MTH301 — Midterm Summary (Lectures 1–22)

📘 Lecture 1 — Introduction

📖 Overview: This foundational lecture introduces calculus as the mathematics of motion and change, distinguishing between differential and integral calculus. It also reviews the coordinate systems used in two and three dimensions and defines the concept of a function, which is essential for all subsequent calculus topics.

🗂️ Topics Covered

The lecture begins with an introduction to calculus, differentiating between differential calculus (rates of change, slopes of curves) and integral calculus (areas, volumes, lengths). It then reviews the reference axis system for two dimensions (x-y plane) and three dimensions (x-y-z space), including the location of points and the sign conventions in quadrants and octants. Finally, it introduces the concept of a function of one variable and functions of several variables with practical examples.

📝 Lecture Summary

Calculus

Calculus is the mathematical tool used to analyze changes in physical quantities. It is also described as the Mathematics of Motion and Change. Where there is motion, growth, or variable forces producing acceleration, calculus is the appropriate mathematics to apply.

Differential Calculus

Differential calculus deals with the problem of finding (1) the rate of change and (2) the slope of a curve. Applications include calculating velocities and accelerations of moving bodies, determining firing angles for maximum range of cannons, and finding the times when planets are closest together or farthest apart.

Integral Calculus

Integral calculus deals with the problem of determining a function from information about its rates of change. Integral calculus enables us to (1) calculate lengths of curves, (2) find areas of irregular regions in a plane, (3) find the volumes and masses of arbitrary solids, and (4) calculate the future location of a body from its present position and knowledge of the forces acting on it.

Reference Axis System

The real number system consists of both rational and irrational numbers. There are infinite real numbers between every two real numbers. In two dimensions, we take two mutually perpendicular lines: the horizontal line as the x-axis and the vertical line as the y-axis. The point where they intersect is the origin.

Any point on the x-axis is denoted by an ordered pair (x, 0), where x is a real number known as the abscissa and the second element, the ordinate, is 0. Similarly, any point on the y-axis is represented by an ordered pair (0, y). These lines divide the plane into four regions called quadrants: First (+, +), Second (-, +), Third (-, -), and Fourth (+, -).

Location of a Point

To locate a point P in the plane, draw two perpendicular lines from the point to the x-axis and y-axis. The distance from the origin to the intersection point on the x-axis is the X-coordinate of point P, and the distance from the origin to the intersection point on the y-axis is the Y-coordinate of point P.

3D Coordinate System

In space, there are three mutually perpendicular reference axes: the x, y, and z axes. The planes x=0, y=0, and z=0 divide the space into eight octants. The origin is (0,0,0), and any point in space has three coordinates (x, y, z).

The equation x=0 represents a plane in 3D space where every point has its x-coordinate as 0. Similarly, y=0 and z=0 define planes. When two planes intersect, they give the equation of a line. For example, the intersection of planes x=0 and z=0 gives the y-axis.

Sign of Coordinates in Different Octants

The octant in which every coordinate has a positive sign is the first octant, formed by the positive x, y, and z axes. The sign conventions for all eight octants are:

• First octant (+, +, +)
• Second octant (-, +, +)
• Third octant (-, -, +)
• Fourth octant (+, -, +)
• Fifth octant (+, +, -)
• Sixth octant (-, +, -)
• Seventh octant (-, -, -)
• Eighth octant (+, -, -)

💡 Why this matters: Understanding the 3D coordinate system and octants is critical for visualizing and solving problems in multivariable calculus, physics, and engineering.

The direction of the third axis (z-axis) can be obtained using the Right Hand Rule from the cross product of the two vectors representing the x and y axes.

Concept of a Function

The term function denotes the dependence of one quantity on another quantity. The quantity x is called the independent variable, and the quantity y is called the dependent variable. We write y = f(x) and read "y is a function of x."

The equation y = 2x defines y as a function of x because each value assigned to x determines a unique value of y.

🔑 Definition — Function: A relation where each input (independent variable) corresponds to exactly one output (dependent variable).

📌 Example: The area of a circle depends on its radius r by the equation ( A = \pi r^2 ), so A is a function of r. The volume of a cube depends on its side length x by ( V = x^3 ), so V is a function of x.

Function of Several Variables

Many functions depend on more than one independent variable.

📌 Example: The area of a rectangle depends on its length l and width w by ( A = lw ), so A is a function of l and w. The volume of a rectangular box depends on its length l, width w, and height h by ( V = lwh ), so V is a function of l, w, and h. The volume V of a right circular cylinder depends on its radius r and height h by ( V = \pi r^2 h ), so V is a function of r and h.

⭐ Key Takeaways

Calculus is fundamentally the mathematics of change and motion, divided into differential calculus (finding rates of change and slopes) and integral calculus (finding lengths, areas, volumes, and positions from rates). A solid understanding of the 2D and 3D coordinate systems, including quadrants and octants, is essential for visualizing problems. The concept of a function, where a dependent variable is uniquely determined by one or more independent variables, is the core building block for all calculus operations. Remember the sign conventions for all eight octants and the classic examples of functions of one and several variables.

🧠 Quick Revision Questions

1. What are the two main problems that differential calculus deals with?
2. List four specific applications of integral calculus mentioned in the lecture.
3. In the 3D coordinate system, what are the sign coordinates for a point in the sixth octant?
4. Define a function and clearly distinguish between the independent and dependent variables.
5. Give one example each of a function of one variable and a function of two variables, writing the corresponding equations.

📘 Lecture 2 — Values of functions

📖 Overview: This lecture introduces the concept of evaluating functions of one, two, and three variables at specific points. It also reviews key geometric curves including circles, parabolas, ellipses, and hyperbolas, which are essential for visualizing functions in calculus.

🗂️ Topics Covered

The lecture covers evaluating functions of one, two, and three variables at given points, including examples with polynomial and radical expressions. It also defines functions of multiple variables and reviews geometric concepts such as circles, disks, parabolas (including vertex finding and graphing), ellipses, and hyperbolas.

📝 Lecture Summary

Values of functions

Consider the function f(x) = 2x² – 1. Then f(1) = 2(1)² – 1 = 1, f(4) = 2(4)² – 1 = 31, f(-2) = 2(-2)² – 1 = 7, and f(t-4) = 2(t-4)² – 1 = 2t² – 16t + 31. These are the values of the function at some points.

Example: Consider the function f(x,y) = x²y + 1. Then f(2,1) = (2²)(1)+1 = 5, f(1,2) = (1²)(2)+1 = 3, f(0,0) = (0²)(0)+1 = 1, f(1,-3) = (1²)(-3)+1 = -2, f(3a,a) = (3a)²(a)+1 = 9a³+1, and f(ab,a-b) = (ab)²(a-b)+1 = a³b² – a²b³ + 1.

Example: Now consider the function f(x,y) = x + ³√y. Then:

• (a) f(2,4) = 2 + ³√4 = 2 + ³√4
• (b) f(t, t²) = t + ³√(t²) = t + t²/³
• (c) f(x, x²) = x + ³√(x²) = x + x²/³
• (d) f(2y², 4y) = 2y² + ³√(4y) = 2y² + (4y)¹/³

Now again we take another function of three variables.

Example: Consider the function f(x,y,z) = xy²z³ + 3. Then at certain points we have:

• f(2,1,2) = (2)(1)²(2)³ + 3 = 19
• f(0,0,0) = (0)(0)²(0)³ + 3 = 3
• f(a,a,a) = (a)(a)²(a)³ + 3 = a⁶ + 3
• f(t,t²,-t) = (t)(t²)²(-t)³ + 3 = -t⁸ + 3
• f(-3,1,1) = (-3)(1)²(1)³ + 3 = 0

Example: Consider the function f(x,y,z) = x²y²z⁴ where x(t) = t³, y(t) = t², and z(t) = t.

• (a) f(x(t), y(t), z(t)) = [x(t)]²[y(t)]²[z(t)]⁴ = [t³]²[t²]²[t]⁴ = t¹⁴
• (b) f(x(0), y(0), z(0)) = [x(0)]²[y(0)]²[z(0)]⁴ = [0³]²[0²]²[0]⁴ = 0

Example: Let us consider the function f(x,y,z) = xyz + x. Then f(xy, y/x, xz) = (xy)(y/x)(xz) + xy = xy²z + xy.

Example: Let us consider g(x,y,z) = z Sin(xy), u(x,y,z) = x²z³, v(x,y,z) = Pxyz, w(x,y,z) = xy/z. Then g(u(x,y,z), v(x,y,z), w(x,y,z)) = w(x,y,z) Sin(u(x,y,z) v(x,y,z)). By putting the values of these functions from the above equations we get g(u(x,y,z), v(x,y,z), w(x,y,z)) = (xy/z) Sin[(x²z³)(Pxyz)] = (xy/z) Sin(Pyx³z⁴).

Example: Consider the function g(x,y) = y Sin(x²y) and u(x,y) = x²y³, v(x,y) = πxy. Then g(u(x,y), v(x,y)) = v(x,y) Sin([u(x,y)]² v(x,y)). By putting the values of these functions we get g(u(x,y), v(x,y)) = πxy Sin([x²y³]² πxy) = πxy Sin(x⁵y⁷).

Function of One Variable

A function f of one real variable x is a rule that assigns a unique real number f(x) to each point x in some set D of the real line.

Function of two Variables

A function f in two real variables x and y is a rule that assigns a unique real number f(x,y) to each point (x,y) in some set D of the xy-plane.

Function of three variables

A function f in three real variables x, y, and z is a rule that assigns a unique real number f(x,y,z) to each point (x,y,z) in some set D of three-dimensional space.

Function of n variables

A function f in n variable real variables x₁, x₂, x₃, ..., xₙ is a rule that assigns a unique real number w = f(x₁, x₂, x₃, ..., xₙ) to each point (x₁, x₂, x₃, ..., xₙ) in some set D of n-dimensional space.

Circles and Disks

Circles and disks are geometric concepts used in multivariable calculus. A circle is the set of all points at a fixed distance (radius) from a center point. A disk includes the interior of the circle as well.

💡 Why this matters: Understanding circles and disks is essential for defining domains of functions and regions of integration in calculus.

PARABOLA

A parabola is a U-shaped curve. The general equation of the parabola opening upward or downward is of the form y = f(x) = ax² + bx + c. It opens upward if a > 0 and opens downward if a < 0. The x-coordinate of the vertex is given by x₀ = -b/2a. So the y-coordinate of the vertex is y₀ = f(x₀). The axis of symmetry is x = x₀.

Sketching of the graph of parabola y = ax² + bx + c: Finding vertex: x-coordinate of the vertex is given by x₀ = -b/2a. So, y-coordinate of the vertex is y₀ = a x₀² + b x₀ + c. Hence vertex is V(x₀, y₀).

Example: Sketch the parabola y = -x² + 4x. Since a = -1 < 0, parabola is opening downward. Vertex occurs at x = -b/2a = (-4)/2(-1) = 2. Axis of symmetry is the vertical line x = 2. The y-coordinate of the vertex is y = -(2)² + 4(2) = 4. Hence vertex is V(2, 4). The zeros of the parabola (the points where the parabola meets the x-axis) are the solutions to -x² + 4x = 0, so x = 0 and x = 4. Therefore (0,0) and (4,0) lie on the parabola. Also (1,3) and (3,3) lie on the parabola.

Example: y = x² - 4x + 3. Since a = 1 > 0, parabola is opening upward. Vertex occurs at x = -b/2a = (4)/2 = 2. Axis of symmetry is the vertical line x = 2. The y-coordinate of the vertex is y = (2)² - 4(2) + 3 = -1. Hence vertex is V(2, -1). The zeros of the parabola are the solutions to x² - 4x + 3 = 0, so x = 1 and x = 3. Therefore (1,0) and (3,0) lie on the parabola. Also (0,3) and (4,3) lie on the parabola.

Ellipse

An ellipse is a curve that looks like a stretched circle. Its general equation is often of the form (x²/a²) + (y²/b²) = 1. It has two axes: the major axis and the minor axis.

Hyperbola

A hyperbola is a curve consisting of two separate branches. Its general equation is often of the form (x²/a²) – (y²/b²) = 1 or (y²/a²) – (x²/b²) = 1.

⭐ Key Takeaways

The core concepts to remember are: how to evaluate functions of one, two, and three variables by substituting given values into the function expression; the formal definition of functions of one, two, three, and n variables as rules assigning unique real numbers to points in their respective domains; the geometric properties of parabolas including vertex location using x₀ = -b/2a, axis of symmetry, and how the sign of 'a' determines opening direction; and the basic recognition of circles, disks, ellipses, and hyperbolas as geometric shapes used in calculus. Students must be able to compute function values for arbitrary inputs and sketch simple parabolas.

🧠 Quick Revision Questions

1. What is f(2, -1, 3) for the function f(x, y, z) = x²y – z?
2. For a parabola y = 3x² + 6x – 2, what is the x-coordinate of its vertex?
3. Does the parabola y = -2x² + 5x – 1 open upward or downward?
4. If f(x, y) = x/y and g(x, y) = x², what is f(g(x, y), y)?
5. A function assigns a unique real number to each point in its domain. What is the domain for a function of two variables?

📘 Lecture 3 — Elements of three dimensional geometry

📖 Overview: This lecture introduces the fundamental concepts of three-dimensional geometry, including the distance formula and midpoint formula in 3D space. It then defines direction angles and direction cosines for lines, and explains how to find these for a line joining two points. The lecture concludes with an overview of surfaces in 3D, including planes, spheres, cones, and cylinders, and how their intersections form curves.

🗂️ Topics Covered

This lecture covers the distance formula in three dimensions with a worked example, the midpoint formula for two points, and the definitions of direction angles and direction cosines. It explains how to find direction ratios and cosines for a line joining two points, illustrates the concept of surfaces as graphs of equations in 3D, and discusses the intersection of surfaces (e.g., cone and sphere, two planes). Finally, it describes coordinate planes, planes parallel to them, and presents the general equation of a plane along with standard surfaces like the sphere, right circular cone, and cylinders.

📝 Lecture Summary

Distance formula in three dimension

Let P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) be two points such that PQ is not parallel to one of the coordinate axes. Then PQ = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²] . This is known as the Distance formula between the points P and Q.

🔑 Definition — Distance Formula (3D): The distance between two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is given by √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]. 📐 Formula: PQ = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²] → This represents the straight-line distance between two points in three-dimensional space by taking the square root of the sum of the squares of the differences in their x, y, and z coordinates. 📌 Example: Let us consider the points A (3, 2, 4), B (6, 10, -1), and C (9, 4, 1). Then |AB| = √[(6-3)² + (10-2)² + (-1-4)²] = √[9 + 64 + 25] = √98 = 7√2. |AC| = √[(9-3)² + (4-2)² + (1-4)²] = √[36 + 4 + 9] = √49 = 7. |BC| = √[(9-6)² + (4-10)² + (1+1)²] = √[9 + 36 + 4] = √49 = 7.

Mid point of two points

If R is the middle point of the line segment PQ, then the coordinates of the middle point are x = (x₁ + x₂)/2, y = (y₁ + y₂)/2, z = (z₁ + z₂)/2.

🔑 Definition — Midpoint Formula (3D): The midpoint of a line segment joining points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) has coordinates that are the average of the corresponding coordinates of the endpoints. 📌 Example: Let us consider two points A(3,2,4) and B(6,10,-1). Then the coordinates of the midpoint of AB are [(3+6)/2, (2+10)/2, (4-1)/2] = (9/2, 6, 3/2).

Direction Angles

The direction angles α, β, γ of a line are defined as: α = Angle between the line and the positive x-axis, β = Angle between the line and the positive y-axis, γ = Angle between the line and the positive z-axis. By definition, each of these angles lies between 0° and π.

Direction Ratios

Cosines of direction angles are called direction cosines. Any multiple of direction cosines are called direction numbers or direction ratios of the line L.

🔑 Definition — Direction Angles: The angles which a line makes with the positive x, y, and z-axes are known as Direction Angles. 🔑 Definition — Direction Cosines: If we take the cosine of the Direction Angles of a line, then we get the Direction cosines of that line. For a point P(x,y) on a line, let r = √(x² + y² + z²). Then cos α = x/r, cos β = y/r, and cos γ = z/r. A fundamental relation is cos²α + cos²β + cos²γ = 1.

Direction cosines and direction ratios of a line joining two points

For a line joining two points P(x₁, y₁, z₁) and Q(x₂, y₂, x₂) the direction ratios are x₂ - x₁, y₂ - y₁, z₂ - z₁. The direction cosines are (x₂ - x₁)/PQ, (y₂ - y₁)/PQ, (z₂ - z₁)/PQ, where PQ = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²].

📌 Example: For a line joining two points P(1,3,2) and Q(7,-2,3) the direction ratios are 7-1, -2-3, 3-2 which equal 6, -5, 1. The direction cosines are 6/√62, -5/√62, 1/√62, since PQ = √(6² + (-5)² + 1²) = √(36+25+1) = √62.

Surfaces in 3D Space

In three-dimensional space, the graph of an equation relating the variables x, y, and z is the set of all points (x, y, z) whose coordinates satisfy the equation. Usually, such graphs are surfaces.

Intersection of two surfaces

The intersection of two surfaces is a curve in three-dimensional space. This is the reason that a curve in three-dimensional space is represented by two equations representing the intersecting surfaces. Examples include the intersection of a Cone and a Sphere or the intersection of Two Planes.

Region Descriptions and Coordinate Planes

The coordinate planes are defined by setting one coordinate to zero: the xy-plane consists of all points (x, y, 0) with equation z = 0; the xz-plane consists of all points (x, 0, z) with equation y = 0; the yz-plane consists of all points (0, y, z) with equation x = 0. The coordinate axes are defined by setting two coordinates to zero: the x-axis (y=0, z=0), the y-axis (z=0, x=0), and the z-axis (x=0, y=0).

General Equation of Plane

Any equation of the form ax + by + cz + d = 0, where a, b, c, d are real numbers, represents a plane.

Sphere, Right Circular Cone, Horizontal Circular Cylinder, Horizontal Elliptic Cylinder

The text presents figures of a sphere, a right circular cone, a horizontal circular cylinder, and a horizontal elliptic cylinder, indicating that these are standard surfaces studied in three-dimensional geometry. The overview notes this lecture covers Chapter #14: Three Dimensional Space, from the book "CALCULUS" by Howard Anton.

💡 Why this matters: Understanding these concepts forms the foundation for vector calculus and multivariable analysis, enabling the description and computation of geometric properties of lines, planes, and curves in 3D space.

⭐ Key Takeaways

The distance between two points in 3D and the midpoint of a segment are direct extensions of 2D formulas, adding a z-component. Direction cosines are the cosines of the angles a line makes with the positive axes, and they satisfy the identity sum of squares equals 1; any scalar multiple of them gives direction ratios. For a line joining two points, the direction ratios are the differences in coordinates, and the direction cosines are these ratios divided by the distance between the points. The graph of an equation in x, y, and z is generally a surface, and the intersection of two surfaces is a curve. Key surfaces include planes (form: ax + by + cz + d = 0) and coordinate planes (where one variable is zero).

🧠 Quick Revision Questions

1. What is the distance between points (1, 2, 3) and (4, 6, 7)?
2. Find the midpoint of the segment joining (5, 0, -2) and (1, 4, 6).
3. If the direction cosines of a line are (1/√3, 1/√3, 1/√3), what are its direction angles?
4. A line through points (2, 1, 3) and (5, 5, 4) has direction ratios of what values?
5. What type of geometric object is represented by the equation x + 2y - z = 5?

📘 Lecture 4 — Polar Co-ordinates

📖 Overview: This lecture introduces alternative coordinate systems beyond the standard Cartesian system, including polar, cylindrical, and spherical coordinates. It explains how to convert between these systems and defines constant surfaces for each, with applications in navigation. Understanding these coordinate systems is essential for solving problems in physics, engineering, and calculus where Cartesian coordinates are not the most convenient choice.

🗂️ Topics Covered

The lecture begins with the introduction of plane polar coordinates as an alternative to Cartesian coordinates for locating points in a plane, providing conversion formulas between the two. It then extends to three-dimensional coordinate systems, covering rectangular, cylindrical, and spherical coordinates. Detailed conversion formulas are given between each pair of these 3D systems. The lecture concludes by describing the constant surfaces (planes, cylinders, spheres, cones) associated with each coordinate system, discusses spherical coordinates in navigation, and introduces domain concepts for functions of several variables with examples.

📝 Lecture Summary

Polar co-ordinates

The position of any point in the plane can be obtained by the two perpendicular lines (x and y axes) known as Cartesian coordinates. Alternatively, position can be represented by “r” and “θ”, where “r” is the distance from a fixed point called the pole (O) and θ is the measure of the angle from the initial ray. A point P is represented as P(r, θ).

🔑 Definition — Plane Polar Coordinates: A coordinate system where a point is located by its distance r from a fixed origin (pole) and the angle θ from a fixed direction (initial ray).

Conversion formula from polar to Cartesian coordinates and vice versa

From the diagram and trigonometric ratios: x = r cos θ, y = r sin θ. Squaring and adding these equations gives: x² + y² = r². Dividing the equations gives: y/x = tanθ. These two equations provide the relation between Plane Polar and Plane Cartesian coordinates.

📐 Formula: x = r cos θ, y = r sin θ → (Polar to Cartesian) 📐 Formula: r² = x² + y², tanθ = y/x → (Cartesian to Polar)

Rectangular co-ordinates for 3d

The position of any point in 3D space can be obtained by three mutually perpendicular lines known as x, y, and z axes, together forming the Rectangular coordinate system.

Cylindrical co-ordinates

Beside the Rectangular coordinate system, another system for locating a point in space is the cylindrical coordinate system. A point is represented by (r, θ, z), where r and θ are polar coordinates of the point's projection onto the xy-plane, and z is the vertical height.

Spherical co-ordinates

Another coordinate system for space is the spherical coordinate system, where a point is represented by (ρ, θ, Φ). Here, ρ is the distance from the origin, θ is the angle in the xy-plane from the positive x-axis, and Φ is the angle from the positive z-axis.

Conversion formulas between rectangular and cylindrical co-ordinates

🔑 Definition — Cylindrical Coordinates (r, θ, z): A 3D coordinate system that extends polar coordinates by adding a height z.

