{
  "index": "1942-A-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "4. Find the orthogonal trajectories of the family of conics \\( (x+2 y)^{2} \\) \\( =a(x+y) \\). At what angle do the curves of one family cut the curves of the other family at the origin?",
  "solution": "Solution. The given family is a family of parabolas all tangent to the line \\( x+y=0 \\) at the origin. For \\( a=0 \\) the parabola degenerates to the double line \\( (x+2 y)^{2}=0 \\) which should be viewed as two degenerate parabolas, the ray in the fourth quadrant being the limiting case as \\( a \\) goes to zero through positive values and the ray in the second quadrant being the limiting parabola as \\( a \\) goes to zero through negative values.\n\nTo find the differential equation of the family we differentiate the given equation and eliminate \\( a \\) between the original equation and its derivative.\n\\[\n\\begin{array}{c}\n(x+2 y)^{2}=a(x+y) \\\\\n2(x+2 y)\\left(1+2 y^{\\prime}\\right)=a\\left(1+y^{\\prime}\\right)\n\\end{array}\n\\]\n\nWe get\n\\[\n2(x+y)(x+2 y)\\left(1+2 y^{\\prime}\\right)=(x+2 y)^{2}\\left(1+y^{\\prime}\\right)\n\\]\nwhich simplifies to\n\\[\n(3 x+2 y) y^{\\prime}+x=0\n\\]\n\nThe factor \\( x+2 y \\) that was cancelled reflects the degeneracy along the line \\( x+2 y=0 \\).\n\nThis differential equation is defined along the line \\( x+y=0 \\) (where the original family of parabolas has no members), so in effect the line \\( x+y=0 \\) is another degenerate member of the family corresponding to the case \\( a=\\infty \\).\n\nThe orthogonal trajectories are obtained by integrating the differential equation\n\\[\nx y^{\\prime}=3 x+2 y\n\\]\n\nWe write this as\n\\[\n\\frac{d}{d x}(y+3 x)=\\frac{2(y+3 x)}{x}\n\\]\n\nThe solution is\n\\[\ny+3 x=k x^{2}\n\\]\nwhere \\( k \\) is an arbitrary constant. This is a new family of parabolas, with a unique member through every point of the plane except for points on the \\( y \\)-axis. The \\( y \\)-axis is an integral curve of differential equation (4) rewritten in the form\n\\[\nx=(3 x+2 y) \\frac{d x}{d y}\n\\]\nso that the \\( y \\)-axis also belongs to the family of orthogonal trajectories.\nAll the curves in the new family (5) are tangent to the line \\( y+3 x=0 \\) at the origin (except the degenerate double parabola made by the \\( y \\)-axis).\n\nThe angle between the two families at the origin is then the angle \\( \\theta \\) between the two lines \\( x+y=0 \\) and \\( 3 x+y=0 \\).\n\nUsing the slopes we get\n\\[\n\\tan \\theta=\\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}=\\frac{-1-(-3)}{(1+(-1)(-3)}=\\frac{2}{4}=\\frac{1}{2}\n\\]\n\nHence \\( \\theta=\\arctan \\left(\\frac{1}{2}\\right) \\).\nRemark. If the degenerate cases are allowed, the answer is not unique and there will be other angles. For example, the angle between the \\( y \\)-axis (degenerate member of the orthogonal family) and \\( x+2 y=0 \\) (degenerate member of the second family) is \\( \\arctan 2 \\).",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "a",
    "k",
    "m_1",
    "m_2",
    "\\\\theta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscis",
        "y": "ordinate",
        "a": "shapeprm",
        "k": "scaleprm",
        "m_1": "slopeone",
        "m_2": "slopetwo",
        "\\theta": "angleval"
      },
      "question": "4. Find the orthogonal trajectories of the family of conics \\( (abscis+2\\,ordinate)^{2}=shapeprm(abscis+ordinate) \\). At what angle do the curves of one family cut the curves of the other family at the origin?",
      "solution": "Solution. The given family is a family of parabolas all tangent to the line \\( abscis+ordinate=0 \\) at the origin. For \\( shapeprm=0 \\) the parabola degenerates to the double line \\( (abscis+2\\,ordinate)^{2}=0 \\) which should be viewed as two degenerate parabolas, the ray in the fourth quadrant being the limiting case as \\( shapeprm \\) goes to zero through positive values and the ray in the second quadrant being the limiting parabola as \\( shapeprm \\) goes to zero through negative values.\n\nTo find the differential equation of the family we differentiate the given equation and eliminate \\( shapeprm \\) between the original equation and its derivative.\n\\[\n\\begin{array}{c}\n(abscis+2\\,ordinate)^{2}=shapeprm(abscis+ordinate) \\\\\n2(abscis+2\\,ordinate)\\left(1+2\\,ordinate^{\\prime}\\right)=shapeprm\\left(1+ordinate^{\\prime}\\right)\n\\end{array}\n\\]\n\nWe get\n\\[\n2(abscis+ordinate)(abscis+2\\,ordinate)\\left(1+2\\,ordinate^{\\prime}\\right)=(abscis+2\\,ordinate)^{2}\\left(1+ordinate^{\\prime}\\right)\n\\]\nwhich simplifies to\n\\[\n(3\\,abscis+2\\,ordinate)\\,ordinate^{\\prime}+abscis=0\n\\]\n\nThe factor \\( abscis+2\\,ordinate \\) that was cancelled reflects the degeneracy along the line \\( abscis+2\\,ordinate=0 \\).