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1 files changed, 55 insertions, 55 deletions

diff --git a/source/know/concept/curvature/index.md b/source/know/concept/curvature/index.md
index bf8ae25..a3aec08 100644
--- a/source/know/concept/curvature/index.md
+++ b/source/know/concept/curvature/index.md
@@ -7,9 +7,9 @@ categories:
 layout: "concept"
 ---
 
-Given a curve or surface, its **curvature** $\kappa$
+Given a curve or surface, its **curvature** $$\kappa$$
 describes how sharply it is bending at a given point.
-It is defined as the inverse of the **radius of curvature** $R$,
+It is defined as the inverse of the **radius of curvature** $$R$$,
 which is the radius of the tangent circle
 that **osculates** (i.e. best approximates)
 the curve/surface at that point:
@@ -18,7 +18,7 @@ $$\begin{aligned}
     \kappa = \frac{1}{R}
 \end{aligned}$$
 
-Typically, $\kappa$ is positive for convex curves/surfaces,
+Typically, $$\kappa$$ is positive for convex curves/surfaces,
 and negative for concave ones, although this distinction is somewhat arbitrary.
 Below, we calculate the curvature in several general cases.
 
@@ -34,13 +34,13 @@ to find the curvature.
 This approach relies on the fact that a circle
 has the highest area-perimeter ratio of any 2D shape,
 and a sphere has the highest volume-surface ratio of any 3D body.
-By the definition of curvature, these shapes have constant $\kappa$.
+By the definition of curvature, these shapes have constant $$\kappa$$.
 
 We will thus minimize the perimeter/surface while keeping the area/volume fixed,
 which will give us a shape with constant curvature,
-and from that we can extrapolate an expression for $\kappa$.
+and from that we can extrapolate an expression for $$\kappa$$.
 
-In 2D, for a single-variable height function $h(x)$,
+In 2D, for a single-variable height function $$h(x)$$,
 the length of a small segment of the curve is:
 
 $$\begin{aligned}
@@ -49,7 +49,7 @@ $$\begin{aligned}
     = \dd{x} \sqrt{1 + h_x^2}
 \end{aligned}$$
 
-Which leads us to define the following Lagrangian $\mathcal{L}$
+Which leads us to define the following Lagrangian $$\mathcal{L}$$
 describing the "energy cost" of the curve:
 
 $$\begin{aligned}
@@ -66,8 +66,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 By putting these things together,
-we arrive at the following energy functional $E[h]$,
-where $\kappa$ is an ominously-named [Lagrange multiplier](/know/concept/lagrange-multiplier/):
+we arrive at the following energy functional $$E[h]$$,
+where $$\kappa$$ is an ominously-named [Lagrange multiplier](/know/concept/lagrange-multiplier/):
 
 $$\begin{aligned}
     E[h]
@@ -83,7 +83,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 We evaluate the terms of this equation
-to arrive at an expression for the curvature $\kappa$:
+to arrive at an expression for the curvature $$\kappa$$:
 
 $$\begin{aligned}
     \boxed{
@@ -92,15 +92,15 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-In this optimization problem, $\kappa$ is a constant,
+In this optimization problem, $$\kappa$$ is a constant,
 but in fact the statement above is valid for variable curvatures too,
-in which case $\kappa$ is a function of $x$.
+in which case $$\kappa$$ is a function of $$x$$.
 
 
 ## 2D in general
 
 We can parametrically describe an arbitrary plane curve
-as a function of the arc length $s$:
+as a function of the arc length $$s$$:
 
 $$\begin{aligned}
     \big( x(s), y(s) \big)
@@ -108,9 +108,9 @@ $$\begin{aligned}
     \dd{s}^2 = \dd{x}^2 + \dd{y}^2
 \end{aligned}$$
 
-If we choose the horizontal $x$-axis as a reference,
-we can furthermore define the **elevation angle** $\theta(s)$
-as the angle between the reference and the curve's tangent vector $\vu{t}$:
+If we choose the horizontal $$x$$-axis as a reference,
+we can furthermore define the **elevation angle** $$\theta(s)$$
+as the angle between the reference and the curve's tangent vector $$\vu{t}$$:
 
 $$\begin{aligned}
     \vu{t}
@@ -118,9 +118,9 @@ $$\begin{aligned}
     = \big( \cos\theta(s), \sin\theta(s) \big)
 \end{aligned}$$
 
-Where $x_s(s) = \idv{x}{s}$.
-The curvature $\kappa$ is defined as
-the $s$-derivative of this elevation angle:
+Where $$x_s(s) = \idv{x}{s}$$.
+The curvature $$\kappa$$ is defined as
+the $$s$$-derivative of this elevation angle:
 
 $$\begin{aligned}
     \kappa
@@ -128,10 +128,10 @@ $$\begin{aligned}
     = \theta_s(s)
 \end{aligned}$$
 
