Applications of Trigonometry

Transcription

1 5144_Demana_Ch06pp /11/06 9:31 PM Page 501 CHAPTER 6 Applications of Trigonometr 6.1 Vectors in the Plane 6. Dot Product of Vectors 6.3 Parametric Equations and Motion 6.4 Polar Coordinates 6.5 Graphs of Polar Equations 6.6 De Moivre s Theorem and nth Roots Young salmon migrate from the fresh water the are born in to salt water and live in the ocean for several ears. When it s time to spawn, the salmon return from the ocean to the river s mouth, where the follow the organic odors of their homestream to guide them upstream. Researchers believe the fish use currents, salinit, temperature, and the magnetic field of the Earth to guide them. Some fish swim as far as 3500 miles upstream for spawning. See a related problem on page

2 5144_Demana_Ch06pp /11/06 9:31 PM Page CHAPTER 6 Applications of Trigonometr JAMES BERNOULLI ( ) The first member of the Bernoulli famil (driven out of Holland b the Spanish persecutions and settled in Switzerland) to achieve mathematical fame, James defined the numbers now known as Bernoulli numbers. He determined the form (the elastica) taken b an elastic rod acted on at one end b a given force and fied at the other end. Chapter 6 Overview We introduce vectors in the plane, perform vector operations, and use vectors to represent quantities such as force and velocit. Vector methods are used etensivel in phsics, engineering, and applied mathematics. Vectors are used to plan airplane flight paths. The trigonometric form of a comple number is used to obtain De Moivre s theorem and find the nth roots of a comple number. Parametric equations are studied and used to simulate motion. One of the principal applications of parametric equations is the analsis of motion in space. Polar coordinates another of Newton s inventions, although James Bernoulli usuall gets the credit because he published first are used to represent points in the coordinate plane. Planetar motion is best described with polar coordinates. We convert rectangular coordinates to polar coordinates, polar coordinates to rectangular coordinates, and stud graphs of polar equations. 6.1 Vectors in the Plane What ou ll learn about Two-Dimensional Vectors Vector Operations Unit Vectors Direction Angles Applications of Vectors... and wh These topics are important in man real-world applications, such as calculating the effect of the wind on an airplane s path. OBJECTIVE Students will be able to appl the arithmetic of vectors and use vectors to solve real-world problems. MOTIVATE Discuss the difference between the statements: Jose lives 3 miles awa from Mar and Jose lives 3 miles west of Mar. LESSON GUIDE Da 1: Two-Dimensional Vectors; Vector Operations Da : Unit Vectors, Direction Angles; Applications of Vectors Two-Dimensional Vectors Some quantities, like temperature, distance, height, area, and volume, can be represented b a single real number that indicates magnitude or size. Other quantities, such as force, velocit, and acceleration, have magnitude and direction. Since the number of possible directions for an object moving in a plane is infinite, ou might be surprised to learn that two numbers are all that we need to represent both the magnitude of an object s velocit and its direction of motion. We simpl look at ordered pairs of real numbers in a new wa. While the pair (a, b) determines a point in the plane, it also determines a directed line segment (or arrow ) with its tail at the origin and its head at (a, b) (Figure 6.1). The length of this arrow represents magnitude, while the direction in which it points represents direction. Because in this contet the ordered pair (a, b) represents a mathematical object with both magnitude and direction, we call it the position vector of (a, b), and denote it as a, b to distinguish it from the point (a, b). O (a) (a, b) O FIGURE 6.1 The point represents the ordered pair (a, b). The arrow (directed line segment) represents the vector a, b. a, b (b) (a, b)

3 5144_Demana_Ch06pp /11/06 9:31 PM Page 503 SECTION 6.1 Vectors in the Plane 503 IS AN ARROW A VECTOR? While an arrow represents a vector, it is not a vector itself, since each vector can be represented b an infinite number of equivalent arrows. Still, it is hard to avoid referring to the vector PQ in practice, and we will often do that ourselves. When we sa the vector u PQ, we reall mean the vector u represented b PQ. DEFINITION Two-Dimensional Vector A two-dimensional vector v is an ordered pair of real numbers, denoted in component form as a, b. The numbers a and b are the components of the vector v. The standard representation of the vector a, b is the arrow from the origin to the point (a, b). The magnitude of v is the length of the arrow, and the direction of v is the direction in which the arrow is pointing. The vector 0 0, 0, called the zero vector, has zero length and no direction. S( 1, 6) P(3, 4) It is often convenient in applications to represent vectors with arrows that begin at points other than the origin. The important thing to remember is that an two arrows with the same length and pointing in the same direction represent the same vector. In Figure 6., for eample, the vector 3, 4 is shown represented b RS, an arrow with initial point R and terminal point S, as well as b its standard representation OP. Two arrows that represent the same vector are called equivalent. R( 4, ) O(0, 0) The quick wa to associate arrows with the vectors the represent is to use the following rule. FIGURE 6. The arrows RS and OP both represent the vector 3, 4, as would an arrow with the same length pointing in the same direction. Such arrows are called equivalent. Head Minus Tail (HMT) Rule If an arrow has initial point 1, 1 and terminal point,, it represents the vector 1, 1. EXAMPLE 1 Showing Arrows are Equivalent Show that the arrow from R ( 4, ) to S ( 1, 6) is equivalent to the arrow from P (, 1) to Q (5, 3) (Figure 6.3). S( 1, 6) R( 4, ) Q(5, 3) O P(, 1) FIGURE 6.3 The arrows RS and PQ appear to have the same magnitude and direction. The Head Minus Tail Rule proves that the represent the same vector (Eample 1). SOLUTION Appling the HMT rule, we see that RS represents the vector 1 ( 4), 6 3, 4, while PQ represents the vector 5, 3 ( 1) 3, 4. Although the have different positions in the plane, these arrows represent the same vector and are therefore equivalent. Now tr Eercise 1.

4 5144_Demana_Ch06pp /11/06 9:31 PM Page CHAPTER 6 Applications of Trigonometr P( 1, 1 ) Q(, ) FIGURE 6.4 The magnitude of v is the length of the arrow PQ,which is found using the distance formula: v 1 1. EXPLORATION 1 Vector Archer See how well ou can direct arrows in the plane using vector information and the HMT Rule. 1. An arrow has initial point (, 3) and terminal point (7, 5). What vector does it represent? 5,. An arrow has initial point (3, 5) and represents the vector 3, 6. What is the terminal point? 0, If P is the point (4, 3) and PQ represents, 4, find Q. 6, 7 4. If Q is the point (4, 3) and PQ represents, 4, find P., 1 If ou handled Eploration 1 with relative ease, ou have a good understanding of how vectors are represented geometricall b arrows. This will help ou understand the algebra of vectors, beginning with the concept of magnitude. The magnitude of a vector v is also called the absolute value of v, so it is usuall denoted b v. (You might see v in some tetbooks.) Note that it is a nonnegative real number, not a vector. The following computational rule follows directl from the distance formula in the plane (Figure 6.4). WHAT ABOUT DIRECTION? You might epect a quick computational rule for direction to accompan the rule for magnitude, but direction is less easil quantified. We will deal with vector direction later in the section. Magnitude If v is represented b the arrow from 1, 1 to,, then v 1 1. If v a, b, then v a b. P( 3, 4) Q( 5, ) v O(0, 0) (, ) EXAMPLE Finding Magnitude of a Vector Find the magnitude of the vector v represented b PQ,where P ( 3, 4) and Q ( 5, ). SOLUTION Working directl with the arrow, v ( 5 ( 3)) ( 4). Or, the HMT Rule shows that v,, so v ( ) ) (. (See Figure 6.5.) Now tr Eercise 5. FIGURE 6.5 The vector v of Eample. Vector Operations The algebra of vectors sometimes involves working with vectors and numbers at the same time. In this contet we refer to the numbers as scalars. The two most basic algebraic operations involving vectors are vector addition (adding a vector to a vector) and scalar multiplication (multipling a vector b a number). Both operations are easil represented geometricall, and both have immediate applications to man real-world problems.

5 5144_Demana_Ch06pp /11/06 9:3 PM Page 505 SECTION 6.1 Vectors in the Plane 505 WHAT ABOUT VECTOR MULTIPLICATION? There is a useful wa to define the multiplication of two vectors in fact, there are two useful was, but neither one of them follows the simple pattern of vector addition. (You ma recall that matri multiplication did not follow the simple pattern of matri addition either, and for similar reasons.) We will look at the dot product in Section 6.. The cross product requires a third dimension, so we will not deal with it in this course. DEFINITION Vector Addition and Scalar Multiplication Let u u 1, u and v v 1, v be vectors and let k be a real number (scalar). The sum (or resultant ) of the vectors u and v is the vector u v u 1 v 1, u v. The product of the scalar k and the vector u is ku k u 1, u ku 1, ku. The sum of the vectors u and v can be represented geometricall b arrows in two was. In the tail-to-head representation, the standard representation of u points from the origin to u 1, u. The arrow from u 1, u to u 1 v 1, u v represents v (as ou can verif b the HMT Rule). The arrow from the origin to u 1 v 1, u v then represents u v (Figure 6.6a). In the parallelogram representation, the standard representations of u and v determine a parallelogram, the diagonal of which is the standard representation of u v (Figure 6.6b). v u u + v u v u + v (a) (b) FIGURE 6.6 Two was to represent vector addition geometricall: (a) tail-to-head, and (b) parallelogram. u u (1/)u u FIGURE 6.7 Representations of u and several scalar multiples of u. The product ku of the scalar k and the vector u can be represented b a stretch (or shrink) of u b a factor of k. If k > 0, then ku points in the same direction as u; if k < 0, then ku points in the opposite direction (Figure 6.7). EXAMPLE 3 Performing Vector Operations Let u 1, 3 and v 4, 7. Find the component form of the following vectors: (a) u v (b) 3u (c) u ( 1)v SOLUTION Using the vector operations as defined, we have: (a) u v 1, 3 4, 7 1 4, 3 7 3, 10 (b) 3u 3 1, 3 3, 9 (c) u ( 1)v 1, 3 ( 1) 4, 7, 6 4, 7 6, 1 Geometric representations of u v and 3u are shown in Figure 6.8 on the net page. continued

6 5144_Demana_Ch06pp /11/06 9:3 PM Page CHAPTER 6 Applications of Trigonometr (3, 10) v 3u = 3, 9 ( 1, 3) u + v u u = 1, 3 (a) (b) FIGURE 6.8 Given that u 1, 3 and v 4, 7, we can (a) represent u v b the tail-to-head method, and (b) represent 3u as a stretch of u b a factor of 3. Now tr Eercise 13. A WORD ABOUT VECTOR NOTATION Both notations, a, b and ai bj, are designed to conve the idea that a single vector v has two separate components. This is what makes a twodimensional vector two-dimensional. You will see both a, b, c and ai bj ck used for three-dimensional vectors, but scientists stick to the notation for dimensions higher than three. Unit Vectors A vector u with length u 1 is a unit vector. If v is not the zero vector 0, 0, then the vector v 1 u v v v is a unit vector in the direction of v. Unit vectors provide a wa to represent the direction of an nonzero vector. An vector in the direction of v, or the opposite direction, is a scalar multiple of this unit vector u. EXAMPLE 4 Finding a Unit Vector Find a unit vector in the direction of v 3,, and verif that it has length 1. SOLUTION v 3, 3 1 3, so v 1 3, v 1 3 3, The magnitude of this vector is 3, ( 1 3 ) 3 ( ) Thus, the magnitude of v v is 1. Its direction is the same as v because it is a positive scalar multiple of v. Now tr Eercise 1.

7 5144_Demana_Ch06pp /11/06 9:3 PM Page 507 SECTION 6.1 Vectors in the Plane 507 bj FIGURE 6.9 The vector v is equal to ai bj. v sin θ ai v θ v = a, b v cos θ FIGURE 6.10 The horizontal and vertical components of v. The two unit vectors i 1, 0 and j 0, 1 are the standard unit vectors. An vector v can be written as an epression in terms of the standard unit vectors: v a, b a, 0 0, b a 1, 0 b 0, 1 ai bj Here the vector v a, b is epressed as the linear combination ai bj of the vectors i and j. The scalars a and b are the horizontal and vertical components, respectivel, of the vector v. See Figure 6.9. Direction Angles You ma recall from our applications in Section 4.8 that direction is measured in different was in different contets, especiall in navigation. A simple but precise wa to specif the direction of a vector v is to state its direction angle, the angle that v makes with the positive -ais, just as we did in Section 4.3. Using trigonometr (Figure 6.10), we see that the horizontal component of v is v cos and the vertical component is v sin. Solving for these components is called resolving the vector. Resolving the Vector If v has direction angle, the components of v can be computed using the formula v v cos, v sin. v = a, b 6 O 115 FIGURE 6.11 The direction angle of v is 115. (Eample 5) From the formula above, it follows that the unit vector in the direction of v is v u cos, sin. v EXAMPLE 5 Finding the Components of a Vector Find the components of the vector v with direction angle 115 and magnitude 6 Figure SOLUTION If a and b are the horizontal and vertical components, respectivel, of v, then v a, b 6 cos 115, 6 sin 115. So, a 6 cos and b 6 sin Now tr Eercise 9.

