nyuuzyou/wikis huggingface dataset:
Equalizer as Limit

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $f_1, f_2: C_1 \to C_2$ be morphisms of $\mathbf C$.
Let their equalizer $e: E \to C_1$ exist in $\mathbf C$.
Then $\struct {E, e}$ is the limit of the diagram $D: \mathbf J \to \mathbf C$ defined by:
::$\begin{xy}\xymatrix@+1em@L+3px{
 \mathbf{J}:
&
 \ast
 \ar[r]<2pt>
 \ar[r]<-2pt>
&
 \star
}\end{xy}$
::$\begin{xy}\xymatrix@+1em@L+3px{
 D:
&
 C_1
 \ar[r]<2pt>^*+{f_1}
 \ar[r]<-2pt>_*+{f_2}
&
 C_2
}\end{xy}$
== Proof ==
== Sources ==
* : $\S 5.4$: Example $5.18$
Category:Category Theory
Equalizer is Monomorphism

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $e: E \to C$ be the equalizer of two morphisms $f, g: C \to D$.
Then $e$ is a monomorphism.
== Proof ==
Suppose that for morphisms $x,y: Z \to E$, it holds that:
:$e \circ y = e \circ x$
Putting $z = e \circ x$, the following commutative diagram applies:
$\quad\quad \begin{xy}\xymatrix{
E
\ar[r]^*{e} 
&
C
\ar[r]<2pt>^*{f}
\ar[r]<-2pt>_*{g} 
&
D 
\\
Z
\ar[u]<2pt>^*{x}
\ar[u]<-2pt>_*{y}
\ar[ur]_*{z}
}\end{xy}$
It follows that $f \circ z = g \circ z$.
Since $e$ is an equalizer, there thus exists a unique $u: Z \to E$ with:
:$z = e \circ u$
Hence $x = u = y$, and it follows that $e$ is a monomorphism.
== Sources ==
* : $\S 3.3$: Proposition $3.16$
Category:Category Theory
Equating Coefficients/Examples/Quadratic

== Example of Equating Coefficients ==
Consider the quadratic equation:
:$(1): \quad x^2 + a x + b = 0$
Let the roots of $(1)$ be $\alpha$ and $\beta$.
Thus:
Hence by equating coefficients:
== Sources ==
* 
* 
Category:Examples of Equating Coefficients
Equation for Line through Two Points in Complex Plane

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $L$ be a straight line through $z_1$ and $z_2$ in the complex plane.
=== Formulation 1 ===
=== Parametric Form $1$ ===
=== Parametric Form $2$ ===
=== Symmetric Form ===
== Examples ==
Category:Geometry of Complex Plane
Category:Equations of Straight Lines in Plane
Category:Equation for Line through Two Points in Complex Plane
Equation for Line through Two Points in Complex Plane/Examples/2+i, 3-2i

== Example of Use of Equation for Line through Two Points in Complex Plane ==
Let $L$ be a straight line through $2 + i$ and $3 - 2 i$ in the complex plane.
Then $L$ can be expressed as an equation in the following ways:
==== Parametric Form: $1$ ====
==== Parametric Form: $2$ ====
==== Standard Form ====
Category:Examples of Use of Equation for Line through Two Points in Complex Plane
Equation for Line through Two Points in Complex Plane/Examples/2+i, 3-2i/Parametric Form 1

== Example of Use of Equation for Line through Two Points in Complex Plane ==
Let $L$ be a straight line through $2 + i$ and $3 - 2 i$ in the complex plane.
Then $L$ can be expressed by the equation:
:$z - \paren {2 + i} = t \paren {1 - 3 i}$
== Proof ==
From Equation for Line through Two Points in Complex Plane: Parametric Form $1$, a straight line $L$ passing through $2$ points $z_1$ and $z_2$ has the equation:
:$z = z_1 + t \paren {z_2 - z_1}$
Hence:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $70 \ \text {(a)}$
Category:Examples of Use of Equation for Line through Two Points in Complex Plane
Equation for Line through Two Points in Complex Plane/Examples/2+i, 3-2i/Parametric Form 2

