Tangent Planes | Tony's Intertwingled Memex

Core Definition

Let $z = f(x, y)$ define a surface in $\mathbb{R}^3$ , and let $f(x, y)$ have continuous first-order partial derivatives near the base point $(x_0, y_0)$ , where $z_0 = f(x_0, y_0)$ . The tangent plane to the surface at the point $P_0(x_0, y_0, z_0)$ is the unique flat plane that contains the tangent lines to all surface trace curves passing through $P_0$ :

Explicit Tangent Plane Equation: $z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$
Point-Normal Form: $f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) - (z - z_0) = 0$
Surface Normal Vectors ( $\vec{n}$ ): $\vec{n}_{\text{down}} = \langle f_x(x_0, y_0), f_y(x_0, y_0), -1 \rangle = f_x\hat{i} + f_y\hat{j} - \hat{k}$ $\vec{n}_{\text{up}} = \langle -f_x(x_0, y_0), -f_y(x_0, y_0), 1 \rangle = -f_x\hat{i} - f_y\hat{j} + \hat{k}$

Concept Explanation

Single-Variable Analogy to Multivariable Extension

In single-variable calculus, the derivative $f'(x_0)$ provides the slope of the unique tangent line to the curve $y = f(x)$ at $x_0$ :

y - y_0 = f'(x_0)(x - x_0) \implies y = y_0 + f'(x_0)(x - x_0)

In multivariable calculus, a function $z = f(x, y)$ represents a curved two-dimensional surface living in three-dimensional space $\mathbb{R}^3$ . At any point $P_0(x_0, y_0, z_0)$ on a smooth surface, there are infinitely many different directions in which one can move away from $P_0$ , generating infinitely many different directional tangent lines. When the surface is differentiable, all of these directional tangent lines lie within a single, unique flat plane: the tangent plane.

                   ^ z
                   |        _.-""-._   Surface z = f(x, y)
                   |      .'  / \   '.
                   |     /   / P_0\   \
                   |    |---/------\---|  <--- Tangent Plane
                   |     \  \  Tx  /  /
                   |      '. \_Ty_/ .'
                   +-------------------------> y
                  /
                 /
                v x

Derivation from Coordinate Traces and Cross Products

The tangent plane can be constructed by analyzing the two fundamental coordinate cross-sections through $P_0(x_0, y_0, z_0)$ :

The $x$ -Trace Curve ( $C_1$ ): Slicing the surface with the vertical plane $y = y_0$ produces the trace curve $z = f(x, y_0)$ . The slope of this curve at $x_0$ is the partial derivative $f_x(x_0, y_0)$ . Moving $\Delta x = 1$ in the $x$ -direction produces $\Delta y = 0$ and $\Delta z = f_x(x_0, y_0)$ . This gives the tangent vector $\vec{T}_x$ :
$\vec{T}_x = \langle 1, 0, f_x(x_0, y_0) \rangle = \hat{i} + f_x(x_0, y_0)\hat{k}$
The $y$ -Trace Curve ( $C_2$ ): Slicing the surface with the vertical plane $x = x_0$ produces the trace curve $z = f(x_0, y)$ . The slope of this curve at $y_0$ is the partial derivative $f_y(x_0, y_0)$ . Moving $\Delta y = 1$ in the $y$ -direction produces $\Delta x = 0$ and $\Delta z = f_y(x_0, y_0)$ . This gives the tangent vector $\vec{T}_y$ :
$\vec{T}_y = \langle 0, 1, f_y(x_0, y_0) \rangle = \hat{j} + f_y(x_0, y_0)\hat{k}$
Computing the Surface Normal Vector: Because both $\vec{T}_x$ and $\vec{T}_y$ lie directly in the tangent plane, their Cross Product produces a normal vector $\vec{n}$ perpendicular to the entire tangent plane:
$\vec{n} = \vec{T}_x \times \vec{T}_y = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & f_x \\ 0 & 1 & f_y \end{vmatrix} = \hat{i}(0 - f_x) - \hat{j}(f_y - 0) + \hat{k}(1 - 0) = \langle -f_x, -f_y, 1 \rangle$ Alternatively, multiplying by $-1$ yields the downward-pointing normal: $\vec{n} = \langle f_x(x_0, y_0), f_y(x_0, y_0), -1 \rangle$
Constructing the Equation: Substituting $\vec{n} = \langle f_x, f_y, -1 \rangle$ and $P_0(x_0, y_0, z_0)$ into the Point-Normal form defined in Equations of Planes in 3D:
$f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) - 1(z - z_0) = 0$ Solving for $z$ yields the standard explicit formula: $z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$

Geometric Meaning of the Normal Vector Components

The normal vector $\vec{n} = \langle f_x, f_y, -1 \rangle$ tells an engineer exactly how the surface tilts in 3D space:

$f_x$ measures the inclination tilt along the $x$ -axis.
$f_y$ measures the inclination tilt along the $y$ -axis.
The constant $-1$ reflects that the surface height $z$ changes proportionally to variations in the inputs $x$ and $y$ .

