{
  "index": "1938-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "3. If a particle moves in the plane, we may express its coordinates \\( x \\) and \\( y \\) as functions of the time \\( t \\). If \\( x=t^{3}-t \\) and \\( y=t^{4}+t \\), show that the curve has a point of inflection at \\( t=0 \\) and that the velocity of the moving particle has a maximum at \\( t=0 \\).",
  "solution": "First Solution. If the velocity vector at time \\( t \\) is of length \\( v \\) and has direction \\( \\theta \\), then \\( \\dot{x}=v \\cos \\theta, \\dot{y}=v \\sin \\theta \\), and \\( \\ddot{x}=\\dot{v} \\cos \\theta-v \\dot{\\theta} \\sin \\theta \\), \\( \\ddot{y}=\\dot{v} \\sin \\theta+\\nu \\dot{\\theta} \\cos \\theta \\). Thus \\( \\dot{x} \\dot{y}-\\dot{y} \\ddot{x}=v^{2} \\dot{\\theta} \\) and \\( \\nu^{2}=\\dot{x}^{2}+\\dot{y}^{2} \\). From the given parametric data,\n\\[\n\\begin{aligned}\nv^{2} \\dot{\\theta} & =\\left(3 t^{2}-1\\right) 12 t^{2}-\\left(4 t^{3}+1\\right) 6 t=6 t\\left(2 t^{3}-2 t-1\\right) \\\\\nv^{2} & =16 t^{6}+9 t^{4}+8 t^{3}-6 t^{2}+2 .\n\\end{aligned}\n\\]\n\nFrom the \\( \\nu^{2} \\dot{\\theta} \\) relation, \\( \\dot{\\theta} \\) changes sign as \\( t \\) passes through 0 . To rule out the possibility of a cusp point, one notes that \\( v^{2} \\neq 0 \\) at \\( t=0 \\). Hence the curve has an inflection point at \\( t=0 \\). Also\n\\[\n\\begin{aligned}\n\\frac{d\\left(v^{2}\\right)}{d t} & =96 t^{5}+36 t^{4}+24 t^{2}-12 t \\\\\n\\left.\\frac{d\\left(v^{2}\\right)}{d t}\\right]_{t}=0 & =0\n\\end{aligned}\n\\]\nand\n\\[\n\\left.\\frac{d^{2}\\left(v^{2}\\right)}{d t^{2}}\\right]_{t=0}=-12 .\n\\]\n\nSo \\( v^{2} \\) has a (local) maximum at \\( t=0 \\). Hence the speed \\( v \\) also has a local maximum.",
  "vars": [
    "x",
    "y",
    "t",
    "v",
    "\\\\theta",
    "\\\\nu"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscis",
        "y": "ordinate",
        "t": "timevar",
        "v": "speedval",
        "\\theta": "angledir",
        "\\nu": "greeknu"
      },
      "question": "3. If a particle moves in the plane, we may express its coordinates \\( abscis \\) and \\( ordinate \\) as functions of the time \\( timevar \\). If \\( abscis=timevar^{3}-timevar \\) and \\( ordinate=timevar^{4}+timevar \\), show that the curve has a point of inflection at \\( timevar=0 \\) and that the velocity of the moving particle has a maximum at \\( timevar=0 \\).",
