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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 ′

    ′=

     

    ( )   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  

    ( )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 

    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 

    ′=∴

    =

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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

    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

    ∫   ++−=0

    2221cossin

    dus

    ∫= 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 

    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=

    +

    =

    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 aV k a N 

    ×==

      2

    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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