Session 10 - Tangent Planes and Linear Approximations

📅 Micro-Lecture

1. Defining Flat Planes in Three-Dimensional Space

Mathematical Foundation: In single-variable calculus, defining a line in 2D space requires a known point and a directional slope. In three-dimensional space $\mathbb{R}^3$ , defining a flat surface—a plane—requires a known base point $P_0(x_0, y_0, z_0)$ and a direction that is completely perpendicular to the entire surface: the normal vector $\vec{n} = \langle a, b, c \rangle = a\hat{i} + b\hat{j} + c\hat{k}$ . Any displacement vector $\vec{r} - \vec{r}_0$ connecting $P_0$ to an arbitrary point $P(x, y, z)$ on the plane must be orthogonal to $\vec{n}$ . Applying the dot product orthogonality condition gives the point-normal equation $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ , which simplifies to the general linear equation $ax + by + cz = d$ , as explored in Equations of Planes in 3D.
Engineering Context: Robotic trajectory-planning systems and Building Information Modeling (BIM) software use planar equations to establish virtual safety barriers, collision envelopes, and structural floor slab elevations.

2. Tangent Planes to Curved Multivariable Surfaces

Mathematical Foundation: In single-variable calculus, the derivative $f'(x_0)$ gives the slope of the tangent line to the curve $y = f(x)$ . In multivariable calculus, a function $z = f(x, y)$ defines a curved 2D surface living in $\mathbb{R}^3$ . As established in Tangent Planes, slicing the surface with coordinate planes yields vertical trace curves whose tangent slopes are given by the partial derivatives $f_x(x_0, y_0)$ and $f_y(x_0, y_0)$ . Taking the cross product of these trace tangent vectors produces the surface normal vector $\vec{n} = \langle f_x(x_0, y_0), f_y(x_0, y_0), -1 \rangle$ . The unique plane containing all directional tangent lines is the tangent plane: $z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ .
Engineering Context: Aerospace aerodynamicists extract surface normal vectors from tangent planes on curved fuselage nose cones and wings to compute local angles of attack, shock wave incidence, and dynamic pressure distributions.

3. Linear Approximations and Total Differentials

Mathematical Foundation: Because smooth differentiable surfaces appear locally flat when viewed under high magnification, the tangent plane serves as the optimal linear model for the surface near the point of tangency. As detailed in Linear Approximations, the linearization function $L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)\Delta x + f_y(x_0, y_0)\Delta y$ replaces complex non-linear functions with rapid linear arithmetic. Furthermore, as shown in Differentials and Error Propagation, the total differential $dz = f_x dx + f_y dy$ quantifies how small uncertainties in independent input measurements propagate into the overall output calculation.
Engineering Context: Embedded microcontrollers in autonomous quadcopters use real-time linear approximations to adjust propulsion thrust during wind gusts without overburdening CPU resources, while manufacturing quality engineers use total differentials to set dimensional tolerances on machined components.

Segment 1 of 8

Skill Block 1

Real-Life Engineering Applications

🤖

Robotics Engineering

Robotic Safety Zone Boundary (Robotics & Computer Engineering)

A robotic manufacturing arm operates inside an automated assembly cell adjacent to an inclined conveyor belt. To prevent the robotic end-effector from colliding with the moving conveyor structure, a safety automation engineer must define the conveyor's flat face as a virtual planar boundary barrier in the robot's trajectory-planning controller.

Metrology calibration instruments establish that the conveyor's surface is perpendicular to an anchor calibration normal vector $\vec{n} = \langle 1, -2, 2 \rangle = \hat{i} - 2\hat{j} + 2\hat{k}$ and passes through a fixed structural mounting pivot located at Cartesian coordinate $P_0(3, 1, 4)$ meters relative to the robot's base origin.

State the point-normal vector formula for a plane in 3D space, identifying the components of the normal vector $\vec{n}$ and base point $P_0$ .
Formulate the geometric equation of the safety boundary plane in point-normal form.
Simplify the equation into standard general linear form ( $ax + by + cz = d$ ).
Explain how the robot's real-time motion controller evaluates the scalar expression $f(x, y, z) = x - 2y + 2z - 9$ to trigger emergency collision-avoidance braking for a tool tip located at $(x_t, y_t, z_t)$ .

💡 Show Worked Derivation

A robotic manufacturing arm operates inside an automated assembly cell adjacent to an inclined conveyor belt. To prevent the robotic end-effector from colliding with the moving conveyor structure, a safety automation engineer must define the conveyor's flat face as a virtual planar boundary barrier in the robot's trajectory-planning controller.

Metrology calibration instruments establish that the conveyor's surface is perpendicular to an anchor calibration normal vector $\vec{n} = \langle 1, -2, 2 \rangle = \hat{i} - 2\hat{j} + 2\hat{k}$ and passes through a fixed structural mounting pivot located at Cartesian coordinate $P_0(3, 1, 4)$ meters relative to the robot's base origin.

State the point-normal vector formula for a plane in 3D space, identifying the components of the normal vector $\vec{n}$ and base point $P_0$ .
Formulate the geometric equation of the safety boundary plane in point-normal form.
Simplify the equation into standard general linear form ( $ax + by + cz = d$ ).
Explain how the robot's real-time motion controller evaluates the scalar expression $f(x, y, z) = x - 2y + 2z - 9$ to trigger emergency collision-avoidance braking for a tool tip located at $(x_t, y_t, z_t)$ .

Step-by-Step Solution:

State the Point-Normal Formula:
- As defined in Equations of Planes in 3D, the equation of a plane with normal vector $\vec{n} = \langle a, b, c \rangle$ passing through base point $P_0(x_0, y_0, z_0)$ is: $\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0 \iff a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$
- Extract given parameters:
  - Normal vector: $a = 1, b = -2, c = 2$
  - Base point coordinates: $x_0 = 3, y_0 = 1, z_0 = 4$
Formulate Point-Normal Form:
- Substitute the parameters into the governing formula: $1(x - 3) - 2(y - 1) + 2(z - 4) = 0$
Simplify to General Form:
- Expand the products: $x - 3 - 2y + 2 + 2z - 8 = 0$ $x - 2y + 2z - 9 = 0$
- Isolate the constant term on the right-hand side: $x - 2y + 2z = 9$
Engineering Interpretation & Collision Avoidance Logic:
- The linear equation $x - 2y + 2z = 9$ represents every point on the physical face of the conveyor belt.
- In robot firmware, the controller computes the signed scalar distance function: $D(x_t, y_t, z_t) = \frac{x_t - 2y_t + 2z_t - 9}{\|\vec{n}\|} = \frac{x_t - 2y_t + 2z_t - 9}{\sqrt{1^2 + (-2)^2 + 2^2}} = \frac{x_t - 2y_t + 2z_t - 9}{3}$
- When $D > 0$ , the tool tip resides safely within the clear operational envelope. As $D \to 0$ , the tip approaches the plane. If $D \le D_{\text{threshold}}$ (e.g., $0.05\text{ m}$ safety margin), the kinematic controller engages dynamic braking to protect machinery and personnel.

🏗️

Civil Engineering

BIM Architectural Facade Alignment (Civil & Construction Engineering)

A structural engineer designs an exterior curtain-wall facade system for a 20-story commercial tower in downtown Las Vegas using Building Information Modeling (BIM) software. Prefabricated glass and aluminum composite facade panels must be installed on the building frame at an inclined angle to deflect high solar heat gain.

Structural engineering specifications dictate that each facade panel plane must be strictly orthogonal to a structural strut anchor vector pointing along $\vec{n} = \langle 3, 4, 12 \rangle = 3\hat{i} + 4\hat{j} + 12\hat{k}$ . Furthermore, the panel plane must intersect a primary structural column node located at spatial grid coordinate $P_0(10, 5, 2)$ feet.

State the scalar equation of a plane in point-normal form, referencing Equations of Planes in 3D.
Determine the standard Cartesian linear equation ( $ax + by + cz = d$ ) of the architectural facade plane.
Compute the unit normal vector $\hat{n} = \frac{\vec{n}}{\|\vec{n}\|}$ that BIM software uses to render directional sun-glare reflection vectors.
If a second facade section has the equation $6x + 8y + 24z = 75$ , determine whether the two facade panels are parallel, perpendicular, or intersecting at an oblique angle.

💡 Show Worked Derivation

A structural engineer designs an exterior curtain-wall facade system for a 20-story commercial tower in downtown Las Vegas using Building Information Modeling (BIM) software. Prefabricated glass and aluminum composite facade panels must be installed on the building frame at an inclined angle to deflect high solar heat gain.

Structural engineering specifications dictate that each facade panel plane must be strictly orthogonal to a structural strut anchor vector pointing along $\vec{n} = \langle 3, 4, 12 \rangle = 3\hat{i} + 4\hat{j} + 12\hat{k}$ . Furthermore, the panel plane must intersect a primary structural column node located at spatial grid coordinate $P_0(10, 5, 2)$ feet.

State the scalar equation of a plane in point-normal form, referencing Equations of Planes in 3D.
Determine the standard Cartesian linear equation ( $ax + by + cz = d$ ) of the architectural facade plane.
Compute the unit normal vector $\hat{n} = \frac{\vec{n}}{\|\vec{n}\|}$ that BIM software uses to render directional sun-glare reflection vectors.
If a second facade section has the equation $6x + 8y + 24z = 75$ , determine whether the two facade panels are parallel, perpendicular, or intersecting at an oblique angle.

