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Math Help - parametric equations of a curve

  1. #1
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    parametric equations of a curve

    A curve: x=(1+t)^0.5, y=(1-t)^-0.5

    After plotting the curve I don't see any turning points. I can see a vertical asymptote at x=2^0.5.

    I have calculated that there is a stationary point at t=-1 and the point (0,2^-0.5). I don't see it on the graph though. At that point the second derivative is 0.25*(2^0.5) and hence greater than 0 and there is a minimum turning point.

    The answers give the point (0,2^0.5) and says it is a minimum end point.
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  2. #2
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    Re: parametric equations of a curve

    Quote Originally Posted by Stuck Man View Post
    I don't see it on the graph though.
    The point, or it looking like a starionary point? It's there.

    Quote Originally Posted by Stuck Man View Post
    The answers give the point (0,2^0.5)
    Must be a typo.

    Quote Originally Posted by Stuck Man View Post
    and says it is a minimum end point.
    i.e. would be a minimum turning point if it wasn't an end point.
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  3. #3
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    Re: parametric equations of a curve

    Left of the supposed stationary point the graph slope down slightly as if there is a horizontal asymptote. The point is not an end point.
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    Re: parametric equations of a curve

    parametric equations of a curve-capture-1.jpg
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  5. #5
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    Re: parametric equations of a curve

    Quote Originally Posted by Stuck Man View Post
    A curve: x=(1+t)^0.5, y=(1-t)^-0.5

    After plotting the curve I don't see any turning points. I can see a vertical asymptote at x=2^0.5.

    I have calculated that there is a stationary point at t=-1 and the point (0,2^-0.5). I don't see it on the graph though. At that point the second derivative is 0.25*(2^0.5) and hence greater than 0 and there is a minimum turning point.

    The answers give the point (0,2^0.5) and says it is a minimum end point.
    1. You certainely have determined that -1 \leq t < 1 ~\implies~ \left\lbrace \begin{array}{rcl}0\leq x \leq \sqrt{2} \\ y \ge \frac1{\sqrt{2}}\end{array}\right.

    2. Your curve is the right branch (drawn in red) of the graph of \displaystyle{f(x)=\frac1{\sqrt{2-x^2}}} (drawn in green)

    3. Obviously the left end point is a minimum with a horizontal tangent.
    Attached Thumbnails Attached Thumbnails parametric equations of a curve-parametric_curve.png  
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  6. #6
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    Re: parametric equations of a curve

    You can see that \left( {0,\frac{1}{{\sqrt 2 }}} \right) is a minimum.
    Attached Thumbnails Attached Thumbnails parametric equations of a curve-untitled.gif  
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  7. #7
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    Re: parametric equations of a curve

    My graphing program is displaying an incorrect line on the left of x=0. It is GraphCalc from 2003. Is there any others you know about?
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    Re: parametric equations of a curve

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