hcju/proofwikiqa huggingface dataset:
Let $H$ be embedded in a cartesian plane with its center at the origin and its cusps positioned on the axes. :400px By symmetry, it is sufficient to evaluate the area shaded yellow and to multiply it by $4$. By Equation of Astroid: :$\begin{cases} x & = a \cos^3 \theta \\ y & = a \sin^3 \theta \end{cases}$ Thus: {{begin-eqn}} {{eqn | l = \AA       | r = 4 \int_0^a y \rd x       | c =  }} {{eqn | r = 4 \int_{x \mathop = 0}^{x \mathop = a} y \frac {\d x} {\d \theta} \rd \theta       | c =  }} {{eqn | r = 4 \int_{x \mathop = 0}^{x \mathop = a} a \sin^3 \theta \, 3 a \cos^2 \theta \sin \theta \paren {-\sin \theta} \rd \theta       | c = differentiating $a \cos^3 \theta$ {{WRT|Differentiation}} $\theta$ }} {{eqn | r = 4 \int_{\theta \mathop = \pi / 2}^{\theta \mathop = 0} a \sin^3 \theta \, 3 a \cos^2 \theta \sin \theta \paren {-\sin \theta} \rd \theta       | c = $x = 0$ when $\theta = \pi / 2$, $x = a$ when $\theta = 0$ }} {{eqn | r = 12 a^2 \int_0^{\pi / 2} \sin^4 \theta \cos^2 \theta \rd \theta       | c = simplifying }} {{end-eqn}} Simplifying the integrand: {{begin-eqn}} {{eqn | l = \sin^4 \theta \cos^2 \theta       | r = \frac {\paren {2 \sin \theta \cos \theta}^2} 4 \frac {2 \sin^2 \theta} 2       | c =  }} {{eqn | r = \frac {\sin^2 2 \theta} 4 \frac {2 \sin^2 \theta} 2       | c = Double Angle Formula for Sine }} {{eqn | r = \frac {\sin^2 2 \theta} 4 \frac {1 - \cos 2 \theta} 2       | c = Square of Sine }} {{eqn | r = \frac {\sin^2 2 \theta - \sin^2 2 \theta \cos 2 \theta} 8       | c =  }} {{eqn | r = \frac {1 - \cos 4 \theta} {16} - \frac {\sin^2 2 \theta \cos 2 \theta} 8       | c = Square of Sine }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \AA       | r = 12 a^2 \int_0^{\pi / 2} \paren {\frac {1 - \cos 4 \theta} {16} - \frac {\sin^2 2 \theta \cos 2 \theta} 8} \rd \theta       | c =  }} {{eqn | r = \frac 3 4 a^2 \int_0^{\pi / 2} \paren {1 - \cos 4 \theta} \rd \theta - \frac 3 2 a^2 \int_0^{\pi / 2} \sin^2 2 \theta \cos 2 \theta \rd \theta       | c = Linear Combination of Integrals }} {{eqn | r = \frac 3 4 a^2 \intlimits {\theta - \frac {\sin 4 \theta} 4} 0 {\pi / 2} - \frac 3 2 a^2 \int_0^{\pi / 2} \sin^2 2 \theta \cos 2 \theta \rd \theta       | c = Primitive of $\cos a x$ }} {{eqn | r = \frac 3 4 a^2 \intlimits {\theta - \frac {\sin 4 \theta} 4} 0 {\pi / 2} - \frac 3 2 a^2 \intlimits {\frac {\sin^3 2 \theta} 6} 0 {\pi / 2}       | c = Primitive of $\sin^n a x \cos a x$ }} {{eqn | r = \frac {3 \pi a^2} 8 - \frac {3 a^2} {16} \sin 2 \pi  - \frac {3 a^2} {12} \sin^3 \pi       | c = evaluating limits of integration }} {{eqn | r = \frac {3 \pi a^2} 8       | c = $\sin 2 \pi = 0$ and $\sin^3 \pi = 0$ }} {{end-eqn}} {{qed}}
Let $H$ be embedded in a cartesian plane with its center at the origin and its cusps positioned on the axes. :400px We have that $\LL$ is $4$ times the length of one arc of the astroid. From Arc Length for Parametric Equations: :$\ds \LL = 4 \int_{\theta \mathop = 0}^{\theta \mathop = \pi/2} \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$ where, from Equation of Astroid: :$\begin{cases} x & = a \cos^3 \theta \\ y & = a \sin^3 \theta \end{cases}$ We have: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d \theta}       | r = -3 a \cos^2 \theta \sin \theta       | c =  }} {{eqn | l = \frac {\d y} {\d \theta}       | r = 3 a \sin^2 \theta \cos \theta       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2}       | r = \sqrt {9 a^2 \paren {\sin^4 \theta \cos^2 \theta + \cos^4 \theta \sin^2 \theta} }       | c =  }} {{eqn | r = 3 a \sqrt {\sin^2 \theta \cos^2 \theta \paren {\sin^2 \theta + \cos^2 \theta} }       | c =  }} {{eqn | r = 3 a \sqrt {\sin^2 \theta \cos^2 \theta}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 3 a \sin \theta \cos \theta       | c =  }} {{eqn | r = \frac {3 a \sin 2 \theta} 2       | c = Double Angle Formula for Sine }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \LL       | r = 4 \frac {3 a} 2 \int_0^{\pi / 2} \sin 2 \theta \rd \theta       | c =  }} {{eqn | r = 6 a \intlimits {\frac {-\cos 2 \theta} 2} 0 {\pi / 2}       | c = Primitive of $\sin a x$ }} {{eqn | r = 6 a \paren {-\frac {\cos \pi} 2 + \frac {\cos 0} 2}       | c = evaluating limits of integration }} {{eqn | r = 6 a \frac {-\paren {-1} + 1} 2       | c =  }} {{eqn | r = 6 a       | c =  }} {{end-eqn}} {{qed}}
By symmetry, it is sufficient to calculate the surface of revolution of $H$ for $0 \le x \le a$. From Area of Surface of Revolution, this surface of revolution is given by: :$\displaystyle \mathcal S = 2 \int_0^{\pi / 2} 2 \pi y \, \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$ From Equation of Astroid: :$\begin{cases} x & = a \cos^3 \theta \\ y & = a \sin^3 \theta \end{cases}$ so: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d \theta}       | r = -3 a \cos^2 \theta \sin \theta       | c =  }} {{eqn | l = \frac {\d y} {\d \theta}       | r = 3 a \sin^2 \theta \cos \theta       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2}       | r = \sqrt {9 a^2 \paren {\sin^4 \theta \cos^2 \theta + \cos^4 \theta \sin^2 \theta} }       | c =  }} {{eqn | r = 3 a \sqrt {\sin^2 \theta \cos^2 \theta \paren {\sin^2 \theta + \cos^2 \theta} }       | c =  }} {{eqn | r = 3 a \sqrt {\sin^2 \theta \cos^2 \theta}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 3 a \sin \theta \cos \theta       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \mathcal S       | r = 2 \int_0^{\pi / 2} 2 \pi a \sin^3 \theta \, 3 a \sin \theta \cos \theta \rd \theta       | c =  }} {{eqn | r = 12 \pi a^2 \int_0^{\pi / 2} \sin^4 \theta \cos \theta \rd \theta       | c =  }} {{eqn | r = 12 \pi a^2 \intlimits {\frac {\sin^5 \theta} 5} 0 {\pi / 2}       | c = Primitive of $\sin^n a x \cos a x$ }} {{eqn | r = \frac {12 \pi a^2} 5 \paren {\sin^5 \theta \frac \pi 2 - \sin^5 \theta 0}       | c = evaluating limits of integration }} {{eqn | r = \frac {12 \pi a^2} 5 \paren {1 - 0}       | c =  }} {{eqn | r = \frac {12 \pi a^2} 5       | c =  }} {{end-eqn}} {{qed}}