Step-by-Step Solution:

State Point-Normal Form Formula:
- As defined in Equations of Planes in 3D, the point-normal equation is: $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$
- Extract parameters:
  - Normal vector components: $a = 3, b = 4, c = 12$
  - Column anchor point: $x_0 = 10, y_0 = 5, z_0 = 2$
Compute the General Linear Equation:
- Substitute parameters into the equation: $3(x - 10) + 4(y - 5) + 12(z - 2) = 0$
- Expand and collect constant terms: $3x - 30 + 4y - 20 + 12z - 24 = 0$ $3x + 4y + 12z - 74 = 0$ $3x + 4y + 12z = 74$
Compute the Unit Normal Vector:
- Calculate the Euclidean norm of $\vec{n}$ : $\|\vec{n}\| = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$
- Normalize the vector: $\hat{n} = \frac{\vec{n}}{\|\vec{n}\|} = \left\langle \frac{3}{13}, \frac{4}{13}, \frac{12}{13} \right\rangle = \frac{3}{13}\hat{i} + \frac{4}{13}\hat{j} + \frac{12}{13}\hat{k}$
Classify Geometric Relationship with Second Facade:
- Extract normal vector of the second facade: $\vec{n}_2 = \langle 6, 8, 24 \rangle$
- Compare with the original normal $\vec{n}_1 = \langle 3, 4, 12 \rangle$ : $\vec{n}_2 = \langle 2(3), 2(4), 2(12) \rangle = 2\vec{n}_1$
- Because $\vec{n}_2$ is a direct scalar multiple of $\vec{n}_1$ ( $k = 2$ ), the two normal vectors point in identical directions. Consequently, the two facade panels are strictly parallel.
- Furthermore, dividing $6x + 8y + 24z = 75$ by $2$ gives $3x + 4y + 12z = 37.5 \neq 74$ , confirming that the planes are distinct parallel planes separated by a non-zero physical offset.

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

Equations of Planes in 3D Space

General Overview
Skill Block 1 covers the geometric and algebraic formulation of planes in three-dimensional Cartesian space. Students learn how to extract normal vectors from equations, convert between point-normal and general forms, construct planes passing through three non-collinear points using vector cross products, and determine geometric relationships between intersecting, parallel, and orthogonal planes.

Conceptual Framework

Normal Vector Orientation: $\vec{n} = \langle a, b, c \rangle = a\hat{i} + b\hat{j} + c\hat{k}$ (Defined in Equations of Planes in 3D.)
Point-Normal (Scalar) Equation of a Plane: Given base point $P_0(x_0, y_0, z_0)$ and normal vector $\vec{n} = \langle a, b, c \rangle$ : $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$
General Linear Equation of a Plane: $ax + by + cz = d, \quad \text{where } d = ax_0 + by_0 + cz_0 = \vec{n} \cdot \vec{r}_0$
Three-Point Plane Construction: Given non-collinear points $A, B, C$ , form displacement vectors $\vec{u} = \vec{AB}$ and $\vec{v} = \vec{AC}$ . The normal vector is obtained via the cross product: $\vec{n} = \vec{AB} \times \vec{AC}$

Key Equations & Structural Rules

Orthogonality Condition: Two planes are perpendicular if $\vec{n}_1 \cdot \vec{n}_2 = 0$ .
Parallelism Condition: Two planes are parallel if $\vec{n}_1 = k\vec{n}_2$ for non-zero scalar $k$ .
Dihedral Angle Formula: $\cos\theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{\|\vec{n}_1\| \|\vec{n}_2\|}$ (see Equations of Planes in 3D).
Zero-Component Rule: If $a = 0$ , the plane $by + cz = d$ has normal $\langle 0, b, c \rangle \perp \hat{i}$ , making the plane parallel to the $x$ -axis.

Worked Example A: Plane Equation with a Zero Normal Component

Find the equation of the plane that passes through the point $P(2, 5, 6)$ and has normal vector $\vec{n} = \langle 0, 2, 6 \rangle = 2\hat{j} + 6\hat{k}$ . Express the final answer in simplest general form.

💡 Reveal Worked Solution

Identify Given Components and State Governing Formula: State the Point-Normal Form of a Plane (defined in Equations of Planes in 3D):
$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$
- Normal vector components: $a = 0, b = 2, c = 6$ .
- Known point coordinates: $x_0 = 2, y_0 = 5, z_0 = 6$ .
Substitute Components into Point-Normal Form:
$0(x - 2) + 2(y - 5) + 6(z - 6) = 0$
Expand and Collect Constant Terms:
$0 + 2y - 10 + 6z - 36 = 0$ $2y + 6z - 46 = 0$ $2y + 6z = 46$
Reduce to Simplest Integer Coefficients: Divide the entire equation by the common factor of $2$ :
$y + 3z = 23$
- Geometric Insight: Because the coefficient of $x$ is zero ( $a = 0$ ), the variable $x$ does not appear in the equation. The normal vector $\vec{n} = \langle 0, 2, 6 \rangle$ is perpendicular to the $x$ -axis ( $\vec{n} \cdot \hat{i} = 0$ ), meaning the physical plane is completely parallel to the $x$ -axis.

Practice Problem 1: Concept & Reading Check

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

(a) In the geometric derivation of the plane equation, why must the dot product $\vec{n} \cdot (\vec{r} - \vec{r}_0)$ equal zero for every point $P(x, y, z)$ on the plane?
(b) In Worked Example A, explain why the variable $x$ does not appear in the simplified equation $y + 3z = 23$ . What does this indicate about the plane's spatial orientation relative to the $x$ -axis?
(c) In Robotic Safety Zone Boundary (Robotics & Computer Engineering), how does the automated trajectory controller use the planar scalar equation $x - 2y + 2z - 9 = 0$ to prevent robotic arm collisions with the conveyor belt?

Practice Space Solve this problem on paper or tablet.

Review the Conceptual Framework and the solutions to Worked Example A and the linked robotics safety application. (Defined in Equations of Planes in 3D.)

Answer: * (a) Because the displacement vector $\vec{r} - \vec{r}_0$ lies entirely within the plane, making it strictly perpendicular to the normal vector $\vec{n}$ .

(b) The normal vector component $a = 0$ , so $\vec{n} \perp \hat{i}$ , meaning the plane is parallel to the $x$ -axis.
(c) The controller evaluates the signed distance function; if the value approaches zero or becomes negative, the robot halts to avoid a collision.

Step-by-Step Execution

(a) As established in Equations of Planes in 3D, let $P_0(x_0, y_0, z_0)$ be a fixed base point on the plane and $P(x, y, z)$ be any arbitrary point on the plane. The displacement vector connecting them is $\vec{P_0P} = \vec{r} - \vec{r}_0 = \langle x - x_0, y - y_0, z - z_0 \rangle$ . Because this displacement vector lies entirely within the flat plane surface, it must be orthogonal to the plane's perpendicular normal vector $\vec{n}$ . By the fundamental definition of the dot product, two non-zero vectors are perpendicular if and only if their dot product vanishes: $\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0$ .
(b) In Worked Example A, the given normal vector is $\vec{n} = \langle 0, 2, 6 \rangle$ . The component along the $x$ -axis is $a = 0$ . When expanding the point-normal form, $0(x - 2) = 0$ , eliminating $x$ from the equation entirely. Geometrically, this means the normal vector has zero tilt along the $x$ -direction ( $\vec{n} \cdot \hat{i} = 0$ ), which forces the plane itself to run infinitely parallel to the $x$ -axis.
(c) In Robotic Safety Zone Boundary (Robotics & Computer Engineering), the conveyor belt surface is defined by $f(x, y, z) = x - 2y + 2z - 9 = 0$ . The robot's trajectory-tracking firmware computes the signed orthogonal distance from its moving tool tip $T(x_t, y_t, z_t)$ to the boundary plane. When $f(x_t, y_t, z_t) > 0$ , the tool is safely in the operational envelope; if $f(x_t, y_t, z_t) \le \epsilon$ , the tool has reached the safety margin, triggering emergency braking before physical impact occurs.

Practice Problem 2

A plane in 3D space has normal vector $\vec{n} = \langle 12, -4, 7 \rangle = 12\hat{i} - 4\hat{j} + 7\hat{k}$ originating from the anchor point $P(0, 7, 4)$ .

Write the equation of the plane in point-normal form.
Simplify the equation into standard general linear form ( $ax + by + cz = d$ ).

Practice Space Solve this problem on paper or tablet.

State the Point-Normal Form of a Plane: $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ . Substitute $a=12, b=-4, c=7$ and $(x_0, y_0, z_0) = (0, 7, 4)$ . (Defined in Equations of Planes in 3D.)

Answer:

12x - 4y + 7z = 0

Step-by-Step Execution

Formulate Point-Normal Form: State the Point-Normal Equation of a Plane:
$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ Substitute normal vector components $a = 12, b = -4, c = 7$ and anchor coordinates $x_0 = 0, y_0 = 7, z_0 = 4$ : $12(x - 0) - 4(y - 7) + 7(z - 4) = 0$
Simplify to General Form: Expand the products:
$12x - 4y + 28 + 7z - 28 = 0$ Notice that the constant terms cancel identically ( $28 - 28 = 0$ ): $12x - 4y + 7z = 0$ Geometric Observation: Because $d = 0$ , the plane passes directly through the Cartesian origin $(0, 0, 0)$ .

Practice Problem 3

Find the equation of the plane that contains the three non-collinear points $A(1, 2, -1)$ , $B(3, -1, 2)$ , and $C(0, 1, 1)$ . Express your final equation in standard general form with positive leading integer coefficients.

Practice Space Solve this problem on paper or tablet.

Form displacement vectors $\vec{u} = \vec{AB}$ and $\vec{v} = \vec{AC}$ . Compute the surface normal vector using the cross product $\vec{n} = \vec{u} \times \vec{v}$ , then apply the point-normal form with point $A, B$ , or $C$ . (Defined in Equations of Planes in 3D.)