\n\nThis differential equation is defined along the line \\( abscis+ordinate=0 \\) (where the original family of parabolas has no members), so in effect the line \\( abscis+ordinate=0 \\) is another degenerate member of the family corresponding to the case \\( shapeprm=\\infty \\).\n\nThe orthogonal trajectories are obtained by integrating the differential equation\n\\[\nabscis\\,ordinate^{\\prime}=3\\,abscis+2\\,ordinate\n\\]\n\nWe write this as\n\\[\n\\frac{d}{d\\,abscis}(ordinate+3\\,abscis)=\\frac{2(ordinate+3\\,abscis)}{abscis}\n\\]\n\nThe solution is\n\\[\nordinate+3\\,abscis=scaleprm\\,abscis^{2}\n\\]\nwhere \\( scaleprm \\) is an arbitrary constant. This is a new family of parabolas, with a unique member through every point of the plane except for points on the ordinate-axis. The ordinate-axis is an integral curve of differential equation (4) rewritten in the form\n\\[\nabscis=(3\\,abscis+2\\,ordinate)\\frac{d\\,abscis}{d\\,ordinate}\n\\]\nso that the ordinate-axis also belongs to the family of orthogonal trajectories.\nAll the curves in the new family (5) are tangent to the line \\( ordinate+3\\,abscis=0 \\) at the origin (except the degenerate double parabola made by the ordinate-axis).\n\nThe angle between the two families at the origin is then the angle \\( angleval \\) between the two lines \\( abscis+ordinate=0 \\) and \\( 3\\,abscis+ordinate=0 \\).\n\nUsing the slopes we get\n\\[\n\\tan angleval=\\frac{slopeone-slopetwo}{1+slopeone slopetwo}=\\frac{-1-(-3)}{1+(-1)(-3)}=\\frac{2}{4}=\\frac{1}{2}\n\\]\n\nHence \\( angleval=\\arctan\\left(\\frac{1}{2}\\right) \\).\n\nRemark. If the degenerate cases are allowed, the answer is not unique and there will be other angles. For example, the angle between the ordinate-axis (degenerate member of the orthogonal family) and \\( abscis+2\\,ordinate=0 \\) (degenerate member of the second family) is \\( \\arctan 2 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "orangetree",
        "y": "sapphire",
        "a": "partridge",
        "k": "lamplight",
        "m_1": "riverbed",
        "m_2": "cobblestone",
        "\\theta": "pendulum"
      },
      "question": "4. Find the orthogonal trajectories of the family of conics \\( (orangetree+2 sapphire)^{2} \\) \\( =partridge(orangetree+sapphire) \\). At what angle do the curves of one family cut the curves of the other family at the origin?",
      "solution": "Solution. The given family is a family of parabolas all tangent to the line \\( orangetree+sapphire=0 \\) at the origin. For \\( partridge=0 \\) the parabola degenerates to the double line \\( (orangetree+2 sapphire)^{2}=0 \\) which should be viewed as two degenerate parabolas, the ray in the fourth quadrant being the limiting case as \\( partridge \\) goes to zero through positive values and the ray in the second quadrant being the limiting parabola as \\( partridge \\) goes to zero through negative values.\n\nTo find the differential equation of the family we differentiate the given equation and eliminate \\( partridge \\) between the original equation and its derivative.\n\\[\n\\begin{array}{c}\n(orangetree+2 sapphire)^{2}=partridge(orangetree+sapphire) \\\\\n2(orangetree+2 sapphire)\\left(1+2 sapphire^{\\prime}\\right)=partridge\\left(1+sapphire^{\\prime}\\right)\n\\end{array}\n\\]\n\nWe get\n\\[\n2(orangetree+sapphire)(orangetree+2 sapphire)\\left(1+2 sapphire^{\\prime}\\right)=(orangetree+2 sapphire)^{2}\\left(1+sapphire^{\\prime}\\right)\n\\]\nwhich simplifies to\n\\[\n(3 orangetree+2 sapphire) sapphire^{\\prime}+orangetree=0\n\\]\n\nThe factor \\( orangetree+2 sapphire \\) that was cancelled reflects the degeneracy along the line \\( orangetree+2 sapphire=0 \\).\n\nThis differential equation is defined along the line \\( orangetree+sapphire=0 \\) (where the original family of parabolas has no members), so in effect the line \\( orangetree+sapphire=0 \\) is another degenerate member of the family corresponding to the case \\( partridge=\\infty \\).