== Example of Use of Equation for Line through Two Points in Complex Plane ==
Let $L$ be a straight line through $2 + i$ and $3 - 2 i$ in the complex plane.
Then $L$ can be expressed by the equations:
:$x = 2 + t, y = 1 - 3 t$
== Proof ==
From Equation for Line through Two Points in Complex Plane: Parametric Form $2$, a straight line $L$ passing through $2$ points $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$ has the equations:
Hence:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $70 \ \text {(a)}$
Category:Examples of Use of Equation for Line through Two Points in Complex Plane
Equation for Line through Two Points in Complex Plane/Examples/2+i, 3-2i/Standard Form

== Example of Use of Equation for Line through Two Points in Complex Plane ==
Let $L$ be a straight line through $2 + i$ and $3 - 2 i$ in the complex plane.
Then $L$ can be expressed by the equation:
:$3 x + y = 7$
== Proof ==
From Two-Point Form of Equation of Straight Line in Plane, a straight line $L$ passing through $2$ points $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$ has the equation:
:$\dfrac {x - x_1} {x_2 - x_1} = \dfrac {y - y_1} {y_2 - y_1}$
Hence:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $70 \ \text {(a)}$
Category:Examples of Use of Equation for Line through Two Points in Complex Plane
Equation for Line through Two Points in Complex Plane/Formulation 1

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $L$ be a straight line through $z_1$ and $z_2$ in the complex plane.
$L$ can be expressed by the equation:
:$\map \arg {\dfrac {z - z_1} {z_2 - z_1} } = 0$
== Proof ==
Let $z$ be a point on the $L$.
Then:
:$z - z_1 = b \paren {z - z_2}$
where $b$ is some real number.
Then:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $137$
Category:Equation for Line through Two Points in Complex Plane
Equation for Line through Two Points in Complex Plane/Parametric Form 1

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $L$ be a straight line through $z_1$ and $z_2$ in the complex plane.
$L$ can be expressed by the equation:
:$z = z_1 + t \paren {z_2 - z_1}$
or:
:$z = \paren {1 - t} z_1 + t z_2$
This form of $L$ is known as the parametric form, where $t$ is the parameter.
== Proof ==
Let $z_1$ and $z_2$ be represented by the points $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ respectively in the complex plane.
Let $z$ be an arbitrary point on $L$ represented by the point $P$.
:
From Geometrical Interpretation of Complex Addition:
As $AP$ and $AB$ are collinear:
:$AP = t AB$
and so:
:$z - z_1 = t \paren {z_2 - z_1}$
The given expressions follow after algebra.
== Sources ==
* : $1$: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $11$
Category:Equation for Line through Two Points in Complex Plane
Equation for Line through Two Points in Complex Plane/Parametric Form 2

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $L$ be a straight line through $z_1$ and $z_2$ in the complex plane.
$L$ can be expressed by the equations:
These are the parametric equations of $L$, where $t$ is the parameter.
== Proof ==
From Equation for Line through Two Points in Complex Plane: Parametric Form 1:
:$z = z_1 + t \paren {z_2 - z_1}$
Letting:
the parametric equations follow by equating real parts and imaginary parts.
== Sources ==
* : $1$: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $11$
Category:Equation for Line through Two Points in Complex Plane
Equation for Line through Two Points in Complex Plane/Symmetric Form

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $L$ be a straight line through $z_1$ and $z_2$ in the complex plane.
$L$ can be expressed by the equation:
:$z = \dfrac {m z_1 + n z_2} {m + n}$
This form of $L$ is known as the symmetric form.
== Proof ==
Let $z_1$ and $z_2$ be represented by the points $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ respectively in the complex plane.
Let $z$ be an arbitrary point on $L$ represented by the point $P$.
:
As $AP$ and $AB$ are collinear:
:$m AP = n PB$
and so:
:$m \paren {z - z_1} = n \paren {z_2 - z_1}$
The result follows.
== Sources ==
* : $1$: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $11$
Category:Equation for Line through Two Points in Complex Plane
Equation for Perpendicular Bisector of Two Points