Answer:

3x + 7y + 5z = 12

Step-by-Step Execution

Construct Two In-Plane Displacement Vectors: State the Displacement Vector Formula:
$\vec{u} = \vec{AB} = \langle x_B - x_A, y_B - y_A, z_B - z_A \rangle = \langle 3 - 1, -1 - 2, 2 - (-1) \rangle = \langle 2, -3, 3 \rangle$ $\vec{v} = \vec{AC} = \langle x_C - x_A, y_C - y_A, z_C - z_A \rangle = \langle 0 - 1, 1 - 2, 1 - (-1) \rangle = \langle -1, -1, 2 \rangle$
Compute the Normal Vector via Cross Product: State the Cross Product Determinant Formula:
$\vec{n} = \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 3 \\ -1 & -1 & 2 \end{vmatrix}$ $\vec{n} = \hat{i}[(-3)(2) - (3)(-1)] - \hat{j}[(2)(2) - (3)(-1)] + \hat{k}[(2)(-1) - (-3)(-1)]$ $\vec{n} = \hat{i}[-6 + 3] - \hat{j}[4 + 3] + \hat{k}[-2 - 3] = -3\hat{i} - 7\hat{j} - 5\hat{k} = \langle -3, -7, -5 \rangle$ Multiplying by $-1$ gives the equivalent outward normal vector: $\vec{n} = \langle 3, 7, 5 \rangle$ .
Substitute into Point-Normal Form Using Point $C(0, 1, 1)$ :
$3(x - 0) + 7(y - 1) + 5(z - 1) = 0$ $3x + 7y - 7 + 5z - 5 = 0$ $3x + 7y + 5z = 12$
Verify Coordinates of Points $A$ and $B$ :
- Point $A(1, 2, -1)$ : $3(1) + 7(2) + 5(-1) = 3 + 14 - 5 = 12$ . (Verified)
- Point $B(3, -1, 2)$ : $3(3) + 7(-1) + 5(2) = 9 - 7 + 10 = 12$ . (Verified)

Practice Problem 4

Find the standard equation of the plane that passes through the point $P(-2, 3, 1)$ and is parallel to the plane $4x - 2y + 5z = 19$ .

Practice Space Solve this problem on paper or tablet.

State the Parallel Planes Condition: Two planes are parallel if their normal vectors are scalar multiples. Extract $\vec{n} = \langle 4, -2, 5 \rangle$ from the given plane and apply point-normal form at $P$ . (Defined in Equations of Planes in 3D.)

Answer:

4x - 2y + 5z = -9

Step-by-Step Execution

Extract the Normal Vector: State the Linear Coefficient Rule for Normal Vectors: For any plane written in general form $ax + by + cz = d$ , the normal vector is $\vec{n} = \langle a, b, c \rangle$ . From the given reference plane $4x - 2y + 5z = 19$ , the normal vector is:
$\vec{n} = \langle 4, -2, 5 \rangle = 4\hat{i} - 2\hat{j} + 5\hat{k}$
Apply the Parallelism Principle: Two planes are parallel if and only if they share the same normal vector direction. Therefore, the required plane has normal vector $\vec{n} = \langle 4, -2, 5 \rangle$ .
Construct the Equation with Point $P(-2, 3, 1)$ : State the Point-Normal Form:
$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ Substitute $a = 4, b = -2, c = 5$ and $x_0 = -2, y_0 = 3, z_0 = 1$ : $4(x - (-2)) - 2(y - 3) + 5(z - 1) = 0$ $4(x + 2) - 2(y - 3) + 5(z - 1) = 0$ $4x + 8 - 2y + 6 + 5z - 5 = 0$ $4x - 2y + 5z + 9 = 0$ $4x - 2y + 5z = -9$

Segment 2 of 8

📅 Regroup 1

Key Takeaway: A plane in 3D is completely determined by a single normal vector $\vec{n} = \langle a, b, c \rangle$ and a single anchor point $P_0(x_0, y_0, z_0)$ .
Common Misconception: Students often try to find the "slope" of a plane as a single scalar number. Emphasize that planes do not have a single slope; their 3D spatial orientation is dictated entirely by their 3-component normal vector $\vec{n}$ .
Cross Product Check: Remind students that when finding a plane through three points, reversing the cross product order ( $\vec{AC} \times \vec{AB}$ ) produces $-\vec{n}$ , which yields the exact same geometric plane after multiplying both sides of the equation by $-1$ .

Segment 3 of 8

Skill Block 2

Real-Life Engineering Applications

🚀

Aerospace Engineering

Supersonic Nose Cone Aerodynamic Pressure (Aerospace & Mechanical Engineering)

An aerospace aerodynamicist evaluates the supersonic airflow across the nose cone of a high-speed atmospheric flight test vehicle. In wind tunnel validation studies, the upper curved aerodynamic profile of the fuselage nose cone is modeled by the paraboloid of revolution:

z = 20 - 0.05x^2 - 0.1y^2

where coordinates

x, y, z

are measured in centimeters (

\text{cm}

).

During supersonic Mach 2 flight, a pitot-static pressure sensor probe is installed at the surface location $(x_0, y_0) = (10, 5)\text{ cm}$ .

Compute the exact height $z_0 = f(10, 5)$ of the fuselage at the sensor installation point.
Determine the general analytical expressions for the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ .
Evaluate the partial derivatives at $(10, 5)$ and state the equation of the tangent plane to the fuselage surface at this location in standard form.
Extract the outward-pointing unit normal vector $\hat{n}$ to the fuselage surface at $(10, 5)$ , and explain why aerodynamicists use $\hat{n}$ to compute the local static pressure force vector $\vec{F}_p = -p \hat{n} \, dA$ .

💡 Show Worked Derivation

An aerospace aerodynamicist evaluates the supersonic airflow across the nose cone of a high-speed atmospheric flight test vehicle. In wind tunnel validation studies, the upper curved aerodynamic profile of the fuselage nose cone is modeled by the paraboloid of revolution:

z = 20 - 0.05x^2 - 0.1y^2

where coordinates

x, y, z

are measured in centimeters (

\text{cm}

).

During supersonic Mach 2 flight, a pitot-static pressure sensor probe is installed at the surface location $(x_0, y_0) = (10, 5)\text{ cm}$ .

Compute the exact height $z_0 = f(10, 5)$ of the fuselage at the sensor installation point.
Determine the general analytical expressions for the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ .
Evaluate the partial derivatives at $(10, 5)$ and state the equation of the tangent plane to the fuselage surface at this location in standard form.
Extract the outward-pointing unit normal vector $\hat{n}$ to the fuselage surface at $(10, 5)$ , and explain why aerodynamicists use $\hat{n}$ to compute the local static pressure force vector $\vec{F}_p = -p \hat{n} \, dA$ .

Step-by-Step Solution:

Compute Fuselage Height at $(10, 5)$ :
$z_0 = f(10, 5) = 20 - 0.05(10)^2 - 0.1(5)^2$ $z_0 = 20 - 0.05(100) - 0.1(25) = 20 - 5.0 - 2.5 = 12.5\text{ cm}$
Compute General Partial Derivatives:
- Differentiating with respect to $x$ (holding $y$ constant): $\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(20 - 0.05x^2 - 0.1y^2) = -0.10x$
- Differentiating with respect to $y$ (holding $x$ constant): $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(20 - 0.05x^2 - 0.1y^2) = -0.20y$
Evaluate and Construct Tangent Plane Equation:
- At $(x_0, y_0) = (10, 5)$ : $f_x(10, 5) = -0.10(10) = -1.0$ $f_y(10, 5) = -0.20(5) = -1.0$
- State the Tangent Plane Formula (defined in Tangent Planes): $z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$
- Substitute values: $z = 12.5 - 1.0(x - 10) - 1.0(y - 5)$ $z = 12.5 - x + 10 - y + 5$ $z = -x - y + 27.5$
- In standard linear form: $x + y + z = 27.5$
Extract Normal Vector and Aerodynamic Application:
- From Tangent Planes, the upward outward normal vector is: $\vec{n}_{\text{up}} = \langle -f_x, -f_y, 1 \rangle = \langle -(-1.0), -(-1.0), 1 \rangle = \langle 1.0, 1.0, 1.0 \rangle = \hat{i} + \hat{j} + \hat{k}$
- Compute its magnitude: $\|\vec{n}\| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$
- The outward unit normal vector is: $\hat{n} = \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle$
- Aerodynamic Application: Air pressure acts purely compressive and strictly normal to any solid immersed boundary. Aerodynamicists integrate $-p \hat{n} \, dA$ over the surface to decompose the resultant aerodynamic load into total vehicle lift (vertical $+z$ component) and wave drag (streamwise $-x$ component).

⚙️

Mechanical Engineering

Hydraulic Excavator Bracket Stress Analysis (Mechanical Engineering)

A mechanical design engineer analyzes the localized contact interface on a cast-steel hydraulic pivot bracket for a heavy-duty mining excavator. The curved load-bearing shoulder of the bracket is modeled mathematically by the logarithmic surface:

z = \ln(x^2 + y)

where

x, y, z

represent spatial coordinates in inches (

\text{in}

).

During heavy bucket digging, a hydraulic actuator piston pushes a spherical pin directly against the bracket shoulder at the localized contact coordinate $(x_0, y_0) = (1, e - 1)\text{ in}$ .

