hcju/proofwikiqa huggingface dataset:
From Forced Vibration of Cart attached to Wall by Spring, the equation of motion of $C$ is: :$(1): \quad m \dfrac {\d^2 \mathbf x} {\d t^2} + c \dfrac {\d \mathbf x} {\d t} + k \mathbf x = \mathbf F_0 \cos \omega t$ From Position of Underdamped Cart attached to Wall by Spring under Forced Vibration, the general solution to $(1)$ is: :$\mathbf x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t} + \dfrac {\mathbf F_0} {\sqrt {\paren {k^2 - \omega^2 m^2}^2 + \omega^2 c^2} } \map \cos {\omega t - \phi}$ where: :$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$ :$\alpha = \sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} }$ :$\phi = \map \arctan {\dfrac {\omega c} {k - \omega^2 m} }$ We have that $(1)$ is in the form: :$(2): \quad \dfrac {\d^2 y} {\d x^2} + 2 b \dfrac {\d y} {\d x} + a^2 x = K \cos \omega x$ where: :$K \in \R: k > 0$ :$a, b \in \R_{>0}: b < a$ whose general solution from Linear Second Order ODE: $y'' + 2 b y' + a^2 y = K \cos \omega x$ where $b < a$ is: :$(1): \quad y = e^{-b x} \paren {C_1 \cos \alpha x + C_2 \sin \alpha x} + \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$ where: :$\alpha = \sqrt {a^2 - b^2}$ :$\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$ By definition, the steady-state component of the general solution to $(2)$ is: :$\dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$ Thus by definition and comparison, the steady-state component of the general solution to $(1)$ is: :$\mathbf x_s = \dfrac {\mathbf F_0} {\sqrt {\paren {k^2 - \omega^2 m^2}^2 + \omega^2 c^2} } \map \cos {\omega t - \phi}$ where: :$\phi = \map \arctan {\dfrac {\omega c} {k - \omega^2 m} }$ {{qed}}
From Linear Second Order ODE: $y'' + 2 b y' + a^2 y = K \cos \omega x$ where $b < a$ the general solution of $(1)$ is: :$(2): \quad y = e^{-b x} \paren {C_1 \cos \alpha x + C_2 \sin \alpha x} + \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$ where: :$\alpha = \sqrt {a^2 - b^2}$ :$\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$ Consider the steady-state component of $(2)$: :$y_s = \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$ By definition, the amplitude of $y_s$ is: :$A = \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} }$ We have that both $a$ and $b$ are greater than $0$. It is also assumed that $\omega$ is also greater than $0$. Thus: :$4 b^2 \omega^2 > 0$ :$a^2 > 0$ :$\omega^2 > 0$ Then $y_s$ is at a maximum when: :$\map f \omega = 4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2$ is at a minimum. Differentiating $\map f \omega$ {{WRT|Differentiation}} $\omega$: {{begin-eqn}} {{eqn | l = \map {f'} \omega       | r = 8 b^2 \omega + 2 \paren {a^2 - \omega^2} \paren {-2 \omega}       | c =  }} {{eqn | r = 8 b^2 \omega - 4 a^2 \omega + 4 \omega^3       | c =  }} {{eqn | r = 4 \omega \paren {2 b^2 - a^2 + \omega^2}       | c =  }} {{end-eqn}} Setting $\map {f'} \omega = 0$: {{begin-eqn}} {{eqn | l = 4 \omega \paren {2 b^2 - a^2 + \omega^2}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \omega       | r = 0       | c =  }} {{eqn | lo= \lor       | l = \omega^2       | r = a^2 - 2 b^2       | c =  }} {{end-eqn}} As $\omega > 0$ the first of these cannot apply. If $a^2 < 2 b^2$ then $a^2 - 2 b^2 < 0$ and so $\omega^2$ has no real roots. Hence the resonance frequency is given by: :$\omega = \sqrt {a^2 - 2 b^2}$ {{iff}} $a^2 > 2 b^2$. {{qed}}
