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<!DOCTYPE html>
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<h1><a href="/">In principio erat Verbum </a></h1>
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<li><a href="/category/expository.html">Expository</a></li>
<li class="active"><a href="/category/math-212.html">Math 212</a></li>
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<h1 class="entry-title">
<a href="/curves.html" rel="bookmark"
title="Permalink to Review Notes: Curves">Review Notes: Curves</a></h1>
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<div class="entry-content">
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<abbr class="published" title="2015-09-04T16:59:00-04:00">
Published: Fri 04 September 2015
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<br />
<abbr class="modified" title="2015-09-05T22:24:00-04:00">
Updated: Sat 05 September 2015
</abbr>
<address class="vcard author">
By <a class="url fn" href="/author/tingran-gao.html">Tingran Gao</a>
</address>
<p>In <a href="/category/math-212.html">Math 212</a>.</p>
<p>tags: <a href="/tag/math-212.html">Math 212</a> <a href="/tag/notes.html">Notes</a> </p>
</footer><!-- /.post-info --> <h2 id="lines-in-mathbbr3">Lines in <span class="math">\(\mathbb{R}^3\)</span></h2>
<p>A line in <span class="math">\(\mathbb{R}^3\)</span> can be uniquely determined by a point <span class="math">\(P_0=\left(x_0, y_0, z_0\right)\)</span> and a direction <span class="math">\(\vec{v} = \left(v_x, v_y, v_z\right)\)</span>, as
</p>
<div class="math">$$\ell:\left\{\overrightarrow{OP_0}+t\,\vec{v}\mid t\in\mathbb{R}\right\}.$$</div>
<p>
Equivalently, any point <span class="math">\(P=\left(x,y,z\right)\)</span> on <span class="math">\(\ell\)</span> must satisfy the <strong>parametric equation</strong>
</p>
<div class="math">\begin{equation}\label{eq:parametricLine}
\left\{
\begin{array}{rl}
x &= x_0+t\,v_x,\\
y &= y_0+t\,v_y,\\
z &= z_0+t\,v_z.
\end{array} \right.
\end{equation}</div>
<div style="text-align:center">
<p><img alt="Parametric Line" src="/images/parametricLine.png" /></p>
<p><em>Figure 1. the parametric equation of a line</em></p>
</div>
<p>If <span class="math">\(v_x\neq 0, v_y\neq 0, v_z\neq 0\)</span>, we can also put \eqref{eq:parametricLine} into the following <strong>symmetric equations</strong>:
</p>
<div class="math">\begin{equation}\label{eq:symmetricEquation}
\frac{x-x_0}{v_x}=\frac{y-y_0}{v_y}=\frac{z-z_0}{v_z}.
\end{equation}</div>
<p>Alternatively, a line in <span class="math">\(\mathbb{R}^3\)</span> can also be determined by two <em>distinct</em> points <span class="math">\(P_0=\left(x_0, y_0, z_0\right)\)</span>, <span class="math">\(P_1=\left(x_1, y_1, z_1\right)\)</span>. In this case, set the direction vector <span class="math">\(\vec{v}\)</span> in \eqref{eq:parametricLine} as
</p>
<div class="math">$$\vec{v}=\overrightarrow{P_0P_1}=\overrightarrow{OP_1}-\overrightarrow{OP_0}=\left(x_1-x_0, y_1-y_0, z_1-z_0\right),$$</div>
<p>
thus any point <span class="math">\(P=\left(x,y,z\right)\)</span> on <span class="math">\(\ell\)</span> must satisfy
</p>
<div class="math">\begin{equation}
\left\{
\begin{array}{rl}
x &= x_0+t\left(x_1-x_0\right),\\
y &= y_0+t\left(y_1-y_0\right),\\
z &= z_0+t\left(z_1-z_0\right).
\end{array} \right.
\end{equation}</div>
<hr />
<h2 id="parametric-curves-in-mathbbr3">Parametric Curves in <span class="math">\(\mathbb{R}^3\)</span></h2>
<p>In the parametric equation \eqref{eq:parametricLine}, coordiantes <span class="math">\(x,y,z\)</span> are functions of <span class="math">\(t\)</span>. We can denote the tuple <span class="math">\(\left(x,y,z\right)\)</span> as a <strong>vector-valued function</strong> of <span class="math">\(t\)</span>:
</p>
<div class="math">$$\vec{r}\left(t\right)=\left(x_0+t\,v_x,y_0+t\,v_y,z_0+t\,v_z\right)=:\left(x\left(t\right),y\left(t\right),z\left(t\right)\right).$$</div>
<p>Taking one step further, <span class="math">\(x\left(t\right), y\left(t\right), z\left(t\right)\)</span> can be much more general functions than linear ones. Each vector-valued funciton of a single variable gives rise to a <strong>parametric curve</strong>
</p>
<div class="math">\begin{equation}\label{eq:parametricCurve}
\vec{r}\left(t\right)=\left(x\left(t\right),y\left(t\right),z\left(t\right)\right).
\end{equation}</div>
<div style="float: right">
<p><img alt="Helix" src="/images/helix.png" /></p>
</div>
<p><strong>Example.</strong> Let
</p>
<div class="math">$$x\left(t\right)=\cos 10t,\quad y\left(t\right)=\sin 10t,\quad z\left(t\right)=t$$</div>
<p>
where <span class="math">\(t\in\left(-\infty,\infty\right)\)</span>. Then
</p>
<div class="math">$$\vec{r}\left(t\right)=\left(\cos 10t, \sin 10t, t\right)$$</div>
<p>
is a <em>helix</em> in <span class="math">\(\mathbb{R}^3\)</span> (see the figure on the right).</p>
<p><strong>Remark.</strong> It is worth noting that the parametrizations of curves are not unique. For instance, the helix in the example above can be equivalently parametrized as
</p>
<div class="math">$$\vec{r}\left(t\right)=\left(\cos t,\sin t, \frac{t}{10}\right)$$</div>
<p>
where <span class="math">\(t\in\left(-\infty,\infty\right)\)</span>. One way to understand this is to consider parametrization as a physical motion, where <span class="math">\(t\)</span> stands for time and <span class="math">\(\vec{r}\left(t\right)\)</span> is the location of a point mass. The curve is the trace of <span class="math">\(\vec{r}\left(t\right)\)</span> in time, but the same trace can be generated from an infinite number of possible motions. For instance, the point mass may move with constant speed, or go back and forth, or stop moving every five minutes......</p>
<p><strong>Example.</strong> The following two sets of parametrization equations
</p>
<div class="math">\begin{align*}
\ell_1:\quad &\vec{r}_1\left(t\right) = \left(1,0,0\right)+t\left(1,2,1\right),\\
\ell_2:\quad &\vec{r}_2\left(t\right) = \left(2,2,1\right)+t\left(2,4,2\right).
