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Sprial Cartesian equation

by Rohitpi
Tags: cartesian, equation, sprial
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Rohitpi
#1
Jan6-13, 06:22 AM
P: 2
I'm volunteering in a summer school for year 12 students in my area, and have to teach them a few topics. I've been struggling to get the parametric equations from this.

Sketch: [tex]|z| = \arg(z)[/tex]

So I thought that the obvious way to explain it to them would be to say: "that as the magnitude of z increases (ie. distance from the origin) the greater the angle becomes, thus producing a spiral" and I can draw it on the whiteboard.

Upon attempting the Cartesian equation, I got a bit stuck:
[tex]|z| = \arg(z)[/tex]
[tex]\sqrt{x^2 + y^2} = \tan^{-1}(\frac{y}{x})[/tex]
[tex]\tan{(\sqrt{x^2 + y^2)}}=\frac{y}{x}[/tex]
[tex]y=x\tan{({\sqrt{x^2 + y^2}})}[/tex]

And that is where I get stuck unfortunately. Any thoughts/solutions on how to proceed in finding a Cartesian and/or parametric equations?

(sorry if this doesn't make sense, I haven't done any maths since year 12 haha as I study med atm :/ I miss maths!)

Thanks :)
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HallsofIvy
#2
Jan6-13, 07:59 AM
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P: 39,327
You titled this "Cartesian equations" and refer to the Cartesian equation in the body but also ask for parametric equations. Which do you want?

You haven't said what parameter you want to use. I would recommend the angle [itex]\theta= arg(z)[/itex]. Then [itex]x= r cos(\theta)[/itex] and [itex]y= rsin(\theta)[/itex] while [itex]r= |z|= arg(z)= \theta[/itex] so the parametric equations are just [itex]x= \theta cos(\theta)[/itex], [itex]y= \theta sin(\theta)[/itex].

As for the Cartesian equation, I think you have it about a simple as you are going to get it.
Rohitpi
#3
Jan6-13, 08:22 AM
P: 2
Ah, thanks! Sorry about the title, I should have included both Parametric and Cartesian. The issue I had with the parametric was as you alluded to, I actually didn't know where to start (hence the query was vague).

Thanks a lot! :)


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