I am having trouble identifying the best manner to define my problem. More specifically, suppose one has a three dimensional, spherical surface charge density. The density is uniform, meaning it is just a constant value $\rho$ everywhere on the surface $S$. This surface can easily be defined by the following parametrization:
$$ S = R\sin\theta\cos\phi \hat{a}_x + R\sin\theta\sin\phi \hat{a}_y + R\cos\theta\hat{a}_z $$
Thus, we can simply state the following to define the surface charge density:
$$ \rho(\mathbf{r})= \begin{cases} \rho,\text{for }(x,y,z)\in S\\ 0,\text{otherwise} \end{cases} $$
However, now suppose that we wish to oscillate the sphere up and down sinusoidally in the $z$ direction according to: $$ M(t)=\Delta z \cos(\omega t) $$ In other words, the surface becomes: $$ S = R\sin\theta\cos\phi \hat{a}_x + R\sin\theta\sin\phi \hat{a}_y + (R\cos\theta+M(t))\hat{a}_z $$ The surface charge density is still defined as shown above in the piecewise function, only now $S$ has changed form a bit.
So now here is the main point of the question. Let's suppose I want to calculate $\frac{\partial \rho(\mathbf{r})}{\partial t}$. $\rho$ isn't actually an explicit function of the coordinates or time other than how its defined in the piecewise function, so how can I calculate the temporal derivative? I thought maybe I can use Heaviside step functions but it is tricky with an infinitesimally thin surface. I then thought perhaps use Dirac delta functions such as:
$$ \rho(\mathbf{r},t) = \rho\left( \delta(x-R\sin\theta \cos\phi)\delta(y-R\sin\theta\sin\phi)\delta(z-R\cos\theta-M(t)) \right) $$
But this doesn't really seem correct , as it seems like a sloppy use of the Dirac Delta (i.e., treating it like the Kronecker delta when it is more nuanced than that).
If I write out the total derivative, it seems like all of the terms go to zero because it necessarily involves derivatives of $\rho$ with respect to the spatial coordinates, which is zero because it is simply equal to a constant value within the specified domain.
Some points are never reached by the oscillating sphere. Here the charge density is always zero, and so you can say the time derivative is zero at these points. Other points, such as $(0,0,R+ \frac{\Delta z}2)$ are reached periodically. At one instant at such a point, the charge density is $\rho,$ while immediately after its zero. The charge density at such points is discontinuous in time, and hence the time derivative doesn't exist.
Added: For example, for a spherical shell with inner radius $r_1$ and outer radius $r_2$, define the function $f: \mathbb{R}^4 \to \mathbb{R}$ by $$f(x,y,z,t) = \begin{cases} \rho \Psi(x^2+y^2+(z-M(t))^2), &r_1^2< x^2+y^2+(z-M(t))^2 < r_2^2\\ 0 &\text{otherwise},\end{cases}$$ where $\Psi$ is a bump function with domain $(r_1^2,r_2^2)$ as on the wiki page. In short, it is a function which is zero outside $(r_1^2,r_2^2)$ which quickly goes to $1$ inside $(r_1^2,r_2^2)$, in such a way that the function is smooth (infinitely differentiable). Try to verify that this function represents a oscillating spherical shell, and try to intuitively see that it should be differentiable as $\Psi$ is differentiable. If you find it worth the time, you could try to write out a rigorous argument that $f$ is time-differentiable.