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prising because, as we see from Figure 9, the side opposite the angle of is also the side 3 π 3 s and 2s. Similarly, cos  _  3 π π π  and cos  , so sin  _ _ _  are exactly the same ratio of the same two sides, √ adjacent to 3 6 6 π and sin  _  are also the same ratio using the same two sides, s and 2s. 6 π _ The interrelationship between the sines and cosines of 6 triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of π _ the other. Since the three angles of a triangle add to π, and the right angle is , the remaining two angles must also 2 π π _ _ add up to —in other words, any . That means that a right triangle can be formed with any two angles that add to 2 2 also holds for the two acute angles in any right π _ and 3 two complementary angles. So we may state a cofunction identity: If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure 10. — β α sin α = cos β sin β = cos α Figure 10 Cofunction identity of sine and cosine of complementary angles SECTION 5.4 right triangle trigonometry 491 Using this identity, we can state without calculating, for instance, that the sine of equals the cosine of equals the cosine of . We can also state that if, for a certain angle t, cos t = π _ 12 π _ 12 5π _ , 12 , then 5 _ 13 and that the sine of π sin  _ − t  = 2 5 _ 13 as well. 5π _ 12 cofunction identities The cofunction identities in radians are listed in Table 1. π __ − t  cos t = sin  2 π __ − t  tan t = cot  2 π __ − t  sec t = csc  2 π __ − t  sin t = cos  2 π __ − t  cot t = tan  2 π __ − t  csc t = sec  2 Table 1 How To… Given the sine and cosine of an angle, find the sine or cosine of its complement. 1. To find the sine of the complementary angle, find the cosine of the original angle. 2. To find the cosine of the complementary angle, find the sine of the original angle. Example 4 Using Cofunction Identities If sin t = 5 _ 12 π _ − t  . , find cos  2 Solution According to the cofunction identities for sine and cosine, π __ − t  . sin t = cos  2 π __ − t  = cos  2 5 __ . 12 So Try It #4 π π If csc   = 2, find sec  _ _  . 3 6 Using Trigonometric Functions In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides. How To… Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides. 1. For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator. 2. Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides. 3. Using the value of the trigonometric function and the known side length, solve for the missing side length. 49 2 CHAPTER 5 trigonometric Functions Example 5 Finding Missing Side Lengths Using Trigonometric Ratios Find the unknown sides of the triangle in Figure 11. 30° a c 7 Figure 11 Solution We know the angle and the opposite side, so we can use the tangent to find the adjacent side. We rearrange to solve for a. We can use the sine to find the hypotenuse. Again, we rearrange to solve for c. 7 __ tan(30°) = a a = 7 ______ tan(30°) ≈ 12.1 7 __ sin(30°) = c c = 7 ______ sin(30°) ≈ 14 Try It #5 π _ A right triangle has one angle of and a hypotenuse of 20. Find the unknown sides and angle of the triangle. 3 Using Right Triangle Trigonometry to Solve Applied Problems Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. See Figure 12. SECTION 5.4 right triangle trigonometry 493 Angle of depression Angle of elevation Figure 12 How To… Given a tall object, measure its height indirectly. 1. Make a sketch of the problem situation to keep track of known and unknown information. 2. Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible. 3. At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal. 4. Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight. 5. Solve the equation for the unknown height. Example 6 Measuring a Distance Indirectly To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of 57° between a line of sight to the top of the tree and the ground, as shown in Figure 13. Find the height of the tree. 57° 30 feet Figure 13 Solution We know that the angle of elevation is 57° and the adjacent side is 30 ft long. The opposite side is the unknown height. The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of 57°, letting h be the unknown height. tan θ = opposite _ adjacent tan(57°) = h __ 30 Solve for h. h = 30tan(57°) Multiply. h ≈ 46.2 Use a calculator. The tree is approximately 46 feet tall. 49 4 CHAPTER 5 trigonometric Functions Try It #6 How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of with the ground? Round to the nearest foot. 5π _ 12 Access these online resources for additional instruction and practice with right triangle trigonometry. • Finding Trig Functions on Calculator (http://openstaxcollege.org/l/findtrigcal) • Finding Trig Functions Using a Right Triangle (http://openstaxcollege.org/l/trigrttri) • Relate Trig Functions to Sides of a Right Triangle (http://openstaxcollege.org/l/reltrigtri) • Determine Six Trig Functions from a Triangle (http://openstaxcollege.org/l/sixtrigfunc) • Determine length of Right Triangle Side (http://openstaxcollege.org/l/rttriside) SECTION 5.4 section exercises 495 5.4 SeCTIOn exeRCISeS VeRBAl 1. For the given right triangle, label the adjacent side, opposite side, and hypotenuse for the indicated angle. 2. When a right triangle with a hypotenuse of 1 is placed in the unit circle, which sides of the triangle correspond to the x- and y-coordinates? 3. The tangent of an angle compares which sides of the 4. What is the relationship between the two acute right triangle? 5. Explain the cofunction identity. AlGeBRAIC angles in a right triangle? For the following exercises, use cofunctions of complementary angles. 6. cos(34°) = sin(_____°) π __  = sin(______) 7. cos  3 8. csc(21°) = sec(_____°) π __  = cot(_____) 9. tan  4 For the following exercises, find the lengths of the missing sides if side a is opposite angle A, side b is opposite angle B, and side c is the hypotenuse. 10. cos B = 4 __ , a = 10 5 11. sin B = 1 __ , a = 20 2 13. tan A = 100, b = 100 12. tan A = , b = 6 5 __ 12 14. sin √ GRAPHICAl 15. a = 5, ∡ A = 60° 16. c = 12, ∡ A = 45° For the following exercises, use Figure 14 to evaluate each trigonometric function of angle A. A 4 10 Figure 14 17. sin A 19. tan A 21. sec A 18. cos A 20. csc A 22. cot A For the following exercises, use Figure 15 to evaluate each trigonometric function of angle A. 23. sin A 25. tan A 27. sec A 24. cos A 26. csc A 28. cot A 10 A 8 Figure 15 49 6 CHAPTER 5 trigonometric Functions For the following exercises, solve for the unknown sides of the given triangle. 29. B 7 30. 31. A 10 A c a 60° c b 30° 45° TeCHnOlOGY For the following exercises, use a calculator to find the length of each side to four decimal places. 32. A 35. B a 33. B 7 a 62° 10 c 12 b 10° c b 36. 35° 34. B a 10 b 65° A b c 16.5 81° 37. b = 15, ∡ B = 15° 38. c = 200, ∡ B = 5° 39. c = 50, ∡ B = 21° 40. a = 30, ∡ A = 27° 41. b = 3.5, ∡ A = 78° exTenSIOnS 42. Find x. 43. Find x. 82 63° 39° x 85 36° 50° x SECTION 5.4 section exercises 497 44. Find x. 45. Find x. 115 56° 35° x 119 70° 26° x 46. A radio tower is located 400 feet from a building. 47. A radio tower is located 325 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is 36°, and that the angle of depression to the bottom of the tower is 23°. How tall is the tower? From a window in the building, a person determines that the angle of elevation to the top of the tower is 43°, and that the angle of depression to the bottom of the tower is 31°. How tall is the tower? 48. A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 15°, and that the angle of depression to the bottom of the tower is 2°. How far is the person from the monument? 49. A 400-foot tall monument is located in
the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 18°, and that the angle of depression to the bottom of the monument is 3°. How far is the person from the monument? 50. There is an antenna on the top of a building. From a location 300 feet from the base of the building, the angle of elevation to the top of the building is measured to be 40°. From the same location, the angle of elevation to the top of the antenna is measured to be 43°. Find the height of the antenna. 51. There is lightning rod on the top of a building. From a location 500 feet from the base of the buil ding, the angle of elevation to the top of the building is measured to be 36°. From the same location, the angle of elevation to the top of the lightning rod is measured to be 38°. Find the height of the lightning rod. ReAl-WORlD APPlICATIOnS 52. A 33-ft ladder leans against a building so that the angle between the ground and the ladder is 80°. How high does the ladder reach up the side of the building? 53. A 23-ft ladder leans against a building so that the angle between the ground and the ladder is 80°. How high does the ladder reach up the side of the building? 54. The angle of elevation to the top of a building in 55. The angle of elevation to the top of a building in New York is found to be 9 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building. Seattle is found to be 2 degrees from the ground at a distance of 2 miles from the base of the building. Using this information, find the height of the building. 56. Assuming that a 370-foot tall giant redwood grows vertically, if I walk a certain distance from the tree and measure the angle of elevation to the top of the tree to be 60°, how far from the base of the tree am I? 49 8 CHAPTER 5 trigonometric Functions CHAPTeR 5 ReVIeW Key Terms adjacent side in a right triangle, the side between a given angle and the right angle angle the union of two rays having a common endpoint angle of depression the angle between the horizontal and the line from the object to the observer’s eye, assuming the object is positioned lower than the observer angle of elevation the angle between the horizontal and the line from the object to the observer’s eye, assuming the object is positioned higher than the observer angular speed the angle through which a rotating object travels in a unit of time arc length the length of the curve formed by an arc area of a sector area of a portion of a circle bordered by two radii and the intercepted arc; the fraction multiplied by the θ _ 2π area of the entire circle cosecant the reciprocal of the sine function: on the unit circle, csc t = 1 cosine function the x-value of the point on a unit circle corresponding to a given angle cotangent the reciprocal of the tangent function: on the unit circle, cot t = x coterminal angles description of positive and negative angles in standard position sharing the same terminal side degree a unit of measure describing the size of an angle as one-360th of a full revolution of a circle hypotenuse the side of a right triangle opposite the right angle identities statements that are true for all values of the input on which they are defined initial side the side of an angle from which rotation begins linear speed the distance along a straight path a rotating object travels in a unit of time; determined by the arc length measure of an angle the amount of rotation from the initial side to the terminal side negative angle description of an angle measured clockwise from the positive x-axis opposite side in a right triangle, the side most distant from a given angle period the smallest interval P of a repeating function f such that f (x + P) = f (x) positive angle description of an angle measured counterclockwise from the positive x-axis Pythagorean Identity a corollary of the Pythagorean Theorem stating that the square of the cosine of a given angle plus the square of the sine of that angle equals 1 quadrantal angle an angle whose terminal side lies on an axis radian the measure of a central angle of a circle that intercepts an arc equal in length to the radius of that circle radian measure the ratio of the arc length formed by an angle divided by the radius of the circle ray one point on a line and all points extending in one direction from that point; one side of an angle