Linear independence of derivatives at a point implies trivial intersection of image Let $f$ and $g$ be differentiable paths from $\mathbb{R}$ to $\mathbb{R^{n}}$ such that $f(a)=g(b)=p$ for some pair $a,b \in \mathbb{R}$.
If $f'(a)$ and $g'(b)$ are linear independent then $\exists k>0$ such that $f(B(a,k))\cap g(B(b,k))=${$p$}.
Here $B(x,k)$ stands for the open ball with radidus $k$ and center at $x$.
It's easy to see that the images of $(a,a+k)$ and $(b,b+k)$ shouldn't have a point in commom because the paths are "growing" to different directions so the inclination may guarantee what I need,still I don't know how to go further.
 A: Without loss of generality suppose $a=b$. If the statement is not true, for all $n\in\mathbb{N}$ there are $a_{n},b_{n}\in B(a,1/n)$ such that $f(a_{n})=g(b_{n})\not=p$. Put $c_{n}=\frac{b_{n}-b}{a_{n}-a}$. Then
$$ \frac{f(a_{n})-f(a)}{a_{n}-a}=c_{n}\frac{g(b_{n})-g(b)}{b_{n}-b} $$
Since $\{f'(a),g'(a)\}$ is linearly independent, some coordinate of $g'(a)$ is not zero. This implies that $c_{n}$ converges to, say, $c$. It follows that
$$f'(a)=cg'(a) $$
which is a contradiction.
A: Exercise: If $u, v \in \mathbb R^n$ are linearly independent, then there exists a constant $c>0$ such that
$$|su+tv|\ge c \sqrt {s^2+t^2}\,\, \text { for } s,t\in \mathbb R.$$
Because $f(a)$ cancels with $g(b),$ the definition of the derivative shows
$$\tag 1 f(a+s)-g(b+t) = sf'(a) - tg'(b) + o(s) + o(t)$$ $$ = sf'(a) - tg'(b) + o(\sqrt {s^2+t^2}).$$
In absolute value $(1)$ is at least
$$|sf'(a) - tg'(b)| - |o(\sqrt {s^2+t^2})|.$$
By the exercise, this is at least $c\sqrt {s^2+t^2}- |o(\sqrt {s^2+t^2})|.$ This is positive for $(s,t)$ close to, but not equal to, $(0,0).$ This is the desired conclusion.
