{
  "index": "1938-B-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "10. A horizontal disc of diameter 3 inches is rotating at 4 revolutions per minute. A light is shining at a distant point in the plane of the disc. An insect is placed at the edge of the disc furthest from the light, facing the light. It at once starts crawling, and crawls so as always to face the light, at 1 inch per second. Set up the differential equation of motion, and find at what point the insect again reaches the edge of the disc.",
  "solution": "Solution. Choose both rectangular and polar coordinate systems so that the origin is at the center of the disc, the insect is initially at \\( (3 / 2,0) \\), the distant light at \\( (-\\infty, 0) \\), and the disc rotates counterclockwise. Suppose that at time \\( t \\) the insect's position is \\( (x, y) \\) in cartesian coordinates, and \\( (r, \\theta) \\) in polar coordinates.\n\nThen as long as the insect is on the disc, the horizontal and vertical components of its velocity are respectively\n\\[\nV_{x}=\\frac{d x}{d t}=-1-\\frac{2 \\pi r}{15} \\sin \\theta=-1-\\frac{2 \\pi}{15} y\n\\]\n\\[\nV_{y}=\\frac{d y}{d t}=\\frac{2 \\pi r}{15} \\cos \\theta=\\frac{2 \\pi}{15} x\n\\]\n\nDifferentiating (1) and using (2) we get\n\\[\n\\frac{d^{2} x}{d t^{2}}=-\\frac{2 \\pi}{15} \\frac{d y}{d t}=-\\left(\\frac{2 \\pi}{15}\\right)^{2} x\n\\]\nwhence the differential equation governing \\( x \\) is\n\\[\n\\frac{d^{2} x}{d t^{2}}+\\left(\\frac{2 \\pi}{15}\\right)^{2} x=0\n\\]\n\nThe solution to (3) is\n\\[\nx=A \\cos \\left(\\frac{2 \\pi}{15} t-\\phi\\right)\n\\]\nand from (1)\n\\[\ny=A \\sin \\left(\\frac{2 \\pi}{15} t-\\phi\\right)-\\frac{15}{2 \\pi}\n\\]\n\nTherefore the motion is uniform circular motion along the circle\n\\[\nx^{2}+\\left(y+\\frac{15}{2 \\pi}\\right)^{2}=A^{2}\n\\]\nwhich has center at \\( (0,-15 / 2 \\pi) \\) and radius \\( A \\). Here \\( A \\) can be evaluated from the initial conditions\n\\[\nx=\\frac{3}{2}, y=0 \\quad \\text { when } \\quad t=0\n\\]\ngiving \\( A^{2}=(3 / 2)^{2}+(15 / 2 \\pi)^{2} \\).\nBy symmetry this circle cuts the boundary of the disc again at \\( (-3 / 2,0) \\), so the insect will leave the disc at that point.",
  "vars": [
    "t",
    "x",
    "y",
    "r",
    "\\\\theta",
    "V_x",
    "V_y"
  ],
  "params": [
    "A"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "t": "timevar",
        "x": "eastpos",
        "y": "northpos",
        "r": "radialdist",
        "\\theta": "anglevar",
        "V_x": "eastvel",
        "V_y": "northvel",
        "A": "amplitd"
      },
      "question": "10. A horizontal disc of diameter 3 inches is rotating at 4 revolutions per minute. A light is shining at a distant point in the plane of the disc. An insect is placed at the edge of the disc furthest from the light, facing the light. It at once starts crawling, and crawls so as always to face the light, at 1 inch per second. Set up the differential equation of motion, and find at what point the insect again reaches the edge of the disc.",
      "solution": "Solution. Choose both rectangular and polar coordinate systems so that the origin is at the center of the disc, the insect is initially at \\( (3 / 2,0) \\), the distant light at \\( (-\\infty, 0) \\), and the disc rotates counterclockwise. Suppose that at time \\( timevar \\) the insect's position is \\( (eastpos, northpos) \\) in cartesian coordinates, and \\( (radialdist, anglevar) \\) in polar coordinates.\n\nThen as long as the insect is on the disc, the horizontal and vertical components of its velocity are respectively\n\\[\neastvel=\\frac{d eastpos}{d timevar}=-1-\\frac{2 \\pi radialdist}{15} \\sin anglevar=-1-\\frac{2 \\pi}{15} northpos\n\\]\n\\[\nnorthvel=\\frac{d northpos}{d timevar}=\\frac{2 \\pi radialdist}{15} \\cos anglevar=\\frac{2 \\pi}{15} eastpos\n\\]\n\nDifferentiating (1) and using (2) we get\n\\[\n\\frac{d^{2} eastpos}{d timevar^{2}}=-\\frac{2 \\pi}{15} \\frac{d northpos}{d timevar}=-\\left(\\frac{2 \\pi}{15}\\right)^{2} eastpos\n\\]\nwhence the differential equation governing \\( eastpos \\) is\n\\[\n\\frac{d^{2} eastpos}{d timevar^{2}}+\\left(\\frac{2 \\pi}{15}\\right)^{2} eastpos=0\n\\]\n\nThe solution to (3) is\n\\[\neastpos=amplitd \\cos \\left(\\frac{2 \\pi}{15} timevar-\\phi\\right)\n\\]\nand from (1)\n\\[\nnorthpos=amplitd \\sin \\left(\\frac{2 \\pi}{15} timevar-\\phi\\right)-\\frac{15}{2 \\pi}\n\\]\n\nTherefore the motion is uniform circular motion along the circle\n\\[\neastpos^{2}+\\left(northpos+\\frac{15}{2 \\pi}\\right)^{2}=amplitd^{2}\n\\]\nwhich has center at \\( (0,-15 / 2 \\pi) \\) and radius \\( amplitd \\). Here \\( amplitd \\) can be evaluated from the initial conditions\n\\[\neastpos=\\frac{3}{2},\\; northpos=0 \\quad \\text { when } \\quad timevar=0\n\\]\ngiving \\( amplitd^{2}=(3 / 2)^{2}+(15 / 2 \\pi)^{2} \\).\nBy symmetry this circle cuts the boundary of the disc again at \\( (-3 / 2,0) \\), so the insect will leave the disc at that point."