== Theorem ==
Let $\tuple {x_1, y_1}$ and $\tuple {y_1, y_2}$ be two points in the cartesian plane.
Let $L$ be the perpendicular bisector of the straight line through $z_1$ and $z_2$ in the complex plane.
$L$ can be expressed by the equation:
:$y - \dfrac {y_1 + y_2} 2 = \dfrac {x_1 - x_2} {y_2 - y_1} \paren {x - \dfrac {x_1 + x_2} 2}$
== Proof ==
Let $M$ be the straight line passing through $z_1$ and $z_2$.
Let $Q$ be the midpoint of $M$.
We have that:
:$Q = \tuple {\dfrac {x_1 + x_2} 2, \dfrac {y_1 + y_2} 2}$
The slope of $M$ is $\dfrac {y_2 - y_1} {x_2 - x_1}$.
As $L$ is perpendicular to the $M$, its slope is $\dfrac {x_1 - x_2} {y_2 - y_1}$.
Thus by Equation of Straight Line in Plane: Point-Slope Form, the equation for $L$ is:
:$y - \dfrac {y_1 + y_2} 2 = \dfrac {x_1 - x_2} {y_2 - y_1} \paren {x - \dfrac {x_1 + x_2} 2}$
Category:Straight Lines
Category:Perpendiculars
Equation for Perpendicular Bisector of Two Points in Complex Plane

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $L$ be the perpendicular bisector of the straight line through $z_1$ and $z_2$ in the complex plane.
=== Parametric Form $1$ ===
=== Parametric Form $2$ ===
=== Standard Form ===
== Examples ==
Category:Equation for Perpendicular Bisector of Two Points in Complex Plane
Category:Perpendicular Bisectors
Category:Geometry of Complex Plane
Category:Equations of Straight Lines in Plane
Equation for Perpendicular Bisector of Two Points in Complex Plane/Examples/2+i, 3-2i

== Examples of Use of Equation for Perpendicular Bisector of Two Points in Complex Plane ==
Let $L$ be $L$ be the perpendicular bisector of the straight line through $2 + i$ and $3 - 2 i$ in the complex plane.
Then $L$ can be expressed as an equation in the following ways:
==== Parametric Form: $1$ ====
==== Parametric Form: $2$ ====
==== Standard Form ====
Category:Examples of Use of Equation for Perpendicular Bisector of Two Points in Complex Plane
Equation for Perpendicular Bisector of Two Points in Complex Plane/Examples/2+i, 3-2i/Parametric Form 1

== Example of Use of Equation for Perpendicular Bisector of Two Points in Complex Plane ==
Let $L$ be the perpendicular bisector of the straight line through $2 + i$ and $3 - 2 i$ in the complex plane.
Then $L$ can be expressed by the equation:
:$z - \paren {\dfrac 5 2 - \dfrac i 2} = t \paren {3 + i}$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $70 \ \text {(b)}$
Category:Examples of Use of Equation for Perpendicular Bisector of Two Points in Complex Plane
Equation for Perpendicular Bisector of Two Points in Complex Plane/Examples/2+i, 3-2i/Parametric Form 2

== Example of Use of Equation for Perpendicular Bisector of Two Points in Complex Plane ==
Let $L$ be the perpendicular bisector of the straight line through $2 + i$ and $3 - 2 i$ in the complex plane.
Then $L$ can be expressed by the equations:
:$x = 3 t + \dfrac 5 2, y = t - \dfrac 1 2$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $70 \ \text {(b)}$
Category:Examples of Use of Equation for Perpendicular Bisector of Two Points in Complex Plane
Equation for Perpendicular Bisector of Two Points in Complex Plane/Examples/2+i, 3-2i/Standard Form

