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24. 25. a. 0a ! b. 0a ! 0 2 0b ! a 2 0 2 c. 4 0a ! 0b ! 0 0 2 0 2 26. Case 1 If and are collinear, then 2b ! is also collinear with both and . But is perpendicular to and c ! c ! 4c ! b ! 9 0b ! c ! 0 2 , so a ! a ! is perpendicular to . Case 2 If b ! and c ! 2b ! b ! b ! 4c ! are not collinear, then by spanning sets, and span a plane in R 3 , and is in that plane. If a ! 2b ! b ! 4c ! c ! is perpendicular to b ! and c ! , then it is perpendicular to the plane and all vectors in the plane. So, is perpendicular to 2b ! 4c ! a ! . Chapter 6 Test, p. 348 1. Let P be the tail of and let Q be the head of The vector sums 3a ! 1b ! c ! a ! c ! . 24 and 31a ! b ! 2 c ! 4 can be depicted as in the diagram below, using the triangle law of addition. We see that 1a ! b ! PQ ! 2 c ! a ! 1b ! c ! 2 . This is the associative property for vector addition. P b a (a + b) (b + c) c PQ= (a + b) + c = a + (b + c) 2. a. (8, 4, 8) b. 12 c. a 2 3 , 1 3 , 2 3 b 3. 19 4. a. x ! 2b ! 3a ! , y ! 3b ! 5a ! b. a , , 5. a. a ! 1 and b ! b 5 c 11 span R 2 , because any vector (x, y) in R 2 can be written as a linear combination of a ! and b ! . These two vectors are not multiples of each other. b. p 2, q 3 6. a. 11, 12, 292 213, 1, 42 711, b. r ! 2,32 cannot be written as a linear combination of and In other words, r ! p ! q ! . does not lie in the plane determined by p ! and q ! . 7. 13, u 3.61; 73.9° relative to x b Q 8. DE ! DE ! CE ! b ! a ! CD ! Also, BA ! CA ! CB ! BA ! 2b ! 2a ! Thus, DE ! 1 2 BA! Chapter 7 Review of Prerequisite Skills, p. 350 1. v 806 km> h N 7.1° E 2. 15.93 units W 32.2° N 3. z x 4. a. b. c. 13, 2, 72; l 7.87 19, 3, 142; l 16.91 11, 1, 02; l 1.41 d. 12, 0, 92; l 9.22 c. (0, y, z) 6. a. i ! b. 6i ! 7j ! c. 8i ! 2j ! d. 4i ! 7. a. i ! 6j ! 11j ! b. 5i ! 3j ! c. 12i ! j ! k ! 8k ! 3k ! j ! k ! 2k ! 5. a. (x, y, 0) b. (x, 0,z) Section 7.1, pp. 362–364 1. a. 10 N is a melon, 50 N is a chair, 100 N is a computer b. Answers will vary. 2. a. 10 N 30 N C(–2, 0, 1) B(–3, 2, 0) y A(0, 1, 0) D(0, 2, –3) 20 N b. 180° 3. a line along the same direction 4. For three forces to be in equilibrium, they must form a triangle, which is a planar figure. 5. a. The resultant is 13 N at an angle of N 22.6° W. The equilibrant is 13 N at an angle of S 22.6° W. b. The resultant is 15 N at an angle of S 36.9° W. The equilibrant is 15 N at N 36.9° E. 6. a. yes b. yes c. no d. yes 7. Arms 90 cm apart will yield a resultant with a smaller magnitude than at 30 cm apart. A resultant with a smaller magnitude means less force to counter your weight, hence a harder chin-up. 8. The resultant would be 12.17 N at 34.7° from the 6 N force toward the 8 N force. The equilibrant would be 12.17 N at 145.3° from the 6 N force away from the 8 N force. 9. 9.66 N 15° from given force, 2.95 N perpendicular to 9.66 N force 10. 49 N directed up the ramp 11. a. 7 N b. 60° 12. approximately 7.1 N 45° south of east 13. a. 7 b. The angle between f 1 and the resultant is The angle between f ! 16.3°. 1 and the equilibrant is 163.7° . 14. a. 1 N 60° 8 N 60° 1 N 13 N 1 N 60° For these three equal forces to be in equilibrium, they must form an equilateral triangle. Since the resultant will lie along one of these lines, and since all angles of an equilateral triangle are 60°, the resultant will be at a 60° angle with the other two vectors. 670 Answers NEL
. Since the equilibrant is directed opposite the resultant, the angle between the equilibrant and the other two vectors is 180° 60° 120°. ! ! 15. 7.65 N, 67.5° from f 2 toward f 3 16. 45° rope: 175.73 N 30° rope: 143.48 17. 24 cm string: approximately 39.2 N, 32 cm string: approximately 29.4 N 18. 8.5° to the starboard side 19. a. magnitude for resultant and equilbrant 13.75 N b. u 5N 111.3°, u 8N 125.6°, u 10N 136.7° 20. We know that the resultant of these two forces is equal in magnitude and angle to the diagonal line of the ! ! parallelogram formed with f 1 and f 2 as ! legs and ! has diagonal length @ f 1 f2 @. We also know from the ! ! cosine rule that @ f 1 f2 @ 2 ! @ f 1 @ 2 ! @ f 2 @ 2 ! ! 2 @ f 1 @@f2 @ cos f, where f is the supplement to u in, our parallelogram. Since we know f 180 u, then cos f cos 1180 u2 cos u. Thus, ! we ! have @ f 1 f2 @ 2 @ f 1 ! @ 2 @ f 2 ! @ 2 2 @ f 1 ! @@f2 ! @ cos f @ f 1 ! f2 ! @ 2 @ f 1 ! @ 2 @ f 2 ! @ 2 2 @ f 1 ! @@f2 ! @ cos f @ f 1 ! f2 ! @ @ f 1 ! @ 2 @ f 2 ! @ 2 2 @ f 1 ! @@f2 ! @cos u Section 7.2, pp. 369–370 1. a. 84 km> h in the direction of the train’s movement b. 76 km> h in the direction of the train’s movement 2. a. 500 km> h north b. 700 km> h north 3. 304.14, W 9.5° S 4. 60° upstream 5. a. 2 m> s forward b. 22 m> s in the direction of the car 6. 13 m> s, N 37.6° W 7. a. 732.71 km> h, N 5.5° W b. about 732.71 km 8. a. about 1383 km b. about 12.5° east of north 9. a. about 10.4° south of west b. 2 h, 53.1° downstream to the bank 10. a. 5 km> h b. about 0.67 km c. 20 min 11. a. about 18.4° west of north b. about 108 km> h 12. 