal Vector
Find the N(normal vector) and the B ( binormal vector)
given . (The point is satisfied
when t = 1). .
)1,3
2,1(,,
3
2,)( 32 ttttr
Solution:
a)
1,2,2
12
1
12
1,2,2
)('
)(')(
)('
)(')( 2
22
2
tttt
tt
tr
trtTwhere
tT
tTtN
so that at t = 1,
1,2,23
1)1( T
)('
)(')(
tT
tTtN
Where
3
2,2,1)1(
12
2,2,12)(
,12
4,4,24)('
2
2
22
2
N
t
ttttN
thent
ttttT
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b) The Binormal Vector at t = 1 is
.1
3
2,
3
1,
3
2)1(
2,2,13
11,2,2
3
1)1(
)1()1()1(
)()()(
twhen
B
B
NTB
tNtTtB
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Practice TaskFind the curvature and TNB given the
following:
1..1,,2,)( 2 twhenttttr
2. .0,2sin,,2cos3)( twhenttttr
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Analyze the curvature of a straight line what will it be?
Compare the curvature between a big circle and a small circle. What can you say about it?
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Practice More!!!
Solve the following problems on page 904-905
Nos. 1, 9, 13 , 17, 21, 39 and 40.
Motion in Space
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Motion in Space: Velocity and Acceleration
A vector function r(t) may represent as the position of a particle in space at time t.
v(t) = r’(t) is the velocity vector and is the
speed
a(t)=r”(t) is the acceleration vector.
and
)(' tr
Thus, dttvtr )()( dttatv )()(
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Find the velocity and position at time t of a particle whose
position is i+j, initial velocity is j + k, and the acceleration
is a(t) = 12 t i + 2 kInitially, (t = 0) jirkjvo )0(;
Answers:
kttjtittr
ktjittv
)()1()12()(
)12(6)(23
2
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Example
Example
What force is required in order for a 5 kg mass to be pushed
in such a way that, at time t, its position is
?23)( 342 ktjtittr
Use F(t)=m a(t)
Answer: ktjtitatF 606030)(5)( 2
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Example
A projectile is fired with a velocity of 200 m / sec at an inclination of 30 degree from a point 10 meters from the ground. Find the vector function that describes the path of motion.
10m
y
x
Use a(t)= - g j
jgt
tittr
jtgitv
jijiC
CvtAt
Cjtgtv
)102
100(3100)(
)100(3100)(
100310030sin20030cos200
;,0
)(
2
00
0
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Acceleration Determined by Unit Tangent Vector and Unit Normal Vector
The acceleration can be found by the combination of T(t) and N(t)
'
"'
'
"':
)()(
r
rraand
r
rrawhere
tNatTaa
NT
NT
Or
'
"''
'
''')(
),('
2
32
r
rrr
r
rrcurvatureisva
vspeedisva
N
T
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vv
Example:
Find the tangential and the normal components of acceleration for
.)( 342 ktjtittr
Answers:
24
24
24
24
9164
649362;
9164
18484
tt
ttta
tt
tta NT
UTP/JBJ 32
'
"'
'
"':
)()(
r
rraand
r
rrawhere
tNatTaa
NT
NT
Use This!
Practice More!!!
Do the following problems on page 914-915
Nos. 5, 9, 11, 13, 16, 18, and 35
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