avewright/proofwiki-math huggingface dataset:
Theorem: Slope of Lemniscate of Bernoulli at Origin

Consider the lemniscate of Bernoulli $M$ embedded in a Cartesian plane such that its foci are at $\tuple {a, 0}$ and $\tuple {-a, 0}$ respectively.
Let $O$ denote the origin.
The tangents to $M$ at $O$ are at an angle of $45 \degrees = \dfrac \pi 4$ to the $x$-axis.
:630px

Proof:
Recall the parametric definition of lemniscate of Bernoulli:
:$\begin {cases} x = \dfrac {a \sqrt 2 \cos t} {\sin^2 t + 1} \\ \\ y = \dfrac {a \sqrt 2 \cos t \sin t} {\sin^2 t + 1} \end {cases}$
Note that $\tuple {x, y} = \tuple {0, 0}$ {{iff}} $\cos t = 0$.
At these points we have $\sin t = \pm \sqrt {1 - \cos^2 t} = \pm 1$.
We have:
{{begin-eqn}}
{{eqn | l = \frac {\d x} {\d t}
      | r = a \sqrt 2 \paren {\frac {-\sin t \paren {\sin^2 t + 1} - \cos t \paren {2 \sin t \cos t} } {\paren {\sin^2 t + 1}^2} }
      | c = Quotient Rule for Derivatives
}}
{{eqn | r = \frac {a \sqrt 2} {\paren {\sin^2 t + 1}^2} \paren {-\sin^3 t - 2 \sin t \cos^2 t - \sin t}
      | c = 
}}
{{eqn | l = \frac {\d y} {\d t}
      | r = a \sqrt 2 \paren {\frac {\paren {\cos^2 t - \sin^2 t} \paren {\sin^2 t + 1} - \cos t \sin t \paren {2 \sin t \cos t} } {\paren {\sin^2 t + 1}^2} }
      | c = Quotient Rule for Derivatives
}}
{{eqn | r = \frac {a \sqrt 2} {\paren {\sin^2 t + 1}^2} \paren {\cos^2 t \sin^2 t - \sin^4 t + \cos^2 t - \sin^2 t - 2 \cos^2 t \sin^2 t}
      | c = 
}}
{{eqn | r = \frac {a \sqrt 2} {\paren {\sin^2 t + 1}^2} \paren {- \cos^2 t \sin^2 t - \sin^4 t + \cos^2 t - \sin^2 t}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d y} {\d x}
      | r = \frac {\d y} {\d t} \times \frac {\d t} {\d x}
      | c = Chain Rule for Derivatives
}}
{{eqn | r = \frac {-\cos^2 t \sin^2 t - \sin^4 t + \cos^2 t - \sin^2 t} {-\sin^3 t - 2 \sin t \cos^2 t - \sin t}
}}
{{end-eqn}}
By substituting $\cos t = 0$ and $\sin t = \pm 1$, we have:
{{begin-eqn}}
{{eqn | l = \valueat {\frac {\d y} {\d x} } {\tuple {x, y} \mathop = \tuple {0, 0} }
      | r = \frac {0 - \paren {\pm 1}^4 + 0 - \paren {\pm 1}^2} {-\paren {\pm 1}^3 - 0 - \paren {\pm 1} }
}}
{{eqn | r = \frac {-2} {\mp 2}
}}
{{eqn | r = \pm 1
}}
{{end-eqn}}
As $\map \tan {\pm 45 \degrees} = \pm 1$, these tangents at $O$ are at an angle of $45 \degrees$ to the $x$-axis.
{{qed}}
Theorem: Area of Lobe of Lemniscate of Bernoulli

Consider the lemniscate of Bernoulli $M$ embedded in a Cartesian plane such that its foci are at $\tuple {a, 0}$ and $\tuple {-a, 0}$ respectively.
Let $O$ denote the origin.
The area of one lobe of $M$ is $a^2$.

