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8x 52. cos 8x cos24x sin24x 65. 66. 67. 68. 69. 70. sin x sin 3x cos x cos 3x tan x sin x sin 3x cos x cos 3x cot 2x sin 4x sin 6x cos 4x cos 6x cot x cos 8x cos 4x cos 8x cos 4x cot 6x cot 2x sin x sin y cos x cos y cot x y 2 b a sin x sin y cos x cos y tan x y 2 b a 71. a. Prove that 1 cos x sin x sin x 1 cos x. b. Use part a and the half-angle identity proved in the text to prove that tan x 2 sin x 1 cos x. 72. If x cos 2t and y sin t, find the relation between x and y by eliminating t. 602 Chapter 9 Trigonometric Identities and Proof 73. The horizontal range of a projectile R is given by the equation R vt cos a, where v is the initial velocity of the projectile, t is is the angle between the the time of flight, and t 2v sin a line of fire and the horizontal. If a, g where g is acceleration due to gravity, show that R v2 sin 2a. g 74. The expression sin p 2 of reflection of light waves. Show that this expression can be written as 1 2 sin2u. 2u a b occurs in the theory 75. The expression 1 2 the theory of the motion of a pendulum. Show that this equation can be written as a sin2 sin2 a cos u cos a. is used in 1 2 u 2 b 76. A batter hits a baseball that is caught by a fielder. If the ball leaves the bat at an angle of radians to the horizontal, with an initial velocity of v feet per second, then the approximate horizontal distance d traveled by the ball is given by u d v2 sin u cos u 16. a. If the initial velocity is 90 ft/sec, find the horizontal distance traveled by the ball when u 0.5 radian. b. Use an identity to show that d v2 sin 2u radian and when u 0.75. 32 9.4 Using Trigonometric Identities Objectives • Use identities to solve trigonometric equations Recall that the basic identities are used to simplify expressions and to algebraically solve trigonometric equations. The trigonometric identities introduced in |
Section 9.3 can also be used with the techniques shown in Section 8.3, where equations were rewritten into a basic form and then solved. Example 1 Use Double-Angle Identities Solve 5 cos x 3 cos 2x 3. Solution cos2x. 5 cos x 3 Use a double-angle identity to rewrite cos 2x in terms of 5 cos x 3 cos 2x 3 2 cos2x 1 3 5 cos x 6 cos2x 3 3 6 cos2x 5 cos x 6 0 0 or 2 1 2 cos x 3 0 3 cos x 2 2 cos x 3 2 1 2 1 double-angle identity factor the quadratic expression cos x 3 2 3 cos x 2 0 cos x 2 3 Section 9.4 Using Trigonometric Identities 603 The equation cos x 3 2 has no solutions because cos x always lies between 1 and 1. A calculator shows that the solutions of cos x 2 3 are x 0.8411 2kp and x 0.8411 2kp for any integer k. ■ Example 2 Use Double-Angle Identities Solve the equation sin x cos x 1. Solution Use the double-angle identity to rewrite sin x cos x. 2 sin x cos x sin 2x sin x cos x 1 2 sin 2x Replace sin x cos x with 1 2 sin 2x and multiply both sides by 2. 1 2 sin 2x 1 sin 2x 2 Because the sine of any number must be between and 1, there is no solution to the last equation. Therefore, there is no solution to the original equation. 1 ■ Example 3 Use Double-Angle Identities Find exact solutions of cos2x sin2x 1 2 Solution Because cos2x sin2x cos 2x, the equation can be rewritten cos 2x 1 2 2x cos 2x p 3 x p 6 2pk or pk 1 1 2 2x 5p 3 x 5p 6 2pk pk for any integer k. ■ 604 Chapter 9 Trigonometric Identities and Proof Example 4 Use Addition Identities Find the exact solutions of sin 2x cos x cos 2x sin x 1. Solution The left side of the equation is similar to the right side of the addition identity for sine. sin 1 x y 2 sin x cos y cos x sin y. Substitute 2x for x and x for y. sin 2x cos x cos 2x sin x sin 1 2x x 1 2 sin 3x |
1 For any integer k, 3x sin 3x p 2 x p 6 11 2pk 2pk 3 Example 5 Use Half-Angle Identities Find the solutions of sin x sin, where 0 x 2p. x 2 Solution CAUTION Squaring both sides of an equation may introduce extraneous solutions. Be sure to check all solutions in the original equation. 1 sin x sin x 2 1 cos x sin x ± B 2 sin2x 1 cos x 1 cos2x 1 cos x 2 2 cos2x 1 cos x 2 2 half-angle identity square both sides Pythagorean identity 2 cos2x cos x 1 0 0 cos x 1 2 cos x 1 2 1 2 cos x 1 0 2 or cos x 1 0 cos x 1 cos x 1 2 For any integer k, x 2p 3 2pk x 4p 3 2pk x 0 2pk For 0 x 2p, x 0, 2p 3, or 2p. ■ ■ Section 9.4 Using Trigonometric Identities 605 Solving a sin x b cos x c (Optional) Equations of the form a sin x b cos x c occur often. For the case when c 0, the equations can be rewritten as a sin x b cos x sin x b a cos x tan x b a The last equation can be solved by methods discussed in Section 8.3. For the case when solutions to c 0, a very different approach is needed to find the The procedure involves rewriting the equation as a sin x b cos x c 2 is the angle whose terminal side contains the point (a, b), and where then using the addition identity for the sine function. a 1 sin x a k, a To find and b, where a lies on the positive x-axis and tex at the origin., construct a right triangle in the coordinate plane with sides a is the angle with its ver- a Begin by writing the equation so that the coefficient of sin x, a, is positive. The position of the point (a, b) depends on whether b is positive or negative. If b is positive, the point is in the first quadrant; if b is negative, the point is in the fourth quadrant. Both possibilities are shown in Figures 9.4-1 and 9.4-2. a 2a2 b2 2a2 b2 Divide each side of the original equation by to obtain a sin x a 2a2 b2 sin x b cos |
x b 2a2 b2 cos x c c 2a2 b2 [2] Substitute the equivalent expressions from [1] into equation [2]. cos a sin x sin a cos x c 2a2 b2 Use the addition identity for sine to rewrite the left side of the equation. sin 1 x a 2 c 2a2 b2 The last equation can be solved by using the methods from Section 8.3. The steps of the procedure are summarized in the following box. Positive b In both cases, sin a b 2a2 b2 and cos a. [1] y y x x (a, b) b a2 + b2 α a Figure 9.4-1 a α a2 + b2 b (a, b) Negative b Figure 9.4-2 606 Chapter 9 Trigonometric Identities and Proof a sin x Solving b cos x c, where c 0 Let a, b, and c be nonzero real numbers. To solve a sin x b cos x c 1. Multiply by 1, if needed, to make a positive. 2. Plot the point (a, b) position that contains A and let be the angle in standard on its terminal side. (a, b) 3. Find • the length of the hypotenuse of the reference triangle • expressions that represent • the measure of A sin A and cos A 4. Divide each side of the equation by 2a2 b2 a 2a2 b2 sin x b 2a2 b2 cos x yielding c 2a2 b2 5. Use the addition identity for sine to rewrite the equation. sin(x A) c 2a2 b2 6. Solve the equation using techniques previously discussed. Example 6 Solve a sin x b cos x c Solve the equation 23 sin x cos x 23. Solution Step 1 Make the coefficient of 1. of the equation by sin x positive by multiplying both sides 23 sin x cos x 23 so a 23 Step 2 Sketch a diagram of the angle a that has nal side. See Figure 9.4-3. A b 1 and 23, 1 on its termi- B Step 3 The length of the hypotenuse is a Find sin and cos 3 A from the figure. 23 2 B 1 1 2 2 2. a cos a 23 2 and sin a 1 2 Find a. a p 6 or a 11p 6 Step 4 Divide both sides of the equation in Step 1 |
by the hypotenuse, 23 3 A B 2 1 1 2 2. 2 23 2 sin x 1 2 cos x 23 2 y 1 −1 0 −1 −2 3 α 2 x 1 ( 3, −1) Figure 9.4-3 Section 9.4 Using Trigonometric Identities 607 Step 5 Rewrite the equation by substituting cos 11p 6 for 23 2 and sin 11p 6 for 1 2 and then use the addition identity for sine. CAUTION When substituting sin for, the 11p 6 1 2 sign between the terms changes to. cos 11p 6 sin x sin 11p 6 x 11p 6 b cos x 23 2 23 2 sin a Step 6 Solve the equation. x 11p 6 sin p x 11p 3 6 x p 3 x 3p 2 2pk 11p 6 2pk or 2pk x p 2 2pk for any integer k. Maxima and Minima of 23 1 2 b a x 11p 2p 3 6 x 2p 3 x 7p 6 x 5p 6 2pk 11p 6 2pk 2pk 2pk ■ f(x) a sin x b cos x a sin x b cos x, maximum and miniFor functions of the form mum values can be found by using a technique similar to that described in the algorithm to solve equations of the form a sin x b cos x c. x f 1 2 Example 7 Maximum and Minimum of f(x) a sin x b cos x Find the maximum and minimum of the function 3 sin x 4 cos x. x f 1 2 Solution Note that 232 42 225 5. Let a cos 1 or equivalently a sin 1 4 5, and write the function in the form sin a NOTE The value of not needed to find the maximum or minimum of the function. is x f 1 2 5 3 5 sin x 4 5 a cos x b 5 cos a sin x sin a cos x 2 1 Because the sine function varies between 5 sin is 5 and the minimum is x a 5. 1 2 x a 1 2 and 1, the maximum of ■ 1 5 sin 608 Chapter 9 Trigonometric Identities and Proof To find the values of x that produce the maximum or minimum values of the function, solve the equation a sin x b cos x c, where c is the maximum or minimum value. In Example 7, the maximum value of 5 occurs when 3 sin x 4 cos x 5 3 sin x 4 5 |
5 x a cos x 1 1 sin 1 2 x a sin 11 p 2 x p 2 a 1.5708 0.9273 a cos 1 3 5 0.9273 0.6435 In one revolution, the maximum occurs at approximately 0.6435. Where the minimum value occurs is found in a similar manner. Exercises 9.4 In Exercises 1–27, find all solutions of the equation in the interval [0, 2P]. 1. sin2x 3 cos2x 0 2. sin 2x cos x 0 3. cos 2x sin x 1 4. sin x 2 1 cos x 5. 4 sin2 x 2 b a cos2x 2 6. sin 4x sin 2x 0 7. sin x sin x 1 cos x 1 2 8. sin 2t cos t cos 2t sin t 0 9. sin 2x sin x cos x 0 10. cos x cos 1 2 x (Check for extraneous solutions.) 11. sin 2x cos 2x 0 12. cos 4x cos x sin 4x sin x 0 13. sin 4x cos 2x 14. cos 2x sin2 x 0 15. 2 cos2x 2 cos 2x 1 16. cos x p 2 b a sin x 1 17. sin x p 2 b a cos x 1 18. sin x 23 cos x 0 19. sin x cos x 0 20. sin 2x cos x 0 21. cos 2x cos x 0 22. 1 sin x cos x 2 1 2 23. sin x cos x 1 2 0 24. sin2 x 2 b a cos x 0 25. csc2 x 2 2 sec x 26. 22 sin x 22 cos x 1 27. 2 sin x 2 cos x 22 In Exercises 28–31, find the solution to each equation in the interval [P, P]. 28. sin x cos x 1 29. sin2x cos 2x 1 30. sin 2x cos x 0 31. cos x 23 sin x 1 In Exercises 32–35, solve each equation in [0, 2P). 32. sin 4x sin 2x sin x Section 9.4 Using Trigonometric Identities 609 33. sin 2x sin 1 2 x 34. cos 3x cos x 0 35. sin 3x sin x 0 In Exercises 36–40, 2 1 f x k sin a. Express each function in the form x a b. Find the maximum value |
that c. Find all values of x in x maximum value of 0, 2p. can assume. that give the 36. 37. 38. 39. 40 23 sin x cos x sin x 23 cos x 2 sin x 2 cos x sin x cos x 4 sin x 3 cos Important Concepts Section 9.1 Basic trigonometry identities................ 574 Strategies for proving trigonometric identities... 574 Section 9.2 Addition and subtraction identities • for sine and cosine..................... 582 • for tangent........................... 584 Cofunction identities...................... 586 Section 9.2.A Angle of inclination....................... 589 Angle between two lines................... 591 Section 9.3 Section 9.4 cos 2x Double-angle identities..................... 593 Forms of........................... 595 Power-reducing identities................... 595 Half-angle identities................... 596–598 Product-to-sum identities................... 599 Sum-to-product identities................... 599 a sin x b cos x c Solving Maxima and minima of a sin x b cos x x f 1 2................. 606................... 607 Important Facts and Formulas Addition and Subtraction |
Identities sin sin cos cos tan tan sin x cos y cos x sin y sin x cos y cos x sin y cos x cos y sin x sin y cos x cos y sin x sin y tan x tan y 1 tan x tan y tan x tan y 1 tan x tan y 610 Chapter Review 611 Cofunction Identities sin x cos p 2 a x b cos x sin p 2 a x b tan x cot sec x csc cot x tan a csc x sec a p 2 p 2 x b x b Double-Angle Identities sin 2x 2 sin x cos x cos 2x cos2x sin2x 2 cos2x 1 1 2 sin2x tan 2x 2 tan x 1 tan2x Half-Angle Identities sin cos tan cos x 2 1 cos x B 2 1 cos x sin x sin x 1 cos x Review Exercises In Exercises 1–4, simplify the given expression. Section 9.1 1. 2. sin2t 1 tan2t 2 tan t 4 3 tan2t 3 tan t 2 cos2t sec2t csc t csc2t sec t 3. tan2x sin2x sec2x sin x cos x 4. 1 sin x cos x 21 sin2x 2 1 In Exercises 5–11, determine graphically whether the equation could not possibly be an identity, or write a proof showing that it is. 5. sin4t cos4t 2 sin2t 1 6. 1 2 cos2t cos4t sin4t 7. 9. sin t 1 cos t 1 cos t sin t 8. sin2t cos2t 1 1 cos2t cos2 sin2 sin2t 10. tan x cot x sec x csc x 11. 1 sin x cos x 2 2 sin 2x 1 612 Chapter Review In Exercises 12–16, prove the given identity. 12. 14. sec x 1 tan x tan x sec x 1 1 tan2x tan2x csc2x 16. tan2x sec2x cot2x csc2x 13. cos4x sin4x 1 tan4x cos4x 15. sec x cos x sin x tan x 17. B a. c.? 1 cos2x 1 sin2x tan x 0 B 0 1 sin2x 1 cos2x e. undefined 18. 1 2 1 csc x? 2 sec2x 1 |
cos2x 1 1 tan2x 21 2 sin x 1 sin x 1 1 tan3x 21 1 a. c. e. b. 0 cot x 0 d. sec x b. sin x sin3x d. sin x 1 1 tan2x 2 Section 9.2 In Exercises 19–20, prove the given identity. 19. 20. cos x y 1 2 x y cos 2 cos x cos y 1 cos x y 2 1 cos2x sin2y 1 tan x tan y 21. Evaluate the following in exact form, where the angles and a b satisfy the conditions: sin a 4 5 for p 2 6 a 6 p tan b 7 24 a b b a c. cos 2 2 1 cot y 5 12 b. 2 tan 1 and p 6 x 6 3p 2, and for p 6 b 6 3p 2, with 3p 2 6 y 6 2p, find with p 6 x 6 3p 2, and sec y 13 12 with 3p 2 6 y 6 2p, find a. 22. If sin 23. If cos 24. If b a sin 1 tan x 4 3 x y. 2 1 sin x 12 13 x y. 2 1 sin x 1 4 and 0 6 x 6 p 2, then sin p 3 a x b? 25. If sin x 2 5 and 3p 2 6 x 6 2p, then cos p 4 a x b? 26. Find the exact value of sin 5p 12. Chapter Review 613 27. Express sec x p 2 1 in terms of sin x and cos x. Section 9.2.A 28. Find the angle of inclination of the straight line through the points (2, 6) and 1 2, 2. 2 29. Find one of the angles between the line L through the points (5, 1) and the line M, which has slope 2. 3, 2 2 1 and Section 9.3 30. Evaluate the following in exact form, where the angles and a b satisfy the conditions: sin a 44 125 for p 2 6 a 6 p tan b 15 112 for 3p 2 6 b 6 2p a. sin b 2 c. tan 2a b. cos 2a d. cos a b cos 1 2 1 a b 2 In Exercises 31–34, prove the given identity. 31. 1 cos 2x tan x sin 2x 32. tan x sin x 2 tan x sin2 x 2 33. |
2 cos x 2 cos3 x sin x sin 2x 34. sin 2x 1 tan x cot 2x 35. If tan x 5 12 and sin x 7 0, find sin 2x. 36. If cos x 15 17 and 0 6 x 6 p 2, find sin x 2. 37. If sin x 0, is it true that sin 2x 0? Justify your answer. 38. If cos x 0, is it true that cos 2x 0? Justify your answer. 39. Show 32 23 22 26 2 by computing cos p 12 in two ways, using the half-angle identity and the subtraction identity for cosine. 40. True or false: 2 sin x sin 2x. Justify your answer. 41. If sin x 0.6 and 0 6 x 6 p 2, find sin 2x. 42. If sin x 0.6 and 0 6 x 6 p 2, find sin x 2. Section 9.4 Solve the equation. Find exact solutions when possible and approximate ones otherwise. 43. 5 tan x 2 sin 2x 44. cos 2x cos x 45. 2 cos x sin x 0 46. sin 2x cos Rates of Change in Trigonometry As discussed in Section 3.7, the difference quotient represents the average rate of change of a function over the interval from x to x h average rate of change As the value of h becomes smaller and smaller, the average rate of change approaches the instantaneous rate of change at x as discussed in the Chapter 3 Can Do Calculus. Another way to represent the instantaneous rate of change is with limit notation, shown in the Can Do Calculus in Chapter 8. That is, instantaneous rate of change of a function f at x is the limit of the difference quotient. f 1 x h 2 h lim hS0 f x 1 2 instantaneous rate of change The instantaneous rate of change of a particular function is given by the expression in x that is the simplified form of the limit given above. Example 1 Instantaneous Rate of Change of f (x) sin x Find an expression for the instantaneous rate of change of f sin x. x 1 2 Solution The difference quotient of addition identity for the sine function. x f 1 2 sin x can be simplified by using the Let y h. sin sin 1 1 x y x h 2 2 sin x cos y cos x sin y sin x cos h cos x sin h Therefore, the difference quotient for the sine function can be |
simplified as follows. (See Example 4 of Section 9.2 for details of the simplification.) sin 1 x h 2 h sin x sin x cos h 1 h a b cos x sin h a h b The expression that represents the instantaneous rate of change is found by finding the limit of sin x a cos h 1 h cos x b sin h a h b as h approaches 0. lim hS0 a sin x cos h 1 h a b cos x sin h a h bb instantaneous rate of change 614 NOTE See Chapter 14 for a complete discussion of limits. The two fractional expressions in the limit above were evaluated in the Chapter 8 Can Do Calculus as follows: lim hS0 sin h h 1 and lim hS0 cos h 1 h 0 Substitute the values above into the expression for instantaneous rate of change to find a simpler expression. lim hS0 a sin x cos h 1 h a b cos x sin h a h bb sin x 1 cos x 0 2 21 cos x 1 2 21 1 instantaneous rate of change ■ The expression for the instantaneous rate of change for can be used to find the instantaneous rate of change at any particular value of x. x f 1 2 sin x Example 2 Finding Instantaneous Rate of Change at x k Find the instantaneous rate of change of f sin x when x 1 2 x p 3. Inter- pret the result. Solution Because the instantaneous rate of change of cos x, x g 1 2 1 2 the instantaneous rate of change at x p 3 1 2, or When. sin x is changing unit per unit increase in x. sin x x f 2 1 x p 3 is given by p cos 3 b p 3, is g a Exercises 1 2 x sin x f each result. x 0 a. c. x p 2 1. What is the instantaneous rate of change of 3. What is the instantaneous rate of change of for the following values of x. Interpret for the following values of x? x cos x f Interpret each result. 1 2 b. d. x p 4 x p 6 a. c. x 0 x p 2 b. x p 4 d. x p 6 2. Find the expression for the instantaneous rate of change of f x 1 2 cos x. ■ 615 C H A P T E R 10 Trigonometric Applications A Bridge over Troubled Waters When planning a bridge or building, architects and engineers must determine the stress on cables and other parts of the structure to be sure that all |
parts are adequately supported. Problems like these can be modeled and solved by using vectors. See Exercise 50 in Section 10.6. 616 Chapter Outline 10.1 The Law of Cosines 10.2 The Law of Sines 10.3 The Complex Plane and Polar Form for Complex Numbers 10.4 DeMoivre’s Theorem and nth Roots of Complex Numbers 10.5 Vectors in the Plane 10.6 Applications of Vectors in the Plane 10.6.A Excursion: The Dot Product Chapter Review can do calculus Euler’s Formula Interdependence of Sections 10.1 10.2 > 10.3 10.5 > > 10.4 10.6 Trigonometry has a variety of useful applications in geometry, alge- bra, and the physical sciences. Several applications are discussed in this chapter. 10.1 The Law of Cosines Objectives • Solve oblique triangles by using the Law of Cosines. Sections 6.1 and 6.2 presented right triangle trigonometry and its applications. In this section, the solutions to oblique triangles, ones that do not contain a right angle, are considered. A c B b a Figure 10.1-1 Standard notation of triangles, which is used in this section and the next, is shown in Figure 10.1-1 and is described below. C Each vertex is labeled with a capital letter, and the length of the side opposite that vertex is denoted by the same letter in lower case. The letter A will also be used to label the angle at vertex A, and similarly for B and C. Thus, statements such as A 37° or cos B 0.326 will be made. The first fact needed to solve oblique triangles is the Law of Cosines, whose proof is given at the end of this section. Law of Cosines In any triangle ABC, with lengths a, b, c, as in Figure 10.1-1, a2 b2 c2 2bc cos A b2 a2 c2 2ac cos B c2 a2 b2 2ab cos C 617 618 Chapter 10 Trigonometric Applications You need only memorize one of these equations since each of them provides essentially the same information: the square of the length of one side of a triangle is given in terms of the angle opposite it and the other two sides. Solving the first equation in the Law of Cosines for cos A results in the alternate form of the Law |
of Cosines, given below. In any triangle ABC, with sides of lengths a, b, and c, as in Figure 10.1-1, cos A b2 c2 a2 2bc. The other two equations in the Law of Cosines can be similarly rewritten in an alternate form. In this form, the Law of Cosines provides a description of each angle of a triangle in terms of the three sides. Consequently, the Law of Cosines can be used to solve triangles in the following cases. 1. Two sides and the angle between them are known (SAS) 2. Three sides are known (SSS) Example 1 Solve a Triangle with SAS Information Solve triangle ABC shown in Figure 10.1-2. Solution Because c is the unknown quantity, use the third equation in the Law of Cosines. c2 a2 b2 2ab cos C c2 162 102 2 10 21 c2 356 320 cos 110° c 2356 320 cos 110° c 21.6 16 2 1 cos 110° Take the square root of each side. Use the alternate form of the Law of Cosines to find the measure of angle A. cos A b2 c2 a2 2bc cos A 102 21.62 162 2 1 cos A 0.7172 A cos A 44.2° 1 0.7172 21.6 10 21 2 1 COS Use the calculator in degree mode. key with the Alternate Form: Law of Cosines NOTE When C is a right angle, then c is the hypotenuse and cos C cos 90° 0, so that the third equation in the Law of Cosines becomes the Pythagorean theorem. c2 a2 b2 C 110° 10 16 A c B Figure 10.1-2 NOTE Throughout this chapter, no rounding is done in the actual computation until the final quantity is obtained. Section 10.1 The Law of Cosines 619 Because the sum of the angle measures in a triangle is 180°, B 180° Thus, c 21.6, A 44.2°, and 44.2° 110° 1 B 25.8°. 2 25.8°. ■ A 15 8.3 Example 2 Solve a Triangle with SSS Information Find the angles of triangle ABC shown in Figure 10.1-3. C 20 B Figure 10.1-3 Solution and c 8.3. To find angles, use a 20, b 15, The given information is the alternate form of the Law |