5 Since her swimming speed is a maximum of 4 km> h, this is her maximum resultant magnitude, which is also the hypotenuse of the triangle formed by her and the river’s velocity vector. Since one of these legs is 5 km> h, we have a triangle with a leg larger than its hypotenuse, which is impossible. 13. a. about 68 m b. 100 s 14. a. about 58.5° , upstream b. about 58.6 s 15. 35 h Section 7.3, pp. 377–378 1. To be guaranteed that the two vectors are perpendicular, the vectors must be nonzero. 2. a ! ! # b is a scalar, and a dot product is only defined for vectors. 3. Answers may vary. For example, let ! ! a ! but 4. a ! ! # ! because c ! a ! b b # ! ! a ! c ! a ! # ! ! a ! î, b ! j ! , c i î # b # b 0, # c! # ! c 0, . a b # ! c 5. 1 6. a. 16 b. 6.93 c. 0 d. 1 e. 0 f. 26.2 7. a. 30° b. 80° c. 53° d. 127° e. 60° f. 120° 8. 22.5 9. a. 2 0a ! 0 2 15 @b ! @ 2 7a ! ! # b b. 10. 0 11. 1 4 6 0x ! 0 2 19x ! # ! ! y 3 @y @ 2 12. a. 1a ! b. 1a ! b ! # ! a ! ! a ! b ! @b ! @ 2 0a ! 0 2 2a ! @b ! @ 2 a ! ! 0a ! b ! # ! a ! ! a ! b ! 2 # ! ! 1a b 2 # ! a # b a 0 # b ! # b 2 # ! ! 1a b 2 # ! a # b a 13. a. 0a ! 0a ! 0 2 @b ! @b ! @ 2 a ! ! # b @ 2 0 2 a ! # ! 1b ! a c ! ! 2 # ! 1b c 2 b. b ! # ! c 0 , vectors are perpendicular. Therefore 0a ! 0 2 @b ! @ 2 0c ! 2 0 , which is the Pythagorean theory. 14. 14 15. 0u ! v ! 0 ! 2 0u ! v ! 2 0 ! # ! ! 1u v 2 1u v 2 ! ! # ! ! 1u v 2 1u v 2 2 0u ! 0 2 2 0v ! 0v ! 2 2u ! 0u ! 2 0v ! 2 0u ! 0 2 2u ! # ! v 0 0 # ! v 0 2 0 16. 3 17. 7 18. d ! c ! b ! b ! d ! c ! # ! ! c ! a 11b # ! ! a 2a 2 # ! a c ! # a ! 1b ! # a ! 21a! # a ! 2 b ! # ! a is a scalar c ! # ! ! a 1b # ! ! a 20a 0 2 c ! # ! ! ! a 1d c 2 # ! a c ! # ! ! d ! a d # ! a c! # ! a # ! a 0 because because 0a ! 0 1 Section 7.4, pp. 385–387 1. Any vector of the form is perpendicular to a ! 1c, c2 . Therefore, there are infinitely many vectors perpendicular to a ! . Answers may vary. For example: 11, 12, 12, 22, 13, 32. 2. a. 0; 90° b. 34 7 0; acute c. 3 6 0; obtuse 3. Answer may vary. For example: a. 10, 0, 12 is perpendicular to every vector in the xy-plane. b. 10, 1, 02 is perpendicular to every vector in the xz-plane. c. 11, 0, 02 is perpendicular to every vector in the yz-plane. 4. a. 11, 2, 12 and 14, 3, 102; b. no 2 a ! # b ! b ! # b ! 0a ! 0 2 a ! # b ! b ! # b ! @b ! @ 2 2b ! # c ! 0c ! 0 2 14, 5, 62 and a 5, 3, 5 6 b NEL Answers 671
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Chapter 1 INTRODUCTION TO CALCULUS
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y x 3 1 , if 3 6 t 6 0 t 6. A func
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What Is Calculus? Two simple geomet
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If a and b were radicals, squaring
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Exercise 1.1 PART A 1. Write the co
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Consider a curve y f 1x2 and a poi
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EXAMPLE 2 Tech Support For help gra
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The slope of the tangent at 13, 42
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INVESTIGATION 4 Tech Support For he
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4. Simplify each of the following d
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18. Copy the following figures. Dra
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A. Determine the average velocity o
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Instantaneous Velocity The velocity
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Solution a. C11002 10V100 1000 1
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Exercise 1.3 C PART A 1. The veloci
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f 1x2 f 1a2 15. Use the alternate
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. Use your results for part a to ap
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The graph shown on your calculator
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limits because, in each case, only
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11. For each function, sketch the g
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EXAMPLE 2 EXAMPLE 3 EXAMPLE 4 Using
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INVESTIGATION Here is an alternate
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In this section, we learned the pro
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A 11. Jacques Charles (1746-1823) d
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A second condition for continuity a
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d. The function F is not continuous
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x 3, if x 3 12. g1x2 e 2 k, if
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Key Concepts Review We began our in
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c. What is the present rate of chan
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16. Evaluate the limit of each diff
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Chapter 2 DERIVATIVES Imagine a dri
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2 4 5. Expand, and collect like te