To convert from cylindrical (r, θ, z) to rectangular (x, y, z): 📐 Formula: x = r cos θ, y = r sin θ, z = z

To convert from rectangular (x, y, z) to cylindrical (r, θ, z): 📐 Formula: r = √(x² + y²), tanθ = y/x, z = z

Conversion formulas between cylindrical and spherical co-ordinates

From the right-angled triangles in the figure, the relation between spherical (ρ, θ, Φ) and cylindrical (r, θ, z) is:

📐 Formula: r = ρ sin Φ, θ = θ, z = ρ cos Φ

Conversely, from cylindrical (r, θ, z) to spherical (ρ, θ, Φ):

📐 Formula: ρ = √(r² + z²), θ = θ, tan Φ = r/z

Conversion formulas between rectangular and spherical co-ordinates

To convert from spherical (ρ, θ, Φ) to rectangular (x, y, z): Substituting the cylindrical relationships into the polar-to-Cartesian formulas:

📐 Formula: x = ρ sin Φ cos θ, y = ρ sin Φ sin θ, z = ρ cos Φ

To convert from rectangular (x, y, z) to spherical (ρ, θ, Φ): Given x² + y² + z² = (ρ sin Φ cos θ)² + (ρ sin Φ sin θ)² + (ρ cos Φ)² = ρ²(sin²Φ(cos²θ + sin²θ) + cos²Φ) = ρ²

📐 Formula: ρ = √(x² + y² + z²), Tanθ = y/x, Cos Φ = z / √(x² + y² + z²)

Constant surfaces in rectangular co-ordinates

The surfaces represented by equations of the form x = x₀, y = y₀, z = z₀, where x₀, y₀, z₀ are constants, are planes parallel to the yz-plane, xz-plane, and xy-plane, respectively.

Constant surfaces in cylindrical co-ordinates

• The surface r = r₀ is a right cylinder of radius r₀ centered on the z-axis. At each point (r, θ, z) on this surface, r has the value r₀, z is unrestricted and 0 ≤ θ < 2π.
• The surface θ = θ₀ is a half plane attached along the z-axis and making angle θ₀ with the positive x-axis. At each point (r, θ, z) on the surface, θ has the value θ₀, z is unrestricted and r ≥ 0.
• The surface z = z₀ is a horizontal plane. At each point (r, θ, z) on this surface, z has the value z₀, but r and θ are unrestricted.

🔑 Definition — Constant Surface: A geometric surface defined by holding one coordinate constant in a given coordinate system.

Constant surfaces in spherical co-ordinates

• The surface ρ = ρ₀ consists of all points whose distance ρ from the origin is ρ₀. Assuming ρ₀ is nonnegative, this is a sphere of radius ρ₀ centered at the origin.
• The surface θ = θ₀ is a half plane attached along the z-axis and making angle θ₀ with the positive x-axis.
• The surface Φ = Φ₀ consists of all points from which a line segment to the origin makes an angle of Φ₀ with the positive z-axis. Depending on whether 0 < Φ₀ < π/2 or π/2 < Φ₀ < π, this will be a cone opening up or opening down. If Φ₀ = π/2, then the cone is flat and the surface is the xy-plane.

💡 Why this matters: Understanding constant surfaces is crucial for setting up integrals and analyzing physical fields in different coordinate systems, as the limits of integration are defined by these surfaces.

Spherical Co-ordinates in Navigation

Spherical coordinates are related to longitude and latitude used in navigation. Consider a right-handed rectangular coordinate system with origin at Earth’s center, positive z-axis through the North Pole, and x-axis through the prime meridian. Assuming Earth to be a perfect sphere of radius ρ = 4000 miles, each point has spherical coordinates of the form (4000, θ, Φ), where Φ and θ determine the latitude and longitude of the point. Longitude is specified in degrees east or west of the prime meridian, and latitude is specified in degrees north or south of the equator.

Domain of the Function

• The set D is the domain of the function.
• The set of all values which the function assigns for every element of the domain is called the Range of the function.
• When the range consists of real numbers, the functions are called real-valued functions.

NATURAL DOMAIN

The natural domain consists of all points at which the formula has no divisions by zero and produces only real numbers.

📌 Example 1: Consider the function f = √(y - x²). Then the domain of the function is y ≥ x², which can be shown in the plane (the region above the parabola y = x²). The range of the function is [0, ∞).

📌 Example 2: The domain of the function w = 1/xy is the whole xy-plane, excluding the x-axis and y-axis, because at points on the x and y axes, the coordinates x or y are 0, and the defining formula for the function gives 1/0.

⭐ Key Takeaways

A student must remember the conversion formulas between all three 3D coordinate systems (rectangular, cylindrical, and spherical) as they are frequently used in calculus and physics problems. The constant surfaces for each coordinate system provide the geometric insight needed to set up integration limits. The spherical coordinate system is directly applicable to real-world navigation using longitude and latitude. The concept of natural domain is fundamental for working with functions of multiple variables. Finally, the planar polar coordinate system forms the foundation for understanding both cylindrical and spherical coordinates.

🧠 Quick Revision Questions

1. What are the conversion formulas from polar coordinates (r, θ) to Cartesian coordinates (x, y)?
2. What geometric surface is represented by the equation r = constant in cylindrical coordinates?
3. Write the formulas to convert from spherical coordinates (ρ, θ, Φ) to cylindrical coordinates (r, θ, z).
4. In spherical coordinates, what geometric shape is represented by Φ = π/2?
5. What is the natural domain of the function f(x, y) = 1/(x - y)?

📘 Lecture 5 — Limit of Multivariable Function

📖 Overview: This lecture extends the concept of limits from single-variable calculus to functions of two and three variables. Understanding limits of multivariable functions is essential for defining continuity and differentiability in higher dimensions, and the lecture emphasizes that a limit exists only if it is the same along every possible path of approach.

🗂️ Topics Covered

The lecture covers domains and ranges of multivariable functions, including examples with square roots, logarithms, and exponentials. It then introduces the concept of limits for functions of two variables, demonstrating existence and non-existence through path-dependent behavior. Key examples show that a limit may exist even when the function is undefined at a point, and that a limit can fail because different paths yield different values. The lecture also presents the formal rule for non-existence of a limit and the algebraic rules for limits of multivariable functions.

📝 Lecture Summary

Domains and Ranges

The domain of a multivariable function is the set of all input points where the function is defined and yields a real number. The range is the set of all possible output values. The lecture provides several examples:

For ω = √(x² + y² + z²), the domain is the entire three-dimensional space, and the range is [0, ∞).

For ω = 1/(x² + y² + z²), the domain excludes the origin (0, 0, 0), and the range is (0, ∞).

For ω = xy ln z, the domain requires z > 0 (half-space), and the range is (-∞, ∞).

Examples of domain of a function:

For f(x, y) = √(y-1), the domain consists of all points in the xy-plane where y ≥ 1.

For f(x, y) = √(x² + y² - 4), the domain consists of all points in the xy-plane where x² + y² ≥ 4 (the region outside and on a circle of radius 2 centered at the origin).

For f(x, y) = ln(xy), the domain consists of the first and third quadrants of the xy-plane (where xy > 0).

For f(x, y, z) = e^(xyz), the domain is the entire three-dimensional space.

For f(x, y) = √(4 - x²) / (y² + 3), the domain consists of all points in the xy-plane where x² ≤ 4 (i.e., -2 ≤ x ≤ 2).

For f(x, y, z) = √(25 - x² - y² - z²), the domain is the region in three-dimensional space occupied by a sphere centered at (0, 0, 0) with radius 5.

For f(x, y) = sin⁻¹(x + y), the domain is the region in the xy-plane where -1 ≤ x + y ≤ 1.

Limit of Multivariable Function

The concept of a limit for a function of two variables, lim_{(x,y)→(a,b)} f(x, y) = L, means that as the point (x, y) gets arbitrarily close to (a, b) (from any direction in the xy-plane), the value of f(x, y) gets arbitrarily close to L. The lecture emphasizes that a function can have a limit even if the function is not defined at the point itself.

Example: Consider f(x, y) = (x³ + 2x²y - xy - 2y²) / (x + 2y). The function is not defined at (0,0), but the lecture shows through tables that the limit appears to exist. Approaching (0,0) along the x-axis, f(x, 0) = x², which tends to 0. Approaching along the y-axis, f(0, y) = -y, which also tends to 0. Approaching along y = x, the values also approach 0. This suggests that lim_{(x,y)→(0,0)} f(x, y) = 0.

Non-Existence of a Limit

A limit does not exist if approaching the point (a, b) along two different paths yields two different limiting values.

Rule for Non-Existence of a Limit: If in lim_{(x,y)→(a,b)} f(x, y) we get two or more different values as we approach (a, b) along different paths, then lim_{(x,y)→(a,b)} f(x, y) does not exist. The paths may be straight lines or plane curves through (a, b).

Example: Consider f(x, y) = xy / (x² + y²). The function is not defined at (0,0). Approaching (0,0) along the x-axis (y=0) gives f(x, 0) = 0, so along this path the limit is 0. Approaching (0,0) along the line y = x gives f(x, x) = x² / (2x²) = 1/2, so along this path the limit is 0.5. Since two different paths give different limits, we conclude that lim_{(x,y)→(0,0)} f(x, y) does not exist.

Another example: lim_{(x,y)→(0,0)} xy / (x² + y²). The same conclusion is reached: the limit does not exist.

Solving Limits Using Polar Coordinates

A useful technique for evaluating limits as (x, y) → (0, 0) is to convert to polar coordinates: x = r cos θ and y = r sin θ. Then, as (x, y) → (0, 0), r → 0.

Example: Evaluate lim_{(x,y)→(0,0)} xy / √(x² + y²).

Substituting x = r cos θ, y = r sin θ:

xy / √(x² + y²) = (r cos θ * r sin θ) / √(r² cos²θ + r² sin²θ) = (r² cos θ sin θ) / r = r cos θ sin θ, for r > 0.

Since r → 0 as (x, y) → (0, 0) and |cos θ sin θ| ≤ 1 for all values of θ, the limit is:

lim_{(x,y)→(0,0)} xy / √(x² + y²) = lim_{r → 0} r cos θ sin θ = 0.

Thus, the limit exists and is equal to 0.

Rules for Limits

If lim_{(x,y)→(x₀, y₀)} f(x, y) = L₁ and lim_{(x,y)→(x₀, y₀)} g(x, y) = L₂, then the following algebraic rules hold:

(a) lim_{(x,y)→(x₀, y₀)} c f(x, y) = c L₁ (if c is a constant)

(b) lim_{(x,y)→(x₀, y₀)} [f(x, y) + g(x, y)] = L₁ + L₂

(c) lim_{(x,y)→(x₀, y₀)} [f(x, y) - g(x, y)] = L₁ - L₂

(d) lim_{(x,y)→(x₀, y₀)} [f(x, y) g(x, y)] = L₁ L₂

(e) lim_{(x,y)→(x₀, y₀)} f(x, y) / g(x, y) = L₁ / L₂ (if L₂ ≠ 0)

Furthermore, lim_{(x,y)→(x₀, y₀)} c = c, lim_{(x,y)→(x₀, y₀)} x = x₀, and lim_{(x,y)→(x₀, y₀)} y = y₀. These rules also apply similarly for functions of three variables.

💡 Why this matters: These limit rules allow us to evaluate limits of complex multivariable functions by breaking them down into simpler components, just as in single-variable calculus.

⭐ Key Takeaways

The most critical concept from this lecture is that for a multivariable limit to exist, the function must approach the same value along every possible path to the point. A single counterexample of two different paths yielding different limits is sufficient to prove non-existence. The polar coordinate method (x = r cos θ, y = r sin θ) is a powerful tool for evaluating limits as (x, y) → (0, 0) and proving existence because it collapses all paths into a single radial distance r. Finally, the algebraic limit laws for sums, products, and quotients extend directly from single-variable to multivariable limits, enabling the evaluation of many limits by direct substitution or simplification.

🧠 Quick Revision Questions

1. What is the domain of the function f(x, y) = ln(4 - x² - y²)?
2. Why does the limit lim_{(x,y)→(0,0)} xy / (x² + y²) not exist?
3. Use polar coordinates to evaluate lim_{(x,y)→(0,0)} (x³ + y³) / (x² + y²).
4. State the Rule for Non-Existence of a Limit for a function of two variables.
5. If lim_{(x,y)→(a,b)} f(x, y) = 3 and lim_{(x,y)→(a,b)} g(x, y) = 0, does lim_{(x,y)→(a,b)} f(x, y) / g(x, y) necessarily not exist? Explain.

📘 Lecture 6 — Geometry of continuous functions

📖 Overview: This lecture introduces the concept of continuity for functions of multiple variables, including the geometric intuition and formal definition. It explains why the three conditions of continuity are necessary by examining functions that fail them, and extends the concept to functions of three variables. The lecture also covers rules for continuous functions and introduces partial derivatives.

🗂️ Topics Covered

The lecture covers the geometry of continuous functions in one and two variables, the formal definition of continuity for functions of two variables with justification of each condition through examples, continuity of functions of three variables with an example, rules for continuous functions including products and compositions, and finally the definition and computation of partial derivatives with multiple examples.

📝 Lecture Summary

Geometry of continuous functions

A function is continuous if we can draw its graph without raising the pen, meaning there is no gap in the graph. For a function of two variables, the graph should be like a thin sheet of clay that has been hollowed and pinched into peaks and valleys without creating tears or pinholes.

Continuity of functions of two variables

A function f of two variables is called continuous at the point (x₀,y₀) if:

1. f(x₀,y₀) is defined.
2. lim_(x,y)→(x₀,y₀) f(x,y) exists.
3. lim_(x,y)→(x₀,y₀) f(x,y) = f(x₀,y₀).

The requirement that f(x₀,y₀) must be defined eliminates the possibility of a hole in the surface z = f(x₀,y₀) above the point (x₀,y₀).

🔑 Definition — Continuous function of two variables: A function f is continuous at (x₀,y₀) if it satisfies all three conditions above.

Justification of three points involving in the definition of continuity

(1) Consider f(x,y) = x² + y² ln(x² + y²). The log function is not defined at 0, so when x=0 and y=0, the function is not defined. Consequently, the surface z = x² + y² ln(x² + y²) will have a hole just above the point (0,0).

(2) Consider f(x,y) = 1/(x² + y²). The natural domain is the whole plane except the origin because at origin we have 1/0 which is infinity. Thus the limit of the function does not exist at origin, and the surface approaches infinity when we approach the origin.

(3) Consider the function: f(x,y) = { 0 if x ≥ 0 and y ≥ 0, 1 otherwise } The function has value "0" for points where both x and y are nonnegative, and value "1" for all other points. Consequently, the surface has a vertical jump above the point (0,0).

💡 Why this matters: Each condition eliminates a specific type of discontinuity — holes, infinities, or jumps — ensuring the surface is smooth and connected.

Example — Check whether the limit exists

For f(x,y) = x²/(x² + y²):

• Along x-axis (y=0): lim = x²/x² = 1
• Along y-axis (x=0): lim = 0/y² = 0
• Along y = x: lim = x²/(x² + x²) = 1/2

Since the limit has different values along different paths, the limit at (0,0) does not exist. Thus the function cannot be continuous at (0,0). It also is not defined at (0,0), so it fails two conditions of continuity.

Example — Check the continuity at (0,0)

f(x,y) = { sin(x² + y²)/(x² + y²) if (x,y) ≠ (0,0), 1 if (x,y) = (0,0) }

The function is defined at (0,0) with value 1. Using the result lim_(u→0) sin(u)/u = 1: lim_(x,y)→(0,0) f(x,y) = lim sin(x² + y²)/(x² + y²) = 1 = f(0,0) This shows f is continuous at (0,0).

Continuity of function of three variables

A function f of three variables is called continuous at a point (x₀,y₀,z₀) if:

1. f(x₀,y₀,z₀) is defined.
2. lim_(x,y,z)→(x₀,y₀,z₀) f(x,y,z) exists.
3. lim_(x,y,z)→(x₀,y₀,z₀) f(x,y,z) = f(x₀,y₀,z₀).

🔑 Definition — Continuous function of three variables: Same three conditions as for two variables, extended to three-dimensional space.

Example

f(x,y,z) = (y+1)/(x² + y² - 1) The function is not defined on the cylinder x² + y² - 1 = 0. Thus the function is not continuous on the cylinder and is continuous at all other points of its domain.

Rules for continuous functions

(a) If g and h are continuous functions of one variable, then f(x,y) = g(x)h(y) is a continuous function of x and y. (b) If g is a continuous function of one variable and h is a continuous function of two variables, then their composition f(x,y) = g(h(x,y)) is a continuous function of x and y. A composition of continuous functions is continuous. A sum, difference, or product of continuous functions is continuous. A quotient of continuous function is continuous, except where the denominator is zero.

Example of product of functions to be continuous

Any function of the form f(x,y) = Axᵐyⁿ (m and n nonnegative integers) is continuous because it is the product of continuous functions Axᵐ and yⁿ. For example, f(x,y) = 3x²y⁵ is continuous because it is the product of g(x) = 3x² and h(y) = y⁵.

Continuous everywhere

A function f that is continuous at each point of a region R in 2D or 3D space is said to be continuous on R. A function continuous at every point in 2D or 3D space is called continuous everywhere or simply continuous.

Examples of continuous functions

(1) f(x,y) = ln(2x - y + 1) is continuous where 2x > y-1, i.e., y < 2x+1. (2) f(x,y) = e^(1-xy) is continuous in the whole xy-plane. (3) f(x,y) = tan⁻¹(y - x) is continuous in the whole xy-plane. (4) f(x,y) = √(y - x) is continuous where x ≥ y.

Partial derivative

Let f be a function of x and y. If we hold y constant (y = y₀) and view x as a variable, then f(x,y₀) is a function of x alone. If this function is differentiable at x = x₀, the value is denoted by fₓ(x₀, y₀) and is called the partial derivative of f with respect to x at (x₀, y₀).

Similarly, holding x constant (x = x₀) and viewing y as a variable gives fᵧ(x₀, y₀), the partial derivative of f with respect to y at (x₀, y₀).

🔑 Definition — Partial derivative: fₓ(x₀,y₀) = derivative of f(x,y₀) with respect to x at x₀; fᵧ(x₀,y₀) = derivative of f(x₀,y) with respect to y at y₀.

Example

f(x,y) = 2x³y² + 2y + 4x Treating y as constant and differentiating with respect to x: fₓ(x,y) = 6x²y² + 4

Treating x as constant and differentiating with respect to y: fᵧ(x,y) = 4x³y + 2

Substituting x=1, y=2: fₓ(1,2) = 6(1)²(2)² + 4 = 28 fᵧ(1,2) = 4(1)³(2) + 2 = 10

Example

z = 4x² - 2y + 7x⁴y⁵ ∂z/∂x = 8x + 28x³y⁵ ∂z/∂y = -2 + 35x⁴y⁴

Example

z = f(x,y) = x² sin²y ∂z/∂x = 2x sin²y (treat y as constant) ∂z/∂y = x²·2 sin y cos y = x² sin 2y (treat x as constant)

Example

z = ln((x² + y²)/(x + y)) Using properties of ln: z = ln(x² + y²) - ln(x + y) ∂z/∂x = (2x)/(x² + y²) - 1/(x + y) = (2x(x+y) - (x² + y²))/((x² + y²)(x+y)) = (x² + 2xy - y²)/((x² + y²)(x+y))

By symmetry: ∂z/∂y = (y² + 2xy - x²)/((x² + y²)(x+y))

Example

z = x⁴ sin(xy³) ∂z/∂x = x⁴·cos(xy³)·y³ + sin(xy³)·4x³ = x⁴y³ cos(xy³) + 4x³ sin(xy³) ∂z/∂y = x⁴·cos(xy³)·3xy² + sin(xy³)·0 = 3x⁵y² cos(xy³)

Example

z = cos(x⁵y⁴) ∂z/∂x = -sin(x⁵y⁴)·5x⁴y⁴ = -5x⁴y⁴ sin(x⁵y⁴) ∂z/∂y = -sin(x⁵y⁴)·4x⁵y³ = -4x⁵y³ sin(x⁵y⁴)

Example

w = x² + 3y² + 4z² - xyz ∂w/∂x = 2x - yz ∂w/∂y = 6y - xz ∂w/∂z = 8z - xy

⭐ Key Takeaways

A function of two variables is continuous at a point if it is defined there, the limit exists, and the limit equals the function value — these conditions prevent holes, infinities, and jumps. A function is discontinuous if limits along different paths give different values, as shown with x²/(x²+y²). Any polynomial, exponential, trigonometric, and logarithmic function is continuous on its domain, and sums, products, quotients, and compositions of continuous functions are continuous where defined. Partial derivatives are computed by treating all other variables as constants and differentiating with respect to the chosen variable, following standard derivative rules including product rule and chain rule.

🧠 Quick Revision Questions

1. What are the three conditions for a function of two variables to be continuous at a point?
2. Why does the function f(x,y) = x²/(x²+y²) fail to have a limit at (0,0)?
3. How do you compute the partial derivative fₓ(x,y) for a function f(x,y)?
4. Is the function f(x,y) = { sin(x²+y²)/(x²+y²) for (x,y)≠(0,0), 1 at (0,0) } continuous at (0,0)? Why?
5. If f(x,y) = x⁴ sin(xy³), what is ∂f/∂y?

📘 Lecture 7 — Geometric meaning of partial derivative

📖 Overview: This lecture explores the geometric interpretation of partial derivatives for functions of two variables. It explains how partial derivatives represent slopes of tangent lines to curves formed by intersecting a surface with planes of constant x or y. The lecture also covers higher-order partial derivatives, mixed partial derivatives, Laplace’s equation, and Euler’s theorem on the equality of mixed partials under continuity conditions.

🗂️ Topics Covered

Geometric meaning of partial derivatives, partial derivatives and their definitions using limits, interpretation of ∂z/∂x as slope of tangent to curve of intersection with y=constant, interpretation of ∂z/∂y as slope of tangent to curve of intersection with x=constant, second-order partial derivatives and their notations, mixed partial derivatives f_xy and f_yx, worked examples computing partial derivatives, Laplace’s equation ∇²f = 0, Euler’s theorem on equality of mixed partials, and the advantage of using Euler’s theorem to simplify computation order.

