{
  "index": "1952-B-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "2. Find the surface generated by the solutions of\n\\[\n\\frac{d x}{y z}=\\frac{d y}{z x}=\\frac{d z}{x y}\n\\]\nwhich intersects the circle \\( y^{2}+z^{2}=1, x=0 \\).",
  "solution": "Solution. The given differential equations mean that at the point \\( (x, y, z) \\) we assign a line element having the direction numbers \\( y z, z x \\), and \\( x y \\), and we seek curves having at each point the assigned line element as tangent. At points of the coordinate axes, however, no line element is assigned, since \\( 0,0,0 \\) is not a set of direction numbers. Furthermore, the direction field cannot be extended to the coordinate axes by continuity, since the line elements assigned on a coordinate plane are always perpendicular to that plane. On the rest of space, however, the equations assign an evidently smooth family of line elements, and so through each point of this domain passes a unique maximal solution curve. We shall determine these curves.\n\nMultiply the given equations by \\( 2 x y z \\) to get\n\\[\n2 x d x=2 y d y=2 z d z\n\\]\n\nThen integration yields\n\\[\nx^{2}=y^{2}+c_{1}=z^{2}+c_{2}\n\\]\n\nIn general, these equations represent the intersection of two hyperbolic cylinders which falls into four connected smooth curves, each of which is a maximal solution curve of the differential system. For example, with the choice \\( c_{1}=-\\alpha^{2}, c_{2}=-\\beta^{2}, \\alpha \\beta \\neq 0 \\), the surfaces (1) pass through the point \\( (0, \\alpha, \\beta) \\) and have the form\n\\[\ny^{2}=x^{2}+\\alpha^{2}, \\quad z^{2}=x^{2}+\\beta^{2}\n\\]\n\nTheir four curves of intersection are given by\n\\[\ny= \\pm \\sqrt{x^{2}+\\alpha^{2}}, \\quad z= \\pm \\sqrt{x^{2}+\\beta^{2}}\n\\]\n\nOne of these curves passes through \\( (0, \\alpha, \\beta) \\). If \\( \\alpha=0 \\) but \\( \\beta \\neq 0 \\), two of these curves pass through each of the points \\( (0,0, \\pm \\beta) \\). Since these points are not in the domain of the original differential system, in this case each of the four curves (3) breaks into two maximal solution curves after removing these points.\nNow we determine the surface formed by the solution curves which intersect the circle \\( C: y^{2}+z^{2}=1, x=0 \\). These solutions are evidently given by (2) with \\( \\alpha^{2}+\\beta^{2}=1, \\alpha \\beta \\neq 0 \\), and therefore they lie on the surface given by\n\\[\ny^{2}+z^{2}=2 x^{2}+1\n\\]\n\nThis is the equation of a hyperboloid of one sheet rotationally symmetric about the \\( x \\)-axis. Since four points of \\( C \\) must be excluded from consideration, we expect that only part of this hyperboloid is generated by the solution curves through \\( C \\). Indeed, if \\( (x, y, z) \\) lies on (2) with \\( \\alpha^{2} \\) and \\( \\beta^{2} \\) positive, then\n\\[\ny^{2}>x^{2}, \\quad z^{2}>x^{2} .\n\\]\n\nConversely, if ( \\( x, y, z \\) ) satisfies (4) and (5), then positive numbers \\( \\alpha \\) and \\( \\beta \\) satisfying (2) and \\( \\alpha^{2}+\\beta^{2}=1 \\) can be chosen. With the appropriate choice of signs, the point \\( (x, y, z) \\) will lie on the curve (3), which meets the circle \\( C \\) at one of the points \\( (0, \\pm \\alpha, \\pm \\beta) \\). Thus the required surface is given by the equation (4) and the inequalities (5).\n\nThe four quadrants of \\( C \\) each generate a portion of the required surface and these portions are bounded by the exceptional curves\n\\[\ny= \\pm x, z= \\pm \\sqrt{x^{2}+1} \\text { and } y= \\pm \\sqrt{x^{2}+1}, z= \\pm x\n\\]\n\nHyperboloid: \\( y^{2}+z^{2}=2 x^{2}+1 \\)\n\\( A(0,1,0) \\) and \\( B(0,0,1) \\) are excluded points. Curves \\( A C \\) and \\( B D \\) are exceptional solution curves. Curves \\( E F \\), \\( G H \\), etc., are ordinary solution curves.\n\nThe solution curves through the illustrated quadrant of \\( C \\) generate that portion of the hyperboloid that lies between the exceptional solution curves.",