      "solution": "First Solution. If the velocity vector at time \\( timevar \\) is of length \\( speedval \\) and has direction \\( angledir \\), then \\( \\dot{abscis}=speedval \\cos angledir, \\dot{ordinate}=speedval \\sin angledir \\), and \\( \\ddot{abscis}=\\dot{speedval} \\cos angledir-speedval \\dot{angledir} \\sin angledir \\), \\( \\ddot{ordinate}=\\dot{speedval} \\sin angledir+greeknu \\dot{angledir} \\cos angledir \\). Thus \\( \\dot{abscis} \\dot{ordinate}-\\dot{ordinate} \\ddot{abscis}=speedval^{2} \\dot{angledir} \\) and \\( greeknu^{2}=\\dot{abscis}^{2}+\\dot{ordinate}^{2} \\). From the given parametric data,\n\\[\n\\begin{aligned}\nspeedval^{2} \\dot{angledir} & =\\left(3 timevar^{2}-1\\right) 12 timevar^{2}-\\left(4 timevar^{3}+1\\right) 6 timevar=6 timevar\\left(2 timevar^{3}-2 timevar-1\\right) \\\\\nspeedval^{2} & =16 timevar^{6}+9 timevar^{4}+8 timevar^{3}-6 timevar^{2}+2 .\n\\end{aligned}\n\\]\n\nFrom the \\( greeknu^{2} \\dot{angledir} \\) relation, \\( \\dot{angledir} \\) changes sign as \\( timevar \\) passes through 0 . To rule out the possibility of a cusp point, one notes that \\( speedval^{2} \\neq 0 \\) at \\( timevar=0 \\). Hence the curve has an inflection point at \\( timevar=0 \\). Also\n\\[\n\\begin{aligned}\n\\frac{d\\left(speedval^{2}\\right)}{d timevar} & =96 timevar^{5}+36 timevar^{4}+24 timevar^{2}-12 timevar \\\\\n\\left.\\frac{d\\left(speedval^{2}\\right)}{d timevar}\\right]_{timevar}=0 & =0\n\\end{aligned}\n\\]\nand\n\\[\n\\left.\\frac{d^{2}\\left(speedval^{2}\\right)}{d timevar^{2}}\\right]_{timevar=0}=-12 .\n\\]\n\nSo \\( speedval^{2} \\) has a (local) maximum at \\( timevar=0 \\). Hence the speed \\( speedval \\) also has a local maximum."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sunflower",
        "y": "chocolate",
        "t": "waterfall",
        "v": "toothbrush",
        "\\\\theta": "suitcase",
        "\\\\nu": "blueberry"
      },
      "question": "3. If a particle moves in the plane, we may express its coordinates \\( sunflower \\) and \\( chocolate \\) as functions of the time \\( waterfall \\). If \\( sunflower=waterfall^{3}-waterfall \\) and \\( chocolate=waterfall^{4}+waterfall \\), show that the curve has a point of inflection at \\( waterfall=0 \\) and that the velocity of the moving particle has a maximum at \\( waterfall=0 \\).",
      "solution": "First Solution. If the velocity vector at time \\( waterfall \\) is of length \\( toothbrush \\) and has direction \\( suitcase \\), then \\( \\dot{sunflower}=toothbrush \\cos suitcase, \\dot{chocolate}=toothbrush \\sin suitcase \\), and \\( \\ddot{sunflower}=\\dot{toothbrush} \\cos suitcase-toothbrush \\dot{suitcase} \\sin suitcase \\), \\( \\ddot{chocolate}=\\dot{toothbrush} \\sin suitcase+blueberry \\dot{suitcase} \\cos suitcase \\). Thus \\( \\dot{sunflower} \\dot{chocolate}-\\dot{chocolate} \\ddot{sunflower}=toothbrush^{2} \\dot{suitcase} \\) and \\( blueberry^{2}=\\dot{sunflower}^{2}+\\dot{chocolate}^{2} \\). From the given parametric data,\n\\[\n\\begin{aligned}\ntoothbrush^{2} \\dot{suitcase} & =(3 \\,waterfall^{2}-1)\\,12\\,waterfall^{2}-(4\\,waterfall^{3}+1)\\,6\\,waterfall=6\\,waterfall\\left(2\\,waterfall^{3}-2\\,waterfall-1\\right) \\\\\ntoothbrush^{2} & =16\\,waterfall^{6}+9\\,waterfall^{4}+8\\,waterfall^{3}-6\\,waterfall^{2}+2 .