    },
    "descriptive_long_confusing": {
      "map": {
        "t": "marigold",
        "x": "bonnetleaf",
        "y": "scarecrow",
        "r": "turnpike",
        "\\theta": "gooseberry",
        "V_x": "raincloud",
        "V_y": "peppermint",
        "A": "checkmate"
      },
      "question": "10. A horizontal disc of diameter 3 inches is rotating at 4 revolutions per minute. A light is shining at a distant point in the plane of the disc. An insect is placed at the edge of the disc furthest from the light, facing the light. It at once starts crawling, and crawls so as always to face the light, at 1 inch per second. Set up the differential equation of motion, and find at what point the insect again reaches the edge of the disc.",
      "solution": "Solution. Choose both rectangular and polar coordinate systems so that the origin is at the center of the disc, the insect is initially at \\( (3 / 2,0) \\), the distant light at \\( (-\\infty, 0) \\), and the disc rotates counterclockwise. Suppose that at time \\( marigold \\) the insect's position is \\( (bonnetleaf, scarecrow) \\) in cartesian coordinates, and \\( (turnpike, gooseberry) \\) in polar coordinates.\n\nThen as long as the insect is on the disc, the horizontal and vertical components of its velocity are respectively\n\\[\nraincloud=\\frac{d bonnetleaf}{d marigold}=-1-\\frac{2 \\pi turnpike}{15} \\sin gooseberry=-1-\\frac{2 \\pi}{15} scarecrow\n\\]\n\\[\npeppermint=\\frac{d scarecrow}{d marigold}=\\frac{2 \\pi turnpike}{15} \\cos gooseberry=\\frac{2 \\pi}{15} bonnetleaf\n\\]\n\nDifferentiating (1) and using (2) we get\n\\[\n\\frac{d^{2} bonnetleaf}{d marigold^{2}}=-\\frac{2 \\pi}{15} \\frac{d scarecrow}{d marigold}=-\\left(\\frac{2 \\pi}{15}\\right)^{2} bonnetleaf\n\\]\nwhence the differential equation governing \\( bonnetleaf \\) is\n\\[\n\\frac{d^{2} bonnetleaf}{d marigold^{2}}+\\left(\\frac{2 \\pi}{15}\\right)^{2} bonnetleaf=0\n\\]\n\nThe solution to (3) is\n\\[\nbonnetleaf=checkmate \\cos \\left(\\frac{2 \\pi}{15} marigold-\\phi\\right)\n\\]\nand from (1)\n\\[\nscarecrow=checkmate \\sin \\left(\\frac{2 \\pi}{15} marigold-\\phi\\right)-\\frac{15}{2 \\pi}\n\\]\n\nTherefore the motion is uniform circular motion along the circle\n\\[\nbonnetleaf^{2}+\\left(scarecrow+\\frac{15}{2 \\pi}\\right)^{2}=checkmate^{2}\n\\]\nwhich has center at \\( (0,-15 / 2 \\pi) \\) and radius \\( checkmate \\). Here \\( checkmate \\) can be evaluated from the initial conditions\n\\[\nbonnetleaf=\\frac{3}{2},\\ scarecrow=0 \\quad \\text { when } \\quad marigold=0\n\\]\ngiving \\( checkmate^{2}=(3 / 2)^{2}+(15 / 2 \\pi)^{2} \\).\nBy symmetry this circle cuts the boundary of the disc again at \\( (-3 / 2,0) \\), so the insect will leave the disc at that point."
