I was given the following theorems:
let $M$ be a $k$ dimension manifold defined by $M=\{a\in \mathbb{R}^n:F(x)=0\}$ when $F\in C^1$ and Rank$(D_f(x))$ is maximal for all $a\in M$ so $T_a(M)={v\in \mathbb{R}^n:D_F(x)v=0}$
Let $M$ be a $k$ dimension manifold with atlas $A$ so for all $a\in M$ $T_a(M)=t_1\frac{\partial \phi}{\partial x_1}(x)+...t_k\frac{\partial \phi}{\partial x_k}(x)$ where $t_1,t_2,...,t_k\in \mathbb{R}$, $\phi\in A$ such that $\phi(x)=a$
Now I am trying to find the tangent plane of using those theorems:
a. $S=\{(x,y): x^2+y^2=R^2\}$ at $(\frac{R}{\sqrt{2}},-\frac{R}{\sqrt{2}})$
b. $z=x^2+y^2+5$ at $(1,1,7)$
a by 1: First we set $F=x^2+y^2$ now we take $D_F=\begin{pmatrix} \frac{\partial F}{\partial x} & \frac{\partial F}{\partial y} \end{pmatrix}=\begin{pmatrix} 2x & 2y \\ \end{pmatrix}$
(we have to check that that the Rank of $D_F$ is maximal for all $a\in M$? what if it is not?)
For any $(a_1,a_2)=a\in x^2+y^2$ the Rank of $(2a_{1} , 2a_{2})$ will be at most $1$ as else we had $S=(0,0)$
now we take the dot product $\begin{pmatrix} 2x & 2y \end{pmatrix} \begin{pmatrix} \frac{R}{\sqrt{2}} \\ -\frac{R}{\sqrt{2}} \end{pmatrix}=0\iff=R\sqrt{2}x-R\sqrt{2}y=0\iff R\sqrt{2}(x-y)=0$ so $x=y$?
b by 1: $F=z-x^2-y^2$ now $D_F=\begin{pmatrix} \frac{\partial F}{\partial x} & \frac{\partial F}{\partial y} & \frac{\partial F}{\partial z} \end{pmatrix}=\begin{pmatrix} -2x & -2y & 1 \\ \end{pmatrix}$ So $\begin{pmatrix} -2x & -2y & 1 \\ \end{pmatrix}\begin{pmatrix} 1 \\ 1 \\ 7 \end{pmatrix}=0\iff -2x-2y+7=0\iff 2x+2y=7$
Is it correct? How can I use method $2$ to find the tangent plane?
If the smooth manifold is given by $\vec{F}(X) = \vec{0}$, then the tangent space to the manifold at point $A$ is the null space (or kernel) of $D_{X}F(A)$.
In your case, $X = \begin{bmatrix}x \\ y \\ z \end{bmatrix}$ , $A=\begin{bmatrix}1 \\ 1 \\ 7 \end{bmatrix}$, $F(X) = z-x^2-y^2$, and the manifold is given by $F(X)=0$.
$D_{X}F(A) = [-2*1, -2*1, 1*7] = [-2,-2,7]$.
The tangent space at $A$ is the solution $\vec{\dot{X}} = \begin{bmatrix}\dot{x} \\ \dot{y} \\ \dot{z} \end{bmatrix}$ to $[-2,-2,7]\vec{\dot{X}} = 0$
To get the points on the tangent plane at A, add $\vec{a}$ to the vectors $\vec{\dot{X}}$ belonging to tangent space.
Hope that helps!