\end{align*}</div>
<p>
To see this, notice that <span class="math">\(\ell_1\)</span> is parallel to the vector <span class="math">\(\left(1,2,1\right)\)</span> and <span class="math">\(\ell_2\)</span> is parallel to the vector <span class="math">\(\left(2,4,2\right)=2\cdot\left(1,2,1\right)\)</span>, which means that <span class="math">\(\ell_1,\ell_2\)</span> are parallel to each other. Moreover, notice that point <span class="math">\(\left(2,2,1\right)\)</span> lies on both <span class="math">\(\ell_1\)</span> and <span class="math">\(\ell_2\)</span>
</p>
<div class="math">\begin{align*}
\ell_1\left(1\right) &= \left(1,0,0\right)+\left(1,2,1\right)=\left(2,2,1\right),\\
\ell_2\left(0\right) &= \left(2,2,1\right)+\left(0,0,0\right)=\left(2,2,1\right),
\end{align*}</div>
<p>
thus <span class="math">\(\ell_1,\ell_2\)</span> are the same line.</p>
<hr />
<p>All standard differential and integral operations can be defined on vector-valued functions; one just need to apply the operations to each component. Assume <span class="math">\(\vec{r}\left(t\right)=\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)\)</span>, then the derivative of <span class="math">\(\vec{r}\left(t\right)\)</span> is
</p>
<div class="math">$$\frac{d}{dt}\vec{r}\left(t\right)=\left(\frac{d}{dt}x\left(t\right),\frac{d}{dt}y\left(t\right),\frac{d}{dt}z\left(t\right)\right).$$</div>
<p>
When the context is clear, one can also write the derivative of <span class="math">\(\vec{r}\left(t\right)\)</span> with the prime notation:
</p>
<div class="math">$$\vec{r}'\left(t\right)=\left(x'\left(t\right),y'\left(t\right),z'\left(t\right)\right).$$</div>
<p>
Major rules of derivatives from single-variable calculus still holds for vector-values functions; it often suffices to replace scalar products with dot or cross products. In particular,
</p>
<div class="math">\begin{align*}
&\left(\vec{u}\left(t\right)+\vec{u}\left(t\right)\right)'=\vec{u}'\left(t\right)+\vec{v}'\left(t\right),\\
&\left(c\vec{u}\left(t\right)\right)'=c\vec{u}'\left(t\right),\\
&\left(c\left(t\right)\vec{u}\left(t\right)\right)'=c'\left(t\right)\vec{u}\left(t\right)+c\left(t\right)\vec{u}'\left(t\right),\\
&\left(\vec{u}\left(t\right)\cdot \vec{v}\left(t\right)\right)'=\vec{u}'\left(t\right)\cdot\vec{v}\left(t\right)+\vec{u}\left(t\right)\cdot\vec{v}'\left(t\right),\\
&\left(\vec{u}\left(t\right)\times \vec{v}\left(t\right)\right)'=\vec{u}'\left(t\right)\times\vec{v}\left(t\right)+\vec{u}\left(t\right)\times\vec{v}'\left(t\right).
\end{align*}</div>
<p><strong>Example.</strong>
</p>
<div class="math">\begin{align*}
\left(\vec{u}\left(t\right)\cdot\left(\vec{v}\left(t\right)\times\vec{w}\left(t\right)\right)\right)'&=\vec{u}'\left(t\right)\cdot\left(\vec{v}\left(t\right)\times\vec{w}\left(t\right)\right)+\vec{u}\left(t\right)\cdot\left(\vec{v}\left(t\right)\times\vec{w}\left(t\right)\right)'\\
&=\vec{u}'\left(t\right)\cdot\left(\vec{v}\left(t\right)\times\vec{w}\left(t\right)\right)+\vec{u}\left(t\right)\cdot\left(\vec{v}'\left(t\right)\times\vec{w}\left(t\right)+\vec{v}\left(t\right)\times\vec{w}'\left(t\right)\right)\\
&=\vec{u}'\left(t\right)\cdot\left(\vec{v}\left(t\right)\times\vec{w}\left(t\right)\right)+\vec{u}\left(t\right)\cdot\left(\vec{v}'\left(t\right)\times\vec{w}\left(t\right)\right)+\vec{u}\left(t\right)\cdot\left(\vec{v}\left(t\right)\times\vec{w}'\left(t\right)\right).
