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Math Help - [SOLVED] Parametric Representation of a Curve

  1. #1
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    [SOLVED] Parametric Representation of a Curve

    Hey,

    Find the equation in x and y for the curve given parametrically
    by x = sec(t) and y = tan(t), where \frac{-\pi}{2} < t < \frac{\pi}{2}. What type of
    curve is it?

    -----
    So I've written that x = \frac{1}{cos(t)} and y = \frac{sin(t)}{cos(t)}.

    From that t= cos^{-1} \frac{1}{x}

    If I substitute that in  y = \frac{sin (cos^{-1} \frac{1}{x})}{\frac{1}{x}}.

    But that's where I get stuck..I'm not sure if I've even started correctly. What do you with the fact they've told you that \frac{-\pi}{2} < t < \frac{\pi}{2}?

    Any help would be greatly appreciated!!
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  2. #2
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    Quote Originally Posted by Solo View Post
    Hey,

    Find the equation in x and y for the curve given parametrically
    by x = sec(t) and y = tan(t), where \frac{-\pi}{2} < t < \frac{\pi}{2}. What type of
    curve is it?

    -----
    So I've written that x = \frac{1}{cos(t)} and y = \frac{sin(t)}{cos(t)}.

    From that t= cos^{-1} \frac{1}{x}

    If I substitute that in  y = \frac{sin (cos^{-1} \frac{1}{x})}{\frac{1}{x}}.

    But that's where I get stuck..I'm not sure if I've even started correctly. What do you with the fact they've told you that \frac{-\pi}{2} < t < \frac{\pi}{2}?

    Any help would be greatly appreciated!!
    If you do this, t= cos^{-1} \frac{1}{x}, you're going to get into trouble real fast.

    From x = \frac{1}{cos(t)} and y = \frac{sin(t)}{cos(t)}, then

    x^2 - y^2 = ? Think identities!
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  3. #3
    Senior Member Spec's Avatar
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    \left\{\begin{array}{ll}x=\frac{1}{\cos t} \Longleftrightarrow t = \arccos \frac{1}{x}\\y=\tan t \Longleftrightarrow t = \arctan y\end{array}\right. \Longleftrightarrow \arccos \frac{1}{x}=\arctan y \Longleftrightarrow y=\tan\left(\arccos\frac{1}{x}\right)=\frac{\sqrt{  1-\frac{1}{x^2}}}{\frac{1}{x}}= \sqrt{x^2-1}
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  4. #4
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    Ah yes!

    x^2 - y^2 = \frac{1}{cos^2(t)} - \frac{sin^2(t)}{cos^2(t)} = \frac{1-sin^2(t)}{cos^2(t)} = \frac{cos^2(t)}{cos^2(t)} = 1

    So Cartesian equation: x^2 - y^2 = 1

    And the curve is a hyperbola?

    As a side note, I have a few questions:

    Isn't the equation for a hyperbola \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1?

    So in this case a = b = 1?

    What implications does that have on the curve when drawn?
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  5. #5
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    Quote Originally Posted by Spec View Post
    \left\{\begin{array}{ll}x=\frac{1}{\cos t} \Longleftrightarrow t = \arccos \frac{1}{x}\\y=\tan t \Longleftrightarrow t = \arctan y\end{array}\right. \Longleftrightarrow \arccos \frac{1}{x}=\arctan y \Longleftrightarrow y=\tan\left(\arccos\frac{1}{x}\right)=\frac{\sqrt{  1-\frac{1}{x^2}}}{\frac{1}{x}}= \sqrt{x^2-1}
    Hmm I see this gives the same result but I don't see how this was deduced:  y=\tan\left(\arccos\frac{1}{x}\right)=\frac{\sqrt{  1-\frac{1}{x^2}}}{\frac{1}{x}}= \sqrt{x^2-1}
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  6. #6
    Senior Member Spec's Avatar
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    List of trigonometric identities - Wikipedia, the free encyclopedia

    It's not a correct solution though (one of the equivalence signs are wrong), since we lost all the negative y values.
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