Session 09 - Partial Derivatives | Tony's Intertwingled Memex

📅 Micro-Lecture

1. The Fundamental Principle of Partial Differentiation

Mathematical Foundation: In single-variable calculus, the derivative $\frac{df}{dx}$ measures the slope of a curve in the $xy$ -plane. In multivariable calculus, a function $z = f(x, y)$ defines a curved two-dimensional surface in 3D space $\mathbb{R}^3$ . Because inputs can vary in infinite directions across the domain plane, we isolate instantaneous rates of change along the coordinate axes using Partial Derivatives. The partial derivative $\frac{\partial f}{\partial x}$ evaluates variation along the $x$ -axis while holding $y$ strictly constant; conversely, $\frac{\partial f}{\partial y}$ evaluates variation along the $y$ -axis while holding $x$ strictly constant.
Engineering Context: Real physical systems depend on multiple simultaneous inputs: thermodynamic pressure $P(V, T)$ depends on chamber volume and temperature; aircraft lift depends on airspeed and angle of attack. Engineers use partial derivatives to isolate how a system responds to a single variable at a time ("fine-tuning" system controls).

2. Geometric Meaning: Cross-Sectional Trace Tangents

Mathematical Foundation: As explored in Geometric Interpretation of Partial Derivatives, slicing the 3D surface $z = f(x, y)$ with a vertical plane $y = y_0$ produces a 2D cross-sectional curve $C_1$ . The partial derivative $f_x(x_0, y_0)$ is the slope of the tangent line $T_1$ to this curve at $(x_0, y_0, f(x_0, y_0))$ . Similarly, slicing with $x = x_0$ produces curve $C_2$ , where $f_y(x_0, y_0)$ is the slope of tangent line $T_2$ .
Engineering Context: Geotechnical engineers evaluate terrain grades along cardinal directions (due east and due north) to design stable building foundations and retaining walls against gravitational shear failure.

3. Visual Estimation from Contour Maps and Sensor Grids

Mathematical Foundation: As detailed in Visual Estimation of Partial Derivatives from Contour Maps, partial derivatives can be read directly from level curve contour plots. Taking a step to the right ( $+x$ , holding $y$ fixed) across higher contour labels indicates $f_x > 0$ , while crossing into lower labels indicates $f_x < 0$ . If stepping parallel to a coordinate axis is tangent to a contour line, the instantaneous rate of change is zero ( $f_x = 0$ or $f_y = 0$ ). Furthermore, closely spaced contour lines represent a steep slope (large derivative magnitude), whereas widely spaced contours indicate gradual, mild slopes.
Engineering Context: Environmental engineers inspect topographic contour lines to determine stormwater runoff velocity and pool locations; materials engineers inspect photoelastic strain fringe densities to locate impending micro-fracture sites near geometric notches.

Segment 1 of 8

Skill Block 1

Real-Life Engineering Applications

🚀

Aerospace Engineering

Thermal Protection Tile Heat Flux (Aerospace Engineering)

An aerospace engineer designs a ceramic thermal protection tile for a hypersonic atmospheric re-entry vehicle. During peak atmospheric deceleration, intense aerodynamic friction generates non-uniform surface heating across the outer tile face.

The temperature distribution $T(x, y)$ (in degrees Celsius, $^\circ\text{C}$ ) across a square test section of the tile is modeled by the Gaussian heat distribution function:

T(x, y) = 800 e^{-0.1(x^2 + 2y^2)}

where

x

and

y

represent spatial coordinates in centimeters (

\text{cm}

) measured from the tile center.

State the partial differentiation rules required to compute $\frac{\partial T}{\partial x}$ and $\frac{\partial T}{\partial y}$ , treating the alternate coordinate as a constant coefficient.
Determine the analytical expressions for the spatial temperature gradients $\frac{\partial T}{\partial x}$ and $\frac{\partial T}{\partial y}$ .
Evaluate the instantaneous temperature gradients at the specific monitoring sensor location $(x, y) = (2, 1)\text{ cm}$ . State the units of each derivative and interpret the physical thermal behavior along the $x$ and $y$ directions.
According to Fourier's Law of Heat Conduction, conductive heat flux across the surface is given by $\vec{q} = -k \nabla T = -k \langle \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \rangle$ , where $k > 0$ is thermal conductivity. Determine the direction of conductive heat flow at $(2, 1)$ .

💡 Show Worked Derivation

An aerospace engineer designs a ceramic thermal protection tile for a hypersonic atmospheric re-entry vehicle. During peak atmospheric deceleration, intense aerodynamic friction generates non-uniform surface heating across the outer tile face.

The temperature distribution $T(x, y)$ (in degrees Celsius, $^\circ\text{C}$ ) across a square test section of the tile is modeled by the Gaussian heat distribution function:

T(x, y) = 800 e^{-0.1(x^2 + 2y^2)}

where

x

and

y

represent spatial coordinates in centimeters (

\text{cm}

) measured from the tile center.

State the partial differentiation rules required to compute $\frac{\partial T}{\partial x}$ and $\frac{\partial T}{\partial y}$ , treating the alternate coordinate as a constant coefficient.
Determine the analytical expressions for the spatial temperature gradients $\frac{\partial T}{\partial x}$ and $\frac{\partial T}{\partial y}$ .
Evaluate the instantaneous temperature gradients at the specific monitoring sensor location $(x, y) = (2, 1)\text{ cm}$ . State the units of each derivative and interpret the physical thermal behavior along the $x$ and $y$ directions.
According to Fourier's Law of Heat Conduction, conductive heat flux across the surface is given by $\vec{q} = -k \nabla T = -k \langle \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \rangle$ , where $k > 0$ is thermal conductivity. Determine the direction of conductive heat flow at $(2, 1)$ .

Step-by-Step Solution:

State the Required Differentiation Rules:
- Exponential Chain Rule: $\frac{\partial}{\partial x}[e^{u(x, y)}] = e^{u(x, y)} \cdot \frac{\partial u}{\partial x}$
- Constant Isolation Rule: When differentiating with respect to $x$ , treat $y$ as a constant; when differentiating with respect to $y$ , treat $x$ as a constant. (Defined in Partial Derivatives.)
Compute the Analytical Expressions:
- Let $u(x, y) = -0.1(x^2 + 2y^2) = -0.1x^2 - 0.2y^2$ .
- Differentiating with respect to $x$ : $\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(-0.1x^2 - 0.2y^2) = -0.2x$ $\frac{\partial T}{\partial x} = 800 e^{-0.1(x^2 + 2y^2)} \cdot (-0.2x) = -160x e^{-0.1(x^2 + 2y^2)}$
- Differentiating with respect to $y$ : $\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(-0.1x^2 - 0.2y^2) = -0.4y$ $\frac{\partial T}{\partial y} = 800 e^{-0.1(x^2 + 2y^2)} \cdot (-0.4y) = -320y e^{-0.1(x^2 + 2y^2)}$
Evaluate at the Point $(2, 1)\text{ cm}$ :
- Compute the exponent at $(2, 1)$ : $u(2, 1) = -0.1(2^2 + 2(1)^2) = -0.1(4 + 2) = -0.6$ $e^{-0.6} \approx 0.54881$
- Evaluate $\frac{\partial T}{\partial x}(2, 1)$ : $\left.\frac{\partial T}{\partial x}\right|_{(2, 1)} = -160(2) e^{-0.6} = -320(0.54881) \approx -175.62^\circ\text{C/cm}$
- Evaluate $\frac{\partial T}{\partial y}(2, 1)$ : $\left.\frac{\partial T}{\partial y}\right|_{(2, 1)} = -320(1) e^{-0.6} = -320(0.54881) \approx -175.62^\circ\text{C/cm}$
- Interpretation: The units are degrees Celsius per centimeter ( $^\circ\text{C/cm}$ ). Because both derivatives are negative, moving outward away from the center (increasing $x$ or $y$ ) causes surface temperature to drop at an instantaneous rate of approximately $175.6^\circ\text{C/cm}$ .
Determine Conductive Heat Flow Direction:
- State Fourier's Law of Heat Conduction: $\vec{q} = -k \nabla T = -k \left\langle \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \right\rangle$
- Substituting our negative spatial gradients: $\vec{q} = -k \langle -175.62, -175.62 \rangle = k \langle 175.62, 175.62 \rangle$
- Because $k > 0$ , the conductive heat flux vector points in the direction of $\langle 1, 1 \rangle$ (into the positive first quadrant, toward decreasing temperatures). Heat naturally diffuses from hot zones (tile center) toward cold outer edges.

🧪

Chemical Engineering

Viscous Polymer Pipeline Flow (Chemical & Petroleum Engineering)

A chemical engineer monitors the laminar flow of a highly viscous polymer solution through a rectangular processing channel in a refinery plant. Due to no-slip boundary conditions at the solid duct walls, fluid velocity drops to zero at the channel perimeter and peaks along the central core line.

The axial velocity $v(x, y)$ (in meters per second, $\text{m/s}$ ) across the rectangular cross-section $0 \le x \le 4\text{ cm}$ and $0 \le y \le 2\text{ cm}$ is modeled by the separable polynomial profile:

v(x, y) = 1.5(4x - x^2)(2y - y^2)

where

x

is horizontal distance from the left wall (in

\text{cm}

) and

y

is vertical distance from the duct floor (in

\text{cm}

).

Explain how to compute $\frac{\partial v}{\partial x}$ and $\frac{\partial v}{\partial y}$ by treating terms involving the other coordinate as constant scaling factors.
Determine the analytical expressions for the transverse velocity gradients $\frac{\partial v}{\partial x}$ and $\frac{\partial v}{\partial y}$ .
In Newtonian fluid mechanics, shear stress across fluid layers is proportional to the velocity gradient: $\tau_{zx} = \mu \frac{\partial v}{\partial x}, \quad \tau_{zy} = \mu \frac{\partial v}{\partial y}$ Evaluate $\frac{\partial v}{\partial x}$ and $\frac{\partial v}{\partial y}$ at the cross-sectional evaluation point $(x, y) = (1, 0.5)\text{ cm}$ . State their units and identify which coordinate axis experiences greater internal fluid shear stress.

💡 Show Worked Derivation

A chemical engineer monitors the laminar flow of a highly viscous polymer solution through a rectangular processing channel in a refinery plant. Due to no-slip boundary conditions at the solid duct walls, fluid velocity drops to zero at the channel perimeter and peaks along the central core line.

The axial velocity $v(x, y)$ (in meters per second, $\text{m/s}$ ) across the rectangular cross-section $0 \le x \le 4\text{ cm}$ and $0 \le y \le 2\text{ cm}$ is modeled by the separable polynomial profile:

v(x, y) = 1.5(4x - x^2)(2y - y^2)

where

x

is horizontal distance from the left wall (in

\text{cm}

) and

y

is vertical distance from the duct floor (in

\text{cm}

).