\end{align*}</div>
<p>Similar to the derivative, the integral and anti-derivative of <span class="math">\(\vec{r}\left(t\right)\)</span> are also defined component-wise:
</p>
<div class="math">\begin{align*}
\int_a^b\vec{r}\left(t\right)dt&=\left(\int_a^bx\left(t\right)dt,\int_a^by\left(t\right)dt,\int_a^bz\left(t\right)dt\right),\quad\textrm{(integral)}\\
\int\vec{r}\left(t\right)dt&=\left(\int x\left(t\right)dt,\int y\left(t\right)dt,\int z\left(t\right)dt\right).\quad\textrm{(anti-derivative)}
\end{align*}</div>
<p>
Recall the difference between integral and anti-derivative in one-variable calculus. If <span class="math">\(X\left(t\right),Y\left(t\right),Z\left(t\right)\)</span> are real-valued functions satisfying
</p>
<div class="math">$$X'\left(t\right)=x\left(t\right),\quad Y'\left(t\right)=y\left(t\right),\quad Z'\left(t\right)=z\left(t\right)$$</div>
<p>
then
</p>
<div class="math">\begin{align*}
\int\vec{r}\left(t\right)dt&=\left(\int x\left(t\right)dt,\int y\left(t\right)dt,\int z\left(t\right)dt\right)\\
&=\left(X\left(t\right)+C_1, Y\left(t\right)+C_2, Z\left(t\right)+C_3\right)\\
&=\left(X\left(t\right), Y\left(t\right), Z\left(t\right)\right)+\left(C_1, C_2, C_3\right)
\end{align*}</div>
<p>
where <span class="math">\(C_1, C_2, C_3\)</span> are arbitrary constants. Moreover, it is important that the <strong>Fundamental Theorem of Calculus</strong> (FTC) holds for vector-valued functions:
</p>
<div class="math">\begin{align*}
\int_a^b\vec{r}\left(t\right)dt&=\left(X\left(b\right)-X\left(a\right), Y\left(b\right)-Y\left(a\right), Z\left(b\right)-Z\left(a\right)\right)\\
&=\left(X\left(b\right),Y\left(b\right),Z\left(b\right)\right)-\left(X\left(a\right),Y\left(a\right),Z\left(a\right)\right).
\end{align*}</div>
<p><strong>Example.</strong> Write <span class="math">\(\vec{r}'\left(t\right)=v\left(t\right)\)</span>, <span class="math">\(\vec{v}'\left(t\right)=\vec{a}\left(t\right)\)</span>. Suppose
</p>
<div class="math">$$\vec{a}\left(t\right)=\left(\cos t,\sin t\right),\quad \vec{v}\left(0\right)=\left(1,2\right), \quad\vec{r}\left(0\right)=\left(3,4\right),$$</div>
<p>
let's compute <span class="math">\(\vec{r}\left(t\right)\)</span>. First, note that
</p>
<div class="math">$$\vec{v}\left(t\right)=\int\vec{a}\left(t\right)dt=\left(\int\cos t\, dt, \int\sin t\, dt\right)=\left(\sin t,-\cos t\right)+\left(C_1,C_2\right)$$</div>
<p>
and
</p>
<div class="math">$$\left(1,2\right)=\vec{v}\left(0\right)=\left(0,-1\right)+\left(C_1,C_2\right)$$</div>
<p>
thus
</p>
<div class="math">$$\left(C_1,C_2\right)=\left(1,2\right)-\left(0,-1\right)=\left(1,3\right)$$</div>
<p>
from which one has
</p>
<div class="math">$$\vec{v}\left(t\right)=\left(\sin t,-\cos t\right)+\left(1,3\right)=\left(\sin t+1, -\cos t+3\right).$$</div>
<p>
Next,
</p>
<div class="math">$$\vec{r}\left(t\right)=\int\vec{v}\left(t\right)dt=\left(\int\left(\sin t + 1\right) dt, \int\left(-\cos t + 3\right) dt\right)=\left(-\cos t + t, -\sin t +3t\right)+\left(C_3,C_4\right),$$</div>
<p>
and we can find <span class="math">\(C_3, C_4\)</span> using the initial condition
</p>
<div class="math">$$\left(3,4\right)=\vec{v}\left(0\right)=\left(-1,0\right)+\left(C_3,C_4\right)\quad\Rightarrow\quad \left(C_3,C_4\right)=\left(4,4\right).$$</div>
<p>
Thus
</p>
<div class="math">$$\vec{v}\left(t\right)=\left(4+t-\cos t, 4+3t-\sin t\right).$$</div>
<hr />
<h2 id="arc-length-parametrization">Arc-Length Parametrization</h2>
<p>We saw from last section that the parametrization of a curve is not unique. In fact, every curve can be parametrized in inifinitely many ways: given a vector-valued function <span class="math">\(\vec{r}\left(t\right)\)</span>, we can pick any smooth monotonic function <span class="math">\(u\left(\tau\right)\)</span> and compose <span class="math">\(\vec{r}\left(t\right)\)</span> with it, obtaining
</p>
<div class="math">$$\vec{r}\left(\tau\right):=\vec{r}\left(u\left(\tau\right)\right),$$</div>
<p>
which is just a different parametrization of the same curve.</p>
<p>Despite the abundant choices of parametrizations available, <a href="https://en.wikipedia.org/wiki/Erlangen_program"><em>geometry</em></a> is all about properties of things that hold <em>invariant</em> regardless of parametrization. <strong>Arc length</strong> of a curve is one such example. Intuitively this makes sense: different parametrizations are like different ways to "draw" the curve, but arc length has nothing to do with the "drawing" process; in fact, it only depends on what is actually "drawn".
<div style="float: left">
<img src="/images/arclength.png", alt="compute arc length", style="width: 450px; height: auto">
</div>
Mathematically, the arc length of a curve <span class="math">\(\vec{v}\left(t\right)\)</span> from time <span class="math">\(t_0\)</span> to time <span class="math">\(t\)</span> can be computed using integrals. Let
</p>
<div class="math">$$t_0<t_1<t_2<\cdots<t_{n-1}<t_n=:t$$</div>
<p>
be a partition of the interval <span class="math">\(\left[t_0, t\right]\)</span>, such that
</p>
<div class="math">$$\Delta t = t_{j+1}-t_j$$</div>
<p>
is the same for all <span class="math">\(j=0,\cdots,n-1\)</span>. When <span class="math">\(\Delta t\)</span> is sufficiently small, the arc length between <span class="math">\(\vec{r}\left(t_{j}\right)\)</span> and <span class="math">\(\vec{r}\left(t_{j+1}\right)\)</span> can be well approximated by the length of the line segment connecting these two points, i.e.,
</p>
<div class="math">\begin{align*}
&\textrm{arc length between } \vec{r}\left(t_{j}\right) \textrm{ and } \vec{r}\left(t_{j+1}\right)\\
&\qquad \approx \left|\vec{r}\left(t_{j}\right) - \vec{r}\left(t_{j+1}\right)\right|.
