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Parabola segment

  1. Jan 28, 2008 #1
    I have a parabola.

    Now, i need to get the length of aprt of a parabolic segment--that is a quarter of a parabola or 1/10 of a parabola, 3/4 of a parabola or anything

    How is this possible?

    Please help
     
  2. jcsd
  3. Jan 28, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    ?? What do you mean by "a quarter of a parabola or 1/10 of a parabola, 3/4 of a parabola"? A parabola has infinite length. If you have a parabola opening downward and are looking at the part above y= 0, please say so.
     
    Last edited: Feb 2, 2008
  4. Feb 2, 2008 #3
    Sorry---I will put it more clearly,

    I have a parabola whose directrix is not parallel to x axis.

    The equation of this parabola is:

    x^2 + 2xy + y^2 - 16y + 32 = 0


    I know three points which lie on the parabola [start point, end point, vertex].

    Now, I want the length of the parabola
     
  5. Feb 2, 2008 #4

    Gib Z

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    Homework Helper

    What exactly makes you think that equation yields a parabola?
     
  6. Feb 2, 2008 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Whatever makes him thinks so, he is right. The quadratic part is [itex]x^2+ 2xy+ y^2= (x+ y)^2[/itex] Let u= x+ y. Then y= u- x and the equation becomes, in terms of u, [itex]u^2- 16(u- x)- 32= 0[/itex] or [itex]u^2- 16u- 16x- 32= 0[/itex] and, finally, [itex]x= (1/16)u^2- u- 2[/itex]. That is a parabola at 45 degrees to the coordinate axes. There is no simple formula for arclength of a parabola. You would have to integrate [tex]\sqrt{1+ ((1/8)u- 1)^2}du[/itex] between the u values of the endpoints.
     
  7. Feb 2, 2008 #6
    To explain in detail what I want:

    Now,

    I have a parabola whose 3 points (x , y) are defined as:

    Point 1 : (-9, -1)

    Point 2 : (-6,6) [VERTEX POINT]

    Point 3: (4,-9)

    I need to get the length of the parabola.

    Now, the directrix of this parabola is not parallel to x axis.OK?

    So, I introduce a transformed coordinate system in which the directrix
    is parallel to an axis:

    Now, always if I draw a tangent to the parabola at the vertex, it will be parallel to the directrix, right?

    So, in the transformed coordinate system , the inclination of directrix would be:

    tan(inverse) of 6/-6 [SEE THE VERTEX POINT].Right?

    that is : -45 degrees

    Now,

    A transformed coordinate axis which
    is rotated about the origin through an angle T is given by:

    [ x' ] [ Cos T Sin T ][ x ] [ x Cos T + y Sin T ]
    | | = | || | = | |
    [ y' ] [ -Sin T Cos T ][ y ] [ -x Sin T + y Cos T ]

    So, I get,

    x ' = (x+y) (-cos 45)

    y' = (y - x) (-cos45)

    Right?

    So, the three points with respect to the rotated coordinate axis would be:

    Point 1 = ( -7.071 , -5.6568)

    Point 2 = (0, 8.4852)

    Point 3 = (3.5355 , 9.1923)

    Solving the equation,

    y = Ax^2 + Bx +C

    I get,

    A = 0.5091
    B = 3.1999
    C = -8.4852

    y = 0.5091x^2 + 3.1999x +(-8.4852)

    Right?

    If i want to find the length of the parabolic arc I use:

    1+ dy/dx^2)^1/2

    and integrate between -7.071 and 3.5355.

    I egt answer 33.3338488

    But correct answer is : 26.603

    Where am I going wrong?

    Please help
     
  8. Feb 2, 2008 #7
    Now,

    I have a parabola whose 3 points (x , y) are defined as:

    Point 1 : (-9, -1)

    Point 2 : (-6,6) [VERTEX POINT]

    Point 3: (4,-9)

    I need to get the length of the parabola.

    Now, the directrix of this parabola is not parallel to x axis.OK?

    So, I introduce a transformed coordinate system in which the directrix
    is parallel to an axis:

    Now, always if I draw a tangent to the parabola at the vertex, it will be parallel to the directrix, right?

    So, in the transformed coordinate system , the inclination of directrix would be:

    tan(inverse) of 6/-6 [SEE THE VERTEX POINT].Right?

    that is : -45 degrees

    Now,

    A transformed coordinate axis which
    is rotated about the origin through an angle T is given by:

    [ x' ] [ Cos T Sin T ][ x ] [ x Cos T + y Sin T ]
    | | = | || | = | |
    [ y' ] [ -Sin T Cos T ][ y ] [ -x Sin T + y Cos T ]

    So, I get,

    x ' = (x+y) (-cos 45)

    y' = (y - x) (-cos45)

    Right?

    So, the three points with respect to the rotated coordinate axis would be:

    Point 1 = ( -7.071 , -5.6568)

    Point 2 = (0, 8.4852)

    Point 3 = (3.5355 , 9.1923)

    Solving the equation,

    y = Ax^2 + Bx +C

    I get,

    A = 0.5091
    B = 3.1999
    C = -8.4852

    y = 0.5091x^2 + 3.1999x +(-8.4852)

    Right?

    If i want to find the length of the parabolic arc I use:

    1+ dy/dx^2)^1/2

    and integrate between -7.071 and 3.5355.

    I egt answer 33.3338488

    But correct answer is : 26.603

    Where am I going wrong?

    Please help
     
  9. Feb 2, 2008 #8
    To explain in detail what i want:

    Now,

    I have a parabola whose 3 points (x , y) are defined as:

    Point 1 : (-9, -1)

    Point 2 : (-6,6) [VERTEX POINT]

    Point 3: (4,-9)

    I need to get the length of the parabola.

    Now, the directrix of this parabola is not parallel to x axis.OK?

    So, I introduce a transformed coordinate system in which the directrix
    is parallel to an axis:

    Now, always if I draw a tangent to the parabola at the vertex, it will be parallel to the directrix, right?

    So, in the transformed coordinate system , the inclination of directrix would be:

    tan(inverse) of 6/-6 [SEE THE VERTEX POINT].Right?

    that is : -45 degrees

    Now,

    A transformed coordinate axis which
    is rotated about the origin through an angle T is given by:

    [ x' ] [ Cos T Sin T ][ x ] [ x Cos T + y Sin T ]
    | | = | || | = | |
    [ y' ] [ -Sin T Cos T ][ y ] [ -x Sin T + y Cos T ]

    So, I get,

    x ' = (x+y) (-cos 45)

    y' = (y - x) (-cos45)

    Right?

    So, the three points with respect to the rotated coordinate axis would be:

    Point 1 = ( -7.071 , -5.6568)

    Point 2 = (0, 8.4852)

    Point 3 = (3.5355 , 9.1923)

    Solving the equation,

    y = Ax^2 + Bx +C

    I get,

    A = 0.5091
    B = 3.1999
    C = -8.4852

    y = 0.5091x^2 + 3.1999x +(-8.4852)

    Right?

    If i want to find the length of the parabolic arc I use:

    1+ dy/dx^2)^1/2

    and integrate between -7.071 and 3.5355.

    I egt answer 33.3338488

    But correct answer is : 26.603

    Where am I going wrong?

    Please help
     
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