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Math Help - Parametric curves

  1. #1
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    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!!!!
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  2. #2
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    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 ...

    x^2 = \tan^2{\theta} = \sec^2{\theta} - 1 = y^2 - 1<br />

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

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

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


    parametrically ...

    x = \tan{\theta}

    1 = \tan{\theta}

    y = \sec{\theta}

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


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

    \frac{dy}{d\theta} = \sec{\theta}\tan{\theta} = \sqrt{2} \cdot 1 = \sqrt{2}<br />

    \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\sqrt{2}}{2}<br />
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