Compute the baseline height $z_0 = f(1, e - 1)$ on the bracket shoulder.
Apply the multivariable chain rule to compute the partial derivatives $f_x(x, y)$ and $f_y(x, y)$ .
Evaluate the partial derivatives at $(1, e - 1)$ and formulate the equation of the tangent plane to the bracket contact zone in both explicit ( $z = \dots$ ) and general ( $ax + by + cz = d$ ) forms.
If the hydraulic piston applies a normal compressive force of magnitude $F_N = 50\text{ kips}$ , explain how finite element analysis (FEA) software uses the tangent plane's normal vector to resolve the force into axial and shear stresses along machine frame axes.

💡 Show Worked Derivation

A mechanical design engineer analyzes the localized contact interface on a cast-steel hydraulic pivot bracket for a heavy-duty mining excavator. The curved load-bearing shoulder of the bracket is modeled mathematically by the logarithmic surface:

z = \ln(x^2 + y)

where

x, y, z

represent spatial coordinates in inches (

\text{in}

).

During heavy bucket digging, a hydraulic actuator piston pushes a spherical pin directly against the bracket shoulder at the localized contact coordinate $(x_0, y_0) = (1, e - 1)\text{ in}$ .

Compute the baseline height $z_0 = f(1, e - 1)$ on the bracket shoulder.
Apply the multivariable chain rule to compute the partial derivatives $f_x(x, y)$ and $f_y(x, y)$ .
Evaluate the partial derivatives at $(1, e - 1)$ and formulate the equation of the tangent plane to the bracket contact zone in both explicit ( $z = \dots$ ) and general ( $ax + by + cz = d$ ) forms.
If the hydraulic piston applies a normal compressive force of magnitude $F_N = 50\text{ kips}$ , explain how finite element analysis (FEA) software uses the tangent plane's normal vector to resolve the force into axial and shear stresses along machine frame axes.

Step-by-Step Solution:

Compute Baseline Coordinate Height:
$z_0 = f(1, e - 1) = \ln\left(1^2 + e - 1\right) = \ln(1 + e - 1) = \ln(e) = 1\text{ in}$ The pin contact point is $P_0(1, e - 1, 1)$ .
Compute General Partial Derivatives:
- Let $u(x, y) = x^2 + y$ . State the Logarithmic Chain Rule: $\frac{\partial}{\partial x}[\ln(u)] = \frac{1}{u} \frac{\partial u}{\partial x}, \quad \frac{\partial}{\partial y}[\ln(u)] = \frac{1}{u} \frac{\partial u}{\partial y}$
- Partial derivatives: $f_x(x, y) = \frac{1}{x^2 + y} \cdot (2x) = \frac{2x}{x^2 + y}$ $f_y(x, y) = \frac{1}{x^2 + y} \cdot (1) = \frac{1}{x^2 + y}$
Evaluate at $(1, e - 1)$ and Formulate Tangent Plane:
- Note that $x^2 + y = 1^2 + (e - 1) = e$ .
- Partial derivative evaluations: $f_x(1, e - 1) = \frac{2(1)}{e} = \frac{2}{e}$ $f_y(1, e - 1) = \frac{1}{e}$
- State the Tangent Plane Equation (defined in Tangent Planes): $z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$
- Substitute parameters: $z = 1 + \frac{2}{e}(x - 1) + \frac{1}{e}(y - (e - 1))$ $z = 1 + \frac{2}{e}x - \frac{2}{e} + \frac{1}{e}y - 1 + \frac{1}{e}$ $z = \frac{2}{e}x + \frac{1}{e}y - \frac{1}{e}$
- General Form (clearing the denominator $e$ ): $2x + y - ez = 1$
FEA Contact Mechanics Application:
- The surface normal vector at the contact point is $\vec{n} = \langle 2, 1, -e \rangle$ .
- The unit normal vector is: $\hat{n} = \frac{\langle 2, 1, -e \rangle}{\sqrt{2^2 + 1^2 + (-e)^2}} = \frac{\langle 2, 1, -e \rangle}{\sqrt{5 + e^2}}$
- In FEA boundary conditions, the $50\text{ kip}$ force is applied along $\vec{F} = -50 \hat{n}\text{ kips}$ . FEA solvers project this force vector onto the local element coordinate axes to calculate von Mises yield stresses and verify that peak contact stresses do not exceed the steel's yield strength ( $\sigma_y = 70\text{ ksi}$ ).

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

Tangent Planes to Curved Multivariable Surfaces

General Overview
Skill Block 2 bridges flat planes in space to differential calculus on curved surfaces $z = f(x, y)$ . Students learn how first-order partial derivatives $f_x$ and $f_y$ yield the slopes of coordinate trace curves, how their cross product generates the surface normal vector $\vec{n} = \langle f_x, f_y, -1 \rangle$ , and how to construct tangent planes for polynomial, radical, trigonometric, and exponential surfaces.

Conceptual Framework

The Tangent Plane Equation: Let $z = f(x, y)$ be differentiable at $(x_0, y_0)$ , with $z_0 = f(x_0, y_0)$ : $z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ (Defined in Tangent Planes.)
Point-Normal Form of the Tangent Plane: $f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) - (z - z_0) = 0$
Surface Normal Vectors:
- Downward-pointing normal: $\vec{n}_{\text{down}} = \langle f_x(x_0, y_0), f_y(x_0, y_0), -1 \rangle$
- Upward-pointing normal: $\vec{n}_{\text{up}} = \langle -f_x(x_0, y_0), -f_y(x_0, y_0), 1 \rangle$
Trace Tangent Vectors: The tangent plane contains the tangent vectors to the $x$ -trace and $y$ -trace curves: $\vec{T}_x = \langle 1, 0, f_x(x_0, y_0) \rangle = \hat{i} + f_x\hat{k}$ $\vec{T}_y = \langle 0, 1, f_y(x_0, y_0) \rangle = \hat{j} + f_y\hat{k}$ Their cross product gives $\vec{T}_x \times \vec{T}_y = \langle -f_x, -f_y, 1 \rangle$ .

Key Equations & Structural Rules

Surface Tangent Formula: $z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$
Horizontal Tangent Plane Condition: If $f_x(x_0, y_0) = 0$ and $f_y(x_0, y_0) = 0$ , the tangent plane is horizontal: $z = z_0$ .

Worked Example A: Tangent Plane to an Oscillating Trigonometric Surface

Find the equation of the tangent plane to the surface $z = \sin(xy)$ at the coordinate $(x_0, y_0) = (0, \pi)$ , where $z_0 = 0$ .

💡 Reveal Worked Solution

Verify the Surface Height at the Given Coordinate:
$z_0 = f(0, \pi) = \sin(0 \cdot \pi) = \sin(0) = 0$
Compute General First-Order Partial Derivatives: Apply the Trigonometric Chain Rule (treating the alternate variable as a constant coefficient):
$f_x(x, y) = \frac{\partial}{\partial x}[\sin(xy)] = \cos(xy) \cdot \frac{\partial}{\partial x}(xy) = y\cos(xy)$ $f_y(x, y) = \frac{\partial}{\partial y}[\sin(xy)] = \cos(xy) \cdot \frac{\partial}{\partial y}(xy) = x\cos(xy)$
Evaluate the Partial Derivatives at $(0, \pi)$ :
$f_x(0, \pi) = \pi \cdot \cos(0 \cdot \pi) = \pi \cdot \cos(0) = \pi(1) = \pi$ $f_y(0, \pi) = 0 \cdot \cos(0 \cdot \pi) = 0 \cdot \cos(0) = 0(1) = 0$
Substitute Components into the Tangent Plane Formula: State the Tangent Plane Equation (defined in Tangent Planes):
$z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ Substitute $z_0 = 0, x_0 = 0, y_0 = \pi, f_x(0, \pi) = \pi, f_y(0, \pi) = 0$ : $z = 0 + \pi(x - 0) + 0(y - \pi)$ $z = \pi x$
Express in General Linear Form:
$\pi x - z = 0$
- Geometric Insight: Because $f_y(0, \pi) = 0$ , slicing the surface parallel to the $y$ -axis produces an instantaneous slope of zero. The tangent plane tilts upward in the positive $x$ -direction with slope $\pi$ , but remains perfectly flat along the $y$ -direction.

Practice Problem 5: Concept & Reading Check

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

(a) Explain how the surface normal vector $\vec{n} = \langle f_x(x_0, y_0), f_y(x_0, y_0), -1 \rangle$ is derived from the cross product of the two coordinate trace tangent vectors $\vec{T}_x$ and $\vec{T}_y$ . Where does the constant component $-1$ come from?
(b) In Worked Example A, explain why the $y$ -term vanishes from the tangent plane equation $z = \pi x$ . What does this indicate about the slope of the surface along the cross-section $x = 0$ ?
(c) In Supersonic Nose Cone Aerodynamic Pressure (Aerospace & Mechanical Engineering), why is finding the unit normal vector $\hat{n}$ to the tangent plane essential for calculating the aerodynamic pressure force $\vec{F}_p = -p \hat{n} \, dA$ ?

Practice Space Solve this problem on paper or tablet.

Review the Conceptual Framework and the solutions to Worked Example A and the linked aerospace application. (Defined in Tangent Planes.)

Answer: * (a) $\vec{T}_x = \langle 1, 0, f_x \rangle$ and $\vec{T}_y = \langle 0, 1, f_y \rangle$ ; their cross product yields $\langle -f_x, -f_y, 1 \rangle$ , or $\langle f_x, f_y, -1 \rangle$ when multiplied by $-1$ .

(b) Because $f_y(0, \pi) = 0$ , the surface has zero slope along the $y$ -trace at that point.
(c) Gas pressure acts purely perpendicular to any immersed solid boundary; the unit normal gives the exact directional vector of this compressive load.

