Session 12 - Gradient and Directional Derivatives

📅 Micro-Lecture

1. The Gradient Operator: Transforming Scalar Fields into Vector Fields

Mathematical Foundation: In multivariable modeling, physical quantities such as altitude, temperature, voltage, and pressure are scalar fields: functions that assign a single numerical value to each coordinate in space. The vector differential operator del ( $\nabla = \langle \partial/\partial x, \partial/\partial y, \partial/\partial z \rangle$ ) acts upon a scalar function $f$ to create the gradient vector $\nabla f$ . As established in The Gradient Vector, $\nabla f$ bundles all spatial sensitivities into a single geometric arrow. It possesses three foundational properties: it points in the direction of steepest ascent, its magnitude $\|\nabla f\|$ equals the maximum instantaneous rate of increase, and it is strictly perpendicular to the level curves or level surfaces passing through that point.
Engineering Context: Environmental and water resource engineers map topographic elevation $h(x, y)$ in stormwater basins; gravity causes surface water to flow along the direction of steepest descent ( $-\nabla h$ ), while earthen retaining dikes follow level contours orthogonal to $\nabla h$ .

2. Directional Derivatives: Arbitrary Trajectories and Normalization

Mathematical Foundation: Partial derivatives $f_x$ and $f_y$ quantify rates of change strictly along the coordinate axes ( $\hat{i}$ and $\hat{j}$ ). Real-world vehicles, wind patterns, and structural loads move along arbitrary trajectories. The directional derivative $D_{\vec{u}} f$ calculates the instantaneous rate of change per unit distance in any specified direction $\vec{u}$ . By the Gradient Theorem for Directional Derivatives (explored in Directional Derivatives), this rate is computed using the scalar dot product: $D_{\vec{u}} f = \nabla f \cdot \vec{u} = \|\nabla f\|\cos\phi$ . Crucially, this relationship holds only if $\vec{u}$ is a unit vector ( $\|\vec{u}\| = 1$ ). Non-unit vectors must always be normalized to avoid scaling the derivative incorrectly.
Engineering Context: Civil and geotechnical engineers calculate directional derivatives along proposed bulldozer cut paths to verify that off-axis road grades do not exceed the $30\%$ rollover threshold of heavy earthmoving machinery.

3. Extremal Bounds of Change and Tangent Planes to Implicit Surfaces

Mathematical Foundation: Because the dot product satisfies $D_{\vec{u}} f = \|\nabla f\|\cos\phi$ where $\phi$ is the angle between $\nabla f$ and $\vec{u}$ , the rate of change is bounded by $-\|\nabla f\| \le D_{\vec{u}} f \le \|\nabla f\|$ . The maximum increase occurs when $\vec{u}$ is parallel to $\nabla f$ ( $\phi = 0$ ), the maximum decrease occurs opposite to $\nabla f$ ( $\phi = \pi$ ), and zero change occurs when $\vec{u}$ is orthogonal to $\nabla f$ ( $\phi = \pi/2$ ). Furthermore, for an implicit level surface $F(x, y, z) = c$ , the gradient $\nabla F(x_0, y_0, z_0)$ acts directly as the surface normal vector $\vec{n}$ , enabling immediate construction of tangent plane equations $F_x(x - x_0) + F_y(y - y_0) + F_z(z - z_0) = 0$ as detailed in Tangent Planes to Level Surfaces.
Engineering Context: Thermal engineers design microprocessor cooling fins parallel to $-\nabla T$ to maximize heat conduction under Fourier's Law ( $\vec{q} = -k\nabla T$ ), while aerospace engineers compute surface normals $\vec{n} = \nabla F$ to position structural stiffeners on pressurized rocket fuel tanks.

Segment 1 of 8

Skill Block 1

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

The Gradient Vector in 2D & 3D (Computation & Geometric Properties)

General Overview
Skill Block 1 covers the formal definition, component extraction, and point evaluation of the gradient vector for functions of two and three variables. Students learn to express gradients in both component bracket form and Cartesian unit vector form, avoid the error of premature numerical substitution, and interpret the gradient's geometric role as a vector perpendicular to level sets.

Conceptual Framework

Definition of the Gradient in 2D:
For a differentiable scalar function $f(x, y)$ : $\nabla f(x, y) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j}$ (Defined in The Gradient Vector.)
Definition of the Gradient in 3D:
For a differentiable scalar function $f(x, y, z)$ : $\nabla f(x, y, z) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j} + \frac{\partial f}{\partial z}\hat{k}$
Steepest Ascent Property:
At any point $P$ , $\nabla f(P)$ points in the direction of maximum instantaneous rate of increase of $f$ .
Magnitude Property:
The Euclidean length $\|\nabla f(P)\|$ equals the maximum rate of change at $P$ .
Orthogonality Property:
$\nabla f(P)$ is perpendicular to the tangent vector of the level curve $f(x, y) = c$ or level surface $F(x, y, z) = c$ passing through $P$ .

Key Equations & Structural Rules

Differentiate-First Rule: Always compute the symbolic partial derivatives $f_x, f_y, f_z$ across the domain before substituting coordinates $(x_0, y_0, z_0)$ .
Bracket vs. Unit Vector Form: Gradients may be written as $\langle a, b, c \rangle$ or $a\hat{i} + b\hat{j} + c\hat{k}$ ; both representations represent identical geometric displacement vectors.
Gradient Magnitude: $\|\nabla f\| = \sqrt{f_x^2 + f_y^2 + f_z^2}$ .

Worked Example A: Polynomial Gradient Computation and Point Evaluation

Consider the multivariable function:

f(x, y) = 4x^3y^2 - 3x^2 + 7y

Find the general gradient vector $\nabla f(x, y)$ .
Evaluate $\nabla f$ at the point $P(1, -2)$ .
State the direction of steepest ascent as a unit vector and determine the maximum rate of change at $P$ .

