# Polar equation to rectangular equation

Poetria

## 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?

Homework Helper
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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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Poetria
Poetria
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.

Homework Helper
Gold Member
2021 Award
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?

Poetria
Poetria
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).

Poetria
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.

Mentor

## 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.

Poetria
Poetria
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.

Mentor
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.

Poetria
Poetria
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. :)

Mentor
Poetria
Poetria
@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. :)