avewright/proofwiki-math huggingface dataset:
Theorem: Reals are Isomorphic to Dedekind Cuts

Let $\mathscr D$ be the set of all Dedekind cuts of the totally ordered set $\struct {\Q, \le}$.
Define a mapping $f: \R \to \mathscr D$ as:
:$\forall x \in \R: \map f x = \set {y \in \Q: y < x}$
Then $f$ is a bijection.

Proof:
First, we will prove that:
:$\forall x \in \R: \map f x \in \mathscr D$
Let $x \in \R$.
It is to be proved that $\map f x$ is a proper subset of $\Q$ such that:
:$(1): \quad \forall z \in \map f x: \forall y \in \Q: y < z \implies y \in \map f x$
:$(2): \quad \forall z \in \map f x: \exists y \in \map f x: z < y$
We have that:
:$x \notin \map f x$
Therefore by definition $\map f x$ is a proper subset of $\Q$.
$(1): \quad$ Let $z \in \map f x, y \in \Q$ such that:
:$y < z$
By definition of $\map f x$:
:$z < x$
Then:
:$y < x$
Thus by definition of $\map f x$:
:$y \in \map f x$
{{qed|lemma}}
$(2): \quad$ Let $z \in \map f x$.
By definition of $\map f x$:
:$z < x$
By Between two Real Numbers exists Rational Number:
:$\exists r \in \Q: z < r < x$
Then by definition of $\map f x$:
:$r \in \map f x$
Thus:
:$\exists r \in \map f x: z < r$
{{qed|lemma}}
By definition of bijection it suffices to prove that $f$ is an injection and a surjection.
We will show by definition that $f: \R \to \mathscr D$ is an injection.
Let $x_1, x_2 \in \R$ such that
:$\map f {x_1} = \map f {x_2}$
{{AimForCont}} $x_1 \ne x_2$.
{{WLOG}} suppose $x_1 < x_2$.
By Between two Real Numbers exists Rational Number:
:$\exists r \in \Q: x_1 < r < x_2$
Then by definition of $\map f x$:
:$r \notin \map f {x_1}$
and
:$r \in \map f {x_2}$
This contradicts $\map f {x_1} = \map f {x_2}$.
We will prove by definition that $f: \R \to \mathscr D$ is a surjection.
Let $L \in \mathscr D$.
By definition of Dedekind cut:
:$L$ is a proper subset of $\Q$.
By definition of proper subset:
:$\exists r \in \Q: r \notin L$
By definition of Dedekind cut:
:$(3): \quad \forall x \in L: \forall y \in \Q: y < x \implies y \in L$
Then
:$\forall x \in L: r \not < x \land r \ne x$
Hence
:$\forall x \in L: r > x$
Then $L$ is bounded above by definition.
By definition of supremum:
:$\map \sup L \le r$
Hence:
:$\map \sup L \in \R$
By definition of supremum:
:$\map \sup L$ is an upper bound of $L$.
Then by definition of upper bound:
:$\forall x \in L: x < \map \sup L$
We will prove that:
:$\forall x \in \Q: x < \map \sup L \implies x \in L$
Let $x \in \Q$ such that:
:$x < \map \sup L$
{{AimForCont}} $x \notin L$.
By $(3)$:
:$\forall x \in L: r \ge x$
By definition:
:$r$ is an upper bound of $L$.
By definition of supremum:
:$r \ge \map \sup L$
This contradicts $x < \map \sup L$.
Thus:
:$L = \map f {\map \sup L}$
{{qed}}
Category:Real Numbers
Category:Dedekind Cuts
bhsd67hmhv8tincb4nhxxv2r2bg01s6
Theorem: Slope of Orthogonal Curves

Let $C_1$ and $C_2$ be curves in a cartesian plane.
Let $C_1$ and $C_2$ intersect each other at $P$.
Let the slope of $C_1$ and $C_2$ at $P$ be $m_1$ and $m_2$.
Then $C_1$ and $C_2$ are orthogonal {{iff}}:
:$m_1 = -\dfrac 1 {m_2}$