8 5144_Demana_Ch06pp /11/06 9:3 PM Page CHAPTER 6 Applications of Trigonometr EXAMPLE 6 Finding the Direction Angle of a Vector Find the magnitude and direction angle of each vector: (a) u 3, (b) v, 5 β u α u = 3, SOLUTION See Figure 6.1. (a) u If is the direction angle of u, then u 3, u cos, u sin. 3 u cos Horizontal component of u v v =, 5 FIGURE 6.1 The two vectors of Eample cos u cos cos ( 1 3 ) is acute. (b) v 5 9. If is the direction angle of v, then v, 5 v cos, v sin. v cos Horizontal component of v 5 cos v ( ) ) ( 5 9 cos 360 cos ( 1 ) Now tr Eercise 33. Applications of Vectors The velocit of a moving object is a vector because velocit has both magnitude and direction. The magnitude of velocit is speed mph FIGURE 6.13 The airplane s path (bearing) in Eample 7. TEACHING NOTE Encourage students to draw pictures to analze the geometr of various situations. v EXAMPLE 7 Writing Velocit as a Vector A DC-10 jet aircraft is fling on a bearing of 65 at 500 mph. Find the component form of the velocit of the airplane. Recall that the bearing is the angle that the line of travel makes with due north, measured clockwise see Section 4.1, Figure 4.. SOLUTION Let v be the velocit of the airplane. A bearing of 65 is equivalent to a direction angle of 5. The plane s speed, 500 mph, is the magnitude of vector v; that is, v 500. See Figure The horizontal component of v is 500 cos 5 and the vertical component is 500 sin 5, so v 500 cos 5 i 500 sin 5 j 500 cos 5, 500 sin , The components of the velocit give the eastward and northward speeds. That is, the airplane travels about mph eastward and about mph northward as it travels at 500 mph on a bearing of 65. Now tr Eercise 41.

9 5144_Demana_Ch06pp /11/06 9:3 PM Page 509 SECTION 6.1 Vectors in the Plane 509 A tpical problem for a navigator involves calculating the effect of wind on the direction and speed of the airplane, as illustrated in Eample 8. EXAMPLE 8 Calculating the Effect of Wind Velocit Pilot Megan McCart s flight plan has her leaving San Francisco International Airport and fling a Boeing 77 due east. There is a 65-mph wind with the bearing 60. Find the compass heading McCart should follow, and determine what the airplane s ground speed will be assuming that its speed with no wind is 450 mph. A 60 C 65 mph θ 450 mph FOLLOW-UP v B D FIGURE 6.14 The -ais represents the flight path of the plane in Eample 8. Have students discuss wh it does not make sense to add a scalar to a vector. ASSIGNMENT GUIDE Da 1: E. 3 7, multiples of 3, 39, 40 Da : E. 9, 3, 34, 37, 4, 43, 45, 46, 49 COOPERATIVE LEARNING Group Activit: E NOTES ON EXERCISES E are problems that students would tpicall encounter in a phsics course. E provide practice with standardized tests. E. 6 and 64 demonstrate connections between vectors and geometr. ONGOING ASSESSMENT Self-Assessment: E. 1, 5, 13, 1, 9, 33, 41, 43, 47 Embedded Assessment: E. 45, 46, 6 SOLUTION See Figure Vector AB represents the velocit produced b the airplane alone, AC represents the velocit of the wind, and is the angle DAB. Vector v AD represents the resulting velocit, so v AD AC AB. We must find the bearing of AB and v. Resolving the vectors, we obtain AC 65 cos 30, 65 sin 30 AB 450 cos, 450 sin AD AC AB 65 cos cos, 65 sin sin Because the plane is traveling due east, the second component of AD must be zero. 65 sin sin 0 sin ( 1 65 sin30 45 ) Thus, the compass heading McCart should follow is Bearing 90 The ground speed of the airplane is v AD 6 5 c o s c o s 0 65 cos cos Using the unrounded value of. McCart should use a bearing of approimatel The airplane will travel due east at approimatel mph. Now tr Eercise 43.

10 5144_Demana_Ch06pp /11/06 9:3 PM Page CHAPTER 6 Applications of Trigonometr D w A EXAMPLE 9 Finding the Effect of Gravit A force of 30 pounds just keeps the bo in Figure 6.15 from sliding down the ramp inclined at 0. Find the weight of the bo. 0 0 C B FIGURE 6.15 The force of gravit AB has a component AC that holds the bo against the surface of the ramp, and a component AD CB that tends to push the bo down the ramp. (Eample 9) SOLUTION We are given that AD 30. Let AB w; then sin 0 C B 3 0. w w Thus, 30 w sin 0 The weight of the bo is about pounds. Now tr Eercise 47. CHAPTER OPENER PROBLEM (from page 501) PROBLEM: During one part of its migration, a salmon is swimming at 6 mph, and the current is flowing downstream at 3 mph at an angle of 7 degrees. How fast is the salmon moving upstream? SOLUTION: Assume the salmon is swimming in a plane parallel to the surface of the water. A θ current salmon swimming in still water B salmon net velocit C In the figure, vector AB represents the current of 3 mph, is the angle CAB,which is 7 degrees, the vector CA represents the velocit of the salmon of 6 mph, and the vector CB is the net velocit at which the fish is moving upstream. So we have AB 3 cos 83, 3 sin ,.98 CA 0, 6 Thus CB CA AB 3 cos 83, 3 sin , 3.0 The speed of the salmon is then CB mph upstream.

