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

    Using this I did the following;

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

    Next;

    ((1-t)^2) = t^2 - b

    1-2t-t^2=t^2 - b

    1-2t=-b

    b=2t-1 Substituting t=1 (Junction point)

    b=1

    Which shows that Case 3 is the correct case, and the values of a, b, and c that make (x(t),y(t)) continuous over [0,2) are;
    a = 1
    b = 1
    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 .
    \lim_{t \to 1^-} (at^2+bt+c \, , \, (1-t)^2) = (a+b+c \, , \, 0)<br />

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

    1 - b = 0 ... b = 1

    a+1+c = 1

    a+c = 0

    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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