\n\nThe orthogonal trajectories are obtained by integrating the differential equation\n\\[\norangetree\\, sapphire^{\\prime}=3 orangetree+2 sapphire\n\\]\n\nWe write this as\n\\[\n\\frac{d}{d orangetree}(sapphire+3 orangetree)=\\frac{2(sapphire+3 orangetree)}{orangetree}\n\\]\n\nThe solution is\n\\[\nsapphire+3 orangetree=lamplight\\, orangetree^{2}\n\\]\nwhere \\( lamplight \\) is an arbitrary constant. This is a new family of parabolas, with a unique member through every point of the plane except for points on the \\( sapphire \\)-axis. The \\( sapphire \\)-axis is an integral curve of differential equation (4) rewritten in the form\n\\[\norangetree=(3 orangetree+2 sapphire) \\frac{d orangetree}{d sapphire}\n\\]\nso that the \\( sapphire \\)-axis also belongs to the family of orthogonal trajectories.\nAll the curves in the new family (5) are tangent to the line \\( sapphire+3 orangetree=0 \\) at the origin (except the degenerate double parabola made by the \\( sapphire \\)-axis).\n\nThe angle between the two families at the origin is then the angle \\( pendulum \\) between the two lines \\( orangetree+sapphire=0 \\) and \\( 3 orangetree+sapphire=0 \\).\n\nUsing the slopes we get\n\\[\n\\tan pendulum=\\frac{riverbed-cobblestone}{1+riverbed\\, cobblestone}=\\frac{-1-(-3)}{(1+(-1)(-3))}=\\frac{2}{4}=\\frac{1}{2}\n\\]\n\nHence \\( pendulum=\\arctan \\left(\\frac{1}{2}\\right) \\).\nRemark. If the degenerate cases are allowed, the answer is not unique and there will be other angles. For example, the angle between the \\( sapphire \\)-axis (degenerate member of the orthogonal family) and \\( orangetree+2 sapphire=0 \\) (degenerate member of the second family) is \\( \\arctan 2 \\ )."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticality",
        "y": "horizontality",
        "a": "mutability",
        "k": "impermanence",
        "m_1": "flatnessone",
        "m_2": "flatnesstwo",
        "\\theta": "straightness"
      },
      "question": "4. Find the orthogonal trajectories of the family of conics \\( (verticality+2 horizontality)^{2} \\) \\( =mutability(verticality+horizontality) \\). At what angle do the curves of one family cut the curves of the other family at the origin?",
      "solution": "Solution. The given family is a family of parabolas all tangent to the line \\( verticality+horizontality=0 \\) at the origin. For \\( mutability=0 \\) the parabola degenerates to the double line \\( (verticality+2 horizontality)^{2}=0 \\) which should be viewed as two degenerate parabolas, the ray in the fourth quadrant being the limiting case as \\( mutability \\) goes to zero through positive values and the ray in the second quadrant being the limiting parabola as \\( mutability \\) goes to zero through negative values.\n\nTo find the differential equation of the family we differentiate the given equation and eliminate \\( mutability \\) between the original equation and its derivative.\n\\[\n\\begin{array}{c}\n(verticality+2 horizontality)^{2}=mutability(verticality+horizontality) \\\\\n2(verticality+2 horizontality)\\left(1+2 horizontality^{\\prime}\\right)=mutability\\left(1+horizontality^{\\prime}\\right)\n\\end{array}\n\\]\n\nWe get\n\\[\n2(verticality+horizontality)(verticality+2 horizontality)\\left(1+2 horizontality^{\\prime}\\right)=(verticality+2 horizontality)^{2}\\left(1+horizontality^{\\prime}\\right)\n\\]\nwhich simplifies to\n\\[\n(3 verticality+2 horizontality) horizontality^{\\prime}+verticality=0\n\\]\n\nThe factor \\( verticality+2 horizontality \\) that was cancelled reflects the degeneracy along the line \\( verticality+2 horizontality=0 \\).\n\nThis differential equation is defined along the line \\( verticality+horizontality=0 \\) (where the original family of parabolas has no members), so in effect the line \\( verticality+horizontality=0 \\) is another degenerate member of the family corresponding to the case \\( mutability=\\infty \\).\n\nThe orthogonal trajectories are obtained by integrating the differential equation\n\\[\nverticality\\ horizontality^{\\prime}=3 verticality+2 horizontality\n\\]\n\nWe write this as\n\\[\n\\frac{d}{d verticality}(horizontality+3 verticality)=\\frac{2(horizontality+3 verticality)}{verticality}\n\\]\n\nThe solution is\n\\[\nhorizontality+3 verticality=impermanence\\ verticality^{2}\n\\]\nwhere \\( impermanence \\) is an arbitrary constant. This is a new family of parabolas, with a unique member through every point of the plane except for points on the \\( horizontality \\)-axis. The \\( horizontality \\)-axis is an integral curve of differential equation (4) rewritten in the form\n\\[\nverticality=(3 verticality+2 horizontality) \\frac{d verticality}{d horizontality}\n\\]\nso that the \\( horizontality \\)-axis also belongs to the family of orthogonal trajectories.