⇪
MATH 20C: CALCULUS AND ANALYTIC GEOMETRY FOR SCIENCE AND ENGINEERING
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
TABLE OF CONTENTS
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
PART A: VECTOR GEOMETRY AND CALCULUS OF VECTOR-VALUED FUNCTIONS
⇒ Lecture 1: Vectors in 2D and 3D [2차원 및 3차원 벡터] ⇐
⇒ Lecture 2: Dot Product [내적] ⇐
⇒ Lecture 3: Cross Product [외적] ⇐
⇒ Lecture 4: Equations of Lines in 2D and 3D [2차원 및 3차원 직선의 방정식] ⇐
⇒ Lecture 5: Equations of Planes Pt. A [평면의 방정식 파트 A] ⇐
⇒ Lecture 6: Equations of Planes Pt. B [평면의 방정식 파트 B] ⇐
⇒ Lecture 7: Introduction of Vector-Valued Functions [벡터 값 함수의 개론] ⇐
⇒ Lecture 8: Calculus of Vector-Valued Functions [벡터 값 함수의 미적분학] ⇐
⇒ Lecture 9: Arc Length [호의 길이] ⇐
PART B: DIFFERENTIATION IN SEVERAL VARIABLES
⇒ Lecture 10: Function in Multiple Variables [다변수 함수] ⇐
⇒ Lecture 11: Limit of Multivariate Functions Pt. A [다변수 함수의 극한 파트 A] ⇐
⇒ Lecture 12: Limit of Multivariate Functions Pt. B [다변수 함수의 극한 파트 B] ⇐
⇒ Lecture 13: Partial Derivatives [편도함수] ⇐
⇒ Lecture 14: Tangent Plane and Linear Approximation [접평면과 선형 근사] ⇐
⇒ Lecture 15: Gradient and Directional Derivative [기울기와 방향도함수] ⇐
⇒ Lecture 16: Direction of Greatest Increase/Decrease + Gradient and Level Curves [최대 증가/감소 방향 + 기울기 및 등고선] ⇐
⇒ Lecture 17: Chain Rules [연쇄법칙] ⇐
⇒ Lecture 18: Implicit Differentiation [암묵적 미분] ⇐
⇒ Lecture 19: Optimization - Critical Points and Second Derivative Test [최적화 - 임계점과 이차 도함수 판정법] ⇐
⇒ Lecture 20: Method of Lagrange's Multiplier Pt. A [라그랑주 승수법 파트 A] ⇐
⇒ Lecture 21: Method of Lagrange's Multiplier Pt. B [라그랑주 승수법 파트 B] ⇐
PART C: MULTIPLE INTEGRATION
⇒ Lecture 22: Riemann Approximation for Volumes [체적에 대한 리만 근사] ⇐
⇒ Lecture 23: Double Integral Over a Rectangular Domain [직사각형 영역에서 이중적분] ⇐
⇒ Lecture 24: Double Integral Over Generalized Regions [일반 영역에서의 이중적분] ⇐
⇒ Lecture 25: Changing the Order of Integration [적분 순서의 변환] ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 1: VECTORS IN 2D AND 3D [2차원 및 3차원 벡터]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
EVALUATE VECTOR GIVEN POINTS [ON EXAM - PRACTICED]
⇒ Given - Points P (x, y, z): let P = (2, 1, 5), Q = (3, 5, 7), R = (1, -3, -2), S = (2, 1, 0) and find components of PQ and RS ⇐
⇒ Step 1 - Evaluate Vector PQ: PQ = ⟨3 - (2), 5 - (1), 7 - (5)⟩ = ⟨1, 4, 2⟩ ⇐
⇒ Step 2 - Evaluate Vector RS: RS = ⟨2 - (1), 1 - (-3), 0 - (-2)⟩ = ⟨1, 4, 2⟩ ⇐
VECTOR OPERATIONS [ON EXAM - PRACTICED]
⇒ Operation 1. Constant c Scalar Multiplication: c · v = (c · x, c · y, c · z) [스칼라 곱셈] ⇐
⇒ Operation 2. Addition or Subtraction: v ± w = (x₁ ± x₂, y₁ ± y₂, z₁ ± z₂) [덧셈과 뺄셈]⇐
⇒ Operation 3. Standard Basis Vectors: i = ⟨1,0,0⟩, j = ⟨0,1,0⟩, k = ⟨0,0,1⟩ [표준기저벡터]⇐
⇒ Operation 4. Magnitude Formula (i.e., Length): ||v|| = √ (x² + y² + z²) [벡터 크기 공식] ⇐
⇒ Operation 5. Normalization and Deriving Unit Vector: û = v / ||v|| [정규화와 단위벡터] ⇐
EVALUATE MAGNITUDE [ON EXAM - PRACTICED]
⇒ Given - Point P (x, y, z) : find ||PQ|| where P = (-1, -2, 0) and Q = (4, -10, -2) ⇐
⇒ Step 1 - Calculate Vector PQ Using Point P and Point Q: PQ = ⟨4 - (-1), -10 - (-2), -2 - (0)⟩ = ⟨5, -8, -2⟩ ⇐
⇒ Step 2 - Evaluate Magnitude ||v|| = √(x² + y² + z²) : ||PQ|| = √ [(5)^2 + (-8)^2 + (-2)^2] = √ [25 + 64 + 4] = √93 ⇐
VECTOR OF SAME DIRECTION WITH LENGTH OF C [ON EXAM - PRACTICED]
⇒ Given - [i] Vector v [ii] Length c: find vector of length 4 pointing in the same direction as PQ ⇐
⇒ Step 1 - Normalize PQ to Create a Unit Vector û = v / ||v|| : û = ⟨5, -8, -2⟩ / ||⟨5, -8, -2⟩|| = ⟨5 / √93 , -8 / √93, -2 / √93⟩ ⇐
⇒ Step 2 - Multiply Length to Unit Vector: scalar multiple 4 · ⟨5 / √93 , -8 / √93, -2 / √93⟩ = ⟨20 / √93 , -32 / √93, -8 / √93⟩ ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 2: DOT PRODUCT [내적]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
DOT PRODUCT OF TWO VECTORS [ON EXAM - PRACTICED]
⇒ Given - [i] Vector ⟨x, y, z⟩ [ii] Standard Basis Vector i + j + k: let v = ⟨-3, -2, 1⟩ and w = 6i + 4j - 2k, find v · w ⇐
⇒ Step 1 - Convert Standard Basis Vector Into ⟨x, y, z⟩: using i = ⟨1, 0, 0⟩, j = ⟨0, 1, 0⟩, k = ⟨0, 0, 1⟩, convert w = 6i + 4j - 2k = ⟨6, 4, -2⟩ ⇐
⇒ Step 2 - Conduct Dot Product Based on x₁x₂ + y₁y₂ + z₁z₂: ⟨-3, -2, 1⟩ · ⟨6, 4, -2⟩ = [(-3) · (6) + (-2) · (4) + (1) · (-2)] = -18 - 8 - 2 = -28 ⇐
EVALUATE ANGLE BETWEEN VECTORS USING DOT PRODUCT FORMULA [ON EXAM - PRACTICED]
⇒ Given - [i] Vector ⟨x, y, z⟩ [ii] Standard Basis Vector i + j + k: let v = ⟨-3, -2, 1⟩ and w = 6i + 4j - 2k, find the angle between v and w ⇐
⇒ Step 1 - Convert Standard Basis Vector Into ⟨x, y, z⟩: using i = ⟨1, 0, 0⟩, j = ⟨0, 1, 0⟩, k = ⟨0, 0, 1⟩, convert w = 6i + 4j - 2k = ⟨6, 4, -2⟩ ⇐
⇒ Step 2 - Conduct Dot Product Based on x₁x₂ + y₁y₂ + z₁z₂: ⟨-3, -2, 1⟩ · ⟨6, 4, -2⟩ = [(-3) · (6) + (-2) · (4) + (1) · (-2)] = -18 - 8 - 2 = -28 ⇐
⇒ Step 3A - Set-up Formula v · w = ||v|| ||w|| cos θ: formulate -28 = ||v|| ||w|| cos θ which will be used to solve for θ ⇐
⇒ Step 3B - Calculate ||v|| and ||w||: get magnitudes ||v|| = ||⟨-3, -2, 1⟩|| = sqrt (14) and ||w|| = ||⟨6, 4, -2⟩|| = 2 sqrt (4) ⇐
⇒ Step 3C - Plug-in ||v|| and ||w|| Into v · w = ||v|| ||w|| cos θ: -28 = (√14) (2√14) (cos θ) becoming cos θ = -28/28 = -1 ⇐
⇒ Step 3D - Conduct arccos to Evaluate θ: θ = arccos (-1) = π = 180 ° because cos (π) = -1, therefore angle is 180 ° ⇐
CONFIRMING ORTHOGONAL VECTOR [PRACTICED]
⇒ Given - [i] Vector v = ⟨x, y, z⟩ [ii] Vector w = ⟨x, y, z⟩: are the vectors v = ⟨-5, 2, 2⟩ and w = ⟨2, 3, -1⟩ orthogonal? ⇐
⇒ Step 1 - Conduct Dot Product: ⟨-5, 2, 2⟩ · ⟨2, 3, -1⟩ = [(-5) · (2) + (2) · (3) + (2) · (-1)] = -10 + 6 - 2 = -12 ⇐
⇒ Step 2 - If v · w = 0, Two Vectors Are Orthogonal, i.e., 90°: since v · w = -12 ≠ 0, vectors not orthogonal ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 3: CROSS PRODUCT [외적]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
CROSS PRODUCT OF TWO VECTORS [ON EXAM - PRACTICED]
⇒ Given - [i] Vector v = ⟨x, y, z⟩ [ii] Vector w = ⟨x, y, z⟩ let v = ⟨2, 4, 6⟩ and w = ⟨3, 2, -5⟩, find cross product v × w ⇐
⇒ Step 1 - Set-up 3 × 3 Determinant Using i, j, k: v × w = first row (i, j, k), second row (v1 = 2, v2 = 4, v3 = 6), third row (w1 = 3, w2 = 2, w3 = -5) ⇐
⇒ Step 2 - Evaluate Using v × w = (v2w3 - v3w2) i - (v1w3 - v3w1) j + (v1w2 - v2w1) k: (-20i + 18j + 4k) + (-18i + 10j -12k) = -32i + 28j - 8k = ⟨-32, 28, -8⟩ ⇐
