polar-parametric transformation

[Moore1879]

Is it possible to transform a parametric "equation" into a polar equation? If so how would I go about it?

Thanks for reading.

[dextercioby]

Is it possible to transform a parametric "equation" into a polar equation? If so how would I go about it?
Thanks for reading.

If the fomulae relating various coordinate systems are diffemorphisms,then why not??Bring an example.A (plane) curve in parametric coordinates.And tell us what u gave to do to express it in (plane) polar coordinates.

Daniel.

[Moore1879]

If I'm given the parametric equations for a hypocycloid:
x(t)=(a/n)[(n-1)cos(t)-cos[(n-1)t]
y(t)-(a/n)[(n-1)sin(t)+sin[(n-1)t]

how would I go about putting it into a function form [tex]r(\theta)[\tex].
There has to be some way to do it. What would it be?

Thanks

[Moore1879]

If I'm given the parametric equations for a hypocycloid:
x(t)=(a/n)[(n-1)cos(t)-cos[(n-1)t]
y(t)-(a/n)[(n-1)sin(t)+sin[(n-1)t]

how would I go about putting it into a function form r(\theta).
There has to be some way to do it. What would it be?

Thanks

[dextercioby]

That's a horrible curve... :yuck: Anyway'ill let u do the calculations of eliminating the parameter. :biggrin:
\rho (t)=\sqrt{x^{2}(t)+y^{2}(t)}
\theta(t)=\arctan({\frac{y(t)}{x(t)}})

Express t(\theta) and plug it into \rho(t) .

Daniel.

PS.My advice:GIVE UP!!It's enough to know that it's possible. :wink:

[Moore1879]

Thanks Kurt? I assume that is your name. That is all I needed. Oh, and I'm not going to give it up. :wink:

[dextercioby]

Thanks Kurt? I assume that is your name. That is all I needed. Oh, and I'm not going to give it up. :wink:

:rofl: :rofl: :rofl: :rofl: :rofl: My name is Daniel.I write it all the time.
That is a "signature".It's edited from the "USER CP" box.Kurt Lewin was a theorist and i loved his idea and decided to quote him.

Daniel.Really,no bull****.

[arildno]

It's much easier to find t(\rho) than t(\theta)
Then, you might invert \theta(\rho) into \rho(\theta)
the inversion is practically impossible to perform, so I concur with Daniel's advice.