Proof:
By the definition of the lemniscate of Bernoulli, we have that the polar equation of $M$ is:
:$r^2 = 2 a^2 \cos 2 \theta$
Let $\AA$ denote the area of one lobe of $M$.
The boundary of the right hand lobe of $M$ is traced out where $-\dfrac \pi 2 \le 2 \theta \le \dfrac \pi 2$.
Thus:
{{begin-eqn}}
{{eqn | l = \AA
      | r = \int_{-\pi / 4}^{\pi / 4} \dfrac {\map {r^2} \theta} 2 \rd \theta
      | c = Area between Radii and Curve in Polar Coordinates
}}
{{eqn | r = \int_{-\pi / 4}^{\pi / 4} \dfrac {2 a^2 \cos 2 \theta} 2 \rd \theta
      | c = {{Defof|Lemniscate of Bernoulli|subdef = Polar Definition}}
}}
{{eqn | r = a^2 \int_{-\pi / 4}^{\pi / 4} \cos 2 \theta \rd \theta
      | c = simplifying
}}
{{eqn | r = a^2 \intlimits {\dfrac {\sin 2 \theta} 2} {-\pi / 4} {\pi / 4}
      | c = Primitive of $\cos a x$
}}
{{eqn | r = \dfrac {a^2} 2 \paren {\sin \dfrac \pi 2 - \map \sin {-\dfrac \pi 2} }
      | c = 
}}
{{eqn | r = \dfrac {a^2} 2 \paren {1 - \paren {-1} }
      | c = Sine of Right Angle etc.
}}
{{eqn | r = a^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Equation of Cardioid/Polar

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:
:525px
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:
{{begin-eqn}}
{{eqn | l = r
      | r = OR + RS + SP
      | c = 
}}
{{eqn | r = a \cos \theta + 2 a + a \cos \theta
      | c = 
}}
{{eqn | r = 2 a \paren {1 + \cos \theta}
      | c = 
}}
{{end-eqn}}
and the result follows.
{{qed}}
Theorem: Equation of Cardioid/Parametric

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:
:525px
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.
{{qed}}
Theorem: Area inside Cardioid

Consider the cardioid $C$ embedded in a polar plane given by its polar equation:
:$r = 2 a \paren {1 + \cos \theta}$
The area inside $C$ is $6 \pi a^2$.

Proof:
Let $\AA$ denote the area inside $C$.
The boundary of $C$ is traced out where $-\pi \le \theta \le \pi$.
Thus:
{{begin-eqn}}
{{eqn | l = \AA
      | r = \int_{-\pi}^\pi \dfrac {\map {r^2} \theta} 2 \rd \theta
      | c = Area between Radii and Curve in Polar Coordinates
}}
{{eqn | r = \int_{-\pi}^\pi \dfrac {\paren {2 a \paren {1 + \cos \theta} }^2} 2 \rd \theta
      | c = Equation of Cardioid
}}
{{eqn | r = 2 a^2 \int_{-\pi}^\pi \paren {1 + 2 \cos \theta + \cos^2 \theta} \rd \theta
      | c = simplifying
}}
{{eqn | r = 2 a^2 \intlimits {\theta + 2 \sin \theta + \frac \theta 2 + \frac {\sin 2 \theta} 4} {-\pi} \pi
      | c = Primitive of $\cos a x$, Primitive of Square of Cosine Function
}}
{{eqn | r = 2 a^2 \paren {\paren {\pi + 2 \sin \pi + \dfrac \pi 2 + \frac {\sin 2 \pi} 4} - \paren {-\pi + 2 \, \map \sin {-\pi} + \dfrac {-\pi} 2 + \frac {\map \sin {-2 \pi} } 4} }
      | c = evaluation between $-\pi$ and $\pi$
}}
{{eqn | r = 2 a^2 \paren {\pi + \dfrac \pi 2 - \paren {-\pi} - \paren {\dfrac {-\pi} 2} }
      | c = Sine of Multiple of Pi
}}
{{eqn | r = 6 \pi a^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Length of Perimeter of Cardioid

Consider the cardioid $C$ embedded in a polar plane given by its polar equation:
:$r = 2 a \paren {1 + \cos \theta}$
where $a > 0$.
The length of the perimeter of $C$ is $16 a$.