  "vars": [
    "x",
    "y",
    "z"
  ],
  "params": [
    "c_1",
    "c_2",
    "\\\\alpha",
    "\\\\beta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "horizcoord",
        "y": "vertical",
        "z": "depthaxis",
        "c_1": "constone",
        "c_2": "consttwo",
        "\\alpha": "shapealpha",
        "\\beta": "shapebeta"
      },
      "question": "2. Find the surface generated by the solutions of\n\\[\n\\frac{d horizcoord}{vertical\\, depthaxis}=\\frac{d vertical}{depthaxis\\, horizcoord}=\\frac{d depthaxis}{horizcoord\\, vertical}\n\\]\nwhich intersects the circle \\( vertical^{2}+depthaxis^{2}=1,\\; horizcoord=0 \\).",
      "solution": "Solution. The given differential equations mean that at the point \\( (horizcoord, vertical, depthaxis) \\) we assign a line element having the direction numbers \\( vertical\\, depthaxis,\\; depthaxis\\, horizcoord \\), and \\( horizcoord\\, vertical \\), and we seek curves having at each point the assigned line element as tangent. At points of the coordinate axes, however, no line element is assigned, since \\( 0,0,0 \\) is not a set of direction numbers. Furthermore, the direction field cannot be extended to the coordinate axes by continuity, since the line elements assigned on a coordinate plane are always perpendicular to that plane. On the rest of space, however, the equations assign an evidently smooth family of line elements, and so through each point of this domain passes a unique maximal solution curve. We shall determine these curves.\n\nMultiply the given equations by \\( 2\\, horizcoord\\, vertical\\, depthaxis \\) to get\n\\[\n2\\, horizcoord\\, d horizcoord = 2\\, vertical\\, d vertical = 2\\, depthaxis\\, d depthaxis\n\\]\n\nThen integration yields\n\\[\nhorizcoord^{2}=vertical^{2}+constone=depthaxis^{2}+consttwo\n\\]\n\nIn general, these equations represent the intersection of two hyperbolic cylinders which falls into four connected smooth curves, each of which is a maximal solution curve of the differential system. For example, with the choice \\( constone=-shapealpha^{2},\\; consttwo=-shapebeta^{2},\\; shapealpha\\, shapebeta \\neq 0 \\), the surfaces (1) pass through the point \\( (0, shapealpha, shapebeta) \\) and have the form\n\\[\nvertical^{2}=horizcoord^{2}+shapealpha^{2}, \\quad depthaxis^{2}=horizcoord^{2}+shapebeta^{2}\n\\]\n\nTheir four curves of intersection are given by\n\\[\nvertical= \\pm \\sqrt{horizcoord^{2}+shapealpha^{2}}, \\quad depthaxis= \\pm \\sqrt{horizcoord^{2}+shapebeta^{2}}\n\\]\n\nOne of these curves passes through \\( (0, shapealpha, shapebeta) \\). If \\( shapealpha=0 \\) but \\( shapebeta \\neq 0 \\), two of these curves pass through each of the points \\( (0,0, \\pm shapebeta) \\). Since these points are not in the domain of the original differential system, in this case each of the four curves (3) breaks into two maximal solution curves after removing these points.\nNow we determine the surface formed by the solution curves which intersect the circle \\( C: vertical^{2}+depthaxis^{2}=1,\\; horizcoord=0 \\). These solutions are evidently given by (2) with \\( shapealpha^{2}+shapebeta^{2}=1,\\; shapealpha\\, shapebeta \\neq 0 \\), and therefore they lie on the surface given by\n\\[\nvertical^{2}+depthaxis^{2}=2\\, horizcoord^{2}+1\n\\]\n\nThis is the equation of a hyperboloid of one sheet rotationally symmetric about the \\( horizcoord \\)-axis. Since four points of \\( C \\) must be excluded from consideration, we expect that only part of this hyperboloid is generated by the solution curves through \\( C \\). Indeed, if \\( (horizcoord, vertical, depthaxis) \\) lies on (2) with \\( shapealpha^{2} \\) and \\( shapebeta^{2} \\) positive, then\n\\[\nvertical^{2}>horizcoord^{2}, \\quad depthaxis^{2}>horizcoord^{2}.