reference angle the measure of the acute angle formed by the terminal side of the angle and the horizontal axis secant the reciprocal of the cosine function: on the unit circle, sec t = 1 sine function the y-value of the point on a unit circle corresponding to a given angle _ x , x ≠ 0 standard position the position of an angle having the vertex at the origin and the initial side along the positive x-axis y _ x , x ≠ 0 tangent the quotient of the sine and cosine: on the unit circle, tan t = terminal side the side of an angle at which rotation ends unit circle a circle with a center at (0, 0) and radius 1. vertex the common endpoint of two rays that form an angle CHAPTER 5 review 499 Key equations arc length s = rθ angular speed area of a sector 1 __ A = θr2 2 θ __ t s _ v = t linear speed related to angular speed v = rω linear speed ω = cosine sine cos t = x sin t = y Pythagorean Identity cos2 t + sin2 t = 1 tangent function secant function cosecant function cotangent function cofunction identities Key Concepts 5.1 Angles = sec t = csc t = cot t = tan t = sin t _ cos t 1 _ cos t 1 _ sin t cos t 1 _ _ tan t sin t π __ − t  cos t = sin  2 π __ − t  sin t = cos  2 π __ − t  tan t = cot  2 π __ − t  cot t = tan  2 π __ − t  sec t = csc  2 π __ − t  csc t = sec  2 • An angle is formed from the union of two rays, by keeping the initial side fixed and rotating the terminal side. The amount of rotation determines the measure of the angle. • An angle is in standard position if its vertex is at the origin and its initial side lies along the positive x-axis. A positive angle is measured counterclockwise from the initial side and a negative angle is measured clockwise. • To draw an angle in standard position, draw the initial side along the positive x-axis and then place the terminal side according to the fraction of a full rotation the angle represents. See Example 1. • In addition to degrees, the measure of an angle can be described in radians. See Example 2. • To convert between degrees and radians, use the proportion θ _ 180 = θR _ π . See Example 3 and Example 4. • Two angles that have the same terminal side are called coterminal angles. • We can find coterminal angles by adding or subtracting 360° or 2π. See Example 5 and Example 6. • Coterminal angles can be found using radians just as they are for degrees. See Example 7. • The length of a circular arc is a fraction of the circumference of the entire circle. See Example 8. 500 CHAPTER 5 trigonometric Functions • The area of sector is a fraction of the area of the entire circle. See Example 9. • An object moving in a circular path has both linear and angular speed. • The angular speed of an object traveling in a circular path is the measure of the angle through which it turns in a unit of time. See Example 10. • The linear speed of an object traveling along a circular path is the distance it travels in a unit of time. See Example 11. 5.2 Unit Circle: Sine and Cosine Functions • Finding the function values for the sine and cosine begins with drawing a unit circle, which is centered at the origin and has a radius of 1 unit. • Using the unit circle, the sine of an angle t equals the y-value of the endpoint on the unit circle of an arc of length t whereas the cosine of an angle t equals the x-value of the endpoint. See Example 1. • The sine and cosine values are most directly determined when the corresponding point on the unit circle falls on an axis. See Example 2. • When the sine or cosine is known, we can use the Pythagorean Identity to find the other. The Pythagorean Identity is also useful for determining the sines and cosines of special angles. See Example 3. • Calculators and graphing software are helpful for finding sines and cosines if the proper procedure for entering information is known. See Example 4. • The domain of the sine and cosine functions is all real numbers. • The range of both the sine and cosine functions is [−1, 1]. • The sine and cosine of an angle have the same absolute value as the sine and cosine of its reference angle. • The signs of the sine and cosine are determined from the x- and y-values in the quadrant of the original angle. • An angle’s reference angle is the size angle, t, formed by the terminal side of the angle t and the horizontal axis. See Example 5. • Reference angles can be used to find the sine and cosine of the original angle. See Example 6. • Reference angles can also be used to find the coordinates of a point on a circle. See Example 7. 5.3 The Other Trigonometric Functions • The tangent of an angle is the ratio of the y-value to the x-value of the corresponding point on the unit circle. • The secant, cotangent, and cosecant are all reciprocals of other functions. The secant is the reciprocal of the cosine function, the cotangent is the reciprocal of the tangent function, and the cosecant is the reciprocal of the sine function. • The six trigonometric functions can be found from a point on the unit circle. See Example 1. • Trigonometric functions can also be found from an angle. See Example 2. • Trigonometric functions of angles outside the first quadrant can be determined using reference angles. See Example 3. • A function is said to be even if f (−x) = f (x) and odd if f (−x) = −f (x). • Cosine and secant are even; sine, tangent, cosecant, and cotangent are odd. • Even and odd properties can be used to evaluate trigonometric functions. See Example 4. • The Pythagorean Identity makes it possible to find a cosine from a sine or a sine
from a cosine. • Identities can be used to evaluate trigonometric functions. See Example 5 and Example 6. • Fundamental identities such as the Pythagorean Identity can be manipulated algebraically to produce new identities. See Example 7. • The trigonometric functions repeat at regular intervals. • The period P of a repeating function f is the smallest interval such that f (x + P) = f (x) for any value of x. • The values of trigonometric functions of special angles can be found by mathematical analysis. • To evaluate trigonometric functions of other angles, we can use a calculator or computer software. See Example 8. CHAPTER 5 review 501 5.4 Right Triangle Trigonometry • We can define trigonometric functions as ratios of the side lengths of a right triangle. See Example 1. • The same side lengths can be used to evaluate the trigonometric functions of either acute angle in a right triangle. See Example 2. • We can evaluate the trigonometric functions of special angles, knowing the side lengths of the triangles in which they occur. See Example 3. • Any two complementary angles could be the two acute angles of a right triangle. • If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa. See Example 4. • We can use trigonometric functions of an angle to find unknown side lengths. • Select the trigonometric function representing the ratio of the unknown side to the known side. See Example 5. • Right-triangle trigonometry permits the measurement of inaccessible heights and distances. • The unknown height or distance can be found by creating a right triangle in which the unknown height or distance is one of the sides, and another side and angle are known. See Example 6. 502 CHAPTER 5 trigonometric Functions CHAPTeR 5 ReVIeW exeRCISeS AnGleS For the following exercises, convert the angle measures to degrees. 1. π _ 4 2. − 5π ___ 3 For the following exercises, convert the angle measures to radians. 3. −210° 5. Find the length of an arc in a circle of radius 7 meters subtended by the central angle of 85°. 4. 180° 6. Find the area of the sector of a circle with diameter 3π _ radians. 5 32 feet and an angle of For the following exercises, find the angle between 0° and 360° that is coterminal with the given angle. 7. 420° 8. −80° For the following exercises, find the angle between 0 and 2π in radians that is coterminal with the given angle. 9. − 20π ____ 11 10. 14π ____ 5 For the following exercises, draw the angle provided in standard position on the Cartesian plane. 11. −210° 5π ___ 4 13. 12. 75° 14. − π _ 3 15. Find the linear speed of a point on the equator of the earth if the earth has a radius of 3,960 miles and the earth rotates on its axis every 24 hours. Express answer in miles per hour. 16. A car wheel with a diameter of 18 inches spins at the rate of 10 revolutions per second. What is the car’s speed in miles per hour? UnIT CIRCle: SIne AnD COSIne FUnCTIOnS π _ 17. Find the exact value of sin . 3 π _ 18. Find the exact value of cos . 4 19. Find the exact value of cos π. 20. State the reference angle for 300°. 21. State the reference angle for 3π _ . 4 22. Compute cosine of 330°. 23. Compute sine of 5π _ . 4 24. State the domain of the sine and 25. State the range of the sine and cosine functions. cosine functions. THe OTHeR TRIGOnOMeTRIC FUnCTIOnS For the following exercises, find the exact value of the given expression. π _ 26. cos 6 π _ 27. tan 4 π _ 28. csc 3 π _ 29. sec 4 For the following exercises, use reference angles to evaluate the given expression. 30. sec 11π _ 3 31. sec 315° 32. If sec(t) = −2.5 , what is the sec(−t)? 33. If tan(t) = −0.6, what is the tan(−t)? CHAPTER 5 review 503 1 _ 34. If tan(t) = , find tan(t − π). 3 35. If cos(t) = , find sin(t + 2π). — 2 √ _ 2 36. Which trigonometric functions are even? 37. Which trigonometric functions are odd? RIGHT TRIAnGle TRIGOnOMeTRY For the following exercises, use side lengths to evaluate. π _ 38. cos 4 π _ 39. cot 3 π _  = sin(_____°) 41. cos  2 42. csc(18°) = sec(_____°) π _ 40. tan 6 For the following exercises, use the given information to find the lengths of the other two sides of the right triangle. 3 __ , a = 6 43. cos B = 5 5 __ , b = 6 44. tan A = 9 For the following exercises, use Figure 1 to evaluate each trigonometric function. A 6 B 11 Figure 1 45. sin A 46. tan B For the following exercises, solve for the unknown sides of the given triangle. 47. b 4√2 45º a 48. b 5 a 30º 49. A 15-ft ladder leans against a building so that the angle between the ground and the ladder is 70°. How high does the ladder reach up the side of the building? 50. The angle of elevation to the top of a building in Baltimore is found to be 4 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building. 504 CHAPTER 5 trigonometric Functions CHAPTeR 5 PRACTICe TeST 1. Convert radians to degrees. 5π _ 6 2. Convert −620° to radians. 3. Find the length of a circular arc with a radius 12 centimeters subtended by the central angle of 30°. 5. Find the angle between 0° and 360° that is coterminal with 375°. 7. Draw the angle 315° in standard position on the Cartesian plane. 9. A carnival has a Ferris wheel with a diameter of 80 feet. The time for the Ferris wheel to make one revolution is 75 seconds. What is the linear speed in feet per second of a point on the Ferris wheel? What is the angular speed in radians per second? 4. Find the area of the sector with radius of 8 feet and an angle of radians. 5π _ 4 6. Find the angle between 0 and 2π in radians that is coterminal with − 4π _ . 7 8. Draw the angle − π _ in standard position on the 6 Cartesian plane. π _ 10. Find the exact value of sin . 6 11. Compute sine of 240°. 12. State the domain of the sine and cosine functions. 13. State the range of the sine and cosine functions. π _ 15. Find the exact value of tan . 3 π _ 14. Find the exact value of cot . 4 16. Use reference angles to evaluate csc 7π _ . 4 17. Use reference angles to evaluate tan 210°. 18. If csc t = 0.68, what is the csc(−t)? 19. If cos t = , find cos(t − 2π). — 3 √ _ 2 20. Which trigonometric functions are even? π _  = sin(_____) 21. Find the missing angle: cos  6 22. Find the missing sides of the triangle ABC : 3 _ , c = 12 sin B = 4 23. Find the missing sides of the triangle. 