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  1. #1
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    Unhappy Help!

    A curve is defined by the parametric equations x = tē and y = 2t + 1
    Find the equation of the tangent at the point where t = 2.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Kiwigirl
    A curve is defined by the parametric equations x = tē and y = 2t + 1
    Find the equation of the tangent at the point where t = 2.
    Let the equation of the tangent at the required point be:

    y=mx+c.

    We need to find the slope of the tangent, which is dx/dy, and
    to find this we need to use the result that:

    <br />
\frac{dy}{dx}=\frac{dy}{dt}\times  \frac{dt}{dx}=2\times \frac{1}{2t}=1/t  <br />

    So at t=2 the slope of the tangent is 1/2.

    So now we know that the equation of the tangent is of the form:

    y=x/2+c,

    and that at t=2, x=4 and y=5, so the tangent passes through the point
    x=4, y=5.

    Hence:

    <br />
5=4/2 +c<br />
,

    so:

    <br />
c=3<br />
,

    and the equation of the tangent is:

    y=2x-3.

    RonL
    Attached Thumbnails Attached Thumbnails Help!-parametric.jpg  
    Last edited by CaptainBlack; April 9th 2006 at 09:21 PM.
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  3. #3
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    Quote Originally Posted by Kiwigirl
    A curve is defined by the parametric equations x = tē and y = 2t + 1
    Find the equation of the tangent at the point where t = 2.
    Hello,

    as you may know the derivation of parametric(?) functions is done with
    \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}

    So you need the coordinates of the tangent point, which is (4, 5) and the slope, which you calculate with the formula given above:
    \frac{dy}{dx}=\frac{2}{2t} . With t = 2 you get a slope of 1/2.
    Now use the point slope formula of a line:
    \frac{y-5}{x-4}=\frac{1}{2} \Longleftrightarrow y=\frac{1}{2} x +3

    Greetings

    EB
    Last edited by earboth; April 9th 2006 at 09:26 PM. Reason: mistake
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  4. #4
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    Wow, thank you both for your incredible help, especially for the graph CaptainBlack. I am very impressed
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