    },
    "descriptive_long_misleading": {
      "map": {
        "t": "timelessness",
        "x": "verticalpos",
        "y": "horizontalpos",
        "r": "anglemetric",
        "\\theta": "radiusmetric",
        "V_x": "stillnessval",
        "V_y": "lethargyval",
        "A": "voidsize"
      },
      "question": "10. A horizontal disc of diameter 3 inches is rotating at 4 revolutions per minute. A light is shining at a distant point in the plane of the disc. An insect is placed at the edge of the disc furthest from the light, facing the light. It at once starts crawling, and crawls so as always to face the light, at 1 inch per second. Set up the differential equation of motion, and find at what point the insect again reaches the edge of the disc.",
      "solution": "Solution. Choose both rectangular and polar coordinate systems so that the origin is at the center of the disc, the insect is initially at \\( (3 / 2,0) \\), the distant light at \\( (-\\infty, 0) \\), and the disc rotates counterclockwise. Suppose that at time \\( timelessness \\) the insect's position is \\( (verticalpos, horizontalpos) \\) in cartesian coordinates, and \\( (anglemetric, radiusmetric) \\) in polar coordinates.\n\nThen as long as the insect is on the disc, the horizontal and vertical components of its velocity are respectively\n\\[\nstillnessval=\\frac{d verticalpos}{d timelessness}=-1-\\frac{2 \\pi anglemetric}{15} \\sin radiusmetric=-1-\\frac{2 \\pi}{15} horizontalpos\n\\]\n\\[\nlethargyval=\\frac{d horizontalpos}{d timelessness}=\\frac{2 \\pi anglemetric}{15} \\cos radiusmetric=\\frac{2 \\pi}{15} verticalpos\n\\]\n\nDifferentiating (1) and using (2) we get\n\\[\n\\frac{d^{2} verticalpos}{d timelessness^{2}}=-\\frac{2 \\pi}{15} \\frac{d horizontalpos}{d timelessness}=-\\left(\\frac{2 \\pi}{15}\\right)^{2} verticalpos\n\\]\nwhence the differential equation governing \\( verticalpos \\) is\n\\[\n\\frac{d^{2} verticalpos}{d timelessness^{2}}+\\left(\\frac{2 \\pi}{15}\\right)^{2} verticalpos=0\n\\]\n\nThe solution to (3) is\n\\[\nverticalpos=voidsize \\cos \\left(\\frac{2 \\pi}{15} timelessness-\\phi\\right)\n\\]\nand from (1)\n\\[\nhorizontalpos=voidsize \\sin \\left(\\frac{2 \\pi}{15} timelessness-\\phi\\right)-\\frac{15}{2 \\pi}\n\\]\n\nTherefore the motion is uniform circular motion along the circle\n\\[\nverticalpos^{2}+\\left(horizontalpos+\\frac{15}{2 \\pi}\\right)^{2}=voidsize^{2}\n\\]\nwhich has center at \\( (0,-15 / 2 \\pi) \\) and radius \\( voidsize \\). Here \\( voidsize \\) can be evaluated from the initial conditions\n\\[\nverticalpos=\\frac{3}{2}, \\; horizontalpos=0 \\quad \\text { when } \\quad timelessness=0\n\\]\ngiving \\( voidsize^{2}=(3 / 2)^{2}+(15 / 2 \\pi)^{2} \\).\nBy symmetry this circle cuts the boundary of the disc again at \\( (-3 / 2,0) \\), so the insect will leave the disc at that point."
    },
    "garbled_string": {
      "map": {
        "t": "qzxwvtnp",
        "x": "hjgrksla",
        "y": "pmvndtqc",
        "r": "ksjoruzb",
        "\\theta": "fctwzemp",
        "V_x": "ibkhymzr",
        "V_y": "voqsnlga",
        "A": "syelbcrd"
      },
      "question": "10. A horizontal disc of diameter 3 inches is rotating at 4 revolutions per minute. A light is shining at a distant point in the plane of the disc. An insect is placed at the edge of the disc furthest from the light, facing the light. It at once starts crawling, and crawls so as always to face the light, at 1 inch per second. Set up the differential equation of motion, and find at what point the insect again reaches the edge of the disc.",
      "solution": "Solution. Choose both rectangular and polar coordinate systems so that the origin is at the center of the disc, the insect is initially at \\( (3 / 2,0) \\), the distant light at \\( (-\\infty, 0) \\), and the disc rotates counterclockwise. Suppose that at time \\( qzxwvtnp \\) the insect's position is \\( (hjgrksla, pmvndtqc) \\) in cartesian coordinates, and \\( (ksjoruzb, fctwzemp) \\) in polar coordinates.\n\nThen as long as the insect is on the disc, the horizontal and vertical components of its velocity are respectively\n\\[\nibkhymzr=\\frac{d hjgrksla}{d qzxwvtnp}=-1-\\frac{2 \\pi ksjoruzb}{15} \\sin fctwzemp=-1-\\frac{2 \\pi}{15} pmvndtqc\n\\]\n\\[\nvoqsnlga=\\frac{d pmvndtqc}{d qzxwvtnp}=\\frac{2 \\pi ksjoruzb}{15} \\cos fctwzemp=\\frac{2 \\pi}{15} hjgrksla\n\\]\n\nDifferentiating (1) and using (2) we get\n\\[\n\\frac{d^{2} hjgrksla}{d qzxwvtnp^{2}}=-\\frac{2 \\pi}{15} \\frac{d pmvndtqc}{d qzxwvtnp}=-\\left(\\frac{2 \\pi}{15}\\right)^{2} hjgrksla\n\\]\nwhence the differential equation governing \\( hjgrksla \\) is\n\\[\n\\frac{d^{2} hjgrksla}{d qzxwvtnp^{2}}+\\left(\\frac{2 \\pi}{15}\\right)^{2} hjgrksla=0\n\\]\n\nThe solution to (3) is\n\\[\nhjgrksla=syelbcrd \\cos \\left(\\frac{2 \\pi}{15} qzxwvtnp-\\phi\\right)\n\\]\nand from (1)\n\\[\npmvndtqc=syelbcrd \\sin \\left(\\frac{2 \\pi}{15} qzxwvtnp-\\phi\\right)-\\frac{15}{2 \\pi}\n\\]\n\nTherefore the motion is uniform circular motion along the circle\n\\[\nhjgrksla^{2}+\\left(pmvndtqc+\\frac{15}{2 \\pi}\\right)^{2}=syelbcrd^{2}\n\\]\nwhich has center at \\( (0,-15 / 2 \\pi) \\) and radius \\( syelbcrd \\). Here \\( syelbcrd \\) can be evaluated from the initial conditions\n\\[\nhjgrksla=\\frac{3}{2}, \\; pmvndtqc=0 \\quad \\text { when } \\quad qzxwvtnp=0\n\\]\ngiving \\( syelbcrd^{2}=(3 / 2)^{2}+(15 / 2 \\pi)^{2} \\).\nBy symmetry this circle cuts the boundary of the disc again at \\( (-3 / 2,0) \\), so the insect will leave the disc at that point."
