Welcome to TiddlyWiki created by Jeremy Ruston, Copyright © 2007 UnaMesa Association
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To get started with this blank [[TiddlyWiki]], you'll need to modify the following tiddlers:
* [[SiteTitle]] & [[SiteSubtitle]]: The title and subtitle of the site, as shown above (after saving, they will also appear in the browser title bar)
* [[MainMenu]]: The menu (usually on the left)
* [[DefaultTiddlers]]: Contains the names of the tiddlers that you want to appear when the TiddlyWiki is opened
You'll also need to enter your username for signing your edits: <<option txtUserName>>
These [[InterfaceOptions]] for customising [[TiddlyWiki]] are saved in your browser
Your username for signing your edits. Write it as a [[WikiWord]] (eg [[JoeBloggs]])
<<option txtUserName>>
<<option chkSaveBackups>> [[SaveBackups]]
<<option chkAutoSave>> [[AutoSave]]
<<option chkRegExpSearch>> [[RegExpSearch]]
<<option chkCaseSensitiveSearch>> [[CaseSensitiveSearch]]
<<option chkAnimate>> [[EnableAnimations]]
----
Also see [[AdvancedOptions]]
Consider the vector functions
$$
{\bf{r_1}} (t) = (t, t^2, t^3), ~1 \leq t \leq 2
$$
$$
{\bf{r_2}} (t) = (e^u, e^{2u}, e^{3u}), ~ 0 \leq u \leq \ln 2.
$$
These vector functions trace out the same geometric object but do so with ''different parametrizations''.
A ''canonical'' or ''standard'' way to parametrize is to use the arc length of the curve. Let's define the ''arc length function''
$$ s(t) = \int_0^t |{\bf{r}}'(u)| du = \int_0^t \sqrt{ (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2 } dt
$$
with
$$
\frac{ds}{dt} = |{\bf{r}}' (t) |.
$$
''Example:'' Reparametrize the circular helix with respect to arc length measured from $(1,0,0)$ in the direction of increasing $t$.
Recall that the circular helix is given by the vector function $t \longmapsto {\bf{r}} (t) = (\cos(t), \sin(t), t)$.
We have calculated
$$
\frac{ds}{dt} = |{\bf{r}}' (t) | = \sqrt{2}
$$
so we have
$$
s(t) = \int_0^t |{\bf{r}}' (t) | dt = \sqrt{2} t.
$$
Therefore $t = t(s) = \frac{s}{\sqrt{2}}$ and the circular helix can be reexpressed as ${\bf{r}}(t(s))$. We obtain
$$
{\bf{r}} (t(s)) = (\cos(\frac{s}{\sqrt{2}}), \sin(\frac{s}{\sqrt{2}}), \frac{s}{\sqrt{2}}).
$$
Let's turn our attention to our next topic: [[Curvature]].
''Question:'' Find the arc length of the circular helix
$$ t \longmapsto ( \cos(t), \sin(t), t)$$
from $(1,0,0)$ to $(1,0, 2 \pi)$.
[img[Helix|http://www.web3d.org/x3d/specifications/vrml/ISO-IEC-14772-VRML97/Images/helix.gif]][This helix goes around 5 times.]
''Answer:'' We calculate
$$
|{\bf{r}}' (t) | = \sqrt{ ( - \sin(t))^2 + (\cos(t))^2 + 1^2} = \sqrt{2}.
$$
Therefore, the arc length of this piece of the circular helix is
$$
L = \int_0^{2 \pi} \sqrt{2} dt = 2 \sqrt{2} \pi.
$$
We can use arc length to build a [[standard parametrization|Arc Length Parametrization]] of curves given by vector functions.
*We have seen that parametric curves in the plane given by $[x = f(t), y = g(t)]$ have ''arc length'' given by the formula
$$
L = \int_a^b \sqrt{ (f'(t))^2 + (g'(t))^2} dt = \int_a^b \sqrt{ (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt.
$$
*For curves ${\bf{r}}(t) = (f(t), g(t), h(t))$ in space, we have a similar formula
$$
L = \int_a^b \sqrt{ (f'(t))^2 + (g'(t))^2 + (h'(t))^2} dt = \int_a^b \sqrt{ (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2 } dt.
$$
*Notice that $|{\bf{r}}' (t) | = \sqrt{ (f'(t))^2 + (g'(t))^2 + (h'(t))^2} $ so the arc length formula may also be expressed
$$
L = \int_a^b | {\bf{r}}' (t) | dt.
$$
Let's see an [[example|Arc Length of Circular Helix]].
[>img[Isaac Newton in 1689 at age 46 (1643-1727)|http://upload.wikimedia.org/wikipedia/commons/thumb/3/39/GodfreyKneller-IsaacNewton-1689.jpg/225px-GodfreyKneller-IsaacNewton-1689.jpg]]
Isaac Newton is the most influential scientist ever. He lived 84 years from 1643 to 1727.
* Newton is regarded as the [[2nd most influential human being in history|http://en.wikipedia.org/wiki/The_100:_A_Ranking_of_the_Most_Influential_Persons_in_History]].
* 1661 Entered Trinity College, Cambridge.