NORMAL VECTOR OF A PLANE USING CROSS PRODUCT [ON EXAM - PRACTICED]
⇒ Given - [i] Vector v = ⟨x, y, z⟩ [ii] Vector w = ⟨x, y, z⟩ let v = ⟨2, 4, 6⟩ and w = ⟨3, 2, -5⟩, find cross product v × w ⇐
⇒ Step 1 - Set-up 3 × 3 Determinant Using i, j, k: v × w = first row (i, j, k), second row (v1 = 2, v2 = 4, v3 = 6), third row (w1 = 3, w2 = 2, w3 = -5) ⇐
⇒ Step 2 - Evaluate Using v × w = (v2w3 - v3w2) i - (v1w3 - v3w1) j + (v1w2 - v2w1) k: (-20i + 18j + 4k) + (-18i + 10j -12k) = -32i + 28j - 8k = ⟨-32, 28, -8⟩ ⇐
⇒ Step 3- Answer is ⟨-32, 28, -8⟩: since cross product of two vectors on a plane results to be orthogonal to both vectors, meaning that it is normal vector to a plane. ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 4: EQUATIONS OF LINES IN 2D AND 3D [2차원 및 3차원 직선의 방정식]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
EQUATION OF LINE IN 3D IN ALL THREE FORMS GIVEN POINT AND VECTOR [ON EXAM - PRACTICED]
⇒ Given - ____________________________________: __________________________________________ ⇐
⇒ Step 1 - Vector Form: r = r₀ + tv, r₀ = ⟨x₀, y₀, z₀⟩, v = ⟨a, b, c⟩: _________________________________________ ⇐
⇒ Step 2 - Parametric Equation: x = x₀ + at, y = y₀ + bt, z = z₀ + ct: __________________________________________ ⇐
⇒ Step 3 - Symmetric Form: (x − x₀) / a = (y − y₀) / b = (z − z₀) / c: __________________________________________ ⇐
EQUATION OF LINE IN 3D IN ALL THREE FORMS GIVEN TWO POINT [ON EXAM - PRACTICED]
⇒ Given - ____________________________________: __________________________________________ ⇐
⇒ Step 1 - ____________________________________: __________________________________________ ⇐
⇒ Step 2 - Vector Form: r = r₀ + tv = ⟨x₀, y₀, z₀⟩ + t ⟨a, b, c⟩: _________________________________________________ ⇐
⇒ Step 3 - Parametric Equation: x = x₀ + at, y = y₀ + bt, z = z₀ + ct: __________________________________________ ⇐
⇒ Step 4 - Symmetric Form: (x − x₀) / a = (y − y₀) / b = (z − z₀) / c: __________________________________________ ⇐
FIND POINT OF INTERSECTION BETWEEN LINES [ON EXAM - PRACTICED]
⇒ Given - ____________________________________: __________________________________________ ⇐
⇒ Step 1 - ____________________________________: __________________________________________ ⇐
⇒ Step 2 - ____________________________________: __________________________________________ ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 5: EQUATIONS OF PLANES PT. A [평면의 방정식 파트 A]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
EQUATION OF PLANE GIVEN POINT AND NORMAL VECTOR [ON EXAM - PRACTICED]
⇒ Given - ____________________________________: __________________________________________ ⇐
⇒ Step 1 - ____________________________________: __________________________________________ ⇐
⇒ Step 2 - ____________________________________: __________________________________________ ⇐
⇒ Step 3 - ____________________________________: __________________________________________ ⇐
⇒ Step 4 - ____________________________________: __________________________________________ ⇐
⇒ Normal Vector: if ax + by + cz + d = 0, then n = ⟨a, b, c⟩ [법선 벡터] ⇐
⇒ Equation of Plane in 3D: a(x − x₀) + b(y − y₀) + c(z − z₀) = 0 [3차원 평면의 방정식] ⇐
⇒ Equation of Plane in 3D - Normal Form: ax + by + cz + d = 0 [3차원 평면의 법선형 방정식] ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 6: EQUATIONS OF PLANES PT. B [평면의 방정식 파트 B]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
CASE 1: EQUATION OF PLANE GIVEN THREE NON-COLINEAR POINTS ON PLANE [ON EXAM - PRACTICED]
⇒ Given - ____________________________________: __________________________________________ ⇐
⇒ Step 1 - ____________________________________: __________________________________________ ⇐
⇒ Step 2 - ____________________________________: __________________________________________ ⇐
⇒ Step 3 - ____________________________________: __________________________________________ ⇐
⇒ Step 4 - ____________________________________: __________________________________________ ⇐
⇒ Case A - Three Non Colinear Points: n = v₁ × v₂ = ⟨a, b, c⟩, formulate an equation [세 점이 한 직선 위에 있지 않은 경우] ⇐
CASE 2: EQUATION OF PLANE GIVEN TWO PARELLEL LINES ON THE PLANE [ON EXAM - PRACTICED]
⇒ Given - ____________________________________: __________________________________________ ⇐
⇒ Step 1 - ____________________________________: __________________________________________ ⇐
⇒ Step 2 - ____________________________________: __________________________________________ ⇐
⇒ Step 3 - ____________________________________: __________________________________________ ⇐
⇒ Step 4 - ____________________________________: __________________________________________ ⇐
⇒ Case B - Two Parallel Lines: n = (p₂ − p₁) × v, where p are two points and v is vector of lines [두 평행한 직선이 주어진 경우] ⇐
CASE 3: EQUATION OF PLANE GIVEN TWO INTERSECTING LINES OF THE PLANE [ON EXAM - PRACTICED]
⇒ Given - ____________________________________: __________________________________________ ⇐
⇒ Step 1 - ____________________________________: __________________________________________ ⇐
⇒ Step 2 - ____________________________________: __________________________________________ ⇐
⇒ Step 3 - ____________________________________: __________________________________________ ⇐
⇒ Step 4 - ____________________________________: __________________________________________ ⇐
⇒ Case C - Two Intersecting Lines: n = v₁ × v₂ = ⟨a, b, c⟩, only if two lines proven to intersect [두 직선이 한 점에서 만나는 경우] ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 7: INTRODUCTION OF VECTOR-VALUED FUNCTIONS [벡터 값 함수의 개론]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
CURVE PARAMETRIZED IN 2D [PRACTICED]
⇒ Curve Parametrized by Paths: a curve traced by r(t) as t varies represents the path of a moving point [경로로 매개화된 곡선] ⇐
⇒ Circle Parametrized: ⟨sin(x), cos(x)⟩ starts from (0, 1); ⟨cos(x), sin(x)⟩ starts from (1, 0); -t draws backward; bigger t is quicker [원의 매개변수화] ⇐
⇒ Ellipse Parametrized: for r cos(x), r equals to length in x-axis; for r sin(x), r equals length in y-axis; properties same as circles [타원의 매개변수화] ⇐
⇒ Point of Origin for the Curve: for cos(x) + c, c is the x-axis of the point of origin and for sin(x) + c, c is the y-axis of the point of origin [곡선의 원점] ⇐
