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Math Help - Motion in the xy-plane

  1. #1
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    Motion in the xy-plane

    I need help with this problem.. it asks me for the accleration and velocity and i already know to find those i need to take the first and second derivatives. but the part i don't get is "Find an equation in x and y whose graph is the path of the particle."

    r(t) = (t+1)i + ((t^2)-1)j
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by aeubz View Post
    I need help with this problem.. it asks me for the accleration and velocity and i already know to find those i need to take the first and second derivatives. but the part i don't get is "Find an equation in x and y whose graph is the path of the particle."

    r(t) = (t+1)i + ((t^2)-1)j
    The x component of r(t) would be x(t), so x(t) = t + 1. Similarly y(t) = t^2 - 1. This is a parametric set of equations describing the curve. Do you know how to eliminate the parameter?

    -Dan
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    Hello, aeubz!

    Find an equation in x and y whose graph is the path of the particle:
    . . . r(t) \;= \;(t+1)i + (t^2-1)j

    We have parametric equations: . \begin{Bmatrix}x &=& t + 1 & {\color{blue}[1]}\\ y &=& t^2-1 & {\color{blue}[2]} \end{Bmatrix}
    Eliminate the parameter.


    Solve [1] for t\!:\;\;t \:=\:x-1

    Substitute into [2]: . x-1)^2 + 1" alt="y \:=\x-1)^2 + 1" />


    Therefore: . y \;=\;x^2 - 2x + 2

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    thanks! you mean y = x^2 - 2x , but thank you for the help!
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by aeubz View Post
    thanks! you mean y = x^2 - 2x , but thank you for the help!
    No he didn't.
    (x - 1)^2 + 1 = (x^2 - 2x + 1) + 1 = x^2 - 2x + 2

    -Dan
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