Ring of mixed characteristic: Difference between revisions

Revision as of 15:24, 20 January 2013

In geometry, a radiodrome is the path followed by a point which is pursuing another point. The term is derived from the Latin word radius (beam) and the Greek word dromos (running). The classical (and best-known) form of a radiodrome is known as the "dog curve"; this is the path a dog follows when it swims across a stream with a current after food it has spotted on the other side. Because the dog drifts downwards with the current, it will have to change its heading; it will also have to swim further than if it had computed the optimal heading. This case was described by Pierre Bouguer in 1732.

It is also the name of a weekly radio show hosted by internet personality Brad Jones and Josh Hadley.

A radiodrome may alternatively be described as the path a dog follows when chasing a hare, assuming that the hare runs in a straight line at a constant velocity. It is illustrated by the following figure:

Graph of a radiodrome, also known as a dog curve — The path of a dog chasing a hare running along a vertical straight line at a constant speed. The dog runs towards the momentary position of the hare, and will have to change his heading continuously. The speed of the dog is 20% faster than the speed of the hare.

Mathematical analysis

Introduce a coordinate system with origin at the position of the dog at time zero and with y-axis in the direction the hare is running with the constant speed $V_{t}$ . The position of the hare at time zero is $(A_{x}, A_{y})$ and at time $t$ it is

Template:NumBlk

The dog runs with the constant speed $V_{d}$ towards the momentary position of the hare. The differential equation corresponding to the movement of the dog, $(x (t), y (t))$ , is consequently

It is possible to obtain a closed form analytical expression $y = f (x)$ for the motion of the dog

From (Template:EquationNote) and (Template:EquationNote) follows that

Template:NumBlk

Multiplying both sides with $T_{x} - x$ and taking the derivative with respect to $x$ using that

Template:NumBlk

one gets

Template:NumBlk

or

Template:NumBlk

From this relation follows that

Template:NumBlk

where $B$ is the constant of integration that is determined by the initial value of $y^{'}$ at time zero, i.e.

Template:NumBlk

From (Template:EquationNote) and (Template:EquationNote) follows after some computations that

Template:NumBlk

If now $V_{t} \neq V_{d}$ this relation is integrated as

Template:NumBlk

where $C$ is the constant of integration.

If $V_{t} = V_{d}$ one gets instead

Template:NumBlk

If $V_{t} < V_{d}$ one gets from (Template:EquationNote) that

Template:NumBlk

In the case illustrated in the figure above $\frac{V_{t}}{V_{d}} = \frac{1}{1.2}$ and the chase starts with the hare at position $(A_{x}, - 0.6 A_{x})$ what means that $y^{'} (0) = - 0.6$ . From (Template:EquationNote) one therefore gets hat the hare is caught at position $(A_{x}, 1.21688 A_{x})$ and consequently that the hare will run the total distance $(1.21688 + 0.6) A_{x}$ before being caught.

If $V_{t} \geq V_{d}$ one gets from (Template:EquationNote) and (Template:EquationNote) that $lim_{x \to A_{x}} y (x) = \infty$ what means that the hare never will be caught whenever the chase starts.

