Line 11: Line 11:
 
[[slecture_differential_geometry_Boutin_Black_Wu_background|'''1. Background and Motivation for Learning Differential Geometry''']]
 
[[slecture_differential_geometry_Boutin_Black_Wu_background|'''1. Background and Motivation for Learning Differential Geometry''']]
  
== 2. On the Geometry of Smooth Curves in Two-dimensional Space==
+
'''2. On the Geometry of Smooth Curves in Two-dimensional Space'''
  
:2.1 Curve Parameterization
+
:[[slecture_differential_geometry_Boutin_Black_Wu_curves_in_plane_parameterization| 2.1 Curve Parameterization]]
 
:2.2
 
:2.2
 
:2.3
 
:2.3
  
 
==3. ==
 
==3. ==
-----
+
----
= Notation  =
+
 
+
For a two-dimensional (2D) space <math>\mathbb{R}^2</math>, for example the <span class="texhtml">''X''</span>-<span class="texhtml">''Y''</span> plane, we can represent a smooth curve, <span class="texhtml">''C''</span>, in it as following: <math>
+
C=\left\{ p(t):\big( x(t),y(t)\big) |t\in I\right\}
+
</math>
+
 
+
, where <span class="texhtml">''I''</span> is some open interval of real number set <math>\mathbb{R}</math>. The smooth curve indicates that the curve is differentiable everywhere (at least 3 times). Imaging we are drawing the this curve <span class="texhtml">''C''</span> in that plane with a pen, and then <span class="texhtml">''p''(''t'')</span> is the position of the pen at time t. Fig.1 and Figs.2 are two examples. However, the same curve <span class="texhtml">''C''</span> can have many different expressions. For example, the straight line that passes through the origin with slope 2, i.e. the straight in Fig.3, can be written as (a) <math>\left\{\begin{array}{l}x(t)=t \\y(t)=2t\end{array}\right.</math>, (b) <math>\left\{\begin{array}{l}x(t)=t^2 \\y(t)=2t^2\end{array}\right.</math>, or (c) <math>\left\{\begin{array}{l}x(t)=\cos (t) \\y(t)=2\cos (t)\end{array}\right.</math>. Therefore, the spacing in time has nothing to do with the spacing in space.
+
 
+
[[Image:Cw df 1 notation 1.gif|thumb|left|400px|Fig 1: Example of curve parameterization]]
+
 
+
[[Image:Cw df 1 notation 2 P.png|thumb|left|300px|Fig 2.a: The points in $X$-$Y$ plane]]
+
[[Image:Cw df 1 notation 2 X.png|thumb|left|300px|Fig 2.b: X(t)]]
+
[[Image:Cw df 1 notation 2 Y.png|thumb|left|300px|Fig 2.c: Y(t)]]
+
 
+
[[Image:Cw df 1 notation 4.png|thumb|left|200px|Fig 3: Straight line go through origin with slope 2]]
+
 
+
In order to make the formulas for the same curve unanimous, we are going to introduce the concept of "arclength". First, we need to define the speed vector of the curve. For the curve C defined in above equation, its speed vector is defined as:
+
 
+
<math>
+
p^{\prime}(t)=\frac{d}{dt}p(t)=\left( \frac{d}{dt}x(t),\frac{d}{dt}y(t)\right).
+
</math>
+
 
+
The speed, i.e. the magnitude of this speed vector, is:
+
 
+
<math>
+
\| p^{\prime}(t)\| =\sqrt{\left( \frac{d}{dt}x(t)\right)^2+\left( \frac{d}{dt}y(t)\right)^2}.
+
</math>
+
 
+
For the same expressions of the straight line <span class="texhtml">''y'' = 2''x''</span> above, we know the corresponding speeds are: (a) <math>\|p^{\prime} (t)\| =\| (1,2)\|=\sqrt{5}</math>, (b) <math>\|p^{\prime} (t)\| =\| (2t,4t)\| =\sqrt{20} |t|</math>, and (c) <math>\| p^{\prime} (t)\| =\| \big( -\sin(t),-2\text{sin} (t) \big)\| =\sqrt{5} |\sin(t)|</math>. From these examples, we can see that the speed is not equal to the slope of the straight line.
+
 
+
Now, we will uniquely parameterize the curve with respect to the "arclength" <span class="texhtml">''s''</span>, such that <math>\| \frac{d}{ds}p(s)\| =1</math>, <math>\forall s</math>, i.e. the unit speed parameterization. Hence, with any given parameter <span class="texhtml">''t''</span>, we can compute the derivative of <span class="texhtml">''p''(''t'')</span> with respect to the arclength <span class="texhtml">''s''</span> as following: <math>
+
\frac{d}{ds}p(t)=\frac{p^{\prime}(t)}{\| p^{\prime}(t)\|}=\frac{\big( x^{\prime}(t),y^{\prime}(t)\big)}{\sqrt{\big( x^{\prime}(t)\big) ^2+\big( y^{\prime}(t)\big) ^2}}.
+
</math>
+
 
+
In other hand, we can also compute the same derivative with the chain rule:
+
 
+
<math>
+
\frac{d}{ds}p(t)=\frac{d}{dt}p(t)\cdot\frac{dt}{ds}=p^{\prime}(t)\cdot\frac{dt}{ds}.
+
</math>
+
 
+
Comparing above two equations, we can derive that:
+
 
+
<math>
+
\frac{dt}{ds}=\frac{1}{\sqrt{\big( x^{\prime}(t)\big) ^2+\big( y^{\prime}(t)\big) ^2}}\Rightarrow
+
ds=\sqrt{\big( x^{\prime}(t)\big) ^2+\big( y^{\prime}(t)\big) ^2}dt=\|p^{\prime}(t)\| dt,
+
</math>
+
 
+
since <math>\| \frac{d}{ds}p(t)\| =1</math>. Then, we can easily compute the length of the curve <span class="texhtml">''p''(''t'')</span> from <span class="texhtml">''t'' = ''a''</span> to <span class="texhtml">''t'' = ''b''</span>:
+
<math>
+
\int^{s(b)}_{s(a)}\|p^{\prime}(s)\| ds=\int^{s(b)}_{s(a)}1\cdot ds=\int^{b}_{a}\|p^{\prime}(t)\| dt=s(b)-s(a),
+
</math>
+
 
+
where <span class="texhtml">''s''(''t'')</span> is the parameter substitution function from <span class="texhtml">''t''</span> to <span class="texhtml">''s''</span>. For example, if <span class="texhtml">''p''(''s'')</span> is parameterized by arclength, the length of the curve from point <span class="texhtml">''p''(2)</span> to <span class="texhtml">''p''(17.5)</span> is <span class="texhtml">17.5 − 2 = 15.5</span>.
+
 
+
 
----
 
----

Revision as of 17:56, 5 March 2015


The Boutin Lectures on Introductory Differential Geometry

SLectures by Chyuan-Tyng "Roger" Wu and Will Black

© 2013, 2015


1. Background and Motivation for Learning Differential Geometry

2. On the Geometry of Smooth Curves in Two-dimensional Space
2.1 Curve Parameterization
2.2
2.3

3.



Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn