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I'm with a problem in an exercise form Do Carmo's Differential Geometry of Curves and Surfaces; it is number 9 section 1.6.

I have a differentiable real function $k(s)$, $s\in I$, and I need to show that the planar curve given by $$ \alpha (s)=\left( \int \cos\theta (s)ds +a,\int \sin\theta (s)ds +b\right) $$ where $$ \theta (s) = \int k(s)ds+\phi $$ has $k(s)$ as its curvature (there is an explanation for this curvature not being exclusively positive in the same book). Both a,b and $\phi$ are constants so they are only important for the second part of the exercise (show that the curve is determined uniquely except of translations and rotations by a $\phi$-angle).

So what I need to show that the curvature of $\alpha$ is $k(s)$. I have already solved exercise 12.d from the same section of the book which gives me

$$ k(s)=\dfrac{x'y''-x''y'}{|\alpha '(s) |^2} $$ Using the definition of $\alpha$ and the Fundamental Theorem of Calculus, I have: $$ \alpha '(s)=(\cos \theta (s)k(s),\sin \theta (s)k(s))=(x'(s),y'(s)) $$ and $$ \alpha ''(s)=(-\sin \theta (s) k(s)^2+\cos \theta(s) k'(s), \cos \theta(s) k(s)^2+\sin \theta(s) k'(s) )

This also gives me $|\alpha '(s)|= |k(s)|$. Substituting on the formula for $k(s)$ above (which should give me k(s)=k(s) ) returns me the value $$ \dfrac{k(s)}{|k(s)|} $$ which is $\pm$ and also very frustrating (shouldn't end up with any restrictions over $k$ ). I can't find where I'm making the mistake. I have tried to do this with arc lenght parametrization, which gave me the same result (by the way, that's how I obtained the formula for $k(s)$).

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The derivate of $\int\cos(\theta(s))$ is simply $\cos(\theta(s))$. So you should find $|\alpha'(s)|=1$. –  Xoff Aug 28 '12 at 18:23
... and therefore we have an example of a really simple problem that got complicated because of a wrong usage of the Calculus Fundamental Theorem. Thanks @Xoff ! That was is accurate and correct. –  Marra Aug 28 '12 at 18:33
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