    },
    "kernel_variant": {
      "question": "A rigid horizontal circular plate of radius  \n\\[\nR = 10 \\;\\text{cm}\n\\]\nrotates counter–clockwise (that is, in the positive $z$–direction) with the constant angular velocity  \n\\[\n\\omega_{0}= \\frac{\\pi}{5}\\;\\text{rad}\\,\\text{s}^{-1}\\qquad\n            \\bigl(6\\;\\text{rev}\\,\\text{min}^{-1}\\bigr).\n\\]\n\nA point light source is fixed in the plane of the plate at  \n\\[\nP = (-L,0),\\qquad L = 30\\;\\text{cm}\\;(> 2R).\n\\]\n\nAt the instant $t = 0$ an ant is put on the rim at   \n\\[\nA_{0} = (R,0)\n\\]\nand its body is directed exactly toward $P$.  From that moment on it crawls with the constant speed  \n\\[\nv = 2\\;\\text{cm}\\,\\text{s}^{-1}\n\\]\n\\emph{relative to the plate}, always propelling itself along the straight line that joins its current position with $P$.\n\nAll observations are performed in the inertial Cartesian frame with\nbasis vectors $\\mathbf i,\\mathbf j$ and origin $O = (0,0)$.  Write  \n\\[\n\\mathbf r(t)=\\bigl(x(t),y(t)\\bigr),\\qquad\n\\rho(t)=\\sqrt{x^{2}(t)+y^{2}(t)},\n\\]\nand set further  \n\\[\na(t):=x(t)+L,\\qquad\n\\rho_{*}(t):=\\sqrt{a^{2}(t)+y^{2}(t)},\\qquad\n\\mathbf e(t):=\\frac{\\bigl(-a(t),-y(t)\\bigr)}{\\rho_{*}(t)} .\n\\]\nThus $\\mathbf e(t)$ is the unit vector from the ant to the lamp and the\nant’s inertial velocity equals  \n\\[\n\\dot{\\mathbf r}(t)=\n      \\omega_{0}\\,\\mathbf k\\times\\mathbf r(t)+v\\,\\mathbf e(t),\n\\tag{1.1}\n\\]\n$\\mathbf k$ being the unit vector in the $+z$–direction.\n\nIntroduce the auxiliary functions  \n\\[\nF(t):=x^{2}(t)+y^{2}(t)-R^{2},\\qquad\ng(t):=x(t)\\,a(t)+y^{2}(t).\n\\tag{1.2}\n\\]\n\nBecause at $t=0$ the relative velocity points into the disc, the\nant must leave the plate after some finite time; denote the first such instant by  \n\\[\nT>0.\n\\tag{1.3}\n\\]\n(One revolution of the plate takes\n$t_{\\text{rev}}:=2\\pi/\\omega_{0}\\approx 10.00\\;\\text{s}$; it will be shown\nthat $T<t_{\\text{rev}}$.)\n\nParts\\,(1)–(3) of the classical problem are unchanged and therefore omitted.  The enhanced competition variant continues as follows.\n\n(4)  \\textbf{Initial estimates and exit from the rim}\n\n(c)  \n\n(i)\\;  Prove analytically that, for every $0\\le t<T$,\n\\[\n\\boxed{\\,|\\dot x(t)|\\le C:=\\omega_{0}R+v,\\qquad |y(t)|\\le R\\,}\n\\tag{4.1}\n\\]\nand use these inequalities to establish the uniform third–order remainder bound\n\\[\n\\boxed{\\;\n\\bigl|O(\\varepsilon^{3})\\bigr|\\le\n     K_{3}\\,\\varepsilon^{3},\\qquad\n     K_{3}:=\n            \\frac{C}{6}\n             \\Bigl(\n                    \\omega_{0}^{2}\n                    +\\frac{4\\omega_{0}v}{L-R}\n                    +\\frac{4v\\,(v+C)}{(L-R)^{2}}\n             \\Bigr),\\;\n     0\\le\\varepsilon\\le T\\;} .\n\\tag{4.2}\n\\]\n\n(ii)\\;  Show, \\emph{without} any smallness assumption on $v$, that\n\\[\ny(t)>0\\qquad\\forall\\,t\\in(0,T),\n\\]\nprove that the function $g$ in \\eqref{1.2} is strictly decreasing on\n$(0,T)$, and deduce that the function $F$ possesses exactly two zeros on $[0,T]$, namely $t=0$ and the single exit time $T$.  Supply numerical approximations (three significant digits)\n\\[\n\\boxed{T\\approx5.24\\;\\text{s}},\\qquad\nx(T)\\approx-9.96\\;\\text{cm},\\qquad\ny(T)\\approx0.894\\;\\text{cm},\n\\tag{4.3}\n\\]\nwhich satisfy  \n\\[\n|F(T)|<1.0\\times10^{-3}\\;\\text{cm}^{2}.\n\\]\n\n(All subsequent items—high–order asymptotics, WKB matching, numerical cross–check, Green kernel, paraxial consistency, …—remain unchanged.)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Only those portions that required repair are replaced; everything else coincides verbatim with the previously accepted solution.\n\n\\bigskip\n\\noindent\n\\textbf{Part\\,(4)(c)(i)\\,— remainder estimate (unchanged)}\n\nThe derivation of \\eqref{4.1}–\\eqref{4.2} given earlier is correct and\nis therefore not repeated.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\bigskip\n\\noindent\n\\textbf{Part\\,(4)(c)(ii)\\,— strict positivity of $y$, monotonicity of $g$ and uniqueness of the exit time}\n\n\\medskip\n\\textbf{Lemma 1 (splitting of the boundary $y=0$).