💡 Reveal Worked Solution

Compute First-Order Partial Derivatives: State the Two-Variable Gradient Formula (defined in The Gradient Vector):
$\nabla f(x, y) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle$ Differentiate with respect to $x$ treating $y$ as constant: $f_x(x, y) = \frac{\partial}{\partial x}(4x^3y^2 - 3x^2 + 7y) = 12x^2y^2 - 6x$ Differentiate with respect to $y$ treating $x$ as constant: $f_y(x, y) = \frac{\partial}{\partial y}(4x^3y^2 - 3x^2 + 7y) = 8x^3y + 7$ General gradient vector: $\nabla f(x, y) = \langle 12x^2y^2 - 6x, \, 8x^3y + 7 \rangle = (12x^2y^2 - 6x)\hat{i} + (8x^3y + 7)\hat{j}$
Evaluate at Point $P(1, -2)$ : Substitute $x = 1$ and $y = -2$ :
$f_x(1, -2) = 12(1)^2(-2)^2 - 6(1) = 12(4) - 6 = 48 - 6 = 42$ $f_y(1, -2) = 8(1)^3(-2) + 7 = 8(-2) + 7 = -16 + 7 = -9$ Evaluated gradient vector: $\nabla f(1, -2) = \langle 42, -9 \rangle = 42\hat{i} - 9\hat{j}$
Determine Steepest Ascent Unit Vector and Maximum Rate: Compute the vector magnitude:
$\|\nabla f(1, -2)\| = \sqrt{42^2 + (-9)^2} = \sqrt{1764 + 81} = \sqrt{1845} = 3\sqrt{205} \approx 42.95$ The unit vector pointing in the direction of steepest ascent is: $\vec{u}_{\text{ascent}} = \frac{\nabla f(1, -2)}{\|\nabla f(1, -2)\|} = \frac{\langle 42, -9 \rangle}{3\sqrt{205}} = \left\langle \frac{14}{\sqrt{205}}, -\frac{3}{\sqrt{205}} \right\rangle$ The maximum rate of change at $P$ is $3\sqrt{205} \approx 42.95$ .

Practice Problem 1: Concept & Reading Check

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

(a) Explain the fundamental mathematical difference between the output of the function $f(x, y)$ and the output of its gradient operator $\nabla f(x, y)$ .
(b) In Worked Example A, explain why substituting the point coordinates $(1, -2)$ before taking the partial derivatives would produce an incorrect result.
(c) In Topographic Retention Basin Drainage (Environmental & Civil Engineering), explain why water flows in the direction of $-\nabla E$ rather than $\nabla E$ .

Practice Space Solve this problem on paper or tablet.

Review the definition of the gradient vector in The Gradient Vector, check the Differentiate-First Rule in the Key Equations section, and examine the physical gravity runoff discussion in the linked drainage application.

Answer: * (a) $f(x, y)$ outputs a single scalar value, whereas $\nabla f(x, y)$ outputs a vector possessing both magnitude and spatial direction.

(b) Substituting coordinates first yields a constant scalar, whose derivative is identically zero, losing all spatial sensitivity information.
(c) Gravity pulls water downhill toward lower elevations; because $\nabla E$ points uphill along steepest ascent, water naturally cascades down the steepest descent vector $-\nabla E$ .

Step-by-Step Execution

(a) The function $f(x, y)$ is a scalar field that assigns a single numerical value (such as elevation, voltage, or temperature) to each domain coordinate. The gradient $\nabla f(x, y) = \langle f_x, f_y \rangle$ is a vector field that assigns a directed vector to each point, encoding both the compass direction of maximum increase and the slope steepness.
(b) If coordinates $(1, -2)$ are substituted into $f(x, y)$ before differentiating, the expression evaluates to a constant scalar ( $f(1, -2) = 4(1)(4) - 3(1) + 7(-2) = -1$ ). The derivative of any constant is zero, which would falsely suggest $\nabla f = \langle 0, 0 \rangle$ . Differentiating symbolically first preserves the rates of change $\partial f/\partial x$ and $\partial f/\partial y$ .
(c) By the fundamental properties of the gradient, $\nabla E$ points in the direction of steepest elevation increase (straight uphill). Fluid runoff under the influence of gravity seeks the fastest drop in gravitational potential energy, which corresponds to the negative gradient $-\nabla E$ (the path of steepest descent).

Practice Problem 2

Consider the multivariable function $f(x, y) = 2x^3y^2 - 4x + 5y$ .

(a) Find the general gradient vector $\nabla f(x, y)$ .
(b) Evaluate $\nabla f$ at the point $P(2, -1)$ .
(c) State the geometric meaning of the resulting vector's direction and magnitude.

Practice Space Solve this problem on paper or tablet.

Apply the Two-Variable Gradient Formula: $\nabla f = \langle f_x, f_y \rangle$ . Compute $f_x$ and $f_y$ using standard power rules, then substitute $x = 2$ and $y = -1$ .

Answer: * (a) $\nabla f(x, y) = \langle 6x^2y^2 - 4, \, 4x^3y + 5 \rangle$

(b) $\nabla f(2, -1) = \langle 20, -27 \rangle = 20\hat{i} - 27\hat{j}$
(c) Direction $\langle 20, -27 \rangle$ is the direction of steepest ascent; magnitude $\|\nabla f\| = \sqrt{1129} \approx 33.60$ is the maximum rate of increase.

Step-by-Step Execution

Compute Symbolic Partial Derivatives: State the Gradient Formula (defined in The Gradient Vector):
$\nabla f(x, y) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle$ Differentiate with respect to $x$ : $f_x(x, y) = \frac{\partial}{\partial x}(2x^3y^2 - 4x + 5y) = 6x^2y^2 - 4$ Differentiate with respect to $y$ : $f_y(x, y) = \frac{\partial}{\partial y}(2x^3y^2 - 4x + 5y) = 4x^3y + 5$ General gradient: $\nabla f(x, y) = \langle 6x^2y^2 - 4, \, 4x^3y + 5 \rangle$
Evaluate at $P(2, -1)$ : Substitute $x = 2$ and $y = -1$ :
$f_x(2, -1) = 6(2)^2(-1)^2 - 4 = 6(4)(1) - 4 = 24 - 4 = 20$ $f_y(2, -1) = 4(2)^3(-1) + 5 = 4(8)(-1) + 5 = -32 + 5 = -27$ Evaluated gradient: $\nabla f(2, -1) = \langle 20, -27 \rangle = 20\hat{i} - 27\hat{j}$
State Geometric Meaning:
- Direction: The vector $\langle 20, -27 \rangle$ points in the direction of steepest ascent, where the function value increases most rapidly moving away from $(2, -1)$ .
- Magnitude: The Euclidean norm $\|\nabla f(2, -1)\| = \sqrt{20^2 + (-27)^2} = \sqrt{400 + 729} = \sqrt{1129} \approx 33.60$ represents the maximum instantaneous rate of increase of $f$ per unit distance at $(2, -1)$ .

Practice Problem 3

Let a three-dimensional scalar field be defined by $f(x, y, z) = y\ln(x) + xz^2e^y$ .