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+{{Underlinked|date=March 2013}}
+In [[geometry]], a '''radiodrome''' is the [[path]] followed by a point which is pursuing another point. The term is derived from the Latin word ''radius'' (beam) and the Greek word ''dromos'' (running). The classical (and best-known) form of a radiodrome is known as the "dog curve"; this is the path a dog follows when it swims across a stream with a current after food it has spotted on the other side. Because the dog drifts downwards with the current, it will have to change its heading; it will also have to swim further than if it had computed the optimal heading. This case was described by [[Pierre Bouguer]] in 1732.
-voyance par telephone - [http://jplusfn.gaplus.kr/xe/?document_srl=2483747 http://jplusfn.gaplus.kr/xe/?document_srl=2483747]. What does it mean to have psychic gifts? In fact, what exactly is a psychic?<br><br>Have you ever given it much thought? After all, it has become a rather common thing to discuss.<br>Perhaps, you or someone you know may have had tangible experiences of seeing beyond the so-called machinal realm. Those who have not had such experiences may find it rather strange and mysterious.<br><br>But, I wonder if we really understand what it means to be psychic. By definition, it is considered to be a transcendental thing of the mind and not necessarily spiritual.<br>A psychic is thought to be a person who has the ability to perceive things beyond the common senses. Such as, clairaudience, clairvoyance, telepathy, seeing the dead and so on.<br><br>To terrain too much emphasis on such things may have little to no meaning at all in your everyday life. For some, it has a great deal of meaning and feels as if it is a show of spiritual prowess. However, that which is of the mind is still a fantastique thing and not the ultimate reality.<br><br>
+It is also the name of a weekly radio show hosted by internet personality Brad Jones and Josh Hadley.
-What is fondamental is to go beyond all psychic phenomenon and discover what spirituality is. That which is psychic is a thing of the mind, which can be distorted and corrupted. To be spiritual is to ajouter the délicate.<br><br>So, to go beyond psychic powers and all such things, is of obligatoire pouvoir. It is up to you to see that spirituality is not a thing of the mind. Spirituality transcends mind as thought and stillness.<br>The superficial stillness of the mind can be cultivated by various means and techniques. That stillness is formulated by sollicitation. It is not the genuine intervalle that is sacred. It is up to you to discover the sacred chut beyond the space of thought.<br><br>Actually, you may develop certifié psychic abilities along your spiritual journey. Realize, all such things are by-products and not to be taken hors circuit of context. To exercice unnecessary emphasis on psychical things leads to delusions and not spiritual understanding.<br><br>In spirituality, you must transcend the limitations of the mind and all its so-called extrasensory sentiment. It is not an attempt to deny such a thing heurt to go beyond it and discover the ultimate reality.<br><br>So, we are not concerned with what the religions have said or new age people have said over the years. Not even the scientists with their limited views on life and its mysteries. What is premier is that you learn to see life clearly for yourself.<br>Therefore, one who is truly interested in truth does not come embout it through any religion, philosophy, or psychic ability. There is nothing for your conscious self to proportion.<br><br>Instead, through the negation of all your limitations, this unknown truth is then revealed within you. That which awakens within cannot be touched or corrupted by any means whatsoever.<br>To realize this charmante truth, the conscious idea of self must end. All you can do is repose in this eternal nothingness. This is true humility that transcends the mind and all psychic powers. So, it is up to you to awaken to this living mystery, which is beyond all things.
+A radiodrome may alternatively be described as the path a dog follows when chasing a hare, assuming that the hare runs in a straight line at a constant velocity.
+It is illustrated by the following figure:
