Consider the smooth vector field on $\mathbb{R}^3$ given by
$$V=-y \frac{\partial }{\partial y}+z \frac{\partial }{\partial z}$$
and let $M=\{(x,y,z)\in \mathbb{R}^3 \ | \ x^2+yz=1\}$.
a) Show that $M$ is an embedded submanifold of $\mathbb{R}^3$.
b) Show that $V$ is tangent to $M$.
I know how to do a), I just need help with b). I am thinking of
Let $f:\mathbb{R}^3\to \mathbb{R}$ to be $f(x,y,z)=x^2+yz$. Then compute the kernel of $df$. However, I’m a little bit stuck at this. Any help would be appreciated.
Thank you.
Given a smooth function $f : \mathbb{R}^3 \to \mathbb{R}$, the derivative $df$ defined on the tangent space of each $(x, y, z) \in \mathbb{R}^3$ is speficied by the matrix
$$ d_{(x, y, z)} f = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \end{bmatrix} = \begin{bmatrix} 2x & z & y \end{bmatrix} $$
and the matrix is with respect to the basis $\{ \frac{\partial}{\partial x} , \frac{\partial}{\partial y} , \frac{\partial}{\partial z} \}$ of the tangent space at $(x, y, z)$. See that the vector field $V$ at each point on the manifold is in the kernel of the above transformation.