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    4.0

    APPLICATIONS OF

    DIFFERENTIATION

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    4.1 TANGENT AND NORMAL

    EQUATION

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    At the end of this topic, students

    should be able :

    a) To find the equations of tangent andnormal to a curve including the parametric

    equation

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    1 . is the gradient to the function ofdxdyxf )('

    )(xfy

    2. At point ( ) on the curve )(xfy the equation of tangent is

    )( 111 xxmyy

    11, yx

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    where

    is the gradient of the tangent to

    the curve at the point

    dxdym 1

    ),( 11 yx

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    )(xfy

    normaltangent

    11, yx

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    EXAMPLE

    Find the tangent and normal equation to

    the curve at x= 2322 xy

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    Solution,

    If

    Differentiate

    so the gradient for tangent at x = 2 is

    4(2) = 8

    xdx

    dy

    4

    32 2 xy

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    If x = 2

    The tangent equation at point (2, 11) is

    113)2(22

    y

    )2(811 xy

    58

    16811

    xy

    xy

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    tangent is perpendicular to the

    normal thus,

    121 mm

    2m8

    1

    1)(8 2 m

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    normal equation at point (2,11)

    9082888

    8

    2

    8

    111

    )2(8

    111

    xyxy

    xy

    xy

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    32 2 xy

    908 xynormaltangent

    58 xy

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    EXAMPLE

    Find the tangent and normal equation to

    the curve at P(1,1)

    Solution,

    Differentiate by using the implicit derivative

    method,

    223 xyyx

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    )3()2(

    02)1(3

    223

    223

    yxyxyxdx

    dy

    dxdyxyyyx

    dxdyx

    xyxyxy

    dxdy

    2)3( 3

    22

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    the gradient for tangent at P(1,1) is

    _

    so tangent equation at P(1,1) is

    743

    3

    4

    3

    41

    )1(3

    41

    xy

    xy

    xy

    3

    4

    )1)(1(21

    ))1()1(31(3

    22

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    gradient for normal at P(1,1) is

    _

    4

    3

    13

    4

    2

    2

    m

    m

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    normal equation at P(1,1)

    134

    4

    3

    4

    31

    )1(4

    31

    xy

    xy

    xy

    E i Fi d th t t d

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    Exercise: Find the tangent and

    normal equation to the curve xy = 8

    at P(4,2)

    Answer;

    Tangent equation 2y = -x + 8Normal equation y = 2x 6

    xy

    dxdy

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    EXAMPLE

    Find the equation of normal to the curve

    given parametrically by

    and

    at point where t= 1

    tx 2 13

    2 ty

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    so

    when t=1

    the gradient for tangent is -3(1) = -3

    32

    3)

    2

    (6 tt

    t

    dx

    dy

    3

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    the normal equation at (2,2) is

    43

    3

    2

    3

    12

    )2(3

    12

    xy

    xy

    xy

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    EXAMPLE

    Given parametric equation

    and wheret 1

    Find the tangent equation which is parallelto line 3y = x

    1

    3

    tx2)1(

    2

    ty

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    Solution

    2)1(

    3

    tdtdx

    3)1(

    4

    tdtdy

    )1(34

    3

    1

    )1(

    4 2

    3

    t

    t

    tdx

    dt

    dt

    dy

    dx

    dy

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    from 3y = x the gradient for the line is

    So

    4 = 1+ t

    t = 3

    3

    1

    31

    )1(34

    tdx

    dy

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    when t = 3

    tangent equation is

    43

    133

    x81

    )31(2

    2 y

    3824

    123

    318

    )4

    3(

    3

    1

    8

    1

    xy

    xy

    xy

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    Exercise:

    Given parametric equation x =

    and y =

    Find the tangent equation when t= 0

    23tt

    2

    42

    t

    t

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    Answer:

    3y = 4x - 6