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/22. 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?

 

[kuruman]

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.

 

[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.

 

[kuruman]

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]

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.

 

[Mark44]

Homework Statement

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

r=e^(a*theta), where -pi/2<theta<pi/22. 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]

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.

 

[Mark44]

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]

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. :)

 

[Mark44]

@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 -- $\ln(x^2 + y^2) = 2a\theta$

 

[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. :)