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Tony's Intertwingled Memex • Reference Manual

Tangent Planes to Level Surfaces

Formulas & Definitions
Core Definition

Let SSS be a smooth level surface in R3\mathbb{R}^3R3 defined implicitly by the equation:

F(x,y,z)=cF(x, y, z) = cF(x,y,z)=c
where FFF has continuous partial derivatives and ∇F(x0,y0,z0)≠0⃗\nabla F(x_0, y_0, z_0) \neq \vec{0}∇F(x0​,y0​,z0​)=0 at a base point P0(x0,y0,z0)P_0(x_0, y_0, z_0)P0​(x0​,y0​,z0​) on SSS.

Because the gradient vector ∇F\nabla F∇F is always geometrically orthogonal to the level surface, it serves directly as the surface normal vector:

n⃗=∇F(x0,y0,z0)=⟨Fx(x0,y0,z0), Fy(x0,y0,z0), Fz(x0,y0,z0)⟩\vec{n} = \nabla F(x_0, y_0, z_0) = \left\langle F_x(x_0, y_0, z_0), \, F_y(x_0, y_0, z_0), \, F_z(x_0, y_0, z_0) \right\ranglen=∇F(x0​,y0​,z0​)=⟨Fx​(x0​,y0​,z0​),Fy​(x0​,y0​,z0​),Fz​(x0​,y0​,z0​)⟩

  1. Tangent Plane Equation:

    Fx(x0,y0,z0)(x−x0)+Fy(x0,y0,z0)(y−y0)+Fz(x0,y0,z0)(z−z0)=0F_x(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = 0Fx​(x0​,y0​,z0​)(x−x0​)+Fy​(x0​,y0​,z0​)(y−y0​)+Fz​(x0​,y0​,z0​)(z−z0​)=0

  2. Normal Line Equation (Symmetric / Parametric):

    r⃗(t)=⟨x0,y0,z0⟩+t∇F(x0,y0,z0)  ⟹  {x(t)=x0+Fxty(t)=y0+Fytz(t)=z0+Fzt\vec{r}(t) = \langle x_0, y_0, z_0 \rangle + t\nabla F(x_0, y_0, z_0) \implies \begin{cases} x(t) = x_0 + F_x t \\ y(t) = y_0 + F_y t \\ z(t) = z_0 + F_z t \end{cases}r(t)=⟨x0​,y0​,z0​⟩+t∇F(x0​,y0​,z0​)⟹⎩⎨⎧​x(t)=x0​+Fx​ty(t)=y0​+Fy​tz(t)=z0​+Fz​t​


Concept Explanation

Derivation via the Multivariable Chain Rule

To understand why ∇F\nabla F∇F is perpendicular to the tangent plane of the level surface F(x,y,z)=cF(x, y, z) = cF(x,y,z)=c, consider any smooth space curve CCC parameterized by r⃗(t)=⟨x(t),y(t),z(t)⟩\vec{r}(t) = \langle x(t), y(t), z(t) \rangler(t)=⟨x(t),y(t),z(t)⟩ that lies entirely on the surface SSS and passes through P0P_0P0​ at t=t0t = t_0t=t0​.

Because every point on CCC satisfies the surface equation:

F(x(t),y(t),z(t))=cfor all tF(x(t), y(t), z(t)) = c \quad \text{for all } tF(x(t),y(t),z(t))=cfor all t
Differentiating both sides with respect to ttt using the Multivariable Chain Rule:
∂F∂xdxdt+∂F∂ydydt+∂F∂zdzdt=0\frac{\partial F}{\partial x}\frac{dx}{dt} + \frac{\partial F}{\partial y}\frac{dy}{dt} + \frac{\partial F}{\partial z}\frac{dz}{dt} = 0∂x∂F​dtdx​+∂y∂F​dtdy​+∂z∂F​dtdz​=0
In vector notation, this is precisely the dot product:
∇F(P0)⋅r⃗′(t0)=0\nabla F(P_0) \cdot \vec{r}'(t_0) = 0∇F(P0​)⋅r′(t0​)=0
Since r⃗′(t0)\vec{r}'(t_0)r′(t0​) is the velocity vector tangent to the curve CCC at P0P_0P0​, and this relation holds for every curve lying on the surface through P0P_0P0​, the gradient vector ∇F(P0)\nabla F(P_0)∇F(P0​) must be orthogonal to every tangent vector. Consequently, ∇F(P0)\nabla F(P_0)∇F(P0​) is perpendicular to the entire tangent plane.