Let $C_1$ be a circle of radius $b$ rolling without slipping around the inside of a circle $C_2$ of radius $a$. 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 $H$ be the hypocycloid formed by the locus of $P$. From Number of Cusps of Hypocycloid from Integral Ratio of Circle Radii we have that $H$ will have $2$ cusps {{iff}}: :$a = 2 b$ By Equation of Hypocycloid a hypocycloid can be expressed in parametric form as: :$x = \paren {a - b} \cos \theta + b \map \cos {\paren {\dfrac {a - b} b} \theta}$ :$y = \paren {a - b} \sin \theta - b \map \sin {\paren {\dfrac {a - b} b} \theta}$ Hence: {{begin-eqn}} {{eqn | l = x       | r = \paren {2 b - b} \cos \theta + b \map \cos {\paren {\dfrac {2 b - b} b} \theta}       | c =  }} {{eqn | r = b \cos \theta + b \cos \theta       | c =  }} {{eqn | r = 2 b \cos \theta        | c =  }} {{end-eqn}} Thus the $x$ coordinate of the $2$ cusp hypocycloid has a range $\closedint {-b} b$. Similarly: {{begin-eqn}} {{eqn | l = y       | r = \paren {2 b - b} \sin \theta - b \map \sin {\paren {\dfrac {2 b - b} b} \theta}       | c =  }} {{eqn | r = b \sin \theta - b \sin \theta       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} Thus the $y$ coordinate of the $2$ cusp hypocycloid is fixed at $y = 0$. Thus the $2$ cusp hypocycloid consists of the line segment: :$x \in \closedint {-b} b, y = 0$. which is a diameter of the containing circle. {{qed}}
By definition, an astroid is a hypocycloid with $4$ cusps. :400px 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 rotor $C_1$ of radius $\dfrac 1 4$ the radius of the stator. 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: :$\begin{cases} x & = 4 b \cos^3 \theta = a \cos^3 \theta \\ y & = 4 b \sin^3 \theta = a \sin^3 \theta \end{cases}$ {{qed}}
By definition, an astroid is a hypocycloid with $4$ cusps. :400px 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: {{begin-eqn}} {{eqn | l = x^{2/3} + y^{2/3}       | r = a^{2/3} \paren {\cos^2 \theta + \sin^2 \theta}       | c =  }} {{eqn | r = a^{2/3}       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} {{qed}}
=== Sufficient Condition === Let :$x \preceq y$ By definition of meet: :$x \wedge y = \inf \left\{ {x, y}\right\}$ By definitions of lower bound and reflexivity: :$x$ is lower bound for $\left\{ {x, y}\right\}$ and :$\forall z \in S: z$ is lower bound for $\left\{ {x, y}\right\} \implies z \preceq x$ Thus by definition of infimum: :$x = \inf \left\{ {x, y}\right\} = x \wedge y$ {{qed|lemma}} === Necessary Condition === Let :$x \wedge y = x$ Thus by Meet Precedes Operands: :$x \preceq y$ {{qed}}
By definition, a deltoid is a hypocycloid with $3$ cusps. :400px 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 rotor $C_1$ of radius $\dfrac 1 3$ the radius of the stator. Thus $a = 3 b$ and the equation of $H$ is now given by: :$\begin{cases} x & = 2 b \cos \theta + b \cos 2 \theta \\ y & = 2 b \sin \theta - b \sin 2 \theta \end{cases}$ {{qed}}
:600px Let $C_1$ have rolled so that the line $OC$ through the radii of $C_1$ and $C_2$ is at angle $\theta$ to the $x$-axis. Let $C_1$ have turned through an angle $\phi$ to reach that point. By definition of sine and cosine, $P = \tuple {x, y}$ is defined by: :$x = \paren {a + b} \cos \theta - b \map \cos {\phi + \theta}$ :$y = \paren {a + b} \sin \theta - b \map \sin {\phi + \theta}$ The arc of $C_1$ between $P$ and $B$ is the same as the arc of $C_2$ between $A$ and $B$. Thus by Arc Length of Sector: :$a \theta = b \phi$ Thus: :$\phi + \theta = \paren {\dfrac {a + b} b} \theta$ whence the result. {{qed}}
