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We just started this chapter and our prof doesn't like text books so he just has lectures. We had one lecture on this and I'm still 100% lost and I'm trying to practice.

So far it's been going okay, but I stumbled upon this question:

$$f(x,y) = \frac{x^2+y^2}{xy}$$

I'm having a lot of trouble sketching this, can anyone point me in the right direction?

So far all I've gathered is $x,y ≠ 0$

And for $C=0$, the equation turns into $x^2+y^2=0$ which won't work because of the restriction.

This doesn't really help me graphing wise, so could someone point me in the right direction.

Also for most of these questions, when I google them, the typical answer is "You should be able to know what these are by looking at them, like $x^2+y^2=z$ is a circle" Is there a website where it will show me the basic level curves and their functions and map like the one above?

Thanks so much!

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This one isn't so easy. But if you rewrite $$\frac{x^2+y^2}{xy} = c \quad\text{as}\quad x^2-cxy+y^2 = 0,$$ then you're on the right track.

Can you see what to do if $c=\pm 2$?

In general, when $|c|<2$, try completing the square, and you should see that when the only solution is $x=y=0$, which as you pointed out is not in the domain of the function. So, all the level curves are ... nothing ... the empty set. When $|c|>2$, completing the square results in a difference of squares $(x-\frac c2y)^2-(\frac{c^2}4-1)y^2 = 0$, which — as Doug M. suggested — will give the union of two lines. Perhaps the graphics software will help you see this "before your very own eyes."

(I don't think I'm fond of this exercise.)

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  • $\begingroup$ Me neither,seriously. That's a nasty problem for a beginner. It seems to have no other purpose but to confuse the students. This is the kind of problem where graphing software is very helpful for checking your answer. $\endgroup$ – Mathemagician1234 Sep 27 '17 at 19:29
  • $\begingroup$ Part b) of the question says "Use the contour plotting command on the functions in part (a) and compare with the pictures you produced in part (a)" Maybe this one was supposed to confuse me? But I highly doubt I'm supposed to leave it blank :\ $\endgroup$ – Hello Mellow Sep 27 '17 at 19:30
  • $\begingroup$ The contour plots should be empty for every $c$ here. I suspect this is a multi-part problem the level curves should be more interesting for other parts. Yes, you should definitely recognize things like lines, parabolas, ellipses, hyperbolas, and so on, but this was truly more of a trick question to see if your algebra is up to snuff. $\endgroup$ – Ted Shifrin Sep 27 '17 at 19:37
  • $\begingroup$ when $|c|\ge 2, x^2 - cxy + y^2 = (x - \frac {c+\sqrt{c^2-4}}{2}y)(x- \frac {c-\sqrt{c^2-4}}{2}y)=0$ $\endgroup$ – Doug M Sep 27 '17 at 19:50
  • $\begingroup$ Ah, right, I was sloppy. I'll edit. $\endgroup$ – Ted Shifrin Sep 27 '17 at 19:53
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Converting to polar

$x = r\cos\theta\\y=r\sin\theta\\f(r,\theta) = \frac {r^2}{r^2\sin\theta\cos\theta} = 2\csc2\theta$

The function is undefined at $(0,0)$

The contours will be lines though the origin (although undefined at the origin).

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  • $\begingroup$ Good approach. And, since $|\csc u|\ge 1$, we have no solution for $|c|<2$. $\endgroup$ – Ted Shifrin Sep 27 '17 at 20:00

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