Let us consider the parametric surface $F(u,v)$, now say we consider the curve $F(u,v_0)$. In my book, they always show this curve as a curve that is projected on the say, $xy$ plane creates a straight line. This is not always the case, right? It is a line in the "domain" of $F(u,v)$ but depending on the parametrization, it could be a very crazy squiggly line that in no way resembles a line no matter how we project it onto whichever plane. If what I am saying is true, then given a point and taking the partial derivatives of the parametric equation, there is no guarantee that these tangent vectors will be perpendicular at all. Yet, my book shows, in all the diagrams, that that is the case. Is it just an unfortunate illustration problem, or am I missing something?
Thank you.
Correct. The partial velocities $F_u$ and $F_v$ (where $F = (x,y,z)$ parametrizes a surface) need not be perpendicular. That special condition $F_u \cdot F_v =0$ makes the parametrization nice and often we form $F$ by freezing on of a triple of an orthonormal coordinate system so the $F$ has this property. For example, $$ F(\phi, \theta) = (R\cos \theta \sin \phi, R\sin \theta \sin \phi, R \cos \phi) $$ for sphere of radius $R$ freezes $\rho = R$, or $$ F(\theta, \rho) = (\rho\cos \theta \sin \phi_o, \rho\sin \theta \sin \phi_o, \rho\cos \phi_o) $$ parametrizes the cone $\phi = \phi_o$. We can construct half-planes parametrized by $\rho, \phi$ by freezing $\theta = \theta_o$ in the same fashion. Then cylindrical coordinates allow easy parametrizations of cylinders in much the same fashion. All of this said, it's easy to give $F$ which are not like this: $$ F(u,v) = (u+v,u,u+v) $$ has $F_u = (1,1,1)$ and $F_v=(1,0,1)$ and $F_u \cdot F_v \neq 0$ is a parametrization of a plane with tangent vectors $\langle 1,1,1 \rangle$ and $\langle 1,0,1 \rangle$. Since $\langle 1,0,-1 \rangle$ is perpendicular to both tangent vectors we discern it is normal to the plane $x=z$ (in retrospect, seeing this is the Cartesian equation of the plane is fairly obvious).