\end{align*}</div>
<p>
Denoting <span class="math">\(s=s\left(t\right)\)</span> for the arc length from <span class="math">\(\vec{r}\left(t_0\right)\)</span> to <span class="math">\(\vec{r}\left(t_n\right)=\vec{r}\left(t\right)\)</span>, we have
</p>
<div class="math">\begin{align*}
s=s\left(t\right)&=\sum_{j=1}^{n-1}\textrm{arc length between } \vec{r}\left(t_{j}\right) \textrm{ and } \vec{r}\left(t_{j+1}\right)\\
&\approx \sum_{j=1}^{n-1}\left|\vec{r}\left(t_{j}\right) - \vec{r}\left(t_{j+1}\right)\right|\\
&\approx\sum_{j=1}^{n-1}\left|\vec{r}'\left(t_j\right)\Delta t\right|\\
&=\sum_{j=1}^{n-1}\left|\vec{r}'\left(t_j\right)\right|\Delta t\longrightarrow \int_{t_0}^t\left|\vec{r}'\left(\tau\right)\right|d\tau\quad\textrm{as }\Delta t\rightarrow 0.
\end{align*}</div>
<p>
Therefore, we have the following formula for the arc length computation:
</p>
<div class="math">\begin{equation}\label{eq:arclength}
s\left(t\right)=\int_{t_0}^t\left|\vec{r}'\left(\tau\right)\right|d\tau.
\end{equation}</div>
<p>
Arc length is the very first <em>geometric quantity</em> among the many we are about to see in this course. Suppose <span class="math">\(\vec{r}\left(\tau\left(\xi\right)\right)\)</span> is another parametrization of the same curve, then the beginning and ending points of the curve should be fixed and thus
</p>
<div class="math">$$t_0=\tau\left(\xi_0\right),\quad t=\tau\left(\xi\right).$$</div>
<p>
Following \eqref{eq:arclength}, the arc length under this new parametrization is
</p>
<div class="math">$$s\left(\xi\right)=\int_{\xi_0}^{\xi}\left|\vec{r}'\left(\xi\right)\right|d\xi.$$</div>
<p>
However, by the chain rule
</p>
<div class="math">$$\vec{r}'\left(\xi\right)=\vec{r}'\left(\tau\right)\tau\left(\xi\right),$$</div>
<p>
thus by the <a href="https://en.wikipedia.org/wiki/Integration_by_substitution">change-of-variables formula</a>
</p>
<div class="math">\begin{align*}
s\left(\xi\right)&=\int_{\xi_0}^{\xi}\left|\vec{r}'\left(\xi\right)\right|d\xi\\
&=\int_{\xi_0}^{\xi}\left|\vec{r}'\left(\tau\right)\right|\tau\left(\xi\right)d\xi\quad\textrm{(chain rule)}\\
&=\int_{t_0}^{t}\left|\vec{r}'\left(\tau\right)\right|d\tau\quad\textrm{(change of variable)}\\
&=s\left(t\right).
\end{align*}</div>
<p>
In other words, \eqref{eq:arclength} defines a quantity that is <em>independent of the parametrization</em> of the curve! Conventionally, we refer to such "parametrization independent" quantities as <em>geometric quantities</em>.</p>
<p>Seeing as arc length is a geometric quantity, it is canonically chosen as the "golden standard" for curve parametrization. We say that a curve is <em>arc-parametrized</em> if it can be written as a vector-valued function of a single variable — the arc length. The <em>arc-length parametrization</em> can be constructed from any parametrization* of a curve, following these steps:</p>
<ol>
<li>Starting with an arbitrary parametrization <span class="math">\(\vec{r}\left(t\right)\)</span> of the curve, compute <span class="math">\(\vec{r}'\left(t\right)\)</span> and <span class="math">\(\left|\vec{r}'\left(t\right)\right|\)</span>;</li>
<li>Compute <span class="math">\(s=s\left(t\right)\)</span> by \eqref{eq:arclength}, and find the inverse function <span class="math">\(t=t\left(s\right)\)</span>;</li>
<li>Compose <span class="math">\(\vec{r}\left(t\right)\)</span> with <span class="math">\(t=t\left(s\right)\)</span> to obtain the arc-length parametrization of the curve <span class="math">\(\vec{r}\left(s\right)=\vec{r}\left(t\left(s\right)\right)\)</span>.</li>
</ol>
<p><strong>Example.</strong> Find the arc-length parametrization of the curve
</p>
<div class="math">$$\vec{r}\left(t\right)=\left(\cos 3t,\sin 3t, t\right),\quad t\in\left[\frac{3\pi}{2},\frac{7\pi}{2}\right].$$</div>
<p>
<em>Solution.</em> Note that
</p>
<div class="math">$$\vec{r}'\left(t\right)=\left(-3\sin 3t, 3\cos 3t, 1\right),\quad \left|\vec{r}'\left(t\right)\right|=\sqrt{9\sin^23t+9\cos^23t+1}=\sqrt{10}$$</div>
<p>
thus the arc length of <span class="math">\(\vec{r}\left(t\right)\)</span> is
</p>
<div class="math">$$s = s\left(t\right)=\int_{\frac{3\pi}{2}}^t\left|\vec{r}'\left(\tau\right)\right|d\tau=\sqrt{10}\left(t-\frac{3\pi}{2}\right).$$</div>
<p>
Inversely,
</p>
<div class="math">$$t = \frac{s}{\sqrt{10}}+\frac{3\pi}{2}.$$</div>
<p>
Thus the arc-length parametrization of <span class="math">\(\vec{r}\left(t\right)\)</span> is
</p>
<div class="math">$$\vec{r}\left(s\right)=\vec{r}\left(t\left(s\right)\right)=\vec{r}\left(\frac{s}{\sqrt{10}}+\frac{3\pi}{2}\right)=\left(\cos\left(\frac{3s}{\sqrt{10}}+\frac{9\pi}{2}\right), \sin\left(\frac{3s}{\sqrt{10}}+\frac{9\pi}{2}\right), \frac{s}{\sqrt{10}}+\frac{3\pi}{2}\right).$$</div>
<p>
A major advantage of arc-length parametrization is that it normalizes the speed of the curve.</p>
<p><strong>Theorem 1.</strong> If <span class="math">\(\vec{r}\left(s\right)\)</span> is an arc-length parametrized curve, then <span class="math">\(\left|\vec{r}'\left(s\right)\right|=1\)</span>.</p>
<p><strong>Proof.</strong> Differentiating both sides of \eqref{eq:arclength}, we see that <span class="math">\(s'\left(t\right)=\left|\vec{r}'\left(t\right)\right|\)</span>. Since <span class="math">\(t=t\left(s\right)\)</span> is the inverse function of <span class="math">\(s=s\left(t\right)\)</span>, there holds
</p>
<div class="math">$$t'\left(s\right)=\frac{1}{s'\left(t\right)}=\frac{1}{\left|\vec{r}'\left(t\right)\right|}.$$</div>
<p>
It then follows from the <a href="https://en.wikipedia.org/wiki/Chain_rule"><em>chain rule</em></a> that
</p>
<div class="math">\begin{align*}
\left|\vec{r}'\left(s\right)\right|=\left|\vec{v}'\left(t\right)\,t'\left(s\right)\right|=\left|\vec{v}'\left(t\right)\cdot\frac{1}{\left|\vec{v}'\left(t\right)\right|}\right|=\frac{1}{\left|\vec{v}'\left(t\right)\right|}\cdot \left|\vec{v}'\left(t\right)\right|=1.