Step-by-Step Execution

(a) As derived in Tangent Planes, slicing the surface $z = f(x, y)$ with the vertical plane $y = y_0$ produces an $x$ -trace curve whose tangent vector is $\vec{T}_x = \langle 1, 0, f_x(x_0, y_0) \rangle$ . Slicing with $x = x_0$ produces a $y$ -trace curve whose tangent vector is $\vec{T}_y = \langle 0, 1, f_y(x_0, y_0) \rangle$ . Because both vectors lie within the tangent plane, their cross product is perpendicular to the plane: $\vec{T}_x \times \vec{T}_y = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & f_x \\ 0 & 1 & f_y \end{vmatrix} = \langle -f_x, -f_y, 1 \rangle$ Multiplying by $-1$ produces the downward-pointing normal $\vec{n} = \langle f_x, f_y, -1 \rangle$ . The constant $-1$ arises directly from rearranging the explicit surface equation $z - f(x, y) = 0$ into point-normal form.
(b) In Worked Example A, evaluating $f_y(x, y) = x\cos(xy)$ at $(0, \pi)$ gives $f_y(0, \pi) = 0 \cdot \cos(0) = 0$ . In the tangent plane equation $z = z_0 + f_x(x - x_0) + f_y(y - y_0)$ , the term $0(y - \pi) = 0$ , leaving $z = \pi x$ . Geometrically, this means that along the vertical trace slice $x = 0$ , the surface has zero instantaneous inclination (a horizontal crest), so moving along $y$ produces no vertical elevation change on the tangent plane.
(c) In Supersonic Nose Cone Aerodynamic Pressure (Aerospace & Mechanical Engineering), physical fluid mechanics dictates that static gas pressure exerts compressive stress strictly perpendicular to any immersed solid surface. The unit normal vector $\hat{n} = \frac{\vec{n}}{\|\vec{n}\|}$ extracted from the tangent plane defines the exact spatial direction of incoming air molecule momentum transfer. Aerodynamicists integrate $-p \hat{n} \, dA$ over the nose cone to resolve resultant forces into vehicle lift and drag.

Practice Problem 6

Find the equation of the tangent plane to the elliptic paraboloid $z = x^2 + y^2$ at the point $(1, 2, 5)$ . Write the final result in general linear form ( $ax + by + cz = d$ ).

Practice Space Solve this problem on paper or tablet.

State the Tangent Plane Equation: $z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ . Compute $f_x = 2x$ and $f_y = 2y$ , evaluate at $(1, 2)$ , and rearrange. (Defined in Tangent Planes.)

Answer:

2x + 4y - z = 5

Step-by-Step Execution

Verify the Point on the Surface:
$z_0 = f(1, 2) = 1^2 + 2^2 = 1 + 4 = 5$ The point of tangency is $P_0(1, 2, 5)$ .
Compute General Partial Derivatives: State the Power Rule for Partial Derivatives:
$f_x(x, y) = \frac{\partial}{\partial x}(x^2 + y^2) = 2x$ $f_y(x, y) = \frac{\partial}{\partial y}(x^2 + y^2) = 2y$
Evaluate Derivatives at $(1, 2)$ :
$f_x(1, 2) = 2(1) = 2$ $f_y(1, 2) = 2(2) = 4$
Substitute into Tangent Plane Formula: State the Tangent Plane Equation:
$z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ $z = 5 + 2(x - 1) + 4(y - 2)$ $z = 5 + 2x - 2 + 4y - 8$ $z = 2x + 4y - 5$
Convert to Standard General Form:
$2x + 4y - z = 5$
- Normal Vector Note: The surface normal vector is $\vec{n} = \langle 2, 4, -1 \rangle = 2\hat{i} + 4\hat{j} - \hat{k}$ .

Practice Problem 7

Find the equation of the tangent plane to the radical surface $z = \sqrt{x + y}$ at the point $(3, 1, 2)$ .

Practice Space Solve this problem on paper or tablet.

Apply the Generalized Power Rule: $\frac{\partial}{\partial x}[u^{1/2}] = \frac{1}{2\sqrt{u}} \frac{\partial u}{\partial x}$ . Evaluate $f_x$ and $f_y$ at $(3, 1)$ and substitute into the tangent plane equation. (Defined in Tangent Planes.)

Answer:

x + y - 4z = -4

Step-by-Step Execution

Verify the Surface Height:
$z_0 = f(3, 1) = \sqrt{3 + 1} = \sqrt{4} = 2$
Compute General Partial Derivatives: Let $u(x, y) = x + y$ . Applying the Chain Rule:
$f_x(x, y) = \frac{1}{2}(x + y)^{-1/2} \cdot \frac{\partial}{\partial x}(x + y) = \frac{1}{2\sqrt{x + y}}$ $f_y(x, y) = \frac{1}{2}(x + y)^{-1/2} \cdot \frac{\partial}{\partial y}(x + y) = \frac{1}{2\sqrt{x + y}}$
Evaluate at the Point $(3, 1)$ :
$f_x(3, 1) = \frac{1}{2\sqrt{3 + 1}} = \frac{1}{2(2)} = \frac{1}{4}$ $f_y(3, 1) = \frac{1}{2\sqrt{3 + 1}} = \frac{1}{2(2)} = \frac{1}{4}$
Substitute into Tangent Plane Formula: State the Tangent Plane Equation:
$z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ $z = 2 + \frac{1}{4}(x - 3) + \frac{1}{4}(y - 1)$ $z = 2 + \frac{1}{4}x - \frac{3}{4} + \frac{1}{4}y - \frac{1}{4}$ $z = \frac{1}{4}x + \frac{1}{4}y + 1$
Clear Fractions to Obtain General Form: Multiply the entire equation by $4$ :
$4z = x + y + 4 \implies x + y - 4z = -4$

Practice Problem 8

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

Practice Space Solve this problem on paper or tablet.

Use the Product Rule for $f_x$ and the Chain Rule for $f_y$ . Evaluate the partial derivatives at $(x_0, y_0) = (2, 0)$ and state the tangent plane in standard form. (Defined in Tangent Planes.)

Answer:

x + 4y - z = 0

Step-by-Step Execution

Verify the Point on the Surface:
$z_0 = f(2, 0) = 2 e^{2 \cdot 0} = 2 e^0 = 2(1) = 2$ The point of tangency is $P_0(2, 0, 2)$ .
Compute General Partial Derivatives:
- For $f_x(x, y)$ : Apply the Multivariable Product Rule because $x$ appears both as a linear factor and in the exponential exponent: $f_x(x, y) = \frac{\partial}{\partial x}[x] \cdot e^{xy} + x \cdot \frac{\partial}{\partial x}[e^{xy}] = 1 \cdot e^{xy} + x(y e^{xy}) = e^{xy}(1 + xy)$
- For $f_y(x, y)$ : The leading factor $x$ is treated as a constant coefficient. Apply the Exponential Chain Rule: $f_y(x, y) = x \cdot \frac{\partial}{\partial y}[e^{xy}] = x(x e^{xy}) = x^2 e^{xy}$
Evaluate Partial Derivatives at $(2, 0)$ : Notice that the exponent is $xy = 2(0) = 0$ , so $e^0 = 1$ :
$f_x(2, 0) = e^0(1 + 2 \cdot 0) = 1(1 + 0) = 1$ $f_y(2, 0) = 2^2 e^0 = 4(1) = 4$
Substitute into Tangent Plane Formula: State the Tangent Plane Equation:
$z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ $z = 2 + 1(x - 2) + 4(y - 0)$ $z = 2 + x - 2 + 4y \implies z = x + 4y$
Convert to Standard General Form:
$x + 4y - z = 0$
- Normal Vector Note: The surface normal vector is $\vec{n} = \langle 1, 4, -1 \rangle = \hat{i} + 4\hat{j} - \hat{k}$ .

Segment 4 of 8

📅 Regroup 2

Key Takeaway: The tangent plane to $z = f(x, y)$ at $(x_0, y_0)$ has the explicit formula $z = z_0 + f_x(x - x_0) + f_y(y - y_0)$ . Its normal vector is always $\vec{n} = \langle f_x, f_y, -1 \rangle$ .
Common Misconception: Students often confuse $z_0$ with zero. Emphasize that $z_0 = f(x_0, y_0)$ must always be calculated first unless explicitly provided in the problem prompt.
Product Rule Alert: When calculating partial derivatives of functions like $x e^{xy}$ , remind students to check whether the differentiated variable appears multiple times in the expression, requiring the product rule.

Segment 5 of 8

Skill Block 3

Real-Life Engineering Applications

🤖

Robotics Engineering

Autonomous Drone Thrust State Estimation (Robotics & Computer Engineering)

An autonomous quadcopter drone's flight guidance computer must continuously calculate its required propulsion thrust percentage ( $T$ ) based on forward pitch angle ( $x$ in radians) and vertical climb velocity ( $y$ in meters per second, $\text{m/s}$ ).

Aerodynamic flight dynamics establish the non-linear thrust response function:

T(x, y) = \ln(x^2 + e^y)

During steady-state hovering, the drone operates at a clean nominal base state $(x_0, y_0) = (0, 1)$ , yielding an exact baseline thrust of $T(0, 1) = \ln(0^2 + e^1) = \ln(e) = 1.0$ ( $100\%$ nominal thrust). Suddenly, a crosswind gust perturbs the drone into a dynamic state of $(x, y) = (0.10, 1.05)$ .

Because evaluating non-linear logarithmic functions on an embedded 32-bit microcontroller consumes excessive clock cycles and battery power, the flight controller executes a local linear approximation $L(x, y)$ centered at $(0, 1)$ to update thrust commands in real time.