of Cosines. cos A b2 c2 a2 cos A 152 8.32 202 15 2 21 1 cos A 0.4261 A 115.2° 2bc 8.3 2 2ac cos B a2 c2 b2 cos B 202 8.32 152 2 1 cos B 0.7346 B 42.7° 8.3 20 21 2 Use the sum of angle measures in a triangle to find the third angle. Thus, A 115.2°, B 42.7°, C 180° 115.2° 42.7° C 22.1°. 1 and 22.1° 2 ■ Example 3 The Distance Between Two Vehicles 125° Two trains leave a station on different tracks. The tracks make an angle of with the station as the vertex. The first train travels at an average speed of 100 kilometers per hour, and the second train travels at an average speed of 65 kilometers per hour. How far apart are the trains after 2 hours? Solution The first train, A, traveling at 100 kilometers per hour for 2 hours, goes a kilometers. The second train, B, travels a disdistance of kilometers. The situation is shown in Figure 10.1-4. tance of 100 2 200 65 2 130 B 130 125° C Station c 200 A Figure 10.1-4 By the Law of Cosines: c2 a2 b2 2ab cos C c2 1302 2002 2 200 130 c2 56,900 52,000 cos 125° c 256,900 52,000 cos 125° c 294.5 21 2 1 cos 125° The trains are about 294.5 kilometers apart after 2 hours. ■ 620 Chapter 10 Trigonometric Applications Example 4 Find Angles with the Horizontal A 100-foot tall antenna tower is to be placed on a hillside that makes an with the horizontal. It is to be anchored by two cables from angle of the top of the tower to points 85 feet uphill and 95 feet downhill from the base. How much cable is needed? 12° Solution The situation is shown in Figure 10.1-5, where AB represents the tower and AC and AD represent the cables. A b 100 k 85 D C 95 12° B E Figure 10.1-5 If the hillside makes an angle of BEC, angle E is a right angle and angle C measures the angle measures in the triangle. with the horizontal, then in triangle Use the sum of 12°. 12° mCBE 180° 90° 12 |
° 78° 1 As shown in Figure 10.1-5, adjacent angles ABC and CBE form a straight angle, which measures 180°. 2 mABC 180° 78° 102° Using the SAS information in triangle ABC, apply the Law of Cosines. b2 a2 c2 2ac cos B b2 952 1002 2 95 2 b2 19,025 19,000 cos 102° b 219,025 19,000 cos 102° b 151.58 100 21 1 cos 102° The length of the downhill cable is about 151.58 feet. To find the length of the uphill cable, notice that adjacent angles ABC and ABD form a straight angle. mABD 180° mABC 180° 102° 78° Section 10.1 The Law of Cosines 621 Using the SAS information in triangle ABD, apply the Law of Cosines. 100 k2 1002 852 2 85 2 k2 17,225 17,000 cos 78° k 217,225 17,000 cos 78° k 117.01 21 1 cos 78° The length of the uphill cable is about 117.01 feet. The amount of cable needed is the sum of the lengths of the uphill and downhill cables. 151.58 117.01 268.59 Therefore, the length of the cable needed is about 268.59 feet. ■ Proof of the Law of Cosines Given triangle ABC, position it on a coordinate plane so that angle A is in standard position with initial side c and terminal side b. Depending on the size of angle A, there are two possibilities, as shown in Figure 10.1-6. y b C(x, y) a B x A c y a C(x, y) b x A c B Figure 10.1-6 The coordinates of B are (c, 0). Let (x, y) be the coordinates of C. Now C is a point on the terminal side of angle A, and the distance from C to the origin A is b. Therefore, according to the definitions of sine and cosine, the following statements are true. x b y b cos A, or equivalently, x b cos A sin A, or equivalently, y b sin A 2 1 2 b cos cos A c 2 b sin A 0 2 Use the distance formula to find a, the distance from C to B. a 2 1 a 2 a2 1 a2 b2 cos2 A 2bc cos A c2 b2 sin2 A |
a2 b2 cos2 A b2 sin2 A c2 2bc cos A a2 b2 c2 2bc cos A cos2 A sin2 A 2 a2 b2 c2 2bc cos A Rearrange terms. Square each side. Factor out b sin A Simplify. 2 2 b2. 1 2 2 2 1 1 2 Pythagorean identity Substitute for x and y. 622 Chapter 10 Trigonometric Applications This proves the first equation in the Law of Cosines. Similar arguments beginning with angles B or C in standard position prove the other two equations. Exercises 10.1 Standard notation for triangle ABC is used throughout. Use a calculator and round your answers to one decimal place at the end of each computation. In Exercises 1–16, solve the triangle ABC under the given conditions. 1. A 20°, b 10, c 7 2. B 40°, a 12, c 20 3. C 118°, a 6, b 10 4. C 52.5°, a 6.5, b 9 5. A 140°, b 12, c 14 6. B 25.4°, a 6.8, c 10.5 7. C 78.6°, a 12.1, b 20.3 8. A 118.2°, b 16.5, c 10.7 9. a 7, b 3, c 5 10. a 8, b 5, c 10 11. a 16, b 20, c 32 12. a 5.3, b 7.2, c 10 13. a 7.2, b 6.5, c 11 14. a 6.8, b 12.4, c 15.1 15. a 12, b 16.5, c 21.3 16. a 5.7, b 20.4, c 16.8 17. Find the angles of the triangle whose vertices are 1, 4 5, 2 and 0 18. Find the angles of the triangle whose vertices are 2, 1 3, 4, and 6, 1,. 19. Two trains leave a station on different tracks. The tracks make a 112° angle with the station as vertex. The first train travels at an average speed of 90 kilometers per hour and the second at an average speed of 55 kilometers per hour. How far apart are the trains after 2 hours and 45 minutes? 20. One plane flies west from Cleveland at 350 miles per hour. A second plane leaves |
Cleveland at the same time and flies southeast at 200 miles per hour. How far apart are the planes after 1 hour and 36 minutes? 21. The pitcher’s mound on a standard baseball diamond (which is actually a square) is 60.5 feet from home plate. How far is the pitcher’s mound from first base? 2nd base 90 ft 90 ft Pitcher's mound 3rd base 1st base 90 ft 60.5 ft 90 ft Home plate 22. If the straight-line distance from home plate over second base to the center field wall in a baseball stadium is 400 feet, how far is it from first base to the same point in center field? Adapt the figure above. 23. A stake is located 10.8 feet from the end of a closed gate that is 8 feet long. The gate swings open, and its end hits the stake. Through what angle does the gate swing? 8 10.8 24. The distance from Chicago to St. Louis is 440 kilometers, from St. Louis to Atlanta 795 kilometers, and from Atlanta to Chicago 950 kilometers. What are the angles in the triangle with these three cities as vertices? 25. A boat runs in a straight line for 3 kilometers, 45° then makes a 6 kilometers. How far is the boat from its starting point? turn and goes for another 45° 3 6 Start 26. A plane flies in a straight line at 400 miles per 15° hour for 1 hour and 12 minutes. It makes a turn and flies at 375 miles per hour for 2 hours and 27 minutes. How far is it from its starting point? 27. The side of a hill makes an angle of with the horizontal. A wire is to be run from the top of a 175-foot tower on the top of the hill to a stake located 120 feet down the hillside from the base of the tower. What length of wire is needed? 12° 28. Two ships leave port, one traveling in a straight course at 22 miles per hour and the other traveling a straight course at 31 miles per hour. Their courses diverge by How far apart are they after 3 hours? 38°. 29. An engineer wants to measure the width CD of a sinkhole. He places a stake B and determines the Section 10.1 The Law of Cosines 623 measurements shown in the figure below. How wide is the sinkhole? B 103° 1 2 0 ft 7 4 ft C D 30. A straight tunnel is to be dug through a hill. |
Two people stand on opposite sides of the hill where the tunnel entrances are to be located. Both can see a stake located 530 meters from the first person and 755 meters from the second. The angle determined by the two people and the stake (the vertex) is How long must the tunnel be? 77°. 31. One diagonal of a parallelogram is 6 centimeters long, and the other is 13 centimeters long. They form an angle of with each other. How long are the sides of the parallelogram? Hint: The diagonals of a parallelogram bisect each other. 42° 32. A parallelogram has diagonals of lengths 12 and How 15 inches that intersect at an angle of long are the sides of the parallelogram? 63.7°. 33. A ship is traveling at 18 miles per hour from 22° Corsica to Barcelona, a distance of 350 miles. To avoid bad weather, the ship leaves Corsica on a route below). After 7 hours the bad weather has been bypassed. Through what angle should the ship now turn to head directly to Barcelona? south of the direct route (see the figure Barcelona Angle of turn Corsica 22° 34. In aerial navigation, directions are given in degrees clockwise from north. Thus east is 180°, and so on, as shown in the following figure. A plane leaves South Bend for Buffalo, 400 miles away, intending to fly a straight course in the south is 90°, 624 Chapter 10 Trigonometric Applications 70°. After flying 180 miles, the pilot direction realizes that an error has been made and that he has actually been flying in the direction 55°. 0° North 270° West 90° East 180° South a. At that time, how far is the plane from Buffalo? b. In what direction should the plane now go to reach Buffalo? 35. Assume that the earth is a sphere of radius 3980 miles. A satellite travels in a circular orbit around the earth, 900 miles above the equator, making one full orbit every 6 hours. If it passes directly over a tracking station at 2 P.M., what is the distance from the satellite to the tracking station at 2:05 P.M.? 36. Two planes at the same altitude approach an airport. One plane is 16 miles from the control tower and the other is 22 miles from the tower. The angle determined by the planes and the tower, with the tower as the vertex, is far apart are the planes? |
11°. How 37. Assuming that the circles in the following figure are mutually tangent, find the lengths of the sides and the measures of the angles in triangle ABC. 8.23 C B 11.27 13 A 38. Assuming that the circles in the following figure are mutually tangent, find the lengths of the sides and the measures of the angles in triangle ABC. A 20.62 7.35 C B 8.04 39. Critical Thinking A rope is attached at points A and B and taut around a pulley whose center is at C, as shown in the following figure. The rope lies on the pulley from D to E and the radius of the pulley is 1 meter. How long is the rope 40. Critical Thinking Use the Law of Cosines to prove that the sum of the squares of the lengths of the two diagonals of a parallelogram equals the sum of the squares of the lengths of the four sides. Section 10.2 The Law of Sines 625 10.2 The Law of Sines Objectives • Solve oblique triangles by using the Law of Sines • Use area formulas to find areas of triangles Law of Sines In Section 10.1, the Law of Cosines was used to solve oblique triangles when SAS or SSS information was given. When different information is given about the triangle, the Law of Cosines may not be sufficient to solve them. In this case, the Law of Sines may be used. In any triangle ABC (in standard notation) a sin A b sin B c sin C. Proof Position triangle ABC on a coordinate plane so that angle C is in standard position, with initial side b and terminal side a, as shown in Figure 10.2-1. 90° < C < 180° y 0° < C < 90 Figure 10.2-1 a B h b D c x A In each case, sin C can be computed by using the point B on the terminal side of angle C. The second coordinate of B is h, and the distance from B to the origin is a. Therefore, by the definition of sine, sin C h a, or equivalently, h a sin C. In each case, right triangle ADB shows that sin A opposite hypotenuse h c, or equivalently, h c sin A. Combine these two expressions for h. a sin C c sin A Because angles in a triangle are nonzero, sin each side of the last equation by (sin A)( |
sin C). A 0 and sin C 0. Divide a sin A c sin C 626 Chapter 10 Trigonometric Applications This proves one proportion in the Law of Sines. Similar arguments beginning with angles A or B in standard position prove the other proportions. The Law of Sines can be used to solve triangles in the following cases. 1. Two angles and one side are known (AAS) 2. Two sides and the angle opposite one of them are known (SSA) Example 1 Solve a Triangle with AAS Information B 20°, C 31°, If lengths. See Figure 10.2-2. and b 210, find the other angle measure and side C 31° 210 a 20° A c B Figure 10.2-2 Solution Because the sum of the angle measures of a triangle is A 180° 20° 31° 129°. 180°, 1 2 In order to find a, notice that three of the four quantities in one of the Law of Sines proportions are known. a sin A a sin 129° b sin B 210 sin 20° a 210 sin 129° a 477.2 sin 20° Multiply each side by sin 129°. c is found in a similar manner. Use a Law of Sines proportion involving c and the three known quantities. c sin C c sin 31° b sin B 210 sin 20° c 210 sin 31° c 316.2 sin 20° Multiply each side by sin 31°. Therefore, A 129°, a 477.2, and c 316.2. ■ Section 10.2 The Law of Sines 627 The Ambiguous Case Given AAS information, such as in Example 1, there is exactly one triangle satisfying the given data (see note). But when two sides of a triangle and the angle opposite one of them are known, as in the SSA case, there may be one, two, or no triangles that satisfy the given data. This is called the ambiguous case. NOTE The four Triangle Congruence Theorems—AAS, SAS, SSS, and ASA—state that a unique triangle can be formed that is congruent to a given triangle. However, SSA is not a triangle congruence theorem because a unique triangle congruent to the given one is not guaranteed; zero, one, or two triangles can be formed from the side-side-angle information. To see why the ambiguous case occurs, suppose sides a and b and angle A are given |
. Place angle A in standard position with terminal side b. If angle A is less than then there are four possibilities for side a. 90°, Ambiguous Case: SSA Information with A 66 90 (i) a b, and side a is too short to reach the third side: no solution. (ii) a b, and side a just reaches the third side and is perpendicular to it: one solution Figure 10.2-3 Figure 10.2-4 (iii) a b, and an arc of radius a (iv) meets the third side at 2 points to the right of A: two solutions. a b, so that an arc of radius a meets the third side at just one point to the right of A: one solution Figure 10.2-5 A B Figure 10.2-6 628 Chapter 10 Trigonometric Applications If angle A is greater than 90°, then there are only two possibilities. Ambiguous Case: SSA Information with A 7 90° (i) so that side a is too a b, short to reach the third side: no solution. so that an arc of radius a 7 b, a meets the third side at just one point to the right of A: one solution. (ii Figure 10.2-7 Figure 10.2-8 Recall from Section 8.3 that when finding all solutions of a trigonometric equation involving the sine function, the identity is used. This same identity, stated in terms of degrees rather than radians, is used to deal with the ambiguous SSA case. sin x sin p x 2 1 Supplementary Angle Identity If 0 U 90, then sin U sin(180 U). D h E y r θ 180° − θ O x Figure 10.2-9 Proof Place the angle in standard position and choose a point D on its terminal side. Let r be the distance from D to the origin. The situation is shown in Figure 10.2-9. 180° u Because h is the second coordinate of D, then sin 180° u 1 h r. 2 Right tri- angle OED shows that sin u opposite hypotenuse h r sin 1 180° u. 2 Example 2 Solve a Triangle with SSA Information Given a possible triangle ABC with a 6, b 7, and A 65°, find angle B. Solution Use a proportion of the Law of Sines involving angle B and three known quantities. Section 10.2 The Law of Sines 629 a sin A |
6 b sin B 7 sin B sin 65° sin B 7 sin 65° 6 sin B 1.06 There is no angle B whose sine is greater than 1. Therefore, there is no triangle satisfying the given data. ■ Example 3 Solve a Triangle with SSA Information An airplane A takes off from carrier B and flies in a straight line for 12 kilometers. At that instant, an observer on destroyer C, located 5 kilometers from the carrier, notes that the angle determined by the carrier, the destroyer (the vertex), and the plane is How far is the plane from the destroyer? 37°. Solution The given information is organized in Figure 10.2-10. B 5 37° C 12 b A Figure 10.2-10 Because there is no proportion of the Law of Sines using B that can be solved, use a proportion involving A to find a second angle of the triangle. This will provide you with enough information to find angle B and hence allowing you to use the Law of Sines to find b. c sin C 12 a sin A 5 sin 37° sin A sin A 5 sin 37° 12 sin A 0.2508 A 14.5° or A 180° 14.5° 165.5° 630 Chapter 10 Trigonometric Applications C 37°, Because that is impossible, then the sum of angles A, B, and C would be is the only possible 14.5° A 165.5° If and greater than 180°. measure of angle A. Therefore, B 180° 37° 14.5° 1 2 128.5°. All of the angles are known, and b can be found using either the Law of Cosines or the Law of Sines. b sin B b sin 128.5° c sin C 12 sin 37° b 12 sin 128.5° b 15.6 sin 37° The plane is approximately 15.6 kilometers from the destroyer. ■ Example 4 Solve a Triangle with SSA Information Solve triangle ABC when a 7.5, b 12, and A 35°. Solution Use a proportion of the Law of Sines involving the known quantities. a sin A 7.5 b sin B 12 sin B sin B 12 sin 35° 7.5 sin B 0.9177 sin 35° B 66.6° or B 180° 66.6° 113.4° Because the sum of angles A and B is less than are two possible cases, as shown in Figure 10.2-11. 180 |
° in each case, there C 12 7.5 7.5 35° A c 113.4° B c Figure 10.2-11 66.6° B Section 10.2 The Law of Sines 631 B 66.6° Case 1 C 180° 35° 66.6° 78.4° 2 1 B 113.4° Case 2 C 180° 35° 113.4° 31.6° 2 1 Use the Law of Sines. Use the Law of Sines. c sin C c sin 78.4° a sin A 7.5 sin 35° c 7.5 sin 78.4° c 12.8 sin 35° c sin C c sin 31.6° a sin A 7.5 sin 35° c 7.5 sin 31.6° c 6.9 sin 35° Thus, in Case 1, B 113.4°, C 31.6°, B 66.6°, C 78.4°, c 6.9. and and c 12.8; and in Case 2, ■ Example 5 Solve a Triangle with ASA Information A plane flying in a straight line parallel to the ground passes directly over point A and later directly over point B, which is 3 miles from A. A few minutes after the plane passes over B, the angle of elevation from A to the plane is How high is the plane at that moment? and the angle of elevation from B to the plane is 67°. 43° Solution If C represents the plane, then the situation is represented in Figure 10.2-12. The height of the plane is h. C h a 67° 43° A 3 B D Figure 10.2-12 Angle ABC measures 180° 67° 113° mBCA 180° 1. So 43° 113° 24°. 2 632 Chapter 10 Trigonometric Applications Use the Law of Sines to find side a of triangle ABC. a sin 43° 3 sin 24° a 3 sin 43° sin 24° a 5.03 Now use sin 67° h a to find h in right triangle CBD. sin 67° h 5.03 h 5.03 sin 67° h 4.63 The plane is about 4.63 miles high. ■ The Area of a Triangle The proof of the Law of Sines leads to the following formula for the area of a triangle. Area of a Triangle The area of a triangle containing an angle C with adjacent sides of lengths a and b is 1 2 � |
� ab sin C. y c B h a D C b Figure 10.2-13 x A Proof Place the vertex of angle C at the origin, with side b on the positive x-axis, as in Figure 10.2-13. Then b is the base and h is the height of the triangle. area of triangle ABC 1 2 base height 1 2 bh. The proof of the Law of Sines shows that area of triangle ABC 1 2 h a sin C. bh 1 2 ab sin C. Therefore, Figure 10.2-13 is the case when C is greater than C is less than is similar. 90° 90°; the argument when Example 6 Find Area with SAS Information Find the area of the triangle shown in Figure 10.2-14. Section 10.2 The Law of Sines 633 8 cm 130° 13 cm Figure 10.2-14 Solution Use the new formula for the area of a triangle. 1 2 ab sin C 1 2 1 8 13 2 21 sin 130° 39.83 Thus, the area is about 39.83 square centimeters. ■ An alternate formula for the area of a triangle, Heron’s formula, gives the area in terms of its sides. Heron’s Formula The area of a triangle with sides a, b, and c is 2s(s a)(s b)(s c), where s 1 2 (a b c). This formula is proved in Exercise 62. Example 7 Find Area with SSS Information Find the area of the triangle whose sides have lengths 7, 9, and 12. Solution Let a 7, b 9, and c 12. To use the Heron’s formula, first find s 12 14 2 Now, use Heron’s formula. s a s b 2s s c 1 21 21 2 14 7 214 21 1 2980 31.3 14 9 14 12 2 21 The area is about 31.3 square units. ■ 634 Chapter 10 Trigonometric Applications Exercises 10.2 Standard notation for triangle ABC is used throughout. Use a calculator and round off your answers to one decimal place at the end of each computation. In Exercises 1–8, solve triangle ABC under the given conditions. 1. A 48°, B 22°, a 5 2. B 33°, C 46°, b 4 3. A 116°, C 50°, a 8 4. A 105°, B 27°, b 10 5. |