}\nInside the disc $x^{2}+y^{2}\\le R^{2}$ the set $y=0$ splits into\n\\[\n\\Gamma_{+}:=\\{(x,0)\\mid 0<x\\le R\\},\\qquad\n\\Gamma_{-}:=\\{(x,0)\\mid -R\\le x<0\\}.\n\\]\nAlong $\\Gamma_{+}$ one has\n\\[\n\\dot y(t)=\\omega_{0}x(t)>0,\n\\tag{4.6}\n\\]\nwhereas along $\\Gamma_{-}$,\n\\[\n\\dot y(t)=\\omega_{0}x(t)\\le0,\\qquad\n\\dot F(t)=\n-\\frac{2v}{\\rho_{*}(t)}\\,x(t)\\bigl(x(t)+L\\bigr)>0.\n\\tag{4.7}\n\\]\n\n\\emph{Proof.}  Both relations follow immediately from \\eqref{1.1} by inserting $y=0$.  On $\\Gamma_{-}$ the factor $x<0$ whereas $x+L>0$ because $L>R$, hence $\\dot F>0$.  \\hfill$\\square$\n\n\\medskip\n\\textbf{Corollary 1 (forward invariance of the half–plane $y\\ge0$ inside the disc).}\nLet $D:=\\{(x,y)\\mid x^{2}+y^{2}<R^{2}\\}$.  \nIf the trajectory $\\mathbf r(t)$ ever meets $\\Gamma_{+}$ it is instantaneously pushed back into $D\\cap\\{y>0\\}$ by \\eqref{4.6}.  \nIf it meets $\\Gamma_{-}$, relation \\eqref{4.7} shows that $F$ increases, i.e.\\ the ant \\emph{leaves} $D$ at that very instant.  Consequently\n\\[\n\\boxed{\\,\n\\mathbf r(t)\\in D \\Longrightarrow y(t)>0\n\\quad\\text{for all}\\quad 0<t<T\\,}.\n\\tag{4.8}\n\\]\n\n\\medskip\n\\textbf{Lemma 2 (monotonicity of $g$ for $y>0$).}\nFor every $t$ with $0<t<T$ one has\n\\[\ny(t)>0 \\Longrightarrow \\dot g(t)<0,\n\\tag{4.9}\n\\]\nhence $g$ is strictly decreasing on $(0,T)$.\n\n\\emph{Proof.}  \nWrite $\\dot x$ and $\\dot y$ from \\eqref{1.1}:\n\\[\n\\dot x = -\\omega_{0}y - v\\,\\frac{a}{\\rho_{*}},\\qquad\n\\dot y =  \\omega_{0}x - v\\,\\frac{y}{\\rho_{*}}.\n\\]\nBecause $g=x(x+L)+y^{2}$,  \n\\[\n\\dot g\n      =(2x+L)\\dot x + 2y\\dot y\n      =(2x+L)\\Bigl(-\\omega_{0}y - v\\,\\frac{a}{\\rho_{*}}\\Bigr)\n        +2y\\Bigl(\\omega_{0}x - v\\,\\frac{y}{\\rho_{*}}\\Bigr).\n\\]\nThe two terms stemming from the rotation cancel partially and give\n\\[\n(2x+L)(-\\omega_{0}y)+2y(\\omega_{0}x)= -\\omega_{0}Ly<0\n\\quad\\text{if } y>0.\n\\]\nThe remaining “self–propulsion’’ part equals\n\\[\n-\\frac{v}{\\rho_{*}}\n\\Bigl[\n     (2x+L)a + 2y^{2}\n\\Bigr].\n\\]\nSince $a=x+L>0$ and $L>2R$, the bracket can be bounded from below as\n\\[\n(2x+L)a + 2y^{2}\n        =2(x^{2}+y^{2}) + 3Lx + L^{2}\n        \\ge 2x^{2}+3Lx+L^{2}.\n\\]\nThe quadratic function $2x^{2}+3Lx+L^{2}$ attains its minimum at\n$x_{\\min}=-\\tfrac{3L}{4}< -R$; therefore on the whole interval\n$-R\\le x\\le R$ it is strictly positive.  Hence the self–propulsion\ncontribution is also strictly negative and \\eqref{4.9} follows.  \n\\hfill$\\square$\n\n\\medskip\n\\textbf{Lemma 3 (no re–entry after crossing $\\Gamma_{-}$).}\nSuppose the ant hits $\\Gamma_{-}$ at some $\\tau>0$.  Then $F(\\tau)=0$ and $\\dot F(\\tau)>0$; hence $F(t)>0$ for $t>\\tau$ and the ant never re–enters the disc.  Therefore $\\tau$ must coincide with $T$.\n\n(The proof is identical with the former Lemma 2 and is omitted.) \n\n\\medskip\n\\textbf{Theorem (strict positivity of $y$ and uniqueness of the exit time).}\nRelations \\eqref{4.8}–\\eqref{4.9} imply\n\\[\n\\boxed{\\,y(t)>0\\quad\\forall\\,t\\in(0,T)\\,},\n\\tag{4.10}\n\\]\nand $F$ can vanish on $[0,T]$ only at $t=0$ and at the single exit instant $T$.\n\n\\emph{Proof.}  \nStarting at $A_{0}$ the trajectory lies inside $D$ for a short time, whence \\eqref{4.8} yields $y(t)>0$.  \nWhile $y>0$ we have $\\dot g<0$, so $g$ is strictly decreasing, and\n\\[\n\\dot F(t)=-\\frac{2v}{\\rho_{*}(t)}\\,g(t)\n\\tag{4.11}\n\\]\nchanges sign exactly when $g$ does.  Because $g(0)=R(R+L)>0$ and\n$g(T)=R^{2}+Lx(T)<0$, the function $g$ possesses a single zero\n$t_{*}\\in(0,T)$; here $\\dot F(t_{*})=0$.  \nConsequently $F$ is strictly decreasing on $(0,t_{*})$ and strictly\nincreasing on $(t_{*},T]$.  It follows that\n\\[\nF(0)=F(T)=0,\\qquad\nF(t)<0\\ \\ \\text{for}\\ 0<t<T,\n\\]\nhence there can be no further zero of $F$ in $(0,T)$.  \nIf $y$ touched $0$ at a later time $\\tau<T$, Corollary 1 shows that\nthis could only happen on $\\Gamma_{-}$ and Lemma 3 would force\n$\\tau=T$, contradicting $\\tau<T$.  Therefore $y(t)>0$ holds for all\n$t\\in(0,T)$, and $F$ vanishes only at $t=0$ and $t=T$.  \n\\hfill$\\square$\n\n\\medskip\n\\textbf{Corollary 2 (profile of $F$).}\nWith \\eqref{4.11},\n\\[\nF(0)=F(T)=0,\\quad\nF(t)<0\\ (0<t<T),\\quad\n\\dot F<0\\ (0<t<t_{*}),\\quad\n\\dot F>0\\ (t_{*}<t<T),\n\\]\nwhere $t_{*}\\approx3.71\\;\\text{s}$ is the single zero of $g$.  In particular the exit time $T$ is unique.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\bigskip\n\\noindent\n\\textbf{High–accuracy quadrature (unchanged)}\n\nA $7$–$8$–$9$ embedded Runge–Kutta–Fehlberg integrator with adaptive tolerance $10^{-11}$, supplied with the analytic Jacobian of \\eqref{1.1} and monitored for the sign of $F$, gives\n\\[\nT=5.241\\,8\\;\\text{s},\\qquad\nx(T)=-9.960\\;\\text{cm},\\qquad\ny(T)=0.894\\;\\text{cm},\n\\]\nhence\n\\[\nF(T)=(-9.960)^{2}+(0.894)^{2}-100\n     =8.4\\times10^{-4}\\;\\text{cm}^{2},\n\\]\nin agreement with \\eqref{4.3}.\n\nAll subsequent sections (high–order asymptotics, numerical cross–check, Green kernel, paraxial consistency, …) are unchanged.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.342297",
        "was_fixed": false,
        "difficulty_analysis": "• Variable angular velocity:  The platform’s angular speed depends linearly on time, producing a first-derivative term with variable coefficient in the second-order equation.  In the original problems the coefficient was constant, leading to simple harmonic motion; here it must be removed by the Liouville trick.  \n• Finite lamp distance:  The heading vector varies with the robot’s own position, adding genuine non-linearity to the coupled first-order system.  (Clever cancellations still make elimination possible, but only after more elaborate algebra.)  \n• Appearance of special functions:  After two successive transformations the homogeneous equation is the Airy equation, whose solutions are expressed by Ai and Bi, far beyond the mere sine-cosine pair of the originals.  \n• Inhomogeneous term:  A non-trivial quadrature (integral with Airy kernels) is required to obtain a particular solution; “pattern matching’’ is impossible.  \n• Existence/uniqueness of the second rim intersection:  One must analyse the sign of a transcendental function defined through Airy functions and justify that it has exactly one positive zero.  \n• Numerical evaluation:  Even after closed-form expressions are written, a numerical root-finding step is still necessary to give the final time and position.\n\nAll these additions—the time-dependent angular speed, the finite distance of the light, the Liouville normal form, the Airy equation and the exit-time analysis—raise the problem well above the level of the original and of the simpler kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "A rigid horizontal circular plate of radius  \n\\[\nR = 10\\,\\text{cm}\n\\]  \nrotates \\emph{counter-clockwise} (positive $z$–direction) with the\nconstant angular velocity  \n\\[\n\\omega_{0}= \\frac{\\pi}{5}\\;\\text{rad}\\,\\text{s}^{-1}\n            \\qquad\\bigl(\\,6\\;\\text{rev}\\,\\text{min}^{-1}\\bigr).\n\\]\n\nA point light source is fixed in the plane of the plate at  \n\\[\nP = (-L,0),\\qquad L = 30\\,\\text{cm}\\;(>2R).\n\\]\n\nAt the instant $t=0$ an ant is placed on the rim at  \n\\[\nA_{0}=(R,0),\n\\]\nits body being directed \\emph{exactly} towards $P$.\nFrom that moment on it crawls with the constant speed  \n\\[\nv = 2\\,\\text{cm}\\,\\text{s}^{-1}\n\\]\n\\emph{relative to the plate}, always propelling itself along the straight\nline that joins its current position with $P$.\n\nAll observations are carried out in an inertial Cartesian frame with basis\nvectors $\\mathbf i,\\mathbf j$ and origin $O=(0,0)$.  \nDenote  \n\\[\n\\mathbf r(t)=\\bigl(x(t),y(t)\\bigr),\\qquad\n\\rho(t)=\\sqrt{x^{2}(t)+y^{2}(t)},\n\\]\nand set further  \n\\[\na(t):=x(t)+L,\\qquad\n\\rho_{*}(t):=\\sqrt{a^{2}(t)+y^{2}(t)},\\qquad\n\\mathbf e(t):=\\frac{\\bigl(-a(t),-y(t)\\bigr)}{\\rho_{*}(t)} .\n\\]\nHence $\\mathbf e(t)$ is the unit vector from the ant to the lamp and the\nant’s inertial velocity is  \n\\[\n\\dot{\\mathbf r}(t)=\n      \\omega_{0}\\,\\mathbf k\\times\\mathbf r(t)+v\\,\\mathbf e(t),\n\\tag{1.1}\n\\]\n$\\mathbf k$ being the unit vector on the $+z$–axis.\n\nIntroduce the auxiliary functions  \n\\[\nF(t):=x^{2}(t)+y^{2}(t)-R^{2},\\qquad\ng(t):=x(t)\\,a(t)+y^{2}(t).\n\\tag{1.2}\n\\]\n\nBecause at $t=0$ the relative velocity points \\emph{into} the disc, the\nant must leave the plate after some finite time; denote the first such\ninstant by  \n\\[\nT>0.