- #1

BraedenP

- 96

- 0

## Homework Statement

I am trying to prove that the length of a helix can be represented by [itex]2\pi=\sqrt{a^2+b^2}[/itex]

## Homework Equations

## The Attempt at a Solution

I have the following so far:

If the helix can be represented by [itex]h(t)=a\cdot cos(t)+a\cdot sin(t)+b(t)[/itex]

Then the length is:

[tex]\int_{0}^{2\pi}\sqrt{(-a\cdot sin(t))^2+(a\cdot cos(t))^2+b^2}\;\: dt[/tex]

My problem comes when integrating this. If I use the stuff in the root as u and do u-substitution, then

*du*equals

*0dt*:

[tex]u=a^2sin^(t)+a^2cos^2(t)+b^2[/tex]

[tex]du=(a^2sin(2t)-a^2sin(2t))dt=0dt[/tex]

My logic fails me when figuring out how to continue from there. I need to somehow represent 1dt. How do I do this?

Help would be awesome!