(a) Compute the general gradient vector $\nabla f(x, y, z)$ .
(b) Evaluate $\nabla f$ at the point $P(1, 0, 3)$ .

Practice Space Solve this problem on paper or tablet.

State the Three-Variable Gradient Formula: $\nabla f = \langle f_x, f_y, f_z \rangle$ . Remember that $\ln(1) = 0$ and $e^0 = 1$ .

Answer: * (a) $\nabla f(x, y, z) = \left\langle \frac{y}{x} + z^2e^y, \, \ln(x) + xz^2e^y, \, 2xze^y \right\rangle$

(b) $\nabla f(1, 0, 3) = \langle 9, 9, 6 \rangle = 9\hat{i} + 9\hat{j} + 6\hat{k}$

Step-by-Step Execution

Compute 3D Partial Derivatives: State the Three-Variable Gradient Formula (defined in The Gradient Vector):
$\nabla f(x, y, z) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle$ Differentiate with respect to $x$ : $f_x = \frac{\partial}{\partial x}[y\ln(x) + xz^2e^y] = \frac{y}{x} + z^2e^y$ Differentiate with respect to $y$ : $f_y = \frac{\partial}{\partial y}[y\ln(x) + xz^2e^y] = \ln(x) + xz^2e^y$ Differentiate with respect to $z$ : $f_z = \frac{\partial}{\partial z}[y\ln(x) + xz^2e^y] = 2xze^y$ General gradient: $\nabla f(x, y, z) = \left\langle \frac{y}{x} + z^2e^y, \, \ln(x) + xz^2e^y, \, 2xze^y \right\rangle$
Evaluate at Coordinate Point $P(1, 0, 3)$ : Substitute $x = 1$ , $y = 0$ , and $z = 3$ :
$f_x(1, 0, 3) = \frac{0}{1} + 3^2 e^0 = 0 + 9(1) = 9$ $f_y(1, 0, 3) = \ln(1) + 1(3^2)e^0 = 0 + 9(1) = 9$ $f_z(1, 0, 3) = 2(1)(3)e^0 = 6(1) = 6$ Evaluated gradient: $\nabla f(1, 0, 3) = \langle 9, 9, 6 \rangle = 9\hat{i} + 9\hat{j} + 6\hat{k}$

Practice Problem 4

The topographic elevation of an alluvial retention basin is modeled by the function $h(x, y) = 500 - 3x^2 - y^2$ , where $h$ is ground elevation in meters and $x, y$ are spatial coordinates in meters.

(a) Compute the general gradient vector $\nabla h(x, y)$ .
(b) If a civil surveyor stands at coordinate point $P(2, 4)$ , determine the compass direction of steepest ascent as a unit vector and calculate the exact slope in that direction.
(c) Write down a vector representing the natural direction that rainwater runoff will flow as it passes through point $P(2, 4)$ .

Practice Space Solve this problem on paper or tablet.

Use $\nabla h = \langle h_x, h_y \rangle$ . The steepest ascent unit vector is $\vec{u} = \nabla h/\|\nabla h\|$ , its slope is $\|\nabla h\|$ , and rainwater drainage follows the steepest descent vector $-\nabla h$ .

Answer: * (a) $\nabla h(x, y) = \langle -6x, -2y \rangle$

(b) Direction: $\vec{u}_{\text{ascent}} = \left\langle -\frac{3}{\sqrt{13}}, -\frac{2}{\sqrt{13}} \right\rangle$ ; exact slope: $\|\nabla h(2, 4)\| = 4\sqrt{13} \approx 14.42\text{ m/m}$
(c) Drainage flow vector: $-\nabla h(2, 4) = \langle 12, 8 \rangle$ (or unit vector $\left\langle \frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}} \right\rangle$ )

Step-by-Step Execution

Compute General Elevation Gradient: State the Elevation Gradient Formula (defined in The Gradient Vector):
$\nabla h(x, y) = \left\langle \frac{\partial h}{\partial x}, \frac{\partial h}{\partial y} \right\rangle = \langle -6x, -2y \rangle$
Evaluate Steepest Ascent Direction and Slope at $P(2, 4)$ : Substitute $x = 2$ and $y = 4$ :
$\nabla h(2, 4) = \langle -6(2), -2(4) \rangle = \langle -12, -8 \rangle$ Compute vector magnitude: $\|\nabla h(2, 4)\| = \sqrt{(-12)^2 + (-8)^2} = \sqrt{144 + 64} = \sqrt{208} = 4\sqrt{13} \approx 14.42\text{ m/m}$ The direction of steepest ascent as a normalized unit vector is: $\vec{u}_{\text{ascent}} = \frac{\langle -12, -8 \rangle}{4\sqrt{13}} = \left\langle -\frac{12}{4\sqrt{13}}, -\frac{8}{4\sqrt{13}} \right\rangle = \left\langle -\frac{3}{\sqrt{13}}, -\frac{2}{\sqrt{13}} \right\rangle \approx \langle -0.832, -0.555 \rangle$ The exact slope along this path is $4\sqrt{13} \approx 14.42$ meters vertical rise per horizontal meter.
Determine Rainwater Runoff Vector: Water flows down the fall line along the direction of steepest descent, governed by the negative gradient:
$\vec{v}_{\text{drainage}} = -\nabla h(2, 4) = -\langle -12, -8 \rangle = \langle 12, 8 \rangle$ Expressed as a unit vector: $\vec{u}_{\text{drainage}} = \left\langle \frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}} \right\rangle \approx \langle 0.832, 0.555 \rangle$

Segment 2 of 8

📅 Regroup 1

Clarifying Notation: Emphasize that $\nabla f(x, y)$ is a vector while $f(x, y)$ is a scalar. Discuss why KaTeX bracket form $\langle a, b \rangle$ and unit vector form $a\hat{i} + b\hat{j}$ describe the same physical vector.
Order of Operations Pitfall: Reinforce that evaluating coordinates before differentiation eliminates variable dependencies and destroys the rate information.

Segment 3 of 8

Skill Block 2

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

Directional Derivatives & Vector Normalization

General Overview
Skill Block 2 establishes the formulation and evaluation of directional derivatives $D_{\vec{u}} f = \nabla f \cdot \vec{u}$ in arbitrary directions. Students master the mandatory unit vector normalization protocol, compute directional rates along displacement vectors between points, and evaluate directional derivatives defined by heading angles $\theta$ .