11 5144_Demana_Ch06pp /11/06 9:3 PM Page 511 SECTION 6.1 Vectors in the Plane 511 QUICK REVIEW 6.1 (For help, go to Sections 4.3 and 4.7.) In Eercises 1 4, find the values of and ; ; 4.5 (, ) (, ) ; ; In (, ) (, ) Eercises 5 and 6, solve for in degrees. 5. sin ( 1 3 ) cos ( 1 1 ) In Eercises 7 9, the point P is on the terminal side of the angle. Find the measure of if P 5, P 5, P, tan 1 (5 ) A naval ship leaves Port Norfolk and averages 4 knots nautical mph traveling for 3 hr on a bearing of 40 and then 5 hr on a course of 15. What is the boat s bearing and distance from Port Norfolk after 8 hr? Distance: naut mi.; Bearing: SECTION 6.1 EXERCISES In Eercises 1 4, prove that RS and PQ are equivalent b showing that the represent the same vector. 1. R 4, 7, S 1, 5, O 0, 0, and P 3,. R 7, 3, S 4, 5, O 0, 0, and P 3, 3. R, 1, S 0, 1, O 1, 4, and P 1, 4. R, 1, S, 4, O 3, 1, and P 1, 4 In Eercises 5 1, let P,, Q 3, 4, R, 5, and S, 8. Find the component form and magnitude of the vector. 5. PQ 5, ; 9 7. QR 5, 1 ; 6 9. QS, 4 ; QR PS 11, 7 ; RS 4, 13 ; PS 4, 10 ; PR 0, 3 ; 3 1. PS 3PQ 11, 16 ; 377 In Eercises 13 0, let u 1, 3, v, 4, and w, 5. Find the component form of the vector. 13. u v 1, u 1 v 3, u w 3, v 6, u 3w 4, u 4v 10, u 3v 4, u v 1, 7 In Eercises 1 4, find a unit vector in the direction of the given vector. 1. u, 4. v 1, 1 3. w i j 4. w 5i 5j In Eercises 5 8, find the unit vector in the direction of the given vector. Write our answer in (a) component form and (b) as a linear combination of the standard unit vectors i and j. 5. u, 1 6. u 3, 7. u 4, 5 8. u 3, 4 In Eercises 9 3, find the component form of the vector v , , v i 0.89j. 0.71i 0.71j i 0.89j i 0.71j v

12 5144_Demana_Ch06pp /11/06 9:3 PM Page CHAPTER 6 Applications of Trigonometr , ,.9 v 47 In Eercises 33 38, find the magnitude and direction angle of the vector , 4 5; , 5 ; i 4j 5; i 5j 3 4 ; cos 135 i sin 135 j 7; cos 60 i sin 60 j ; 60 In Eercises 39 and 40, find the vector v with the given magnitude and the same direction as u. 39. v, u 3, v 5, u 5, Navigation An airplane is fling on a bearing of 335 at 530 mph. Find the component form of the velocit of the airplane. 3.99, Navigation An airplane is fling on a bearing of 170 at 460 mph. Find the component form of the velocit of the airplane , Flight Engineering An airplane is fling on a compass heading bearing of 340 at 35 mph. A wind is blowing with the bearing 30 at 40 mph. (a) Find the component form of the velocit of the airplane. (b) Find the actual ground speed and direction of the plane. 44. Flight Engineering An airplane is fling on a compass heading bearing of 170 at 460 mph. A wind is blowing with the bearing 00 at 80 mph. (a) Find the component form of the velocit of the airplane. (b) Find the actual ground speed and direction of the airplane. 45. Shooting a Basketball A basketball is shot at a 70 angle with the horizontal direction with an initial speed of 10 m sec. (a) Find the component form of the initial velocit. (b) Writing to Learn Give an interpretation of the horizontal and vertical components of the velocit. 46. Moving a Heav Object In a warehouse a bo is being pushed up a 15 inclined plane with a force of.5 lb, as shown in the figure..5 lb v (a) Find the component form of the force..41, 0.65 (b) Writing to Learn Give an interpretation of the horizontal and vertical components of the force. v Moving a Heav Object Suppose the bo described in Eercise 46 is being towed up the inclined plane, as shown in the figure below. Find the force w needed in order for the component of the force parallel to the inclined plane to be.5 lb. Give the answer in component form..0, Combining Forces Juana and Diego Gonzales, ages si and four respectivel, own a strong and stubborn pupp named Corporal. It is so hard to take Corporal for a walk that the devise a scheme to use two leashes. If Juana and Diego pull with forces of 3 lb and 7 lb at the angles shown in the figure, how hard is Corporal pulling if the pupp holds the children at a standstill? about lb 3 lb lb w In Eercises 49 and 50, find the direction and magnitude of the resultant force. 49. Combining Forces A force of 50 lb acts on an object at an angle of 45. A second force of 75 lb acts on the object at an angle of 30. F lb and Combining Forces Three forces with magnitudes 100, 50, and 80 lb, act on an object at angles of 50, 160, and 0, respectivel. F lb and Navigation A ship is heading due north at 1 mph. The current is flowing southwest at 4 mph. Find the actual bearing and speed of the ship ; 9.6 mph 5. Navigation A motor boat capable of 0 mph keeps the bow of the boat pointed straight across a mile-wide river. The current is flowing left to right at 8 mph. Find where the boat meets the opposite shore. 0.4 mi downstream 53. Group Activit A ship heads due south with the current flowing northwest. Two hours later the ship is 0 miles in the direction 30 west of south from the original starting point. Find the speed with no current of the ship and the rate of the current mph; 7.07 mph 54. Group Activit Epress each vector in component form and prove the following properties of vectors. (a) u v v u (b) u v w u v w (c) u 0 u, where 0 0, 0

13 5144_Demana_Ch06pp /11/06 9:3 PM Page 513 SECTION 6.1 Vectors in the Plane 513 (d) u u 0, where a, b a, b (e) a u v au av (f) a b u au bu (g) ab u a bu (h) a0 0, 0u 0 (i) 1 u u, 1 u u (j) au a u Standardized Test Questions 55. True or False If u is a unit vector, then u is also a unit vector. Justif our answer. 56. True or False If u is a unit vector, then 1 u is also a unit vector. Justif our answer. False. 1/u is not a vector. In Eercises 57 60, ou ma use a graphing calculator to solve the problem. 57. Multiple Choice Which of the following is the magnitude of the vector, 1? D (A) 1 (B) 3 (C) 5 5 (D) 5 (E) Multiple Choice Let u, 3 and v 4, 1. Which of the following is equal to u v? E (A) 6, 4 (B), (C), (D) 6, (E) 6, Multiple Choice Which of the following represents the vector v shown in the figure below? A O 30 3 (A) 3 cos 30, 3 sin 30 (B) 3 sin 30, 3 cos 30 (C) 3 cos 60, 3 sin 60 (D) 3 cos 30, 3 sin 30 (E) 3 cos 30, 3 sin Multiple Choice Which of the following is a unit vector in the direction of v i 3j? C (A) i j (B) i j (C) i 3 10 j 1 (D) 10 i 3 10 j (E) 1 3 i j 8 8 v Eplorations 61. Dividing a Line Segment in a Given Ratio Let A and B be two points in the plane, as shown in the figure. B (a) Prove that BA OA OB,where O is the C origin. A (b) Let C be a point on the line segment BA which divides the segment in the ratio : where 1. That is, B C C A. O Show that OC OA OB. 6. Medians of a Triangle Perform the following steps to use vectors to prove that the medians of a triangle meet at a point O which divides each median in the ratio 1 :. M 1, M,and M 3 are midpoints of the sides of the triangle shown in the figure. A (a) Use Eercise 61 to prove that OM 1 1 OA 1 OB OM 1 OC 1 OB OM 1 3 OA 1 OC (b) Prove that each of OM 1 OC,OM OA,OM3 OB is equal to OA OB OC. (c) Writing to Learn Eplain wh part b establishes the desired result. Etending the Ideas C M 3 M O M 1 B 63. Vector Equation of a Line Let L be the line through the two points A and B. Prove that C, is on the line L if and onl if OC toa 1 t OB,where t is a real number and O is the origin. 64. Connecting Vectors and Geometr Prove that the lines which join one verte of a parallelogram to the midpoints of the opposite sides trisect the diagonal. 55. True. u and u have the same length but opposite directions. Thus, the length of u is also 1.

