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1

Divergence at a point measures the flux flowing out from an infinitesimal volume around that point. Suppose you have a volume $v$ in vector field $\mathbf{h}$. Divide it into two parts: $$v = v_1 +v _2$$. Now the flux out of volume $v$ is given by : $\int \mathbf{h}\cdot da_v = \int \mathbf{h} \cdot da_{v_1} + \int \mathbf{h}\cdot da_{v_2}$. Start dividing ...

1

Think about it one more time. If $\vec{F}$ has continuous partial derivatives, then $$\vec\nabla\cdot\vec{F}=\sum_i \frac{\partial F_i}{\partial x_i}$$ is also continuous. If a function is continuous, it's approximately constant on sufficiently small volumes: that's pretty much the definition of continuity! So your original understanding was just fine. ...

2

If you have the position vector along a path $\vec{r}(q)$ parametrized by $q$, where $q$ can be time, angle, distance, or whatever then the derivatives are: $$\vec{v}(q,\dot{q}) = \frac{\partial \vec{r}(q)}{\partial q} \dot{q}$$ $$v = \| \vec{v} \|$$ $$\vec{e} = \frac{ \vec{v}}{v}$$ $$\vec{a}(q,\dot{q},\ddot{q}) = \frac{\partial ... 2 \newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}Elaborating on Mikael's answer, note that equations like \vec v = \dv{\vec r}{t} are sort of shorthand notations to make the life of a physicist easier. Note that there are two (three in 3d) equations in this condensed notation. What we mean by such equations is simply the following:$$v_x = ...

2

You first write your position vector as $\vec r = (x_0 -t\cdot10~\mathrm{m/s},y_0)$ and then take the derivative of that. This produces $$\vec v = \frac{d\vec r}{dt}=(-10,0)~\mathrm{ms^{-1}}=-10\hat i\,~\mathrm{ms^{-1}}$$ which is what you want.

1

If after falling some distance $y$ the rain drop has radius $r(y)$ and cross-sectional area $A(y)=\pi (r(y))^2$when it falls an additional small distance $\delta y$ it sweeps out an additional volume $\delta V_v\approx A(y)\delta y= \pi (r(y))^2\delta y$. Then taking limits as $\delta y \to 0$ gives: $$\frac{dV_c}{dy}=\pi (r(y))^2$$

2

Ain't nothing quite like a picture to explain this: The drop falls a distance $v$ in one second, so the volume of the cylinder is $\pi r^2 v$. Of course, the drop grows as it picks up all the moisture in the gray region - so $r$ will be a function of distance fallen, and the volume swept will increase accordingly (right hand diagram). In essence the ...

0

Sourisse's comment answers your question, but just for the record I'll expand on it here as a Wiki answer. Note that this is a physicist's answer - any mathemticians present would be wise to avert their gaze now. Remember that when we say that the volume element is: $$dV = 4\pi r^2 dr \tag{1}$$ We are talking about the limit in which $dr \rightarrow 0$. ...

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