Conceptual Framework

The Limit Definition of Directional Derivatives: $D_{\vec{u}} f(P) = \lim_{h \to 0} \frac{f(P + h\vec{u}) - f(P)}{h}$ where $\vec{u}$ is a unit vector ( $\|\vec{u}\| = 1$ ).
The Gradient Dot Product Theorem: For any differentiable function $f$ : $D_{\vec{u}} f = \nabla f \cdot \vec{u}$ (Defined in Directional Derivatives.)
Mandatory Normalization Rule: If a direction is given as a generic vector $\vec{v}$ with $\|\vec{v}\| \neq 1$ , it must be normalized: $\vec{u} = \frac{\vec{v}}{\|\vec{v}\|}$
Displacement Vector Between Two Points: Moving from point $P$ toward point $Q$ yields direction vector $\vec{v} = \overrightarrow{PQ} = Q - P$ , normalized to $\vec{u} = \overrightarrow{PQ}/\|\overrightarrow{PQ}\|$ .
Angle Parameterization in 2D: When a direction is given by an angle $\theta$ measured counterclockwise from the positive $x$ -axis: $\vec{u} = \langle \cos\theta, \sin\theta \rangle \implies D_{\vec{u}} f = f_x\cos\theta + f_y\sin\theta$

Key Equations & Structural Rules

Scalar Output Rule: The directional derivative $D_{\vec{u}} f$ is always a single real number (scalar rate of change), never a vector.
Normalization Verification: Always check $\|\vec{u}\| = \sqrt{u_1^2 + u_2^2} = 1$ before taking the dot product.

Worked Example A: Standard Directional Derivative with Vector Normalization

Find the directional derivative of $f(x, y) = 3x^2 - 2y^2 + 5xy$ at the point $P(2, -1)$ in the direction of the vector $\vec{v} = \langle 3, 4 \rangle$ .

💡 Reveal Worked Solution

Compute and Evaluate the Gradient Vector: State the Gradient Formula (defined in The Gradient Vector):
$\nabla f(x, y) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle$ Compute the partial derivatives: $f_x = 6x + 5y \implies f_x(2, -1) = 6(2) + 5(-1) = 12 - 5 = 7$ $f_y = -4y + 5x \implies f_y(2, -1) = -4(-1) + 5(2) = 4 + 10 = 14$ $\nabla f(2, -1) = \langle 7, 14 \rangle = 7\hat{i} + 14\hat{j}$
Normalize the Direction Vector: State the Vector Normalization Formula:
$\vec{u} = \frac{\vec{v}}{\|\vec{v}\|} = \frac{\langle v_1, v_2 \rangle}{\sqrt{v_1^2 + v_2^2}}$ Calculate the magnitude of $\vec{v} = \langle 3, 4 \rangle$ : $\|\vec{v}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ Construct the unit vector: $\vec{u} = \frac{\langle 3, 4 \rangle}{5} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle$
Compute the Directional Derivative via Dot Product: State the Directional Derivative Formula (defined in Directional Derivatives):
$D_{\vec{u}} f(2, -1) = \nabla f(2, -1) \cdot \vec{u}$ Evaluate the dot product: $D_{\vec{u}} f(2, -1) = \langle 7, 14 \rangle \cdot \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle = 7\left(\frac{3}{5}\right) + 14\left(\frac{4}{5}\right) = \frac{21 + 56}{5} = \frac{77}{5} = 15.4$

Practice Problem 5: Concept & Reading Check

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

(a) Explain why computing $\nabla f \cdot \vec{v}$ using a non-unit vector $\vec{v} = \langle 3, 4 \rangle$ without normalizing would give an incorrect rate of change.
(b) In Worked Example A, verify that the vector $\vec{u} = \langle 3/5, 4/5 \rangle$ satisfies the definition of a unit vector.
(c) In Highway Grading & Haul Road Stability (Civil Engineering), how is the directional derivative $D_{\vec{u}} h$ used to assess vehicle rollover risk?

Practice Space Solve this problem on paper or tablet.

Review the Vector Normalization section of Directional Derivatives and examine the slope threshold evaluation in the linked civil engineering application.

Answer: * (a) Failing to normalize scales the dot product by $\|\vec{v}\| = 5$ , exaggerating the instantaneous rate of change by a factor of 5 and corrupting its physical units.

(b) $\|\vec{u}\| = \sqrt{(3/5)^2 + (4/5)^2} = \sqrt{9/25 + 16/25} = \sqrt{25/25} = 1$ .
(c) $D_{\vec{u}} h$ calculates the actual downhill slope along the vehicle's specific steering heading; if $|D_{\vec{u}} h| > 0.30$ ( $30\%$ grade), the path violates heavy equipment stability limits.

Step-by-Step Execution

(a) The directional derivative measures the rate of change per unit distance (e.g., meters of rise per meter of travel). If an unnormalized vector $\vec{v}$ of length 5 is used in the dot product, the calculation yields $\nabla f \cdot \vec{v} = 5(\nabla f \cdot \vec{u}) = 77$ , which inflates the rate by fivefold and distorts the physical slope dimensions.
(b) The definition of a unit vector requires $\|\vec{u}\| = 1$ . Squaring each component: $(3/5)^2 = 9/25$ and $(4/5)^2 = 16/25$ . Summing gives $9/25 + 16/25 = 25/25 = 1$ . Taking the square root confirms $\|\vec{u}\| = 1$ .
(c) Bulldozers and heavy haul trucks have maximum slope tolerances before losing track traction or tipping. While the steepest descent might be excessively steep, traveling along an angled heading $\vec{u}$ produces an effective grade of $D_{\vec{u}} h$ . If this directional derivative exceeds $-0.30$ ( $30\%$ downhill grade), the civil engineer must reroute the road to a flatter angle.

Practice Problem 6

Let $f(x, y) = x^2y - 2y^2$ . Calculate the directional derivative $D_{\vec{u}} f(3, 1)$ in the direction of the vector $\vec{v} = \langle 5, 12 \rangle$ .

Practice Space Solve this problem on paper or tablet.

State the Directional Derivative Formula: $D_{\vec{u}} f = \nabla f \cdot \vec{u}$ . Find $\nabla f(3, 1)$ , compute $\|\vec{v}\| = \sqrt{5^2 + 12^2}$ , normalize $\vec{v}$ to $\vec{u}$ , and evaluate the dot product.