\end{align*}</div>
<p>
<span class="math">\(\blacksquare\)</span></p>
<hr />
<h2 id="curvature-and-principal-normal-vectors">Curvature and Principal Normal Vectors</h2>
<p>As one might infer from the name, <strong>curvature</strong> measures the "curviness" of a curve (how much it deviates from a straight line). Since the direction of tangent vectors along a line stay constant, it is natural to quantify "curviness" using the change in the direction of tangent vectors. The direction of tangent vectors is fully captured in their "unit" counterpart.</p>
<p><strong>Definition.</strong> The <strong>unit tangent vector</strong> along a curve <span class="math">\(\vec{r}\left(t\right)\)</span> is
</p>
<div class="math">$$\vec{T}\left(t\right)=\frac{\vec{r}'\left(t\right)}{\left|\vec{r}'\left(t\right)\right|}.$$</div>
<p><strong>Proposition 1.</strong> Under arc-length parametrization, <span class="math">\(\vec{T}\left(s\right)=\vec{r}'\left(s\right)\)</span>.</p>
<p><strong>Proof.</strong> Recall from Theorem 1 that <span class="math">\(\left|\vec{r}'\left(s\right)\right|=1\)</span>. <span class="math">\(\blacksquare\)</span></p>
<p>We define curvature as the magnitude of the change of the unit tangent vectors under arc-length parametrization.</p>
<p><strong>Definition.</strong> The <strong>curvature</strong> of a curve <span class="math">\(\vec{r}\left(t\right)\)</span> is
</p>
<div class="math">$$\kappa\left(s\right)=\left|\frac{d\vec{T}\left(s\right)}{ds}\right|.$$</div>
<p>
There are 2 key points in this definition: (1) If one needs to compute <span class="math">\(\kappa\)</span> from its definition, then the curve has to be first cast into arc-length parametrization (there are also other ways to compute <span class="math">\(\kappa\)</span>, as the textbook indicates); (2) The curvature of a curve is always non-negative (this is no longer true for the curvature of surfaces, as we shall see later in this course). According to Proposition 1, under arc-length parametrization
</p>
<div class="math">$$\kappa\left(s\right)=\left|\vec{r}''\left(s\right)\right|.$$</div>
<p>
In words, for a curve under arc-length parametrization, its curvature is the magnitude of the second order derivative of the vector-valued function that represents the curve.</p>
<p><strong>Example.</strong> Let us compute the curvature of the curve
</p>
<div class="math">$$\vec{r}\left(t\right)=\left(\cos 3t,\sin 3t, t\right),\quad t\in\left[\frac{3\pi}{2},\frac{7\pi}{2}\right].$$</div>
<p>
Note that this is exactly the same curve for which we computed arc-length parametrization in the previous section. The arc-length parametrization is
</p>
<div class="math">$$\vec{r}\left(s\right)=\left(\cos\left(\frac{3s}{\sqrt{10}}+\frac{9\pi}{2}\right), \sin\left(\frac{3s}{\sqrt{10}}+\frac{9\pi}{2}\right), \frac{s}{\sqrt{10}}+\frac{3\pi}{2}\right),$$</div>
<p>
thus
</p>
<div class="math">\begin{align*}
\vec{r}'\left(s\right)&=\left(-\frac{3}{\sqrt{10}}\sin\left(\frac{3s}{\sqrt{10}}+\frac{9\pi}{2}\right), \frac{3}{\sqrt{10}}\cos\left(\frac{3s}{\sqrt{10}}+\frac{9\pi}{2}\right), \frac{1}{\sqrt{10}}\right),\\
\vec{r}''\left(s\right)&=\left(-\frac{9}{10}\cos\left(\frac{3s}{\sqrt{10}}+\frac{9\pi}{2}\right), -\frac{9}{10}\sin\left(\frac{3s}{\sqrt{10}}+\frac{9\pi}{2}\right), 0\right),
\end{align*}</div>
<p>
and we have
</p>
<div class="math">$$\kappa\left(s\right)=\left|\vec{r}''\left(s\right)\right|=\frac{9}{10}.$$</div>
<p>
<strong>A Computational Trick.</strong> Finding the arc-length parametrization involves computing <span class="math">\(\vec{r}'\left(t\right)\)</span> and <span class="math">\(\left|\vec{r}'\left(t\right)\right|\)</span>, which essentially gives <span class="math">\(\vec{T}\left(t\right)\)</span>. But this is not good enough for curvature — we have to take the derivative of <span class="math">\(\vec{T}\left(s\right)\)</span>, not <span class="math">\(\vec{T}\left(t\right)\)</span>! When we move on to computing <span class="math">\(\kappa\left(s\right)\)</span> under the arc-length parametrization, it seems too much repeated work that we have to take derivatives of each component of <span class="math">\(\vec{r}'\left(s\right)\)</span> all over again. A computational trick is that one can reuse <span class="math">\(\vec{T}\left(t\right)\)</span> for computing <span class="math">\(\vec{T}\left(s\right)\)</span>, as long as one recalls from chain rule
