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Transcript of math 281- unit 1-1f
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CALCULUS 3: MATH 281
Mrs. McWilliam & Dr. Kounta (COB)
Vector-Valued Functions
February 4th, 2016
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Syllabus
Recall: 3 Dimensional Coordinate System
Vector-Valued Functions
Arc Length in R 3
Motion in Space
i. Position
ii. Velocity and Acceleration
Curvature
Tangent and Normal Vectors
Tangential and Normal Component of Acceleration
Parametric Surfaces
Outline
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TANGENTAND
NORMAL VECTORS
Mrs. McWilliam & Dr. Kounta (COB)
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Unit Tangent Vectors
Let r(t) be a smooth curve, then the unit tangentvector to the curve , T (t), is given by
( ) ( )
( )t r t r
t
T ′
′=
( ) vector gent theishg f t r Where tan,,, ′′′=′
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Ex. Let r(t) = t i + t 2 j, find the unit tangent vectorto the curve at t = 1.
t r 2,1=
241 t r +=′
22241
2,41
141
2,1t
t t t
t T ++
=+
=
( )5
2,
5
11 =T
SOLUTION
( ) t t r 21+=′
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Ex. Find T (t) and the parametric equations of thetangent line to
r(t) = 2cost i + 2sint j + t k at t = .4π
SOLUTION
1,cos2,sin2 t t r −=′
( ) 1,2,24
−=′ π r
( ) 51224 =++=′ π
r
( )5
1
5
2
5
2
4 ,,−=π T
( ) 4,2,24π
π
=r
+=
+=
−=
t z
t y
t x
4
22
22
π
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We know that
||T(t )|| = 1 for all t.
We saw last class that if ||u|| = c, then u ∙ u′ = 0.
So T ∙ T′ = 0 for all t .
T′ is orthogonal to T for all t .
( ) ( )
( )t r t r
t T ′
′=
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The principal unit normal vector:
( ) ( )( )t T t T
t N ′′=
Unit Tangent Vectors
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T shows the direction that the curve is goingN shows the direction that the curve is turning
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Ex. Find the principal unit normal vector of
( ) cos ,sin ,r t t t t =
t t t r ,cos,sin−=′
( ) ( ) 21cossin 22
=++−=′ t t r
2
1,
2
cos,
2
sin t t T
−=
2
1
2
sin
2
cos 22
=+= t t
T
0,2
sin,2
cos t t T −−=′
0,sin,cos t t N −−=
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CURVATURE
Mrs. McWilliam & Dr. Kounta (COB)
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Curvature
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The curvature, κ , of a curve describes how sharply
it bends.
big κ
small κ
Curvature
The curvature, κ , measures how fast a curve ischanging direction at a given point.
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Thm. The curvature of a curve is
where T is the unit tangent vector and s is the arclength parameter.
( )sT
dsT d
′=∴
=
k
k
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Ex. Find the curvature of
Hint:
( ) 3 33cos ,3sins sr s =
( )22
33cos3,
33sin3
sssssr
−⋅
−⋅−=′
2
3
sk =
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Arc Length
We saw the length of a curve given by
on the interval a ≤ t ≤ b is
( ) ( ) ( )2 2 2
b
a
s f t g t h t dt ′ ′ ′ = + + ∫
( ) ( ) ( ) ( ), ,r t f t g t h t =
( )
vector gent thehg f r note
dt t r L
b
a
tan,,,: ′′′=′
′= ∫
L
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Ex. Find the length of the curve r(t) = cos t i + sin t j + t k from (1,0,0) to (1,0,2π )
( ) dt t r Lb
a
∫ ′=
dt ∫=π 2
0
2
π 2
02t =
22π =
↑
= 0t
↑
= π 2t
SOLUTION
We first need the tangent vector and its magnitude
The length is then,
( ) 1,cos,sin t t t r −=′
( ) ( ) ( ) ( ) 21cossin 222 =++−=′ t t t r
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These vector functions are the same:
They are just re-parameterizations of thesame curve.