14 5144_Demana_Ch06pp /11/06 9:3 PM Page CHAPTER 6 Applications of Trigonometr 6. Dot Product of Vectors What ou ll learn about The Dot Product Angle Between Vectors Projecting One Vector onto Another Work... and wh Vectors are used etensivel in mathematics and science applications such as determining the net effect of several forces acting on an object and computing the work done b a force acting on an object. The Dot Product Vectors can be multiplied in two different was, both of which are derived from their usefulness for solving problems in vector applications. The cross product (or vector product or outer product) results in a vector perpendicular to the plane of the two vectors being multiplied, which takes us into a third dimension and outside the scope of this chapter. The dot product (or scalar product or inner product) results in a scalar. In other words, the dot product of two vectors is not a vector but a real number! It is the important information conveed b that number that makes the dot product so worthwhile, as ou will see. Now that ou have some eperience with vectors and arrows, we hope we won t confuse ou if we occasionall resort to the common convention of using arrows to name the vectors the represent. For eample, we might write u PQ as a shorthand for u is the vector represented b PQ. This greatl simplifies the discussion of concepts like vector projection. Also, we will continue to use both vector notations, a, b and ai bj, so ou will get some practice with each. DEFINITION Dot Product The dot product or inner product of u u 1, u and v v 1, v is u v u 1 v 1 u v. DOT PRODUCT AND STANDARD UNIT VECTORS (u 1 i u j) (v 1 i v j) u 1 v 1 u v OBJECTIVE Students will be able to calculate dot products and projections of vectors. MOTIVATE Ask students to guess the meaning of a projection of one vector onto another. LESSON GUIDE Da 1: The Dot Product; Angle Between Vectors Da : Projecting One Vector Onto Another; Work Dot products have man important properties that we make use of in this section. We prove the first two and leave the rest for the Eercises. Properties of the Dot Product Let u, v, and w be vectors and let c be a scalar. 1. u v v u 4. u v w u v u w. u u u u v w u w v w 3. 0 u 0 5. cu v u cv c u v Proof Let u u 1, u and v v 1, v. Propert 1 u v u 1 v 1 u v Definition of u v v 1 u 1 v u v u Commutative propert of real numbers Definition of u v Propert u u u 1 u u 1 u u Definition of u u Definition of u

15 5144_Demana_Ch06pp /11/06 9:3 PM Page 515 SECTION 6. Dot Product of Vectors 515 TEACHING NOTE If ou do not plan to cover Chapter 8 and ou want to cover vectors in threedimensional space, ou can cover the relevant parts of Section 8.6 after ou finish Section 6.. EXAMPLE 1 Finding Dot Products Find each dot product. (a) 3, 4 5, (b) 1, 4, 3 (c) i j 3i 5j SOLUTION (a) 3, 4 5, (b) 1, 4, (c) i j 3i 5j Now tr Eercise 3. DOT PRODUCTS ON CALCULATORS It is reall a waste of time to compute a simple dot product of two-dimensional vectors using a calculator, but it can be done. Some calculators do vector operations outright, and others can do vector operations via matrices. If ou have learned about matri multiplication alread, ou will know wh the matri v [ 1 ] v product [u 1, u ] ields the dot product u 1, u v 1, v as a 1-b-1 matri. (The same trick works with vectors of higher dimensions.) This book will cover matri multiplication in Chapter 7. Propert of the dot product gives us another wa to find the length of a vector, as illustrated in Eample. EXAMPLE Using Dot Product to Find Length Use the dot product to find the length of the vector u 4, 3. SOLUTION It follows from Propert that u u u. Thus, 4, 3 4, 3 4, Now tr Eercise 9. Angle Between Vectors Let u and v be two nonzero vectors in standard position as shown in Figure The angle between u and v is the angle,0 or The angle between an two nonzero vectors is the corresponding angle between their respective standard position representatives. We can use the dot product to find the angle between nonzero vectors, as we prove in the net theorem. v v u u FIGURE 6.16 The angle between nonzero vectors u and v. THEOREM Angle Between Two Vectors If is the angle between the nonzero vectors u and v, then u v cos u v and cos ( 1 u u ) v v

16 5144_Demana_Ch06pp /11/06 9:3 PM Page CHAPTER 6 Applications of Trigonometr Proof We appl the Law of Cosines to the triangle determined b u, v, and v u in Figure 6.16, and use the properties of the dot product. v u u v u v cos v u v u u v u v cos v v v u u v u u u v u v cos v u v u u v u v cos u v u v cos u v cos u v cos ( 1 u u ) v v v =, 5 u =, 3 θ (a) u =, 1 θ EXAMPLE 3 Finding the Angle Between Vectors Find the angle between the vectors u and v. (a) u, 3, v, 5 (b) u, 1, v 1, 3 SOLUTION (a) See Figure 6.17a. Using the Angle Between Two Vectors Theorem, we have u v, 3, 5 11 cos. u v, 3, So, cos (b) See Figure 6.17b. Again using the Angle Between Two Vectors Theorem, we have u v, 1 1, cos. u v, 1 1, So, cos Now tr Eercise 13. v = 1, 3 (b) FIGURE 6.17 The vectors in (a) Eample 3a and (b) Eample 3b. If vectors u and v are perpendicular, that is, if the angle between them is 90, then u v u v cos 90 0 because cos DEFINITION Orthogonal Vectors The vectors u and v are orthogonal if and onl if u v 0.