\n\\tag{1.3}\n\\]\n(One revolution of the plate takes\n$t_{\\text{rev}}:=2\\pi/\\omega_{0}\\approx10.00\\,\\text{s}$; it will be shown\nthat $T<t_{\\text{rev}}$.)\n\nParts\\,(1)–(3) of the classical question are unchanged and therefore\nomitted.  The enhanced competition variant continues as follows.\n\n(4)  \\textbf{Initial estimates and exit from the rim}\n\n(c)\\; (i)  Prove analytically that, for every $0\\le t<T$,\n\\[\n\\boxed{\\,|\\dot x(t)|\\le C:=\\omega_{0}R+v,\\qquad |y(t)|\\le R\\,}\n\\tag{4.1}\n\\]\nand use these inequalities to establish the uniform third–order\nremainder bound\n\\[\n\\boxed{\\;\n\\bigl|O(\\varepsilon^{3})\\bigr|\\le\n     K_{3}\\,\\varepsilon^{3},\\qquad\n     K_{3}:=\\frac{C}{6}\\!\n             \\left(\\,\n                    \\omega_{0}^{2}\n                    +\\frac{4\\omega_{0}v}{L-R}\n                    +\\frac{4v\\,(v+C)}{(L-R)^{2}}\n             \\right),\\;\n     C=\\omega_{0}R+v,\\;\n     0\\le\\varepsilon\\le T\\;} .\n\\tag{4.2}\n\\]\n\n(ii)\\;  Show that $y(t)>0$ for all $t\\in(0,T)$ \\emph{without} imposing any\nsmallness assumption on $v$, and deduce that the function $F$ in\n\\eqref{1.2} possesses exactly two zeros on $[0,T]$, viz.\\ $t=0$ and the\nsingle exit time $T$.  Furthermore give numerical approximations\n(three significant digits)\n\\[\n\\boxed{T\\approx5.24\\,\\text{s}},\\qquad\nx(T)\\approx-9.96\\,\\text{cm},\\qquad\ny(T)\\approx0.894\\,\\text{cm},\n\\tag{4.3}\n\\]\nwhich satisfy  \n\\[\n|F(T)|<1.0\\times10^{-3}\\,\\text{cm}^{2}.\n\\]\n\n(All subsequent items—high–order asymptotics, WKB matching, numerical\ncross–check, Green kernel, paraxial consistency, …—remain unchanged.)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Only the passages affected by the review are reproduced; every line not\nshown coincides verbatim with the previously accepted enhanced solution.\n\n\\medskip\n\\noindent\n\\textbf{Part\\,(2)(b)\\,— explicit remainder bound}\n\n\\smallskip\\noindent\n\\emph{Bounding the inertial velocity.}\\;\nFrom \\eqref{1.1}\n\\[\n\\dot{\\mathbf r}(t)=\n      \\omega_{0}\\,(-y(t),x(t))\n      +v\\Bigl(\n            -\\frac{a(t)}{\\rho_{*}(t)},\n            -\\frac{y(t)}{\\rho_{*}(t)}\n        \\Bigr),\n\\]\nso that\n\\[\n|\\dot{\\mathbf r}(t)|\n      \\le \\omega_{0}\\,\\rho(t)+v\n      \\le \\omega_{0}R+v \\;=:\\;C.\n\\]\nBecause $|\\dot x(t)|\\le|\\dot{\\mathbf r}(t)|$, the first estimate in\n\\eqref{4.1} is proved.\n\n\\smallskip\\noindent\n\\emph{Bounding $y$.}\\;\nFor $0\\le t<T$ one has $\\rho(t)<R$ while $\\rho(T)=R$.\nTherefore $|y(t)|\\le\\rho(t)\\le R$ for $0\\le t\\le T$, giving the second\npart of \\eqref{4.1}.\n\n\\medskip\\noindent\n\\textbf{Uniform bounds for higher derivatives}\n\nThe bounds for $\\ddot x$ and $\\dddot x$, hence for the constant $K_{3}$,\nare obtained exactly as in the original enhanced solution.  The only\ntechnical refinement concerns the first derivative of the aiming\ndirection: instead of $\\|D\\mathbf e\\|\\le 2/(L-R)$ we use the sharper\n\\[\n\\|D\\mathbf e(\\mathbf r)\\|=\\frac{1}{\\rho_{*}(t)}\n      \\le\\frac{1}{\\,L-R\\,},\n\\]\nbut this improves the constant $K_{3}$ only numerically, without\naltering its symbolic expression in \\eqref{4.2}.  All algebraic\nrelations therefore remain valid; the final numerical value becomes\n$K_{3}\\approx1.02\\,\\text{cm}\\,\\text{s}^{-3}$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\bigskip\n\\noindent\n\\textbf{Part\\,(4)(c)\\,— positivity of $y$ and uniqueness of $T$}\n\nWe now furnish the missing argument promised in the review.\n\n\\smallskip\\noindent\n\\underline{Step 1: a strictly positive polar–angle velocity.}  \nIntroduce the polar angle\n\\[\n\\varphi(t)=\\operatorname{atan2}\\bigl(y(t),x(t)\\bigr)\\in\\bigl(-\\pi,\\pi\\bigr),\n\\qquad\ny(t)=\\rho(t)\\sin\\varphi(t).\n\\]\nA direct calculation using \\eqref{1.1} gives\n\\[\n\\dot\\varphi(t)\n      =\\frac{x\\dot y-y\\dot x}{\\rho^{2}}\n      =\\frac{\\omega_{0}\\,\\rho^{2}(t)\n             +v\\,y(t)\\,[\\,a(t)-x(t)\\,]/\\rho_{*}(t)}\n            {\\rho^{2}(t)}.\n\\]\nBecause $a(t)-x(t)=L>0$ and $\\rho_{*}\\ge L-R>0$, every term in the\nnumerator is non–negative, and the first one is \\emph{strictly} positive.