First Fundamental Form and angle between curves in surface I'm reading this page about the first fundamental form, and in the part of angle between two curves $r_1$ and $r_2$ on a surface $r(u,v)$ where
$$r_1 = r(u_1(t), v_1(t))$$
$$r_2 = r(u_2(t), v_2(t))$$
Then the angle of two curves gives the equation 3.17, but I don't know how to arrive there. It says that I must take the inner product between the tangent vectors of $r_1$ and $r_2$, but I should take the derivative with respect to what? $u,v$ or $t$? How to arrive at that?
What are those $du_1, du_2$? There's no integral, how can there be differentials?
 A: Linear algebra background:
Let $V$ be a vector space over field $F$ and $[e_1 \dots e_n]$ a base for $V$.
Bilinear form on $V$ is a function $f:V^2\rightarrow F$ such that for every $\alpha_1,\alpha_2\in F$


*

*for every $x_1,x_2,y\in V\ f(\alpha_1x_1+\alpha_2x_2,y)=f(\alpha_1x_1,y)+f(\alpha_2x_2,y)$

*for every $x,y_1,y_2\in V\ f(x,\alpha_1y_1+\alpha_2y_2)=f(x,\alpha_1y_1)+f(x,\alpha_2y_2)$


Matrix of bilinear form $f$, in the base $[e_1 \dots e_n]$ is:
$$\begin{bmatrix}f(e_1,e_1) & f(e_1,e_2) & \dots & f(e_1,e_n)\\
\vdots & \vdots & & \vdots\\
f(e_n,e_1) & f(e_n,e_2) & \dots & f(e_n,e_n)
\end{bmatrix}$$
For every $a,b\in V,\ a=\alpha_1e_1+ \dots +\alpha_ne_n,\ b=\beta_1e_1+ \dots +\beta_ne_n$:
$$f(a,b)=\begin{bmatrix}\alpha_1 & \dots & \alpha_n \end{bmatrix}\begin{bmatrix}f(e_1,e_1) & f(e_1,e_2) & \dots & f(e_1,e_n)\\
\vdots & \vdots & & \vdots\\
f(e_n,e_1) & f(e_n,e_2) & \dots & f(e_n,e_n)
\end{bmatrix}\begin{bmatrix}\beta_1\\ \vdots\\ \beta_n\end{bmatrix}$$
Geometry:
You should know that for every curve $\alpha(t)=r(u(t),v(t))$ on the surface $r$ $\alpha'(t)=u'(t)r_u+v'(t)r_v$. That means that for every $p=(u,v)\in\mathbb{R}$x$\mathbb{R}$ vectors $r_u(u,v)$ and $r_v(u,v)$ make a base for the tangent space $T_p$ in $(u,v)$.
First fundamental form of the surface is a restriction of the standard dot product in $\mathbb{R}^3$ on the tangent space.
That means that first fundamental form is a bilinear form on the tangent space, and it has it's own matrix. However, in this case, as a base we use vectors $r_u$ and $r_v$, since they are the most convenient.
$$I:T_p^2 \rightarrow \mathbb{R},\ a=k_a^1r_u+k_a^2r_v,\ b=k_b^1r_u+k_b^2r_v,\\ I(a,b)=\begin{bmatrix}k_a^1 & k_a^2\end{bmatrix}\begin{bmatrix}
E & F\\
F & G
\end{bmatrix}\begin{bmatrix} k_b^1\\k_b^2\end{bmatrix}$$,
where $$E=<r_u,r_u>\\F=<r_u,r_v>\\G=<r_v,r_v>\\$$ are coefficients of the first fundamental form.
Just like with the standard dot product, where $cos\angle(a,b)=\frac{<a,b>}{\lVert a\rVert\lVert b\rVert}$, it's true that: $cos\angle(a,b)=\frac{I(a,b)}{\sqrt{I(a,a)I(b,b)}}$.
Finding the angle between two curves is equivalent to finding the angle between their tangent vectors. If $\alpha_1=r(u_1(t),v_1(t))$ and $\alpha_2=r(u_2(t),v_2(t))$, then:
$$I(\alpha_1',\alpha_2')=\begin{bmatrix}u_1' & v_1'\end{bmatrix}\begin{bmatrix}
E & F\\
F & G
\end{bmatrix}\begin{bmatrix} u_2'\\v_2'\end{bmatrix}$$
and
$$\lVert \alpha_1' \rVert=\sqrt{I(\alpha_1',\alpha_1')},\\
I(\alpha_1',\alpha_1')=\begin{bmatrix}u_1' & v_1'\end{bmatrix}\begin{bmatrix}
E & F\\
F & G
\end{bmatrix}\begin{bmatrix} u_1'\\v_1'\end{bmatrix}
$$, and similarily for $I(\alpha_2',\alpha_2')$
