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Thread: Parametric Equation Continuity;

  1. #1
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    Parametric Equation Continuity;

    Hey I got a question here involving solving for values of 3 variables that would make a parametric function contuous over a given interval. I have come up with an answer, but I want to make sure my rationale (/answer) is correct!

    The question is as follows;

    Find all values for $\displaystyle a, b, c$ such that
    $\displaystyle (x(t),y(t))=\left\{\begin{array}{cc}(at^2+bt+c),(( 1-t)^2)&\mbox{ if }
    0 \leq t < 1\\(t^2,t^2-b) & \mbox{ if } 1 \leq t < 2\end{array}\right. $
    is a continuous curve on $\displaystyle [0,2)$.
    Off the top, the parametric function will be continuous at $\displaystyle t=1$ where the first function $\displaystyle x(t)$ over $\displaystyle 0 \leq t < 1$ is equal to the second function $\displaystyle x(t)$ over $\displaystyle 1 \leq t < 2$, right? Ditto for the $\displaystyle y(t)'s$.

    Using this I did the following;

    $\displaystyle at^2 + bt +c = t^2$
    Which gives us three cases,
    $\displaystyle 1) a=1, b=0, c=0$
    $\displaystyle 2) a=1, b=-1, c=-1$
    $\displaystyle 3) a=1, b=1, c=-1$

    Next;

    $\displaystyle ((1-t)^2) = t^2 - b$

    $\displaystyle 1-2t-t^2=t^2 - b$

    $\displaystyle 1-2t=-b$

    $\displaystyle b=2t-1 $ Substituting $\displaystyle t=1$ (Junction point)

    $\displaystyle b=1$

    Which shows that Case 3 is the correct case, and the values of a, b, and c that make $\displaystyle (x(t),y(t))$ continuous over $\displaystyle [0,2)$ are;
    $\displaystyle a = 1$
    $\displaystyle b = 1$
    $\displaystyle c = -1$

    Does this look on the up and up? Thanks.
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  2. #2
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    Find all values for such that

    is a continuous curve on .
    $\displaystyle \lim_{t \to 1^-} (at^2+bt+c \, , \, (1-t)^2) = (a+b+c \, , \, 0)
    $

    $\displaystyle \lim_{t \to 1^+} (t^2 \, , \, t^2 - b) = (1 \, , \, 1-b)$

    $\displaystyle 1 - b = 0$ ... $\displaystyle b = 1$

    $\displaystyle a+1+c = 1$

    $\displaystyle a+c = 0$

    $\displaystyle a = -c$
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  3. #3
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    Beauty, I gotta remember to look for all values rather than find a valid case and stick to it. Thanks Skeeter.
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