I am self studying Differntial Geometry with Do Carmo and I am attempting problem 9 of section 2-4.
Specifically I am confused about the rotation statement.
After not understanding that part I had to peek at the solution in Slader:
But I am still confused.
If we define $u=u_0$ then the coordinate remains fixed. This in turn means that that normal, defined as:
$<\sin(u_0)a, -\cos(u_0)a, v>$ is not rotating.
For it to rotate I would need to fix $v$ and let $u$ be the one that varies, i.e. $<\sin(u)a, -\cos(u)a, v_0>$.
I may be misreading the question too. The question, for those that do not have access to the book is:
Given the surface $X(u,v)=<v\cos(u), v\sin(u), au>, a\neq 0$ Compute it's normal $N(u,v)$ and show that along the coordinate line $u=u_0$ the tangent plane of $x$ rotates about this line in such a way that the tangent of its angle with the $z$axis is proportional to its distance $v=\sqrt{x^2 + y^2}$ of the point $X(u_0, v)$ to the $z$ axis.
What am I misreading about the question?