Turns of closed path $\lambda$ around origin is related to area of paralelogram $\lambda(t), \lambda'(t)$ 
Given the closed path $\lambda:[a,b]\to\mathbb{R^2}-\{0\}$, $C^1$,
  call $A(t)$ the 'oriented area' of the paralelogram determined by the
  vectors $\lambda(t)$ and $\lambda'(t)$ for each $t\in [a,b]$. Show
  that the number of turns that $\lambda$ does arount he origin is:
$$n(\lambda, 0) = \frac{1}{2\pi} \int_a^b \frac{A(t)}{|\lambda(t)|^2}\
 dt$$



*

*The oriented area of a paralelogram such that its sides are given by vectros $v,w\in \mathbb{R^2}$, in this order, is positive when the rotation is from $v\to w$ by the least anti-clockwise angle, negative otherwise, and $0$ is the vectors are linearly dependent


As I understood, for each $t$, we have one $\lambda(t)$ and $\lambda'(t)$. We must consider the paralelogram formed by these $2$ vectors for each $t$ and call $A(t)$ its area in the point $t$. But I honestly can't see a relation from this to turns around the origin. 
 A: The key here is to realize that, if we express $\lambda(t)$ in polar coordinates $(r(t), \theta(t))$, then
$\theta'(t) = \dfrac{d\theta}{dt} = \dfrac{A(t)}{\vert \lambda(t) \vert^2}. \tag 1$
We now demonstrate this amazing fact.  In so doing, it will be convenient to introduce the standard cylidnrical coordinate system $(r, \theta, z)$ on $\Bbb R^3$, and to consider the planar polars $(r, \theta)$ to be embedded in the cylidnrical system in the conventional manner.
The vector $\lambda(t) = (x(t), y(t))$, when expressed in polar coordinates, is
$\lambda(t) = (r(t) \cos \theta(t), r(t) \sin \theta(t)) = r(t)(\cos \theta(t), \sin \theta(t)); \tag 2$
thus, in the cylindrical system, we have
$\lambda(t) = (r(t) \cos \theta(t), r(t) \sin \theta(t), 0) = r(t)(\cos \theta(t), \sin \theta(t), 0); \tag 3$
in these terms,
$\vert \lambda(t) \vert = \sqrt{r^2(t) \sin^2 \theta(t) + r^2(t) \cos^2 \theta(t)} = r(t); \tag 4$
also, from (3), 
$\lambda'(t) = r'(t)(\cos \theta(t), \sin \theta(t), 0) + r(t) (\cos \theta(t), \sin \theta(t), 0)'$
$= r'(t)(\cos \theta(t), \sin \theta(t), 0) + r(t)\theta'(t)(-\sin \theta(t), \cos \theta(t), 0). \tag 5$
Now the signed area $A(t)$ of the parallelogram whose sides are $\lambda(t)$ and $\lambda'(t)$ is
$A(t) = (\lambda(t) \times \lambda'(t)) \cdot \mathbf k, \tag 6$
where "$\times$" is the ordinary vector cross product in $\Bbb R^3$, and $\mathbf k$ is the usual unit normal vector to the $x$-$y$ plane, which is spanned by the usual unit vectors $\mathbf i$ and $\mathbf j$; see this widipedia article on the cross product.  We compute $(\lambda(t) \times \lambda'(t))\cdot \mathbf k$ based upon (2) and (5):
$\lambda(t) \times \lambda'(t)$
$= r(t)(\cos \theta(t), \sin \theta(t), 0) \times (r'(t)(\cos \theta(t), \sin \theta(t), 0) + r(t)\theta'(t)(-\sin \theta(t), \cos \theta(t), 0))$
$= r(t)r'(t)((\cos \theta(t), \sin \theta(t), 0) \times (\cos \theta(t), \sin \theta(t), 0))$
$+ r^2(t) \theta'(t)((\cos \theta(t), \sin \theta(t), 0) \times (-\sin \theta(t), \cos \theta(t), 0)); \tag 7$
now
$(\cos \theta(t), \sin \theta(t), 0) \times (\cos \theta(t), \sin \theta(t), 0) = 0, \tag 8$
and
$(\cos \theta(t), \sin \theta(t), 0) \times (-\sin \theta(t), \cos \theta(t), 0) = \mathbf k; \tag 9$
bringing (8) and (9) into (7) yields
$\lambda(t) \times \lambda'(t) = r^2(t)\theta'(t) \mathbf k; \tag{10}$
thus,
$A(t) = (\lambda(t) \times \lambda'(t)) \cdot \mathbf k = r^2(t)\theta'(t) \mathbf k \cdot \mathbf k = r^2(t) \theta'(t), \tag{11}$
or, via (4),
$A(t) = \vert \lambda(t) \vert^2 \theta'(t), \tag{12}$
whence
$\theta'(t) = \dfrac{A(t)}{\vert \lambda(t) \vert^2}, \tag{13}$
as was claimed.
With formula (13) in hand it is easy to dispense with the present question. Since $\lambda(t)$ is a closed path, we must have
$\lambda(a) = \lambda(b) \tag{14}$
or, in the polar system,
$(r(a)\cos \theta(a), r(a) \sin \theta(a)) = (r(b)\cos \theta(b), r(b) \sin \theta(b)), \tag{15}$
whence
$r(a) = r(b), \tag{16}$
$\cos \theta(a) = \cos \theta(b), \tag{17}$
$\sin \theta(a) = \sin \theta(b); \tag{18}$
the only way (17)-(18) can simultaneously bind is if
$\theta(b) = \theta(a) + 2\pi n \tag {19}$
for some $n \in \Bbb Z$ is clearly the number of times $\lambda(t)$ encircles the origin 'twixt $t = a$ and $t = b$; therefore we may write
$n(\lambda, 0) = \dfrac{1}{2\pi}(\theta(b) - \theta(a)) = \dfrac{1}{2\pi} \displaystyle \int_a^b \theta'(s)ds = \dfrac{1}{2\pi}\int_a^b \dfrac{A(s)}{\vert \lambda(s) \vert^2} ds, \tag{20}$
as was to be shown.