</p>
<div class="math">$$\vec{T}'\left(s\right)=\vec{T}'\left(t\right)\cdot t'\left(s\right).$$</div>
<p>
<strong>Example.</strong> Find the curvature of the curve
</p>
<div class="math">$$\vec{r}\left(t\right)=\left(\cos t+t\,\sin t, \sin t-t\,\cos t\right),\quad t\in \left[0,\infty\right).$$</div>
<p>
<em>Solution.</em> Note that
</p>
<div class="math">$$\vec{r}'\left(t\right)=\left(-\sin t+\sin t+t\,\cos t, \cos t-\cos t +t\,\sin t\right)=\left(t\,\cos t, t\,\sin t\right),\quad \left|\vec{r}'\left(t\right)\right|=t,$$</div>
<p>
thus
</p>
<div class="math">$$\vec{T}\left(t\right)=\frac{\vec{r}'\left(t\right)}{\left|\vec{r}'\left(t\right)\right|}=\frac{1}{t}\cdot \left(t\,\cos t, t\,\sin t\right)=\left(\cos t,\sin t\right).$$</div>
<p>
The arc length is
</p>
<div class="math">$$s=s\left(t\right)=\int_0^t\left|\vec{r}'\left(\tau\right)\right|d\tau=\int_0^t\tau d\tau=\frac{t^2}{2}.$$</div>
<p>
Inversely,
</p>
<div class="math">$$t=t\left(s\right)=\sqrt{2s},\quad\Rightarrow\quad t'\left(s\right)=\frac{1}{\sqrt{2s}}.$$</div>
<p>
Therefore,
</p>
<div class="math">$$\kappa\left(s\right)=\left|\vec{T}'\left(s\right)\right|=\left|\vec{T}'\left(t\right)\cdot t'\left(s\right)\right|=\left|\left(-\sin t, \cos t\right)\cdot \frac{1}{\sqrt{s}}\right|=\frac{1}{\sqrt{2s}}.$$</div>
<p>As curvature is the magnitude of <span class="math">\(\vec{T}'\left(s\right)\)</span>, what does the direction of <span class="math">\(\vec{T}'\left(s\right)\)</span> tell us? This motivates the following definition of <em>principal normal vectors</em>.</p>
<p><strong>Definition.</strong> On a curve <span class="math">\(\vec{r}\left(s\right)\)</span> (arc-length parametrized) where <span class="math">\(\kappa\left(s\right)\neq 0\)</span>, the <strong>principal normal vector</strong> is defined as
</p>
<div class="math">$$\vec{N}\left(s\right)=\frac{\vec{T}'\left(s\right)}{\left|\vec{T}'\left(s\right)\right|}=\frac{1}{\kappa\left(s\right)}\vec{T}'\left(s\right).$$</div>
<p>
<strong>Remark.</strong> Note that at points <span class="math">\(\vec{r}\left(s\right)\)</span> where <span class="math">\(\kappa\left(s\right)=0\)</span> the principal normal curvature is not defined (since <span class="math">\(\kappa\left(s\right)\)</span> has to appear in the denominator).
<div style="float: right">
<img src="/images/principalNormal.png", alt="compute arc length", style="width: 350px; height: auto">
</div>
<strong>Proposition.</strong> Let <span class="math">\(s\)</span> denote the arc-length parameter. Wherever <span class="math">\(\vec{N}\left(s\right)\)</span> is defined, there holds
</p>
<div class="math">$$\vec{N}\left(s\right)\cdot\vec{T}\left(s\right)=0.$$</div>
<p>
<strong>Proof.</strong> Since <span class="math">\(\left|\vec{T}\left(s\right)\right|=1\)</span>, we have
</p>
<div class="math">$$\vec{T}\left(s\right)\cdot\vec{T}\left(s\right)=1.$$</div>
<p>
Differentiating both sides of the equality, we see that
</p>
<div class="math">$$2\vec{T}'\left(s\right)\cdot\vec{T}\left(s\right)=0.$$</div>
<p>
Since
</p>
<div class="math">$$\vec{T}'\left(s\right)=\kappa\left(s\right)\vec{N}\left(s\right),$$</div>
<p>
we conclude that
</p>
<div class="math">$$\kappa\left(s\right)\vec{N}\left(s\right)\cdot\vec{T}\left(s\right)=0.$$</div>
<p>
Recall that <span class="math">\(\kappa\left(s\right)\neq 0\)</span> wherever <span class="math">\(\vec{N}\left(s\right)\)</span> is defined, thus
</p>
<div class="math">$$\vec{N}\left(s\right)\cdot\vec{T}\left(s\right)=0.$$</div>
<p>
<span class="math">\(\blacksquare\)</span></p>
<hr />
<h2 id="decompose-accerleration">Decompose Accerleration</h2>
<p>For a curve <span class="math">\(\vec{r}\left(t\right)\)</span>, <span class="math">\(\vec{v}\left(t\right)=\vec{r}'\left(t\right)\)</span> and <span class="math">\(\vec{a}\left(t\right)=\vec{r}''\left(t\right)\)</span> are often referred to as the <em>velocity</em> and the <em>accerleration</em> vector, respectively. According to the definition, the unit tangent vector <span class="math">\(\vec{T}\left(t\right)\)</span> is always in the same direction as <span class="math">\(\vec{r}'\left(t\right)\)</span>. The direction of an acceleration vector is less straightforward; however, it is easy to derive that <span class="math">\(\vec{r}''\left(t\right)\)</span> always lives in the plane spanned by <span class="math">\(\vec{T}\)</span> and <span class="math">\(\vec{N}\)</span>.</p>
