# Length of a 3D parametric function

1. Jan 1, 2014

### applestrudle

1. The problem statement, all variables and given/known data

find the length of a circular helix expressed in parametric form x= cos(t), y=sin(t) and z = t
from t = 0 to t =2pi

2. Relevant equations

L = integrate ds

(ds)^2 = (dx)^2+(dy)^2+(dz)^2

3. The attempt at a solution

I got to ds = (1 + (dt)^2)^0.5

but I can't integrate as the dt is squared and inside the square root.

:(

2. Jan 1, 2014

### HallsofIvy

It shouldn't be! You have x= cos(t) so dx= -sin(t)dt, y= sin(t) so dy= cos(t)dt, and z= t so dz= dt. Looks to me like you forgot the "dt" in both dx and dy.
$$ds= \sqrt{dx^2+ dy^2+ dz^2}= \sqrt{sin^2(t)dt^2+ cos^2(t)dt^2+ dt^2}= \sqrt{2}dt$$.