Explain how to compute $\frac{\partial v}{\partial x}$ and $\frac{\partial v}{\partial y}$ by treating terms involving the other coordinate as constant scaling factors.
Determine the analytical expressions for the transverse velocity gradients $\frac{\partial v}{\partial x}$ and $\frac{\partial v}{\partial y}$ .
In Newtonian fluid mechanics, shear stress across fluid layers is proportional to the velocity gradient: $\tau_{zx} = \mu \frac{\partial v}{\partial x}, \quad \tau_{zy} = \mu \frac{\partial v}{\partial y}$ Evaluate $\frac{\partial v}{\partial x}$ and $\frac{\partial v}{\partial y}$ at the cross-sectional evaluation point $(x, y) = (1, 0.5)\text{ cm}$ . State their units and identify which coordinate axis experiences greater internal fluid shear stress.

Step-by-Step Solution:

State the Constant Scaling Principle:
- When differentiating with respect to $x$ , the entire vertical term $(2y - y^2)$ behaves as a constant multiplier and can be factored outside the derivative operator: $\frac{\partial v}{\partial x} = 1.5(2y - y^2) \frac{\partial}{\partial x}(4x - x^2)$
- Similarly, when differentiating with respect to $y$ , $(4x - x^2)$ behaves as a constant multiplier: $\frac{\partial v}{\partial y} = 1.5(4x - x^2) \frac{\partial}{\partial y}(2y - y^2)$ (Defined in Partial Derivatives.)
Compute Analytical Expressions:
- Differentiate the polynomial factors using the single-variable power rule: $\frac{\partial}{\partial x}(4x - x^2) = 4 - 2x = 2(2 - x)$ $\frac{\partial}{\partial y}(2y - y^2) = 2 - 2y = 2(1 - y)$
- Substitute back into the partial derivatives: $\frac{\partial v}{\partial x} = 1.5(2y - y^2)(4 - 2x) = 3(2 - x)(2y - y^2)$ $\frac{\partial v}{\partial y} = 1.5(4x - x^2)(2 - 2y) = 3(1 - y)(4x - x^2)$
Evaluate at $(x, y) = (1, 0.5)\text{ cm}$ :
- Compute coordinate factors at $(1, 0.5)$ : $4x - x^2 = 4(1) - (1)^2 = 3$ $2 - x = 2 - 1 = 1$ $2y - y^2 = 2(0.5) - (0.5)^2 = 1 - 0.25 = 0.75$ $1 - y = 1 - 0.5 = 0.5$
- Evaluate $\frac{\partial v}{\partial x}$ : $\left.\frac{\partial v}{\partial x}\right|_{(1, 0.5)} = 3(1)(0.75) = 2.25\text{ (m/s)/cm} = 225\text{ s}^{-1}$
- Evaluate $\frac{\partial v}{\partial y}$ : $\left.\frac{\partial v}{\partial y}\right|_{(1, 0.5)} = 3(0.5)(3) = 4.5\text{ (m/s)/cm} = 450\text{ s}^{-1}$
- Units and Interpretation: The physical units are meters per second per centimeter ( $(\text{m/s})/\text{cm}$ , or $100\text{ s}^{-1}$ in SI base units).
- Internal Shear Comparison: Because $\left|\frac{\partial v}{\partial y}\right| = 4.5 > \left|\frac{\partial v}{\partial x}\right| = 2.25$ , the velocity gradient in the vertical direction is twice as large as the horizontal gradient. Since shear stress $\tau = \mu \frac{\partial v}{\partial s}$ , the fluid experiences significantly greater internal shear stress in the vertical direction ( $\tau_{zy} = 2\tau_{zx}$ ) at this location.

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Computing First-Order Partial Derivatives in Two and Three Variables

General Overview
Skill Block 1 covers the formal limit definition of partial derivatives, notation conventions ( $f_x, \frac{\partial f}{\partial x}, \partial_x f$ ), the fundamental constant-variable rule, and the algebraic application of single-variable differentiation rules (product, quotient, power, chain rule) to functions of two and three variables.

Conceptual Framework

Formal Limit Definition of Partial Derivatives: $\frac{\partial f}{\partial x}(x, y) = \lim_{h \to 0} \frac{f(x+h, y) - f(x, y)}{h}$ $\frac{\partial f}{\partial y}(x, y) = \lim_{h \to 0} \frac{f(x, y+h) - f(x, y)}{h}$ (Defined in Partial Derivatives.)
The Core Differentiation Rule:
- When finding $\frac{\partial f}{\partial x}$ , treat $y$ (and $z$ ) as constant numbers.
- When finding $\frac{\partial f}{\partial y}$ , treat $x$ (and $z$ ) as constant numbers.
- When finding $\frac{\partial f}{\partial z}$ , treat $x$ and $y$ as constant numbers. (Detailed in Partial Derivatives of Functions of Three or More Variables.)

Key Equations & Structural Rules

Constant Multiplier Rule: $\frac{\partial}{\partial x}[c(y) \cdot g(x)] = c(y) \cdot g'(x)$
Multivariable Product Rule: $\frac{\partial}{\partial x}[u \cdot v] = \frac{\partial u}{\partial x} v + u \frac{\partial v}{\partial x}$
Multivariable Quotient Rule: $\frac{\partial}{\partial x}\left[\frac{u}{v}\right] = \frac{\frac{\partial u}{\partial x} v - u \frac{\partial v}{\partial x}}{v^2}$
Multivariable Chain Rule: $\frac{\partial}{\partial x}[g(u(x, y))] = g'(u(x, y)) \cdot \frac{\partial u}{\partial x}$

Worked Example A: Mixed Rational and Trigonometric Function

Compute $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ for the function:

f(x, y) = \frac{x^2 y + \sin(xy)}{y^2 + 1}

💡 Reveal Worked Solution

Compute $\frac{\partial f}{\partial x}$ : State the Constant Multiple Rule for Partial Derivatives: When differentiating with respect to $x$ , the entire denominator $(y^2 + 1)$ depends only on $y$ and is treated as a constant scalar factor. Factor it outside the derivative:
$\frac{\partial f}{\partial x} = \frac{1}{y^2 + 1} \cdot \frac{\partial}{\partial x}\left[x^2 y + \sin(xy)\right]$
- Differentiate $x^2 y$ with respect to $x$ (treating $y$ as a constant): $\frac{\partial}{\partial x}(x^2 y) = 2xy$
- Differentiate $\sin(xy)$ with respect to $x$ using the Chain Rule: $\frac{\partial}{\partial x}[\sin(xy)] = \cos(xy) \cdot \frac{\partial}{\partial x}(xy) = \cos(xy) \cdot y = y\cos(xy)$ Combine the terms: $\frac{\partial f}{\partial x} = \frac{2xy + y\cos(xy)}{y^2 + 1}$
Compute $\frac{\partial f}{\partial y}$ : State the Quotient Rule for Partial Derivatives: Because $y$ appears in both the numerator and the denominator, apply $\frac{\partial}{\partial y}\left[\frac{u}{v}\right] = \frac{\frac{\partial u}{\partial y} v - u \frac{\partial v}{\partial y}}{v^2}$ with $u = x^2 y + \sin(xy)$ and $v = y^2 + 1$ .
- Differentiate numerator $u$ with respect to $y$ (treating $x$ as a constant): $\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(x^2 y) + \frac{\partial}{\partial y}[\sin(xy)] = x^2(1) + \cos(xy) \cdot x = x^2 + x\cos(xy)$
- Differentiate denominator $v$ with respect to $y$ : $\frac{\partial v}{\partial y} = \frac{\partial}{\partial y}(y^2 + 1) = 2y$ Substitute into the Quotient Rule: $\frac{\partial f}{\partial y} = \frac{(x^2 + x\cos(xy))(y^2 + 1) - (x^2 y + \sin(xy))(2y)}{(y^2 + 1)^2}$

Worked Example B: Three-Variable Function with Logarithmic Composition

Compute $\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}$ , and $\frac{\partial g}{\partial z}$ for the three-variable function:

g(x, y, z) = \ln(x^2 + y^2 + z^2)

💡 Reveal Worked Solution

Compute $\frac{\partial g}{\partial x}$ : State the Natural Logarithm Chain Rule:
$\frac{\partial}{\partial x}[\ln(u)] = \frac{1}{u} \cdot \frac{\partial u}{\partial x}$
where $u(x, y, z) = x^2 + y^2 + z^2$ .
- Holding $y$ and $z$ constant, their derivatives are zero: $\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(x^2 + y^2 + z^2) = 2x + 0 + 0 = 2x$ Combine components: $\frac{\partial g}{\partial x} = \frac{2x}{x^2 + y^2 + z^2}$
Compute $\frac{\partial g}{\partial y}$ and $\frac{\partial g}{\partial z}$ by Symmetry: Notice that the function $g(x, y, z)$ is algebraically symmetric with respect to all three variables. Holding $x$ and $z$ constant when differentiating with respect to $y$ , and holding $x$ and $y$ constant when differentiating with respect to $z$ , yields:
$\frac{\partial g}{\partial y} = \frac{2y}{x^2 + y^2 + z^2}$ $\frac{\partial g}{\partial z} = \frac{2z}{x^2 + y^2 + z^2}$

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) State the core operational rule of partial differentiation when computing $\frac{\partial f}{\partial x}$ . How are all other variables treated?
(b) In Worked Example A, explain why the Quotient Rule was mandatory when calculating $\frac{\partial f}{\partial y}$ , but was unnecessary when calculating $\frac{\partial f}{\partial x}$ .
(c) In Thermal Protection Tile Heat Flux (Aerospace Engineering), why does conductive heat flux depend directly on the directional spatial gradients $\frac{\partial T}{\partial x}$ and $\frac{\partial T}{\partial y}$ ?

Practice Space Solve this problem on paper or tablet.

Review the Core Operational Rule and the solutions to Worked Example A and the linked aerospace application. (Defined in Partial Derivatives.)

Answer: * (a) Differentiate with respect to $x$ while treating all other variables ( $y, z$ ) as constant numbers.

(b) Because $y$ is present in both numerator and denominator; $x$ is only in the numerator, so $(y^2+1)$ acts as a constant scalar multiplier.
(c) Fourier's law states heat flux is proportional to negative temperature gradients: $\vec{q} = -k \langle \partial T/\partial x, \partial T/\partial y \rangle$ .