Proof:
Let the slopes of $C_1$ and $C_2$ at $P$ be defined by the vectors $\mathbf v_1$ and $\mathbf v_2$ represented as column matrices:
:$\mathbf v_1 = \begin{bmatrix} x_1 \\ y_1 \end{bmatrix} , \mathbf v_2 = \begin{bmatrix} x_2 \\ y_2 \end{bmatrix}$
By Non-Zero Vectors are Orthogonal iff Perpendicular:
:$\mathbf v_1 \cdot \mathbf v_2 = 0$ {{iff}} $C_1$ is orthogonal to $C_2$
where $\mathbf v_1 \cdot \mathbf v_2$ denotes the dot product of $C_1$ and $C_2$.
Thus:
{{begin-eqn}}
{{eqn | l = \mathbf v_1 \cdot \mathbf v_2
      | r = 0
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = x_1 x_2 + y_1 y_2
      | r = 0
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \frac {y_1} {x_1} + \frac {x_2} {y_2}
      | r = 0
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \frac {x_1} {y_1} 
      | r = -\frac 1 {\paren {\dfrac {y_2} {x_2} } }
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = m_1
      | r = -\frac 1 {m_2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Analytic Geometry
dpqhwj7fxb28utoh8ryh8noad4iz4k7
Theorem: Orthogonal Trajectories/Examples/Concentric Circles

Consider the one-parameter family of curves:
:$(1): \quad x^2 + y^2 = c$
Its family of orthogonal trajectories is given by the equation:
:$y = c x$
:500px

Proof:
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ {{WRT|Differentiation}} $x$ gives:
:$2 x + 2 y \dfrac {\d y} {\d x} = 0$
from which:
:$\dfrac {\d y} {\d x} = -\dfrac x y$
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
:$\dfrac {\d y} {\d x} = \dfrac y x$
By Solution to Separable Differential Equation:
:$\ds \int \frac {\d y} y = \int \frac {\d x} x$
which by Primitive of Reciprocal gives:
:$\ln y = \ln x + \ln c$
or:
:$y = c x$
Hence the result.
{{qed}}
Theorem: Angle of Tangent to Radius in Polar Coordinates

Let $C$ be a curve embedded in a plane defined by polar coordinates.
Let $P$ be the point at $\polar {r, \theta}$.
Then the angle $\psi$ made by the tangent to $C$ at $P$ with the radial coordinate is given by:
:$\tan \psi = r \dfrac {\d \theta} {\d r}$

Proof:
:400px
{{ProofWanted}}
Theorem: Orthogonal Trajectories/Examples/Circles Tangent to Y Axis

Consider the one-parameter family of curves:
:$(1): \quad x^2 + y^2 = 2 c x$
which describes the loci of circles tangent to the $y$-axis at the origin.
Its family of orthogonal trajectories is given by the equation:
:$x^2 + y^2 = 2 c y$
which describes the loci of circles tangent to the $x$-axis at the origin.
:600px