== Example of Use of Equation for Perpendicular Bisector of Two Points in Complex Plane ==
Let $L$ be the perpendicular bisector of the straight line through $2 + i$ and $3 - 2 i$ in the complex plane.
Then $L$ can be expressed by the equation:
:$x - 3 y = 4$
== Proof ==
Let $z_1 = 2 + i$ and $z_2 = 3 - 2 i$.
From Equation for Perpendicular Bisector of Two Points in Complex Plane/Standard Form, $L$ can be expressed by the equation:
:$\map \Re {z_2 - z_1} x + \map \Im {z_2 - z_1} y = \dfrac {\cmod {z_2}^2 - \cmod {z_1}^2} 2$
We have:
:$z_2 - z_1 = 1 - 3 i$
:$\cmod z_1^2 = 2^2 + 1^2 = 5$
:$\cmod z_2^2 = 3^2 + 2^2 = 13$
and hence we have the equation:
:$x - 3 y = \dfrac {13 - 5} 2 = 4$
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $70 \ \text {(b)}$
Category:Examples of Use of Equation for Perpendicular Bisector of Two Points in Complex Plane
Equation for Perpendicular Bisector of Two Points in Complex Plane/Parametric Form 1

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $L$ be the perpendicular bisector of the straight line through $z_1$ and $z_2$ in the complex plane.
$L$ can be expressed by the equation:
:$z - \dfrac {z_1 + z_2} 2 = i t\paren {z_2 - z_1}$
or:
:$z = \dfrac {z_1 + z_2} 2 + i t\paren {z_2 - z_1}$
This form of $L$ is known as the parametric form, where $t$ is the parameter.
== Proof ==
Let $z_1$ and $z_2$ be represented by the points $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ respectively in the complex plane.
Let $z$ be an arbitrary point on $L$ represented by the point $P$.
:
We have that $L$ passes through the point:
:$\dfrac {z_1 + z_2} 2$
and is perpendicular to the straight line:
:$z = z_1 + t \paren {z_2 - z_1}$
By Multiplication by Imaginary Unit is Equivalent to Rotation through Right Angle:
:the vector $i \paren {z_2 - z_1}$ is perpendicular to the vector $\paren {z_2 - z_1}$
Hence the perpendicular bisector can be written as:
:$z = \dfrac {z_1 + z_2} 2 + t i \paren {z_2 - z_1}$
Category:Equation for Perpendicular Bisector of Two Points in Complex Plane
Equation for Perpendicular Bisector of Two Points in Complex Plane/Standard Form

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $L$ be the perpendicular bisector of the straight line through $z_1$ and $z_2$ in the complex plane.
$L$ can be expressed by the equation:
:$\map \Re {z_2 - z_1} x + \map \Im {z_2 - z_1} y = \dfrac {\cmod {z_2}^2 - \cmod {z_1}^2} 2$
== Proof ==
Let $z_1$ and $z_2$ be represented by the points $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ respectively in the complex plane.
By Equation for Perpendicular Bisector of Two Points, the equation of their perpendicular bisector can be expressed as:
Category:Equation for Perpendicular Bisector of Two Points in Complex Plane
Equation of Astroid

== Theorem ==
Let $H$ be the astroid generated by the epicycle $C_1$ of radius $b$ rolling without slipping around the inside of a deferent $C_2$ of radius $a = 4 b$.
Let $C_2$ be embedded in a cartesian plane with its center $O$ located at the origin.
Let $P$ be a point on the circumference of $C_1$.
Let $C_1$ be initially positioned so that $P$ is its point of tangency to $C_2$, located at point $A = \tuple {a, 0}$ on the $x$-axis.
Let $\tuple {x, y}$ be the coordinates of $P$ as it travels over the plane.
=== Parametric Form ===
=== Cartesian Form ===
Category:Astroids
Equation of Astroid/Cartesian Form