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Section 2.1—The Derivative Functi
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For the Parabola f1x2 x 2 The slop
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Note that f 1t2 Vt is defined for
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EXAMPLE 6 Reasoning about different
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Exercise 2.1 PART A 1. State the do
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T 15. Match each function in graphs
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EXAMPLE 1 Applying the constant fun
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We conclude this section with the s
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EXAMPLE 6 Connecting the derivative
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x 1 8. Determine the slope of the
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Section 2.3—The Product Rule In t
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EXAMPLE 3 Selecting an efficient st
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EXAMPLE 6 Selecting a strategy to d
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PART B dy 5. Determine the value of
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9. Determine the equation of the ta
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EXAMPLE 1 Applying the quotient rul
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Exercise 2.4 PART A 1. What are the
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Section 2.5—The Derivatives of Co
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with respect to x is equal to the p
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EXAMPLE 7 Combining derivative rule
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Exercise 2.5 PART A 1. Given f 1x2
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Technology Extension: Derivatives o
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Key Concepts Review Now that you ha
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9. For what values of x does the cu
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24. At a manufacturing plant, produ
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Chapter 3 DERIVATIVES AND THEIR APP
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4. Determine the area of each figur
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Section 3.1—Higher-Order Derivati
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Motion on a Straight Line An object
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d. The object moves in a positive d
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EXAMPLE 5 Analyzing motion under gr
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Exercise 3.1 C K PART A 1. Explain
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T 12. A dragster races down a 400 m
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H. Repeat part D for the following
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Only t 3 is in the given interval,
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IN SUMMARY Key Ideas • The maximu
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T PART B 4. Using the algorithm for
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Mid-Chapter Review 1. Determine the
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Section 3.3—Optimization Problems
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For extreme values, set V¿1x2 0.
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Exercise 3.3 C PART A 1. A piece of
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a. Determine the dimensions that sh
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The revenue function is R1x2 17 0
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The minimal cost is approximately $
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T C 8. The fuel cost per hour for r
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CAREER LINK WRAP-UP Investigate and
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10. For each of the following cost
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s1t2 t 26. Find the absolute maxi
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Chapter 4 CURVE SKETCHING If you ar
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5. Determine the derivative of each
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Section 4.1—Increasing and Decrea
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dy Since this is a polynomial funct
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IN SUMMARY Key Ideas • A function
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9. Each of the following graphs rep
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Solution dy First determine dx . dy
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Note that there is no value of x fo
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EXAMPLE 4 Graphing the derivative g
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4. Find the x- and y-intercepts of
