Math in Mechanics

Question

I thought $\frac{d}{dx}, \frac{\partial }{\partial x}$ are basically the same thing and whenever it occurs in a problem I used just $\left(x\right)^{'}$. But recently my professor has been using $\int \:$ for $\frac{d}{dx}$. Why is that so? What fundamental misunderstandings have I? Or is is something else? Specific example that my professer used: First: $$\frac{dV}{dr}=\frac{r}{2a}\cdot \frac{\partial p}{\partial s}=\frac{r}{2a}\cdot \frac{dp}{ds}$$ (side question: is it ok to just change $\frac{\partial p}{\partial s}$ to $\frac{dp}{ds}$ like he did? or am I still missing something here) After: $$V\left(r\right)=\frac{r^2}{4a}\cdot \frac{dp}{ds}+C$$ P.S. It's laminar pipe flow example. $V$-Volume, $r$-pipe radius, $p$-pressure, $a$-liquid viscosity, $s$-depth.

Answer 1

Take a look at these examples: $f = f(x)$ is a notation for a function of the variable $x$, $g = g(x,y)$ for a function of the variables $x,y$, and again $h = h(x,y(x))$ is a notation for a function of two variables $x,y$, where $y$ in turn depends on $x$. When you differentiate a function of one variable like $f$ you usually write $$f' = \frac{df}{dx}$$ whereas when the function depends on two variables (like $g$ above) you use the notation of partial derivation: $$\frac{\partial g}{\partial x} \quad \text{ and } \quad \frac{\partial g}{\partial y}.$$ The notation $\frac{\partial }{\partial x}$ (resp. $\frac{\partial}{\partial y}$) has the meaning of differentiating with respect to the first (resp. second) variable! To see what I mean, check this: suppose you want to differentiate $h$ as above with respect to the variable $x$. In that case you still use the notation $\frac{d}{dx}$ as follows:

$$\frac{d}{dx}h(x,y(x)) = \frac{\partial h}{\partial x}(x,y)+ \frac{\partial h}{\partial y}(x,y)\frac{dy}{dx}(x).$$ This is sometimes called 'total derivative'. Observe how the notation changes and that this derivation is fundamentally different from the partial derivation with respect to the first $\mbox{variable, $\frac{\partial }{\partial x}$.}$

Finally, about your side question: if $\rho$ depends only on $s$ yes, you should change your notation.