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Diffstat (limited to 'source/know/concept/curvature')
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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 |