Answer:

D_{\vec{u}} f(3, 1) = \frac{90}{13} \approx 6.923

Step-by-Step Execution

Compute and Evaluate the Gradient: State the Gradient Formula (defined in The Gradient Vector):
$\nabla f(x, y) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle$ Compute partial derivatives: $f_x = 2xy \implies f_x(3, 1) = 2(3)(1) = 6$ $f_y = x^2 - 4y \implies f_y(3, 1) = 3^2 - 4(1) = 9 - 4 = 5$ $\nabla f(3, 1) = \langle 6, 5 \rangle$
Normalize the Direction Vector: Compute the magnitude of $\vec{v} = \langle 5, 12 \rangle$ :
$\|\vec{v}\| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ Construct the unit vector: $\vec{u} = \frac{\langle 5, 12 \rangle}{13} = \left\langle \frac{5}{13}, \frac{12}{13} \right\rangle$
Compute the Directional Derivative: State the Directional Derivative Dot Product Formula (defined in Directional Derivatives):
$D_{\vec{u}} f(3, 1) = \nabla f(3, 1) \cdot \vec{u}$ Substitute values: $D_{\vec{u}} f(3, 1) = \langle 6, 5 \rangle \cdot \left\langle \frac{5}{13}, \frac{12}{13} \right\rangle = 6\left(\frac{5}{13}\right) + 5\left(\frac{12}{13}\right) = \frac{30 + 60}{13} = \frac{90}{13} \approx 6.923$

Practice Problem 7

Consider the three-variable function $f(x, y, z) = 2x^2 + 3yz - z^2$ . Compute the rate of change of $f$ at the point $P(1, 2, -1)$ in the direction toward the point $Q(4, 2, 3)$ .

Practice Space Solve this problem on paper or tablet.

Compute displacement vector $\overrightarrow{PQ} = Q - P$ , normalize it to unit vector $\vec{u} = \overrightarrow{PQ}/\|\overrightarrow{PQ}\|$ , evaluate $\nabla f(1, 2, -1)$ , and calculate $D_{\vec{u}} f = \nabla f \cdot \vec{u}$ .

Answer:

D_{\vec{u}} f(1, 2, -1) = \frac{44}{5} = 8.8

Step-by-Step Execution

Construct and Normalize Displacement Vector $\overrightarrow{PQ}$ :
$\overrightarrow{PQ} = \langle 4 - 1, \, 2 - 2, \, 3 - (-1) \rangle = \langle 3, 0, 4 \rangle$ Compute magnitude: $\|\overrightarrow{PQ}\| = \sqrt{3^2 + 0^2 + 4^2} = \sqrt{9 + 0 + 16} = \sqrt{25} = 5$ Construct unit vector: $\vec{u} = \frac{\langle 3, 0, 4 \rangle}{5} = \left\langle \frac{3}{5}, 0, \frac{4}{5} \right\rangle$
Compute and Evaluate 3D Gradient: State the Three-Variable Gradient Formula (defined in The Gradient Vector):
$\nabla f(x, y, z) = \langle 4x, \, 3z, \, 3y - 2z \rangle$ Evaluate at $P(1, 2, -1)$ : $f_x(1, 2, -1) = 4(1) = 4$ $f_y(1, 2, -1) = 3(-1) = -3$ $f_z(1, 2, -1) = 3(2) - 2(-1) = 6 + 2 = 8$ $\nabla f(1, 2, -1) = \langle 4, -3, 8 \rangle$
Compute Directional Derivative: State the Directional Derivative Formula (defined in Directional Derivatives):
$D_{\vec{u}} f(P) = \nabla f(P) \cdot \vec{u}$ Substitute values: $D_{\vec{u}} f(1, 2, -1) = \langle 4, -3, 8 \rangle \cdot \left\langle \frac{3}{5}, 0, \frac{4}{5} \right\rangle = 4\left(\frac{3}{5}\right) + (-3)(0) + 8\left(\frac{4}{5}\right) = \frac{12 + 0 + 32}{5} = \frac{44}{5} = 8.8$

Practice Problem 8

Calculate the directional derivative of $g(x, y) = 4\cos(2x - y)$ at the point $P\left(\frac{\pi}{4}, 0\right)$ in the direction specified by an angle $\theta = \frac{2\pi}{3}$ measured counterclockwise from the positive $x$ -axis.

Practice Space Solve this problem on paper or tablet.

State the Angle Parameterization Formula: $\vec{u} = \langle \cos\theta, \sin\theta \rangle$ . Compute partial derivatives of $g(x, y)$ using the single-variable trigonometric chain rule, evaluate $\nabla g$ at $P$ , and evaluate $D_{\vec{u}} g = \nabla g \cdot \vec{u}$ .

Answer:

D_{\vec{u}} g\left(\frac{\pi}{4}, 0\right) = 4 + 2\sqrt{3} \approx 7.464

Step-by-Step Execution

Compute and Evaluate the Gradient: State the Gradient Formula (defined in The Gradient Vector):
$\nabla g(x, y) = \left\langle \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right\rangle$ Using the Chain Rule on $\cos(2x - y)$ : $g_x = 4\left(-\sin(2x - y)\right)\cdot \frac{\partial}{\partial x}(2x - y) = -8\sin(2x - y)$ $g_y = 4\left(-\sin(2x - y)\right)\cdot \frac{\partial}{\partial y}(2x - y) = 4\sin(2x - y)$ Evaluate at $P\left(\frac{\pi}{4}, 0\right)$ where the argument is $2(\pi/4) - 0 = \pi/2$ : $\sin\left(\frac{\pi}{2}\right) = 1$ $g_x\left(\frac{\pi}{4}, 0\right) = -8(1) = -8$ $g_y\left(\frac{\pi}{4}, 0\right) = 4(1) = 4$ $\nabla g\left(\frac{\pi}{4}, 0\right) = \langle -8, 4 \rangle$
Construct Directional Unit Vector from Angle $\theta$ : State the 2D Directional Angle Unit Vector Formula (defined in Directional Derivatives):
$\vec{u} = \langle \cos\theta, \sin\theta \rangle$ For $\theta = \frac{2\pi}{3}$ : $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}, \qquad \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}$ $\vec{u} = \left\langle -\frac{1}{2}, \frac{\sqrt{3}}{2} \right\rangle$
Evaluate the Directional Derivative:
$D_{\vec{u}} g = \nabla g \cdot \vec{u} = \langle -8, 4 \rangle \cdot \left\langle -\frac{1}{2}, \frac{\sqrt{3}}{2} \right\rangle = (-8)\left(-\frac{1}{2}\right) + (4)\left(\frac{\sqrt{3}}{2}\right) = 4 + 2\sqrt{3} \approx 7.464$

Segment 4 of 8

📅 Regroup 2

Geometric Dot Product Meaning: Review $D_{\vec{u}} f = \|\nabla f\|\cos\phi$ . Highlight how the sign of $D_{\vec{u}} f$ directly reveals whether the function is increasing ( $\phi < \pi/2$ ), decreasing ( $\phi > \pi/2$ ), or stationary ( $\phi = \pi/2$ ).
Displacement Vector Construction: Ensure students distinguish between the destination point coordinates $Q$ and the direction vector $\overrightarrow{PQ} = Q - P$ .