{{begin-eqn}} {{eqn | l = x       | r = \paren {a - \paren {-b} } \cos \theta + \paren {-b} \map \cos {\paren {\dfrac {a - \paren {-b} } {\paren {-b} } } \theta}       | c = putting $-b$ for $b$ }} {{eqn | r = \paren {a + b} \cos \theta - b \map \cos {-\paren {\dfrac {a + b} b} \theta}       | c =  }} {{eqn | r = \paren {a + b} \cos \theta - b \map \cos {\paren {\dfrac {a + b} b} \theta}       | c = Cosine Function is Even }} {{end-eqn}} {{begin-eqn}} {{eqn | l = y       | r = \paren {a - \paren {-b} } \sin \theta - \paren {-b} \map \sin {\paren {\dfrac {a - \paren {-b} } {\paren {-b} } } \theta}       | c = putting $-b$ for $b$ }} {{eqn | r = \paren {a + b} \sin \theta + b \map \sin {-\paren {\dfrac {a + b} b} \theta}       | c =  }} {{eqn | r = \paren {a + b} \sin \theta - b \map \sin {\paren {\dfrac {a + b} b} \theta}       | c = Sine Function is Odd }} {{end-eqn}} {{qed}}
By definition, a nephroid is an epicycloid with $2$ cusps. :600px By Equation of Epicycloid, the equation of $H$ is given by: :$\begin{cases} x & = \paren {a + b} \cos \theta - b \map \cos {\paren {\dfrac {a + b} b} \theta} \\ y & = \paren {a + b} \sin \theta - b \map \sin {\paren {\dfrac {a + b} b} \theta} \end{cases}$ When $a = 2 b$ the equation of $H$ is now given by: :$\begin{cases} x & = 3 b \cos \theta - b \cos 3 \theta \\ y & = 3 b \sin \theta - b \sin 3 \theta \end{cases}$ {{qed}}
By definition of image of set: :$\map {f^\to} D = \set {x \wedge y: \tuple {x, y} \in D}$ By definition of subset: :$\map {f^\to} D \subseteq \set {x \wedge y: x \in \map {\pr_1^\to} D, y \in \map {\pr_2^\to} D}$ By Up-Complete Product/Lemma 2: :$D_1 := \map {\pr_1^\to} D$ is directed and :$D_2 := \map {\pr_2^\to} D$ is directed. By Meet of Directed Subsets is Directed: :$\set {x \wedge y: x \in D_1, y \in D_2}$ is directed. By definition of up-complete: :$\set {x \wedge y: x \in D_1, y \in D_2}$ admits a supremum. By Meet is Increasing: :$f$ is increasing mapping. By Image of Directed Subset under Increasing Mapping is Directed: :$\map {f^\to} D$ is directed. By definition of up-complete: :$\map {f^\to} D$ admits a supremum. By Supremum of Subset: :$\map \sup {\map {f^\to} D} \preceq \sup \set {x \wedge y: x \in D_1, y \in D_2}$ We will prove that :$\set {x \wedge y: x \in D_1, y \in D_2} \subseteq \paren {\map {f^\to} D}^\preceq$ where :$\paren {\map {f^\to} D}^\preceq$ denotes the lower closure of $\map {f^\to} D$. Let $z \in \set {x \wedge y: x \in D_1, y \in D_2}$. Then :$\exists x \in D_1, y \in D_2: z = x \wedge y$ By definition of image of set: :$\exists \tuple {a, b} \in D: \map {\pr_1} {a, b} = x$ and :$\exists \tuple {c, d} \in D: \map {\pr_2} {c, d} = y$ By definition of first projection and second projection: :$a = x$ and $d = y$ By definition of directed subset: :$\exists \tuple {g, h} \in D: \tuple {x, b} \precsim \tuple {g, h} \land \tuple {c, y} \precsim \tuple {g, h}$ By definition of Cartesian product of ordered sets: :$x \preceq g$ and $y \preceq h$ By Meet Semilattice is Ordered Structure: :$x \wedge y \preceq g \wedge h \in \set {x \wedge y: \tuple {x, y} \in D}$ Thus by definition of lower closure: :$z \in \paren {\map {f^\to} D}^\preceq$ {{qed|lemma}} By Supremum of Lower Closure of Set: :$\paren {\map {f^\to} D}^\preceq$ admits a supremum and :$\map \sup {\map {f^\to} D}^\preceq = \map \sup {\map {f^\to} D}$ By Supremum of Subset: :$\sup \set {x \wedge y: x \in D_1, y \in D_2} \preceq \map \sup {\map {f^\to} D}$ Thus by definition of antisymmetry: :$\map \sup {\map {f^\to} D} = \sup \set {x \wedge y: x \in D_1, y \in D_2}$ {{qed}}
