Equating Acceleration Terms for a Parametric Surface Suppose we have a smooth 2D surface embedded in 3D Euclidean space defined parametrically by,
$$
r : \mathbb{R^2} \to \mathbb{R}^3\ ,\ \ \ r(x,y) := \begin{bmatrix} \bar{x}(x,y) \\ \bar{y}(x,y) \\ \bar{z}(x,y) \end{bmatrix}
$$
We also have its Jacobian and higher partial derivatives:
$$
J(x,y) := \begin{bmatrix} \frac{\partial r}{\partial x}(x,y) & \frac{\partial r}{\partial y}(x,y) \end{bmatrix} \in \mathbb{R}^{3 \times 2}
$$
Consider a 1D trajectory through the parametric space:
$$
q : \mathbb{R} \to \mathbb{R}^2\ ,\ \ \ q(t) := \begin{bmatrix} x(t) \\ y(t) \end{bmatrix}
$$
By chain-rule (and over-dot notation) the ambient velocity is,
\begin{align}
\frac{dr}{dt} &= \frac{dr}{dq} \frac{dq}{dt}\\
\dot{r} &= J \dot{q}
\end{align}
By product-rule the ambient acceleration is,
$$
\ddot{r} = J \ddot{q} + \dot{J} \dot{q}
$$
A similar two-term expression for the acceleration can be derived by expressing the velocity as a product of its magnitude $v := ||\dot{r}||$ and direction $\hat{\tau} := \dot{r} / v$.
\begin{align}
\dot{r} &= v\hat{\tau}\\
\implies\ \ddot{r} &= \dot{v}\hat{\tau} + v\dot{\hat{\tau}}
\end{align}
Obviously we can equate these $\ddot{r}$ expansions, but I am wondering about whether the following red/blue terms are individually equatable:
$$
\ddot{r} = \color{red}{J \ddot{q}} + \color{blue}{\dot{J} \dot{q}} = \color{red}{\dot{v}\hat{\tau}} + \color{blue}{v\dot{\hat{\tau}}}
$$
The term $\color{red}{J \ddot{q}}$ is always in the tangent plane because it is a linear combination of tangent vectors:
$$
\color{red}{J \ddot{q}} = \ddot{x} \frac{\partial r}{\partial x} + \ddot{y} \frac{\partial r}{\partial y}
$$
The same is true for $\color{red}{\dot{v}\hat{\tau}}$ since it is in the direction of the velocity. This inclines me to write a stronger pair of equalities:
\begin{gather}
\color{red}{J \ddot{q} \overset{?}{=} \dot{v}\hat{\tau}} \tag{1}\\
\color{blue}{\dot{J} \dot{q} \overset{?}{=} v\dot{\hat{\tau}}} \tag{2}
\end{gather}
To actually prove this, I think I would need to show that $\color{blue}{\dot{J} \dot{q}}$ and $\color{blue}{v\dot{\hat{\tau}}}$ are orthogonal to the tangent plane. Edit: nevermind, that isn't true, but also isn't necessary.
I have been unable to show this in general (though I might just be missing a fact that could allow me to simplify things). How should I conclude on conjectures (1) and (2)? And if they are false, can you shed some light on the geometry of the relationship between the two different $\ddot{r}$ expansions?
Thanks!
 A: The conjectures are false. To see why, first lets equate our expressions for $\dot{r}$.
$$
\dot{r} = J\dot{q} = v\hat{\tau}
$$
Non-degeneracy of the surface-coordinates imposes that $\frac{\partial r}{\partial x}$ is linearly independent of $\frac{\partial r}{\partial y}$. Therefore, $J$ is full-column-rank and admits a left-pseudoinverse $\ \tilde{J} := (J^\intercal J)^{-1} J^\intercal\ $ such that $\ \tilde{J}J = I$.
$$
\dot{q} = v\tilde{J}\hat{\tau}
$$
Taking another $t$-derivative we have,
$$
\ddot{q} = \dot{v}\tilde{J}\hat{\tau} + v\dot{\tilde{J}}\hat{\tau} + v\tilde{J}\dot{\hat{\tau}}
$$
Lets simplify this with the following definition:
\begin{gather}
h := \tilde{J}\hat{\tau}\ \implies\ \dot{h} = \dot{\tilde{J}}\hat{\tau} + \tilde{J}\dot{\hat{\tau}}\\[4pt]
\therefore\ \ddot{q} = \dot{v}\tilde{J}\hat{\tau} + v\dot{h}
\end{gather}
In general, $J\tilde{J} \neq I$. However, consider that $\tilde{J}$ acts to orthogonally project ambient vectors onto the local tangent-plane. Therefore, vectors that already live in the tangent-plane like $\hat{\tau}$ have the property $J\tilde{J}\hat{\tau} = \hat{\tau}$. (This isn't the case for $\dot{\hat{\tau}}$ which can have a normal component). Thus, left-multiplying our last equation by $J$ yields the corrected red equation:
$$
\color{red}{J\ddot{q} = \dot{v}\hat{\tau} + vJ\dot{h}}
$$
Substituting this into the original acceleration equation yields the corrected blue equation:
\begin{align}
J\ddot{q} + \dot{J}\dot{q} &= \dot{v}\hat{\tau} + v\dot{\hat{\tau}}\\[4pt]
\dot{v}\hat{\tau} + vJ\dot{h} + \dot{J}\dot{q} &= \dot{v}\hat{\tau} + v\dot{\hat{\tau}}\\[4pt]
\color{blue}{\dot{J}\dot{q}} &\color{blue}{=} \color{blue}{v\dot{\hat{\tau}} - vJ\dot{h}}\\[4pt]
\end{align}
Since in general $v \neq 0$ and $J \neq 0$, the question of whether the original conjectures were true is a question of whether $\dot{h} \overset{?}{=} 0$. Examples can show this is false in general, but there is insight to be had now that we've isolated the correction.
One can geometrically interpret $h = \tilde{J}\hat{\tau} = \dot{q}/v$ as the 2D "heading" vector that an ant on the surface could express its direction of travel with, all in the tangent-plane basis. Whenever the ant turns or the tangent-plane basis-vectors change interior angle, $\dot{h}$ will be nonzero and the original conjectures won't hold. These ideas are put to use here.