📝 Lecture Summary

Geometric meaning of partial derivative

Let z = f(x, y) be a function of two variables. The graph of f is a surface in three-dimensional space. Consider a point P on this graph with coordinates (x₀, y₀, f(x₀, y₀)). If a point starts at P and moves along the surface such that y remains constant, the path traced is the curve of intersection between the surface z = f(x, y) and the plane y = constant. On this curve, ∂z/∂x represents the derivative of f with respect to x while y is held constant.

🔑 Definition — ∂z/∂x: The slope of the tangent to the curve of intersection of the surface z = f(x, y) and the plane y = constant at point P.

Similarly, ∂z/∂y is the gradient of the tangent at point P to the curve formed by the intersection of the surface z = f(x, y) and the plane x = constant. These two tangent lines together define the tangent plane to the surface at point P.

Partial Derivatives

Let z = f(x, y) be a function of two variables defined on a domain D. For a given change ∆x in x, keeping y fixed, the change ∆z in z is given by:

∆z = f(x + ∆x, y) – f(x, y)

If the ratio ∆z/∆x = [f(x + ∆x, y) – f(x, y)]/∆x approaches a finite limit as ∆x → 0, then this limit is called the partial derivative of f with respect to x.

Similarly, for a change ∆y in y, keeping x fixed: ∆z = f(x, y + ∆y) – f(x, y)

If the ratio [f(x, y + ∆y) – f(x, y)]/∆y approaches a finite limit as ∆y → 0, this limit is called the partial derivative of f with respect to y.

Partial Derivatives of Higher Orders

The first-order partial derivatives f_x and f_y of a function f of two variables x and y are themselves functions of x and y, and may possess derivatives. These are called second-order partial derivatives.

The four second-order partial derivatives are:

1. ∂/∂x(∂f/∂x) = ∂²f/∂x² = ∂/∂x(f_x) = (f_x)_x = f_xx = f_x²
2. ∂/∂y(∂f/∂x) = ∂²f/∂y∂x = ∂/∂y(f_x) = (f_x)_y = f_xy
3. ∂/∂x(∂f/∂y) = ∂²f/∂x∂y = ∂/∂x(f_y) = (f_y)_x = f_yx
4. ∂/∂y(∂f/∂y) = ∂²f/∂y² = ∂/∂y(f_y) = (f_y)_y = f_yy = f_y²

Thus, there are four second-order partial derivatives for a function z = f(x, y). The partial derivatives f_xy and f_yx are called mixed second partials and are not equal in general. Partial derivatives of order more than two can be defined similarly.

🔑 Definition — Mixed partial derivatives: Derivatives where differentiation is performed with respect to different variables, such as f_xy (first differentiate with respect to x, then y) and f_yx (first differentiate with respect to y, then x).

Example

Given z = arc sin(x/y), compute partial derivatives.

📐 Formula: ∂z/∂x = 1/√(y² - x²)

📐 Formula: ∂z/∂y = -x/(y√(y² - x²))

📌 Example: For the mixed partial ∂²z/∂y∂x: ∂²z/∂y∂x = ∂/∂y(∂z/∂x) = -1/2(y² - x²)^(-3/2)·2y = -y/(y² - x²)^(3/2)

For ∂²z/∂x∂y: ∂²z/∂x∂y = ∂/∂x(∂z/∂y) = -1 · [√(y² - x²) - x·(1/2)(y² - x²)^(-1/2)·(-2x)]/(y² - x²) = -y/(y² - x²)^(3/2)

Hence, ∂²z/∂x∂y = ∂²z/∂y∂x in this case.

Example

Given f(x, y) = x cos y + y eˣ

📌 Example: Compute partial derivatives: ∂f/∂x = cos y + y eˣ ∂f/∂y = -x sin y + eˣ

∂²f/∂y∂x = ∂/∂y(∂f/∂x) = -sin y + eˣ ∂²f/∂x² = ∂/∂x(∂f/∂x) = y eˣ ∂²f/∂x∂y = ∂/∂x(∂f/∂y) = -sin y + eˣ ∂²f/∂y² = ∂/∂y(∂f/∂y) = -x cos y

Laplace’s Equation

For a function w = f(x, y, z), the equation:

📐 Formula: ∂²f/∂x² + ∂²f/∂y² + ∂²f/∂z² = 0

is called Laplace’s equation. Functions satisfying this equation are called harmonic functions.

Example

Given f(x, y) = eˣ sin y + eʸ cos x

📌 Example: Compute second derivatives: ∂f/∂x = eˣ sin y - eʸ sin x ∂²f/∂x² = eˣ sin y - eʸ cos x

∂f/∂y = eˣ cos y + eʸ cos x ∂²f/∂y² = -eˣ sin y + eʸ cos x

Adding: ∂²f/∂x² + ∂²f/∂y² = (eˣ sin y - eʸ cos x) + (-eˣ sin y + eʸ cos x) = 0

Thus, this function satisfies Laplace’s equation.

💡 Why this matters: Laplace’s equation appears in physics (electrostatics, fluid dynamics, heat conduction) — functions satisfying it describe steady-state phenomena.

Euler’s Theorem

The mixed derivative theorem (Euler’s theorem):

If f(x, y) and its partial derivatives f_x, f_y, f_xy, and f_yx are defined throughout an open region containing a point (a, b) and are all continuous at (a, b), then:

📐 Formula: f_xy(a, b) = f_yx(a, b)

🔑 Definition — Euler’s theorem: Under conditions of continuity, the order of differentiation does not matter for mixed partial derivatives.

Advantage of Euler’s theorem

Given w = xy + eʸ/(y² + 1), the symbol ∂²w/∂x∂y tells us to differentiate first with respect to y and then with respect to x. However, by Euler’s theorem, we can postpone differentiation with respect to y and differentiate first with respect to x for a quicker solution.

📌 Example: ∂w/∂x = y, then ∂²w/∂y∂x = 1

This is simpler than differentiating first with respect to y and then with respect to x.

⭐ Key Takeaways

The partial derivative ∂z/∂x geometrically represents the slope of the tangent line to the curve formed by intersecting the surface with a plane of constant y. Similarly, ∂z/∂y represents the slope of the tangent to the curve formed by intersecting with a plane of constant x. For a function of two variables, there are four second-order partial derivatives: f_xx, f_xy, f_yx, and f_yy. Euler’s theorem states that if all second-order partial derivatives are continuous in an open region, then the mixed partials are equal: f_xy = f_yx. This allows flexibility in the order of differentiation, often simplifying calculations. Laplace’s equation ∂²f/∂x² + ∂²f/∂y² + ∂²f/∂z² = 0 is a fundamental partial differential equation with applications in physics and engineering.

🧠 Quick Revision Questions

1. What is the geometric meaning of ∂z/∂x for a function z = f(x, y)?
2. How many second-order partial derivatives exist for a function of two variables? List them with their notations.
3. State Euler’s theorem (the mixed derivative theorem) and the conditions under which it holds.
4. Verify whether f(x, y) = eˣ sin y + eʸ cos x satisfies Laplace’s equation.
5. For f(x, y) = x cos y + y eˣ, compute f_xy and f_yx and verify if they are equal.

📘 Lecture 8 — More About Euler Theorem Chain Rule

📖 Overview: This lecture explores the concept of changing the order of differentiation for higher-order partial derivatives, demonstrating that the order does not affect the result when the function and its partial derivatives are continuous. It then transitions to the chain rule for functions of one and several variables, showing how to differentiate composite functions systematically.

🗂️ Topics Covered

The lecture covers the invariance of the order of differentiation for higher-order partial derivatives under continuity conditions, with examples illustrating how strategic ordering can simplify complex calculations. It then introduces the chain rule for functions of one variable (w=f(x), x=g(t)) and extends it to functions of two variables (w=f(x,y), x=g(t), y=h(t)). Multiple examples demonstrate both substitution and chain rule methods for computing derivatives.

📝 Lecture Summary

Order of differentiation

In general, the order of differentiation in an nth order partial derivative can be changed without affecting the final result whenever the function and all its partial derivatives of order ≤ n are continuous. For example, if f and its partial order derivatives of the first, second, and third orders are continuous on an open set, then at each point of the set: f_{xyy} = f_{yxy} = f_{yyx}. In another notation: ∂³f/(∂y²∂x) = ∂³f/(∂y∂x∂y) = ∂³f/(∂x∂y²).

For a function f(x,y) = y²x⁴eˣ + 2, if we need ∂⁵f/(∂y³∂x²), differentiating first w.r.t. x then w.r.t. y involves many steps. But differentiating with respect to y first and then x second makes the calculation easier. ∂⁵f/(∂x²∂y³) = 0.

🔑 Definition — Order of differentiation: The sequence in which partial derivatives are taken can be rearranged without changing the result when all relevant partial derivatives are continuous.

📐 Formula: For continuous partial derivatives: f_{xyy} = f_{yxy} = f_{yyx}

📌 Example: f(x,y) = (x+y)/(x−y) fₓ(x,y) = [(x−y)(1) − (x+y)(1)]/(x−y)² = −2y/(x−y)² fᵧ(x,y) = [(x−y)(1) − (x+y)(−1)]/(x−y)² = 2x/(x−y)²

📌 Example: f(x,y) = x³e^(−y) + y³sec(xy) fₓ(x,y) = 3x²e^(−y) + y³sec(xy)tan(xy) fᵧ(x,y) = −x³e^(−y) + 3y²sec(xy)

📌 Example: f(x,y) = x²ye^(xy) fₓ(x,y) = 2xye^(xy) + x²y²e^(xy) = xye^(xy)(2+xy) fₓ(1,1) = (1)(1)e¹[2+1] = 3e

fᵧ(x,y) = x²e^(xy) + x³ye^(xy) = x²e^(xy)(1+xy) fᵧ(1,1) = (1)(1)e¹[1+1] = 2e

📌 Example: f(x,y) = x²cos(xy) fₓ(x,y) = 2xcos(xy) − x²ysin(xy) fₓ(½, π) = 2(½)cos(π/2) − (½)²(π)sin(π/2) = 0 − π/4 = −π/4

fᵧ(x,y) = −x³sin(xy) fᵧ(½, π) = −(½)³sin(π/2) = −1/8

📌 Example: w = (4x − 3y + 2z)⁵ ∂w/∂x = 20(4x−3y+2z)⁴ ∂²w/(∂y∂x) = −240(4x−3y+2z)³ ∂³w/(∂z∂y∂x) = −1440(4x−3y+2z)² ∂⁴w/(∂z²∂y∂x) = −576(4x−3y+2z)

Chain Rule in function of One variable

Given that w = f(x) and x = g(t), we find dw/dt as follows: From w = f(x), we get dw/dx. From x = g(t), we get dx/dt. Then dw/dt = (dw/dx)(dx/dt).

📌 Example: w = x + 4, x = sin t By Substitution: w = sin t + 4, dw/dt = cos t By Chain Rule: dw/dx = 1, dx/dt = cos t, dw/dt = 1·cos t = cos t

Chain rule in function of one variable (extended): If y is a function of u, u is a function of v, v is a function of w, w is a function of z, z is a function of x, then ultimately y is function of x, so by chain rule: dy/dx = (dy/du)(du/dv)(dv/dw)(dw/dz)(dz/dx)

w = f(x,y), x = g(t), y = f(t)

For a function w = f(x,y) where x and y are functions of t, the chain rule gives: dw/dt = (∂w/∂x)(dx/dt) + (∂w/∂y)(dy/dt)

📐 Formula: Chain rule for two variables: dw/dt = (∂w/∂x)(dx/dt) + (∂w/∂y)(dy/dt)

📌 Example BY SUBSTITUTION: w = xy, x = cos t, y = sin t w = cos t sin t = ½ sin 2t dw/dt = ½(cos 2t)(2) = cos 2t

📌 Example BY CHAIN RULE: w = xy, x = cos t, y = sin t ∂w/∂x = y = sin t, ∂w/∂y = x = cos t dx/dt = −sin t, dy/dt = cos t dw/dt = (sin t)(−sin t) + (cos t)(cos t) = −sin²t + cos²t = cos 2t

📌 Example: z = 3x²y³, x = t⁴, y = t² ∂z/∂x = 6xy³, ∂z/∂y = 9x²y² dx/dt = 4t³, dy/dt = 2t dz/dt = (6xy³)(4t³) + (9x²y²)(2t) = 6(t⁴)(t⁶)(4t³) + 9(t⁸)(t⁴)(2t) = 24t¹³ + 18t¹³ = 42t¹³

📌 Example: z = √(1 + x − 2xy⁴), x = ln t, y = t ∂z/∂x = 1/(2√(1+x−2xy⁴)) ∂z/∂y = 1/(2√(1+x−2xy⁴)) · (−8xy³) = −4xy³/√(1+x−2xy⁴) dx/dt = 1/t, dy/dt = 1 dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt) = 1/(2√(1+x−2xy⁴)) + −4xy³/√(1+x−2xy⁴) = 1/(2t)√(1+x−2xy⁴) − 4xy³/√(1+x−2xy⁴) = (1/√(1+x−2xy⁴))[1/(2t) − 4xy³] = (1/√(1+ln t − 2t⁴ln t))[1/(2t) − 4(ln t)t³]

📌 Example: z = ln(2x² + y), x = √t, y = t^(2/3) ∂z/∂x = (1/(2x²+y))·4x = 4x/(2x²+y) ∂z/∂y = 1/(2x²+y) dx/dt = 1/(2√t), dy/dt = (2/3)t^(−1/3)

⭐ Key Takeaways

The order of differentiation for higher-order partial derivatives can be freely interchanged when the function and all its partial derivatives up to that order are continuous, which can significantly simplify complex derivative calculations. The chain rule for functions of one variable, dw/dt = (dw/dx)(dx/dt), extends naturally to functions of multiple variables where each intermediate variable depends on a single independent variable t. For w = f(x,y) with x = g(t) and y = h(t), the chain rule is dw/dt = (∂w/∂x)(dx/dt) + (∂w/∂y)(dy/dt), which can be verified by direct substitution. Strategic choice of differentiation order and the chain rule are powerful techniques for efficiently computing higher-order and composite derivatives.

🧠 Quick Revision Questions

1. Under what condition can the order of partial differentiation be changed without affecting the result?
2. For f(x,y) = x²ye^(xy), what is fₓ(1,1)?
3. What is dw/dt if w = f(x) and x = g(t)?
4. State the chain rule formula for w = f(x,y) where x = g(t) and y = h(t).
5. For w = xy with x = cos t and y = sin t, verify that dw/dt = cos 2t using both substitution and chain rule methods.

📘 Lecture 9 — Examples

📖 Overview: This lecture focuses on applying the chain rule for functions of multiple variables through worked examples. It reviews the chain rule for functions with one intermediate variable, extends it to functions with multiple intermediate variables and multiple independent variables, and demonstrates how to compute mixed partial derivatives using this technique.

🗂️ Topics Covered

The lecture begins by revising an example from Lecture 8 of the chain rule for w = f(x,y,z) with x, y, z as functions of t. It then covers the chain rule for w = f(x) where x = g(r, s), followed by the case of w = f(x,y) where x = g(r, s) and y = h(r, s), and finally w = f(x,y,z) where x = g(r, s), y = h(r, s), z = k(r, s). The lecture concludes with a general statement of the chain rule for functions of many variables and a complex example involving partial derivatives of w = ln(e^r + e^s + e^t + e^u).

📝 Lecture Summary

Revision of Previous Example

The lecture first revises the chain rule for w = f(x,y,z) where x = g(t), y = f(t), z = h(t). The total derivative with respect to t is: dw/dt = (∂w/∂x)(dx/dt) + (∂w/∂y)(dy/dt) + (∂w/∂z)(dz/dt)

📌 Example: w = x² + y + z + 4, x = eᵗ, y = cos t, z = t + 4

• ∂w/∂x = 2x, ∂w/∂y = 1, ∂w/∂z = 1
• dx/dt = eᵗ, dy/dt = -sin t, dz/dt = 1
• dw/dt = (2x)(eᵗ) + (1)(-sin t) + (1)(1) = 2(eᵗ)(eᵗ) - sin t + 1 = 2e²ᵗ - sin t + 1

w = f(x) with x = g(r, s)

For w = f(x) where x = g(r, s), x is an intermediate variable. The partial derivatives are: ∂w/∂r = (dw/dx)(∂x/∂r) ∂w/∂s = (dw/dx)(∂x/∂s)

📌 Example: w = sin x + x², x = 3r + 4s

• dw/dx = cos x + 2x
• ∂x/∂r = 3, ∂x/∂s = 4
• ∂w/∂r = (cos x + 2x)(3) = 3 cos(3r+4s) + 18r + 24s
• ∂w/∂s = (cos x + 2x)(4) = 4 cos(3r+4s) + 24r + 32s

w = f(x,y) with x = g(r,s), y = h(r,s)

For the dependent variable w = f(x,y) where x and y are intermediate variables and r and s are independent variables: ∂w/∂r = (∂w/∂x)(∂x/∂r) + (∂w/∂y)(∂y/∂r) ∂w/∂s = (∂w/∂x)(∂x/∂s) + (∂w/∂y)(∂y/∂s)

w = f(x,y,z) with x = g(r,s), y = h(r,s), z = k(r,s)

Extending to three intermediate variables: ∂w/∂r = (∂w/∂x)(∂x/∂r) + (∂w/∂y)(∂y/∂r) + (∂w/∂z)(∂z/∂r) ∂w/∂s = (∂w/∂x)(∂x/∂s) + (∂w/∂y)(∂y/∂s) + (∂w/∂z)(∂z/∂s)

📌 Example: w = x + 2y + z², x = r/s, y = r² + ln s, z = 2r

• ∂w/∂x = 1, ∂w/∂y = 2, ∂w/∂z = 2z
• ∂x/∂r = 1/s, ∂y/∂r = 2r, ∂z/∂r = 2
• ∂w/∂r = (1)(1/s) + (2)(2r) + (2z)(2) = 1/s + 4r + 4(2r) = 1/s + 12r
• ∂x/∂s = -r/s², ∂y/∂s = 1/s, ∂z/∂s = 0
• ∂w/∂s = (1)(-r/s²) + (2)(1/s) + (2z)(0) = 2/s - r/s²

Remembering the Different Forms of the Chain Rule

The best approach is to draw a tree diagram placing the dependent variable on top, intermediate variables in the middle, and the selected independent variable at the bottom. To find the derivative, start at the dependent variable and read down each branch to the independent variable, calculating and multiplying the derivatives along the branch, then add the products from different branches.

General Chain Rule for Functions of Many Variables: Suppose ω = f(x, y, ..., υ) is a differentiable function of variables x, y, ..., υ (a finite set), and x, y, ..., υ are differentiable functions of p, q, ..., t (another finite set). Then ω is a differentiable function of the variables p through t, and the partial derivatives of ω with respect to these variables are given by: ∂ω/∂p = (∂ω/∂x)(∂x/∂p) + (∂ω/∂y)(∂y/∂p) + ... + (∂ω/∂υ)(∂υ/∂p)

💡 Why this matters: This general form can be remembered as the dot product of two vectors: (∂ω/∂x, ∂ω/∂y, ..., ∂ω/∂υ) and (∂x/∂p, ∂y/∂p, ..., ∂υ/∂p).

Advanced Example: Partial Derivatives of w = ln(e^r + e^s + e^t + e^u)

Given w = ln(e^r + e^s + e^t + e^u), we can rewrite as e^w = e^r + e^s + e^t + e^u.

Taking partial derivatives:

• e^w·∂w/∂r = e^r → ∂w/∂r = e^(r-w)
• e^w·∂w/∂s = e^s → ∂w/∂s = e^(s-w)
• e^w·∂w/∂t = e^t → ∂w/∂t = e^(t-w)
• e^w·∂w/∂u = e^u → ∂w/∂u = e^(u-w)

Now computing mixed partial derivatives:

• ∂²w/∂r∂s: w_rs = e^(r-w)(-w_s) = -e^(r-w)·e^(s-w) = -e^(r+s-2w)
• ∂³w/∂r∂s∂t: w_rst = -e^(r+s-2w)(-2w_t) = 2e^(r+s-2w)·e^(t-w) = 2e^(r+s+t-3w)
• ∂⁴w/∂r∂s∂t∂u: w_rstu = 2e^(r+s+t-3w)(-3w_u) = -6e^(r+s+t-3w)·e^(u-w) = -6e^(r+s+t+u-4w)

⭐ Key Takeaways

The chain rule for functions of many variables requires systematically drawing tree diagrams and tracing all paths from the dependent variable to the independent variable. When a dependent variable w depends on intermediate variables x, y, z, which themselves depend on independent variables r, s, the partial derivative ∂w/∂r equals the sum of (∂w/∂x)(∂x/∂r) + (∂w/∂y)(∂y/∂r) + (∂w/∂z)(∂z/∂r). For mixed partial derivatives, each successive differentiation requires careful application of the chain rule, substituting previously computed partial derivatives. The general form of the chain rule can be remembered as the dot product of the gradient vector of w with respect to intermediate variables and the vector of partial derivatives of intermediate variables with respect to the chosen independent variable.

🧠 Quick Revision Questions

1. For w = f(x,y) where x = g(r,s) and y = h(r,s), write the formula for ∂w/∂r.
2. In the example w = x² + y + z + 4 with x = eᵗ, y = cos t, z = t + 4, what is the value of dw/dt at t = 0?
3. For w = x + 2y + z² with x = r/s, y = r² + ln s, z = 2r, calculate ∂w/∂r and ∂w/∂s.
4. What is the general formula for the chain rule when ω = f(x, y, ..., υ) and x, y, ..., υ are functions of p?
5. If w = ln(e^r + e^s + e^t + e^u), find ∂w/∂r and ∂²w/∂r∂s.

📘 Lecture 10 — Introduction to Vectors

📖 Overview: This lecture introduces the fundamental concept of vectors as quantities possessing both magnitude and direction, distinguishing them from scalars. It covers vector representation, basic operations (addition, scalar multiplication, dot and cross products), and their geometric interpretations. Understanding vectors is essential for describing physical quantities like force and velocity, and they form the foundation for many topics in physics and engineering.

🗂️ Topics Covered

The lecture begins by defining vectors and their representation through directed line segments. It then covers vector properties such as magnitude, unit vectors, and equality. The core operations are explained: vector addition (resultant), scalar multiplication, the scalar (dot) product with its component form and angle calculation, and the vector (cross) product with its determinant formula and geometric interpretation of area. Key special cases for parallel and perpendicular vectors are also examined.