B 44°, C 48°, b 12 6. A 67°, C 28°, a 9 7. A 102.3°, B 36.2°, a 16 8. B 97.5°, C 42.5°, b 7 In Exercises 9–16, find the area of triangle ABC under the given conditions. 9. a 4, b 8, C 27° 10. b 10, c 14, A 36° 11. c 7, a 10, B 68° 12. a 9, b 13, C 75° 13. a 11, b 15, c 18 14. a 4, b 12, c 14 15. a 7, b 9, c 11 16. a 17, b 27, c 40 In Exercises 17 – 36, solve the triangle. The Law of Cosines may be needed in Exercises 27–36. 17. b 15, c 25, B 47° 18. b 30, c 50, C 60° 19. a 12, b 5, B 20° 20. b 12.5, c 20.1, B 37.3° 21. a 5, c 12, A 102° 22. a 9, b 14, B 95° 23. b 11, c 10, C 56° 24. a 12.4, c 6.2, A 72° 25. A 41°, B 67°, a 10.5 26. a 30, b 40, A 30° 27. b 4, c 10, A 75° 28. a 50, c 80, C 45° 29. a 6, b 12, c 16 30. B 20.67°, C 34°, b 185 31. a 16.5, b 18.2, C 47° 32. a 21, c 15.8, B 71° 33. b 17.2, c 12.4, B 62.5° 34. b 24.1, c 10.5, C 26.3° 35. a 10.1, b 18.2, A 50.7° 36. b 14.6, c 7.8, B 40.4° In Exercises 37 and 38, find the area of the triangle with the given vertices. 37. 38. 1 1, 0, 0 2 4, 2 2, 5 5, 1 2 3, 0 2 In Exercises 39 and 40, find the area of the polygonal |
region. Hint: Divide the region into triangles. 39. 120 55 89° 96° 103° 68.4 72° 40. 40 135 80° 20 120° 135° 23 1 3 75° 130° 30 30 41. A surveyor marks points A and B 200 meters apart on one bank of a river. She sights a point C on the opposite bank and determines the angles shown in the figure below. What is the distance from A to C? C 57° A 42° B 42. A forest fire is spotted from two fire towers. The triangle determined by the two towers and the fire has angles of and If the towers are 3000 meters apart, which one is closer to the fire? at the tower vertices. 28° 37° 43. A visitor to the Leaning Tower of Pisa observed that the tower’s shadow was 40 meters long and that the angle of elevation from the tip of the The shadow to the top of the tower was tower is now 54 meters tall, measured from the ground to the top along the center line of the a tower (see the figure). Approximate the angle that the center line of the tower makes with the vertical. 57°. α Section 10.2 The Law of Sines 635 angle of elevation from the end of the pole’s shadow to the top of the pole is the pole? 53°. How long is 45. A side view of a bus shelter is shown in the following figure. The brace d makes an angle of with the back and an angle of with 37.25° the top of the shelter. How long is the brace? 34.85° d 8 ft 5 ft 46. A straight path makes an angle of with the 6° horizontal. A statue at the higher end of the path casts a 6.5-meter shadow straight down the path. The angle of elevation from the end of the shadow to the top of the statue is the statue? How tall is 32°. 47. A vertical statue 6.3 meters high stands on top of a hill. At a point on the side of the hill 35 meters from the statue’s base, the angle between the hillside and a line from the top of the statue is 10°. What angle does the side of the hill make with the horizontal? 48. A fence post is located 36 feet from one corner of a building and 40 feet from the adjacent corner. Fences are put up between the post and the building corners to form a triangular |
garden area. The 40-foot fence makes a building. What is the area of the garden? angle with the 58° 57° 49. Two straight roads meet at an angle of 40° in Harville, one leading to Eastview and the other to Wellston (see the figure on the next page). Eastview is 18 kilometers from Harville and 20 kilometers from Wellston. What is the distance from Harville to Wellston? 44. A pole tilts at an angle 9° from the sun, and casts a shadow 24 feet long. The from the vertical, away 636 Chapter 10 Trigonometric Applications Harville 40° Eastview Wellston class. He has a long tape measure, but no way to measure angles. While pondering what to do, he paces along the side of the river using the five paths joining points A, B, C, and D (see the following figure). If he does not determine the width of the river, he will not pass the course. 50. Each of two observers 400 feet apart measures the angle of elevation to the top of a tree that sits on the straight line between them. These angles are 51° How far is the base of its trunk from each observer? respectively. How tall is the tree? and 65° 51. From the top of the 800-foot-tall Cartalk Tower, Tom sees a plane; the angle of elevation is 67°. the same instant, Ray, who is on the ground 1 mile from the building, notes that his angle of elevation to the plane is elevation to the top of Cartalk Tower is Assume that Tom, Ray, and the airplane are in a plane perpendicular to the ground. (See the following figure.) How high is the airplane? and that his angle of 8.6°. 81° At E A C B D a. Save Charlie from disaster by explaining how he can determine the width AE simply by measuring the lengths AB, AC, AD, BC, and BD and using trigonometry. b. Charlie determines that AD 90 AC 25 BD 22 BC 80 feet, feet. How wide is the river between A and E? feet, feet, and AB 75 feet, 54. A plane flies in a direction of 85° from Chicago. [Note: Aerial navigation directions are explained in Exercise 34 of Section 10.1.] It then turns and flies in the direction of 200° then 195 miles from its starting point. How far did the plane fly in the direction of for 150 miles. |
It is 85°? 81° 67° 800 ft 8.6° 1 mile 52. A plane flies in a direction of 105° from airport A. 55. A hinged crane makes an angle of 50° with the ground. A malfunction causes the lock on the hinge to fail and the top part of the crane swings down (see the figure). How far from the base of the crane does the top hit the ground? [Note: Aerial navigation directions are explained in Exercise 34 of Section 10.1.] After a time, it turns and proceeds in a direction of it lands at airport B, 120 miles directly south of airport A. How far has the plane traveled? 267°. Finally, 53. Charlie is afraid of water; he can’t swim and refuses to get in a boat. However, he must measure the width of a river for his geography 14.6 m 19 m 50° Section 10.3 The Complex Plane and Polar Form for Complex Numbers 637 56. A triangular lot has sides of 120 feet and 160 feet. The angle between these sides is Adjacent to this lot is a rectangular lot whose longest side is 200 feet and whose shortest side is the same length as the shortest side of the triangular lot. What is the total area of both lots? 42°. 57. If a gallon of paint covers 400 square feet, how many gallons are needed to paint a triangular deck with sides of 65 feet, 72 feet, and 88 feet? 58. Critical Thinking Find the volume of the pyramid in the figure below. The volume is given by the formula V 1 3 Bh, where B is the area of the base and h is the height. h 34° 36° 46° 10 59. Critical Thinking A rigid plastic triangle ABC rests on three vertical rods, as shown in the figure. What is its area Horizontal plane 60. Critical Thinking Prove that the area of triangle ABC, in standard notation is given by a2 sin B sin C 2 sin A. 61. Critical Thinking What is the area of a triangle whose sides have lengths 12, 20, and 36? Hint: Drawing a diagram may be helpful. 62. Critical Thinking Use the area formula 1 ab sin C 2 sin2C 1 cos2C ab sin C and the Pythagorean identity to show that 1 2 Then use the Law of Cosines to show that 1 s c 2 1 cos C 1 cos C s 1 2, where s ab ab ab cos and ab 1 2 1 cos C 2 s |
a 1 2 s b 2 21. Combine the facts to prove Heron’s Formula. 10.3 The Complex Plane and Polar Form for Complex Numbers* Objectives • Graph a complex number in the complex plane • Find the absolute value of a complex number • Express a complex number in polar form • Perform polar multiplication The real number system is represented geometrically by the number line. The complex number system can be represented geometrically by the coordinate plane: The complex number a bi corresponds to the point (a, b) in the plane. For example, the point (2, 3) shown in Figure 10.3-1 is labeled by The other points shown are labeled similarly. 2 3i. and division *Section 4.5 is a prerequisite for this section. 638 Chapter 10 Trigonometric Applications Technology Tip Recall that complex numbers are entered by using the special i key on the TI keyboard or in the CPLX submenu of the Casio OPTN menu. i 2 + 3i −6 + 2.3i 2i = 0 + 2i −5 − 3i Figure 10.3-1 real 5.5 = 5.5 + 0i 4 − 3i When the coordinate plane is used to graph complex numbers in this way, it is called the complex plane. Each real number corresponds on the horizontal axis; so this axis is called the real axis. to the point The vertical axis is called the imaginary axis because every imaginary on the vertical axis. number corresponds to the point bi 0 bi a a 0i 0, b a, 0 2 1 1 2 is defined to be the distance from The absolute value of a real number c is the distance from c to 0 on the number line. So the absolute value (or modulus) of the complex number a bi to the origin in the complex plane. a bi 2 represents the distance from, which is given by a bi 2 2a2 b2. 0 a 0 b 0 0, 0 2 a, b to 2 1 2 1 0 1 2 1 2 Absolute Value of a Complex Number Technology Tip Use the ABS key to find the absolute value of a complex number. It is in the NUM submenu of TI and in the CPLX submenu of the Casio OPTN menu. The absolute value (or modulus) of the complex number a bi is a bi 00 00 2a2 b2. Example 1 Find Absolute Value of a Complex Number Find the absolute value of each complex number. |
a. 3 2i b. 4 5i Solution 3 2i a. 0 232 22 213 0 b. 0 4 5i 0 242 5 1 2 241 ■ 2 Absolute values and trigonometry lead to a useful way of representing be a nonzero complex number and denote complex numbers. Let a bi by r. Then r is the length of the line segment joining (a, b) and 0 (0, 0) in the plane. Let be the angle in standard position with this line segment as terminal side, as shown in Figure 10.3-2. a bi u 0 Section 10.3 The Complex Plane and Polar Form for Complex Numbers 639 i θ (a, b) r real Figure 10.3-2 NOTE It is customary to sin u place i in front of rather than after it. Some books abbreviate cos u i sin u r as r cis u. 1 2 Polar Form of a Complex Number Using the definitions of sine and cosine, the coordinates a and b can be expressed in terms of r and u. cos u a r a r cos u and sin u b r b r sin u Consequently, a bi r cos u 1 a bi r sin u i r 1 2 cos u i sin u. 2 is written in this way, it is said to be in When a complex number u polar form or trigonometric form. The angle is called the argument and a bi is called is usually expressed in radian measure. The number the modulus (plural, moduli). The number 0 can also be written in polar notation by letting and be any angle. r 0 r u 0 0 Every complex number can be written in polar form a bi r(cos U i sin U) 2a2 b2, a r cos U, and b r sin U. where r a bi 00 00 When a complex number is written in polar form, the argument uniquely determined because conditions in the box. is not and so on, all satisfy the u ± 4p, u ± 2p, u, u Example 2 Find Polar Form Express 23 i in polar form. Solution In this case, a 23 r 2a2 b2 2 and b 1. Therefore, 13 1 2 2 12 23 1 2. The angle must satisfy the following two conditions. u cos u a r 23 2 and sin u b r 1 2 13 i is represented by the Because point in the complex plane, it lies in the second quadrant, as shown in Figure 10. |
3-3. Therefore, must be a second- 13, 1 A u B quadrant angle. So, conditions. u 5p 6 satisfies these – 3 + i i 1 2 3 5π 6 real Thus, 23 i 2 cos a 5p 6 i sin 5p 6 b. Figure 10.3-3 ■ 640 Chapter 10 Trigonometric Applications Example 3 Find Polar Form Express 2 5i in polar form. Solution In this case, a 2 and b 5. r 2a2 b2 2 Therefore, 2 1 2 2 52 229. The angle must satisfy u cos u a r 2 229 and sin u b r 5 229, so that i −2 + 5i 29 θ real Figure 10.3-4 Polar Multiplication and Division tan u sin u cos u 5 129 2 129 5 2 2.5. lies in the second quadrant (see Figure 10.3-4), u lies Because 2 5i p 2 solution to between p. and Using the tan u 2.5 TAN u 1.1903. is 1 key, the calculator indicates that a Because that angle is in the fourth quadrant, the only solution between p 2 and p is Thus, 2 5i 229 1 u 1.1903 p 1.9513. cos 1.9513 i sin 1.9513. 2 ■ Multiplication and Division of Complex Numbers z1 i sin U1) r1(cos U1 If two complex numbers, then r1r2[cos(U1 z1z2 U2) i sin(U1 U2)] and z2 r2(cos U2 i sin U2) are any and z1 z2 r1 r2 [cos(U1 U2) i sin(U1 U2)], z2 0. That is, given two complex numbers written in polar form, • to multiply the two numbers multiply the moduli and add the arguments. • to divide the two numbers divide the moduli and subtract the arguments. Section 10.3 The Complex Plane and Polar Form for Complex Numbers 641 The proof of the multiplication statement, which is given at the end of this section, uses the addition identities for sine and cosine. Example 4 Multiplication of Numbers in Polar Form Find z1z2 when z1 2 cos a 5p 6 i sin 5p 6 b and z2 3 cos a 7p 4 i sin 7p 4 b. NOTE A complex number written in |
polar form can be written in rectangular form by evaluating each term and simplifying. For example, Solution In this case, r1 2, u1 5p 6, r2 3, and u2 z1z2 r1r23 2 3 1 21 cos u1 1 cos 2 c 2 cos a 5p 6 i sin 2 23 2 a 23 i 5p cos cos 10p a 12 31p 12 u1 1 i sin. 7p 4 u22 4 5p 6 10p 12 a a i sin a i sin 7p 4 b u22 5p 6 21p 12 b i sin 31p 12 b Therefore, 7p 4 b d 21p 12 b d Example 5 Division of Numbers in Polar Form z1 Find when z2 10 z1 p 3 cos Q i sin p 3 R and z2 2 cos a p 4 i sin p 4 b. Solution In this case, r1 10, u1 p 3, r2 2, and u2 p 4. Therefore, z1 z2 10 2 c cos p 3 a p 4 b i sin p 3 a p 4 b d 5 cos a p 12 i sin p 12b ■ ■ z1z2 Proof of the Polar Multiplication Rule r21 Let cos u1 and z2 r11 z1 r11 3 r1r21 r1r21 r1r23 1 i sin u12 i sin u12 4 3 i sin u121 cos u1 cos u1 cos u1 cos u2 cos u1 cos u2 cos u2 r21 cos u2 i sin u1 cos u2 sin u1 sin u22 cos u2 i sin u22 4 i sin u22 i sin u22. i sin u2 cos u1 i sin u1 cos u2 1 i2 sin u1 sin u22 sin u2 cos u12 4 Recall from Section 9.2 that u22 cos u1 1 cos u1 cos u2 sin u1 sin u2 642 Chapter 10 Trigonometric Applications and Therefore, sin u1 1 u22 sin u1 cos u2 cos u1 sin u2. z1z2 r1r23 1 r1r23 cos u1 cos u2 u22 cos sin u1 sin u22 i sin u22 4 u1 1 This completes the proof of the multiplication rule. The division rule is proved similarly. See Exercise 51. sin u2 cos u12 4 sin u1 cos u2 i u1 1 |
1 Exercises 10.3 In Exercises 1–8, plot the point in the complex plane that corresponds to each number. 23. 1. 3 2i 2. 7 6i 4. 22 7i 7. 2i 3 5 a 2 i b 5. 8. 1 i 1 i 2 21 1 4 3 i 1 6 3i 2 3. 6. 8 3 5 3 i 2 i 1 21 1 2i 2 In Exercises 9–14, find each absolute value. 9. 12. 5 12i 0 2 3i 0 0 0 10. 13. 2i 0 12i 0 0 0 11. 14. 0 0 1 i22 0 i 7 0 15. Give an example of complex numbers z and w z such that z w w. 0 0 0 0 16. If z 3 4i, find z 2 and where zz z is the 0 0 0 0 conjugate of z. See Section 4.5 for the definition of a complex conjugate. In Exercises 17–24, sketch the graph of the equation in the complex plane (z denotes a complex number of the form a bi ). 17. 18. 19. 20. 22. 4 z 0 lie 4 units from the origin. 0 Hint: The graph consists of all points that z 1 0 0 z 1 0 complex plane. What does the equation say about the distance from z to 1? Hint: 1 corresponds to (1, 0) in the 10 0 z 3 1 0 z 3i 2 z 2 3i 0 1 0 0 0 3 2 0 21. 0 z 2i 4 0 Hint: Rewrite it as 3. 1 2 z Re 2 z a bi denoted Re(z).] [The real part of the complex number is defined to be the number a and is 24. Im 5 2 z 2 1 [The imaginary part of z a bi is defined to be the number b (not bi) and is denoted Im(z).] In Exercises 25–32, express each number in polar form. 25. 3 4i 26. 4 3i 28. 27 3i 29. 1 2i 27. 5 12i 30. 3 5i 31. 5 2 7 2 i 32. 25 i211 In Exercises 33–38, perform the indicated multiplication or division. Express your answer in both recr(cos U i sin U). tangular form and polar form a bi 33. 3 cos a |
34. 3 cos a p 12 p 8 i sin p 12b 2 cos a 7p 12 i sin 7p 12 b i sin p 8 b 12 cos a 3p 8 i sin 3p 8 b 7 2 a cos p 4 i sin p 4 b 35. 12 cos a 11p 12 i sin 11p 12 b 36. 37. 8 cos a 5p 18 cos 4 a p 9 6 cos a 7p 20 4 cos a p 10 i sin i sin i sin i sin 5p 18 b p 9 b 7p 20 b p 10b Section 10.3 The Complex Plane and Polar Form for Complex Numbers 643 254 38. cos a 26 cos a 9p 4 7p 12 i sin 9p 4 b i sin 7p 12 b In Exercises 39–46, convert to polar form and then multiply or divide. Express your answer in polar form. 1 i23 B A B 39. 41. 1 i A 1 i 1 i 43. 3i A 223 2i B 40. 42. 44. 3 3i 2 1 i 21 1 2 2i 1 i 4i 23 i 45. 46. i 1 i 1 1 1 i 23 i 21 223 2i 2 21 4 4i23 2 21 47. Explain what is meant by saying that multiplying z r a complex number 1 amounts to rotating z 90° around the origin. Hint: Express i and iz in polar form; what are their relative positions in the complex plane? cos u i sin u counterclockwise by i 2 48. Describe what happens geometrically when you multiply a complex number by 2. a bi 49. Critical Thinking The sum of two distinct complex c di, can be found a bi in the complex plane and form the and numbers, geometrically by means of the so-called parallelogram rule: Plot the points c di parallelogram, three of whose vertices are 0, and vertex of the parallelogram is the point whose coordinate is the sum a bi as in the figure below. Then the fourth c di and a bi 1 2 1 c di a c 1 2 1 2 b d i. 2 a + bi i 0 c + di real i 0 c + di a + bi real Complete the following proof of the parallelogram c 0. rule when a. Find the slope of the line K from 0 to a bi. Hint: K contains the points (0, 0) and (a, b). a 0 |
and c di. b. Find the slope of the line N from 0 to c. Find the equation of the line L, through a bi and parallel to line N of part b. Hint: The point (a, b) is on L; find the slope of L by using part b and facts about the slope of parallel lines. d. Find the equation of the line M, through c di and parallel to line K of part a. e. Label the lines K, L, M, and N in each figure. f. Show by using substitution that the point 2 2 satisfies both the equation of line a c, b d 1 L and the equation of line M. Therefore, a c, b d lies on both L and M. Because 1 the only point on both L and M is the fourth vertex of the parallelogram, this vertex must be coordinate a c Hence, this vertex has 2 b d a c, b d c di a bi i. 1. 1 2 1 50. Critical Thinking Let number and denote its conjugate 2 z z. Prove that z 0 0 2 1 z a bi 2 1 2 be a complex z. a bi by and cos u1 Let z2 z1 r21 z1 z2 51. Critical Thinking Proof of the polar division rule. i sin u12. i sin u12 i sin u22 i sin u12 i sin u22 a. Multiply the denominators and use the i sin u22 cos u1 cos u2 cos u1 cos u2 r11 cos u2 r11 r21 r11 r21 cos u2 cos u2 i sin u2 i sin u2 Pythagorean identity to show that it is the number r2. b. Multiply the numerators; use the subtraction identities for sine and cosine (Section 9.2) to show that it is cos u1 1 u22 i sin u1 1 u22 4. r1 3 Therefore, z1 z2 r1 r2b 3 a cos u1 1 u22 i sin u1 1 u22 4. r a. If cos u i sin u 52. Critical Thinking s explain 1 must be true. Hint: Think distance. why r explain, 2 1 Hint: See why Property 5 of the complex numbers in Section 4.5. cos b i sin b s r cos b i sin b cos b cos u cos u i sin u r 1 sin b sin u |
. and b. If, 1 2 2 2 644 Chapter 10 Trigonometric Applications c. If cos b cos u b u angles and same terminal side. Hint: cos u, sin u 1 2 sin b sin u, and show that in standard position have the 1 are points on the unit circle. 2 cos b, sin b and d. Use parts a–c to prove this equality rule for polar form: cos b i sin b s r cos u i sin u 2 1 for some exactly when integer k. Hint: Angles with the same terminal side must differ by an integer multiple of 2p. b u 2kp 2 and s r 1 10.4 DeMoivre’s Theorem and n th Roots of Complex Numbers Objectives • Calculate powers and roots of complex numbers • Find and graph roots of unity Polar form provides a convenient way to calculate both powers and roots cos u i sin u, of complex numbers. If then the multiplication for2 mula from Section 10.3 shows the following: z2 z z r r i sin z r u u 1 cos u u cos 2u i sin 2u 2 1 3 1 2 4 and r2 1 z3 z2 z r2 r cos 2u u i sin 2u u 1 2 4 3 2 1 cos 3u i sin 3u r3 1 2 2 and so on. Repeated application of the multiplication formula proves DeMoivre’s Theorem. DeMoivre’s Theorem For any complex number positive integer n, z r(cos U i sin U) and any zn r n(cos nU i sin nU). NOTE 1 tan 1 23 5p 6 Example 1 Find Powers of Complex Numbers Evaluate 23 i A B 5. Solution First express the complex number 2 of Section 10.3.) 23 i in polar form. (See Example 23 i 2 cos a 5p 6 i sin 5p 6 b Section 10.4 DeMoivre’s Theorem and nth Roots of Complex Numbers 645 Apply DeMoivre’s Theorem. 23 i 5 25 B cos c A 5 5p 6 b a 25p 6 i sin 5 5p a 25p 6 b d 32 cos a i sin 6 b Because 25p 6 p 6 24p 6 p 6 4p, p 6 can be substituted for 25p 6. 23 i A B i sin 25p 6 b 5 32 32 cos a |
