keizhacarbajal58 keizhacarbajal58
  • 10-10-2019
  • Mathematics
contestada

Find the length of the arc of the circular helix with vector equation r(t) = 2 cos t i + 2 sin t j + tk from the point (2, 0, 0) to the point (2, 0, 2π).

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LammettHash
LammettHash LammettHash
  • 10-10-2019

[tex]\vec r(t)=2\cos t\,\vec\imath+2\sin t\,\vec\jmath+t\,\vec k[/tex]

[tex]\implies\vec r'(t)=-2\sin t\,\vec\imath+2\cos t\,\vec\jmath+\vec k[/tex]

[tex]\implies\|\vec r'(t)\|=\sqrt{(-2\sin t)^2+(2\cos t)^2+1^2}=\sqrt5[/tex]

Then the length of the arc is

[tex]\displaystyle\int_0^{2\pi}\|\vec r'(t)\|\,\mathrm dt=\sqrt5\int_0^{2\pi}\mathrm dt=2\sqrt5\,\pi[/tex]

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