# Thread: Help with arc length of parametric curves

1. ## Help with arc length of parametric curves

x(t) = 2t - 2sin(t)
y(t) = 2 - 2cos(t)

t is from 0 to 2Pi

Okay, I am able to set up the problem correctly:

x'(t) = 2 - 2cos(t)
y'(t) = 2sin(t)

Length = Integral from 0 to 2Pi of (Sqrt[[2 - 2cos(t)]^2 + [2sin(t)]^2])
= Integral from 0 to 2Pi of (Sqrt[4[1-cos(t)]^2 + [2sin(t)]^2])

I know I can pull the 4 out (making it a 2), but I'm stuck after that. What good does making a substitution of 1 - cos(t) = 2sin(t/2)^2 do?

Help would be greatly appreciated!

2. Dear Drumiester,

next step: bring out 4 from root
next step: break down the bracket under the root and simplify
next step: you can bring out again 2 from root
you get $\sqrt{1-cos(t)}$

Now one trick: $cos(t) = cos(2*t/2) = cos^2(t/2) - sin^2(t/2)$
and
$1 = cos^2(t/2) + sin^2(t/2)$
So
$\sqrt{1-cos(t)} = \sqrt{cos^2(t/2)+sin^2(t/2) - cos^2(t/2) + sin^2(t/2)} = \sqrt{2}*sin(t/2)$

Now it's not so wild, isn't it?

3. Thanks man! I just discovered this site, so I'll be sure to come here for help