Engineering Applications

Aerodynamics & Hydrodynamics: Determining the normal vector to an aircraft fuselage or submarine hull to compute the angle of attack of oncoming fluid streamlines, surface skin-friction drag, and dynamic pressure distribution.
Mechanical Contact Mechanics: Modeling contact interfaces between gear teeth, ball bearings, and cam followers where localized crushing stresses depend on the tangent plane of contact.
Solar Energy Engineering: Orienting dual-axis solar photovoltaic arrays so that their normal vector directly aligns with the sun's position vector, maximizing energy capture.
Computer Graphics Shading: Calculating normal vectors across curved 3D polygonal models for Phong and Gouraud specular reflection rendering.

Example Problems

Example 1: Elliptic Paraboloid

Find the equation of the tangent plane to the surface $z = x^2 + y^2$ at the point $(1, 2, 5)$ .

Solution

Verify Base Point: $z_0 = f(1, 2) = 1^2 + 2^2 = 1 + 4 = 5$
Compute Partial Derivatives: $f_x(x, y) = \frac{\partial}{\partial x}(x^2 + y^2) = 2x$ $f_y(x, y) = \frac{\partial}{\partial y}(x^2 + y^2) = 2y$
Evaluate at $(1, 2)$ : $f_x(1, 2) = 2(1) = 2$ $f_y(1, 2) = 2(2) = 4$
State Tangent Plane Formula and Substitute: $z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ $z = 5 + 2(x - 1) + 4(y - 2)$ $z = 5 + 2x - 2 + 4y - 8 \implies z = 2x + 4y - 5$
Standard Form: $2x + 4y - z = 5$

Example 2: Trigonometric Surface

Find the equation of the tangent plane to $z = \sin(xy)$ at $(x_0, y_0) = (0, \pi)$ , where $z_0 = 0$ .

Solution

Compute Partial Derivatives via Chain Rule: $f_x(x, y) = \frac{\partial}{\partial x}[\sin(xy)] = y\cos(xy)$ $f_y(x, y) = \frac{\partial}{\partial y}[\sin(xy)] = x\cos(xy)$
Evaluate at $(0, \pi)$ : $f_x(0, \pi) = \pi \cos(0 \cdot \pi) = \pi \cos(0) = \pi(1) = \pi$ $f_y(0, \pi) = 0 \cos(0 \cdot \pi) = 0(1) = 0$
Construct Tangent Plane Equation: $z = 0 + \pi(x - 0) + 0(y - \pi)$ $z = \pi x \implies \pi x - z = 0$ Geometric Note: Because $f_y = 0$ , moving parallel to the $y$ -axis produces zero slope at this point; the tangent plane is horizontal in the $y$ -direction.

Example 3: Product and Exponential Surface

Find the equation of the tangent plane to $z = x e^{xy}$ at the point $(2, 0, 2)$ .

Solution

Verify Base Point: $z_0 = f(2, 0) = 2 e^{2 \cdot 0} = 2 e^0 = 2$
Compute Partial Derivatives:
- For $f_x$ : Apply the Product Rule: $f_x(x, y) = \frac{\partial}{\partial x}[x] \cdot e^{xy} + x \cdot \frac{\partial}{\partial x}[e^{xy}] = 1 \cdot e^{xy} + x(y e^{xy}) = e^{xy}(1 + xy)$
- For $f_y$ : Treat $x$ as a constant multiplier: $f_y(x, y) = x \cdot \frac{\partial}{\partial y}[e^{xy}] = x(x e^{xy}) = x^2 e^{xy}$
Evaluate at $(2, 0)$ : $f_x(2, 0) = e^0(1 + 0) = 1$ $f_y(2, 0) = 2^2 e^0 = 4$
Substitute into Tangent Plane Formula: $z = 2 + 1(x - 2) + 4(y - 0)$ $z = 2 + x - 2 + 4y \implies z = x + 4y$ General Form: $x + 4y - z = 0$ .

Connections & References

Parent Concepts: Partial Derivatives | Geometric Interpretation of Partial Derivatives
Related Notes: Equations of Planes in 3D | Linear Approximations | Differentials and Error Propagation | The Gradient Vector | Tangent Planes to Level Surfaces
Course Sessions: Calc-3 Session 10 | Calc-3 Session 12