\nAll the curves in the new family (5) are tangent to the line \\( horizontality+3 verticality=0 \\) at the origin (except the degenerate double parabola made by the \\( horizontality \\)-axis).\n\nThe angle between the two families at the origin is then the angle \\( straightness \\) between the two lines \\( verticality+horizontality=0 \\) and \\( 3 verticality+horizontality=0 \\).\n\nUsing the slopes we get\n\\[\n\\tan straightness=\\frac{flatnessone-flatnesstwo}{1+flatnessone flatnesstwo}=\\frac{-1-(-3)}{(1+(-1)(-3)}=\\frac{2}{4}=\\frac{1}{2}\n\\]\n\nHence \\( straightness=\\arctan \\left(\\frac{1}{2}\\right) \\).\nRemark. If the degenerate cases are allowed, the answer is not unique and there will be other angles. For example, the angle between the \\( horizontality \\)-axis (degenerate member of the orthogonal family) and \\( verticality+2 horizontality=0 \\) (degenerate member of the second family) is \\( \\arctan 2 \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "a": "fbsnqryw",
        "k": "lmpdtjra",
        "m_1": "rvgxkzse",
        "m_2": "teofwlaq",
        "\\theta": "wopranty"
      },
      "question": "4. Find the orthogonal trajectories of the family of conics \\( (qzxwvtnp+2 hjgrksla)^{2} \\) \\( =fbsnqryw(qzxwvtnp+hjgrksla) \\). At what angle do the curves of one family cut the curves of the other family at the origin?",
      "solution": "Solution. The given family is a family of parabolas all tangent to the line \\( qzxwvtnp+hjgrksla=0 \\) at the origin. For \\( fbsnqryw=0 \\) the parabola degenerates to the double line \\( (qzxwvtnp+2 hjgrksla)^{2}=0 \\) which should be viewed as two degenerate parabolas, the ray in the fourth quadrant being the limiting case as \\( fbsnqryw \\) goes to zero through positive values and the ray in the second quadrant being the limiting parabola as \\( fbsnqryw \\) goes to zero through negative values.\n\nTo find the differential equation of the family we differentiate the given equation and eliminate \\( fbsnqryw \\) between the original equation and its derivative.\n\\[\n\\begin{array}{c}\n(qzxwvtnp+2 hjgrksla)^{2}=fbsnqryw(qzxwvtnp+hjgrksla) \\\\\n2(qzxwvtnp+2 hjgrksla)\\left(1+2 hjgrksla^{\\prime}\\right)=fbsnqryw\\left(1+hjgrksla^{\\prime}\\right)\n\\end{array}\n\\]\n\nWe get\n\\[\n2(qzxwvtnp+hjgrksla)(qzxwvtnp+2 hjgrksla)\\left(1+2 hjgrksla^{\\prime}\\right)=(qzxwvtnp+2 hjgrksla)^{2}\\left(1+hjgrksla^{\\prime}\\right)\n\\]\nwhich simplifies to\n\\[\n(3 qzxwvtnp+2 hjgrksla) hjgrksla^{\\prime}+qzxwvtnp=0\n\\]\n\nThe factor \\( qzxwvtnp+2 hjgrksla \\) that was cancelled reflects the degeneracy along the line \\( qzxwvtnp+2 hjgrksla=0 \\).\n\nThis differential equation is defined along the line \\( qzxwvtnp+hjgrksla=0 \\) (where the original family of parabolas has no members), so in effect the line \\( qzxwvtnp+hjgrksla=0 \\) is another degenerate member of the family corresponding to the case \\( fbsnqryw=\\infty \\).\n\nThe orthogonal trajectories are obtained by integrating the differential equation\n\\[\nqzxwvtnp hjgrksla^{\\prime}=3 qzxwvtnp+2 hjgrksla\n\\]\n\nWe write this as\n\\[\n\\frac{d}{d qzxwvtnp}(hjgrksla+3 qzxwvtnp)=\\frac{2(hjgrksla+3 qzxwvtnp)}{qzxwvtnp}\n\\]\n\nThe solution is\n\\[\nhjgrksla+3 qzxwvtnp=lmpdtjra qzxwvtnp^{2}\n\\]\nwhere \\( lmpdtjra \\) is an arbitrary constant. This is a new family of parabolas, with a unique member through every point of the plane except for points on the \\( hjgrksla \\)-axis. The \\( hjgrksla \\)-axis is an integral curve of differential equation (4) rewritten in the form\n\\[\nqzxwvtnp=(3 qzxwvtnp+2 hjgrksla) \\frac{d qzxwvtnp}{d hjgrksla}\n\\]\nso that the \\( hjgrksla \\)-axis also belongs to the family of orthogonal trajectories.\nAll the curves in the new family (5) are tangent to the line \\( hjgrksla+3 qzxwvtnp=0 \\) at the origin (except the degenerate double parabola made by the \\( hjgrksla \\)-axis).\n\nThe angle between the two families at the origin is then the angle \\( wopranty \\) between the two lines \\( qzxwvtnp+hjgrksla=0 \\) and \\( 3 qzxwvtnp+hjgrksla=0 \\).\n\nUsing the slopes we get\n\\[\n\\tan wopranty=\\frac{rvgxkzse-teofwlaq}{1+rvgxkzse teofwlaq}=\\frac{-1-(-3)}{(1+(-1)(-3)}=\\frac{2}{4}=\\frac{1}{2}\n\\]\n\nHence \\( wopranty=\\arctan \\left(\\frac{1}{2}\\right) \\).\nRemark. If the degenerate cases are allowed, the answer is not unique and there will be other angles. For example, the angle between the \\( hjgrksla \\)-axis (degenerate member of the orthogonal family) and \\( qzxwvtnp+2 hjgrksla=0 \\) (degenerate member of the second family) is \\( \\arctan 2 \\)."