\n\\end{aligned}\n\\]\n\nFrom the \\( blueberry^{2} \\dot{suitcase} \\) relation, \\( \\dot{suitcase} \\) changes sign as \\( waterfall \\) passes through 0. To rule out the possibility of a cusp point, one notes that \\( toothbrush^{2} \\neq 0 \\) at \\( waterfall=0 \\). Hence the curve has an inflection point at \\( waterfall=0 \\). Also\n\\[\n\\begin{aligned}\n\\frac{d\\left(toothbrush^{2}\\right)}{d waterfall} & =96\\,waterfall^{5}+36\\,waterfall^{4}+24\\,waterfall^{2}-12\\,waterfall \\\\\n\\left.\\frac{d\\left(toothbrush^{2}\\right)}{d waterfall}\\right]_{waterfall}=0 & =0\n\\end{aligned}\n\\]\nand\n\\[\n\\left.\\frac{d^{2}\\left(toothbrush^{2}\\right)}{d waterfall^{2}}\\right]_{waterfall=0}=-12 .\n\\]\n\nSo \\( toothbrush^{2} \\) has a (local) maximum at \\( waterfall=0 \\). Hence the speed \\( toothbrush \\) also has a local maximum."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "farpoint",
        "y": "lowpoint",
        "t": "spacemeas",
        "v": "slowness",
        "\\\\theta": "straightdir",
        "\\\\nu": "stillness"
      },
      "question": "3. If a particle moves in the plane, we may express its coordinates \\( farpoint \\) and \\( lowpoint \\) as functions of the time \\( spacemeas \\). If \\( farpoint=spacemeas^{3}-spacemeas \\) and \\( lowpoint=spacemeas^{4}+spacemeas \\), show that the curve has a point of inflection at \\( spacemeas=0 \\) and that the velocity of the moving particle has a maximum at \\( spacemeas=0 \\).",
      "solution": "First Solution. If the velocity vector at time \\( spacemeas \\) is of length \\( slowness \\) and has direction \\( straightdir \\), then \\( \\dot{farpoint}=slowness \\cos straightdir, \\dot{lowpoint}=slowness \\sin straightdir \\), and \\( \\ddot{farpoint}=\\dot{slowness} \\cos straightdir-slowness \\dot{straightdir} \\sin straightdir \\), \\( \\ddot{lowpoint}=\\dot{slowness} \\sin straightdir+stillness \\dot{straightdir} \\cos straightdir \\). Thus \\( \\dot{farpoint} \\dot{lowpoint}-\\dot{lowpoint} \\ddot{farpoint}=slowness^{2} \\dot{straightdir} \\) and \\( stillness^{2}=\\dot{farpoint}^{2}+\\dot{lowpoint}^{2} \\). From the given parametric data,\n\\[\n\\begin{aligned}\nslowness^{2} \\dot{straightdir} & =\\left(3 spacemeas^{2}-1\\right) 12 spacemeas^{2}-\\left(4 spacemeas^{3}+1\\right) 6 spacemeas=6 spacemeas\\left(2 spacemeas^{3}-2 spacemeas-1\\right) \\\\\nslowness^{2} & =16 spacemeas^{6}+9 spacemeas^{4}+8 spacemeas^{3}-6 spacemeas^{2}+2 .\n\\end{aligned}\n\\]\n\nFrom the \\( stillness^{2} \\dot{straightdir} \\) relation, \\( \\dot{straightdir} \\) changes sign as \\( spacemeas \\) passes through 0 . To rule out the possibility of a cusp point, one notes that \\( slowness^{2} \\neq 0 \\) at \\( spacemeas=0 \\). Hence the curve has an inflection point at \\( spacemeas=0 \\). Also\n\\[\n\\begin{aligned}\n\\frac{d\\left(slowness^{2}\\right)}{d spacemeas} & =96 spacemeas^{5}+36 spacemeas^{4}+24 spacemeas^{2}-12 spacemeas \\\\\n\\left.