Let $\left({x, y}\right)$, $\left({z, t}\right) \in S \times S$ such that :$\left({x, y}\right) \precsim \left({z, t}\right)$ By definition of Cartesian product of ordered sets: :$x \preceq z$ and $y \preceq t$ By Meet Semilattice is Ordered Structure: :$x \wedge y \preceq z \wedge t$ By definition of $f$: :$f\left({x, y}\right) \preceq f\left({z, t}\right)$ Thus by definition :$f$ is increasing mapping. {{qed}}
By Meet Precedes Operands: :$\paren {\sup D} \wedge x \preceq \sup D$ By assumption: :$\paren {\sup D} \wedge x \preceq \sup \set {\paren {\sup D} \wedge x \wedge d: d \in D}$ By definition of supremum: :$\forall d \in D: d \preceq \sup D$ By Preceding iff Meet equals Less Operand: :$\forall d \in D: d \wedge \sup D = d$ By Meet is Associative and Meet is Commutative: :$\paren {\sup D} \wedge x \preceq \sup \set {d \wedge x: d \in D}$ By Meet Semilattice is Ordered Structure: :$\forall d \in D: d \wedge x \preceq \paren {\sup D} \wedge x$ By definition: :$\paren {\sup D} \wedge x$ is upper bound for $\set {d \wedge x: d \in D}$ By definition of supremum: :$\sup \set {d \wedge x: d \in D} \preceq \paren {\sup D} \wedge x$ if $\set {d \wedge x: d \in D}$ admits a supremum We will prove that: :$\set {d \wedge x: d \in D}$ is directed Let $y, z \in \set {d \wedge x: d \in D}$. Then :$\exists d_1 \in D: y = d_1 \wedge x$ and :$\exists d_2 \in D: z = d_2 \wedge x$ By definition of directed subset: :$\exists d \in D: d_1 \preceq d \land d_2 \preceq d$ By Meet Semilattice is Ordered Structure: :$y \preceq d \wedge x$ and $z \preceq d \wedge x$ Thus :$d \wedge x \in \set {d \wedge x: d \in D}$ Hence by definition: :$\set {d \wedge x: d \in D}$ is directed. {{qed|lemma}} By definition of up-complete: :$\set {d \wedge x: d \in D}$ admits a supremum Thus by definition of antisymmetry: :$\paren {\sup D} \wedge x = \sup \set {d \wedge x: d \in D}$ {{qed}}
By Up-Complete Product: :$\struct {S \times S, \precsim}$ is up-complete. By definition of up-complete: :$D$ admits a supremum. By definition of Cartesian product: :$\exists d_1, d_2 \in S: \sup D = \tuple {d_1, d_2}$ By Up-Complete Product/Lemma 2: :$D_1 := \map {\pr_1^\to} D$ is directed and :$D_2 := \map {\pr_2^\to} D$ is directed. By definition of up-complete: :$D_1$ admits a supremum and :$D_2$ admits a supremum We will prove that :$d_2$ is upper bound for $D_2$ Let $x \in D_2$. By definition of image of set: :$\exists \tuple {a, b} \in D: \map {\pr_2} {a, b} = x$ By definition of second projection: $b = x$ By definition of supremum: :$\tuple {d_1, d_2}$ is upper bound for $D$. By definition of upper bound: :$\tuple {a, x} \precsim \tuple {d_1, d_2}$ Thus by definition of Cartesian product of ordered sets: :$x \preceq d_2$ {{qed|lemma}} Analogically we have that :$d_1$ is upper bound for $D_1$ By definition of supremum: :$\sup D_1 \preceq d_1$ and $\sup D_2 \preceq d_2$ By definition of Cartesian product of ordered sets: :$\tuple {\sup D_1, \sup D_2} \precsim \sup D$ By Up-Complete Product/Lemma 1: :$D_1 \times D_2$ is directed. By definition of up-complete: :$D_1 \times D_2$ admits a supremum. By definition of subset: :$D \subseteq D_1 \times D_2$ By Supremum of Subset: :$\sup D \precsim \map \sup {D_1 \times D_2}$ By Supremum of Cartesian Product: :$\sup D \precsim \tuple {\sup D_1, \sup D_2}$ Thus by definition of antisymmetry: :$\sup D = \tuple {\sup D_1, \sup D_2}$ {{qed}}
