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Tony's Intertwingled Memex • Reference Manual

Geometric Interpretation of Partial Derivatives

Formulas & Definitions
Core Definition

Let SSS be the surface defined by z=f(x,y)z = f(x, y)z=f(x,y), and let P(x0,y0,z0)P(x_0, y_0, z_0)P(x0​,y0​,z0​) be a point on SSS where z0=f(x0,y0)z_0 = f(x_0, y_0)z0​=f(x0​,y0​).

  • Slicing the surface SSS with the vertical plane y=y0y = y_0y=y0​ produces a 2D cross-sectional curve C1C_1C1​. The partial derivative fx(x0,y0)f_x(x_0, y_0)fx​(x0​,y0​) is the slope of the tangent line T1T_1T1​ to the curve C1C_1C1​ at point PPP.
  • Slicing the surface SSS with the vertical plane x=x0x = x_0x=x0​ produces a 2D cross-sectional curve C2C_2C2​. The partial derivative fy(x0,y0)f_y(x_0, y_0)fy​(x0​,y0​) is the slope of the tangent line T2T_2T2​ to the curve C2C_2C2​ at point PPP.

Concept Explanation

Visualizing Cross-Sectional Trace Slices

A single real-valued multivariable function z=f(x,y)z = f(x, y)z=f(x,y) generates a curved surface in three-dimensional space R3\mathbb{R}^3R3. To understand the directional nature of partial derivatives geometrically, imagine cutting through this 3D topography with vertical planar slices:

                      z ^
                        |              Plane y = y_0 (Vertical slice)
                        |             +-----------------------+
                        |            /|                      /|
                        |           / |    Trace Curve C_1  / |
                        |          +--|-------(P)----------+  |
                        |          |  |      /   \         |  |
                        |          |  |    Tangent T_1     |  |
                        |          |  +---|----------------|--+-----> y
                        |          | /    |                | /
                        |          +------|----------------+
                        |                /
                        +--------------/---------------------> x
  1. The xxx-Direction Slope (fx(x0,y0)f_x(x_0, y_0)fx​(x0​,y0​)):

    • Intersect the surface z=f(x,y)z = f(x, y)z=f(x,y) with the vertical plane y=y0y = y_0y=y0​ (parallel to the xzxzxz-plane).
    • This intersection is the trace curve z=f(x,y0)z = f(x, y_0)z=f(x,y0​).
    • The value fx(x0,y0)f_x(x_0, y_0)fx​(x0​,y0​) gives the instantaneous slope ΔzΔx\frac{\Delta z}{\Delta x}ΔxΔz​ of the tangent line T1T_1T1​ resting in the plane y=y0y = y_0y=y0​.
    • Tangible Meaning: If you stand at point PPP and step directly east (along the positive xxx-axis) while holding your latitude (y=y0y = y_0y=y0​) fixed, fx(x0,y0)f_x(x_0, y_0)fx​(x0​,y0​) is the grade of the incline you ascend or descend.
  2. The yyy-Direction Slope (fy(x0,y0)f_y(x_0, y_0)fy​(x0​,y0​)):

    • Intersect the surface z=f(x,y)z = f(x, y)z=f(x,y) with the vertical plane x=x0x = x_0x=x0​ (parallel to the yzyzyz-plane).
    • This intersection is the trace curve z=f(x0,y)z = f(x_0, y)z=f(x0​,y).
    • The value fy(x0,y0)f_y(x_0, y_0)fy​(x0​,y0​) gives the instantaneous slope ΔzΔy\frac{\Delta z}{\Delta y}ΔyΔz​ of the tangent line T2T_2T2​ resting in the plane x=x0x = x_0x=x0​.
    • Tangible Meaning: If you stand at point PPP and step directly north (along the positive yyy-axis) while holding your longitude (x=x0x = x_0x=x0​) fixed, fy(x0,y0)f_y(x_0, y_0)fy​(x0​,y0​) is the grade of the incline you ascend or descend.