This follows by mutatis mutandis of the proof of Directed in Join Semilattice. {{qed}}
From Forced Vibration of Cart attached to Wall by Spring, the equation of motion of $C$ is: :$(1): \quad m \dfrac {\d^2 \mathbf x} {\d t^2} + c \dfrac {\d \mathbf x} {\d t} + k \mathbf x = \mathbf F_0 \cos \omega t$ Let: :$a^2 = \dfrac k m$ :$2 b = \dfrac c m$ :$K = \dfrac {\mathbf F_0} m$ This is in the form: :$(2): \quad \dfrac {\d^2 y} {\d x^2} + 2 b \dfrac {\d y} {\d x} + a^2 x = K \cos \omega x$ where: :$K \in \R: k > 0$ :$a, b \in \R_{>0}: b < a$ From Condition for Resonance in Forced Vibration of Underdamped System, $C$ is in resonance when: :$\omega = \sqrt {a^2 - 2 b^2}$ Thus, substituting back for $a$ and $b$: {{begin-eqn}} {{eqn | l = \omega       | r = \sqrt {\dfrac k m - 2 \paren {\dfrac c {2 m} }^2}       | c =  }} {{eqn | r = \sqrt {\dfrac k m - \dfrac {c^2} {2 m^2} }       | c =  }} {{end-eqn}} By definition of frequency of simple harmonic motion: :$\nu_R = \dfrac 1 {2 \pi} \sqrt {\dfrac k m - \dfrac {c^2} {2 m^2} }$ {{qed}}
=== Sufficient Condition === Let us assume that :$H$ is directed. Let $A$ be a non-empty finite subset of $H$. By Directed iff Finite Subsets have Upper Bounds: :$\exists h \in H: \forall a \in A: a \preceq h$ By definition :$z$ is upper bound of $A$ By Existence of Non-Empty Finite Suprema in Join Semilattice: :$\sup A$ exists in $\left({S, \preceq}\right)$ By definition of supremum: :$\sup A \preceq h$ Thus by definition of lower set: :$\sup A \in H$ {{qed|lemma}} === Necessary Condition === Let us assume that :for every non-empty finite subset $A$ of $H$, $\sup A \in H$ Let $x, y \in H$. :$A := \left\{ {x, y}\right\}$ is a non-empty finite subset of $H$ By assumption: :$\sup \left\{ {x, y}\right\} \in H$ By definition of supremum: :$\sup A$ is upper bound for $\left\{ {x, y}\right\}$ Thus :$x \preceq \sup A \land y \preceq \sup A$ Thus by definition :$H$ is directed. {{qed}}
From Natural Frequency of Underdamped System, the natural frequency of $(2)$ is: :$\nu = \dfrac {\sqrt {a^2 - b^2} } {2 \pi}$ From Condition for Resonance in Forced Vibration of Underdamped System, the resonance frequency of $S$ is: :$\omega = \sqrt {a^2 - 2 b^2}$ :$\nu_R = \dfrac {\sqrt {a^2 - 2 b^2} } {2 \pi}$ We have that: {{begin-eqn}} {{eqn | l = \dfrac \nu {\nu_R}       | r = \dfrac {\sqrt {a^2 - b^2} } {\sqrt {a^2 - 2 b^2} }       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\nu^2} { {\nu_R}^2}       | r = \dfrac {a^2 - b^2} {a^2 - 2 b^2}       | c =  }} {{eqn | r = \dfrac {a^2 - 2 b^2 + b^2} {a^2 - 2 b^2}       | c =  }} {{eqn | r = 1 + \dfrac {b^2} {a^2 - 2 b^2}       | c =  }} {{end-eqn}} Thus $\nu_R < \nu$ {{iff}} $a^2 - 2 b^2 > 0$. But from Condition for Resonance in Forced Vibration of Underdamped System, in order for there to be resonance, $a^2 - 2 b^2 > 0$. The two conditions are compatible, and so hence the result. {{qed}}
This follows by mutatis mutandis of the proof of Directed in Join Semilattice with Finite Suprema. {{qed}}