State the multivariable linear approximation formula $L(x, y)$ referencing Linear Approximations.
Compute the analytical partial derivatives $T_x(x, y)$ and $T_y(x, y)$ , and evaluate them at the hovering base state $(0, 1)$ .
Construct the explicit linearization formula $L(x, y)$ for the quadcopter's thrust controller.
Estimate the required thrust $T(0.10, 1.05)$ using $L(x, y)$ , and compute the absolute percentage error relative to the true numerical value $T_{\text{true}} = \ln(0.01 + e^{1.05}) \approx 1.0535$ .

💡 Show Worked Derivation

An autonomous quadcopter drone's flight guidance computer must continuously calculate its required propulsion thrust percentage ( $T$ ) based on forward pitch angle ( $x$ in radians) and vertical climb velocity ( $y$ in meters per second, $\text{m/s}$ ).

Aerodynamic flight dynamics establish the non-linear thrust response function:

T(x, y) = \ln(x^2 + e^y)

During steady-state hovering, the drone operates at a clean nominal base state $(x_0, y_0) = (0, 1)$ , yielding an exact baseline thrust of $T(0, 1) = \ln(0^2 + e^1) = \ln(e) = 1.0$ ( $100\%$ nominal thrust). Suddenly, a crosswind gust perturbs the drone into a dynamic state of $(x, y) = (0.10, 1.05)$ .

Because evaluating non-linear logarithmic functions on an embedded 32-bit microcontroller consumes excessive clock cycles and battery power, the flight controller executes a local linear approximation $L(x, y)$ centered at $(0, 1)$ to update thrust commands in real time.

State the multivariable linear approximation formula $L(x, y)$ referencing Linear Approximations.
Compute the analytical partial derivatives $T_x(x, y)$ and $T_y(x, y)$ , and evaluate them at the hovering base state $(0, 1)$ .
Construct the explicit linearization formula $L(x, y)$ for the quadcopter's thrust controller.
Estimate the required thrust $T(0.10, 1.05)$ using $L(x, y)$ , and compute the absolute percentage error relative to the true numerical value $T_{\text{true}} = \ln(0.01 + e^{1.05}) \approx 1.0535$ .

Step-by-Step Solution:

State the Linearization Formula:
- As defined in Linear Approximations, the local linear approximation of a two-variable function at base point $(x_0, y_0)$ is: $L(x, y) = T(x_0, y_0) + T_x(x_0, y_0)(x - x_0) + T_y(x_0, y_0)(y - y_0)$
Compute Analytical Partial Derivatives:
- Differentiating $T(x, y) = \ln(x^2 + e^y)$ via the Logarithmic Chain Rule: $T_x(x, y) = \frac{1}{x^2 + e^y} \cdot \frac{\partial}{\partial x}(x^2 + e^y) = \frac{2x}{x^2 + e^y}$ $T_y(x, y) = \frac{1}{x^2 + e^y} \cdot \frac{\partial}{\partial y}(x^2 + e^y) = \frac{e^y}{x^2 + e^y}$
- Evaluate at nominal base state $(x_0, y_0) = (0, 1)$ : $x_0^2 + e^{y_0} = 0^2 + e^1 = e$ $T_x(0, 1) = \frac{2(0)}{e} = 0$ $T_y(0, 1) = \frac{e^1}{e} = 1$
- Baseline thrust value: $T(0, 1) = \ln(e) = 1.0$ .
Construct the Linearization Formula:
- Substitute parameters into the formula: $L(x, y) = 1.0 + 0(x - 0) + 1.0(y - 1)$ $L(x, y) = 1.0 + (y - 1) = y$
- Physical Insight: Because $T_x(0, 1) = 0$ , first-order variations in pitch angle $x$ around hover have zero linear sensitivity. In the immediate vicinity of hover, thrust commands depend almost exclusively on vertical climb rate $y$ .
Estimate Perturbed Thrust and Compute Error:
- Perturbed state: $(x, y) = (0.10, 1.05)$ .
- Evaluate linearization: $L(0.10, 1.05) = 1.05 \quad (105\% \text{ thrust})$
- Compare with true value $T_{\text{true}} = \ln(0.01 + e^{1.05}) = \ln(0.01 + 2.85765) = \ln(2.86765) \approx 1.05349$ : $\text{Absolute Error} = |1.05349 - 1.05000| = 0.00349$ $\text{Percentage Error} = \frac{0.00349}{1.05349} \times 100\% \approx 0.33\%$
- The microcontroller achieves a computation speedup of orders of magnitude while introducing less than $0.35\%$ error, well within actuator tolerance bands.

⚡

Electrical Engineering

Semiconductor Transistor Current Fluctuation (Electrical Engineering)

In integrated circuit design, an advanced field-effect transistor (FET) regulates output current based on gate-to-source voltage ( $V_g$ ) and drain-to-source voltage ( $V_d$ ). In the saturation regime, the electrical drain current $I$ (in milliamperes, $\text{mA}$ ) is modeled by the non-linear constitutive equation:

I(V_g, V_d) = \sqrt{V_g^3 + V_d^2}

The transistor's nominal operating bias point (Q-point) is established at $V_g = 2.0\text{ V}$ and $V_d = 4.0\text{ V}$ , providing a baseline current of:

I(2, 4) = \sqrt{2^3 + 4^2} = \sqrt{8 + 16} = \sqrt{24} \approx 4.8990\text{ mA}

During high-frequency switching, power rail switching noise causes the terminal voltages to shift slightly to an off-nominal state of $V_g = 2.02\text{ V}$ and $V_d = 3.95\text{ V}$ .

State the multivariable total differential formula $dI$ for the transistor current in terms of voltage differentials $dV_g$ and $dV_d$ .
Compute the analytical partial derivatives $\frac{\partial I}{\partial V_g}$ and $\frac{\partial I}{\partial V_d}$ , and evaluate them at the nominal bias point $(2, 4)$ .
Use the linear approximation $L(V_g, V_d)$ centered at $(2, 4)$ to estimate the perturbed current $I(2.02, 3.95)\text{ mA}$ .
If manufacturer power rail tolerances guarantee $|dV_g| \le 0.05\text{ V}$ and $|dV_d| \le 0.10\text{ V}$ , compute the maximum worst-case current fluctuation $|dI|_{\max}$ .

💡 Show Worked Derivation

In integrated circuit design, an advanced field-effect transistor (FET) regulates output current based on gate-to-source voltage ( $V_g$ ) and drain-to-source voltage ( $V_d$ ). In the saturation regime, the electrical drain current $I$ (in milliamperes, $\text{mA}$ ) is modeled by the non-linear constitutive equation:

I(V_g, V_d) = \sqrt{V_g^3 + V_d^2}

The transistor's nominal operating bias point (Q-point) is established at $V_g = 2.0\text{ V}$ and $V_d = 4.0\text{ V}$ , providing a baseline current of:

I(2, 4) = \sqrt{2^3 + 4^2} = \sqrt{8 + 16} = \sqrt{24} \approx 4.8990\text{ mA}

During high-frequency switching, power rail switching noise causes the terminal voltages to shift slightly to an off-nominal state of $V_g = 2.02\text{ V}$ and $V_d = 3.95\text{ V}$ .

State the multivariable total differential formula $dI$ for the transistor current in terms of voltage differentials $dV_g$ and $dV_d$ .
Compute the analytical partial derivatives $\frac{\partial I}{\partial V_g}$ and $\frac{\partial I}{\partial V_d}$ , and evaluate them at the nominal bias point $(2, 4)$ .
Use the linear approximation $L(V_g, V_d)$ centered at $(2, 4)$ to estimate the perturbed current $I(2.02, 3.95)\text{ mA}$ .
If manufacturer power rail tolerances guarantee $|dV_g| \le 0.05\text{ V}$ and $|dV_d| \le 0.10\text{ V}$ , compute the maximum worst-case current fluctuation $|dI|_{\max}$ .

Step-by-Step Solution:

State the Total Differential Formula:
- As defined in Differentials and Error Propagation, the total differential $dI$ is: $dI = \frac{\partial I}{\partial V_g} dV_g + \frac{\partial I}{\partial V_d} dV_d$
Compute Analytical Partial Derivatives:
- Let $u(V_g, V_d) = V_g^3 + V_d^2$ . Applying the Generalized Power Rule: $\frac{\partial I}{\partial V_g} = \frac{1}{2}(V_g^3 + V_d^2)^{-1/2}(3V_g^2) = \frac{3V_g^2}{2\sqrt{V_g^3 + V_d^2}}$ $\frac{\partial I}{\partial V_d} = \frac{1}{2}(V_g^3 + V_d^2)^{-1/2}(2V_d) = \frac{V_d}{\sqrt{V_g^3 + V_d^2}}$
- Evaluate at $(V_g, V_d) = (2, 4)$ , where $\sqrt{V_g^3 + V_d^2} = \sqrt{24} = 2\sqrt{6}$ : $\left.\frac{\partial I}{\partial V_g}\right|_{(2, 4)} = \frac{3(2^2)}{2\sqrt{24}} = \frac{12}{2\sqrt{24}} = \frac{6}{\sqrt{24}} = \frac{3}{\sqrt{6}} \approx 1.2247\text{ mA/V}$ $\left.\frac{\partial I}{\partial V_d}\right|_{(2, 4)} = \frac{4}{\sqrt{24}} = \frac{2}{\sqrt{6}} \approx 0.8165\text{ mA/V}$
Construct Linear Approximation and Estimate Current:
- Formulate the linearization function centered at $(2, 4)$ : $L(V_g, V_d) = I(2, 4) + \left.\frac{\partial I}{\partial V_g}\right|_{(2, 4)}(V_g - 2) + \left.\frac{\partial I}{\partial V_d}\right|_{(2, 4)}(V_d - 4)$ $L(V_g, V_d) = \sqrt{24} + \frac{6}{\sqrt{24}}(V_g - 2) + \frac{4}{\sqrt{24}}(V_d - 4)$
- Calculate offsets for target point $(2.02, 3.95)$ : $\Delta V_g = 2.02 - 2.0 = +0.02\text{ V}$ $\Delta V_d = 3.95 - 4.0 = -0.05\text{ V}$
- Substitute offsets: $L(2.02, 3.95) = \sqrt{24} + \frac{6(0.02) + 4(-0.05)}{\sqrt{24}} = \sqrt{24} + \frac{0.12 - 0.20}{\sqrt{24}} = \sqrt{24} - \frac{0.08}{\sqrt{24}}$ $L(2.02, 3.95) \approx 4.89898 - \frac{0.08}{4.89898} \approx 4.89898 - 0.01633 \approx 4.8827\text{ mA}$
Compute Worst-Case Current Fluctuation:
- Applying the Triangle Inequality for maximum error bound: $|dI|_{\max} \le \left|\frac{\partial I}{\partial V_g}\right| |dV_g| + \left|\frac{\partial I}{\partial V_d}\right| |dV_d|$ $|dI|_{\max} \le (1.2247)(0.05) + (0.8165)(0.10) = 0.06124 + 0.08165 = 0.1429\text{ mA}$
- The maximum current deviation due to power rail noise is bounded by approximately $\pm 0.143\text{ mA}$ (or $\frac{0.1429}{4.8990} \times 100\% \approx 2.92\%$ ).

