hcju/proofwikiqa huggingface dataset:
By No Bijection from Set to its Power Set: : there exist no bijections $X \to \mathcal P \left({X}\right)$ Then by definition of set equivalence: :$X \not\sim \mathcal P \left({X}\right)$ Hence by definition of cardinality: :$(1): \quad \left\vert{X}\right\vert \ne \left\vert{\mathcal P \left({X}\right)}\right\vert$ By Cardinality of Set of Singletons: :$(2): \quad \left\vert{\left\{{\left\{{x}\right\}: x \in X}\right\}}\right\vert = \left\vert{X}\right\vert$ By definition of subset: :$\forall x \in X: \left\{{x}\right\} \subseteq X$ Then by definition of power set: :$\forall x \in X: \left\{{x}\right\} \in \mathcal P \left({X}\right)$ Hence by definition of subset: :$\left\{{\left\{{x}\right\}: x \in X}\right\} \subseteq \mathcal P \left({X}\right)$ Then by Subset implies Cardinal Inequality and $(2)$: :$\left\vert{X}\right\vert \leq \left\vert{\mathcal P \left({X}\right)}\right\vert$ Thus by $(1)$: :$\left\vert{X}\right\vert < \left\vert{\mathcal P \left({X}\right)}\right\vert$ {{qed}}
:500px Consider a rectangle bounded by the lines: :$y = 0$ :$x = \xi$ :$x = \xi + \delta x$ :$y = \map f x$ Consider the cylinder generated by revolving it about the $x$-axis. By Volume of Cylinder, the volume of this cylinder is: :$V_\xi = \pi \paren {\map f x}^2 \delta x$ {{finish|Needs finishing off, needs a rigorous treatment.}}
By Reals are Isomorphic to Dedekind Cuts there exists bijection: :$f: \R \to \mathscr D$ where: :$\mathscr D$ denotes the set of all Dedekind cuts of $\struct {\Q, \le}$. Dedekind's cuts are subsets of $\Q$. Therefore by definition of power set: :$\mathscr D \subseteq \powerset \Q$ By Subset implies Cardinal Inequality: :$\card {\mathscr D} \le \card {\powerset \Q}$ By Rational Numbers are Countably Infinite: :$\Q$ is countably infinite. Then by definition of countably infinite there exists a bijection: :$g: \Q \to \N$ By definition of set equivalence: :$\Q \sim \N$ Hence by definition of cardinality: :$\card \Q = \card \N$ Then by Cardinality of Power Set is Invariant: :$\card {\powerset \Q} = \card {\powerset \N}$ By definition of set equivalence: :$\R \sim \mathscr D$ Hence by definition of cardinality: :$\card \R = \card {\mathscr D}$ Thus: :$\mathfrak c \le \card {\powerset \N}$ Define a mapping $h: \map {\operatorname {Fin} } \N \times \powerset \N \to \R^+$: :$\forall F \in \map {\operatorname {Fin} } \N, A \in \powerset \N: \map h {F, A} = \displaystyle \sum_{i \mathop \in F} 2^i + \sum_{i \mathop \in A} \paren {\frac 1 2}^i$ where $\map {\operatorname {Fin} } \N$ denotes the set of all finite subsets of $\N$. A pair $\tuple {F, A}$ corresponds to binary denotation of a real number $\map h {F, A}$. It means that $h$ is a surjection. By Surjection iff Cardinal Inequality: :$\card {\map {\operatorname {Fin} } \N \times \powerset \N} \le \card {\R^+}$ By definition of subset: :$\map {\operatorname {Fin} } \N \subseteq \powerset \N$ Then by Subset implies Cardinal Inequality: :$\card {\map {\operatorname {Fin} } \N} \le \card {\powerset \N}$ {{begin-eqn}} {{eqn | l = \card {\map {\operatorname {Fin} } \N \times \powerset \N}       | r = \max \set {\card {\map {\operatorname {Fin} } \N}, \card {\powerset \N} }       | c = Cardinal Product Equal to Maximum }} {{eqn | r = \card {\powerset \N} }} {{end-eqn}} Because $\R^+ \subseteq \R$, we have by Subset implies Cardinal Inequality: :$\card {\R^+} \le \card \R$ Thus: :$\card {\powerset \N} \le \mathfrak c$ Hence the result: :$\mathfrak c = \card {\powerset \N}$ {{qed}}
By definition, the acceleration of a body $M$ is defined as the first derivative of the velocity $\mathbf v$ of $M$ relative to a given point of reference with respect to time: :$\mathbf a = \dfrac {\d \mathbf v} {\d t}$ Also by definition, the velocity of $M$ is defined as the first derivative of the displacement of $M$ from a given point of reference with respect to time: :$\mathbf v = \dfrac {\d \mathbf s} {\d t}$ That is: :$\mathbf a = \map {\dfrac \d {\d t} } {\dfrac {\d \mathbf s} {\d t} }$ Hence the result by definition of the second derivative. {{qed}}
By definition of cardinality: :$X \sim Y$ where $\sim$ denotes the set equivalence. Then by definition of set equivalence: : there exists a bijection $f: X \to Y$ By definition of bijection :$f$ is an injection and a surjection. By Direct Image Mapping of Injection is Injection: :the direct image mapping $\map {f^\to}: \powerset X \to \powerset Y$ is an injection. By Direct Image Mapping of Surjection is Surjection: :$f^\to$ is a surjection. Then by definition of bijection: :$f^\to: \powerset X \to \powerset Y$ is a bijection. Hence by definition of set equivalence: :$\powerset X \sim \powerset Y$ Thus the result by definition of cardinality: :$\card {\powerset X} = \card {\powerset Y}$ {{qed}}