    },
    "kernel_variant": {
      "question": "Let  \n\n  u = x + 6y + 2z ,  v = x + 2y + z ,      \n  A = ∇u = (1,6,2) ,  B = ∇v = (1,2,1).                                         (†)  \n\nand consider the one–parameter family of algebraic surfaces  \n\n  (x + 6y + 2z)³ = a (x + 2y + z) ,    a ∈ ℝ.                                (★)\n\nThroughout Parts A–C assume v = x + 2y + z ≠ 0 unless stated otherwise, so that the right-hand side of (★) does not vanish.\n\nA.  Put F(x,y,z,a)=u³–av and show that simultaneous elimination of the parameter a from  \n\n  F = 0 and dF = 0  \n\nforces every non-degenerate member of (★) to possess the vector field  \n\n    C(x,y,z) = 3v A – u B = (3v–u, 18v–2u, 6v–u)                         (1)\n\nas a field of normals (i.e. C is everywhere perpendicular to the surface).  \nEquivalently, if Φ(x,y,z) is any potential of the family (★) one must have  \n\n    ∇Φ × C = 0,                                                       (2)\n\nor, written as two independent first–order linear PDE’s,  \n\n    (18v – 2u) Φ_x – (3v – u) Φ_y = 0 ,                             (2a)  \n    ( 6v –  u) Φ_x – (3v – u) Φ_z = 0 .                             (2b)\n\n(i) Show that in passing from (★) to (2) the entire plane  \n\n    Π : v = x + 2y + z = 0                                      (3)\n\nis lost, although Π is still an integral surface of the Pfaffian system generated by C.  \n\n(ii) Give the precise geometric reason for the loss, and verify explicitly that C is everywhere normal to Π (hence Π also satisfies ∇Φ × C = 0 when the potential is chosen as Φ = v).\n\nB.  A smooth surface Σ is called an orthogonal trajectory of the family (★) if, at each intersection point, the normals of Σ and the corresponding member of (★) are perpendicular.\n\n(i) For a trajectory written implicitly as G(x,y,z)=g(u,v)=0, prove that the orthogonality condition C·∇G = 0 reduces to the autonomous linear PDE  \n\n    (123 − 15 u/v) g_u + (45 − 6 u/v) g_v = 0 .                     (4)\n\n(ii) Solve (4) by the method of characteristics and show that every orthogonal trajectory can be expressed in closed form as  \n\n    (41v² − 20uv + 2u²)^{1/2} ·  \n    [(41v − (10 + √18)u)/(41v − (10 − √18)u)]^{5/√72} = C ,      (5)\n\n   where C is a non-zero real constant (either branch of the logarithm may be used).\n\nC.  Consider the concrete member  \n\n    S : (x + 6y + 2z)³ = x + 2y + z                            (the case a = 1)\n\nwhich passes through the point  \n\n    P = (1,0,0)   (u(P) = v(P) = 1).\n\nPutting u = v = 1 in (5) gives  \n\n    C₀ = √23 · [(31 − √18)/(31 + √18)]^{5/√72}.                   (6)\n\nDenote by  \n\n    Σ : (41v² − 20uv + 2u²)^{1/2} ·  \n      [(41v − (10 + √18)u)/(41v − (10 − √18)u)]^{5/√72} = C₀     (7)\n\nthe unique orthogonal trajectory that contains P.\n\n(i) Show directly that S and Σ meet orthogonally at P and state the angle of intersection.\n\n(ii) Let γ be the intersection curve S ∩ Σ with γ(0)=P, parameterised by arc length s and oriented so that γ′(0) points in the direction (2,1,–4).  Compute the curvature κ(0) and torsion τ(0) of γ at P.\n\n\n\n───────────────────────────────────────────────────────────────",
      "solution": "Pre-computed scalar products of the constant vectors in (†):  \n\n |A|² = 41 , A·B = 15 , |B|² = 6 , A×B = (2,1,−4).                       (8)\n\n\n\n--------------------------------------------------------------------  \nA.  Differential system of the family  \n--------------------------------------------------------------------  \nLet  \n\n  F(x,y,z,a)=u³−av.                                                       (9)\n\nDifferentiating,\n\n  dF = 3u² du − a dv.                                                   (10)\n\nBecause v ≠ 0 we may eliminate a by a = u³/v, obtaining  \n\n  3u² du − (u³/v) dv = 0 ⇔ 3v du − u dv = 0.                        (11)\n\nWriting du = A·dx and dv = B·dx gives  \n\n  (3v A − u B)·dx = 0,                                               (12)\n\nso the non-degenerate members of (★) have the normal field  \n\n  C = 3v A − u B,                                                    (1)\n\nand consequently any potential Φ describing the family satisfies  \n\n  ∇Φ × C = 0, equivalently (2a),(2b).\n\n(i)  Lost surface.  The algebraic step a = u³/v is legitimate only for v ≠ 0; therefore the surface Π : v = 0 is discarded when (11) is derived even though Π satisfies (★) with a = 0.\n\n(ii) Geometric reason.  On Π one has v = 0, whence  \n\n  C|_{Π} = 3·0·A − u B = −u B.                                      (13)\n\nSince the normal of Π is B (because ∇v = B), the vector C|_{Π} is everywhere parallel to the normal of Π; hence Π is an integral surface in the sense that its normal field coincides (up to scale) with C.  Choosing the potential Φ = v we obtain ∇Φ = B and  \n\n  ∇Φ × C = B × (−u B) = 0   on Π,                                  (14)\n\nconfirming that Π satisfies (2).  The surface is lost solely because the algebraic elimination required division by v.