length of path travelled on $(t, \cos t, \sin t)$ from times $t = 0$ and $t = 2\pi$ Let the position of a particle in three dimensional space at time t be
$(t, \cos t, \sin t)$. Then the length of the path traversed by the particle between
the times $t = 0$ and $t = 2\pi$ is
(A) $2\pi$. 
(B) $2\sqrt{2}\pi$
(C) $\sqrt{2}\pi$ 
(D) none of the above.
My thoughts: $ds = \sqrt{(dx)^2 + (dy)^2 + (dz)^2}dt=\sqrt{(1 + \sin^2t + \cos^2t)} dt$
= $\sqrt{2} dt$
Integrating between  $t = 0$ and $t = 2\pi$, I get $2\sqrt{2}\pi$.
However, If I try to visualise this :
$\cos^2t +\sin^2t = 1$.
so, $x^2+y^2 =1$
Hence, the particle describes a circle in yz plane.
and in xy plane it is a cosine wave, in xz plane it is a sine wave.
combining these three, it seems that for each $2\pi$ time it completes one revolution in xy plane.
Each revolution is the perimeter of the circle $x^2+y^2 =1$, so it should be $2\pi$.
I am getting conflicting answers.
I am not sure if I have done this correctly. Can someone help me confirm this ?
Thanks.
 A: Here is a graph which may gives you a clearer idea

Here is another view

A: By definition, the arc-length of a curve $\gamma: [a,b] \to \Bbb R^n$ is: $$L[\gamma] = \int_a^b \|\gamma'(t)\|\,{\rm d}t.$$ Here, $\gamma: [0,2\pi] \to \Bbb R^3$ is given by $\gamma(t) = (t,\cos t, \sin t)$, so $\gamma'(t) = (1,-\sin t, \cos t)$ and so: $$\|\gamma'(t)\| = \sqrt{2} \implies L[\gamma] = \int_0^{2\pi}\sqrt{2}\,{\rm dt} = 2\sqrt{2} \pi.$$

Each revolution is the perimeter of the circle $x^2+y^2=1$, so it should
   be $2π$.

It is not $2\pi$. The vertical displacement adds a bit of length there. Not even the $x^2+y^2=1$ circle, actually.. it would be the $y^2+z^2=1$. And if you think in $3D$, these equations by themselves describe cylinders, not circles.
A: The following way to visualize the curve works in this situation, but it is not reliable in general.
If you look at the $yz$-plane, the curve traces out a circle of length $2\pi$.  If you look in the $x$-axis, the curve traces out a line of length $2\pi$.  If you think of the final curve as a combination of these two curves, you could "apply" the Pythagorean theorem to get a length of $\sqrt{(2\pi)^2+(2\pi)^2}=2\sqrt{2}\pi$.
Warning!  This will not always lead to the right answer - it is just a way to think about this problem.
A: Making a paper model of the cylinder in question and then flattening it out you will see that the curve appears as diagonal of a square with side length $2\pi$. Its length therefore is $\sqrt{2}\cdot2\pi$.
