Find the points of self-intersection for the curve x(t) = t^2-2t, y(t) = t^3 - 3t^2 + 2t, t∈ℝ and
then find the area of the loop.

Question

Find the points of self-intersection for the curve x(t) = t^2-2t, y(t) = t^3 - 3t^2 + 2t, t∈ℝ and
then find the area of the loop.

anonymous · Answer

How does a curve exactly self intersect? >.> .

anonymous · Answer

Well, neither one will intersect with itself, but x(t) will intersect y(t). The system is a loop if you're talking about a parametric equation.

anonymous · Answer

Yeah, it's parametric.

anonymous · Answer

How would I exactly find this point? >.> .

anonymous · Answer

Whenever you have $x(t_1) = x(t_2)$

anonymous · Answer

$y(t_1) = y(t_2)$

anonymous · Answer

Umm... Okay, Just any random value of t should work right?

anonymous · Answer

It's two values of $t$ though...

anonymous · Answer

I am thinking of setting them equal to each other?

anonymous · Answer

THis is a tricky question.

anonymous · Answer

One thing you can do is just try graphing it.

anonymous · Answer

I would normally say that you should try to isolate 't' and substitute into the other equation. However, this does not happen nicely in this case.

anonymous · Answer

Make a table of $t$, $x$, and $y$ values.

anonymous · Answer

@qpHalcy0n If you know a parametrized curve self-intersects, you know it's going to be hard to de-parametrize.

anonymous · Answer

There isn't an elegant solution for this.

anonymous · Answer

I mean there isn't an elegant solution that I know of for solving self intersection among parametrized curves that I know of.

anonymous · Answer

I have to do the entire questione by hand :( .

anonymous · Answer

|dw:1364967099507:dw|

anonymous · Answer

Try filling this out a bit and see what happens.

anonymous · Answer

Ew... Alright. I just need a visual of this I guess. No harm in plotting then right? >.> .

anonymous · Answer

Okay, how about putting $x(t)$ in vertex form?

anonymous · Answer

Completing the square?

anonymous · Answer

Yes, if it is in vertex form then you know the line of symmetry.

anonymous · Answer

And you can find a general equation for $t_2$ given $t_1$ such that $x(t_1)=x(t_2)$

anonymous · Answer

Well if I complete the square I get x(t) = (t-1)^2 - 1

anonymous · Answer

Okay so the vertex is at $(1, -1)$

anonymous · Answer

Don't you mean (-1,0) ?

anonymous · Answer

Why?

anonymous · Answer

No never mind. My mistake.

anonymous · Answer

Anyway, basically if I give you $t_1$ how would you get $t_2$?
You reflect it over the axis of symmetry.

anonymous · Answer

Here is a visual:

|dw:1364967834035:dw|