Answer to Question #351032 in Differential Geometry | Topology for Anna

Question #351032

Calculate T ; N; B; "\\kappa" ; "\\tau" of the curve x(t) = (t; t 2 ; t 4) at the point (1; 1; 1).


1
Expert's answer
2022-06-26T17:19:42-0400
"\\vec{x}(t)=\\langle t,t^2,t^4\\rangle"

"\\vec{x}'(t)=\\langle 1,2t,4t^3\\rangle"

"|\\vec{x}'(t)|=\\sqrt{(1)^2+(2t)^2+(4t^3)^2}"

"\\vec{x}'(1)=\\langle 1,2,4\\rangle"

"|\\vec{x}'(1)|=\\sqrt{1^2+2^2+4^2}=\\sqrt{21}"

a.


"\\vec{T}(t)=\\dfrac{\\vec{x}'(t)}{|\\vec{x}'(t)|}"

"\\vec{T}(1)=\\dfrac{\\vec{x}'(1)}{|\\vec{x}'(1)|}=\\langle \\dfrac{1}{\\sqrt{21}},\\dfrac{2}{\\sqrt{21}},\\dfrac{4}{\\sqrt{21}}\\rangle"



b.


"\\vec{N}(t)=\\dfrac{\\vec{T}'(t)}{|\\vec{T}'(t)|}"

"\\vec{T}'(t)=\\langle 0,2,12t^2\\rangle"

"|\\vec{T}'(t)|=\\sqrt{(0)^2+(2)^2+(12t^2)^2}"

"\\vec{T}'(1)=\\langle 0,2,12\\rangle"

"|\\vec{T}'(1)|=\\sqrt{(0)^2+(2)^2+(12)^2}=2\\sqrt{37}"

"\\vec{N}(1)=\\dfrac{\\vec{T}'(1)}{|\\vec{T}'(1)|}=\\langle 0,\\dfrac{1}{\\sqrt{37}},\\dfrac{6}{\\sqrt{37}}\\rangle"

c.


"\\vec{B}(t)=\\vec{T}(t)\\times \\vec{N}(t)"

"\\vec{B}(1)=\\vec{T}(1)\\times \\vec{N}(1)=\\begin{vmatrix}\n \\vec{i} & \\vec{j} & \\vec{k} \\\\\n \\dfrac{1}{\\sqrt{21}} & \\dfrac{2}{\\sqrt{21}} & \\dfrac{4}{\\sqrt{21}} \\\\\n 0 & \\dfrac{1}{\\sqrt{37}} & \\dfrac{6}{\\sqrt{37}} \n\\end{vmatrix}"

"= \\dfrac{8}{\\sqrt{777}}\\vec{i}- \\dfrac{6}{\\sqrt{777}}\\vec{j} + \\dfrac{1}{\\sqrt{777}}\\vec{k}"

"\\vec{B}(1)=\\langle \\dfrac{8}{\\sqrt{777}},- \\dfrac{6}{\\sqrt{777}}, \\dfrac{1}{\\sqrt{777}}\\rangle"

d.


"\\kappa=\\dfrac{|\\vec{T}'(t)|}{|\\vec{x}'(t)|}"

"t=1"

"\\kappa=\\dfrac{|\\vec{T}'(1)|}{|\\vec{x}'(1)|}=\\dfrac{2\\sqrt{37}}{\\sqrt{21}}"


e.


"\\vec{x}''(t)=\\langle 0,2,12t^2\\rangle"

"\\vec{x}''(1)=\\langle 0,2,12\\rangle"

"\\vec{x'}(1)\\times \\vec{x''}(1)=\\begin{vmatrix}\n \\vec{i} & \\vec{j} & \\vec{k} \\\\\n 1 & 2& 4\\\\\n 0 & 2 & 12 \n\\end{vmatrix}"

"=16\\vec{i}- 12\\vec{j} +2\\vec{k}"

"\\vec{x'}(1)\\times \\vec{x''}(1)=\\langle 16,-12,2\\rangle"

"|\\vec{x'}(1)\\times \\vec{x''}(1)|=\\sqrt{(16)^+(-12)^2+(2)^2}=2\\sqrt{101}"



"\\vec{x}'''(t)=\\langle 0,0,24t\\rangle"

"\\vec{x}'''(1)=\\langle 0,0,24\\rangle"

"(\\vec{x'}(1)\\times \\vec{x''}(1))\\cdot\\vec{x}'''(1)=0-0+48=48"

"\\tau=\\dfrac{(\\vec{x'}(t)\\times \\vec{x''}(t))\\cdot\\vec{x}'''(t)}{|\\vec{x'}(t)\\times \\vec{x''}(t)|^2}"

"t=1"


"\\tau=\\dfrac{(\\vec{x'}(1)\\times \\vec{x''}(1))\\cdot\\vec{x}'''(1)}{|\\vec{x'}(1)\\times \\vec{x''}(1)|^2}"

"=\\dfrac{48}{404}=\\dfrac{12}{101}"

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