Let the position of $S$ be described in the canonical form: :$(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \cos \left({\alpha t - \theta}\right)$ where: :$\alpha = \sqrt {a^2 - b^2}$. :$\theta = \arctan \left({\dfrac b \alpha}\right)$ From Period of Oscillation of Underdamped System is Regular, the period of oscillation $T$ is given by: :$T = \dfrac {2 \pi} {a^2 - b^2}$ Differentiating {{{WRT}} $t$: {{begin-eqn}} {{eqn | l = x'       | r = -b \dfrac {x_0 \, a} \alpha e^{-b t} \cos \left({\alpha t - \theta}\right) - \dfrac {x_0 \, a} \alpha e^{-b t} \alpha \sin \left({\omega t - \theta}\right)       | c =  }} {{eqn | r = -\dfrac {x_0 \, a} \alpha e^{-b t} \left({b \cos \left({\alpha t - \theta}\right) + \alpha \sin \left({\alpha t - \theta}\right)}\right)       | c =  }} {{end-eqn}} From Derivative at Maximum or Minimum, the local maxima and local minima of $x$ occur at $x' = 0$: {{begin-eqn}} {{eqn | l = x'       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = -\dfrac {x_0 \, a} \alpha e^{-b t} \left({b \cos \left({\alpha t - \theta}\right) + \alpha \sin \left({\alpha t - \theta}\right)}\right)       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = b \cos \left({\alpha t - \theta}\right)       | r = -\alpha \sin \left({\alpha t - \theta}\right)       | c =  }} {{eqn | ll= \leadsto       | l = \tan \left({\alpha t - \theta}\right)       | r = -\frac b \alpha       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\tan \alpha t - \tan \theta} {1 + \tan \alpha t \tan \theta}       | r = -\frac b \alpha       | c = Tangent of Difference }} {{eqn | ll= \leadsto       | l = \frac {\tan \alpha t - \dfrac b \alpha} {1 + \tan \alpha t \dfrac b \alpha}       | r = -\frac b \alpha       | c = Definition of $\theta$ }} {{eqn | ll= \leadsto       | l = \tan \alpha t - \dfrac b \alpha        | r = -\frac b \alpha - \tan \alpha t \dfrac {b^2} {\alpha^2}       | c =  }} {{eqn | ll= \leadsto       | l = \tan \alpha t \left({1 + \dfrac {b^2} {\alpha^2} }\right)       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \tan \alpha t       | r = 0       | c = as $1 + \dfrac {b^2} {\alpha^2} > 0$ }} {{eqn | ll= \leadsto       | l = \alpha t       | r = 0, \pi, 2 \pi, \ldots       | c = Tangent of Zero, Tangent of Angle plus Straight Angle }} {{end-eqn}} It remains to be determined which of these points at which $x' = 0$ are local maxima. This occurs when $x > 0$. From Cosine of Angle plus Full Angle: :$\cos \left({x}\right) = \cos \left({2 \pi + x}\right)$ We have that at $x = x_0$ at $t = 0$. It is given that $x_0 > 0$. So at $t = 0, 2 \pi, 4 \pi, \ldots$ we have that: :$\cos \alpha t > 0$ Similarly, from Cosine of Angle plus Straight Angle: :$\cos \left({x}\right) = -\cos \left({\pi + x}\right)$ So at $t = \pi, 3 \pi, 5 \pi, \ldots$ we have that: :$\cos \alpha t < 0$ Thus we have that: :$\alpha L = 2 \pi$ where $L$ is the value of $t$ between consecutive local maxima of $x$. Thus: :$L = \dfrac {2 \pi} {\alpha} = \dfrac {2 \pi} {a^2 - b^2} = T$ as required. {{qed}}