CURVE PARAMETRIZED IN 3D [PRACTICED]
⇒ Parellel to a Plane: if ⟨cos(x), sin(x), c⟩, then the curve is parallel to the xy plane [평면의 평행] ⇐
⇒ Radius of a Circle: if ⟨5 cos(x), 9, 3 + 5 sin (x)⟩, the radius of a circle is equal to 5 [원의 반지름] ⇐
⇒ Point of a Origin for the Curve: for ⟨4 + cos(x), 9, 3 + sin(x)⟩, the origin is (5, 9, 3) [곡선의 원전] ⇐
⇒ Intersection of Plane and Sphere: plug in plane y = 5 into the sphere formula [평면과 구의 교선] ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 8: CALCULUS OF VECTOR-VALUED FUNCTIONS [벡터 값 함수의 미분]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
CONVERT POSITION VECTOR TO ACCELERATION VECTOR [ON EXAM - PRACTICED]
⇒ Given - ____________________________________: __________________________________________ ⇐
⇒ Step 1 - ____________________________________: __________________________________________ ⇐
⇒ Step 2 - ____________________________________: __________________________________________ ⇐
⇒ Step 3 - ____________________________________: __________________________________________ ⇐
⇒ Step 4 - ____________________________________: __________________________________________ ⇐
CONVERT ACCELERATION VECTOR TO VELOCITY VECTOR [ON EXAM - PRACTICED]
⇒ Given - ____________________________________: __________________________________________ ⇐
⇒ Step 1 - ____________________________________: __________________________________________ ⇐
⇒ Step 2 - ____________________________________: __________________________________________ ⇐
⇒ Step 3 - ____________________________________: __________________________________________ ⇐
⇒ Step 4 - ____________________________________: __________________________________________ ⇐
EVALUATE SPEED AT TIME GIVEN POSITION VECTOR [ON EXAM - PRACTICED]
⇒ Given - ____________________________________: __________________________________________ ⇐
⇒ Step 1 - ____________________________________: __________________________________________ ⇐
⇒ Step 2 - ____________________________________: __________________________________________ ⇐
⇒ Step 3 - ____________________________________: __________________________________________ ⇐
⇒ Step 4 - ____________________________________: __________________________________________ ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 9: ARC LENGTH [호의 길의]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
EQUATION OF TANGENT LINE TO PATH FUNCTION AT GIVEN POINT [ON EXAM - PRACTICED]
⇒ Given - ____________________________________: __________________________________________ ⇐
⇒ Step 1 - ____________________________________: __________________________________________ ⇐
⇒ Step 2 - ____________________________________: __________________________________________ ⇐
⇒ Step 3 - ____________________________________: __________________________________________ ⇐
⇒ Step 4 - ____________________________________: __________________________________________ ⇐
⇒ Tangent Line to a Path: line(t) = r(t₀) + (t - t₀) · r′(t₀) [경로의 접선] ⇐
EVALUATE AN ESTIMATE OF LOCATION AROUND THE POINT OF TANGENT [ON EXAM - PRACTICED]
⇒ Given - ____________________________________: __________________________________________ ⇐
⇒ Step 1 - ____________________________________: __________________________________________ ⇐
⇒ Step 2 - ____________________________________: __________________________________________ ⇐
⇒ Step 3 - ____________________________________: __________________________________________ ⇐
ARC LENGTH OF CURVE TRACED BY PATH FUNCTIONS BETWEEN TWO POINTS [ON EXAM - PRACTICED]
⇒ Given - ____________________________________: __________________________________________ ⇐
⇒ Step 1 - ____________________________________: __________________________________________ ⇐
⇒ Step 2 - ____________________________________: __________________________________________ ⇐
⇒ Step 3 - ____________________________________: __________________________________________ ⇐
⇒ Step 4 - ____________________________________: __________________________________________ ⇐
⇒ Arc Length: L = ∫ₐᵇ speed dt = ∫ₐᵇ ||r(t)|| dt [호의 길이] ⇐
⇒ Problem Approach: solve for speed ^ 2, then measure length [문제풀이 방식] ⇐
LOCATION AT TIME T WHEN TRAVELLED BY THE ARC LENGTH [ON EXAM - PRACTICED]
⇒ Given - ____________________________________: __________________________________________ ⇐
⇒ Step 1 - ____________________________________: __________________________________________ ⇐
⇒ Step 2 - ____________________________________: __________________________________________ ⇐
⇒ Step 3 - ____________________________________: __________________________________________ ⇐
⇒ Step 4 - ____________________________________: __________________________________________ ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 10: FUNCTION IN MULTIPLE VARIABLES [다변수 함수]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
DOMAIN AND RANGE OF MULTIVARIABLE FUNCTION [PRACTICED]
⇒ Given - Multivariate Equation f (x, y): find the domain and range of f (x,y) ⇐
⇒ Step 1 - Check Function Type: is the function fraction, square root, or something else? ⇐
⇒ Step 2 - Check Undefinable Values: in which points does the function results undefined? ⇐
⇒ Step 3 - Define Domain: set of all points (x, y) in R2, [i] under possible condition or [ii] except impossible condition. ⇐
⇒ Step 4- Define Range: set of all possible real numbers which result from the restraint conditions of the defined domain ⇐
⇒ Fraction: [domain] denominator should not be 0 [range] possible output is set of all real numbers [분모가 0이 되지 않는 도메인] ⇐
⇒ Square Root: [domain] value in square root should not be negative [range] possible output, e.g., [0, 3] [루트가 음수가 아닌 도메인] ⇐
SKETCH THE GRAPH [PRACTICED]
⇒ Given - An Equation of f (x, y): sketch the graph of function f (x, y) = 9 - x^2 - y^2 ⇐
⇒ Step 1 - Set f (x, y) = z = c: make the function z = f (x, y) = 9 - x^2 - y^2 as c = 9 - x^2 - y^2 ⇐
⇒ Step 2 - Identify Level Curve as Circle or Straight Line: [circle] x^2 + y^2 = 5 - c [line] y = 5x + 5c ⇐
⇒ Step 3 - Search for Values of C Where The Equation Becomes Undefined: [circle] c = 5 is dot at origin and c > 5 is undefined ⇐
⇒ Step 4 - Find Equation of Level Curve for Possible Values of c: figure out a few equations where c does not result in undefined equation ⇐