\\frac{d\\left(slowness^{2}\\right)}{d spacemeas}\\right]_{spacemeas}=0 & =0\n\\end{aligned}\n\\]\nand\n\\[\n\\left.\\frac{d^{2}\\left(slowness^{2}\\right)}{d spacemeas^{2}}\\right]_{spacemeas=0}=-12 .\n\\]\n\nSo \\( slowness^{2} \\) has a (local) maximum at \\( spacemeas=0 \\). Hence the speed \\( slowness \\) also has a local maximum."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "t": "dmcvbryk",
        "v": "npslfqiw",
        "\\theta": "rjqmzgeb",
        "\\nu": "plxrtbsa"
      },
      "question": "3. If a particle moves in the plane, we may express its coordinates \\( qzxwvtnp \\) and \\( hjgrksla \\) as functions of the time \\( dmcvbryk \\). If \\( qzxwvtnp=dmcvbryk^{3}-dmcvbryk \\) and \\( hjgrksla=dmcvbryk^{4}+dmcvbryk \\), show that the curve has a point of inflection at \\( dmcvbryk=0 \\) and that the velocity of the moving particle has a maximum at \\( dmcvbryk=0 \\).",
      "solution": "First Solution. If the velocity vector at time \\( dmcvbryk \\) is of length \\( npslfqiw \\) and has direction \\( rjqmzgeb \\), then \\( \\dot{qzxwvtnp}=npslfqiw \\cos rjqmzgeb, \\dot{hjgrksla}=npslfqiw \\sin rjqmzgeb \\), and \\( \\ddot{qzxwvtnp}=\\dot{npslfqiw} \\cos rjqmzgeb-npslfqiw \\dot{rjqmzgeb} \\sin rjqmzgeb \\), \\( \\ddot{hjgrksla}=\\dot{npslfqiw} \\sin rjqmzgeb+plxrtbsa \\dot{rjqmzgeb} \\cos rjqmzgeb \\). Thus \\( \\dot{qzxwvtnp} \\dot{hjgrksla}-\\dot{hjgrksla} \\ddot{qzxwvtnp}=npslfqiw^{2} \\dot{rjqmzgeb} \\) and \\( plxrtbsa^{2}=\\dot{qzxwvtnp}^{2}+\\dot{hjgrksla}^{2} \\). From the given parametric data,\n\\[\n\\begin{aligned}\nnpslfqiw^{2} \\dot{rjqmzgeb} & =\\left(3 dmcvbryk^{2}-1\\right) 12 dmcvbryk^{2}-\\left(4 dmcvbryk^{3}+1\\right) 6 dmcvbryk=6 dmcvbryk\\left(2 dmcvbryk^{3}-2 dmcvbryk-1\\right) \\\\\nnpslfqiw^{2} & =16 dmcvbryk^{6}+9 dmcvbryk^{4}+8 dmcvbryk^{3}-6 dmcvbryk^{2}+2 .\n\\end{aligned}\n\\]\n\nFrom the \\( plxrtbsa^{2} \\dot{rjqmzgeb} \\) relation, \\( \\dot{rjqmzgeb} \\) changes sign as \\( dmcvbryk \\) passes through 0. To rule out the possibility of a cusp point, one notes that \\( npslfqiw^{2} \\neq 0 \\) at \\( dmcvbryk=0 \\). Hence the curve has an inflection point at \\( dmcvbryk=0 \\). Also\n\\[\n\\begin{aligned}\n\\frac{d\\left(npslfqiw^{2}\\right)}{d dmcvbryk} & =96 dmcvbryk^{5}+36 dmcvbryk^{4}+24 dmcvbryk^{2}-12 dmcvbryk \\\\\n\\left.\\frac{d\\left(npslfqiw^{2}\\right)}{d dmcvbryk}\\right]_{dmcvbryk}=0 & =0\n\\end{aligned}\n\\]\nand\n\\[\n\\left.\\frac{d^{2}\\left(npslfqiw^{2}\\right)}{d dmcvbryk^{2}}\\right]_{dmcvbryk=0}=-12 .\n\\]\n\nSo \\( npslfqiw^{2} \\) has a (local) maximum at \\( dmcvbryk=0 \\). Hence the speed \\( npslfqiw \\) also has a local maximum."