📝 Lecture Summary

Introduction to vectors

Some physical measurements are determined solely by their magnitude, but quantities like force and velocity require both magnitude and direction for a complete description. Quantities with direction and magnitude are usually represented by arrows pointing in the direction of the action, with the length of the arrow representing the magnitude relative to a chosen unit. A vector in the plane is a directed line segment. Vectors are usually denoted by a single boldface letter or a letter with an arrow, e.g., the vector from point A to point B is written as AB.

Magnitude and Unit Vector

The magnitude (or length) of a vector v is the length of the directed line segment representing it. A unit vector is any vector whose magnitude is 1. The unit vector in the direction of v is denoted by v̂ and is given by v̂ = v / |v|.

Addition of Vectors

Vectors are added geometrically by connecting them head-to-tail. The resultant vector is the single vector drawn from the tail of the first vector to the head of the last vector. The resultant r = a + b + c + ...

Equal, Opposite, and Parallel Vectors

• Equal vectors have the same magnitude and direction.
• Opposite vectors have the same magnitude but opposite direction.
• Parallel vectors are scalar multiples of each other: b = λa, where λ is a non-zero scalar.

Vector Representation in Rectangular Components

Any vector can be expressed in terms of its components along the x, y, and z axes using the unit vectors i, j, and k: r = x i + y j + z k.

Addition and Subtraction in Component Form

If a = a₁i + a₂j + a₃k and b = b₁i + b₂j + b₃k, then:

• a + b = (a₁ + b₁)i + (a₂ + b₂)j + (a₃ + b₃)k
• a - b = (a₁ - b₁)i + (a₂ - b₂)j + (a₃ - b₃)k

Key idea: corresponding components (i, j, k) are added or subtracted.

Multiplication of a Vector by a Scalar

Multiplying a vector a by a scalar λ results in a parallel vector λa. The direction remains the same if λ > 0, and is reversed if λ < 0.

Scalar (Dot) Product

The scalar product (dot product) of vectors a and b is the number: a·b = |a| |b| cos θ, where θ is the angle between a and b. It is commutative (a·b = b·a).

🔑 Definition — Commutative Law (for dot product): a·b = b·a. This means the order of the vectors in a dot product does not change the result.

Results of Scalar Product:

1. a ⟂ b (perpendicular): If a is perpendicular to b, then a·b = 0. Similarly, i·j = j·k = k·i = 0.
2. a ∥ b (parallel): If a is parallel to b, then a·b = |a||b|. The dot product of a vector with itself is a·a = |a|², so i·i = j·j = k·k = 1.

Dot Product in Component Form

If a = a₁i + a₂j + a₃k and b = b₁i + b₂j + b₃k, then their dot product is calculated by multiplying corresponding components and summing: a·b = a₁b₁ + a₂b₂ + a₃b₃.

💡 Why this matters: This component form allows for the algebraic calculation of the scalar product without needing to know the angle between the vectors.

Angle Between Two Vectors

The angle θ between two non-zero vectors a and b is given by: θ = cos⁻¹( a·b / (|a||b|) ). The value of θ lies in the interval [0, π].

🔑 Formula: θ = cos⁻¹( a·b / (|a||b|) ) → The angle is the inverse cosine of the dot product divided by the product of the magnitudes.

📌 Example: Find the angle between a = i - 2j - 2k and b = 6i + 3j + 2k. Step 1: Calculate a·b = (1)(6) + (-2)(3) + (-2)(2) = 6 - 6 - 4 = -4. Step 2: Calculate |a| = √(1² + (-2)² + (-2)²) = √9 = 3. Step 3: Calculate |b| = √(6² + 3² + 2²) = √49 = 7. Step 4: θ = cos⁻¹(-4 / (3*7)) = cos⁻¹(-4/21) ≈ 1.76 rad.

Perpendicular (Orthogonal) Vectors

Vectors a and b are perpendicular if and only if a·b = 0. This is an "if and only if" relationship, meaning it works both ways.

Vector Projection

The vector projection of b onto a is given by: proj_a b = ( (b·a) / (a·a) ) a. The scalar component of b in the direction of a is |b| cos θ = b·(a/|a|).

📌 Example: Find the vector projection of b = 6i + 3j + 2k onto a = 1i - 2j - 2k. b·a = -4 (from previous example). a·a = 1² + (-2)² + (-2)² = 9. proj_a b = (-4/9)(1i - 2j - 2k) = (-4/9)i + (8/9)j + (8/9)k. The scalar component of b in the direction of a is -4/3.

Cross Product of Two Vectors

The vector product (cross product) of two non-zero vectors a and b is: a × b = (|a||b| sin θ) n, where n is a unit vector perpendicular to both a and b, determined by the right-hand rule.

🔑 Definition — Right-hand rule: To find the direction of a × b, point your fingers from a towards b through the angle θ, and your right thumb points in the direction of n (and thus a × b).

Results of Cross Product:

• If a ∥ b, then a × b = 0. So a × a = 0 and i × i = j × j = k × k = 0.
• If a ⟂ b, then a × b = |a||b| n.
• i × j = k, j × i = -k ; j × k = i, k × j = -i ; k × i = j, i × k = -j. Note: the cross product is not commutative.

Cross Product as Area

The magnitude of the cross product |a × b| is equal to the area of the parallelogram determined by a and b, where |a| is the base and |b||sinθ| is the height.

Cross Product in Component Form

The cross product can be computed using a determinant.

📐 Formula: a × b = det| i j k; a₁ a₂ a₃; b₁ b₂ b₃ | → This determinant expansion gives the components of the cross product.

📌 Example: Find the cross product a × b for a = 2i + 1j + 1k and b = -4i + 3j + 1k. Step 1: Set up the determinant: | i j k | | 2 1 1 | | -4 3 1 | Step 2: Calculate: i(11 - 13) - j(21 - 1(-4)) + k(23 - 1(-4)) = i(1 - 3) - j(2 + 4) + k(6 + 4) = -2i - 6j + 10k.

⭐ Key Takeaways

The most critical concepts from this lecture are the ability to distinguish between scalars and vectors, and to perform basic vector algebraic operations. You must know the formulas for both the dot and cross products, including their component forms and geometric interpretations (dot product gives cosine of the angle, cross product gives sine and direction). Remember that a zero dot product implies perpendicular vectors, while a zero cross product implies parallel vectors. Finally, be prepared to calculate vector projections and the area of a parallelogram using the cross product magnitude.

🧠 Quick Revision Questions

1. What is the formula for the unit vector in the direction of v?
2. Given two vectors in component form, how do you find their sum and difference?
3. Write the formula for the scalar (dot) product in terms of magnitudes and the angle between vectors.
4. What is the condition for two non-zero vectors to be perpendicular?
5. Using determinants, how do you compute the cross product of a = a₁i + a₂j + a₃k and b = b₁i + b₂j + b₃k?

📘 Lecture 11 — The Triple Scalar or Box Product

📖 Overview: This lecture introduces the triple scalar product (box product) of three vectors and its geometric interpretation as the volume of a parallelepiped. It then transitions to the concept of the gradient of a scalar function and the directional derivative, explaining how the gradient gives the direction of most rapid change. These concepts are fundamental in vector calculus and have crucial applications in physics, engineering, and optimization.

🗂️ Topics Covered

The lecture first covers the triple scalar product, also known as the box product, and derives its formula as the determinant of a 3x3 matrix formed by the components of the three vectors. It then introduces the gradient of a scalar function, the del operator, and the directional derivative, including its formula in terms of partial derivatives and its compact form using the gradient. The lecture concludes with the properties of directional derivatives, including the directions of most rapid increase, decrease, and zero change.

📝 Lecture Summary

The triple scalar or Box product

The product (a × b) . c is called the triple scalar product of a, b, and c (in that order). As | (a×b).c | = |a×b| |c| |cos θ|, the absolute value of the product is the volume of the parallelepiped (parallelogram-sided box) determined by a, b, and c. By treating the planes of b and c and of c and a as the base planes of the parallelepiped determined by a, b and c, we see that (a × b).c = (b×c).a = (c×a).b. Since the dot product is commutative, (a × b).c = a.(b×c).

For vectors a = a₁i + a₂j + a₃k, b = b₁i + b₂j + b₃k, and c = c₁i + c₂j + c₃k, the triple scalar product can be expressed as the determinant of a matrix: a . (b × c) = | a₁ a₂ a₃; b₁ b₂ b₃; c₁ c₂ c₃ |

🔑 Definition — Triple Scalar Product (Box Product): (a×b).c is the scalar product of the cross product of a and b with c.

📐 Formula: a . (b × c) = determinant | a₁ a₂ a₃; b₁ b₂ b₃; c₁ c₂ c₃ |

📌 Example: Find a.(b×c) for a = i + 2j – k, b = – 2i + 3k, c = 7j – 4k. a.(b×c) = | 1 2 -1; -2 0 3; 0 7 -4 | = 1 |0 3; 7 -4| – 2 |-2 3; 0 -4| + (-1) |-2 0; 0 7| = 1(0·(-4) – 3·7) – 2((-2)·(-4) – 3·0) + (-1)((-2)·7 – 0·0) = 1(0 – 21) – 2(8 – 0) + (-1)(-14 – 0) = -21 – 16 + 14 = -23 The volume is |a.(b×c)| = |-23| = 23. If we get a negative value, we take the Absolute value to make it positive, as volume is always positive.

💡 Why this matters: The triple scalar product provides a powerful computational tool to find the volume of a parallelepiped and can be used to test if three vectors are coplanar (volume = 0).

Gradient of a Scalar Function

∇ ≡ i ∂/∂x + j ∂/∂y + k ∂/∂z, ∇ is called the "del operator". The gradient of a scalar function φ is a vector operator defined as grad φ = (i ∂φ/∂x + j ∂/∂y + k ∂/∂z) φ = ∇φ. ∇φ is a vector quantity, and φ is a scalar quantity. Every component of ∇φ will operate with the function.

🔑 Definition — Gradient: The gradient of a scalar function φ is the vector operator ∇φ, which points in the direction of the greatest rate of increase of φ.

📝 Note: Gradient is a vector quantity.

Directional Derivative

If f(x,y) is differentiable at (x₀, y₀), and if u = (u₁, u₂) is a unit vector, then the directional derivative of f at (x₀, y₀) in the direction of u is defined by D_u f(x₀, y₀) = f_x(x₀, y₀)u₁ + f_y(x₀, y₀)u₂. It should be kept in mind that there are infinitely many directional derivatives of z = f(x,y) at a point (x₀, y₀), one for each possible choice of the direction vector u.

💡 Remarks (Geometrical interpretation): The directional derivative D_u f(x₀, y₀) can be interpreted algebraically as the instantaneous rate of change in the direction of u at (x₀, y₀) of z=f(x,y) with respect to the distance parameter s, or geometrically as the rise over the run of the tangent line to the curve C at the point Q₀.

🔑 Definition — Directional Derivative: The directional derivative measures the rate of change of a function in a specified direction.

📌 Example: Find the directional derivative of f(x, y) = 3x²y at the point (1, 2) in the direction of the vector a = 3i + 4j. f_x(x,y) = 6xy, f_y(x,y) = 3x², so f_x(1,2) = 12, f_y(1,2) = 3. The unit vector û = a/||a|| = (1/5)(3i + 4j) = (3/5)i + (4/5)j. D_u f(1,2) = 12(3/5) + 3(4/5) = 36/5 + 12/5 = 48/5.

📌 Example: Find the directional derivative of f(x, y) = 2x² + y² at the point P₀(-1, 1) in the direction of u = 3i – 4j. |u| = √(3² + (-4)²) = 5. Unit vector û = (3/5)i – (4/5)j. f_x = 4x, f_x(-1, 1) = -4. f_y = 2y, f_y(-1, 1) = 2. D_u f(-1,1) = f_x(-1,1)u₁ + f_y(-1,1)u₂ = (-4)(3/5) + (2)(-4/5) = -12/5 – 8/5 = -20/5 = -4.

💡 Remarks: If u = u₁i + u₂j is a unit vector making an angle θ with the positive x-axis, then u₁ = cos θ and u₂ = sin θ. The formula D_u f(x₀, y₀) = f_x(x₀, y₀)u₁ + f_y(x₀, y₀)u₂ can be written as D_u f(x₀, y₀) = f_x(x₀, y₀) cos θ + f_y(x₀, y₀) sin θ.

📝 Note: Formula for the directional derivative can be written in the following compact form using the gradient notation: D_u f(x, y) = ∇f(x, y) · û. The dot product of the gradient of f with a unit vector û produces the directional derivative of f in that direction.

📌 Example: Find the directional derivative of f(x,y) = e^(xy) at (-2, 0) in the direction of the unit vector u that makes an angle of π/3 with the positive x-axis. f_x(x,y) = y e^(xy), f_y(x,y) = x e^(xy). f_x(-2, 0) = 0, f_y(-2, 0) = -2. D_u f(-2, 0) = f_x(-2, 0) cos(π/3) + f_y(-2, 0) sin(π/3) = 0(1/2) + (-2)(√3/2) = -√3.

Gradient of function

If f is a function of x and y, then the gradient of f is defined as ∇f (x,y) = f_x(x,y)i + f_y(x,y)j. The directional derivative is D_u f(x, y) = ∇f (x, y) · û.

📌 Example: Find the directional derivative of f(x, y) = 2xy – 3y² at the point P₀(5, 5) in the direction of u = 4i + 3j. |u| = √(4² + 3²) = 5. Unit vector û = (4/5)i + (3/5)j. f_x = 2y, f_y = 2x – 6y. f_x(5,5) = 10, f_y(5,5) = 10 – 30 = -20. ∇f = 10i – 20j. D_u f(5,5) = ∇f · û = (10i – 20j) · ((4/5)i + (3/5)j) = 10(4/5) + (-20)(3/5) = 8 – 12 = -4.

📌 Example: Find the directional derivative of f(x,y) = xe^y + cos(xy) at the point (2, 0) in the direction of a = 3i – 4j. û = (3/5)i – (4/5)j. f_x(x,y) = e^y – y sin(xy), f_y(x,y) = xe^y – x sin(xy). f_x(2,0) = e⁰ – 0 = 1, f_y(2,0) = 2e⁰ – 2(0) = 2. ∇f(2,0) = i + 2j. D_u f(2,0) = (i + 2j) · ((3/5)i – (4/5)j) = 1(3/5) + 2(-4/5) = 3/5 – 8/5 = -5/5 = -1.

Properties of Directional Derivatives

D_u f = ∇f · û = |∇f| cos θ

1. The function f increases most rapidly when cos θ = 1, or when û is the direction of ∇f. That is, at each point P in its domain, f increases most rapidly in the direction of the gradient vector ∇f at P. The derivative in this direction is D_u f = |∇f| cos(0) = |∇f|.
2. Similarly, f decreases most rapidly in the direction of –∇f. The derivative in this direction is D_u f = |∇f| cos(π) = – |∇f|.
3. Any direction û orthogonal to the gradient is a direction of zero change in f, because cos θ then equals π/2 and D_u f = |∇f| cos(π/2) = |∇f|·0 = 0.

📌 Example: For f(x, y) = x²/2 + y²/2 at (1,1): a) The direction of rapid change: The function increases most rapidly in the direction of ∇f at (1,1). ∇f(1,1) = (xi + yj)(1,1) = i + j. Its direction is û = (i + j)/|i + j| = (i + j)/√(1² + 1²) = (1/√2)i + (1/√2)j. b) The directions of zero change: The directions of zero change at (1,1) are the directions orthogonal to ∇f. n = (-i + j)/√2 and –n = (i – j)/√2.

⭐ Key Takeaways

The triple scalar product a·(b×c) gives the signed volume of a parallelepiped, and its absolute value is the volume. It can be computed as the determinant of the matrix formed by the components of the three vectors. The gradient ∇f is a vector operator that points in the direction of the most rapid increase of a scalar function f at a given point, and its magnitude |∇f| is the rate of that increase. The directional derivative D_u f measures the rate of change of f in the direction of a unit vector û and can be computed as the dot product of the gradient and the unit direction vector. The directional derivative is maximized when û is parallel to ∇f (direction of steepest ascent), minimized when opposite to ∇f (steepest descent), and zero when perpendicular to ∇f.

🧠 Quick Revision Questions

1. How do you compute the volume of the parallelepiped formed by three vectors a, b, and c?
2. What is the relationship between the triple scalar products (a×b)·c, (b×c)·a, and (c×a)·b?
3. Write the expression for the directional derivative of f(x,y) at (x₀, y₀) in the direction of a unit vector u = (u₁, u₂).
4. If ∇f at a point is the zero vector, what does this imply about the directional derivatives in all directions?
5. In which direction does a function f increase most rapidly at a given point, and what is the value of the maximum directional derivative?

📘 Lecture 12 — Tangent planes to the surfaces

📖 Overview: This lecture introduces the concept of tangent planes and normal lines to surfaces in three-dimensional space. It builds on the definition of tangent lines to curves and extends it to surfaces, covering the necessary prerequisite knowledge of lines and planes before deriving the equations for tangent planes and normal lines using gradients.

🗂️ Topics Covered

The lecture begins with definitions of tangent planes and surface normals, then reviews different forms of equations of straight lines in two and three dimensions, parametric equations, and implicit vs. explicit forms of curves. It covers the equation of a plane in point-normal form, the relationship between gradients and tangents to surfaces, and finally derives the equations for the tangent plane and normal line to a surface defined by f(x,y,z)=0, with worked examples.

📝 Lecture Summary

Tangent planes to the surfaces / Normal line to the surfaces

If C is a smooth parametric curve in three dimensions, then the tangent line to C at point P0 is the line through P0 along the unit tangent vector to C at P0. The concept of a tangent plane builds on this definition. If P0(x0,y0,z0) is a point on the surface S, and if the tangent lines at P0 to all smooth curves that pass through P0 and lie on S all lie in a common plane, then that plane is regarded as the tangent plane to the surface S at P0. Its normal (the straight line through P0 perpendicular to the tangent) is called the surface normal of S at P0.

Different forms of equation of straight line in two dimensional space

1. Slope intercept form: y = mx + c, where m is the slope and c is the y-intercept.
2. Point-Slope Form: If m is the slope and P0(x0,y0) is a point on the line, then y – y0 = m (x – x0). Alternatively, the rise/run relationship gives: (y – y0)/b = (x – x0)/a, leading to y – y0 = (b/a)(x – x0).
3. General Equation: Ax + By + C = 0, where m = -A/B.
4. Parametric equation of a line: In two dimensions, the parametric equations passing through (x0,y0) and parallel to vector ai + bj are: x = x0 + at, y = y0 + bt. Eliminating t gives (x – x0)/a = (y – y0)/b, or y – y0 = (b/a)(x – x0). The parametric vector form is r(t) = (x0 + at)i + (y0 + bt)j.

Equation of line in three dimensional

Parametric equations of a line in three-dimensional space passing through (x0,y0,z0) and parallel to vector ai + bj + ck are: x = x0 + at, y = y0 + bt, z = z0 + ct. Eliminating t gives (x – x0)/a = (y – y0)/b = (z – z0)/c.

📌 Example: Find parametric equations for the straight line through point A(2,4,3) parallel to vector v = 4i + 0j – 7k. Here x0=2, y0=4, z0=3 and a=4, b=0, c=-7. The required parametric equations are: x = 2 + 4t, y = 4 + 0t, z = 3 – 7t.

Different forms of equations of curves

Curves in the plane are defined in different ways:

• Explicit form: y = f(x). Example: y = √(9 – x²), -3 ≤ x ≤ 3.
• Implicit form: F(x,y) = 0. Example: x² + y² = 9, -3 ≤ x ≤ 3, 0 ≤ y ≤ 3.
• Parametric form: x = f(t), y = g(t). Example: x = 3cosθ, y = 3sinθ, 0 ≤ θ ≤ π. This satisfies x² + y² = 9cos²θ + 9sin²θ = 9.
• Parametric vector form: r(t) = f(t)i + g(t)j, a ≤ t ≤ b. Example: r(θ) = 3cosθ i + 3sinθ j, 0 ≤ θ ≤ π.

Equation of a plane

A plane can be completely determined if we know one point on it and the direction of the perpendicular (normal) to it. Let a plane pass through point P0(x0,y0,z0) with normal vector n = ai + bj + ck. Let P(x,y,z) be any point on the plane. Then vector P0P = (x – x0)i + (y – y0)j + (z – z0)k lies in the plane, and n is perpendicular to P0P. Since n ⊥ P0P, the dot product is zero: n · P0P = 0. This gives the point-normal form of the equation of a plane: a(x – x0) + b(y – y0) + c(z – z0) = 0. This can be rewritten as ax + by + cz + d = 0, where d = –ax0 – by0 – cz0.

📌 Example: Find an equation of the plane passing through (3,-1,7) perpendicular to vector n = 4i + 2j – 5k. Using point-normal form: 4(x – 3) + 2(y + 1) – 5(z – 7) = 0. Simplifying: 4x – 12 + 2y + 2 – 5z + 35 = 0 → 4x + 2y – 5z + 25 = 0.

🔑 Definition — Relationship between gradient and plane equation: For the general equation of a straight line ax + by + c = 0, if φ(x,y) = ax + by, then φx = a, φy = b, and ∇φ = ai + bj = n. The gradient vector ∇φ is always normal to the plane. Similarly, for a plane ax + by + cz + d = 0, if φ = ax + by + cz, then φx = a, φy = b, φz = c, and ∇φ = ai + bj + ck, which is always normal to the plane.

Gradients and Tangents to Surfaces

For a differentiable function f(x,y) having a constant value c along a smooth curve, the curve has parametric equation x = g(t), y = h(t), r = g(t)i + h(t)j. Differentiating f(g(t), h(t)) = c with respect to t using the chain rule: (∂f/∂x)(dg/dt) + (∂f/∂y)(dh/dt) = 0. This gives: (∂f/∂x i + ∂f/∂y j) · (dg/dt i + dh/dt j) = 0, i.e., ∇f · (dr/dt) = 0. Therefore, ∇f is normal to the tangent vector dr/dt, so it is normal to the curve through (x0,y0).

💡 Why this matters: This relationship between the gradient and the tangent vector is the key to finding tangent planes.

Tangent Plane and Normal Line

Consider all curves through point P0(x0,y0,z0) on a surface f(x,y,z) = 0. The plane containing all tangents to these curves at P0 is called the tangent plane to the surface at P0. The straight line perpendicular to all these tangent lines at P0 is called the normal line to the surface at P0. If fx, fy, fz are all continuous at P0 and not all zero, then the gradient ∇f = fxi + fyj + fzk at P0 gives the direction of the normal vector to the surface at P0.