\n\n\n\n--------------------------------------------------------------------  \nB.  Orthogonal trajectories  \n--------------------------------------------------------------------  \n(i)  Let G(x,y,z)=g(u,v).  Then ∇G = g_u A + g_v B.  Orthogonality with C means  \n\n  C·∇G = (3v A − u B)·(g_u A + g_v B) = 0.                        (15)\n\nUsing (8),\n\n  (3v·41 − u·15) g_u + (3v·15 − u·6) g_v = 0  \n\n  ⇔ (123 v − 15 u) g_u + (45 v − 6 u) g_v = 0.                     (16)\n\nDividing by v (again v≠0) yields the autonomous linear PDE (4).\n\n(ii)  Characteristic integration.  \nIntroduce w = v/u (so v = wu).  From (16) the characteristic ODE is  \n\n  dv/du = (45v − 6u)/(123v − 15u) = (45w − 6)/(123w − 15).         (17)\n\nBecause v = wu, we have dv/du = w + u dw/du, so\n\n  w + u dw/du = (45w − 6)/(123w − 15).                              (18)\n\nSolving for dw/du and separating variables gives  \n\n  u dw/du = −(41w² − 20w + 2)/(41w − 5).                            (19)\n\nHence  \n\n  ∫ (41w − 5)/(41w² − 20w + 2) dw = −∫ du/u.                       (20)\n\nThe left integral equals  \n\n  ½ ln(41w² − 20w + 2) + (5/√72) ln[(41w − 10 − √18)/(41w − 10 + √18)].  \n\nExponentiating and replacing w by v/u one obtains the first integral  \n\n  u (41w² − 20w + 2)^{1/2}  \n   · [(41w − 10 − √18)/(41w − 10 + √18)]^{5/√72} = const,          (21)\n\nwhich, after restoring w = v/u, is exactly the closed form (5).  Each non-zero constant C determines one orthogonal trajectory and every trajectory arises in this way.\n\n\n\n--------------------------------------------------------------------  \nC.  Geometry at P = (1,0,0) (u = v = 1)  \n--------------------------------------------------------------------  \nConstant of the trajectory.  Setting u = v = 1 in (5) gives (6); hence the trajectory through P is Σ given by (7).\n\nNormals at P.  \n• For S (a = 1) we have F(x,y,z) = u³ − v, so  \n\n  ∇F = 3u² A − B; at P: N_S = 3A − B = (2,16,5).                    (22)\n\n• For Σ write  \n\n  ψ(u,v) = ½ ln(41v² − 20uv + 2u²)  \n      + k[ln(41v − 10 − √18 u) − ln(41v − 10 + √18 u)], k = 5/√72. (23)\n\nEquation (7) is ψ = ln C₀, so  \n\n  N_Σ = ∇ψ = ψ_u A + ψ_v B.                                          (24)\n\n(i)  Orthogonality at P.  Substituting u = v = 1 in the autonomous relation (4) satisfied by ψ gives  \n\n  (123 − 15) ψ_u + (45 − 6) ψ_v = 0 ⇒ 108 ψ_u + 39 ψ_v = 0.        (25)\n\nUsing (8),\n\n  N_S·N_Σ = (3|A|² − A·B) ψ_u + (3A·B − |B|²) ψ_v  \n       = 108 ψ_u + 39 ψ_v = 0.                                  (26)\n\nHence N_S ⟂ N_Σ and the two surfaces meet at a right angle, i.e. the angle of intersection is 90°.\n\n(ii)  Curvature and torsion of γ = S ∩ Σ at P.  \nThe unit tangent t of γ is parallel to A×B, hence  \n\n  t = (2,1,−4)/√21, with A·t = B·t = 0.                            (27)\n\nSecond derivatives.  \n• Hess F = 6u A Aᵀ ⇒ tᵀ Hess F t = 6u (A·t)² = 0.  \n• Hess G = Hess ψ = ψ_{uu} A Aᵀ + 2ψ_{uv} A Bᵀ + ψ_{vv} B Bᵀ, and again tᵀ Hess G t = 0 because A·t = B·t = 0.\n\nLet a = γ″(0).  Since γ is arc-length parametrised, t·a = 0, so write  \n\n  a = β N_S + γ N_Σ.                                                (28)\n\nThe second-derivative constraints F(γ)=0 and G(γ)=0 yield  \n\n  N_S·a = −tᵀ Hess F t = 0, N_Σ·a = −tᵀ Hess G t = 0.             (29)\n\nBecause N_S ⟂ N_Σ, equations (29) force β = γ = 0 and hence a = 0.  Therefore  \n\n  κ(0) = |a| = 0,  τ(0) may be taken as 0 (the standard convention when κ = 0). (30)\n\nThus the intersection curve γ has zero curvature at P; to second order it coincides with its tangent line in the direction (2,1,−4).\n\n\n\n───────────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.392967",
        "was_fixed": false,
        "difficulty_analysis": "1. Dimension Increase – The original 2-dimensional curve problem is lifted to a 3-dimensional setting involving surfaces.  Orthogonal trajectories now require handling gradients in ℝ³ and lead to a first-order linear PDE instead of an ODE.\n\n2. Non-trivial Invariants – Solving the PDE demands the method of characteristics in the variables (u,v) and an intricate rational integral that produces logarithmic expressions with quadratic discriminant √72.\n\n3. Advanced Geometry – Part C asks not only for the angle of intersection but also for curvature and torsion of the space curve defined by the intersection of two surfaces.  The solution uses the Frenet-Serret framework, properties of Hessians, and the implicit-function method—concepts absent from the original exercise.\n\n4. Hidden Rank-One Structure – Recognising that Hess F has rank 1 (because F is a cubic in a linear form) is essential for the surprisingly zero curvature; overlooking this would lead to lengthy but futile calculations.\n\n5. Multiple Interacting Concepts – Linear algebra (rank arguments), differential geometry (normals, curvature, torsion), multivariable calculus (gradient, Hessian, method of characteristics) and algebraic manipulation all interact, making the exercise markedly more sophisticated than the original."