Finding a function with level curves

Question

I'm asked to find a function f(x,y) defined on R2, whose level curves include the two following curves:

(x(t), y(t)) = (cos(t) + 1, sin(t) - 1), t ∈ R

and

(x(t), y(t)) = (1/2cos(t) + 1, 1/2sin(t) - 1), t ∈ R

I'm not sure where to start. I understand how to find level curves of a function, but how do I do the reverse? I have no doubt I am overthinking this

Answer 1

I would first eliminate the parametric equations. The first one:

$$x=\cos(t) + 1$$ $$\cos(t) = x - 1$$ And using the right triangle trick: $$\sin(t)=\sqrt{2x-x^2}$$ So, $y=\sqrt{2x-x^2}-1$ is the equation for the first parametric equation. Cleaning this up a little:

$$y+1 = \sqrt{2x-x^2}$$ $$y^2+2y+1 = 2x-x^2$$ $$\boxed{y^2+2y + x^2-2x = -1}$$

Second parametric equation is done similarly:

$$x=\frac{1}{2}\cos(t) + 1$$ $$\cos(t) = 2x - 2$$ And using the right triangle trick: $$\sin(t)=\sqrt{-4x^2+8x-3}$$ So, $y=\frac{1}{2}\sqrt{-4x^2+8x-3}-1$ is the equation for the second parametric equation. Cleaning this up a little:

$$y+1 = \frac{1}{2}\sqrt{-4x^2+8x-3}$$ $$y^2+2y+1 = -x^2+2x-\frac{3}{4}$$ $$\boxed{y^2+2y + x^2-2x = -\frac{7}{4}}$$

So, looking at the two boxed equations, the left sides are the same and the right sides are two different constants. The left side is your $z=y^2+2y + x^2-2x$ and the two constants on the right are the two values of $z$ that are your level curves.

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One final thing. Mathematicians like things to be pretty. All that matters is that the left sides are the same. They can include a constant too. So lets complete the squares. In this case though, we add one for the y's and subtract one for the x's.

The boxed equations become:

$$(y+1)^2 + (x-1)^2=-1$$ $$(y+1)^2 + (x-1)^2=-\frac{7}{4}$$

So: $$z=(y+1)^2 + (x-1)^2$$

Answer 2

Think about this geometrically. If you think about graphing these curves, or if you recognize the parametric equations for a circle, you realize that both your level curves are circles centered at the point $(1, -1)$. Here is a graph:

So your function could simply be the distance from that center point:

$$f(x,y) = \sqrt{(x-1)^2+(y+1)^2}$$

Or it could be the square of the distance, which would remove the square-root symbol and be simpler. Or it could be almost any function of the distance at all.

As a reminder: Parametric equations for a circle of radius $r$ centered at $(h,k)$ are:

$$x=r\cos(t)+h,\qquad y=r\sin(t)+k, \qquad 0\le t\le 2\pi$$