Question on the notation $ds^2=dx^2+dy^2$ : what does it really mean? I know that if a curve $C$ is parametrize as $$s(t)=(x(t),y(t)),\quad  t\in [a,b]$$
then $$\ell(C)=\int ds=\int_a^b \sqrt{\dot x^2(t)+\dot y^2(t)}dt.\tag{*}$$
In the french wikipedia, it's written that $(*)$ can be written as $$ds^2=dx^2+dy^2.$$
I really don't understand the logical. Could someone explain ? What is the link between $(*)$ and $ds^2=dx^2+dy^2$ ? I tried $$ds^2=dx^2+dy^2\implies ds=\sqrt{dx^2+dy^2}\implies \int ds=\int\sqrt{dx^2+dy^2},$$
but I can't give an interpretation of the RHS... So it should be wrong.
 A: The object $ds^2$ is a tensor field. It assigns to each point a tensor. The formula $ds^2=dx^2+dy^2$ tells you how this tensor field is defined.
The terminology looks fancy, but it simply does the following:
Write the canonical inner product on $\Bbb R^2$ as $\langle\cdot,\cdot\rangle:\Bbb R^2\times\Bbb R^2\to\Bbb R$. This function is a bilinear function. A tensor is simply a real-valued function that takes multiple vector as inputs, such that this function is linear in each argument. So inner product is itself a tensor. The tensor field $ds^2=dx^2+dy^2$ tells you that it assigns to each point the tensor $\langle\cdot,\cdot\rangle$, i.e. $ds^2$ is a function that sends each point $p\in\Bbb R^2$ to $\langle\cdot,\cdot\rangle$. For a point $p\in\Bbb R^2$, $ds^2(p)$ is a function that takes two vector entries (the inner product). So generically you can write $ds^2(p)(v,w)=\langle v,w\rangle$ for two vectors $v,w\in\Bbb R^2$.
What this tensor field does is the following:
A smooth curve $$\gamma(t)=(\gamma_1(t),\gamma_2(t)),\quad  t\in [a,b]$$ is given. (I don't want to use $s$, to indicate that $ds^2$ should not be associated to any curve. The tensor is defined independently from curve.) You can differentiate this curve at a point $t$, to obtain a vector $\gamma'(t)\in\Bbb R^2$.
We can define a quantity from this curve $\gamma$ and the tensor field $ds^2$, as follows:
$$\int_a^b\sqrt{ds^2(\gamma(t))(\gamma'(t),\gamma'(t))}\ dt.$$
If you look carefully enough, you would see $$ds^2(\gamma(t))(\gamma'(t),\gamma'(t))=\langle\gamma'(t),\gamma'(t)\rangle=\dot\gamma_1^2(t)+\dot\gamma_2^2(t)$$
So the above quantity is the same as $$\int_a^b \sqrt{\dot \gamma_1^2(t)+\dot \gamma_2^2(t)}\ dt.$$
There is one thing I have not quite justified yet, and that is, how $ds^2=dx^2+dy^2$ tells us the tensor field $ds^2$ assigns to each point the inner product $\langle\cdot,\cdot\rangle$ and not something else? Well, for the moment, take the following as a definition:
A general tensor field on $\Bbb R^2$ that assigns to each point a bilinear function is of the following general form $$p\mapsto f(p)dx\otimes dx+g(p)dx\otimes dy+h(p)dy\otimes dx+k(p)dy\otimes dy$$ for four real-valued functions $f,g,h,k$. The whole thing on the right is a bilinear function. By "definition", it takes two vector inputs $$\vec a=\begin{bmatrix}a_1\\a_2\end{bmatrix},\vec b=\begin{bmatrix}b_1\\b_2\end{bmatrix}\in\Bbb R^2$$ and outputs the following number: $$f(p)dx\otimes dx+g(p)dx\otimes dy+h(p)dy\otimes dx+k(p)dy\otimes dy(\vec a,\vec b)=\begin{bmatrix}a_1&a_2\end{bmatrix}\begin{bmatrix}f(p)&g(p)\\h(p)&k(p)\end{bmatrix}\begin{bmatrix}b_1\\b_2\end{bmatrix}.$$
(It is not really a definition, because it can be derived from some more fundamental definitions. I won't go into it here. If you want to know what the symbols $dx\otimes dy$ etc. mean, learn tensors from a book on analysis on $\Bbb R^n$ or differential geometry, and ask another question about these symbols.)
Now, $dx^2$ is a shorthand for $dx\otimes dx$ and $dy^2$ is a shorthand for $dy\otimes dy$. When the tensor field is $dx^2+dy^2$ (i.e. when $f=k=$ constant $1$ function, $g=h=$ constant zero function), you should recognise that $$\begin{bmatrix}a_1&a_2\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}b_1\\b_2\end{bmatrix}=\begin{bmatrix}a_1&a_2\end{bmatrix}\begin{bmatrix}b_1\\b_2\end{bmatrix}$$ is just the canonical inner product of $\vec a$ and $\vec b$.
Here is my comment on your manipulation of $ds^2$:
You should NOT take square root of $ds^2$ to become $ds$. The first reason is that this symbol $ds^2$ is NOT the square of some other symbol $ds$. The notation $$\int ds$$ is just an abuse of notation, to denote the total length of the curve. There is actually no such object $ds$.
A: Formally we have
$$
\int \sqrt{\dot x^2(t)+\dot y^2(t)} \, dt
= \int \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} \, dt \\
= \int \sqrt{\left[\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2\right]dt^2}
= \int \sqrt{dx^2+dy^2}
$$
It's not really stranger than this.
A: On your curve the expression
$$
dx = \dot{x}(t)dt
$$
says (informally) that if you change time by the "infinitesimal" amount $dt$ the $x$ value change $dx$ in $x$ will be given by that formula. Similarly for $dy$. 
You are interested in the amount $ds$ of arclength this change causes. To find that, you calculate the hypotenuse of the right triangle with sides $dx$ and $dy$. 
To find the length of the curve you add up those infinitesimal changes - that is, you integrate. 
Then $$\int\sqrt{dx^2+dy^2}$$ just rewrites (*) informally to remind you about what the integrand means geometrically.