<p><strong>Theorem 2.</strong> The accerleration vector can be decomposed as
</p>
<div class="math">\begin{equation}\label{eq:accerlerationDecomposition}
\vec{a}\left(t\right)=a_T\left(t\right)\vec{T}\left(t\right)+a_N\left(t\right)\vec{N}\left(t\right),
\end{equation}</div>
<p>
where
</p>
<div class="math">$$a_T\left(t\right)=\frac{d}{dt}\left|\vec{r}'\left(t\right)\right|,\quad a_N\left(t\right)=\kappa\left(t\right)\left|\vec{r}'\left(t\right)\right|^2.$$</div>
<p>
<strong>Proof.</strong> Direct computation yields
</p>
<div class="math">\begin{align*}
\vec{a}\left(t\right)&=\vec{r}''\left(t\right)=\frac{d}{dt}\left(\left|\vec{r}'\left(t\right)\right|\cdot \frac{\vec{r}'\left(t\right)}{\left|\vec{r}'\left(t\right)\right|}\right)=\frac{d}{dt}\left(\left|\vec{r}'\left(t\right)\right|\cdot\vec{T}\left(t\right)\right)\\
&=\left(\frac{d}{dt}\left|\vec{r}'\left(t\right)\right|\right)\vec{T}\left(t\right)+\left|\vec{r}'\left(t\right)\right|\frac{d}{dt}\vec{T}\left(t\right)\\
&=\left(\frac{d}{dt}\left|\vec{r}'\left(t\right)\right|\right)\vec{T}\left(t\right)+\left|\vec{r}'\left(t\right)\right|\frac{d}{ds}\vec{T}\left(s\right)s'\left(t\right)\\
&=\left(\frac{d}{dt}\left|\vec{r}'\left(t\right)\right|\right)\vec{T}\left(t\right)+\left|\vec{r}'\left(t\right)\right|\kappa\left(s\right)\vec{N}\left(s\right)\cdot\left|\vec{r}'\left(t\right)\right|\\
&=\left(\frac{d}{dt}\left|\vec{r}'\left(t\right)\right|\right)\vec{T}\left(t\right)+\kappa\left(s\right)\left|\vec{r}'\left(t\right)\right|^2\vec{N}\left(s\right).
\end{align*}</div>
<p>
The conclusion follows as we define
</p>
<div class="math">$$a_T\left(t\right):=\frac{d}{dt}\left|\vec{r}'\left(t\right)\right|,$$</div>
<div class="math">$$a_N\left(t\right):=\kappa\left(s\right)\left|\vec{r}'\left(t\right)\right|^2=\kappa\left(s\left(t\right)\right)\left|\vec{r}'\left(t\right)\right|^2=\kappa\left(t\right)\left|\vec{r}'\left(t\right)\right|^2,$$</div>
<p>
and conventionally write <span class="math">\(\vec{N}\left(t\right):=\vec{N}\left(s\left(t\right)\right)=\vec{N}\left(s\right)\)</span>.
<span class="math">\(\blacksquare\)</span></p>
<p><strong>Remark.</strong> Note that <span class="math">\(a_N\left(t\right)\geq 0\)</span> always holds, since <span class="math">\(\kappa\left(t\right)\geq0\)</span> and <span class="math">\(\left|\vec{r}'\left(t\right)\right|^2\geq 0\)</span>.</p>
<p>Typical questions for the decomposition of accerleration vectors involves computing <span class="math">\(\vec{T},\vec{N},a_T,a_N,\kappa\)</span> given curve <span class="math">\(\vec{r}\left(t\right)\)</span>. We summarize these computations in the following "road map".</p>
<ol>
<li>Compute <span class="math">\(\vec{r}'\left(t\right)\)</span> and <span class="math">\(\vec{r}''\left(t\right)\)</span>. Both of them are straightforward derivative computations on each component of <span class="math">\(\vec{r}\left(t\right)\)</span>.</li>
<li>The unit tangent vector <span class="math">\(\vec{T}\)</span> can be computed from its definition
<div class="math">$$\vec{T}\left(t\right)=\frac{\vec{r}'\left(t\right)}{\left|\vec{r}'\left(t\right)\right|}.$$</div>
</li>
<li>The "tangent component" of <span class="math">\(\vec{a}\left(t\right)\)</span>, or <span class="math">\(a_T\)</span>, can be computed from the decomposition formula \eqref{eq:accerlerationDecomposition}. In fact, dot product both sides of \eqref{eq:accerlerationDecomposition} with <span class="math">\(\vec{T}\left(t\right)\)</span> and using the fact that <span class="math">\(\vec{T}\cdot\vec{N}=0\)</span>, one has
<div class="math">$$\vec{a}\left(t\right)\cdot\vec{T}\left(t\right)=a_T\left(t\right)\vec{T}\left(t\right)\cdot\vec{T}\left(t\right)=a_T\left(t\right),$$</div>
or equivalently
<div class="math">\begin{equation}\label{eq:tangentComp}
a_T\left(t\right)=\vec{r}''\left(t\right)\cdot\vec{T}\left(t\right)=\frac{\vec{r}''\left(t\right)\cdot\vec{r}'\left(t\right)}{\left|\vec{r}'\left(t\right)\right|}.