17 5144_Demana_Ch06pp /11/06 9:3 PM Page 517 SECTION 6. Dot Product of Vectors 517 EXPLORATION EXTENSIONS Now suppose B(, ) is a point that is not on the given circle. If a,what can ou sa about u v? If a, what can ou sa about u v? The terms perpendicular and orthogonal almost mean the same thing. The zero vector has no direction angle, so technicall speaking, the zero vector is not perpendicular to an vector. However, the zero vector is orthogonal to ever vector. Ecept for this special case, orthogonal and perpendicular are the same. EXAMPLE 4 Proving Vectors are Orthogonal Prove that the vectors u, 3 and v 6, 4 are orthogonal. SOLUTION We must prove that their dot product is zero. u v, 3 6, The two vectors are orthogonal. Now tr Eercise 3. A( a, 0) θ B(, ) C(a, 0) FIGURE 6.18 The angle ABC inscribed in the upper half of the circle a. (Eploration 1) EXPLORATION 1 Angles Inscribed in Semicircles Figure 6.18 shows ABC inscribed in the upper half of the circle a. 1. For a, find the component form of the vectors u BA and v BC.,,,. Find u v. What can ou conclude about the angle between these two vectors? Repeat parts 1 and for arbitrar a. Answers will var P u Q R FIGURE 6.19 The vectors u PQ, v PS, and the vector projection of u onto v, PR proj v u. FOLLOW-UP Ask students to name a pair of vectors that are orthogonal but not perpendicular. ASSIGNMENT GUIDE Da 1: E. 1 1, multiples of 3, 30 4, multiples of 3 Da : E. 7 51, multiples of 3, COOPERATIVE LEARNING Group Activit: E. 58, 59 v S Projecting One Vector onto Another The vector projection of u PQ onto a nonzero vector v PS is the vector PR determined b dropping a perpendicular from Q to the line PS (Figure 6.19). We have resolved u into components PR and RQ u PR RQ with PR and RQ perpendicular. The standard notation for PR,the vector projection of u onto v, is PR proj v u. With this notation, RQ u projv u. We ask ou to establish the following formula in the Eercises (see Eercise 58). Projection of u onto v If u and v are nonzero vectors, the projection of u onto v is proj v u u v v v.

18 5144_Demana_Ch06pp /11/06 9:3 PM Page CHAPTER 6 Applications of Trigonometr u v = 5, 5 FIGURE 6.0 The vectors u 6,, v 5, 5, u 1 proj v u, and u u u 1. (Eample 5) 45 u = 6, u u proj v u FIGURE 6.1 If we pull on a bo with force u, the effective force in the direction of v is proj v u, the vector projection of u onto v. F 1 FIGURE 6. The sled in Eample 6. v F EXAMPLE 5 Decomposing a Vector into Perpendicular Components Find the vector projection of u 6, onto v 5, 5. Then write u as the sum of two orthogonal vectors, one of which is proj v u. SOLUTION We write u u 1 u where u 1 proj v u and u u u 1 (Figure 6.0). v u 1 proj v u u v v 0 5, 5, 50 u u u 1 6,, 4, 4 Thus, u 1 u, 4, 4 6, u. Now tr Eercise 5. If u is a force, then proj v u represents the effective force in the direction of v (Figure 6.1). We can use vector projections to determine the amount of force required in problem situations like Eample 6. EXAMPLE 6 Finding a Force Juan is sitting on a sled on the side of a hill inclined at 45. The combined weight of Juan and the sled is 140 pounds. What force is required for Rafaela to keep the sled from sliding down the hill? (See Figure 6..) SOLUTION We can represent the force due to gravit as F 140j because gravit acts verticall downward. We can represent the side of the hill with the vector v cos 45 i sin 45 j i j. The force required to keep the sled from sliding down the hill is F 1 proj v F ( F v v ) v F v v because v 1. So, F 1 F v v 140 ( ) v 70 i j. The magnitude of the force that Rafaela must eert to keep the sled from sliding down the hill is pounds. Now tr Eercise 45. Work If F is a constant force whose direction is the same as the direction of AB, then the work W done b F in moving an object from A to B is W F AB.

19 5144_Demana_Ch06pp /11/06 9:3 PM Page 519 SECTION 6. Dot Product of Vectors 519 NOTES ON EXERCISES E can be completed b using dot products or b using common sense. Encourage students to tr both methods. E involve work done b a force that is not parallel to the direction of motion. E provide practice with standardized tests. ONGOING ASSESSMENT Self-Assessment: E. 3, 9, 13, 3, 5, 45, 53 Embedded Assessment: E. 67, 68 UNITS FOR WORK Work is usuall measured in footpounds or Newton-meters. One Newton-meter is commonl referred to as one Joule. If F is a constant force in an direction, then the work W done b F in moving an object from A to B is W F AB F AB cos where is the angle between F and AB. Ecept for the sign, the work is the magnitude of the effective force in the direction of AB times AB. EXAMPLE 7 Finding Work Find the work done b a 10 pound force acting in the direction 1, in moving an object 3 feet from 0, 0 to 3, 0. SOLUTION The force F has magnitude 10 and acts in the direction 1,, so 1, 10 F 10 1,. 1, 5 The direction of motion is from A 0, 0 to B 3, 0, so AB 3, 0. Thus, the work done b the force is F AB , 3, foot-pounds. 5 5 Now tr Eercise 53. QUICK REVIEW 6. (For help, go to Section 6.1.) In Eercises 1 4, find u. 1. u, u 3i 4j 5 3. u cos 35 i sin 35 j 1 4. u cos 75 i sin 75 j In Eercises 5 8, the points A and B lie on the circle 4. Find the component form of the vector AB. 5. A, 0, B 1, 3 6. A, 0, B 1, 3 3, 3 1, 3 7. A, 0, B 1, 3 1, 3 8. A, 0, B 1, 3 3, 3 In Eercises 9 and 10, find a vector u with the given magnitude in the direction of v. 9. u, v, u 3, v 4i 3j 4 6, , SECTION 6. EXERCISES In Eercises 1 8, find the dot product of u and v. 1. u 5, 3, v 1, 4 7. u 5,, v 8, u 4, 5, v 3, u, 7, v 5, u 4i 9j, v 3i j u i 4j, v 8i 7j u 7i, v i 5j u 4i 11j, v 3j 33 In Eercises 9 1, use the dot product to find u. 9. u 5, u 8, u 4i 4 1. u 3j 3