* 1665 Discovered Binomial Series
* 1665 Great Plague. Cambridge shuts down. Newton discovers the keys to the universe.
* 1669 Became Lucasian Professor of Mathematics.
* 1687 Publishes [[Principia Mathematica|Newton's Principia Mathematica]].
**[[Newton's Laws of Motion]]
**[[Newton's Law of Universal Gravitation]]
* 1699-1716 [[Newton vs. Leibniz Calculus Controversy|http://en.wikipedia.org/wiki/Newton_v._Leibniz_calculus_controversy]]
French mathematician ~Joseph-Louis Lagrange often said that Newton was the greatest genius who ever lived, and once added that Newton was also "the most fortunate, for we cannot find more than once a system of the world to establish."
[<img[Johannes Kepler on Wikipedia|http://upload.wikimedia.org/wikipedia/commons/thumb/d/d4/Johannes_Kepler_1610.jpg/225px-Johannes_Kepler_1610.jpg]][[Wikipedia>>Johannes Kepler|http://en.wikipedia.org/wiki/Johannes_Kepler]]
Kepler lived 1571-1630. He had an enticing idea called [[Mysterium Cosmographicum|http://en.wikipedia.org/wiki/Mysterium_cosmographicum]] to explain the structure of the solar system. The idea was wrong but his pursuit of this vision led to his discovery of [[three laws of planetary motion|Kepler's Laws]]. You can read more about Kepler and his high drama with [[Tycho Brahe|http://en.wikipedia.org/wiki/Tycho_Brahe]].
[<img[Mysterium Cosmographicum|http://upload.wikimedia.org/wikipedia/commons/thumb/2/2b/Tycho_Brahe.JPG/220px-Tycho_Brahe.JPG]]
[>img[Mysterium Cosmographicum|http://upload.wikimedia.org/wikipedia/commons/thumb/1/19/Kepler-solar-system-1.png/250px-Kepler-solar-system-1.png]]
[[TiddlerToddlerHelp|http://tiddlertoddler.tiddlyspot.com]]
[[Tutorial|http://www.blogjones.com/TiddlyWikiTutorial.html]]
PageTemplate
|>|>|SiteTitle - SiteSubtitle|
|MainMenu|DefaultTiddlers<br><br><br><br>ViewTemplate<br><br>EditTemplate|SideBarOptions|
|~|~|OptionsPanel|
|~|~|AdvancedOptions|
|~|~|<<tiddler Configuration.SideBarTabs>>|
''StyleSheet:'' StyleSheetColors - StyleSheetLayout - StyleSheetPrint
SiteUrl
*A parametrization $t \longmapsto {\bf{r}}(t)$ is ''smooth on an interval $I$'' if ${\bf{r}}'$ is continuous AND for all $t \in I$,
$$
{\bf{r}}' (t) \neq 0.
$$
*A curve $\mathcal{C}$ is called ''smooth'' if it has a smooth parametrization.
If $\mathcal{C}$ is a smooth curve defined by the vector function ${\bf{r}}(t)$ then the ''unit tangent vector'' is well-defined:
$$
{\bf{T}} (t) = \frac{{\bf{r}}'(t)}{|{\bf{r}}'(t)|}
$$
''Definition:'' The ''curvature'' of a curve is
$$
\kappa = | \frac{d {\bf{T}}}{ds} |.
$$
where ${\bf{T}}$ is the unit tangent vector and $s$ is the arc length.
''Reexpress curvature formula?''
Calculating the curvature with this formula is a bit of a hassle. If we are given ${\bf{r}}(t)$, we have to reparametrize it with arc length and then calculate the time derivative of the tangent vector. Wouldn't it would be more convenient if we could calculate the curvature directly from the formula for ${\bf{r}}(t)$?
*First Simplification of curvature formula:
Observe that
$$
\frac{d {\bf{T}}}{ds} = \frac{d {\bf{T}}}{dt} \frac{dt}{ds}= \frac{ \frac{d {\bf{T}}}{dt} }{\frac{ds}{dt}}
$$
and therefore
$$
\kappa = \frac{|{\bf{T}}' (t)|}{|{\bf{r}}' (t)|}.
$$
This is better than the formula given by the definition but still requires us to find ${\bf{T}}$ and calculate ${\bf{T}}'$.
''Example:'' Show that the curvature of a circle of radius $a$ is $\frac{1}{a}$.
The circle is given parametrically as
$${\bf{r}}(t) = (a \cos(t), a \sin(t)).$$
We calculate the derivative to see
$${\bf{r}}'(t) = (-a \sin(t), a \cos(t)), ~|{\bf{r}}'(t)| = a.$$
So, we can continue and calculate
$${\bf{T}}(t) = \frac{ {\bf{r}}'(t)}{|{\bf{r}}'(t)|} = \frac{ {\bf{r}}'(t)}{a}.
$$
Therefore, we see that ${\bf{T'}}(t)$ has length 1 since
$${\bf{T}}' (t) = \frac{1}{a} {\bf{r}}'' (t) = (- \cos (t), -\sin(t)).$$
Using the formula obtained using the chain rule, we find
$$
\kappa = \frac{1}{a}.
$$
We'd still like a [[better formula|Curvature in Terms of Vector Function]] for calculating the curvature in terms of ${\bf{r}}(t)$.
