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Let $f(x,y) = (x^2+y^2-1)(2x+y-1)$.

Then how can I sketch the level curves of $f(x,y)$?

Thank you for your help in advance.

math2357
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    Welcome to math.SE. Do you have any thoughts on the problem? It will make our job easier and more likely to answer you. – Don Larynx Dec 19 '13 at 19:16
  • Are you looking for curves defined by $f(x,y)=z$, for any given $z$, or to the specific case $z=0$, that is $f(x,y)=0$? The latter is almost trivial (union of the unit circle and a line), while the former is less obvious. – Jean-Claude Arbaut Dec 19 '13 at 19:23

2 Answers2

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According to the definition of level curves, if we are given a function of two variables $z=f(x, y)$,the cross-section between the surface and a horizontal plane is called a level curve or a contour curve. Thus, level curves have algebraic equations of the form: $$f(x, y) =k$$ for all possible values of $k$. Now let's do this goal by using a mathematical software like Mathematica or Maple. I did it by Maple 16 for you:

enter image description here

Mikasa
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  • the level curves should be in the plane? – ILoveMath Dec 19 '13 at 19:50
  • @AtahualpaInca: If we assume that $z$ is the same as $f(x,y)$, so they are some surfaces as you noted in 3d. – Mikasa Dec 19 '13 at 19:52
  • I still think they should live in the plane – ILoveMath Dec 19 '13 at 19:56
  • @AtahualpaInca: I got what you mean. Let me correct it accordingly to be out of $\mathbb R^4$. – Mikasa Dec 19 '13 at 19:58
  • Thank you, But I was wondering I could get those level sets without using computer. Just by hand. When k=0, it is easy. but other cases such as k=-1, 1, 2 etc.... that's my original question.. – math2357 Dec 19 '13 at 20:07
  • @math2357: Yes, it would be hard without a machine. If this post doesn't satisfy you, tell me to remove it. Indeed, without a software, sketching these curves, seems very difficult. Sorry if I could't help u. – Mikasa Dec 19 '13 at 20:13
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    no, it helps to me. you helped me a lot. Thanks! – math2357 Dec 19 '13 at 20:21
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You have a function $f: \mathbb{R}^2 \to \mathbb{R}$. The level curves of $f$ is the set $$ \{ (x,y) \in \mathbb{R}^2 : f(x,y) = K, K \in \mathbb{R} \}$$. So, in order to find the level curves of your function, just set it equal to a constant K, and try different values of $K$. For instance

$$f(x,y) = (x^2 + y^2 -1)(2x + y -1) = K$$

Now, test values foe $K$, say $K=-1,-2,0,1,2,3$, and graph it in each different scenario.

ILoveMath
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  • I know, but can I compute this without computer? For example, when K=0, it is easy to get the curve. But K=1, K=2, or K=-1, then it seems very hard to figure out the whole level curves... – math2357 Dec 19 '13 at 20:05
  • look at B.S. Answer. – ILoveMath Dec 19 '13 at 20:06
  • @ILoveMath What if I don't have the formula for the function, but I have formulas for the components of its gradient vector field instead? Is there a way to get back the original function from its gradient? – SasQ Jun 07 '20 at 20:20