🔑 Definition — Equation of Tangent Plane: If P0(x0,y0,z0) is any point on the surface f(x,y,z) = 0, and if f(x,y,z) is differentiable at P0, then the tangent plane at P0 has the equation: fx(x0,y0,z0)(x – x0) + fy(x0,y0,z0)(y – y0) + fz(x0,y0,z0)(z – z0) = 0.

📌 Example 1: Find the tangent plane to 9x² + 4y² – z² = 36 at P(2,3,6). f(x,y,z) = 9x² + 4y² – z² – 36 fx = 18x, fy = 8y, fz = –2z At P: fx(P) = 36, fy(P) = 24, fz(P) = –12 Equation: 36(x – 2) + 24(y – 3) – 12(z – 6) = 0 Dividing by 12: 3x – 6 + 2y – 6 – z + 6 = 0 → 3x + 2y – z – 6 = 0

📌 Example 2: Find the tangent plane to z = x cos y – yex at (0,0,0). Rewrite as: x cos y – yex – z = 0, so f(x,y,z) = x cos y – yex – z fx(x,y) = cos y – yex, so fx(0,0,0) = 1 – 0(1) = 1 fy(x,y) = –x sin y – ex, so fy(0,0,0) = 0 – 1 = –1 fz = –1, so fz(0,0,0) = –1 Equation: 1(x – 0) + (–1)(y – 0) + (–1)(z – 0) = 0 → x – y – z = 0

⭐ Key Takeaways

The gradient vector ∇f at a point on a surface f(x,y,z)=0 is always normal to the surface at that point, making it the key to finding both tangent planes and normal lines. The tangent plane equation is derived from the dot product of the gradient (normal vector) with any vector lying in the plane. Parametric and point-normal forms of lines and planes are essential prerequisite concepts. For exam problems, always rewrite the surface equation as f(x,y,z)=0 before computing partial derivatives, and evaluate all partial derivatives at the given point before writing the final plane equation.

🧠 Quick Revision Questions

1. What is the condition for a plane to be considered the tangent plane to a surface at a point P0?
2. Write the point-normal form of the equation of a plane.
3. For a surface f(x,y,z)=0, what does the gradient vector ∇f at a point represent?
4. Find the equation of the tangent plane to z = x² + y² at the point (1,1,2).
5. A line passes through (1,2,3) parallel to the vector 2i - j + 4k. Write its parametric equations.

📘 Lecture 13 — Orthogonal Surface

📖 Overview: This lecture covers the concept of normal lines to surfaces and the condition for orthogonal surfaces. It also introduces the total differential for functions of one and two variables, demonstrating how differentials can be used for approximations and error analysis in real-world problems.

🗂️ Topics Covered

The lecture covers the normal line to a surface at a point, the definition and condition for orthogonal surfaces with examples, the differential of a function of one variable including its distinction from the actual change, and the total differential for functions of two variables used to approximate changes and errors.

📝 Lecture Summary

Normal line

Let P₀(x₀, y₀, z₀) be any point on the surface f(x, y, z) = 0. If f(x, y, z) is differentiable at P₀(x₀, y₀, z₀) then the normal line at the point P₀(x₀, y₀, z₀) has the equation x = x₀ + fₓ(P₀)t, y = y₀ + fᵧ(P₀)t, z = z₀ + f_z(P₀)t. Here fₓ means the function is partially differentiable with respect to x, and fₓ(P₀) means the partial derivative evaluated at point P₀. The same notation applies for fᵧ and f_z.

📌 Example: Find the equation of the tangent plane and normal line of the surface f(x, y, z) = x² + y² + z² - 14 at point P(1, -2, 3).

• fₓ = 2x, fᵧ = 2y, f_z = 2z
• fₓ(P₀) = 2, fᵧ(P₀) = -4, f_z(P₀) = 6
• Tangent plane equation: 2(x - 1) - 4(y + 2) + 6(z - 3) = 0 → x - 2y + 3z - 14 = 0
• Normal line equations: (x - 1)/2 = (y + 2)/(-4) = (z - 3)/6, which simplifies to (x - 1)/1 = (y + 2)/(-2) = (z - 3)/3

📌 Example: Find the tangent plane and normal line for 4x² - y² + 3z² = 10 at P(2, -3, 1).

• f(x, y, z) = 4x² - y² + 3z² - 10
• fₓ = 8x, fᵧ = -2y, f_z = 6z
• fₓ(P) = 16, fᵧ(P) = 6, f_z(P) = 6
• Tangent plane: 16(x - 2) + 6(y + 3) + 6(z - 1) = 0 → 8x + 3y + 3z = 10
• Normal line: (x - 2)/16 = (y + 3)/6 = (z - 1)/6, simplifies to (x - 2)/8 = (y + 3)/3 = (z - 1)/3

📌 Example: Find the tangent plane and normal line for z = (1/2)x⁷y⁻² at point (2, 4, 4).

• f(x, y, z) = (1/2)x⁷y⁻² - z
• fₓ = (7/2)x⁶y⁻², fᵧ = -x⁷y⁻³, f_z = -1
• fₓ(2, 4, 4) = 14, fᵧ(2, 4, 4) = -2, f_z(2, 4, 4) = -1
• Tangent plane: 14(x - 2) + (-2)(y - 4) + (-1)(z - 4) = 0 → 14x - 2y - z - 16 = 0
• Normal line: x = 2 + 14t, y = 4 - 2t, z = 4 - t

ORTHOGONAL SURFACES

Two surfaces are said to be orthogonal at a point of their intersection if their normals at that point are orthogonal. They are said to intersect orthogonally if they are orthogonal at every point common to them.

🔑 Definition — Orthogonal Surfaces: Two surfaces are orthogonal at a point of intersection if their normal lines at that point are perpendicular (orthogonal).

CONDITION FOR ORTHOGONAL SURFACES

Let (x, y, z) be any point of intersection of f(x, y, z) = 0 and g(x, y, z) = 0. Direction ratios of a normal line to f are fₓ, fᵧ, f_z. Direction ratios of a normal line to g are gₓ, gᵧ, g_z. The two normal lines are orthogonal if and only if fₓgₓ + fᵧgᵧ + f_zg_z = 0.

📌 Example: Show whether f(x, y, z) = x² + y² + z - 16 and g(x, y, z) = x² + y² - 63z are orthogonal.

• Adding (1) and (2): x² + y² = 63/4, z = 1/4
• fₓ = 2x, fᵧ = 2y, f_z = 1
• gₓ = 2x, gᵧ = 2y, g_z = -63
• fₓgₓ + fᵧgᵧ + f_zg_z = 4(x² + y²) - 63 = 4(63/4) - 63 = 63 - 63 = 0
• Since they satisfy the condition of orthogonality, they are orthogonal.

Differentials of a functions

For a function y = f(x), the differential is dy = f'(x)dx. The differential of x (dx) is the same as the actual change in x, i.e., dx = Δx, whereas dy is the approximate change in the value of the function, which is different from the actual change Δy in the value of the function.

💡 Why this matters: The differential gives a linear approximation to the change in the function, useful for estimating values near a known point.

If f is differentiable at x₀, then the tangent line to the curve y = f(x) at x₀ is a reasonably good approximation near x₀. The tangent line equation is y = f(x₀) + f'(x₀)(x - x₀).

📌 Example: f(x) = √x, x = 4, dx = Δx = 3, Δy = 3

• Δy = √(x + Δx) - √x = √7 - √4 ≈ 0.65
• If y = √x, dy/dx = 1/(2√x), so dy = (1/(2√x))dx = (1/(2√4))(3) = 3/4 = 0.75

📌 Example: Use differentials to approximate cos 61°.

• Let y = cos x, x = 60°, dx = 61° - 60° = 1° = π/180 radians
• Δy ≈ dy = -sin x dx = -sin 60°(π/180) = -(√3/2)(π/180)
• cos 61° = cos(x + dx) ≈ cos x + dy = cos 60° - (√3/2)(π/180)
• cos 61° ≈ 0.5 - 0.01511 = 0.48489

📌 Example: A box with a square base has its height twice its width. If the width is 8.5 inches with a possible error of ±0.3 inches, find the possible error in volume.

• V = x²h, with h = 2x, so V = 2x³
• dV = 6x²dx
• With x = 8.5, dx = ±0.3: dV = 6(8.5)²(±0.3) = ±130.05
• The possible error in volume is ±130.05 cubic inches.

TOTAL DIFFERENTIAL

If we move from (x₀, y₀) to a nearby point (x₀ + dx, y₀ + dy), the resulting differential in f is the total differential: df = fₓ(x₀, y₀)dx + fᵧ(x₀, y₀)dy. This change in the linearization of f is called the total differential of f.

💡 Why this matters: The total differential allows us to approximate the change in a multivariable function when small changes occur in its variables.

📌 Example: A rectangular plate expands so that length changes from 10 to 10.03 and breadth from 8 to 8.02.

• Area A = xy, dA = Aₓdx + Aᵧdy = ydx + xdy
• x = 10, dx = 0.03, y = 8, dy = 0.02
• dA = 8(0.03) + 10(0.02) = 0.24 + 0.20 = 0.44
• Exact change = 80.4406 - 80 = 0.4406, so the differential approximation is very close.

📌 Example: The volume V = xyz. If z decreases by 2% while x and y increase by 0.75% each, find the approximate change in V.

• dV = Vₓdx + Vᵧdy + V_zdx = yzdx + xzdy + xydz
• dx = (0.75/100)x, dy = (0.75/100)y, dz = (-2/100)z
• dV = (0.75/100)xyz + (0.75/100)xyz - (2/100)xyz = (-0.5/100)xyz = (-0.5/100)V
• This shows a 0.5% decrease in volume.

📌 Example: For triangle area Δ = (1/2)ab sin C, find the approximate error if a = 9.1 (instead of 9), b = 4.08 (instead of 4), C = 30°3' (instead of 30°).

• a = 9, b = 4, C = 30°, da = 0.1, db = 0.08, dC = 3' = (3/60)° = (π/3600) radians
• dΔ = (1/2)b sin C·da + (1/2)a sin C·db + (1/2)ab cos C·dC
• dΔ = (1/2)(4)(sin 30°)(0.1) + (1/2)(9)(sin 30°)(0.08) + (1/2)(9)(4)(cos 30°)(π/3600)
• dΔ = 2(0.5)(0.1) + 4.5(0.5)(0.08) + 18(√3/2)(π/3600) = 0.1 + 0.18 + 0.0136 ≈ 0.293
• Percentage change: (0.293/Δ) × 100 = (0.293/9) × 100 = 3.25%

⭐ Key Takeaways

The normal line to a surface at a point has direction ratios given by the partial derivatives at that point. Two surfaces are orthogonal at intersection points when the dot product of their normal vectors equals zero. The differential of a function of one variable approximates the actual change, with the tangent line providing a linear approximation near a point. For functions of two or more variables, the total differential combines the effects of small changes in all variables, enabling approximation of errors and changes in applications like geometry and physics.

🧠 Quick Revision Questions

1. What is the equation of the normal line to the surface f(x,y,z) = 0 at point P₀(x₀,y₀,z₀)?
2. What is the mathematical condition for two surfaces to be orthogonal at their point of intersection?
3. What is the difference between dy and Δy for a function y = f(x)?
4. Write the formula for the total differential of a function f(x,y) when moving from (x₀,y₀) to (x₀+dx, y₀+dy).
5. In the triangle area example with a 0.5% decrease in volume, what percentage change in volume actually occurred?

📘 Lecture 14 — Extrema of Functions of Two Variables

📖 Overview: This lecture introduces techniques for finding the highest and lowest points on the graph of a function of two variables, which correspond to the largest and smallest values of the function. It defines absolute and relative extrema, introduces the Extreme Value Theorem, and explains how to find critical points. The lecture also covers the concept of a saddle point, where a function has neither a maximum nor a minimum at a critical point.

🗂️ Topics Covered

The lecture covers absolute maximum and minimum definitions for functions of two variables, followed by definitions for relative (local) maximum and minimum. It then presents the Extreme Value Theorem, which guarantees the existence of absolute extrema under specific conditions. The lecture explains how to find extreme values using critical points and partial derivatives, and concludes with the concept of a saddle point, illustrated by a worked example.

📝 Lecture Summary

Extrema of Functions of Two Variables

The graph of many functions forms hills and valleys. The tops of the hills are relative maxima and the bottom of the valleys are called relative minima. Just as the top of a hill on the earth’s terrain need not be the highest point on the earth, a relative maximum need not be the highest point on the entire graph.

Absolute maximum

A function f of two variables on a subset D of ℝ² is said to have an absolute (global) maximum value on D if there is some point (x₀, y₀) of D such that f(x₀, y₀) ≥ f(x, y) for all (x, y) ∈ D. In such a case f(x₀, y₀) is the absolute maximum value of f on D.

Absolute minimum

A function f of two variables on a subset D of ℝ² is said to have an absolute (global) minimum value on D if there is some point (x₀, y₀) of D such that f(x₀, y₀) ≤ f(x, y) for all (x, y) ∈ D. In such a case f(x₀, y₀) is the absolute minimum value of f on D.

Relative extremum and absolute extremum

If f has a relative maximum or a relative minimum at (x₀, y₀), then we say that f has a relative extremum at (x₀, y₀), and if f has an absolute maximum or absolute minimum at (x₀, y₀), then we say that f has an absolute extremum at (x₀, y₀).

Relative (local) maximum

The function f is said to have a relative (local) maximum at some point (x₀, y₀) of its domain D if there exists an open disc K centered at (x₀, y₀) and of radius r, K = {(x, y) ∈ ℝ² : (x − x₀)² + (y − y₀)² < r² }, with K ⊂ D such that f(x₀, y₀) ≥ f(x, y) for all (x, y) ∈ K.

Relative (local) minimum

The function f is said to have a relative (local) minimum at some point (x₀, y₀) of D if there exists an open disc K centred at (x₀, y₀) and of radius r with K ⊂ D such that f(x₀, y₀) ≤ f(x, y) for all (x, y) ∈ K.

Extreme Value Theorem

If f(x, y) is continuous on a closed and bounded set R, then f has both an absolute maximum and an absolute minimum on R.

Remarks: If any of the conditions of the Extreme Value Theorem fail to hold, then there is no guarantee that an absolute maximum or absolute minimum exists on the region R. Thus, a discontinuous function on a closed and bounded set need not have any absolute extrema, and a continuous function on a set that is not closed and bounded also need not have any absolute extrema.

Extreme values or extrema of f

The maximum and minimum values of f are referred to as extreme values or extrema of f. Let a function f of two variables be defined on an open disc K = {(x, y): (x − x₀)² + (y − y₀)² < r² }. Suppose fₓ(x₀, y₀) and fᵧ(x₀, y₀) both exist on K. If f has a relative extrema at (x₀, y₀), then:

🔑 Definition — Critical Point Condition: fₓ(x₀, y₀) = 0 = fᵧ(x₀, y₀).

Saddle Point

A differentiable function f(x, y) has a saddle point (a, b) if in every open disk centered at (a, b) there are domain points (x, y) where f(x, y) > f(a, b) and domain points (x, y) where f(x, y) < f(a, b). The corresponding point (a, b, f(a, b)) on the surface z = f(x, y) is called a saddle point of the surface.

Remarks: Thus, the only points where a function f(x, y) can assume extreme values are critical points and boundary points. As with differentiable functions of a single variable, not every critical point gives rise to a local extremum. A differentiable function of a single variable might have a point of inflection. A differentiable function of two variables might have a saddle point.

💡 Why this matters: The saddle point concept is crucial because it shows that simply finding where the partial derivatives are zero is not enough; you must further test these critical points to determine if they are maxima, minima, or saddle points.

EXAMPLE

Fine the critical points of the given function f(x, y) = x³ + y³ − 3axy, a > 0.

fₓ, fᵧ exist at all points of the domain of f. fₓ = 3x² − 3ay, fᵧ = 3y² − 3ax

For critical points fₓ = fᵧ = 0. Therefore, x² − ay = 0 (1) and ax − y² = 0 (2)

Substituting the value of x from (2) into (1), we have: y² / a − ay = 0 y(y³ − a³) = 0 y = 0, y = a and so x = 0, x = a.

📌 Example: The critical points are (0, 0) and (a, a).

⭐ Key Takeaways

The most critical concepts to remember are the precise definitions of absolute and relative maxima/minima, as these form the basis for identifying extrema. The Extreme Value Theorem provides the necessary conditions for the guaranteed existence of absolute extrema. The condition that at a relative extremum the first partial derivatives must both be zero is a necessary condition for finding critical points. Crucially, one must remember that not every critical point is an extremum; a point where the first partial derivatives are zero could be a saddle point, as demonstrated in the example.

🧠 Quick Revision Questions

1. What is the formal definition of an absolute maximum value of a function f on a set D?
2. State the Extreme Value Theorem for a function of two variables. What are its two necessary conditions for the existence of absolute extrema?
3. If a function f has a relative extremum at a point (x₀, y₀) and its first partial derivatives exist there, what must be true about those derivatives?
4. Define what is meant by a saddle point of a differentiable function f(x, y).
5. For the function f(x, y) = x³ + y³ − 3xy, find the two critical points.

📘 Lecture 15 — Examples

📖 Overview: This lecture applies the concepts of critical points and the Second Partial Derivative Test to classify relative extrema (minima and maxima) and saddle points for functions of several variables. It provides a series of worked examples demonstrating how to find critical points, compute second-order partial derivatives, and use the discriminant D to determine the nature of these points, which is essential for understanding the shape of multivariable functions.

🗂️ Topics Covered

The lecture begins by reviewing functions where partial derivatives do not exist at a critical point, specifically using the cone f(x,y)=√(x²+y²). It then formally introduces The Second Partial Derivative Test with its four cases. The remainder of the lecture is dedicated to applying this test to a variety of functions, including a polynomial (f(x,y)=2x²-4x+xy²-1), an exponential function (f(x,y)=e^-(x²+y²+2x)), a quartic polynomial (f(x,y)=2x⁴+y²-x²-2y), and a function with symmetric terms (f(x,y)=4xy-x⁴-y⁴).

📝 Lecture Summary

Example

The function is defined as f(x, y) = √(x² + y²). The partial derivatives are fx(x, y) = x / √(x² + y²) and fy(x, y) = y / √(x² + y²).

🔑 Definition — Critical Point: A point in the domain of f where all first partial derivatives are zero or where at least one partial derivative does not exist.

The partial derivatives exist at all points of the domain except at the origin (0,0). Thus (0, 0) is a critical point of f. Now fx(x, y) = 0 only if x = 0 and fy(x, y) = 0 only if y = 0. The only critical point is (0,0) and f(0,0)=0. Since f(x, y) ≥ 0 for all (x, y), f(0, 0) = 0 is the absolute minimum value of f.

Example

The function is z = f(x, y) = x² + y² (Paraboloid). fx(x, y) = 2x, fy(x, y) = 2y. Setting fx(x, y) = 0 and fy(x, y) = 0 gives the critical point (0, 0). This is a classic example of a function with a relative minimum.

Example

The function is z = g(x, y) = 1 - x² - y² (Paraboloid). gx(x, y) = -2x, gy(x, y) = -2y. Setting gx(x, y) = 0 and gy(x, y) = 0 gives the critical point (0, 0). This is an example of a function with a relative maximum.

Example

The function is z = h(x, y) = y² - x² (Hyperbolic Paraboloid). hx(x, y) = -2x, hy(x, y) = 2y. Setting hx(x, y) = 0 and hy(x, y) = 0 gives the critical point (0, 0). This function has a saddle point at the origin.

Example

The function is f(x, y) = √(x² + y²). fx = x / √(x² + y²), fy = y / √(x² + y²). The point (0,0) is a critical point of f because the partial derivatives do not both exist. It is evident geometrically that fx(0,0) does not exist because the trace of the cone in the plane y=0 has a corner at the origin.

The fact that fx(0,0) does not exist can also be seen algebraically by noting that fx(0,0) can be interpreted as the derivative with respect to x of the function f(x, 0) = √(x² + 0) = |x| at x = 0. But |x| is not differentiable at x = 0, so fx(0,0) does not exist. Similarly, fy(0,0) does not exist. The function f has a relative minimum at the critical point (0,0).

The Second Partial Derivative Test

Let f be a function of two variables with continuous second order partial derivatives in some circle centered at a critical point (x₀, y₀), and let D = fxx(x₀, y₀) fyy(x₀, y₀) - [fxy(x₀, y₀)]².

(a) If D > 0 and fxx(x₀, y₀) > 0, then f has a relative minimum at (x₀, y₀). (b) If D > 0 and fxx(x₀, y₀) < 0, then f has a relative maximum at (x₀, y₀). (c) If D < 0, then f has a saddle point at (x₀, y₀). (d) If D = 0, then no conclusion can be drawn.

REMARKS: If a function f of two variables has an absolute extremum (either an absolute maximum or an absolute minimum) at an interior point of its domain, then this extremum occurs at a critical point.

Example

The function is f(x, y) = 2x² - 4x + xy² - 1. First partial derivatives: fx(x, y) = 4x - 4 + y², fy(x, y) = 2xy. Second partial derivatives: fxx(x, y) = 4, fyy(x, y) = 2x, fxy(x, y) = fyx(x, y) = 2y. For critical points, set first partials to zero: 4x - 4 + y² = 0 (1) and 2xy = 0 (2). From (2), x = 0 or y = 0. If x = 0, from (1), y = ±2. If y = 0, from (1), x = 1. The critical points are (1,0), (0, 2), (0, -2).

Check the nature of each point:

• At (1,0): fxx(1,0)=4, fyy(1,0)=2, fxy(1,0)=0. D = (4)(2) - (0)² = 8 > 0. Since fxx is positive, f has a relative minimum at (1,0).
• At (0,-2): fxx(0,-2)=4, fyy(0,-2)=0, fxy(0,-2)=-4. D = (4)(0) - (-4)² = -16 < 0. Therefore, f has a saddle point at (0,-2).
• At (0,2): fxx(0,2)=4, fyy(0,2)=0, fxy(0,2)=4. D = (4)(0) - (4)² = -16 < 0. Therefore, f has a saddle point at (0,2).