''Theorem:'' The curvature of a curve given by the vector function ${\bf{r}} (t) $ is
$$
\kappa (t) = \frac{| {\bf{r}}' (t) \times {\bf{r}}'' (t)|}{| {\bf{r}}' (t) |^3}.
$$
''Proof:'' By definition and our manipulations above,
$$ {\bf{T}} = \frac{{\bf{r}}'}{|{\bf{r}}'|} ~{\mbox{and}}~ |{\bf{r}}'| = \frac{ds}{dt}.
$$
We can manipulate this to write
$$
{\bf{r}}' = |{\bf{r}}'| {\bf{T}} = \frac{ds}{dt} {\bf{T}}.
$$ and differentiating again we obtain
$$
{\bf{r}}'' = \frac{d^2 s}{dt^2} {\bf{T}} +\frac{ds}{dt} {\bf{T}}'.
$$
Notice that ${\bf{T}} \times {\bf{T}} = {\bf{0}}$ (''Why?'') so
$$
{\bf{r}}' \times {\bf{r}}'' = (\frac{ds}{dt})^2 ( {\bf{T}} \times {\bf{T}}').
$$
Now $|{\bf{T}}(t)| = 1$ so ${\bf{T}}$ and ${\bf{T}}'$ are orthogonal (''Why?''). Therefore we obtain
$$
|{\bf{r}}' \times {\bf{r}}'' |= (\frac{ds}{dt})^2 | {\bf{T}} \times {\bf{T}}'| = (\frac{ds}{dt})^2 | {\bf{T}} | |{\bf{T}}'|.
$$
Since $ | {\bf{T}} | = 1$, we obtain
$$
|{\bf{T}}'| = \frac{|{\bf{r}}' \times {\bf{r}}'' |}{(\frac{ds}{dt})^2} = \frac{|{\bf{r}}' \times {\bf{r}}'' |}{|{\bf{r}}'|^2 }.
$$
But, by definition,
$$
\kappa = \frac{ |{\bf{T}}'| }{|{\bf{r}}'|} = \frac{|{\bf{r}}' \times {\bf{r}}'' |}{|{\bf{r}}'|^3 }. QED
$$
Let's turn our attention to the [[Frenet Frame]] to a curve given by a vector function.
[[Derivative of Vector Functions]]
[[Background on Johannes Kepler]]
The definite integral of a vector function is defined using Riemann sums as
$$
\int_a^b {\bf{r}} (t) dt = \lim_{\|P \| \rightarrow 0} \sum_{i = 1}^n {\bf{r}} ( t_i^*) (t_i - t_{i-1}).
$$
By breaking down ${\bf{r}}(t) = (f(t), g(t), h(t))$ we see that
$$
\int_a^b {\bf{r}} (t) dt = ( \int_a^b f(t) dt, \int_a^b g(t) dt, \int_a^b h(t) dt)
$$
So, integrals of vector functions are done component by component.
With the integral now defined, we can use it to define the [[arc length|Arc Length of Parametric Curves given by Vector Functions]].
This list of topics have been labeled with the tag Definition.
!Combining Newton's laws to specify the two body problem
We demonstrate Kepler's first law (The orbit of every planet is an ellipse with the Sun at a focus.) is a consequence of Newton's laws:
* ${\bf{F}} = m {\bf{a}}$
* $ {\bf{F}} = - G \frac{m_1 m_2}{r^3} {\bf{r}} = - G \frac{m_1 m_2}{r^2} {\bf{u}}$
The ${\bf{F}} = m {\bf{a}}$ law tells us how an object's motion reacts to the application of a force. The other law specifies the graviational force between any two objects.
We combine these laws to treat the ''idealized two body problem'' consisting of a big homogeneous spherical Sun and a tiny homogeneous spherical Earth. Since these objects are round and homogenous, a theorem of Newton (that we will not prove) allows us to collapse the objects to point masses located at their respective centres.
Equating ${\bf{F}}$ and ${\bf{F}}$, and cancelling $m$, we obtain
$$
{\bf{a}} = - \frac{GM}{r^3} {\bf{r}}.
$$
!The planet moves in a plane
This tells us that ${\bf{F}}$ and ${\bf{a}}$ are parallel. Therefore, $ {\bf{r}} \times {\bf{a}}= 0.$ We can use this!
$$\frac{d}{dt} ({\bf{r}} \times {\bf{v}}) = {\bf{r}}' \times {\bf{v}} + {\bf{r}}\times {\bf{v}}' = {\bf{v}} \times {\bf{v}} + {\bf{r}}\times {\bf{a}} = {\bf{0}}.
$$
Therefore ${\bf{r}} \times {\bf{v}}$ must be a constant vector. Let's call that vector ${\bf{h}}$ so
$$
{\bf{r}} \times {\bf{v}} = {\bf{h}}.
$$
We can also assume that ${\bf{v}}$ and ${\bf{r}}$ are not parallel (''Why?''). This ensures that $ {\bf{h}} \neq {\bf{0}}$.
