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Suppose I sketched the domain of a function $f(x, y) = \sqrt{(x-1)^2 + (y-2)^2 - 1}$ on the x-y plane. There's no problem with that. But suppose on that same set of x-y axes I wanted to add (by hand) some level curves of the function for $z = \sqrt{3}, \sqrt{7}$ and $\sqrt{12}$ (Feel free to use any $\sqrt{}$ values).

If we allow $ z = k = (x-1)^2 + (y-2)^2 ≥ 1$ I could maybe manoeuvre $x = y-1$ which yields $2(y-2)^2 ≥ 1$

If this is correct, what steps are required next to input and graph the z-values?

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I'm not sure if I'm answering your question, but drawing level curves by hand for this function is convenient by considering the square of the function $$z^2=(x-1)^2 + (y-2)^2 - 1$$ So, we have a constant $c=r^2=z^2+1$ where $r$ is the radius of a circular level curve in the $xy$-plane with center $(x,y)=(1,2)$. Hence, for $z=\sqrt{3}$ we have $c=4$ corresponding to a radius $r=2$.

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Why not use ubiquitous plotting programs rather than "by hand"?

enter image description here

Such a 3D plot shows the structure extremely well.

But if you're required to lose the information (such as the relative heights of contours), you can easily simplify such graphs:

enter image description here

All the above took no more than one minute in software. I wonder how long the OP will work on doing such plots "by hand" (and not making a mistake) and whether all that hand work really helps.

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  • $\begingroup$ To demonstrate understanding in regards to the function I suppose $\endgroup$
    – DuncanK3
    Jan 5, 2021 at 2:53
  • $\begingroup$ I, frankly, don't believe that is the way to understanding. See this: ted.com/talks/… $\endgroup$ Jan 5, 2021 at 3:08
  • $\begingroup$ Nevertheless, it's required to be within the x-y plane only $\endgroup$
    – DuncanK3
    Jan 5, 2021 at 5:08

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