Proof:
Let $\LL$ denote the length of the perimeter of $C$.
The boundary of the $C$ is traced out where $-\pi \le \theta \le \pi$.
From Arc Length for Polar Curve:
:$\ds \LL = \int_{\theta \mathop = -\pi}^{\theta \mathop = \pi} \sqrt {r^2 + \paren {\frac {\d r} {\d \theta} }^2} \rd \theta$
where:
:$r = 2 a \paren {1 + \cos \theta}$
Note that we have: 
{{begin-eqn}}
{{eqn | l = \frac {\d r} {\d \theta}
      | r = 2 a \frac \d {\d \theta} \paren {1 + \cos \theta}
}}
{{eqn | r = -2 a \sin \theta
      | c = Sum Rule for Derivatives, Derivative of Cosine Function
}}
{{end-eqn}}
We therefore have: 
{{begin-eqn}}
{{eqn | l = \LL
      | r = \int_{-\pi}^\pi \sqrt {4 a^2 \paren {1 + \cos \theta}^2 + 4 a^2 \sin^2 \theta} \rd \theta
}}
{{eqn | r = 2 a \int_{-\pi}^\pi \sqrt {1 + 2 \cos \theta + \cos^2 \theta + \sin^2 \theta} \rd \theta
      | c = extracting a factor of $\sqrt {4 a^2} = 2 a$, noting that $a \ge 0$
}}
{{eqn | r = 2 a \int_{-\pi}^\pi \sqrt {2 + 2 \cos \theta} \rd \theta
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 4 a \int_0^\pi \sqrt {4 \cos^2 \frac \theta 2} \rd \theta
      | c = Definite Integral of Even Function, {{Corollary|Double Angle Formula for Cosine|1}}
}}
{{eqn | r = 8 a \int_0^\pi \size {\cos \frac \theta 2} \rd \theta
}}
{{eqn | r = 8 a \int_0^\pi \cos \frac \theta 2 \rd \theta
      | c = {{Defof|Absolute Value}}, noting that $\cos \theta \ge 0$ for $0 \le \theta \le \dfrac \pi 2$
}}
{{eqn | r = 16 a \intlimits {\sin \frac \theta 2} 0 \pi
      | c = Primitive of $\cos a x$
}}
{{eqn | r = 16 a \paren {\sin \frac \pi 2 - \sin 0}
}}
{{eqn | r = 16 a
      | c = Sine of Right Angle, Sine of Zero is Zero
}}
{{end-eqn}}
{{qed}}
Theorem: Arc Length for Polar Curve

Let $a$ and $b$ be real numbers.
Let $\CC$ be a simple curve continuous on $\closedint a b$ and continuously differentiable on $\openint a b$.
Let $\CC$ be described by the parametric equations:
:$\begin {cases} x & = r \cos \theta \\ y & = r \sin \theta \end {cases}$
where:
:$r$ is a function of $\theta$
:$\theta \in \closedint a b$.
Then the length $s$ of $\CC$ is given by:
:$\ds s = \int_a^b \sqrt {r^2 + \paren {\frac {\d r} {\d \theta} }^2} \rd \theta$

Proof:
Note that $\CC$ satisfies the conditions of Arc Length for Parametric Equations.
So: 
:$\ds s = \int_a^b \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$
We have: 
{{begin-eqn}}
{{eqn	| l = \frac {\d x} {\d \theta}
	| r = \frac \d {\d \theta} \paren {r \cos \theta}
}}
{{eqn	| r = \frac {\d r} {\d \theta} \cos \theta + r \frac \d {\d \theta} \paren {\cos \theta}
	| c = Product Rule for Derivatives
}}
{{eqn	| r = \frac {\d r} {\d \theta} \cos \theta - r \sin \theta
	| c = Derivative of Cosine Function
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn	| l = \frac {\d y} {\d \theta}
	| r = \frac \d {\d \theta} \paren {r \sin \theta}
}}
{{eqn	| r = \frac {\d r} {\d \theta} \sin \theta + r \frac \d {\d \theta} \paren {\sin \theta}
	| c = Product Rule for Derivatives
}}
{{eqn	| r = \frac {\d r} {\d \theta} \sin \theta + r \cos \theta
	| c = Derivative of Sine Function
}}
{{end-eqn}}
We then have:
{{begin-eqn}}
{{eqn	| l = s
	| r = \int_a^b \sqrt {\paren {\frac {\d r} {\d \theta} \cos \theta - r \sin \theta}^2 + \paren {\frac {\d r} {\d \theta} \sin \theta + r \cos \theta}^2} \rd \theta
}}
{{eqn	| r = \int_a^b \sqrt {\paren {\frac {\d r} {\d \theta} }^2 \cos^2 \theta - 2 r \frac {\d r} {\d \theta} \sin \theta \cos \theta + r^2 \sin^2 \theta + \paren {\frac {\d r} {\d \theta} }^2 \sin^2 \theta + 2 r \frac {\d r} {\d \theta} \sin \theta \cos \theta + r^2 \cos^2 \theta} \rd \theta
	| c = Square of Sum
}}
{{eqn	| r = \int_a^b \sqrt {\paren {\frac {\d r} {\d \theta} }^2 \paren {\sin^2 \theta + \cos^2 \theta} + r^2 \paren {\sin^2 \theta + \cos^2 \theta} } \rd \theta
}}
{{eqn	| r = \int_a^b \sqrt {r^2 + \paren {\frac {\d r} {\d \theta} }^2} \rd \theta
	| c = Sum of Squares of Sine and Cosine
}}
{{end-eqn}}
{{qed}}
Theorem: Number of Petals of Odd Index Rhodonea Curve