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 $\Q$. Ad. $(1)$: 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 $\map f x$: :$y \in \map f x$ Ad. $(2)$: 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: :$\exists r \in \map f x: z < r$ 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 rf1bi2w5yjvrm65r9a92qd7eo5dskgu
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 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 nq4m6foku7g4yfj2punw324qg09tx4k
:500px 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: :$\dfrac {\d y} {\d x} = \dfrac y x$ Using the technique of Separation of Variables: :$\displaystyle \int \frac {\d y} y = \int \frac {\d x} x$ which by Primitive of Reciprocal gives: :$\ln y = \ln x + \ln c$ or: :$y = c x$ Hence the result. {{qed}}
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: :$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$ 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}}
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 Separation of Variables: :$\displaystyle \int \frac {\d r} r = \int \dfrac {\cos \theta} {\sin \theta} \rd \theta$ 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}}
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 = Separation of Variables }} {{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}}
We use the technique of formation of ordinary differential equation by elimination. Differentiating $(1)$ {{WRT}} $x$ 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 = Separation of Variables }} {{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}}
We use the technique of formation of ordinary differential equation by elimination. Differentiating $(1)$ {{WRT}} $r$ 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 = Separation of Variables }} {{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}}
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 = Separation of Variables }} {{eqn | ll= \leadsto       | l = y^2       | r = -2 x + c       | c =  }} {{end-eqn}} Hence the result. {{qed}}
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}}
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}}
By definition of weight there exists a basis $\mathcal B$ of $T$: :$w \left({T}\right) = \left\vert{\mathcal B}\right\vert$ where $\left\vert{\mathcal B}\right\vert$ denotes the cardinality of $\mathcal B$. By Axiom of Choice define a mapping $f: \left\{{U \in \mathcal B: U \ne \varnothing}\right\} \to S$: :$\forall U \in \mathcal B: U \ne \varnothing \implies f \left({U}\right) \in U$ We will prove that :$\forall U \in \tau: U \ne \varnothing \implies U \cap \operatorname{Im} \left({f}\right) \ne \varnothing$ where $\operatorname{Im} \left({f}\right)$ denotes the image of $f$. Let $U \in \tau$ such that :$U \ne \varnothing$ By definition of empty set: :$\exists x: x \in U$ Then by definition of basis: :$\exists V \in \mathcal B: x \in V \subseteq U$ By definition of image: :$f \left({V}\right) \in \operatorname{Im} \left({f}\right)$ By definition of $f$: :$f \left({V}\right) \in V$ By definition of subset: :$f \left({V}\right) \in U$ Then by definition of intersection: :$f \left({V}\right) \in U \cap \operatorname{Im} \left({f}\right)$ Thus by definition of empty set: :$U \cap \operatorname{Im} \left({f}\right) \ne \varnothing$  Then: :$\forall x \in S: \forall U \in \tau: x \in U \implies U \cap \operatorname{Im} \left({f}\right) \ne \varnothing$ Hence by Condition for Point being in Closure: :$\forall x \in S: x \in \left( {\operatorname{Im} \left({f}\right)}\right)^-$ where $^-$ denotes the topological closure. Then by definition of subset: :$S \subseteq \left( {\operatorname{Im} \left({f}\right)}\right)^- \subseteq S$ Thus by definition of set equality: :$S = \left( {\operatorname{Im} \left({f}\right)}\right)^-$ Thus by definition: :$\operatorname{Im} \left({f}\right)$ is dense By definition of density: :$d \left({T}\right) \leq \left\vert{ \operatorname{Im} \left({f}\right) }\right\vert$ $f$ as $\left\{{U \in \mathcal B: U \ne \varnothing} \right\} \to \operatorname{Im} \left({f}\right)$ by definition is surjection. Therefore by Surjection iff Cardinal Inequality: :$\left\vert{\operatorname{Im} \left({f}\right)}\right\vert \leq \left\vert{\left\{{U \in \mathcal B: U \ne \varnothing} \right\}}\right\vert$ By definition of subset: :$\left\{{U \in \mathcal B: U \ne \varnothing} \right\} \subseteq \mathcal B$ Then by Subset implies Cardinal Inequality: :$\left\vert{\left\{{U \in \mathcal B: U \ne \varnothing} \right\}}\right\vert \leq \left\vert{\mathcal B}\right\vert$ Thus: :$d \left({T}\right) \leq w \left({T}\right)$ {{qed}}
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}}
From Body under 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}}
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}}
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}}
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}}