⇒ Step 5 - Identify and Sketch Graph Combining Level Curves: figure out what kind of shape the level curves make, e.g., sphere, cone, paraboloid ⇐
SKETCH THE LEVEL CURVE [PRACTICED]
⇒ Given - [i] An Equation of f (x, y) [ii] c = number: sketch the level curve of f (x, y) = x^2 + y^2 for c = 81 ⇐
⇒ Step 1 - Set f (x, y) = z = c: make the function z = f (x, y) = x^2 + y^2 as c = x^2 + y^2 ⇐
⇒ Step 2 - Plug-in the Value of c to the Original Equation: plug in c = 81 to set an equation 81 = x^2 + y^2 ⇐
⇒ Step 3A - Identify Type of Level Curve: [circle] x^2 + y^2 = c [line] y = mx + c [hyperbola] c = xy [hyperbola] c = x^2 - y^2 ⇐
⇒ Step 4 - Sketch the Level Curve When c = number: plug in possible numbers in x to get (x, y) points, then draw identified type involving set of points ⇐
⇒ Circle z = x^2 + y^2: to simply find the radius of the circle as 9 and then draw a circle ⇐
⇒ Hyperbola z = xy: to identify the shape quickly by plugging in x = 1/2, 1, 2, -1/2, -1, -2 ⇐
⇒ Hyperbola z = y^2 - x^2: to set y = sqrt (x^2 - 9) and plug in 3, -3, 5, -5; make sure y can both be negative and positive for all x plug-ins ⇐
⇒ Hyperbola z = x^2 + y^2: to set x = sqrt (y^2 - 9) and plug in 3, -3, 5, -5; make sure x can both be negative and positive for all y plug-ins ⇐
EQUTION OF THE LEVEL CURVE WITH C [PRACTICED]
⇒ Given - [i] An Equation of f (x, y) [ii] c = number: identify the level curve of function f (x, y) for c = 2 ⇐
⇒ Step 1 - Set f (x, y) = z = c: make the function z = f (x, y) = x^2 + y^2 as c = x^2 + y^2 ⇐
⇒ Step 2 - Plug-in Value of c to Above Equation: plug in c = 81 to set an equation 81 = x^2 + y^2 ⇐
⇒ Step 3 - Write Equation in Terms of y as a Function of x: convert the equation as y = sqrt (-x^2 + 81) ⇐
EQUATION OF THE LEVEL CURVE WITH POINT [PRACTICED]
⇒ Given - [i] An Equation of f (x, y) [ii] P (-3, 3): identify the level curve of function f (x, y) that passes through the point P (-3, 3) ⇐
⇒ Step 1 - Plug-in Point (-3, 3) to f (x, y) to Evaluate C: to the equation f (x, y) = 3 - 4x^2 - y^2, plug in f (-3, 3) = 3 - 4(-3)^2 - (3^2) = -42 ⇐
⇒ Step 2 - Plug-in Value of c to Above Equation: plug in c = -42 to the original equation f (x, y) = 3 - 4x^2 - y^2 to get an equation y^2 + 4x^2 = 45 ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 11: LIMIT OF MULTIVARIABLE FUNCTIONS PT. A [다변수 함수의 극한 파트 A]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
EVALUATE THE LIMIT WHERE (X, Y) GOES TO (A, B) [PRACTICED]
⇒ Given - Limit of Function Where (x, y) → (a, b): find the limit given lim (x, y) → (−4, −4) [ x^2 / (x^4 + y^4) ] ⇐
⇒ Step 1 - Plug-in (a, b) to the function f (x): (-4)^2 / [(-4)^4 + (-4)^4] = 16/512 = 1/32 ⇐
⇒ Step 2 - Phrase an Answer: lim (x, y) → (−4, −4) [ x^2 / (x^4 + y^4) ] is equal to 1/32 ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 12: LIMIT OF MULTIVARIABLE FUNCTIONS PT. B [다변수 함수의 극한 파트 B]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LIMIT WHERE (X, Y) GOES TO (0, 0) - DIFFERENT PATHS TO PROVE D.N.E. [ON EXAM - PRACTICED]
⇒ Given - Limit of Function Where (x, y) → (0, 0): show that lim (x, y) → (0, 0) [ 7xy / (x^2 + y^2) ] is undefined ⇐
⇒ Step 1 - Test First Path Where y = 0, i.e., lim (x, 0) → (0, 0): lim (x, 0) → (0, 0) [ 7x(0) / x^2 + (0)^2) ] = lim (x, 0) → (0, 0) lim 0 / x^2 = 0 ⇐
⇒ Step 2 - Test Second Path Where x = 0, i.e., lim (0, y) → (0, 0): lim (0, y) → (0, 0) [ 7(0)y / (0)^2 + y^2 ] = lim (0, y) → (0, 0) lim 0 / y^2 = 0 ⇐
⇒ Step 3 - Test Third Path Where y = x, i.e., lim (x, x) → (0, 0): lim (x, x) → (0, 0) [ 7x(x) / (x)^2 + y^2) ] = lim (x, x) → (0, 0) lim 7x^2 / 2x^2 = 7/2 ⇐
⇒ Step 4 - Determine D.N.E. With Different Limits From Other Paths: since lim (x, 0) → (0, 0) and lim (0, y) → (0, 0) do not match with lim (x, x) → (0, 0) ⇐
LIMIT WHERE (X, Y, Z) GOES TO (0, 0, 0) - DIFFERENT PATHS TO PROVE D.N.E. [ON EXAM - PRACTICED]
⇒ Given - Limit of Function Where (x, y, z) → (0, 0, 0): show that lim (x, y, z) → (0, 0, 0) [ (x^2 - y^2 - z^2) / (x^2 + y^2 - z^2) ] is undefined ⇐
⇒ Step 1 - Test First Path Where y = 0 and z = 0, i.e., lim (x, 0, 0) → (0, 0, 0) : lim (x, 0, 0) → (0, 0, 0) [ x^2 / x^2] = lim (x, 0, 0) → (0, 0, 0) [1] = 1 ⇐
⇒ Step 2 - Test Second Path Where x = 0 and z = 0, i.e., lim (0, y, 0) → (0, 0, 0) : lim (0, y, 0) → (0, 0, 0) [ -y^2 / y^2] = lim (0, y, 0) → (0, 0, 0) [-1] = -1 ⇐
⇒ Step 3 - Test Third Path Where x = 0 and y = 0, i.e., lim (0, 0, z) → (0, 0, 0) : lim (0, 0, z) → (0, 0, 0) [ -z^2 / -z^2] = lim (0, 0, z) → (0, 0, 0) [1] = 1 ⇐
⇒ Step 4 - Determine D.N.E. Due to Different Limits From Paths: since lim (x, 0, 0) and lim (0, 0, z) do not match with lim (0, y, 0), the limit is D.N.E. ⇐
LIMIT WHERE (X, Y, Z) GOES TO (0, 0, 0) - POLAR COORDINATES TO EVALUATE LIMIT [ON EXAM - PRACTICED]
⇒ Given - _____________: _____________________ ⇐
⇒ Step 1 - _____________: _____________________ ⇐
⇒ Step 2 - _____________: _____________________ ⇐
⇒ Step 3 - _____________: _____________________ ⇐
⇒ Only Needs x^2 + y^2 = r^2: since sqrt (x^2 + y^2) = sqrt (r^2) = r, convert the function to sin (r) / r, resulting in 1 ⇐
⇒ Needs All Three Polar Conversions - Limit Exists: when θ exists in converted function, but also multiplier of r, then limit is 0 ⇐
⇒ Needs All Three Polar Conversions - Limit D.N.E.: when θ exists in converted function without r, limit is D.N.E. as answer depends on θ ⇐
LIMIT WHERE (X, Y, Z) GOES TO (0, 0, 0) - SQUEEZE THEOREM TO EVALUATE LIMIT [ON EXAM - PRACTICED]
⇒ Given - _____________: _____________________ ⇐
⇒ Step 1 - _____________: _____________________ ⇐
⇒ Step 2 - _____________: _____________________ ⇐
⇒ Step 3 - Isolate x Inside Absolute Value to be Either + or -: _____________________ ⇐
⇒ Convert Addition in Denominator: when denominator is x^2 + y^2, make y^2 to compare | 2xy^2 / (x^2 + y^2) | ≤ | 2xy^2 / (x^2 | ⇐
⇒ Convert Addition in Denominator + Plug-in sin (x * y) to sin (0 * 0): make sin (xy) as sin (0) = 1 to simplify making l (x, y) and u (x, y) ⇐
⇒ Plug-in tan^-1 (1 / e^xy) to tan ^-1 (1): tan (π / 2) = 1 and (tan^-1 (1) = π / 2, then do the squeeze theorem to (- π / 2) | xy | ≤ f (x) ≤ (π / 2) | xy | ⇐
DETERMINE PARAMETER VALUES TO MAKE PIECE-WISE FUNCTION CONTINUOUS [ON EXAM - PRACTICED]