Let $n$ be an odd positive integer.
Let $R$ be a '''rhodonea curve''' defined by one of the polar equations:
{{begin-eqn}}
{{eqn | l = r
      | r = a \cos n \theta
}}
{{eqn | l = r
      | r = a \sin n \theta
}}
{{end-eqn}}
Then $R$ has $n$ petals.

Proof:
Let $R$ be defined by $r = a \cos n \theta$.
The tips of each of the petals of $R$ occur when $\cos n \theta = \pm 1$.
This happens whenever $n \theta \in \set {0, \pi}$.
Thus for $0 \le \theta < 2 \pi$, the tips occur at:
:$\theta \in \set {\dfrac {2 k \pi} {2 n}: k \in \set {0, 1, \ldots, 2 n - 1} }$
Let $k < n$.
Then each of $\dfrac {2 k \pi} {2 n}$ is a distinct value between $0$ and $\pi$.
Hence there are at least $n$ petals, one for each $k \in \set {0, 1, \ldots, n - 1}$.
Now consider $\theta \in \hointr \pi {2 \pi}$.
From Cosine of Angle plus Straight Angle, we have that:
:$\cos \theta = -\map \cos {\theta - \pi}$
But as $\theta$ is on the opposite side of the circle to $\theta - \pi$, the points they describe are exactly the same.
Hence the petals for $\theta \ge \pi$ are in the same place as that for $\theta < \pi$.
So for $\pi \le \theta < 2 \pi$ the pattern will be traced out again.
Hence the result.
{{qed}}
{{handwaving|Anyone care to make this a bit more rigorous?}}
Theorem: Equation of Trochoid

Consider a circle $C$ of radius $a$ rolling without slipping along the x-axis of a cartesian plane.
Consider the point $P$ on on the line of a radius of $C$ at a distance $b$ from the center of $C$.
Let $P$ be on the y-axis when the center of $C$ is also on the y-axis.
Consider the trochoid traced out by the point $P$.
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 equations:
:$x = a \theta  - b \sin \theta$
:$y = a - b \cos \theta$

Proof:
Let $C$ have rolled so that the radius to the point $P = \tuple {x, y}$ is at angle $\theta$ to the vertical.
The center of $C$ is at $\tuple {a \theta, a}$.
Then it follows from the definition of sine and cosine that:
:$x = a \theta - b \sin \theta$
:$y = a - b \cos \theta$
whence the result.
{{qed}}
Theorem: Equation of Witch of Agnesi/Parametric Form

The equation of the Witch of Agnesi can be presented in parametric form as:
:$\begin {cases} x = 2 a \cot \theta \\ y = a \paren {1 - \cos 2 \theta} \end {cases}$

Proof:
Let $P = \tuple {x, y}$ and $A = \tuple {d, y}$.
Let $\theta$ be the angle that $ON$ makes with the horizontal.
We have by definition of cotangent:
:$\dfrac {OM} {MN} = \dfrac {2 a} x = \cot \theta$
By Thales' Theorem $\angle OAM$ is a right angle.
Hence $\angle OMA = \theta$ and so:
:$OA = 2 a \cos \theta$
Thus:
:$2 a - y = 2 a \cos^2 \theta$
{{begin-eqn}}
{{eqn | l = 2 a - y
      | r = 2 a \cos^2 \theta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = 2 a \paren {1 - \dfrac {\cos 2 \theta + 1} 2}
      | c = 
}}
{{eqn | r = a \paren {2 - \paren {\cos 2 \theta + 1} }
      | c = 
}}
{{eqn | r = a \paren {1 - \cos 2 \theta}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Area of Loop of Folium of Descartes

