MAT1300 Final Review. Pieter Hofstra. December 4, 2009

Transcription

1 December 4, 2009

2 Sections from the book to study (8th Edition) Chapter 0: 0.1: Real line and Order 0.2: Absolute Value and Distance 0.3: Exponents and Radicals 0.4: Factoring Polynomials (you may omit the part on the rational zero theorem, pp. 23) 0.5: Fractions and Rationalization

3 Sections from the book to study (8th Edition) Chapter 1: 1.1: Cartesian Plane and Distance (you may skip the part on translating points, p. 39) 1.2: Graphs of Equations 1.3: Lines and Slope (may skip linear depreciation) 1.4: Functions 1.5: Limits 1.6: Continuity (may skip applications)

4 Sections from the book to study (8th Edition) Chapter 2: 2.1: Derivative and Slope of a Graph (you may skip the part on translating points, p. 39) 2.2: Rules of Differentiation 2.3: Rates of Change: velocity and marginals 2.4: Product and Quotient Rules 2.5: Chain Rule 2.7: Implicit Differentation 2.8: Related Rates

5 Sections from the book to study (8th Edition) Chapter 3: 3.1: Increasing and Decreasing Functions 3.2: Extrema and the First Derivative Test 3.3: Concavity and the Second Derivative Test (may skip diminishing returns) 3.4: Optimization Problems 3.5: Business and Economics Applications 3.6: Asymptotes

6 Sections from the book to study (8th Edition) Chapter 4: 4.1: Exponential Functions 4.2: Natural Exponential Functions (may skip Example 4) 4.3: Derivatives of Exponential Functions 4.4: Logarithmic Functions 4.5: Derivatives of Logarithmic Functions 4.6: Exponential Growth and Decay

7 Sections from the book to study (8th Edition) Chapter 5: 5.1 Antiderivatives and Indefinite Integrals 5.2 Substitution and General Power Rule (skip p. 371) 5.3 Exponential and Logarithmic Integrals 5.4 Area and the Fundamental Theorem (up to p. 387) 5.5 Area Bounded by Two Graphs

8 Sections from the book to study (8th Edition) Chapter 6: 6.1 Integration by Parts and Present Value 6.5 Improper Integrals (skip the vertical asymptote examples)

9 Chapter 7: 7.1 The 3d Coordinate System 7.2 Surfaces in Space (only first 2 pages) 7.3 Functions of Several Variables (we only do 2 variables, up to p. 499) 7.4 Partial Derivatives (may skip pp ) 7.5 Extrema of Functions of Two Variables (up to p. 520)

10 Key concepts from Chapter 1: 1 Graphing equations; finding intercepts 2 Break-even analysis; finding a break-even point 3 Slope of a line; finding equation y = mx + b for a line 4 Functions: domain, range; finding domain and range of a function 5 Composite and inverse: finding the inverse of a function, horizontal line test 6 Limits: one- and two-sided; evaluating the limit 7 Operations on limits; calculating limits of polynomials and rational functions 8 Continuity; understand definition and apply to given functions

11 Key concepts from Chapter 2: 1 Definition of derivative: Use definition to calculate derivative of standard functions 2 Interpretation of derivative as slope of tangent line: finding equation of tangent line, demand functions 3 Differentiability: understanding why certain functions are not differentiable at some points 4 Rules for differentiation: be able to use all the rules (and combine them as needed) 5 Implicit differentiation: finding derivative of implicitly defined functions 6 Related Rates; Identifying dependent and independent variables, using the chain rule, solving for unknown rate.

12 Key concepts from Chapter 3: 1 Increasing/decreasing behaviour of a function; using the derivative to find where a function is increasing or decreasing 2 Critical points of a function; finding critical points 3 Local extrema; classifying the critical points using first- or second derivative test 4 Absolute extrema; finding the absolute min or max of a function on a given interval 5 Optimization; solving optimization problems, both geometrical (volume, area, etc.) and economical (cost, revenue, profit) 6 Elasticity: computing price elasticity of demand; elastic vs. inelastic 7 Asymptotes; finding horizontal and vertical asymptotes of rational functions

13 Key concepts from Chapter 4: 1 Exponential functions; natural base e; know graphs, calculation rules 2 Compound interest: know formulas for compound interest and continuous compounding and know how to apply these 3 Logarithms; know graphs, calculation rules; apply logarithms to solve exponential equations, e.g. in compound interest problems 4 Derivatives of exponential functions; know calculation rules 5 Derivatives of logarithms: know calculation rules 6 Exponential growth and decay; setting up a formula, using given data to solve for unknowns; solve problems similar to the examples studied in class (population growth, compound interest, exponential decay