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

Linear Approximations and Total Differentials

General Overview
Skill Block 3 focuses on using tangent planes as operational computational tools. Students learn how to construct the multivariable linearization function $L(x, y)$ , select appropriate clean base points, approximate difficult decimal evaluations, analyze approximation error, and compute total differentials $dz = f_x dx + f_y dy$ to quantify measurement error propagation in engineering systems.

Conceptual Framework

Multivariable Linearization Function $L(x, y)$ : $L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ (Defined in Linear Approximations.)
Local Approximation Principle: $f(x, y) \approx L(x, y) \quad \text{for } (x, y) \approx (x_0, y_0)$
Total Differential Equation: $dz = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy = f_x(x, y) dx + f_y(x, y) dy$ (Defined in Differentials and Error Propagation.)
True Increment vs. Total Differential:
- True change along surface: $\Delta z = f(x_0 + \Delta x, y_0 + \Delta y) - f(x_0, y_0)$
- Linearized change along tangent plane: $dz = f_x \Delta x + f_y \Delta y$
- Approximation: $\Delta z \approx dz$

Key Equations & Structural Rules

Linearization Equation: $L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)\Delta x + f_y(x_0, y_0)\Delta y$
Total Differential: $dz = f_x dx + f_y dy$
Worst-Case Absolute Error Bound: $|dz|_{\max} \le |f_x| \delta_x + |f_y| \delta_y$
Relative Error: $\frac{dz}{z}$
Percentage Error: $\frac{dz}{z} \times 100\%$

Worked Example A: Estimating a Radical Expression via Linear Approximation

Use a linear approximation at the base point $(2, 1)$ to estimate the value of $f(2.1, 0.9)$ for the function $f(x, y) = \sqrt{x^2 + y}$ .

💡 Reveal Worked Solution

Evaluate the True Function at the Base Point $(2, 1)$ :
$f(2, 1) = \sqrt{2^2 + 1} = \sqrt{4 + 1} = \sqrt{5}$
Compute General First-Order Partial Derivatives: Apply the Generalized Power Rule to $f(x, y) = (x^2 + y)^{1/2}$ :
$f_x(x, y) = \frac{1}{2}(x^2 + y)^{-1/2}(2x) = \frac{x}{\sqrt{x^2 + y}}$ $f_y(x, y) = \frac{1}{2}(x^2 + y)^{-1/2}(1) = \frac{1}{2\sqrt{x^2 + y}}$
Evaluate the Partial Derivatives at $(2, 1)$ :
$f_x(2, 1) = \frac{2}{\sqrt{2^2 + 1}} = \frac{2}{\sqrt{5}}$ $f_y(2, 1) = \frac{1}{2\sqrt{2^2 + 1}} = \frac{1}{2\sqrt{5}}$
Construct the Linearization Function $L(x, y)$ : State the Linear Approximation Formula (defined in Linear Approximations):
$L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ $L(x, y) = \sqrt{5} + \frac{2}{\sqrt{5}}(x - 2) + \frac{1}{2\sqrt{5}}(y - 1)$
Evaluate at the Perturbed Target Point $(2.1, 0.9)$ : Calculate coordinate displacements:
$\Delta x = x - x_0 = 2.1 - 2 = 0.1$

$\Delta y = y - y_0 = 0.9 - 1 = -0.1$
Substitute into $L(x, y)$ :
$L(2.1, 0.9) = \sqrt{5} + \frac{2}{\sqrt{5}}(0.1) + \frac{1}{2\sqrt{5}}(-0.1)$

$L(2.1, 0.9) = \sqrt{5} + \frac{0.2}{\sqrt{5}} - \frac{0.05}{\sqrt{5}} = \sqrt{5} + \frac{0.15}{\sqrt{5}}$
Using the numerical approximation $\sqrt{5} \approx 2.23607$ :
$L(2.1, 0.9) \approx 2.23607 + \frac{0.15}{2.23607} \approx 2.23607 + 0.06708 = 2.30315$
- Accuracy Check: The exact value is $f(2.1, 0.9) = \sqrt{2.1^2 + 0.9} = \sqrt{4.41 + 0.9} = \sqrt{5.31} \approx 2.30434$ . The linear approximation error is only $|2.30434 - 2.30315| = 0.00119$ ( $0.05\%$ relative error).

Practice Problem 9: Concept & Reading Check

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

(a) What is the mathematical and geometric connection between the Tangent Plane equation and the Linearization function $L(x, y)$ ?
(b) In Worked Example A, explain why $(2, 1)$ was chosen as the base point rather than evaluating $f(2.1, 0.9)$ directly. What two criteria should guide the selection of a base point $(x_0, y_0)$ ?
(c) In Autonomous Drone Thrust State Estimation (Robotics & Computer Engineering), why does the quadcopter's flight computer use a linear approximation rather than evaluating the non-linear equation $T(x, y) = \ln(x^2 + e^y)$ in real-time?

Practice Space Solve this problem on paper or tablet.

Review the Conceptual Framework and the solutions to Worked Example A and the linked drone flight control application. (Defined in Linear Approximations.)

Answer: * (a) The graph of the linearization function $z = L(x, y)$ is identically the tangent plane to the surface at $(x_0, y_0, z_0)$ .

(b) $(2, 1)$ produces clean integer evaluations for $f$ and its derivatives; base points must evaluate cleanly and lie close to the target coordinate.
(c) Linearization executes in microseconds on embedded microcontrollers, saving CPU clock cycles and battery power while maintaining high accuracy.

Step-by-Step Execution

(a) As defined in Linear Approximations, the equation $z = L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ is algebraically and geometrically identical to the point-normal equation of the tangent plane. The linearization function $L(x, y)$ treats the flat tangent plane as an explicit mathematical function whose outputs approximate the curved surface $z = f(x, y)$ in the immediate vicinity of the base point.
(b) In Worked Example A, evaluating $f(2.1, 0.9) = \sqrt{5.31}$ requires a high-precision radical root extraction algorithm that cannot be done by hand. In contrast, the base point $(2, 1)$ yields $2^2 + 1 = 5$ , giving clean analytical derivatives. In general, an optimal base point $(x_0, y_0)$ must satisfy two criteria: (1) Clean Evaluation: $f(x_0, y_0)$ and its partial derivatives must evaluate to simple, known numbers; and (2) Proximity: the displacements $\Delta x = x - x_0$ and $\Delta y = y - y_0$ must be small so that higher-order curvature errors remain negligible.
(c) In Autonomous Drone Thrust State Estimation (Robotics & Computer Engineering), the quadcopter's 32-bit flight controller updates motor commands at $400\text{ Hz}$ ( $2.5\text{ ms}$ loop cycle). Computing floating-point transcendental functions (logarithms and exponentials) consumes substantial clock cycles and battery energy. By executing the linear approximation $L(x, y) = y$ , the microcontroller executes a single register read operation in nanoseconds while introducing less than $0.35\%$ error, well within motor actuator tolerances.

Practice Problem 10

Use the linear approximation of $f(x, y) = x^2 + y^2$ at the base point $(1, 1)$ to estimate $f(1.1, 0.9)$ . Compare your estimate with the exact value and calculate the approximation error.

Practice Space Solve this problem on paper or tablet.

State the Linear Approximation Formula: $L(x, y) = f(1, 1) + f_x(1, 1)(x - 1) + f_y(1, 1)(y - 1)$ . Compute $f(1, 1) = 2$ and partial derivatives $f_x = 2x, f_y = 2y$ . (Defined in Linear Approximations.)