By definition, the evolute of $C$ is the locus of the centers of curvature of each point on $C$ {{WLOG}}, take the circle $C$ of radius $a$ whose center is positioned at the origin of a cartesian plane. From Equation of Circle, $C$ has the equation: :$x^2 + y^2 = a^2$ From the definition of curvature in cartesian form: :$k = \dfrac {y''} {\left({1 + y'^2}\right)^{3/2} }$ Here we have: {{begin-eqn}} {{eqn | l = 2 x + 2 y \frac {\mathrm d y} {\mathrm d x}       | r = 0       | c = differentiating {{WRT|Differentiation}} $x$ }} {{eqn | ll= \leadsto       | l = \frac {\mathrm d y} {\mathrm d x}       | r = -\frac x y       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\mathrm d^2 y} {\mathrm d x^2}       | r = \frac {x \frac {\mathrm d y} {\mathrm d x} - y} {y^2}       | c = Quotient Rule for Derivatives }} {{eqn | r = \frac {y - x \left({-\frac x y}\right) } {y^2}       | c = substituting for $\dfrac {\mathrm d y} {\mathrm d x}$ }} {{eqn | r = \frac {y^2 + x^2} {y^3}       | c = simplifying }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = k       | r = \dfrac {\frac {y^2 + x^2} {y^3} } {\left({1 + \left({-\frac x y}\right)^2}\right)^{3/2} }       | c =  }} {{eqn | r = \dfrac {y^2 + x^2} {y^3 \left({1 + \frac {x^2} {y^2} }\right)^{3/2} }       | c =  }} {{eqn | r = \dfrac {y^2 + x^2} {y^3 \left({\frac {y^2 + x^2} {y^2} }\right)^{3/2} }       | c =  }} {{eqn | r = \dfrac {y^2 + x^2} {\left({y^2 + x^2}\right)^{3/2} }       | c =  }} {{eqn | r = \dfrac 1 {\left({y^2 + x^2}\right)^{1/2} }       | c =  }} {{eqn | r = \dfrac 1 a       | c = as $x^2 + y^2 = a^2$ }} {{end-eqn}} Thus the curvature of $C$ is constant. The radius of curvature of $C$ is likewise constant: :$\rho = a$ From Radius at Right Angle to Tangent, the normal to $C$ at all points on $C$ passes through the center of $C$. We have that $a$ is the distance from $C$ to the center of $C$. Thus it follows that the center of curvature of $C$ is the center of $C$ at all points. Hence the result by definition of evolute. {{qed}}
Consider the curvature of a curve $C$ at a point $P$ expressed as a Whewell equation: :$\kappa = \dfrac {\d \psi} {\d s}$ where: :$\psi$ is the turning angle of $C$ :$s$ is the arc length of $C$. The derivative of the tangent of the turning angle $\psi$ at a point $P$ {{WRT|Differentiation}} $\psi$ is also the derivative of the tangent to $C$ at $P$, again {{WRT|Differentiation}} $\psi$. That is: {{begin-eqn}} {{eqn | l = \frac {\d} {\d \psi} \tan \psi       | r = \frac {\d} {\d \psi} \frac {\d y} {\d x}       | c =  }} {{eqn | ll= \leadsto       | l = \sec^2 \psi       | r = \frac {\d y'} {\d \psi}       | c = Derivative of Tangent Function }} {{eqn | ll= \leadsto       | l = 1 + \tan^2 \psi       | r = \frac {\d y'} {\d \psi}       | c = Difference of Squares of Secant and Tangent }} {{eqn | ll= \leadsto       | l = 1 + y'^2       | r = \frac {\d y'} {\d \psi}       | c =  }} {{eqn | ll= \leadsto       | l = \frac 1 {1 + y'^2}       | r = \frac {\d \psi} {\d y'}       | c =  }} {{end-eqn}} {{explain|Clarify the relationship between the tangent and turning angle. The latter is still inadequately defined.