⇒ Given - _____________: _____________________ ⇐
⇒ Step 1 - _____________: _____________________ ⇐
⇒ Step 2 - _____________: _____________________ ⇐
⇒ Step 3 - _____________: _____________________ ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 13: PARTIAL DERIVATIVE [편도함수]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
FIRST ORDER PARTIAL DERIVATIVE [ON EXAM - PRACTICED]
⇒ Given - _____________: _____________________ ⇐
⇒ Step 1 - _____________: _____________________ ⇐
⇒ Step 2A - Multiplication: for e^x sin (y) (multiplication of different terms) and for ∂f / ∂x, only do derivative to x term ⇐
⇒ Step 2B - Addition: for x + y + xy (addition of terms), only calculate derivative of x term for ∂f / ∂x and y term for ∂f / ∂y ⇐
⇒ Step 2C - Product Rule: for e^x sin (xy) (multiplication of same terms), use product rule while leaving y derivative undone ⇐
SECOND ORDER PARTIAL DERIVATIVE [ON EXAM - PRACTICED]
⇒ Given - _____________: _____________________ ⇐
⇒ Step 1 - _____________: _____________________ ⇐
⇒ Step 2 - _____________: _____________________ ⇐
⇒ Step 3 - _____________: _____________________ ⇐
CLAIRAUT'S THEOREM FOR C^2 FUNCTIONS [ON EXAM - PRACTICED]
⇒ Given - _____________: _____________________ ⇐
⇒ Step 1 - _____________: _____________________ ⇐
⇒ Step 2 - _____________: _____________________ ⇐
⇒ Step 3 - _____________: _____________________ ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 14: TANGENT PLANE AND LINEAR APPROXIMATION [접평면과 선형 근사]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
NORMAL VECTOR TO TANGENT PLANE [PRACTICED]
⇒ Given - [i] Equation z = f (x, y) [ii] P (a, b, c): find normal vector to tangent plane of z = 6e^(x^2 - 2y) at P (8, 16, 6) ⇐
⇒ Step 1 - Set F (x, y, z): convert f (x, y) = z = 7e^(x^2 - 2y) into F (x, y, z) = 7e^(x^2 - 2y) - z = 0 to be used for step 2 and step 3 ⇐
⇒ Step 2 - Evaluate ⟨∂F / ∂x, ∂F / ∂y, ∂F / ∂z⟩: evaluate ∂F / ∂x, ∂F / ∂y, ∂F / ∂z to make vector ⟨12xe^(x^2-4y), -24e^(x^2-4y), -1⟩ ⇐
⇒ Step 3 - Plug in P (4, 8, 7) Into F (x, y, z): plug-in P (4, 8, 7) to an above vector to figure out the unit vector which is ⟨96, -24, -1⟩ ⇐
EQUATION OF TANGENT PLANE [PRACTICED]
⇒ Given - [i] Equation z = f (x, y) [ii] P (a, b, c): find tangent plane to the equation of z = 5x^2 + 2y^2 - 3y at P (3, -3, 72) ⇐
⇒ Step 1 - Set F (x, y, z): convert f (x, y) = z = 5x^2 + 2y^2 - 3y into F (x, y, z) = 5x^2 + 2y^2 - 3y - z = 0 ⇐
⇒ Step 2 - Evaluate ⟨∂F / ∂x, ∂F / ∂y, ∂F / ∂z⟩: evaluate ∂F / ∂x, ∂F / ∂y, ∂F / ∂z to make vector ⟨10x, 4y - 3, -1⟩ ⇐
⇒ Step 3 - Plug-in P (3, -3, 72) to ⟨∂F / ∂x, ∂F / ∂y, ∂F / ∂z⟩: plug-in P (3, -3, 72) to complete unit vector ⟨30, -15, -1⟩ ⇐
⇒ Step 4 - With Unit Vector and Point, Set Equation: with ⟨30, -15, -1⟩ and P (3, -3, 72), set up 0 = 30 (x - 3) - 15 (y + 3) - 1 (z - 72) ⇐
⇒ Step 5 - Evaluate to Make Equation of z: evaluate current equation to set 0 = 30x - 15y - z - 63, then convert the equation to z = 30x - 15y - 63 ⇐
LINEAR APPROXIMATION [PRACTICED]
⇒ Given - [i] Equation z = f (x, y) [ii] Decimal P (x, y) : use a tangent plane to approximate value of f (x, y) at P (3.1, -3.1) ⇐
⇒ Step 1 - Evaluate Partial Derivatives ∂F / ∂x and ∂F / ∂y: _________________________________ ⇐
⇒ Step 2 - Set Formula f (x, y) = f (a, b) + fx (a, b) (x - a) + fy (a, b) (y - b): _____________________ ⇐
⇒ Step 3 - Calculate f (a, b), fx (a, b), fy (a, b), (x - a), and (y - b): ______________________________ ⇐
⇒ Step 4 - Plug-in to Formula and Complete Linear Approximation: ____________________________ ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 15: GRADIENT AND DIRECTIONAL DERIVATIVE [기울기와 방향도함수]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
CALCULATE GRADIENT [ON EXAM - PRACTICED]
⇒ Given - [i] Equation f (x, y) [ii] P (a, b) : calculate the gradient of f (x, y) = xy at P (-1, 2) ⇐
⇒ Step 1 - Evaluate Partial Derivatives ∂f / ∂x and ∂f / ∂y: _____________________ ⇐
⇒ Step 2 - Form Gradient ∇ f (x, y) by Plugging-in ⟨∂f / ∂x, ∂f / ∂y⟩: _____________________ ⇐
⇒ Step 3 - Plug-in P (a, b) to Above Gradient Vector and Evaluate Answer: _____________________ ⇐
CALCULATE DIRECTIONAL DERIVATIVE, I.E., RATE OF CHANGE [ON EXAM - PRACTICED]
⇒ Given - [i] Equation f (x, y) [ii] Direction v = ⟨a, b⟩ [iii] P (x, y): _____________________ ⇐
⇒ Step 1 - Evaluate Partial Derivatives ∂f / ∂x and ∂f / ∂y: _____________________ ⇐
⇒ Step 2 - Form Gradient ∇ f (x, y) by Plugging-in ⟨∂f / ∂x, ∂f / ∂y⟩: _____________________ ⇐
⇒ Step 3 - Plug-in P (a, b) to Above Gradient Vector and Evaluate Answer: _____________________ ⇐
⇒ Step 4 - Evaluate Unit Vector With Formula û = v / ||v|| Based on Direction: _____________________ ⇐
⇒ Step 5 - Evaluate Directional Derivative With Dot Product (Dûf) (x0) = ∇ f (x0) * û: _____________________ ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 16: DIRECTION OF GREATEST INCREASE/DECREASE + GRADIENT AND LEVEL CURVES [최대 증가/감소 방향 + 기울기 및 등고선]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
MAXIMUM/MINIMUM RATE OF CHANGE AT POINT [ON EXAM - PRACTICED]
⇒ Given - [i] Equation f (x, y) [ii] P (a, b): _____________________ ⇐
⇒ Step 1 - Evaluate Partial Derivatives ∂f / ∂x and ∂f / ∂y: _____________________ ⇐
⇒ Step 2 - Form Gradient ∇ f (x, y) by Plugging-in ⟨∂f / ∂x, ∂f / ∂y⟩: _____________________ ⇐
⇒ Step 3 - Plug-in P (a, b) to Above Gradient Vector and Evaluate Answer: _____________________ ⇐
⇒ Step 4A - With Gradient at P (a, b), Evaluate Maximum Rate of Change ||∇ f (a, b)||: _____________________ ⇐
⇒ Step 4B - With Gradient at P (a, b), Evaluate Minimum Rate of Change - ||∇ f (a, b)||: _____________________ ⇐
⇒ Step 4C - With Gradient at P (a, b), Evaluate Minimum Rate of Change 0.5 ||∇ f (a, b)||: _____________________ ⇐
DIRECTION IN WHICH FUNCTION INCREASE/DECREASE FASTEST [ON EXAM - PRACTICED]
⇒ Given - [i] Equation f (x, y) [ii] P (a, b): find direction in which maximum rate of change occurs for f (x, y) = 5x sin (xy) at P (4, 2) ⇐
⇒ Step 1 - Evaluate Partial Derivatives ∂f / ∂x and ∂f / ∂y: _____________________ ⇐
⇒ Step 2 - Form Gradient ∇ f (x, y) by Plugging-in ⟨∂f / ∂x, ∂f / ∂y⟩: _____________________ ⇐
⇒ Step 3 - Plug-in P (a, b) to Above Gradient Vector and Evaluate Answer: _____________________ ⇐