20 5144_Demana_Ch06pp /11/06 9:3 PM Page CHAPTER 6 Applications of Trigonometr In Eercises 13, find the angle between the vectors. 13. u 4, 3, v 1, u,, v 3, u, 3, v 3, u 5,, v 6, u 3i 3j, v i 3 j u i, v 5j u ( cos 4 ) ( i sin 4 ) ( j, v cos 3 ) ( i sin 3 ) j u ( cos 3 ) ( i sin 3 ) ( j, v 3 cos 5 6 ) ( i 3 sin 5 6 ) j (8, 5) 5 ( 3, 4) 4 3 v 1 u ( 3, 8) ( 1, 9) 9 10 In Eercises 3 4, prove that the vectors u and v are orthogonal. 3. u, 3, v 3, 1 4. u 4, 1, v 1, 4 In Eercises 5 8, find the vector projection of u onto v. Then write u as a sum of two orthogonal vectors, one of which is proj v u. 5. u 8, 3, v 6, 6. u 3, 7, v, 6 7. u 8, 5, v 9, 8. u, 8, v 9, 3 In Eercises 9 and 30, find the interior angles of the triangle with given vertices. 9. 4, 5, 1, 10, 3, , 1, 1, 6, 5, 1 In Eercises 31 and 3, find u v satisfing the given conditions where is the angle between u and v , u 3, v 8 3., u 1, v 40 3 In Eercises 33 38, determine whether the vectors u and v are parallel, orthogonal, or neither. 33. u 5, 3, v 1 0, 3 4 Parallel 34. u, 5, v 1 0, Neither 35. u 15, 1, v 4, 5 Neither 36. u 5, 6, v 1, 10 Orthogonal 37. u 3, 4, v 0, 15 Orthogonal 38. u, 7, v 4, 14 Parallel In Eercises 39 4, find (a) the -intercept A and -intercept B of the line. (b) the coordinates of the point P so that AP is perpendicular to the line and AP 1. (There are two answers.) In Eercises 43 and 44, find the vector(s) v satisfing the given conditions. 43. u, 3, u v 10, v u, 5, u v 11, v Sliding Down a Hill Ojemba is sitting on a sled on the side of a hill inclined at 60. The combined weight of Ojemba and the sled is 160 pounds. What is the magnitude of the force required for Mandisa to keep the sled from sliding down the hill? 46. Revisiting Eample 6 Suppose Juan and Rafaela switch positions. The combined weight of Rafaela and the sled is 15 pounds. What is the magnitude of the force required for Juan to keep the sled from sliding down the hill? pounds 47. Braking Force A 000 pound car is parked on a street that makes an angle of 1 with the horizontal (see figure). 1 (a) Find the magnitude of the force required to keep the car from rolling down the hill pounds (b) Find the force perpendicular to the street pounds

21 5144_Demana_Ch06pp /11/06 9:3 PM Page 51 SECTION 6. Dot Product of Vectors Effective Force A 60 pound force F that makes an angle of 5 with an inclined plane is pulling a bo up the plane.the inclined plane makes an 18 angle with the horizontal (see 5 figure). What is the magnitude of the effective force pulling the bo up the plane? pounds Work Find the work done lifting a 600 pound car 5.5 feet. 14,300 foot-pounds 50. Work Find the work done lifting a 100 pound bag of potatoes 3 feet. 300 foot-pounds 51. Work Find the work done b a force F of 1 pounds acting in the direction 1, in moving an object 4 feet from 0, 0 to 4, foot-pounds 5. Work Find the work done b a force F of 4 pounds acting in the direction 4, 5 in moving an object 5 feet from 0, 0 to 5, foot-pounds 53. Work Find the work done b a force F of 30 pounds acting in the direction, in moving an object 3 feet from 0, 0 to a point in the first quadrant along the line Work Find the work done b a force F of 50 pounds acting in the direction, 3 in moving an object 5 feet from 0, 0 to a point in the first quadrant along the line. 55. Work The angle between a 00 pound force F and AB i 3j is 30. Find the work done b F in moving an object from A to B foot-pounds 56. Work The angle between a 75 pound force F and AB is 60, where A 1, 1 and B 4, 3. Find the work done b F in moving an object from A to B foot-pounds 57. Properties of the Dot Product Let u, v, and w be vectors and let c be a scalar. Use the component form of vectors to prove the following properties. (a) 0 u 0 (b) u v w u v u w (c) u v w u w v w (d) cu v u cv c u v 58. Group Activit Projection of a Vector Let u and v be nonzero vectors. Prove that (a) proj v u ( u v v ) v (b) u proj v u proj v u Group Activit Connecting Geometr and Vectors Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. 60. If u is an vector, prove that we can write u as u u i i u j j. Standardized Test Questions 61. True or False If u v 0, then u and v are perpendicular. Justif our answer. 6. True or False If u is a unit vector, then u u 1. Justif our answer. True. u u u (1) 1 In Eercises 63 66, ou ma use a graphing calculator to solve the problem. 63. Multiple Choice Let u 1, 1 and v 1, 1. Which of the following is the angle between u and v? D (A) 0 (B) 45 (C) 60 (D) 90 (E) Multiple Choice Let u 4, 5 and v, 3. Which of the following is equal to u v? C (A) 3 (B) 7 (C) 7 (D) 3 (E) Multiple Choice Let u 3, 3 and v, 0. Which of the following is equal to proj v u? A (A) 3, 0 (B) 3, 0 (C) 3, 0 (D) 3, 3 (E) 3, Multiple Choice Which of the following vectors describes a 5 lb force acting in the direction of u 1, 1? B 5 (A) 5 1, 1 (B) 1, 1 (C) 5 1, 1 5 (D) 1, 1 Eplorations (E) 5 1, Distance from a Point to a Line Consider the line L with equation 5 10 and the point P 3, 7. (a) Verif that A 0, and B 5, 0 are the - and -intercepts of L. (b) Find w 1 proj AB AP and w AP proj AB AP. (c) Writing to Learn Eplain wh w is the distance from P to L. What is this distance? (d) Find a formula for the distance of an point P 0, 0 to L. (e) Find a formula for the distance of an point P 0, 0 to the line a b c. Etending the Ideas 68. Writing to Learn Let w cos t u sin t v where u and v are not parallel. (a) Can the vector w be parallel to the vector u? Eplain. (b) Can the vector w be parallel to the vector v? Eplain. (c) Can the vector w be parallel to the vector u v? Eplain. 69. If the vectors u and v are not parallel, prove that au bv cu dv a c, b d.