For all values of $t$, we have therefore learned that ${\bf{r}}(t)$ is always perpendicular to a fixed vector ${\bf{h}}$.
Our first significant conclusion is: ''The motion of the planet takes place in a plane through the origin perpendicular to ${\bf{h}}$.''
!Vector Analysis
Let's try to understand this vector ${\bf{h}}$ a bit better.
$${\bf{h}} = {\bf{r}} \times {\bf{v}}= {\bf{r}} \times {\bf{r}}' = r {\bf{u}} \times {r\bf{u}}'.
$$
Continuing the calculation by applying the product rule, we get
$$
{\bf{h}} = r {\bf{u}} \times [ r {\bf{u}}' + r' {\bf{u}}] = r^2 {\bf{u}} \times {\bf{u}}' .
$$
Let's take the cross product of ${\bf{h}}$ with ${\bf{a}}$.
$$
{\bf{a}} \times {\bf{h}} = \frac{-GM}{r^2} {\bf{u}} \times [ r^2 {\bf{u}} \times {\bf{u}}' ] = -GM {\bf{u}} \times ( {\bf{u}} \times {\bf{u}}' ).
$$
We use the vector identity
$$
{\bf{a}} \times ( {\bf{b}} \times {\bf{c}}) = {\bf{a}} \cdot {\bf{c}} {\bf{b}}- {\bf{a}}\cdot {\bf{b}} {\bf{c}}
$$
to get
$$
{\bf{a}} \times {\bf{h}} = -GM [({\bf{u}} \cdot {\bf{u}}') {\bf{u}} - ({\bf{u}}\cdot {\bf{u}}) {\bf{u}}']
$$
Therefore, we observe that
$$
{\bf{a}} \times {\bf{h}} = GM {\bf{u}}'.
$$
Let's go backwards from this to get information on ${\bf{v}}$ by observing
$$
({\bf{v}} \times {\bf{h}})' = {\bf{v}}' \times {\bf{h}} = {\bf{a}} \times {\bf{h}} = GM {\bf{u}}'.
$$
Integrating both sides, we find that
$$
({\bf{v}} \times {\bf{h}}) = GM {\bf{u}} + {\bf{c}},
$$
where $ {\bf{c}}$ is some constant vector.
!Introduce Coordinates to Understand Dynamics Better
Let's choose axes so that ${\bf{h}}$ points up along the $z$ axis. We then observe that the planet moves in the $x,y$-plane. Since ${\bf{v}} \times {\bf{h}}$ and ${\bf{u}}$ are both perpendicular to ${\bf{h}}$, we have some information about which way the vector ${\bf{c}}$ points. In particular, ${\bf{c}}$ must lie in the $x,y$-plane. Let's choose the $x$ axis so that it points along the vector ${\bf{c}}$.
Let $\theta$ denote the angle between ${\bf{c}}$ and ${\bf{r}}$. Then $(r, \theta)$ are polar coordinates of the planet. We have
$$
{\bf{r}} \cdot ( {\bf{v}} \times {\bf{h}}) = {\bf{r}} \cdot (GM {\bf{u}} + {\bf{c}}) = GM {\bf{r}} \cdot {\bf{u}} + {\bf{r}} \cdot {\bf{c}}.
$$
Since ${\bf{r}} \cdot {\bf{u}} = r$ and ${\bf{r}} \cdot {\bf{c}} = |{\bf{r}}| {\bf{c}} \cos \theta = r c \cos \theta.$
Therefore, upon rewriting things, we have
$$
r = \frac{ {\bf{r}} \cdot ( {\bf{v}} \times {\bf{h}})}{GM + c \cos \theta}.
$$
But, we can calculate the numerator:
$$
{ {\bf{r}} \cdot ( {\bf{v}} \times {\bf{h}})} = ({\bf{r}} \times {\bf{v}} ) \cdot {\bf{h}} = {\bf{h}} \cdot {\bf{h}} = h^2.
$$
So, we have now found that
$$
r = \frac{\frac{h^2}{GM}}{1 + {\frac{c}{GM}} \cos \theta}.
$$
This is the equation of a conic section with focus at the origin and eccentricity $e = \frac{c}{GM}$. Since we know that the orbit of the earth is a closed curve, this conic section must be an ''ellipse''.
Entries tagged with Derivative.
We can describe a curve ${\mathcal{C}} \subset {\mathbb{R}}^3$ using a vector function
$$t \rightarrow {\bf{r}} (t) = (f(t), g(t), h(t)).$$
Let's calculate the derivative of this function:
$$
{\bf{r}}' (t) = \lim_{h \rightarrow 0} \frac{ {\bf{r}}(t+h) - {\bf{r}}(t) }{h} = ( f'(t), g'(t), h'(t)).
$$
So, derivatives of vector functions are calculated component by component.
If ${\bf{r}}' (t) \neq 0$ then we can define the (unit length) tangent vector to the curve:
$$
{\bf{T}} (t) = \frac{ {\bf{r}}' (t) }{|{\bf{r}}' (t) |}.
$$
Let's see an [[example|Example: Derivative Calculation for Elliptic Helix]].