Proof:
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ {{WRT|Differentiation}} $x$ gives:
:$2 x + 2 y \dfrac {\d y} {\d x} = 2 c$
from which:
:$\dfrac {\d y} {\d x} = \dfrac {y^2 - x^2} {2 x y}$
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
:$\dfrac {\d y} {\d x} = \dfrac {2 x y} {x^2 - y^2}$
Let:
:$\map M {x, y} = 2 x y$
:$\map N {x, y} = x^2 - y^2$
Put $t x, t y$ for $x, y$:
{{begin-eqn}}
{{eqn | l = \map M {t x, t y}
      | r = 2 t x t y
      | c = 
}}
{{eqn | r = t^2 \paren {2 x y}
      | c = 
}}
{{eqn | r = t^2 \, \map M {x, y}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map N {t x, t y}
      | r = \paren {t x}^2 - \paren {t y}^2
      | c = 
}}
{{eqn | r = t^2 \map N {x^2 - y^2}
      | c = 
}}
{{eqn | r = t \map N {x, y}
      | c = 
}}
{{end-eqn}}
Thus both $M$ and $N$ are homogeneous functions of degree $2$.
Thus, by definition, $(1)$ is a homogeneous differential equation.
By Solution to Homogeneous Differential Equation, its solution is:
:$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
:$\map f {x, y} = \dfrac {2 x y} {x^2 - y^2}$
Thus:
{{begin-eqn}}
{{eqn | l = \ln x
      | r = \int \frac {\d z} {\dfrac {2 z} {1 - z^2} - z} + C_1
      | c = 
}}
{{eqn | r = \int \frac {1 - z^2} {z \paren {1 + z^2} } \rd z + C_1
      | c = 
}}
{{eqn | r = \int \frac {\d z} {z \paren {1 + z^2} } \rd z - \int \frac z {\paren {1 + z^2} } \rd z + C_1
      | c = 
}}
{{eqn | r = \frac 1 2 \map \ln {\frac {z^2} {z^2 + 1} } - \frac 1 2 \map \ln {z^2 + 1} + C_1
      | c = 
}}
{{eqn | r = \frac 1 2 \map \ln {\frac {z^2} {\paren {z^2 + 1}^2} } + C_1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = C_2 x^2
      | r = \frac {z^2} {\paren {z^2 + 1}^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = C_3 x
      | r = \frac {y / x} {\paren {y / x}^2 + 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x^2 + y^2
      | r = 2 C y
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Orthogonal Trajectories/Examples/Circles Tangent to Y Axis

Consider the one-parameter family of curves:
:$(1): \quad x^2 + y^2 = 2 c x$
which describes the loci of circles tangent to the $y$-axis at the origin.
Its family of orthogonal trajectories is given by the equation:
:$x^2 + y^2 = 2 c y$
which describes the loci of circles tangent to the $x$-axis at the origin.
:600px

Proof:
We use the technique of formation of ordinary differential equation by elimination.
Expressing $(1)$ in polar coordinates, we have:
:$(2): \quad r = 2 c \cos \theta$
Differentiating $(1)$ {{WRT|Differentiation}} $\theta$ gives:
:$(3): \quad \dfrac {\d r} {\d \theta} = -2 c \sin \theta$
Eliminating $c$ from $(2)$ and $(3)$:
:$r \dfrac {\d \theta} {\d r} = -\dfrac {\cos \theta} {\sin \theta}$
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
:$r \dfrac {\d \theta} {\d r} = \dfrac {\sin \theta} {\cos \theta}$
Using the technique of Solution to Separable Differential Equation:
:$\ds \int \frac {\d r} r = \int \dfrac {\cos \theta} {\sin \theta} \rd \theta$
which by Primitive of Reciprocal and various others gives:
:$\ln r = \map \ln {\sin \theta} + \ln 2 c$
or:
:$r = 2 c \sin \theta$
This can be expressed in Cartesian coordinates as:
:$x^2 + y^2 = 2 c y$
Hence the result.
{{qed}}
Theorem: Orthogonal Trajectories/Examples/Rectangular Hyperbolas

Consider the one-parameter family of curves of rectangular hyperbolas:
:$(1): \quad x y = c$
Its family of orthogonal trajectories is given by the equation:
:$x^2 - y^2 = c$
:600px

Proof:
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ {{WRT|Differentiation}} $x$ gives:
:$x \dfrac {\d y} {\d x} + y = 0$
{{begin-eqn}}
{{eqn | l = x \frac {\d y} {\d x} + y
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d y} {\d x}
      | r = -\frac y x
      | c = 
}}
{{end-eqn}}
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
:$\dfrac {\d y} {\d x} = \dfrac x y$
So:
{{begin-eqn}}
{{eqn | l = \frac {\d y} {\d x}
      | r = \frac x y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int x \rd x
      | r = \int y \rd y
      | c = Solution to Separable Differential Equation
}}
{{eqn | ll= \leadsto
      | l = x^2
      | r = y^2 + c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x^2 - y^2
      | r = c
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Orthogonal Trajectories/Examples/Parabolas Tangent to X Axis