25p 6 p 6 23 1 2 2 1623 16i cos a 32 a i sin p 6 b polar form i b rectangular form ■ NOTE tan 1 1 p 4 Example 2 Find Powers of Complex Numbers Evaluate 1 i 10. 2 1 Solution Express the complex number 1 i in polar form. 1 i 22 cos a p 4 i sin p 4 b Apply DeMoivre’s Theorem. 1 i 2 1 10 22 10p 4 i sin 10p 4 b 10 B cos a 5p cos a 2 0 i 2 A 32 32 1 32i i sin 5p 2 b polar form rectangular form ■ NOTE With complex a bi, numbers, it is not possible to choose the positive root as the nth root of as is done with real numbers, because “positive” and “negative” are not meaningful terms in the complex numbers. For 3 2i instance, should called positive or negative? be Nth Roots a bi is a complex number and n is a positive integer, the equation If zn a bi may have n different solutions in the complex numbers. Furthermore, there is no obvious way to designate one of these solutions as the nth root of (see note). Consequently, any solution of the equazn a bi tion a bi is called an nth root of a bi. Every real number is a complex number. When the definition of nth root of a complex number is applied to a real number, the terminology for real numbers no longer applies. For instance, in the complex numbers, 16 has 2, 2, 2i, z4 16, four fourth roots because each of whereas in the real numbers, 2 is the fourth root of 16. is a solution of 2i and 646 Chapter 10 Trigonometric Applications tan NOTE p 3 is in 823 1 8 8 8i˛23 Because Quadrant II of the complex plane, the angle is p p 2p 3 3. Although nth roots are not unique in the complex numbers, the radical symbol will be used only for nonnegative real numbers and will have the same meaning as before. That is, if r is a nonnegative real number, then 2n denotes the unique nonnegative real number whose nth power is r. r Example 3 Find Roots of Complex Numbers Find the fourth roots of 8 8i23. Solution Express the complex number 8 8i23 in polar form. 8 8i23 16 cos a 2p 3 i sin 2p 3 b To solve z s |
z4 16 cos a cos b i sin b 2 1 2p 3 i sin 2p 3 b, find s and b such that is a solution. In other words, find s and b such that cos b i sin b s 1 3 4 16 2 4 cos a 2p 3 i sin 2p 3 b. Use DeMoivre’s Theorem to rewrite the left side. cos 4b i sin 4b s4 1 16 2 cos a 2p 3 i sin 2p 3 b The equality rules for complex numbers (proved in Exercise 52 of Section 10.3) show that the above equation is true if s4 16 and s 24 16 2 2kp 4b 2p 3 kp 2 b p 6 Therefore, the solutions of z4 16 cos a 2p 3 i sin 2p 3 b are z 2 cos c p a 6 i sin kp 2 b k 0, 1, 2, p a 6 kp 2 b d, and 3 produces four distinct where k is any integer. Letting solutions. k 0: z 2 cos a p 6 i sin p 6 b 23 i k 1: z 2 cos c 2 cos a k 2: z 2 cos c 2 cos a p a 6 2p 3 p a 6 7p 6 p 2 b i sin i sin p 6 a p 2 b d 2p 3 b 1 i23 p b i sin p 6 a p b d i sin 7p 6 b 23 i Section 10.4 DeMoivre’s Theorem and nth Roots of Complex Numbers 647 k 3: z 2 cos c 2 cos a 3p 2 b i sin p a 6 5p 3 i sin p 6 a 3p 2 b d 5p 3 b 1 i 23 Any other value of k produces an angle with the same terminal side as one of the four angles already used and is the same solution. For instance, b when k 4, then b p 6 4p 2 p 6 2p, so has the same terminal side b as p 6. Therefore, all fourth roots of 8 8i23 have been found. ■ zn r cos u i sin u The general equation same method used in Example 3: substitute n for 4, r for 16, and 2p 3 can be solved exactly by the for as follows. A solution is a number cos b i sin b such that: u s, 2 1 2 cos b i sin b 2 4 1 cos nb i |
sin nb 1 n r r 2 cos u i sin u cos u i sin u 2 2 1 1 s 3 sn 1 Therefore, sn r s 2n r and nb u 2kp b u 2kp n where k is any integer. Letting angles b. This is stated in the following formula for nth roots. k 0, 1, 2, p, n 1 produces n distinct Formula for nth Roots For each positive integer n, the nonzero complex number r(cos U i sin U) has exactly n distinct nth roots. They are given by U 2kP n U 2kP n i sin cos 2n r a, b d c b where k 0, 1, 2, 3, p, n 1. a Example 4 Find Roots of Complex Numbers Find the fifth roots of 4 4i. Solution Express the complex number 4 4i in polar form. 4 4i 422 cos a p 4 i sin p 4 b 648 Chapter 10 Trigonometric Applications Apply the root formula with n 5, r 422, u p 4, and k 0, 1, 2, 3, and 4. 2n r 25 412 422 1 5 1 2 222 1 5 2 The fifth roots have the following form. 1 A B 1 5 2 5 10 2 1 2 22 5 2 2 1 2 22 scos a p 4 2kp 5 b i sin p 4 a 2kp 5 bt, for k 0, 1, 2, 3, and 4 Therefore, the five distinct roots are as follows. p 20 b a i sin p 20 bt a 9p 20 b a i sin 9p 20 b T a 17p 20 b a i sin 17p 20 b T a 5p 4 b a i sin 5p 4 b T a k 0: 22 scos k 1: 22 scos a a p 4 5 b p 4 i sin p 4 5 bt a 2p 5 i sin b k 2: 22 scos a p 4 4p 5 b i sin k 3: 22 scos a p 4 6p 5 b i sin k 4: 22 scos a p 4 8p 5 b i sin Roots of Unity 22 scos 2p 5 p 4 bt a 22 cos a 22 bt cos p 4 p 4 S 4p 5 S 6p 5 a 22 bt cos p 4 S 8p 5 a 22 bt 33p 20 b a cos S i sin 33p 20 b T a |
■ The n distinct nth roots of 1 (the solutions of roots of unity. Because number 1 is u 0 produces a formula for roots of unity. cos 0 i sin 0. sin 0 0, Applying the root formula with ) are called the nth the polar form of the and cos 0 1 r 1 and zn 1 Formula for Roots of Unity For each positive integer n, there are n distinct nth roots of unity, which have the following form. cos 2kp n i sin 2kp n, for k 0, 1, 2, p, n 1. Section 10.4 DeMoivre’s Theorem and nth Roots of Complex Numbers 649 Example 5 Find Roots of Unity Find the cube roots of unity. Solution Apply the formula for roots of unity with k 0: cos 0 i sin 0 1 2p 2p 3 3 4p 4p 3 3 k 2: cos k 1: cos i sin i sin n 3 and k 0, 1, and 2. 1 2 1 2 23 2 23 2 i i ■ All roots of unity can be found from the first nonreal root. Let the first nonreal cube root of unity obtained in Example 5 be denoted by v. v cos 2p 3 i sin 2p 3 Using DeMoivre’s Theorem to find roots of unity. v2 and produces the other two cube v3 v2 cos a 2p 3 i sin 2 2p 3 b cos 4p 3 i sin 4p 3 v3 cos a 2p 3 i sin 3 2p 3 b cos 6p 3 i sin 6p 3 cos 2p i sin 2p 1 In other words, all the cube roots of unity are powers of true in the general case, as stated below. v. The same is All Roots of Unity Let n be a positive integer with n 77 1. Then the number z cos 2P n i sin 2P n is an nth root of unity and all the nth roots of unity are z, z2, z3, z4, p, zn1, zn 1. The nth roots of unity have an interesting geometric interpretation. Every nth root of unity has absolute value of 1: 2kp n cos ` i sin 2kp n ` cos B a 2 2kp n b 2 2kp n b sin a 2kp n sin2 2kp n cos2 B 21 1 650 Chapter 10 Trigon |
ometric Applications Therefore, in the complex plane, every nth root of unity is exactly 1 unit from the origin. That is, the nth roots of unity all lie on the unit circle in the complex plane. Example 6 Find nth Roots of Unity Find the fifth roots of unity. Solution The fifth roots of unity have the following form. cos 2kp 5 i sin 2kp 5, for k 0, 1, 2, 3, and 4 Therefore, the five roots of unity are i sin k 1: cos k 2: cos k 0: cos 0 i sin 0 1 2p 2p 5 5 4p 4p 5 5 6p 6p 5 5 8p 8p 5 5 k 3: cos k 4: cos i sin i sin i sin ■ These five roots can be plotted in the complex plane by starting at 1 1 0i, and moving counterclockwise around the unit circle, moving through an angle of 2p 5 at each step, as shown in Figure 10.4-1. If you connect these five roots, they form the vertices of a regular pentagon, as shown in Figure 10.4-2. cos 4π 5 + i sin 4π 5 i 1 y cos 2π 5 + i sin 2π 5 i 2π 5 2π 5 2π 5 2π 5 2π 5 cos 0 + i sin 0 real 1 cos 6π 5 + i sin 6π 5 cos 8π 5 + i sin 8π 5 Figure 10.4-1 Figure 10.4-2 real Section 10.4 DeMoivre’s Theorem and nth Roots of Complex Numbers 651 NOTE On wide screen calculators, you may 2 x 2 choose to use so that or the unit circle looks like a circle. 1.7 x 1.7 Graphing Exploration With your calculator in parametric graphing mode, use the following window settings. 0 t 2p t-step 0.067 2.2 x 2.2 1.5 y 1.5 Graph the unit circle, whose parametric equations are x cos t and y sin t Reset the t-step to be 2p 5, and graph again. Your screen should now look exactly like the solid lines in Figure 10.4-2 because the calculator plotted only the five points corresponding to t 0, 2p 5, 4p 5, 6p 5, and 8p 5, and connected them with the shortest possible segments. Use the trace feature to move |
along the graph. The cursor will jump from vertex to vertex, that is, from one fifth root of unity to the next. Example 7 Graph Roots of Unity Graph the tenth roots of unity, and estimate the two tenth roots of unity in the first quadrant. Solution With a graphing calculator in parametric mode, set the range values as in the graphing exploration above. Because n 10, 2p 10 ular decagon whose vertices are the tenth roots of unity. By using the trace feature, you can approximate each tenth root of unity. and graph. The result is a reg- reset the t-step to p 5 1.5 1.5 2.2 2.2 2.2 2.2 1.5 Figure 10.4-3 1.5 Figure 10.4-4 Figures 10.4-3 and 10.4-4 show the two approximate tenth roots of unity in the first quadrant. 0.8090 0.5878i and 0.3090 0.9511i. ■ 5910ac10_616-687 9/21/05 2:31 PM Page 652 652 Chapter 10 Trigonometric Applications Exercises 10.4 In Exercises 1 – 10, calculate each given product and express your answer in the form a bi. 1. cos a p 12 i sin 6 p 12 b 2. cos a p 5 i sin 20 p 5 b 3. 3 cos a c 7p 30 i sin 5 7p 30 b d 4. 5. 23 4 c cos a 7p 36 i sin 12 7p 36 b d 12 1 i 1 theorem. 2 Hint: Use polar form and DeMoivre’s 6. 1 2 2i 8 2 8. 1 a 2 20 23 2 i b 10. 1 1 23 i 8 2 7. 9. 23 2 a 1 2 i b 10 1 22 a 14 i 22 b In Exercises 11 and 12, find all indicated roots of unity and express your answers in the form a bi. 11. fourth roots of unity 12. sixth roots of unity In Exercises 13–22, find the nth roots of each given number in polar form. 13. 64 cos a p 5 i sin p 5 b ; n 3 14. 8 cos a p 10 i sin p 10 b ; n 3 15. 81 cos a p 12 i sin p 12 b ; n 4 16. 16 cos a p 7 i sin p 7 b ; n |
5 29. x4 1 i 23 30. x4 8 8i 23 In Exercises 31–35, represent the roots of unity graphto obtain ically. Then use feature approximations of the form for each root (round to four places). trace a bi the 31. seventh roots of unity 32. fifth roots of unity 33. eighth roots of unity 34. twelfth roots of unity 35. ninth roots of unity 36. Solve the equation x3 x2 x 1 0. find the quotient when then consider solutions of x4 1 0. x4 1 is divided by Hint: First x 1, 37. Solve x5 x4 x3 x2 x 1 0 Hint: Consider x6 1 and x 1 and see Exercise 36. 38. What are the solutions of xn1 xn2 p x3 x2 x 1 0? Exercises 36 and 37.) (See 39. Critical Thinking In the complex plane, the unit 0 1. circle consists of all numbers (points) z such that Suppose v and w are two points z 0 (numbers) that move around the unit circle in such a way that made one complete trip around the circle, how many trips has v made? Hint: Think polar and DeMoivre. at all times. When w has v w12 40. Critical Thinking Suppose u is an nth root of unity. Show that 1 u is also an nth root of unity. Hint: Use the definition, not polar form. 41. Critical Thinking Let u1, u2, p, un be the distinct nth roots of unity and suppose v is a nonzero cos u i sin u solution of the equation vu1, vu2, p,vun are n distinct solutions Show that of the equation. Hint: Each ui is a solution of xn 1. zn r. 2 1 17. 1; n 5 18. 1; n 7 19. i; n 5 42. Critical Thinking Use the formula for nth roots and 20. i; n 6 21. 1 i; n 2 22. 1 i23; n 3 the identities In Exercises 23 – 30, solve the given equation in the complex number system. 23. x6 1 24. x6 64 0 25. x3 i 26. x4 i 27. x3 27i 0 28. x6 729 0 cos 1 x p 2 cos x sin x p 2 1 |
sin x cos u i sin u to show that the nonzero complex number r has two square roots and that 1 these square roots are negatives of each other. 2 Section 10.5 Vectors in the Plane 653 10.5 Vectors in the Plane Objectives • Find the components and magnitude of a vector • Perform scalar multiplication of vectors, vector addition, and vector subtraction Once a unit of measure has been agreed upon, quantities such as area, length, time, and temperature can be described by a single number. Other quantities, such as an east wind of 10 miles per hour, require two numbers to describe them because they involve both magnitude and direction. Quantities that have magnitude and direction are called vectors and are represented geometrically by a directed line segment or arrow, as shown in Figures 10.5-1 and 10.5-2. u Q v w P Figure 10.5-1 Figure 10.5-2 When a vector extends from a point P to a point Q, as in Figure 10.5-1, P is called the initial point of the vector, and Q is called the terminal point, and the vector is written Its length is denoted by!. PQ! PQ. When the endpoints are not specified, as in Figure 10.5-2, vectors are denoted by boldface letters such as u, v, and w. The length of a vector u is denoted by and is called the magnitude of u. u If u and v are vectors with the same magnitude and direction, the vectors u and v are said to be equivalent, written Some examples and nonexamples are shown in Figure 10.5-3. u v same magnitude, different directions u ≠ v same direction, different magnitudes u ≠ v different directions, different magnitudes Figure 10.5-3 654 Chapter 10 Trigonometric Applications Example 1 Confirm Equivalent Vectors P Let Show that 1 1, 2, Q 5, 4!! 1 2 OR PQ., O 2 0, 0, 2 1 and R 4, 2, 2 1 as in Figure 10.5-4. y Solution 5 4 3 2 1 Q (5, 4) P (1, 2) R (4, 2) x O 1 2 3 4 5 Figure 10.5-4 Equivalent Vectors y P (x1, y1) O Q (x2, y2) R (x2 – x |
1, y2 – y1) x Figure 10.5-5 The distance formula shows that and! OR have the same length.! PQ 4 2 2 0! 2 PQ! OR 2 5 1 1 4 0 1! PQ 2 2 and 2 2! OR 1 1 2 242 22 220 2 242 22 220 2 2 have the same slope: The lines containing › 4 2 5 1! OR and! OR ‹ PQ slope! PQ! PQ Because slope, and 2 4 1 2 ‹ slope OR › 2 0 4 0 2 4 1 2 both point to the upper right on lines of the same have the same direction. Therefore,! PQ! OR. ■ According to the definition of equivalence, if two vectors are equivalent, then one of the vectors may be moved from one location to another, provided that its magnitude and direction are not changed, and the two vectors will remain equivalent.! PQ Every vector point at the origin. If is equivalent to a vector! PQ! OR, P (x1, y1) and where R (x2! OR Q (x2, y2), x1, y2 y1). with initial then and! OR Proof The proof is similar to the one used in Example 1. It follows! PQ have the same length: from the fact that! 2 y12 x12 OR x2 3 1 x12 2 x2! 1 PQ 0 4 y2 0 2 1 2 4 2 y2 3 1 y12 2! OR and either the lines containing the same slope:! PQ and are both vertical or they have slope ‹ OR 0 0 y12 y2 › 1 x12 x2 1 y1 y2 x1 x2 ‹ slope ›, PQ as shown in Figure 10.5-5. Every vector can be written as a vector with the origin as its initial point. The magnitude and direction of a vector with the origin as its initial point Section 10.5 Vectors in the Plane 655 are completely determined by the coordinates of its terminal point. Consequently, the vector with initial point (0, 0) and terminal point (a, b) is The numbers a and b are called the components of the denoted by a, b. vector I a, b. I H H a, b is the distance from (0, 0) to (a, b), Because the length of the vector I the distance formula gives its magnitude, |
which is also called its norm. H Magnitude The magnitude (or norm) of the vector 2a2 b2. v 7 7 v a, b I H is CAUTION Example 2 Find Components and Magnitude of a Vector The order in which the coordinates of the initial point and terminal point are subtracted to a, b obtain I H cant. For the points x2, y22 Q P and : x1, y2 y1I x2, y1 y2I x1, y12 1! PQ x2! H QP x1 H is signifi- 1 Scalar Multiplication and! PQ 3: y2! OR, Find the components and the magnitude of the vector with initial point P and terminal point 4, 3 2, 6 Q. 1 2 1 2 Solution According to the properties of equivalent vectors, where x2 4, x1 2, y1 6, where, 9. 2 1 2 In other words,! PQ! OR 6, 9 H. I Therefore, the magnitude is! PQ! OR 262 9 1 2 2 236 81 2117 ■ Vector Arithmetic Vectors can be added, can be subtracted, and can be multiplied in three different ways. Addition, subtraction, and one type of multiplication are discussed in this section. Another type of multiplication is presented in the Excursion 10.6.A. Scalar Multiplication When dealing with vectors, it is customary to refer to ordinary real numbers as scalars. Scalar multiplication is an operation in which a scalar k is “multiplied” by a vector v to produce another vector denoted by kv. If k is a real number and is a vector, then v a, b HH II kv is the vector ka, kb. II HH The vector kv is called a scalar multiple of v. 656 Chapter 10 Trigonometric Applications Example 3 Perform Scalar Multiplication Find the components of 3v and 2v. Let v 3, 1 H I. Solution 3v 3 3, 1 I H 3 3, 3 1 H 9, 3 H I I 2v 2 3, 1 H I 2 3, 2 1 H 6, 2 H I I ■ The graphs of v, 3v, and illustrates that 3v has the same direction as v and direction. from Example 3, shown in Figure 10.5-6, has the opposite 2v 2v y 〈3, 1 |
〉 v 〈9, 3〉 3v x −2v 〈−6, −2〉 Figure 10.5-6 Also note that 2v 7 7 Therefore, v 7 7 6, 2 H I 3, 1 I H 2 232 12 210 6 1 2 2 22 240 2210 3v Similarly, it can be verified that 7 an illustration of the following facts. 7 2v 7 7 2210. Figure 10.5-6 is Geometric Interpretation of Scalar Multiplication The magnitude of the vector kv is is, 00 k 00 times the length of v, that kv k v. 00 The direction of kv is the same as that of v when k is positive and opposite that of v when k is negative. 7 7 7 7 00 Section 10.5 Vectors in the Plane 657 Vector Addition Vector addition is an operation in which two vectors u and v are added, u v resulting in a new vector denoted. Vector Addition If u a, b II HH and v then, c, d II HH u v a c, b d. II HH Example 4 Perform Vector Addition Let u 5, 2 H I and v 3, 1. I H Find the components of u v. Solution u v H 5, 2 3, 1 I I H 5 3, 2 1 H 2, 3 H I I ■ Geometric Interpretation of u v The graph of u v from Example 4 is shown in Figure 10.5-7. y 〈−5, 2〉 〈−2, 35 –2 –1 〈3, 1〉 x v 1 3 Figure 10.5-7 Figure 10.5-7 illustrates the following geometric interpretation of vector addition. Geometric Interpretation of Vector Addition 1. If u and v are vectors with the same initial point P, then is the diagonal of the where! u v, PQ parallelogram with adjacent sides u and v. is the vector! PQ 2. If the vector v is moved without changing its magnitude or direction so that its initial point lies on the endpoint of the vector u, then is the vector with the same initial point P as u and the same terminal point Q as v. u v 658 Chapter 10 Trigonometric Applications Exercise 33 asks for the proof of the geometric interpretation of vector addition stated above. Vector Subtraction The negative (or opposite) of |
a vector c, d c, d tor H is defined using the negative of a vector as follows. v and is denoted c, d H v 1 1, I 2H I I 2 1 1 is defined to be the vecv. Vector subtraction Vector Subtraction If u a, b II HH and v, c, d II HH u (v) then u v a, b II HH a c, b d HH II is the vector c, d II HH Example 5 Perform Vector Subtraction Let u 2, 5 I H and v 6, 1 H. I Find the components of u v. Solution NOTE The vectors v and v have the same magnitude, and lines that contain them have the same slope, but v and v have opposite directions. u v I 6, 1 2, 5 I H H 2 6, 5 1 H 4, 4 H I I ■ Geometric Interpretation of The graph of u v u v from Example 5 is shown in Figure 10.5-8. 〈−4, 4〉 u y 〈2, 5〉 u − v 1 u −v −1 1 Figure 10.5-8 〈6, 1〉 x v The Zero Vector The vector 0, 0 H I is called the zero vector and is denoted 0. Example 6 Perform Combined Vector Operations Let u 1, and w 2, h 5 2 i. Find the components of each vector. Section 10.5 Vectors in the Plane 659 a. 2u 3v b. 4w 2u Solution a. 2u 3v 2 I 3 2 1, 6 h H 3 2, 12 2, 12 I H 2 2, 12 12 H 0. i 4w 2u 4 I I 2, 2 5 1, 6 h H 2 i 2, 12 8, 10 H 8 2, 10 12 H 10, 2 H I I H I I ■ Properties of Vector Addition and Scalar Multiplication Vector Properties Operations on vectors share many of the same properties as arithmetical operations on numbers. For any vectors u, v, and w and any scalars r and s, 1. 2. 3. 4. 5. 6. 7. 8. 9. u (v w) (u vv) 0 r(u v) ru rv (r s)v rv sv (rs)v r(sv) s(rv) 1v v |
0v 0 and r0 0 associative for addition commutative additive identity additive inverse distributive distributive associative for multiplication multiplicative identity multiplication by 0 u v v u Proof of u a, b Let and I H therefore, v c, d H Addition of real numbers is commutative;. I H u v c, d a, b I I H a c, b d H I c a, d b H I c, d a, b H I H I v u The other properties are proved similarly. See Exercises 26–31. 660 Chapter 10 Trigonometric Applications Exercises 10.5 In Exercises 1–4, find the magnitude of the vector! PQ. force and find an additional force v, which, if added to the system, produces equilibrium. 2 7, 11 2 4, 5 1 1 24. u1 25. u1 2, 5, u2 I 3, 7, u2 I H H 2 6, 1 I H 8, 2 H I, u3, u3 I 4, 8 H 9, 0 H, u4 I 5, 4 H I 1. P 2. P 3. P 4. P 1 1 1 1, Q 5, 9 2, 3 2 3, 5 1, Q 7, 0, Q 2 2 30, 12 2, Q 25, 5 1 2 In Exercises 5–10, find a vector equivalent to the vector with its initial point at the origin.! PQ 5. P 6. P 7. P 8. P 1 1 1 1 7, 11 2 2, 9, Q 1, 5 2 1, Q 2, 7 2 4, 8 1, Q 2 10, 2 2, Q 1 7, 9 2 1 5, 6 2 9 17 5 a 2, 12 5 b 10. P 22, 4 A, Q B A 23,1 B In Exercises 11–15, find u v, u v, and 3u 2v. 11. u 12. u 13. u I 4, 0 2, 4, v 6, 1 H I 1, 3, v H I, v 3, 312 I H H H H I 412, 1 I 14, h 19 3 i 15. u 2 2, 12 I In Exercises 26 – 31, let w stated property holds. e, f HH, II and HH and let r and s be scalars. Prove that the c |
