parametrization question

[logistic_guy]

here is the question

i need to parametrize the curve defined by $\displaystyle y = \frac{x^3 + y^3}{3x}$

i tried $\displaystyle x = t$ then $\displaystyle y = \frac{t^3 + y^3}{3t}$

$\displaystyle y3t = t^3 + y^3$

$\displaystyle y3t - y^3 = t^3$

$\displaystyle y(3t - y^2) = t^3$

can i say $\displaystyle y = t^3$ and $\displaystyle 3t - y^2 = t^3$

this give $\displaystyle y = \sqrt{3t - t^3}$ and $\displaystyle y = -\sqrt{3t - t^3}$

i call the curve $\displaystyle \pmb{\alpha}(t) = (t, t^3)$ or $\displaystyle \pmb{\alpha}(t) = (t, \sqrt{3t - t^3})$ or $\displaystyle \pmb{\alpha}(t) = (t, -\sqrt{3t - t^3})$

 

[Dr.Peterson]

$\displaystyle y(3t - y^2) = t^3$

can i say $\displaystyle y = t^3$ and $\displaystyle 3t - y^2 = t^3$

this give $\displaystyle y = \sqrt{3t - t^3}$ and $\displaystyle y = -\sqrt{3t - t^3}$

i call the curve $\displaystyle \pmb{\alpha}(t) = (t, t^3)$ or $\displaystyle \pmb{\alpha}(t) = (t, \sqrt{3t - t^3})$ or $\displaystyle \pmb{\alpha}(t) = (t, -\sqrt{3t - t^3})$

No, you can't.

Can you tell us why not?

 

[logistic_guy]

No, you can't.

Can you tell us why not?

because i need the curve? not parts of the curve?

 

[BigBeachBanana]

because i need the curve? not parts of the curve?

Review the domain of the function. Which rule does it violate?

 

[Otis]

$\displaystyle y = \frac{x^3 + y^3}{3x}$
$\displaystyle x=t$

In that case, we have $\displaystyle y = \frac{t^3 + y^3}{3t}$

… can i say $\displaystyle y = t^3$

If you change $\displaystyle \frac{t^3 + y^3}{3t}$ to $\displaystyle t^3$, then you change the original curve (it becomes the curve of a cubic polynomial).

i call the curve $\displaystyle \pmb{\alpha}(t) = (t, t^3)$ or $\displaystyle \pmb{\alpha}(t) = (t, \sqrt{3t - t^3})$ or $\displaystyle \pmb{\alpha}(t) = (t, -\sqrt{3t - t^3})$

I'm not sure what you're thinking here, but, if a parameterization contains three functions, then it's a surface, not a curve.

Did you try using software to solve the original equation for y?
$\;$

 

[logistic_guy]

Review the domain of the function. Which rule does it violate?

this remind me by my first question in this forum. i learn from it. domain not have zero in the denomator. $\displaystyle x = 0$ not in the domain. correct?

 

[logistic_guy]

In that case, we have $\displaystyle y = \frac{t^3 + y^3}{3t}$


If you change $\displaystyle \frac{t^3 + y^3}{3t}$ to $\displaystyle t^3$, then you change the original curve (it becomes the curve of a cubic polynomial).


I'm not sure what you're thinking here, but, if a parameterization contains three functions, then it's a surface, not a curve.

in my previeous post, BigBeachBanana and Dr.Peterson say i have the freedome to choose any parametrization. also Dr.Peterson say we choose and try, if it doesn't work, we choose something else. this look messy but it is my first try

Did you try using software to solve the original equation for y?
$\;$

i don't knwo about this. you mean there is a software that can parametrize functions for me???

 

[Otis]

[others] say i have the [freedom] to choose any [parameterization]

The context is any valid parameterization. Changing the original curve to y=x^3 is not valid.

you mean there is a software that can parametrize functions for me?

There probably is, but that's not what I mean. If we solve the original equation for Real values of y, then we get an expression for the y-coordinate in terms of x. (My prior question is whether or not you've tried solving the original equation for y, using software. That question does not contain the word 'parameterize'.)

Replacing symbol x with parameter t provides the y-coordinate's function in terms of t. The x-coordinate's function is simply X(t)=t. Don't forget to specify a domain for t.

Points plotted using coordinates $\big(X(t),Y(t)\big)$ shown below.


$\;$

 

[Dr.Peterson]

No, you can't.

Can you tell us why not?