⇒ Step 4 - Create a Unit Vector With Gradient Vector û = ∇ f (a, b) / ||∇ f (a, b)||: _____________________ ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 17: CHAIN RULES [연쇄법칙]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
FUNCTION OF 2 VARIABLES + 1 INDEPENDENT VARIABLE [ON EXAM - PRACTICED]
⇒ Given - [i] Equation z (x, y) [ii] Equatin of x to t [ii] Equation of x to t: calculate dz / dt by letting z (x, y) = x^8 * y where x = t^4 and y = t^9 ⇐
⇒ Step 1 - Sketch Dependency Graph From f to x and y to t: _____________________ ⇐
⇒ Step 2 - Calculate ∂z / ∂x, ∂z / ∂y, dx / dt, dy / dt to be Plugged Next: _____________________ ⇐
⇒ Step 3 - Plug-in to Formula dz / dt = (∂z / ∂x) * (dx / dt) + (∂z / ∂y) * (dy / dt): _____________________ ⇐
⇒ Step 4 - Replace all x terms and y terms in formula by plugging in x = t^4 and y = t^9: _____________________ ⇐
FUNCTION OF 2 VARIABLES + 2 INDEPENDENT VARIABLES [ON EXAM - PRACTICED]
⇒ Given - [i] Equation z (x, y) [ii] x to s and t [ii] x to s and t: calculate dz / ds and dz / dt where z (x, y) = -3x^2 + 9y^2 where x = 9s - 7t and y = -2s + 5t ⇐
⇒ Step 1 - Sketch Dependency Graph From f to x and y to s or t: _____________________ ⇐
⇒ Step 2 - Calculate ∂z / ∂x, ∂z / ∂y, ∂x / ∂s, ∂y / ∂s, ∂x / ∂t, ∂y / ∂t, to be Plugged Next: _____________________ ⇐
⇒ Step 3A - Plug-in to Formula for ∂z / ∂s = (∂z / ∂x) * (∂x / ∂s) + (∂z / ∂y) * (∂y / ∂s): _____________________ ⇐
⇒ Step 3B - Replace all x and y terms by plugging in x = 9s - 7t and y = -2s + 5t: _____________________ ⇐
⇒ Step 4A - Plug-in to Formula for ∂z / ∂t = (∂z / ∂x) * (∂x / ∂t) + (∂z / ∂y) * (∂y / ∂t): _____________________ ⇐
⇒ Step 4B - Replace all x and y terms by plugging in x = 9s - 7t and y = -2s + 5t: _____________________ ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 18: IMPLICIT DIFFERENTIATION [암묵적 미분]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
IMPLICIT DIFFERENTIATION [EXCLUDED FROM EXAM]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 19: OPTIMIZATION - CRITICAL POINTS AND SECOND DERIVATIVE TEST [최적화 - 임계점과 이차 도함수 판정법]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
FIND VALUES OF X, Y, Z CORRESPONDING TO CRITICAL POINT [PRACTICED]
⇒ Given - Equation z = f (x, y): find values of x, y, z which correspond to the critical point of f (x, y) = 5x^2 - 7x + 4y + 4y^2 ⇐
⇒ Step 1 - Evaluate Partial Derivatives ∂f / ∂x and ∂f / ∂y: _____________________ ⇐
⇒ Step 2 - Format Partial Derivatives to Piece-Wise Function: _____________________ ⇐
⇒ Step 3 - Solve for All Values of x and y Satisfying ∇ f (x, y) = ⟨0, 0⟩: ____________________ ⇐
⇒ Step 4 - Plug-in Values of x and y Found into f (x, y) to Find Value of z: _____________________ ⇐
FIND CRITICAL POINTS OF A FUNCTION IN 2 VARIABLES [ON EXAM - PRACTICED]
⇒ Given - Equation z = f (x, y): evaluate all critical points of function f (x, y) = 8xy^2 + 12x^2 - 24xy ⇐
⇒ Step 1 - Evaluate Partial Derivatives ∂f / ∂x and ∂f / ∂y: _____________________ ⇐
⇒ Step 2 - Format Partial Derivatives to Piece-Wise Function: _____________________ ⇐
⇒ Step 3 - Solve for All Values of x and y Satisfying ∇ f (x, y) = ⟨0, 0⟩: ____________________ ⇐
⇒ Step 4 - Define All Critical Points P (x, y) Satisfying Values Found Above: _____________________ ⇐
⇒ Substitution Needed to Find x and y: if each equation in a piece-wise function only has one x and y, then simple substitution works, e.g., 2x - 3y + 5 = 0 ⇐
⇒ Simplification Needed to Find Multiple x and y: if equation in piece-wise function has more than one x or y, then simplify equation, e.g., x ( 16y - 24) = 0 ⇐
⇒ Simplification Result Still Includes x and y Inside: plug in x = 6y to the first equation to find out y-value, then plug y to x = 6y to find x, e.g., x (x - 6y) = 0 ⇐
CLASSIFY CRITICAL POINTS USING SECOND DERIVATIVE TEST [ON EXAM - PRACTICED]
⇒ Given - Equation z = f (x, y): find and classify the critical points of function f (x, y) = 8xy^2 + 12x^2 - 24xy ⇐
⇒ Step 1 to 4 - Solving for Critical Points: after going through step 1 to 4, the critical points include (0, 0), (0,3), (3/4, 3/2) ⇐
⇒ Step 5 - Solve for fxx (x, y), fyy (x, y), and fxy (x, y) to be plugged-in to a formula in next step: _____________________ ⇐
⇒ Step 6 - Set up formula by plugging fxx, fyy, fxy in D (x, y) = fxx (x, y) * fyy (x, y) - (fxy (x, y))^2: _____________________ ⇐
⇒ Step 7 - Calculate Discriminant D (x, y) by Plugging-in Critical Point Into Above Formula: _____________________ ⇐
⇒ Step 8A - If D (x, y) < 0, Then (x0, y0) Is a Saddle Point: _____________________ ⇐
⇒ Step 8B - If D (x, y) > 0, Then Check Local Min or Local Max: _____________________ ⇐
⇒ Step 8C - If fxx (x, y) or fyy (x, y) > 0, Then (x0, y0) Is a Local Minimum: _____________________ ⇐
⇒ Step 8D - If fxx (x, y) or fyy (x, y) < 0, Then (x0, y0) Is a Local Maximum: _____________________ ⇐
⇒ Step 9 - Repeat Step 7 to 8 for All Critical Points: ____________________________________________ ⇐
FIND GLOBAL EXTREMA OF A FUNCTION IN 2 VARIABLES OVER A COMPACT DOMAIN [ON EXAM - PRACTICED]
⇒ Given - [i] Equation z = f (x, y) [ii] Region U: find the global extrema of f (x, y) = (x^2)(y) - x - 3xy^2 over region defined by U = {(x, y): 0 ≤ x ≤ 5, 0 ≤ y ≤ x} ⇐
⇒ Step 1 to 4 - Solving for Critical Points on Interior of U: after going through step 1 to 4, critical point possibilities include (2, 1/3) and (-2, -1/3) ⇐
⇒ Step 5 - Check All Critical Point Whether They Belong Inside Defined Region: while (2, 1/3) satisfies U, (-2, -1/3) does not satisfy U, so drop (-2, -1/3) ⇐
⇒ Step 3 - To Solve for Critical Points on Boundary of U, Set Sides of Boundary U Equations: [i] y = 0 & 0 ≤ x ≤ 5 [ii] x = 5 & 0 ≤ x ≤ 5 [iii] x = y & 0 ≤ x ≤ 5 ⇐
⇒ Step 4A - Alongside Side I, Plug-in P (x, 0) and Evaluate: for the side y = 0 & 0 ≤ x ≤ 5, evaluate f (x, 0) = (x^2) (0) - x - 3(0)y^2 = - x and therefore f (x, 0) = -x ⇐
⇒ Step 4B - Alongside Side I, Convert f (x, 0) to g (x) and Evaluate g' (x) to Check Points of g' (x) = 0: f (x, 0) = g (x) = -x and g' (x) = -1, so g' (x) = 0 is impossible ⇐
⇒ Step 4C - Alongside Side I, Since g'(x) = 0 Nowhere, Critical Points Are Only End Points: critical points for side y = 0 & 0 ≤ x ≤ 5 are endpoints (0, 0) and (0, 5) ⇐