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\n  u = x + 6y + 2z ,  v = x + 2y + z ,      \n  A = ∇u = (1,6,2) ,  B = ∇v = (1,2,1).                                         (†)  \n\nand consider the one–parameter family of algebraic surfaces  \n\n  (x + 6y + 2z)³ = a (x + 2y + z) ,    a ∈ ℝ.                                (★)\n\nThroughout Parts A–C assume v = x + 2y + z ≠ 0 unless stated otherwise, so that the right-hand side of (★) does not vanish.\n\nA.  Put F(x,y,z,a)=u³–av and show that simultaneous elimination of the parameter a from  \n\n  F = 0 and dF = 0  \n\nforces every non-degenerate member of (★) to possess the vector field  \n\n    C(x,y,z) = 3v A – u B = (3v–u, 18v–2u, 6v–u)                         (1)\n\nas a field of normals (i.e. C is everywhere perpendicular to the surface).  \nEquivalently, if Φ(x,y,z) is any potential of the family (★) one must have  \n\n    ∇Φ × C = 0,                                                       (2)\n\nor, written as two independent first–order linear PDE’s,  \n\n    (18v – 2u) Φ_x – (3v – u) Φ_y = 0 ,                             (2a)  \n    ( 6v –  u) Φ_x – (3v – u) Φ_z = 0 .                             (2b)\n\n(i) Show that in passing from (★) to (2) the entire plane  \n\n    Π : v = x + 2y + z = 0                                      (3)\n\nis lost, although Π is still an integral surface of the Pfaffian system generated by C.  \n\n(ii) Give the precise geometric reason for the loss, and verify explicitly that C is everywhere normal to Π (hence Π also satisfies ∇Φ × C = 0 when the potential is chosen as Φ = v).\n\nB.  A smooth surface Σ is called an orthogonal trajectory of the family (★) if, at each intersection point, the normals of Σ and the corresponding member of (★) are perpendicular.\n\n(i) For a trajectory written implicitly as G(x,y,z)=g(u,v)=0, prove that the orthogonality condition C·∇G = 0 reduces to the autonomous linear PDE  \n\n    (123 − 15 u/v) g_u + (45 − 6 u/v) g_v = 0 .                     (4)\n\n(ii) Solve (4) by the method of characteristics and show that every orthogonal trajectory can be expressed in closed form as  \n\n    (41v² − 20uv + 2u²)^{1/2} ·  \n    [(41v − (10 + √18)u)/(41v − (10 − √18)u)]^{5/√72} = C ,      (5)\n\n   where C is a non-zero real constant (either branch of the logarithm may be used).\n\nC.  Consider the concrete member  \n\n    S : (x + 6y + 2z)³ = x + 2y + z                            (the case a = 1)\n\nwhich passes through the point  \n\n    P = (1,0,0)   (u(P) = v(P) = 1).\n\nPutting u = v = 1 in (5) gives  \n\n    C₀ = √23 · [(31 − √18)/(31 + √18)]^{5/√72}.                   (6)\n\nDenote by  \n\n    Σ : (41v² − 20uv + 2u²)^{1/2} ·  \n      [(41v − (10 + √18)u)/(41v − (10 − √18)u)]^{5/√72} = C₀     (7)\n\nthe unique orthogonal trajectory that contains P.\n\n(i) Show directly that S and Σ meet orthogonally at P and state the angle of intersection.\n\n(ii) Let γ be the intersection curve S ∩ Σ with γ(0)=P, parameterised by arc length s and oriented so that γ′(0) points in the direction (2,1,–4).  Compute the curvature κ(0) and torsion τ(0) of γ at P.\n\n\n\n───────────────────────────────────────────────────────────────",
      "solution": "Pre-computed scalar products of the constant vectors in (†):  \n\n |A|² = 41 , A·B = 15 , |B|² = 6 , A×B = (2,1,−4).                       (8)\n\n\n\n--------------------------------------------------------------------  \nA.  Differential system of the family  \n--------------------------------------------------------------------  \nLet  \n\n  F(x,y,z,a)=u³−av.                                                       (9)\n\nDifferentiating,\n\n  dF = 3u² du − a dv.                                                   (10)\n\nBecause v ≠ 0 we may eliminate a by a = u³/v, obtaining  \n\n  3u² du − (u³/v) dv = 0 ⇔ 3v du − u dv = 0.                        (11)\n\nWriting du = A·dx and dv = B·dx gives  \n\n  (3v A − u B)·dx = 0,                                               (12)\n\nso the non-degenerate members of (★) have the normal field  \n\n  C = 3v A − u B,                                                    (1)\n\nand consequently any potential Φ describing the family satisfies  \n\n  ∇Φ × C = 0, equivalently (2a),(2b).\n\n(i)  Lost surface.  The algebraic step a = u³/v is legitimate only for v ≠ 0; therefore the surface Π : v = 0 is discarded when (11) is derived even though Π satisfies (★) with a = 0.\n\n(ii) Geometric reason.  On Π one has v = 0, whence  \n\n  C|_{Π} = 3·0·A − u B = −u B.                                      (13)\n\nSince the normal of Π is B (because ∇v = B), the vector C|_{Π} is everywhere parallel to the normal of Π; hence Π is an integral surface in the sense that its normal field coincides (up to scale) with C.  Choosing the potential Φ = v we obtain ∇Φ = B and  \n\n  ∇Φ × C = B × (−u B) = 0   on Π,                                  (14)\n\nconfirming that Π satisfies (2).  The surface is lost solely because the algebraic elimination required division by v.\n\n\n\n--------------------------------------------------------------------  \nB.  