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
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- $\begingroup$ The thing about curves is that they also have an inherent velocity (chosen when you chose a parametrization). So your tangent vector will reflect this, a longer tangent coming from a faster curve. So yes, there are many vectors that point in the same "correct" direction, but you also care about how fast, and that is what dictates the true tangent vector and the direction of the resulting line (along inertia). $\endgroup$Randall– Randall2018-10-04 17:15:14 +00:00Commented Oct 4, 2018 at 17:15
- $\begingroup$ It is the direction of $\mathbf{T}$ that has me baffled. Many other tangent lines could be drawn to point $P$ that go all of the other directions. $\endgroup$Narlin– Narlin2018-10-04 17:20:59 +00:00Commented Oct 4, 2018 at 17:20
- $\begingroup$ I might understand what you mean by velocity, since that has to have both magnitude and direction. Why this direction? $\endgroup$Narlin– Narlin2018-10-04 17:22:21 +00:00Commented Oct 4, 2018 at 17:22
- $\begingroup$ The answer below is getting at it. When you ride the curve and slam on the breaks (no seatbelt), your body wants to travel a certain direction at a certain speed. That's the data in your tangent vector. $\endgroup$Randall– Randall2018-10-04 17:30:02 +00:00Commented Oct 4, 2018 at 17:30
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1 Answer
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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 $T_1=T_2$.
