How to know if the curve $x^6 + y^6 = x^4y$ is closed ? And how to know that the curve $y^2 = x^2\frac{1 - x}{ 1 + x}$ has a loop ? PS: When I ask this two questions, I'm not considering the usage of a computer program. Another consideration is that I used the "Multivariable-Calculus" tag for the questions but I think it's necessary more than that to answer them.
Asked
Active
Viewed 54 times
2
-
maybe some polar? That may clarify the second curve I'm not sure about the first one – Teh Rod Nov 20 '16 at 22:45
2 Answers
2
In polar coordinates, $x = r \cos(\theta)$, $y = r \sin(\theta)$, the first curve becomes $$r^6 (\cos^6(\theta) + \sin^6(\theta)) = r^5 \cos^4(\theta) \sin(\theta)$$ One solution is $r = 0$ (the origin, an isolated point); otherwise $$ r = \dfrac{\cos^4(\theta) \sin(\theta)}{\cos^6(\theta) + \sin^6(\theta)}$$ Note that the denominator is bounded away from $0$, so this is a closed curve.
The second curve is symmetric about the $x$ axis: $$ y = \pm x \sqrt{\frac{1-x}{1+x}}$$ for those $x$ such that $(1-x)/(1+x) \ge 0$, i.e. $-1 < x \le 1$. We have $y=0$ at $x=0$ and $x=1$. The loop consists of $y = +x \sqrt{(1-x)/(1+x)}$ in the upper half plane and $y = -x \sqrt{(1-x)/(1+x)}$ in the lower half plane, for $0 \le x \le 1$.
Robert Israel
- 448,999
-
Your aswer helped a lot! But I have some questions in relation to the first part. What do you mean by "bounded away from zero" ? Why being the denominator bounded away from zero makes the curve closed ? Searching for that expression, it seemed to me that it comes from Real Analysis. Is this true ? Do I need some background in Real Analysis to understand what you did ? – danilocn94 Nov 20 '16 at 23:24
1
I'm trying to comment on the other answer but I am unable to. To answer our question in the comments, you only need to understand the terminology in the definition, you don't really need a full course in real analysis to understand what he is saying. Here is an answer that I think will help. (Feel free to delete this since it doesn't adequately answer the post. )
-
I still don't understand why being the denominator of the polar equation bounded away from zero makes the curve closed. I don't find this fact anywhere. Could you give me references(or books that talk about that) ? Or could you explain in more details, giving references preferably ? – danilocn94 Nov 21 '16 at 03:06
-
My assumption is that the author of the other solution said "the denominator is bounded away from zero" in the sense that this is a well defined solution. In other words the denominator will never be zero, which is something we like. Aside from that, the reason this shows the curve is close is because we have two solutions for r. One where r=0 and the other when r is a quite interesting looking function but is periodic. – El Spiffy Nov 21 '16 at 04:31