Results 1 to 2 of 2

Thread: Parametric curves

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    13

    Parametric curves

    The question is:

    Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter.
    x = tan(θ)
    y = sec(θ)

    (1, sqrt(2))


    I am unsure as to what the question is referring to. Maybe I'm being silly and it's actually very easy, I don't know.

    Please help!!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    Quote Originally Posted by DarthPipsqueak View Post
    The question is:

    Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter.
    x = tan(θ)
    y = sec(θ)

    (1, sqrt(2))


    I am unsure as to what the question is referring to. Maybe I'm being silly and it's actually very easy, I don't know.

    Please help!!!!
    eliminating the parameter ...

    $\displaystyle x^2 = \tan^2{\theta} = \sec^2{\theta} - 1 = y^2 - 1
    $

    $\displaystyle \frac{d}{dx}(x^2 = y^2 - 1)$

    $\displaystyle 2x = 2y \cdot \frac{dy}{dx}$

    $\displaystyle \frac{dy}{dx} = \frac{x}{y} = \frac{1}{\sqrt{2}}$


    parametrically ...

    $\displaystyle x = \tan{\theta}$

    $\displaystyle 1 = \tan{\theta}$

    $\displaystyle y = \sec{\theta} $

    $\displaystyle \sqrt{2} = \sec{\theta}$


    $\displaystyle \frac{dx}{d\theta} = \sec^2{\theta} = 2$

    $\displaystyle \frac{dy}{d\theta} = \sec{\theta}\tan{\theta} = \sqrt{2} \cdot 1 = \sqrt{2}
    $

    $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\sqrt{2}}{2}
    $
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help on Parametric Curves please
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Dec 10th 2010, 12:02 PM
  2. Parametric Curves
    Posted in the Calculus Forum
    Replies: 9
    Last Post: Oct 20th 2009, 06:14 PM
  3. Parametric Curves
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 6th 2009, 10:34 PM
  4. parametric curves
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Aug 6th 2008, 10:35 PM
  5. Parametric curves
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 27th 2008, 10:11 PM

Search Tags


/mathhelpforum @mathhelpforum