                   ^ z
                   |        ^ n = grad F (Surface Normal)
                   |        |
                   |      .'|'.   Level Surface F(x, y, z) = c
                   |     /  *--\-----> r'(t) (Tangent vector)
                   |    |  P_0  |
                   |     \     /
                   |      '...'  Tangent Plane: n . (r - r_0) = 0
                   +-------------------------> y
                  /
                 /
                v x

Unifying Explicit and Implicit Tangent Planes

Any explicitly defined surface z=f(x,y)z = f(x, y)z=f(x,y) can be reformulated as a level surface of an implicit three-variable function:

F(x,y,z)=f(x,y)−z=0F(x, y, z) = f(x, y) - z = 0F(x,y,z)=f(x,y)−z=0
Taking the partial derivatives of FFF:
Fx=fx,Fy=fy,Fz=−1F_x = f_x, \qquad F_y = f_y, \qquad F_z = -1Fx​=fx​,Fy​=fy​,Fz​=−1
Evaluating the gradient gives:
∇F=⟨fx,fy,−1⟩\nabla F = \langle f_x, f_y, -1 \rangle∇F=⟨fx​,fy​,−1⟩
Substituting into the implicit plane formula yields:
fx(x−x0)+fy(y−y0)−1(z−z0)=0  ⟹  z−z0=fx(x−x0)+fy(y−y0)f_x(x - x_0) + f_y(y - y_0) - 1(z - z_0) = 0 \implies z - z_0 = f_x(x - x_0) + f_y(y - y_0)fx​(x−x0​)+fy​(y−y0​)−1(z−z0​)=0⟹z−z0​=fx​(x−x0​)+fy​(y−y0​)
This perfectly reproduces the explicit tangent plane formula developed in Tangent Planes, demonstrating that the gradient-level surface formulation is the overarching, general framework.


Engineering Applications

  • Aerospace & Mechanical Propulsion: In supersonic nozzle and pressure vessel design, internal combustion surfaces are modeled implicitly by quadric or polynomial pressure envelopes F(x,y,z)=PburstF(x, y, z) = P_{\text{burst}}F(x,y,z)=Pburst​. Structural engineers compute ∇F\nabla F∇F to orient external reinforcement ribs exactly parallel to the normal vectors n⃗\vec{n}n.
  • Optical & Acoustical Engineering: Curved parabolic and ellipsoidal reflectors are defined by implicit quadric shells. The normal vector n⃗=∇F\vec{n} = \nabla Fn=∇F dictates the angle of incidence for incoming laser beams or acoustic wavefronts according to Snell's Law and specular reflection (v⃗reflected=v⃗−2(v⃗⋅n^)n^\vec{v}_{\text{reflected}} = \vec{v} - 2(\vec{v} \cdot \hat{n})\hat{n}vreflected​=v−2(v⋅n^)n^).
  • Computer Graphics & CAD Modeling: 3D rendering engines and finite-element mesh generators utilize ∇F\nabla F∇F to compute vertex shading normals for smooth Phong shading and collision detection algorithms.

Example Problems

Example 1: Tangent Plane to a Hyperboloid Level Surface

Find the equation of the tangent plane and the parametric equations of the normal line to the hyperboloid of one sheet:

x2+y2−z2=4x^2 + y^2 - z^2 = 4x2+y2−z2=4
at the point P0(3,2,3)P_0(3, 2, 3)P0​(3,2,3).

Solution
  1. Verify point and compute gradient:

    F(3,2,3)=32+22−32=9+4−9=4✓F(3, 2, 3) = 3^2 + 2^2 - 3^2 = 9 + 4 - 9 = 4 \quad \checkmarkF(3,2,3)=32+22−32=9+4−9=4✓
    ∇F(x,y,z)=⟨2x,2y,−2z⟩\nabla F(x, y, z) = \langle 2x, 2y, -2z \rangle∇F(x,y,z)=⟨2x,2y,−2z⟩

  2. Evaluate normal vector n⃗\vec{n}n at P0(3,2,3)P_0(3, 2, 3)P0​(3,2,3):

    n⃗=∇F(3,2,3)=⟨2(3),2(2),−2(3)⟩=⟨6,4,−6⟩\vec{n} = \nabla F(3, 2, 3) = \langle 2(3), 2(2), -2(3) \rangle = \langle 6, 4, -6 \ranglen=∇F(3,2,3)=⟨2(3),2(2),−2(3)⟩=⟨6,4,−6⟩

