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Polar equation to rectangular equation

  • Thread starter Poetria
  • Start date
  • #1
183
4

Homework Statement


[/B]
a - a fixed non-zero real number

r=e^(a*theta), where -pi/2<theta<pi/2


2. The attempt at a solution

r^2=(e^(a*theta))^2
x^2 + y^2 = e^(2*a*theta)
ln(x^2 + y^2) = 2*a*theta
ln(x^2 + y^2) = 2*a*(pi+arctan(y/x))

Is this OK?

 

Answers and Replies

  • #2
kuruman
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Are you trying to convert r(θ) into y(x)? Your title seems to indicate otherwise.

Mod note: I changed the title to reflect what the problem is asking.
 
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  • #3
183
4
Are you trying to convert r(θ) into y(x)? Your title seems to indicate otherwise.
Oh dear, of course. There were problems the other way round too.
 
  • #4
kuruman
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Is this OK?
It is not OK because you have not found y as a function of x. Both x and y appear on both sides of the equation. Is this the question that you are asked to solve or is it part of a solution that you reached following the algebra and you don't know how to continue?
 
  • #5
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It is not OK because you have not found y as a function of x. Both x and y appear on both sides of the equation. Is this the question that you are asked to solve or is it part of a solution that you reached following the algebra and you don't know how to continue?
I see. This is a multi-option problem (there are several options given and I have to choose). I have tried to do some algebra to arrive at the correct one. Well, in every option there are x and y on both sides. :(

e.g. ln(x^2+y^2)=a*tan(y/x).
 
  • #6
183
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I see. This is a multi-option problem (there are several options given and I have to choose). I have tried to do some algebra to arrive at the correct one. Well, in every option there are x and y on both sides. :(

e.g. ln(x^2+y^2)=a*tan(y/x).
I could transform it of course, but there is no option with y on one side.
 
  • #7
33,075
4,776

Homework Statement


[/B]
a - a fixed non-zero real number

r=e^(a*theta), where -pi/2<theta<pi/2


2. The attempt at a solution

r^2=(e^(a*theta))^2
x^2 + y^2 = e^(2*a*theta)
ln(x^2 + y^2) = 2*a*theta
ln(x^2 + y^2) = 2*a*(pi+arctan(y/x))

Is this OK?
The next-to-last equation looks fine to me. Why did you add ##\pi## in the last equation? After all, since ##\frac y x = \tan(\theta)##, then ##\theta = \tan^{-1}(\frac y x) = \arctan(\frac y x)##.

Also, the problem might not require an equation with y explicitly in terms of x. In your equation, the relationship between x and y is an implicit one.
 
  • #8
183
4
The next-to-last equation looks fine to me. Why did you add ##\pi## in the last equation? After all, since ##\frac y x = \tan(\theta)##, then ##\theta = \tan^{-1}(\frac y x) = \arctan(\frac y x)##.

Also, the problem might not require an equation with y explicitly in terms of x. In your equation, the relationship between x and y is an implicit one.
I have added pi because -pi/2<theta<pi/2 -
oh wait I think I am wrong.

Yeah, this problem does not require y explicitly in terms of x.
 
  • #9
33,075
4,776
I have added pi because -pi/2<theta<pi/2 -
This just specifies the interval for ##\theta##, which determines endpoints for the graph (which by the way is some sort of spiral).

oh wait I think I am wrong.
Yes -- you shouldn't add the ##\pi## term.
 
  • #10
183
4
This just specifies the interval for ##\theta##, which determines endpoints for the graph (which by the way is some sort of spiral).

Yes -- you shouldn't add the ##\pi## term.
Many thanks. I got it. :)
 
  • #11
33,075
4,776
  • #12
183
4
@Poetria, take a look at our tutorial in using LaTeX to format equations and such -- https://www.physicsforums.com/help/latexhelp/. In just a few minutes you could go from writing this -- ln(x^2 + y^2) = 2*a*theta

to this -- ##\ln(x^2 + y^2) = 2a\theta##
Oh yes, I will. My way of writing is difficult to read.
Thank you so much for the link. :)
 

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