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$s(t)=(\frac{2}{t^2+1},\frac{2t}{t^2+1})$

I need to calculate a line integral along this path. But I have trouble understanding what it is. I did some googling and it looks that it is a parabola, but I am not sure. And if it is can someone give me the parameter change function to parametrize it in more simple way for solving the above mentioned integral. I mean something like $s(x)=(x,x^{2})$

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2 Answers 2

up vote 1 down vote accepted

In fact, the trajectory is a circle. To see why, let $t = \tan \theta / 2$ where $-\pi < \theta < \pi$. Then since

$$ \frac{1-t^2}{1+t^2} = \cos \theta \quad \text{and} \quad \frac{2t}{1+t^2} = \sin \theta, $$

we have

$$ s(t) = \left( 1 + \frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2} \right) = (1 + \cos\theta, \sin \theta). $$

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Thankyou @sos440. More cleare impossible. –  Mykolas Nov 17 '12 at 10:11

If $$x=\frac 2{t^2+1},y=\frac{2t}{t^2+1}$$

On division, $\frac yx=t$

So, $$2=x(t^2+1)=x\{\left(\frac yx\right)^2+1\}\implies 2x=x^2+y^2\implies (x-1)^2+y^2=(1)^2$$

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