  3. Assemble tangent plane equation:

    6(x−3)+4(y−2)−6(z−3)=06(x - 3) + 4(y - 2) - 6(z - 3) = 06(x−3)+4(y−2)−6(z−3)=0
    6x−18+4y−8−6z+18=0  ⟹  6x+4y−6z=86x - 18 + 4y - 8 - 6z + 18 = 0 \implies 6x + 4y - 6z = 86x−18+4y−8−6z+18=0⟹6x+4y−6z=8
    Dividing by 2 gives the simplified standard form:
    3x+2y−3z=43x + 2y - 3z = 43x+2y−3z=4

  4. Parametric equations of normal line:

    {x(t)=3+3ty(t)=2+2tz(t)=3−3t\begin{cases} x(t) = 3 + 3t \\ y(t) = 2 + 2t \\ z(t) = 3 - 3t \end{cases}⎩⎨⎧​x(t)=3+3ty(t)=2+2tz(t)=3−3t​


Example 2: Tangent Plane to a Mixed Implicit Polynomial Surface

Find the equation of the tangent plane to the surface x2y+yz+z3=11x^2y + yz + z^3 = 11x2y+yz+z3=11 at the point P0(2,1,2)P_0(2, 1, 2)P0​(2,1,2).

Solution
  1. Compute partial derivatives:

    Fx=∂∂x[x2y+yz+z3]=2xyF_x = \frac{\partial}{\partial x}[x^2y + yz + z^3] = 2xyFx​=∂x∂​[x2y+yz+z3]=2xy
    Fy=∂∂y[x2y+yz+z3]=x2+zF_y = \frac{\partial}{\partial y}[x^2y + yz + z^3] = x^2 + zFy​=∂y∂​[x2y+yz+z3]=x2+z
    Fz=∂∂z[x2y+yz+z3]=y+3z2F_z = \frac{\partial}{\partial z}[x^2y + yz + z^3] = y + 3z^2Fz​=∂z∂​[x2y+yz+z3]=y+3z2

  2. Evaluate at P0(2,1,2)P_0(2, 1, 2)P0​(2,1,2):

    Fx(2,1,2)=2(2)(1)=4F_x(2, 1, 2) = 2(2)(1) = 4Fx​(2,1,2)=2(2)(1)=4
    Fy(2,1,2)=22+2=4+2=6F_y(2, 1, 2) = 2^2 + 2 = 4 + 2 = 6Fy​(2,1,2)=22+2=4+2=6
    Fz(2,1,2)=1+3(2)2=1+12=13F_z(2, 1, 2) = 1 + 3(2)^2 = 1 + 12 = 13Fz​(2,1,2)=1+3(2)2=1+12=13
    n⃗=∇F(2,1,2)=⟨4,6,13⟩\vec{n} = \nabla F(2, 1, 2) = \langle 4, 6, 13 \ranglen=∇F(2,1,2)=⟨4,6,13⟩

  3. Write and simplify tangent plane equation:

    4(x−2)+6(y−1)+13(z−2)=04(x - 2) + 6(y - 1) + 13(z - 2) = 04(x−2)+6(y−1)+13(z−2)=0
    4x−8+6y−6+13z−26=0  ⟹  4x+6y+13z=404x - 8 + 6y - 6 + 13z - 26 = 0 \implies 4x + 6y + 13z = 404x−8+6y−6+13z−26=0⟹4x+6y+13z=40


Connections & References

  • Prerequisites:
    • The Gradient Vector — Orthogonality of ∇F\nabla F∇F to level sets.
    • Tangent Planes — Explicit tangent plane formula z−z0=fx(x−x0)+fy(y−y0)z - z_0 = f_x(x - x_0) + f_y(y - y_0)z−z0​=fx​(x−x0​)+fy​(y−y0​).
    • Equations of Planes in 3D — Point-normal plane geometry n⃗⋅(r⃗−r⃗0)=0\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0n⋅(r−r0​)=0.
  • Direct Successors & Extensions:
    • Session 13 - Optimization with Constraints — Lagrange multipliers relying on parallel gradients ∇f=λ∇g\nabla f = \lambda \nabla g∇f=λ∇g.
  • Course Sessions:
    • Session 12 - Gradient and Directional Derivatives