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Explain why $\sqrt{x^2-y^2}+\arcsin(x/y)=0$ does not define $y$ as an implicit function of $x$.

Quite confused by this, mainly because I do not fully understand really what it means for an equation to define $y$ as an implicit function of $x$ even though I kind of get the idea.

I think the reason is something to do with the fact that the first term requires $|x|\ge|y|$ and the second term the opposite but I would appreciate if someone would explain this to me a bit more clearly, thanks.

Edit: Sorry I messed up the question it was meant to be a minus in the root.

zipirovich
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Carlos Bacca
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  • Tip: Use $\arcsin \left( \frac xy \right)$ to obtain $\arcsin \left( \frac xy \right)$. – For the love of maths Feb 02 '19 at 02:52
  • I must confess that I do not either understand the question. I could be a problem of formal terminology. Use a contour plot for $\sqrt{x^2 + y^2}+\arcsin \left( \frac xy \right)=0$ for $-1 \leq x \leq 1$ and $-1 \leq y \leq 1$ and see the result. – Claude Leibovici Feb 02 '19 at 02:56
  • i plotted it and it looks like it defines y as an implicit function of x to me, but then again i dont really know what im looking for – Carlos Bacca Feb 02 '19 at 03:45
  • I think its because the restrictions on the first and second term in my post mean that the only function you can replace y with is x and this does not give a true inequality which you need. Say if you replaced inverse sin with inverse cos then it would define y as an implicit function of x and that function would be g(x)=x. I think thats why... – Carlos Bacca Feb 02 '19 at 04:09

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The question sounds a little strange to me… But I agree with what you said in a comment above — essentially, the reason is that this equation has no solutions at all, so it doesn't define anything. Geometrically, its graph is the empty set in the plane.

And yes, just as you said, it has a lot to do with the fact that the first term requires $|x|\ge|y|$ and the second term requires $|x|\le|y|$, along with $y\ne0$. The two inequalities together imply that $|x|=|y|$, so $y=\pm x$. In either case, $\sqrt{x^2-y^2}=0$, so we're left with $\arcsin(x/y)=\arcsin(\pm1)=\pm\pi/2\ne0$.

zipirovich
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