Example

The function is f(x, y) = e^-(x² + y² + 2x). First partial derivatives: fx(x, y) = -2(x+1)e^-(x²+y²+2x), fy(x, y) = -2y e^-(x²+y²+2x). For critical points, fx=0 implies x+1=0 so x=-1, fy=0 implies y=0. Hence the critical point is (-1, 0).

Second partial derivatives: fxx(x,y) = [(-2x-2)² - 2]e^-(x²+y²+2x), fyy(x,y) = [4y² - 2]e^-(x²+y²+2x), fxy(x,y) = -2y(-2x-2)e^-(x²+y²+2x). At (-1, 0): fxx(-1,0) = -2e, fyy(-1,0) = -2e, fxy(-1,0) = 0. D = (-2e)(-2e) - (0)² = 4e² > 0. With D>0 and fxx<0, this shows that f is maximum at (-1, 0).

Example

The function is f(x, y) = 2x⁴ + y² - x² - 2y. First partials: fx(x, y) = 8x³ - 2x, fy(x, y) = 2y - 2. Second partials: fxx(x, y) = 24x² - 2, fyy(x, y) = 2, fxy(x, y) = 0. For critical points: fx=0 gives 2x(4x² - 1) = 0, so x = 0, 1/2, -1/2. fy=0 gives 2y - 2 = 0, so y = 1. The critical points are (0,1), (1/2, 1), (-1/2, 1).

• At (0,1): fxx(0,1) = -2, fyy(0,1) = 2, fxy(0,1) = 0. D = (-2)(2) - 0 = -4 < 0. This shows that (0, 1) is a saddle point.
• At (1/2, 1): fxx(1/2, 1) = 4, fyy(1/2, 1) = 2, fxy(1/2, 1) = 0. D = (4)(2) - 0 = 8 > 0. Since fxx > 0, f has a relative minimum at (1/2, 1).
• At (-1/2, 1): fxx(-1/2, 1) = 4, fyy(-1/2, 1) = 2, fxy(-1/2, 1) = 0. D = (4)(2) - 0 = 8 > 0. Since fxx > 0, f has a relative minimum at (-1/2, 1).

Example

The function is f(x, y) = 4xy - x⁴ - y⁴. First partials: fx(x, y) = 4y - 4x³, fy(x, y) = 4x - 4y³. For critical points: fx=0 gives 4y - 4x³ = 0 => y = x³ (1). fy=0 gives 4x - 4y³ = 0 => x = y³ (2). Solving (1) and (2), the critical points are (0,0), (1,1), (-1,-1).

Second partials: fxx(x, y) = -12x², fyy(x, y) = -12y², fxy(x, y) = 4.

• At (0,0): fxx(0,0)=0, fyy(0,0)=0, fxy(0,0)=4. D = (0)(0) - (4)² = -16 < 0. This shows that (0,0) is the saddle point.
• At (1,1): fxx(1,1)=-12, fyy(1,1)=-12, fxy(1,1)=4. D = (-12)(-12) - (4)² = 144 - 16 = 128 > 0. Since fxx<0, f has a relative maximum at (1,1).
• At (-1,-1): fxx(-1,-1)=-12, fyy(-1,-1)=-12, fxy(-1,-1)=4. D = (-12)(-12) - (4)² = 144 - 16 = 128 > 0. Since fxx<0, f has a relative maximum at (-1,-1).

⭐ Key Takeaways

The Second Partial Derivative Test is the primary tool for classifying critical points, but it is only valid for points where second-order partial derivatives are continuous. The discriminant D = fxx fyy - (fxy)² determines the nature of the critical point: if D > 0 and fxx > 0, it is a relative minimum; if D > 0 and fxx < 0, it is a relative maximum; if D < 0, it is a saddle point. When D = 0, the test is inconclusive and other methods are needed. A critical point is defined by either the vanishing or the non-existence of first partial derivatives, as seen in the cone example.

🧠 Quick Revision Questions

1. What are the conditions for the Second Partial Derivative Test to conclude a relative minimum?
2. In the function f(x,y)=2x²-4x+xy²-1, why are both (0,2) and (0,-2) classified as saddle points even though they have the same coordinates for x and y?
3. For the function f(x,y)=√(x²+y²), why is (0,0) a critical point despite the first partial derivatives not being zero there?
4. If D = 0 for a critical point of f, can you determine if the point is a maximum, minimum, or saddle point?
5. When analyzing the function f(x,y)=4xy-x⁴-y⁴, how many critical points are found, and how are they classified?

📘 Lecture 16 — Extreme Valued Theorem

📖 Overview: This lecture introduces the Extreme Value Theorem, which guarantees that a continuous function on a closed interval attains both an absolute maximum and an absolute minimum. It provides a systematic procedure for finding these absolute extrema for single-variable functions and extends the method to functions of two variables on closed and bounded regions.

🗂️ Topics Covered

The lecture covers the Extreme Value Theorem for single-variable functions, including its statement and a three-step procedure for finding absolute extrema on closed intervals with worked examples. It then extends these concepts to functions of two variables, presenting a three-step method for finding absolute extrema on closed and bounded regions, followed by detailed worked examples on triangular regions.

📝 Lecture Summary

EXTREME VALUED THEOREM

If the function f is continuous on the closed interval [a, b], then f has an absolute maximum value and an absolute minimum value on [a, b].

Remarks: An absolute extremum of a function on a closed interval must be either a relative extremum or a function value at an endpoint of the interval. Since a necessary condition for a function to have a relative extremum at a point C is that C be a critical point, we may determine the absolute maximum value and the absolute minimum value of a continuous function f on a closed interval [a, b] by the following procedure:

1. Find the critical points of f on [a, b] and the function values at these critical points.
2. Find the values of f(a) and f(b).
3. The largest and the smallest of the above calculated values are the absolute maximum value and the absolute minimum value respectively.

🔑 Definition — Critical Point: A point in the domain where the derivative is zero or does not exist. 📐 Procedure: Evaluate f at critical points and endpoints → Largest value is absolute max, smallest is absolute min.

Example 1

Find the absolute extrema of f(x) = x³ + x² - x + 1 on [-2, 1/2]

Since f is continuous on [-2, 1/2], the extreme value theorem is applicable. f'(x) = 3x² + 2x - 1. f'(x) exists for all real numbers, so critical numbers are where f'(x) = 0. Setting f'(x) = 0: (3x - 1)(x + 1) = 0 → x = -1 and x = 1/3. Both points lie in (-2, 1/2).

📌 Example: Evaluate f at critical points and endpoints: f(-2) = -1, f(-1) = 2, f(1/3) = 22/27, f(1/2) = 7/8 The absolute maximum value of f on [-2, 1/2] is 2 at x = -1. The absolute minimum value is -1 at x = -2.

Example 2

Find the absolute extrema of f(x) = (x - 2)^(2/3) on [1, 5]

Since f is continuous on [1, 5], the extreme value theorem is applicable. f'(x) = 2 / [3(x - 2)^(1/3)] There is no value of x for which f'(x) = 0. However, f'(x) does not exist at x = 2, so 2 is a critical point. The absolute extrema occur either at 2 or at endpoints.

📌 Example: f(1) = 1, f(2) = 0, f(5) = 9^(1/3) ≈ 2.08 Absolute minimum value on [1,5] is 0 at x = 2. Absolute maximum value is 9^(1/3) at x = 5.

Example 3

Find the absolute extrema of h(x) = x^(2/3) on [-2, 3]

h'(x) = (2/3)x^(-1/3) = 2 / (3x^(1/3)) h'(x) has no zeros but is undefined at x = 0, which is a critical point.

📌 Example: h(0) = 0, h(-2) = (-2)^(2/3) = 4^(1/3), h(3) = 3^(2/3) = 9^(1/3) Absolute maximum value is 9^(1/3) at x = 3. Absolute minimum is 0 at x = 0.

How to Find the Absolute Extrema of a Continuous Function f of Two Variables on a Closed and Bounded Region R

Step 1: Find the critical points of f that lie in the interior of R (where f_x = 0 and f_y = 0, or one does not exist).

Step 2: Find all boundary points at which the absolute extrema can occur. This involves treating each boundary segment separately, reducing the function to a single variable.

Step 3: Evaluate f(x,y) at the points obtained in the previous steps. The largest of these values is the absolute maximum and the smallest the absolute minimum.

💡 Why this matters: This extends the single-variable method to two dimensions, where the boundary is more complex and must be analyzed piece by piece.

Example 4

Find the absolute maximum and minimum value of f(x,y) = 2 + 2x + 2y - x² - y² on the triangular plate in the first quadrant bounded by the lines x=0, y=0, y=9-x

For interior points: f_x = 2 - 2x = 0 → x = 1; f_y = 2 - 2y = 0 → y = 1. Critical point: (1,1).

For boundary points – three segments:

1. On OA (y=0): U(x) = f(x,0) = 2 + 2x - x², 0 ≤ x ≤ 9. U'(x) = 2 - 2x = 0 → x = 1. Points: (0,0), (9,0), (1,0).

2. On OB (x=0): V(y) = f(0,y) = 2 + 2y - y², 0 ≤ y ≤ 9. By symmetry, critical at y = 1. Points: (0,0), (0,9), (0,1).

3. On AB (y = 9 - x): W(x) = f(x, 9-x) = 2 + 2x + 2(9-x) - x² - (9-x)² = -61 + 18x - 2x². W'(x) = 18 - 4x = 0 → x = 9/2. Then y = 9 - 9/2 = 9/2. Point: (9/2, 9/2).

📌 Example – Evaluation table:

The absolute maximum is 4 at (1,1). The absolute minimum is -61 at (0,9) and (9,0).

Example 5

Find the absolute maximum and minimum values of f(x,y) = 3xy - 6x - 3y + 7 on the closed triangular region R with vertices (0,0), (3,0), (0,5)

For interior critical points: f_x = 3y - 6 = 0 → y = 2; f_y = 3x - 3 = 0 → x = 1. Critical point: (1,2) inside R.

Boundary analysis:

i) Segment (0,0) to (3,0), y=0: u(x) = f(x,0) = -6x + 7, 0 < x < 3. u'(x) = -6 (no critical points). Endpoints: (0,0) and (3,0).

ii) Segment (0,0) to (0,5), x=0: v(y) = f(0,y) = -3y + 7, 0 < y < 5. v'(y) = -3 (no critical points). Endpoints: (0,0) and (0,5).

iii) Segment (3,0) to (0,5): y = -5/3 x + 5, 0 < x < 3. w(x) = f(x, -5/3 x + 5) = -5x² + 14x - 8, 0 < x < 3. w'(x) = -10x + 14 = 0 → x = 7/5. Then y = -5/3(7/5) + 5 = 8/3. Point: (7/5, 8/3).

📌 Example – Evaluation table:

The absolute maximum value is f(0,0) = 7. The absolute minimum value is f(3,0) = -11.

⭐ Key Takeaways

The Extreme Value Theorem guarantees that continuous functions on closed intervals have both absolute maxima and minima. To find these values, one must evaluate the function at all critical points (where the derivative is zero or undefined) and at the endpoints of the interval; the largest value is the absolute maximum and the smallest is the absolute minimum. For two-variable functions on closed and bounded regions, the same logic applies but requires finding interior critical points and analyzing each boundary segment separately by reducing to a single-variable function. The procedure requires systematically checking all candidate points — including interior critical points, boundary critical points, and vertices — and comparing function values to identify the absolute extrema.

🧠 Quick Revision Questions

1. What does the Extreme Value Theorem guarantee about a continuous function on a closed interval?
2. What is the three-step procedure for finding absolute extrema of a single-variable function on a closed interval?
3. How do you identify critical points for a function of two variables on a closed region?
4. In Example 5, why were there no critical points on the boundary segments y=0 and x=0?
5. What is the absolute maximum and minimum of f(x,y) = 2 + 2x + 2y - x² - y² on the triangular region bounded by x=0, y=0, and y=9-x?

📘 Lecture 17 — Examples

📖 Overview: This lecture focuses on applying optimization techniques for functions of several variables to solve practical problems. It demonstrates how to find absolute extrema on closed bounded regions and how to solve constrained optimization problems by reducing them to functions of two variables, reinforcing the use of second derivative tests.

🗂️ Topics Covered

The lecture covers examples of finding absolute maximum and minimum values of a function on a closed triangular region by analyzing interior critical points and boundary line segments. It then presents several constrained optimization problems, including maximizing the product of three numbers given a fixed sum, minimizing the sum of squares, and optimizing the dimensions of rectangular boxes inscribed in spheres or made from materials with different costs. The lecture also covers finding maximum values of functions on planes in the first octant.

📝 Lecture Summary

EXAMPLE

Find the absolute maximum and minimum values of f(x,y) = xy - x - 3y on the closed triangular region R with vertices (0, 0), (0, 4), and (5, 0).

To find the absolute extrema on a closed bounded region, we must check both critical points in the interior and points on the boundary.

First, find the critical points of f(x,y) = xy - x - 3y.

• f_x(x,y) = y - 1
• f_y(x,y) = x - 3

Setting f_x = 0 and f_y = 0 gives y = 1 and x = 3. Thus, (3, 1) is the only critical point in the interior of R.

Next, analyze the three boundary line segments: (i) Line segment between (0,0) and (5,0): On this segment, y = 0, so the function simplifies to u(x) = f(x,0) = -x for 0 ≤ x ≤ 5. u'(x) = -1 is never zero, so extreme values occur at the endpoints x = 0 (point (0,0)) and x = 5 (point (5,0)).

(ii) Line segment between (0,0) and (0,4): On this segment, x = 0, so the function simplifies to v(y) = f(0,y) = -3y for 0 ≤ y ≤ 4. v'(y) = -3 is never zero, so extreme values occur at endpoints y = 0 (point (0,0)) and y = 4 (point (0,4)).

(iii) Line segment between (5,0) and (0,4): The equation of this line is y = -(4/5)x + 4 for 0 ≤ x ≤ 5. Substituting into f gives: w(x) = f(x, -(4/5)x + 4) = -(4/5)x² + (27/5)x - 12. Setting w'(x) = -(8/5)x + (27/5) = 0 gives x = 27/8. This critical point corresponds to the point (27/8, 13/10).

Finally, evaluate f(x,y) at all candidate points: (0,0), (5,0), (0,4), (27/8, 13/10), and (3,1).

• f(0,0) = 0
• f(5,0) = -5
• f(0,4) = -12
• f(27/8, 13/10) = -231/80
• f(3,1) = -3

📌 Example: The absolute maximum value of f is f(0,0) = 0, and the absolute minimum value is f(0,4) = -12.

EXAMPLE

Find three positive numbers whose sum is 48 and such that their product is as large as possible.

Let the numbers be x, y, and z. The constraint is x + y + z = 48, so z = 48 - x - y. The product function to maximize is: P(x, y) = xy(48 - x - y) = 48xy - x²y - xy².

Find critical points:

• P_x = 48y - 2xy - y²
• P_y = 48x - 2xy - x²

Setting P_x = 0 and P_y = 0, we solve the system:

• y(48 - 2x - y) = 0
• x(48 - x - 2y) = 0 Since positive numbers are required (x, y > 0), we have 48 - 2x - y = 0 and 48 - x - 2y = 0. Solving these gives x = 16, y = 16, and therefore z = 48 - 16 - 16 = 16.

Apply the Second Derivative Test:

• P_xx(x,y) = -2y ⇒ P_xx(16,16) = -32
• P_yy(x,y) = -2x ⇒ P_yy(16,16) = -32
• P_xy(x,y) = 48 - 2x - 2y ⇒ P_xy(16,16) = -16

Compute D = P_xx(16,16)·P_yy(16,16) - [P_xy(16,16)]² = (-32)(-32) - (-16)² = 1024 - 256 = 768 > 0. Since P_xx(16,16) = -32 < 0, the function has a maximum at (16,16).

📌 Example: The required three positive numbers are 16, 16, 16.

EXAMPLE

Find three positive numbers whose sum is 27 and such that the sum of their squares is as small as possible.

Let the numbers be x, y, and z. The constraint is x + y + z = 27, so z = 27 - x - y. The function to minimize is the sum of squares: S(x, y) = x² + y² + (27 - x - y)² = x² + y² + 729 - 54x - 54y + 2xy + x² + y² = 2x² + 2y² - 54x - 54y + 2xy + 729.

Find critical points:

• S_x = 4x + 2y - 54
• S_y = 2x + 4y - 54

Setting S_x = 0 and S_y = 0, we solve the system:

• 4x + 2y = 54
• 2x + 4y = 54 Solving gives x = 9, y = 9, and therefore z = 27 - 9 - 9 = 9.

Apply the Second Derivative Test:

• S_xx(x,y) = 4
• S_yy(x,y) = 4
• S_xy(x,y) = 2

D = (4)(4) - (2)² = 16 - 4 = 12 > 0. Since S_xx = 4 > 0, the function has a minimum at (9,9).

📌 Example: The required three positive numbers are 9, 9, 9.

EXAMPLE

Find the dimensions of the rectangular box of maximum volume that can be inscribed in a sphere of radius 4.

Let the dimensions of the box be x, y, and z. The volume is V = xyz. Since the box is inscribed in a sphere of radius 4, the box's vertices lie on the sphere x² + y² + z² = 16. From this, we get z = √(16 - x² - y²).

The volume function to maximize is V(x, y) = xy√(16 - x² - y²).

Find critical points by setting V_x = 0 and V_y = 0: V_x = y√(16 - x² - y²) + xy * [(-2x) / (2√(16 - x² - y²))] = y√(16 - x² - y²) - (x²y) / √(16 - x² - y²)

Setting V_x = 0: y [ (16 - x² - y²) - x² ] / √(16 - x² - y²) = 0 ⇒ y(16 - 2x² - y²) = 0. Since y ≠ 0 for a nonzero box, we have 2x² + y² = 16 ... (a)

V_y = x√(16 - x² - y²) - (xy²) / √(16 - x² - y²) Setting V_y = 0: x(16 - x² - 2y²) = 0 ⇒ x² + 2y² = 16 ... (b)

Solving (a) and (b): Subtract (b) from (a): (2x² + y²) - (x² + 2y²) = 0 ⇒ x² - y² = 0 ⇒ x = y. Substituting into (b): x² + 2x² = 16 ⇒ 3x² = 16 ⇒ x = 4/√3. So y = 4/√3. Then z = √(16 - (16/3) - (16/3)) = √(16/3) = 4/√3.

The Second Derivative Test confirms this gives a maximum. 💡 Why this matters: This proves that for a sphere of any radius, a cube inscribed in it has maximum volume.

📌 Example: The dimensions of the box of maximum volume are x = 4/√3, y = 4/√3, z = 4/√3.

EXAMPLE

A closed rectangular box with volume of 16 ft³ is made from two kinds of materials. The top and bottom are made of material costing Rs. 10 per square foot and the sides from material costing Rs.5 per square foot. Find the dimensions of the box so that the cost of materials is minimized.

Let length = x, width = y, height = z. The cost function is: Cost = Cost of top/bottom + Cost of four sides Top and bottom area = 2(xy), cost = 10 * 2xy = 20xy. Four sides area = 2(xz) + 2(yz), cost = 5 * (2xz + 2yz) = 10xz + 10yz. So, C = 20xy + 10xz + 10yz. This is the cost function to minimize.

The constraint is the constant volume: xyz = 16 ⇒ z = 16 / (xy). Substitute into the cost function: C(x, y) = 20xy + 10x * (16/xy) + 10y * (16/xy) = 20xy + 160/y + 160/x.

Find critical points:

• C_x = 20y - 160/x²
• C_y = 20x - 160/y²

Setting C_x = 0 and C_y = 0:

• 20y = 160/x² ⇒ y = 8/x²
• 20x = 160/y² ⇒ x = 8/y²

Substitute y = 8/x² into x = 8/y²: x = 8 / (8/x²)² = 8 / (64/x⁴) = x⁴/8. This gives x⁴ = 8x, so x(x³ - 8) = 0. Since x > 0, x³ = 8 ⇒ x = 2. Then y = 8/(2)² = y = 2. Finally, z = 16 / (2 * 2) = z = 4.

Apply the Second Derivative Test:

• C_xx = 320/x³ ⇒ C_xx(2,2) = 320/8 = 40
• C_yy = 320/y³ ⇒ C_yy(2,2) = 320/8 = 40
• C_xy = 20 ⇒ C_xy(2,2) = 20

D = (40)(40) - (20)² = 1600 - 400 = 1200 > 0. Since C_xx = 40 > 0, the function has a minimum at (2,2).

📌 Example: The dimensions that minimize cost are 2 ft × 2 ft × 4 ft.

EXAMPLE

Find the dimensions of the rectangular box of maximum volume that can be inscribed in a sphere of radius a.

This is a generalization of the earlier sphere problem where the radius was 4. The sphere equation is x² + y² + z² = a², so z = √(a² - x² - y²).

The volume function is V(x, y) = xy√(a² - x² - y²). Following the same procedure as before:

Setting V_x = 0 gives y(a² - 2x² - y²) = 0 ⇒ 2x² + y² = a² ... (a) Setting V_y = 0 gives x(a² - x² - 2y²) = 0 ⇒ x² + 2y² = a² ... (b)

Solving (a) and (b): Subtracting gives x² - y² = 0 ⇒ x = y. Substituting into (b): x² + 2x² = a² ⇒ 3x² = a² ⇒ x = a/√3. So y = a/√3, and z = √(a² - (a²/3) - (a²/3)) = √(a²/3) = a/√3.

The Second Derivative Test confirms this gives a maximum value for the volume. 💡 Why this matters: This demonstrates that regardless of the sphere's radius, the box of maximum volume inscribed within it is always a cube whose side length is the sphere's radius divided by √3.

📌 Example: The dimensions of the box are x = a/√3, y = a/√3, z = a/√3.

EXAMPLE

Find the points on the plane x + y + z = 5 in the first octant at which f(x,y,z) = xy²z² has maximum value.

Solution Method 1: Reduce to a function of y and z. From the plane equation, x = 5 - y - z. Substitute into f: u(y,z) = f(5-y-z, y, z) = (5 - y - z)y²z² = 5y²z² - y³z² - y²z³.

Find critical points:

• u_y = 10yz² - 3y²z² - 2yz³ = yz²(10 - 3y - 2z)
• u_z = 10y²z - 2y³z - 3y²z² = y²z(10 - 2y - 3z)

Since we are in the first octant (y>0, z>0), set (10 - 3y - 2z) = 0 and (10 - 2y - 3z) = 0. Solving this system: subtract the first equation from the second: (10-2y-3z) - (10-3y-2z) = 0 ⇒ y - z = 0 ⇒ y = z. Then 10 - 3y - 2y = 0 ⇒ 10 - 5y = 0 ⇒ y = 2, z = 2. Then x = 5 - 2 - 2 = 1.

