1
$\begingroup$

I came across the function, $$z=\frac{4xy(x^2-y^2)}{x^2+y^2}$$ and I had to draw the level curve for this equation. Clearly the equation is already explicitly expressed in the form $z=f(x,y)$ and therefore for every value of $z$, the "shape" of the curve would be the same. We assume $z$ to be some constant $c$ and proceed. The main problem arises when we try to graph the following function which is now implicit in $x$ and $y$: $$c=\frac{4xy(x^2-y^2)}{x^2+y^2}$$


MY ATTEMPT: Firstly I tried proceeding normally. The function does not seem to be defined at the origin $(0,0)$. Now I tried checking behaviour at infinity. $$c=\lim_{x\to\infty}\frac{4xy(x^2-y^2)}{x^2+y^2} $$ Now the very first question that arises is how do we deal with $y$? We do not know if the limit of $y$ as $x$ approaches $\infty$ is $0$, $\infty$ or finite. Therefore we cannot comment on how the function behaves at either infinity. The next thing that we do is look at the derivative. The derivative here is: $$y'= \frac{12xy^2+4y^3-12x^2y+2cx}{4x^3-2cy}$$ Clearly this derivative is also difficult to handle because of the implicit nature of the function. The only takeaway from it is the fact that there exists a vertical asymptote at those points where $$y=\frac{2x^3}{c}\;\;\;\;\;\;...(1)$$ which again being tedious gives this: $$64x^8+(8c-8c^2)x^4+c^3=0$$ which anyone would discard at first look. If we just had to solve equation $(1)$ and our curve we could have tried approximating and playing around with graphs but that does not work here because the graph is what we actually want. Clearly I have ventured the wrong way, correct it is, but too rigorous to be solved.


MY 2nd ATTEMPT: Another train of thought that arises is to try polar coordinates, which, by the way, was also suggested in the book. Doing so, we get the curve to be: $$c= r^2sin(4\theta)$$ Now the problem here is that the polar system is a lot less intuitive compared to the Cartesian coordinate system as the variables deal with distance from the origin and angle with the x-axis. I still do not see a clear way out.


So how do we deal with this? I know that a graphing calculator is the simplest solution, but then I am not a fan, especially in a question that deals specifically with "sketching" a graph. The book is old enough that there weren't any easily available graphing calculators around like we have today. So, could someone describe the technique to sketch this curve or is it that there isn't any?

P.S: I have seen some Stack Exchange questions and browsed the internet but those were simpler ones like $x=f(y)$ or $y^2=f(x)$ type, so I won't be linking them here.

$\endgroup$
1
$\begingroup$

As mentioned in the hint, you can write this as

$$ c = r^2 \sin(4\theta)$$

For $c=0$, $r=0$ or $\sin(4\theta) = 0$ when $\theta$ is a multiple of $\pi/4$. That is, the level curve for $z=0$ consists of the $x$ and $y$ axes and the lines $y=x$ and $y=-x$.

For $c \ne 0$, $r = \sqrt{c/\sin(4\theta)}$. Of the eight intervals $(0, \pi/4), (\pi/4, \pi/2), \ldots, (7\pi/4, 2\pi)$, there are four where $\sin(4\theta)$ has the same sign as $c$. This gives you curves that are closest to the origin at the middle of the interval, and go off to $\infty$ as you approach the endpoints. The curves are symmetric under rotation by $\pi/2$.

enter image description here

$\endgroup$
1
  • $\begingroup$ Sir, how did you come to the conclusion that in the level curves where $z\ne0$ the curves are closest to the origin at the mid point of the interval? $\endgroup$ – Tesla's Coil Oct 28 '20 at 18:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.