14 Key concepts from Chapter 5: 1 Antiderivatives; general versus particular solutions; finding antiderivatives of polynomial functions; finding a particular solution using initial data 2 Power Rule and Substitution: using these rules to solve integrals; rewriting integrals so that these rules become applicable 3 Exponential and logarithmic integrals; basic integration rules; combining these with substitution rule 4 FTC, Area under graph; finding area under a graph of a function; difference between area and definite integral 5 Area between two curves; finding intersection points; locating area; calculating area using FTC; consumer and producer surplus

15 Key concepts from Chapter 6: 1 Integration by parts; using formula for IBP, recognizing when to use this (as opposed to, say substitution); present value 2 Improper Integrals; definition using limit; divergence vs. convergence; evaluating improper integrals; present value

16 Key concepts from Chapter 7: 1 3d coordinates; x,y,z axis, coordinate planes, general planes 2 Surfaces; skim this section, it helps you understand the rest 3 Functions; evaluating functions; finding the domain of a function; level curves 4 Partial Derivatives; finding partial derivatives using differentiation; geometric interpretation; second partial derivatives 5 Extrema; critical points; finding critical points by solving two equations simultaneously; using second derivative test to classify critical points

17 of the exam: The final exam consists of: 10 MC questions (worth 30% in total) 4 on derivatives and applications 4 on integrals and applications 2 on functions of 2 variables 5 long answer questions (worth 70% in total) one related rates problem one area problem one indefinite integral problem The exam is 3 hours long. No books, notes or calculators are allowed.

18 Problem 1 Suppose that for a certain product, the demand function is given by D(x) = 11 x 2 and the supply function is given by S(x) = 2x + 3. Calculate the producer surplus.

19 Problem 1 Suppose that for a certain product, the demand function is given by D(x) = 11 x 2 and the supply function is given by S(x) = 2x + 3. Calculate the producer surplus. Solution. Find the equilibrium price: D(x) = S(x) 2x + 3 = 11 x 2 x 2 + 2x 8 = 0 (x 2)(x + 4) = 0 x = 2, x = 4

20 Problem 1 Suppose that for a certain product, the demand function is given by D(x) = 11 x 2 and the supply function is given by S(x) = 2x + 3. Calculate the producer surplus. Solution. Find the equilibrium price: D(x) = S(x) 2x + 3 = 11 x 2 x 2 + 2x 8 = 0 (x 2)(x + 4) = 0 x = 2, x = 4 We only consider x = 2. The equilibrium price is p = 7.

21 Problem 1 CS = x 2 7dx = x 2 dx = 4x x = = 16 3 PS = (2x + 3)dx = xdx = 4x x = 8 4 = 4

22 Problem 2 If f (x) is a function such that f (x) = e 2x and f (ln(3)) = 5, find f (0).

23 Problem 2 If f (x) is a function such that f (x) = e 2x and f (ln(3)) = 5, find f (0). Solution. We have f (x) = e 2x dx = 1 2 e2x + C.

24 Problem 2 If f (x) is a function such that f (x) = e 2x and f (ln(3)) = 5, find f (0). Solution. We have f (x) = e 2x dx = 1 2 e2x + C.Use the initial data f (ln(3)) = 5 to solve for C: Thus 5 = 1 2 e2 ln(3) + C = C = C C = 1 2 f (x) = 1 2 e2x + 1 2

25 Problem 3. Calculate: 1 dx 3 x.

26 Problem 3. Calculate: 1 Solution. dx 3 x. We have 1 dx 3 x = lim b b 1 x 1/3 dx = lim b [3 2 x 2/3 b 1] = lim b [3 2 b2/3 ( 3 2 1)] =

27 Problem 3. Calculate: 1 Solution. dx 3 x. We have 1 dx 3 x = lim b b 1 x 1/3 dx = lim b [3 2 x 2/3 b 1] = lim b [3 2 b2/3 ( 3 2 1)] = The limit does not exist, so the integral diverges.

28 Problem 4. Consider the function of two variables f (x, y) = 2x 2 4xy + 4y 2 5x + 9y + 3. Calculate the first-order partial derivatives. Find all critical points. Identify what type of critical points they are (local max, local min or saddle point).