Apply the Second Derivative Test:

• u_yy = 10z² - 6yz² - 2z³ ⇒ u_yy(2,2) = 40 - 48 - 16 = -24
• u_zz = 10y² - 2y³ - 6y²z ⇒ u_zz(2,2) = 40 - 16 - 48 = -24
• u_yz = 20yz - 6y²z - 6yz² ⇒ u_yz(2,2) = 80 - 48 - 48 = -16

D = (-24)(-24) - (-16)² = 576 - 256 = 320 > 0. Since u_yy = -24 < 0, this is a maximum.

📌 Example: The point at which f has its maximum value is (1, 2, 2).

Solution Method 2: Reduce to a function of x and y. From the plane, z = 5 - x - y. f(x,y) = xy²(5 - x - y)².

Find critical points by setting f_x = 0 and f_y = 0. The calculations are more involved but lead to the same solution: x = 1, y = 2, z = 2.

⭐ Key Takeaways

To solve optimization problems for functions of two or three variables, you must first reduce the problem to a function of two variables using any given constraints. The critical points of this reduced function are then found by setting its first partial derivatives to zero. The Second Derivative Test, which involves computing D = f_xx·f_yy - (f_xy)², is essential to classify these critical points as maxima, minima, or saddle points, where D > 0 and f_xx < 0 indicates a maximum and D > 0 and f_xx > 0 indicates a minimum. For absolute extrema on a closed bounded region, you must also evaluate the function on all boundary segments, including their endpoints.

🧠 Quick Revision Questions

1. What are the steps to find the absolute maxima and minima of a function on a closed triangular region?
2. For the product maximization problem with sum 48, what system of equations must be solved to find the critical point, and what is the final result?
3. For the cost minimization problem of a box with fixed volume, how is the constraint used to reduce the cost function to two variables?
4. For a rectangular box inscribed in a sphere of radius 'a', what is the relationship between the dimensions at the maximum volume?
5. For the function f(x,y,z) = xy²z² on the plane x+y+z=5, what is the single critical point that gives a maximum, and what is D at that point?

📘 Lecture 18 — Revision of Integration

📖 Overview: This lecture provides a comprehensive review of integration concepts, focusing on double integrals over rectangular regions. It demonstrates how to evaluate double integrals using iterated or repeated integrals and establishes the key theorem that the order of integration does not matter for continuous functions over a rectangle, which simplifies computation.

🗂️ Topics Covered

Revision of single integration with respect to x and y separately, introduction to double integrals for finding volume under a surface, evaluation of double integrals using iterated integrals in both orders (dxdy and dydx), the theorem showing order of integration does not affect the result for continuous functions over rectangles, and worked examples including exponential functions using properties of e and ln.

📝 Lecture Summary

Revision of Integration

The lecture begins by revisiting single integration with respect to one variable while treating the other as constant. For example, integrating (xy + y²) with respect to x from 0 to 1 treats y as constant, yielding y/2 + y². Integrating the same expression with respect to y from 0 to 1 treats x as constant, yielding x/2 + 1/3.

📌 Example 1: Consider ∫₀¹ (xy + y²)dx. Here y is treated as constant, so ∫₀¹ xy dx + ∫₀¹ y² dx = y(1/2) + y². Result: y/2 + y²

📌 Example 2: Consider ∫₀¹ (xy + y²)dy. Here x is treated as constant, so x∫₀¹ y dy + ∫₀¹ y² dy = x(1/2) + 1/3. Result: x/2 + 1/3

Double Integral

Symbolically, the double integral of two variables x and y over a certain region R of the plane is denoted by ∬ᵣ f(x, y) dxdy.

📌 Example: Use a double integral to find the solid bounded above by the plane Z = 4 – x – y and below by the rectangle R = {(x, y): 0 ≤ x ≤ 1, 0 ≤ y ≤ 2}.

Solution: The volume V = ∬ᵣ (4 – x – y) dA

Evaluating: V = ∫₀² ∫₀¹ (4 – x – y) dx dy

First integrate with respect to x: ∫₀¹ (4 – x – y) dx = [4x – x²/2 – xy] from x=0 to x=1 = 4(1) – (1)²/2 – 1(y) – [0] = 4 – 1/2 – y = 7/2 – y

Then integrate with respect to y: ∫₀² (7/2 – y) dy = [7y/2 – y²/2] from y=0 to y=2 = (7(2)/2 – (2)²/2) – 0 = (14/2 – 4/2) = 7 – 2 = 5

Result: V = 5

📌 Example: Evaluate the double integral ∫₀¹ ∫₀¹ (xy + y²) dx dy

First integrate with respect to x: ∫₀¹ (xy + y²) dx = [x²y/2 + xy²] from x=0 to x=1 = (1²y/2 + 1y²) – 0 = y/2 + y²

Then integrate with respect to y: ∫₀¹ (y/2 + y²) dy = [y²/4 + y³/3] from y=0 to y=1 = (1/4 + 1/3) – 0 = (3/12 + 4/12) = 7/12

Iterated or Repeated Integral

The expression ∫ᶜᵈ [∫ₐᵇ f(x, y) dx] dy is called an iterated or repeated integral. Often the brackets are omitted and this expression is written as ∫ᶜᵈ ∫ₐᵇ f(x, y) dx dy. Here, ∫ₐᵇ f(x, y) dx yields a function of y, which is then integrated over the interval c ≤ y ≤ d.

Similarly, ∫ₐᵇ ∫ᶜᵈ f(x, y) dy dx = ∫ₐᵇ [∫ᶜᵈ f(x, y) dy] dx. Here, ∫ᶜᵈ f(x, y) dy yields a function of x, which is then integrated over the interval a ≤ x ≤ b.

📌 Example: Evaluate the integral ∫₀¹ ∫₀² (x + 3) dy dx

First integrate with respect to y: ∫₀² (x + 3) dy = (x + 3)[y] from y=0 to y=2 = (x + 3)(2 – 0) = 2(x + 3)

Then integrate with respect to x: ∫₀¹ 2(x + 3) dx = 2[x²/2 + 3x] from x=0 to x=1 = 2[(1/2 + 3) – 0] = 2(1/2 + 3) = 2(7/2) = 7

Now change the order: ∫₀² ∫₀¹ (x + 3) dx dy

First integrate with respect to x: ∫₀¹ (x + 3) dx = [x²/2 + 3x] from x=0 to x=1 = (1/2 + 3) – 0 = 7/2

Then integrate with respect to y: ∫₀² (7/2) dy = (7/2)[y] from y=0 to y=2 = (7/2)(2 – 0) = 7

Note: The value remains the same when the order of integration is changed.

Theorem

Let R be the rectangle defined by the inequalities a < x < b and c < y < d. If f(x, y) is continuous on this rectangle, then:

∬ᵣ f(x, y) dA = ∫ᶜᵈ ∫ₐᵇ f(x, y) dx dy = ∫ₐᵇ ∫ᶜᵈ f(x, y) dy dx

💡 Why this matters: This powerful theorem enables us to evaluate a double integral over a rectangle by calculating an iterated integral. Moreover, the theorem tells us that the order of integration in the iterated integral does not matter — we can choose whichever order is easier.

📌 Example: Evaluate the integral ∫₀^(ln 2) ∫₀^(ln 3) e^(x+y) dx dy

Note that e^(x+y) = eˣ · eʸ

First integrate with respect to x: ∫₀^(ln 3) eˣ·eʸ dx = eʸ[eˣ] from x=0 to x=ln 3 = eʸ(e^(ln 3) – e⁰) = eʸ(3 – 1) = 2eʸ (Since e^(ln 3) = 3)

Then integrate with respect to y: ∫₀^(ln 2) 2eʸ dy = 2[eʸ] from y=0 to y=ln 2 = 2(e^(ln 2) – e⁰) = 2(2 – 1) = 2 (Since e^(ln 2) = 2)

Result: 2

📌 Example: Evaluate the integral ∫₀^(ln 3) ∫₀^(ln 2) e^(x+y) dy dx (order changed)

First integrate with respect to y: ∫₀^(ln 2) eˣ·eʸ dy = eˣ[eʸ] from y=0 to y=ln 2 = eˣ(e^(ln 2) – e⁰) = eˣ(2 – 1) = eˣ

Then integrate with respect to x: ∫₀^(ln 3) eˣ dx = [eˣ] from x=0 to x=ln 3 = e^(ln 3) – e⁰ = 3 – 1 = 2

Note that in both cases the integral has the same value of 2.

⭐ Key Takeaways

A double integral ∬ᵣ f(x, y) dA over a rectangle R can be evaluated as an iterated integral by integrating with respect to one variable at a time while treating the other as constant. The order of integration (dx dy or dy dx) does not matter for continuous functions over a rectangle, due to the powerful Fubini-type theorem demonstrated. The properties of exponential and natural logarithmic functions as inverse functions (e^(ln k) = k) can simplify integrals involving e^(x+y). The evaluation of double integrals requires careful application of single-variable integration techniques and proper substitution of limits.

🧠 Quick Revision Questions

1. What is the difference between a double integral and an iterated integral?
2. If you integrate f(x, y) with respect to x first (holding y constant), what kind of function do you get before integrating with respect to y?
3. Evaluate the double integral ∫₀³ ∫₀¹ (2x + y) dx dy.
4. Why does the order of integration not matter for a continuous function over a rectangle?
5. Explain why e^(x+y) can be written as eˣ·eʸ and show how this property is used in double integration.

📘 Lecture 19 — Use Of Integrals

📖 Overview: This lecture demonstrates the application of integrals to compute areas and volumes, extending from single-variable calculus to double integrals over rectangular and non-rectangular regions. It establishes the fundamental relationship between double integrals and volume, and introduces Type I and Type II regions for evaluating double integrals over more complex domains.

🗂️ Topics Covered

The lecture begins by showing area as an anti-derivative and volume as a double integral over a rectangle. It then lists properties of double integrals analogous to single-variable definite integrals. Two worked examples compute volumes under surfaces over rectangular regions. The concept of computing volume using cross-sectional areas perpendicular to the y-axis and x-axis is developed, leading to iterated integrals. Finally, the lecture defines and explains double integrals over Type I and Type II non-rectangular regions, providing formulas for their evaluation.

📝 Lecture Summary

Area as anti-derivatives

The lecture first reviews that the definite integral can compute area under a curve. For example, the area under y = 2x from x=0 to x=4 is given by the definite integral ∫₀⁴ 2x dx = [x²]₀⁴ = 16. This matches the geometric formula for the area of a triangle: ½ × base × altitude = ½ (4)(8) = 16. This demonstrates the anti-derivative property of the integral for area.

volume as anti-derivative

The concept extends to volume using a double integral over a rectangular region. For a rectangular solid defined by 0 ≤ x ≤ 2, 0 ≤ y ≤ 3, 0 ≤ z ≤ 5, the volume is the double integral of z=5 over the rectangle in the xy-plane: ∫₀² ∫₀³ 5 dy dx. Evaluating this gives ∫₀² [5y]₀³ dx = ∫₀² 15 dx = [15x]₀² = 30, which is equivalent to the geometric volume 2 × 3 × 5 = 30.

Properties of Double Integrals

Double integrals over a region R have properties analogous to single integrals. For constants c and functions f and g:

• Constant Multiple Rule: ∬ᵣ c f(x,y) dA = c ∬ᵣ f(x,y) dA
• Sum Rule: ∬ᵣ [f(x,y) + g(x,y)] dA = ∬ᵣ f(x,y) dA + ∬ᵣ g(x,y) dA
• Difference Rule: ∬ᵣ [f(x,y) - g(x,y)] dA = ∬ᵣ f(x,y) dA - ∬ᵣ g(x,y) dA

Example: Volume under z = 3x³ + 3x²y over a rectangle

Find the volume under the surface z = 3x³ + 3x²y and over the rectangle R = {(x,y): 1 ≤ x ≤ 3, 0 ≤ y ≤ 2}. Volume = ∫₀² ∫₁³ (3x³ + 3x²y) dx dy First, integrate with respect to x, treating y as constant: ∫₁³ (3x³ + 3x²y) dx = [3x⁴/4 + x³y]₁³ = [3(3)⁴/4 + (3)³y] - [3(1)⁴/4 + (1)³y] = [243/4 + 27y] - [3/4 + y] = 60 + 26y Then, integrate with respect to y: ∫₀² (60 + 26y) dy = [60y + 13y²]₀² = (120 + 52) - 0 = 172

📐 Formula: Volume = ∫ᵧₘᵢₙ^ᵧₘₐₓ ∫ₓₘᵢₙ^ₓₘₐₓ f(x,y) dx dy → The volume under the surface z = f(x,y) over a rectangle.

Example: Volume under z = x² in the first octant

Find the volume of the solid in the first octant enclosed by the surface z = x² and the planes x=2, y=0, y=3, and z=0. The region R in the xy-plane is a rectangle: 0 ≤ x ≤ 2, 0 ≤ y ≤ 3. Volume = ∫₀² ∫₀³ x² dy dx First, integrate with respect to y: ∫₀³ x² dy = [x²y]₀³ = 3x² Then, integrate with respect to x: ∫₀² 3x² dx = [x³]₀² = 8

Properties: Nonnegativity and Additivity

• If f(x,y) ≥ 0 on R, then ∬ᵣ f(x,y) dA ≥ 0.
• If f(x,y) ≥ g(x,y) on R, then ∬ᵣ f(x,y) dA ≥ ∬ᵣ g(x,y) dA.
• For a nonnegative function f on a region R that is subdivided into two regions R₁ and R₂, the double integral is additive: ∬ᵣ f(x,y) dA = ∬ᵣ₁ f(x,y) dA + ∬ᵣ₂ f(x,y) dA.

Volume Using Cross Sections Perpendicular to the y-axis

The volume of a solid S can be obtained using cross-sections perpendicular to the y-axis. The volume is Vol(S) = ∫ᶜᵈ A(y) dy (1), where A(y) is the area of the cross-section perpendicular to the y-axis at point y. For each fixed y in the interval c ≤ y ≤ d, A(y) = ∫ₐᵇ f(x, y) dx. Substituting this into (1) gives: Vol(S) = ∫ᶜᵈ [∫ₐᵇ f(x, y) dx] dy = ∫ᶜᵈ ∫ₐᵇ f(x, y) dx dy

🔑 Definition — A(y): The area of a cross-section of the solid, perpendicular to the y-axis, at a given value of y.

Volume Using Cross Sections Perpendicular to the x-axis

Similarly, the volume can be obtained using cross-sections perpendicular to the x-axis. The volume is Vol(S) = ∫ₐᵇ A(x) dx (3), where A(x) is the area of the cross-section perpendicular to the x-axis at point x. For each fixed x in the interval a ≤ x ≤ b, A(x) = ∫ᶜᵈ f(x, y) dy. Substituting this into (3) gives: Vol(S) = ∫ₐᵇ [∫ᶜᵈ f(x, y) dy] dx = ∫ₐᵇ ∫ᶜᵈ f(x, y) dy dx

From these two derivations, we have the equivalence of iterated integrals: ∬ᵣ f(x,y) dA = ∫ᶜᵈ ∫ₐᵇ f(x,y) dx dy = ∫ₐᵇ ∫ᶜᵈ f(x,y) dy dx 💡 Why this matters: This shows that the order of integration (dx dy or dy dx) does not matter for a rectangular region, as long as the limits are correctly applied.

Double integral for non-rectangular region

The lecture now extends double integrals to non-rectangular regions, defining Type I and Type II regions.

Type I Region

A Type I region is bounded left and right by vertical lines x=a and x=b, and bounded below and above by continuous curves y = g₁(x) and y = g₂(x) where g₁(x) ≤ g₂(x) for a ≤ x ≤ b. If R is a Type I region and f(x, y) is continuous, then: ∬ᵣ f(x,y) dA = ∫ₐᵇ ∫{g₁(x)}^{g₂(x)} f(x,y) dy dx (1) This is derived from the cross-sectional method where **A(x) = ∫{g₁(x)}^{g₂(x)} f(x, y) dy** is the area of the cross-section at a fixed x. Substituting this into the volume formula Vol(S) = ∫ₐᵇ A(x) dx yields the iterated integral.

📐 Formula: ∬ᵣ f(x,y) dA = ∫ₐᵇ ∫_{g₁(x)}^{g₂(x)} f(x,y) dy dx → For a Type I region, integrate first with respect to y from the lower curve to the upper curve, then with respect to x from a to b.

Type II Region

A Type II region is bounded below and above by horizontal lines y=c and y=d, and bounded left and right by continuous curves x = h₁(y) and x = h₂(y) where h₁(y) ≤ h₂(y) for c ≤ y ≤ d. If R is a Type II region and f(x, y) is continuous, then: ∬ᵣ f(x,y) dA = ∫ᶜᵈ ∫_{h₁(y)}^{h₂(y)} f(x,y) dx dy (2) In this case, the partial definite integral ∫ᶜᵈ f(x,y) dy produces a function of x when holding x fixed and integrating with respect to y.

📐 Formula: ∬ᵣ f(x,y) dA = ∫ᶜᵈ ∫_{h₁(y)}^{h₂(y)} f(x,y) dx dy → For a Type II region, integrate first with respect to x from the left curve to the right curve, then with respect to y from c to d.

⭐ Key Takeaways

1. Double integrals over a rectangle compute the volume under a surface z = f(x,y) and can be evaluated by iterated integration in either order (dx dy or dy dx) using constant limits.
2. Properties of double integrals, including linearity and additivity over subregions, are directly analogous to those for single-variable definite integrals.
3. For non-rectangular regions, the limits of integration are variable, depending on the boundaries of the region, which are classified as Type I (bounded by functions of x) or Type II (bounded by functions of y).
4. The cross-sectional method provides a conceptual bridge from single-variable integrals to double integrals, where A(x) or A(y) represents the area of a slice of the solid.
5. The correct setup of a double integral requires careful identification of the region's boundaries and choosing the appropriate order of integration to ensure the inner integral's limits are expressed in terms of the outer variable.

🧠 Quick Revision Questions

1. What is the volume under the surface z = 3x² over the rectangle 0 ≤ x ≤ 4, 0 ≤ y ≤ 2?
2. State the formula for evaluating a double integral over a Type I region.
3. In the cross-sectional method for volume perpendicular to the y-axis, what does A(y) represent, and how is it computed?
4. For a Type II region bounded by x = h₁(y) and x = h₂(y), what is the order of integration in the iterated integral?
5. How does the additivity property of double integrals apply when a region R is subdivided into R₁ and R₂?

📘 Lecture 20 — Double integral for non-rectangular region

📖 Overview: This lecture extends double integration from rectangular regions to more general, non-rectangular regions. It introduces two types of regions—Type I and Type II—and demonstrates how to set up and evaluate double integrals over them, including techniques for reversing the order of integration when necessary. This is crucial for solving real-world problems where integration regions are bounded by curves, not just straight lines.

🗂️ Topics Covered

The lecture covers double integrals over Type I regions (bounded by vertical lines and curves y = g₁(x), y = g₂(x)), Type II regions (bounded by horizontal lines and curves x = h₁(y), x = h₂(y)). It includes examples of sketching regions, writing double integrals from sketches, and evaluating integrals. A key technique is reversing the order of integration from Type I to Type II (or vice versa) to simplify calculations, especially when the integrand lacks an elementary antiderivative in the original order.

📝 Lecture Summary

Double integral for non-rectangular region

The lecture defines two standard types of non-rectangular regions for double integration.

🔑 Definition — Type I region: A region bounded on the left and right by vertical lines x = a and x = b, and bounded below and above by continuous curves y = g₁(x) and y = g₂(x) where g₁(x) ≤ g₂(x) for a ≤ x ≤ b. 📐 Formula: ∬R f(x,y) dA = ∫{x=a}^{b} ∫_{y=g₁(x)}^{g₂(x)} f(x,y) dy dx 💡 Why this matters: This formula allows integration over regions where the y-limits are functions of x, not constants.

🔑 Definition — Type II region: A region bounded below and above by horizontal lines y = c and y = d, and bounded on the left and right by continuous curves x = h₁(y) and x = h₂(y) where h₁(y) ≤ h₂(y) for c ≤ y ≤ d. 📐 Formula: ∬R f(x,y) dA = ∫{y=c}^{d} ∫_{x=h₁(y)}^{h₂(y)} f(x,y) dx dy 💡 Why this matters: This formula is useful when the region is more naturally described with x as a function of y.

The lecture provides sketches of regions and asks to write the double integral. For example, one region yields: ∫{y=1}^{ln8} ∫{x=0}^{ln y} f(x,y) dx dy. Another yields: ∫{y=0}^{ln8} ∫{x=e^x}^{ln8} f(x,y) dx dy. Another: ∫{y=0}^{1} ∫{x=0}^{y²} f(x,y) dx dy, and: ∫{x=0}^{1} ∫{y=0}^{√x} f(x,y) dy dx.

EXAMPLE

The lecture gives an example of reversing the order of integration to evaluate an integral.

Problem: Draw the region and evaluate an equivalent integral with the order of integration reversed: ∫{x=0}^{2} ∫{y=x²}^{2x} (4x + 2) dy dx

Region of integration: Given by inequalities x² ≤ y ≤ 2x and 0 ≤ x ≤ 2. The region is bounded below by y = x² (a parabola opening upward) and above by y = 2x (a straight line) for x between 0 and 2.

To reverse the order, express the region as a Type II region. The curves intersect where x² = 2x → x² - 2x = 0 → x(x-2) = 0 → x = 0, x = 2. For a given y, x ranges from the left curve (x = y/2) to the right curve (x = √y). y ranges from 0 to 4.

Reversed integral: ∫{y=0}^{4} ∫{x=y/2}^{√y} (4x + 2) dx dy

Evaluation: First integrate with respect to x: Inner integral: ∫{x=y/2}^{√y} (4x + 2) dx = [2x² + 2x]{x=y/2}^{√y} = (2(√y)² + 2√y) - (2(y/2)² + 2(y/2)) = (2y + 2√y) - (2(y²/4) + y) = 2y + 2√y - (y²/2 + y) = y + 2√y - y²/2

Now integrate with respect to y: Final result = ∫{y=0}^{4} (y + 2√y - y²/2) dy = [y²/2 + (4/3)y^(3/2) - y³/6]{0}^{4} = (16/2 + (4/3)(8) - 64/6) - (0) = 8 + 32/3 - 32/3 = 8

📌 Example: Evaluate ∫{y=0}^{4} ∫{x=√y}^{2} y cos(x⁵) dx dy by reversing the order.

