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;

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 $\displaystyle y(t)'s$.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 $\displaystyle [0,2)$.

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.