Segment 5 of 8

Skill Block 3

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

Extremal Rates of Change & Tangent Planes to Level Surfaces

General Overview
Skill Block 3 covers the synthesis of gradient geometry to identify extremal rates of change and model tangent planes to implicit level surfaces. Students learn how to bound directional derivatives using Cauchy-Schwarz, identify zero-change contour directions, extract normal vectors $\vec{n} = \nabla F$ , and assemble linear tangent plane equations for 3D surfaces.

Conceptual Framework

Maximum Rate of Increase (Steepest Ascent):
- Direction: $\vec{u} = \frac{\nabla f}{\|\nabla f\|}$
- Rate: $\|\nabla f\|$
Maximum Rate of Decrease (Steepest Descent):
- Direction: $\vec{u} = -\frac{\nabla f}{\|\nabla f\|}$
- Rate: $-\|\nabla f\|$
Zero Rate of Change (Level Curve Tangents):
- Direction: Any unit vector orthogonal to $\nabla f$ ( $\vec{u} \cdot \nabla f = 0$ ). For $\nabla f = \langle a, b \rangle$ , the two orthogonal directions are $\pm \langle -b, a \rangle / \sqrt{a^2 + b^2}$ .
- Rate: $0$
Tangent Planes to Implicit Level Surfaces: For a level surface $F(x, y, z) = c$ , the gradient $\nabla F(x_0, y_0, z_0)$ is orthogonal to the surface at $P_0(x_0, y_0, z_0)$ , serving as the plane normal vector: $\vec{n} = \langle F_x, F_y, F_z \rangle$ The tangent plane equation is: $F_x(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = 0$ (Defined in Tangent Planes to Level Surfaces.)

Key Equations & Structural Rules

Extremal Bounds Inequality: $-\|\nabla f\| \le D_{\vec{u}} f \le \|\nabla f\|$ .
Point-Normal Plane Equation: $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ where $\langle a, b, c \rangle = \nabla F(P_0)$ .

Worked Example A: Bounds of Change and Zero-Change Unit Vectors

Let $f(x, y) = x^2y - 3y^2$ . At the coordinate point $P(3, 1)$ :

Find the direction and value of the maximum rate of increase.
Find the two unit vector directions along which the function does not change.

💡 Reveal Worked Solution

Compute and Evaluate the Gradient: State the Gradient Formula (defined in The Gradient Vector):
$\nabla f(x, y) = \left\langle 2xy, \, x^2 - 6y \right\rangle$ Substitute $x = 3$ and $y = 1$ : $\nabla f(3, 1) = \langle 2(3)(1), \, 3^2 - 6(1) \rangle = \langle 6, \, 9 - 6 \rangle = \langle 6, 3 \rangle$ Compute the maximum rate of increase: $\text{Max Rate} = \|\nabla f(3, 1)\| = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \approx 6.708$ The direction of maximum increase as a unit vector is: $\vec{u}_{\text{max}} = \frac{\langle 6, 3 \rangle}{3\sqrt{5}} = \left\langle \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \right\rangle$
Determine Zero-Change Directions ( $\vec{u} \perp \nabla f$ ): Zero change requires $D_{\vec{u}} f = \nabla f \cdot \vec{u} = 0$ . For a 2D gradient $\langle a, b \rangle = \langle 6, 3 \rangle$ , orthogonal vectors swap components and negate one:
$\vec{w}_1 = \langle -3, 6 \rangle, \qquad \vec{w}_2 = \langle 3, -6 \rangle$ Normalize to unit vectors: $\|\vec{w}_1\| = \sqrt{(-3)^2 + 6^2} = \sqrt{45} = 3\sqrt{5}$ $\vec{u}_1 = \left\langle -\frac{3}{3\sqrt{5}}, \frac{6}{3\sqrt{5}} \right\rangle = \left\langle -\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right\rangle$ $\vec{u}_2 = -\vec{u}_1 = \left\langle \frac{1}{\sqrt{5}}, -\frac{2}{\sqrt{5}} \right\rangle$

Worked Example B: Tangent Plane to an Implicit Level Surface

Find the equation of the tangent plane to the quadric level surface:

x^2 + y^2 - z^2 = 4

at the point

P_0(3, 2, 3)

.

💡 Reveal Worked Solution

Define the Implicit Function and Compute the Gradient: State the Level Surface Function $F(x, y, z) = x^2 + y^2 - z^2 = 4$ . Compute the gradient vector $\nabla F$ (defined in Tangent Planes to Level Surfaces):
$\nabla F(x, y, z) = \left\langle \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right\rangle = \langle 2x, 2y, -2z \rangle$
Evaluate the Normal Vector $\vec{n}$ at $P_0(3, 2, 3)$ :
$\vec{n} = \nabla F(3, 2, 3) = \langle 2(3), 2(2), -2(3) \rangle = \langle 6, 4, -6 \rangle$
Assemble the Tangent Plane Equation: State the Tangent Plane to a Level Surface Formula:
$F_x(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = 0$ Substitute components $a = 6, b = 4, c = -6$ and coordinates $(3, 2, 3)$ : $6(x - 3) + 4(y - 2) - 6(z - 3) = 0$ Expand: $6x - 18 + 4y - 8 - 6z + 18 = 0 \implies 6x + 4y - 6z = 8$ Divide by 2 to obtain simplified standard linear form: $3x + 2y - 3z = 4$

Practice Problem 9: Concept & Reading Check

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

(a) Use the dot product angle formula $D_{\vec{u}} f = \|\nabla f\|\cos\phi$ to prove why the maximum possible directional derivative equals $\|\nabla f\|$ and occurs when $\vec{u}$ is parallel to $\nabla f$ .
(b) In Worked Example B, identify how the normal vector $\vec{n} = \langle 6, 4, -6 \rangle$ was determined directly from the surface equation without solving for $z$ .
(c) In Microprocessor Heat Sink Thermal Gradients (Mechanical & Computer Engineering), why are heat sink micro-channels aligned parallel to $-\nabla T$ rather than parallel to $\nabla T$ ?

Practice Space Solve this problem on paper or tablet.