    },
    "kernel_variant": {
      "question": "Let a particle move in space according to the vector–valued function  \n\n  γ(t)= (x(t),y(t),z(t)), where  \n  x(t)=t−3t³−5t⁵+7t⁷,  \n  y(t)=t−3t³−4t⁵,  \n  z(t)=2t⁴−t⁶,            (−∞<t<∞).\n\nDefine  \n\n  v(t)=‖γ′(t)‖   (the speed),  \n  κ(t)=‖γ′(t)×γ″(t)‖ / ‖γ′(t)‖³ (the curvature),  \n  τ(t)= [ (γ′×γ″)·γ‴ ] / ‖γ′×γ″‖² (the torsion).\n\n(a)  Show that t=0 is an inflection point of the spatial curve, i.e. κ(0)=0 while κ(t)>0 for every t≠0 sufficiently close to 0.  Compute the first non-vanishing term of κ(t) and prove that  \n\n  κ(t)=12\\,t²+O(t³).  \n\n(b)  Show that the torsion has the finite, non–zero limit  \n\n  lim_{t→0} τ(t)=−5/12.  \n\n(c)  Verify that the speed v(t) attains a strict local maximum of second order at t=0, whereas the magnitude of the acceleration  \n\n  a(t)=‖γ″(t)‖  \n\nattains a strict local minimum there.  Moreover prove the asymptotic expansion  \n\n  ‖γ″(t)‖=18√2\\,|t|+O(t³).  \n\n(d)  Denote by  \n\n  T(t)=γ′/‖γ′‖ (unit tangent),   \n  N(t)=T′/‖T′‖ (unit principal normal),   \n  B(t)=T×N   (unit binormal).  \n\nDetermine the limits  \n\n  lim_{t→0}T(t), lim_{t→0}N(t), lim_{t→0}B(t)  \n\nand describe explicitly the osculating plane and the binormal direction at the inflection point.",
      "solution": "Throughout we put  \n\n v(t)=γ′(t), s(t)=‖v(t)‖, T(t)=v/s, N(t)=T′/‖T′‖, B(t)=T×N.\n\nStep 1.  Low–order expansions of derivatives.  \nDifferentiate three times:\n\nv(t)=γ′(t)=⟨1−9t²−25t⁴+49t⁶, 1−9t²−20t⁴, 8t³−6t⁵⟩,\n\nγ″(t)=⟨−18t−100t³+294t⁵, −18t−80t³, 24t²−30t⁴⟩,\n\nγ‴(t)=⟨−18−300t²+1470t⁴, −18−240t², 48t−120t³⟩.\n\nAt t=0  \n\n v(0)=⟨1,1,0⟩, γ″(0)=0, γ‴(0)=⟨−18,−18,0⟩, s(0)=√2.\n\nStep 2.  The cross–product v×γ″.  \nRetaining terms up to order t⁴ we obtain  \n\nv×γ″ = ⟨24t²−102t⁴, −(24t²−102t⁴), 20t³⟩+O(t⁵).  (2.1)\n\nHence  \n\n‖v×γ″‖² = (24t²)²+(−24t²)²+(20t³)²+O(t⁶)=1152t⁴+O(t⁶), (2.2)  \n‖v×γ″‖  = 24√2\\,|t|²+O(t³).                (2.3)\n\nStep 3.  The speed.  \nBecause v(t)=⟨1,1,0⟩+O(t²),\n\n s(t)=√2+O(t²), s³(t)=2√2+O(t²).            (3.1)\n\nStep 4.  Curvature.  \nBy definition,\n\nκ(t)=‖v×γ″‖/s³(t)= (24√2\\,|t|²+O(t³))/(2√2+O(t²))  \n    = 12\\,t²+O(t³).                  (4.1)\n\nThus κ(0)=0 and κ(t)>0 for small t≠0; the curve has an inflection at t=0.\n\nStep 5.  Torsion.  \nCompute the triple product to order t⁴:\n\n(v×γ″)·γ‴  \n = ⟨24t²−102t⁴, −24t²+102t⁴, 20t³⟩·⟨−18−300t², −18−240t², 48t⟩  \n = −480t⁴+O(t⁶).                    (5.1)\n\nConsequently  \n\nτ(t)= (v×γ″)·γ‴ / ‖v×γ″‖² = (−480t⁴+O(t⁶))/(1152t⁴+O(t⁶))  \n   → −5/12  (t→0).                 (5.2)\n\nStep 6.  Extremal behaviour of speed and acceleration.\n\n(a) Speed.  Since s²(t)=v·v,\n\n (s²)′(0)=2v·γ″(0)=0.                (6.1)\n\nA second derivative gives (s²)″(t)=2(‖γ″‖²+v·γ‴).  \nAt t=0, v·γ‴(0)=⟨1,1,0⟩·⟨−18,−18,0⟩=−36 and ‖γ″(0)‖=0, hence  \n\n (s²)″(0)=−72<0, so s(t) has a strict quadratic local maximum at t=0.\n\n(b) Acceleration.  From the components of γ″,\n\n ‖γ″(t)‖² = 648t²+O(t⁴) ⇒ ‖γ″(t)‖ = 18√2\\,|t|+O(t³). (6.2)\n\nTherefore ‖γ″‖ has a simple zero at t=0 and increases linearly away from it; the acceleration magnitude attains a strict local minimum there.\n\nStep 7.  Limit of the Frenet frame (corrected).\n\nFirst, T(t)=v/s and (3.1) gives  \n\n lim_{t→0}T(t)= (1/√2)⟨1,1,0⟩.            (7.1)\n\nBecause κ(0)=0, direct use of N=T′/‖T′‖ must be handled carefully.  \nIt is more convenient to employ B(t)=(v×γ″)/‖v×γ″‖.  \nUsing (2.1)–(2.3),\n\nB(t)= ⟨1/√2,−1/√2,(20t³)/(24√2\\,|t|²)⟩+O(t²)  \n  = ⟨1/√2,−1/√2,(5t)/(6√2)⟩+O(t²),                        \n\nso  \n\n lim_{t→0}B(t)=⟨1/√2,−1/√2,0⟩.            (7.2)\n\nFinally,\n\nN(t)=B(t)×T(t)→⟨0,0,1⟩.                (7.3)\n\nOsculating plane and binormal.  \nThe osculating plane at t=0 is spanned by the limiting T and N, i.e. by the vectors  \n\n (1/√2,1/√2,0) and (0,0,1).\n\nIts normal is the limiting binormal B(0)=⟨1/√2,−1/√2,0⟩, so the plane has equation  \n\n (x−y)=0.\n\nThus the osculating plane is the vertical plane x=y through the z–axis, and the binormal direction lies horizontally along ⟨1/√2,−1/√2,0⟩.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.334584",
        "was_fixed": false,
        "difficulty_analysis": "1. Dimensional upgrade – the problem moves from planar motion to a fully-fledged space curve, forcing the competitor to handle curvature, torsion and the full Frenet apparatus.  \n2. Higher algebraic complexity – seventh-degree polynomials in x and y combined with a sixth-degree z-component generate lengthy derivative chains; extracting the leading asymptotics requires careful bookkeeping rather than routine differentiation.  \n3. Subtle cancellations – to locate the precise orders of vanishing of κ and of the triple product that enters τ, one must arrange (and track) cancellations up to order 4, well beyond the usual linear check that suffices for a planar inflection.  \n4. Interacting extrema – simultaneously analysing maxima of speed and minima of acceleration adds an independent layer that combines vector-calculus identities with one-variable extremum tests.  \n5. Frenet-frame limits – deciding the limits of T, N, B at a point where κ vanishes but τ does not requires working with asymptotic rather than pointwise definitions of the frame.