Answer:

L(1.1, 0.9) = 2.0, \quad f(1.1, 0.9) = 2.02, \quad \text{Error} = 0.02

Step-by-Step Execution

Evaluate Baseline Value at Base Point $(1, 1)$ :
$f(1, 1) = 1^2 + 1^2 = 1 + 1 = 2$
Compute General Partial Derivatives and Evaluate at $(1, 1)$ :
$f_x(x, y) = 2x \implies f_x(1, 1) = 2(1) = 2$ $f_y(x, y) = 2y \implies f_y(1, 1) = 2(1) = 2$
Construct the Linearization Function $L(x, y)$ : State the Linearization Equation:
$L(x, y) = f(1, 1) + f_x(1, 1)(x - 1) + f_y(1, 1)(y - 1)$ $L(x, y) = 2 + 2(x - 1) + 2(y - 1)$
Evaluate at Target Point $(1.1, 0.9)$ : Identify displacements: $\Delta x = 1.1 - 1 = 0.1$ and $\Delta y = 0.9 - 1 = -0.1$ .
$L(1.1, 0.9) = 2 + 2(0.1) + 2(-0.1) = 2 + 0.2 - 0.2 = 2.0$
Compare with Exact Value and Calculate Error:
$f(1.1, 0.9) = (1.1)^2 + (0.9)^2 = 1.21 + 0.81 = 2.02$

$\text{Approximation Error} = |f(1.1, 0.9) - L(1.1, 0.9)| = |2.02 - 2.00| = 0.02$
- Curvature Insight: The first-order linear contributions canceled exactly ( $+0.2 - 0.2 = 0$ ). The remaining discrepancy of $0.02$ represents second-order surface curvature: $(\Delta x)^2 + (\Delta y)^2 = (0.1)^2 + (-0.1)^2 = 0.01 + 0.01 = 0.02$ .

Practice Problem 11

Use a linear approximation at the base point $(4, 0)$ to estimate $f(4.1, -0.1)$ for the radical function $f(x, y) = \sqrt{x + y + 4}$ .

Practice Space Solve this problem on paper or tablet.

Compute $f(4, 0) = \sqrt{8} = 2\sqrt{2}$ . Evaluate $f_x$ and $f_y$ at $(4, 0)$ , set up $L(x, y)$ , and substitute $\Delta x = 0.1, \Delta y = -0.1$ . (Defined in Linear Approximations.)

Answer:

L(4.1, -0.1) = 2\sqrt{2} \approx 2.8284

Step-by-Step Execution

Evaluate Baseline Value at Base Point $(4, 0)$ :
$f(4, 0) = \sqrt{4 + 0 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2.82843$
Compute General Partial Derivatives: Applying the Chain Rule to $f(x, y) = (x + y + 4)^{1/2}$ :
$f_x(x, y) = \frac{1}{2\sqrt{x + y + 4}}, \quad f_y(x, y) = \frac{1}{2\sqrt{x + y + 4}}$
Evaluate Derivatives at $(4, 0)$ :
$f_x(4, 0) = \frac{1}{2\sqrt{4 + 0 + 4}} = \frac{1}{2\sqrt{8}} = \frac{1}{4\sqrt{2}} = \frac{\sqrt{2}}{8}$ $f_y(4, 0) = \frac{1}{2\sqrt{4 + 0 + 4}} = \frac{1}{2\sqrt{8}} = \frac{1}{4\sqrt{2}} = \frac{\sqrt{2}}{8}$
Construct Linearization and Evaluate at $(4.1, -0.1)$ : State the Linear Approximation Formula:
$L(x, y) = f(4, 0) + f_x(4, 0)(x - 4) + f_y(4, 0)(y - 0)$

$L(x, y) = 2\sqrt{2} + \frac{1}{4\sqrt{2}}(x - 4) + \frac{1}{4\sqrt{2}}y$
Substitute $\Delta x = 4.1 - 4 = 0.1$ and $\Delta y = -0.1 - 0 = -0.1$ :
$L(4.1, -0.1) = 2\sqrt{2} + \frac{1}{4\sqrt{2}}(0.1) + \frac{1}{4\sqrt{2}}(-0.1)$

$L(4.1, -0.1) = 2\sqrt{2} + \frac{0.1 - 0.1}{4\sqrt{2}} = 2\sqrt{2} + 0 = 2\sqrt{2} \approx 2.82843$
- Remarkable Geometric Note: Notice that $\Delta x + \Delta y = 0.1 - 0.1 = 0$ , so the exact evaluation is $f(4.1, -0.1) = \sqrt{4.1 - 0.1 + 4} = \sqrt{8} = 2\sqrt{2}$ . Because the level curves of this surface are parallel lines $x + y = C$ , the linear approximation yielded the exact true value.

Practice Problem 12

The storage capacity of a cylindrical industrial chemical tank is given by $V(r, h) = \pi r^2 h$ . During construction, field measurements record a radius of $r = 2.0\text{ m}$ with a maximum measurement uncertainty of $dr = \pm 0.02\text{ m}$ and a height of $h = 5.0\text{ m}$ with an uncertainty of $dh = \pm 0.05\text{ m}$ .

Compute the total differential $dV$ in terms of $r, h, dr,$ and $dh$ .
Use $dV$ to estimate the maximum worst-case absolute error in the calculated volume.
Determine the maximum relative error $\frac{dV}{V}$ and percentage error in the storage volume.

Practice Space Solve this problem on paper or tablet.

State the Total Differential Equation: $dV = \frac{\partial V}{\partial r} dr + \frac{\partial V}{\partial h} dh$ . Use the Triangle Inequality $|dV|_{\max} \le |V_r| |dr| + |V_h| |dh|$ . (Defined in Differentials and Error Propagation.)

Answer:

\text{(a) } dV = 2\pi r h \, dr + \pi r^2 \, dh, \quad \text{(b) } |\Delta V| \lesssim 0.60\pi \approx 1.885\text{ m}^3, \quad \text{(c) } \pm 3.0\%

Step-by-Step Execution

Compute Partial Derivatives and State Total Differential: State the Total Differential Formula (defined in Differentials and Error Propagation):
$dV = \frac{\partial V}{\partial r} dr + \frac{\partial V}{\partial h} dh$ Differentiate $V(r, h) = \pi r^2 h$ : $\frac{\partial V}{\partial r} = 2\pi r h, \quad \frac{\partial V}{\partial h} = \pi r^2$ $dV = 2\pi r h \, dr + \pi r^2 \, dh$
Estimate Maximum Absolute Error at $(r, h) = (2.0, 5.0)$ : Evaluate sensitivity coefficients:
$\frac{\partial V}{\partial r}(2, 5) = 2\pi(2.0)(5.0) = 20\pi$ $\frac{\partial V}{\partial h}(2, 5) = \pi(2.0)^2 = 4\pi$ Apply worst-case tolerances $|dr| \le 0.02\text{ m}$ and $|dh| \le 0.05\text{ m}$ using the Triangle Inequality: $|dV|_{\max} \le \left|\frac{\partial V}{\partial r}\right| |dr| + \left|\frac{\partial V}{\partial h}\right| |dh|$ $|dV|_{\max} \le (20\pi)(0.02) + (4\pi)(0.05) = 0.40\pi + 0.20\pi = 0.60\pi \approx 1.8850\text{ m}^3$
Compute Relative and Percentage Error: The nominal tank volume is $V(2, 5) = \pi(2.0)^2(5.0) = 20\pi\text{ m}^3 \approx 62.8319\text{ m}^3$ . State the Relative Error Formula:
$\frac{dV}{V} = \frac{2\pi r h \, dr + \pi r^2 \, dh}{\pi r^2 h} = 2\frac{dr}{r} + \frac{dh}{h}$
Substitute relative measurement uncertainties:
$\frac{dr}{r} = \frac{0.02}{2.0} = 0.01 \quad (1.0\%), \quad \frac{dh}{h} = \frac{0.05}{5.0} = 0.01 \quad (1.0\%)$

$\frac{dV}{V} = 2(0.01) + 0.01 = 0.02 + 0.01 = 0.030$

$\text{Percentage Error} = 0.030 \times 100\% = 3.0\%$
- Engineering Sensitivity Takeaway: Because the radius is squared in the volume formula ( $r^2$ ), its relative measurement uncertainty contributes with a weight factor of $2$ , while height contributes with a weight factor of $1$ . Precision control on radius is twice as critical to tank capacity accuracy as height.

Segment 6 of 8

📅 Regroup 3

Key Takeaway: The total differential $dz = f_x dx + f_y dy$ approximates the true change $\Delta z$ by measuring vertical displacement along the flat tangent plane rather than the curved surface.
Common Misconception: Students often forget to take absolute values when estimating maximum worst-case error. If $dr = \pm 0.02$ and $dh = \pm 0.05$ , the worst-case scenario occurs when both errors push the output in the same direction, requiring $|f_x| |dx| + |f_y| |dh|$ .
Relative Error Shortcut: In power products of the form $f(x, y) = C x^m y^n$ , logarithmic differentiation reveals that relative error is simply $\frac{df}{f} = m \frac{dx}{x} + n \frac{dy}{y}$ , providing an instantaneous mental check for engineering tolerance studies.

Segment 7 of 8

📅 Synthesis Wrap-up

Core Takeaways from Session 10:
1. Planes in 3D Space: Uniquely oriented by a normal vector $\vec{n} = \langle a, b, c \rangle$ . The scalar equation $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$ expresses that every displacement vector in the plane is orthogonal to $\vec{n}$ .
2. Tangent Planes to Surfaces: For any differentiable multivariable surface $z = f(x, y)$ , the surface normal vector is $\vec{n} = \langle f_x, f_y, -1 \rangle$ , yielding the tangent plane $z = z_0 + f_x(x-x_0) + f_y(y-y_0)$ .
3. Linearization & Differentials: The tangent plane acts as a local computational proxy $L(x, y)$ . The total differential $dz = f_x dx + f_y dy$ provides a rapid first-order tool for propagating measurement tolerances and analyzing system sensitivities.
Overview of Knewton Alta Homework:
- Tangent Planes and Linear Approximations of Functions of Several Variables: Focus on calculating partial derivatives accurately, converting equations between point-normal and standard general forms, evaluating linearizations at nearby offsets $\Delta x, \Delta y$ , and bounding total differentials in physical measurement applications.

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