[{"@context":"http:\/\/schema.org\/","@type":"BlogPosting","@id":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/dog-curve-wikipedia\/#BlogPosting","mainEntityOfPage":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/dog-curve-wikipedia\/","headline":"Dog curve – Wikipedia","name":"Dog curve – Wikipedia","description":"before-content-x4 Curve followed by a dog pursuing his master who runs in a straight line. The dog curve is the","datePublished":"2021-11-09","dateModified":"2021-11-09","author":{"@type":"Person","@id":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/author\/lordneo\/#Person","name":"lordneo","url":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/author\/lordneo\/","image":{"@type":"ImageObject","@id":"https:\/\/secure.gravatar.com\/avatar\/44a4cee54c4c053e967fe3e7d054edd4?s=96&d=mm&r=g","url":"https:\/\/secure.gravatar.com\/avatar\/44a4cee54c4c053e967fe3e7d054edd4?s=96&d=mm&r=g","height":96,"width":96}},"publisher":{"@type":"Organization","name":"Enzyklop\u00e4die","logo":{"@type":"ImageObject","@id":"https:\/\/wiki.edu.vn\/wiki4\/wp-content\/uploads\/2023\/08\/download.jpg","url":"https:\/\/wiki.edu.vn\/wiki4\/wp-content\/uploads\/2023\/08\/download.jpg","width":600,"height":60}},"image":{"@type":"ImageObject","@id":"https:\/\/upload.wikimedia.org\/wikipedia\/commons\/thumb\/f\/fd\/Radiodrome-simple-y-bw.png\/220px-Radiodrome-simple-y-bw.png","url":"https:\/\/upload.wikimedia.org\/wikipedia\/commons\/thumb\/f\/fd\/Radiodrome-simple-y-bw.png\/220px-Radiodrome-simple-y-bw.png","height":"327","width":"220"},"url":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/dog-curve-wikipedia\/","wordCount":15270,"articleBody":"     (adsbygoogle = window.adsbygoogle || []).push({});before-content-x4 Curve followed by a dog pursuing his master who runs in a straight line. The dog curve is the curve described by a dog seeking to join his master by constantly orienting his trajectory in the direction of the latter. We assume their constant speeds.        (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4The model actually applies to any situation of prosecution and interception of a target and we also call this curve “pursuit curve” or “interception curve” [ first ] . It has applications in guide systems.          (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4 First page of Pierre Bouguer’s article, Memoirs of the Royal Science Academy, 1732 This classic plane curve was studied from the XVIII It is century. Paul J. Nahin indicates that the often repeated anecdote according to which Leonardo da Vinci has studied such curves in his notebooks is apparently baseless [ 2 ] . The curve has been studied in particular by Pierre Bouguer [ 3 ] Who associates it with the problem of pursuing one ship by another. Table of Contents       (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4Case where the prosecuted trajectory is rectilinear [ modifier | Modifier and code ] Problem modeling [ modifier | Modifier and code ] Equation [ modifier | Modifier and code ] Mathematical treatment of the equation [ modifier | Modifier and code ] Starting problem solution [ modifier | Modifier and code ] The prosecution curve [ modifier | Modifier and code ] Case where the prosecuted trajectory is circular [ modifier | Modifier and code ] Problem statement [ modifier | Modifier and code ] Elements for solving the problem [ modifier | Modifier and code ] Case where the prosecuted trajectory is rectilinear [ modifier | Modifier and code ] When the prosecuted trajectory, also called line or “leakage curve”, is a right, pursuing it catches it if and only if its speed is higher. In the event of equal speed, the leak line is an asymptote for the prosecution curve. The problem can be modeled by a differential equation. Problem modeling [ modifier | Modifier and code ] It is a question of establishing the equation of the trajectory covered by a dog when his master, distant, calls it while moving on a right perpendicular to the right segment separating them at the moment. We will agree that at this moment, the master is at a distance a {displaystyle a} of the dog ; It moves uniformly at speed in {displaystyle v} . The dog runs at a speed k in {displaystyle kv} And so as to be, at any time, directed towards his master. We can therefore choose an orthonormal benchmark, such as the dog is at point o of coordinates (0.0) and the master at point A of coordinates ( a , 0) At the start, the right described by the master being that of equation x = a {displaystyle x=a} . We designate by ( x , and ) {displaystyle (x,y)} Dog coordinates after time t {displaystyle t} . Equation [ modifier | Modifier and code ]  The scheme describes the situation when the dog has arrived at point C following the curve (in black) from the start. At the end of the time t {displaystyle t} , the master is in M {displaystyle M} , such as A M = in t {displaystyle AM=vt} . If we note l {displaystyle l} The trajectory arc O C {displaystyle OC} traveled by the dog at speed k in {displaystyle kv} , then the slope to the point C {displaystyle C} is worth: dydx= dissolve \u2061 ( a ) = BMBC{displaystyle {frac {{mathrm {d} }y}{{mathrm {d} }x}}=tan(alpha )={frac {BM}{BC}},} either : dydx= vt\u2212ya\u2212x{displaystyle {frac {{mathrm {d} }y}{{mathrm {d} }x}}={frac {vt-y}{a-x}},} On the other hand, in time t {displaystyle t} , the dog has traveled a length l = k in t {displaystyle l=kvt} , Which give : ( a – x ) dydx+ and = lk{displaystyle (a-x){frac {{mathrm {d} }y}{{mathrm {d} }x}}+y={frac {l}{k}},} with x \u2a7d a . {displaystyle xleqslant a.}  By deriving in relation to x {displaystyle x} , we obtain : [(a\u2212x)dydx]\u2032 = – dydx+ ( a – x ) d2ydx2{displaystyle left[(a-x){frac {{mathrm {d} }y}{{mathrm {d} }x}}right]’=-{frac {{mathrm {d} }y}{{mathrm {d} }x}}+(a-x){frac {{mathrm {d} ^{2}}y}{{mathrm {d} }x^{2}}},} Either : – dydx+ ( a – x ) d2ydx2+ dydx= 1kdldx{displaystyle -{frac {{mathrm {d} }y}{{mathrm {d} }x}}+(a-x){frac {{mathrm {d} ^{2}}y}{{mathrm {d} }x^{2}}}+{frac {{mathrm {d} }y}{{mathrm {d} }x}}={frac {1}{k}}{frac {{mathrm {d} }l}{{mathrm {d} }x}},} or ( a – x ) d2ydx2= first k dldx{displaystyle (a-x){frac {{mathrm {d} ^{2}}y}{{mathrm {d} }x^{2}}}={frac {1}{k}}{frac {{mathrm {d} }l}{{mathrm {d} }x}},}  Using the fact that d l 2 = d x 2 + d and 2 {displaystyle {mathrm {d} }l^{2}={mathrm {d} }x^{2}+{mathrm {d} }y^{2}} , and by posing and \u2032 = dydx{displaystyle y’={frac {{mathrm {d} }y}{{mathrm {d} }x}}} , we finally obtain the following equation [ 4 ] :  ( a – x ) dy\u2032dx= first k first + y\u20322{displaystyle (a-x){frac {{mathrm {d} }y’}{{mathrm {d} }x}}={frac {1}{k}}{sqrt {1+y’^{2}}},}  Mathematical treatment of the equation [ modifier | Modifier and code ] The theory of differential equations can then be implemented to obtain the trajectory. Calculation details From the equation:  ( a – x ) dy\u2032dx= 1k1+y\u20322{displaystyle (a-x){frac {{mathrm {d} }y’}{{mathrm {d} }x}}={frac {1}{k}}{sqrt {1+y’^{2}}},}  We can separate the variables, hence the differential equation of the second order: dy\u20321+y\u20322=1kdxa\u2212x{displaystyle {frac {{mathrm {d} }y’}{sqrt {1+y’^{2}}}}={frac {1}{k}}{frac {{mathrm {d} }x}{a-x}},} By integrating we obtain: ln\u2061(y\u2032+1+y\u20322)=\u22121kln\u2061(a\u2212x)+C{displaystyle ln left(y’+{sqrt {1+y’^{2}}}right)=-{frac {1}{k}}ln(a-x)+C,} Just now t = 0 {displaystyle t=0} , the dog is at point o and its trajectory is tangent to the right of the abscissa, therefore y\u2032= 0 {displaystyle y’=0} And x = 0 {displaystyle x=0} , which gives for the constant: C=1kln\u2061(a){displaystyle C={frac {1}{k}}ln(a),} He comes then: ln\u2061(y\u2032+1+y\u20322)=\u22121kln\u2061(a\u2212x)+1kln\u2061(a)=ln\u2061(aa\u2212xk){displaystyle ln left(y’+{sqrt {1+y’^{2}}}right)=-{frac {1}{k}}ln(a-x)+{frac {1}{k}}ln(a)=ln left({sqrt[{k}]{frac {a}{a-x}}}right),} or : y\u2032+1+y\u20322=aa\u2212xk.{displaystyle y’+{sqrt {1+y’^{2}}}={sqrt[{k}]{frac {a}{a-x}}}.} From the equation to the logarithms above, we also deduce the equality of the reverse, and therefore: y\u2032\u22121+y\u20322=\u2212a\u2212xak{displaystyle y’-{sqrt {1+y’^{2}}}=-{sqrt[{k}]{frac {a-x}{a}}},} Then, by making the sum of the last two equations it comes: y\u2032\u22121+y\u20322+y\u2032+1+y\u20322=\u2212a\u2212xak+aa\u2212xk{displaystyle y’-{sqrt {1+y’^{2}}}+y’+{sqrt {1+y’^{2}}}=-{sqrt[{k}]{frac {a-x}{a}}}+{sqrt[{k}]{frac {a}{a-x}}},} The equation giving y\u2032{displaystyle y’} is then : y\u2032=12(\u2212a\u2212xak+aa\u2212xk){displaystyle y’={frac {1}{2}}(-{sqrt[{k}]{frac {a-x}{a}}}+{sqrt[{k}]{frac {a}{a-x}}}),} We can then integrate: y=12(\u222b\u2212a\u2212xakdx+\u222baa\u2212xkdx).{displaystyle y={frac {1}{2}}left(int {-{sqrt[{k}]{frac {a-x}{a}}}}{mathrm {d} }x+int {sqrt[{k}]{frac {a}{a-x}}}{mathrm {d} }xright).} And k {displaystyle k} is different from 1, we obtain, using the fact that a primitive of xN{displaystyle x^{N}} East xN+1N+1{Displaystyle {frac {x^{n+1}} {n+1}}} , y=k(a\u2212x)2(1k+1(a\u2212xa)1k+11\u2212k(a\u2212xa)\u22121k)+D{displaystyle y={frac {k(a-x)}{2}}left({frac {1}{k+1}}left({frac {a-x}{a}}right)^{frac {1}{k}}+{frac {1}{1-k}}left({frac {a-x}{a}}right)^{-{frac {1}{k}}}right)+D,} For x = 0; y = 0 we have D=kak2\u22121{Displaystyle d = {frac {ka} {k^{2} -1}},} , The equation of the dog’s trajectory is therefore, for k{displaystyle k} different from 1: and = k(a\u2212x)2( 1k+1( a\u2212xa)1k+ 11\u2212k( a\u2212xa)\u22121k) + kak2\u22121{Displaystyle y = {frac {k (A-X)} {2}} ({frac {1} {k+1}} ({Frac {A-X} {A}})^{Frac {1}} {k}}++ {Frac {1} {1-K}} ({Frac {A-X} {A}})^{-{Frac {1} {k}}})+{frac {ka} {k^{2} -11 }},} .  And k = first {displaystyle k=1} , the integration of y\u2032= 12( – a\u2212xa+ aa\u2212x) {Displaystyle y ‘= {frac {1} {2}} (-{Frac {A-X} {A}}+{Frac {A} {A-X}})} women and = (a\u2212x)24a– a2ln \u2061 ( a – x ) + D , {displaystyle y={frac {(a-x)^{2}}{4a}}-{frac {a}{2}}ln(a-x)+D,}  And we determine as before the constant D based on the initial value x=0,y=0{displaystyle x=0,y=0} ,  providing : D=a2ln\u2061a\u2212a4{displaystyle D={frac {a}{2}}ln a-{frac {a}{4}}} Finally: y=(a\u2212x)24a\u2212a2ln\u2061(a\u2212x)+a2ln\u2061a\u2212a4,{displaystyle y={frac {(a-x)^{2}}{4a}}-{frac {a}{2}}ln(a-x)+{frac {a}{2}}ln a-{frac {a}{4}},} or : y=x2\u22122ax4a+a2ln\u2061aa\u2212x.