== Theorem ==
Let $H$ be the astroid generated by the epicycle $C_1$ of radius $b$ rolling without slipping around the inside of a deferent $C_2$ of radius $a = 4 b$.
Let $C_2$ be embedded in a cartesian plane with its center $O$ located at the origin.
Let $P$ be a point on the circumference of $C_1$.
Let $C_1$ be initially positioned so that $P$ is its point of tangency to $C_2$, located at point $A = \tuple {a, 0}$ on the $x$-axis.
Let $\tuple {x, y}$ be the coordinates of $P$ as it travels over the plane.
The point $P = \tuple {x, y}$ is described by the equation:
:$x^{2/3} + y^{2/3} = a^{2/3}$
== Proof ==
By definition, an astroid is a hypocycloid with $4$ cusps.
:
From the parametric form of the equation of an astroid, $H$ can be expressed as:
:$\begin{cases}
x & = 4 b \cos^3 \theta = a \cos^3 \theta \\
y & = 4 b \sin^3 \theta = a \sin^3 \theta
\end{cases}$
Squaring, taking cube roots and adding:
== Sources ==
* : $\S 11$: Special Plane Curves: Hypocycloid with Four Cusps: $11.8$
* 
* : Chapter $\text {B}.21$: The Cycloid
* 
* : $\S 9$: Special Plane Curves: Hypocycloid with Four Cusps: $9.8.$
Category:Astroids
Equation of Astroid/Parametric Form

== Theorem ==
Let $H$ be the astroid generated by the epicycle $C_1$ of radius $b$ rolling without slipping around the inside of a deferent $C_2$ of radius $a = 4 b$.
Let $C_2$ be embedded in a cartesian plane with its center $O$ located at the origin.
Let $P$ be a point on the circumference of $C_1$.
Let $C_1$ be initially positioned so that $P$ is its point of tangency to $C_2$, located at point $A = \tuple {a, 0}$ on the $x$-axis.
Let $\tuple {x, y}$ be the coordinates of $P$ as it travels over the plane.
The point $P = \tuple {x, y}$ is described by the parametric equation:
:$\begin{cases}
x & = a \cos^3 \theta \\
y & = a \sin^3 \theta
\end{cases}$
where $\theta$ is the angle between the $x$-axis and the line joining the origin to the center of $C_1$.
== Proof ==
By definition, an astroid is a hypocycloid with $4$ cusps.
:
By Equation of Hypocycloid, the equation of $H$ is given by:
:$\begin{cases}
x & = \paren {a - b} \cos \theta + b \map \cos {\paren {\dfrac {a - b} b} \theta} \\
y & = \paren {a - b} \sin \theta - b \map \sin {\paren {\dfrac {a - b} b} \theta}
\end{cases}$
From Number of Cusps of Hypocycloid from Integral Ratio of Circle Radii, this can be generated by a epicycle $C_1$ of radius $\dfrac 1 4$ the radius of the deferent.
Thus $a = 4 b$ and the equation of $H$ is now given by:
:$\begin{cases}
x & = 3 b \cos \theta + b \cos 3 \theta \\
y & = 3 b \sin \theta - b \sin 3 \theta
\end{cases}$
From Triple Angle Formula for Cosine:
:$\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$
and from Triple Angle Formula for Sine:
:$\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$
Thus $H$ can be expressed as:
:$\begin{cases}
x & = 4 b \cos^3 \theta = a \cos^3 \theta \\
y & = 4 b \sin^3 \theta = a \sin^3 \theta
\end{cases}$
== Sources ==
* : $\S 11$: Special Plane Curves: Hypocycloid with Four Cusps: $11.9$
* 
* : Chapter $\text {B}.21$: The Cycloid
* 
* : $\S 9$: Special Plane Curves: Hypocycloid with Four Cusps: $9.9.$
* 
Category:Astroids
Equation of Cardioid/Parametric

