Tangent plane
= Tangent plane
For an embedded <regular surface> $M\subseteq\mathbb R^3$ and $p\in M$, the affine <tangent plane> is
$$
p+T_pM=\{p+v:v\in T_pM\}.
$$
It is a set of points through $p$, whereas the <tangent space> $T_pM$ is a <vector space> of displacements with <zero vector>. Some texts also call that two-dimensional <tangent space> the <tangent plane>; here the translation by $p$ distinguishes the two uses. For a regular <parametric surface> $X(u,v)$ with $p=X(u_0,v_0)$,
$$
T_pM=\operatorname{span}\{X_u,X_v\},\qquad
p+T_pM=\{p+aX_u+bX_v:a,b\in\mathbb R\},
$$
where the <partial derivatives> are evaluated at $(u_0,v_0)$. Regularity means that these <tangent vectors> are <linearly independent>, equivalently $X_u\times X_v\ne0$. Their <cross product> is a <normal vector>. https://www.math.uga.edu/sites/default/files/inline-files/ShifrinDiffGeo.pdf[Theodore Shifrin, Differential Geometry: A First Course in Curves and Surfaces, Chapter 2, §1: regular parametrizations and tangent planes]{ref}
For an implicit <regular surface> $F(x,y,z)=0$, require a <smooth function> $F$ with $\nabla F(p)\ne0$. Then
$$
T_pM=\ker DF_p=\{v:\nabla F(p)\cdot v=0\},\qquad
p+T_pM=\{r:\nabla F(p)\cdot(r-p)=0\}.
$$
Differentiating $F(\gamma(t))=0$ proves that every surface <velocity> lies in the displayed <kernel of a linear map>; the <implicit function theorem> gives a local graph with two independent coordinates, proving equality. The <gradient> is the <normal vector> with respect to the ambient <dot product>. https://www.math.toronto.edu/mein/teaching/LectureNotes/Man.pdf[Eckhard Meinrenken, Differentiable Manifolds, Theorem 1.15: regular level sets]{ref}
For the <sphere> $F(r)=r\cdot r-R^2$, $\nabla F(p)=2p\ne0$, and the affine <tangent plane> has equation $p\cdot r=R^2$. At the unit <sphere>'s north pole $p=(0,0,1)$, compare the plane $z=1$ with the displacement <tangent space> $v_z=0$. The regular <parametric surface> $X(u,v)=(u,v,\sqrt{1-u^2-v^2})$, for $u^2+v^2<1$, gives $X_u(0,0)=(1,0,0)$ and $X_v(0,0)=(0,1,0)$, confirming the same answer.
\Image[/tangent-vectors-spaces-planes-sphere-plane.png]
{title=The sphere's tangent plane passes through the base point; its tangent space contains displacements}
This analytical illustration shows a unit <sphere> and a vertical cross-section. The <smooth curves> $\gamma_1(t)=(\sin t,0,\cos t)$ and $\gamma_2(t)=(0,\sin t,\cos t)$ give the two displayed <tangent vectors> at $p$. The straight point $p+tv$ for any nonzero $v\in T_pS^2$ is outside the <sphere> when $t\ne0$, since $\|p+tv\|^2=1+t^2\|v\|^2$. A <tangent plane> therefore captures first-order motion, not an exact patch of the <regular surface>.
The same idea makes local approximation useful. For the <graph of a function> $z=f(x,y)$ with $f$ a <smooth function>, the affine <tangent plane> is $z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$. https://openstax.org/books/calculus-volume-3/pages/4-4-tangent-planes-and-linear-approximations[Gilbert Strang and Edwin Herman, OpenStax Calculus Volume 3, §4.4: tangent planes and linear approximations]{ref} With continuous second <partial derivatives>, the vertical error is $O((x-a)^2+(y-b)^2)$ in <Big O notation>. On the unit <sphere> near its north pole, $\sqrt{1-u^2-v^2}=1-\tfrac12(u^2+v^2)+O((u^2+v^2)^2)$: the <tangent plane> loses quadratic bending while preserving all first-order <velocities>.
A failed regularity test needs interpretation. Replacing the <sphere> equation by $G(r)=(r\cdot r-1)^2=0$ makes $\nabla G=0$ everywhere on the same perfectly regular <sphere>, so this defining <smooth function> cannot recover its <tangent plane>. Similarly, the usual spherical <parametric surface> loses <matrix rank> at the poles, while the graph parametrization above remains regular there. A vanishing <gradient> or degenerate parametrization invalidates that calculation; it alone does not prove that the underlying geometry has no <tangent plane>.