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

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

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

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

Then the length of the arc is

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