\n\\]\n\nConversely, if \\( (horizcoord, vertical, depthaxis) \\) satisfies (4) and (5), then positive numbers \\( shapealpha \\) and \\( shapebeta \\) satisfying (2) and \\( shapealpha^{2}+shapebeta^{2}=1 \\) can be chosen. With the appropriate choice of signs, the point \\( (horizcoord, vertical, depthaxis) \\) will lie on the curve (3), which meets the circle \\( C \\) at one of the points \\( (0, \\pm shapealpha, \\pm shapebeta) \\). Thus the required surface is given by the equation (4) and the inequalities (5).\n\nThe four quadrants of \\( C \\) each generate a portion of the required surface and these portions are bounded by the exceptional curves\n\\[\nvertical= \\pm horizcoord, \\quad depthaxis= \\pm \\sqrt{horizcoord^{2}+1} \\text { and } \\quad vertical= \\pm \\sqrt{horizcoord^{2}+1}, \\quad depthaxis= \\pm horizcoord\n\\]\n\nHyperboloid: \\( vertical^{2}+depthaxis^{2}=2\\, horizcoord^{2}+1 \\)\n\\( A(0,1,0) \\) and \\( B(0,0,1) \\) are excluded points. Curves \\( A C \\) and \\( B D \\) are exceptional solution curves. Curves \\( E F \\), \\( G H \\), etc., are ordinary solution curves.\n\nThe solution curves through the illustrated quadrant of \\( C \\) generate that portion of the hyperboloid that lies between the exceptional solution curves."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "tangerine",
        "y": "elephant",
        "z": "pinecone",
        "c_1": "marigold",
        "c_2": "butterfly",
        "\\alpha": "sunflower",
        "\\beta": "chocolate"
      },
      "question": "2. Find the surface generated by the solutions of\n\\[\n\\frac{d tangerine}{elephant pinecone}=\\frac{d elephant}{pinecone tangerine}=\\frac{d pinecone}{tangerine elephant}\n\\]\nwhich intersects the circle \\( elephant^{2}+pinecone^{2}=1, tangerine=0 \\).",
      "solution": "Solution. The given differential equations mean that at the point \\( (tangerine, elephant, pinecone) \\) we assign a line element having the direction numbers \\( elephant pinecone, pinecone tangerine \\), and \\( tangerine elephant \\), and we seek curves having at each point the assigned line element as tangent. At points of the coordinate axes, however, no line element is assigned, since \\( 0,0,0 \\) is not a set of direction numbers. Furthermore, the direction field cannot be extended to the coordinate axes by continuity, since the line elements assigned on a coordinate plane are always perpendicular to that plane. On the rest of space, however, the equations assign an evidently smooth family of line elements, and so through each point of this domain passes a unique maximal solution curve. We shall determine these curves.\n\nMultiply the given equations by \\( 2 tangerine elephant pinecone \\) to get\n\\[\n2 tangerine d tangerine=2 elephant d elephant=2 pinecone d pinecone\n\\]\n\nThen integration yields\n\\[\ntangerine^{2}=elephant^{2}+marigold=pinecone^{2}+butterfly\n\\]\n\nIn general, these equations represent the intersection of two hyperbolic cylinders which falls into four connected smooth curves, each of which is a maximal solution curve of the differential system. For example, with the choice \\( marigold=-sunflower^{2}, butterfly=-chocolate^{2}, sunflower chocolate \\neq 0 \\), the surfaces (1) pass through the