Let the position of $S$ be described in the canonical form: :$(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \cos \left({\alpha t - \theta}\right)$ where: :$\alpha = \sqrt {a^2 - b^2}$. :$\theta = \arctan \left({\dfrac b \alpha}\right)$ From Period of Oscillation of Underdamped System is Regular, the period of oscillation $T$ is given by: :$T = \dfrac {2 \pi} {a^2 - b^2}$ From Interval between Local Maxima for Underdamped Free Vibration, the local maxima of $x$ occur at: :$t = 0, T, 2 T, \ldots$ Let $x_1$ occur at $n T$. Thus $x_2$ occurs at $\left({n + 1}\right) T$. Differentiating {{{WRT}} $t$: {{begin-eqn}} {{eqn | l = x_1       | r = \dfrac {x_0 \, a} \alpha e^{-b n T} \cos \left({\alpha n T - \theta}\right)       | c =  }} {{eqn | l = x_2       | r = \dfrac {x_0 \, a} \alpha e^{-b \left({n + 1}\right) T} \cos \left({\alpha \left({n + 1}\right) T - \theta}\right)       | c =  }} {{end-eqn}} As $T = \dfrac {2 \pi} {a^2 - b^2} = \dfrac {2 \pi} \alpha$, we have: {{begin-eqn}} {{eqn | l = \cos \left({\alpha \left({n + 1}\right) T - \theta}\right)       | r = \cos \left({\alpha \left({n + 1}\right) \dfrac {2 \pi} \alpha - \theta}\right)       | c =  }} {{eqn | r = \cos \left({\left({n + 1}\right) 2 \pi - \theta}\right)       | c =  }} {{eqn | r = \cos \left({n 2 \pi - \theta}\right)       | c =  }} {{eqn | r = \cos \left({\alpha n \dfrac {2 \pi} \alpha  - \theta}\right)       | c =  }} {{eqn | r = \cos \left({\alpha n T  - \theta}\right)       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \dfrac {x_1} {x_2}       | r = \dfrac {\dfrac {x_0 \, a} \alpha e^{-b n T} \cos \left({\alpha n T - \theta}\right)} {\dfrac {x_0 \, a} \alpha e^{-b \left({n + 1}\right) T} \cos \left({\alpha \left({n + 1}\right) T - \theta}\right)}       | c =  }} {{eqn | r = \dfrac {e^{-b n T} \cos \left({\alpha n T - \theta}\right)} {e^{-b \left({n + 1}\right) T} \cos \left({\alpha n T - \theta}\right)}       | c =  }} {{eqn | r = \dfrac {e^{-b n T} } {e^{-b \left({n + 1}\right) T} }       | c =  }} {{eqn | r = e^{-b n T + b \left({n + 1}\right) T}       | c =  }} {{eqn | r = e^{b T}       | c =  }} {{end-eqn}} as required. {{qed}}
Let the radius vector $\mathbf r$ from the origin to $p$ be expressed as: :$(1): \quad \mathbf r = r \mathbf u_r$ From Velocity Vector in Polar Coordinates, the velocity $\mathbf v$ of $p$ can be expressed in vector form as: :$\mathbf v = r \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \mathbf u_r$ From Acceleration Vector in Polar Coordinates, the the acceleration $\mathbf a$ of $p$ can be expressed as: :$\mathbf a = \left({r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} + 2 \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} }\right) \mathbf u_\theta + \left({\dfrac {\mathrm d^2 r} {\mathrm d t^2} - r \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2}\right) \mathbf u_r$ We have: :$\mathbf F = F_r \mathbf u_r + F_\theta \mathbf u_\theta$ and from Newton's Second Law of Motion: :$\mathbf F = m \mathbf a$ Hence: :$\mathbf F = m \left({r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} + 2 \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} }\right) \mathbf u_\theta + m \left({\dfrac {\mathrm d^2 r} {\mathrm d t^2} - r \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2}\right) \mathbf u_r$ Equating components: {{begin-eqn}} {{eqn | l = F_r \mathbf u_r       | r = m \left({\dfrac {\mathrm d^2 r} {\mathrm d t^2} - r \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2}\right) \mathbf u_r       | c =  }} {{eqn | l = F_\theta \mathbf u_\theta       | r =  m \left({r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} + 2 \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} }\right) \mathbf u_\theta       | c =  }} {{end-eqn}} Hence the result: {{begin-eqn}} {{eqn | l = F_\theta       | r =  m \left({r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} + 2 \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} }\right)       | c =  }} {{eqn | l = F_r       | r = m \left({\dfrac {\mathrm d^2 r} {\mathrm d t^2} - r \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2}\right)       | c =  }} {{end-eqn}} {{qed}}