Consider the one-parameter family of curves of parabolas which are tangent to the $x$-axis at the origin:
:$(1): \quad y = c x^2$
Its family of orthogonal trajectories is given by the equation:
:$x^2 + 2 y^2 = c$
:600px

Proof:
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ {{WRT|Differentiation}} $x$ gives:
:$x \dfrac {\d y} {\d x} + y = 0$
{{begin-eqn}}
{{eqn | n = 2
      | l = \frac {\d y} {\d x}
      | r = 2 c x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d y} {\d x}
      | r = -\frac {2 y} x
      | c = eliminating $c$ between $(1)$ and $(2)$
}}
{{end-eqn}}
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
:$\dfrac {\d y} {\d x} = -\dfrac x {2 y}$
So:
{{begin-eqn}}
{{eqn | l = \frac {\d y} {\d x}
      | r = -\dfrac x {2 y}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int x \rd x
      | r = -2 \int y \rd y
      | c = Solution to Separable Differential Equation
}}
{{eqn | ll= \leadsto
      | l = x^2
      | r = -2 y^2 + c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x^2 + 2 y^2
      | r = c
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Orthogonal Trajectories/Examples/Cardioids

Consider the one-parameter family of curves of cardioids given in polar form as:
:$(1): \quad r = c \paren {1 + \cos \theta}$
Its family of orthogonal trajectories is given by the equation:
:$r = c \paren {1 - \cos \theta}$
:600px

Proof:
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ {{WRT|Differentiation}} $r$ gives:
{{begin-eqn}}
{{eqn | n = 2
      | l = \frac {\d r} {\d \theta}
      | r = - c \sin \theta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d r} {\d \theta}
      | r = -\frac {r \sin \theta} {1 + \cos \theta}
      | c = eliminating $c$ between $(1)$ and $(2)$
}}
{{eqn | ll= \leadsto
      | l = r \frac {\d \theta} {\d r}
      | r = -\frac {1 + \cos \theta} {\sin \theta}
      | c = 
}}
{{end-eqn}}
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
:$r \dfrac {\d \theta} {\d r} = \dfrac {\sin \theta} {1 + \cos \theta}$
So:
{{begin-eqn}}
{{eqn | l = r \dfrac {\d \theta} {\d r}
      | r = \frac {\sin \theta} {1 + \cos \theta}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d r} r
      | r = \int \frac {1 + \cos \theta} {\sin \theta} \rd \theta
      | c = Solution to Separable Differential Equation
}}
{{eqn | r = \int \paren {\csc \theta + \cot \theta} \rd \theta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \ln r
      | r = \ln \size {\csc \theta - \cot \theta} + \ln \size {\sin \theta} + c
      | c = 
}}
{{eqn | r = \ln \size {\paren {\csc \theta - \cot \theta} \sin \theta} + c
      | c = 
}}
{{eqn | r = \ln \size {1 - \cos \theta} + c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = r
      | r = c \paren {1 - \cos \theta}
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Orthogonal Trajectories/Examples/Exponential Functions

Consider the one-parameter family of curves of graphs of the exponential function:
:$(1): \quad y = c e^x$
Its family of orthogonal trajectories is given by the equation:
:$y^2 = -2 x + c$
:600px