Step-by-Step Execution

(a) As defined in Partial Derivatives, the fundamental rule states that to compute $\frac{\partial f}{\partial x}$ , you perform ordinary single-variable differentiation with respect to $x$ while holding all other variables ( $y, z$ ) strictly fixed as if they were constant numerical coefficients.
(b) In Worked Example A, the denominator is $y^2 + 1$ . When differentiating with respect to $x$ , this denominator contains zero instances of $x$ , making $\frac{1}{y^2 + 1}$ a constant coefficient that factors cleanly out of the derivative. When differentiating with respect to $y$ , the variable $y$ appears in both the numerator and the denominator, making the Quotient Rule mandatory.
(c) In Thermal Protection Tile Heat Flux (Aerospace Engineering), Fourier's Law of Heat Conduction states that heat energy flows spontaneously from high temperature to low temperature at a rate proportional to the spatial temperature gradient: $\vec{q} = -k \nabla T = -k \langle \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \rangle$ . The partial derivatives isolate heat diffusion rates along orthogonal structural axes.

Practice Problem 2

Compute $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ for the radical function:

f(x, y) = \sqrt{x^2 + y^2}

Practice Space Solve this problem on paper or tablet.

State the Power Chain Rule: $\frac{\partial}{\partial x}[u^{1/2}] = \frac{1}{2} u^{-1/2} \frac{\partial u}{\partial x}$ , where $u = x^2 + y^2$ . (Defined in Partial Derivatives.)

Answer:

\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}, \quad \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}}

Step-by-Step Execution

Differentiate with respect to $x$ : State the Generalized Power Rule:
$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left[(x^2 + y^2)^{1/2}\right] = \frac{1}{2}(x^2 + y^2)^{-1/2} \cdot \frac{\partial}{\partial x}(x^2 + y^2)$ Holding $y$ constant, $\frac{\partial}{\partial x}(x^2 + y^2) = 2x + 0 = 2x$ : $\frac{\partial f}{\partial x} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot (2x) = \frac{x}{\sqrt{x^2 + y^2}}$
Differentiate with respect to $y$ : Holding $x$ constant, $\frac{\partial}{\partial y}(x^2 + y^2) = 0 + 2y = 2y$ :
$\frac{\partial f}{\partial y} = \frac{1}{2}(x^2 + y^2)^{-1/2} \cdot (2y) = \frac{y}{\sqrt{x^2 + y^2}}$

Practice Problem 3

Compute $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ for the exponential function:

f(x, y) = e^{xy}

Practice Space Solve this problem on paper or tablet.

State the Exponential Chain Rule: $\frac{\partial}{\partial x}[e^u] = e^u \frac{\partial u}{\partial x}$ , where $u = xy$ . (Defined in Partial Derivatives.)

Answer:

\frac{\partial f}{\partial x} = y e^{xy}, \quad \frac{\partial f}{\partial y} = x e^{xy}

Step-by-Step Execution

Differentiate with respect to $x$ : State the Exponential Chain Rule:
$\frac{\partial f}{\partial x} = e^{xy} \cdot \frac{\partial}{\partial x}(xy)$ Holding $y$ constant, $y$ acts as a constant scalar coefficient on $x$ : $\frac{\partial}{\partial x}(xy) = y(1) = y$ : $\frac{\partial f}{\partial x} = y e^{xy}$
Differentiate with respect to $y$ : Holding $x$ constant, $x$ acts as a constant scalar coefficient on $y$ : $\frac{\partial}{\partial y}(xy) = x(1) = x$ :
$\frac{\partial f}{\partial y} = x e^{xy}$

Practice Problem 4

Let $f(x, y) = x^2 + 3xy$ . Compute $\frac{\partial f}{\partial x}(2, 1)$ using the formal limit definition of a partial derivative:

\frac{\partial f}{\partial x}(2, 1) = \lim_{h \to 0} \frac{f(2+h, 1) - f(2, 1)}{h}

Practice Space Solve this problem on paper or tablet.

State the Limit Definition of Partial Derivative with Respect to $x$ : hold $y = 1$ fixed throughout the difference quotient and evaluate the single-variable limit as $h \to 0$ . (Defined in Partial Derivatives.)

Answer:

\frac{\partial f}{\partial x}(2, 1) = 7

Step-by-Step Execution

State the Formal Limit Definition:
$\frac{\partial f}{\partial x}(a, b) = \lim_{h \to 0} \frac{f(a+h, b) - f(a, b)}{h}$ For $(a, b) = (2, 1)$ , we hold $y = 1$ constant.
Evaluate the Difference Quotient Terms:
- At $(2, 1)$ : $f(2, 1) = (2)^2 + 3(2)(1) = 4 + 6 = 10$
- At $(2+h, 1)$ : $f(2+h, 1) = (2+h)^2 + 3(2+h)(1) = (4 + 4h + h^2) + (6 + 3h) = 10 + 7h + h^2$
Simplify and Evaluate the Limit:
$\frac{f(2+h, 1) - f(2, 1)}{h} = \frac{(10 + 7h + h^2) - 10}{h} = \frac{7h + h^2}{h} = \frac{h(7 + h)}{h}$ For $h \ne 0$ , cancel $h$ : $\lim_{h \to 0} (7 + h) = 7 + 0 = 7$ (Verification: $\frac{\partial f}{\partial x} = 2x + 3y \implies f_x(2, 1) = 2(2) + 3(1) = 4 + 3 = 7$ .)

Practice Problem 5

Compute $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ for the mixed product function:

f(x, y) = x^2 y \cos(x + 2y)

Practice Space Solve this problem on paper or tablet.

Apply the Multivariable Product Rule: $\frac{\partial}{\partial x}[u \cdot v] = \frac{\partial u}{\partial x}v + u\frac{\partial v}{\partial x}$ . Identify which terms contain the variable of differentiation. (Defined in Partial Derivatives.)

Answer:

\frac{\partial f}{\partial x} = 2xy \cos(x + 2y) - x^2 y \sin(x + 2y)

\frac{\partial f}{\partial y} = x^2 \cos(x + 2y) - 2x^2 y \sin(x + 2y)

Step-by-Step Execution

Compute $\frac{\partial f}{\partial x}$ : Let $u(x, y) = x^2 y$ and $v(x, y) = \cos(x + 2y)$ .
- Holding $y$ constant: $\frac{\partial u}{\partial x} = 2xy$ $\frac{\partial v}{\partial x} = -\sin(x + 2y) \cdot \frac{\partial}{\partial x}(x + 2y) = -\sin(x + 2y) \cdot (1 + 0) = -\sin(x + 2y)$
- By the Product Rule: $\frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} v + u \frac{\partial v}{\partial x} = (2xy)\cos(x + 2y) + (x^2 y)(-\sin(x + 2y))$ $\frac{\partial f}{\partial x} = 2xy \cos(x + 2y) - x^2 y \sin(x + 2y)$
Compute $\frac{\partial f}{\partial y}$ : With $u(x, y) = x^2 y$ and $v(x, y) = \cos(x + 2y)$ :
- Holding $x$ constant: $\frac{\partial u}{\partial y} = x^2(1) = x^2$ $\frac{\partial v}{\partial y} = -\sin(x + 2y) \cdot \frac{\partial}{\partial y}(x + 2y) = -\sin(x + 2y) \cdot 2 = -2\sin(x + 2y)$
- By the Product Rule: $\frac{\partial f}{\partial y} = (x^2)\cos(x + 2y) + (x^2 y)(-2\sin(x + 2y))$ $\frac{\partial f}{\partial y} = x^2 \cos(x + 2y) - 2x^2 y \sin(x + 2y)$

Segment 2 of 8

📅 Regroup 1

Constant Extraction Strategy: When differentiating with respect to $x$ , mentally replace every $y$ with a constant number (like $5$ ) to confirm whether terms differentiate to zero (additive constants) or factor through (multiplicative constants).
Quotient vs Constant Multiplier: If the variable of differentiation does not appear in the denominator, never use the Quotient Rule—simply pull the denominator out as a constant scalar factor.

Segment 3 of 8

Skill Block 2

Real-Life Engineering Applications

🏗️

Civil Engineering

Stress Concentration at Bolt Hole (Mechanical & Civil Engineering)

A structural engineer analyzes stress distribution around a circular bolt fastening hole in a structural steel gusset plate under heavy tensile loading. Geometric discontinuities cause severe localized stress risers that can initiate micro-cracking and brittle fatigue fracture.

From high-resolution Finite Element Analysis (FEA) simulation, the Von Mises mechanical stress $\sigma(x, y)$ (in megapascals, $\text{MPa}$ ) near the bolt hole boundary is described by:

\sigma(x, y) = \frac{400}{x^2 + 2y^2 + 1}

where

x

and

y

represent planar spatial coordinates in millimeters (

\text{mm}

) measured from the center of the fastening aperture.

State the Quotient Rule and Reciprocal Power Rule representations used to compute $\frac{\partial \sigma}{\partial x}$ and $\frac{\partial \sigma}{\partial y}$ .
Determine general algebraic expressions for the spatial stress gradients $\frac{\partial \sigma}{\partial x}$ and $\frac{\partial \sigma}{\partial y}$ .
Evaluate the instantaneous stress gradients at the critical monitoring coordinate $(x, y) = (1, 1)\text{ mm}$ .
State the physical units of these derivatives. Geometrically interpret what these values mean if an inspector moves away from the bolt hole along the $+x$ vs $+y$ directions.

💡 Show Worked Derivation

A structural engineer analyzes stress distribution around a circular bolt fastening hole in a structural steel gusset plate under heavy tensile loading. Geometric discontinuities cause severe localized stress risers that can initiate micro-cracking and brittle fatigue fracture.

From high-resolution Finite Element Analysis (FEA) simulation, the Von Mises mechanical stress $\sigma(x, y)$ (in megapascals, $\text{MPa}$ ) near the bolt hole boundary is described by:

\sigma(x, y) = \frac{400}{x^2 + 2y^2 + 1}

where

x

and

y

represent planar spatial coordinates in millimeters (

\text{mm}

) measured from the center of the fastening aperture.

State the Quotient Rule and Reciprocal Power Rule representations used to compute $\frac{\partial \sigma}{\partial x}$ and $\frac{\partial \sigma}{\partial y}$ .
Determine general algebraic expressions for the spatial stress gradients $\frac{\partial \sigma}{\partial x}$ and $\frac{\partial \sigma}{\partial y}$ .
Evaluate the instantaneous stress gradients at the critical monitoring coordinate $(x, y) = (1, 1)\text{ mm}$ .
State the physical units of these derivatives. Geometrically interpret what these values mean if an inspector moves away from the bolt hole along the $+x$ vs $+y$ directions.