point \\( (0, sunflower, chocolate) \\) and have the form\n\\[\nelephant^{2}=tangerine^{2}+sunflower^{2}, \\quad pinecone^{2}=tangerine^{2}+chocolate^{2}\n\\]\n\nTheir four curves of intersection are given by\n\\[\nelephant= \\pm \\sqrt{tangerine^{2}+sunflower^{2}}, \\quad pinecone= \\pm \\sqrt{tangerine^{2}+chocolate^{2}}\n\\]\n\nOne of these curves passes through \\( (0, sunflower, chocolate) \\). If \\( sunflower=0 \\) but \\( chocolate \\neq 0 \\), two of these curves pass through each of the points \\( (0,0, \\pm chocolate) \\). Since these points are not in the domain of the original differential system, in this case each of the four curves (3) breaks into two maximal solution curves after removing these points.\nNow we determine the surface formed by the solution curves which intersect the circle \\( C: elephant^{2}+pinecone^{2}=1, tangerine=0 \\). These solutions are evidently given by (2) with \\( sunflower^{2}+chocolate^{2}=1, sunflower chocolate \\neq 0 \\), and therefore they lie on the surface given by\n\\[\nelephant^{2}+pinecone^{2}=2 tangerine^{2}+1\n\\]\n\nThis is the equation of a hyperboloid of one sheet rotationally symmetric about the \\( tangerine \\)-axis. Since four points of \\( C \\) must be excluded from consideration, we expect that only part of this hyperboloid is generated by the solution curves through \\( C \\). Indeed, if \\( (tangerine, elephant, pinecone) \\) lies on (2) with \\( sunflower^{2} \\) and \\( chocolate^{2} \\) positive, then\n\\[\nelephant^{2}>tangerine^{2}, \\quad pinecone^{2}>tangerine^{2} .\n\\]\n\nConversely, if ( \\( tangerine, elephant, pinecone \\) ) satisfies (4) and (5), then positive numbers \\( sunflower \\) and \\( chocolate \\) satisfying (2) and \\( sunflower^{2}+chocolate^{2}=1 \\) can be chosen. With the appropriate choice of signs, the point \\( (tangerine, elephant, pinecone) \\) will lie on the curve (3), which meets the circle \\( C \\) at one of the points \\( (0, \\pm sunflower, \\pm chocolate) \\). Thus the required surface is given by the equation (4) and the inequalities (5).\n\nThe four quadrants of \\( C \\) each generate a portion of the required surface and these portions are bounded by the exceptional curves\n\\[\nelephant= \\pm tangerine, pinecone= \\pm \\sqrt{tangerine^{2}+1} \\text { and } elephant= \\pm \\sqrt{tangerine^{2}+1}, pinecone= \\pm tangerine\n\\]\n\nHyperboloid: \\( elephant^{2}+pinecone^{2}=2 tangerine^{2}+1 \\)\n\\( A(0,1,0) \\) and \\( B(0,0,1) \\) are excluded points. Curves \\( A C \\) and \\( B D \\) are exceptional solution curves. Curves \\( E F \\), \\( G H \\), etc., are ordinary solution curves.\n\nThe solution curves through the illustrated quadrant of \\( C \\) generate that portion of the hyperboloid that lies between the exceptional solution curves."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "lateralaxis",
        "z": "depthaxis",
        "c_1": "rigidconstant",
        "c_2": "stiffconstant",
        "\\alpha": "omegamarker",
        "\\beta": "zetamarker"
      },
      "question": "2. Find the surface generated by the solutions of\n\\[\n\\frac{d verticalaxis}{lateralaxis\\, depthaxis}=\\frac{d lateralaxis}{depthaxis\\, verticalaxis}=\\frac{d depthaxis}{verticalaxis\\, lateralaxis}\n\\]\nwhich intersects the circle \\( lateralaxis^{2}+depthaxis^{2}=1, verticalaxis=0 \\).",
