Length of a 3D parametric function

applestrudle
Messages
64
Reaction score
0

Homework Statement



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


Homework Equations



L = integrate ds

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

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.

:(
 
Physics news on Phys.org
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.
 
  • Like
Likes 1 person
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top