Step-by-Step Solution:

State the Differentiation Rules:
- Writing $\sigma(x, y) = 400(x^2 + 2y^2 + 1)^{-1}$ , apply the Generalized Power Chain Rule: $\frac{\partial}{\partial x}[c \cdot u^{-1}] = -c \cdot u^{-2} \cdot \frac{\partial u}{\partial x} = -\frac{c}{u^2} \frac{\partial u}{\partial x}$ (Defined in Partial Derivatives and Geometric Interpretation of Partial Derivatives.)
Compute the General Algebraic Expressions:
- Let $u(x, y) = x^2 + 2y^2 + 1$ .
- Differentiating with respect to $x$ (holding $y$ constant): $\frac{\partial u}{\partial x} = 2x$ $\frac{\partial \sigma}{\partial x} = -\frac{400}{(x^2 + 2y^2 + 1)^2} \cdot (2x) = -\frac{800x}{(x^2 + 2y^2 + 1)^2}$
- Differentiating with respect to $y$ (holding $x$ constant): $\frac{\partial u}{\partial y} = 4y$ $\frac{\partial \sigma}{\partial y} = -\frac{400}{(x^2 + 2y^2 + 1)^2} \cdot (4y) = -\frac{1600y}{(x^2 + 2y^2 + 1)^2}$
Evaluate at the Point $(1, 1)\text{ mm}$ :
- Evaluate denominator $u(1, 1)$ : $u(1, 1) = 1^2 + 2(1)^2 + 1 = 1 + 2 + 1 = 4$ $u^2 = 4^2 = 16$
- Evaluate $\frac{\partial \sigma}{\partial x}(1, 1)$ : $\left.\frac{\partial \sigma}{\partial x}\right|_{(1, 1)} = -\frac{800(1)}{16} = -50\text{ MPa/mm}$
- Evaluate $\frac{\partial \sigma}{\partial y}(1, 1)$ : $\left.\frac{\partial \sigma}{\partial y}\right|_{(1, 1)} = -\frac{1600(1)}{16} = -100\text{ MPa/mm}$
Physical Units and Geometric Interpretation:
- Units: Megapascals per millimeter ( $\text{MPa/mm}$ ).
- Geometric Interpretation: At $(1, 1)$ , moving in the positive $x$ -direction (holding $y = 1\text{ mm}$ ) causes localized stress to drop at an instantaneous rate of $50\text{ MPa/mm}$ . Moving in the positive $y$ -direction (holding $x = 1\text{ mm}$ ) causes stress to plunge twice as rapidly, at $-100\text{ MPa/mm}$ .
- Engineering Consequence: The stress gradient is significantly steeper along the $y$ -axis because of the coefficient $2$ in the ellipse geometry. Crack initiation risk diminishes twice as rapidly as you step along $y$ away from the notch than along $x$ .

🏗️

Civil Engineering

Hillside Foundation Slope Stability (Civil & Construction Engineering)

A geotechnical and civil engineering firm assesses hillside terrain stability prior to constructing a reinforced concrete retaining wall and foundation slab for a commercial development in Henderson, Nevada.

A drone-based LiDAR survey generates a continuous digital elevation model $E(x, y)$ (in meters, $\text{m}$ ) across the building pad:

E(x, y) = 50 - 0.05x^3 + 0.1xy + 0.02y^2

where

x

represents the distance eastward (in meters) and

y

represents the distance northward (in meters) relative to an anchored geodetic survey monument.

State how cross-sectional vertical cuts $y = y_0$ and $x = x_0$ define directional slopes on the terrain surface.
Determine the algebraic expressions for the directional slopes $\frac{\partial E}{\partial x}$ and $\frac{\partial E}{\partial y}$ .
Evaluate the slopes $\frac{\partial E}{\partial x}$ and $\frac{\partial E}{\partial y}$ at the proposed foundation center location $(x, y) = (4, 10)\text{ m}$ .
State whether the ground slopes uphill or downhill when walking due east ( $+x$ ) and due north ( $+y$ ) from the center location, and state the angle of inclination along each cardinal direction in degrees ( $\theta = \arctan(m)$ ).

💡 Show Worked Derivation

A geotechnical and civil engineering firm assesses hillside terrain stability prior to constructing a reinforced concrete retaining wall and foundation slab for a commercial development in Henderson, Nevada.

A drone-based LiDAR survey generates a continuous digital elevation model $E(x, y)$ (in meters, $\text{m}$ ) across the building pad:

E(x, y) = 50 - 0.05x^3 + 0.1xy + 0.02y^2

where

x

represents the distance eastward (in meters) and

y

represents the distance northward (in meters) relative to an anchored geodetic survey monument.

State how cross-sectional vertical cuts $y = y_0$ and $x = x_0$ define directional slopes on the terrain surface.
Determine the algebraic expressions for the directional slopes $\frac{\partial E}{\partial x}$ and $\frac{\partial E}{\partial y}$ .
Evaluate the slopes $\frac{\partial E}{\partial x}$ and $\frac{\partial E}{\partial y}$ at the proposed foundation center location $(x, y) = (4, 10)\text{ m}$ .
State whether the ground slopes uphill or downhill when walking due east ( $+x$ ) and due north ( $+y$ ) from the center location, and state the angle of inclination along each cardinal direction in degrees ( $\theta = \arctan(m)$ ).

Step-by-Step Solution:

State the Cross-Sectional Trace Principle:
- Slicing the 3D elevation surface with a plane $y = y_0$ isolates an east-west cross-section where longitude $y$ is constant; the derivative $\frac{\partial E}{\partial x}$ is the instantaneous slope $\frac{\Delta E}{\Delta x}$ along this cut.
- Slicing with a plane $x = x_0$ isolates a north-south cross-section where latitude $x$ is constant; the derivative $\frac{\partial E}{\partial y}$ is the instantaneous slope $\frac{\Delta E}{\Delta y}$ along this cut. (Defined in Geometric Interpretation of Partial Derivatives.)
Compute the Algebraic Slope Expressions:
- Differentiating with respect to $x$ (holding $y$ constant): $\frac{\partial E}{\partial x} = \frac{\partial}{\partial x}(50) - 0.05 \frac{\partial}{\partial x}(x^3) + 0.1y \frac{\partial}{\partial x}(x) + 0.02 \frac{\partial}{\partial x}(y^2)$ $\frac{\partial E}{\partial x} = 0 - 0.05(3x^2) + 0.1y(1) + 0 = -0.15x^2 + 0.1y$
- Differentiating with respect to $y$ (holding $x$ constant): $\frac{\partial E}{\partial y} = \frac{\partial}{\partial y}(50) - 0.05 \frac{\partial}{\partial y}(x^3) + 0.1x \frac{\partial}{\partial y}(y) + 0.02 \frac{\partial}{\partial y}(y^2)$ $\frac{\partial E}{\partial y} = 0 - 0 + 0.1x(1) + 0.02(2y) = 0.1x + 0.04y$
Evaluate at $(x, y) = (4, 10)\text{ m}$ :
- For the east-west slope: $\left.\frac{\partial E}{\partial x}\right|_{(4, 10)} = -0.15(4)^2 + 0.1(10) = -0.15(16) + 1.0 = -2.40 + 1.0 = -1.40\text{ m/m}$
- For the north-south slope: $\left.\frac{\partial E}{\partial y}\right|_{(4, 10)} = 0.1(4) + 0.04(10) = 0.40 + 0.40 = 0.80\text{ m/m}$
Interpret Cardinal Behavior and Angles of Inclination:
- Eastward Direction ( $+x$ ): Since $\frac{\partial E}{\partial x} = -1.40 < 0$ , walking due east is walking downhill at a $-140\%$ grade. $\theta_x = \arctan(-1.40) \approx -54.46^\circ$ (A steep drop of $54.5^\circ$ below the horizontal).
- Northward Direction ( $+y$ ): Since $\frac{\partial E}{\partial y} = 0.80 > 0$ , walking due north is walking uphill at an $+80\%$ grade. $\theta_y = \arctan(0.80) \approx 38.66^\circ$ (A climb of $38.7^\circ$ above the horizontal).
- Engineering Conclusion: The steep drop to the east mandates an engineered retaining wall system to prevent catastrophic landslide slippage along the cut slope.

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Evaluating and Interpreting Partial Derivatives at Points & Geometric Traces

General Overview
Skill Block 2 focuses on evaluating partial derivatives at specific numerical coordinates, interpreting them as slopes of tangent lines to vertical trace curves ( $x = x_0$ and $y = y_0$ ), constructing 3D tangent vectors, and analyzing directional rates of change in physical scalar fields.

Conceptual Framework

Evaluation at Coordinate Points: $\left.\frac{\partial f}{\partial x}\right|_{(x_0, y_0)} = f_x(x_0, y_0), \quad \left.\frac{\partial f}{\partial y}\right|_{(x_0, y_0)} = f_y(x_0, y_0)$
Geometric Interpretation on Vertical Cross-Sections:
- $f_x(x_0, y_0)$ is the slope of the tangent line to the surface trace in the vertical slicing plane $y = y_0$ .
- $f_y(x_0, y_0)$ is the slope of the tangent line to the surface trace in the vertical slicing plane $x = x_0$ . (Detailed in Geometric Interpretation of Partial Derivatives.)

Key Equations & Structural Rules

Tangent Line Direction Vectors in 3D Space: $\vec{v}_1 = \langle 1, 0, f_x(x_0, y_0) \rangle = \hat{i} + f_x(x_0, y_0)\hat{k}$ $\vec{v}_2 = \langle 0, 1, f_y(x_0, y_0) \rangle = \hat{j} + f_y(x_0, y_0)\hat{k}$
Three-Variable Directional Rate of Change: In a 3D scalar field $w = f(x, y, z)$ , $f_x, f_y, f_z$ represent the instantaneous rates of change of $w$ per unit distance traveled parallel to the $x$ , $y$ , and $z$ axes.

Worked Example A: Evaluating and Interpreting Slopes on a Cubic-Quadratic Surface

For the surface defined by:

f(x, y) = x^2 y - y^3

Compute the general partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ .
Evaluate $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ at the point $(1, 2)$ .
Interpret the physical and geometric meaning of these numerical values in terms of surface slopes along vertical cross-sections.