\end{equation}</div>
</li>
<li>The "normal component" of <span class="math">\(\vec{a}\left(t\right)\)</span>, or <span class="math">\(a_N\)</span>, can also be computed from the decomposition formula \eqref{eq:accerlerationDecomposition}. This time we cross product both sides of \eqref{eq:accerlerationDecomposition} with <span class="math">\(\vec{T}\left(t\right)\)</span> and use the facts <span class="math">\(\vec{T}\cdot\vec{N}=0\)</span>, <span class="math">\(\vec{T}\left(t\right)\times\vec{T}\left(t\right)=0\)</span>, and
<div class="math">$$\left|\vec{N}\left(t\right)\times\vec{T}\left(t\right)\right|=\left|\vec{N}\left(t\right)\right|\left|\vec{T}\left(t\right)\right|\sin\left(\frac{\pi}{2}\right)=1$$</div>
to conclude that
<div class="math">$$\left|\vec{a}\left(t\right)\times\vec{T}\left(t\right)\right|=\left|a_N\left(t\right)\vec{N}\left(t\right)\times\vec{T}\left(t\right)\right|=\left|a_N\left(t\right)\right|\stackrel{(*)}{=}a_N\left(t\right),$$</div>
and thus
<div class="math">\begin{equation}\label{eq:normalComp}
a_N\left(t\right)=\left|\vec{a}\left(t\right)\times\vec{T}\left(t\right)\right|=\frac{\left|\vec{r}''\left(t\right)\times\vec{r}'\left(t\right)\right|}{\left|\vec{r}'\left(t\right)\right|}.
\end{equation}</div>
Note that at equality <span class="math">\((*)\)</span> we used the fact that <span class="math">\(a_N\left(t\right)\geq0\)</span> to drop the absolute value bars.</li>
<li>Once <span class="math">\(a_T,a_N,\vec{T}\)</span> are obtained, <span class="math">\(\vec{N}\)</span> can be very easily computed from the decomposition formula \eqref{eq:accerlerationDecomposition} as
<div class="math">\begin{equation}\label{eq:normalVector}
\vec{N}\left(t\right)=\frac{\vec{a}\left(t\right)-a_T\left(t\right)\vec{T}\left(t\right)}{a_N\left(t\right)}=\frac{1}{a_N\left(t\right)}\left(\vec{r}''\left(t\right)-a_T\left(t\right)\frac{\vec{r}'\left(t\right)}{\left|\vec{r}'\left(t\right)\right|}\right).
\end{equation}</div>
</li>
<li>By Theorem 2, <span class="math">\(a_N\left(t\right)=\kappa\left(t\right)\left|\vec{r}'\left(t\right)\right|^2\)</span>, thus the curvature <span class="math">\(\kappa\)</span> can also be easily computed from <span class="math">\(a_N\)</span> as
<div class="math">\begin{equation}\label{eq:curvatureFromAccerleration}
\kappa\left(t\right)=\frac{a_N\left(t\right)}{\left|\vec{r}'\left(t\right)\right|^2}=\frac{\left|\vec{r}''\left(t\right)\times\vec{r}'\left(t\right)\right|}{\left|\vec{r}'\left(t\right)\right|^3}.
\end{equation}</div>
</li>
</ol>
<hr />
<h2 id="plane-curves-a-special-case">Plane Curves: A Special Case</h2>
<p>Everything we developed so far applies equally well to plane (2D) curves. The only trick is to "embed" a 2D curve into 3D by trivially attaching a <span class="math">\(0\)</span> at the end of the 2D vector that represents the plane curve, making the curve 3D. If one needs to find <span class="math">\(\vec{T}\)</span> and <span class="math">\(\vec{N}\)</span> for a plane curve using this trick, remember to drop the artificially attached <span class="math">\(z\)</span>-component (thus bringing them back to 2D) at the end of the computation.</p>
<p><strong>Example.</strong> Compute the curvature of the plane curve
</p>
<div class="math">$$t\mapsto\left(x\left(t\right),y\left(t\right)\right).$$</div>
<p><em>Solution.</em> Let <span class="math">\(\vec{r}\left(t\right)=\left(x\left(t\right),y\left(t\right),0\right)\)</span>. Then
</p>
<div class="math">\begin{align*}
\vec{r}'\left(t\right)&=\left(x'\left(t\right),y'\left(t\right),0\right),\\
\left|\vec{r}'\left(t\right)\right|&=\sqrt{\left(x'\left(t\right)\right)^2+\left(y'\left(t\right)\right)^2},\\
\vec{r}''\left(t\right)&=\left(x''\left(t\right),y''\left(t\right),0\right),\\
\vec{r}''\left(t\right)\times\vec{r}'\left(t\right)&=\begin{vmatrix}
\vec{i} & \vec{j} & \vec{k} \\
x'\left(t\right) & y'\left(t\right) & 0 \\
x''\left(t\right) & y''\left(t\right) & 0 \notag
\end{vmatrix}
=\left(x'\left(t\right)y''\left(t\right)-y'\left(t\right)x''\left(t\right)\right)\vec{k}.
\end{align*}</div>
<p>
The curvature can thus be computed from \eqref{eq:curvatureFromAccerleration} as
</p>
<div class="math">\begin{equation}\label{eq:curvaturePlaneCurve}
\kappa\left(t\right)=\frac{\left|\vec{r}''\left(t\right)\times\vec{r}'\left(t\right)\right|}{\left|\vec{r}'\left(t\right)\right|^3}=\frac{\left|x'\left(t\right)y''\left(t\right)-y'\left(t\right)x''\left(t\right)\right|}{\left[\left(x'\left(t\right)\right)^2+\left(y'\left(t\right)\right)^2\right]^{\frac{3}{2}}}.
\end{equation}</div>
<p>
Note that this is exactly the same formula as in the textbook for plane curves (\$12.6 Equation (12)).</p>
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<h2>useful links</h2>
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<li><a href="http://www.math.duke.edu/">Department of Mathematics</a></li>
<li><a href="http://www.math.duke.edu/courses/math_everywhere/">Math Everywhere @ Duke</a></li>
<li><a href="http://www.wolframalpha.com/">Wolfram Alpha</a></li>
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<h2>social</h2>
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<li><a href="https://github.com/trgao10">GitHub</a></li>
<li><a href="https://www.linkedin.com/pub/tingran-gao/89/8a8/a8a">LinkedIn</a></li>
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