24. The angle of elevation to the top of a building in Chicago is found to be 9 degrees from the ground at a distance of 2,000 feet from the base of the building. Using this information, find the height of the building. A 9 60º C B Periodic Functions 6 Figure 1 (credit: "Maxxer_", Flickr) CHAPTeR OUTlIne 6.1 Graphs of the Sine and Cosine Functions 6.2 Graphs of the Other Trigonometric Functions 6.3 Inverse Trigonometric Functions Introduction Each day, the sun rises in an easterly direction, approaches some maximum height relative to the celestial equator, and sets in a westerly direction. The celestial equator is an imaginary line that divides the visible universe into two halves in much the same way Earth’s equator is an imaginary line that divides the planet into two halves. The exact path the sun appears to follow depends on the exact location on Earth, but each location observes a predictable pattern over time. The pattern of the sun’s motion throughout the course of a year is a periodic function. Creating a visual representation of a periodic function in the form of a graph can help us analyze the properties of the function. In this chapter, we will investigate graphs of sine, cosine, and other trigonometric functions. 505 506 CHAPTER 6 periodic Functions leARnInG OBjeCTIVeS In this section, you will: • Graph variations of y = sin(x ) and y = cos(x ). • Use phase shifts of sine and cosine curves. 6.1 GRAPHS OF THe SIne AnD COSIne FUnCTIOnS Figure 1 light can be separated into colors because of its wavelike properties. (credit: "wonderferret"/ Flickr) White light, such as the light from the sun, is not actually white at all. Instead, it is a composition of all the colors of the rainbow in the form of waves. The individual colors can be seen only when white light passes through an optical prism that separates the waves according to their wavelengths to form a rainbow. Light waves can be represented graphically by the sine function. In the chapter on Trigonometric Functions, we examined trigonometric functions such as the sine function. In this section, we will interpret and create graphs of sine and cosine functions. Graphing Sine and Cosine Functions Recall that the sine and cosine functions relate real number values to the x- and y-coordinates of a point on the unit circle. So what do they look like on a graph on a coordinate plane? Let’s start with the sine function. We can create a table of values and use them to sketch a graph. Table 1 lists some of the values for the sine function on a unit circle. x sin(x Table 1 π _ 2 1 2π _ 3 3π 5π _ 6 1 _ 2 π 0 Plotting the points from the table and continuing along the x-axis gives the shape of the sine function. See Figure 2. y 2 1 –1 –2 y = sin (x) π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 x 2π Figure 2 The sine function SECTION 6.1 graphs oF the sine and cosine Functions 507 Notice how the sine values are positive between 0 and π, which correspond to the values of the sine function in quadrants I and II on the unit circle, and the sine values are negative between π and 2π, which correspond to the values of the sine function in quadrants III and IV on the unit circle. See Figure 3. y = sin (x1 Now let’s take a similar look at the cosine function. Again, we can create a table of values and use them to sketch a graph. Table 2 lists some of the values for the cosine function on a unit circle. Figure 3 Plotting values of the sine function x cos(x Table 2 2π _ 3 3π _ 4 5π 1 — 3 √ _ 2 As with the sine function, we can plots points to create a graph of the cosine function as in Figure 4. y 1 0 –1 y = cos
(x) 2π 3 3π 4 5π 5π 4 3π 2 7π 4 x 2π Figure 4 The cosine function Because we can evaluate the sine and cosine of any real number, both of these functions are defined for all real numbers. By thinking of the sine and cosine values as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval [−1, 1]. In both graphs, the shape of the graph repeats after 2π, which means the functions are periodic with a period of 2π. A periodic function is a function for which a specific horizontal shift, P, results in a function equal to the original function: f (x + P) = f (x) for all values of x in the domain of f . When this occurs, we call the smallest such horizontal shift with P > 0 the period of the function. Figure 5 shows several periods of the sine and cosine functions. y = sin (x) y 1 1 period y = cos (x) 1 period y 1 –3π –2π –π π 2π x 3π –3π –2π –π π 2π 3π x –1 –1 Figure 5 508 CHAPTER 6 periodic Functions Looking again at the sine and cosine functions on a domain centered at the y-axis helps reveal symmetries. As we can see in Figure 6, the sine function is symmetric about the origin. Recall from The Other Trigonometric Functions that we determined from the unit circle that the sine function is an odd function because sin(−x) = −sin x. Now we can clearly see this property from the graph. y 2 1 –1 –2 y = sin (x) π x 2π –2π –π Figure 7 shows that the cosine function is symmetric about the y-axis. Again, we determined that the cosine function is an even function. Now we can see from the graph that cos(−x) = cos x. Figure 6 Odd symmetry of the sine function y 2 1 –1 –2 –2π –π π x 2π y = cos (x) Figure 7 even symmetry of the cosine function characteristics of sine and cosine functions The sine and cosine functions have several distinct characteristics: • They are periodic functions with a period of 2π. • The domain of each function is (−∞, ∞) and the range is [−1, 1]. • The graph of y = sin x is symmetric about the origin, because it is an odd function. • The graph of y = cos x is symmetric about the y- axis, because it is an even function. Investigating Sinusoidal Functions As we can see, sine and cosine functions have a regular period and range. If we watch ocean waves or ripples on a pond, we will see that they resemble the sine or cosine functions. However, they are not necessarily identical. Some are taller or longer than others. A function that has the same general shape as a sine or cosine function is known as a sinusoidal function. The general forms of sinusoidal functions are y = Asin(Bx − C) + D and y = Acos(Bx − C) + D SECTION 6.1 graphs oF the sine and cosine Functions 509 Determining the Period of Sinusoidal Functions Looking at the forms of sinusoidal functions, we can see that they are transformations of the sine and cosine functions. We can use what we know about transformations to determine the period. 2π _ . If ∣ B ∣ > 1, then the period is less than 2π and the function In the general formula, B is related to the period by P = ∣ B ∣ undergoes a horizontal compression, whereas if ∣ B ∣ < 1, then the period is greater than 2π and the function undergoes a horizontal stretch. For example, f (x) = sin(x), B = 1, so the period is 2π, which we knew. If f (x) = sin(2x), then B = 2, x 1 _ _  , then B = so the period is π and the graph is compressed. If f (x) = sin  , so the period is 4π and the graph is 2 2 stretched. Notice in Figure 8 how the period is indirectly related to ∣ B ∣ . y f (x) = sin (x) 1 –1 f (x) = sin( )x 2 x 2π π 2 π 3π 2 f (x) = sin (2x) Figure 8 period of sinusoidal functions If we let C = 0 and D = 0 in the general form equations of the sine and cosine functions, we obtain the forms The period is 2π _ . ∣ B ∣ y = Asin(Bx) y = Acos(Bx) Identifying the Period of a Sine or Cosine Function Example 1 π _ Determine the period of the function f (x) = sin  x  . 6 Solution Let’s begin by comparing the equation to the general form y = Asin(Bx). π _ In the given equation, B = , so the period will be 6 P = = 2π _ ∣ B ∣ 2π _ π _ 6 6 _ = 2π ⋅ π = 12 Try It #1 x __ Determine the period of the function g(x) =  cos  . 3 Determining Amplitude Returning to the general formula for a sinusoidal function, we have analyzed how the variable B relates to the period. Now let’s turn to the variable A so we can analyze how it is related to the amplitude, or greatest distance from rest. A represents the vertical stretch factor, and its absolute value ∣ A ∣ is the amplitude. The local maxima will be a distance ∣ A ∣ above the vertical midline of the graph, which is the line x = D ; because D = 0 in this case, the midline is the x-axis. The local minima will be the same distance below the midline. If ∣ A ∣ > 1, the function is stretched. For example, the amplitude of f (x) = 4sin x is twice the amplitude of f (x) = 2sin x. If ∣ A ∣ < 1, the function is compressed. Figure 9 compares several sine functions with different amplitudes. 510 CHAPTER 6 periodic Functions – 9π 2 – 5π 2 –11π 2 – 7π 2 – 3π 2 f (x) = 4 sin(x) f (x) = 3 sin(x) f (x) = 2 sin(x) f (x) = 1 sin(x) 11π 2 3π 2 7π 2 5π 2 9π 2 y 4 3 2 1 –1 –2 –3 –4 Figure 9 amplitude of sinusoidal functions If we let C = 0 and D = 0 in the general form equations of the sine and cosine functions, we obtain the forms y = Asin(Bx) and y = Acos(Bx) The amplitude is A, and the vertical height from the midline is ∣ A ∣ . In addition, notice in the example that 1 ∣ maximum − minimum ∣ __ ∣ A ∣ = amplitude = 2 Example 2 Identifying the Amplitude of a Sine or Cosine Function What is the amplitude of the sinusoidal function f (x) = −4sin(x)? Is the function stretched or compressed vertically? Solution Let’s begin by comparing the function to the simplified form y = Asin(Bx). In the given function, A = −4, so the amplitude is ∣ A ∣ = ∣ −4 ∣ = 4. The function is stretched. Analysis The negative value of A results in a reflection across the x-axis of the sine function, as shown in Figure 10. y – 3π 2 – π 2 f (x) = −4 sin x π 2 3π 2 x 4 3 2 –1 –2 –3 –4 Figure 10 Try It #2 1 __ sin(x)? Is the function stretched or compressed vertically ? What is the amplitude of the sinusoidal function f (x) = 2 Analyzing Graphs of Variations of y = sin x and y = cos x Now that we understand how A and B relate to the general form equation for the sine and cosine functions, we will explore the variables C and D. Recall the general form: y = Asin(Bx − C) + D and y = Acos(Bx − C) + D C C __ __   + D   + D and y = Acos  B  x − y = Asin  B  x − B B or SECTION 6.1 graphs oF the sine and cosine Functions 511 C __ B The value for a sinusoidal function is called the phase shift, or the horizontal displacement of the basic sine or cosine function. If C > 0, the graph shifts to the right. If C < 0, the graph shifts to the left. The greater the value of ∣ C ∣ , the more the graph is shifted. Figure 11 shows that the graph of f (x) = sin(x − π) shifts to the right by π units, which π π __ __ is more than we see in the graph of f (x) = sin  x −  , which shifts to the right by units. 4 4 y 1 f (x) = sin(x) f (x) = sin x − π 4 f (x) = sin(x − π) π 2 π 3π 2 2π 5π 2 x 3π While C relates to the horizontal shift, D indicates the vertical shift from the midline in the general formula for a sinusoidal function. See Figure 12. The function y = cos(x) + D has its midline at y = D. Figure 11 y y = A sin(x) + D Midline π 2π 3π y = D x Figure 12 Any value of D other than zero shifts the graph up or down. Figure 13 compares f (x) = sin x with f (x) = sin x + 2, which is shifted 2 units up on a graph. y 3 2 1 –1 f (x) = sin(x) + 2 f (x) = sin(x) π 2 π 3π 2 2π 5π 2 x 3π Figure 13 variations of sine and cosine functions C __ Given an equation in the form f (x) = Asin(Bx − C) + D or f (x) = Acos(Bx − C) + D, B D is the vertical shift. is the phase shift and Example 3 Identifying the Phase Shift of a Function π __  − 2. Determine the direction and magnitude of the phase shift for f (x) = sin  x + 6 Solution Let’s begin by comparing the equation to the general form y = Asin(Bx − C) + D. 512 CHAPTER 6 periodic Functions π __ In the given equation, notice that B = 1 and C = − . So the phase shift is 6 C __ B = − π __ 6 _ 1 π __ = − 6 π __ or units to the left. 6 Analysis We must pay attention to the sign in the equation for the general form of a sinusoidal function. The equation π __ π __   − 2.  − 2 can be rewritten as f (x) = sin  x −  − shows a minus sign before C. Therefore f (x) = sin  x + 6 6 If the value of C is negative, the shift is to the left. Try It #3 π Determine the direction and magnitude of the phase shift for f (x) = 3cos  x − _  . 2 Example 4 Identifying the Vertical Shift of a Function Determine the direction and magnitude of the vertical shift for f (x) = cos(x) − 3. Solution Let’s begin by comparing the equation to the general form y = Acos(Bx − C) + D. In the given equation, D = −3 so the shift is 3 units downward. Try It #4 Determine the direction and magnitude of the vertical shift for f (x) = 3sin(x) + 2. How To… Given a sinusoidal function in the form f (x) = Asin(Bx − C) + D, identify the midline, amplitude, period, and phase shift. 1. Determine the amplitude as ∣ A ∣ . 2π _ 2. Determine the period as P = . ∣ B ∣ C __ . B 3. Determine the phase shift as 4. Determine the midline as y = D. Example 5 Identifying the Variations of a Sinusoidal Function from an Equation Determine the midline, amplitude, period, and phase shift of the function y = 3sin(2x) + 1. Solution Let’s begin by comparing the equation to the general form y = Asin(Bx − C) + D. A = 3, so the amplitude is ∣ A ∣ = 3. Next, B = 2, so the period is P = 2π __ 2 C __ There is no added constant inside the parentheses, so C = 0 and the phase shift is B 2π _ ∣ B ∣ = π. = 0 __ = 0. = 2 Finally, D = 1, so the midline is y = 1. Analysis Figure 14. Inspecting the graph, we can determine that the period is π, the midline is y = 1, and the amplitude is 3. See SECTION 6.1 graphs oF the sin
