From 2153e89b1fbe03f390313844a2953a7ede645bf1 Mon Sep 17 00:00:00 2001
From: Maciej Topyla <m.m.topyla@student.tudelft.nl>
Date: Sat, 3 Sep 2022 17:36:11 +0000
Subject: [PATCH] Update src/2_coordinates.md

---
 src/2_coordinates.md | 45 ++++++++++++++++++++++++--------------------
 1 file changed, 25 insertions(+), 20 deletions(-)

diff --git a/src/2_coordinates.md b/src/2_coordinates.md
index 780b939..d55bb13 100644
--- a/src/2_coordinates.md
+++ b/src/2_coordinates.md
@@ -29,7 +29,7 @@ and $(x'_1, x'_2, \ldots, x'_n)$ can be quickly computed using a general formula
 
 $$\Delta s^2 = (x'_1 - x_1)^2 + (x'_2 - x_2)^2 + \ldots + (x'_n - x_n)^2.$$
 
-(A space with such a distance definition is called a *Euclidean
+(A space with such a distance definition is called an *Euclidean
 space*.)
 
 In mathematics, we are often dealing with so-called *infinitesimally* small
@@ -195,7 +195,7 @@ 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
-of the [inverse relations](#The-inverse-relation). In fact, we have been rather sloppy in our notation above,
+of [the inverse relations](#the-inverse-relation). In fact, we have been rather sloppy in our notation above,
 as the functions $f(x,y)$ and $f(r, \varphi)$ do not mean that I substitute $x=r$
 and $y=\varphi$! It is more precise to state that there are two diferent
 functions $f_\text{cart}(x,y)$ and $f_\text{polar}(r, \varphi)$ that are equivalent,
@@ -214,22 +214,22 @@ $$ \frac{\partial}{\partial x} f(r, \varphi) =
 $$
 
 and it is now a matter of (tedious) calculus to arrive at the right result.
-This is the task of exercises 3 and 4, which finally compute the Laplacian
+This is the task of exercises 3 and 4, which lead you to compute the Laplacian
 in polar coordinates.
 
-!!! warning Inverse function theorem
-    In this calculation one might be tempted to use the inverse
+!!! warning "Inverse function theorem"
+    In this calculation, one might be tempted to use the inverse
     function theorem to compute derivatives like
     $\frac{\partial \varphi}{\partial x}$ from the much simpler
-    $\frac{\partial x}{\partial \varphi}$. Note though that here we
-    are dealing with functions depending on several variables, so the
-    *Jacobian* has to be used (see [Wikipedia](https://en.wikipedia.org/wiki/Inverse_function_theorem)). A direct calculation is in this particular case more easy.
+    $\frac{\partial x}{\partial \varphi}$. However, note that here we
+    are dealing with functions depending on several variables, so an appropriate
+    *Jacobian* has to be used (see [Wikipedia](https://en.wikipedia.org/wiki/Inverse_function_theorem)). A direct calculation is in this particular case considerably easier.
 
-Note that this procedure also carries over to other coordinate systems,
-although the calculations can become quite tedious. In these cases,
-it's usually best to look up the correct form.
+Note that this procedure also applies to transformations to other coordinate systems,
+although the calculations can become quite tedious. In conventional cases,
+it is usually advised to look up the correct form.
 
-## Coordinate systems in 3D
+## 2.3. Coordinate systems in 3D
 
 <iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/VjUbnZN1BvA?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
 
@@ -239,10 +239,10 @@ it's usually best to look up the correct form.
 Three dimensional systems may have axial symmetry. An example is an
 electrically charged wire of which we would like to calculate the
 electric field, or a current-carrying wire for which we would like to
-calculate the magnetic field. For such problems the most convenient
-coordinates are *cylindrical coordinates*. For convenience, we choose
-the symmetry-axis as the $z$-axis. Note that this can be done as we can
-choose the coordinate system ourselves - this is not imposed by the
+calculate the magnetic field. For such problems, the most convenient
+coordinates are *cylindrical coordinates*. For a further convenience, we choose
+the symmetry-axis as the $z$-axis. Note that this allowed, because we may
+choose the coordinate system ourselves - it is not imposed by the
 problem.
 
 Cylindrical coordinates are defined straightforwardly: we use polar
@@ -250,12 +250,15 @@ coordinates $r$ and $\varphi$ in the $xy$ plane, and the distance $z$
 along the symmetry-axis as the third coordinate. If the axis system is
 chosen in physical space, we have two coordinates which have the
 dimension of a distance: $r$ and $z$. The other coordinate,
-$\varphi$, is of course dimensionless.
+$\varphi$ is of course dimensionless.
 
 What is the distance travelled along a path when we express this in
-cylindrical coordinates? Let’s consider an example (Figure).
+cylindrical coordinates? Let’s consider an example shown in the figure below.
 
-![image](figures/Coordinates_13_0.svg)
+<figure markdown>
+  ![image](figures/Coordinates_13_0.svg)
+  <figcaption></figcaption>
+</figure>
 
 We want to find the length of the (small) red segment $d s$. By
 inspecting the figure, we see that the horizontal (i.e. parallel to the
@@ -265,7 +268,9 @@ segment in the $xy$ plane, expressed in polar coordinates, we
 immediately find:
 $$d s^2 = d l^2 + d z^2 = d r^2 + r^2 d \varphi^2 + d z^2.$$
 The volume element is consequently given as
-$$dV = r dr d\varphi dz.$$
+
+!!! info "Volume element in cylindrical coordinates"
+    $$dV = r dr d\varphi dz.$$
 
 ### Spherical coordinates
 
-- 
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