Plot the level curve of sine function in multiple variables I'm very confused about how I could go about this, as it seems that the question cannot be done using only the information given.
The question is: plot the level curve for $f(x,y) = \sin(k^2x^2 + y^2) =$ $\frac{1}{\sqrt{2}}$ for some (unknown) fixed $k \in \mathbb{R}$.
I cannot see how you could plot this without knowing the value of $k$. This was an exam question and the paper asked for it to be drawn during the exam today. However, can you actually draw it when you don't have a value of $k$ to go by? For example, when I replace $k$ with $3$, Wolfram Alpha gives this: http://bit.ly/1Knjdo1 .
Would this be realistic to sketch on paper during an exam? If someone could help explain this, that would help greatly, thanks.
 A: First, $\sin\theta = \frac{1}{\sqrt{2}}$ if and only if $\theta = \frac{\pi}{4} + 2n\pi$ or $\theta = \frac{3\pi}{4} + 2n\pi$ for some integer $n$. Consequently,
$$
\sin(k^{2}x^{2} + y^{2}) = \tfrac{1}{\sqrt{2}}
$$
if and only if $k^{2}x^{2} + y^{2} = \frac{\pi}{4} + 2n\pi$ or $k^{2}x^{2} + y^{2} = \frac{3\pi}{4} + 2n\pi$ for some integer $n$.
If $c < 0$, the equation $k^{2}x^{2} + y^{2} = c$ has no (real) solutions, so you may assume the integer $n$ above is positive. If $c > 0$, the curve $k^{2}x^{2} + y^{2} = c$ is either


*

*An axis-oriented ellipse centered at the origin, with $x$-intercepts $\pm\frac{1}{k}\sqrt{c}$ and $y$-intercepts $\pm\sqrt{c}$ (if $k \neq 0$), or

*A pair of horizontal lines, $y = \pm \sqrt{c}$ (if $k = 0$).
To summarize, if $k \neq 0$, the level curve is a family of nested ellipses
$$
\left.
\begin{aligned}
  k^{2}x^{2} + y^{2} &= \tfrac{\pi}{4} + 2n\pi, \\
  k^{2}x^{2} + y^{2} &= \tfrac{3\pi}{4} + 2n\pi, \\
\end{aligned}\right\}\qquad\text{for some integer $n > 0$.}
$$
If $k = 0$, the level curve is the family of horizontal lines obtained by setting $k = 0$ in the preceding.
Note that plotting software has trouble giving an accurate picture: The function $f$ "oscillates rapidly", and quickly exceeds the sampling resolution of a bitmap graphic. All the non-elliptical bits and pieces in the linked graph are artifacts, not genuine features of the level curve.
This is fairly challenging for a timed test, but I'd say it's "reasonable"; your instructor may be hoping to see what productive strategies you find under pressure, rather than looking for a complete solution. And in any case, it should be fairly routine to get started: To see the level curves are ellipses (and perhaps to notice the need to separate the case $k = 0$), and to see how the qualitative shape depends on $k$.
A: let k = 1, then x^2+y^2 = c^2, where c is constant.
then (x, y) which satisfy above equation will form circle with radius c.
then level curve will be circular for f(x, y) because any (x,y) on unit circle evaluate to same f(x,y)
k scale x, causing circle to become elliptic.  so level curve become elliptic.
analytically, you can solve value of k, which satisfy f(x,y) equation by:
solve for sine(z)=1/(2)^1/2, there will be multiple solutions. the smallest positive solution is pie/4.
then k^2*x^2 + y^2 = w.  you will find elliptic satisfying this equation to be your level curve
edit note:
i have been thinking about the weird contour plot using mathematica.
to help solve the issue, i tried following:
set k = 1, which should not matter cuz k simply scale x to make circle into elliptic
set contour to equal 0, instead of 1/(2)^1/2.  make math easier. sine(w)=0 are 0, 1*pi, 2*pi, 3*pi ... do not worry about negative w, cuz squared value can never be negative
also change 3d to 2d problem: sine(x^2+y^2) -> sine(x^2), which mean set y=0.
now plot sine(x^2).  you will get something similar to 
plot from -6*pi, to 6*pi.  you will notice the roots of sine(x^2) become closer as x increase.  
now 2d->3d, this is effectively showing contour plot will be increasingly closer to each other as x & y increase for sine(x^2+y^2)
i think the weird effect you see from mathematica may be due to numerical issue, due to contour plot infinitely becoming closer
