Length of a 3D parametric function

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SUMMARY

The discussion focuses on calculating the length of a circular helix defined by the parametric equations x = cos(t), y = sin(t), and z = t, over the interval from t = 0 to t = 2π. The correct formula for the differential arc length, ds, is derived as ds = √(sin²(t)dt² + cos²(t)dt² + dt²), simplifying to ds = √2 dt. The integration of ds over the specified interval yields the total length of the helix, which is 2√2π.

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  • Understanding of parametric equations in 3D space
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of differential geometry concepts
  • Ability to manipulate trigonometric identities
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  • Study the process of integrating parametric equations in 3D
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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.

:(
 
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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.
 
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