, d II, v a, b II u HH 26. v 0 v 0 v 28. 30. r 1 u v 2 v r rs 2 1 ru rv sv 2 1 s rv 2 1 27. 29. 31. v v 0 2 1 r s v rv sv 2 1 1v v and 0v 0 32. Let v be the vector with initial point x1, y12 and let k be any real 1 and 1 x2, y22 terminal point number. a. Find the component form of v and kv. b. Calculate and c. Use the fact that v 7 7 kv. 7 7 2k2 kv 7 7 k 0 0 to verify the v k 0 0 7 7 following equation: 33. Let u a, b I H and v c, d H I. Verify the accuracy of the two geometric interpretations of vector addition given on page 657 as follows: a. Show that the distance from to is the same as a c, b d 2 b. Show that the distance from (c, d) to a c, b d is the same as u. c. Show that the line through (a, b) and a, b d 2 1 they have the same slope. is parallel to v by showing that d. Show that the line through (c, d) and a c, b d is parallel to u. 2 1 1 1 In Exercises 16 – 23, let w. 3, 1 HH Find the magnitude of each vector. II II, 8, 4 HH, v u and 7 34. Let u u v 7 6, 2 HH u v II 16. 18. 3u v 20. 22 17. u v 19. v w 21. 23. 2 2 w 2u 1 v 2 3 7 6 v u1, u2, p, uk act on an object at the origin, the If forces resultant force is the sum The forces are said to be in equilibrium if their resultant force is 0. In Exercises 24 and 25, find the resultant p uk. u2 u1 a, b I H w and. 7 7 v c, d H. I Show that y (a – b, c – d) (a, b) u – v u –v w x v (c, d) Section 10.6 Applications of Vectors in the Plane 661 10.6 Applications of Vectors in the Plane Objectives • Perform |
operations with linear combinations of vectors • Determine the direction angle of a vector • Determine resultant forces in physical applications In the previous section, vectors were introduced and vector arithmetic was defined. In this section, vectors are applied to real-world situations. Unit Vectors A vector with length 1 is called a unit vector. For instance, 3 5, 4 5 i h is a unit vector because 5b 2 4 5b a B a 9 25 16 25 B 25 25 B 1. Example 1 Unit Vectors Find a unit vector u with the same direction as the vector v 5, 12 H. I Solution Multiplying vector v by a scalar that is the reciprocal of its length produces a unit vector. The length of v is v 7 7 H 5, 12 252 122 2169 13. I Let u 1 13 v 1 13 5, 12 H I 5 13, 12 13 i. h The vector u 5 13, 12 13 i h is a unit vector because u 1 13 v g g 1 13 ` ` v 1 13 13 1 ■ Multiplying a vector by a positive scalar produces a vector with the same direction. Thus, u the vector v 5, 12 H h. I 5 13, 12 13 i is a unit vector with the same direction as Multiplying a vector by the reciprocal of its length to produce a unit vector, as in Example 1, works in the general case, as stated below. Unit Vectors If v is a nonzero vector, then same direction as v. 1 v 7 7 v is a unit vector with the 662 Chapter 10 Trigonometric Applications Alternate Vector Notation I u 7 H 0, 1 H I, I then 5i 7j. i It can be verified that the vectors are unit vectors. The vectors i and j play a special role because they lead to a useful alternate notation for vectors. For example, if 0, 1 H 5, 7 1, 0 H and j I Similarly, if 0, 7 H 5 1, 0 H I I is any vector, then I u v 5, 0 H a, b I H v a, b a, 0 H I H v ai bj I 0, b I H a 1, 0 H b 0, 1 H is said to be a linear combination of i and j. When The vector vectors are written as linear combinations of i and j, then the properties of vector addition and scalar multiplication, given in Section 10.5, can be used |
to write the rules for vector addition and scalar multiplication in terms of i and j. ai bj. I I ai bj 1 2 1 ci dj and cai cbj ai bj c 1 2 Example 2 Perform Operations with Linear Combinations If u 2i 6j and v 5i 2j, find 3u 2v. Solution 3u 2v 3 2i 6j 2 5i 2j 1 2 6i 18j 10i 4j 16i 22j 1 2 ■ Direction Angles ai bj v H a, b I If u determined by the standard position angle between terminal side is v, as shown in Figure 10.6-1. is a vector, then the direction of v is completely whose 360°, and 0° b v y θ x 〈a, b〉 a Figure 10.6-1 Section 10.6 Applications of Vectors in the Plane 663 The angle definitions of the trigonometric functions, u is called the direction angle of the vector v. According to the cos u a v and sin u b v 7 7. Rewriting each of these equations gives the following fact. If v a, b II HH ai bj, then a v cos U and b 7 is the direction angle of v. 7 sin U v 7 7 where U Example 3 Find Velocity Vectors Find the component form of the vector that represents the velocity of an airplane at the instant its wheels leave the ground, if the plane is going angle with the 60 miles per hour and the body of the plane makes a horizontal. 7° Solution The velocity vector u 7°, v ai bj as shown in Figure 10.6-2. Therefore, has magnitude 60 and direction angle 7 i v cos u v 1 7 i 60 cos 7° 1 59.55i 7.31j 59.55, 7.31 H 2 2 I 7 sin u v 1 7 60 sin 7° 1 j 2 j 2 ■ Components of the Direction Angle y 10 60 7° v x 10 30 50 Figure 10.6-2 If v ai bj is a nonzero vector with direction angle u, then tan u sin u cos u b v a v b a. This fact provides a convenient way to find the direction angle of a vector. Example 4 Find Direction Angles Find the direction angle of each vector. v 10i 7j a. u 5i 13j b. x Solution |
a. The direction angle of u satisfies u tan u b a 13 5 2.6. y u θ 16 12 8 4 〈5, 13〉 13 4 5 8 a Figure 10.6-3 664 Chapter 10 Trigonometric Applications TAN1 Using the vector u is shown in Figure 10.6-3. b. The direction angle of v satisfies u key on a calculator indicates that u 68.96°. The y tan u b a 7 10 0.7. 〈–10, 7〉 8 4 v Because v lies in the second quadrant, must be between 180°. approximately equal to 90° has a tangent that is so u t 34.99° The period of tangent is t 180° A calculator shows that 0.7. tan t tan 180°, and 1 2 145.01° x for every t. Therefore, –12 –8 –4 –34.99° b Figure 10.6-4 u 34.99° 180° 145.01° is the angle between v is shown in Figure 10.6-4. 90° and 180° such that tan u 0.7. The vector ■ Vector Applications A common application of vectors is in modeling a system of forces acting on an object. Every force has direction and magnitude, therefore, each can be represented by a vector. The sum of all the forces acting on an object is called the resultant force. Example 5 Resultant Force An object at the origin is acted upon by two forces. A 150-pound force with the positive x-axis, and the other force of 100 makes an angle of with the positive x-axis, as shown in Figpounds makes an angle of ure 10.6-5. Find the direction and magnitude of the resultant force. 20° 70° R y 100 Q 50 70° P 20° x O 50 100 150 Figure 10.6-5 Section 10.6 Applications of Vectors in the Plane 665 Solution The forces acting upon the object are:! OP! OQ! OR 1 1 150 cos 20° 100 cos 70° 2 2 is the sum of The resultant force! OR! OR 150 cos 20° 1 175.16i 145.27j i 2 1 150 sin 20° The magnitude of the resultant force i i! OP j 2 150 sin 20° 100 sin 70° j j 2 2 1 1 and! OQ. 100 cos 70° 1 i 2 1 |
100 sin 70° j 2! OR 2175.162 145.272 227.56. is! OR 7 7 The direction angle of the resultant force satisfies u tan u 145.27 175.16 0.8294 A calculator in degree mode shows that u 39.67°. ■ Example 6 Resultant Force A 200-pound box lies on a ramp that makes an angle of with the horizontal. A rope is tied to the box from a post at the top of the ramp to keep it in position (see Figure 10.6-6). Ignoring friction, how much force is being exerted on the rope by the box? 24° P α T 24° C S θ Q R Figure 10.6-6 Solution! TR Because of gravity, the box exerts a 200-pound weight straight down (vec! TR ). As Figure 10.6-6 shows, The force tor! TP, the vector of the force pulling the box on the rope is represented by is the sum of! TQ.! TP and 666 Chapter 10 Trigonometric Applications down the ramp, and 7! TP TSC, a 24° 90°, 7 represents the magnitude of the force. In right TRP, a u 90°. triangle Therefore, The box weighs 200 pounds, so and in right triangle Use sin u to find! TP 7. 7 7 200. a u a 24° u 24°! TR! TP! TR! TP 200 7 7 7 7 sin u 7 7 sin 24° 7! TP 7 7 200 sin 24° 81.35 The force on the rope is about 81.35 pounds. ■ Example 7 Resultant Force 50° with an air speed of 300 miles An airplane is flying in the direction 50° per hour. If there was no wind, the course of the airplane would be. 120°, However, there is a 35-mile-per-hour wind from the direction as represented by the vectors p and w in Figures 10.6-7, which shows the angles using aerial navigation orientation. Find the course and ground speed of the plane (that is, its direction and speed relative to the ground taking the effect of the wind into consideration). y 240 120 50° w 120° –60 p x 240 Figure 10.6-7 y p + w p Solution 40° x 30° 240 Figure 10.6-8 shows p, w, and p w tion angle of the vector p w. p w. The course of the plane is the dire |
c, and its ground speed is the magnitude of 240 120 60° w –60 Figure 10.6-8 The direction angle of p (the angle it makes with the positive x-axis) is 90° 50° 40°. The angle that w makes with the positive y-axis is Section 10.6 Applications of Vectors in the Plane 667 so the direction angle of w, as measured from the pos- 180° 120° 60°, itive x-axis, is p w p w 60° 90° 150°. Therefore, j 300 sin 40° 2 j 35 sin 150° 2 300 sin 40° 300 cos 40° 2 35 cos 150° 2 300 cos 40° 1 1 3 1 2 300 cos 40° 35 cos 150° 2 35 sin 150 199.50i 210.34j p w is The direction angle of j 4 3 1 35 cos 150° i 2 i 2 1 300 sin 40° 35 sin 150° j 2 tan u 210.34 199.50 tan u 1.0543 u 46.5° The course of the plane is the angle between 90° 46.5° 43.5° p w and true north. The ground speed of the plane is p w. p w 2199.502 210.342 289.9 Thus, the plane’s course is about miles per hour. 43.5° and its ground speed is about 289.9 ■ Exercises 10.6 In Exercises 1–5, find u v, u v, and 3u 2v. 1. u i j, v 2i j In Exercises 12–19, find the component form of the vector v whose magnitude and direction angle are given. U 2. 3. u 8i, v 2 3i 2j 1 i j 2, v 3i u 4 2 1 2i. u 5. u 22j, v 23i 12. 14. 16. 18. 4, u 0° 10, u 225° v v 7 7 7 7 v 6, u 40° 7 7 v 1 2, u 250° 13. 15. 17. 19, u 30° 20, u 120° 8, u 160° 3, u 310° In Exercises 6–11, find the components of the given w 4i j. vector, where u i 2j, v 3i j, and In Exercises 20–27, find the magnitude and direction angle of the vector v. |
6. u 2w w 8. 10. 1 2 1 4 7. 1 2 1 3v w 2 9. 2u 3v 20. 22. 24. v 4, 4 H v I 8, 0 H v 6j 21. v 5, 523 H I I 23. 25. v 4, 5 H v 4i 8j I 8u 4v w 2 1 11. 3 1 u 2v 2 6w 26. v 2i 8j 27. v 15i 10j 668 Chapter 10 Trigonometric Applications In Exercises 28–31, find a unit vector that has the same direction as the given vector. 28. 30. 4, 5 H 5i 10j I 29. 7i 8j 31. 3i 9j In Exercises 32–35, an object at the origin is acted upon Uv, by two forces u and v, with direction angle respectively. Find the direction and magnitude of the resultant force. and Uu 32. u 30 pounds, uu 0°; v 90 pounds, uv 60° 33. u 6 pounds, uu 45°; v 6 pounds, uv 120° 34. 35. u 12 kilograms, uu uv 250° u 30 kilograms, uu uv 40° 130°; v 20 kilograms, 300°; v 80 kilograms, 36. Two ropes are tied to a wagon. A child pulls one with a force of 20 pounds, while another child pulls the other with a force of 30 pounds. See the figure. If the angle between the two ropes is 28°, how much force must be exerted by a third child, standing behind the wagon, to keep the wagon from moving? Hint: Assume the wagon is at the origin and one rope runs along the positive x-axis. Proceed as in Example 5 to find the resultant force on the wagon from the ropes. The third child must use the same amount in the opposite direction. 2 0 l b 28˚ 30 lb 37. Two circus elephants, Bessie and Maybelle, are dragging a large wagon, as shown in the figure. If Bessie pulls with a force of 2200 pounds and Maybelle with a force of 1500 pounds and the wagon moves along the dashed line, what is angle u? Bessie 24° θ Maybelle Exercises 38 – 41 deal with an object on an inclined plane. The situation is similar to that in Figure 10.6 |
-6! is the component of the of Example 6, where TP 7 weight of the object parallel to the plane and is the component of the weight perpendicular to the plane.! TQ 7 7 7 38. An object weighing 50 pounds lies on an inclined angle with the horizontal. plane that makes a Find the components of the weight parallel and perpendicular to the plane. Hint: Solve an appropriate triangle. 40° 39. Do Exercise 38 when the object weighs 200 20° pounds and the inclined plane makes a with the horizontal. angle 40. If an object on an inclined plane weighs 150 pounds and the component of the weight perpendicular to the plane is 60 pounds, what angle does the plane make with the horizontal? 41. A force of 500 pounds is needed to pull a cart up 15° a ramp that makes a Assuming that no friction is involved, find the weight of the cart. Hint: Draw a picture similar to Figure 10.6-6; the 500-pound force is parallel to the ramp. angle with the ground. In Exercises 42–47, find the course and ground speed of the plane under the given conditions. See Example 7. All angle measurements are given as aerial navigation directions. See Exercise 55 of Section 6.2. 42. air speed 250 miles per hour in the direction 60°; wind speed 40 miles per hour from the direction 330° 43. air speed 400 miles per hour in the direction 150°; wind speed 30 miles per hour from the direction 60° 44. air speed 300 miles per hour in the direction 300°; wind speed 50 miles per hour in (not from) the direction 30° 5910ac10_616-687 9/21/05 1:59 PM Page 669 45. air speed 500 miles per hour in the direction wind speed 70 miles per hour in the direction 180°; 40° Section 10.6 Applications of Vectors in the Plane 669 46. The course and ground speed of a plane are 70° and 400 miles per hour respectively. There is a 60-mile-per-hour wind blowing from the south. Find the approximate direction and air speed of the plane. 47. A plane is flying in the direction 200° with an air u speed of 500 miles per hour. Its course and and 450 miles per hour, ground speed are respectively. What is the direction and speed of the wind? 210° 48. A river flows from east to west. A swimmer on the south bank wants to swim to |
a point on the opposite shore directly north of her starting point. She can swim at 2.8 miles per hour, and there is a 1-mile-per-hour current in the river. In what direction should she swim in order to travel directly north (that is, what angle should the swimmer make with the south bank of the river)? 49. A river flows from west to east. A swimmer on the north bank swims at 3.1 miles per hour along a line that makes a angle with the north bank of the river and reaches the south bank at a point directly south of his starting point. How fast is the current in the river? 75° 50. A 400-pound weight is suspended by two cables (see the following figure). What is the force (tension) on each cable? Hint: Imagine that the weight is at the origin and that the dashed line is the x-axis. Then cable v is represented by the vector c cos 65° i c sin 65° j 1 2 1 which has magnitude c. (Why?) Represent cable u similarly, denoting its magnitude as d. Use the fact that of two equations in the unknowns c and d. (why?) to set up a system u v 0i 400j 2 32° v 65° 400 51. A 175-pound high-wire artist stands balanced on a tightrope, which sags slightly at the point where he is standing. The rope in front of him makes a angle with the horizontal and the rope behind 6° him makes a force on each end of the rope. Hint: Use a picture and procedure similar to that in Exercise 50. angle with the horizontal. Find the 4° 52. Let v be the vector with initial point x1, y12 and let k be any real 1 and terminal point number. a. Show that x2, y22 tan u tan b, 1 where u is the direction b is the direction angle of kv. angle of v and t tan Use the fact that tan conclude that v and kv have either the same or opposite directions. t 180° to 1 2 b. Use the fact that (c, d) and c, d lie on the 1 2 same straight line on opposite sides of the origin to verify that v and kv have the same and opposite directions if direction if k 6 0. k 7 0 53. Let 7 and u v In Exercise 34 of was shown (see the u v a, b c, |
d. I H I H w Section 10.5, 7 figure with Exercise 34 of Section 10.5). Show that u v is equivalent to the vector w with initial point (c, d) and terminal point (a, b) by now showing that and w have the same direction. u v 7 7 670 Chapter 10 Trigonometric Applications 10.6.A Excursion: The Dot Product Objectives • Find the dot product of two vectors and the angle between two vectors • Determine projection and component vectors and use them in physical applications Dot Product NOTE The dot product of two vectors is found by multiplying corresponding components and finding the sum of the products. Properties of the Dot Product Unlike multiplication of real numbers where there is only one type of multiplication, vector operations include three types of multiplication: scalar multiplication, dot products, and cross products. This section discusses the vector operation called the dot product. Unlike scalar multiplication of vectors, the dot product is not a vector. The dot product of two vectors is a real number. The dot product of vectors ci dj is denoted ac bd. Thus, uv u ai bj c, d II HH and is defined to be the real number a, b II HH v and uv ac bd. Example 1 Find Dot Product of Two Vectors for the given vectors u and v. a. u v 2, 6 Find the dot product and v u I u 4i 2j and v 3i j u 5, 3 H b. c. H I and v I 2, 4 H 6, 3 H I a. Solution u v u v u v b. c. 2, 6 5, 3 H I H 4i 2j 2 2, 4 H 1 6, 3 5 I 3i 21 0 1 1 21 2 2 14 2 ■ The dot product has a number of useful properties. If u, v, and w are vectors, and k is a real number, then: 1. 2. 3v w) u v u w ku v k(u v) u kv 4. 5. 0 u 0 commutative distributive Section 10.6.A Excursion: The Dot Product 671 Proof Let a, b, c, and d be real numbers. 1. If 2. If 7 a u, u then a, b a, b I H I H v a, b and I H a, b I H c, d H I 2a2 b2. |
Therefore, a b b a2 b2 2 2, I 1 then 1 c, d H ac bd ca db 2a2 b2 a 2 b 2. u 7 7 c, d H I a, b I H v u. v θ u Figure 10.6.A-1 The proofs of the last three statements are asked for in the exercises. Angles Between Vectors u v I u c, d H a, b I H are any nonzero vectors, then the angle between If and u and v is the smallest angle formed by these two vectors, as shown in Figure 10.6.A-1. The clockwise or counterclockwise rotation is ignored, and the angle between v and u is considered to be the same as the angle between u and v. Thus, the radian measure of. 4 Nonzero vectors u and v are said to be parallel if the angle between them radians. In other words, u and v are parallel if they lie on is either 0 or the same straight line through the origin and have either the same or opposite directions. The zero vector 0 is considered to be parallel to every vector. is in the interval 0, p p u 3 Any scalar multiple of u is parallel to u because it lies on the same straight line as u. Conversely, if v is parallel to u, it is easy to show that v must be a scalar multiple of u. This is shown in the exercises. Parallel Vectors Vectors u and v are parallel exactly when v ku, for some real number k. Example 2 Determine Parallel Vectors Determine whether the vectors u 2, 3 H I and v 8, 12 H I are parallel. Solution Vector v is a scalar multiple of u. I Thus, vectors u and v are parallel. v 8, 12 H 4 2, 3 H I 4u ■ The angle between nonzero vectors u and v is closely related to their dot product. 672 Chapter 10 Trigonometric Applications Angle Theorem If u is the angle between the nonzero vectors u and v, then u v u v 7 7 7 7 cos U, or equivalently, cos a, b) x u θ v (c, d) Figure 10.6.A-2 Proof Let a, b, c, and d be real numbers, and suppose that, v p, u a, b is not 0 or 7 I H then u and v form two sides of |
a triangle, as shown in Figure 10.6.A-2. and the angle 0 and 0. c, d H u If, I v u 7 7 7 The lengths of two sides of the triangle are 2c2 d2. v 7 side (opposite angle Cosines produces the following result. The distance formula shows that the length of the third b d Therefore, the Law of a c u ) is 2 2 2. u 1 1 2 2 7 7 7 2a2 b2 and c2 d2 a2 b2 a2 2ac c2 b2 2bd d2 a2 b2 c2 d2 2 2ac 2bd cos u cos u cos u cos ac bd ac bd cos u 7 7 The proof when u 7 is 0 or 7 7 7 is exercise 41. p v 7 2 u cos u u 7 v 7 7 7 7 7 cos u v 7 7 cos u u v ac bd y 〈–3, 1〉 2 1 u v 〈5, 2〉 –3 –1 1 5 Figure 10.6.A-3 Example 3 Find the Angle Between Vectors x Find the angle between the vectors in Figure 10.6.A-3. u 3, 1 H I and 5, 2 H I, which are shown Solution Apply the formula from the Angle Theorem with v u 3, 1 H I and. 5, 2 I H cos u u v u v COS1 Using the 3 5 21 2 2 12 1 3 1 2 1 2 252 22 2 1 1 key shows that 2 21 13 210 229 13 2290 2 u 2.4393 radians, or 139.76°. ■ The Angle Theorem has several useful consequences. For instance, taking u v and using the fact the absolute value of each side of cos u, u v 7 7 7 7 Section 10.6.A Excursion: The Dot Product 673 v 7 7 7 (because u v 7 7 7 7 is always positive), produces the 7 0 v u u that 7 7 7 0 7 following results. u v u 0 0 For any angle v 7 7 0 u, 7 7 cos u 0 0 u v u cos u 7 7 0 7 0 1; therefore, u v 7 This proves the Schwarz inequality. 7 7 7 0 0 0 cos cos u 0 ‘ u ‘ ‘ v ‘ cos u 0 0 Schwarz Inequality For any vectors u and |