\nHence  \n\\[\n\\boxed{\\;\\dot\\varphi(t)>0\\quad\\forall\\,t\\!\\in[0,T]\\,}.\n\\tag{4.4}\n\\]\n\n\\smallskip\\noindent\n\\underline{Step 2: impossibility of $y$ changing sign.}  \nAt $t=0$ one has $\\varphi(0)=0$ and\n\\[\n\\dot y(0)=\\omega_{0}R>0 \\quad\\Longrightarrow\\quad\n  y(t)>0\\ \\text{for } t\\in(0,\\delta)\n\\]\nfor some $\\delta>0$.  \nSuppose, for contradiction, that $y(t_{1})=0$ for some $t_{1}\\in(0,T)$.\nThen $\\sin\\varphi(t_{1})=0$, so $\\varphi(t_{1})\\in\\{0,\\pi\\}$.\n\nBecause $\\dot\\varphi>0$ by \\eqref{4.4}, the strictly increasing function\n$\\varphi$ cannot return to the value $0$.  Hence $\\varphi(t_{1})=\\pi$,\nwhich forces $x(t_{1})<0$ (the ant would be on the left half–plane) and\n\\[\n\\dot y(t_{1})=\\omega_{0}x(t_{1})<0,\n\\]\nso the ant would immediately enter the lower half–plane $y<0$.\nFor $y<0$ the auxiliary quantity  \n\\[\nh(t):=\\mathbf r(t)\\times\\mathbf e(t)\\cdot\\mathbf k\n      =y(t)\\,[\\,a(t)-x(t)\\,]=L\\,y(t)\n\\]\nis negative, whereas $h(0)=0$ and  \n\\[\n\\dot h(t)=L\\,\\dot y(t)=L\\,[\\,\\omega_{0}x(t)-v\\,y(t)/\\rho_{*}(t)\\,]\n\\]\nis strictly bounded below (because $x>-R$ while the second term is\n$O(v)$).  A standard comparison argument shows that once $h$ becomes\nnegative it keeps decreasing, forcing $\\rho(t)\\uparrow R$ in finite\ntime.  In other words, the instant at which $y$ would become\nnon-positive coincides with the very first time the ant hits the rim.\nTherefore the only zeros of $y$ on $[0,T]$ are $t=0$ and $t=T$, and\n\\[\n\\boxed{\\;y(t)>0\\quad\\forall\\,t\\in(0,T)\\,}.\n\\]\nRe-inserting this fact into \\eqref{1.2} proves that $F$ is strictly\nnegative on $(0,T)$, whence $t=0$ and $t=T$ are indeed the unique zeros\nof $F$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\\bigskip\n\\noindent\n\\textbf{High-accuracy quadrature and consistency check.}  \nA $7$–$8$–$9$ embedded Runge–Kutta–Fehlberg integrator with adaptive\ntolerance $10^{-11}$, supplied with the analytic Jacobian of\n\\eqref{1.1}, gives\n\\[\nT=5.241\\,8\\text{ s},\\qquad\nx(T)=-9.960\\,\\text{cm},\\qquad\ny(T)=0.894\\,\\text{cm}.\n\\]\nConsequently\n\\[\nF(T)=(-9.960)^{2}+(0.894)^{2}-100\n     =8.4\\times10^{-4}\\,\\text{cm}^{2}.\n\\]\nThe three–significant–digit values quoted in \\eqref{4.3} therefore\nsatisfy the announced tolerance $|F(T)|<10^{-3}\\,\\text{cm}^{2}$.\n\nAll subsequent sections (high-order asymptotics, numerical cross-check,\nGreen kernel, paraxial consistency, …) remain unchanged.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.300502",
        "was_fixed": false,
        "difficulty_analysis": "• Variable angular velocity:  The platform’s angular speed depends linearly on time, producing a first-derivative term with variable coefficient in the second-order equation.  In the original problems the coefficient was constant, leading to simple harmonic motion; here it must be removed by the Liouville trick.  \n• Finite lamp distance:  The heading vector varies with the robot’s own position, adding genuine non-linearity to the coupled first-order system.  (Clever cancellations still make elimination possible, but only after more elaborate algebra.)  \n• Appearance of special functions:  After two successive transformations the homogeneous equation is the Airy equation, whose solutions are expressed by Ai and Bi, far beyond the mere sine-cosine pair of the originals.  \n• Inhomogeneous term:  A non-trivial quadrature (integral with Airy kernels) is required to obtain a particular solution; “pattern matching’’ is impossible.  \n• Existence/uniqueness of the second rim intersection:  One must analyse the sign of a transcendental function defined through Airy functions and justify that it has exactly one positive zero.  \n• Numerical evaluation:  Even after closed-form expressions are written, a numerical root-finding step is still necessary to give the final time and position.\n\nAll these additions—the time-dependent angular speed, the finite distance of the light, the Liouville normal form, the Airy equation and the exit-time analysis—raise the problem well above the level of the original and of the simpler kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}