-We have two ways of writing $\vu{t}$:
-using the derivatives $x_s$ and $y_s$,
-or the elevation angle $\theta$.
-Now, let us take the $s$-derivative of both expressions,
+We have two ways of writing $$\vu{t}$$:
+using the derivatives $$x_s$$ and $$y_s$$,
+or the elevation angle $$\theta$$.
+Now, let us take the $$s$$-derivative of both expressions,
 and equate them:
 
 $$\begin{aligned}
@@ -147,15 +147,15 @@ $$\begin{aligned}
     y_{ss} = \kappa x_s
 \end{aligned}$$
 
-We multiply these equation by $y_s$ and $x_s$, respectively,
+We multiply these equation by $$y_s$$ and $$x_s$$, respectively,
 and subtract the first from the last:
 
 $$\begin{aligned}
     y_{ss} x_s - x_{ss} y_s = \kappa x_s^2 + \kappa y_s^2
 \end{aligned}$$
 
-Isolating this for $\kappa$ and using the fact that $x_s^2 + y_s^2 = 1$
-thanks to $s$ being the arc length:
+Isolating this for $$\kappa$$ and using the fact that $$x_s^2 + y_s^2 = 1$$
+thanks to $$s$$ being the arc length:
 
 $$\begin{aligned}
     \kappa
@@ -165,8 +165,8 @@ $$\begin{aligned}
 
 While this result is correct,
 we would like to generalize it to cases where the curve
-is parametrized by some other $t$, not necessarily the arc length.
-Let prime denote the $t$-derivative:
+is parametrized by some other $$t$$, not necessarily the arc length.
+Let prime denote the $$t$$-derivative:
 
 $$\begin{aligned}
     x_s
@@ -182,7 +182,7 @@ $$\begin{aligned}
     = y'' t_s^2 + x' t_{ss}
 \end{aligned}$$
 
-By inserting these expression into the earlier formula for $\kappa$, we find:
+By inserting these expression into the earlier formula for $$\kappa$$, we find:
 
 $$\begin{aligned}
     \kappa
@@ -194,9 +194,9 @@ $$\begin{aligned}
     &= t_s^3 (x' y'' - y' x'')
 \end{aligned}$$
 
-Since $x_s^2 + y_s^2 = 1$, we know that $(x')^2 + (y')^2 = 1 / t_s^2$,
+Since $$x_s^2 + y_s^2 = 1$$, we know that $$(x')^2 + (y')^2 = 1 / t_s^2$$,
 which leads us to the following general expression for
-the curvature $\kappa$ of a plane curve:
+the curvature $$\kappa$$ of a plane curve:
 
 $$\begin{aligned}
     \boxed{
@@ -205,13 +205,13 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-If the curve happens to be a height function, i.e. $y(x)$,
-then $x' = 1$ and $x'' = 0$, and we arrive at our previous result again.
+If the curve happens to be a height function, i.e. $$y(x)$$,
+then $$x' = 1$$ and $$x'' = 0$$, and we arrive at our previous result again.
 
 
 ## 3D height functions
 
-The generalization to a 3D height function $h(x, y)$ is straightforward:
+The generalization to a 3D height function $$h(x, y)$$ is straightforward:
 the cost of an infinitesimal portion of the surface is as follows,
 using the same reasoning as before:
 
@@ -220,8 +220,8 @@ $$\begin{aligned}
     = \sqrt{1 + h_x^2 + h_y^2}
 \end{aligned}$$
 
-Keeping the volume $V$ constant,
-we get the following energy functional $E$ to minimize:
+Keeping the volume $$V$$ constant,
+we get the following energy functional $$E$$ to minimize:
 
 $$\begin{aligned}
     E[h]
@@ -229,14 +229,14 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Which gives us an Euler-Lagrange equation
-involving the Lagrange multiplier $\lambda$:
+involving the Lagrange multiplier $$\lambda$$:
 
 $$\begin{aligned}
     0
     = \pdv{\mathcal{L}}{h} - \dv{}{x}\Big( \pdv{\mathcal{L}}{h_x} \Big) - \dv{}{y}\Big( \pdv{\mathcal{L}}{h_y} \Big) + \lambda
 \end{aligned}$$
 
