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Tangent Planes

Formulas & Definitions
Core Definition

Let z=f(x,y)z = f(x, y) define a surface in R3\mathbb{R}^3, and let f(x,y)f(x, y) have continuous first-order partial derivatives near the base point (x0,y0)(x_0, y_0), where z0=f(x0,y0)z_0 = f(x_0, y_0). The tangent plane to the surface at the point P0(x0,y0,z0)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 P0P_0:

  1. Explicit Tangent Plane Equation:
    z=z0+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)
  2. Point-Normal Form:
    fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)−(z−z0)=0f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) - (z - z_0) = 0
  3. Surface Normal Vectors (n⃗\vec{n}):
    n⃗down=⟨fx(x0,y0),fy(x0,y0),−1⟩=fxi^+fyj^−k^\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}
    n⃗up=⟨−fx(x0,y0),−fy(x0,y0),1⟩=−fxi^−fyj^+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′(x0)f'(x_0) provides the slope of the unique tangent line to the curve y=f(x)y = f(x) at x0x_0:

y−y0=f′(x0)(x−x0)  ⟹  y=y0+f′(x0)(x−x0)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)z = f(x, y) represents a curved two-dimensional surface living in three-dimensional space R3\mathbb{R}^3. At any point P0(x0,y0,z0)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 P0P_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 P0(x0,y0,z0)P_0(x_0, y_0, z_0):

  1. The xx-Trace Curve (C1C_1): Slicing the surface with the vertical plane y=y0y = y_0 produces the trace curve z=f(x,y0)z = f(x, y_0). The slope of this curve at x0x_0 is the partial derivative fx(x0,y0)f_x(x_0, y_0). Moving Δx=1\Delta x = 1 in the xx-direction produces Δy=0\Delta y = 0 and Δz=fx(x0,y0)\Delta z = f_x(x_0, y_0). This gives the tangent vector T⃗x\vec{T}_x:

    T⃗x=⟨1,0,fx(x0,y0)⟩=i^+fx(x0,y0)k^\vec{T}_x = \langle 1, 0, f_x(x_0, y_0) \rangle = \hat{i} + f_x(x_0, y_0)\hat{k}

  2. The yy-Trace Curve (C2C_2): Slicing the surface with the vertical plane x=x0x = x_0 produces the trace curve z=f(x0,y)z = f(x_0, y). The slope of this curve at y0y_0 is the partial derivative fy(x0,y0)f_y(x_0, y_0). Moving Δy=1\Delta y = 1 in the yy-direction produces Δx=0\Delta x = 0 and Δz=fy(x0,y0)\Delta z = f_y(x_0, y_0). This gives the tangent vector T⃗y\vec{T}_y:

    T⃗y=⟨0,1,fy(x0,y0)⟩=j^+fy(x0,y0)k^\vec{T}_y = \langle 0, 1, f_y(x_0, y_0) \rangle = \hat{j} + f_y(x_0, y_0)\hat{k}

  3. Computing the Surface Normal Vector: Because both T⃗x\vec{T}_x and T⃗y\vec{T}_y lie directly in the tangent plane, their Cross Product produces a normal vector n⃗\vec{n} perpendicular to the entire tangent plane:

    n⃗=T⃗x×T⃗y=∣i^j^k^10fx01fy∣=i^(0−fx)−j^(fy−0)+k^(1−0)=⟨−fx,−fy,1⟩\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 −1-1 yields the downward-pointing normal:
    n⃗=⟨fx(x0,y0),fy(x0,y0),−1⟩\vec{n} = \langle f_x(x_0, y_0), f_y(x_0, y_0), -1 \rangle

  4. Constructing the Equation: Substituting n⃗=⟨fx,fy,−1⟩\vec{n} = \langle f_x, f_y, -1 \rangle and P0(x0,y0,z0)P_0(x_0, y_0, z_0) into the Point-Normal form defined in Equations of Planes in 3D:

    fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)−1(z−z0)=0f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) - 1(z - z_0) = 0
    Solving for zz yields the standard explicit formula:
    z=z0+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)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 n⃗=⟨fx,fy,−1⟩\vec{n} = \langle f_x, f_y, -1 \rangle tells an engineer exactly how the surface tilts in 3D space:

  • fxf_x measures the inclination tilt along the xx-axis.
  • fyf_y measures the inclination tilt along the yy-axis.
  • The constant −1-1 reflects that the surface height zz changes proportionally to variations in the inputs xx and yy.