💡 Reveal Worked Solution

Find General Partial Derivatives: State the Constant-Variable Rule:
- Differentiating with respect to $x$ (treating $y$ as a constant): $\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2 y) - \frac{\partial}{\partial x}(y^3) = 2xy - 0 = 2xy$
- Differentiating with respect to $y$ (treating $x$ as a constant): $\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2 y) - \frac{\partial}{\partial y}(y^3) = x^2(1) - 3y^2 = x^2 - 3y^2$
Evaluate at $(x_0, y_0) = (1, 2)$ : Substitute $x = 1$ and $y = 2$ :
$\left.\frac{\partial f}{\partial x}\right|_{(1, 2)} = 2(1)(2) = 4$ $\left.\frac{\partial f}{\partial y}\right|_{(1, 2)} = (1)^2 - 3(2)^2 = 1 - 3(4) = 1 - 12 = -11$
Geometric Interpretation:
- $x$ -Direction Slope ( $m_x = 4$ ): Slicing the surface with the vertical plane $y = 2$ produces the parabolic trace curve $z = f(x, 2) = 2x^2 - 8$ . The tangent line to this trace at $(1, 2, -6)$ has slope $4$ . If an engineer stands at $(1, 2, -6)$ and walks parallel to the positive $x$ -axis, the surface rises at an instantaneous rate of $4$ units of height per unit of horizontal distance.
- $y$ -Direction Slope ( $m_y = -11$ ): Slicing the surface with the vertical plane $x = 1$ produces the cubic trace curve $z = f(1, y) = y - y^3$ . The tangent line to this trace at $(1, 2, -6)$ has slope $-11$ . If the engineer walks parallel to the positive $y$ -axis, the surface drops steeply at a rate of $11$ units of depth per unit of distance.

Practice Problem 6: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Slicing the surface $z = f(x, y)$ with the vertical plane $y = y_0$ produces a 2D trace curve. How is $\frac{\partial f}{\partial x}(x_0, y_0)$ related to this curve?
(b) In Worked Example A, describe what a walker at $(1, 2, -6)$ experiences if they step in the $+x$ direction versus the $+y$ direction.
(c) In Stress Concentration at Bolt Hole (Mechanical & Civil Engineering), why are the spatial derivatives $\frac{\partial \sigma}{\partial x}$ and $\frac{\partial \sigma}{\partial y}$ negative at $(1, 1)\text{ mm}$ ?

Practice Space Solve this problem on paper or tablet.

Review the Geometric Interpretation on Vertical Cross-Sections and the solution to Worked Example A. (Defined in Geometric Interpretation of Partial Derivatives.)

Answer: * (a) It is the slope of the tangent line to the trace curve at $(x_0, y_0, f(x_0, y_0))$ inside the slicing plane $y = y_0$ .

(b) Stepping in $+x$ climbs an uphill slope of $+4$ ; stepping in $+y$ plunges down a steep downhill drop of $-11$ .
(c) Moving outward away from the center of the bolt hole reduces stress concentration; stress decreases with distance.

Step-by-Step Execution

(a) As detailed in Geometric Interpretation of Partial Derivatives, slicing $z = f(x, y)$ with the plane $y = y_0$ produces the cross-sectional trace curve $C_1: z = f(x, y_0)$ . The partial derivative $\frac{\partial f}{\partial x}(x_0, y_0)$ is precisely the instantaneous slope $\frac{dz}{dx}$ of the tangent line $T_1$ to $C_1$ at the point $P(x_0, y_0, f(x_0, y_0))$ .
(b) In Worked Example A, the partial derivatives at $(1, 2)$ are $f_x = 4$ and $f_y = -11$ . A person standing at $(1, 2, -6)$ walking parallel to the positive $x$ -axis climbs upward at an incline grade of $+4$ ( $+400\%$ ). Walking parallel to the positive $y$ -axis plunges steeply downward at an incline grade of $-11$ ( $-1100\%$ ).
(c) In Stress Concentration at Bolt Hole (Mechanical & Civil Engineering), the stress distribution is $\sigma(x, y) = \frac{400}{x^2 + 2y^2 + 1}$ . Peak stress occurs at the aperture edge $(0, 0)$ . As an inspector moves away from the hole (increasing $x$ or $y$ ), the denominator increases, causing stress to diminish. Negative derivatives indicate decreasing stress concentration away from the defect.

Practice Problem 7

For the rational surface:

f(x, y) = \frac{x}{y + 1}

Determine the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ .
Evaluate $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ at the point $(2, 1)$ .
Interpret these values geometrically in terms of trace curve slopes.

Practice Space Solve this problem on paper or tablet.

Notice that when differentiating with respect to $x$ , $\frac{1}{y+1}$ is a constant scalar. When differentiating with respect to $y$ , use the reciprocal rule $\frac{d}{dy}[(y+1)^{-1}] = -(y+1)^{-2}$ . (Defined in Partial Derivatives and Geometric Interpretation of Partial Derivatives.)

Answer: 1. $\frac{\partial f}{\partial x} = \frac{1}{y+1}, \quad \frac{\partial f}{\partial y} = -\frac{x}{(y+1)^2}$ 2. $\frac{\partial f}{\partial x}(2, 1) = \frac{1}{2}, \quad \frac{\partial f}{\partial y}(2, 1) = -\frac{1}{2}$ 3. Slopes of trace tangent lines: $+0.5$ in the $x$ -direction and $-0.5$ in the $y$ -direction.

Step-by-Step Execution

Find General Partial Derivatives:
- With respect to $x$ : $\frac{\partial f}{\partial x} = \frac{1}{y+1} \cdot \frac{\partial}{\partial x}(x) = \frac{1}{y+1}$
- With respect to $y$ : $\frac{\partial f}{\partial y} = x \cdot \frac{\partial}{\partial y}\left[(y+1)^{-1}\right] = x\left[-(y+1)^{-2}(1)\right] = -\frac{x}{(y+1)^2}$
Evaluate at $(2, 1)$ :
$\left.\frac{\partial f}{\partial x}\right|_{(2, 1)} = \frac{1}{1+1} = \frac{1}{2}$ $\left.\frac{\partial f}{\partial y}\right|_{(2, 1)} = -\frac{2}{(1+1)^2} = -\frac{2}{4} = -\frac{1}{2}$
Geometric Interpretation:
- In the vertical plane $y = 1$ , the trace is a straight line $z = \frac{1}{2}x$ . Its slope is a constant $+1/2$ everywhere.
- In the vertical plane $x = 2$ , the trace is a hyperbola $z = \frac{2}{y+1}$ . At $y = 1$ , the tangent line has slope $-1/2$ , meaning the surface decreases at rate $0.5$ in the positive $y$ -direction.

Practice Problem 8

For the inverse trigonometric surface:

f(x, y) = \arctan(xy)

Determine $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ .
Evaluate $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ at the point $(0, 1)$ .
Explain the geometric significance of the value obtained for $\frac{\partial f}{\partial y}(0, 1)$ .

Practice Space Solve this problem on paper or tablet.

State the Arctangent Chain Rule: $\frac{\partial}{\partial x}[\arctan(u)] = \frac{1}{1+u^2} \frac{\partial u}{\partial x}$ , where $u = xy$ . (Defined in Partial Derivatives.)

Answer: 1. $\frac{\partial f}{\partial x} = \frac{y}{1+x^2 y^2}, \quad \frac{\partial f}{\partial y} = \frac{x}{1+x^2 y^2}$ 2. $\frac{\partial f}{\partial x}(0, 1) = 1, \quad \frac{\partial f}{\partial y}(0, 1) = 0$ 3. $f_y(0, 1) = 0$ means the surface is momentarily completely flat (zero slope) along the vertical cross-section $x = 0$ .

Step-by-Step Execution

Find General Partial Derivatives: State the Arctangent Derivative Rule:
$\frac{\partial f}{\partial x} = \frac{1}{1 + (xy)^2} \cdot \frac{\partial}{\partial x}(xy) = \frac{y}{1 + x^2 y^2}$ $\frac{\partial f}{\partial y} = \frac{1}{1 + (xy)^2} \cdot \frac{\partial}{\partial y}(xy) = \frac{x}{1 + x^2 y^2}$
Evaluate at $(0, 1)$ :
$\left.\frac{\partial f}{\partial x}\right|_{(0, 1)} = \frac{1}{1 + (0)^2(1)^2} = \frac{1}{1} = 1$ $\left.\frac{\partial f}{\partial y}\right|_{(0, 1)} = \frac{0}{1 + (0)^2(1)^2} = \frac{0}{1} = 0$
Geometric Significance:
- Along the plane $x = 0$ (the $yz$ -plane), $f(0, y) = \arctan(0) = 0$ for all $y$ . The trace curve is the horizontal line $z = 0$ .
- Because this trace is perfectly horizontal, its slope $\frac{\partial f}{\partial y}(0, 1)$ is identically zero. The surface is tangent to a horizontal line in the $y$ -direction at $(0, 1, 0)$ .

Practice Problem 9

For the three-variable scalar field:

f(x, y, z) = x^2 + y^2 - z^2

Determine $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}$ , and $\frac{\partial f}{\partial z}$ .
Evaluate all three partial derivatives at the point $(1, -1, 2)$ .
Interpret these values in terms of directional rates of change within 3D space.

Practice Space Solve this problem on paper or tablet.

Hold two variables constant at a time. Evaluate the resulting expressions at $(x, y, z) = (1, -1, 2)$ . (Defined in Partial Derivatives of Functions of Three or More Variables.)

Answer: 1. $\frac{\partial f}{\partial x} = 2x, \quad \frac{\partial f}{\partial y} = 2y, \quad \frac{\partial f}{\partial z} = -2z$ 2. $f_x(1, -1, 2) = 2, \quad f_y(1, -1, 2) = -2, \quad f_z(1, -1, 2) = -4$ 3. Moving in $+x$ increases $f$ at rate $2$ ; moving in $+y$ decreases $f$ at rate $2$ ; moving in $+z$ decreases $f$ rapidly at rate $4$ .

Step-by-Step Execution

Find General Partial Derivatives:
- Treating $y, z$ as constants: $\frac{\partial f}{\partial x} = 2x + 0 - 0 = 2x$ .
- Treating $x, z$ as constants: $\frac{\partial f}{\partial y} = 0 + 2y - 0 = 2y$ .
- Treating $x, y$ as constants: $\frac{\partial f}{\partial z} = 0 + 0 - 2z = -2z$ .
Evaluate at $(1, -1, 2)$ :
$\left.\frac{\partial f}{\partial x}\right|_{(1, -1, 2)} = 2(1) = 2$ $\left.\frac{\partial f}{\partial y}\right|_{(1, -1, 2)} = 2(-1) = -2$ $\left.\frac{\partial f}{\partial z}\right|_{(1, -1, 2)} = -2(2) = -4$
Physical Interpretation in 3D Space: In this 3D scalar field:
- Stepping parallel to the positive $x$ -axis (holding $y = -1, z = 2$ ) increases $f$ at an instantaneous rate of $2$ units per unit distance.
- Stepping parallel to the positive $y$ -axis (holding $x = 1, z = 2$ ) decreases $f$ at an instantaneous rate of $2$ units per unit distance.
- Stepping parallel to the positive $z$ -axis (holding $x = 1, y = -1$ ) causes $f$ to decrease rapidly at a rate of $4$ units per unit distance.