Entries with the label Example.
''Question:'' Find parametric equations for the tangent line to the helix
$$t \longmapsto (2 \cos (t), \sin(t), t)
$$
at the point $(0,1, \frac{\pi}{2})$.
''Answer:'' We see that $z = t$ and when $t=0, 2 \cos(t) = 0$ and $\sin(t) =1$ so this point is indeed on the helix.
$$
{\bf{r}}' (t) = ( - 2 \sin (t), \cos(t), 1)
$$
At time $t = \frac{\pi}{2}$, we see that ${\bf{r}}' (t) = (-2, 0, 1)$. Our task is to describe a line passing through the point $(0, 1, \frac{\pi}{2})$ directed along the tangent vector $(-2, 0 , 1)$. The line can be described using the vector function
$$
s \longmapsto {\bf{l}} (s) = (0,1,\frac{\pi}{2}) + s ( -2, 0 , 1).
$$
Let's turn our attention next to the [[integral of a vector function|Definite Integral of Vector Function]]
!Normal Plane
Consider the curve $\mathcal{C}$ given by the vector function $t \longmapsto {\bf{r}}(t) \in {\mathbb{R}}^3$. Along the direction of the curve, we have identified the (unit) tangent vector ${\bf{T}}(t)$. At time $t$, the particle moving along this curve is passing along the tangent vector exactly perpindicular to the ''normal plane'' to the curve. The geometry of the curve determines an intrinsic coordinate system on the normal plane.
The (principal unit) ''Normal Vector'' ${\bf{N}} (t)$ to the curve is defined by
$$
{\bf{N}} (t) = \frac{ {\bf{T}}'(t)}{|{\bf{T}}' (t) |}.
$$
The vector
$${\bf{B}}(t) = {\bf{T}}(t) \times {\bf{N}} (t)
$$
is perpindicular to both $ {\bf{T}}(t)$ and to ${\bf{N}} (t) $. This vector is called the (unit) ''binormal vector''.Therefore ${\bf{B}}$ is the direction in the normal plane which is also perpindicular to the principal unit normal vector ${\bf{N}} (t) $.
The normal plane passes through ${\bf{r}}(t)$ and is spanned by the vectors ${\bf{N}}(t)$ and ${\bf{B}}(t)$. The tangent vector ${\bf{B}}(t)$ is orthogonal to the normal plane.
${\bf{T}}$ orthogonally penetrates the normal plane while bending toward the principal normal ${\bf{N}}$. In the other direction, orthogonal to the plane generated by ${\bf{T}}$ and ${\bf{N}}$, points the binormal ${\bf{B}}$.
!Osculating Plane
Consider the curve $\mathcal{C}$ given by the vector function ${\bf{r}}(t)$. At a given point in time, we have the tangent vector ${\bf{T}}$ which represents the direction of the curve. The curve is bending toward the unit normal ${\bf{N}}$ so the best plane for approximating the motion at this instant is the plane spanned by ${\bf{T}}$ and ${\bf{N}}$. This plane is called the ''osculating (kissing) plane'' to the curve. How fast is the curve bending toward the normal vector ${\bf{N}}$. This is measured by the curvature which can be understood as the reciprocal of the radius of the circle in the osculating plane which best approximates the curve. This circle is called the ''osculating circle'' to the curve. The direction orthogonal to the osculating plane is the ''binormal direction''.
The osculating plane passes through ${\bf{r}}(t)$ and is spanned by the vectors ${\bf{T}}(t)$ and ${\bf{N}}(t)$. The binormal vector ${\bf{B}}(t)$ is orthogonal to the osculating plane.
''Example:'' The unit circle $S^1$ is parametrized by $t \longmapsto {\bf{r}}(t) = ( \cos(t), \sin(t)) \in {\mathbb{R}^2_{x,y}}$. We can think of this circle sitting inside its normal plane, say $z = 0$, if we want to view it in 3 dimensions. The tangent vector is perpindicular to ${\bf{r}}(t) $. The principal unit normal is $-{\bf{r}}(t)$. The binormal is along the $z$ axis away from the plane containing the circle. Other curves can twist out of their normal plane but they linger in their normal plane.
The ''Frenet frame'' is the orthogonal system of coordinates defined by the vectors $({\bf{T}}(t), {\bf{N}}(t), {\bf{B}}(t))$ along the curve described parametrically by ${\bf{r}}(t)$.
*[[Nice Java Application for visualizing the Frenet frame|http://www.math.tu-berlin.de/geometrie/lab/curvesnsurfaces.shtml#FrenetFrames]]
*[[ShapeTape by Measurand Corporation|http://www.measurand.com/products/ShapeTape.html]] provides an intuitive introduction to the Frenet frame. Here is a [[nice video|http://www.youtube.com/watch?v=wAeow7Yq0l4]] demonstrating ShapeTape in a computer graphics application.