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Section 4.3—Vertical and Horizont
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The behaviour of the graph can be i
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EXAMPLE 2 Expressing a polynomial f
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When lim f 1x2 k or lim f 1x2 k,
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For horizontal asymptotes, 3x f 1x2
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Now let's consider the straight lin
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Exercise 4.3 PART A 1. State the eq
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12. Use the features of each functi
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11. a. What does f ¿1x2 7 0 imply
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From these investigations, we can m
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Setting dy dx 0, we obtain 3 1x 2
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The second derivative of f 1x2 is f
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Exercise 4.4 K PART A 1. For each f
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Section 4.5—An Algorithm for Curv
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Therefore, by the second derivative
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Interval 1, 12 11, 0.82 10.8, 22 12
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K A 1x PART B 4. Use the algorithm
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Key Concepts Review In this chapter
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5. For each of the following, check
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17. If f 1x2 x3 8 , determine the
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Chapter 5 DERIVATIVES OF EXPONENTIA
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Radian Measure A radian is the meas
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5. Convert the following angles to
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Section 5.1—Derivatives of Expone
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From the investigation, you should
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EXAMPLE 4 Connecting the derivative
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A C PART B 6. Determine the equatio
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Section 5.2—The Derivative of the
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In general, for the exponential fun
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Solution a. January 1, 1900, is exa
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Section 5.3—Optimization Problems
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The average revenue to the company
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Exercise 5.3 PART A 1. Use graphing
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12. Find the maximum and minimum va
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9. The rapid growth in the number o
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INVESTIGATION 2 A.Using your graphi
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With practice, you will learn how t
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We evaluate f 1x2 at the critical n
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A T 6. Determine the absolute extre
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Derivatives of Composite Functions
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Review Exercise 1. Differentiate ea
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18. The position of a particle is g
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Chapter 5 Test dy 1. Determine the
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8. Use algebraic methods to evaluat
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f 1x2 142e 5x1 24. For each functi
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Review of Prerequisite Skills In th
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CAREER LINK Investigate CHAPTER 6:
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What is a Vector? A vector is a mat
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EXAMPLE 1 B A E D C Connecting vect
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A K B 5. Given the vector AB ! as s
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Section 6.2—Vector Addition In th
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the overall magnitude would be equa
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Solution c b a a - b + c a - b c a
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W N S E The vectors are drawn so th
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Exercise 6.2 PART A 1. The vectors
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10. a. In ! the ! example ! involvi
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0 0 The previous example illustrate
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Solution a. 30˚ y x 2x-y 2x -y 30
