Question

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).

Answer 1

`\vec r(t)=2\cos t\,\vec\imath+2\sin t\,\vec\jmath+t\,\vec k`

`\implies\vec r'(t)=-2\sin t\,\vec\imath+2\cos t\,\vec\jmath+\vec k`

`\implies\|\vec r'(t)\|=√((-2\sin t)^2+(2\cos t)^2+1^2)=\sqrt5`

Then the length of the arc is

`\displaystyle\int_0^(2\pi)\|\vec r'(t)\|\,\mathrm dt=\sqrt5\int_0^(2\pi)\mathrm dt=2\sqrt5\,\pi`