Answer to Question #289747 in Differential Geometry | Topology for Abi

Question #289747

find the curvature and torsion of the curve z=u,y=1+u/u,z=1-u^2/u


1
Expert's answer
2022-01-25T06:51:29-0500

a)


"r(u)=\\langle u, \\dfrac{1+u}{u}, \\dfrac{1-u^2}{u}\\rangle"

"r'(u)=\\langle1, -\\dfrac{1}{u^2},-\\dfrac{1}{u^2}-1\\rangle"

"r''(u)=\\langle0, \\dfrac{2}{u^3}, \\dfrac{2}{u^3}\\rangle"

"|r'(u)|=\\sqrt{(1)^2+(-\\dfrac{1}{u^2})^2+(-\\dfrac{1}{u^2}-1)^2}"

"=\\dfrac{\\sqrt{2u^4+2u^2+2}}{u^2}"

"r'(u)\\times r''(u)=\\begin{vmatrix}\n i & j & k \\\\\n\\\\\n 1 & -\\dfrac{1}{u^2} & -\\dfrac{1}{u^2}-1 \\\\ \\\\\n 0 & \\dfrac{2}{u^3}&\\dfrac{2}{u^3}\n\\end{vmatrix}"

"=i\\begin{vmatrix}\n -\\dfrac{1}{u^2} & -\\dfrac{1}{u^2}-1 \\\\ \\\\\n \\dfrac{2}{u^3} & \\dfrac{2}{u^3}\n\\end{vmatrix}-j\\begin{vmatrix}\n 1 & -\\dfrac{1}{u^2}-1 \\\\ \\\\\n 0& \\dfrac{2}{u^3}\n\\end{vmatrix}"

"+k\\begin{vmatrix}\n 1 & -\\dfrac{1}{u^2} \\\\ \\\\\n 0 & \\dfrac{2}{u^3}\n\\end{vmatrix}=-\\dfrac{2}{u^3}i-\\dfrac{2}{u^3}j+\\dfrac{2}{u^3}k"

"|r'(t)\\times r''(t)|=\\sqrt{(-\\dfrac{2}{u^3})^2+(-\\dfrac{2}{u^3})^2+(\\dfrac{2}{u^3})^2}"

"=\\dfrac{2\\sqrt{3}}{u^2|u|}"

Find curvature


"\\kappa(t)=\\dfrac{|r'(t)\\times r''(t)|}{(|r'(t)|)^{3}}"

"=\\dfrac{\\dfrac{2\\sqrt{3}}{u^2|u|}}{(\\dfrac{\\sqrt{2u^4+2u^2+2}}{u^2})^{3}}"

"=\\dfrac{\\sqrt{6}u^2|u|}{2(u^4+u^2+1)^{3\/2}}"

"\\kappa(u)=\\dfrac{\\sqrt{6}u^2|u|}{2(u^4+u^2+1)^{3\/2}}"

b)


"r'''(u)=\\langle0, -\\dfrac{6}{u^4}, -\\dfrac{6}{u^4}\\rangle"

"(r'(u)\\times r''(u))\\cdot r'''(u)=0-\\dfrac{2}{u^3}(-\\dfrac{6}{u^4})+\\dfrac{2}{u^3}(-\\dfrac{6}{u^4})"

"=0"


"\\tau(u)=\\dfrac{(r'(u)\\times r''(u))\\cdot r'''(u)}{(|r'(t)\\times r''(t)|)^2}=0"


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