e and cosine Functions 513 y 4 3 2 –1 –2 Amplitude: \|A\| = 3 y = 3 sin (2x) + 1 Midline: y = 1 π 2 π 3π 2 x 2π Period = π Figure 14 Try It #5 – π x 1 __ __ __ Determine the midline, amplitude, period, and phase shift of the function y =  . cos  3 3 2 Example 6 Identifying the Equation for a Sinusoidal Function from a Graph Determine the formula for the cosine function in Figure 15. y 1 0.5 –2π –π π 2π 3π 4π x Figure 15 Solution To determine the equation, we need to identify each value in the general form of a sinusoidal function. y = Asin(Bx − C) + D y = Acos(Bx − C) + D The graph could represent either a sine or a cosine function that is shifted and/or reflected. When x = 0, the graph has an extreme point, (0, 0). Since the cosine function has an extreme point for x = 0, let us write our equation in terms of a cosine function. Let’s start with the midline. We can see that the graph rises and falls an equal distance above and below y = 0.5. This value, which is the midline, is D in the equation, so D = 0.5. The greatest distance above and below the midline is the amplitude. The maxima are 0.5 units above the midline and the minima are 0.5 units below the midline. So ∣ A ∣ = 0.5. Another way we could have determined the amplitude is 1 __ by recognizing that the difference between the height of local maxima and minima is 1, so ∣ A ∣ = = 0.5. Also, the 2 graph is reflected about the x-axis so that A = −0.5. The graph is not horizontally stretched or compressed, so B = 1; and the graph is not shifted horizontally, so C = 0. Putting this all together, g(x) = −0.5cos(x) + 0.5 Try It #6 Determine the formula for the sine function in Figure 16. y 3 2 1 –2π –π π 2π x Figure 16 514 CHAPTER 6 periodic Functions Example 7 Identifying the Equation for a Sinusoidal Function from a Graph Determine the equation for the sinusoidal function in Figure 17. y 1 –5 –3 –1 1 3 5 7 x –1 –2 –3 –4 –5 Figure 17 Solution With the highest value at 1 and the lowest value at −5, the midline will be halfway between at −2. So D = −2. The distance from the midline to the highest or lowest value gives an amplitude of ∣ A ∣ = 3. The period of the graph is 6, which can be measured from the peak at x = 1 to the next peak at x = 7, or from the distance between the lowest points. Therefore, P = = 6. Using the positive value for B, we find that 2π _ ∣ B ∣ 2π __ = P B = π 2π __ __ = 6 3 π π __ __ x − C  − 2. For the shape and shift, we have more x − C  − 2 or y = 3cos  So far, our equation is either y = 3sin  3 3 than one option. We could write this as any one of the following: • a cosine shifted to the right • a negative cosine shifted to the left • a sine shifted to the left • a negative sine shifted to the right While any of these would be correct, the cosine shifts are easier to work with than the sine shifts in this case because they involve integer values. So our function becomes π π π __ __ __  − 2 or y = −3cos  x + x − y = 3cos  3 3 3 2π __  −2 3 Again, these functions are equivalent, so both yield the same graph. Try It #7 Write a formula for the function graphed in Figure 18. y 8 6 4 2 –9 –7 –5 –3 –1 1 3 5 7 9 11 x Figure 18 SECTION 6.1 graphs oF the sine and cosine Functions 515 Graphing Variations of y = sin x and y = cos x Throughout this section, we have learned about types of variations of sine and cosine functions and used that information to write equations from graphs. Now we can use the same information to create graphs from equations. Instead of focusing on the general form equations we will let C = 0 and D = 0 and work with a simplified form of the equations in the following examples. y = Asin(Bx − C) + D and y = Acos(Bx − C) + D, How To… Given the function y = Asin(Bx), sketch its graph. 1. Identify the amplitude, ∣ A ∣ . 2π ___ 2. Identify the period, P = . ∣ B ∣ 3. Start at the origin, with the function increasing to the right if A is positive or decreasing if A is negative. π ____ 2 ∣ B ∣ 4. At x = there is a local maximum for A > 0 or a minimum for A < 0, with y = A. 5. The curve returns to the x-axis at x = π ___ . ∣ B ∣ 6. There is a local minimum for A > 0 (maximum for A < 0 ) at x = 7. The curve returns again to the x-axis at x = π ____ . 2 ∣ B ∣ 3π ____ 2 ∣ B ∣ with y = −A. Example 8 Graphing a Function and Identifying the Amplitude and Period Sketch a graph of f (x) = −2sin  Solution Let’s begin by comparing the equation to the form y = Asin(Bx). πx __  . 2 Step 1. We can see from the equation that A = −2, so the amplitude is 2. π __ Step 2. The equation shows that B = , so the period is 2 ∣ A ∣ = 2 P = 2π _ π __ 2 2 __ = 2π ⋅ π = 4 Step 3. Because A is negative, the graph descends as we move to the right of the origin. Step 4–7. The x-intercepts are at the beginning of one period, x = 0, the horizontal midpoints are at x = 2 and at the end of one period at x = 4. The quarter points include the minimum at x = 1 and the maximum at x = 3. A local minimum will occur 2 units below the midline, at x = 1, and a local maximum will occur at 2 units above the midline, at x = 3. Figure 19 shows the graph of the function. y πx y = f (x) = −2sin 2 2 1 –2 –1 1 2 3 4 5 6 x –1 –2 Figure 19 516 CHAPTER 6 periodic Functions Try It #8 Sketch a graph of g(x) = −0.8cos(2x). Determine the midline, amplitude, period, and phase shift. How To… Given a sinusoidal function with a phase shift and a vertical shift, sketch its graph. 1. Express the function in the general form y = Asin(Bx − C) + D or y = Acos(Bx − C) + D. 2. Identify the amplitude, ∣ A ∣ . 2π ___ . ∣ B ∣ C __ . B 3. Identify the period, P = 4. Identify the phase shift, C __ 5. Draw the graph of f (x) = Asin(Bx) shifted to the right or left by B and up or down by D. Example 9 Graphing a Transformed Sinusoid π π __ __ x − Sketch a graph of f (x) = 3sin   . 4 4 Solution π π __ __ x − Step 1. The function is already written in general form: f (x) = 3sin   . This graph will have the shape of a sine 4 4 function, starting at the midline and increasing to the right. Step 2. The amplitude is 3. π π  = Step 3. Since ∣ B ∣ =  __ __ , we determine the period as follows. 4 4 P = 2π ___ ∣ B ∣ = 4 _ π = 8 = 2π ⋅ 2π _ π __ 4 The period is 8. π __ Step 4. Since C = , the phase shift is 4 The phase shift is 1 unit. Step 5. Figure 20 shows the graph of the function. = = 1. C __ x) = 3 sin 1 –1 –2 –3 –7 –5 –3 Figure 20 A horizontally compressed, vertically stretched, and horizontally shifted sinusoid SECTION 6.1 graphs oF the sine and cosine Functions 517 Try It #9 π π __ __ Draw a graph of g(x) = −2cos   . Determine the midline, amplitude, period, and phase shift. x + 6 3 Example 10 Identifying the Properties of a Sinusoidal Function π __ x + π  + 3, determine the amplitude, period, phase shift, and horizontal shift. Then graph the Given y = −2cos  2 function. Solution Begin by comparing the equation to the general form and use the steps outlined in Example 9. y = Acos(Bx − C) + D Step 1. The function is already written in general form. Step 2. Since A = −2, the amplitude is ∣ A ∣ = 2. 2π π _ __ Step 3. ∣ B ∣ = , so the period is P = π 2 __ 2 2π ___ ∣ B ∣ = 2 __ = 4. The period is 4. = 2π ⋅ π C __ Step 4. C = −π, so we calculate the phase shift as B = 2 __ = −2. The phase shift is −2. = −π ⋅ π −π _ π __ 2 Step 5. D = 3, so the midline is y = 3, and the vertical shift is up 3. Since A is negative, the graph of the cosine function has been reflected about the x-axis. Figure 21 shows one cycle of the graph of the function. y = −2 cos 2 π x + π + 3 –6 –4 –2 y Amplitude = 2 Midline: y = 3 Period = 4 2 x 5 4 3 2 1 –1 Using Transformations of Sine and Cosine Functions Figure 21 We can use the transformations of sine and cosine functions in numerous applications. As mentioned at the beginning of the chapter, circular motion can be modeled using either the sine or cosine function. Example 11 Finding the Vertical Component of Circular Motion A point rotates around a circle of radius 3 centered at the origin. Sketch a graph of the y-coordinate of the point as a function of the angle of rotation. Solution Recall that, for a point on a circle of radius r, the y-coordinate of the point is y = r sin(x), so in this case, we get the equation y(x) = 3 sin(x). The constant 3 causes a vertical stretch of the y-values of the function by a factor of 3, which we can see in the graph in Figure 22. y(x) = 3 sin x π 2 3π 2 5π 2 7π 2 x y 3 2 1 –1 –2 –3 Figure 22 518 CHAPTER 6 periodic Functions Analysis Notice that the period of the function is still 2π ; as we travel around the circle, we return to the point (3, 0) for x = 2π, 4π, 6π, ... Because the outputs of the graph will now oscillate between –3 and 3, the amplitude of the sine wave is 3. Try It #10 What is the amplitude of the function f (x) = 7cos(x)? Sketch a graph of this function. Example 12 Finding the Vertical Component of Circular Motion A circle with radius 3 ft is mounted with its center 4 ft off the ground. The point closest to the ground is labeled P, as shown in Figure 23. Sketch a graph of the height above the ground of the point P as the circle is rotated; then find a function that gives the height in terms of the angle of rotation. 3 ft P Figure 23 4 ft Solution Sketching the height, we note that it will start 1 ft above the ground, then increase up to 7 ft above the ground, and continue to oscillate 3 ft above and below the center value of 4 ft, as shown in Figure 243 cos x + 4 Midline: y = 4 π 2π 3π 4π x Figure 24 Although we could use a transformation of either the sine or cosine function, we start by looking for characteristics that would make one function easier to use than the other. Let’s use a cosine function because it starts at the highest or lowest value, while a sine function starts at the middle value. A standard cosine starts at the highest value, and this graph starts at the lowest value, so we need to incorporate a vertical reflection. Second, we see that the graph oscillates 3 above and below the center, while a basi
