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Math Help - Help with finding Cartesian Equations from Parametric Equations using Chain Rule

  1. #1
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    Help with finding Cartesian Equations from Parametric Equations using Chain Rule

    Hi!

    This is my first post here

    I have encountered some trouble in my math homework.

    A curve is defined parametrically by the equations:
    x= t^3 - 6t + 4 and y= t - 3 + 2/t

    Find:
    i. the equations of the normals to the curve at the points where the curve meets the x-axis.
    ii. the coordinates of their point of intersection.

    I am simply stumped...
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  2. #2
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    Hello, macho_maggot2001!

    Welcome aboard!


    A curve is defined parametrically by the equations: . \begin{array}{ccc}x & = & t^3 - 6t + 4\\<br />
y & = & t - 3 + \frac{2}{t}\end{array}
    Find:

    1) the equations of the normals to the curve at the points
    . . .where the curve meets the x-axis.

    2) the coordinates of their point of intersection.

    1) The curve meets the x-axis when y = 0
    . . t - 3 + \frac{2}{t}\:=\:0\quad\Rightarrow\quad t^2-3t + 2 \:=\:0\quad\Rightarrow\quad (t-1)(t-2) \:=\:0\quad\Rightarrow\quad t\,=\,1,\,2

    Then: . x\,=\,-1,\,0
    . . The x-intercepts are: . (-1,\,0),\;(0,\,0)


    Then we need the derivative, \frac{dy}{dx}, to find the slope of the tangents.

    We have: . \begin{Bmatrix}\frac{dy}{dt} & = & 1 - \frac{2}{t^2} \\ \\<br />
\frac{dx}{dt} & = & 3t^2-6\end{Bmatrix}\quad\Rightarrow\quad<br />
\frac{dy}{dx}\;=\;\frac{1-\frac{2}{t^2}}{3t^2-6} \;=\;\frac{t^2-2}{t^2(3t^2-6)}


    When t = 1\!:\;\frac{dy}{dx} \:=\:\frac{1}{3}
    When t = 2\!:\;\frac{dy}{dx} \:=\:\frac{1}{12}


    One normal has point (-1,0) and m = -3
    . . Its equation is: . y - 0 \:=\:-3(x + 1)\quad\Rightarrow\quad\bf{{\color{blue} y \:=\:-3x-3}}

    The other has point (0,0) and , = -12
    . . Its equation is: . y - 0 \:=\:-12(x-0)\quad\Rightarrow\quad\bf{{\color{blue} y \:=\:-12x}}


    2) They intersect when: . -3x - 3 \:=\:-12x\quad\Rightarrow\quad9x\:=\:3

    . . . x \,=\,\frac{1}{3}\quad\Rightarrow\quad y \,=\,-4

    Their intersection is: . \bf{{\color{blue}\left(\frac{1}{3},\,-4\right)}}

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