Consider the folium of Descartes $F$, given in parametric form as:
:$\begin {cases} x = \dfrac {3 a t} {1 + t^3} \\ y = \dfrac {3 a t^2} {1 + t^3} \end {cases}$
The area $\AA$ of the loop of $F$ is given as:
:$\AA = \dfrac {3 a^2} 2$

Proof:
From Behaviour of Parametric Equations for Folium of Descartes according to Parameter we have that the loop is traversed for $0 \le t < +\infty$.
We convert the parametric equation to polar form:
{{begin-eqn}}
{{eqn | l = r^2
      | r = x^2 + y^2
      | c = 
}}
{{eqn | r = \dfrac {\paren {3 a t}^2} {\paren {1 + t^3}^2} + \dfrac {\paren {3 a t^2}^2} {\paren {1 + t^3}^2}
      | c = 
}}
{{eqn | r = \dfrac {\paren {3 a t}^2 \paren {1 + t^2} } {\paren {1 + t^3}^2}
      | c = 
}}
{{eqn | l = \tan \theta
      | r = \dfrac y x
      | c = 
}}
{{eqn | r = \dfrac {3 a t^2} {1 + t^3} \dfrac {1 + t^3} {3 a t}
      | c = 
}}
{{eqn | r = t
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \theta
      | r = \arctan t
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d \theta} {\d t}
      | r = \dfrac 1 {1 + t^2}
      | c = Derivative of Arctangent Function
}}
{{end-eqn}}
Then we have:
{{begin-eqn}}
{{eqn | l = \AA
      | r = \dfrac 1 2 \int_{t \mathop = 0}^{t \mathop \to \infty} r^2 \rd \theta
      | c = Area between Radii and Curve in Polar Coordinates
}}
{{eqn | r = \dfrac 1 2 \int_{t \mathop = 0}^{t \mathop \to \infty} \dfrac {\paren {3 a t}^2 \paren {1 + t^2} } {\paren {1 + t^3}^2} \dfrac 1 {1 + t^2} \rd t
      | c = Integration by Substitution
}}
{{eqn | r = \dfrac {3 a^2} 2 \int_{t \mathop = 0}^{t \mathop \to \infty} \dfrac {3 t^2 \rd t} {\paren {1 + t^3}^2}
      | c = Integration by Substitution
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = u
      | r = 1 + t^3
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d u} {\d t}
      | r = 3 t^2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = t
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = u
      | r = 1
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = t
      | o = \to
      | r = +\infty
      | c = 
}}
{{eqn | ll= \leadsto
      | l = u
      | o = \to
      | r = +\infty
      | c = 
}}
{{end-eqn}}
which leads us to:
{{begin-eqn}}
{{eqn | l = \AA
      | r = \dfrac {3 a^2} 2 \int_{t \mathop = 0}^{t \mathop \to \infty} \dfrac {3 t^2 \rd t} {\paren {1 + t^3}^2}
      | c = 
}}
{{eqn | r = \dfrac {3 a^2} 2 \int_{u \mathop = 1}^{u \mathop \to \infty} \dfrac {\d u} {u^2} 
      | c = Integration by Substitution
}}
{{eqn | r = \dfrac {3 a^2} 2 \intlimits {-\dfrac 1 u} 1 {+\infty}
      | c = Primitive of Power
}}
{{eqn | r = \dfrac {3 a^2} 2 \paren {-0 - \paren {-1} }
      | c = evaluating limits: $u \to +\infty \implies \dfrac 1 u \to 0$
}}
{{eqn | r = \dfrac {3 a^2} 2
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Asymptote to Folium of Descartes

Consider the folium of Descartes $F$, given in parametric form as:
:$\begin {cases} x = \dfrac {3 a t} {1 + t^3} \\ y = \dfrac {3 a t^2} {1 + t^3} \end {cases}$
The straight line whose equation is given by:
:$x + y + a = 0$
is an asymptote to $F$.

