MATH-283 (Calculus III) Curriculum

Session 09 - Partial Derivatives

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

By the end of this session, you will be able to:

Compute first-order partial derivatives ( $f_x, f_y, f_z$ ) for multivariable functions using algebraic differentiation rules (product, quotient, power, chain rule) and the formal limit definition, treating non-differentiated variables as constant coefficients.
Evaluate and geometrically interpret partial derivatives at specified points $(x_0, y_0)$ or $(x_0, y_0, z_0)$ as instantaneous slopes of tangent lines to vertical trace curves ( $y = y_0$ and $x = x_0$ ) and directional rates of change.
Visually estimate the signs, relative magnitudes, and zero-slope tangencies of partial derivatives from 2D contour maps (level curve diagrams) and tabular sensor grids.

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.

Skill Block 1

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}

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)

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 $\frac{\partial T}{\partial y}$ ?

Solution Key:

(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.

Show your steps and calculations here.

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}

Solution Key:

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, $\frac{\partial}{\partial y}(x^2 + y^2) = 0 + 2y = 2y$ :
$\frac{\partial f}{\partial y} = \frac{1}{2}(x^2 + y^2)^{-1/2} \cdot (2y) = \frac{y}{\sqrt{x^2 + y^2}}$

Show your steps and calculations here.

Practice Problem 3

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

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

Solution Key:

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 $y$ : $\frac{\partial}{\partial y}(xy) = x(1) = x$ :
$\frac{\partial f}{\partial y} = x e^{xy}$

Show your steps and calculations here.

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}

Solution Key:

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: $\frac{\partial f}{\partial x} = 2x + 3y \implies f_x(2, 1) = 2(2) + 3(1) = 4 + 3 = 7$ .)

Show your steps and calculations here.

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)

Solution Key:

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

Show your steps and calculations here.

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.

Skill Block 2

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.

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 $(1, 1)\text{ mm}$ ?

Solution Key:

(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.

Show your steps and calculations here.

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.

Solution Key:

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.

Show your steps and calculations here.

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 $\frac{\partial f}{\partial y}(0, 1)$ .

Solution Key:

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 $(0, 1, 0)$ .

Show your steps and calculations here.

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.

Solution Key:

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.

Show your steps and calculations here.

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.

Skill Block 3

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 $(1, 0)$ .

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?

Solution Key:

(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.

Show your steps and calculations here.

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 $\frac{\partial f}{\partial y}(2, 0)$ .

Solution Key:

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:
$\frac{\partial f}{\partial x} = 2x \implies f_x(2, 0) = 2(2) = 4$ $\frac{\partial f}{\partial y} = 2y \implies f_y(2, 0) = 2(0) = 0$

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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.

Solution Key:

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 $1 \cdot \cos(45^\circ) = \frac{\sqrt{2}}{2}$ .

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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?

Solution Key:

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.

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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.

Solution Key:

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.

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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 $\mathcal{O}(\Delta x^2)$ .

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.