v Vectors u and v are said to be orthogonal (or perpendicular) if the angle between them is p 2 fact about orthogonal vectors follows. radians 90°, 1 2 or if at least one of them is 0. The key Orthogonal Vectors Let u and v be vectors. Then u and v are orthogonal exactly when uv 0. u v 0. Proof If u or v is 0, then vectors, then by the Angle Theorem: If u and v are nonzero orthogonal u v u v cos u u v cos Conversely, if u and v are vectors such that asks for a proof that u and v are orthogonal. u v p 2 u v 0, 0 0 1 2 then Exercise 42 Example 4 Find Orthogonal Vectors Determine whether the given vectors are orthogonal. a. u 2,6 H and v I 9, 3 H I b. u 1 2 i 5j and v 10i j Solution a. u v 2,6 H I 9, 3 H I b. 2 9 6 1 2 2 18 18 5j b a 1 1 2 5 5 0 10 5 2 10i j 1 2 1 2 1 Vectors u and v are orthogonal. Vectors u and v are orthogonal. ■ 674 Chapter 10 Trigonometric Applications Projections and Components u If u and v are nonzero vectors, and is the angle between them, construct the perpendicular line segment from the terminal point P of u to the straight line on which v lies. This perpendicular segment intersects the line at a point Q. The three possibilities are shown in Figure 10.6.A-4. is called the projection of u onto v and is denoted projvu.! OQ The vector A useful description of P u v Q O projvu projvu follows. P u Q O v projvu Figure 10.6.A-4 P u v O Q projvu Projection of u onto v If u and v are nonzero vectors, then the projection of u onto v is the vector projvu u v v2 b v. a projvu projvu kv Proof Because and v lie on the same straight line, they are parallel. for some real number k. Construct the orthogonal Therefore, vector from a point Q on v through point P, the terminal point of u. Let w be the vector with its |
initial point at the origin and the same length and direction as as in the two cases shown in Figure 10.6.A-5.! QP, P u v Q projvu w O P u w v Q O projvu Figure 10.6.A-5 Because w is parallel to 10.6.A-5, the dot product: u projvu w kv w. it is orthogonal to v. As shown in Figure Consequently, by the properties of! QP Section 10.6.A Excursion: The Dot Product 675 2 kv w u v 1 kv 1 2 v v k kv2 substitution distributive v v v2 But w v 0 because w and v are orthogonal. So, u v kv2 or equivalently, k u v v2. Finally, multiplying both sides of the last statement by v, and substituting projvu for kv, the desired result is proved. u v v2 b projvu kv v a Example 5 Find Projection Vectors If u 8i 3j and v 4i 2j, find projvu and projuv. Solution v2 v v 42 1 2 u v 8 2 26 v u 2 20 and u2 u u 82 32 73 3 2 4 1 1 2 2 Therefore, and projvu u v v2 b a v 26 20 a b 1 4i 2j 26 5 i 13 5 j 2 projuv v u u2 b a u 26 73 a b 1 8i 3j 208 73 i 78 73 j 2 ■ The projection vectors from Example 5 are shown in Figure 10.6.A-6. y projuv v projvu u x Figure 10.6.A-6 676 Chapter 10 Trigonometric Applications Projections and Components Recall from Section 10.6 that 1 v v is a unit vector in the direction of v. Then, projvu can be expressed as a scalar multiple of this unit vector. projvu u v v2 The scalar compvu. is called the component of u along v, and is denoted projvu compvu 1 v v b a Because 1 v v is a unit vector, it can be used to find the length of projvu. projvu compvu g 1 v v b g a 0 compvu 1 v v b g 0 0 g a compvu. 0 Also, because u v u v cos |
u, is the angle between u and v, u where compvu u v v u v cos u v u cos u This result is stated formally as follows. Projections and Components If u and v are nonzero vectors, and them, then u is the angle between and compvu u v v u cos U projvu compvu. 00 00 Example 6 Find Component Vectors If u 2i 3j and v 5i 2j, find compvu and compuv. Solution v 2 5 1 2 u v 2 2 22 229 and u 222 32 213 Section 10.6.A Excursion: The Dot Product 677 Therefore, compvu u v v 4 229 and compuv v u u 4 213 ■ Applications Vectors and the dot product can be used to solve a variety of problems. Example 7 Find Forces Due to Gravity A 4000-pound automobile is on an inclined ramp that makes a angle with the horizontal. Find the force required to keep it from rolling down the ramp, assuming that the only force that must be overcome is that due to gravity. 15° Solution The situation is shown in Figure 10.6.A-7, where the coordinate system is chosen so that the car is at the origin, the vector F representing the downward force of gravity is on the y-axis, and v is a unit vector from the origin down the ramp. 90° 15° 75°. The angle between v Because the car weighs 4000 pounds, and F is is the force pulling the car The vector down the ramp, so a force of the same magnitude in the opposite direction is needed to keep the car motionless. F 4000j. projvF projvF compvF F cos 75° 0 4000 cos 75° 1035.3 0 0 0 Therefore, a force of 1035.3 pounds is required to hold the car in place. ■ If a constant force F is applied to an object, pushing or pulling it a distance d in the direction of the force, as shown in Figure 10.6.A-8, the amount of work done by the force is defined to be the product of the magnitude of the force and the distance. W 1 magnitude of force distance 2 1 F d 1 2 2 If the magnitude of F is measured in pounds and d in feet, then the units for W are foot-pounds. For example, if you push a car for 35 feet along a level driveway by exerting a constant force of 110 pounds, the amount of |
foot-pounds. work done is 3850 110 35 1 2 When a force F moves an object in the direction of a vector d rather than in the direction of F, as shown in Figure 10.6.A-9, then the motion of the which is a force object can be considered as the result of the vector in the same direction as d. projdF, projvF 15° y v 75° F R a m p 15° x Figure 10.6.A-7 F d Figure 10.6.A-8 F θ d projdF Figure 10.6.A-9 678 Chapter 10 Trigonometric Applications NOTE This formula cos u cos 0 1, reduces to the previous one when F and d have the same direction because in that case, so that W F d of force times distance moved. magnitude Therefore, the amount of work done by F is the same as the amount of work done by projdF, as shown below. W projdF d 0 d d 2 compdF cos u 0 F 1 F d See note. Consequently, work can be described as follows. Work The work W done by a constant force F as its point of application moves along the vector d is W 00 compdF 00 d or equivalently, W F d. Example 8 Compute Work How much work is done by a child who pulls a sled 100 feet over level 45° ground by exerting a constant 20-pound force on a rope that makes a angle with the ground? F 45° d Figure 10.6.A-10 Solution The situation is shown in Figure 10.6.A-10, where force F on the rope has magnitude 20, and the sled moves along vector d of length 100. W F d F d cos u 22 2 20 100 100022 1414.2 Therefore, the work done is 1414.2 foot-pounds. ■ Exercises 10.6.A In Exercises 1–6, find u v, u u, and v v. 1. u, v 3, 4 II HH 5, 2 II HH 2. u, v 1, 6 II HH hh 4, 1 3 ii 3. u 2i j, v 3i 4. u i j, v 5j 5. u 3i 2j, v 2i 3j 6. u 4i j, v i 2j In Exercises 7–12, find the |
dot product when u w and, II 2, 5 HH u, v 4, 3 HH v w II 2 v w 1 u v 2 3u v 1 1 2 1 2w 7. 9. 11. 8. 10. 12. 2 2, 1 HH u. II v w 1 u v 2 u 4v 1 1 1 1 2 2 u v 2 2u w 2 In Exercises 13–18, find the angle between vectors u and v. 13. 14. 15. H u, v 4, 3 u 1, 2 H 0, 5 H I, v I u 2i 3j, v i 2, 4 H I I 16. u 2j, v 4i j 17. u 22 i 22 j, v i j 18. u 3i 5j, v 2i 3j In Exercises 19–24, determine whether the vectors u and v are parallel, orthogonal, or neither. 19. u 20. u 21. u I 2, 6 3, 3 I, v H 9, 6 2, 6 I H 6, 4 H I H I 22. u i 2j, v 2i 4j 23. u 2i 2j, v 5i 8j 24. u 6i 4j, v 2i 3j Section 10.6.A Excursion: The Dot Product 679 In Exercises 25–28, find a real number k such that vectors u and v are orthogonal. 25. u 2i 3j, v 3i kj 26. u 3i j, v 2ki 4j 27. u i j, v ki 22 j 28. u 4i 5j, v 2i 2kj In Exercises 29–32, find projuv and projvu. 29. u 3i 5j, v 6i 2j 30. u 2i 3j, v i 2j 31. u i j, v i j 32. u 5i j, v 2i 3j In Exercises 33–36, find compvu. 33. u 10i 4j, v 3i 2j 34. u i 2j, v 3i j 35. u 3i 2j, v i 3j 36. u i j, v 3i 2j In Exercises 37 – 39, let and w Verify that the given property of dot products is valid by calculating the quantities on each side of the equal sign. r |
, s II HH c, d II, v a, b II u HH HH. 37. 38. u v w u v u w 1 ku v k 2 u v 1 u kv 2 39. 0 u 0 40. Suppose u parallel vectors. c 0, a. If a, b I H and v c, d H I are nonzero show that u and v lie on the same nonvertical straight line through the origin. v c show that a 0, (that is, v is a scalar b. If a u multiple of u). Hint: The equation of the line on y mx which u and v lie is m (why?), which implies that d mc. c 0 Hint: If (otherwise u 0, show that v is a scalar multiple of u. c 0 b 0 for some constant b ma (why?) and so then. a 0 and c. If 2 680 Chapter 10 Trigonometric Applications 41. Prove the Angle Theorem in the case when u is 0 or p. 42. If u and v are nonzero vectors such that show that u and v are orthogonal. Hint: If angle between u and v, what is does this say about cos u u and what u? u v 0, is the 43. Show that, 1, 2, 1 2 right triangle by considering the sides of the triangle as vectors. are the vertices of a 3, 4 5, 2 2 1 2 1 44. Find a number x such that the angle between the vectors 1, 1 H I and x, 1 H I is p 4 radians. 45. Find nonzero vectors u, v, and u v u w, orthogonal to u. v w w, and neither v nor w is such that 46. If u and v are nonzero vectors, show that the uv vu, uv vu are orthogonal. vectors 30° 47. A 600-pound trailer is on an inclined ramp that makes a angle with the horizontal. Find the force required to keep it from rolling down the ramp, assuming that the only force that must be overcome is due to gravity. 48. In Example 7, find the vector that represents the force necessary to keep the car motionless. In Exercises 49–52, find the work done by a constant force F as the point of application of F moves along the vector! PQ. 52. F 5i j, P 1 |
, 2 2 1, Q 1 4, 3 2 53. A lawn mower handle makes an angle of with the ground. A woman pushes on the handle with a force of 30 pounds. How much work is done as she moves the lawn mower a distance of 75 feet on level ground? 60° 54. A child pulls a wagon along a level sidewalk by exerting a force of 18 pounds on the wagon handle, which makes an angle of horizontal. How much work is done as she pulls the wagon 200 feet? with the 25° 20° 55. A 40-pound cart is pushed 100 feet up a ramp that makes a angle with the horizontal. How much work is done against gravity? Hint: The amount of work done against gravity is the negative of the amount of work done by gravity. Position the cart on a coordinate plane so that the cart is at the origin. Then the cart moves along vector i d downward force of gravity is and the F 0i 40j. 100 cos 20° 100 sin 20° j 2 1 1 2 20° 49. F 2i 5j, P 1 0, 0 50. 51. F i 2j, P 0, 0 1 1, Q F 2i 3j, P! component form of PQ. 2, 1 2 1 5, 2 2 5, 9 1 2 Hint: Find the 56. Suppose the child in Exercise 54 is pulling the wagon up a hill that makes an angle of the horizontal, and all other conditions remain the same. How much work is done in pulling the wagon 150 feet? with 20° C H A P T E R 10 R E V I E W Important Concepts Section 10.1 Section 10.2 Section 10.3 Section 10.4 Section 10.5 Standard notation for triangles.............. 617 Law of Cosines.......................... 617 Law of Sines............................ 625 Ambiguous case......................... 627 Area formulas for triangles................. 632 Complex plane.......... |
................ 638 Real axis............................... 638 Imaginary axis........................... 638 Absolute value of a complex number......... 638 Argument.............................. 639 Modulus............................... 639 Polar form.............................. 639 Polar multiplication and division............ 640 DeMoivre’s Theorem...................... 644 Formula for nth roots..................... 647 Roots of unity........................... 648 Vector................................. 653 Magnitude.............................. 655 Components............................ 655 Scalar multiplication...................... 655 Vector addition and subtraction............. 657 Section 10.6 Unit vector............................. 661 Linear combination of i and j......... |
...... 662 Direction angle of a vector................. 663 Section 10.6.A Dot product............................. 670 Angle between vectors.................... 671 Parallel vectors.......................... 671 Angle theorem.......................... 672 Schwarz inequality....................... 673 Orthogonal vectors....................... 673 Projection of u on v....................... 674 Component of u along v................... 676 Work.................................. 678 681 682 Chapter Review Important Facts and Formulas Law of Cosines a2 b2 c2 2bc cos A Law of Cosines, Alternate Form: cos A b2 c2 a2 2bc Law of Sines a sin A b sin B c sin C Area of triangle ABC: ab sin C 2 Heron’s Formula Area of triangle a b c where. s 1 2 1 2 ABC 2s s a s b s c, 2 21 21 1 2a2 b2 a bi 0 0 a bi r 1 cos u i sin u, where 2 cos u1 r11 i sin u12 r 2a2 b2, a r cos u, b r sin u i sin u22 r1r23 u22 cos r21 cos u2 u1 1 i sin u1 1 u22 4 cos u1 cos u2 r11 r21 i sin u12 i sin u22 r1 r2 3 DeMoivre’s Theorem cos u1 |
1 u22 i sin u1 1 u22 4 cos u i sin u r 3 The distinct nth roots of 1 r 3 2 4 n rn cos u i sin u u 2kp n b d a 1 i sin 2n r cos a c u 2kp n b cos nu i sin nu 4 are: 2 k 0, 1, 2, p, n 1 1 2 The distinct nth roots of unity are: cos 2kp n i sin 2kp n 1 k 0, 1, 2, p, n 1 Q x2, y22 1, then PQ › x1, y2 x2 H 2 y1I. and x1, y12 1 2a2 b2 If P a, b I H u If a, b I H and k is a scalar, then ku ka, kb. I H If u a, b and I H u v v c, d then, I H a c, b d H I and u v a c, b d H. I Chapter Review 683 Properties of Vector Addition and Scalar Multiplication For any vectors u, v, and w and any scalars r and s: u v w 2 1. 2. 3. 4. 5. 6. 7. 8 rs 2 1 1v v 0v 0 r0 0 2 ru rv 2 v rv sv rv 2 v r s sv 1 1 2 1 2 9. u If a, b I H ai bj, then a u cos u and b u sin u, where u is the direction angle of u. If u a, b I H ai bj and v c, d H u v ac bd. ci dj, then I If u is the angle between nonzero vectors u and v, then u v u v cos u. Schwarz Inequality u v 0 0 u v Vectors u and v are orthogonal exactly when u v 0. projvu u v v2 b a v compvu u v v u cos u, where u is the angle between u and v. Properties of Dot Products If u, v, and w are vectors, and k is a real number, then: 2. 1. u u u2 u v v u u v w 1 ku v k 4. 5. 0 u 0 3 kv 2 684 Chapter Review Review Exercises Note: Standard notation is used for triangles. Section |
10.1 In Exercises 1–6, use the Law of Cosines to solve triangle ABC. 1. a 12, b 10, c 15 2. a 7.5, b 3.2, c 6.4 3. a 10, c 14, B 130° 4. a 7, b 8.6, C 72.4° 5. a 5, c 8, B 76° 6. a 90, b 70, c 40 7. Two trains depart simultaneously from the same station. The angle between their two tracks is 45 miles per hour and the other at 70 miles per hour. How far apart are the trains after 3 hours? One train travels at an average speed of 120°. 8. A 40-foot flagpole sits on the side of a hill. The hillside makes a angle with the horizontal. How long is a wire that runs from the top of the pole to a point 72 feet downhill from the base of the pole? 17° 9. Find angle ABC in the figure below. A B 18 12 C 10 10. A surveyor stakes out points A and B on opposite sides of a building. Point What is C is 300 feet from A and 440 feet from B. Angle ACB measures the distance from A to B? 38°. Section 10.2 In Exercises 11–18, use the Law of Sines to solve triangle ABC. 11. B 124°, C 31°, c 3.5 12. A 96°, B 44°, b 12 13. a 75, c 84, C 62° 14. a 5, c 2.5, C 30° 15. a 3.5, b 4, A 60° 16. a 3.8, c 2.8, C 41° 17. A 48°, B 57°, b 47 18. A 67°, c 125, a 100 19. Find the area of triangle ABC if b 24, c 15, and A 55°. 20. Find the area of triangle ABC if a 10, c 14, and B 75°. 21. A boat travels for 8 kilometers in a straight line from the dock. It is then sighted from a lighthouse which is 6.5 kilometers from the dock. The angle determined by the dock, the lighthouse (vertex), and the boat is far is the boat from the lighthouse? How 25°. 22. A pole tilts 12° from the vertical, away from the sun, and casts a |
34-foot shadow on level ground. The angle of elevation from the end of the 64°. shadow to the top of the pole is How long is the pole? Chapter Review 685 23. Two surveyors, Joe and Alice, are 240 meters apart on a riverbank. Each sights a flagpole on the opposite bank. The angle from the pole to Joe (vertex) to Alice is 54°. The angle from the pole to Alice (vertex) to Joe is How far are Joe and Alice from the pole? 63°. 24. A straight road slopes at an angle of angle of elevation of the sun is 62.5°, road casts a 15-foot shadow downhill, parallel to the road. How high is the telephone pole? with the horizontal. When the 10° a telephone pole at the side of the 25. A woman on the top of a 448-foot building spots a small plane. As she 62°. views the plane, its angle of elevation is At the same instant a man at the ground-level entrance to the building sees the plane and notes that its angle of elevation is a. How far is the woman from the plane? b. How far is the man from the plane? c. How high is the plane? 65°. 26. Use the Law of Sines to prove Engelsohn’s equations given below: For any triangle ABC (standard notation), a b c sin A sin B sin C and a b c sin A sin B sin C. In Exercises 27–30, find the area of the triangle described. 27. angle of 30°, adjacent side lengths 5 and 8 28. angle of 40°, adjacent side lengths 3 and 12 29. side lengths 7, 11, and 14 30. side lengths 4, 8, and 10 Section 10.3 31. Simplify 0 4 2i i 1 3 2i 2 0 3 i 0 1 2i. 0 32. Simplify 0 33. Graph the equation 0 0. in the complex plane. 0 2 z 0 z 3 0 0 34. Graph the equation 1 0 in the complex plane. 35. Express 1 i23 in polar form. 36. Express 4 5i in polar form. In Exercises 37–41, express the given number in the form a bi. 37. 2 cos a p 12 i sin p 12b 4 cos a p 6 i sin 38. 3 cos a p 8 i sin p 8 b 2 cos a 3p 8 |
i sin p 6 b 3p 8 b 686 Chapter Review 12 39. cos a i sin 3 cos a i sin 7p 12 5p 12 7p 12 b 5p 12 b Section 10.4 40. cos a p 12 i sin 18 p 12b 41. 13 3 c cos a 5p 36 i sin 12 5p 36b d In Exercises 42–46, solve the given equation in the complex number system, and express your answers in polar form. 42. x3 i 45. x4 i 43. x6 1 46. x3 1 i 44. x8 23 3i Section 10.5 In Exercises 47–50, let u 3, 2 HH II and v 8, 1 HH II. Find each of the following: 47. u v 48. 3v 49. 2v 4u 50. 3u 1 2 v In Exercises 51–54, let u 2i j and v 3i 4j. Find each of the following: 51. 4u v 52. u 2v 53. u v 54. u v 55. Find the components of the vector v such that v 5 and the direction angle of v is 45°. Section 10.6 56. Find the magnitude and direction angle of 3i 4j. 57. Find a unit vector whose direction is opposite the direction of 3i 6j. 58. An object at the origin is acted upon by a 10-pound force with direction 90° angle magnitude and direction of the resultant force. and a 20-pound force with direction angle 30°. Find the 59. A plane flies in the direction 120°, with an air speed of 300 miles per hour. The wind is blowing from north to south at 40 miles per hour. Find the course and ground speed of the plane. 60. An object weighing 40 pounds lies on an inclined plane that makes a 30° angle with the horizontal. Find the components of the weight parallel and perpendicular to the plane. Section 10.6.A In Exercises 61 –64, let following: u 3,4, v II, and w 2, 5 II HH HH 0, 3 HH. II Find each of the 61. u v 62. u u v v 63. u v 2 1 w 64. u w 1 2 1 w 3v 2 65. What is the angle between the vectors 5i 2j and 3i j? 66. Is 3i 2j orthogonal to 4i 6j |
? Chapter Review 687 In Exercises 67 and 68, let ing: u 4i 3j and v 2i j. Find each of the follow- 67. projvu 68. compuv 69. If u and v have the same magnitude, show that u v and u v are orthogonal. 70. If u and v are nonzero vectors, show that the vector u kv is orthogonal to v, where k u v v2. 71. A 3500-pound automobile is on an inclined ramp that makes a angle with the horizontal. Find the force required to keep it from rolling down the ramp, assuming the only force that must be overcome is due to gravity. 30° 72. A sled is pulled along level ground by a rope that makes a angle with the horizontal. If a force of 40 pounds is used to pull the sled, how much work is done in pulling it 100 feet? 50° C H A P T E R 10 Euler’s Formula One of the most interesting and surprising identities in all of mathematics is one that relates the exponential function to the trigonometric functions sin x and cos x and the imaginary number i. e x ix cos x i sin x e The identity, known as Euler’s formula, is named after the mathematician Leonhard Euler (1707–1783), who discovered it in 1748. The formula is used in many areas of calculus—most notably differential equations. Euler's formula is true for any real number x, and many real numbers produce surprising results. Example 1 Evaluating Euler’s Formula Evaluate eip. Solution Substitute p for x in the formula and simplify. eip cos p i sin p 1 i 1 Therefore, eip 1. 1 0 2 ■ Rewriting the last equation connects the five most common constants of mathematics: e, p, i, 0, and 1. eip 1 0 One of the most surprising aspects of this displayed equation is that raising an irrational number to an irrational power results in an integer. In fact, raising an imaginary number to an imaginary power can also give a real number, as shown in the next example. Example 2 Imaginary Numbers Raised to Imaginary Powers Show that i e p 2 i and find ii. Solution Substitute x p 2 into Euler’s formula and simplify. p 2 i cos e p 2 i sin p 2 0 i i 1 1 2 688 To find ii, raise |
