Results 1 to 4 of 4

Thread: Help!

  1. #1
    Junior Member
    Joined
    Feb 2006
    From
    New Zealand
    Posts
    26

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    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:

    $\displaystyle y=mx+c$.

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

    $\displaystyle
    \frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=2\times \frac{1}{2t}=1/t
    $

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

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

    $\displaystyle y=x/2+c$,

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

    Hence:

    $\displaystyle
    5=4/2 +c
    $,

    so:

    $\displaystyle
    c=3
    $,

    and the equation of the tangent is:

    $\displaystyle y=2x-3$.

    RonL
    Attached Thumbnails Attached Thumbnails Help!-parametric.jpg  
    Last edited by CaptainBlack; Apr 9th 2006 at 09:21 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,854
    Thanks
    138
    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
    $\displaystyle \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:
    $\displaystyle \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:
    $\displaystyle \frac{y-5}{x-4}=\frac{1}{2} \Longleftrightarrow y=\frac{1}{2} x +3$

    Greetings

    EB
    Last edited by earboth; Apr 9th 2006 at 09:26 PM. Reason: mistake
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2006
    From
    New Zealand
    Posts
    26
    Wow, thank you both for your incredible help, especially for the graph CaptainBlack. I am very impressed
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum