Local Flatness of a Sphere in 3-D Euclidean Space

In 3-D Euclidean space, a sphere can be locally approximated as a plane. The proof is as follows.

Notation

Symbol	Type	Explanation
$R$	${\mathbb{R}}^{+}$	radius of the sphere
$x,y,z$	$\mathbb{R}$	coordinates on the sphere
$P(x_0,y_0,z_0)$	${\mathbb{R}}^3$	point on the sphere's surface
$C(a,b,c)$	${\mathbb{R}}^3$	center of the sphere
$x_0,y_0,z_0$	$\mathbb{R}$	coordinates of $P(x_0,y_0,z_0)$
$a,b,c$	$\mathbb{R}$	coordinates of $C(a,b,c)$
$\mathbf{n}$	${\mathbb{R}}^3$	normal vector at $P$
$A,B,C$	$\mathbb{R}$	components of $\mathbf{n}$
$\mathrm{\Delta }x,\mathrm{\Delta }y,\mathrm{\Delta }z$	$\mathbb{R}$	small perturbations at $P$
$\approx$	relation	approximation
$(\mathrm{\Delta }x{)}^2,(\mathrm{\Delta }y{)}^2,(\mathrm{\Delta }z{)}^2$	$\mathbb{R}$	second-order infinitesimals

Proof

1. Sphere in 3-D Euclidean Space

Consider a sphere with center $C(a,b,c)$ and radius $R$ in 3-dimensional Euclidean space. The equation of the sphere is: $$(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=R^2$$ where $(x,y,z)$ are the coordinates of any point on the surface of the sphere.

2. Tangent Plane to the Sphere

Let $P(x_0,y_0,z_0)$ be a point on the sphere's surface. The tangent plane at $P$ is the plane that touches the sphere at exactly one point, $P$, and is perpendicular to the radius of the sphere at $P$.

To find the equation of the tangent plane at $P(x_0,y_0,z_0)$, observe that the vector from the center of the sphere $C(a,b,c)$ to the point $P(x_0,y_0,z_0)$ is given by: $$\mathbf{n}=(x_0-a,y_0-b,z_0-c)$$

This vector is the normal (perpendicular) vector to the tangent plane at $P$. The general equation of a plane with normal vector $\mathbf{n}=(A,B,C)$ passing through a point $(x_0,y_0,z_0)$ is: $$A(x-x_0)+B(y-y_0)+C(z-z_0)=0$$

Substituting the components of the normal vector $\mathbf{n}$ into the equation, the equation of the tangent plane at $P(x_0,y_0,z_0)$ becomes: $$(x_0-a)(x-x_0)+(y_0-b)(y-y_0)+(z_0-c)(z-z_0)=0$$

3. Simplifying the Tangent Plane Equation

Expanding and rearranging the equation, we get: $$(x_0-a)x+(y_0-b)y+(z_0-c)z=(x_0-a)x_0+(y_0-b)y_0+(z_0-c)z_0$$

This is the equation of the tangent plane to the sphere at $P(x_0,y_0,z_0)$.

4. Local Approximation

Consider any point $Q(x,y,z)$ on the sphere around the point $P(x_0,y_0,z_0)$, where $$x=x_0+\mathrm{\Delta }x,y=y_0+\mathrm{\Delta }y,z=z_0+\mathrm{\Delta }z.$$

Since $Q(x,y,z)$ is on the sphere, we get $${(x_0+\mathrm{\Delta }x-a)}^2+{(y_0+\mathrm{\Delta }y-b)}^2+{(z_0+\mathrm{\Delta }z-c)}^2=R^2,$$ which we further expand to $$(x_0-a{)}^2+2(x_0-a)\mathrm{\Delta }x+(\mathrm{\Delta }x{)}^2+(y_0-b{)}^2+2(y_0-b)\mathrm{\Delta }y+(\mathrm{\Delta }y{)}^2+(z_0-c{)}^2+2(z_0-c)\mathrm{\Delta }z+(\mathrm{\Delta }z{)}^2=R^2.$$ Notice that the terms $$(x_0-a{)}^2+(y_0-b{)}^2+(z_0-c{)}^2=R^2$$ because $P(x_0,y_0,z_0)$ lies on the sphere, so these terms cancel out with the $R^2$ on the right-hand side. We are left with: $$2(x_0-a)\mathrm{\Delta }x+2(y_0-b)\mathrm{\Delta }y+2(z_0-c)\mathrm{\Delta }z+(\mathrm{\Delta }x{)}^2+(\mathrm{\Delta }y{)}^2+(\mathrm{\Delta }z{)}^2=0$$ The terms $(\mathrm{\Delta }x{)}^2,(\mathrm{\Delta }y{)}^2$, and $(\mathrm{\Delta }z{)}^2$ are second-order infinitesimals, meaning they become negligible when $\mathrm{\Delta }x,\mathrm{\Delta }y$, and $\mathrm{\Delta }z$ are very small.

Therefore, we can approximate the equation by neglecting these second-order terms: $$2(x_0-a)\mathrm{\Delta }x+2(y_0-b)\mathrm{\Delta }y+2(z_0-c)\mathrm{\Delta }z\approx 0$$ Dividing the above equation by 2 gives the linear approximation: $$(x_0-a)\mathrm{\Delta }x+(y_0-b)\mathrm{\Delta }y+(z_0-c)\mathrm{\Delta }z=0$$

Replace $x$, $y$, $z$ with $x_0+\mathrm{\Delta }x$, $y_0+\mathrm{\Delta }y$, $z_0+\mathrm{\Delta }z$: $$(x_0-a)(x-x_0)+(y_0-b)(y-y_0)+(z_0-c)(z-z_0)=0$$ This equation is precisely the tangent plane equation through the point $P(x_0,y_0,z_0)$.

Therefore, near any point $P(x_0,y_0,z_0)$ on the sphere, the sphere's equation can be approximated by the tangent plane equation through $P(x_0,y_0,z_0)$.