# Basis for tangent space of embedded submanifold

I am studying for a geometry exam, and one of our practice questions is as follows:

Let $f:\mathbb{R}^3\to\mathbb{R}$ be defined by $$f(x,y,z)=2x^3+y^2+z^4-3xz^2.$$ a.) Prove that $M=f^{-1}(\{4\})$ is an embedded submanifold of $\mathbb{R}^3$

b.) Find a basis for the tangent space of the submanifold at $p=(1,2,1)$.

I know that for part a.) we just show that the differential of $f$ is always surjective at every point in M. That is pretty easy to do. What I am having trouble understanding is how to figure out which vectors in $T_p(\mathbb{R}^3)$ are also vectors in $T_p(M)$. I know that as $M$ has codimension $1$, $T_p(M)$ will be a 2 dimensional subspace of $\mathbb{R}^2$, but how do we explicitly compute this subspace?

If $v\in T_pM$, then $v$ is represented by a curve $\gamma : (-\epsilon , \epsilon) \to M$ so that $\gamma (0) = p$ and $\gamma'(0) = v$. As $M = f^{-1}(\{4\})$, $f\circ \gamma$ is a constant function and so
$$0 = (f\circ \gamma)(0)=\langle \nabla f(p) , \gamma'(0)\rangle = \langle \nabla f(p) , v\rangle.$$
Then as $v\in T_pM$ is arbitrary, $T_pM \subset \nabla f(p)^\perp$. By dimension counting, $$T_pM = \nabla f(p)^\perp.$$
Now $\nabla f = (6x^2 - 3z^2, 2y, 4z^3- 6xz)$ and $p = (1, 2, 1)$. So $\nabla f(p) = (3, 4, -2)$. It is easy to see that $$(4, -3, 0), (0, 1, 2)$$ forms a basis for $(3, 4, -2)^\perp$.
• Thank you for your answer. As a follow up, could you also just find the kernel of the differential? I am asking because we also have problems in which the image is, for instance, $\mathbb{R}^2$, in which case the gradient would not be as helpful. – TomGrubb Dec 12 '15 at 16:16
• In this case the kernel is just $\nabla f^\perp$ @bburGsamohT – user99914 Dec 12 '15 at 20:49