 ... ... @@ -67,19 +67,19 @@ $(r,\varphi)$ indicated. From this, we can see that the *Cartesian* coordinates $(x,y)$ of the point are related to the polar ones as follows: $$x = r \cos\varphi;$$ $$y = r \sin \varphi.$$ $$\begin{equation} x = r \cos\varphi; \end{equation}$$ $$\begin{equation} y = r \sin \varphi.\end{equation}$$ ![image](figures/Coordinates_9_0.svg) The inverse relation is given as $$r=\sqrt{x^2 + y^2};$$ $$\varphi=\begin{cases}$$\begin{equation} r=\sqrt{x^2 + y^2}; \label{rxy}\end{equation}\begin{equation} \varphi=\begin{cases} \arctan(y/x) & \text{$x>0$,}\\ \pi + \arctan(y/x) & \text{$x<0$ and $y>0$,}\\ -\pi + \arctan(y/x) & \text{$x<0$ and $y<0$.} \end{cases}$$\end{cases} \label{phixy}\end{equation}$$ The last formula for $\varphi$ warrants a closer explanation: It is easy to see that $\tan(\varphi)=y/x$ - but this is not a unique relation, due to ... ... @@ -133,8 +133,8 @@ understood: the area swept by an angle difference $d\varphi$ We find: $$\int_0^{2\pi} d\varphi \int_0^r_0 r dr =\\ 2 \int_0^{2\pi} d\varphi \int_0^r_0 r dr = 2\pi \int_0^r_0 r dr = 2 \pi \frac{1}{2} r_0^2 = \pi r_0^2,$$ which is indeed the area of a circle with radius 0. ... ... @@ -143,10 +143,10 @@ understood: the area swept by an angle difference $d\varphi$ Often, in physics important equations involve derivatives given in terms of Cartesian coordinates. One prominent example are equations of the form $$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2)\right)$$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right) f(x, y) = \ldots.$$The derivative operator \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2)\right) is so common it has its own name: \frac{\partial^2}{\partial y^2}\right) is so common it has its own name: the Laplacian (here for two-dimensional space). Such an equation is universal, but for particular situations it might be ... ... @@ -162,9 +162,17 @@ involving the chain rule for a function of several variables. Let f be a function of n variables: f(y_1, y_2, \ldots, y_n), as well as g_i(x_1, x_2, \ldots, x_n) for i=1,2,\ldots, n. Then$$\frac{\partial}{\partial{x_i} = \sum_{j=1}^n $$\frac{\partial}{\partial x_i} = \sum_{j=1}^n \fac{partial f}{\partial y_j} \frac{\partial g_j}{\partial x_i}$$ We start by replacing the function $f(x, y)$ by a function in polar coordinates $f(r, \varphi)$, and ask what is $\frac{\partial}{\partial x} f(r, \varphi)$. When we look at this expression, we need to understand what it *means* to take the derivative of a function of $r, \varphi$ in terms of $x$? For this, we need to realize that there are relations between the coordinate systems. In particular, $r = r(x,y)$ and $\varphi = \varphi(x, y)$ as defined in equations \ref{rxy} and \ref{phixy}. # Cylindrical coordinates ... ...