parametrization question

[blamocur]

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

I was disappointed at first by seeing a simpler rational parameterization in there, but then remembered that I had a similar one but modified it because it was not defined for [imath]t=-1[/imath].

 

[BigBeachBanana]

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

The idea behind [imath]y=tx[/imath] is to parameterize the curve by slopes ([imath]t[/imath]) of lines through the origin. Besides the origin and [imath]t=-1[/imath], any line through the origin touches the curve at exactly one point.

 

[Otis]

The symmetric nature of the equation probably helped

Ah, I see. You had first checked for origin symmetry, by replacing (x,y) with (-x,-y)?
[imath]\;[/imath]

 

[blamocur]

Ah, I see. You had first checked for origin symmetry, by replacing (x,y) with (-x,-y)?

No, I wasn't thinking of that, just using [imath]y=tx[/imath]. Just trying all kinds of things without any real system.

 

[Otis]

No, I wasn't thinking of that

Fair enough. I'm curious how you knew the original equation has symmetry.
[imath]\;[/imath]

 

[blamocur]

Fair enough. I'm curious how you knew the original equation has symmetry.
[imath]\;[/imath]

[imath]x^3+y^3 = 3xy[/imath] or [imath]\frac{x^2}{y} +\frac{y^2}{x} = 3[/imath]

 

[Steven G]

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

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

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

Then y(3t-y^2) = t^3*t^3 = t^6. What happened to y(3t-y^2) = t^3?????

If a=c and b=c, then it does NOT follow that ab=c. In fact, ab=c^2

 

[mario99]

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

Well after two to three failures from the OP, I would know that he's fully stuck and it's time to reveal my secret of parameterization. Also, you don't have to wait because professor blamocur has already teleported my discovery in his post #11.

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

I will not pretend to be smart or has the knowledge of famous curves (like the Folium of Descartes). Frankly, I am not both. I just did what any normal student would do: try to find any good parametrization by plugging random [imath]t[/imath]'s in [imath]x[/imath] and [imath]y[/imath]. I failed to find any after 20 minutes, but then suddenly I noticed something.

If you know me enough, Mr. Otis, you will realize that I am the Mario who has solved [imath]1000[/imath] differential equations. I remember a long time ago, I solved this differential equation: [imath]\displaystyle \frac{dy}{dx} = \frac{x^2 - y^2}{2xy}[/imath] which has almost a similar form of the the Folium of Descartes curve. I remember I did this substitution to get rid of one of the variables: [imath]y = vx[/imath] (which led me to [imath]y = tx[/imath])

which turned this:

[imath]\displaystyle \frac{dy}{dx} = \frac{x^2 - y^2}{2xy}[/imath] (not solvable)

into this:

[imath]\displaystyle v + x\frac{dv}{dx} = \frac{x^2 - v^2x^2}{2vx^2} = \frac{1 - v^2}{2v}[/imath] (solvable)

I don't know if you will call this as recollection! Maybe, yes, but indirectly!

 

[logistic_guy]

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

i still don't understand algebra, so graphs and domain is my enemey. i'm learning

There probably is, but that's not what I mean.

if you know please give me. i'm strugle in parametrize

(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'.)
View attachment 38300
[imath]\;[/imath]

i'm still not in the level to draw graphs but thank for this advise. when i see the graph i understand the question better

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".)

algebra is my weaknest. i'm still miss lots of basics. i think i'm understand your explanation

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!

so the secand idea is to stop once the parametrize get messy. i think i'm understand

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'm still not in the level to use technology to help me solve. i think once i know this solving will be easier

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.

me too. i even don't understand what is parapetrization and why we do it

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.)

i hear this word a lot in calculus polar but i still don't get its full idea. i only understand angle appear there

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.

there is not contex or anything. the exercise is exactly as what i give. it want me to parametrize a given function. it's just like warming up to be able to solve other problems because most of them depend on this first step parametrize

I managed to work it out by assuming [imath]y = tx[/imath].

thank

i'll use this parametrize \(\displaystyle y = tx\)

\(\displaystyle y = \frac{x^3 + y^3}{3x} = \frac{x^3 + (tx)^3}{3x} = \frac{x^3 + t^3x^3}{3x} = \frac{x(x^2 + t^3x^2)}{3x} = \frac{(x^2 + t^3x^2)}{3}\)

i think this doesn't help much as i've still \(\displaystyle y\) on the left side

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.

thank

if your parametrize \(\displaystyle y = tx\) i check it, it doesn't help much

Then y(3t-y^2) = t^3*t^3 = t^6. What happened to y(3t-y^2) = t^3?????

If a=c and b=c, then it does NOT follow that ab=c. In fact, ab=c^2

thank

this is similar what Dr.Peterson explain. i think i'm understand now

i'll try to think of other parametrize on paper. if i get any i'll try to post here. thank

 

[blamocur]

i think this doesn't help much as i've still y\displaystyle yy on the left side

[imath]y = tx[/imath]

 

[Dr.Peterson]

there is not contex or anything. the exercise is exactly as what i give.

Nonsense. Everything has context. In particular, I asked if it comes from a book; what book? What topic? Why would they ask you to do this?

But there is also the larger context, namely your own. It is becoming more and more clear that you need to study Algebra 1 before you can ask any questions at a higher level. Please do so; I may start answering you again when you have done that. Until then, it seems useless.

 

[Otis]

i still don't understand algebra, so graphs and domain [are my enemies]

Perhaps, you do not understand my comment. You know enough to not change given information in an exercise, yes?

i'm still not in the level to draw graphs but thank for this advise. when i see the graph i understand the question better

I don't understand "not in the level to draw graphs". Do you mean that you don't know how or that you're trying to learn without graphing or something else? Otherwise, I see no issue with using software to visualize a curve. Well, no issue as long as you're careful, but I was not!

I've assumed that you saw Dr. Peterson tell me the graph I'd posted in reply #8 is not complete. I was careless; I didn't bother to evaluate (x,y) points using test values of t to determine why the curve stopped in the first quadrant — because I've seen that type of behavior before. The graph stopped because the actual curve does not represent a single function of x; it fails the vertical-line test. So, the software's function-plotting command graphed only one piece. As Dr. Peterson noted, it's better to parameterize using polar coordinates.

I regret the misinformation in my post#8, but hopefully we've both learned a lesson here. Just because we (or a computer) can solve an equation for y in terms of x from a given equation in which y appears on both sides does not always mean that there is a single, implicit function y=f(x) hidden in there.
[imath]\;[/imath]