Path-Connected Subsets

Question

Let $S⊆\mathbb R^2$ and $a,b∈S$ be as follows: $S=\left\{(x,y):\frac{1}{4}<x^2+y^2<9\right\}$ and $a=(-1,\sqrt3)$ and $b=\left(\frac{-1}{\sqrt2},\frac{1}{\sqrt2}\right)$. Find a continuous path $r(t): [0,1]\to\mathbb R$ which connects the points $a$ and $b$ and which stays inside $S$.

I'm trying to formulate a general approach. I used polar coordinates to show $\frac{1}{2}<r<3$. Graphing this in the $r-\Theta$ plane, I'm trying to find the straight line that connected these two points $a$ and $b$ within this region via a straight line. So far I have $r(t)-(1-t)a+tb$. Where do I go from here? How do I find this path?

Answer 1

Since $a=2\left(\cos\left(\frac{5\pi}6\right),\sin\left(\frac{5\pi}6\right)\right)$ and $b=\left(\cos\left(\frac{3\pi}4\right),\sin\left(\frac{3\pi}4\right)\right)$, you can define$$r(t)=(2-t)\left(\cos\left((1-t)\frac{5\pi}6+t\frac{3\pi}4\right),\sin\left((1-t)\frac{5\pi}6+t\frac{3\pi}4\right)\right).$$

Answer 2

The following is easy to show:

If the two points $V_1 = (x, y_1)$ and $V_2 = (x, y_2)$ belong to $S$ and are in the same quadrant, then the line segment joining $V_1$ and $V_2$ is contained in $S$.

If the two points $H_1 = (x_1, y)$ and $H_2 = (x_2, y)$ belong to $S$ and are in the same quadrant, then the line segment joining $H_1$ and $H_2$ is contained in $S$.

Now connect the points,

$\quad \left(\frac{-1}{\sqrt2},\frac{1}{\sqrt2}\right) \to \left(-2,\frac{1}{\sqrt2}\right) \to \left(-2,2 \right) \to \left(-2,\sqrt 3\right) \to \left(-1,\sqrt3\right)$

I didn't concern myself with checking if the line segment joining $a$ to $b$ was already contained in $S$ when tackling this problem. I simply wanted to link up to the point $(-2,2)$ since that would provide some 'elbow room'. If that did not work then back to the drawing board (and I would curse whoever came up with the exercise).

The OP can construct the path mapping to trace out these line segments.