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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?

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  • $\begingroup$ Tangent vectors are $dr_1=(du_1,dv_1)$ and $dr_2=(du_2,dv_2)$ all respective to t $\endgroup$ – Djura Marinkov Sep 18 '16 at 20:40
  • $\begingroup$ @DjuraMarinkov I can't understand your notation :( $\endgroup$ – Guerlando OCs Sep 18 '16 at 20:44
  • $\begingroup$ @DjuraMarinkov $r_1$ and $r_2$ are curves, how can I take it's derivative with respect to anything but $t$? $\endgroup$ – Guerlando OCs Sep 18 '16 at 20:50
  • $\begingroup$ You don't find derivative dv/du if you asking that? you need tangent vectors $\endgroup$ – Djura Marinkov Sep 18 '16 at 21:00
  • $\begingroup$ Tangent vectors are on u,v surface, so they have some u,v coordinates. You find them by derivating u1,v1,u2,v2 respective to t. After that you can play with them to find angle between... $\endgroup$ – Djura Marinkov Sep 18 '16 at 21:06
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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 $\alpha_1,\alpha_2\in F$

  1. for every $x_1,x_2,y\in V\ f(\alpha_1x_1+\alpha_2x_2,y)=f(\alpha_1x_1,y)+f(\alpha_2x_2,y)$
  2. 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 $I(\alpha_2',\alpha_2')$

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