Let $\mathbf F$ be expressed as: :$\mathbf F = F_r \mathbf u_r + F_\theta \mathbf u_\theta$ where: :$\mathbf u_r$ is the unit vector in the direction of the radial coordinate of $p$ :$\mathbf u_\theta$ is the unit vector in the direction of the angular coordinate of $p$ :$F_r$ and $F_\theta$ are the magnitudes of the components of $\mathbf F$ in the directions of $\mathbf u_r$ and $\mathbf u_\theta$ respectively. From Motion of Particle in Polar Coordinates, the second order ordinary differential equations governing the motion of $m$ under the force $\mathbf F$ are: {{begin-eqn}} {{eqn | l = F_\theta       | r = m \paren {r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t} }       | c =  }} {{eqn | l = F_r       | r = m \paren {\dfrac {\d^2 r} {\d t^2} - r \paren {\dfrac {\d \theta} {\d t} }^2}       | c =  }} {{end-eqn}} However, we are given that $\mathbf F$ is a central force. :600px Thus: {{begin-eqn}} {{eqn | l = F_\theta       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = r^2 \dfrac {\d^2 \theta} {\d t^2} + 2 r \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t}       | r = 0       | c = multiplying through by $r$ }} {{eqn | ll= \leadsto       | l = \dfrac \d {\d t} \paren {r^2 \dfrac {\d \theta} {\d t} }       | r = 0       | c = Product Rule for Derivatives }} {{eqn | ll= \leadsto       | l = r^2 \dfrac {\d \theta} {\d t}       | r = h       | c = Derivative of Constant }} {{end-eqn}} for some constant $h$. That is: :$\dfrac {\d \theta} {\d t} = \dfrac h {r^2}$ Hence the result, by definition of inverse proportion. {{qed}}
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 0$ :$q = 1$ :$\map R x = K$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$(2): \quad y'' + y = 0$ From Linear Second Order ODE: $y'' + y = 0$, this has the general solution: :$y_g = C_1 \sin x + C_2 \cos x$ We have that: :$\map R x = K$ So from the Method of Undetermined Coefficients for Polynomial: :$y_p = A K$ where $A$ is to be determined. Hence: {{begin-eqn}} {{eqn | l = y_p       | r = A K       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = 0       | c = Derivative of Constant }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = 0       | c = Derivative of Constant }} {{end-eqn}} Substituting into $(1)$: {{begin-eqn}} {{eqn | l = A K       | r = K       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = 1 }} {{end-eqn}} Our work is done. So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 \sin x + C_2 \cos x + K$ {{qed}} Category:Examples of Constant Coefficient LSOODEs Category:Examples of Method of Undetermined Coefficients qowaor0rptc3den7cksl9o1zijqpv4k
