MATH-283 (Calculus III) Curriculum

Session 10 - Tangent Planes and Linear Approximations

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

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

Construct vector, point-normal ( $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$ ), and general ( $ax+by+cz=d$ ) equations of flat planes in 3D space from a point and normal vector or three non-collinear points using cross products.
Formulate equations of tangent planes to smooth surfaces $z = f(x, y)$ at given base points $(x_0, y_0, z_0)$ by evaluating first-order partial derivatives and constructing upward- or downward-pointing surface normal vectors $\vec{n} = \langle f_x, f_y, -1 \rangle$ .
Construct linearizations $L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ to approximate multivariable function values near a base point, and compute total differentials $dz = f_x dx + f_y dy$ to estimate physical propagation of measurement uncertainties and tolerances.

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.

Skill Block 1

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.

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?

Solution Key:

(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.

Show your steps and calculations here.

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 ( $ax + by + cz = d$ ).

Solution Key:

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 $(0, 0, 0)$ .

Show your steps and calculations here.

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.

Solution Key:

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)

Show your steps and calculations here.

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 $4x - 2y + 5z = 19$ .

Solution Key:

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 $x_0 = -2, y_0 = 3, z_0 = 1$ : $4(x - (-2)) - 2(y - 3) + 5(z - 1) = 0$ $4(x + 2) - 2(y - 3) + 5(z - 1) = 0$ $4x + 8 - 2y + 6 + 5z - 5 = 0$ $4x - 2y + 5z + 9 = 0$ $4x - 2y + 5z = -9$

Show your steps and calculations here.

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 $-1$ .

Skill Block 2

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 $z_0 = 0$ .

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 $\vec{F}_p = -p \hat{n} \, dA$ ?

Solution Key:

(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.

Show your steps and calculations here.

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 ( $ax + by + cz = d$ ).

Solution Key:

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 $\vec{n} = \langle 2, 4, -1 \rangle = 2\hat{i} + 4\hat{j} - \hat{k}$ .

Show your steps and calculations here.

Practice Problem 7

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

Solution Key:

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 $4$ :
$4z = x + y + 4 \implies x + y - 4z = -4$

Show your steps and calculations here.

Practice Problem 8

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

Solution Key:

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 $\vec{n} = \langle 1, 4, -1 \rangle = \hat{i} + 4\hat{j} - \hat{k}$ .

Show your steps and calculations here.

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.

Skill Block 3

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

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?

Solution Key:

(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.

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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.

Solution Key:

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: $(\Delta x)^2 + (\Delta y)^2 = (0.1)^2 + (-0.1)^2 = 0.01 + 0.01 = 0.02$ .

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Practice Problem 11

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

Solution Key:

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.

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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.

Solution Key:

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.

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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.

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.