Vector Representations of the Tangent Lines

The slopes define direction vectors for the two fundamental tangent lines at P(x0,y0,z0)P(x_0, y_0, z_0)P(x0​,y0​,z0​):

  • Tangent line in xxx-direction (T1T_1T1​): A unit run Δx=1\Delta x = 1Δx=1 with Δy=0\Delta y = 0Δy=0 produces a vertical rise Δz=fx(x0,y0)\Delta z = f_x(x_0, y_0)Δz=fx​(x0​,y0​):
    v⃗1=⟨1,0,fx(x0,y0)⟩=i^+fx(x0,y0)k^\vec{v}_1 = \langle 1, 0, f_x(x_0, y_0) \rangle = \hat{i} + f_x(x_0, y_0)\hat{k}v1​=⟨1,0,fx​(x0​,y0​)⟩=i^+fx​(x0​,y0​)k^
  • Tangent line in yyy-direction (T2T_2T2​): A unit run Δy=1\Delta y = 1Δy=1 with Δx=0\Delta x = 0Δx=0 produces a vertical rise Δz=fy(x0,y0)\Delta z = f_y(x_0, y_0)Δz=fy​(x0​,y0​):
    v⃗2=⟨0,1,fy(x0,y0)⟩=j^+fy(x0,y0)k^\vec{v}_2 = \langle 0, 1, f_y(x_0, y_0) \rangle = \hat{j} + f_y(x_0, y_0)\hat{k}v2​=⟨0,1,fy​(x0​,y0​)⟩=j^​+fy​(x0​,y0​)k^

These two tangent vectors span the tangent plane to the surface at PPP, establishing the geometric foundation for multivariable differentiability and linear approximations.


Engineering Applications

  • Civil & Transportation Engineering: In road design, fxf_xfx​ and fyf_yfy​ represent roadway superelevation and cross-slope grades on embankments, ensuring proper stormwater shedding without causing vehicle rollovers.
  • Geotechnical Engineering: On hillside construction sites, slopes ∂E∂x\frac{\partial E}{\partial x}∂x∂E​ and ∂E∂y\frac{\partial E}{\partial y}∂y∂E​ identify critical shear slip angles for slope stability.
  • Robotics & Autonomous Navigation: Real-time terrain slope vectors ⟨fx,fy⟩\langle f_x, f_y \rangle⟨fx​,fy​⟩ guide rover traction control and tilt-sensor compensation over uneven terrain.

Example Problems

Example 1: Calculating Slopes and Tangent Vectors on an Elliptic Paraboloid

Consider the paraboloid surface z=f(x,y)=4−x2−2y2z = f(x, y) = 4 - x^2 - 2y^2z=f(x,y)=4−x2−2y2.

  1. Find the slopes of the trace curves passing through the point (1,1,1)(1, 1, 1)(1,1,1).
  2. Write direction vectors for the tangent lines T1T_1T1​ and T2T_2T2​ at this point.
Solution
  1. Compute and Evaluate Partial Derivatives:

    ∂f∂x=−2x  ⟹  fx(1,1)=−2(1)=−2\frac{\partial f}{\partial x} = -2x \implies f_x(1, 1) = -2(1) = -2∂x∂f​=−2x⟹fx​(1,1)=−2(1)=−2
    ∂f∂y=−4y  ⟹  fy(1,1)=−4(1)=−4\frac{\partial f}{\partial y} = -4y \implies f_y(1, 1) = -4(1) = -4∂y∂f​=−4y⟹fy​(1,1)=−4(1)=−4

    • The slope of the tangent line to the trace curve in the plane y=1y = 1y=1 is mx=−2m_x = -2mx​=−2 (the surface drops at rate 222 in the +x+x+x direction).
    • The slope of the tangent line to the trace curve in the plane x=1x = 1x=1 is my=−4m_y = -4my​=−4 (the surface drops at rate 444 in the +y+y+y direction).
  2. Construct Tangent Vectors:

    v⃗1=⟨1,0,−2⟩=i^−2k^\vec{v}_1 = \langle 1, 0, -2 \rangle = \hat{i} - 2\hat{k}v1​=⟨1,0,−2⟩=i^−2k^
    v⃗2=⟨0,1,−4⟩=j^−4k^\vec{v}_2 = \langle 0, 1, -4 \rangle = \hat{j} - 4\hat{k}v2​=⟨0,1,−4⟩=j^​−4k^


Connections & References

  • Foundational Theory: Partial Derivatives
  • Cross-Sectional Foundations: Traces and Cross-Sections of Surfaces
  • Subsequent Topics: Session 10 - Tangent Planes and Linear Approximations, Session 12 - Gradient and Directional Derivatives
  • Course Sessions: Session 09 - Partial Derivatives