Proof:
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ {{WRT|Differentiation}} $x$ gives:
:$\dfrac {\d y} {\d x} = c e^x$
{{begin-eqn}}
{{eqn | n = 2
      | l = \frac {\d y} {\d x}
      | r = c e^x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d y} {\d x}
      | r = y
      | c = eliminating $c$ between $(1)$ and $(2)$
}}
{{end-eqn}}
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
:$\dfrac {\d y} {\d x} = -\dfrac 1 y$
So:
{{begin-eqn}}
{{eqn | l = \frac {\d y} {\d x}
      | r = -\dfrac 1 y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int y \rd y
      | r = -\int \rd x
      | c = Solution to Separable Differential Equation
}}
{{eqn | ll= \leadsto
      | l = y^2
      | r = -2 x + c
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Orthogonal Trajectories/Examples/Parabolas with Focus at Origin

Consider the one-parameter family of curves of parabolas whose focus is at the origin and whose axis is the $x$-axis:
:$(1): \quad y^2 = 4 c \paren {x + c}$
Its family of orthogonal trajectories is given by the equation:
:$y^2 = 4 c \paren {x + c}$
:600px

Proof:
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $(1)$ {{WRT|Differentiation}} $x$ gives:
{{begin-eqn}}
{{eqn | n = 2
      | l = 2 y \frac {\d y} {\d x}
      | r = 4 c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c
      | r = \frac y 2 \frac {\d y} {\d x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y^2
      | r = 2 y \frac {\d y} {\d x} \paren {x + \frac y 2 \frac {\d y} {\d x} }
      | c = substituting for $c$ into $(1)$
}}
{{eqn | r = 2 x y \frac {\d y} {\d x} + y^2 \paren {\frac {\d y} {\d x} }^2
      | c = 
}}
{{end-eqn}}
Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:
{{begin-eqn}}
{{eqn | l = y^2
      | r = -2 x y \frac {\d x} {\d y} + y^2 \paren {-\frac {\d x} {\d y} }^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y^2 \paren {\frac {\d x} {\d y} }^2
      | r = 2 x y \frac {\d x} {\d y} + y^2
      | c = 
}}
{{end-eqn}}
{{Finish}}
Theorem: Half-Life of Radioactive Substance

Let a radioactive element $S$ decay with a rate constant $k$.
Then its half-life $T$ is given by:
:$T = \dfrac {\ln 2} k$
:400px

Proof:
Let $x_0$ be the quantity of $S$ at time $t = 0$.
At time $t = T$ the quantity of $S$ has been reduced to $x = \dfrac {x_0} 2$.
This gives:
{{begin-eqn}}
{{eqn | l = x_0 e^{-k T}
      | r = \frac {x_0} 2
      | c = First-Order Reaction
}}
{{eqn | ll= \leadsto
      | l = e^{k T}
      | r = 2
      | c =
}}
{{eqn | ll= \leadsto
      | l = k T
      | r = \ln 2
      | c =
}}
{{eqn | ll= \leadsto
      | l = T
      | r = \frac {\ln 2} k
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Density not greater than Weight

Let $T = \struct {S, \tau}$ be a topological space.
Then
:$\map d T \le \map w T$
where
:$\map d T$ denotes the density of $T$,
:$\map w T$ denotes the weight of $T$.