}} We also have that: {{begin-eqn}} {{eqn | l = \d s       | r = \sqrt {\d x^2 + \d y^2}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d s} {\d x}       | r = \sqrt {\paren {\frac {\d x} {\d x} }^2 + \paren {\frac {\d y} {\d x} }^2}       | c =  }} {{eqn | r = \sqrt {1 + y'^2}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d s}       | r = \frac 1 {\paren {1 + y'^2}^{1/2} }       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \kappa       | r = \dfrac {\d \psi} {\d s}       | c =  }} {{eqn | r = \dfrac {\d \psi} {\d y'} \dfrac {\d y'} {\d x} \dfrac {\d x} {\d s}       | c = Chain Rule for Derivatives }} {{eqn | r = \frac 1 {1 + y'^2} y'' \frac 1 {\paren {1 + y'^2}^{1/2} }       | c =  }} {{eqn | r = \dfrac {y''} {\paren {1 + y'^2}^{3/2} }       | c =  }} {{end-eqn}} which is the Cartesian form of curvature as required. {{qed}} {{finish|Need to demonstrate the parametric form}} Category:Analytic Geometry Category:Curvature qg409gem0ceaasctp4c1f6l3paok5db
By definition, the radius of curvature $\rho$ is given by: :$\rho = \dfrac 1 {\size \kappa}$ where $\kappa$ is the curvature, given in Cartesian form as: :$\kappa = \dfrac {y''} {\paren {1 + y'^2}^{3/2} }$ As $\paren {1 + y'^2}^{3/2}$ is positive, it follows that: :$\size {\dfrac {y''} {\paren {1 + y'^2}^{3/2} } } = \dfrac {\size {y''} } {\paren {1 + y'^2}^{3/2} }$ Hence the result. {{qed}}
:400px Let $P = \tuple {x, y}$ be a general point on $C$. Let $Q = \tuple {X, Y}$ be the center of curvature of $C$ at $P$. From the above diagram: :$x - X = \pm \rho \sin \psi$ :$Y - y = \pm \rho \cos \psi$ where: :$\rho$ is the radius of curvature of $C$ at $P$ :$\psi$ is the angle between the tangent to $C$ at $P$ and the $x$-axis. Whether the sign is plus or minus depends on whether the curve is convex or concave. By definition of radius of curvature: :$(1): \quad \begin {cases} x - X = \dfrac 1 k \sin \psi \\ Y - y = \dfrac 1 k \cos \psi \end {cases}$ where $k$ is the curvature of $C$ at $P$, given by: :$k = \dfrac {y''} {\paren {1 + y'^2}^{3/2} }$ We have that: :$\sin \psi = \dfrac {\d y} {\d s} = \dfrac {y'} {\sqrt {1 + y'^2} }$ :$\cos \psi = \dfrac {\d x} {\d s} = \dfrac 1 {\sqrt {1 + y'^2} }$ Substituting for $k$ and $\psi$ in $(1)$ gives: {{begin-eqn}} {{eqn | l = x - X       | r = \dfrac {\paren {1 + y'^2}^{3/2} } {y''} \dfrac {y'} {\sqrt {1 + y'^2} }       | c =  }} {{eqn | r = \dfrac {y' \paren {1 + y'^2} } {y''}       | c =  }} {{eqn | ll= \leadsto       | l = X       | r = x - \dfrac {y' \paren {1 + y'^2} } {y''}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = Y - y       | r = \dfrac {\paren {1 + y'^2}^{3/2} } {y''} \dfrac 1 {\sqrt {1 + y'^2} }       | c =  }} {{eqn | r = \dfrac {1 + y'^2} {y''}       | c =  }} {{eqn | ll= \leadsto       | l = Y       | r = y + \dfrac {1 + y'^2} {y''}       | c =  }} {{end-eqn}} {{qed}}