Segment 4 of 8

📅 Regroup 2

Trace Plane Visualization: Always remember that $f_x$ lives in the slicing plane $y = \text{constant}$ and $f_y$ lives in the slicing plane $x = \text{constant}$ .
Vector Bridge: The two tangent vectors $\vec{v}_1 = \langle 1, 0, f_x \rangle$ and $\vec{v}_2 = \langle 0, 1, f_y \rangle$ cross to form the normal vector $\vec{n} = \langle f_x, f_y, -1 \rangle$ , which is the foundation of tangent planes in Session 10.

Segment 5 of 8

Skill Block 3

Real-Life Engineering Applications

🏗️

Civil Engineering

Topographic Retention Basin Drainage (Environmental & Civil Engineering)

An environmental and water resources engineer studies a high-resolution topographic survey contour map of an alluvial valley in the Las Vegas wash basin to design a regional stormwater detention basin. The contour lines represent lines of constant ground surface elevation $E(x, y)$ marked in meters ( $\text{m}$ ) above sea level.

At a prospective retention inlet located at survey coordinate point $P(x_0, y_0)$ :

The contour lines run perfectly horizontal (straight east-to-west, parallel to the $x$ -axis).
Moving vertically northward (in the positive $y$ -direction) from point $P$ , the contour elevation labels decrease steadily: $105\text{ m} \to 100\text{ m} \to 95\text{ m}$ .
The contour lines representing $5\text{ m}$ elevation drops are spaced tightly at $\Delta y = 10\text{ m}$ intervals on the ground.

State the visual rules for determining the sign of partial derivatives $\frac{\partial E}{\partial x}$ and $\frac{\partial E}{\partial y}$ from a contour plot.
Determine the signs and instantaneous numerical values (or approximations) of $\frac{\partial E}{\partial x}$ and $\frac{\partial E}{\partial y}$ at point $P$ .
Explain why the horizontal orientation of the contour line forces $\frac{\partial E}{\partial x} = 0$ .
In hydrological engineering, stormwater runoff velocity $V$ is proportional to the hydraulic ground gradient $S = \sqrt{(\partial E/\partial x)^2 + (\partial E/\partial y)^2}$ . Calculate $S$ at point $P$ and state the direction in which surface water will flow.

💡 Show Worked Derivation

An environmental and water resources engineer studies a high-resolution topographic survey contour map of an alluvial valley in the Las Vegas wash basin to design a regional stormwater detention basin. The contour lines represent lines of constant ground surface elevation $E(x, y)$ marked in meters ( $\text{m}$ ) above sea level.

At a prospective retention inlet located at survey coordinate point $P(x_0, y_0)$ :

The contour lines run perfectly horizontal (straight east-to-west, parallel to the $x$ -axis).
Moving vertically northward (in the positive $y$ -direction) from point $P$ , the contour elevation labels decrease steadily: $105\text{ m} \to 100\text{ m} \to 95\text{ m}$ .
The contour lines representing $5\text{ m}$ elevation drops are spaced tightly at $\Delta y = 10\text{ m}$ intervals on the ground.

State the visual rules for determining the sign of partial derivatives $\frac{\partial E}{\partial x}$ and $\frac{\partial E}{\partial y}$ from a contour plot.
Determine the signs and instantaneous numerical values (or approximations) of $\frac{\partial E}{\partial x}$ and $\frac{\partial E}{\partial y}$ at point $P$ .
Explain why the horizontal orientation of the contour line forces $\frac{\partial E}{\partial x} = 0$ .
In hydrological engineering, stormwater runoff velocity $V$ is proportional to the hydraulic ground gradient $S = \sqrt{(\partial E/\partial x)^2 + (\partial E/\partial y)^2}$ . Calculate $S$ at point $P$ and state the direction in which surface water will flow.

Step-by-Step Solution:

State the Visual Step-and-Observe Rules:
- For $\frac{\partial E}{\partial x}$ : Step to the right (east, $+x$ ) holding $y$ constant. Crossing into higher contours means $\frac{\partial E}{\partial x} > 0$ ; crossing into lower contours means $\frac{\partial E}{\partial x} < 0$ ; traveling tangent along a contour means $\frac{\partial E}{\partial x} = 0$ .
- For $\frac{\partial E}{\partial y}$ : Step upward (north, $+y$ ) holding $x$ constant. Crossing into higher contours means $\frac{\partial E}{\partial y} > 0$ ; crossing into lower contours means $\frac{\partial E}{\partial y} < 0$ . (Defined in Visual Estimation of Partial Derivatives from Contour Maps.)
Evaluate the Signs and Magnitudes:
- East-West Derivative $\frac{\partial E}{\partial x}$ : Moving along the $+x$ axis travels directly along the horizontal contour line. Because elevation does not change along the contour, the instantaneous rate of change is zero: $\frac{\partial E}{\partial x} = 0$
- North-South Derivative $\frac{\partial E}{\partial y}$ : Moving along the $+y$ axis crosses from $100\text{ m}$ to $95\text{ m}$ over a distance of $\Delta y = 10\text{ m}$ . $\Delta E = 95 - 100 = -5\text{ m}$ $\frac{\partial E}{\partial y} \approx \frac{\Delta E}{\Delta y} = \frac{-5\text{ m}}{10\text{ m}} = -0.50\text{ m/m} = -50\%$ Thus, $\frac{\partial E}{\partial y}$ is negative, with a magnitude of $0.50\text{ m/m}$ .
Geometric Reason for Vanishing East-West Derivative:
- A contour line $E(x, y) = c$ represents an isoline of constant height. Because the contour runs east-west, the tangent vector to the contour is $\vec{T} = \langle 1, 0 \rangle = \hat{i}$ .
- Differentiating in the direction of the contour tangent measures zero change in elevation. Hence, the ground is completely level along the east-west axis at point $P$ .
Calculate Hydraulic Gradient and Flow Direction:
- State the Hydraulic Ground Gradient Formula: $S = \sqrt{\left(\frac{\partial E}{\partial x}\right)^2 + \left(\frac{\partial E}{\partial y}\right)^2} = \sqrt{0^2 + (-0.50)^2} = \sqrt{0.25} = 0.50\text{ m/m}$
- Flow Direction: Under gravity, water flows down the steepest descent gradient ( $-\nabla E$ ): $-\nabla E = -\left\langle \frac{\partial E}{\partial x}, \frac{\partial E}{\partial y} \right\rangle = -\langle 0, -0.50 \rangle = \langle 0, 0.50 \rangle$
- Surface runoff flows due north (straight down the fall line into the detention inlet) at a steep $50\%$ drop.

⚙️

Mechanical Engineering

Ceramic Engine Component Thermal Strain (Materials & Mechanical Engineering)

A materials and mechanical reliability engineer evaluates experimental digital image correlation (DIC) full-field strain measurements on a silicon carbide (SiC) ceramic stator vane in an experimental gas turbine engine. Under severe thermal cycling, brittle ceramics fail catastrophically when local strain gradients exceed critical thresholds.

The engineer inspects a 2D contour map of total mechanical strain $\epsilon(x, y)$ (dimensionless microstrain, $\mu\epsilon$ , where $1\mu\epsilon = 10^{-6}\text{ m/m}$ ) around a cooling hole fillet at evaluation point $Q(x_0, y_0)$ :

The contour lines form a family of nested, elongated concentric ellipses centered near the notch edge.
Stepping to the right (positive $x$ -direction) from point $Q$ , the strain contours are separated by microscopic spacing ( $\Delta x = 0.2\text{ mm}$ ) and cross values of $2000\mu\epsilon \to 5000\mu\epsilon \to 9000\mu\epsilon$ .
Stepping vertically upward (positive $y$ -direction) from point $Q$ , the strain contours are widely spaced ( $\Delta y = 2.0\text{ mm}$ ) and cross values of $2000\mu\epsilon \to 2100\mu\epsilon \to 2200\mu\epsilon$ .

State how contour line density (spacing) relates mathematically and physically to the magnitude of partial derivatives.
Calculate numerical approximations for the directional strain gradients $\frac{\partial \epsilon}{\partial x}$ and $\frac{\partial \epsilon}{\partial y}$ at point $Q$ .
Compare the magnitudes of $\left|\frac{\partial \epsilon}{\partial x}\right|$ and $\left|\frac{\partial \epsilon}{\partial y}\right|$ . Explain what this extreme anisotropy reveals about directional mechanical loading.
If the critical fracture criterion for the ceramic material establishes a maximum permissible spatial gradient of $10{,}000\mu\epsilon/\text{mm}$ , determine whether point $Q$ violates the safety threshold and identify along which coordinate axis failure is imminent.

💡 Show Worked Derivation

A materials and mechanical reliability engineer evaluates experimental digital image correlation (DIC) full-field strain measurements on a silicon carbide (SiC) ceramic stator vane in an experimental gas turbine engine. Under severe thermal cycling, brittle ceramics fail catastrophically when local strain gradients exceed critical thresholds.

The engineer inspects a 2D contour map of total mechanical strain $\epsilon(x, y)$ (dimensionless microstrain, $\mu\epsilon$ , where $1\mu\epsilon = 10^{-6}\text{ m/m}$ ) around a cooling hole fillet at evaluation point $Q(x_0, y_0)$ :

The contour lines form a family of nested, elongated concentric ellipses centered near the notch edge.
Stepping to the right (positive $x$ -direction) from point $Q$ , the strain contours are separated by microscopic spacing ( $\Delta x = 0.2\text{ mm}$ ) and cross values of $2000\mu\epsilon \to 5000\mu\epsilon \to 9000\mu\epsilon$ .
Stepping vertically upward (positive $y$ -direction) from point $Q$ , the strain contours are widely spaced ( $\Delta y = 2.0\text{ mm}$ ) and cross values of $2000\mu\epsilon \to 2100\mu\epsilon \to 2200\mu\epsilon$ .

State how contour line density (spacing) relates mathematically and physically to the magnitude of partial derivatives.
Calculate numerical approximations for the directional strain gradients $\frac{\partial \epsilon}{\partial x}$ and $\frac{\partial \epsilon}{\partial y}$ at point $Q$ .
Compare the magnitudes of $\left|\frac{\partial \epsilon}{\partial x}\right|$ and $\left|\frac{\partial \epsilon}{\partial y}\right|$ . Explain what this extreme anisotropy reveals about directional mechanical loading.
If the critical fracture criterion for the ceramic material establishes a maximum permissible spatial gradient of $10{,}000\mu\epsilon/\text{mm}$ , determine whether point $Q$ violates the safety threshold and identify along which coordinate axis failure is imminent.

