Answer to Question #278191 in Differential Geometry | Topology for Rabia qadeer

Question #278191

Find




the Curvature and




torsion of the Curves r = (u, (1 + u)/u, (1 - u ^ 2)/u)

1
Expert's answer
2021-12-13T17:58:06-0500

The curvature of the curve given by the vector function "\\bm r" is


"k(u)=\\dfrac{|\\bm r'(u)\\times\\bm r''(u)|}{|\\bm r'(u)|^3}"

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

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

"\\bm r''(u)=\\langle 0 , \\dfrac{2}{u^3}, \\dfrac{2}{u^3}\\rangle"

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

"=(-2\/u^5+2\/u^3+2\/u^5)\\bm i-(2\/u^3)\\bm j+()2\/u^3\\bm k"

"=(2\/u^3)\\bm i-(2\/u^3)\\bm j+(2\/u^3)\\bm k"

"|\\bm r'(u)|=\\sqrt{1+1\/u^4+1+2\/u^2+1\/u^4}"

"=\\dfrac{\\sqrt{2}}{u^2}\\sqrt{u^4+u^2+1}"


"|\\bm r'(u)\\times\\bm r''(u)|=\\dfrac{\\sqrt{6}}{u^2|u|}"

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

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

The torsion of the curve given by the vector function "\\bm r" is


"\\tau(u)=\\dfrac{(\\bm r'(u)\\times\\bm r''(u))\\cdot\\bm r'''(u)}{|\\bm r'(u)\\times\\bm r''(u)|^2}"

"\\bm r'''(u)=\\langle 0 , -\\dfrac{6}{u^4}, -\\dfrac{6}{u^4}\\rangle"

"(\\bm r'(u)\\times\\bm r''(u))\\cdot\\bm r'''(u)=0+12\/u^7-12\/u^7=0"

"\\tau(u)=0"


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