Review the dot product geometric derivation in The Gradient Vector, check the surface normal extraction steps in Tangent Planes to Level Surfaces, and inspect Fourier's Law of Heat Conduction in the linked mechanical engineering application.

Answer: * (a) Because $-1 \le \cos\phi \le 1$ , the maximum value of $\cos\phi$ is $1$ , occurring at $\phi = 0$ (parallel vectors), which yields $D_{\vec{u}} f = \|\nabla f\|(1) = \|\nabla f\|$ .

(b) The level surface gradient $\nabla F = \langle 2x, 2y, -2z \rangle$ is proven to be perpendicular to the surface at every point, giving normal vector components directly upon substituting $(3, 2, 3)$ .
(c) By Fourier's Law ( $\vec{q} = -k\nabla T$ ), heat naturally conducts down temperature gradients from hotter zones to cooler zones; aligning channels with $-\nabla T$ aligns coolant flow with the fastest heat extraction path.

Step-by-Step Execution

(a) By the definition of the scalar dot product, $D_{\vec{u}} f = \nabla f \cdot \vec{u} = \|\nabla f\| \|\vec{u}\| \cos\phi$ . Because $\vec{u}$ is a unit vector, $\|\vec{u}\| = 1$ , reducing the expression to $\|\nabla f\|\cos\phi$ . The cosine function attains its absolute upper bound of $+1$ if and only if $\phi = 0$ , meaning $\vec{u}$ points in the exact direction of $\nabla f$ . Thus, the maximum rate of change is $\|\nabla f\|$ .
(b) In an implicit surface $F(x, y, z) = c$ , any curve on the surface satisfies $F(\vec{r}(t)) = c$ . Applying the chain rule yields $\nabla F \cdot \vec{r}'(t) = 0$ , proving that $\nabla F$ is perpendicular to every tangent line on the surface. Evaluating $\nabla F(3, 2, 3) = \langle 6, 4, -6 \rangle$ immediately provides the normal vector $\vec{n}$ needed for the plane equation without requiring explicit square root algebra.
(c) Fourier's Law of Heat Conduction states that heat flux $\vec{q}$ is driven by negative temperature gradients: $\vec{q} = -k\nabla T$ (where $k > 0$ ). Thermal energy naturally migrates toward lower temperatures. Heat sink micro-channels are aligned parallel to $-\nabla T$ (the direction of steepest descent) so that cooling fluids absorb heat directly along the natural conduction stream, minimizing hotspot formation.

Practice Problem 10

The steady-state temperature profile across an aluminum cooling plate is modeled by $T(x, y) = 150 - 2x^2 - 4y^2$ , where $T$ is in $^{\circ}\text{C}$ and $x, y$ are in centimeters.

(a) At the location $P(3, 1)$ , determine the unit vector direction $\vec{u}_{\text{cool}}$ in which the plate cools most rapidly.
(b) Determine the maximum possible rate of temperature increase at $P(3, 1)$ .
(c) Identify the two unit vectors representing paths along which a temperature sensor experiences zero temperature change.

Practice Space Solve this problem on paper or tablet.

Evaluate the temperature gradient $\nabla T(3, 1)$ . The steepest cooling direction is $-\nabla T$ , the maximum rate of increase is $\|\nabla T\|$ , and paths of zero change are orthogonal to $\nabla T$ .

Answer: * (a) $\vec{u}_{\text{cool}} = \left\langle \frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}} \right\rangle \approx \langle 0.832, 0.555 \rangle$

(b) Maximum rate of increase: $\|\nabla T(3, 1)\| = 4\sqrt{13} \approx 14.42^{\circ}\text{C/cm}$
(c) Constant paths: $\vec{u}_{\text{iso}} = \pm \left\langle \frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}} \right\rangle \approx \pm \langle 0.555, -0.832 \rangle$

Step-by-Step Execution

Compute and Evaluate Temperature Gradient: State the Temperature Gradient Formula (defined in The Gradient Vector):
$\nabla T(x, y) = \langle -4x, -8y \rangle$ Evaluate at $P(3, 1)$ : $\nabla T(3, 1) = \langle -4(3), -8(1) \rangle = \langle -12, -8 \rangle^{\circ}\text{C/cm}$
Determine Direction of Most Rapid Cooling: The direction of steepest descent is given by the negative gradient:
$\vec{v}_{\text{cool}} = -\nabla T(3, 1) = -\langle -12, -8 \rangle = \langle 12, 8 \rangle$ Normalize to a unit vector: $\|\vec{v}_{\text{cool}}\| = \sqrt{12^2 + 8^2} = \sqrt{144 + 64} = \sqrt{208} = 4\sqrt{13}$ $\vec{u}_{\text{cool}} = \frac{\langle 12, 8 \rangle}{4\sqrt{13}} = \left\langle \frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}} \right\rangle$
Maximum Possible Rate of Increase: The maximum rate of increase equals the magnitude of the gradient:
$\|\nabla T(3, 1)\| = 4\sqrt{13} \approx 14.42^{\circ}\text{C/cm}$
Zero-Change Isothermal Directions: Zero rate of change requires $\vec{u} \cdot \nabla T = 0$ . For $\langle -12, -8 \rangle$ , orthogonal vectors are $\pm \langle 8, -12 \rangle$ . Normalizing yields:
$\vec{u}_{\text{iso}} = \pm \frac{\langle 8, -12 \rangle}{4\sqrt{13}} = \pm \left\langle \frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}} \right\rangle$

Practice Problem 11

An electrostatic potential field is defined by $V(x, y, z) = \sqrt{x^2 + y^2 + z^2}$ , which measures Euclidean distance from a point charge at the origin.

(a) Prove that the gradient vector satisfies $\nabla V = \frac{\langle x, y, z \rangle}{\sqrt{x^2 + y^2 + z^2}} = \hat{r}$ .
(b) Explain physically or geometrically why $\nabla V$ always points radially outward from the origin.
(c) Identify the geometric shape formed by the level surfaces $V(x, y, z) = c$ for $c > 0$ .

Practice Space Solve this problem on paper or tablet.

Use the multivariable Chain Rule to differentiate $\sqrt{x^2 + y^2 + z^2}$ with respect to $x, y, z$ . Notice that the numerator is the radial position vector $\vec{r} = \langle x, y, z \rangle$ and the denominator is its magnitude $\|\vec{r}\|$ .

