# 5th power perameter for parametric equations

1. Dec 29, 2013

### Lebombo

Given parametric equations:

[g(t)= x = $t^{5}-4t^{3}$] and [h(t) = y = $t^{2}$]

Since polynomials can only be solved up to the 4th degree as I've just learned here on PhysicsForums, I guess it's not possible to isolate t in terms of x in the g(t) function and substitute into the h function to create a new function:

[f(x) = y = ...]

Thus the only way to deal with parametric equations of 5th degree and up is to use calculus ...dy/dx = dy/dt/dx/dt.

Correct?

EDIT: P.S. I think its called eliminating the parameter. I just recalled there is a name that describes what I'm trying to do here.

Last edited: Dec 29, 2013
2. Dec 29, 2013

### Jhenrique

It's depend of how is x(t) and y(x). Sometimes is possible make y(x), other times no.

You can do ∫ydt, dy/dt, ∫xdt, dx/dt and combine these calculus for result dy/dx or ∫ydx, like you already noticed. Detail: calculate the area of a parametrized curve can be more easy that calculate by ∫f(x)dx. For example, for calculate ∫√[x²-1]dx you need make a trigonometric substituition or parametrize x(t)=cosh(t), y(t)=sinh(t) and calculate ∫y(t)·dx/dt·dt.

3. Dec 30, 2013

### HallsofIvy

Staff Emeritus
You cannot solve a generic 5th degree equation for t in terms of x but you can go the other way:
$t= \pm\sqrt{y}$, then either $x= y^{5/2}- 4y^{3/2}$ or $x= -y^{5/2}+ 4y^{3/2}$.

4. Dec 30, 2013

### Lebombo

HallsofIvy,

As I'm discovering in another thread, an x(t) function of 5th degree can be solved for t in terms of x if the function or equation is one-to-one.

And to prove if a function is one-to-one, you have to prove f(a) = f(b).

So my question is, do you know of or have you ever come across an equation of 5th degree that happened to also be one-to-one? Is being 1-1 not a common characteristic of 5th degree equations?

P.S.

I'm also exploring this issue in another thread that started as a separate question but lead to the same line of inquiry: https://www.physicsforums.com/showthread.php?t=730479

Don't want to pit you against the information being provided elsewhere, but it happens to be that the other thread mentions that a 5th degree x(t) polynomial may in fact be solvable for t in terms of x if the function is 1-1.

5. Dec 30, 2013

### jbriggs444

This thing which you say that you have learned is not true.

What is true is that a function that is one-to-one is invertible. That is, it has an inverse.

This does not mean that the inverse can be written down as a finite formula using only elementary functions and operations (plus, minus, times, divide, root extraction, exponentiation and trig functions). Such a formula is called "closed form".

One could generate an approximate solution with a series of calculations that converges on the correct answer. But that does not count as "closed form".

A function is "one to one" if no two different inputs produce the same output.

Functions can be "one to one". The concept does not apply to equations. The fifth degree polynomial function f defined by f(x) = x5 is one to one.

Every polynomial function of any non-zero degree can be split up into ranges and will be one to one over each of the ranges. For instance the polynomial function f defined by f(x) = x^2 is one to one over the range {x: x <= 0} and also over the range {x: x >= 0}.

In general, a n'th degree [where n is non-zero] polynomial function can always be split up into ranges where it will be one to one over each. This will take, at most, n ranges. So a fifth degree polynomial could, in principle be divided into no more than five separate pieces. Each piece would be one-to-one, so each piece would be invertible. But there would still be no way, in general, to write down a formula for the inverse of each piece.

[Local extrema of a polynomial function can only occur where its first derivitive is zero. The first derivitive of an n'th degree polynomial is an n-1'st degree polynomial which can have at most n-1 distinct real roots. These roots partition the real numbers into at most n ranges]