\n\nTogether these additions demand deeper theoretical insight and substantially longer computations than either the original textbook exercise or the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let a particle move in space according to the vector–valued function  \n\n  γ(t)= (x(t),y(t),z(t)), where  \n  x(t)=t−3t³−5t⁵+7t⁷,  \n  y(t)=t−3t³−4t⁵,  \n  z(t)=2t⁴−t⁶,            (−∞<t<∞).\n\nDefine  \n\n  v(t)=‖γ′(t)‖   (the speed),  \n  κ(t)=‖γ′(t)×γ″(t)‖ / ‖γ′(t)‖³ (the curvature),  \n  τ(t)= [ (γ′×γ″)·γ‴ ] / ‖γ′×γ″‖² (the torsion).\n\n(a)  Show that t=0 is an inflection point of the spatial curve, i.e. κ(0)=0 while κ(t)>0 for every t≠0 sufficiently close to 0.  Compute the first non-vanishing term of κ(t) and prove that  \n\n  κ(t)=12\\,t²+O(t³).  \n\n(b)  Show that the torsion has the finite, non–zero limit  \n\n  lim_{t→0} τ(t)=−5/12.  \n\n(c)  Verify that the speed v(t) attains a strict local maximum of second order at t=0, whereas the magnitude of the acceleration  \n\n  a(t)=‖γ″(t)‖  \n\nattains a strict local minimum there.  Moreover prove the asymptotic expansion  \n\n  ‖γ″(t)‖=18√2\\,|t|+O(t³).  \n\n(d)  Denote by  \n\n  T(t)=γ′/‖γ′‖ (unit tangent),   \n  N(t)=T′/‖T′‖ (unit principal normal),   \n  B(t)=T×N   (unit binormal).  \n\nDetermine the limits  \n\n  lim_{t→0}T(t), lim_{t→0}N(t), lim_{t→0}B(t)  \n\nand describe explicitly the osculating plane and the binormal direction at the inflection point.",
      "solution": "Throughout we put  \n\n v(t)=γ′(t), s(t)=‖v(t)‖, T(t)=v/s, N(t)=T′/‖T′‖, B(t)=T×N.\n\nStep 1.  Low–order expansions of derivatives.  \nDifferentiate three times:\n\nv(t)=γ′(t)=⟨1−9t²−25t⁴+49t⁶, 1−9t²−20t⁴, 8t³−6t⁵⟩,\n\nγ″(t)=⟨−18t−100t³+294t⁵, −18t−80t³, 24t²−30t⁴⟩,\n\nγ‴(t)=⟨−18−300t²+1470t⁴, −18−240t², 48t−120t³⟩.\n\nAt t=0  \n\n v(0)=⟨1,1,0⟩, γ″(0)=0, γ‴(0)=⟨−18,−18,0⟩, s(0)=√2.\n\nStep 2.  The cross–product v×γ″.  \nRetaining terms up to order t⁴ we obtain  \n\nv×γ″ = ⟨24t²−102t⁴, −(24t²−102t⁴), 20t³⟩+O(t⁵).  (2.1)\n\nHence  \n\n‖v×γ″‖² = (24t²)²+(−24t²)²+(20t³)²+O(t⁶)=1152t⁴+O(t⁶), (2.2)  \n‖v×γ″‖  = 24√2\\,|t|²+O(t³).                (2.3)\n\nStep 3.  The speed.  \nBecause v(t)=⟨1,1,0⟩+O(t²),\n\n s(t)=√2+O(t²), s³(t)=2√2+O(t²).            (3.1)\n\nStep 4.  Curvature.  \nBy definition,\n\nκ(t)=‖v×γ″‖/s³(t)= (24√2\\,|t|²+O(t³))/(2√2+O(t²))  \n    = 12\\,t²+O(t³).                  (4.1)\n\nThus κ(0)=0 and κ(t)>0 for small t≠0; the curve has an inflection at t=0.\n\nStep 5.  Torsion.  \nCompute the triple product to order t⁴:\n\n(v×γ″)·γ‴  \n = ⟨24t²−102t⁴, −24t²+102t⁴, 20t³⟩·⟨−18−300t², −18−240t², 48t⟩  \n = −480t⁴+O(t⁶).                    (5.1)\n\nConsequently  \n\nτ(t)= (v×γ″)·γ‴ / ‖v×γ″‖² = (−480t⁴+O(t⁶))/(1152t⁴+O(t⁶))  \n   → −5/12  (t→0).                 (5.2)\n\nStep 6.  Extremal behaviour of speed and acceleration.\n\n(a) Speed.  Since s²(t)=v·v,\n\n (s²)′(0)=2v·γ″(0)=0.                (6.1)\n\nA second derivative gives (s²)″(t)=2(‖γ″‖²+v·γ‴).  \nAt t=0, v·γ‴(0)=⟨1,1,0⟩·⟨−18,−18,0⟩=−36 and ‖γ″(0)‖=0, hence  \n\n (s²)″(0)=−72<0, so s(t) has a strict quadratic local maximum at t=0.\n\n(b) Acceleration.  From the components of γ″,\n\n ‖γ″(t)‖² = 648t²+O(t⁴) ⇒ ‖γ″(t)‖ = 18√2\\,|t|+O(t³). (6.2)\n\nTherefore ‖γ″‖ has a simple zero at t=0 and increases linearly away from it; the acceleration magnitude attains a strict local minimum there.\n\nStep 7.  Limit of the Frenet frame (corrected).\n\nFirst, T(t)=v/s and (3.1) gives  \n\n lim_{t→0}T(t)= (1/√2)⟨1,1,0⟩.            (7.1)\n\nBecause κ(0)=0, direct use of N=T′/‖T′‖ must be handled carefully.  \nIt is more convenient to employ B(t)=(v×γ″)/‖v×γ″‖.  \nUsing (2.1)–(2.3),\n\nB(t)= ⟨1/√2,−1/√2,(20t³)/(24√2\\,|t|²)⟩+O(t²)  \n  = ⟨1/√2,−1/√2,(5t)/(6√2)⟩+O(t²),                        \n\nso  \n\n lim_{t→0}B(t)=⟨1/√2,−1/√2,0⟩.            (7.2)\n\nFinally,\n\nN(t)=B(t)×T(t)→⟨0,0,1⟩.                (7.3)\n\nOsculating plane and binormal.  \nThe osculating plane at t=0 is spanned by the limiting T and N, i.e. by the vectors  \n\n (1/√2,1/√2,0) and (0,0,1).\n\nIts normal is the limiting binormal B(0)=⟨1/√2,−1/√2,0⟩, so the plane has equation  \n\n (x−y)=0.\n\nThus the osculating plane is the vertical plane x=y through the z–axis, and the binormal direction lies horizontally along ⟨1/√2,−1/√2,0⟩.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.296366",
        "was_fixed": false,
        "difficulty_analysis": "1. Dimensional upgrade – the problem moves from planar motion to a fully-fledged space curve, forcing the competitor to handle curvature, torsion and the full Frenet apparatus.  \n2. Higher algebraic complexity – seventh-degree polynomials in x and y combined with a sixth-degree z-component generate lengthy derivative chains; extracting the leading asymptotics requires careful bookkeeping rather than routine differentiation.  \n3. Subtle cancellations – to locate the precise orders of vanishing of κ and of the triple product that enters τ, one must arrange (and track) cancellations up to order 4, well beyond the usual linear check that suffices for a planar inflection.  \n4. Interacting extrema – simultaneously analysing maxima of speed and minima of acceleration adds an independent layer that combines vector-calculus identities with one-variable extremum tests.  \n5. Frenet-frame limits – deciding the limits of T, N, B at a point where κ vanishes but τ does not requires working with asymptotic rather than pointwise definitions of the frame.\n\nTogether these additions demand deeper theoretical insight and substantially longer computations than either the original textbook exercise or the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}