== Theorem ==
Let $C$ be a cardioid embedded in a Cartesian coordinate plane such that:
:its deferent of radius $a$ is positioned with its center at $\tuple {a, 0}$
:there is a cusp at the origin.
Then $C$ can be expressed by the parametric equation:
:$\begin {cases} x = 2 a \cos t \paren {1 + \cos t} \\ y = 2 a \sin t \paren {1 + \cos t} \end {cases}$
== Proof ==
:
Let $P = \polar {x, y}$ be an arbitrary point on $C$.
From Polar Form of Equation of Cardioid, $C$ is expressed as a polar equation as:
:$r = 2 a \paren {1 + \cos \theta}$
We have that:
:$x = r \cos \theta$
:$y = r \sin \theta$
Replace $\theta$ with $t$ and the required parametric equation is the result.
== Sources ==
* 
Category:Cardioids
Equation of Cardioid/Polar

== Theorem ==
Let $C$ be a cardioid embedded in a polar coordinate plane such that:
:its deferent of radius $a$ is positioned with its center at $\polar {a, 0}$
:there is a cusp at the origin.
The polar equation of $C$ is:
:$r = 2 a \paren {1 + \cos \theta}$
== Proof ==
:
Let $P = \polar {r, \theta}$ be an arbitrary point on $C$.
Let $A$ and $B$ be the centers of the deferent and epicycle respectively.
Let $Q$ be the point where the deferent and epicycle touch.
By definition of the method of construction of $C$, we have that the arc $OQ$ of the deferent equals the arc $PQ$ of the epicycle.
Thus:
:$\angle OAQ = \angle PBQ$
and it follows that $AB$ is parallel to $OP$.
With reference to the diagram above, we have:
and the result follows.
== Also presented as ==
== Sources ==
* : $\S 11$: Special Plane Curves: Cardioid: $11.12$
* 
* 
* : $\S 9$: Special Plane Curves: Cardioid: $9.12.$
Category:Cardioids
Equation of Catenary/Cartesian

== Curve ==
Let a catenary be embedded in a cartesian plane so that the $y$-axis passes through the lowest point of the catenary.
==== Formulation 1 ====
==== Formulation 2 ====
== Proof of Formulation 1 ==
== Proof of Formulation 2 ==
== Historical Note ==
== Linguistic Note ==
== Sources ==
* : Chapter $8$: The System of the World: The English get left behind
Category:Catenary
Equation of Catenary/Cartesian/Formulation 1

== Theorem ==
Consider a catenary.
Let a cartesian plane be arranged so that the y-axis passes through the lowest point of the catenary.
The catenary is described by the equation:
:$y = \dfrac {e^{a x} + e^{-a x} } {2 a} = \dfrac {\cosh a x} a$
where $a$ is a constant.
The lowest point of the catenary is at $\tuple {0, \dfrac 1 a}$.
== Proof ==
== Historical Note ==
== Lingustic Note ==
Category:Catenary
Equation of Catenary/Cartesian/Formulation 1/Proof