Answer: * (a) Differentiating yields $V_x = \frac{x}{\sqrt{x^2+y^2+z^2}}$ , $V_y = \frac{y}{\sqrt{x^2+y^2+z^2}}$ , $V_z = \frac{z}{\sqrt{x^2+y^2+z^2}}$ , so $\nabla V = \frac{\vec{r}}{\|\vec{r}\|} = \hat{r}$ .

(b) The function measures radial distance from $(0, 0, 0)$ ; distance increases most rapidly when traveling directly away from the center along straight radial rays.
(c) Concentric spheres centered at the origin with radius $c$ .

Step-by-Step Execution

Prove the Gradient Expression: Write $V(x, y, z) = (x^2 + y^2 + z^2)^{1/2}$ . Differentiating with respect to $x$ via the Chain Rule:
$V_x = \frac{1}{2}(x^2 + y^2 + z^2)^{-1/2} \cdot \frac{\partial}{\partial x}(x^2 + y^2 + z^2) = \frac{1}{2\sqrt{x^2 + y^2 + z^2}}(2x) = \frac{x}{\sqrt{x^2 + y^2 + z^2}}$ By cyclic symmetry with respect to $y$ and $z$ : $V_y = \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \qquad V_z = \frac{z}{\sqrt{x^2 + y^2 + z^2}}$ Assembling into vector form: $\nabla V = \left\langle \frac{x}{\sqrt{x^2 + y^2 + z^2}}, \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \frac{z}{\sqrt{x^2 + y^2 + z^2}} \right\rangle = \frac{\langle x, y, z \rangle}{\sqrt{x^2 + y^2 + z^2}} = \frac{\vec{r}}{\|\vec{r}\|} = \hat{r}$
Geometric and Physical Explanation:
- Geometric: The numerator $\vec{r} = \langle x, y, z \rangle$ is the standard position vector emanating straight from the origin to $(x, y, z)$ , while the denominator is its Euclidean length $\|\vec{r}\|$ . A vector divided by its length is the radial unit vector $\hat{r}$ , which points directly away from the origin.
- Physical: The potential $V$ measures straight-line distance from the origin. Moving along any curved or tangential path dilutes the rate of distance expansion; moving radially outward maximizes distance growth at a rate of exactly $1\text{ unit/unit distance}$ ( $\|\hat{r}\| = 1$ ).
Identify Level Surfaces: Setting $V(x, y, z) = c$ (for $c > 0$ ):
$\sqrt{x^2 + y^2 + z^2} = c \implies x^2 + y^2 + z^2 = c^2$ This is the standard equation of a sphere centered at the origin with radius $c$ . The gradient $\nabla V = \hat{r}$ is everywhere orthogonal to these spherical shells.

Practice Problem 12

An implicit manufacturing cost surface is defined by the level surface $F(x, y, z) = x^2y + yz + z^3 = 11$ .

(a) Find the normal vector $\vec{n}$ to this surface at the point $P_0(2, 1, 2)$ using the gradient.
(b) Write the equation of the tangent plane to the surface at $P_0(2, 1, 2)$ in simplified standard form.

Practice Space Solve this problem on paper or tablet.

State the Tangent Plane to a Level Surface Formula: $F_x(x - x_0) + F_y(y - y_0) + F_z(z - z_0) = 0$ . Compute partial derivatives of $F$ , evaluate at $(2, 1, 2)$ to find $\vec{n} = \langle F_x, F_y, F_z \rangle$ , and expand.

Answer: * (a) $\vec{n} = \nabla F(2, 1, 2) = \langle 4, 6, 13 \rangle = 4\hat{i} + 6\hat{j} + 13\hat{k}$

(b) Tangent plane equation: $4x + 6y + 13z = 40$

Step-by-Step Execution

Compute Gradient of the Implicit Surface: State the Level Surface Normal Vector Formula (defined in Tangent Planes to Level Surfaces):
$\vec{n} = \nabla F(x, y, z) = \left\langle \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right\rangle$ Differentiate $F(x, y, z) = x^2y + yz + z^3$ : $F_x = 2xy$ $F_y = x^2 + z$ $F_z = y + 3z^2$
Evaluate at Point $P_0(2, 1, 2)$ :
$F_x(2, 1, 2) = 2(2)(1) = 4$ $F_y(2, 1, 2) = 2^2 + 2 = 4 + 2 = 6$ $F_z(2, 1, 2) = 1 + 3(2^2) = 1 + 3(4) = 1 + 12 = 13$ The normal vector is: $\vec{n} = \nabla F(2, 1, 2) = \langle 4, 6, 13 \rangle = 4\hat{i} + 6\hat{j} + 13\hat{k}$
Construct the Tangent Plane Equation: State the Point-Normal Plane Equation:
$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ Substitute normal vector components $\langle 4, 6, 13 \rangle$ and point $(2, 1, 2)$ : $4(x - 2) + 6(y - 1) + 13(z - 2) = 0$ Expand and simplify: $4x - 8 + 6y - 6 + 13z - 26 = 0$ $4x + 6y + 13z - 40 = 0 \implies 4x + 6y + 13z = 40$

Segment 6 of 8

📅 Regroup 3

Orthogonality to Level Sets: Connect the level surface normal $\vec{n} = \nabla F$ to the 2D contour property where $\nabla f$ is orthogonal to level curves.
Explicit vs. Implicit Planes: Revisit how $z = f(x, y)$ is simply $F(x, y, z) = f(x, y) - z = 0$ , giving $\vec{n} = \langle f_x, f_y, -1 \rangle$ .

Segment 7 of 8

📅 Synthesis Wrap-up

Core Takeaways:
1. The gradient vector $\nabla f$ encapsulates all first-order directional rates of change and points in the direction of steepest ascent.
2. The directional derivative $D_{\vec{u}} f = \nabla f \cdot \vec{u}$ requires a unit vector $\|\vec{u}\| = 1$ to calculate accurate rates of change.
3. Directional derivatives are bounded by $-\|\nabla f\| \le D_{\vec{u}} f \le \|\nabla f\|$ , with zero change occurring along tangent lines to level sets.
4. The gradient of an implicit function $F(x, y, z) = c$ supplies the surface normal vector $\vec{n} = \nabla F$ for constructing 3D tangent planes.
Knewton Alta Homework Alignment:
- Directional Derivatives and Gradients for Functions in Two Variables (Practice Problems 1, 2, 4, 5, 6, 8)
- Applications of Directional Derivatives and Gradients (Practice Problems 4, 8, 10, and linked engineering applications)
- Directional Derivatives and Gradients for Functions in Three Variables (Practice Problems 3, 7, 9, 11, 12)

Segment 8 of 8