Material derivatives and the chain rule

I am currently using this webpage to understand more about the material derivative. Specifically I am looking at the example at the bottom of the page involving free fall motion.

The equation describing free fall of an object with initial velocity is as follows \begin{align} y = y_0 + v_o t - {1 \over 2} g \, t^2 \tag 1\\ \end{align}

Differentiating once yields velocity as a function of time :

\begin{align} v = {dy \over dt} = v_o - g \, t \tag 2\\ \end{align}

Solving equation (1) for $t$ and plugging into (2) yields velocity as a function of position $y$

\begin{align} v_y = \pm \sqrt{v_o^{\,2} - 2 g (y - y_o)} \tag 3\\ \end{align}

Now the author claims the following:

\begin{align} a = {D v_y \over D t} = {\partial v_y \over \partial t} + v_y {\partial v_y \over \partial y} \tag 4\\ \end{align}

\begin{align} {\partial v_y \over \partial t}=0 \tag 5 \end{align}

To explain (5) the author states it is "because the velocity, as described here, is not transient. It does not change with time at any given location."

I don't agree with either statement. Regarding statement (4), and regardless of how the equations are arranged, $y$ is still a function of time, and therefore the chain rule should apply as follows:

\begin{align} \frac {d} {dt} v(y(t))=\frac{dv}{dy} \frac {dy}{dt} \tag 6 \end{align}

$$\frac {d} {dt} v(y(t)) = {\mp g (v_o - g \, t) \over \sqrt{v_o^{\,2} - 2 g (y - y_o)}}$$

which, yes, if you substitute (1) back into the above equation will yield a = -g. However I don't understand where equation (4) comes from and why it yields the same result as equation (6). In fact, why are we even taking partial derivatives? This is all scalar valued functions in 1D....

• equation $(4)$ come into play only when the velocity is a function of both position $y$ and time $t.$ this is what happens in the eulerian description of fluid flow. there is also the lagrangian description that follows each particle.
– abel
Jan 12 '17 at 14:55

$$\frac{Dv_y}{Dt}=\frac{\partial v_y}{\partial t}+v_y\frac{\partial v_y}{\partial y}=\frac{dv}{dy}\frac{dy}{dt}$$
since $\frac{dy}{dt}=v_y$. We take partial derivatives since $v_y$ as expressed in $(3)$ is not directly a function of time. You are right that single variable chain rule suffices, as multi-variate chain rule reduces down to it, but, the multivariate version is better. For example, imagine if we replaced $g$ with $a(t)$, then,
$$\frac{\partial v_y}{\partial t}\ne0$$