Let the radius vector $\mathbf r$ from the origin to $p$ be expressed as: :$(1): \quad \mathbf r = r \mathbf u_r$ :600px From Derivatives of Unit Vectors in Polar Coordinates: {{begin-eqn}} {{eqn | n = 2       | l = \dfrac {\mathrm d \mathbf u_r} {\mathrm d \theta}       | r = \mathbf u_\theta       | c =  }} {{eqn | n = 3       | l = \dfrac {\mathrm d \mathbf u_\theta} {\mathrm d \theta}       | r = -\mathbf u_r       | c =  }} {{end-eqn}} The velocity of $p$ is by definition the rate of change in its position: {{begin-eqn}} {{eqn | l = \mathbf v       | r = \dfrac {\mathrm d \mathbf r} {\mathrm d t}       | c =  }} {{eqn | r = r \dfrac {\mathrm d \mathbf u_r} {\mathrm d t} + \mathbf u_r \dfrac {\mathrm d r} {\mathrm d t}       | c = from $(1)$ and Product Rule for Derivatives }} {{eqn | r = r \dfrac {\mathrm d \mathbf u_r} {\mathrm d \theta} \dfrac {\mathrm d \theta} {\mathrm d t} + \mathbf u_r \dfrac {\mathrm d r} {\mathrm d t}       | c = Chain Rule for Derivatives }} {{eqn | r = r \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \mathbf u_r       | c = substituting from $(2)$ and $(3)$ }} {{end-eqn}} {{qed}}
By definition of sine and cosine: {{begin-eqn}} {{eqn | n = 1       | l = \mathbf u_r       | r = \mathbf i \cos \theta + \mathbf j \sin \theta }} {{eqn | n = 2       | l = \mathbf u_\theta       | r = -\mathbf i \sin \theta + \mathbf j \cos \theta }} {{end-eqn}} where $\mathbf i$ and $\mathbf j$ are the unit vectors in the $x$-axis and $y$-axis respectively. :600px Differentiating $(1)$ and $(2)$ {{WRT}} $\theta$ gives: {{begin-eqn}} {{eqn | l = \dfrac {\mathrm d \mathbf u_r} {\mathrm d \theta}       | r = -\mathbf i \sin \theta + \mathbf j \cos \theta       | c =  }} {{eqn | r = \mathbf u_\theta       | c =  }} {{eqn | l = \dfrac {\mathrm d \mathbf u_\theta} {\mathrm d \theta}       | r = -\mathbf i \cos \theta - \mathbf j \sin \theta       | c =  }} {{eqn | r = -\mathbf u_r       | c =  }} {{end-eqn}} {{qed}}
Let the radius vector $\mathbf r$ from the origin to $p$ be expressed as: :$(1): \quad \mathbf r = r \mathbf u_r$ :600px From Derivatives of Unit Vectors in Polar Coordinates: {{begin-eqn}} {{eqn | n = 2       | l = \dfrac {\mathrm d \mathbf u_r} {\mathrm d \theta}       | r = \mathbf u_\theta       | c =  }} {{eqn | n = 3       | l = \dfrac {\mathrm d \mathbf u_\theta} {\mathrm d \theta}       | r = -\mathbf u_r       | c =  }} {{end-eqn}} From Velocity Vector in Polar Coordinates: :$\mathbf v = r \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \mathbf u_r$ where $\mathbf v$ is the velocity of $p$. The acceleration of $p$ is by definition the rate of change in its velocity: {{begin-eqn}} {{eqn | l = \mathbf a       | r = \dfrac {\mathrm d \mathbf v} {\mathrm d t}       | c =  }} {{eqn | r = r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + r \dfrac {\mathrm d \theta} {\mathrm d t} \dfrac {\mathrm d \mathbf u_\theta} {\mathrm d t} + \dfrac {\mathrm d^2 r} {\mathrm d t^2} \mathbf u_r + \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \mathbf u_r} {\mathrm d t}       | c = Product Rule for Derivatives }} {{eqn | r = r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + r \dfrac {\mathrm d \theta} {\mathrm d t} \dfrac {\mathrm d \mathbf u_\theta} {\mathrm d \theta} \dfrac {\mathrm d \theta} {\mathrm d t} + \dfrac {\mathrm d^2 r} {\mathrm d t^2} \mathbf u_r + \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \mathbf u_r} {\mathrm d \theta} \dfrac {\mathrm d \theta} {\mathrm d t}       | c = Chain Rule for Derivatives }} {{eqn | r = r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta - r \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_r \dfrac {\mathrm d \theta} {\mathrm d t} + \dfrac {\mathrm d^2 r} {\mathrm d t^2} \mathbf u_r + \dfrac {\mathrm d r} {\mathrm d t} \mathbf u_\theta \dfrac {\mathrm d \theta} {\mathrm d t}       | c = substituting from $(2)$ and $(3)$ }} {{eqn | r = \left({r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} + 2 \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} }\right) \mathbf u_\theta + \left({\dfrac {\mathrm d^2 r} {\mathrm d t^2} - r \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2}\right) \mathbf u_r       | c = gathering terms }} {{end-eqn}} {{qed}}
Let: :$\mathbf u_r$ be the unit vector in the direction of the radial coordinate of $p$ :$\mathbf u_\theta$ be the unit vector in the direction of the angular coordinate of $p$. By Kinetic Energy of Motion, the kinetic energy of $p$ is: :$K = \dfrac {m v^2} 2$ where $v$ is the magnitude of the velocity of $p$. Thus: {{begin-eqn}} {{eqn | l = K       | r = \dfrac {m \mathbf v \cdot \mathbf v} 2       | c = Dot Product of Vector with Itself }} {{eqn | r = \dfrac 1 2 m \paren {r \dfrac {\d \theta} {\d t} \mathbf u_\theta + \dfrac {\d r} {\d t} \mathbf u_r} \cdot \paren {r \dfrac {\d \theta} {\d t} \mathbf u_\theta + \dfrac {\d r} {\d t} \mathbf u_r}       | c = Velocity Vector in Polar Coordinates }} {{eqn | n = 1       | r = \dfrac 1 2 m \paren {r^2 \paren {\dfrac {\d \theta} {\d t} }^2 + \paren {\dfrac {\d r} {\d t} }^2}       | c = {{Defof|Dot Product}} }} {{end-eqn}} The potential energy $P$ of the system is the negative of the work required to move $p$ to infinity: {{MissingLinks|Definition of Work, the above definition of P.E.}} {{begin-eqn}} {{eqn | l = P       | r = -\int_r^\infty \dfrac {G M m} {r^2} \rd r       | c = Newton's Law of Universal Gravitation }} {{eqn | r = \intlimits {\dfrac {G M m} r} r \infty       | c = Primitive of Power }} {{eqn | r = -\dfrac {G M m} r       | c =  }} {{eqn | n = 2       | r = -\dfrac {k m} r       | c = defining $k = G M$: constant for this system }} {{end-eqn}} By the Principle of Conservation of Energy, the total energy in the system remains constant: $E$, say. So: :$E = \dfrac 1 2 m \paren {r^2 \paren {\dfrac {\d \theta} {\d t} }^2 + \paren {\dfrac {\d r} {\d t} }^2} - \dfrac {k m} r$ {{WLOG}}, let us arrange the polar axis so as to make $r$ a minimum when $\theta = 0$. By Kepler's First Law of Planetary Motion, the position of $p$ in polar coordinates is: :$(3): \quad r = \dfrac {h^2 / k} {1 + e \cos \theta}$ At the instant when $\theta = 0$, we therefore have: :$r = \dfrac {h^2 / k} {1 + e}$ At this point, $r$ is at a local minimum. Hence: :$\dfrac {m r^2} 2 \dfrac {h^2} {r^4} - \dfrac {k m} r = E$ Eliminating $r$ from these gives: :$e = \sqrt {1 + E \paren {\dfrac {2 h^2} {m k^2} } }$ Thus equation $(3)$ for the orbit of $p$ can be written as: :$r = \dfrac {h^2 / k} {1 + \sqrt {1 + E \paren {2 h^2 / m k^2} \cos \theta} }$ Thus from Equation of Conic Section in Polar Form, it can be seen that the orbit is: :an ellipse when $E < 0$ :a parabola when $E = 0$ :a hyperbola when $E > 0$. {{qed}}