-Inserting $\mathcal{L}$ into this and evaluating all the derivatives
+Inserting $$\mathcal{L}$$ into this and evaluating all the derivatives
 yields a result for the (variable) curvature:
 
 $$\begin{aligned}
@@ -247,10 +247,10 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-What are $\kappa_1$ and $\kappa_2$?
+What are $$\kappa_1$$ and $$\kappa_2$$?
 Well, the problem in 3D is that the curvature of an osculating circle
 depends on the orientation of that circle.
-The **principal curvatures** $\kappa_1$ and $\kappa_2$
+The **principal curvatures** $$\kappa_1$$ and $$\kappa_2$$
 are the largest and smallest curvatures at a given point,
 but finding their values and the corresponding **principal directions** is not so easy.
 Fortunately, in practice, we are often only interested in their sum:
@@ -261,7 +261,7 @@ $$\begin{aligned}
     = \frac{1}{R_1} + \frac{1}{R_2}
 \end{aligned}$$
 
-These **principal radii** $R_1$ and $R_2$ are important
+These **principal radii** $$R_1$$ and $$R_2$$ are important
 for e.g. the [Young-Laplace law](/know/concept/young-laplace-law/).
 
 
@@ -269,10 +269,10 @@ for e.g. the [Young-Laplace law](/know/concept/young-laplace-law/).
 
 To find a general expression for the mean curvature of an arbitrary surface,
 we "cut off" a small part of the surface that we can regard as a height function.
-We call the "cutting" reference plane $(x, y)$,
-and the surface it describes $h(x, y)$.
-We then define the unit tangent vectors $\vu{t}_x$ and $\vu{t}_y$
-to be parallel to the $x$-axis and $y$-axis, respectively:
+We call the "cutting" reference plane $$(x, y)$$,
+and the surface it describes $$h(x, y)$$.
+We then define the unit tangent vectors $$\vu{t}_x$$ and $$\vu{t}_y$$
+to be parallel to the $$x$$-axis and $$y$$-axis, respectively:
 
 $$\begin{aligned}
     \vu{t}_x
@@ -290,7 +290,7 @@ $$\begin{aligned}
 
 Since they were chosen to lie along the axes,
 these vectors are not necessarily orthogonal,
-so we need to normalize the resulting normal vector $\vu{n}$:
+so we need to normalize the resulting normal vector $$\vu{n}$$:
 
 $$\begin{aligned}
     \vu{n}
@@ -301,7 +301,7 @@ $$\begin{aligned}
     \end{bmatrix}
 \end{aligned}$$
 
-Let us take a look at the divergence of $\vu{n}$,
+Let us take a look at the divergence of $$\vu{n}$$,
 or to be precise, its *projection* onto the reference plane
 (although this distinction is not really important for our purposes):
 
@@ -310,7 +310,7 @@ $$\begin{aligned}
     = - \dv{}{x}\bigg( \frac{h_x}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg) - \dv{}{y}\bigg( \frac{h_y}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg)
 \end{aligned}$$
 
-Compare this with the expression for $\lambda$ we found earlier,
+Compare this with the expression for $$\lambda$$ we found earlier,
 with the help of variational calculus:
 
 $$\begin{aligned}
@@ -342,8 +342,8 @@ $$\begin{aligned}
     &= \frac{1}{2} a r^2 \cos^2\varphi + \frac{1}{2} b r^2 \sin^2\varphi + c r^2 \cos\varphi \sin\varphi
 \end{aligned}$$
 
-Sufficiently close to the extremum, where $h_x$ and $h_y$ are negligible,
-the curvature along a certain direction $\varphi$ is given by
+Sufficiently close to the extremum, where $$h_x$$ and $$h_y$$ are negligible,
+the curvature along a certain direction $$\varphi$$ is given by
 our earlier formula for a 2D height function:
 
 $$\begin{aligned}
@@ -352,8 +352,8 @@ $$\begin{aligned}
     = a \cos^2\varphi + b \sin^2\varphi + c \sin(2 \varphi)
 \end{aligned}$$
 
-To find the extremes of $\kappa$,
-we differentiate with respect to $\varphi$ and demand that it is zero:
+To find the extremes of $$\kappa$$,
+we differentiate with respect to $$\varphi$$ and demand that it is zero:
 
 $$\begin{aligned}
     0
@@ -370,10 +370,10 @@ $$\begin{aligned}
     = \tan(2 \varphi)
 \end{aligned}$$
 
-Since the $\tan$ function is $\pi$-periodic,
-this has two solutions, $\varphi_0$ and $\varphi_0 + \pi/2$,
+Since the $$\tan$$ function is $$\pi$$-periodic,
+this has two solutions, $$\varphi_0$$ and $$\varphi_0 + \pi/2$$,
 which are clearly orthogonal,
-hence the principal directions are at an angle of $\pi/2$.
+hence the principal directions are at an angle of $$\pi/2$$.
 
 Finally, it is also worth mentioning that
 the principal directions always lie in planes