c cosine has an amplitude of 1, so this graph has been vertically stretched by 3, as in the last example. Finally, to move the center of the circle up to a height of 4, the graph has been vertically shifted up by 4. Putting these transformations together, we find that y = −3cos(x) + 4 SECTION 6.1 graphs oF the sine and cosine Functions 519 Try It #11 A weight is attached to a spring that is then hung from a board, as shown in Figure 25. As the spring oscillates up and down, the position y of the weight relative to the board ranges from −1 in. (at time x = 0) to −7 in. (at time x = π) below the board. Assume the position of y is given as a sinusoidal function of x. Sketch a graph of the function, and then find a cosine function that gives the position y in terms of x. y Figure 25 Example 13 Determining a Rider’s Height on a Ferris Wheel The London Eye is a huge Ferris wheel with a diameter of 135 meters (443 feet). It completes one rotation every 30 minutes. Riders board from a platform 2 meters above the ground. Express a rider’s height above ground as a function of time in minutes. Solution With a diameter of 135 m, the wheel has a radius of 67.5 m. The height will oscillate with amplitude 67.5 m above and below the center. Passengers board 2 m above ground level, so the center of the wheel must be located 67.5 + 2 = 69.5 m above ground level. The midline of the oscillation will be at 69.5 m. The wheel takes 30 minutes to complete 1 revolution, so the height will oscillate with a period of 30 minutes. Lastly, because the rider boards at the lowest point, the height will start at the smallest value and increase, following the shape of a vertically reflected cosine curve. • Amplitude: 67.5, so A = 67.5 • Midline: 69.5, so D = 69.5 π __ 15 • Period: 30, so B = 2π ___ 30 = • Shape: −cos(t) An equation for the rider’s height would be y = − 67.5cos  t  + 69.5 π __ 15 where t is in minutes and y is measured in meters. Access these online resources for additional instruction and practice with graphs of sine and cosine functions. • Amplitude and Period of Sine and Cosine (http://openstaxcollege.org/l/ampperiod) • Translations of Sine and Cosine (http://openstaxcollege.org/l/translasincos) • Graphing Sine and Cosine Transformations (http://openstaxcollege.org/l/transformsincos) • Graphing the Sine Function (http://openstaxcollege.org/l/graphsinefunc) 520 CHAPTER 6 periodic Functions 6.1 SeCTIOn exeRCISeS VeRBAl 1. Why are the sine and cosine functions called periodic functions? 3. For the equation Acos(Bx + C) + D, what constants affect the range of the function and how do they affect the range? 5. How can the unit circle be used to construct the graph of f(t) = sin t? GRAPHICAl 2. How does the graph of y = sin x compare with the graph of y = cos x? Explain how you could horizontally translate the graph of y = sin x to obtain y = cos x. 4. How does the range of a translated sine function relate to the equation y = Asin(Bx + C) + D? For the following exercises, graph two full periods of each function and state the amplitude, period, and midline. State the maximum and minimum y-values and their corresponding x-values on one period for x > 0. Round answers to two decimal places if necessary. 6. f (x) = 2sin x 9. f (x) = 4sin x 1 __ x  12. f (x) = 2sin  2 7. f (x) = 2 __ cos x 3 10. f (x) = 2cos x 13. f (x) = 4cos(πx) 8. f (x) = −3sin x 11. f (x) = cos(2x) 6 __ x  14. f (x) = 3cos  5 15. y = 3sin(8(x + 4)) + 5 16. y = 2sin(3x − 21) + 4 17. y = 5sin(5x + 20) − 2 For the following exercises, graph one full period of each function, starting at x = 0. For each function, state the amplitude, period, and midline. State the maximum and minimum y-values and their corresponding x-values on one period for x > 0. State the phase shift and vertical translation, if applicable. Round answers to two decimal places if necessary. 18. f(t) = 2sin  t − 5π ___  6 π __  + 1 19. f(t) = −cos  t + 3 π __   − 3 20. f (t) = 4cos  2  t + 4 1 __ 21. f(t) = −sin  t + 2 5π ___  3 π _ (x − 3)  + 7 22. f (x) = 4sin  2 23. Determine the amplitude, midline, period, and an equation involving the sine function for the graph shown in Figure 26. 24. Determine the amplitude, period, midline, and an equation involving cosine for the graph shown in Figure 27. f(x) –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x – 3π 2 –π – π 2 f(x) 5 4 3 2 1 –1 –2 –3 –4 –5 π 2 π x 3π 2 Figure 26 Figure 27 SECTION 6.1 section exercises 521 25. Determine the amplitude, period, midline, and an equation involving cosine for the graph shown in Figure 28. 26. Determine the amplitude, period, midline, and an equation involving sine for the graph shown in Figure 29. f(x) f(x) –6 –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 6 x –20 –16 –12 –8 5 4 3 2 1 –4–1 –2 –3 –4 –5 84 12 16 20 x Figure 28 Figure 29 27. Determine the amplitude, period, midline, and an equation involving cosine for the graph shown in Figure 30. 28. Determine the amplitude, period, midline, and an equation involving sine for the graph shown in Figure 31. f(x2 + π 2 –2 –4 f(x) –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x Figure 30 Figure 31 29. Determine the amplitude, period, midline, and an equation involving cosine for the graph shown in Figure 32. 30. Determine the amplitude, period, midline, and an equation involving sine for the graph shown in Figure 33. f(x) f(x) –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x Figure 32 Figure 33 522 CHAPTER 6 periodic Functions AlGeBRAIC For the following exercises, let f (x) = sin x. 31. On [0, 2π), solve f (x) = 0. π __  . 33. Evaluate f  2 32. On [0, 2π), solve f (x) = 1 __ . 2 34. On [0, 2π), f (x) = . Find all values of x. — 2 √ ____ 2 35. On [0, 2π), the maximum value(s) of the function 36. On [0, 2π), the minimum value(s) of the function occur(s) at what x-value(s)? occur(s) at what x-value(s)? 37. Show that f(−x) = −f (x). This means that f (x) = sin x is an odd function and possesses symmetry with respect to ________________. For the following exercises, let f (x) = cos x. 38. On [0, 2π), solve the equation f (x) = cos x = 0. 40. On [0, 2π), find the x-intercepts of f (x) = cos x. 42. On [0, 2π), solve the equation f (x) = — 3 √ ____ . 2 39. On [0, 2π), solve f (x) = 1 __ . 2 41. On [0, 2π), find the x-values at which the function has a maximum or minimum value. TeCHnOlOGY 43. Graph h(x) = x + sin x on [0, 2π]. Explain why the 44. Graph h(x) = x + sin x on [−100, 100]. Did the graph appears as it does. graph appear as predicted in the previous exercise? 45. Graph f (x) = x sin x on [0, 2π] and verbalize how the graph varies from the graph of f (x) = sin x. 46. Graph f (x) = x sin x on the window [−10, 10] and explain what the graph shows. 47. Graph f (x) = on the window [−5π, 5π] and sin x ____ x explain what the graph shows. ReAl-WORlD APPlICATIOnS 48. A Ferris wheel is 25 meters in diameter and boarded from a platform that is 1 meter above the ground. The six o’clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. The function h(t) gives a person’s height in meters above the ground t minutes after the wheel begins to turn. a. Find the amplitude, midline, and period of h(t). b. Find a formula for the height function h(t). c. How high off the ground is a person after 5 minutes? SECTION 6.2 graphs oF the other trigonometric Functions 523 leARnInG OBjeCTIVeS In this section, you will: • Analyze the graph of y = tan x. • Graph variations of y = tan x. • Analyze the graphs of y = sec x and y = csc x. • Graph variations of y = sec x and y = csc x. • Analyze the graph of y = cot x. • Graph variations of y = cot x. 6.2 GRAPHS OF THe OTHeR TRIGOnOMeTRIC FUnCTIOnS We know the tangent function can be used to find distances, such as the height of a building, mountain, or flagpole. But what if we want to measure repeated occurrences of distance? Imagine, for example, a police car parked next to a warehouse. The rotating light from the police car would travel across the wall of the warehouse in regular intervals. If the input is time, the output would be the distance the beam of light travels. The beam of light would repeat the distance at regular intervals. The tangent function can be used to approximate this distance. Asymptotes would be needed to illustrate the repeated cycles when the beam runs parallel to the wall because, seemingly, the beam of light could appear to extend forever. The graph of the tangent function would clearly illustrate the repeated intervals. In this section, we will explore the graphs of the tangent and other trigonometric functions. Analyzing the Graph of y = tan x We will begin with the graph of the tangent function, plotting points as we did for the sine and cosine functions. Recall that tan x = sin x ____ cos x The period of the tangent function is π because the graph repeats itself on intervals of kπ where k is a constant. If we to π __ graph the tangent function on − π __ , we can see the behavior of the graph on one complete cycle. If we look at any 2 2 larger interval, we will see that the characteristics of the graph repeat. We can determine whether tangent is an odd or even function by using the definition of tangent. tan(−x) = Definition of tangent. sin(−x) ______ cos(−x) = −sin x ______ cos x = − sin x ____ cos x Sine is an odd function, cosine is even. The quotient of an odd and an even function is odd. = −tan x Definition of tangent. Therefore, tangent is an odd function. We can further analyze the graphical behavior of the tangent function by looking at values for some of the special angles, as listed in Table 1 tan(x) undefined − √ — 3 −1 − π _ 6 — 3 √ ____ 3 − 0 0 Table 1 π _ 6 — 3 √ ____ undefined These points will help us draw our graph, but we need to determine how the graph behaves where it is undefined. π π π _ _ _ ≈ 1.05 and < x < , we can use a table to look for a trend. Because If w
e look more closely at values when 3 2 3 π _ ≈ 1.57, we will evaluate x at radian measures 1.05 < x < 1.57 as shown in Table 2. 2 524 CHAPTER 6 periodic Functions x tan x 1.3 3.6 1.5 14.1 Table 2 1.55 48.1 1.56 92.6 π _ , the outputs of the function get larger and larger. Because y = tan x is an odd function, we see the As x approaches 2 corresponding table of negative values in Table 3. x tan x −1.3 −3.6 −1.5 −1.55 −1.56 −14.1 −48.1 −92.6 Table 3 We can see that, as x approaches − π _ , the outputs get smaller and smaller. Remember that there are some values of x 2 π 3π _ _  = 0. At these values, the tangent function is undefined,  = 0 and cos  for which cos x = 0. For example, cos  2 2 π 3π _ _ so the graph of y = tan x has discontinuities at x = . At these values, the graph of the tangent has vertical and 2 2 π _ asymptotes. Figure 1 represents the graph of y = tan x. The tangent is positive from 0 to and from π to 2 corresponding to quadrants I and III of the unit circle. 3π _ , 2 y y = tan(x) 5 3 1 –1 –3 –5 π x x = π 2 x = 3π 2 –π x = − 3π 2 x = − π 2 Figure 1 Graph of the tangent function Graphing Variations of y = tan x As with the sine and cosine functions, the tangent function can be described by a general equation. y = Atan(Bx) We can identify horizontal and vertical stretches and compressions using values of A and B. The horizontal stretch can typically be determined from the period of the graph. With tangent graphs, it is often necessary to determine a vertical stretch using a point on the graph. Because there are no maximum or minimum values of a tangent function, the term amplitude cannot be interpreted as it is for the sine and cosine functions. Instead, we will use the phrase stretching/compressing factor when referring to the constant A. features of the graph of y = Atan(Bx) • The stretching factor is ∣ A ∣ . • The period is P = π ___ . ∣ B ∣ • The domain is all real numbers x, where x ≠ • The range is (−∞, ∞). • The asymptotes occur at = Atan(Bx) is an odd function. π _ ∣ B ∣ + π ____ such that k is an integer. k, where k is an integer. SECTION 6.2 graphs oF the other trigonometric Functions 525 Graphing One Period of a Stretched or Compressed Tangent Function We can use what we know about the properties of the tangent function to quickly sketch a graph of any stretched and/ or compressed tangent function of the form f (x) = Atan(Bx). We focus on a single period of the function including the origin, because the periodic property enables us to extend the graph to the rest of the function’s domain if we π P P wish. Our limited domain is then the interval  − P _ __ __ __  and the graph has vertical asymptotes at ± where On  − π _ __ __  , the graph will come up from the left asymptote at x = − , cross through the origin, and continue to , 2 2 2 π _ increase as it approaches the right asymptote at x = . To make the function approach the asymptotes at the correct 2 rate, we also need to set the vertical scale by actually evaluating the function for at least one point that the graph will pass through. For example, we can use P P __ __  = Atan  B  = Atan  B f  4 4 π __ 4B  = A π __  = 1. because tan  4 How To… Given the function f (x) = Atan(Bx), graph one period. 