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Solution Draw a sketch and determin
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8. The two vectors a ! and b ! are
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Section 6.4—Properties of Vectors
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Demonstrating the distributive law
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IN SUMMARY Key Idea • Properties
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Mid-Chapter Review 1. ABCD is a par
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Section 6.5—Vectors in R 2 and R
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Right-Handed System of Coordinates
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Solution a. A16, 0, 02 is a point o
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There is one further observation th
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C A 14. a. What is the equation of
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R 2 EXAMPLE 1 Representing vectors
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EXAMPLE 2 Representing the sum and
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EXAMPLE 5 Calculating the magnitude
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A 10. Parallelogram is determined b
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EXAMPLE 1 Representing vectors in R
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. Using components, a ! b ! c !
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IN SUMMARY Key Ideas • In R 3 , i
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Section 6.8—Linear Combinations a
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In R 2 , it is possible to take any
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Geometrically, the linear combinati
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IN SUMMARY Key Ideas • In R 2 , O
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Review Exercise 1. Determine whethe
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15. A rectangular prism measuring 3
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Chapter 6 Test 1. The vectors a ! ,
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Section 7.1—Vectors as Forces The
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of 6 cm, also pointing east. The ve
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is drawn approximately to scale, an
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The magnitude of the vertical compo
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! T1 sin 105° 1961sin 45°2 T ! 1
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49 34 30 cos CBA 1 cos CBA 2 120
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C T f 3 40 N O 35 N 45° 30 N f 2 2
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Solution We start by drawing positi
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v a v + w w 20 triangles, Solving t
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A T C 9. A small airplane has an ai
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Perhaps the most important observat
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If we look at the right-angled tria
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Solution Consider the following dia
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A T 10. What is the value of the do
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Expanding, we get b 2 1 2a 1 b 1
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. 0 a ! 0 2 @b ! 112 2 122 2 142
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One of the most important propertie
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4. a. From the set of vectors e11,
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Mid-Chapter Review 1. a. If 0 a ! 0
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Section 7.5—Scalar and Vector Pro
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EXAMPLE 1 Reasoning about the chara
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x EXAMPLE 3 z P (2, 1, 4) g b a O(0
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EXAMPLE 4 Calculating a specific di
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IN SUMMARY Key Idea • A projectio
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B C 12. In the diagram shown, ^ ABC
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a 3 b, k = 1 O b 3 a, k = -1 b a is
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If we carry out an identical proced
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EXAMPLE 2 Calculating cross product
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K A C T 4. Calculate the cross prod
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EXAMPLE 1 Using the dot product to
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. We start by constructing position
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EXAMPLE 4 Using the cross product t
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Review Exercise 1. Given that a !
