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HTML Settings

Here you can Sed posuere consectetur est at lobortis. Donec ullamcorper nulla non metus auctor fringilla. Maecenas sed diam eget risus varius blandit sit amet non magna. Donec id elit non mi porta gravida at eget metus. Praesent commodo cursus magna, vel scelerisque nisl consectetur et.

## HTML

            
<h1>Responsive scaling with MathJax</h1>

<p>We scale down math that's too big to fit, much like MathML's <code>overflow=scale</code>. We get a little bit of help from our markup and assume that all our (display) math sits in wrappers with <code>class="formula"</code></p>

<h1>Extract From Struik</h1>

<p>
This procedure is simply a generalization of the method used in Sects. 1-3 and 1-4 to obtain the equations of the osculating plane and the osculating circle. Let $f(u)$ near $P(u=u_0)$ have finite derivatives $f^{(i)}(u_0)$, $i = 1, 2, \ldots, n+1$. Then
if we take $u=u_1$ at $A$ and write $h = u_1 - u_0$, then there exists a Taylor development of $f(u)$ of the form (compare Eq. (1-5)):
</p>
<div class="formula">

$$f(u_1) = f(u_0) + hf'(u_0)+{h^2\over 2!}f''(u_0) + \cdots + {h^{n+1}\over (n+1)!}f^{(n+1)}(u_0) + o(h^{n+1}).$$
</div>

<p>Here, $f(u_0)=0$ since $P$ lies on $\Sigma_2$, and $h$ is of order $AP$ (see theorem Sec. 1-2); $f(u_1)$ is of order $AD$.
<I>Hence necessary and
sufficient conditions that the surface has a contact of order $n$ at $P$
with the curve are that at $P$ the relations hold:</I>
</p>
<div class="formula">
$$f(u) = f'(u) = f''(u) = \cdots = f^{(n)}(u) = 0;\quad f^{(n+1)}(u) \ne 0.$$
</div>

<hr />
<p>
The converse problem is somewhat more complicated: Find the curves which admit a given curve $C$ as involute. Such curves are called
<I>evolutes</I> of $C$ (German:
<I>Evolute;</I> French:
<I>d&eacute;velopp&eacute;es</I>). Their tangents are normal to $C({\bf x})$ and we can therefore write the equation of the evolute ${\bf y}$ (Fig. 1-34):
</p>

<div class="formula">
$${\bf y} = {\bf x} + a_1{\bf n} + a_2{\bf b}.$$
</div>

<p>Hence</p>
<div class="formula">
$${d{\bf y}\over ds} = {\bf t}(1-a_1\kappa) + {\bf n}\left({da_1\over ds}-\tau a_2\right) + {\bf b}\left({da_2\over ds}+\tau a_1\right)$$
</div>
<p>must have the direction of $a_1{\bf n} + a_2{\bf b}$, this tangent to the evolute:</p>
<div class="formula">
$$\kappa = 1/a, \qquad R= a_1,$$
</div>

<p>and</p>

<div class="formula">
$${{da_1\over ds} - \tau a_2\over a_1} = {{da_2\over ds}+\tau a_1\over a_2},$$
</div>
<p>which can be written in the form:</p>
<div class="formula">
$${a_2{dR\over ds} - R{da_2\over ds} \over a_2^2 + R^2} = \tau.$$
</div>
<p>This expression can be integrated:
</p>
<div class="formula">

$$\tan^{-1}{R\over a_2} = \int \tau\,ds + {\rm const},$$
</div>
<p>or</p>
<div class="formula">
$$a_2 = R\left[{\rm cot}\left(\int \tau\,ds + {\rm const}\right)\right].$$

</div>
<p>The equation of the evolute is:
</p>
<div class="formula">
$${\bf y} = {\bf x} + R\left[{\bf n} + {\rm cot}\left(\int \tau\,ds + {\rm const}\right){\bf b}\right].$$

</div>

<hr />
<p>If $P(u,v)$ and $Q(u,v)$ are two functions of $u$ and $v$ on a surface, then according to Green's theorem and the expression in Chapter 2, Eq. (3-4) for the element area:
</p>
<div class="formula">
$$\int_C P\,du + Q\, dv = \int\!\!\!\int_A \left({\partial Q\over \partial u} - {\partial P\over \partial v}\right) {1\over \sqrt{EG-F^2}}\,dA,$$

</div>
<p>where $dA$ is the element of area of the region $R$ enclosed by the curve $C$. With the aid of this theorem we shall evaluate
</p>
<div class="formula">
$$\int_C \kappa_g\,ds,$$

</div>

<p>where $\kappa_g$ is the geodesic curvature of the curve $C$. If $C$ at a point $P$ makes the angle $\theta$ with the coordinate curve $v = {\rm constant}$ and if the coordinate curves are orthogonal, then, according to Liouville's formula (1-13):
</p>
<div class="formula">
$$\kappa_g\,ds = d\theta + \kappa_1(\cos\theta)\,ds + \kappa_2(\sin\theta)\,ds.$$

</div>

<p>Here, $\kappa_1$ and $\kappa_2$ are the geodesic curvatures of the curves $v = {\rm constant}$ and $u = {\rm constant}$ respectively. Since
</p>
<div class="formula">
$$\cos\theta\,ds = \sqrt{E}\,du, \qquad \sin\theta\,ds = \sqrt{G}\,dv,$$
</div>

<p>we find by application of Green's theorem:</p>
<div class="formula">
$$\int_C\kappa_g\,ds = \int_C d\theta + \int\!\!\!\int_A\left({\partial\over\partial u} \left(\kappa_2\sqrt{G}\,\right) - {\partial\over \partial v}\left(\kappa_1\sqrt{E}\,\right)\right)\,du\,dv.$$

</div>

<p>
The Gaussian curvature can be written, according to Chapter 3, Eq. (3-7),
</p>
<div class="formula">
$$K = -{1\over 2\sqrt{EG}} \left[{\partial\over\partial u}{G_u\over \sqrt{EG}} + {\partial\over\partial v}{E_v\over\sqrt{EG}}\right] ={1\over\sqrt{EG}}\left[ -{\partial\over\partial u} \left(\kappa_2\sqrt{G}\,\right) + {\partial\over\partial v} \left(\kappa_1\sqrt{E}\,\right)\right],$$

</div>
<p>so we obtain the formula</p>
<div class="formula">
$$\int_C\kappa_g\,ds = \int_C d\theta - \int\!\!\!\int_A K\,dA.$$
</div>

<p>The integral $\int\!\!\int_A K\,dA$ is known as the
<I>total</I> or
<I>integral curvature</I>, or
<I>curvature integra</I>, of the region $R$, the name by which Gauss introduced it.
</p>


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