Really basic parametric equations question

**Mathstud28** · June 15th 2008, 12:35 PM

Hi, this is really simple but I just wanted to make sure that since it is so simple that I am not missing something,

Consider the parametric equations

$x=f(\xi)$

and $y=g(\xi)$

where x is of degree $N$ and y of degree $n$

then since math $\xi=f^{-1}(x)$ which I believ would have degree $\frac{1}{N}$

and to get our equation into regular form with y as a dependent and x an independent variable we must take

$y=g(f^{-1}(x))$

will our final function in terms of x and y be of degree $\frac{n}{N}$ ?

**Moo** · June 16th 2008, 12:21 AM

Hello,

Originally Posted by Mathstud28

Hi, this is really simple but I just wanted to make sure that since it is so simple that I am not missing something,

Consider the parametric equations

$x=f(\xi)$

and $y=g(\xi)$

where x is of degree $N$ and y of degree $n$

then since math $\xi=f^{-1}(x)$ which I believ would have degree $\frac{1}{N}$

and to get our equation into regular form with y as a dependent and x an independent variable we must take

$y=g(f^{-1}(x))$

will our final function in terms of x and y be of degree $\frac{n}{N}$ ?

Prove that $t \mapsto f^{-1}(t)$ exists ^^

But there's something disturbing me, because the degree of a polynomial is defined as being "the least integer such that the coefficient is null".

$x=f(\xi)=a_N\xi^N+a_{N-1}\xi^{N-1}+\dots+a_0$ (assuming that x is a polynomial because you didn't mention it).

How would you find $f^{-1}(x)$ ? How can you know it's something like the N-th root of x ?

**Mathstud28** · June 16th 2008, 09:07 AM

Originally Posted by Moo

Hello,

Prove that $t \mapsto f^{-1}(t)$ exists ^^

But there's something disturbing me, because the degree of a polynomial is defined as being "the least integer such that the coefficient is null".

$x=f(\xi)=a_N\xi^N+a_{N-1}\xi^{N-1}+\dots+a_0$ (assuming that x is a polynomial because you didn't mention it).

How would you find $f^{-1}(x)$ ? How can you know it's something like the N-th root of x ?

Thanks for responding Moo, this is not in any way a proof just an observation, i am saying that if you solved a quadratic for x you would get a squareroot, a cubic would yield a cubic root. I know this is informal but I think it works loosely. For example if you are given a few graphs and you need to match the parametric equations to it, you look for y(t) composed of x inverse of t which by my observation( which could totally be wrong) would be the degree of y over the degree of x.

So for example if I was given

$x(t)=at+b$
and

$y(t)=cx^3+dx+e$

I would expect

$y(x)$ to be a cubic function

**Moo** · June 16th 2008, 10:47 AM

So for example if I was given

$x(t)=at+b$
and

$y(t)=cx^3+dx+e$

I would expect

$y(x)$ to be a cubic function

It would be a quartic function, huh ?
But yeah, this is true

Originally Posted by Mathstud28

Thanks for responding Moo, this is not in any way a proof just an observation, i am saying that if you solved a quadratic for x you would get a squareroot, a cubic would yield a cubic root.

While dealing with functions which are defined to be polynomials, yep.
But since you were talking about $f^{-1}$ , it just didn't make sense... for me at least..perhaps it is different for someone else >.<

I know this is informal but I think it works loosely. For example if you are given a few graphs and you need to match the parametric equations to it, you look for y(t) composed of x inverse of t which by my observation( which could totally be wrong) would be the degree of y over the degree of x.

I don't really agree with that... Because it would mean that the inverse function is also a polynomial... Prove it ? =)
Plus, like I said, the degree of a polynomial is an integer (it was quoted from Wikipedia)

Polynomial - Wikipedia, the free encyclopedia << they also give examples which are not polynomials.

**Mathstud28** · June 16th 2008, 01:48 PM

Originally Posted by Moo

It would be a quartic function, huh ?
But yeah, this is true

While dealing with functions which are defined to be polynomials, yep.
But since you were talking about $f^{-1}$ , it just didn't make sense... for me at least..perhaps it is different for someone else >.<

I don't really agree with that... Because it would mean that the inverse function is also a polynomial... Prove it ? =)
Plus, like I said, the degree of a polynomial is an integer (it was quoted from Wikipedia)

Polynomial - Wikipedia, the free encyclopedia << they also give examples which are not polynomials.

I never said the inverse was a polynomial, just an expression of degree 1/N

**Moo** · June 16th 2008, 01:57 PM

Originally Posted by Mathstud28

I never said the inverse was a polynomial, just an expression of degree 1/N

Two things

:
- The term "degree" applies, as far as I know, to polynomials
- The degree has to be an integer (I can't find any certain source for it, I read it in the wikipedia and Mathworld Wolfram doesn't mention it)

**Mathstud28** · June 16th 2008, 05:33 PM

Originally Posted by Moo

Two things

:
- The term "degree" applies, as far as I know, to polynomials
- The degree has to be an integer (I can't find any certain source for it, I read it in the wikipedia and Mathworld Wolfram doesn't mention it)

But, no offense I am not asking about the semantics of polynomial terminology (although oh so interesting as it is, that was not sarcastic), I am interested if my observation has any truth?

**Moo** · June 16th 2008, 11:11 PM

Originally Posted by Mathstud28

But, no offense I am not asking about the semantics of polynomial terminology (although oh so interesting as it is, that was not sarcastic), I am interested if my observation has any truth?

Your observation isn't correct since you don't even know if $f^{-1}$ is a polynomial

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