# Thread: paramatised curve and self-intersection points.

1. ## paramatised curve and self-intersection points. Need Extra help!

So i have the curve:

$\displaystyle \gamma(t) =(t^2-t+1, t^2-t)$

And i need to show that it has one self-intersection point.

To me, it is obviously at 0 and 1, as at these points $\displaystyle \gamma$ is equal to zero.

Now, the thing is, i think i must have missed this lecture. I can't find any reference in my differential geometry book, and my google fu is failing me.

Can anyone show me the method on how to show self-intersection points?
I'll solve the question once i know how.

Thus far, i've been unable to figure out my own way. These damn parametric equations upset me so.

so far i've though maybe if
$\displaystyle \gamma(t) = \gamma(u)$
where
$\displaystyle (t^2-t+1, t^3-t) = (u^2-u+1, u^3-u)$
but yeh.

Halp?

2. Hello, paronga!

Your idea is a good one!

So i have the curve: .$\displaystyle \gamma(t) \:=\:(t^2-t+1, t^2-t)$

And i need to show that it has one self-intersection point.

So far, i've thought maybe if : .$\displaystyle \gamma(t) = \gamma(u)$

where: .$\displaystyle (t^2-t+1, t^3-t) \:=\: (u^2-u+1, u^3-u)$ . Yes!

So we have: .$\displaystyle \begin{array}{ccccc}t^2-t+1 &=& u^2 - u + 1 & [1] \\ t^3 - t &=& u^3 - u & [2] \end{array}$

From [1], we have: .$\displaystyle t^2-u^2 - (t - u) \:=\:0$

. . $\displaystyle (t-u)(t+u) - (t - u) \;=\;0 \quad\Rightarrow\quad (t-u)(t+u-1) \:=\:0$

We have: .$\displaystyle t-u \:=\:0 \quad\Rightarrow\quad \rlap{//////}t \:=\:u$
. . . . .$\displaystyle t + u - 1 \:=\:0 \quad\Rightarrow\quad u \:=\:1 - t\;\;[3]$

From [2], we have: .$\displaystyle t^3 - u^3 -t+u\:=\:0$

. . $\displaystyle (t-u)(t^2+tu+u^2) - (t-u)\:=\:0$

. . $\displaystyle (t-u)(t^2+tu+u^2-1) \:=\:0$

. . $\displaystyle t^2 + tu + u^2-1 \:=\:0\;\;[4]$

Substitute [3] into [4]: .$\displaystyle t^2 + t(1-t) + (1-t)^2-1 \:=\:0$

. . which simplifies to: .$\displaystyle t^2-t\:=\:0$

. . . . . . . . . . . . . . . $\displaystyle t(t-1)\:=\:0 \quad\Rightarrow\quad t \:=\:0,\,1$

It turns out that: .$\displaystyle \gamma(0) \:=\:\gamma(1) \:=\:(1,0)$

Therefore, there is one self-intersection point.

3. huuzar!!
I see where i was going wrong.

I thought it was something like that but my simplification was wrong.
I thank you very much.

I did have another question. Maybe i should just post it in this thread instead of making a new one?
It's on the same problem set for the same subject. My weekend if just me doing problems

It's to do with showing a vector function in planar. Once again, i can do it say visually.

Anyway, i've got the unit tangent vector to the curve:

$\displaystyle \gamma(t) = \frac{1+t^2}{t} , t+1 , \frac{1-t}{t}$

with

$\displaystyle \gamma\prime (t) = \frac{t^2-1}{t^2} , 1 , \frac{1}{t^2}$

Which means the unit tangent vector is

$\displaystyle T(t) = \frac{\gamma\prime(t)}{|\gamma\prime(t)|}$

where (for simplification)

$\displaystyle |\gamma\prime(t)| = \frac{\sqrt{2t^4-2t^2+2}}{t^2}$

finally

$\displaystyle T(t) = \frac{t^2-1}{|\gamma\prime(t)|}, \frac{t^2}{|\gamma\prime(t)|} , \frac{1}{|\gamma\prime(t)|}$

So now i have this vector, which i believe to be normal to the whole original curve. how do i show that this is in a plane? or is also normal to a plane. Hmm

The original question says to show that $\displaystyle \gamma(t)$ is planar.
Perhaps i'm on the wrong path?
Any halp?