Step-by-Step Solution:

State the Contour Density Principle:
- On any contour map, the spatial gradient magnitude is inversely proportional to contour spacing: $\left|\frac{\partial f}{\partial s}\right| \approx \frac{|\Delta f|}{\Delta s}$
- Narrow spacing ( $\Delta s \to 0$ ) implies an extremely steep slope and high rate of change. Broad spacing ( $\Delta s \gg 0$ ) represents a gradual, mild slope. (Defined in Visual Estimation of Partial Derivatives from Contour Maps.)
Calculate Directional Strain Gradients:
- Horizontal Strain Gradient $\frac{\partial \epsilon}{\partial x}$ : Using a forward difference from $2000\mu\epsilon$ to $5000\mu\epsilon$ over $\Delta x = 0.2\text{ mm}$ : $\frac{\partial \epsilon}{\partial x} \approx \frac{\Delta \epsilon}{\Delta x} = \frac{5000\mu\epsilon - 2000\mu\epsilon}{0.2\text{ mm}} = \frac{3000\mu\epsilon}{0.2\text{ mm}} = 15{,}000\mu\epsilon/\text{mm}$ (Over the next interval to $9000\mu\epsilon$ , the gradient increases further to $\frac{4000}{0.2} = 20{,}000\mu\epsilon/\text{mm}$ ).
- Vertical Strain Gradient $\frac{\partial \epsilon}{\partial y}$ : Using a forward difference from $2000\mu\epsilon$ to $2100\mu\epsilon$ over $\Delta y = 2.0\text{ mm}$ : $\frac{\partial \epsilon}{\partial y} \approx \frac{\Delta \epsilon}{\Delta y} = \frac{2100\mu\epsilon - 2000\mu\epsilon}{2.0\text{ mm}} = \frac{100\mu\epsilon}{2.0\text{ mm}} = 50\mu\epsilon/\text{mm}$
Compare Magnitudes and Directional Anisotropy:
$\frac{|\partial \epsilon / \partial x|}{|\partial \epsilon / \partial y|} = \frac{15{,}000\mu\epsilon/\text{mm}}{50\mu\epsilon/\text{mm}} = 300$
- The rate of strain escalation along the $x$ -axis is 300 times greater than along the $y$ -axis.
- Physical Rationale: This extreme anisotropy demonstrates severe unconstrained stress concentration along the $x$ -direction (likely perpendicular to a sharp structural notch or cooling hole boundary), whereas the $y$ -direction lies nearly parallel to the principal neutral load axis.
Safety and Failure Assessment:
- Threshold criterion: $\left|\frac{\partial \epsilon}{\partial s}\right| \le 10{,}000\mu\epsilon/\text{mm}$ .
- Along the $x$ -axis, $\frac{\partial \epsilon}{\partial x} \approx 15{,}000\mu\epsilon/\text{mm} > 10{,}000\mu\epsilon/\text{mm}$ .
- Conclusion: Point $Q$ substantially violates the safety threshold in the $x$ -direction. Thermal-mechanical shock will initiate catastrophic brittle micro-cracking propagating along the $y$ -axis perpendicular to this extreme horizontal gradient.

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Interpreting Partial Derivatives Visually from Contour Maps & Grid Data

General Overview
Skill Block 3 teaches students how to extract partial derivative information from 2D contour maps (level curves) and discrete numerical grids without algebraic formulas, determining signs, relative magnitudes, zero-slope tangencies, and finite-difference approximations.

Conceptual Framework

The Step-and-Observe Rule on Contour Maps:
- With respect to $x$ ( $\frac{\partial f}{\partial x}$ ): Move to the right (positive $x$ -direction) holding $y$ constant. Crossing into higher contours means $f_x > 0$ ; crossing into lower contours means $f_x < 0$ .
- With respect to $y$ ( $\frac{\partial f}{\partial y}$ ): Move upward (positive $y$ -direction) holding $x$ constant. Crossing into higher contours means $f_y > 0$ ; crossing into lower contours means $f_y < 0$ .
- Tangency Condition: If the coordinate step is tangent to a level curve, the instantaneous derivative is zero ( $f_x = 0$ or $f_y = 0$ ). (Detailed in Visual Estimation of Partial Derivatives from Contour Maps.)
Magnitude and Contour Density:
- Closely spaced contour lines indicate a large rate of change (steep slope).
- Widely spaced contour lines indicate a small rate of change (gentle slope).
Central Difference Estimation on Discrete Grids: $\frac{\partial f}{\partial x}(x_0, y_0) \approx \frac{f(x_0 + \Delta x, y_0) - f(x_0 - \Delta x, y_0)}{2\Delta x}$ $\frac{\partial f}{\partial y}(x_0, y_0) \approx \frac{f(x_0, y_0 + \Delta y) - f(x_0, y_0 - \Delta y)}{2\Delta y}$

Worked Example A: Visual Estimation on Elliptic Paraboloid Contours

Given the contour plot of $f(x, y) = x^2 + 2y^2$ , whose level curves are concentric ellipses $x^2 + 2y^2 = c$ :

Visually estimate the signs and relative values of $\frac{\partial f}{\partial x}(1, 0)$ and $\frac{\partial f}{\partial y}(1, 0)$ from the contour plot.
Verify your visual estimations by computing the exact analytical partial derivatives at $(1, 0)$ .

💡 Reveal Worked Solution

Visual Geometric Analysis from Contour Map:
- Locate Point $(1, 0)$ : The point sits on the positive $x$ -axis on the ellipse $x^2 + 2y^2 = 1$ ( $c = 1$ ).
- Analyze $\frac{\partial f}{\partial x}(1, 0)$ : Take a small step to the right (in the $+x$ direction) along the line $y = 0$ . As $x$ increases past $1$ , you cross out of the $c = 1$ ellipse and enter regions bounded by $c = 2, 4, 9$ . Because contour values are increasing, $\frac{\partial f}{\partial x}(1, 0) > 0$ (positive slope).
- Analyze $\frac{\partial f}{\partial y}(1, 0)$ : Take a small step straight upward (in the $+y$ direction) along the line $x = 1$ . The ellipse $x^2 + 2y^2 = 1$ has a vertical tangent line at its vertex $(1, 0)$ . Because the vertical line $x = 1$ is tangent to the contour ellipse at this point, taking an infinitesimal step along $y$ remains along the constant height curve. Therefore, $\frac{\partial f}{\partial y}(1, 0) = 0$ .
Analytical Verification:
- Differentiate $f(x, y) = x^2 + 2y^2$ : $\frac{\partial f}{\partial x} = 2x \implies \left.\frac{\partial f}{\partial x}\right|_{(1, 0)} = 2(1) = 2 > 0$ $\frac{\partial f}{\partial y} = 4y \implies \left.\frac{\partial f}{\partial y}\right|_{(1, 0)} = 4(0) = 0$ The analytical calculations confirm the visual observations: $\frac{\partial f}{\partial x}$ is positive ( $+2$ ) and $\frac{\partial f}{\partial y}$ is exactly zero.

Practice Problem 10: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) How can you immediately tell whether $\frac{\partial f}{\partial x}$ is positive, negative, or zero directly from a contour plot?
(b) Why did $\frac{\partial f}{\partial y}(1, 0)$ equal zero in Worked Example A? How is this related to curve tangency?
(c) In Ceramic Engine Component Thermal Strain (Materials & Mechanical Engineering), how does the ratio of contour spacing between the $x$ and $y$ axes inform engineers about directional stress concentrations?

Practice Space Solve this problem on paper or tablet.

Review the Step-and-Observe Rule and the solutions to Worked Example A and the linked materials engineering application. (Defined in Visual Estimation of Partial Derivatives from Contour Maps.)

Answer: * (a) Step right (+x): crossing into higher labels = positive, lower = negative, traveling along contour = zero.

(b) The vertical line $x = 1$ is tangent to the elliptical contour at $(1, 0)$ ; no change in height occurs along the tangent.
(c) Microscopic contour spacing along $x$ versus wide spacing along $y$ indicates an extreme directional strain gradient ( $\partial \epsilon/\partial x \gg \partial \epsilon/\partial y$ ).

Step-by-Step Execution

(a) As established in Visual Estimation of Partial Derivatives from Contour Maps, you apply the Step-and-Observe Rule: start at the point and step to the right along the positive $x$ -direction (holding $y$ constant). If you cross into contour lines with higher numerical values, $\frac{\partial f}{\partial x} > 0$ . If you cross into lower values, $\frac{\partial f}{\partial x} < 0$ . If your step is tangent to the contour line, $\frac{\partial f}{\partial x} = 0$ .
(b) In Worked Example A, the level curves are concentric ellipses centered at the origin. At the point $(1, 0)$ on the $x$ -axis, the vertical line $x = 1$ is perfectly tangent to the elliptical level curve. Stepping in the $y$ -direction moves along the tangent line of the constant-value contour, meaning the instantaneous rate of change is zero.
(c) In Ceramic Engine Component Thermal Strain (Materials & Mechanical Engineering), contour lines of strain are spaced at $\Delta x = 0.2\text{ mm}$ along the $x$ -axis but $\Delta y = 2.0\text{ mm}$ along the $y$ -axis. Since gradient magnitude is inversely proportional to contour spacing, $\frac{\partial \epsilon}{\partial x}$ is 300 times larger than $\frac{\partial \epsilon}{\partial y}$ , indicating severe unconstrained stress concentration across the horizontal axis.

Practice Problem 11

Consider the circular paraboloid $f(x, y) = x^2 + y^2$ .

Describe the geometry of the level curves $f(x, y) = c$ for $c > 0$ .
At the point $(2, 0)$ , determine the visual signs of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ .
Verify by computing the exact analytical values of $\frac{\partial f}{\partial x}(2, 0)$ and $\frac{\partial f}{\partial y}(2, 0)$ .

Practice Space Solve this problem on paper or tablet.

The level curves are circles $x^2 + y^2 = c$ . Consider what happens when you step right ( $+x$ ) and step up ( $+y$ ) from $(2, 0)$ . (Defined in Visual Estimation of Partial Derivatives from Contour Maps.)

Answer: 1. Concentric circles centered at $(0, 0)$ with radius $r = \sqrt{c}$ . 2. $\frac{\partial f}{\partial x}(2, 0) > 0$ and $\frac{\partial f}{\partial y}(2, 0) = 0$ . 3. Analytical values: $f_x(2, 0) = 4, \quad f_y(2, 0) = 0$ .

