Tangent Space to a manifold So, I have a manifold $M=\{\mathbf{x}:\mathbf{\Theta}\left(\mathbf{x}\right)=\mathbf{0}\}$.
I can also write $M=\{\mathbf{x}:\mathbf{F}(\mathbf{x})=\mathbf{c}\}$. Both functions are differentiable. I understand why we can see the Tangent space $T_{\mathbf{x}_0}$ as the $Ker(D\mathbf{\Theta}(\mathbf{x}_0))$.( I think...) But why can we not say the same about $Ker(D\mathbf{F}(\mathbf{x}_0))$?  If we can, is $T_{\mathbf{x}_0}$ tangent to $\mathbf{\Theta}(\mathbf{x}_0)$ the same as $T_{\mathbf{x}_0}$ tangent to $\mathbf{F}(\mathbf{x}_0)$?Why?
To the first question, I would answer yes, and to the second no. Am I correct?
Many Thanks.
 A: Assuming that you are viewing $M$ as the level set of a constant rank function $\mathbf{\Theta}$ defined on a larger ambient manifold . . . 
For your question concerning why we (might) have $T_{\mathbf{x}_0}M = Ker\left(D\mathbf{\Theta}\left(\mathbf{x}_0\right)\right)$, you have to be a bit careful with the condition of $D\mathbf{\Theta}$.  Namely, to ensure that $T_{\mathbf{x}_0}M = Ker\left(D\mathbf{\Theta}\left((\mathbf{x}_0\right)\right)$ you need to know that $D\mathbf{\Theta}$ is surjective at $\mathbf{x}_{0}$. 
For a simple example of this, consider the following situation:
Let $M = \left\{ \mathbf{x}=(x, y) \in \mathbb{R}^{2} \Big\vert\,  x = 0\right\}$.  $M$ is clearly a submanifold of $\mathbb{R}^2$ and it is cut out by the function $\mathbf{\Theta}: \mathbb{R}^{2} \to \mathbb{R}$ defined by $\mathbf{\Theta}\left(\mathbf{x}\right) = x$.  In this instance, $T_{\mathbf{x}_0}M = Ker\left(D\mathbf{\Theta}\left(\mathbf{x}_0\right)\right)$ for all $\mathbf{x}_{0}$ in $M$.  However, you could also define $M$ as the level set of the function $\mathbf{F} : \mathbb{R}^{2} \to \mathbb{R}$ defined by $\mathbf{F}(\mathbf{x}) = x^2$.  In this instance, $\mathbf{F}$ is not a map of constant rank as the rank drops at $\mathbf{x}_{0} = \left(0, 0\right)$ and $T_{\mathbf{x}_0}M \ne Ker\left(D\mathbf{F}\left(\mathbf{x}_0\right)\right)$, but instead $T_{\mathbf{x}_0}M \subset Ker\left(D\mathbf{F}\left(\mathbf{x}_0\right)\right)$.
Regarding the intuition of this situation, there is in fact a beautiful (and wonderfully intuitive) description.  You have defined $M$ as a level set of some function $\mathbf{\Theta}$, which means that $\mathbf{\Theta}$ is constant along $M$.  If $\mathbf{x}_0 \in M$ and $v \in T_{\mathbf{x}_{0}}M$ is tangent to $M$, then one expects $D\mathbf{\Theta}\left(\mathbf{x}_0\right)\left(v\right)$ to be zero precisely because $\mathbf{\Theta}$ is constant on $M$! (Note by the example above, $D\mathbf{\Theta} \left(\mathbf{x}_0\right)\left(v\right)$ being zero is merely a necessary condition for $v$ to be tangent to $M$ and not a sufficient condition for $v$ to be tangent to $M$.)