both sides of the identity p 2 i2 e 2 1 p e 2 i e ii p 2 i e A B p A calculator computes the value of Figure 10.C-1. So ii 0.2079. p 2 i i e 1 2 e to the power i. p 2 to be 0.2078795764, as shown in Figure 10.C-1 Euler’s formula can be used to define complex powers of e, that is, xiy e x e cos y i sin y The equation terms of a real power of e and the cosine and sine of a real number. 2 1 defines a complex power of e in ez e xiy e cos y i sin y iy e e 1 2 x x ■ xiy. e Example 3 Complex Power of e Find the exact value and approximate value of ep2i. Solution Substitute x p and y 2 ep2i ep into the formula cos 2 i sin 2 z e e x 1 1 0.4161 0.9093i 23.14 9.629926 21.041772i 2 1 cos y i sin y. 2 exact value 2 approximate value ■ Exercises Find the exact value and the approximate value of the following powers of e. 1. pie e pi 3. ie e i 3i B 2. e3i A p 4 i 4. e 5. e 1ip 7. e1 p 3 i 6. e 2pi 8. epi 689 C H A P T E R 11 Analytic Geometry You are there! All planets travel in elliptical orbits around the sun, moons travel in elliptical orbits around planets, and satellites follow elliptical paths around the earth. Parabolic reflectors are used in spotlights, radar antennas, and satellite dishes. The long-range navigation system (LORAN) uses hyperbolas to enable a ship to determine its exact location. See Exercise 68 in Section 11.4. 690 Chapter Outline 11.1 Ellipses 11.2 Hyperbolas 11.3 Parabolas 11.4 11.5 11.6 11.7 Translations and Rotations of Conics 11.4.A Excursion: Rotation of Axes Polar Coordinates Polar Equations of Conics Plane Curves and Parametric Equations 11.7.A Excursion: Parameterizations of Conic Sections Chapter Review can do calculus Arc Length of a Polar Graph Interd |
ependence of Sections ⎫ 11.1 ⎪ ⎬ 11.2 ⎪ ⎭ 11.3 11.5 11.7 > 11.4 >> 11.6 Sections 3.1, 3.2, and 3.4 are prerequisites for this chapter. Except for the discussion of standard equations for conics in Sections 11.1–11.4, Chapter 6 (Trigonometry) is also a prerequisite. When a right circular cone is cut by a plane, the intersection is a curve called a conic section, as shown in the figure below. (A point, a line, or two intersecting lines are sometimes called degenerate conic sections.) Conic sections were studied by the ancient Greeks and are still of interest. For example, the orbits of planets are ellipses, parabolic mir- rors are used in telescopes, and certain atomic particles follow hyperbolic paths. P Circle Ellipse Hyperbola Parabola Point Line Two intersecting lines Although the Greeks studied conic sections from a purely geometric point of view, the modern approach is to describe them in terms of the coordinate plane and distance, or as the graphs of certain types of equations. The study of the geometric properties of objects using a coordinate system is called analytic geometry. Circles are discussed in the appendix and used in prior sections. In this chapter, ellipses, hyperbolas, and parabolas are defined in terms of points and distances, and their equations are determined. The standard form of the equation of a conic includes the key information necessary for a rough sketch of its graph. Techniques for graphing conic sections with a calculator are discussed in the sections that define each conic section, and applications are given. 691 692 Chapter 11 Analytic Geometry 11.1 Ellipses Objectives • Define an ellipse • Write the equation of an ellipse • Identify important characteristics of ellipses • Graph ellipses An ellipse is a closed figure that can be thought of as a circle that has been elongated along a line of symmetry through its center. In this section, ellipses are defined in terms of points and distances, and then their equations are derived from the definition. Definition of an Ellipse Let P and Q be points in the plane and k a number greater than the distance from P to Q. The ellipse with foci (singular: focus) P and Q is the set |
of all points X such that the sum of the distance from X to P and the distance from X to Q is k. Written algebraically with X representing a point XP XQ k x, y 1 2 on the ellipse, To draw the ellipse, pin the ends of a string of length k at points P and Q, as shown in Figure 11.1-1. Place a pencil against the string, and keep the string taut while moving the pencil. center major axis P Q P Q Figure 11.1-1 Figure 11.1-2 vertex foci vertex minor axis PQ joining the foci is the center of the ellipse. The midpoint of the segment The points where the line through the foci intercept the ellipse are its vertices. The segment connecting the vertices is the major axis, and the segment through the center of the ellipse perpendicular to the major axis is the minor axis, as shown in Figure 11.1-2. If the points P and Q coincide, the ellipse generated is a circle with radius k 2 Thus, a circle is a special case of an ellipse.. Equation of an Ellipse The simplest case is an ellipse centered at the origin, with its foci on the c, 0 x- or y-axis. Suppose that the foci are on the x-axis at the points P 1 2 Section 11.1 Ellipses 693 Let a k 2, so that k 2a. Then x, y 1 2 is on the and Q c, 0, where 1 2 ellipse exactly when c 7 0. distance from 3 x, y 1 2 to P 4 3 distance from x, y 1 2 to Q 4 k 2a. Written algebraically 2a 2 2 1 1 which can be rewritten as x c 2 1 2 y˛ 2 2a 2 x c 1 2 2 y˛ 2. 2 Square both sides and simplify the result. Square both sides again and simplify. a2 1 x c 2 2 y˛ 2 a˛ 2 cx 2 c˛ a˛ 2 1 2 2 a˛ x˛ a˛ 1 2 c˛ 2 a˛ 2 2 y˛ b 2a˛ 2 c˛ 2 2 a˛ 2 c˛ 2 2 so that 2, To simplify the last equation, let Equation [1] then becomes b˛ [1 |
] 2˛x˛ b˛ 2 a˛ 2˛y˛ 2 a˛ 2 2˛b˛ Dividing both sides by the ellipse satisfy the equation 2 2b˛ a˛ shows that the coordinates of every point on 2 x˛ 2 a˛ 2 y˛ 2 b˛ 1 Standard form of an ellipse Conversely, it can be shown that every point whose coordinates satisfy 0, c the equation is on the ellipse. The equation for an ellipse with foci and on the y-axis is developed similarly. 0, c 2 1 1 2 Standard Equation of an Ellipse Centered at the Origin Let a and b be real numbers such that graph of each of the following equations is an ellipse centered at the origin. 0 66 b 66 a. Then the Foci on the x-axis: y˛ 2 b˛ 2 x˛ 2 a˛ 1 2 Foci on the y-axis: y˛ 2 a˛ 2 x˛ b˛ 1 2 2 y b −b x c a −a −c y a c −b x b −c −a 694 Chapter 11 Analytic Geometry When the equation is in standard form, the x- and y-intercepts are easily found. x-intercepts 2 2 y 0 2 x˛ 2 a˛ 1 1 0˛ 2 b˛ 2 a˛ 2 x˛ x –a y-intercepts 2 2 x 0 2 0˛ 2 a˛ 1 1 y˛ 2 b˛ 2 b˛ 2 y˛ y –b The characteristics of the graph of an ellipse centered at the origin are shown in the following box. Characteristics of Ellipses For 0 66 b 66 a, Foci on the x-axis: y˛ 2 1 2 2 2 x˛ a˛ b˛ Foci on the y-axis: y˛ 2 2 x˛ 2 b˛ 2 a˛ 1 foci −a −c minor axis y b −b center x c a major axis y a c foci x b major axis center −b minor axis −c −a ±±a ±±b x-intercepts: y-intercepts: major axis is on the x-axis vertices foci c 2a� |
� (a, 0) and 2 and (c, 0), (c, 0) 2 b˛ (a, 0) where ±±b ±±a x-intercepts: y-intercepts: major axis is on the y-axis vertices foci c 2a˛ (0, a) and 2 and (0, c), (0, c) 2 b˛ (0, a) where Notice the following facts in both cases. • the foci are within the ellipse • the major axis always contains the foci and is determined by which denominator is larger • the distance between the foci is 2c • the center is the midpoint between the foci and the midpoint between the vertices • the distance between the vertices is 2a • 2 b˛ 2 a˛ c˛ 2 When the equation of an ellipse centered at the origin is in standard form, the denominator of the x term always gives the x-intercepts and the denominator of the y term always gives the y-intercepts. Section 11.1 Ellipses 695 Graphing an Ellipse Example 1 Graph an Ellipse Show that the graph of the equation the foci, the vertices, the major axis, and the minor axis. 25x˛ 2 16y˛ 2 400 is an ellipse. Label y 10 5 major axis vertices −10 −5 0 foci −5 x 5 10 minor axis −10 Figure 11.1-3 Solution Put the equation in standard form by dividing both sides by 400. 2 x˛ 16 2 y˛ 25 1 This is the equation of an ellipse with its center at the origin. The foci are on the y-axis because the denominator of and 2 16, b˛ To graph the ellipse, plot the intercepts and draw the ellipse, as shown in Figure 11.1-3. is larger. Since and the y-intercepts are 2 25 a˛ ± a ± 5. the x-intercepts are ± b ± 4 2 y˛ To locate the foci, note that on the y-axis. Therefore, the foci are c 125 16 19 3 0, 3 0, 3 and and that the foci are. 1 2 2 and the major axis lies on the y-axis 1 The vertices are 1 with endpoints |
at the vertices. and 2 1 0, 5 0, 5, 2 The minor axis lies on the x-axis, with endpoints 4, 0 1 and 4, 0. 2 1 2 ■ Example 2 Graph an Ellipse on a Calculator Graph the ellipse with equation 4x˛ 2 9y˛ 2 36 on a calculator. Solution Solve the equation for y. 3.1 9y˛ 2 36 4x˛ 2 2 36 4x˛ 2 y˛ −4.7 4.7 y ± B 9 36 4x2 9 ± 2 3 29 x˛ 2 The ellipse is defined by the two functions −3.1 Y1 2 3 29 x˛ 2 and Y2 2 3 29 x˛ 2, Figure 11.1-4 whose graphs are shown in Figure 11.1-4. ■ NOTE Graph all conic sections using a square window to see the correct shape. 696 Chapter 11 Analytic Geometry Ellipse Equations Example 3 Find the Equation of an Ellipse Find the equation of the ellipse that has vertices at 0, –216 Then sketch its graph by using the intercepts.. 1 0, –6 and foci at 2 A B Solution The foci of the ellipse lie on the y-axis, and its center is the origin. Thus, the equation has the form 2 x˛ 2 b˛ c 216 2 y˛ 2 a˛ 1 and by using the relationship among Find b by letting the values for an ellipse. a 6 and A x 4 8 Thus, the equation is 216 2 c˛ 2 2 b2 2 2 b˛ 2 a˛ 2 6 B 1 24 36 b˛ 2 12 b˛ b 212 3.5 2 x˛ 112 2 y˛ 2 6˛ 1 or 2 x˛ 12 2 y˛ 36 1 2 B and the intercepts are A 0, ± 6 1 2 and A ± 212, 0 B. See Figure 11.1-5. ■ Applications of Ellipses Elliptical surfaces have interesting reflective properties. A sound wave or light ray that passes through one focus and reflects off an ellipse will always pass through the other focus, as shown in Figure 11.1-6. Example 4 Finding the Foci The Whispering Gallery at the Museum of Science and Industry in Chicago is elliptical in shape, |
with a parabolic dish at each focus. (Parabolas are discussed in Section 11.3.) The shape of the room and two parabolic dishes carry the quietest sound from one focus to the other. The width of the ellipse is 13 feet 6 inches and the length of the ellipse is 47 feet 4 inches. Assume that the ellipse is centered at the origin. Find its equation, sketch its graph, and locate the foci. Solution Because the length of the ellipse is 47 feet 4 inches, the value of a is half that amount, or 23 2 3 feet. Because the width is 13 feet 6 inches, the value y 8 4 −4 −4 −8 −8 Figure 11.1-5 foci Figure 11.1-6 where 2a NOTE The area of an A pab, ellipse is is the length of the major axis and 2b is the length of the minor axis. See Exercise 24 for a method to estimate the circumference of an ellipse. Section 11.1 Ellipses 697 of b is half that, or 6 3 4 feet. Therefore, the equation of the ellipse is 2 x˛ 23 a 2 2 3b 2 x˛ 5041 9 2 2 y˛ 3 6 4b a 2 y˛ 729 16 16y˛ 729 2 1 1 1 2 9x˛ 5041 The distance from the center to each focus is c, which is given by 2 a˛ 2 b˛ c˛ 2 5041 9 729 16 74,095 144 c 74,095 144 B 22.7 y 20 10 F1 F2 x −20 −10 0 10 20 −10 −20 Figure 11.1-7 Therefore, the foci are approximately in Figure 11.1-7. F11 22.7, 0 and F21 2 22.7, 0 2, as shown ■ The planets and many comets have elliptical orbits, with the Sun at one focus. The Moon travels in an elliptical orbit with Earth at one focus, and man-made satellites usually travel in elliptical orbits around Earth. Example 5 Elliptical Orbits Earth’s orbit around the Sun is an ellipse that is almost a circle. The Sun is at one focus, the major axis is 299,190,000 km in length, and the minor axis is 299,148,000 km in length. What |
are the minimum and maximum distances from Earth to the Sun? Solution Choose a coordinate system with the center of the ellipse at the origin and the Sun at the point to get a diagram of the orbit. See Figure 11.1-8a. c, 0 2 1 Earth Sun Figure 11.1-8a Figure 11.1-8b 698 Chapter 11 Analytic Geometry The length of the major axis is 2a and the length of the minor axis is 2b, so 2a 299,190,000 a 149,595,000 2b 299,148,000 b 149,574,000 Therefore, the distance from each focus to the center is given by c 2a2 b2 2,507,000 It can be proved algebraically that the minimum and maximum distances from a focus to a point on the ellipse are at the endpoints of the major axis. That is, the maximum distance is and the minimum distance is See Figure 11.1-8b. a c. a c • minimum distance • maximum distance a c a c 147,088,000 km 91.2 million miles 152,102,000 km 94.3 million miles ■ Exercises 11.1 In Exercises 1–6, find the equation of the ellipse centered at the origin that satisfies the given conditions. 7. 1. foci on x-axis; x-intercepts ± 7, y-intercepts ± 2 2. foci on y-axis; x-intercepts ± 1, y-intercepts ± 8 3. foci on x-axis; major axis of length 12; minor axis of length 8 4. foci on y-axis; major axis of length 20; minor axis of length 18 y 4 2 0 −2 −4 −4 −2 x 2 4 5. endpoints of major and minor axes: 3, 0 1 3, 0, 1 2 2 0, 7 1 0, 7, 2 1, 2 6. vertices 8, 0 2 1 and 1 8, 0 2, minor axis of length 8 In Exercises 7–12, match one of the following equations to the given graph. 2 x˛ 9 2 x˛ 4 2x˛ 2 y˛ 16 2 y˛ 25 2 y˛ 1 1 2 12 2 1 y˛ 2 x˛ 9 16 2 y˛ 2 x˛ 25 4 2 6y˛ 2 |
18 x˛ 1 8. y 4 2 0 −4 −2 −2 −4 x 2 4 9. y 4 2 −4 −2 0 −2 −4 4 2 −4 0 −2 −2 −4 4 2 0 −4 −2 −2 −4 y y y 10. 11. 124 −2 −2 −4 2 4 Section 11.1 Ellipses 699 Calculus can be used to show that the area of the ellipse 2 with equation y˛ 2 b˛ the area of each ellipse in Exercises 17 – 22. Pab. 2 x˛ 2 a˛ 1 is Use this fact to find 2 x˛ 16 2 y˛ 4 1 3x˛ 2 4y˛ 2 12 6x˛ 2 2y˛ 2 14 17. 19. 21. 18. 20. 22. 2 x˛ 9 2 y˛ 5 1 7x˛ 2 5y˛ 2 35 5x˛ 2 y˛ 2 5 23. Washington, D.C. has a park located next to the White House called The Ellipse. Letting the center of the ellipse be at the origin of a coordinate system, the equation that defines the boundary of the park is 2 x˛ 562,500 2 y˛ 409,600 1 Find how many square feet of grass is needed to cover the entire park. 24. The Indian mathematician Ramanujan is credited with developing the following formula that approximates the circumference of an ellipse. If 2a and 2b are the lengths of the major and minor axes of the ellipse, the circumference can be approximated by p 3a 3b 2 a 3b b 3a 2 1 1 Estimate the amount of fencing needed to enclose The Ellipse park described in Exercise 23. 2 2 1 25. Consider the ellipse whose equation is Show that if circle. a b, then the graph is actually a 2 x˛ 2 a˛ 2 y˛ 2 b˛ 1. 26. Complete the derivation of the equation of the ellipse as follows. a. By squaring both sides, show that the equation 2 x c 2 y˛ 2 2a 2 1 2 may be simplified as x c 2 y˛ 2 2 ˛ 1 In Exercises 13–16, find a complete graph of the equation. b. Show |
that the last equation in part a may be further simplified as a2 1 x c 2 y˛ 2 ˛ 2 a˛ 2 cx. 2 x˛ 25 2 y˛ 4 1 14. 2 x˛ 6 2 y˛ 16 1 4x˛ 2 3y˛ 2 12 16. 9x˛ 2 4y˛ 2 72 13. 15. 2 c˛ 2 a˛ 1 x˛ 2 2 a˛ 2 y˛ 2 a˛ 2 c˛ 2 a˛ 2 1. 2 27. Sketch the graph of b 8, b 12, and 2 2 y˛ 4 b 20. 1 x˛ 2 b˛ What happens to the b 2, b 4, for 700 Chapter 11 Analytic Geometry ellipse as b takes larger and larger values? Could the graph ever degenerate into a vertical line? 28. Halley’s Comet has an elliptical orbit with the sun at one focus and a major axis of 1,636,484,848 miles. The closest the comet comes to the sun is 54,004,000 miles. What is the maximum distance from the comet to the sun? 29. The orbit of the Moon around Earth is an ellipse with Earth at one focus. If the length of the major axis of the orbit is 477,736 miles and the length of the minor axis is 477,078 miles, find the minimum and maximum distances from Earth to the Moon. 30. Critical Thinking An arched footbridge over a 100-foot river is shaped like half an ellipse. The maximum height of the bridge over the river is 20 feet. Find the height of the bridge over a point in the river exactly 25 feet from the center of the river. 31. Critical Thinking Find the length of the sides (in terms of a, b, and c) of triangle FOC in the following ellipse. Its equation is 2 x˛ a2 F is one focus. Justify your answer. 2 y˛ 2 b˛ 1 and y C O x F 11.2 Hyperbolas Objectives • Define a hyperbola • Write the equation of a hyperbola • Identify important characteristics of hyperbolas • Graph hyperbolas Like an ellipse, a hyperbola has two foci, two vertices, and a center; but its shape is quite different. Definition |
of a Hyperbola Let P and Q be points in the plane and k be a positive number. The hyperbola with foci P and Q is the set of all points X such that the absolute value of the difference of the distance from X to P and the distance from X to Q is k. That is, 0 XP XQ 0 k, where X represents the point (x, y) and k is called the distance difference. Section 11.2 Hyperbolas 701 As shown in Figure 11.2-1, a hyperbola consists of two separate branches (shown in red). The distances XP and XQ are shown in blue. The dotted straight lines are the asymptotes of the hyperbola. The asymptotes are not part of the hyperbola but are useful in graphing. A hyperbola approaches its asymptotes, but it never touches them. is the center of the The midpoint of the segment joining the foci, hyperbola, and the line through P and Q is called the focal axis. The points where the focal axis intercepts the hyperbola are its vertices. PQ, center X P Q vertices foci Figure 11.2-1 Equation of a Hyperbola The simplest case is a hyperbola centered at the origin with its foci on the x- or y-axis. The equation of a hyperbola is derived by using the distance formula, and it is left as an exercise. Standard Equation of a Hyperbola Centered at the Origin Let a and b be positive real numbers. Then the graph of each of the following equations is a hyperbola centered at the origin. foci on the x-axis: y˛ b˛ 1 x˛ a˛ 2 2 2 2 foci on the y-axis: 2 x˛ 2 b˛ 2 y˛ 2 a˛ 1 b −c −a −b y y = xb a x c a y = − xb a y c a b −b −a −c y = xa b x y = − xa b The characteristics of the graph of a hyperbola centered at the origin are shown in the following list. Notice in both cases that • the hyperbola bends toward the foci • the positive term determines which way the hyperbola opens • the distance between the foci is 2c • the distance between the vertices is 2 |
a • the center is the midpoint between the foci and the midpoint between the vertices 2 a˛ c˛ 2 b˛ 2 • When the equation is in standard form with the x term positive and y term negative, the hyperbola intersects the x-axis and opens left and right. When the x term is negative and the y term is positive, the hyperbola intersects the y-axis and opens up and down. 702 Chapter 11 Analytic Geometry Characteristics of Hyperbolas For positive numbers a and b, Foci on the x-axis: y˛ 2 b˛ 1 x˛ a vertices −c −a a c x foci a x-intercepts: y-intercepts: none focal axis is on the x-axis (a, 0) (a, 0) vertices (c, 0) foci and c 2a2 b2 and (c, 0), where Foci on the y-axis: x˛ b˛ 2 y˛ 2 a˛ 1 2 2 y c a −a −c vertices y = − x a b foci x a y = x b x-intercepts: none ±± a y-intercepts: focal axis is on the y-axis (0, a) (0, a) vertices (0, c) foci and c 2a2 b2 and (0, c), where y b a x and asymptotes: y b a x y a b x and asymptotes: y a b x Graphing a Hyperbola Example 1 Graph a Hyperbola 9x˛ Show that the graph of the equation is a hyperbola. Graph it and its asymptotes. Find the equations of the asymptotes, and label the foci and the vertices. 2 4y˛ 2 36 Solution Put the equation in standard form by dividing both sides by 36 and simplifying. 2 2 2 9x˛ 36 x˛ 4 2 x˛ 22 4y˛ 36 2 y˛ 9 2 y˛ 32 1 1 1 Section 11.2 Hyperbolas 703 b 3 Applying the fact in the box with is a hyperbola centered at the origin with vertices a 2 and shows that the graph and 2, 0 2, 0 and has asymptotes y 3 2 x and y 3 |
2 x. 2 First plot the vertices and sketch 1 2 1 the auxiliary rectangle determined by the vertical lines and y ± b ± 3. the horizontal lines The asymptotes go through the origin and the corners of this rectangle, as shown on the left in Figure 11.2-2. It is then easy to sketch the hyperbola by drawing curves that are asymptotic to the dashed lines. x ± a ± 2 y (0, 3) y = 3 2 x y y = − x3 2 y = x3 2 − x2 4 y2 9 = 1 (–2, 0) (2, 0) (– 13, 0) (–2, 0) (2, 0) ( 13, 0) x x (0, –3) y = − x3 2 Figure 11.2-2 Locate the foci by using the formula c 222 33 14 9 113 3.6 c 2a2 b2 with a 2 and b 3. Therefore, the foci are the right in Figure 11.2-2. A 113, 0 and A B 113, 0 B, as shown on the graph on ■ Example 2 Graph a Hyperbola on a Calculator Identify the graph of 4x˛ 2 9y˛ 2 36, and then graph it on a calculator. Solution Dividing both sides of of a hyperbola. 4x˛ 2 9y˛ 2 36 by 36 shows that it is the equation 2 x˛ 9 2 y˛ 4 1 704 Chapter 11 Analytic Geometry To graph this hyperbola on a calculator, solve the original equation for y. 6.2 −9.4 9.4 9y˛ y˛ 2 4x˛ 2 4x˛ 2 36 2 36 9 y ± 4x˛ 2 36 9 B ± 2 3 2x2 9 The hyperbola is defined by the two functions −6.2 Y1 2 3 2x˛ 2 9 and Y2 2 3 2x˛ 2 9 Figure 11.2-3 whose graphs are shown in Figure 11.2-3. ■ NOTE The two branches of the hyperbola in Figure 11.2-3 do not correspond to the two functions shown in Example 2. One function gives the part above the x-axis and the other gives the part below the x-axis. y (0, 10) (3, 2) 4 2 − |