*[[Wikipedia>>Frenet-Serret Formulas|http://en.wikipedia.org/wiki/Frenet–Serret_formulas]]
To get started with this blank [[TiddlyWiki]], you'll need to modify the following tiddlers:
* [[SiteTitle]] & [[SiteSubtitle]]: The title and subtitle of the site, as shown above (after saving, they will also appear in the browser title bar)
* [[MainMenu]]: The menu (usually on the left)
* [[DefaultTiddlers]]: Contains the names of the tiddlers that you want to appear when the TiddlyWiki is opened
You'll also need to enter your username for signing your edits: <<option txtUserName>>
Entries tagged with Integral.
[>img[Kepler's first and second law|http://upload.wikimedia.org/wikipedia/commons/thumb/c/c9/Kepler-second-law.svg/220px-Kepler-second-law.svg.png]]
1. The orbit of every planet is an ellipse with the Sun at a focus.
2. A line joining a planet and the Sun sweeps out equal areas during equal intervals of time.[1]
3. The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
''Remarks:''
*The [[Kepler Applet|http://www.surendranath.org/Applets/Dynamics/Kepler/Kepler1Applet.html]] provides a nice visualization.
*Kepler obtained these insights by carefully studying astronomical data.
*We will obtain the first law as a consequence of [[Newton's Laws of Motion]] and [[Newton's Law of Universal Gravitation]].
Before we do that, let's look at some [[background on Isaac Newton|Background on Isaac Newton]].
[[GettingStarted]]
[[ConFig]]
[[Definition]]
[[Example]]
[[Derivative]]
[[Integral]]
[[Kepler]]
Just type in LaTeX:
$$\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}6$$
Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between the point masses:
$$
F = G \frac{m_1 m_2}{r^2},
$$
where:
[>img[Newton'a Law of Gravity|http://upload.wikimedia.org/wikipedia/commons/thumb/0/0e/NewtonsLawOfUniversalGravitation.svg/300px-NewtonsLawOfUniversalGravitation.svg.png]]
*$F$ is the magnitude of the gravitational force between the two point masses,
*$G$ is the gravitational constant,
*$m_1$ is the mass of the first point mass,
*$m_2$ is the mass of the second point mass, and
*$r$ is the distance between the two point masses.
Let's turn our attention to a [[Derivation of Keplerian Ellipses from Newton's Laws]].
#In the absence of a net force, the center of mass of a body either is at rest or moves at a constant velocity.
#A body experiencing a force ${\bf{F}}$ experiences an acceleration a related to ${\bf{F}}$ by ${\bf{F}} = m{\bf{a}}$, where $m$ is the mass of the body. Alternatively, force is equal to the time derivative of momentum.
#Whenever a first body exerts a force ${\bf{F}}$ on a second body, the second body exerts a force $-{\bf{F}}$ on the first body. ${\bf{F}}$ and $-{\bf{F}}$ are equal in magnitude and opposite in direction.
[[Newton's Principia Mathematica (1687) on Google Books|http://books.google.com/books?id=Tm0FAAAAQAAJ&pg=PA90#v=onepage&q=&f=false]]
/***
|Name|Plugin: Scientific Notation|
|Created by|BobMcElrath|
|Email|my first name at my last name dot org|
|Location|http://bob.mcelrath.org/tiddlyjsmath-2.0.3.html|
|Version|1.0|
|Requires|[[TiddlyWiki|http://www.tiddlywiki.com]] ≥ 2.0.3, [[jsMath|http://www.math.union.edu/~dpvc/jsMath/]] ≥ 3.0, [[Plugin: jsMath]]|
!Description
This plugin will render numbers expressed in scientific notation, such as {{{3.5483e12}}} using the jsMath plugin to display it in an intuitive way such as 3.5483e12. You may customize the number of significant figures displayed, as well as "normalize" numbers so that {{{47392.387e9}}} is displayed as 47392.387e9.
!Installation
Install the Requirements, above, add this tiddler to your tiddlywiki, and give it the {{{systemConfig}}} tag.
!History
* 1-Feb-06, version 1.0, Initial release
!Code
***/
//{{{
config.formatters.push({
name: "scientificNotation",
match: "\\b[0-9]+\\.[0-9]+[eE][+-]?[0-9]+\\b",
element: "span",
className: "math",
normalize: true, // set to 'true' to convert numbers to X.XXX \times 10^{y}
sigfigs: 3, // with this many digits in the mantissa
handler: function(w) {
var snRegExp = new RegExp("\\b([0-9]+(?:\\.[0-9]+)?)[eE]([-0-9+]+)\\b");
var mymatch = snRegExp.exec(w.matchText);
var mantissa = mymatch[1];
var exponent = parseInt(mymatch[2]);
// normalize the number.
if(this.normalize) {
mantissa = parseFloat(mantissa);
while(mantissa > 10.0) {
mantissa = mantissa / 10.0;
exponent++;
}
while(mantissa < 1.0) {
mantissa = mantissa * 10.0;
exponent--;
}
var sigfigsleft = this.sigfigs;
mantissa = parseInt(mantissa) + "." + (Math.round(Math.pow(10,this.sigfigs-1)*mantissa)+"").substr(1,this.sigfigs-1);
}
var e = document.createElement(this.element);
e.className = this.className;
if(exponent == 0) {
e.appendChild(document.createTextNode(mantissa));
} else {
e.appendChild(document.createTextNode(mantissa + "\\times 10^{" + exponent + "}"));
}
w.output.appendChild(e);
}
});
//}}}
/***
|Name|Plugin: arXiv Links|
|Created by|BobMcElrath|
|Email|my first name at my last name dot org|
|Location|http://bob.mcelrath.org/tiddlyjsmath-2.0.3.html|
|Version|1.0|
|Requires|[[TiddlyWiki|http://www.tiddlywiki.com]] ≥ 2.0.3|
!Description
This formatting plugin will render links to the [[arXiv|http://www.arxiv.org]] preprint system. If you type a paper reference such as hep-ph/0509024, it will be rendered as an external link to the abstract of that paper.