Step-by-Step Execution

Describe Level Curves: The level curves are $x^2 + y^2 = c$ . For each $c > 0$ , this represents a circle centered at $(0, 0)$ with radius $r = \sqrt{c}$ . As $c$ increases, the circles expand outward.
Visual Estimation at $(2, 0)$ :
- Point $(2, 0)$ lies on the circle of radius $2$ ( $c = 4$ ).
- Stepping right ( $+x$ ) along $y = 0$ moves to larger radii ( $r > 2$ ), crossing into outer circles where $c > 4$ . Thus, $\frac{\partial f}{\partial x}(2, 0) > 0$ .
- Stepping upward ( $+y$ ) along $x = 2$ is tangent to the circle $x^2 + y^2 = 4$ at the rightmost vertex $(2, 0)$ . Thus, $\frac{\partial f}{\partial y}(2, 0) = 0$ .
Analytical Verification:
$\frac{\partial f}{\partial x} = 2x \implies f_x(2, 0) = 2(2) = 4$ $\frac{\partial f}{\partial y} = 2y \implies f_y(2, 0) = 2(0) = 0$

Practice Problem 12

Use the contour plot of the circular cone $f(x, y) = \sqrt{x^2 + y^2}$ to estimate and evaluate $\frac{\partial f}{\partial y}(1, 1)$ .

Visually determine the sign of $\frac{\partial f}{\partial y}(1, 1)$ by observing how stepping in the positive $y$ -direction moves relative to the level curves.
Compute the exact analytical value of $\frac{\partial f}{\partial y}(1, 1)$ and compare it to the cone's constant radial slope.

Practice Space Solve this problem on paper or tablet.

The level curves are concentric circles of radius $r = c$ . At $(1, 1)$ , distance from the origin is $r = \sqrt{2}$ . (Defined in Visual Estimation of Partial Derivatives from Contour Maps.)

Answer: 1. Positive ( $\frac{\partial f}{\partial y} > 0$ ). 2. $\frac{\partial f}{\partial y}(1, 1) = \frac{\sqrt{2}}{2} \approx 0.707$ .

Step-by-Step Execution

Visual Analysis: The level curves are $f(x, y) = \sqrt{x^2 + y^2} = c \implies x^2 + y^2 = c^2$ , circles of radius $c$ .
- Point $(1, 1)$ has distance $r = \sqrt{1^2 + 1^2} = \sqrt{2}$ from the origin, lying on contour $c = \sqrt{2} \approx 1.414$ .
- Taking a step upward along $x = 1$ in the positive $y$ -direction increases $y$ . The distance to the origin becomes $\sqrt{1^2 + y^2} > \sqrt{2}$ , crossing into outer circles with higher $c$ values.
- Therefore, $\frac{\partial f}{\partial y}(1, 1)$ must be positive.
Analytical Computation: State the Radical Chain Rule:
$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\left[(x^2 + y^2)^{1/2}\right] = \frac{1}{2\sqrt{x^2 + y^2}} \cdot (2y) = \frac{y}{\sqrt{x^2 + y^2}}$ Evaluate at $(1, 1)$ : $\left.\frac{\partial f}{\partial y}\right|_{(1, 1)} = \frac{1}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \approx 0.707$ Because the cone has a constant generator slope of $1$ radially, moving purely along the $y$ -direction at a $45^\circ$ angle yields a slope of $1 \cdot \cos(45^\circ) = \frac{\sqrt{2}}{2}$ .

Practice Problem 13

Conceptual Contour Analysis: Suppose the level curves of a surface $z = f(x, y)$ are a family of parallel horizontal straight lines $y = c$ .

Determine the value of $\frac{\partial f}{\partial x}$ at any point $(x, y)$ .
What can you conclude about the behavior of $\frac{\partial f}{\partial y}$ ? Can $f(x, y)$ be written as a single-variable function?

Practice Space Solve this problem on paper or tablet.

If level curves are lines of constant height, what happens when you walk along one of those horizontal lines? (Defined in Visual Estimation of Partial Derivatives from Contour Maps.)

Answer: 1. $\frac{\partial f}{\partial x} = 0$ everywhere. 2. $\frac{\partial f}{\partial y} \ne 0$ in general; the function depends purely on $y$ : $f(x, y) = g(y)$ .

Step-by-Step Execution

Analyze the $x$ -Direction: If the level curves are horizontal lines parallel to the $x$ -axis, then along any line $y = y_0$ , the function value $f(x, y_0)$ is strictly constant. Moving in the positive or negative $x$ -direction never crosses into another contour line. Therefore, there is zero change in function value along $x$ :
$\frac{\partial f}{\partial x} = 0 \quad \text{everywhere}$
Analyze the $y$ -Direction: Because the contours vary as you move vertically across the horizontal strips, moving in the $y$ -direction crosses from one contour line to another. Therefore, $\frac{\partial f}{\partial y}$ will generally be non-zero (either positive or negative depending on whether height increases or decreases northward). Since $\frac{\partial f}{\partial x} = 0$ , integrating with respect to $x$ shows that $f(x, y)$ has no $x$ -dependence:
$f(x, y) = g(y)$ The 3D surface is a cylindrical trough or ridge oriented parallel to the $x$ -axis.

Practice Problem 14

A thermal sensor array embedded in a circuit board substrate measures surface temperatures $T(x, y)$ in $^\circ\text{C}$ at $1\text{ cm}$ spatial grid intervals:

$y \backslash x$	$x = 0\text{ cm}$	$x = 1\text{ cm}$	$x = 2\text{ cm}$
$y = 2\text{ cm}$	$85^\circ\text{C}$	$92^\circ\text{C}$	$101^\circ\text{C}$
$y = 1\text{ cm}$	$78^\circ\text{C}$	$84^\circ\text{C}$	$91^\circ\text{C}$
$y = 0\text{ cm}$	$70^\circ\text{C}$	$75^\circ\text{C}$	$80^\circ\text{C}$

Estimate $\frac{\partial T}{\partial x}(1, 1)$ using the central difference quotient formula: $\frac{\partial T}{\partial x}(1, 1) \approx \frac{T(1 + \Delta x, 1) - T(1 - \Delta x, 1)}{2\Delta x}$
Estimate $\frac{\partial T}{\partial y}(1, 1)$ using the central difference quotient formula: $\frac{\partial T}{\partial y}(1, 1) \approx \frac{T(1, 1 + \Delta y) - T(1, 1 - \Delta y)}{2\Delta y}$
State the units of each derivative and interpret which axis exhibits greater thermal gradient across the center sensor.

Practice Space Solve this problem on paper or tablet.

Note that $\Delta x = 1\text{ cm}$ and $\Delta y = 1\text{ cm}$ . Read the neighboring sensor values directly from the table row $y = 1$ and column $x = 1$ . (Defined in Visual Estimation of Partial Derivatives from Contour Maps.)

Answer: 1. $\frac{\partial T}{\partial x}(1, 1) \approx 6.5^\circ\text{C/cm}$ 2. $\frac{\partial T}{\partial y}(1, 1) \approx 8.5^\circ\text{C/cm}$ 3. Units: $^\circ\text{C/cm}$ ; the vertical $y$ -axis exhibits greater thermal gradient ( $8.5 > 6.5$ ).

Step-by-Step Execution

Estimate $\frac{\partial T}{\partial x}(1, 1)$ using Central Difference: State the Central Difference Formula along $x$ : With $\Delta x = 1\text{ cm}$ , the neighbor coordinates along the row $y = 1$ are $(0, 1)$ and $(2, 1)$ :
$T(0, 1) = 78^\circ\text{C}, \quad T(2, 1) = 91^\circ\text{C}$ $\frac{\partial T}{\partial x}(1, 1) \approx \frac{T(2, 1) - T(0, 1)}{2\Delta x} = \frac{91^\circ\text{C} - 78^\circ\text{C}}{2(1\text{ cm})} = \frac{13^\circ\text{C}}{2\text{ cm}} = 6.5^\circ\text{C/cm}$
Estimate $\frac{\partial T}{\partial y}(1, 1)$ using Central Difference: State the Central Difference Formula along $y$ : With $\Delta y = 1\text{ cm}$ , the neighbor coordinates along the column $x = 1$ are $(1, 0)$ and $(1, 2)$ :
$T(1, 0) = 75^\circ\text{C}, \quad T(1, 2) = 92^\circ\text{C}$ $\frac{\partial T}{\partial y}(1, 1) \approx \frac{T(1, 2) - T(1, 0)}{2\Delta y} = \frac{92^\circ\text{C} - 75^\circ\text{C}}{2(1\text{ cm})} = \frac{17^\circ\text{C}}{2\text{ cm}} = 8.5^\circ\text{C/cm}$
Units and Engineering Interpretation:
- Units: Degrees Celsius per centimeter ( $^\circ\text{C/cm}$ ).
- Interpretation: Temperature rises at an average instantaneous rate of $6.5^\circ\text{C/cm}$ moving eastward and $8.5^\circ\text{C/cm}$ moving northward from the central sensor $(1, 1)$ . Because $\frac{\partial T}{\partial y} = 8.5 > \frac{\partial T}{\partial x} = 6.5$ , heat accumulation is more pronounced along the vertical axis, signaling that active heat pipe cooling should prioritize heat extraction along the $y$ -direction.

Segment 6 of 8

📅 Regroup 3

Contour Density Intuition: Tightly spaced contour lines always equal a steep slope (large rate of change); widely spaced contour lines always equal a gentle slope (small rate of change).
Central vs Forward Difference: Central difference quotients are preferred in sensor grid processing because truncation errors cancel, providing a second-order accurate approximation $\mathcal{O}(\Delta x^2)$ .

Segment 7 of 8

📅 Synthesis Wrap-up

Core Session Takeaways:
1. A partial derivative isolates variation along a single coordinate axis, holding all other variables constant.
2. Geometrically, $f_x$ and $f_y$ represent the slopes of tangent lines to vertical trace curves in the slicing planes $y = y_0$ and $x = x_0$ .
3. Visually, partial derivatives can be estimated from contour maps by stepping in the coordinate direction and observing whether height increases ( $> 0$ ), decreases ( $< 0$ ), or remains tangent ( $= 0$ ).
Looking Ahead to Session 10: In Session 10, we will combine the two tangent vectors $\langle 1, 0, f_x \rangle$ and $\langle 0, 1, f_y \rangle$ to construct the Tangent Plane Equation $z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ and establish linear approximations for multivariable engineering systems.
Knewton Alta Homework Scope: Complete adaptive problem sets: Partial Derivatives of a Function and Partial Derivatives of a Function of Three Variables.

Segment 8 of 8