Polar-parametric transformation

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Transforming a parametric equation into a polar equation is possible, and the process involves eliminating the parameter to express the equations in terms of polar coordinates. For a hypocycloid defined by specific parametric equations, one can calculate the radius \( r(t) \) and angle \( \theta(t) \) using \( r(t) = \sqrt{x^2(t) + y^2(t)} \) and \( \theta(t) = \arctan\left(\frac{y(t)}{x(t)}\right) \). The challenge lies in expressing \( t \) as a function of \( \theta \) and subsequently inverting to find \( \rho(\theta) \), which can be complex. While it is feasible to perform these transformations, the inversion process is often impractical. Understanding the possibility of such transformations is crucial, even if the calculations can be daunting.
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Is it possible to transform a parametric "equation" into a polar equation? If so how would I go about it?

Thanks for reading.
 
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Moore1879 said:
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.
 
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)[\tex].<br /> There has to be some way to do it. What would it be?<br /> <br /> Thanks
 
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
 
That's a horrible curve... 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:
 
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Thanks Kurt? I assume that is your name. That is all I needed. Oh, and I'm not going to give it up. :wink:
 
Moore1879 said:
Thanks Kurt? I assume that is your name. That is all I needed. Oh, and I'm not going to give it up. :wink:

:smile: :smile: :smile: :smile: :smile: 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****.
 
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.
 
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