22 5144_Demana_Ch06pp /11/06 9:3 PM Page 5 5 CHAPTER 6 Applications of Trigonometr 6.3 Parametric Equations and Motion What ou ll learn about Parametric Equations Parametric Curves Eliminating the Parameter Lines and Line Segments Simulating Motion with a Grapher... and wh These topics can be used to model the path of an object such as a baseball or a golf ball. t = 0, = 40 t = 4, = 164 t = 1, = 404 t =, = 356 t = 3, = 76 Parametric Equations Imagine that a rock is dropped from a 40-ft tower. The rock s height in feet above the ground t seconds later (ignoring air resistance) is modeled b 16t 40 as we saw in Section.1. Figure 6.3 shows a coordinate sstem imposed on the scene so that the line of the rock s fall is on the vertical line.5. The rock s original position and its position after each of the first 5 seconds are the points.5, 40,.5, 404,.5, 356,.5, 76,.5, 164,.5, 0, which are described b the pair of equations.5, 16t 40, when t 0, 1,, 3, 4, 5. These two equations are an eample of parametric equations with parameter t. As is often the case, the parameter t represents time. Parametric Curves In this section we stud the graphs of parametric equations and investigate motion of objects that can be modeled with parametric equations. t = 5, = 0 [0, 5] b [ 10, 500] FIGURE 6.3 The position of the rock at 0, 1,, 3, 4, and 5 seconds. OBJECTIVE Students will be able to define parametric equations, graph curves parametricall, and solve application problems using parametric equations. MOTIVATE Have students use a grapher to graph the parametric equations t and t for 5 t 5. Have them write the equation for this graph in the form f(). ( ) LESSON GUIDE Da 1: Parametric Equations; Parametric Curves; Eliminating the Parameter; Lines and Line Segments Da : Simulating Motion with a Grapher DEFINITION Parametric Curve, Parametric Equations The graph of the ordered pairs, where f t, g t are functions defined on an interval I of t-values is a parametric curve. The equations are parametric equations for the curve, the variable t is a parameter, and I is the parameter interval. When we give parametric equations and a parameter interval for a curve, we have parametrized the curve. A parametrization of a curve consists of the parametric equations and the interval of t-values. Sometimes parametric equations are used b companies in their design plans. It is then easier for the compan to make larger and smaller objects efficientl b just changing the parameter t. Graphs of parametric equations can be obtained using parametric mode on a grapher. EXAMPLE 1 Graphing Parametric Equations For the given parameter interval, graph the parametric equations t, 3t. (a) 3 t 1 (b) t 3 (c) 3 t 3 continued

23 5144_Demana_Ch06pp /11/06 9:3 PM Page 53 SECTION 6.3 Parametric Equations and Motion 53 NOTES ON EXAMPLES Eample 1 is important because it shows how a parametric graph is affected b the chosen range of t-values. SOLUTION In each case, set Tmin equal to the left endpoint of the interval and Tma equal to the right endpoint of the interval. Figure 6.4 shows a graph of the parametric equations for each parameter interval. The corresponding relations are different because the parameter intervals are different. Now tr Eercise 7. [ 10, 10] b [ 10, 10] (a) [ 10, 10] b [ 10, 10] (b) [ 10, 10] b [ 10, 10] (c) FIGURE 6.4 Three different relations defined parametricall. (Eample 1) TEACHING NOTE If students are not familiar with parametric graphing, it might be helpful to show them the graph of the linear function f() 3 and compare it to one defined parametricall as t and 3t, using a trace ke to show how t,, and are related. Eliminating the Parameter When a curve is defined parametricall it is sometimes possible to eliminate the parameter and obtain a rectangular equation in and that represents the curve. This often helps us identif the graph of the parametric curve as illustrated in Eample. EXAMPLE Eliminating the Parameter Eliminate the parameter and identif the graph of the parametric curve 1 t, t, t. SOLUTION We solve the first equation for t: 1 t t 1 t 1 1 Then we substitute this epression for t into the second equation: t 1 1 [ 10, 5] b [ 5, 5] FIGURE 6.5 The graph of (Eample ) The graph of the equation is a line with slope 0.5 and -intercept 1.5 Figure 6.5. Now tr Eercise 11.

24 5144_Demana_Ch06pp /11/06 9:33 PM Page CHAPTER 6 Applications of Trigonometr EXPLORATION EXTENSIONS Determine the smallest possible range of t-values that produces the graph shown in Figure 6.5, using the given parametric equations. ALERT Man students will confuse range values of t with range values on the function grapher. Point out that while the scale factor does not affect the wa a graph is drawn, the Tstep does affect the wa the graph is displaed. EXPLORATION 1 Graphing the Curve of Eample Parametricall 1. Use the parametric mode of our grapher to reproduce the graph in Figure 6.5. Use for Tmin and 5.5 for Tma.. Prove that the point 17, 10 is on the graph of Find the corresponding value of t that produces this point. t 8 3. Repeat part for the point 3, 10. t 1 4. Assume that a, b is on the graph of Find the corresponding value of t that produces this point. t 1 a b 5. How do ou have to choose Tmin and Tma so that the graph in Figure 6.5 fills the window? Tmin and Tma 5.5 If we do not specif a parameter interval for the parametric equations f t, g t, it is understood that the parameter t can take on all values which produce real numbers for and. We use this agreement in Eample 3. PARABOLAS The inverse of a parabola that opens up or down is a parabola that opens left or right. We will investigate these curves in more detail in Chapter 8. EXAMPLE 3 Eliminating the Parameter Eliminate the parameter and identif the graph of the parametric curve t, 3t. SOLUTION Here t can be an real number. We solve the second equation for t obtaining t 3 and substitute this value for into the first equation. t ( 3 ) 9 9 Figure 6.4c shows what the graph of these parametric equations looks like. In Chapter 8 we will call this a parabola that opens to the right. Interchanging and we can identif this graph as the inverse of the graph of the parabola 9. Now tr Eercise 15. [ 4.7, 4.7] b [ 3.1, 3.1] FIGURE 6.6 The graph of the circle of Eample 4. EXAMPLE 4 Eliminating the Parameter Eliminate the parameter and identif the graph of the parametric curve cos t, sin t, 0 t. SOLUTION The graph of the parametric equations in the square viewing window of Figure 6.6 suggests that the graph is a circle of radius centered at the origin. We confirm this result algebraicall. continued