!Installation
Add this tiddler to your tiddlywiki, and give it the {{{systemConfig}}} tag.
!History
* 1-Feb-06, version 1.0, Initial release
!Code
***/
//{{{
config.formatters.push({
name: "arXivLinks",
match: "\\b(?:astro-ph|cond-mat|hep-ph|hep-th|hep-lat|gr-qc|nucl-ex|nucl-th|quant-ph|(?:cs|math|nlin|physics|q-bio)(?:\\.[A-Z]{2})?)/[0-9]{7}\\b",
element: "a",
handler: function(w) {
var e = createExternalLink(w.output, "http://arxiv.org/abs/"+w.matchText);
e.target = "_blank"; // open in new window
w.outputText(e,w.matchStart,w.nextMatch);
}
});
//}}}
/***
|Name|Plugin: jsMath|
|Created by|BobMcElrath|
|Email|my first name at my last name dot org|
|Location|http://bob.mcelrath.org/tiddlyjsmath.html|
|Version|1.5.1|
|Requires|[[TiddlyWiki|http://www.tiddlywiki.com]] ≥ 2.0.3, [[jsMath|http://www.math.union.edu/~dpvc/jsMath/]] ≥ 3.0|
!Description
LaTeX is the world standard for specifying, typesetting, and communicating mathematics among scientists, engineers, and mathematicians. For more information about LaTeX itself, visit the [[LaTeX Project|http://www.latex-project.org/]]. This plugin typesets math using [[jsMath|http://www.math.union.edu/~dpvc/jsMath/]], which is an implementation of the TeX math rules and typesetting in javascript, for your browser. Notice the small button in the lower right corner which opens its control panel.
!Installation
In addition to this plugin, you must also [[install jsMath|http://www.math.union.edu/~dpvc/jsMath/download/jsMath.html]] on the same server as your TiddlyWiki html file. If you're using TiddlyWiki without a web server, then the jsMath directory must be placed in the same location as the TiddlyWiki html file.
I also recommend modifying your StyleSheet use serif fonts that are slightly larger than normal, so that the math matches surrounding text, and \\small fonts are not unreadable (as in exponents and subscripts).
{{{
.viewer {
line-height: 125%;
font-family: serif;
font-size: 12pt;
}
}}}
If you had used a previous version of [[Plugin: jsMath]], it is no longer necessary to edit the main tiddlywiki.html file to add the jsMath <script> tag. [[Plugin: jsMath]] now uses ajax to load jsMath.
!History
* 11-Nov-05, version 1.0, Initial release
* 22-Jan-06, version 1.1, updated for ~TW2.0, tested with jsMath 3.1, editing tiddlywiki.html by hand is no longer necessary.
* 24-Jan-06, version 1.2, fixes for Safari, Konqueror
* 27-Jan-06, version 1.3, improved error handling, detect if ajax was already defined (used by ZiddlyWiki)
* 12-Jul-06, version 1.4, fixed problem with not finding image fonts
* 26-Feb-07, version 1.5, fixed problem with Mozilla "unterminated character class".
* 27-Feb-07, version 1.5.1, Runs compatibly with TW 2.1.0+, by Bram Chen
!Examples
|!Source|!Output|h
|{{{The variable $x$ is real.}}}|The variable $x$ is real.|
|{{{The variable \(y\) is complex.}}}|The variable \(y\) is complex.|
|{{{This \[\int_a^b x = \frac{1}{2}(b^2-a^2)\] is an easy integral.}}}|This \[\int_a^b x = \frac{1}{2}(b^2-a^2)\] is an easy integral.|
|{{{This $$\int_a^b \sin x = -(\cos b - \cos a)$$ is another easy integral.}}}|This $$\int_a^b \sin x = -(\cos b - \cos a)$$ is another easy integral.|
|{{{Block formatted equations may also use the 'equation' environment \begin{equation} \int \tan x = -\ln \cos x \end{equation} }}}|Block formatted equations may also use the 'equation' environment \begin{equation} \int \tan x = -\ln \cos x \end{equation}|
|{{{Equation arrays are also supported \begin{eqnarray} a &=& b \\ c &=& d \end{eqnarray} }}}|Equation arrays are also supported \begin{eqnarray} a &=& b \\ c &=& d \end{eqnarray} |
|{{{I spent \$7.38 on lunch.}}}|I spent \$7.38 on lunch.|
|{{{I had to insert a backslash (\\) into my document}}}|I had to insert a backslash (\\) into my document|
!Code
***/
//{{{
// AJAX code adapted from http://timmorgan.org/mini
// This is already loaded by ziddlywiki...
