Parameterization for a Straight Line

[nacho-man]

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.

 

[alyafey22]

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.

 

[nacho-man]

Re: parameterization question

Thanks a tonne zaid, that clears up a lot.

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

 

[alyafey22]

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.

 

[blue_raver22]

Hello,

Thank you for reaching out with your question about parameterization. Parameterization is a mathematical technique used to represent a curve or surface in terms of one or more variables called parameters. In the context of your question, it is used to represent a straight line between two points, $z_1$ and $z_2$, using a parameter, $t$.

To answer your first question, it does not matter which end points are assigned to $z_1$ and $z_2$. As long as the parameterization equation is correctly set up, the line will be represented accurately. In the case of the line $y_3$ in the attached image, both $z_1$ and $z_2$ can be assigned to either end point of the line.

To find the correct values for $z_1$ and $z_2$, you can use either of the parameterization equations provided in your question. Using the first equation, $z_1$ can be assigned to the point (1, 0) and $z_2$ can be assigned to the point (3, 2). Using the second equation, $z_1$ can be assigned to the point (3, 2) and $z_2$ can be assigned to the point (1, 0). Both of these assignments will give you the correct parameterization for the line $y_3$.

Regarding the direction of the line, it does not change the parameterization equation. The direction of the line is represented by the value of the parameter, $t$. In this case, since the arrow is pointing from right to left, the parameter is ranging from -2<t<0. If the arrow was pointing from left to right, the parameter would range from 0<t<2. This is because the parameter represents the distance along the line from $z_1$ to $z_2$, and in this case, the direction is opposite to the usual positive direction.

I hope this helps to clarify your confusion about parameterization. If you have any further questions, please don't hesitate to ask.

Best,