29 Problem 4. Consider the function of two variables f (x, y) = 2x 2 4xy + 4y 2 5x + 9y + 3. Calculate the first-order partial derivatives. Find all critical points. Identify what type of critical points they are (local max, local min or saddle point). Solution. For the derivatives, calculate f x = 4x 4y 5, f y = 4x + 8y + 9

30 Problem 4. Consider the function of two variables f (x, y) = 2x 2 4xy + 4y 2 5x + 9y + 3. Calculate the first-order partial derivatives. Find all critical points. Identify what type of critical points they are (local max, local min or saddle point). Solution. For the derivatives, calculate f x = 4x 4y 5, f y = 4x + 8y + 9 To find critical points, we need f x = 0 and f y = 0. Add the equations to get 4y + 4 = 0 y = 1, x = 1 4

31 Problem 4. Consider the function of two variables f (x, y) = 2x 2 4xy + 4y 2 5x + 9y + 3. Calculate the first-order partial derivatives. Find all critical points. Identify what type of critical points they are (local max, local min or saddle point). Solution. For the derivatives, calculate f x = 4x 4y 5, f y = 4x + 8y + 9 To find critical points, we need f x = 0 and f y = 0. Add the equations to get 4y + 4 = 0 y = 1, x = 1 4 Thus there is one CP, namely ( 1 4, 1).

32 Problem 4. We calculate the second derivatives: f xx = 4, f yy = 8, f xy = 4

33 Problem 4. We calculate the second derivatives: Then f xx = 4, f yy = 8, f xy = 4 D = f xx f yy f 2 xy = 4 8 ( 4) 2 = = 16 > 0

34 Problem 4. We calculate the second derivatives: Then f xx = 4, f yy = 8, f xy = 4 D = f xx f yy f 2 xy = 4 8 ( 4) 2 = = 16 > 0 Since D > 0 and f xx > 0, we have a local min at ( 1 4, 1).

35 Problem 5. The profit function of a company is given by P(x) = 2x 2 200x. The current production level is 100 units, and production is decreasing by 2 units per day. At what rate is profit decreasing?

36 Problem 5. The profit function of a company is given by P(x) = 2x 2 200x. The current production level is 100 units, and production is decreasing by 2 units per day. At what rate is profit decreasing? Solution. This is a related rates problem. Identify the three variables: P, x, t.

37 Problem 5. The profit function of a company is given by P(x) = 2x 2 200x. The current production level is 100 units, and production is decreasing by 2 units per day. At what rate is profit decreasing? Solution. This is a related rates problem. Identify the three variables: P, x, t.since x depends on t and P depends on x we have dp dt = dp dx dx dt

38 Problem 5. The profit function of a company is given by P(x) = 2x 2 200x. The current production level is 100 units, and production is decreasing by 2 units per day. At what rate is profit decreasing? Solution. This is a related rates problem. Identify the three variables: P, x, t.since x depends on t and P depends on x we have dp dt = dp dx dx dt We know dx dt dp dp = 2. We need to find dt. So we must calculate dx : dp dx = 4x 200.

39 Problem 5. The profit function of a company is given by P(x) = 2x 2 200x. The current production level is 100 units, and production is decreasing by 2 units per day. At what rate is profit decreasing? Solution. This is a related rates problem. Identify the three variables: P, x, t.since x depends on t and P depends on x we have dp dt = dp dx dx dt We know dx dt dp dp = 2. We need to find dt. So we must calculate dx : dp dx At x = 100, this gives dp dx = 200. = 4x 200.

40 Problem 5. The profit function of a company is given by P(x) = 2x 2 200x. The current production level is 100 units, and production is decreasing by 2 units per day. At what rate is profit decreasing? Solution. This is a related rates problem. Identify the three variables: P, x, t.since x depends on t and P depends on x we have dp dt = dp dx dx dt We know dx dt dp dp = 2. We need to find dt. So we must calculate dx : dp dx = 4x 200. At x = 100, this gives dp dp dx = 200.Thus dt = 200 ( 2) = 400. Therefore profit is decreasing by $400 per day.

41 Problem 6. Consider f (x, y) = xy. Find the domain of f.

42 Problem 6. Consider f (x, y) = xy. Find the domain of f. Solution. For the domain, note that we must have a non-negative value in the square root. Thus we must have xy 0, i.e. xy 0. This means that either x is positive and y is negative or vice versa. Thus the domain is dom(f ) = {(x, y) x 0&y 0} {(x, y) x 0&y 0} (This region corresponds to the union of the second and fourth quadrants in the plane.)

43 Problem 6. Consider f (x, y) = xy. Find the domain of f. Solution. For the domain, note that we must have a non-negative value in the square root. Thus we must have xy 0, i.e. xy 0. This means that either x is positive and y is negative or vice versa. Thus the domain is dom(f ) = {(x, y) x 0&y 0} {(x, y) x 0&y 0} (This region corresponds to the union of the second and fourth quadrants in the plane.)