( ) 2 31 , , 1 2r t t t t t = ≤ ≤
( ) 2 32 , , 0 ln 2u u u
r u e e e u= ≤ ≤
( ) ( ) ( ) ( )2 33 1 , 1 , 1 1 0r w w w w w= − − − − ≤ ≤
( ) ( ) ( )2 3
4 4 4 4, , 4 8t t t r t t = ≤ ≤
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The arc length function is defined as
for t between a and b. We will seehow to re-parameterize a curve using
s.
( ) ( ) ( ) ( )2 2 2
t
a
s t f u g u h u du′ ′ ′ = + + ∫
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Ex. Reparameterize the curve r(t) = cost i + sint j + t k with respect to the arc
length measured from (1,0,0).( ) ( ) ( ) ( ) duuut s
t
∫ ++−=0
2221cossin
dus
t
∫= 0 2t
us0
2=
2t s =
0=t
2
st =
( )( ) ( ) 222 ,sin,cos sss
sr st r ==
With the reparameterization we can now tell
where we are on the curve after we’ve
traveled a distance of s along the curve
Td
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To evaluate a description of the curve interms of the arch length is needed but at times it
may be difficult to obtain.
Chain Rule: ( ) ( )
dt
ds
t T
ds
T d
dt
ds
ds
T d
dt
T d t T
′=⇒•==′
ds
T d
( ) ( ) ( )t r duur dt
d t s
dt
d
dt
ds t
a
′=′== ∫
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An easier equation for curvature would be
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Ex. Find the curvature of a line
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Ex. Find the curvature of the circle with radius b.
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Ex. Find the curvature of a helix
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( ) t t t t r
of curvaturetheisWhat
5,sin4,cos4=
41
4
2516
44=
+
=
k
In this case the curvature is constant. This means that the curve is changing direction atthe same rate at every point along it. This curve is a helix therefore the result makes
sense.
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We can also find curvature using
3v
va×=
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Ex. Find the curvature of the space curve
at any point t.( ) 2 3, ,r t t t t =
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If we’re looking for the curvature of a plane curve y = f ( x), there’s yet another formula:
( )
( )( )
322
1
f x
f x
κ
′′=
′+
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Ex. Find the curvature of the y = x2 at the
points (0,0), (1,1), and (-1,1).
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Alternative formula for the
principal unit normal vector:
( )ds
T d k
dsT d ds
T d
t N
1==
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TANGENTIAL COMPONENTS
ANDNORMAL COMPONENTS
Mrs. McWilliam & Dr. Kounta (COB)
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Back to velocity and acceleration:
||v(t )|| = constant
v ∙ v = c
v ∙ a = 0
So if velocity is constant, then velocity isorthogonal to acceleration
This is not necessarily true for variable velocity
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Because T(t ) and N(t ) are orthogonal, they define a plane.
Thm. If r(t ) is a smooth function for position and ifN(t ) exists, then the acceleration vector a(t ) liesin the plane determined by T(t ) and N(t ).
This means that a(t ) is a linear combination ofT(t ) and N(t ):
N aT aa N T
+=
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In the study of the motion of objects the acceleration is often broken up into a tangential
component, aT , and a normal component, a
N. The tangential component is the part of the
acceleration that is tangential to the curve and the normal component is the part of the
acceleration that is normal (or orthogonal) to the curve. If we do this we can write the
acceleration as, N aT aa N T
+=
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V
V aV k a N
×==
2
V
V a
dt
sd aT
•== 2
2
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( ) ( )( )
tangentialcomponent of
T r t r t a a T
ar t
′ ′′⋅ = = ⋅ =
′
( ) ( )
( )
normalcomponent of N
r t r t a a N a
r t
′ ′′× = = ⋅ =
′
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Ex. Find the tangential and normal components of
acceleration for the position function
( )2 2 3
, ,r t t t t =
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Ex. Let , find T(t), N(t), aT, and aN at
t = 1. [Use a = aT
T + aN
N to find N.]
( )222, ,t r t t t =
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