How to Find the Length of a Parametric Path?

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Homework Statement


Find the length of the path over (2 cos t - cos 2t, 2 sin t - sin 2t) 0<=t<=(pi/2)


Homework Equations


sin^(2) x = (1-cos 2x)/2


The Attempt at a Solution


I have worked my way thru the problem and I have arrived at 2 (sq root 2) antiderivative (sq root (1-cos t)), but I'm not sure how this coverts into the relevant equation above? Thanks in advance!
 
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stau40 said:

Homework Statement


Find the length of the path over (2 cos t - cos 2t, 2 sin t - sin 2t) 0<=t<=(pi/2)


Homework Equations


sin^(2) x = (1-cos 2x)/2


The Attempt at a Solution


I have worked my way thru the problem and I have arrived at 2 (sq root 2) antiderivative (sq root (1-cos t)), but I'm not sure how this coverts into the relevant equation above? Thanks in advance!

\sqrt{1 - cos(t)} = \sqrt{\frac{2(1 - cos(t))}{2}}=\sqrt{2sin^2(\frac{t}{2})}=\sqrt{2}sin(\frac{t}{2})
 
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...
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