4 −2 0 2 x 4 −2 −4 (0, − 10) Figure 11.2-4 Writing the Equation of a Hyperbola Example 3 Find the Equation of a Hyperbola Find the equation of the hyperbola that has vertices at and passes through the point A asymptotes, and label the foci. 0, 1 1 2 Then sketch its graph by using the 3, 12 and 0, 1 B. 1 2 Solution The vertices are on the y-axis and the equation has the form 2 y˛ 2 a˛ 2 x˛ 2 b˛ 1, with a 1. Because 3, 12 A B A 2 B 1 is on the graph, 12 32 b2 12 2 9 b2 b2 9. 1 Therefore, b 3 and the equation of the hyperbola is 2 y˛ 12 2 x˛˛ 32 1 or y˛ 2 2 x˛ 9 1 The asymptotes of the hyperbola are the lines y ± 1 3 x. The foci are on the y-axis c units away from the center, 0, 0 1 2, where Thus, the foci are at c 212 33 11 9 110 0, 110 and, 0, 110 A B B A as shown in Figure 11.2-4. ■ Section 11.2 Hyperbolas 705 Applications of Hyperbolas Applications modeled by hyperbolas occur in science, business, and economics. Unlike Halley’s comet, which as an elliptical orbit, some comets have hyperbolic orbits. These comets pass through the solar system once and never return. Additionally, the reflective properties of hyperbolas are used in the design of camera and telescope lenses. The Hubble Space Telescope incorporates a Cassegrain telescope (invented in 1672), which has both a hyperbolic mirror and a parabolic mirror. If a light ray passes through one focus of a hyperbola and reflects off the hyperbola at a point P, then the reflected ray moves along the straight line determined by P and the other focus, as shown in Figure 11.2-5. focus focus focus P focus P Figure 11.2-5 The next example illustrates the way hyperbolas are used in location systems. Example 4 Determine Locations An explosion was heard on a passenger ship and on a naval ship that are mile apart. Passengers on Ship A heard the sound 1 2 1 2 second |
before sailors on Ship B. The speed of sound in air is approximately 1100 feet per second. Describe the possible locations of the explosion. Solution In 1 2 second, the sound traveled 1 2 1 1100 2, or 550 feet. Therefore, the explo- sion occurred at a point 550 feet closer to ship A than to ship B. That is, the difference between the distance from the explosion to ship B and from the explosion to ship A is 550 feet. Whenever a difference is constant, a hyperbola is usually a good model. Draw a coordinate system and place A and B on the x-axis equidistant from the origin. The locations of the ships are the foci of the hyperbola, and the hyperbola contains all possible locations of the explosion. 706 Chapter 11 Analytic Geometry y 1000 A tB −1000 0 1000 −1000 Figure 11.2-6a of A and B are 1 Because the distance from A to B is mile, or 2640 feet, the coordinates 2 1320, 0 as shown in Figure 11.2-6a. 1 The explosion occurred on one branch of the hyperbola with foci 0 XB XA 0 and 1 hyperbola. 1320, 0 2 for every point X on the 1320, 0 such that 550 1320, 0 and, 1 1 2 2 2 V1 V1 Let to focus A. Because the difference between the distances BV1 be the vertex of the hyperbola closer is on the hyperbola, and is 550 feet. As shown in Figure 11.2-6b, AV1 0 c a 0 1 2 0 2a 0 0 BV1 2 550 550 550 a 275 Because a 7 0, c a AV1 0 1 y c − a A V1 c + a B x a a c c NOTE For every point X on the branch of the hyperbola closer to A, XB XA 550. For every point X on the branch of the hyperbola closer to B, XB XA 550. Thus, a 275, c 1320, and b2 c2 a2 13202 2752 Figure 11.2-6b 1,742,400 75,625 b2 1,666,775 Therefore, the equation of the hyperbola is x2 a2 y2 b2 x2 75,625 y2 1,666,775 1. The explosion occurred somewhere on the branch of the hyperbola |
closer to the passenger ship at A, as illustrated in Figure 11.2-6c. y x2 75,625 − y2 1,666,775 = 1 A B x (−1320, 0) (−275, 0) (1320, 0) (275, 0) Figure 11.2-6c To find the exact location of the explosion, the sound must be detected at a third location that is the focus of another hyperbola that shares one of the two foci given in the original problem. The intersection of the two hyperbolas will identify the precise location of the explosion, but the second hyperbola will not have its center at the origin. See Example 10 in Section 11.4, which gives the procedure for finding the exact location. ■ Section 11.2 Hyperbolas 707 9. y 8 4 x 0 −8 −4 −4 4 8 Exercises 11.2 In Exercises 1–6, find the equation of the hyperbola centered at the origin that satisfies the given conditions. 1. x-intercepts ± 3, asymptote y 2x 2. y-intercepts ± 12, asymptote y 3x 2, passing through A B B 10. A B 213, 6 3. vertex passing through 4, 13, 2, 0 2 1 0, 112 A 3, 0 4. vertex 5. focus 6. focus 1 1 and vertex 2 2, 0 2 1 0, 4 2 and vertex 0, 112 A B In Exercises 7–12, match one of the following equations to the given graph. 2 y˛ 16 2 x˛ 9 1 8x2 y 2 8 16 y˛ 2 2 25x˛ 9 1 11. 12. 2 x˛ 9 2 y˛ 4 1 8x˛ 2 y˛ 2 8 2 x˛ 9 2 y˛ 4 1 8 4 0 −8 −4 −4 −8 8 4 y y 7. 8. 4 8 x t 4 8 −8 0 −4 −4 −8 −8 2 1 −2 0 −1 −1 −2 4 2 y y −4 −2 0 −2 −4 y 4 2 −4 −2 0 −2 − In Exercises 13–18, sketch a complete graph of the equation. Label the foci and the asymptote equations. 2 x˛ 6 2 y˛ 16 1 4x |
˛ 2 y˛ 2 16 14. 16. x2 4 y 2 1 3y˛ 2 5x˛ 2 15 18y˛ 2 8x˛ 2 2 0 18. x˛ 2 2y˛ 2 1 13. 15. 17. 708 Chapter 11 Analytic Geometry In Exercises 19–24, graph each equation using a graphing calculator. 30. y2 8 x2 36 1 at 6, 4 1 2 19. 4x2 y 2 16 21. 2 2y 2 1 x 20. 22. x2 4 x2 6 y2 1 y2 16 1 23. 3y 2 5x 2 15 24. 18y 2 8x 2 2 0 y2 25. Sketch the graph of 4 b 20. b 8, b 12, hyperbola as b takes larger and larger values? Could the graph ever degenerate into a pair of horizontal lines? x2 b2 What happens to the 1 and for b 2, b 4, 26. Sketch the graph of y2 a2 a 0.5. 1 x2 16 What happens to the a 8, a 4, for and a 2, a 1, hyperbola as a takes smaller and smaller values? Could the graph ever degenerate into a pair of horizontal lines? 27. April and Marty, 2 miles apart, are talking on the phone when lightning strikes nearby. They each hear the thunder, but April hears it 2.4 seconds after Marty. Sketch a graph of the locations where the lightning could have struck. [Sound travels at approximately 1100 feet per second.] 28. In Exercise 27, suppose that later in the conversation Marty hears the thunder 3 seconds after April. Sketch a graph of the locations where the lightning could have struck. For Exercises 29–30, write the equation of the tangent line to the given curve at the given point by using the following facts. The slope m of the tangent line to a hyperbola at the point (x, y) is m b2x a2 y m a2x b2 y for for x2 a2 y2 a2 y2 b2 x2 b2 1 1 29. x2 8 y2 4 1 at 4, 2 1 2 31. Show that the difference between the distance from each focus to any point on a hyperbola is equal to the distance between the vertices. 32. Derive the equation of a hyperbola centered at the be |
a point on the hyperbola with origin as follows. a. Let x, y P 2 1 c, 0 F11 foci F1 P 7 F2 P. the distance formula, and Exercise 31, and 2 By the definition of a hyperbola, Assume that F21 c, 0. 2 F1 P F2 2a 2 Show that the last equation simplifies as shown. cx a2 a2 x c 2 1 2 y 2 b. Show that the last equation in part a may be further simplified as shown. c2 a2 2 x 1 a2 y2 a2 c2 a2 2 1 and show that the equation in 2 c. Let b2 c2 a2 b simplifies to the standard form of a hyperbola centered at the origin. 33. Critical Thinking The following hyperbola is centered at the origin with vertex V and the auxiliary rectangle as shown. Use the length of one of the sides of triangle POV to locate the foci. Justify your answer. y P x O V Section 11.3 Parabolas 709 11.3 Parabolas Objectives • Define a parabola • Write the equation of a parabola • Identify important characteristics of parabolas • Graph parabolas Parabolas appeared in Section 3.3 as the graphs of quadratic functions, which are a special case of the following more general definition. Definition of a Parabola Let L be a line in the plane and P be a point not on L. If X is any point not on L, the distance from X to L is defined to be the length of the perpendicular line segment from X to L. The parabola with focus P and directrix L is the set of all points X such that distance from X to P distance from X to L as shown in Figure 11.3-1. L X a xis P focus vertex directrix Figure 11.3-1 The line through P perpendicular to L is called the axis. The intersection of the axis with the parabola, which is the midpoint of the segment of the axis from P to L, is the vertex of the parabola, as shown in Figure 11.3-1. Equation of a Parabola Suppose that the focus is on the y-axis at the point nonzero constant, and that the directrix is the horizontal line x, y 2 1 horizontal line x, p, where p is a y p. |
If x, y to the x, y is the length of the vertical segment from to is any point on the parabola, then the distance from as shown in Figure 11.3-2. y p 0 By the definition of a parabola, y (x, y) (0, p) (0, −p) x (x, −p) Figure 11.3-2 2 1 distance from distance from x, y 1 x, y 1 x 0 to to 2 2 2 0, p 2 1 0, p 2 1 y p 2 1 2 distance from 1 distance from 1 2 2 2 x x x, y x, y to y p x, p to 710 Chapter 11 Analytic Geometry Square both sides of the equation and simplify. x 0 1 2 y˛ x 2py p2 02 y˛ x˛ 2 4py 1 y p 2 2 2 1 2 2 2py p2 standard form of a parabola Conversely, it can be shown that every point whose coordinates satisfy this equation is on the parabola. A similar argument works for the on the x-axis and directrix the vertical line parabola with focus x p, and leads to the following conclusion. p, 0 1 2 Standard Equation of a Parabola with Vertex at the Origin Let p be a nonzero real number. Then the graph of each of the following equations is a parabola with vertex at the origin. focus at directrix: (0, p) y p focus at directrix: (p, 0) x p x2 4py y2 4px The equations of a parabola in standard form can also be y 1 2 4 p x˛ NOTE written as and x 1 4 p y˛ 2. One of the variables is quadratic and the other variable is linear. When the y term is linear, the parabola opens up or down; when the x term is linear, the parabola opens left or right. The characteristics of the graph of a parabola with vertex at the origin are shown in the following box. Notice the following facts in both cases. • the parabola bends toward the focus and away from the directrix • the linear term determines the orientation of the parabola and the axis of symmetry—left/right or up/down • the sign of p determines which way the parabola opens • the distance between the focus and the directrix |
is 2p • the distance from the vertex to the focus and the distance from the vertex to the directrix is p • the vertex is the midpoint of the line segment joining the focus and the directrix Characteristics of Parabolas Section 11.3 Parabolas 711 For a nonzero real number p, parabolas have the following characteristics. x2 4py y2 4px • focus on the y axis at (0, p) • focus on the x-axis at (p, 0) • directrix y p • directrix x p • axis of symmetry is the • axis of symmetry is the y-axis x-axis • If p 77 0, then the parabola: opens up opens right y focus vertex x y directrix focus x vertex directrix • If p 66 0, then the parabola: opens down opens left y directrix vertex x focus y directrix focus vertex x Example 1 Graphing a Parabola Show that the graph of the equation graph, and then find and label its focus and directrix. 2 8y 0 x˛ is a parabola. Draw its Solution Write the equation in standard form: form 2 8y. This equation is of the so the graph is a parabola. To find the value of p, note that 2 4py, x˛ x˛ 4p 8 p 2 712 Chapter 11 Analytic Geometry y 4 y = 2 x 8 4 F(0, −2) −8 −4 0 −4 −8 Therefore, the following statements are true, as shown in Figure 11.3-3. • the focus is 0, p 1 0, 2 2 1 2 y p • the directrix is • the parabola opens downward because p is negative 1 2 2 2 ■ Example 2 Graph a Parabola on a Calculator Figure 11.3-3 On a calculator, graph the parabola y2 12x. 10 1 Solution Solve the equation for y and enter both functions into a calculator. y ± 112x 10 The graphs of Y1 112x and Y2 112x are shown in Figure 11.3-4. ■ 10 Figure 11.3-4 Writing the Equation of a Parabola Example 3 Find the Equation of a Parabola y 8 4 x = 3 F(−3, 0) −8 −4 0 4 x 8 −4 −8 Figure 11.3-5 Find the focus, directrix |
, and equation of the parabola that passes through and has its focus on the x-axis. 0, 0 the point 1 Sketch the graph of the parabola, and label its focus and directrix. has vertex 1, 112 A B,, 2 Solution Because the focus is on the x-axis, the equation has the form Since is on the graph, 1, 112 A B 112 A 1 2 2 4p B 1 12 4p 3 p 2 4˛px. y˛ Therefore, the focus is The equation of the parabola is Figure 11.3-5. 2 1 3, 0 and the directrix is the vertical line x 3. whose graph is shown in 2 12x, y˛ ■ Applications of Parabolas Projectiles follow a parabolic curve, a fact that is used in the design of water slides in which the rider slides down a sharp incline, then up and over a hill, before plunging downward into a pool. At the peak of the hill, the rider shoots up along a parabolic arc several inches above the slide, experiencing a sensation of weightlessness. Section 11.3 Parabolas 713 Certain laws of physics show that sound waves or light rays from a source at the focus of a parabola will reflect off the parabola in rays parallel to the axis of the parabola, as shown in Figure 11.3-6. This is the reason that parabolas are used in automobile headlights and searchlights. focus axis axis focus Figure 11.3-6 Figure 11.3-7 Conversely, a sound wave or light ray coming toward a parabola will be reflected into the focus, as shown in Figure 11.3-7. This fact is used in the design of radar antennas, satellite dishes, and field microphones used at outdoor sporting events to pick up conversations on the field. Example 4 Parabola Application A radio telescope in the Very Large Array at Socorro, New Mexico, shown in Figure 11.3-8a, has the shape of a parabolic dish (a cross section through the center of the dish is a parabola). It is approximately 12 feet deep at the center and has a diameter of 82 feet. How far from the vertex of the parabolic dish should the receiver be placed in order to “catch” all the radio waves that hit the dish? Figure 11.3-8a 714 Chapter 11 Analytic Geometry y 40 20 −40 |
0 −20 Figure 11.3-8b 20 (41, 12) x 40 All radio waves hitting the dish are reflected into the focus, so the receiver should be located there. To find the focus, draw a cross section of the dish, with the vertex at the origin, as shown in Figure 11.3-8b. The equation of Because the point (41, 12) is on the this parabola is of the form parabola, x2 4py. 12 x2 4py 412 4p 1 412 48p p 412 48 p 1681 48 2 35 feet Substitute Simplify Divide both sides by 48 The focus is the point 0, 0 2 fore, the receiver should be placed about 35 feet from the vertex. which is p units from the vertex 0, p, 1 1 2. There- ■ Exercises 11.3 In Exercises 1–6, find the equation of the parabola with vertex at the origin that satisfies the given condition. 7. 1. axis x 0, passing through 2. axis y 0, passing through 2, 12 2, 12 2 2 1 1 3. focus 4. focus 1 1 5, 0 2 0, 3.5 5. directrix 2 x 2 6. directrix y 3 In Exercises 7–10, match one of the following equations to the given graph. 8. 6x y˛ 2 2 4x y˛ y x2 4 2 8y x˛ y 8 4 −8 −4 0 4 −4 −8 8 4 y −8 −4 0 4 −4 −8 x 8 x 8 5910ac11_690-775 9/21/05 2:00 PM Page 715 9. 10. y 8 4 −8 −4 0 4 −4 −8 8 4 y −8 −4 0 4 −4 −8 x 8 x 8 In Exercises 11–14, sketch a complete graph of the equation. 11. x 6y˛ 2 13. 8x 2y˛ 2 12. 1 2 y2 2x 14. 4y x˛ 2 In Exercises 15–18, find the focus and directrix of the parabola. 15. y 3x2 17. y 1 4 x2 16. x 1 2 y2 18. x 6y˛ 2 Section 11.3 Parabolas 715 In Exercises 19–22, find the equation of the parab |
ola with vertex at the origin passing through the given points. 2 2 8x y˛ 19. 21. 1 1 1, 2 and 1, 2 1 2 2 20. 3,3 2 b a and 3 2 b 3, a 1, 2 and 1 2 1, 2 2 22. 1 2, 5 and 2, 5 2 1 23. Find the point on the graph of that is closest to the focus of the parabola. 24. Find the point on the graph of 2 3y x˛ that is closest to the focus of the parabola. 25. The receiver in a parabolic television dish is 2 feet from the vertex and is located at the focus. Find an equation of a cross section of the receiver. (Assume that the dish is directed to the right and that the vertex is at the origin.) 26. The receiver in a parabolic television dish is 1.5 feet from the vertex and is located at the focus. Find an equation of a cross section of the receiver. (Assume that the dish is directed upward and that the vertex is at the origin.) 27. The filament of a flashlight bulb is located at the focus, which is 1 2 inch from the vertex of a parabolic reflector. Find an equation of a cross section of the reflector. (Assume that the flashlight is directed to the left and that the vertex is at the origin.) 28. The filament of a flashlight bulb is located at the focus, which is 1 4 inch from the vertex of a parabolic reflector. Find an equation of a cross section of the reflector. (Assume that the flashlight is directed downward and that the vertex is at the origin.) 716 Chapter 11 Analytic Geometry 11.4 Translations and Rotations of Conics Objectives • Write the equation of a translated conic • Graph a translated conic • Determine the shape of a translated conic without graphing • Apply translated conics to real-world problems Horizontal and Vertical Shifts Now that you are familiar with conic sections centered at the origin, the discussion will be expanded to include both conics centered at other points in the plane and ones with axes that may not be parallel to the coordinate axes. 2 1 x As you saw in Section 3.4, replacing a variable x with y f a function x 5, whereas replacing x with to the left. Similarly, if the rule of a function is given by replacing y with y 4 f is equivalent to t |
ions, a similar result applies. in the rule of shifts the graph of the function 5 units to the right, 5 shifts the graph 5 units y f then shifts the graph 4 units upward, because For equations that are not func- that is, y 4 y f 4 and y with in an equation shifts the graph of the equation: • Let h and k be constant. Replacing x with y k 00 h 00 negative h 00 k 00 negative k • units upward for positive k and downward for units to the right for positive h and to the left for The process of writing the equation of a conic is the same as that discussed in Sections 11.1 through 11.3, except that x and y are replaced with x h y k, where (h, k) is the vertex of a parabola or the center of an ellipse or a hyperbola. and Example 1 Graph a Translated Conic Identify and sketch the graph of 36 2 1 and find its center, major axis, and minor axis. Solution The given equation can be obtained from the equation x2 9 y 2 36 1, whose graph is known to be an ellipse, as follows: y and replace y by replace x by x 5 1 This is the situation described in the previous box with 4 y 4. 2 h 5 and k 4. Therefore, the graph is the ellipse x2 9 y 2 36 1 shifted 5 units to the right Section 11.4 Translations and Rotations of Conics 717 and 4 units downward, as shown in Figure 11.4-1. • The center of the ellipse is 5, 4. 2 1 • The major axis lies on the vertical line • The minor axis is on the horizontal line y 4. x 5. y 8 4 −8 −4 0 4 8 x −4 −8 Figure 11.4-1 ■ Before identifying a conic section and determining its characteristics, the corresponding equation should be rewritten in standard form. Example 2 Identify a Conic Identify and sketch the graph of 4x 2 9y 2 32x 90y 253 0. Solution Rewrite the equation as 253 253 1 4x 4 1 2 32x x2 8x 1 9 2 90y 9y y2 10y 2 2 1 2 10y. y and 2 10y 25 y 2 2 2 8x 9 2 1 2 253?? x Complete the square in 2 8x 16 x 4 1 4 16 64 Be careful here |
: 16 and 25 were not added to the left side of the equation. Actually were added, when the left side is multiplied out. Therefore, to leave the original equation unchanged, 64 and 225 must be added to the right side. 9 25 225 and x 4 1 2 8x 16 2 4 4 1 253 64 225 2 2 36 36 x 4 9 2 10y 25 36 NOTE Review the technique of completing the square in Section 2.2, if needed. 718 Chapter 11 Analytic Geometry The graph of this equation is the ellipse 2 x˛ 9 2 y˛ 4 1 shifted 4 units to the right and 5 units upward. Its center is at (4, 5), its major axis lies on the as horizontal line shown in Figure 11.4-2. and its minor axis lies on the vertical line x 4, y 5, y 8 4 −4 0 4 8 x −4 Figure 11.4-2 ■ Example 3 Writing the Equation of a Translated Conic Find the equation of an ellipse with center at points of its major and minor axes are Find the coordinates of the foci. 5, 2, 1 2 1 1 5, 10 5, 4 2, 1 2 such that the end2, 4 8, 4 and., 1 2 2 Solution The major axis has length 12 and is parallel to the y-axis. The minor axis has length 6 and is parallel to the x-axis. Therefore, and the equation of the ellipse has the form a 6, b 3, 2 1 x h 32 2 1 2 y k 62 2 1 Because it has its center at form 1 5, 4, 2 the equation of the ellipse has the y 8 4 1 −8 −4 0 4 8 x Since a 6 and 2 2 x 5 32 22 1 4 1 62 y 4 36 c 2a2 b2, 1. 2 2 and in an ellipse, b 3, c 262 32 136 9 127 313 −4 −8 The foci of 36 2 1 Figure 11.4-3 on the major axis. That is, the foci are as shown in Figure 11.4-3. A are 313 5, 4 313 units from the center 5, 4 5, 4 313 A 1 2, and B B ■ Section 11.4 Translations and Rotations of Conics 719 Example 4 Identify a Translated Conic Identify and sketch the graph of istics of the conic. |
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