TITLE: Describe tangent and normal bundle to a manifold
QUESTION [7 upvotes]: Consider the set
$X:=\{(x,y,z)\in\mathbb{R}^3 | x^2+3y^2=1+z^2\}$
I have to show that $X$ is a submanifold of $\mathbb{R}^3$ (and this is trivial); then, using on $T\mathbb{R}^3$ the standard coordinates $x,y,z,\partial_x,\partial_y,\partial_z$ I have to find the equations that describe $TX\subseteq T\mathbb{R}^3$ and $NX\subseteq T\mathbb{R}^3$.
Any suggestion? Thanks...

REPLY [3 votes]: The gradient vector $\nabla f = f_x\,\partial_x + f_y\,\partial_y + f_z \, \partial_z$ is at right angles to the zero-level set $f=0$. In your case $f(x,y,z) = x^2+3y^2-z^2-1$ and so:
$$\nabla f = 2x \, \partial_x + 6y \, \partial_y - 2z \, \partial_z \, .$$
A vector belongs to $X$ if it is based at a point of $X$ and is tangent to $X$ at that point. Equivalently, a vector must be based at a point of $X$ and be perpendicular to $\nabla f$. Such a vector must be based at a point $(x,y,z)$ for which $x^2+3y^2-z^2-1=0$ and be of the form $a \, \partial_x + b \, \partial_y + c \, \partial_z$ for whch $2xa + 6yb -2zc = 0$. This last condition means that 
$$a \, \partial_x + b \, \partial_y + c \, \partial_z \perp 2x \, \partial_x + 6y \, \partial_y - 2z \, \partial_z \, .$$
Putting this all together, it follows that:
$$TX = \{ (x,y,z;a,b,c) : x^2+3y^2-z^2 = 1 \ \text{ and } \ xa + 3yb - zc = 0 \} \, . $$
For the normal bundle, you want vectors based on $X$ and parallel to $\nabla f$ at that point. Hence:
$$NX = \left\{ (x,y,z;2kx,6ky,-2kz) : x^2+3y^2-z^2 = 1 \ \text{ and } \ k \in \mathbb{R} \right\} \, . $$