      "solution": "Solution. The given differential equations mean that at the point \\( (verticalaxis, lateralaxis, depthaxis) \\) we assign a line element having the direction numbers \\( lateralaxis\\, depthaxis, depthaxis\\, verticalaxis \\), and \\( verticalaxis\\, lateralaxis \\), and we seek curves having at each point the assigned line element as tangent. At points of the coordinate axes, however, no line element is assigned, since \\( 0,0,0 \\) is not a set of direction numbers. Furthermore, the direction field cannot be extended to the coordinate axes by continuity, since the line elements assigned on a coordinate plane are always perpendicular to that plane. On the rest of space, however, the equations assign an evidently smooth family of line elements, and so through each point of this domain passes a unique maximal solution curve. We shall determine these curves.\n\nMultiply the given equations by \\( 2 verticalaxis\\, lateralaxis\\, depthaxis \\) to get\n\\[\n2 verticalaxis\\, d verticalaxis=2 lateralaxis\\, d lateralaxis=2 depthaxis\\, d depthaxis\n\\]\n\nThen integration yields\n\\[\nverticalaxis^{2}=lateralaxis^{2}+rigidconstant=depthaxis^{2}+stiffconstant\n\\]\n\nIn general, these equations represent the intersection of two hyperbolic cylinders which falls into four connected smooth curves, each of which is a maximal solution curve of the differential system. For example, with the choice \\( rigidconstant=-omegamarker^{2}, stiffconstant=-zetamarker^{2}, omegamarker zetamarker \\neq 0 \\), the surfaces (1) pass through the point \\( (0, omegamarker, zetamarker) \\) and have the form\n\\[\nlateralaxis^{2}=verticalaxis^{2}+omegamarker^{2}, \\quad depthaxis^{2}=verticalaxis^{2}+zetamarker^{2}\n\\]\n\nTheir four curves of intersection are given by\n\\[\nlateralaxis=\\pm \\sqrt{verticalaxis^{2}+omegamarker^{2}}, \\quad depthaxis=\\pm \\sqrt{verticalaxis^{2}+zetamarker^{2}}\n\\]\n\nOne of these curves passes through \\( (0, omegamarker, zetamarker) \\). If \\( omegamarker=0 \\) but \\( zetamarker \\neq 0 \\), two of these curves pass through each of the points \\( (0,0, \\pm zetamarker) \\). Since these points are not in the domain of the original differential system, in this case each of the four curves (3) breaks into two maximal solution curves after removing these points.\nNow we determine the surface formed by the solution curves which intersect the circle \\( C: lateralaxis^{2}+depthaxis^{2}=1, verticalaxis=0 \\). These solutions are evidently given by (2) with \\( omegamarker^{2}+zetamarker^{2}=1, omegamarker zetamarker \\neq 0 \\), and therefore they lie on the surface given by\n\\[\nlateralaxis^{2}+depthaxis^{2}=2 verticalaxis^{2}+1\n\\]\n\nThis is the equation of a hyperboloid of one sheet rotationally symmetric about the \\( verticalaxis \\)-axis. Since four points of \\( C \\) must be excluded from consideration, we expect that only part of this hyperboloid is generated by the solution curves through \\( C \\). Indeed, if \\( (verticalaxis, lateralaxis, depthaxis) \\) lies on (2) with \\( omegamarker^{2} \\) and \\( zetamarker^{2} \\) positive, then\n\\[\nlateralaxis^{2}>verticalaxis^{2}, \\quad depthaxis^{2}>verticalaxis^{2} .\n\\]\n\nConversely, if \\( ( verticalaxis, lateralaxis, depthaxis ) \\) satisfies (4) and (5), then positive numbers \\( omegamarker \\) and \\( zetamarker \\) satisfying (2) and \\( omegamarker^{2}+zetamarker^{2}=1 \\) can be chosen. With the appropriate choice of signs, the point \\( (verticalaxis, lateralaxis, depthaxis) \\) will lie on the curve (3), which meets the circle \\( C \\) at one of the points \\( (0, \\pm omegamarker, \\pm zetamarker) \\). Thus the required surface is given by the equation (4) and the inequalities (5).