I would like to prove the following for the motion of an object on a space curve.
"The intrinsic acceleration of the object is equal to the dot product of the velocity and acceleration vectors, divided by the speed."
This is probably pretty easy, but I'm just not seeing it. Also, I would like to rewrite the formula using math symbolic notation, but I don't know the math code I would need to use. Can you direct me?
I will update the question when I figure this out. Thank you.
Suppose the curve in question is $\gamma(t)$, i.e., $\gamma$ maps some open interval $I \subset \Bbb R$ into $\Bbb R^n$, $n$ a postive integer. Note we are addressing the possibility that $n > 3$, since the essential ideas are the same for all positive $n$. With $\gamma: I \to \Bbb R^n$ at least twice differentiable ($\gamma \in C^m(I, \Bbb R^n)$ with $ m \ge 2$), we have the tangent vector $\dot \gamma(t)$ to $\gamma(t)$, in terms of which the speed $\dot s$, that is, the instantaneous rate of distance along the curve traversed per unit time, is given by
$\dot s(t) = \dfrac{ds(t)}{dt} = (\dot \gamma \cdot \dot \gamma)^{1/2}; \tag 1$
then the acceleration along $\gamma(t)$, in the sense of the rate of change of $\dot s(t)$, is
$\ddot s(t) = \dfrac{d}{dt}(\dot \gamma \cdot \dot \gamma)^{1/2} = \dfrac{1}{2}(\dot \gamma \cdot \dot \gamma)^{-1/2}\dfrac{d}{dt}(\dot \gamma \cdot \dot \gamma) = \dfrac{1}{2}(\dot \gamma \cdot \dot \gamma)^{-1/2}(2(\ddot \gamma \cdot \dot \gamma)) = (\dot \gamma \cdot \dot \gamma)^{-1/2}(\ddot \gamma \cdot \dot \gamma), \tag 2$
which is exactly the formula described in the text of the question.
