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Parameterization question

nacho

Active member
Sep 10, 2013
156
Hi,

Just wanted to clarify with parameterising.

So for a straight line, connected from the point $z_1$ to $z_2$
We can use the formula

$ (1-t)z_{1} + z_{2}t$ or
$ z_{1} + t[(z_1 - z_2)]$

I am just wondering, does it matter which end points you allocate to be z1 and z2?

if so, in which manner do we go about assigning the points?

So if you could look at the attached image, for the line $y_3$
could someone please give me what the correct $z_1$ and $z_2$ would be? I've tried both, but can't seem to get the answer.
I did also notice that the direction of the line is going from right to left, so what does that change for us, is it a negative direction?

I am confused why the parameter is ranging from -2<t< 0, although i suspect it may be because the arrow is pointing from right to left.

Any help is very much appreciated.
 

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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: parameterization question

Hi,

Just wanted to clarify with parameterising.

So for a straight line, connected from the point $z_1$ to $z_2$
We can use the formula

$ (1-t)z_{1} + z_{2}t$ or
$ z_{1} + t[(z_1 - z_2)]$

I am just wondering, does it matter which end points you allocate to be z1 and z2?
Yes, because the curve has a direction it matters which direction you choose actually the two curves will differ by a negative sign.

If you let the following \(\displaystyle \gamma(t) = (1-t)z_1 +z_2 t = z_1+t(z_2-z_1)\)

if you plug \(\displaystyle t=0\) in the first equation you get \(\displaystyle \gamma(0) = z_1\) so the starting point is \(\displaystyle z_1\) and if you plug \(\displaystyle t=1\) you get \(\displaystyle \gamma(1) = z_2\) so the final point is $z_2$ hence the parmatrization of the curve is

\(\displaystyle \int_{\gamma}f(z)\, dz = \int^1_0 f(\gamma(t)) \gamma'(t)\, dt\)


if so, in which manner do we go about assigning the points?
For straight lines it is quite easy $z_1$ is the starting point and $z_2$ is the final point, most of the time it is by trial and error so test it first .

So if you could look at the attached image, for the line $y_3$
could someone please give me what the correct $z_1$ and $z_2$ would be? I've tried both, but can't seem to get the answer.
I did also notice that the direction of the line is going from right to left, so what does that change for us, is it a negative direction?

I am confused why the parameter is ranging from -2<t< 0, although i suspect it may be because the arrow is pointing from right to left.

Any help is very much appreciated.
I understand your confusion because parametrization is not unique so you can get many parametrizations for the same curve but they are actually the same.

In \(\displaystyle \gamma_3\) what is the starting point $z_1$ and what is $z_2$ ? and just plug them in the formula above.

To check that the formula the book used is correct note that $z_3(t) = -t(1+i)$ so we must have $z_3(-2) = 2+2i $ and $z_3(0) =0$ which is indeed correct. since the paratmerization represents the right coordinates on the curve. But that is not enough we have to prove that \(\displaystyle z_3\) is an equation of a line and since a line between two point is unique the parametrization is correct.

Note that another valid parametrizations is $z_3(t) = t(1+i)$ where $2 \leq t \leq 0$ since the starting point and the end points coincide with the curve it is correct.
 
Last edited:

nacho

Active member
Sep 10, 2013
156
Re: parameterization question

Thanks a tonne zaid, that clears up a lot.

i also just remembered that i forgot paramterizations need not be unique
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: parameterization question

Thanks a tonne zaid, that clears up a lot.

i also just remembered that i forgot paramterizations need not be unique
Note that direction IS important. Regardless of what parametrization you use it should EXACTLY represents the same curve with the correct direction.