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I need to find the winding number of the closed curve $c(t)=(a \cos(t),b \sin(t))^T $, where $a,b > 0$ and $c:[0,2\pi) \to\mathbb{R}^2\setminus\{0\}$.

I don't understand how to do this.

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integrate $d\theta$ along the curve? If you get $2\pi$ then the winding number is one. If you get $4\pi$ then the number is $2$. In any event, the answer will be an integer multiple of $2\pi$ since your curve is closed. –  James S. Cook Oct 22 '12 at 2:31

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Integrate $d\theta = d(tan^{-1}(y/x)) = \frac{-y \ dx+x \ dy}{x^2+y^2}$. You have $x = a \cos(t)$ and $y=b\sin(t)$ hence $dx=-a\sin(t) dt$ and $dy=b\cos(t)dt$. We find: $$ d \theta = \frac{abdt}{a^2\cos^2(t)+b^2\sin^2(t)} $$ now integrate... or perhaps there is a lazier argument using homotopy.

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Oh, I think we are supposed to use the formula that the winding number of a closed curve about the origin is (1/2*pi)*the integral of alpha along the curve where alpha equals (-ydx+xdy)/(x^2+y^2). I don't really understand this. –  Nerlbao Nerl Oct 22 '12 at 2:49
    
@Nerlbao Nerl Suppose $x = \cos (t)$ and $y=\sin(t)$ and $t \in [0,4\pi)$ then you can calculate $d\theta = dt$ and so the integral gives $4\pi$. This means the unit-circle is covered twice by the curve. The winding number simply counts how many times you circle the origin in a CCW-fashion. The winding number would be negative if you had a CW-oriented curve. How was "winding number" defined for your context? –  James S. Cook Oct 22 '12 at 5:47
    
Oh, I suppose you are right. –  Nerlbao Nerl Oct 22 '12 at 6:03

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