Show that this implicit equation can be expressed as the vector equation

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Homework Statement


Show that [tex]x^3+y^3=9xy[/tex] can be expressed as
[tex]r(t) = \frac{t}{1+t^3} \hat{i} + \frac{t^2}{1+t^3} \hat{j}[/tex]


The Attempt at a Solution


I only know how to convert parametric/vector equations to Cartesian, not the other way around which is what I'm pretty sure I have to do...thanks

EDIT: Sorry, not sure how to fix up the latex

Fixed the LaTeX for you.
 
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  • #2
SammyS
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Homework Statement


Show that [tex]x^3+y^3=9xy[/tex] can be expressed as [tex]r(t)=\frac{t}{1+t^3}\vec{i}+ (\frac{t^2}{1+t^3})\vec{j}[/tex]

The Attempt at a Solution


I only know how to convert parametric/vector equations to Cartesian, not the other way around which is what I'm pretty sure I have to do...thanks

EDIT: Sorry, not sure how to fix up the latex
You have an extra "{" in front of you last fraction.
 
  • #3
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Let [itex]A=\{(x,y)\inℝ^2:x^3+y^3=9xy\}.[/itex]

Let [itex]B=\{ (\frac{t}{1+t^3},\frac{t^2}{1+t^3})\in ℝ^2:t\in ℝ\}.[/itex]

It is probably easy to show that [itex]B\subset A[/itex], what I'm not sure is if it was intended for you to show that [itex]A\subset B[/itex].
 
  • #4
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Is there a restriction on t? Did you type the problem correctly? Something seems wrong or missing:

[itex]\vec{r}(1)=<\frac{1}{2},\frac{1}{2}>[/itex]

but...

[itex](\frac{1}{2})^{3}+(\frac{1}{2})^{3}=\frac{1}{4}≠9(\frac{1}{2})(\frac{1}{2})=\frac{9}{4}[/itex]
 
  • #5
HallsofIvy
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Indeed, as cjc0117 suggests, the fundamental difficulty here is that the given parametic equation is NOT equivalent to the given Cartesian equation! What you want to show simply is not true.
 
  • #6
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Sorry, there's supposed to be a 9 in the numerator as well. I was posting from memory yesterday
 
  • #7
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Can't seem to edit my post so this is what it should be:

Show that [tex]x^3+y^3=9xy[/tex] can be expressed as [tex]r(t)=\frac{9t}{1+t^3}\vec{i}+(\frac{9t^2}{1+t^3})\vec{j}[/tex]
 
  • #8
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Okay, can you show [itex]B\subset A[/itex] ?
 
  • #9
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^ I honestly have no idea. We haven't covered any of this stuff yet, it's not in my text either. Might not even be in the syllabus (I think this class would be the equivalent of Calculus I with some other stuff - vectors, complex numbers, coordinate geometry etc.)
 
  • #10
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Take a point (x(t),y(t)) from B, and check if it satisfies the condition to be in A.
 
  • #11
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algebrat, that would only prove that the vector equation represents the cartesian equation at one point (x(t0),y(t0)). What OP needs to do is plug r(t)=<x(t),y(t)> into the cartesian equation and see if both sides equal one another:

[itex](\frac{9t}{1+t^{3}})^{3}+(\frac{9t^{2}}{1+t^{3}})^{3}=9(\frac{9t}{1+t^{3}})(\frac{9t^{2}}{1+t^{3}})[/itex]

Simplify this. Is it true?
 
  • #12
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Really? So then I would have to manipulate both sides until they're equal? Is there a systematic method of doing so? How can you tell when (guess is turning into an algebra question) it's in the simplest form (on that note, what exactly is the definition of simplest form?)?

At the moment, I've got

[tex]\frac{729t^3+729t^6}{(1+t^3)^3}=\frac{729t^3}{(1+t^3)^2}[/tex]


Also, how do I tell if a particular equation can be represented by a particular vector equation? Say if I plug in a few values and they're consistent and I tried to do what you showed...is the only way of knowing then would be to recognize that it's impossible to show both sides are equal?
 
  • #13
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I hate denominators, get rid of them. (Cross multiply.)

@cj, the point is arbitrary.

Actually, maybe don't cross multiply yet, don't some factors cancel?
 
  • #14
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^ I think they do, going to try and work it out when I get home. Are you allowed to cross multiply? I was avoiding carrying out any operations because I thought I had to just show that two expressions on either side are equal and not solve for t
 
  • #15
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[STRIKE]Yes, you can cross multiply[/STRIKE]. But first do some factoring as algebrat suggested, and you won't even have to cross multiply anymore.

EDIT: Also, I'm not sure of what the definition of "simplest form" is in a mathematical context. But all we have to do is make it clear that the equality I wrote above, [itex] (\frac{9t}{1+t^{3}})^{3}+(\frac{9t^{2}}{1+t^{3}})^ {3}=9(\frac{9t}{1+t^{3}})(\frac{9t^{2}}{1+t^{3}})[/itex], makes sense/is true for any value t. After factoring and canceling, you should be able to simplify it to 1=1, which is sufficiently simple I think, and makes clear that the original equality is true. But if you got, say, 1=2, clearly that does not make sense and would imply the original equality is false.

EDIT2: Actually, I'm not too sure now. Can someone else clear this up? Is it valid to cross multiply or should you just perform operations separately on both sides until they equal one another?
 
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  • #16
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If you take the left side and cancel a factor of 1+t^3, then you will see both side are equal. And from the very beginning, t cannot equal -1. The set of t for which the statement is true is not effected by this cancellation.

The sort of thing you have to look out for is when the operation changes the set of values for which the statement is true. For instance when picking positive or negative square root, you are decreasing the size of set for which a statement is true.

If you square both sides, often you are changing the set of values for which soemthing is true. When you divid both sides, you can change the set, but we didn't here, since we couldn't have t=-1 in the first place.

x=2, square both sides, x^2=4, now true for both x=2 and x=-2.

x^2-4=x^2+4x+4, true for x=-2. divide both sides by x+2, now it is never true.
 
  • #17
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Damn, I can't work out how to factor out the factor!
 
  • #18
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At the moment, I've got

[tex]\frac{729t^3+729t^6}{(1+t^3)^3}=\frac{729t^3}{(1+t ^3)^2}[/tex]

All you have to do is factor the numerator on the left side. What factors out of both terms in the numerator?
 
  • #19
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Ha, thanks! Had a brain meltdown there...I was just trying to factor out (1+t^3) for some reason and wasn't looking at anything else. Thank you to everyone who chipped in, also the moderator who fixed up my initial post
 

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