I was trying to understand why the unit normal vector is normal to the direction of motion. Note that $\mathbf T(t) = \frac{\mathbf r'(t)}{||\mathbf r'(t)||}$ is the unit tangent vector for some vector valued function $\mathbf r(t)$, and the unit normal vector is defined to be $\mathbf N(t) = \frac{\mathbf T'(t)}{||\mathbf T'(t)||}$My book offered the following proof:
To prove that $\mathbf T(t)$ and $\mathbf T'(t)$ are orthogonal, note that $\mathbf T(t)$ is a unit vector, so $\mathbf T(t) \bullet \mathbf T(t) = 1$.
//this is because $\mathbf u \bullet \mathbf v = (||\mathbf u||$ $ ||\mathbf v|| ) cos\theta$
$\frac{d}{dt}\mathbf T(t)\bullet\mathbf T(t) = 2\mathbf T(t)\bullet\mathbf T'(t) = 0$
This shows that $\mathbf T(t)\bullet\mathbf T'(t) = 0$
That's all well and good, and I understand WHY that proves that $\mathbf N(t)$ is orthogonal to $\mathbf T(t)$, but now I'm extremely confused because I can use the same proof to show that $\mathbf T(t)$ is orthogonal to $\mathbf r(t)$.
$\mathbf r(t) \bullet \mathbf r(t) = c$
$\frac{d}{dt}\mathbf r(t)\bullet\mathbf r(t) = 2\mathbf r(t)\bullet\mathbf r'(t) = 0$
What am I missing?
It's not so much what is being missed here, as it is what is being assumed . . .
The assertion that
$\mathbf r(t) \cdot \mathbf r(t) = c \tag{1}$
only holds under the condition that $\Vert r(t) \Vert = \sqrt{c}$ is a constant, i.e., the curve lies in a sphere of radius $\sqrt{c}$ centered at the origin. In this case,
$\mathbf r(t) \cdot \mathbf r'(t) = 0 \tag{2}$
follows and makes perfect sense: every tangent vector to a sphere is normal to its radial vector. For a general curve,
$\mathbf r(t) \cdot \mathbf r(t) = c(t) \tag{3}$
where $c(t)$ is a time-varying function. Then (2) will not hold though we still have
$\mathbf T(t) \cdot \mathbf T'(t) = 0, \tag{4}$
since
$\mathbf T(t) \cdot \mathbf T(t) = 1, \tag{5}$
where $\mathbf T(t) = \mathbf r(t) / \Vert \mathbf r(t) \Vert$.
Hope this helps. Cheers!
And as ever,
Fiat Lux!!!