for the curve x=a(3u-u³), y=3au², z=a(3u+u³) prove that k=t
"r(u)=\\langle a(3u-u^3),\\ \\ 3au^2,\\ \\ a(3u+u^3)\\rangle"
"r^\\prime (u)=\\langle 3a(1-u^2),\\ \\ 6au,\\ \\ 3a(1+u^2)\\rangle"
"r^{\\prime \\prime}(u)=\\langle -6au,\\ \\ 6a,\\ \\ 6au\\rangle"
"r^{\\prime\\prime\\prime}(u)=\\langle -6a,\\ \\ 0,\\ \\ 6a\\rangle"
"r^\\prime \\times r^{\\prime \\prime }=\\begin{vmatrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\n\\\\\n3a(1-u^2)&6au&3a(1+u^2)\n\\\\\n-6au&6a&6au\n\\end{vmatrix}=18a^2(u^2-1)\\mathbf{i}-36a^2u\\mathbf{j}+18a^2(u^2+1)\\mathbf{k}"
"|r^\\prime\\times r^{\\prime \\prime}|=18a^2\\sqrt{(u^2-1)^2+(-2u)^2+(u^2+1)^2}=18\\sqrt{2}a^2(u^2+1)"
"|r^\\prime|=3a\\sqrt{(1-u^2)^2+(2u)^2+(1+u^2)^2}=3\\sqrt{2}a(u^2+1)"
"(r^\\prime \\times r^{\\prime \\prime})\\cdot r^{\\prime\\prime\\prime}=\\begin{vmatrix}\n3a(1-u^2)&6au&3a(1+u^2)\n\\\\\n-6au&6a&6au\\\\\n-6a&0&6a\n\\end{vmatrix}=216a^3"
"\\kappa =\\frac{|r^\\prime \\times r^{\\prime \\prime }|}{|r^\\prime|^3}=\\frac{18\\sqrt{2}a^2(u^2+1)}{54\\sqrt{2}a^3(u^2+1)^3}=\\frac{1}{3a(u^2+1)^2}"
"\\tau =\\frac{(r^\\prime \\times r^{\\prime \\prime })\\cdot r^{\\prime \\prime\\prime}}{|r^\\prime \\times r^{\\prime \\prime}|^2}=\\frac{216a^3}{648a^4(u^2+1)^2}=\\frac{1}{3a(u^2+1)^2}"
So, "\\kappa =\\tau" .
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