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18. For the vectors m ! 1 V3, 2, 3
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Chapter 7 Test 1. Given the vectors
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Both of these vectors can be multip
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c. If the point Q121, 232 lies on t
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In this section, the vector and par
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C K A T 6. a. If the equation of a
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EXAMPLE 1 Representing the Cartesia
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It is not possible, in this case, t
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AP ! 1x 4, y 1222 1x 4, y 22.
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u cos 1 a 3 12 1 10 21 25 2 b u
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A T 10. For each pair of lines, det
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is its direction vector. If represe
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EXAMPLE 4 Representing the equation
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C A T 6. a. Determine parametric eq
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13. The Cartesian equation of a lin
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s t s(1, 2, 1) t(0, 2, 1), s, tR P
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After deriving vector and parametri
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EXAMPLE 4 Representing the equation
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A K T 7. a. Determine parameters co
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To derive the equation of this plan
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A number of observations can be mad
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When we considered lines in R 2 , w
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IN SUMMARY Key Idea • The Cartesi
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Section 8.6—Sketching Planes in R
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EXAMPLE 2 Representing the graph of
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EXAMPLE 5 Graphing planes whose Car
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IN SUMMARY Key Idea • A sketch of
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Review Exercise 1. Determine vector
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20. Calculate the acute angle that
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Chapter 8 Test 1. a. Given the poin
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4 13 s2 211 4s2 12 8s2 8 0 1
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Method 2: Again, this result can be
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We can now select any two of the th
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IN SUMMARY Key Ideas • Line and p
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A T 11. a. Show that the lines and
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equations. Since the lines would be
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Solution 1: Interchange equations 1
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Substituting into the second equati
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1: Multiply equation 1 by k, and ad
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K C PART B 4. Solve each system of
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Section 9.3—The Intersection of T
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If we had solved the system using e
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then x 3s 1. Now it is a matter o
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Exercise 9.3 C K PART A 1. A system
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Mid-Chapter Review 1. Determine the
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Section 9.4—The Intersection of T
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1: Create two new equations, 4 and
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Thus, 7y 5t 3. Dividing by 7, we g
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Inconsistent Systems for Three Equa
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Therefore, we can choose ! ! ! m 1
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Exercise 9.4 PART A 1. A student is
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9. Solve each system of equations u
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Section 9.5—The Distance from a P
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EXAMPLE 1 Calculating the distance
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In triangle PQR, sin u d @ QP ! @
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Exercise 9.5 PART A 1. Determine th
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Section 9.6—The Distance from a P
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It is also possible to use the dist
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p 1 p 2 U V d L 1 : r = (-2, 1, 0)
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The required distance between the t
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K A T PART B 2. Determine the follo
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Review Exercise 1. The lines 2x y
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14. You are given the lines , and r
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Chapter 9 Test 1. a. Determine the
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P(1, -2, 4) Q P9 2x - 3y - 4z + 66
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29. A line that passes through the
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Solution a. Differentiate both side
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Exercise PART A 1. State the chain
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dA b. To determine differentiate A
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1 16 dh dt Therefore, at the momen
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11. Two cyclists depart at the same
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Solution a. y ln15x2 Using the cha
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x1x 42 0 x 4 or x 0 But x 0 is
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The Derivatives of General Logarith
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The Derivative of a Composite Funct
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To solve this, we take the natural
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Exercise PART A 1. Differentiate ea
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y eliminating unknowns. In Example
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Properties of a Matrix in Row-Echel
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Exercise PART A 1. Write an augment
Page 591 and 592:
12. Determine the equation of the p
Page 593 and 594:
1 0 0 18 4 (row 3) row 1 £ 0 1 0
Page 595 and 596:
Finally, convert the entry in the s
Page 597 and 598:
Review of Technical Skills Appendix
Page 599 and 600:
3 Evaluating a Function 1. Enter th
Page 601 and 602:
7 Making a Table of Differences To
Page 603 and 604:
4. Cursor along the curve to any po
Page 605 and 606:
