How to tell if two vector functions are NOT different parametrizations

[Dmsan]

https://ibb.co/nq1VRz

I know that they are the same but I can't show it
x1=sin(t)
y1=cos(2t)
z1=sin^2(t)

x2=t
y2=1-2t^2
z2=t^2

If somehow t=sin(t), then the two vector functions are the same.
y1 and y2 are related by cos(2t)=1-2sin^2(t)

 

[Dr.Peterson]

I know that they are the same but I can't show it
x1=sin(t)
y1=cos(2t)
z1=sin^2(t)

x2=t
y2=1-2t^2
z2=t^2

If somehow t=sin(t), then the two vector functions are the same.
y1 and y2 are related by cos(2t)=1-2sin^2(t)

The problem asks you to show that these are NOT different parametrizations of the same curve.

Yes, you can express every point of

x1=sin(s)
y1=cos(2s)
z1=sin^2(s)

as a point of

x2=t
y2=1-2t^2
z2=t^2

by letting t = sin(s).

But can you express every point of the second curve as a point of the first? That is, given any t, can you find a corresponding s?

 

[HallsofIvy]

https://ibb.co/nq1VRz

I know that they are the same but I can't show it
x1=sin(t)
y1=cos(2t)
z1=sin^2(t)

x2=t
y2=1-2t^2
z2=t^2

If somehow t=sin(t), then the two vector functions are the same.
y1 and y2 are related by cos(2t)=1-2sin^2(t)

t= sin(t) makes no sense! The difficulty is that you are using the same letter to represent two different parameters.

Instead write the parametric equations as
x1= sin(u)
x2= cos(2u)
x3= sin^2(u)

and

x2= v
y2= 2- 2v^2
z2= v^2


Certainly, if you set v= sin(u) then x2= v= sin(u)= x1 and x3= sin^2(u)= v^2 but what about y1 and y2? y2= 2- 2v^2= 2(1- v^2)= 2(1- sin^2(u))= 2cos^2(u). But that is NOT cos(2u).

(And a kind of obvious point is that the equations for the first point are all periodic while the equations for the second point are not.)

 

[Dmsan]

The problem asks you to show that these are NOT different parametrizations of the same curve.

Yes, you can express every point of

x1=sin(s)
y1=cos(2s)
z1=sin^2(s)

as a point of

x2=t
y2=1-2t^2
z2=t^2

by letting t = sin(s).

But can you express every point of the second curve as a point of the first? That is, given any t, can you find a corresponding s?

t= sin(t) makes no sense! The difficulty is that you are using the same letter to represent two different parameters.

Instead write the parametric equations as
x1= sin(u)
x2= cos(2u)
x3= sin^2(u)

and

x2= v
y2= 2- 2v^2
z2= v^2


Certainly, if you set v= sin(u) then x2= v= sin(u)= x1 and x3= sin^2(u)= v^2 but what about y1 and y2? y2= 2- 2v^2= 2(1- v^2)= 2(1- sin^2(u))= 2cos^2(u). But that is NOT cos(2u).

(And a kind of obvious point is that the equations for the first point are all periodic while the equations for the second point are not.)

Ok so I don't know what I'm doing after all or how to proceed.
Doesn't two vector function being the same parametrization of the same curve means that they are similar to each others when substitutions are made? Since this is not the case for this problem, does that mean that they are different parametrizations of the same curve?

 

[pka]

Ok so I don't know what I'm doing after all or how to proceed.
Doesn't two vector function being the same parametrization of the same curve means that they are similar to each others when substitutions are made? Since this is not the case for this problem, does that mean that they are different parametrizations of the same curve?

The phrase same curve is rather self evident: are they point by point the same?

Did you take note of Prof. Ivy's comment, "(And a kind of obvious point is that the equations for the first point are all periodic while the equations for the second point are not.)"? Did you also consider looking at the derivatives of each curve?

 

[Dmsan]

The phrase same curve is rather self evident: are they point by point the same?

Did you take note of Prof. Ivy's comment, "(And a kind of obvious point is that the equations for the first point are all periodic while the equations for the second point are not.)"? Did you also consider looking at the derivatives of each curve?

Perhaps I am not understanding a fundamental concept behind what the question is asking. The concept of 3d parameterization is still new to me and I didn't follow what the professor was teaching during class.
I think that they are not point by point the same since the equation for the first point is periodic while the second one is not.
As for derivatives, I would have no idea how to use the information from finding the derivatives of each curves.

How would I go to prove that they are indeed not different parameterizations of the same curve?

 

[HallsofIvy]

Ok so I don't know what I'm doing after all or how to proceed.
Doesn't two vector function being the same parametrization of the same curve means that they are similar to each others when substitutions are made? Since this is not the case for this problem, does that mean that they are different parametrizations of the same curve?

You are starting to get me confused now! (I'll admit that's not hard to do, Half the time I confuse myself!) The "same" parameterization would mean exactly the same functions, not "similar". "Different parameterizations of the same curve" mean different formulas that describe the same curve: x= cos(t), y= sin(t), for \(\displaystyle 0\le x\le \pi\) and \(\displaystyle x= t\), \(\displaystyle y= \sqrt{1- t^2}\), for \(\displaystyle -1\le t\le 1\) are very different parameterizations for the same curve- the upper semicircle with center at (0, 0) and radius 1.

A simpler example (wish I thought of it first) would be x= t, y= t, z= t and \(\displaystyle x= t^3\), \(\displaystyle y= t^3\), \(\displaystyle z= t^3\) which are both parameterizations of the three dimensional line x= y= z.

 

[Dr.Peterson]

Perhaps I am not understanding a fundamental concept behind what the question is asking. The concept of 3d parameterization is still new to me and I didn't follow what the professor was teaching during class.
I think that they are not point by point the same since the equation for the first point is periodic while the second one is not.
As for derivatives, I would have no idea how to use the information from finding the derivatives of each curves.

How would I go to prove that they are indeed not different parameterizations of the same curve?

Read carefully what I wrote before. Your work (after changing one of the parameters to s or u to avoid confusion) showed that points satisfying one set of equations will satisfy the other, which is part of showing that they are the same curve. What make them not the same curve is a subtle detail: one is only part of the curve represented by the other. Answer the question I asked, and think about the implications. What values can x1 have? How about x2?

 

[pka]

How would I go to prove that they are indeed not different parameterizations of the same curve?

\(\displaystyle \mathcal{C}_1(t): <\sin(t),\cos(2t),\sin^2(t)>~\&~\mathcal{C}_2(u):<u,1-2u,u^2>\).
It clear to you that \(\displaystyle \mathcal{C}_2(3)=<3,-5,9>?\) Can \(\displaystyle \mathcal{C}_1(t)=<3,-5,9>~?\)

 

[HallsofIvy]

Very good!