{displaystyle y={frac {x^{2}-2ax}{4a}}+{frac {a}{2}}ln {frac {a}{a-x}}.} We finally obtain the equation of the dog’s trajectory, if K is different from 1:  and = k(a\u2212x)2 ( 1k+1(a\u2212xa)1k+ 11\u2212k(a\u2212xa)\u22121k) + kak2\u22121{displaystyle y={frac {k(a-x)}{2}}left({frac {1}{k+1}}left({frac {a-x}{a}}right)^{frac {1}{k}}+{frac {1}{1-k}}left({frac {a-x}{a}}right)^{-{frac {1}{k}}}right)+{frac {ka}{k^{2}-1}},} .  et if k = 1,  and = x2\u22122ax4a+ a 2 ln \u2061 a a\u2212x{displaystyle y={frac {x^{2}-2ax}{4a}}+{frac {a}{2}}ln {frac {a}{a-x}}} .  Starting problem solution [ modifier | Modifier and code ] The dog catches up with his master when x = a , so for and = kak2\u22121{Displaystyle y = {frac {ka} {k^{2} -1}},} , provided that this value of and {displaystyle y} be positive, so if and only if 1″>(That is to say if the dog goes faster than the master). He then catches up in time t = ka(k2\u22121)v. {DisplayStyle t = {frac {ka} {(k^{2} -1) v}}.}  When K = 1, that is to say if the master and the dog have the same speed, the dog does not catch up with the master, the vertical leakage right is asymptote to the trajectory of the dog, which is a leaflet. The prosecution curve [ modifier | Modifier and code ] And k {displaystyle k} is not a rational number, or if k = first {displaystyle k=1} , the prosecution curve is a transcendent curve. When k {displaystyle k} is a different rational number of 1, the prosecution curve is unicursal [ 5 ] . A configuration is given for example by posing:  x = a – t m , and = m 2 t m ( 1m+ntnak+ 1n\u2212maktn) + kak2\u22121, {displaystyle x=a-t^{m},qquad y={frac {m}{2}}t^{m}left({frac {1}{m+n}}{frac {t^{n}}{sqrt[{k}]{a}}}+{frac {1}{n-m}}{frac {sqrt[{k}]{a}}{t^{n}}}right)+{frac {ka}{k^{2}-1}},}  Or k = m n {Displaystyle k = {frac {m} {n}}}} , m {displaystyle m} And n {displaystyle n} Being two first whole between them. Case where the prosecuted trajectory is circular [ modifier | Modifier and code ]  Pursuit curve with circular leakage trajectory. The master moves from “MO” to “M ‘” and the dog following the curve drawn in red and so as to always be oriented towards his master Problem statement [ modifier | Modifier and code ] It is a question of establishing the equation of the prosecution trajectory that a dog travels when his master, distant, calls it while moving on a circular trajectory. It will be appropriate that the master moves uniformly at the speed “V” and in the trigonometric direction. The dog runs at a “KV” speed and so as to be directed to his master. At the end of the time “t” we agree that the master has moved from “Mo” to “M ‘”, that is to say from an angle “\u03a9 = VT\/R” Elements for solving the problem [ modifier | Modifier and code ] A method consists of a graphical resolution of the problem. This article is written from the document of the “College of Redwoods” [ 6 ] . By reasoning on the schematic representation opposite we deduce: OC(\u03c9)\u2032\u2192+ C\u2032M(\u03c9)\u2032\u2192= OM(\u03c9)\u2032\u2192{displaystyle {overrightarrow {OC’_{(omega )}}}+{overrightarrow {C’M’_{(omega )}}}={overrightarrow {OM’_{(omega )}}},} With, by reasoning in the complex plane: O M (\u03c9)\u2032 = x m( oh ) + i and m( oh ) = r cos \u2061 oh + i r sin \u2061 oh {Displayyle Om ‘_ {(omega)} = x_ {m} (omega) + mathrm {i} y_ {m} (omega) = rcos omega + mathrm {i} rsin omega,} O C (\u03c9)\u2032 = x c( oh ) + i and c( oh ) {displaystyle OC’_{(omega )}=x_{c}(omega )+mathrm {i} y_{c}(omega ),} And so : C \u2032 M (\u03c9)\u2032 = OM(\u03c9)\u2032\u2192– OC(\u03c9)\u2032\u2192= r cos \u2061 oh – x c( oh ) + i r sin \u2061 oh – i and c( oh ) {DisplayStyle C’M ‘_ {(omega)} = {OverrightTarrow {OM’ _}}} – {OverrightTarrow {OC ‘_}}} = RCOS OMEGA -X_ {C} (Omega ) + mathrm {i} rsin omega -mathrm {i} y_ {c} (Omega),} After development and considering that the master moves from the Mo point on a unit radius circle then: x m= cos \u2061 oh {displaystyle x_{m}=cos omega ,} and m= sin \u2061 oh {displaystyle y_ {m} = without omega,} The following two relationships are obtained: dxcd\u03c9= k cos\u2061\u03c9\u2212xc(cos\u2061\u03c9\u2212xc)2+(sin\u2061\u03c9\u2212yc)2{Displaystyle {FRAC {{Mathrm {d}} x_ {c}} {{Mathrm {d}} omega}} = K {Cos Omega -x_ {C}} {Sqrt {(COS OMEGA -X_ {C} )^{2}+(singa -y_ {c})^{2}}}},} dycd\u03c9= k sin\u2061\u03c9\u2212yc(cos\u2061\u03c9\u2212xc)2+(sin\u2061\u03c9\u2212yc)2{DisplayStyle {frac {{mathrm {d}} {d}} {}} {d} omega}} = k {frac {sin omega -y_ {c}} {(cos omega -x_ {c} ) ^ {2} + (sin omega -y_ {c}) ^ {2}}}},} The Count of Lautr\u00e9amont refers to the dog’s curve in Les Chants de Maldoror : “The Grand Duke of Virginia, beautiful as a memory on the curve described by a dog running after his master, sank into the crevices of a ruined convent” (Chant V) [ 7 ] . Jules Verne evokes this curve in the jacanda: “With what grace he made this ball described this learned curve to this ball, whose mathematicians may not yet calculate the value, who have determined, however, famous curve “of the dog who follows his master!” ” \u2191 ‘ Remarkable curves \u00bb, Revue du Palais de la D\u00e9couverte , n O Special 8, July 1976 , p. 113  \u2191 (in) Paul j. nahin, Chases and Escapes\u00a0: The Mathematics of Pursuits and Evasion , Princeton, Princeton University Press, 2012 ( first re  ed. 2007), 253 p.  (ISBN\u00a0 978-0-691-12514-5 And 978-0-691-15501-2 , read online )  \u2191 Pierre Boast \u00ab On new curves to which one can give the name of pursuit lines \u00bb, Memoirs of mathematics and physics drawn from the registers of the Royal Academy of Sciences , 1732 , p. 1-15 . The following article, due to Pierre Louis Moreau de Maupertuis, proposes a generalization to all forms of trajectories of the prosecution.  \u2191 Teixera gum 1909, p. 255  \u2191 Teixera gum 1909, p. 256  \u2191 http:\/\/online.redwoods.cc.ca.us\/instruct\/darnold\/deproj\/sp08\/mseverdia\/pursuit.pdf  \u2191 Count of Lautr\u00e9amont , Les Chants de Maldoror , s. l. E. Wittmann, 1874 ( read online ) , “Fifth song”        (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4"},{"@context":"http:\/\/schema.org\/","@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"item":{"@id":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/#breadcrumbitem","name":"Enzyklop\u00e4die"}},{"@type":"ListItem","position":2,"item":{"@id":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/dog-curve-wikipedia\/#breadcrumbitem","name":"Dog curve – Wikipedia"}}]}]