5. Display and analyze the results.
Page 607 and 608:
14 Graphing a Trigonometric Functio
Page 609 and 610:
2. Enter the second equation. Press
Page 611 and 612:
Use either the calculator keypad or
Page 613 and 614:
20 Graphing the Derivative of a Fun
Page 615 and 616:
4. Create a function. Right-click t
Page 617 and 618:
composition: the process of combini
Page 619 and 620:
G geometric vectors: vectors that a
Page 621 and 622:
ationalizing the denominator: the p
Page 623 and 624:
Answers Chapter 1 0 0 0 0 0 Review
Page 625 and 626:
d. about 1 e. about 7 8 f. no tang
Page 627 and 628: d. 13. m 3; b 1 14. a 3, b 2, c
Page 629 and 630: Review Exercise, pp. 56-59 1. a. 3
Page 631 and 632: 19. a. y 7 b. y 5x 5 c. y 18x 9
Page 633 and 634: 20. a. 34.3 m> s b. 39.2 m> s c. 54
Page 635 and 636: 17. a. 100 bacteria b. 1200 bacteri
Page 637 and 638: 11. a. b. 160x y 16 0 60x y 61
Page 639 and 640: 3. 1 2x 4. a. x 2 15x 6 b. 6012x
Page 641 and 642: 14. t 1 s; away 15. a. s1t2 kt 2
Page 643 and 644: 14. a. triangle side length 0.96 cm
Page 645 and 646: Review Exercise, pp. 156-159 11. 20
Page 647 and 648: c. i. ii. iii. d. i. ii. iii. 5 4 3
Page 649 and 650: . local minimum: 11.41, 39.62, loca
Page 651 and 652: 2. increasing: x 6 1 and x 7 2, dec
Page 653 and 654: to a zero of its derivative, the nu
Page 655 and 656: 7. (-2, 10) 10 8 y 10. a. 8 y e. 4
Page 657 and 658: 4. hole at x 2; large and negative
Page 659 and 660: sin x b. 1 sin 2 tan x sec x x L
Page 661 and 662: 12. a. no maximum or minimum value
Page 663 and 664: Review Exercise, pp. 263-265 1. a.
Page 665 and 666: x 2. 5. a. 19 000 fish> year b. 23
Page 667 and 668: Section 6.1, pp. 279-281 6. a. b. c
Page 669 and 670: 17. P G 4. Answers may vary. For ex
Page 671 and 672: 11. a. AG ! a ! b ! c ! , b. AG
Page 673 and 674: d. OD ! 11, 1, 12 (0, 0, 1) (1, 0,
Page 675 and 676: Section 6.7, pp. 332-333 1. a. 1i !
Page 677: 24. 25. a. 0a ! b. 0a ! 0 2 0b !
Page 681 and 682: 7. a. scalar projection: 0, vector
Page 683 and 684: 2. a. 3 b. 7 c. 43 d. 217 e. 5 f. 1
Page 685 and 686: cos A 1 cos B 1 cos C 1 area of t
Page 687 and 688: . r ! 11, 1, 02 t10, 1, 12, tR; x
Page 689 and 690: 2. 12, 3, 42 3. P must lie on plane
Page 691 and 692: 25. A plane has two parameters, bec
Page 693 and 694: coincident with the plane, meaning
Page 695 and 696: Mid-Chapter Review, pp. 518-519 1.
Page 697 and 698: 8. a. b. y 1 z 1 4 , 2 c. x 3t
Page 699 and 700: 10. vector equation: r ! 10, 0, 42
Page 701 and 702: . 23. a. b. 2a 1 2 b a 6 7 , 2 7 ,
Page 703 and 704: 16. 17. 18. 19. 2 3 4 m>min 144 m>m
Page 705 and 706: ln x 1 e. ln 2 4x 1 x ln 13x 2 2
Page 707 and 708: Index A Absolute extrema, 130, 132,
Page 709: normal, 461-469, 486, 512, 524 para
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