== Theorem ==
== Proof ==
Let $\tuple {x, y}$ be an arbitrary point on the chain.
Let $s$ be the length along the arc of the chain from the lowest point to $\tuple {x, y}$.
Let $w_0$ be the linear mass density of the chain, that is, its weight per unit length.
The section of chain between the lowest point and $\tuple {x, y}$ is in static equilibrium under the influence of three forces, as follows:
:The tension $T_0$ at the lowest point
:The tension $T$ at the point $\tuple {x, y}$
:The weight $w_0 s$ of the chain between these two points.
As the chain is (ideally) flexible, the tension $T$ is along the line of the chain, and therefore along a tangent to the chain.
:
We can resolve this system of forces to obtain the horizontal and vertical components:
:$T_0 = T \cos \theta$
:$w_0 s = T \sin \theta$
We divide one by the other to eliminate $T$ and set $a = w_0 / T_0$:
:$\tan \theta = a s = \dfrac {\d y} {\d x}$
Differentiating with respect to $x$:
:$\dfrac {\d^2 y} {\d x^2} = a \dfrac {\d s} {\d x}$
From Derivative of Arc Length, we have:
:$\dfrac {\d s} {\d x} = \sqrt {1 + \paren {\dfrac {\d y} {\d x} }^2}$
So we have this differential equation to solve:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} = a \sqrt {1 + \paren {\dfrac {\d y} {\d x} }^2}$
Let $p = \dfrac {\d y} {\d x}$.
This transforms $(1)$ into:
:$\dfrac {\d p} {\d x} = a \sqrt {1 + p^2}$
This can be solved by Solution to Separable Differential Equation:
:$(2): \quad \ds \int \frac {\d p} {\sqrt {1 + p^2} } = \int a \rd x$
The is worked by using Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$:
::$\ds \int \frac {\d p} {\sqrt {1 + p^2} } = \map \ln {\sqrt {1 + p^2} + p} + c_1$
The is worked by using Primitive of Constant:
::$\ds \int a \rd x = a x + c_2$
So $(2)$ becomes:
:$\map \ln {\sqrt {1 + p^2} + p} = a x + c_3$
When $x = 0$ we have that $\theta = \dfrac {\d y} {\d x} = p = 0$ and so $c_3 = 0$, so:
:$\map \ln {\sqrt {1 + p^2} + p} = a x$
Therefore:
and:
which gives us:
:$p = \dfrac {\d y} {\d x} = \dfrac {e^{a x} - e^{-a x} } 2$
By Derivative of Exponential Function, we get:
:$y = \dfrac {e^{a x} + e^{-a x} } {2 a} + c_4$
We now need to position the coordinate axes so as to make $c_4 = 0$.
So, setting $x = 0$ and $c_4 = 0$:
:$y = \dfrac 1 a$
and the lowest point of the catenary is seen to be at $\tuple {0, \dfrac 1 a}$.
Hence the result:
:$y = \dfrac {e^{a x} + e^{-a x} } {2 a}$
== Historical Note ==
== Lingustic Note ==
== Sources ==
* : $\S 2.12$: The Hanging Chain. Pursuit Curves: Example $(1)$
* : Chapter $\text {B}.11$: The Catenary, or Curve of a Hanging Chain
Category:Catenary
Equation of Catenary/Cartesian/Formulation 2

== Theorem ==
Consider a catenary.
Let a cartesian plane be arranged so that the $y$-axis passes through the lowest point of the catenary.
The catenary is described by the equation:
:$y = \dfrac a 2 \paren {e^{x / a} + e^{-x / a} } = a \cosh \dfrac x a$
where $a$ is a constant.
The lowest point of the chain is at $\tuple {0, a}$.
== Proof ==
== Historical Note ==
== Lingustic Note ==
== Sources ==
* : $\S 11$: Special Plane Curves: Catenary: $11.15$
* 
* 
* : $\S 9$: Special Plane Curves: Catenary: $9.15.$
Category:Catenary
Equation of Catenary/Cartesian/Formulation 2/Proof

== Curve ==
== Proof ==
Take the equation of the catenary according to Formulation 1:
:$y = \dfrac {e^{ax} + e^{-ax}} {2 a}$
Put this in a form which uses the hyperbolic cosine:
:$y = \dfrac {\cosh a x} a$
Replace $a$ with $\dfrac 1 a$:
:$y = a \cosh \dfrac x a$
Hence the result.
== Historical Note ==
== Lingustic Note ==
Category:Catenary
Equation of Catenary/Cesàro/Formulation 1

== Theorem ==
The catenary can be described by the Cesàro equation:
:$a = \kappa \paren {a^2 + s^2}$
where:
:$s$ is the arc length
:$\kappa$ is the curvature
:$a$ is a constant.
== Proof ==
== Historical Note ==
== Lingustic Note ==
Category:Cesàro Equation for Catenary
Category:Catenary
Category:Cesàro Equations