Engineering Applications

Example Problems

Example 1: Elliptic Paraboloid

Find the equation of the tangent plane to the surface z=x2+y2z = x^2 + y^2 at the point (1,2,5)(1, 2, 5).

Solution
  1. Verify Base Point:
    z0=f(1,2)=12+22=1+4=5z_0 = f(1, 2) = 1^2 + 2^2 = 1 + 4 = 5
  2. Compute Partial Derivatives:
    fx(x,y)=∂∂x(x2+y2)=2xf_x(x, y) = \frac{\partial}{\partial x}(x^2 + y^2) = 2x
    fy(x,y)=∂∂y(x2+y2)=2yf_y(x, y) = \frac{\partial}{\partial y}(x^2 + y^2) = 2y
  3. Evaluate at (1,2)(1, 2):
    fx(1,2)=2(1)=2f_x(1, 2) = 2(1) = 2
    fy(1,2)=2(2)=4f_y(1, 2) = 2(2) = 4
  4. State Tangent Plane Formula and Substitute:
    z=z0+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)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 + 2(x - 1) + 4(y - 2)
    z=5+2x−2+4y−8  ⟹  z=2x+4y−5z = 5 + 2x - 2 + 4y - 8 \implies z = 2x + 4y - 5
  5. Standard Form:
    2x+4y−z=52x + 4y - z = 5

Example 2: Trigonometric Surface

Find the equation of the tangent plane to z=sin⁡(xy)z = \sin(xy) at (x0,y0)=(0,π)(x_0, y_0) = (0, \pi), where z0=0z_0 = 0.

Solution
  1. Compute Partial Derivatives via Chain Rule:
    fx(x,y)=∂∂x[sin⁡(xy)]=ycos⁡(xy)f_x(x, y) = \frac{\partial}{\partial x}[\sin(xy)] = y\cos(xy)
    fy(x,y)=∂∂y[sin⁡(xy)]=xcos⁡(xy)f_y(x, y) = \frac{\partial}{\partial y}[\sin(xy)] = x\cos(xy)
  2. Evaluate at (0,π)(0, \pi):
    fx(0,π)=πcos⁡(0⋅π)=πcos⁡(0)=π(1)=πf_x(0, \pi) = \pi \cos(0 \cdot \pi) = \pi \cos(0) = \pi(1) = \pi
    fy(0,π)=0cos⁡(0⋅π)=0(1)=0f_y(0, \pi) = 0 \cos(0 \cdot \pi) = 0(1) = 0
  3. Construct Tangent Plane Equation:
    z=0+π(x−0)+0(y−π)z = 0 + \pi(x - 0) + 0(y - \pi)
    z=πx  ⟹  πx−z=0z = \pi x \implies \pi x - z = 0
    Geometric Note: Because fy=0f_y = 0, moving parallel to the yy-axis produces zero slope at this point; the tangent plane is horizontal in the yy-direction.

Example 3: Product and Exponential Surface

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

Solution
  1. Verify Base Point:
    z0=f(2,0)=2e2⋅0=2e0=2z_0 = f(2, 0) = 2 e^{2 \cdot 0} = 2 e^0 = 2
  2. Compute Partial Derivatives:
    • For fxf_x: Apply the Product Rule:
      fx(x,y)=∂∂x[x]⋅exy+x⋅∂∂x[exy]=1⋅exy+x(yexy)=exy(1+xy)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 fyf_y: Treat xx as a constant multiplier:
      fy(x,y)=x⋅∂∂y[exy]=x(xexy)=x2exyf_y(x, y) = x \cdot \frac{\partial}{\partial y}[e^{xy}] = x(x e^{xy}) = x^2 e^{xy}
  3. Evaluate at (2,0)(2, 0):
    fx(2,0)=e0(1+0)=1f_x(2, 0) = e^0(1 + 0) = 1
    fy(2,0)=22e0=4f_y(2, 0) = 2^2 e^0 = 4
  4. Substitute into Tangent Plane Formula:
    z=2+1(x−2)+4(y−0)z = 2 + 1(x - 2) + 4(y - 0)
    z=2+x−2+4y  ⟹  z=x+4yz = 2 + x - 2 + 4y \implies z = x + 4y
    General Form: x+4y−z=0x + 4y - z = 0.

Connections & References