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Math Help - parameterization of a curve

  1. #1
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    parameterization of a curve

    can someone help please:

    Find what kind of curves are given by the following representations and draw the curves:

    1. r(t) = (2t-5, -3t+1, 4)
    2. r(t) = (0, -cos(t), 3sin(t))
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  2. #2
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    Quote Originally Posted by sonia1 View Post
    can someone help please:

    Find what kind of curves are given by the following representations and draw the curves:

    1. r(t) = (2t-5, -3t+1, 4)
    2. r(t) = (0, -cos(t), 3sin(t))
    The first is a straight line parallel to the xy plane moved 4 units in the z direction and the second, an ellipse in the yz plane.
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    how did you find that out, could u please show me
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    Quote Originally Posted by sonia1 View Post
    how did you find that out, could u please show me
    Sure. In the first case

    x = 2t -5,\;\;y= -3t+1,\;\;z=4

    z isn't changing, x and y are. Eliminating t gives

    \frac{x+5}{2} = \frac{y-1}{-3} a straight line.

    In the second, x = 0, it's not changing and

    y = - \cos t,\;\; z = 3 \sin t

    Eliminating t using the fact that \cos^2 t + \sin^2 t= 1 gives

    y^2 + \frac{z^2}{3^2} = 1, the formula for an ellipse in the yz plane. Hope that helps.
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    yes thanks alot,
    i'm also stuck on a different question on the same topic

    wonder if you can help:

    find a parametric representation of the curve :

    x^2 + y^2= 36 and z= (1/pi) arctan(x/y)

    i.e. find a representation in the form x=x(t), y=y(t), z=z(t)
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    Quote Originally Posted by sonia1 View Post
    yes thanks alot,
    i'm also stuck on a different question on the same topic

    wonder if you can help:

    find a parametric representation of the curve :

    x^2 + y^2= 36 and z= (1/pi) arctan(x/y)

    i.e. find a representation in the form x=x(t), y=y(t), z=z(t)
    You'll notice with the first equation and the identity

    \sin^2t + \cos^2 t = 1 then 36\sin^2t + 36\cos^2 t = 36 so comparing gives

    x = 6 \sin t,\;\;\;y = 6 \cos t. Now sub. these into z and simplify.
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  7. #7
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    thanks..i simplified it and i get z= (1/pi) arc 6tan^2 t, is that correct
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    regarding the before question, could u please show me a drawing of what the ellipse actually looks like.
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    Quote Originally Posted by sonia1 View Post
    thanks..i simplified it and i get z= (1/pi) arc 6tan^2 t, is that correct
    Not 6\tan^2 t but \tan t
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  10. #10
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    r u sure it should be tan^2 t because (6sint/6cost) =tan t

    and then we get 1/pi arc tan^2t right
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  11. #11
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    Here's a picture of the ellipse.
    Attached Thumbnails Attached Thumbnails parameterization of a curve-ellipse.jpg  
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  12. #12
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    thankz for your help,

    did u get tan^2 t then not tant
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  13. #13
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    Quote Originally Posted by sonia1 View Post
    thankz for your help,

    did u get tan^2 t then not tant
    From your earlier post,

    x^2 + y^2= 36 \;\; \text{and}\;\; z= \frac{1}{\pi}\tan^{-1}(\frac{x}{y})

    if x = 6\sin t, y =6\cos t then \frac{x}{y} = \frac{6\sin t}{6\cos t} = \tan t
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  14. #14
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    thankz for your help!
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    Quote Originally Posted by sonia1 View Post
    thankz for your help!
    You can thank danny (and add to his reputation) by clicking the Thanks button (bottom right).
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