Parametrically defined curves in $\mathbb R^3$ and tangent vectors (how to show that tangent vector as defined really is the tangent vector)

Question

If we define smooth curve in $\mathbb R^3$ as an image of a differentiable map $m: [0,1] \to \mathbb R^3$ then a curve is represented as a set of points $m(t)=(x(t),y(t),z(t))$ where $t \in [0,1]$.

Then a tangent vector at a point $(x(t_0),y(t_0),z(t_0))$ is defined as $(x´(t_0),y´(t_0),z´(t_0))$.

But, how can we be sure that really that is the tangent vector, that is, a vector that in some neigbourhood approximates that curve better than any other vector, in some strictly and well-defined sense?

I understand that $x´(t)$ gives us a speed of change of a point with respect to $x$-coordinate, that is, it gives us a speed of change of a projection of a point to the $x$-axis. The same for $y´(t)$ and $z´(t)$.

But... I am still not sure that $(x´(t_0),y´(t_0),z´(t_0))$ really is the tangent vector at point $(x(t_0),y(t_0),z(t_0))$.

I believe that it is but, geometrically, I do not see that.

Can some explain to me this issues I am facing with?

Answer 1

Pick some point $t_0 \in [0,1]$ and let: $$ \phi (t) = m(t_0) + m'(t_0)(t-t_0)$$ We want to show that $\phi$ is the best approximation of $m$ locally. Suppose that there is another tangent vector that is a better approximation, say $v$, so that within some neighbourhood $|t- t_0|<\epsilon$: $$||m(t) - m(t_0) - v(t-t_0)|| \leq ||m(t) - m(t_0) - m'(t_0)(t-t_0)||$$ Dividing by $|t-t_0|$, you have: $$ ||\frac{m(t) - m(t_0)}{|t-t_0|} - v \frac{t-t_0}{|t-t_0|} || \leq ||\frac{m(t) - m(t_0)}{|t-t_0|} - m'(t_0) \frac{t-t_0}{|t-t_0|} ||$$ Taking the limit as $ t\to t_0$ and using the continuity of the norm and the absolute value, and noting that as $m$ is differentiable, the limit exists so we don't have to worry about only having the positive limit in the denominator, then you just get the derivative in the first terms of the above: $$||m'(t_0) - v \frac{t_0}{|t_0|}|| \leq || m'(t_0) - m'(t_0)\frac{t_0}{|t_0|}|| $$ Since $t_0 \in [0,1]$ is non negative, this is just: $$|| m'(t_0) - v|| \leq ||m'(t_0) - m'(t_0)|| = 0$$ So this implies that $m'(t_0) -v =0$, or $v = m'(t_0)$, which shows that this is the best tangent vector you can possible have as it is the best linear approximation you can get.