Hey everyone. I'm having a hard time getting a visual of $\displaystyle \vec{T} $ being perpendicular to $\displaystyle \vec{T'} $. The proof makes sense. Here it is:

Assume the vector function $\displaystyle |\vec{r(t)}| = c $

Thus, we take the dot product of the same vector:

$\displaystyle \vec{r(t)} * \vec{r(t)} = |\vec{r(t)}|^2 = c^2 $

If we were to take the derivative of this equation, we would get:

$\displaystyle 0 = \frac{d}{dt}[\vec{r(t)} * \vec{r(t)}] = \vec{r'(t)} * \vec{r(t)} + \vec{r(t)} * \vec{r'(t)} = 2\vec{r'(t)} * \vec{r(t)}$

Thus, $\displaystyle \vec{r(t)} * \vec{r'(t)} = 0 $

Now this make sense because we are talking about a position vector and its tangent. This is also easy to visualize: a point on a circle. The position vector is like the centripetal force, it points toward the center. So this makes perfect sense.

Now let's talk about $\displaystyle \vec{T(t)} $ and $\displaystyle \vec{T'(t)} $

$\displaystyle |\vec{T(t)}| = 1$ so the proof above applies here too. The problem is that I can't visual this. I mean I can but it contradicts the proof. $\displaystyle \vec{T(t)} $ is not a position vector of $\displaystyle \vec{r(t)} $. It is actually the unit vector of $\displaystyle \vec{r'(t)} $; thus, we're basically saying $\displaystyle \vec{r'(t)} $ is perpendicular to $\displaystyle \vec{r''(t)} $, which is not true. (ie. if f'(x) = x^2, f'"(x) = 2. They are not perpendicular.) HOWEVER, that is in 2D. idk what happens in 3D cuz I am having a hard time visualizing it.

So the question is, can someone please give me an example of where the tangent and the 2nd derivative of a function are perpendicular? Or somehow help me grasp this. Thanks in advance!