Existence of a smooth path connecting two points

Question

Let $U=\{(x,y)\in \mathbb{R}^2 | 1<x^2+y^2<4\}$. Let $p,q\in U$. Show that there is a continuous map $\gamma : [0,1] \to U$ such that $\gamma (1)=q$ and $\gamma (0)=p$ and such that $\gamma$ is differentiable on $(0,1)$

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I really don't know how think this problem $\gamma $ is a continuous function and given that $\gamma (1)=q$ and $\gamma (0)=p$ then how construct $\gamma $ is differentiable on $(0,1)$? Any help plzzz..

Answer 1

Hint

Identify $\mathbb{R}^2$ with the complex plane $\mathbb{C}.$ Write $p=r_1 e^{i\theta_1}$ and $q=r_2 e^{i\theta_2}$.

Consider

$$\gamma(t)=(tr_1+(1-t)r_2)e^{i(t\theta_1+(1-\theta_2)}.$$

Answer 2

Write $p=(a\cos(\alpha),a\sin(\alpha))$ and $q=(b\cos(\beta),b\sin(\beta))$ with $1<a,b<2$. Then define \begin{align} &r(t)=a(1-t)+bt& &\theta(t)=\alpha(1-t)+\beta t \end{align} $$\gamma(t)=(r(t)\cos(\theta(t)),r(t)\sin(\theta(t)))$$