\n\nThe four quadrants of \\( C \\) each generate a portion of the required surface and these portions are bounded by the exceptional curves\n\\[\nlateralaxis=\\pm verticalaxis, \\quad depthaxis=\\pm \\sqrt{verticalaxis^{2}+1} \\text { and } lateralaxis=\\pm \\sqrt{verticalaxis^{2}+1}, \\quad depthaxis=\\pm verticalaxis\n\\]\n\nHyperboloid: \\( lateralaxis^{2}+depthaxis^{2}=2 verticalaxis^{2}+1 \\)\n\\( A(0,1,0) \\) and \\( B(0,0,1) \\) are excluded points. Curves \\( A C \\) and \\( B D \\) are exceptional solution curves. Curves \\( E F \\), \\( G H \\), etc., are ordinary solution curves.\n\nThe solution curves through the illustrated quadrant of \\( C \\) generate that portion of the hyperboloid that lies between the exceptional solution curves."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "z": "mnlpftoe",
        "c_1": "fbquxels",
        "c_2": "rdzvojki",
        "\\alpha": "vsrhkeum",
        "\\beta": "ixmuqzan"
      },
      "question": "2. Find the surface generated by the solutions of\n\\[\n\\frac{d qzxwvtnp}{hjgrksla mnlpftoe}=\\frac{d hjgrksla}{mnlpftoe qzxwvtnp}=\\frac{d mnlpftoe}{qzxwvtnp hjgrksla}\n\\]\nwhich intersects the circle \\( hjgrksla^{2}+mnlpftoe^{2}=1, qzxwvtnp=0 \\).",
      "solution": "Solution. The given differential equations mean that at the point \\( (qzxwvtnp, hjgrksla, mnlpftoe) \\) we assign a line element having the direction numbers \\( hjgrksla mnlpftoe, mnlpftoe qzxwvtnp \\), and \\( qzxwvtnp hjgrksla \\), and we seek curves having at each point the assigned line element as tangent. At points of the coordinate axes, however, no line element is assigned, since \\( 0,0,0 \\) is not a set of direction numbers. Furthermore, the direction field cannot be extended to the coordinate axes by continuity, since the line elements assigned on a coordinate plane are always perpendicular to that plane. On the rest of space, however, the equations assign an evidently smooth family of line elements, and so through each point of this domain passes a unique maximal solution curve. We shall determine these curves.\n\nMultiply the given equations by \\( 2 qzxwvtnp hjgrksla mnlpftoe \\) to get\n\\[\n2 qzxwvtnp d qzxwvtnp = 2 hjgrksla d hjgrksla = 2 mnlpftoe d mnlpftoe\n\\]\n\nThen integration yields\n\\[\nqzxwvtnp^{2} = hjgrksla^{2} + fbquxels = mnlpftoe^{2} + rdzvojki\n\\]\n\nIn general, these equations represent the intersection of two hyperbolic cylinders which falls into four connected smooth curves, each of which is a maximal solution curve of the differential system. For example, with the choice \\( fbquxels = -vsrhkeum^{2}, rdzvojki = -ixmuqzan^{2}, vsrhkeum ixmuqzan \\neq 0 \\), the surfaces (1) pass through the point \\( (0, vsrhkeum, ixmuqzan) \\) and have the form\n\\[\nhjgrksla^{2} = qzxwvtnp^{2} + vsrhkeum^{2}, \\quad mnlpftoe^{2} = qzxwvtnp^{2} + ixmuqzan^{2}\n\\]\n\nTheir four curves of intersection are given by\n\\[\nhjgrksla = \\pm \\sqrt{qzxwvtnp^{2}+vsrhkeum^{2}}, \\quad mnlpftoe = \\pm \\sqrt{qzxwvtnp^{2}+ixmuqzan^{2}}\n\\]\n\nOne of these curves passes through \\( (0, vsrhkeum, ixmuqzan) \\). If \\( vsrhkeum = 0 \\) but \\( ixmuqzan \\neq 0 \\), two of these curves pass through each of the points \\( (0,0, \\pm ixmuqzan) \\). Since these points are not in the domain of the original differential system, in this case each of the four curves (3) breaks into two maximal solution curves after removing these points.\n\nNow we determine the surface formed by the solution curves which intersect the circle \\( C: hjgrksla^{2}+mnlpftoe^{2} = 1, qzxwvtnp = 0 \\). These solutions are evidently given by (2) with \\( vsrhkeum^{2} + ixmuqzan^{2} = 1, vsrhkeum ixmuqzan \\neq 0 \\), and therefore they lie on the surface given by\n\\[\nhjgrksla^{2} + mnlpftoe^{2} = 2 qzxwvtnp^{2} + 1\n\\]\n\nThis is the equation of a hyperboloid of one sheet rotationally symmetric about the \\( qzxwvtnp \\)-axis. Since four points of \\( C \\) must be excluded from consideration, we expect that only part of this hyperboloid is generated by the solution curves through \\( C \\). Indeed, if \\( (qzxwvtnp, hjgrksla, mnlpftoe) \\) lies on (2) with \\( vsrhkeum^{2} \\) and \\( ixmuqzan^{2} \\) positive, then\n\\[\nhjgrksla^{2} > qzxwvtnp^{2}, \\quad mnlpftoe^{2} > qzxwvtnp^{2}.