+[[File:Dog curve.svg|center|frame|The path of a dog chasing a hare running along a vertical straight line at a constant speed. The dog runs towards the momentary position of the hare, and will have to change his heading continuously. The speed of the dog is 20% faster than the speed of the hare.|alt=Graph of a radiodrome, also known as a dog curve]]
+==[[Mathematical analysis]]==
+Introduce a coordinate system with origin at the position of the dog at time
+zero and with y-axis in the direction the hare is running with the constant
+speed <math>V_t</math>. The position of the hare at time zero is <math>(A_x\ ,\ A_y)</math> and at time <math>t</math> it is
+#    {{NumBlk|:|<math>(T_x\ ,\ T_y)\ =\ (A_x\ ,\ A_y+V_t t)</math>|{{EquationRef|1}}}}
+The dog runs with the constant speed <math>V_d</math> towards the momentary position of the hare. The differential equation corresponding to the movement of the dog, <math>(x(t)\ ,\ y(t))</math>, is consequently
+#    {{NumBlk|:|<math> \dot x= V_d\ \frac{T_x-x}{\sqrt{(T_x-x)^2+(T_y-y)^2}}</math>|{{EquationRef|2}}}}
+#    {{NumBlk|:|<math> \dot y= V_d\ \frac{T_y-y}{\sqrt{(T_x-x)^2+(T_y-y)^2}} </math>|{{EquationRef|3}}}}
+It is possible to obtain a closed form analytical expression <math>y=f(x)</math> for the motion of the dog
+From ({{EquationNote|2}}) and ({{EquationNote|3}}) follows that
+#    {{NumBlk|:| <math>y'(x)=\frac{T_y-y}{T_x-x}</math> |{{EquationRef|4}}}}
+Multiplying both sides with <math>T_x-x</math> and taking the derivative with respect to <math>x</math> using that
+#    {{NumBlk|:| <math> \frac{dT_y}{dx}\ =\ \frac{dT_y}{dt}\ \frac{dt}{dx}\ =\ \frac{V_t}{V_d}\ \sqrt{{y'}^2+1} </math> |{{EquationRef|5}}}}
+one gets
+#    {{NumBlk|:| <math> y''=\frac{V_t\ \sqrt{1+{y'}^2}}{V_d(A_x-x)}  </math> |{{EquationRef|6}}}}
+or
+#    {{NumBlk|:| <math> \frac{y''}{\sqrt{1+{y'}^2}}=\frac{V_t}{V_d(A_x-x)} </math> |{{EquationRef|7}}}}
+From this relation follows that
+#    {{NumBlk|:| <math> \sinh^{-1}(y')=B-\frac{V_t}{V_d}\ \ln(A_x-x) </math> |{{EquationRef|8}}}}
+where <math>B</math> is the constant of integration that is determined by the initial value of <math>y'</math> at time zero, i.e.
+#    {{NumBlk|:| <math> B=\frac{V_t}{V_d}\ \ln(A_x)+\ln\left(y'(0)+\sqrt{{y'(0)}^2+1}\right) </math> |{{EquationRef|9}}}}
+From ({{EquationNote|8}}) and ({{EquationNote|9}}) follows after some computations that
+#    {{NumBlk|:| <math> y'= \frac{1}{2}\left(\frac{y'(0)+\sqrt{{y'(0)}^2+1}}{(1-\frac{x}{A_x})^{\frac{V_t}{V_d}}}  - \frac{(1-\frac{x}{A_x})^{\frac{V_t}{V_d}}}{y'(0)+\sqrt{{y'(0)}^2+1}}\right) </math> |{{EquationRef|10}}}}
+If now <math>V_t \neq V_d</math> this relation is integrated as
+#    {{NumBlk|:| <math> y= C - \frac{1}{2}\ A_x\left(
+\frac{(y'(0)+\sqrt{{y'(0)}^2+1})\ (1-\frac{x}{A_x}) ^{1 - \frac{V_t}{V_d}} }{1-\frac{V_t}{V_d}} -
+\frac{ (1-\frac{x}{A_x}) ^{1 + \frac{V_t}{V_d}} }{ (y'(0)+\sqrt{{y'(0)}^2+1})\ (1 + \frac{V_t}{V_d}) }
+\right)</math>|{{EquationRef|11}}}}
+where <math>C</math> is the constant of integration.
+If <math>V_t = V_d</math> one gets instead
+#    {{NumBlk|:| <math> y= C -\frac{1}{2}A_x\ \left(\left(y'(0)+\sqrt{{y'(0)}^2+1}\right)\ \ln(1-\frac{x}{A_x})   -
+\frac{ (1-\frac{x}{A_x})  ^2}{(y'(0)+\sqrt{{y'(0)}^2+1})\ 2}\right) </math> |{{EquationRef|12}}}}
+If <math>V_t < V_d</math> one gets from ({{EquationNote|11}}) that
+#    {{NumBlk|:| <math> \lim_{x \to A_x}y(x) = C = \frac{1}{2}\ A_x\left( \frac{y'(0)+\sqrt{{y'(0)}^2+1} }{1-\frac{V_t}{V_d}} - \frac{1}{ (y'(0)+\sqrt{{y'(0)}^2+1})\ (1 + \frac{V_t}{V_d}) } \right) </math> |{{EquationRef|13}}}}
+In the case illustrated in the figure above <math>\frac{V_t}{V_d} = \frac{1}{1.2}</math> and the chase starts with the hare at position <math>(A_x\ ,\ -0.6\ A_x)</math> what means that <math>y'(0) = -0.6</math>.  From ({{EquationNote|13}}) one therefore gets  hat the hare is caught at position <math>(A_x\ ,\ 1.21688\ A_x)</math> and consequently that the hare will run the total distance <math>(1.21688\ +\ 0.6)\ A_x</math> before being caught.
+If <math>V_t \geq V_d</math> one gets from ({{EquationNote|11}}) and ({{EquationNote|12}}) that <math>\lim_{x \to A_x}y(x) = \infty</math> what means that the hare never will be caught whenever the chase starts.
+[[Category:Curves]]
+[[Category:Differential equations]]
+[[Category:Analytic geometry]]

Ring of mixed characteristic: Difference between revisions

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Mathematical analysis

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