Region: 0 ≤ y ≤ 4, x = √y and x = 2. This is equivalent to: 0 ≤ x ≤ 2, y = x² and y = 4.

Reversed integral (Type I): ∫{x=0}^{2} ∫{y=0}^{x²} y cos(x⁵) dy dx

Evaluate: Inner integral: ∫{y=0}^{x²} y cos(x⁵) dy = cos(x⁵) * [y²/2]{0}^{x²} = cos(x⁵) * (x⁴/2) = (x⁴/2) cos(x⁵)

Now integrate with respect to x: I = ∫{x=0}^{2} (x⁴/2) cos(x⁵) dx = (1/10) ∫{x=0}^{2} 5x⁴ cos(x⁵) dx = (1/10) [sin(x⁵)]_{0}^{2} = (1/10) sin(32)

📌 Example: Evaluate I = ∫{x=0}^{1/2} ∫{y=2x}^{1} e^(y²) dy dx

The integral cannot be evaluated in the given order since e^(y²) has no elementary antiderivative. We change the order of integration.

Region R: 0 ≤ x ≤ 1/2, y = 2x and y = 1. This region is also enclosed by x = 0, x = y/2, and 0 ≤ y ≤ 1.

Reversed integral (Type II): I = ∫{y=0}^{1} ∫{x=0}^{y/2} e^(y²) dx dy

Evaluate: Inner integral: ∫{x=0}^{y/2} e^(y²) dx = e^(y²) * [x]{0}^{y/2} = (y/2) e^(y²)

Now integrate with respect to y: I = ∫{y=0}^{1} (y/2) e^(y²) dy Let u = y², du = 2y dy → y dy = du/2 I = ∫{u=0}^{1} (1/2) e^u (du/2) = (1/4) ∫{0}^{1} e^u du = (1/4) [e^u]{0}^{1} = (1/4)(e - 1)

📌 Example: Evaluate ∫{x=1}^{3} ∫{y=0}^{ln x} x dy dx by reversing the order of integration.

Original region: 1 ≤ x ≤ 3, 0 ≤ y ≤ ln x. The region is bounded below by y = 0, above by y = ln x, and x from 1 to 3. The point (3, ln3) is on the curve.

To reverse: For a given y, x ranges from the left curve (x = e^y) to x = 3. y ranges from 0 to ln3.

Reversed integral: ∫{y=0}^{ln3} ∫{x=e^y}^{3} x dx dy

Evaluate: Inner integral: ∫{x=e^y}^{3} x dx = [x²/2]{e^y}^{3} = (9/2) - (e^(2y)/2)

Now integrate: I = ∫{y=0}^{ln3} (9/2 - e^(2y)/2) dy = (1/2) ∫{0}^{ln3} (9 - e^(2y)) dy = (1/2) [9y - (e^(2y))/2]_{0}^{ln3} = (1/2) [ (9 ln3 - (e^(2 ln3))/2) - (0 - 1/2) ] = (1/2) [ 9 ln3 - (9)/2 + 1/2 ] = (1/2) [ 9 ln3 - 8/2 ] = (1/2) [ 9 ln3 - 4 ]

⭐ Key Takeaways

The most important concepts from this lecture are: (1) Double integrals over non-rectangular regions are evaluated by iterated integrals where one set of limits is a function of the other variable. (2) A Type I region uses vertical slices (y as a function of x), while a Type II region uses horizontal slices (x as a function of y). (3) When an integral is difficult or impossible to evaluate in a given order, reversing the order of integration can simplify the problem drastically. (4) To reverse the order, sketch the region, express its boundaries with the opposite variable as the inner integration variable, and adjust the outer limits accordingly. (5) Always correctly identify the bounding curves and their intersection points to define the new limits equally.

🧠 Quick Revision Questions

1. What are the standard forms for a double integral over a Type I and a Type II non-rectangular region?
2. For a given region bounded by y = x² and y = 2x, how would you set up the double integral in both Type I and Type II orders?
3. Why might reversing the order of integration be necessary when evaluating an integral like ∫∫ e^(y²) dy dx?
4. What is the first step in reversing the order of integration for a double integral over a non-rectangular region?
5. If a region is defined by 1 ≤ x ≤ 3 and 0 ≤ y ≤ ln x, what are the limits for the reversed integral when x is integrated first?

📘 Lecture 21 — Examples

📖 Overview: This lecture presents worked examples of double integrals, focusing on techniques for reversing the order of integration and calculating volumes of solids. These examples are crucial for mastering the practical application of iterated integrals in multivariable calculus.

🗂️ Topics Covered

The lecture covers four main examples: reversing the order of integration to evaluate a double integral, calculating a double integral over a triangular region, evaluating an integral by reversing order when no elementary antiderivative exists, and using double integrals to find volumes of solids bounded by various surfaces including planes and paraboloids.

📝 Lecture Summary

[Lecture No -21 Examples]

Example: Evaluate the double integral (\int_{0}^{4} \int_{0}^{4x} e^{-y^2} dy dx) by reversing the order of integration.

The original integral has an inner integral with respect to y and an outer integral with respect to x. Since the antiderivative of (e^{-y^2}) is not elementary, we reverse the order. The region of integration is bounded by y=0, y=4x, x=0, and x=4. Reversing the order, the region is described as: x varies from 0 to y/4, and y varies from 0 to 16. The new integral is (\int_{0}^{16} \int_{0}^{y/4} e^{-y^2} dx dy).

Evaluating the inner integral gives (\int_{0}^{y/4} e^{-y^2} dx = e^{-y^2} \cdot x \big|{0}^{y/4} = \frac{y}{4} e^{-y^2}). Then the outer integral becomes (\int{0}^{16} \frac{y}{4} e^{-y^2} dy).

Using substitution (u = -y^2), (du = -2y dy), so (y dy = -\frac{1}{2} du), the integral becomes (-\frac{1}{8} \int_{0}^{16} e^{-y^2} (-2y) dy = -\frac{1}{8} \int_{y=0}^{y=16} e^{u} du).

Evaluating: (-\frac{1}{8} [e^{-y^2}]_{0}^{16} = -\frac{1}{8} (e^{-256} - e^{0}) = \frac{1}{8} (1 - e^{-256})).

📐 Formula: Reversing order of integration uses the transformation: (\int_{x=a}^{b} \int_{y=g(x)}^{h(x)} f(x,y) dy dx = \int_{y=c}^{d} \int_{x=p(y)}^{q(y)} f(x,y) dx dy), where the region is described equivalently.

Example: Calculate (\iint_R \frac{\sin x}{x} dA), where R is the triangle bounded by the x-axis, the line y=x, and the line x=1.

We integrate first with respect to y, then x. The region R: x from 0 to 1, y from 0 to x.

The integral is (\int_{0}^{1} \int_{0}^{x} \frac{\sin x}{x} dy dx = \int_{0}^{1} \left[ \frac{\sin x}{x} \cdot y \right]{0}^{x} dx = \int{0}^{1} \frac{\sin x}{x} \cdot x dx = \int_{0}^{1} \sin x dx).

Evaluating: (\int_{0}^{1} \sin x dx = [-\cos x]_{0}^{1} = -\cos(1) - (-\cos(0)) = -\cos(1) + 1 \approx 0.46).

🔑 Definition — Triangle region: defined by the x-axis (y=0), the line y=x, and the line x=1.

Example: Evaluate (\int_{0}^{2} \int_{y/2}^{1} e^{x^2} dx dy).

Since there is no elementary antiderivative of (e^{x^{2}}), the integral cannot be evaluated by performing the x-integration first. We must reverse the order. The region: x from y/2 to 1, y from 0 to 2. This is a triangular region bounded by x=y/2, x=1, y=0.

Reversing to Type I region: y from 0 to 2x, x from 0 to 1. The new integral is (\int_{0}^{1} \int_{0}^{2x} e^{x^2} dy dx).

Evaluating the inner integral: (\int_{0}^{2x} e^{x^2} dy = e^{x^2} \cdot y \big|{0}^{2x} = 2x e^{x^2}). Then the outer integral: (\int{0}^{1} 2x e^{x^2} dx). Let (u = x^2), (du = 2x dx), so (\int_{0}^{1} e^{u} du = [e^{x^2}]_{0}^{1} = e^{1} - e^{0} = e - 1).

Example: Use a double integral to find the volume of the solid bounded above by the plane z=4-x-y and below by the rectangle R = {(x,y): 0 ≤ x ≤ 1, 0 ≤ y ≤ 2}.

Volume V = (\iint_R (4-x-y) dA = \int_{0}^{2} \int_{0}^{1} (4-x-y) dx dy).

Evaluating the inner integral: (\int_{0}^{1} (4-x-y) dx = [4x - \frac{x^2}{2} - xy]_{0}^{1} = 4(1) - \frac{1^2}{2} - 1y - (0) = 4 - \frac{1}{2} - y = \frac{7}{2} - y).

Now the outer integral: (\int_{0}^{2} (\frac{7}{2} - y) dy = [\frac{7}{2}y - \frac{y^2}{2}]_{0}^{2} = \frac{7}{2}(2) - \frac{4}{2} = 7 - 2 = 5).

Example: Use a double integral to find the volume of the tetrahedron bounded by the coordinate planes and the plane z=4-4x-2y.

The tetrahedron is bounded above by the plane z=4-4x-2y and below by the triangular region R. The region R is bounded by the x-axis (y=0), the y-axis (x=0), and the line where z=0: 0=4-4x-2y → y=2-2x.

Volume V = (\iint_R (4-4x-2y) dA), treating R as a Type I region: x from 0 to 1, y from 0 to 2-2x.

The integral: (\int_{0}^{1} \int_{0}^{2-2x} (4-4x-2y) dy dx).

Inner integral: (\int_{0}^{2-2x} (4-4x-2y) dy = [4y - 4xy - y^2]_{0}^{2-2x} = 4(2-2x) - 4x(2-2x) - (2-2x)^2).

Simplify: (8-8x - 8x + 8x^2 - (4 - 8x + 4x^2) = 8 - 16x + 8x^2 - 4 + 8x - 4x^2 = 4 - 8x + 4x^2).

Now the outer integral: (\int_{0}^{1} (4 - 8x + 4x^2) dx = [4x - 4x^2 + \frac{4x^3}{3}]_{0}^{1} = 4(1) - 4(1) + \frac{4}{3} = \frac{4}{3}).

🔑 Definition — Tetrahedron: a solid with four triangular faces, bounded here by the coordinate planes (x=0, y=0, z=0) and the given plane.

Example: Find the volume of the solid bounded by the cylinder (x^2 + y^2 = 4) and the planes y+z=4 and z=0.

The solid is bounded above by the plane z=4-y and below by the region R within the circle (x^2 + y^2 = 4). Volume V = (\iint_R (4-y) dA).

Treating R as a Type I region: x from -2 to 2, y from (-\sqrt{4-x^2}) to (\sqrt{4-x^2}).

The integral: (\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (4-y) dy dx).

Inner integral: (\int (4-y) dy = 4y - \frac{y^2}{2}), evaluated from (-\sqrt{4-x^2}) to (\sqrt{4-x^2}).

At y = (\sqrt{4-x^2}): (4\sqrt{4-x^2} - \frac{(4-x^2)}{2}). At y = (-\sqrt{4-x^2}): (-4\sqrt{4-x^2} - \frac{(4-x^2)}{2}). Difference: (8\sqrt{4-x^2} - 0 = 8\sqrt{4-x^2}).

Now the outer integral: (\int_{-2}^{2} 8\sqrt{4-x^2} dx). This is 8 times the area of a semicircle of radius 2, but we evaluate using trigonometric substitution: (\int \sqrt{4-x^2} dx = \frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) + C).

So (8 \left[ \frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) \right]_{-2}^{2}).

At x=2: (\frac{2}{2}\sqrt{4-4} + 2\sin^{-1}(1) = 0 + 2\cdot \frac{\pi}{2} = \pi). At x=-2: (\frac{-2}{2}\sqrt{4-4} + 2\sin^{-1}(-1) = 0 + 2\cdot (-\frac{\pi}{2}) = -\pi). Difference: (\pi - (-\pi) = 2\pi).

So the result is (8(2\pi) = 16\pi).

Example: Use double integral to find the volume of the solid bounded above by the paraboloid (z = 9x^2 + y^2), below by the plane z=0, and laterally by the planes x=0, y=0, x=3, y=2.

Volume V = (\iint_R (9x^2 + y^2) dA = \int_{0}^{3} \int_{0}^{2} (9x^2 + y^2) dy dx).

Inner integral: (\int_{0}^{2} (9x^2 + y^2) dy = [9x^2 y + \frac{y^3}{3}]_{0}^{2} = 9x^2(2) + \frac{8}{3} - 0 = 18x^2 + \frac{8}{3}).

Now the outer integral: (\int_{0}^{3} (18x^2 + \frac{8}{3}) dx = [6x^3 + \frac{8}{3}x]_{0}^{3} = 6(27) + \frac{8}{3}(3) = 162 + 8 = 170).

⭐ Key Takeaways

Reversing the order of integration is a powerful technique when a function has no elementary antiderivative with respect to one variable, as demonstrated with the Gaussian-type integral. The key steps involve correctly re-describing the region of integration from a Type II (y varies first) to a Type I region (x varies first), or vice versa. When using double integrals for volume, the integrand is always the height function (upper surface minus lower surface), and the domain is the projection of the solid onto the xy-plane. For solids bounded by planes, the region R is often a triangle or rectangle; for cylinders, the region is a circle. Mastering trigonometric substitutions is essential for evaluating integrals involving circular boundaries.

🧠 Quick Revision Questions

1. In the first example, why was it necessary to reverse the order of integration for (\int_{0}^{4} \int_{0}^{4x} e^{-y^2} dy dx)?
2. What are the bounds for the region in the second example (triangle bounded by x-axis, y=x, x=1) when integrating with respect to y first?
3. After reversing order in the third example, what was the new integrand and bounds for evaluating (\int_{0}^{2} \int_{y/2}^{1} e^{x^2} dx dy)?
4. For the tetrahedron volume example, how was the line y=2-2x derived from the plane equation z=4-4x-2y?
5. In the cylinder volume example, why did the inner integral (\int (4-y) dy) simplify to exactly (8\sqrt{4-x^2}) after evaluating the bounds?

📘 Lecture 22 — Examples

📖 Overview: This lecture demonstrates the application of double integrals to solve various geometric problems, including calculating the volume of solids bounded by cylinders and planes, and computing the area of planar regions enclosed by curves. It uses both type I and type II integration orders to illustrate flexibility in problem-solving.

🗂️ Topics Covered

The lecture covers examples of double integrals over trapezoidal regions for function evaluation, computing the volume of a wedge cut from an elliptic cylinder by a plane, finding the volume of the intersection of two perpendicular cylinders, and calculating the area of a region as a double integral with different integration orders.

📝 Lecture Summary

EXAMPLE – Double Integral Over a Trapezium

The first example evaluates (\iint_R xy , dA) where (R) is the trapezium with vertices (1,3), (5,3), (2,1), and (4,1). The region is bounded by lines AD and BC. The slope of AD is (-2) and of BC is (2). The equations of the lines are derived as:

• Line AD (left boundary): (x = -\frac{y-5}{2})
• Line BC (right boundary): (x = \frac{y+7}{2})

The integral is set up with y from 1 to 3 as the outer variable. [ \int_{1}^{3} \int_{-(y-5)/2}^{(y+7)/2} xy , dx , dy ] Evaluating the inner integral with respect to x: [ \int_{1}^{3} \left[ \frac{x^2}{2} y \right]{-(y-5)/2}^{(y+7)/2} dy = \int{1}^{3} \frac{y}{2} \left( \left(\frac{y+7}{2}\right)^2 - \left(-\frac{y-5}{2}\right)^2 \right) dy ] This simplifies to (\int_{1}^{3} (3y^2 + 3y) , dy). Evaluating from 1 to 3 gives: [ \left[ y^3 + \frac{3y^2}{2} \right]_{1}^{3} = (27 + 13.5) - (1 + 1.5) = 40.5 - 2.5 = 38 ]

📌 Example: (\iint_R xy , dA = 38) for the given trapezium.

EXAMPLE – Volume of a Wedge from an Elliptic Cylinder

Find the volume of the wedge cut from the cylinder (4x^2 + y^2 = 9) by the planes (z=0) and (z=y+3).

The cylinder is an ellipse described by (\frac{x^2}{(3/2)^2} + \frac{y^2}{3^2} = 1). The limits for x are (x = \pm \frac{\sqrt{9-y^2}}{2}), and for y the limits are -3 to 3. The volume is: [ V = \int_{-3}^{3} \int_{-\sqrt{9-y^2}/2}^{\sqrt{9-y^2}/2} (y+3) , dx , dy ] Integrating with respect to x: [ \int_{-3}^{3} \left[ (y+3)x \right]{-\sqrt{9-y^2}/2}^{\sqrt{9-y^2}/2} dy = \int{-3}^{3} (y+3) \sqrt{9-y^2} , dy ] This is split into (\int_{-3}^{3} y\sqrt{9-y^2} , dy + 3\int_{-3}^{3} \sqrt{9-y^2} , dy). The first integral evaluates to 0 (odd function over symmetric interval). The second integral is the area of a semicircle of radius 3: [ 3 \cdot \frac{\pi (3)^2}{2} = 3 \cdot \frac{9\pi}{2} = \frac{27\pi}{2} ]

📌 Example: (V = \frac{27\pi}{2}) cubic units.

EXAMPLE – Volume of Intersection of Two Cylinders

Find the volume of the solid common to the cylinders (x^2 + y^2 = 25) and (x^2 + z^2 = 25).

The solid is symmetric. Using 1/8 of the total volume in the first octant: [ V = 8 \int_{0}^{5} \int_{0}^{\sqrt{25-x^2}} dy , dx = 8 \int_{0}^{5} \sqrt{25-x^2} , dx ] Integrate: [ 8 \left[ \frac{x}{2}\sqrt{25-x^2} + \frac{25}{2}\sin^{-1}\left(\frac{x}{5}\right) \right]{0}^{5} = 8 \left(0 + \frac{25}{2} \cdot \frac{\pi}{2} - 0 - 0\right) = 8 \cdot \frac{25\pi}{4} = 50\pi ] Alternatively, using the given calculation in the text: [ 8 \int{0}^{5} (25 - x^2) dx = 8 \left[ 25x - \frac{x^3}{3} \right]_{0}^{5} = 8 \left(125 - \frac{125}{3}\right) = 8 \cdot \frac{250}{3} = \frac{2000}{3} ] 💡 Why this matters: The first method gives (50\pi \approx 157.08), while the second gives (2000/3 \approx 666.67). The correct volume is (50\pi), as the area formula used in the second method is incorrect for this shape.

📌 Example: (V = 50\pi) cubic units.

AREA CALCULATED AS A DOUBLE INTEGRAL

The area of a region R can be computed as: [ \text{Area of } R = \iint_R dA ] This follows from the volume of a solid with constant height 1, where (V = \iint_R 1 , dA = \iint_R dA = \text{Area of base} \times 1).

EXAMPLE – Area Between Parabola and Line (Type I)

Find the area of the region R enclosed between the parabola (y = \frac{1}{2}x^2) and the line (y = 2x).

Set up with x as outer variable from 0 to 4: [ \text{Area} = \int_{0}^{4} \int_{x^2/2}^{2x} dy , dx = \int_{0}^{4} \left(2x - \frac{x^2}{2}\right) dx = \left[ x^2 - \frac{x^3}{6} \right]_{0}^{4} = 16 - \frac{64}{6} = \frac{96 - 64}{6} = \frac{32}{6} = \frac{16}{3} ]

Treating as type II (y from 0 to 8): [ \text{Area} = \int_{0}^{8} \int_{y/2}^{\sqrt{2y}} dx , dy ] Note: The text uses (x = \sqrt{2y}) for the parabola and (x = y/2) for the line. [ \int_{0}^{8} \left( \sqrt{2y} - \frac{y}{2} \right) dy = \left[ \frac{2\sqrt{2}}{3} y^{3/2} - \frac{y^2}{4} \right]_{0}^{8} ] At (y=8): (\frac{2\sqrt{2}}{3}(8\sqrt{8}) - \frac{64}{4} = \frac{2\sqrt{2}}{3}(16\sqrt{2}) - 16 = \frac{64}{3} - 16 = \frac{64 - 48}{3} = \frac{16}{3})

EXAMPLE – Area Enclosed by Parabola and Line

Find the area of the region R enclosed by the parabola (y = x^2) and the line (y = x+2).

Find intersection points: (x^2 = x+2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x-2)(x+1)=0), so x = -1, 2. Set up as type I (x from -1 to 2): [ \text{Area} = \int_{-1}^{2} \int_{x^2}^{x+2} dy , dx = \int_{-1}^{2} (x+2 - x^2) dx ] [ = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2} ] At x=2: (2 + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}) At x=-1: (\frac{1}{2} - 2 + \frac{1}{3} = \frac{3 - 12 + 2}{6} = -\frac{7}{6}) Subtract: (\frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20 + 7}{6} = \frac{27}{6} = \frac{9}{2})

📌 Example: Area = (9/2) square units.

⭐ Key Takeaways

Remember that double integrals can compute volumes and areas. The area of a region is simply (\iint_R dA). When setting up integrals, choose the order of integration (type I or type II) that simplifies the limits. For symmetric solids, use symmetry to integrate over a fraction and multiply. Be careful with algebraic simplification of integrands and evaluating definite integrals accurately. Always check that the limits of integration cover the entire region without overlap.

🧠 Quick Revision Questions

1. How do you derive the x-limits for a region bounded by two non-vertical lines when using type II integration?
2. Why does (\int_{-a}^{a} y\sqrt{a^2 - y^2} , dy) evaluate to 0?
3. What is the volume of intersection of two perpendicular cylinders of equal radius (r)?
4. Why can area be expressed as (\iint_R dA) rather than (\iint_R f(x,y) dA)?
5. How do you find intersection points when a region is bounded by a parabola and a line?