The particular issue I hoped you would see is this, which I called your attention to:

$\displaystyle y(3t - y^2) = t^3$

can i say $\displaystyle y = t^3$ and $\displaystyle 3t - y^2 = t^3$

If you know that AB = C, does it follow that A = C and B = C? No, that's true only if C = 0. (And even then it would be "or", not "and".)

in my previeous post, BigBeachBanana and Dr.Peterson say i have the freedome to choose any parametrization. also Dr.Peterson say we choose and try, if it doesn't work, we choose something else. this look messy but it is my first try

If you were to follow through on your correct statement that $\displaystyle y = \frac{t^3 + y^3}{3t}$, the next step would be to solve for y as a function of t. That isn't easy. (In fact, y is not a function of x!)

So, yes, it is messy. It would have been my first try, too!

Now you have to try something else. I have my own idea, which I got after using Desmos to graph the equation, in order to more quickly see what sort of parameter might work -- that is, as you go along the curve, what about it might change in a consistent way? (Using x = t would force you to move from left to right, and the curve doesn't do that.) But I might have thought of it without first doing that first; in fact, before modern software made it easy to visualize complicated equations, I wouldn't even have had that option.

I don't know of a general method for finding a parametrization, or software that would do it, but I am not up on all the latest methods.

But the idea I considered, which turned out to work, is to first try putting the equation in polar form, in the hope that that might reveal another variable I could use for t. (@Otis's graph is a function of x, and doesn't completely represent the equation.)

Another thing I wanted to ask: What is the context of your question? If it's a problem you were given, please quote it completely, so we can see if there were any implied hints we could point out. Again, parametrization is not an inherently easy thing to do; you have to recognize possibilities, which come with experience as I've said, and often context (that is, why you want to parametrize) provides ideas.

 

[Otis]

(@Otis's graph is a function of x, and doesn't completely represent the equation.)

Ah, I ought to have used the implicitPlot command, before trying anything else. That would have suggested using a polar coordinate system. Oops.
$\;$

 

[blamocur]

I managed to work it out by assuming $y = tx$.

 

[mario99]

I have worked for 20 minutes in this problem and I have discovered a beautiful parametrization. I will wait for the OP to get fully stuck or find something before I reveal it.

FYI: I have not used an AI or any artificial help. I have just discovered it after a few tries by coincidence.

 

[blamocur]

I have worked for 20 minutes in this problem and I have discovered a beautiful parametrization. I will wait for the OP to get fully stuck or find something before I reveal it.

FYI: I have not used an AI or any artificial help. I have just discovered it after a few tries by coincidence.

How beautiful? Mine is pretty hairy: ratios of 3rd and 2nd degree polynomials.

 

[mario99]

How beautiful? Mine is pretty hairy: ratios of 3rd and 2nd degree polynomials.

Well professor blamocur, you have stolen my work

Just kidding but that was exactly my discovery. I could have posted it before you, but I loved to give the OP a chance to sharpen his pencil.

 

[blamocur]

What bugs me that I'd expect a nicer parameterization with hyperbolic trigonometric functions, but I couldn't find one.

 

[Dr.Peterson]

I managed to work it out by assuming $y = tx$.

Yes, that is ultimately equivalent to my idea, and in fact the final form of mine can be rewritten using your t in place of my tangents.

How beautiful? Mine is pretty hairy: ratios of 3rd and 2nd degree polynomials.

That does sound like mine, though the way I wrote it (with reciprocals) didn't look that bad.

 

[Otis]

What bugs me…

At least you have a good reason. I'm bugged without one.

I'd thought that, when a parameterized curve cannot be represented by a single function, the situation would show itself via multiple, Real solutions for y. Clearly, there are exceptions. As I don't consider myself broadly knowledgeable, there's no reason for me to be bugged by such a goof. Yet, I am!

I managed to work it out by assuming $y = tx$.

I'll be interested in learning the train of thought to arrive at that (if it's more than recollection of seeing the situation before).

I will wait for the OP to get fully stuck … before I reveal it.

How will you know when? Seems like we may be there, already, but I'll wait.
$\;$

 

[BigBeachBanana]

At least you have a good reason. I'm bugged without one.

I'd thought that, when a parameterized curve cannot be represented by a single function, the situation would show itself via multiple, Real solutions for y. Clearly, there are exceptions. As I don't consider myself broadly knowledgeable, there's no reason for me to be bugged by such a goof. Yet, I am!


I'll be interested in learning the train of thought to arrive at that (if it's more than recollection of seeing the situation before).


How will you know when? Seems like we may be there, already, but I'll wait.
$\;$

It's a well-known curve. Look up Folium of Descartes.

 

[Otis]

It's a well-known curve. Look up Folium of Descartes

Thanks. I look forward also to how folks obtained the parameterization without recognizing 'Folium of Descartes'.
$\;$

 

[blamocur]

I'll be interested in learning the train of thought to arrive at that (if it's more than recollection of seeing the situation before).

I don't really have any train to report, probably some subconscious recollection. The symmetric nature of the equation probably helped to nudge my brain.