⇒ Step 5A - Alongside Side II, Plug-in P (5, y) and Evaluate: for the side x = 5 & 0 ≤ x ≤ 5, evaluate f (5, y) = 25y - 5 - 15y^2, then therefore f (5, y) = 25y - 5 - 15y^2 ⇐
⇒ Step 5B - Alongside Side II, Convert f (5, y) Into h (x) and Evaluate h' (x) to 0: f (5, y) = h (x) and h' (x) = 25 - 30y to figure out that g' (x) = 0 when y = 25/30 = 5/6 ⇐
⇒ Step 5C - Alongside Side II, Since h'(x) = 0 exists, Three Critical Points Exists: critical points for side x = 5 & 0 ≤ x ≤ 5 are endpoints (5, 0), (5, 5), then also (5, 5/6) ⇐
⇒ Step 6A - Alongside Side III, Plug-in P (x, x) and Evaluate: for the side x = y & 0 ≤ x ≤ 5, evaluate f (x, x) = x^3 - x - 3x^2, then simplify result to f (x, x) = -2x^3 - x ⇐
⇒ Step 6B - Alongside Side III, Convert f (x, x) Into k (x) and Evaluate k' (x) to 0: f (x, x) = k (x) and k' (x) = -6x^2 - 1 to find out k' (x) = 0 is impossible at any point ⇐
⇒ Step 6C - Alongside Side III, Since k'(x) = 0 Nowhere, Critical Points Are Only End Points: critical points for side x = y & 0 ≤ x ≤ 5 are endpoints (0, 0) and (5, 5) ⇐
⇒ Step 7 - List All Critical Points Both on Interior of U and Boundary of U: [i] on three corner endpoints, (0, 0), (5, 0), (5, 5) [ii] side II, (5, 5/6) [iii] interior, (2, 1/3) ⇐
⇒ Step 8 - Plug-in All Previously Listed Critical Points to an Original Function z = f (x, y): f (0, 0) = 0, f (5, 0) = -5, f (5, 5) = -255, f (2, 1/3) = - 4/3, f (5, 5/6) = 65/12 ⇐
⇒ Step 9 - Critical Point With a Biggest Value Is Global Max and Critical Point With a Smallest Value Is Global Min: (5, 5/6) is global max and (5, 5) is global min ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 20: METHOD OF LAGRANGE'S MULTIPLIER PT. A [라그랑주 승수법 파트 A]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LAGRANGE'S MULTIPLIER TO FIND GLOBAL EXTREMA OF A FUNCTION IN 2 VARIABLES [ON EXAM - PRACTICED]
⇒ Given - [i] Equation z = f (x, y) [ii] Constraint Equation g (x, y): use lagrange's multiplier to find extrema of f (x, y) = xy subject to constraint x^2 + y^2 = 1 ⇐
⇒ Step 1A - Formulate System ∇ f (x, y) = λ ∇ g (x, y) and g (x, y): system of ∇ f (x, y) = λ ∇ g (x, y) and g (x, y) = x^2 + y^2 = 1 ⇐
⇒ Step 1B - Evaluate ∇ f (x, y) and ∇ g (x, y): for f (x, y) = xy, ∇ f (x, y) = ⟨y, x⟩ and for g (x, y) = x^2 + y^2, ∇ g (x, y) = ⟨2x, 2y⟩ ⇐
⇒ Step 1C - Formulate Lagrange's Condition System: system of y = λ(2x), x = λ(2y), x^2 + y^2 = 1 by using two above gradients ⇐
⇒ Step 2A - Plug-in x and y into g (x, y): 1 = x^2 + y^2 = (2λx)^2 + (2λy)^2 = 4(λ^2)(x^2) + 4(λ^2)(y^2) = (4λ^2)(y^2 + x^2) ⇐
⇒ Step 2B - Solve for λ Using Plugged-in g (x, y): 1 = (4λ^2)(y^2 + x^2) and since (y^2 + x^2) = 1, 1 = (4λ^2), so λ = ± (1 / 2) ⇐
⇒ Step 3A - Evaluate x When λ = 1 / 2: since y = 2λx, y = 2(1/2)x, so y = x; then use it for 1 = x^2 + y^2 = x^2 + x^2 = 2x^2; x = ± (1 / sqrt (2))⇐
⇒ Step 3B - Critical Point When λ = 1 / 2 and x = 1 / sqrt (2): for x = 1 / sqrt(2), plug-in x and λ values to x = λ(2y) to get (1 / sqrt (2), 1 / sqrt (2) ⇐
⇒ Step 3C - Critical Points When λ = 1 / 2 and x = - 1 / sqrt (2): for x = - 1 / sqrt(2), plug-in x and λ to x = λ(2y) to get (- 1 / sqrt (2), - 1 / sqrt (2)) ⇐
⇒ Step 4A - Evaluate x When λ = - 1 / 2: since y = 2λx, y = 2(-1/2)x, so y = - x; then use it for 1 = x^2 + y^2 = x^2 + (- x)^2 = 2x^2; x = ± (1 / sqrt (2))⇐
⇒ Step 4B - Critical Point When λ = - 1 / 2 and x = 1 / sqrt (2): for x = 1 / sqrt(2), plug-in x and λ values to x = λ(2y) to obtain (1 / sqrt (2), - 1 / sqrt (2) ⇐
⇒ Step 4C - Critical Points When λ = 1 / 2 and x = - 1 / sqrt (2): for x = - 1 / sqrt(2), plug-in x and λ values to x = λ(2y) to get (- 1 / sqrt (2), 1 / sqrt (2)) ⇐
⇒ Step 5 - List Every Critical Points: P (1 / sqrt (2), 1 / sqrt (2)), P (- 1 / sqrt (2), - 1 / sqrt (2)), P (1 / sqrt (2), - 1 / sqrt (2)), P (- 1 / sqrt (2), 1 / sqrt (2)) ⇐
⇒ Step 6 - Plug-in All Critical Points to the Function z = f (x, y): f (1/√2, 1/√2) = 1/2, f (-1/√2, -1/√2) = 1/2, f (1/√2, -1/√2) = -1/2, f (-1/√2, 1/√2) = -1/2 ⇐
⇒ Step 7 - Evaluate Global Max and Global Min: P (1/√2, 1/√2) and P(-1/√2, -1/√2) are global max and (1/√2, -1/√2) and P(-1/√2, 1/√2) are global min ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 21: METHOD OF LAGRANGE'S MULTIPLIER PT. B [라그랑주 승수법 파트 B]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LAGRANGE'S MULTIPLIER TO FIND GLOBAL EXTREMA OF A FUNCTION IN 3 VARIABLES [ON EXAM - PRACTICED]
⇒ Given - [i] Equation f (x, y, z) [ii] Constraint Equation g (x, y, z): find max and min of f (x, y, z) = x - y + 2z subject to constraint x^2 + y^2 + z^2 = 2 ⇐
⇒ Step 1A - Formulate System ∇ f (x, y, z) = λ ∇ g (x, y, z) and g (x, y, z): system of ∇ f (x, y, z) = λ ∇ g (x, y, z) and g (x, y, z) = x^2 + y^2 + z^2 = 2 ⇐
⇒ Step 1B - Evaluate∇ f (x, y, z) and ∇ g (x, y, z): for f (x, y, z), evaluate ∇ f (x, y, z) = ⟨1, -1, 2⟩ and for g (x, y), evaluate∇ g (x, y, z) = ⟨2x, 2y, 2z⟩ ⇐
⇒ Step 1C - Formulate Lagrange's Condition System: system of 1 = λ(2x), -1 = λ(2y), 2 = λ(2y), x^2 + y^2 + z^2 = 2 by using two above gradients ⇐
⇒ Step 1D - Convert Lagrange's Condition System: system of x = 1/(2λ), y = -1/(2λ), z = 1/λ, x^2 + y^2 + z^2 = 2, changing three rows about x, y, z ⇐
⇒ Step 2A - Plug-in x, y, z into g (x, y, z): 2 = (1/2λ)^2 + (-1/2λ)^2 + (1/λ)^2 = (1/(λ^2))(1/4 + 1/4 + 1) = (1/(λ^2))(6/4) = 6/(4λ^2) ⇐
⇒ Step 2B - Solve for λ Using Plugged-in g (x, y, z): 2 = (1/(λ^2))(6/4) and since 1/(λ^2) = 4/3, λ^2 = 3/4, return λ = ± (sqrt (3) / 2) ⇐
⇒ Step 3A - Critical Point P(x, y, z) When λ = (√3)/2: x = 1/(2λ) = 1/[(2)(√3/2)] = 1/√3; y = -1/(2λ) = -1/[(2)(√3/2)] = -1/√3; z = 1/(λ) = 1/(√3/2) = 2/√3 ⇐
⇒ Step 3B - Critical Point P(x, y, z) When λ = - (√3)/2: x = 1/(2λ) = 1/[(2)(-√3/2)] = -1/√3; y = -1/(2λ) = -1/[(2)(-√3/2)] = 1/√3; z = 1/(λ) = 1/(-√3/2) = -2/√3 ⇐
⇒ Step 4 - List Every Critical Points: P (1/√3, -1/√3, 2/√3) and P (-1/√3, 1/√3, -2/√3) are critical points and global extrema candidates ⇐
⇒ Step 5 - Plug-in All Critical Points to the Function f (x, y, z) = x - y + 2z: f (1/√3, -1/√3, 2/√3) = 6/√3 and f (-1/√3, 1/√3, -2/√3) = -6/√3 ⇐