1. Identify the stretching factor, ∣ A ∣ . 2. Identify B and determine the period, P = π _ . ∣ B ∣ and x = P 3. Draw vertical asymptotes at . For A > 0, the graph approaches the left asymptote at negative output values and the right asymptote at positive output values (reverse for A < 0). , A  , (0, 0), and  − P P __ __ , −A  , and draw the graph through these points. 5. Plot reference points at  4 4 Example 1 Sketching a Compressed Tangent π __ Sketch a graph of one period of the function y = 0.5tan  x  . 2 Solution First, we identify A and B. π __ y = 0.5 tan  x  2 ↑ y = Atan(Bx) π __ Because A = 0.5 and B = , we can find the stretching/compressing factor and period. The period is 2 = 2, so the π _ π _ 2 asymptotes are at x = ±1. At a quarter period from the origin, we have f (0.5) = 0.5tan  0.5π ____  2 π __ = 0.5tan   4 = 0.5 This means the curve must pass through the points (0.5, 0.5), (0, 0), and (−0.5, −0.5). The only inflection point is at the origin. Figure 2 shows the graph of one period of the function. 526 CHAPTER 6 periodic Functions πx y = 0.5 tan 2 y 4 2 (0.5, 0.5) x (–0.5, –0.5) –2 –4 x = −1 x = 1 Figure 2 Try It #1 π __ Sketch a graph of f (x) = 3tan  x  . 6 Graphing One Period of a Shifted Tangent Function Now that we can graph a tangent function that is stretched or compressed, we will add a vertical and/or horizontal (or phase) shift. In this case, we add C and D to the general form of the tangent function. f (x) = Atan(Bx − C) + D The graph of a transformed tangent function is different from the basic tangent function tan x in several ways: features of the graph of y = Atan(Bx − C) + D • The stretching factor is ∣ A ∣ . • The period is π ___ . ∣ B ∣ C __ • The domain is x ≠ B + π ___ 2 ∣ B ∣ k, where k is an odd integer. • The range is (−∞, ∞). C __ • The vertical asymptotes occur at x = B + π ____ 2 ∣ B ∣ k, where k is an odd integer. • There is no amplitude. • y = Atan(Bx) is an odd function because it is the quotient of odd and even functions (sine and cosine respectively). How To… Given the function y = Atan(Bx − C) + D, sketch the graph of one period. 1. Express the function given in the form y = Atan(Bx − C) + D. 2. Identify the stretching/compressing factor, ∣ A ∣ . π ___ 3. Identify B and determine the period, P = . ∣ B ∣ C __ . B 4. Identify C and determine the phase shift, C __ 5. Draw the graph of y = Atan(Bx) shifted to the right by B + C __ 6. Sketch the vertical asymptotes, which occur at x = B π ____ 2 ∣ B ∣ and up by D. k, where k is an odd integer. 7. Plot any three reference points and draw the graph through these points. SECTION 6.2 graphs oF the other trigonometric Functions 527 Example 2 Graphing One Period of a Shifted Tangent Function Graph one period of the function y = −2tan(πx + π) − 1. Solution Step 1. The function is already written in the form y = Atan(Bx − C) + D. Step 2. A = −2, so the stretching factor is ∣ A ∣ = 2. Step 3. B = π, so the period is P = = π π __ ___ = 1. ∣ B ∣ π −π C ___ __ = −1. Step 4. C = −π, so the phase shift is = π B Step 5-7. The asymptotes are at x = − 3 and x = − 1 __ __ and the three recommended reference points are (−1.25, 1), 2 2 (−1, −1), and (−0.75, −3). The graph is shown in Figure 3. y = −2 tan (πx + π) − 1 (−1.25, 1) –1 (−1, −1) (−0.75, −3) x = −1.5 x = −0.5 Figure 3 y 3 1 –1 –3 –5 x 0.5 Analysis Note that this is a decreasing function because A < 0. Try It #2 How would the graph in Example 2 look different if we made A = 2 instead of −2? How To… Given the graph of a tangent function, identify horizontal and vertical stretches. 1. Find the period P from the spacing between successive vertical asymptotes or x-intercepts. π __ 2. Write f (x) = Atan  x  . P 3. Determine a convenient point (x, f (x)) on the given graph and use it to determine A. Example 3 Identifying the Graph of a Stretched Tangent Find a formula for the function graphed in Figure 4. x = 4 x = 12 x = −12 x = −4 y 6 4 2 –10 –6 –2 2 6 10 x –2 –4 –6 Figure 4 A stretched tangent function 528 CHAPTER 6 periodic Functions Solution The graph has the shape of a tangent function. Step 1. One cycle extends from –4 to 4, so the period is P = 8. Since P = π __ Step 2. The equation must have the form f (x) = Atan  x  . 8 Step 3. To find the vertical stretch A, we can use the point (2, 2). π ___ ∣ B ∣ π π __ __ , we have B = = . 8 P π __  = 1, A = 2. Because tan  4 π __ This function would have a formula f (x) = 2tan  x  . 8 π π __ __ 2 = Atan  · 2  = Atan   4 8 Try It #3 Find a formula for the function in Figure 5. Analyzing the Graphs of y = sec x and y = csc x x = − 3π = 3π π – π 2 –2 –4 –6 Figure 5 The secant was defined by the reciprocal identity sec x = . Notice that the function is undefined when the cosine π __ , is 0, leading to vertical asymptotes at 2 3π __ , etc. Because the cosine is never more than 1 in absolute value, the secant, 2 1 ____ cos x being the reciprocal, will never be less than 1 in absolute value. We can graph y = sec x by observing the graph of the cosine function because these two functions are reciprocals of one another. See Figure 6. The graph of the cosine is shown as a blue wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function increases. Where the graph of the cosine function increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined. The secant graph has vertical asymptotes at each value of x where the cosine graph crosses the x-axis; we show these in the graph below with dashed vertical lines, but will not show all the asymptotes explicitly on all later graphs involving the secant and cosecant. Note that, because cosine is an even function, secant is also an even function. That is, sec(−x) = sec x. x = − 3π 2 y x = 3π 2 y = cos (x) –2π y = sec (x) x 2π 8 4 –4 –8 Figure 6 Graph of the secant function, f (x) = sec x = 1 ____ cos x As we did for the tangent function, we will again refer to the constant ∣ A ∣ as the stretching factor, not the amplitude. SECTION 6.2 graphs oF the other trigonometric Functions 529 features of the graph of y = Asec(Bx) • The stretching factor is ∣ A ∣ . • The period is π ____ 2 ∣ B ∣ • The domain is x ≠ • The range is (−∞, − ∣ A ∣ ] ∪ [ ∣ A ∣ , ∞). • The vertical asymptotes occur at x = 2π ___ . ∣ B ∣ • There is no amplitude. k, where k is an odd integer. π ____ 2 ∣ B ∣ k, where k is an odd integer. • y = Asec(Bx) is an even function because cosine is an even function. Similar to the secant, the cosecant is defined by the reciprocal identity csc x = undefined when the sine is 0, leading to a vertical asymptote in the graph at 0, π, etc. Since the sine is never more than 1 in absolute
value, the cosecant, being the reciprocal, will never be less than 1 in absolute value. . Notice that the function is 1 ____ sin x We can graph y = csc x by observing the graph of the sine function because these two functions are reciprocals of one another. See Figure 7. The graph of sine is shown as a blue wave so we can see the relationship. Where the graph of the sine function decreases, the graph of the cosecant function increases. Where the graph of the sine function increases, the graph of the cosecant function decreases. The cosecant graph has vertical asymptotes at each value of x where the sine graph crosses the x-axis; we show these in the graph below with dashed vertical lines. Note that, since sine is an odd function, the cosecant function is also an odd function. That is, csc(−x) = −cscx. The graph of cosecant, which is shown in Figure 7, is similar to the graph of secant. x = −2π y x = −π 10 x = π x = 2π y = csc (x) 6 2 –6 –10 y = sin (x) x Figure 7 The graph of the cosecant function, f (x) = cscx = 1 ____ sinx features of the graph of y = Acsc(Bx) • The stretching factor is ∣ A ∣ . • The period is 2π ___ . ∣ B ∣ • The domain is x ≠ k, where k is an integer. π ___ ∣ B ∣ • The range is (−∞, − ∣ A ∣ ] ∪ [ ∣ A ∣ , ∞). • The asymptotes occur at x = π ___ ∣ B ∣ • y = Acsc(Bx) is an odd function because sine is an odd function. k, where k is an integer. Graphing Variations of y = sec x and y = csc x For shifted, compressed, and/or stretched versions of the secant and cosecant functions, we can follow similar methods to those we used for tangent and cotangent. That is, we locate the vertical asymptotes and also evaluate the functions for a few points (specifically the local extrema). If we want to graph only a single period, we can choose the interval for the 530 CHAPTER 6 periodic Functions period in more than one way. The procedure for secant is very similar, because the cofunction identity means that the secant graph is the same as the cosecant graph shifted half a period to the left. Vertical and phase shifts may be applied to the cosecant function in the same way as for the secant and other functions. The equations become the following. y = Asec(Bx − C) + D y = Acsc(Bx − C) + D features of the graph of y = Asec(Bx − C) + D • The stretching factor is ∣ A ∣ . • The period is 2π ___ . ∣ B ∣ C __ • The domain is x ≠ B + π ____ 2 ∣ B ∣ k, where k is an odd integer. • The range is (−∞, − ∣ A ∣ + D] ∪ [ ∣ A ∣ + D, ∞). π C ____ __ • The vertical asymptotes occur at x = 2 ∣ B ∣ B • There is no amplitude. • y = Asec(Bx) is an even function because cosine is an even function. + k, where k is an odd integer. features of the graph of y = Acsc(Bx − C) + D • The stretching factor is ∣ A ∣ . • The period is 2π ___ . ∣ B ∣ C __ • The domain is x ≠ B + π ____ ∣ B ∣ k, where k is an integer. • The range is (−∞, − ∣ A ∣ + D] ∪ [ ∣ A ∣ + D, ∞). π • The vertical asymptotes occur at x = C ___ __ ∣ B ∣ B + k, where k is an integer. • There is no amplitude. • y = Acsc(Bx) is an odd function because sine is an odd function. How To… Given a function of the form y = Asec(Bx), graph one period. 1. Express the function given in the form y = Asec(Bx). 2. Identify the stretching/compressing factor, ∣ A ∣ . 2π ___ 3. Identify B and determine the period, P = . ∣ B ∣ 4. Sketch the graph of y = Acos(Bx). 5. Use the reciprocal relationship between y = cos x and y = sec x to draw the graph of y = Asec(Bx). 6. Sketch the asymptotes. 7. Plot any two reference points and draw the graph through these points. Example 4 Graphing a Variation of the Secant Function Graph one period of f (x) = 2.5sec(0.4x). Solution Step 1. The given function is already written in the general form, y = Asec(Bx). Step 2. A = 2.5 so the stretching factor is 2.5. Step 3. B = 0.4 so P = = 5π. The period is 5π units. Step 4. Sketch the graph of the function g(x) = 2.5cos(0.4x). 2π ___ 0.4 Step 5. Use the reciprocal relationship of the cosine and secant functions to draw the cosecant function. SECTION 6.2 graphs oF the other trigonometric Functions 531 Steps 6–7. Sketch two asymptotes at x = 1.25π and x = 3.75π. We can use two reference points, the local minimum at (0, 2.5) and the local maximum at (2.5π, −2.5). Figure 8 shows the graph. f (x) = 2.5 sec (0.4x) y 6 4 2 –π π 2π 3π 4π x –2 –4 –6 Figure 8 Try It #4 Graph one period of f (x) = −2.5sec(0.4x). Q & A… Do the vertical shift and stretch/compression affect the secant’s range? Yes. The range of f (x) = Asec(Bx − C) + D is (−∞, − ∣ A ∣ + D] ∪ [ ∣ A ∣ + D, ∞). How To… Given a function of the form f (x) = Asec(Bx − C) + D, graph one period. 1. Express the function given in the form y = A sec(Bx − C) + D. 2. Identify the stretching/compressing factor, ∣ A ∣ . 