if(typeof(window["ajax"]) == "undefined") {
ajax = {
x: function(){try{return new ActiveXObject('Msxml2.XMLHTTP')}catch(e){try{return new ActiveXObject('Microsoft.XMLHTTP')}catch(e){return new XMLHttpRequest()}}},
gets: function(url){var x=ajax.x();x.open('GET',url,false);x.send(null);return x.responseText}
}
}
// Load jsMath
jsMath = {
Setup: {inited: 1}, // don't run jsMath.Setup.Body() yet
Autoload: {root: new String(document.location).replace(/[^\/]*$/,'jsMath/')} // URL to jsMath directory, change if necessary
};
var jsMathstr;
try {
jsMathstr = ajax.gets(jsMath.Autoload.root+"jsMath.js");
} catch(e) {
alert("jsMath was not found: you must place the 'jsMath' directory in the same place as this file. "
+"The error was:\n"+e.name+": "+e.message);
throw(e); // abort eval
}
try {
window.eval(jsMathstr);
} catch(e) {
alert("jsMath failed to load. The error was:\n"+e.name + ": " + e.message + " on line " + e.lineNumber);
}
jsMath.Setup.inited=0; // allow jsMath.Setup.Body() to run again
// Define wikifers for latex
config.formatterHelpers.mathFormatHelper = function(w) {
var e = document.createElement(this.element);
e.className = this.className;
var endRegExp = new RegExp(this.terminator, "mg");
endRegExp.lastIndex = w.matchStart+w.matchLength;
var matched = endRegExp.exec(w.source);
if(matched) {
var txt = w.source.substr(w.matchStart+w.matchLength,
matched.index-w.matchStart-w.matchLength);
if(this.keepdelim) {
txt = w.source.substr(w.matchStart, matched.index+matched[0].length-w.matchStart);
}
e.appendChild(document.createTextNode(txt));
w.output.appendChild(e);
w.nextMatch = endRegExp.lastIndex;
}
}
config.formatters.push({
name: "displayMath1",
match: "\\\$\\\$",
terminator: "\\\$\\\$\\n?", // 2.0 compatability
termRegExp: "\\\$\\\$\\n?",
element: "div",
className: "math",
handler: config.formatterHelpers.mathFormatHelper
});
config.formatters.push({
name: "inlineMath1",
match: "\\\$",
terminator: "\\\$", // 2.0 compatability
termRegExp: "\\\$",
element: "span",
className: "math",
handler: config.formatterHelpers.mathFormatHelper
});
var backslashformatters = new Array(0);
backslashformatters.push({
name: "inlineMath2",
match: "\\\\\\\(",
terminator: "\\\\\\\)", // 2.0 compatability
termRegExp: "\\\\\\\)",
element: "span",
className: "math",
handler: config.formatterHelpers.mathFormatHelper
});
backslashformatters.push({
name: "displayMath2",
match: "\\\\\\\[",
terminator: "\\\\\\\]\\n?", // 2.0 compatability
termRegExp: "\\\\\\\]\\n?",
element: "div",
className: "math",
handler: config.formatterHelpers.mathFormatHelper
});
backslashformatters.push({
name: "displayMath3",
match: "\\\\begin\\{equation\\}",
terminator: "\\\\end\\{equation\\}\\n?", // 2.0 compatability
termRegExp: "\\\\end\\{equation\\}\\n?",
element: "div",
className: "math",
handler: config.formatterHelpers.mathFormatHelper
});
// These can be nested. e.g. \begin{equation} \begin{array}{ccc} \begin{array}{ccc} ...
backslashformatters.push({
name: "displayMath4",
match: "\\\\begin\\{eqnarray\\}",
terminator: "\\\\end\\{eqnarray\\}\\n?", // 2.0 compatability
termRegExp: "\\\\end\\{eqnarray\\}\\n?",
element: "div",
className: "math",
keepdelim: true,
handler: config.formatterHelpers.mathFormatHelper
});
// The escape must come between backslash formatters and regular ones.
// So any latex-like \commands must be added to the beginning of
// backslashformatters here.
backslashformatters.push({
name: "escape",
match: "\\\\.",
handler: function(w) {
w.output.appendChild(document.createTextNode(w.source.substr(w.matchStart+1,1)));
w.nextMatch = w.matchStart+2;
}
});
config.formatters=backslashformatters.concat(config.formatters);
window.wikify = function(source,output,highlightRegExp,tiddler)
{
if(source && source != "") {
if(version.major == 2 && version.minor > 0) {
var wikifier = new Wikifier(source,getParser(tiddler),highlightRegExp,tiddler);
wikifier.subWikifyUnterm(output);
} else {
var wikifier = new Wikifier(source,formatter,highlightRegExp,tiddler);
wikifier.subWikify(output,null);
}
jsMath.ProcessBeforeShowing();
}
}
//}}}
Chapter 14 of Stewart's Calculus