Coordinates of a particle (coupled differential equations)

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Homework Help Overview

The problem involves two coupled differential equations describing the motion of a particle in the xy plane, specifically \(\dot{x} + \omega y = 0\) and \(\dot{y} - \omega x = 0\). The task is to express the complex variable \(z = x + iy\) and find its differential, ultimately leading to expressions for \(x\) and \(y\) as functions of time.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the complex variable \(z\) and the original equations, with attempts to express \(\dot{z}\) in terms of \(x\) and \(y\). Questions arise about the derivation of the equation \(\dot{z} - i\omega z = 0\) and the implications of substituting the original equations into this form.

Discussion Status

Some participants have made progress in relating \(\dot{z}\) to the original equations, while others express confusion about the steps involved. There is an ongoing exploration of how to incorporate the phase angle in the solution, with guidance provided regarding the nature of the integration constant.

Contextual Notes

Participants note the challenge of integrating the equations without prior experience in differentials, and there is mention of assumptions regarding the form of the integration constant and its implications for the solution.

Linday12
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Homework Statement


I'm given two equations for coordinates of a certain particle in the xy plane, \dot{x}+ωy=0
and \dot{y}-ωx=0.
Then using the complex variable z=x+iy, find the differential for z, and solve it. Hence give x and y as functions of time.


Homework Equations




The Attempt at a Solution


I'm not sure how this is gotten:
\dot{z}-iωz=0

Any help would be highly appreciated. Thank you!
 
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Hi Linday12! :smile:

Can you write \dot z in terms of x and y?
And then substitute the equations you have?

Can you also write iωz in terms of x and y?
 
Hi. I'm not really sure what to do. As far as I know, taking \dot{z}=\dot{x}+i\dot{y}, and \dot{y}=ωx and \dot{x}=-ωy, which doesn't seem to get me anywhere. I have no idea what I am doing.
 
Linday12 said:
Hi. I'm not really sure what to do. As far as I know, taking \dot{z}=\dot{x}+i\dot{y}, and \dot{y}=ωx and \dot{x}=-ωy, which doesn't seem to get me anywhere. I have no idea what I am doing.

So substitute your 2nd and 3rd equation in the first?
 
Yes. I then get \dot{z}=-ωy+iωx=ω(-y+ix). So I can sort of see the relation here now, except my variables seem to be the wrong way, and I have no idea where the \dot{z}-iωz=o comes from still, because i would be gone if the variables were the proper way and I subbed in z for the x-iy.
 
What is iz in terms of x and y?
 
Wow, interesting. So \dot{z}=iωz → \dot{z}-iωz=0 because iz=ix-y. Thank you! Sorry, one more question. Now that I have that, when solving it I get z=z_{0}e^{ωt}, but the answer has an additional phase angle \phi in it, so I was just wondering if there was a quick explanation for that.

It's been a while since I did any differentials (I haven't had a class in them yet), I think perhaps I'm used to the general formula that includes the phase angle in the constant z_0, but if you were to solve it you would get ln(z)=iωt+c, which is where the phase angle came from, but then where does the z_0 come from? Because the answer is z=z_0e^{ωt+\phi} Sorry, I hope that makes sense.
 
Your z0 is the integration constant that is itself a complex number.
Write z_0=r_0e^{i\phi} and you get:
z=z_0e^{i\omega t}=r_0e^{i\phi}e^{i\omega t}=r_0e^{i(\omega t + \phi)}

Btw, note the extra i and the use of the non-negative real number r0 instead of z0.
 

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