⇒ Step 6 - Evaluate Global Max and Global Min: P (1/√3, -1/√3, 2/√3) is a global maxima and P (-1/√3, 1/√3, -2/√3) is a global minimum ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 22: RIEMANN APPROXIMATION FOR VOLUMES [체적에 대한 리만 근사]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
COMPUTE RIEMANN SUM USING LOWER-LEFT VERTEX SAMPLE POINTS [ON EXAM - PRACTICED]
⇒ Given - [i] Domain R = [A, B] X [C, D] [ii] Riemann Sum S(N,M): let R = [2, 6] X [1, 4] and f (x, y) = (x^2)(y), compute Riemann sum S(2,3) for∬ f (x, y) dA ⇐
⇒ Step 1A - Concert [A, B] X [C, D] into {A ≤ x ≤ B, C ≤ y ≤ D}: convert R = [2, 6] X [1, 4] into {2 ≤ x ≤ 6, 1 ≤ y ≤ 4} and sketch a rectangle bounded by R ⇐
⇒ Step 1B - Subdivision of Rectangular Domain Into Grid of N X M: using S(2,3) meaning N = 2 and M = 3, divide a rectangle into 2 X 3 subdivisions ⇐
⇒ Step 2 - Calculate Δx and Δy With Formula (Total Length / Number of Pieces): using R and S, calculate Δx = (6 - 2) / 2 = 2 and Δy = (4 - 1) / 3 = 1 ⇐
⇒ Step 3A - Find Out All Sample Points Based on Lower-Left Vertex of Subdivided Rectangles: P (2, 1), P (2, 2), P (2, 3), P (4, 1), P (4, 2), P (4, 3) ⇐
⇒ Step 3B - Calculate f (x, y) for All Sample Points: plug-in all six sample points to a function f (x, y) = (x^2)(y) to evaluate 4, 8, 12, 16, 32, 48 ⇐
⇒ Step 3C - Calculate f (sample points) * Δx * Δy for All Sample Points: since Δx = 2, Δy = 1, multiply 2 for previous to get 8, 16, 24, 32, 64, 96 ⇐
⇒ Step 4A - Evaluate Riemann Sum S(n,m) = ∑ ∑ f (Pij) * Δxi * Δyi: for the Riemann sum, add up 8 + 16 + 24 + 32 + 64 + 96 to evaluate 240 ⇐
⇒ Step 4B - Define the Answer: the Riemann sum S(2,3) = 240 is an approximation of the value of double integral over a rectangular domain ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 23: DOUBLE INTEGRAL OVER A RECTANGULAR DOMAIN [직사각형 영역에서 이중적분]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
DOUBLE INTEGRATAL OVER A RECTANGULAR DOMAIN [ON EXAM - PRACTICED]
⇒ Given - [i] Equation z = f (x, y) [ii] Domain R = [A, B] X [C, D]: find volume under paraboloid z = x^2 + y^2 over rectangle [0, 2] X [-1, 1] ⇐
⇒ Step 1 - Convert ∬ f (x, y) dA Into ∫ (B on top of A) [∫ (D on top of C) f (x, y) dy] dx : set-up the volume = ∫ (2 on top of 0) [∫ (1 on top of -1) f (x, y) dy] dx ⇐
⇒ Step 2 - Compute s (x) With Integral of ∫ (D on top of C) f (x, y) dy: ∫ (x^2 + y^2) dy = (x^2)(y) + (y^3)/3 and substitute ∫ f (1) - ∫ f (-1) to get s(x) = 2x^2 + 2/3 ⇐
⇒ Step 3 - Compute Volume V With Integral of ∫ (B on top of C) s (x) dx: ∫ (2x^2 + 2/3) dx = (2x^3)/3 + (2x)/3 and substitute ∫ s (2) - ∫ s (0) to get the volume = 20/3 ⇐
FUBINI'S THEOREM OF SWITCHING ORDER OF INTEGRATION WHEN DOMAIN IS RECTANGLE [PRACTICED]
⇒ Given - [i] Equation z = f (x, y) [ii] Domain R = [A, B] X [C, D]: find volume under paraboloid z = x^2 + y^2 over rectangle [0, 2] X [-1, 1] ⇐
⇒ Step 1 - Convert ∬ f (x, y) dA Into ∫ (D on top of C) ∫ (B on top of A) f (x, y) dx dy : set-up volume = ∫ (1 on top of -1) [∫ (2 on top of 0) f (x, y) dx] dy ⇐
⇒ Step 2 - Compute s (y) With Integral of ∫ (B on top of A) f (x, y) dx: ∫ (x^2 + y^2) dx = (x^2)/3 + xy^2 and substitute ∫ f (2) - ∫ f (0) to get s(y) = 2y^2 + 8/3 ⇐
⇒ Step 3 - Compute Volume v With Integral of ∫ (D on top of C) s(y) dy: ∫ (2y^2 + 8/3) dy = (2y^3)/3 + (8y)/3 and substitute ∫ s (1) - ∫ s (-1) to get volume = 20/3 ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 24: DOUBLE INTEGRAL OVER GENERALIZED REGIONS [일반 영역에서의 이중적분]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
DOUBLE INTEGRAL OVER A NON-RETANGULAR DOMAIN [ON EXAM - PRACTICED]
⇒ Given - [i] Equation z = f (x, y) [ii] Domain D: find volume of solid under paraboloid z = 1 - x^2 - y^2 and above region D = {(x, y): -1 ≤ x ≤ 1, 0 ≤ y ≤ 1 + x^2} ⇐
⇒ Step 1 - Set Outer Integral to Only Include Constants: given -1 ≤ x ≤ 1, outer integral should be set as ∫ (1 on top of -1) s (x) dx ⇐
⇒ Step 2 - Set Inner Integral to Include Variables: given 0 ≤ y ≤ 1 + x^2, inner integral should be ∫ (1 + x^2 on top of 0) f (x, y) dy ⇐
⇒ Step 3 - Compute s (y) With ∫ (1 + x^2 on top of 0) f (x, y) dy: ∫ (1 + x^2 on top of 0) [1 - x^2 - y^2] dy = - x^6 - 2x^3 - x^2 + 2/3 ⇐
⇒ Step 4 - Compute Volume v With Inner Integral ∫ (1 on top of -1) s (x) dx: ∫ (1 on top of -1) [- x^6 - 2x^3 - x^2 + 2/3] dx = - 8 / 25 ⇐
⇒ Rule #1 - Outer Integral: The bounds of outer integral must not include one of the variables, e.g., ∫ (x on top of -1) s (x) dx can not work ⇐
⇒ Rule #2 - Bound Variable: The bound of the integral should never include same variable, e.g., ∫ (x^2 on top of 0) f (x, y) dx can not work ⇐
⇒ Vertically Simple: if D is region given by A ≤ x ≤ B, g1 (x) ≤ y ≤ g2 (x), then D is called vertically simple region expressed as ∫∫ f (x, y) dy dx ⇐
⇒ Horizontally Simple: if D is region given by C ≤ y ≤ D, h1 (y) ≤ x ≤ h2 (y), then D is called horizontally simple region expressed as ∫∫ f (x, y) dx dy ⇐
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
LECTURE 25: CHANGING THE ORDER OF INTEGRATION PT. A [적분 순서의 변환 파트 A]
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
CHANGE ORDER OF INTEGRATION FROM DYDX INTO DXDY [ON EXAM - PRACTICED]
⇒ Given - [i] Equation f (x, y) [ii] Domain D: change the order of integration to evaluate ∫ (1 on top of 0) [∫ (√x on top of 0) xe^(y^5 - 5y) dy] dx ⇐
⇒ Step 1 - Set-up Domain of Current Order of Integration dydx: formulate {D : (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ √x} by using the above bounds of double integral ⇐
⇒ Step 2 - Sketch Region in Graph Based on Defined Domain: draw lines of outer bound x = 0 and y = 0 and inner bound y = 0 and y = √x and interior region ⇐
⇒ Step 3 - Convert Vertically Simple Domain Into Horizontally Simple Region: formulate {D : (x, y) | 0 ≤ y ≤ 1, y^2 ≤ x ≤ 1} using how horizontal lines are bounded ⇐
⇒ Step 4 - Set-up Double Integral With Changed Order Into dxdy: using horizontally simple domain evaluate ∫ (1 on top of 0) [∫ (1 on top of y^2) xe^(y^5 - 5y) dx] dy ⇐
⇒ Step 5 - Compute s (y) Given ∫ (1 on top of y^2) f (x, y) xy: evaluate s (y) which becomes e^(y^5 - 5y) [1/2 - (y^2)/4] ⇐
⇒ Step 6 - Compute Volume With Inner Integral ∫ (1 on top of 0) s (y) dy: evaluate v which becomes 1/10 [-e^(-4) + 1] ⇐
⇒ Important Note: In process of finalizing s (y), fully expand expression to make integral to be applied to each variable. ⇐