Proof:
First we note that from Behaviour of Parametric Equations for Folium of Descartes according to Parameter:
:when $t = 0$ we have that $x = y = 0$
:when $t \to \pm \infty$ we have that $x \to 0$ and $y \to 0$
:when $t \to -1^+$ we have that $1 + t^3 \to 0+$, and so:
::$x \to -\infty$
::$y \to +\infty$
:when $t \to -1^-$ we have that $1 + t^3 \to 0-$, and so:
::$x \to +\infty$
::$y \to -\infty$
We have that:
{{begin-eqn}}
{{eqn | l = x + y
      | r = \dfrac {3 a t} {1 + t^3} + \dfrac {3 a t^2} {1 + t^3}
      | c = 
}}
{{eqn | r = \dfrac {3 a t \paren {1 + t} } {\paren {1 + t} \paren {1 - t + t^2} }
      | c = Sum of Two Cubes
}}
{{eqn | r = \dfrac {3 a t} {1 - t + t^2}
      | c = simplifying
}}
{{end-eqn}}
So setting $t = -1$:
{{begin-eqn}}
{{eqn | l = x + y
      | r = \dfrac {3 a \times \paren {-1} } {1 - \paren {-1} + \paren {-1}^2}
      | c = 
}}
{{eqn | r = \dfrac {-3 a} {1 + 1 + 1}
      | c = 
}}
{{eqn | r = -a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x + y + a
      | r = 0
      | c = 
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Behaviour of Parametric Equations for Folium of Descartes according to Parameter

Consider the folium of Descartes $F$, given in parametric form as:
:$\begin {cases} x = \dfrac {3 a t} {1 + t^3} \\ y = \dfrac {3 a t^2} {1 + t^3} \end {cases}$
Then:
:$F$ has a discontinuity at $t = -1$.
:For $t < -1$, the section in the $4$th quadrant is generated
:For $-1 < t \le 0$, the section in the $2$nd quadrant is generated 
:For $0 \le t$, the section in the $1$st quadrant is generated.

Proof:
:500px
Theorem: Parametric Equation of Involute of Circle

Let $C$ be a circle of radius $a$ whose center is at the origin of a cartesian plane.
The involute $V$ of $C$ can be described by the parametric equation:
:$\begin {cases} x = a \paren {\cos \theta + \theta \sin \theta} \\ y = a \paren {\sin \theta - \theta \cos \theta} \end {cases}$

Proof:
By definition the involute of $C$ is described by the endpoint of a string unwinding from $C$.
Let that endpoint start at $\tuple {a, 0}$ on the circumference of $C$.
:500px
Let $P = \tuple {x, y}$ be an arbitrary point on $V$.
Let $Q$ be the point at which the cord is tangent to $C$.
Then $PQ$ equals the arc of $C$ from which the cord has unwound.
Thus:
:$PQ = a \theta$
where $\theta$ is the angle of $OQ$ to the $x$-axis.
Thus:
{{begin-eqn}}
{{eqn | l = x
      | r = OQ \cos \theta + PQ \, \map \cos {\theta - \dfrac \pi 2}
      | c = 
}}
{{eqn | r = a \cos \theta + a \theta \paren {\cos \theta \cos \dfrac \pi 2 + \sin \theta \sin \dfrac \pi 2}
      | c = Cosine of Difference
}}
{{eqn | r = a \paren {\cos \theta + \theta \sin \theta}
      | c = various trigonometrical identities
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = y
      | r = OQ \sin \theta + PQ \, \map \sin {\theta - \dfrac \pi 2}
      | c = 
}}
{{eqn | r = a \sin \theta + a \theta \paren {\sin \theta \cos \dfrac \pi 2 - \cos \theta \sin \dfrac \pi 2}
      | c = Sine of Difference
}}
{{eqn | r = a \paren {\sin \theta - \theta \cos \theta}
      | c = various trigonometrical identities
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Equation of Ovals of Cassini/Cartesian Form

The Cartesian equation:
:$\paren {x^2 + y^2 + a^2}^2 - 4 a^2 x^2 = b^4$
describes the '''ovals of Cassini'''.

