Results 1 to 6 of 6

Thread: [SOLVED] Parametric Representation of a Curve

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    21

    [SOLVED] Parametric Representation of a Curve

    Hey,

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

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

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

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

    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 $\displaystyle \frac{-\pi}{2} < t < \frac{\pi}{2}$?

    Any help would be greatly appreciated!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    Quote Originally Posted by Solo View Post
    Hey,

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

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

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

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

    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 $\displaystyle \frac{-\pi}{2} < t < \frac{\pi}{2}$?

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

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

    $\displaystyle x^2 - y^2 =$ ? Think identities!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    $\displaystyle \left\{\begin{array}{ll}x=\frac{1}{\cos t} \Longleftrightarrow t = \arccos \frac{1}{x}\\y=\tan t \Longleftrightarrow t = \arctan y\end{array}\right.$ $\displaystyle \Longleftrightarrow \arccos \frac{1}{x}=\arctan y \Longleftrightarrow y=\tan\left(\arccos\frac{1}{x}\right)=\frac{\sqrt{ 1-\frac{1}{x^2}}}{\frac{1}{x}}=$ $\displaystyle \sqrt{x^2-1}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2009
    Posts
    21
    Ah yes!

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

    So Cartesian equation: $\displaystyle x^2 - y^2 = 1$

    And the curve is a hyperbola?

    As a side note, I have a few questions:

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

    So in this case $\displaystyle a = b = 1$?

    What implications does that have on the curve when drawn?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2009
    Posts
    21
    Quote Originally Posted by Spec View Post
    $\displaystyle \left\{\begin{array}{ll}x=\frac{1}{\cos t} \Longleftrightarrow t = \arccos \frac{1}{x}\\y=\tan t \Longleftrightarrow t = \arctan y\end{array}\right.$ $\displaystyle \Longleftrightarrow \arccos \frac{1}{x}=\arctan y \Longleftrightarrow y=\tan\left(\arccos\frac{1}{x}\right)=\frac{\sqrt{ 1-\frac{1}{x^2}}}{\frac{1}{x}}=$ $\displaystyle \sqrt{x^2-1}$
    Hmm I see this gives the same result but I don't see how this was deduced: $\displaystyle y=\tan\left(\arccos\frac{1}{x}\right)=\frac{\sqrt{ 1-\frac{1}{x^2}}}{\frac{1}{x}}=$ $\displaystyle \sqrt{x^2-1}$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Parametric representation
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Apr 21st 2010, 05:05 AM
  2. Parametric representation for the surface
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Nov 20th 2009, 01:44 PM
  3. parametric representation of a curve
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Oct 9th 2008, 12:01 AM
  4. Parametric Representation of a Curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Sep 23rd 2008, 05:24 PM
  5. Parametric Representation of a curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 17th 2008, 03:00 PM

Search Tags


/mathhelpforum @mathhelpforum