Proof:
By definition of weight there exists a basis $\BB$ of $T$:
:$\map w T = \card \BB$
where $\card \BB$ denotes the cardinality of $\BB$.
By Axiom of Choice define a mapping $f: \paren {\BB \setminus \O} \to S$:
:$\forall U \in \BB: U \ne \O \implies f \sqbrk U \in U$
We will prove that
:$\forall U \in \tau: U \ne \O \implies U \cap \Img f \ne \O$
where $\Img f$ denotes the image of $f$.
Let $U \in \tau$ such that
:$U \ne \O$
By definition of empty set:
:$\exists x: x \in U$
Then by definition of basis:
:$\exists V \in \BB: x \in V \subseteq U$
By definition of image:
:$f \sqbrk V \in \Img f$
By definition of $f$:
:$f \sqbrk V \in V$
By definition of subset:
:$f \sqbrk V \in U$
Then by definition of intersection:
:$f \sqbrk V \in U \cap \Img f$
Thus by definition of empty set:
:$U \cap \Img f \ne \O$ 
Then:
:$\forall x \in S: \forall U \in \tau: x \in U \implies U \cap \Img f \ne \O$
Hence by Condition for Point being in Closure:
:$\forall x \in S: x \in \paren {\Img f}^-$
where $^-$ denotes the topological closure.
Then by definition of subset:
:$S \subseteq \paren {\Img f}^- \subseteq S$
Thus by definition of set equality:
:$S = \paren {\Img f}^-$
Thus by definition:
:$\Img f$ is dense
By definition of density:
:$\map d T \le \card {\Img f}$
$f$ as $\paren {\BB \setminus \O} \to \Img f$ by definition is surjection.
Therefore by Surjection iff Cardinal Inequality:
:$\card {\Img f} \le \card {\BB \setminus \O}$
From Set Difference is Subset:
:$\BB \setminus \O \subseteq \BB$
Then by Subset implies Cardinal Inequality:
:$\card {\BB \setminus \O} \le \card \BB$
Thus:
:$\map d T \le \map w T$
{{qed}}
Theorem: Temperature of Body under Newton's Law of Cooling

Let $B$ be a body in an environment whose ambient temperature is $H_a$.
Let $H$ be the temperature of $B$ at time $t$.
Let $H_0$ be the temperature of $B$ at time $t = 0$.
Then:
:$H = H_a - \paren {H_0 - H_a} e^{-k t}$
where $k$ is some positive constant.

Proof:
By Newton's Law of Cooling:
:The rate at which a hot body loses heat is proportional to the difference in temperature between it and its surroundings.
We have the differential equation:
:$\dfrac {\d H} {\d t} \propto - \paren {H - H_a}$
That is:
:$\dfrac {\d H} {\d t} = - k \paren {H - H_a}$
where $k$ is some positive constant.
This is an instance of the Decay Equation, and so has a solution:
:$H = H_a + \paren {H_0 - H_a} e^{-k t}$
{{qed}}
{{Namedfor|Isaac Newton|cat = Newton}}
Theorem: Speed of Body under Free Fall from Height

Let an object $B$ be released above ground from a point near the Earth's surface and allowed to fall freely.
Let $B$ fall a distance $s$.
Then:
:$v = \sqrt {2 g s}$
where:
:$v$ is the speed of $B$ after having fallen a distance $s$
:$g$ is the Acceleration Due to Gravity at the height through which $B$ falls.
It is supposed that the distance $s$ is small enough that $g$ can be considered constant throughout.

Proof:
From Equations of Motion with Constant Acceleration: Velocity after Distance:
:$\mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u + 2 \mathbf g \cdot \mathbf s$
All dot products are between pairs of parallel vectors.
Thus by Cosine Formula for Dot Product:
:$v^2 = u^2 + 2 g s$
Here the body falls from rest, so:
:$\mathbf u = \mathbf 0$
Thus:
:$v^2 = 2 g s$
and so:
:$v = \sqrt {2 g s}$
{{qed}}
Theorem: Speed of Body under Free Fall from Height

Let an object $B$ be released above ground from a point near the Earth's surface and allowed to fall freely.
Let $B$ fall a distance $s$.
Then:
:$v = \sqrt {2 g s}$
where:
:$v$ is the speed of $B$ after having fallen a distance $s$
:$g$ is the Acceleration Due to Gravity at the height through which $B$ falls.
It is supposed that the distance $s$ is small enough that $g$ can be considered constant throughout.

