Why is the tangent vector in $\mathbb{R}^3$ unique

Question

Given a vector function which has a derivative not equal to zero, for example $\mathbf{r}=(t,\,\cos(t)\,\sin(t))$, its derivative is called a tangent vector to the function $\mathbf{r}.$ $$\mathbf{r^\prime}=\left(\begin{array}{c}1\\-\sin(t)\\ \cos(t)\end{array}\right),\qquad 0\le t \le 2\pi.$$ If I graph $\mathbf{r^\prime}$ it just looks like a circle with center $\left(1,0,0\right),$ radius $1,$ and in the plane $x=1.$ However, I can choose some $t$ value and create a point on $\mathbf{r}$ as $P=\mathbf{r}(t)$ and that point will now be a vector of scalars. Indeed, $\mathbf{r^\prime}$ is a direction vector for a single tangent line which is tangent to the curve called $\mathbf{r}$ at the point $P$. Again, for example, if we let $t=\pi/2$, then $$\mathbf{r}\left(\frac{\pi}{2}\right) =P=\left(\begin{array}{c}\frac{\pi}{2}\\0\\1 \end{array}\right)$$ and "the" tangent line is $$\left(\begin{array}{c} x\\ y\\ z \end{array}\right)=P+\lambda\left(\begin{array}{c} 1\\ -\sin\left(\frac{\pi}{2}\right)\\ \cos\left(\frac{\pi}{2}\right) \end{array}\right) \tag{eq 1}$$ Call the tangent vector $\mathbf{\mathcal{T}}$ and note that wlog it could be $\mathbf{T}$ where $$\mathbf{T}=\frac{\mathbf{r^\prime}}{\Vert\mathbf{r^\prime}\Vert}$$ It just so happens that I want $\mathbf{\mathcal{T}}$ or $\mathbf{T}$ to be in the osculating plane, but I feel extraordinarily lucky that it occurred there, given that we could find an infinite number of other lines with corresponding direction vectors that would be tangent to point $P$. I am adding a picture, but my question is $\underline{\text{why does this math just happen to pick the tangent line that I want?}}$ Basically, I have the same question for the unit Normal, but perhaps if I can understand this one, I will figure the other out for myself. In context, I am trying to explain tangents, normals, and binormals to someone else and finally realized that I probably don't know it]1

Answer 1

You feel that you have many choices for the tangent, but that's not the case.

Intuitively, if you think you are "riding" the curve, you can think of the tangent vector as telling exactly what to do with your steering wheel (that would be the unit tangent vector) and the accelerator/brakes (that would be tangent's vector magnitude). So there is a single one.

Mathematically, you have $$ r(t+h)=r(t)+hr'(t)+ o(h^2).$$ If you have two "tangent vectors" $T_1$ and $T_2$, you would have $$ r(t+h)=r(t)+hT_1+o(h^2), \ \ \text{ and } r(t+h)=r(t)+hT_2+o(h^2). $$ This gives you $$ h(T_1-T_2)=o(h^2), $$ that can only happen when $T_1=T_2$.