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I'm a little rusty on this stuff, so bear with me.

Let $F(t) = (x(t), y(t))$ be a closed continuously differentiable curve in $\mathbb{R}^2 \setminus \{(0,0) \}$. How do I compute the following integral?

$$\int_{0}^{2\pi} \frac{x(t)x'(t) + y(t)y'(t)}{x^2(t) + y^2(t)}dt .$$

I believe there's a substitution I can make. That is, if I let $u = \ln(x^2(t) + y^2(t))$, then the integral reduces to $$ \int_{t=0}^{t=2\pi}du, $$ which is rather easy to deal with. But, when I started to compute $du/dt$, I realized that I'm actually not quite sure how to take such a derivative. That is, we clearly have that $$\frac{du}{dt} = \frac{(x^2(t))' + (y^2(t))'}{x^2(t) + y^2(t)} ,$$ but what is, for example, $(x^2(t))'?$ By the chain rule, I assume it is something like $(x(x(t)))' = x'(x(t))x'(t)$. But that doesn't seem right... and isn't quite what I want, which would be $x(t)x'(t)$.

I feel like I'm perhaps over-looking something quite obvious...

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  • $\begingroup$ You have the right idea. Your function is conservative. $\endgroup$ – Doug M May 11 '18 at 20:55
  • $\begingroup$ You don't want $t$-limits on a $u$-integral. $\endgroup$ – Fly by Night May 11 '18 at 21:03
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$$x^2(t)=x(t)\cdot x(t)$$ so $$(x^2(t))'=2x(t)\cdot x'(t)$$ Your mistake was to think that $$x^2(t)=x(x(t))$$

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  • $\begingroup$ Hmm. That certainly gets me what I want, but it's sort of strange since $x$ is a function of $t$. If, for example, I were to see the expression $f^2(t)$, I'd certainly interpret this as $f(f(t))$, and interpret, say, $(f(t))^2$ as $f(t) \cdot f(t)$. I suppose it's simply a contextual nuance that I'm overlooking... $\endgroup$ – thisisourconcerndude May 11 '18 at 20:55
  • $\begingroup$ Have you ever seen $\sin^2x+\cos^2x=1$? It is not $\sin(\sin(x))$ $\endgroup$ – Andrei May 11 '18 at 20:57
  • $\begingroup$ Ah. Yes, that's right. Thanks. $\endgroup$ – thisisourconcerndude May 11 '18 at 21:01
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$$\frac{du}{dt} = \frac{(x^2(t))' + (y^2(t))'}{x^2(t) + y^2(t)}=$$

$$\frac{2x(t)x'(t) + 2y(t)y'(t)}{x^2(t) + y^2(t)}$$

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