3. Identify B and determine the period, 2π ___ . ∣ B ∣ 4. Identify C and determine the phase shift, C __ 5. Draw the graph of y = A sec(Bx) but shift it to the right by B π ____ 2 ∣ B ∣ C __ 6. Sketch the vertical asymptotes, which occur at x = B + C __ . B and up by D. k, where k is an odd integer. Example 5 Graphing a Variation of the Secant Function π π __ __  + 1. x − Graph one period of y = 4sec  2 3 Solution π π __ __ Step 1. Express the function given in the form y = 4sec   + 1. x − 2 3 Step 2. The stretching/compressing factor is ∣ A ∣ = 4. Step 3. The period is Step 4. The phase shift is 2π ___ ∣ B ∣ = = 2π _ π _ 3 3 2π __ ___ · 1 π = 6 π __ 2 _ = π __ 3 π 3 __ __ = · 2 π = 1.5 C __ B 532 CHAPTER 6 periodic Functions C __ Step 5. Draw the graph of y = Asec(Bx), but shift it to the right by B = 1.5 and up by D = 6. Step 6. Sketch the vertical asymptotes, which occur at x = 0, x = 3, and x = 6. There is a local minimum at (1.5, 5) and a local maximum at (4.5, −3). Figure 9 shows the graph. y x = 3 x = 6 10 8 6 4 2 –2 –4 –6 –8 y = 4 sec Figure 9 Try It #5 Graph one period of f (x) = −6sec(4x + 2) − 8. Q & A… The domain of csc x was given to be all x such that x ≠ kπ for any integer k. Would the domain of y = Acsc(Bx − C) + D be x ≠ C + kπ ______ ? B Yes. The excluded points of the domain follow the vertical asymptotes. Their locations show the horizontal shift and compression or expansion implied by the transformation to the original function’s input. How To… Given a function of the form y = Acsc(Bx), graph one period. 1. Express the function given in the form y = Acsc(Bx). 2. Identify the stretching/compressing factor, ∣ A ∣ . 2π ___ 3. Identify B and determine the period, P = . ∣ B ∣ 4. Draw the graph of y = Asin(Bx). 5. Use the reciprocal relationship between y = sin x and y = csc x to draw the graph of y = Acsc(Bx). 6. Sketch the asymptotes. 7. Plot any two reference points and draw the graph through these points. Example 6 Graphing a Variation of the Cosecant Function Graph one period of f (x) = −3csc(4x). Solution Step 1. The given function is already written in the general form, y = Acsc(Bx). Step 2. ∣ A ∣ = ∣ −3 ∣ = 3, so the stretching factor is 3. π π __ __ = Step 3. B = 4, so P = units. . The period is 2 2 Step 4. Sketch the graph of the function g(x) = −3sin(4x). 2π __ 4 Step 5. Use the reciprocal relationship of the sine and cosecant functions to draw the cosecant function. π π __ __ , and x = Steps 6–7. Sketch three asymptotes at x = 0, x = . We can use two reference points, the local maximum 2 4 π __ , −3  and the local minimum at  at  8 3π __ , 3  . Figure 10 shows the graph. 8 SECTION 6.2 graphs oF the other trigonometric Functions 533 y 6 4 2 –2 –4 –(x) = –3 csc (4x) π 8 3π 8 x Figure 10 Try It #6 Graph one period of f (x) = 0.5csc(2x). How To… Given a function of the form f (x) = Acsc(Bx − C) + D, graph one period. 1. Express the function given in the form y = Acsc(Bx − C) + D. 2. Identify the stretching/compressing factor, ∣ A ∣ . 3. Identify B and determine the period, 2π ___ . ∣ B ∣ 4. Identify C and determine the phase shift, C __ 5. Draw the graph of y = Acsc(Bx) but shift it to the right by B π 6. Sketch the vertical asymptotes, which occur at x = C ___ __ ∣ B ∣ B C __ . B + and up by D. k, where k is an integer. Graphing a Vertically Stretched, Horizontally Compressed, and Vertically Shifted Cosecant Example 7 π _ x  + 1. What are the domain and range of this function? Sketch a graph of y = 2csc  2 Solution π __ Step 1. Express the function given in the form y = 2csc  x  + 1. 2 Step 2. Identify the stretching/compressing factor, ∣ A ∣ = 2. 2π _ π __ 2 = 0. Step 4. The phase shift is Step 3. The period is 2 _ π = 4. · 2π ___ ∣ B ∣ 2π _ 1 = = 0 _ π __ 2 Step 5. Draw the graph of y = Acsc(Bx) but shift it up D = 1. Step 6. Sketch the vertical asymptotes, which occur at x = 0, x = 2, x = 4. The graph for this function is shown in Figure 11. y 6 3 –3 –6 –3 y = 2 csc 4 –6 x = −2 x = 2 x = 4 Figure 11 A transformed cosecant function 534 CHAPTER 6 periodic Functions Analysis The vertical asymptotes shown on the graph mark off one period of the function, and the local extrema in this π __ x  + 1, interval are shown by dots. Notice how the graph of the transformed cosecant relates to the graph of f (x) = 2sin  2 shown as the blue wave. Try It #7 π π __ __ x  + 1 on the same axes. x  + 1 shown in Figure 12, sketch the graph of g(x) = 2sec  Given the graph of f (x) = 2cos  2 2 f (x) 6 4 f (x) = 2 cos π x 2 + 1 –5 –3 –1 1 3 5 x –2 –4 –6 Figure 12 Analyzing the Graph of y = cot x The last trigonometric function we need to explore is cotangent. The cotangent is defined by the reciprocal identity cot x = 1 ____ tan x . Notice that the function is undefined when the tangent function is 0, leading to a vertical asymptote in the graph at 0, π, etc. Since the output of the tangent function is all real numbers, the output of the cotangent function is also all real numbers. We can graph y = cot x by observing the graph of the tangent function because these two functions are reciprocals of one another. See Figure 13. Where the graph of the tangent function decreases, the graph of the cotangent function increases. Where the graph of t
he tangent function increases, the graph of the cotangent function decreases. The cotangent graph has vertical asymptotes at each value of x where tan x = 0; we show these in the graph below with dashed lines. Since the cotangent is the reciprocal of the tangent, cot x has vertical asymptotes at all values of x where tan x = 0, and cot x = 0 at all values of x where tan x has its vertical asymptotes. f (x) = cot x y 20 10 x – 3π 2 – π 2 3π 2 π 2 –10 –20 x = −π x = −2π Figure 13 The cotangent function x = 2π x = π features of the graph of y = Acot(Bx) • The stretching factor is ∣ A ∣ . • The period is P = π ___ . ∣ B ∣ π ___ ∣ B ∣ • The range is (−∞, ∞). • The asymptotes occur at x = π ___ ∣ B ∣ • y = Acot(Bx) is an odd function. • The domain is x ≠ k, where k is an integer. k, where k is an integer. SECTION 6.2 graphs oF the other trigonometric Functions 535 Graphing Variations of y = cot x We can transform the graph of the cotangent in much the same way as we did for the tangent. The equation becomes the following. y = Acot(Bx − C) + D features of the graph of y = Acot(Bx − C) + D • The stretching factor is ∣ A ∣ . • The period is π ___ . ∣ B ∣ C __ • The domain is x ≠ B + π ___ ∣ B ∣ k, where k is an integer. • The range is (−∞, ∞). • The vertical asymptotes occur at x = C __ B • There is no amplitude. • y = Acot(Bx) is an odd function because it is the quotient of even and odd functions (cosine and sine, respectively) k, where k is an integer. π ___ ∣ B ∣ + How To… Given a modified cotangent function of the form f (x) = Acot(Bx), graph one period. 1. Express the function in the form f (x) = Acot(Bx). 2. Identify the stretching factor, ∣ A ∣ . 3. Identify the period, P = π ___ . ∣ B ∣ 4. Draw the graph of y = Atan(Bx). 5. Plot any two reference points. 6. Use the reciprocal relationship between tangent and cotangent to draw the graph of y = Acot(Bx). 7. Sketch the asymptotes. Example 8 Graphing Variations of the Cotangent Function Determine the stretching factor, period, and phase shift of y = 3cot(4x), and then sketch a graph. Solution Step 1. Expressing the function in the form f (x) = Acot(Bx) gives f (x) = 3cot(4x). Step 2. The stretching factor is ∣ A ∣ = 3. π __ Step 3. The period is P = . 4 Step 4. Sketch the graph of y = 3tan(4x). Step 5. Plot two reference points. Two such points are  Step 6. Use the reciprocal relationship to draw y = 3cot(4x). π __ 16 , 3  and  3π __ 16 , −3  . π __ Step 7. Sketch the asymptotes, x = 0, x = . 4 The blue graph in Figure 14 shows y = 3tan(4x) and the red graph shows y = 3cot(4x). y = 3 cot (4x) y = 3 tan (4x) 6 4 2 –2 –4 –6 Figure 14 536 CHAPTER 6 periodic Functions How To… Given a modified cotangent function of the form f (x) = Acot(Bx − C) + D, graph one period. 1. Express the function in the form f (x) = Acot(Bx − C) + D. 2. Identify the stretching factor, ∣ A ∣ . 3. Identify the period, P = π ___ . ∣ B ∣ C __ . B C __ 5. Draw the graph of y = Atan(Bx) shifted to the right by B 4. Identify the phase shift, C __ 6. Sketch the asymptotes x = B π ___ ∣ B ∣ 7. Plot any three reference points and draw the graph through these points. k, where k is an integer. + and up by D. Graphing a Modified Cotangent Example 9 π π __ __  − 2. x − Sketch a graph of one period of the function f (x) = 4cot  2 8 Solution Step 1. The function is already written in the general form f (x) = Acot(Bx − C) + D. Step 2. A = 4, so the stretching factor is 4. π __ , so the period is P = Step 3. B = 8 π ___ ∣ B ∣ = π _ π __ 8 = 8. π __ Step 4. C = , so the phase shift is 2 π __ 2 _ π __ 8 π π __ __  − 2. x − Step 5. We draw f (x) = 4tan  2 8 C __ B = = 4. Step 6-7. Three points we can use to guide the graph are (6, 2), (8, −2), and (10, −6). We use the reciprocal relationship of π π __ __  − 2. x − tangent and cotangent to draw f (x) = 4cot  2 8 Step 8. The vertical asymptotes are x = 4 and x = 12. The graph is shown in Figure 15. y 40 30 20 10 –10 –20 –30 –40 f (x) = 4 cot 10 12 14 x x = 4 x = 12 Figure 15 One period of a modified cotangent function Using the Graphs of Trigonometric Functions to Solve Real-World Problems Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a police car and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how do we determine the distance? We can use the tangent function. SECTION 6.2 graphs oF the other trigonometric Functions 537 Example 10 Using Trigonometric Functions to Solve Real-World Scenarios π __ Suppose the function y = 5tan  t  marks the distance in the movement of a light beam from the top of a police car 4 across a wall where t is the time in seconds and y is the distance in feet from a point on the wall directly across from the police car. a. Find and interpret the stretching factor and period. b. Graph on the interval [0, 5]. c. Evaluate f (1) and discuss the function’s value at that input. Solution π __ a. We know from the general form of y = Atan(Bt) that ∣ A ∣ is the stretching factor and is the period. B π __ y = 5 tan  t  4 ↑ ↑ A B Figure 16 We see that the stretching factor is 5. This means that the beam of light will have moved 5 ft after half the period. The period is π 4 __ __ = = 4. This means that every 4 seconds, the beam of light sweeps the wall. The distance · π 1 π _ π __ 4 from the spot across from the police car grows larger as the police car approaches. b. To graph the function, we draw an asymptote at t = 2 and use the stretching factor and period. See Figure 17 y 6 4 2 –2 –4 –6 π t y = 5 tan 4 1 3 4 5 t t = 2 Figure 17 π __ (1)  = 5(1) = 5; after 1 second, the beam of light has moved 5 ft from the spot across c. Period: f (1) = 5tan  4 from the police car. Access these online resources for additional instruction and practice with graphs of other trigonometric functions. • Graphing the Tangent Function (http://openstaxcollege.org/l/graphtangent) • Graphing Cosecant and Secant Functions (http://openstaxcollege.org/l/graphcscsec) • Graphing the Cotangent Function (http://openstaxcollege.org/l/graphcot) 538 CHAPTER 6 periodic Functions 6.2 SeCTIOn exeRCISeS VeRBAl 1. Explain how the graph of the sine function can be used to graph y = csc x. 2. How can the graph of y = cos x be used to construct the graph of y = sec x? 3. Explain why the period of tan x is equal to π. 4. Why are there no intercepts on the graph of 5. How does the period of y = csc x compare with the period of y = sin x? y = csc x? AlGeBRAIC For the following exercises, match each trigonometric function with one of the graphs in Figure 182π –π x –π 2π – π 2 x π π 2 –2π –π π 2π x x = π II Figure 18 III IV 6. f (x) = tan x 8. f (x) = csc x 7. f (x) = sec x 9. f (x) = cot x For the following exercises, find the period and horizontal shift of each of the functions. π 10. f (x) = 2tan(4x − 32) __ 11. h(x) = 2sec  (x + 1)  4 13. If tan x = −1.5, find tan(−x). π __ 12. m(x) = 6csc  x + π  3 14. If sec x = 2, find sec(−x). 15. If csc x = −5, find csc(−x). 16. If xsin x = 2, find (−x)sin(−x). For the following exercises, rewrite each expression such that the argument x is positive. 