The following parametric equations trace out a loop.
x=3-(2/2)t^2
y=(-2/6)t^3+2t+2
Find the t values at which the curve intersects itself:
t = +/- .
What is the total area inside the loop?
Area = .

Question

The following parametric equations trace out a loop.
x=3-(2/2)t^2
y=(-2/6)t^3+2t+2
Find the t values at which the curve intersects itself: 
t = +/-   . 
What is the total area inside the loop? 
Area =  .

blockcolder · Answer

First, you need to find $t_1$ and $t_2$ such that $x(t_1)=x(t_2)$ and $t_1\neq t_2$:
A little clarification: is $x= 3-\frac{2}{2}t^2$ or $x=3-\frac{2}{2t^2}$?

blockcolder · Answer

Well, whichever it is, the t-values you're looking for ate $\large t=\pm\frac{\sqrt{6}}{2}$.

anonymous · Answer

i think its 2/2, it is very unclear, So I have to equate the x equation?

blockcolder · Answer

Yeah, so you'll end up with $3-t_1^2=3-t_2^2 \Rightarrow$ $t_1^2=t_2^2 \Rightarrow$ $|t_1|=|t_2| $. WLOG, let $t_2=-t_1$, since we don't want $t_1=t_2$.
Now, you have to equate $y(t_1)=y(t_2)$ and use the fact that t2=-t1 to find out the value of t1. Let me know what you got.

anonymous · Answer

aaahhhhh!!!! I think I got it, after equating both x and finding out that for t_2 to be equal to t_1, one of them has to be the opposite. Then when I equated y(t1)=y(t2) I can just do -(yt2)=(yt2) and then solve for t, getting t=0 and t= sqrt(6). However the one thing I don't understand is how we're allowed to equate both x.