1. ## 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
$(x(t),y(t))=\left\{\begin{array}{cc}(at^2+bt+c),(( 1-t)^2)&\mbox{ if }
0 \leq t < 1\\(t^2,t^2-b) & \mbox{ if } 1 \leq t < 2\end{array}\right.$

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.

2. 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)
$

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

$1 - b = 0$ ... $b = 1$

$a+1+c = 1$

$a+c = 0$

$a = -c$

3. Beauty, I gotta remember to look for all values rather than find a valid case and stick to it. Thanks Skeeter.