I think you might be interested in the *acceleration* of integral curves of the gradient vector field (aka, "gradient flow curves").

Suppose $f:\mathbb{R}^n\to\mathbb{R}$ is at least $C^2$, with gradient $\nabla f$ and Hessian $\operatorname{H}f$.

If $\gamma$ is an integral curve for $\nabla f$, then for all $t$,
$$ \ddot{\gamma}(t) = \operatorname{H}f_{\gamma(t)} \dot{\gamma}(t) $$
This follows from $\nabla f_q \approx_{q\to p} \nabla f_p + \operatorname{H}f_p (q-p)$ applied to the fact that $\dot{\gamma} = \nabla f$.

Since $f\in C^2$, its Hessian is symmetric, so admits an eigenvalue decomposition. Say a **gradient flow curve of minimal acceleration** is a gradient flow curve $\gamma$ with $\dot{\gamma}$ always lying in a minimum-eigenvalue eigenspace of $\operatorname{H}f$. (This need not always happen.)

Reversing time, so $t\to -\infty$, gradient flow curves will tend toward minimal acceleration curves. Here is a quick qualitative argument: If $e_j, \lambda_j$ is an orthonormal basis of eigenvectors of $\operatorname{H}f$ at point, write in coordinates $\dot{\gamma} = c_1v_1+\ldots +c_nv_n$. Now observe that $c_j(t - dt) \approx c_j(t) / (1 + dt \lambda_j)$, so the larger $\lambda_j$, the faster the component of $\dot{\gamma}$ in the direction of $v_j$ decays as $t\to -\infty$.

This tendency confirms the intuition of a waterslide: as water tumbles downhill (following exactly gradient descent), it accumulates toward a curve of minimal acceleration.

I would expect physicists have studied this concept as it is similar to the Hamiltonian formulation of classical mechanics. But I don't know the keywords to search.

## Examples

Just to check whether this concept tracks with the desired intuition, this section works through the examples shared by Iosif Pinelis above, plus one more to illustrate.

### (A) $f(x,y) = x^2 + y$

The Hessian of $f$ has two eigenvalues, $2$ and $0$, with eigenspaces parallel to the $x$- and $y$-axes respectively. For initial data $c_x, c_y$ the gradient flow curves have the form $ \gamma(t) = \big( c_x \exp(2t), t + c_y \big) $. Thus the minimum acceleration curve is the y-axis and as $t\to -\infty$ the gradient flow paths accumulate to the $y$-axis as expected.

### (B) $f(x,y) = x^2 - y^2$

The Hessian of $f$ has two eigenvalues at each point, $2$ and $-2$, with eigenspaces parallel to the $x$- and $y$-axes respectively. For initial data $c_x, c_y$ the gradient flow curves have the form
$ \gamma(t) = \big( c_x \exp(2t), c_y\exp(-2t) \big) $
and the two minimal acceleration gradient flow curves are the positive and negative y axes: exactly how we would expect water to flow off a saddle.

### (C) $f(x,y) = x + \frac{1}{3}x^3 + y$

This example illustrates that curves of minimal acceleration need not exist. The Hessian of $f$ at $(x,y)$ has two eigenvalues, $2x$ and $0$, with eigenspaces parallel to the $x$-axis and the $y$-axis respectively.

The gradient flow curves have the form $\gamma(t) = \big( \tan(c_x + t), c_y + t \big)$. As $\dot{\gamma}(t) = \big(\sec^2(c_x + t), 1\big)$ we see that no gradient flow curves are ever tangent to an eigenspace of the Hessian. So there are no minimal acceleration gradient flow curves.

### (D) $f(x,y) = \sin(x) + y$

Here's a picture of the graph of $f$:

$\sin(x)+y$" />

Intuition is that the minimal acceleration gradient curves lie along the valleys of this slope. The gradient of $f$ is $\nabla f = \big( \cos(x), 1\big)$ and computing the Hessian of $f$ yields
$$ \operatorname{H}f = \begin{bmatrix} -\sin(x) & 0 \\ 0 & 0 \end{bmatrix} $$

Gradient flow curves for initial data $c_x,c_y$ are
$ \gamma(t) = \big( 2\arctan(\tanh((t + c_x)/2), t + c_y \big)$, except for a degenerate family of paths moving at constant speed parallel to the $y$ axis when $x = \pi/2 + k\pi$ corresponding to the ridges and valleys.

The eigenvalues of the Hessian matrix are $-\sin(x), 0$ with eigenspaces parallel to the $x$ and $y$ axes respectively (away from $x$ a multiple of $2\pi$, in which case there is a single $0$-eigenspace). The only paths that are always tangent to an eigenspace are the degenerate paths $(c_x, t+c_y)$ when $c_x = \pi/2 + k\pi$ ($k\in\mathbb{Z}$).

When $k$ is odd these degenerate paths are gradient flow curves of minimal acceleration (Hessian eigenvalue $0$) and as expected the gradient flow accumulates to these paths as $t\to -\infty$.

As an aside: at $x = \pi/2 + k\pi$ ($k$ odd) the function $f$ is equal to example (A), the "waterslide," to second order (and scaling and addition of a constant). So it is no surprise that the gradient curves behave the same way.

To illustrate, here is a plot of the gradient flow of $f$ superimposed on a contourplot of $f$:
$\sin(x)+y$" />