Proof:
The '''ovals of Cassini''' are the loci of points $M$ in the plane such that:
:$P_1 M \times P_2 M = b^2$
Let $b$ be chosen.
Let $P = \tuple {x, y}$ be an arbitrary point of $M$.
We have:
{{begin-eqn}}
{{eqn | l = P_1 P
      | r = \sqrt {\paren {x - a}^2 + y^2}
      | c = Distance Formula
}}
{{eqn | l = P_2 P
      | r = \sqrt {\paren {x + a}^2 + y^2}
      | c = Distance Formula
}}
{{eqn | ll= \leadsto
      | l = \paren {P_1 P} \paren {P_2 P}
      | r = \sqrt {\paren {x + a}^2 + y^2} \sqrt {\paren {x - a}^2 + y^2}
      | c = {{Defof|Ovals of Cassini}}
}}
{{eqn | r = b^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b^4
      | r = \paren {\paren {x + a}^2 + y^2} \paren {\paren {x - a}^2 + y^2}
      | c = squaring both sides
}}
{{eqn | r = \paren {x^2 + y^2 + a^2 + 2 a x} \paren {x^2 + y^2 + a^2 - 2 a x}
      | c = multiplying out and rearranging
}}
{{eqn | r = \paren {x^2 + y^2 + a^2}^2 - 4 a^2 x^2
      | c = Difference of Two Squares
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Equation of Ovals of Cassini/Polar Form

The polar equation:
:$r^4 + a^4 - 2 a^2 r^2 \cos 2 \theta = b^4$
describes the '''ovals of Cassini'''.

Proof:
The '''ovals of Cassini''' are the loci of points $M$ in the plane such that:
:$P_1 M \times P_2 M = b^2$
Let $b$ be chosen.
Let $P = \tuple {x, y}$ be an arbitrary point of $M$.
Let $d_1 = \size {P_1 P}$ and $d_2 = \size {P_2 P}$.
We have:
{{begin-eqn}}
{{eqn | l = b^2
      | r = d_1 d_2
      | c = {{Defof|Ovals of Cassini}}
}}
{{eqn | r = \sqrt {r^2 + a^2 - 2 a r \cos \theta} \times \sqrt {r^2 + a^2 - 2 a r \, \map \cos {\pi - \theta} }
      | c = Cosine Rule
}}
{{eqn | ll= \leadsto
      | l = \paren {b^2}^2
      | r = \paren {r^2 + a^2 - 2 a r \cos \theta} \paren {r^2 + a^2 + 2 a r \cos \theta}
      | c = Cosine of Supplementary Angle, and squaring throughout
}}
{{eqn | r = \paren {r^2 + a^2}^2 - \paren {2 a r \cos \theta}^2
      | c = Difference of Two Squares
}}
{{eqn | r = \paren {r^2}^2 + 2 a^2 r^2 + \paren {a^2}^2 - 4 a^2 r^2 \cos^2 \theta
      | c = Square of Sum
}}
{{eqn | r = r^4 + a^4 - 2 a^2 r^2 \paren {2 \cos^2 \theta - 1}
      | c = simplifying
}}
{{eqn | r = r^4 + a^4 - 2 a^2 r^2 \cos 2 \theta
      | c = {{Corollary|Double Angle Formula for Cosine|1}}
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Lemniscate of Bernoulli is Special Case of Ovals of Cassini

The lemniscate of Bernoulli is a special case of the ovals of Cassini.

Proof:
The ovals of Cassini can be defined by a Cartesian equation as follows:
{{:Equation of Ovals of Cassini/Cartesian Form}}
The lemniscate of Bernoulli can be defined by a Cartesian equation as follows:
{{:Definition:Lemniscate of Bernoulli/Cartesian Definition}}
Setting $b = a$:
{{begin-eqn}}
{{eqn | l = \paren {x^2 + y^2 + a^2}^2 - 4 a^2 x^2
      | r = a^4
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x^2 + y^2}^2 + 2 \paren {x^2 + y^2} a^2 + a^4 - 4 a^2 x^2
      | r = a^4
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x^2 + y^2}^2
      | r = 4 a^2 x^2 - 2 a^2 x^2 - 2 a^2 y^2
      | c = simplifying and rearranging
}}
{{eqn | r = 2 a^2 \paren {x^2 - y^2}
      | c = further simplification
}}
{{end-eqn}}
It follows that the lemniscate of Bernoulli is an oval of Cassini where $b = a$.
{{qed}}
Theorem: Lemniscate of Bernoulli is Special Case of Ovals of Cassini

The lemniscate of Bernoulli is a special case of the ovals of Cassini.

Proof:
The ovals of Cassini are defined as follows:
{{:Definition:Ovals of Cassini}}
The lemniscate of Bernoulli is defined geometrically as:
{{:Definition:Lemniscate of Bernoulli/Geometric Definition}}
It follows that the lemniscate of Bernoulli is an oval of Cassini where $b = a$.
{{qed}}