\n\\]\n\nConversely, if \\( (qzxwvtnp, hjgrksla, mnlpftoe) \\) satisfies (4) and (5), then positive numbers \\( vsrhkeum \\) and \\( ixmuqzan \\) satisfying (2) and \\( vsrhkeum^{2} + ixmuqzan^{2} = 1 \\) can be chosen. With the appropriate choice of signs, the point \\( (qzxwvtnp, hjgrksla, mnlpftoe) \\) will lie on the curve (3), which meets the circle \\( C \\) at one of the points \\( (0, \\pm vsrhkeum, \\pm ixmuqzan) \\). Thus the required surface is given by the equation (4) and the inequalities (5).\n\nThe four quadrants of \\( C \\) each generate a portion of the required surface and these portions are bounded by the exceptional curves\n\\[\nhjgrksla = \\pm qzxwvtnp, \\; mnlpftoe = \\pm \\sqrt{qzxwvtnp^{2}+1} \\text{ and } \\; hjgrksla = \\pm \\sqrt{qzxwvtnp^{2}+1}, \\; mnlpftoe = \\pm qzxwvtnp\n\\]\n\nHyperboloid: \\( hjgrksla^{2}+mnlpftoe^{2} = 2 qzxwvtnp^{2} + 1 \\)\n\\( A(0,1,0) \\) and \\( B(0,0,1) \\) are excluded points. Curves \\( A C \\) and \\( B D \\) are exceptional solution curves. Curves \\( E F \\), \\( G H \\), etc., are ordinary solution curves.\n\nThe solution curves through the illustrated quadrant of \\( C \\) generate that portion of the hyperboloid that lies between the exceptional solution curves."
    },
    "kernel_variant": {
      "question": "Let the autonomous system of first–order differential equations in ℝ³  \n\n  dx /(yz)=dy /(zx)=dz /(xy)  (S)\n\nbe interpreted as a direction field.  At every point that has at most one vanishing coordinate the three numbers (yz , zx , xy) determine a non–zero tangent direction, whereas on the three coordinate axes \n   x=y=0  (z–axis), y=z=0  (x–axis), z=x=0  (y–axis)\nall three direction numbers are 0 and the field is undefined.\n\nDescribe completely the surface that is swept out by all maximal integral curves of (S) which meet the circle\n\n  𝒞 : x=1, y²+z²=4.\n\n(“Describe completely” means: give explicit equations and/or inequalities that characterise exactly the points that are reached – no more and no less.)",
      "solution": "1.  The vector field and two first integrals\nThe differential system (S) is equivalent to the smooth vector field\n        F(x,y,z)=(yz, zx, xy).\nExcept on the three coordinate axes F never vanishes, so through every point of the domain Ω := ℝ³\\{axes} there passes one and only one maximal integral curve.\n\nFor the two functions\n        Φ₁(x,y,z)=x²−y²,  Φ₂(x,y,z)=x²−z²\nwe have\n        ∇Φ₁·F=(2x,−2y,0)·(yz,zx,xy)=0,\n        ∇Φ₂·F=(2x,0,−2z)·(yz,zx,xy)=0.\nHence both quantities stay constant along each integral curve.  Denoting the (constant) values by a and b we obtain the two first-integral relations\n        x²−y²=a,  x²−z²=b  (1)\nwhich are valid on every maximal solution.\n\nConversely, for every ordered pair (a,b)∈ℝ² the two quadratic cylinders (1) intersect in (up to a change of sign in y and z) the four space curves\n        y=±√(x²−a),  z=±√(x²−b)  (2)\nall of which are integral curves of the system.  Thus (2) with a unique pair (a,b) parametrises every maximal solution.\n\n2.  Parameters of the curves through 𝒞\nA point of the circle 𝒞 has coordinates (1,y₀,z₀) with y₀²+z₀²=4.  Substituting x=1 in (1) gives\n        a=1−y₀²,  b=1−z₀².  (3)\nBecause y₀²+z₀²=4 we have\n        a+b = 1−y₀² + 1−z₀² = −2.  (4)\nMoreover 0≤y₀²,z₀²≤4 implies\n        −3≤a≤1,  −3≤b≤1.  (5)\nConversely, every pair (a,b) satisfying (4) and (5) produces real numbers\n        y₀=±√(1−a), z₀=±√(1−b)\nwith y₀²+z₀²=4, so the corresponding integral curves (2) do meet 𝒞.  Hence the relevant parameters are exactly the points of the line segment\n        L := {(a,b)∈ℝ² | a+b=−2, −3≤a≤1}.  (6)\n\n3.  A first description of the swept surface\nAdding the two equalities in (1) and using (4) we obtain, along every curve with parameters in L,\n        y²+z² = 2x² −(a+b) = 2x² +2.  (7)\nThus all such curves lie on the one-sheeted hyperboloid\n        H : y²+z² = 2x² +2.  (8)\nThe bounds (5) yield further inequalities.  From a=x²−y² we get\n        −3≤x²−y²≤1  ⇔  y² ≥ x²−1.  (9)\nThe analogous statement with b gives\n        z² ≥ x²−1.  (10)\n\n4.  Elimination of the axis points\nConditions (8)–(10) describe the set\n        S₀ := {(x,y,z)∈ℝ³ | y²+z² = 2x²+2, y² ≥ x²−1, z² ≥ x²−1}.  (11)\nAlthough (11) contains all points reached by the required integral curves, it also contains four extra points\n        P₁=(0, 0, √2), P₂=(0, 0,−√2), P₃=(0, √2, 0), P₄=(0,−√2, 0),\nlying on the coordinate axes.  At each of them two coordinates vanish, hence F(Pᵢ)=0 and the direction field is undefined.  An integral curve with parameters (a,b)∈L approaches such a point as x→0, but the maximal curve stops beforehand and never passes through it.  Therefore these four axis points have to be removed.\n\nEquivalently, one may require that if x=0 then y·z≠0, or simply delete the four points listed above.\n\n5.  Exhaustion of the reduced surface\nDefine\n        S := S₀ \\ {P₁,P₂,P₃,P₄}.\nLet (x,y,z)∈S.  Putting\n        a := x²−y²,  b := x²−z²\nwe still have a+b=−2 and −3≤a,b≤1; hence (a,b)∈L, and (2) with the choice of signs matching the signs of y and z gives an integral curve that contains both (x,y,z) and a point of 𝒞.  Consequently every point of S is indeed reached by at least one required integral curve, and by construction no other point is.\n\n6.  Final description\nThe set swept out by all integral curves of (S) that intersect the circle x=1, y²+z²=4 is\n        S = { (x,y,z)∈ℝ³ | y²+z² = 2x²+2,\n                              y² ≥ x²−1,\n                              z² ≥ x²−1 } \\ { (0,0,±√2), (0,±√2,0) }.\nA convenient equivalent formulation is\n        y²+z² = 2x²+2, y² ≥ x²−1, z² ≥ x²−1,\n        and if x=0 then y·z ≠ 0.\nGeometrically: it is that portion of the one-sheeted hyperboloid (8) which lies outside the two coaxial cones y²=x²−1 and z²=x²−1 (both real only for |x|≥1) and which, in the cross-section x=0, omits the two coordinate axes.  The four ruling curves where one of the inequalities becomes an equality form its boundary.",
      "_meta": {
        "core_steps": [
          "Clear denominators to get proportionality 2x dx = 2y dy = 2z dz",
          "Integrate → invariants  x² – y² = c₁  and  x² – z² = c₂",
          "Rewrite integral curves as intersections  y² = x² + α² ,  z² = x² + β²",
          "Use the circle condition at the chosen plane to fix  α² + β² = R²",
          "Eliminate α,β to obtain the hyperboloid  y² + z² = 2x² + R²  (with domain exclusions)"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Radius-squared of the circle that the solution curves must meet",
            "original": "1  (in  y² + z² = 1)"
          },
          "slot2": {
            "description": "Location of the circle along the x-axis (plane in which it lies)",
            "original": "x = 0"
          },
          "slot3": {
            "description": "Scalar factor used when multiplying the differential equations before integration (e.g. 2 in 2xyz)",
            "original": "2"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}