Orthogonal trajectories  \n--------------------------------------------------------------------  \n(i)  Let G(x,y,z)=g(u,v).  Then ∇G = g_u A + g_v B.  Orthogonality with C means  \n\n  C·∇G = (3v A − u B)·(g_u A + g_v B) = 0.                        (15)\n\nUsing (8),\n\n  (3v·41 − u·15) g_u + (3v·15 − u·6) g_v = 0  \n\n  ⇔ (123 v − 15 u) g_u + (45 v − 6 u) g_v = 0.                     (16)\n\nDividing by v (again v≠0) yields the autonomous linear PDE (4).\n\n(ii)  Characteristic integration.  \nIntroduce w = v/u (so v = wu).  From (16) the characteristic ODE is  \n\n  dv/du = (45v − 6u)/(123v − 15u) = (45w − 6)/(123w − 15).         (17)\n\nBecause v = wu, we have dv/du = w + u dw/du, so\n\n  w + u dw/du = (45w − 6)/(123w − 15).                              (18)\n\nSolving for dw/du and separating variables gives  \n\n  u dw/du = −(41w² − 20w + 2)/(41w − 5).                            (19)\n\nHence  \n\n  ∫ (41w − 5)/(41w² − 20w + 2) dw = −∫ du/u.                       (20)\n\nThe left integral equals  \n\n  ½ ln(41w² − 20w + 2) + (5/√72) ln[(41w − 10 − √18)/(41w − 10 + √18)].  \n\nExponentiating and replacing w by v/u one obtains the first integral  \n\n  u (41w² − 20w + 2)^{1/2}  \n   · [(41w − 10 − √18)/(41w − 10 + √18)]^{5/√72} = const,          (21)\n\nwhich, after restoring w = v/u, is exactly the closed form (5).  Each non-zero constant C determines one orthogonal trajectory and every trajectory arises in this way.\n\n\n\n--------------------------------------------------------------------  \nC.  Geometry at P = (1,0,0) (u = v = 1)  \n--------------------------------------------------------------------  \nConstant of the trajectory.  Setting u = v = 1 in (5) gives (6); hence the trajectory through P is Σ given by (7).\n\nNormals at P.  \n• For S (a = 1) we have F(x,y,z) = u³ − v, so  \n\n  ∇F = 3u² A − B; at P: N_S = 3A − B = (2,16,5).                    (22)\n\n• For Σ write  \n\n  ψ(u,v) = ½ ln(41v² − 20uv + 2u²)  \n      + k[ln(41v − 10 − √18 u) − ln(41v − 10 + √18 u)], k = 5/√72. (23)\n\nEquation (7) is ψ = ln C₀, so  \n\n  N_Σ = ∇ψ = ψ_u A + ψ_v B.                                          (24)\n\n(i)  Orthogonality at P.  Substituting u = v = 1 in the autonomous relation (4) satisfied by ψ gives  \n\n  (123 − 15) ψ_u + (45 − 6) ψ_v = 0 ⇒ 108 ψ_u + 39 ψ_v = 0.        (25)\n\nUsing (8),\n\n  N_S·N_Σ = (3|A|² − A·B) ψ_u + (3A·B − |B|²) ψ_v  \n       = 108 ψ_u + 39 ψ_v = 0.                                  (26)\n\nHence N_S ⟂ N_Σ and the two surfaces meet at a right angle, i.e. the angle of intersection is 90°.\n\n(ii)  Curvature and torsion of γ = S ∩ Σ at P.  \nThe unit tangent t of γ is parallel to A×B, hence  \n\n  t = (2,1,−4)/√21, with A·t = B·t = 0.                            (27)\n\nSecond derivatives.  \n• Hess F = 6u A Aᵀ ⇒ tᵀ Hess F t = 6u (A·t)² = 0.  \n• Hess G = Hess ψ = ψ_{uu} A Aᵀ + 2ψ_{uv} A Bᵀ + ψ_{vv} B Bᵀ, and again tᵀ Hess G t = 0 because A·t = B·t = 0.\n\nLet a = γ″(0).  Since γ is arc-length parametrised, t·a = 0, so write  \n\n  a = β N_S + γ N_Σ.                                                (28)\n\nThe second-derivative constraints F(γ)=0 and G(γ)=0 yield  \n\n  N_S·a = −tᵀ Hess F t = 0, N_Σ·a = −tᵀ Hess G t = 0.             (29)\n\nBecause N_S ⟂ N_Σ, equations (29) force β = γ = 0 and hence a = 0.  Therefore  \n\n  κ(0) = |a| = 0,  τ(0) may be taken as 0 (the standard convention when κ = 0). (30)\n\nThus the intersection curve γ has zero curvature at P; to second order it coincides with its tangent line in the direction (2,1,−4).\n\n\n\n───────────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.337846",
        "was_fixed": false,
        "difficulty_analysis": "1. Dimension Increase – The original 2-dimensional curve problem is lifted to a 3-dimensional setting involving surfaces.  Orthogonal trajectories now require handling gradients in ℝ³ and lead to a first-order linear PDE instead of an ODE.\n\n2. Non-trivial Invariants – Solving the PDE demands the method of characteristics in the variables (u,v) and an intricate rational integral that produces logarithmic expressions with quadratic discriminant √72.\n\n3. Advanced Geometry – Part C asks not only for the angle of intersection but also for curvature and torsion of the space curve defined by the intersection of two surfaces.  The solution uses the Frenet-Serret framework, properties of Hessians, and the implicit-function method—concepts absent from the original exercise.\n\n4. Hidden Rank-One Structure – Recognising that Hess F has rank 1 (because F is a cubic in a linear form) is essential for the surprisingly zero curvature; overlooking this would lead to lengthy but futile calculations.\n\n5. Multiple Interacting Concepts – Linear algebra (rank arguments), differential geometry (normals, curvature, torsion), multivariable calculus (gradient, Hessian, method of characteristics) and algebraic manipulation all interact, making the exercise markedly more sophisticated than the original."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}