Proof:
From Acceleration is Second Derivative of Displacement with respect to Time:
:$\mathbf g = \dfrac {\d^2 \mathbf s} {\d t^2}$
Integrating with respect to $t$, and by definition of velocity:
:$\mathbf v = \dfrac {\d \mathbf s} {\d t} = \mathbf g t + \mathbf c_1$
When $t = 0$, we have that $\mathrm c_1$ is the initial velocity $\mathbf v_0$, and so:
:$\mathbf v = \dfrac {\d \mathbf s} {\d t} = \mathbf g t + \mathbf v_0$
Integrating with respect to $t$ again:
:$\mathbf s = \dfrac {\mathbf g t^2} 2 + \mathbf v_0 t + \mathbf c_2$
When $t = 0$, we have that $\mathrm c_2$ is the initial displacement $\mathbf s_0$, and so:
:$\mathbf s = \dfrac {\mathbf g t^2} 2 + \mathbf v_0 t + \mathbf s_0$
We have that $B$ is released at rest starting at $\mathbf s = \mathbf 0$.
Thus $\mathbf v_0 = \mathbf s_0 = \mathbf 0$ and so:
:$\mathbf s = \dfrac {\mathbf g t^2} 2$
:$\mathbf v = \mathbf g t$
By eliminating $t$ and taking the scalar quantities:
:$v = \sqrt {2 g s}$
{{qed}}
Theorem: Speed of Body under Free Fall from Height

Let an object $B$ be released above ground from a point near the Earth's surface and allowed to fall freely.
Let $B$ fall a distance $s$.
Then:
:$v = \sqrt {2 g s}$
where:
:$v$ is the speed of $B$ after having fallen a distance $s$
:$g$ is the Acceleration Due to Gravity at the height through which $B$ falls.
It is supposed that the distance $s$ is small enough that $g$ can be considered constant throughout.

Proof:
From the Principle of Conservation of Energy:
:$K + P = C$
where:
:$K$ is the kinetic energy of $B$
:$P$ is the potential energy of $B$
:$C$ is a constant.
Let the mass of $B$ be $m$.
From Kinetic Energy of Motion:
:$K = \dfrac {m v^2} 2$
where $v$ is the speed of $B$.
From Potential Energy of Position:
:$P = m g s$
where $s$ is the distance fallen by $B$.
Since $B$ falls from rest, its initial kinetic energy is zero.
Having fallen a distance $s$, $B$ has lost potential energy $m g s$.
Therefore:
:$\dfrac {m v^2} 2 = m g s$
from which the result follows by dividing both sides by $m$ and extracting the square root.
{{qed}}
Theorem: Terminal Speed of Body under Fall Retarded Proportional to Speed

Let $B$ be a body falling in a gravitational field.
Let $B$ be falling through a medium which exerts a resisting force $k \mathbf v$ upon $B$ which is proportional to the speed of $B$ relative to the medium.
Then the terminal speed of $B$ is given by:
:$v = \dfrac {g m} k$

Proof:
Let $B$ start from rest.
From Motion of Body Falling through Air, the differential equation governing the motion of $B$ is given by:
:$m \dfrac {\d^2 \mathbf s} {\d t^2} = m \mathbf g - k \dfrac {\d \mathbf s} {\d t}$
Dividing through by $m$ and setting $c = \dfrac k m$ gives:
:$\dfrac {\d^2 \mathbf s} {\d t^2} = \mathbf g - c \dfrac {\d \mathbf s} {\d t}$
By definition of velocity:
:$\dfrac {\d \mathbf v} {\d t} = \mathbf g - c \mathbf v$
and so:
{{begin-eqn}}
{{eqn | l = \int \dfrac {\d \mathbf v} {\mathbf g - c \mathbf v}
      | r = \int \rd t
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\dfrac 1 c \map \ln {\mathbf g - c \mathbf v}
      | r = t + c_1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \mathbf g - c \mathbf v
      | r = \mathbf c_2 e^{-c t}
      | c = 
}}
{{end-eqn}}
When $t = 0$ we have that $\mathbf v = 0$ and so:
:$\mathbf c_2 = \mathbf g$
Hence by taking magnitudes:
:$v = \dfrac g c \paren {1 - e^{-c t} }$
Since $c > 0$ it follows that $v \to \dfrac g c$ as $t \to \infty$.
Thus in the limit:
:$v = \dfrac g c = \dfrac {g m} k$
{{qed}}