17. cot(−x)cos(−x) + sin(−x) 18. cos(−x) + tan(−x)sin(−x) GRAPHICAl For the following exercises, sketch two periods of the graph for each of the following functions. Identify the stretching factor, period, and asymptotes. 19. f (x) = 2tan(4x − 32) π __ 22. j(x) = tan  x  2 π __ 25. f (x) = tan  x +  4 28. f (x) = − 1 __ csc(x) 4 31. f (x) = 7sec(5x) π __  − 2 34. f (x) = −sec  x − 3 π __ (x + 1)  20. h(x) = 2sec  4 π __ 23. p(x) = tan  x −  2 26. f (x) = π tan(πx − π) − π π __ x + π  21. m(x) = 6csc  3 24. f (x) = 4tan(x) 27. f (x) = 2csc(x) 29. f (x) = 4sec(3x) 30. f (x) = −3cot(2x) csc(πx) 32. f (x) = 9 __ 10 π 35. f (x) = 7 __ __ csc  x −  4 5 π __  − 1 33. f (x) = 2csc  x + 4 π __  − 3  36. f (x) = 5  cot  x + 2 SECTION 6.2 section exercises 539 For the following exercises, find and graph two periods of the periodic function with the given stretching factor, ∣ A ∣ , period, and phase shift. π _ 37. A tangent curve, A = 1, period of ; and phase shift 3 π __ (h, k) =  , 2  4 π _ 38. A tangent curve, A = −2, period of , and phase 4 shift (h, k) =  − π __ , −2  4 For the following exercises, find an equation for the graph of each function. f (x) 8 4 –0.5 0.5 –4 –8 x = −1 x 1 x = 1 39. x = −π f (x) x = π 2 40. 10 6 2 –6 –10 x = − π 2 x = π 41. x = − π 4 f (x) x = π 2 x x 10 6 2 –6 –10 π 2 x = − 42. f (x) 43. f (x) x = −2π x = π 10 6 2 – 3π 8 π 8 –6 –10 x 5π 8 10 6 2 –6 –10 x = −π x = 2π x x = π 4 x = − 5π 8 x = − π 8 44. f (x) 45. 10 6 2 –2π –π π 2π x −0.01 –10 x = −0.005 TeCHnOlOGY x = 3π 8 f (x) 4 3 2 1 –1 –2 –3 –4 x 0.01 x = 0.005 For the following exercises, use a graphing calculator to graph two periods of the given function. Note: most graphing calculators do not have a cosecant button; therefore, you will need to input csc x as 1 ____ . sin x 46. f (x) = \|csc(x)\| 48. f (x) = 2csc(x) 47. f (x) = \|cot(x)\| 49. f (x) = csc(x) _____ sec(x) 50. Graph f (x) = 1 + sec2 (x) − tan2 (x). What is the function shown in the graph? 51. f (x) = sec(0.001x) 52. f (x) = cot(100πx) 53. f (x) = sin2 x + cos2 x 540 CHAPTER 6 periodic Functions ReAl-WORlD APPlICATIOnS π 54. The function f (x) = 20tan  __ 10 x  marks the distance in the movement of a light beam from a police car across a wall for time x, in seconds, and distance f (x), in feet. a. Graph on the interval [0, 5]. b. Find and interpret the stretching factor, period, and asymptote. c. Evaluate f (1) and f (2.5) and discuss the function’s values at those inputs. 55. Standing on the shore of a lake, a fisherman sights a boat far in the distance to his left. Let x, measured in radians, be the angle formed by the line of s
ight to the ship and a line due north from his position. Assume due north is 0 and x is measured negative to the left and positive to the right. (See Figure 19.) The boat travels from due west to due east and, ignoring the curvature of the Earth, the distance d(x), in kilometers, from the fisherman to the boat is given by the function d(x) = 1.5sec(x). a. What is a reasonable domain for d(x)? b. Graph d(x) on this domain. c. Find and discuss the meaning of any vertical asymptotes on the graph of d(x). d. Calculate and interpret d  − π __  . Round to the 3 second decimal place. π __  . Round to the e. Calculate and interpret d  6 second decimal place. f. What is the minimum distance between the fisherman and the boat? When does this occur? 56. A laser rangefinder is locked on a comet 57. A video camera is focused on a rocket on a Figure 19 approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x) = 250,000csc  a. Graph g(x) on the interval [0, 35]. π __ 30 x  . b. Evaluate g(5) and interpret the information. c. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? d. Find and discuss the meaning of any vertical asymptotes. launching pad 2 miles from the camera. The angle of elevation from the ground to the rocket after x seconds is π ___ x. 120 a. Write a function expressing the altitude h(x), in miles, of the rocket above the ground after x seconds. Ignore the curvature of the Earth. b. Graph h(x) on the interval (0, 60). c. Evaluate and interpret the values h(0) and h(30). d. What happens to the values of h(x) as x approaches 60 seconds? Interpret the meaning of this in terms of the problem. SECTION 6.3 inverse trigonometric Functions 541 leARnInG OBjeCTIVeS In this section, you will: • Understand and use the inverse sine, cosine, and tangent functions. • Find the exact value of expressions involving the inverse sine, cosine, and tangent functions. • Use a calculator to evaluate inverse trigonometric functions. • Find exact values of composite functions with inverse trigonometric functions. 6.3 InVeRSe TRIGOnOMeTRIC FUnCTIOnS For any right triangle, given one other angle and the length of one side, we can figure out what the other angles and sides are. But what if we are given only two sides of a right triangle? We need a procedure that leads us from a ratio of sides to an angle. This is where the notion of an inverse to a trigonometric function comes into play. In this section, we will explore the inverse trigonometric functions. Understanding and Using the Inverse Sine, Cosine, and Tangent Functions In order to use inverse trigonometric functions, we need to understand that an inverse trigonometric function “undoes” what the original trigonometric function “does,” as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function, and vice versa, as summarized in Figure 1. Trig Functions Domain: Measure of an angle Range: Ratio Inverse Trig Functions Domain: Ratio Range: Measure of an angle Figure 1 For example, if f (x) = sin x, then we would write f −1(x) = sin−1 x. Be aware that sin−1 x does not mean following examples illustrate the inverse trigonometric functions: 1 ___ sinx . The 1 π 1 π __ __ __ __  . = sin−1   = • Since sin  , then 2 6 2 6 • Since cos(π) = −1, then π = cos−1(−1). π π __ __ = tan−1(1).  = 1, then • Since tan  4 4 In previous sections, we evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that, for a oneto-one function, if f(a) = b, then an inverse function would satisfy f −1(b) = a. Bear in mind that the sine, cosine, and tangent functions are not one-to-one functions. The graph of each function would fail the horizontal line test. In fact, no periodic function can be one-to-one because each output in its range corresponds to at least one input in every period, and there are an infinite number of periods. As with other functions that are not one-to-one, we will need to restrict the domain of each function to yield a new function that is one-toone. We choose a domain for each function that includes the number 0. Figure 2 shows the graph of the sine function π limited to  − π __ __  and the graph of the cosine function limited to [0, π]. , 2 2 y (a) – π 2 – π 4 1 –1 y f(x) = sin x x π 4 π 2 (b) 1 –1 f (x) = cos x π 4 π 2 3π 4 x π Figure 2 (a) Sine function on a restricted domain of  − π __ 2 , π __ 2  ; (b) Cosine function on a restricted domain of [0, π] 542 CHAPTER 6 periodic Functions π Figure 3 shows the graph of the tangent function limited to  − π __ __  . , x) = tan x π 4 y 4 3 2 1 –1 –2 –3 –4 Figure 3 Tangent function on a restricted domain of  − π __ 2 , π __  2 These conventional choices for the restricted domain are somewhat arbitrary, but they have important, helpful characteristics. Each domain includes the origin and some positive values, and most importantly, each results in a one-to-one function that is invertible. The conventional choice for the restricted domain of the tangent function also has the useful property that it extends from one vertical asymptote to the next instead of being divided into two parts by an asymptote. On these restricted domains, we can define the inverse trigonometric functions. • The inverse sine function y = sin−1 x means x = sin y. The inverse sine function is sometimes called the arcsine function, and notated arcsinx. π y = sin−1 x has domain [−1, 1] and range  − π __ __  , 2 2 • The inverse cosine function y = cos−1 x means x = cos y. The inverse cosine function is sometimes called the arccosine function, and notated arccos x. y = cos−1 x has domain [−1, 1] and range [0, π] • The inverse tangent function y = tan−1 x means x = tan y. The inverse tangent function is sometimes called the arctangent function, and notated arctan x. π y = tan−1 x has domain (−∞, ∞) and range  − π __ __  , 2 2 The graphs of the inverse functions are shown in Figure 4, Figure 5, and Figure 6. Notice that the output of each of these inverse functions is a number, an angle in radian measure. We see that sin−1 x has domain [−1, 1] and π range  − π __ __  , cos−1 x has domain [−1, 1] and range [0, π], and tan−1 x has domain of all real numbers and range , 2 2  − π π __ __  . To find the domain and range of inverse trigonometric functions, switch the domain and range of the , 2 2 original functions. Each graph of the inverse trigonometric function is a reflection of the graph of the original function about the line y = x. y = sin–1 (x) y = x y = sin (x Figure 4 The sine function and inverse sine (or arcsine) function SECTION 6.3 inverse trigonometric Functions 543 y = cos–1 (x = cos (x) π 2 x π y = tan (x) y = x y = tan–1 (x Figure 5 The cosine function and inverse cosine (or arccosine) function Figure 6 The tangent function and inverse tangent (or arctangent) function relations for inverse sine, cosine, and tangent functions For angles in the interval  − π π  , if sin y = x, then sin−1 x = y. __ __ , 2 2 For angles in the interval [0, π], if cos y = x, then cos−1 x = y. π For angles in the interval  − π __ __  , if tan y = x, then tan−1 x = y. , 2 2 Example 1 Writing a Relation for an Inverse Function 5π __ 12  ≈ 0.96593, write a relation involving the inverse sine. Given sin  Solution Use the relation for the inverse sine. If sin y = x, then sin−1 x = y. In this problem, x = 0.96593, and y = 5π __ . 12 sin−1 (0.96593) ≈ 5π ___ 12 Try It #1 Given cos(0.5) ≈ 0.8776, write a relation involving the inverse cosine. Finding the exact Value of expressions Involving the Inverse Sine, Cosine, and Tangent Functions Now that we can identify inverse functions, we will learn to evaluate them. For most values in their domains, we must evaluate the inverse trigonometric functions by using a calculator, interpolating from a table, or using some other numerical technique. Just as we did with the original trigonometric functions, we can give exact values for the π π π __ __ __ (60°), and their reflections (45°), and (30°), inverse functions when we are using the special angles, specifically 3 4 6 into other quadrants. How To… Given a “special” input value, evaluate an inverse trigonometric function. 1. Find angle x for which the original trigonometric function has an output equal to the given input for the inverse trigonometric function. 2. If x is not in the defined range of the inverse, find another angle y that is in the defined range and has the same sine, cosine, or tangent as x, depending on which corresponds to the given inverse function. Example 2 Evaluating Inverse Trigonometric Functions for Special Input Values Evaluate each of the following. 1 __  a. sin−1  2 2 b. sin−1  − √  ____ 2 — c. cos−1  − — 3 √  ____ 2 d. tan−1(1) 544 Solution CHAPTER 6 periodic Functions 1 1 _ _  is the same as determining the angle that would have a sine value of a. Evaluating sin−1  . In other words, 2 2 1 _ what angle x would satisfy sin(x) = ? There are multiple values that would satisfy this relationship, such 2 π π 5π , but we know we need the angle in the interval  − π 1 _ _ _ _ _  =  , so the answer will be sin− and as 6 Remember that the inverse is a function, so for each input, we will get exactly one output. 2 b. To evaluate sin−1  − √  , we know that _ 2 5π _ 4 π interval  − π _ _  . For that, we need the negative angle coterminal with , 2 2 2 _ 7π both have a sine value of − √ _ , but neither is in the 4 2 — 2 : sin−1  − √ 7π  = − π _ ____ ___ . 2 4 4 and — 3 3  , we are looking for an angle in the interval [0, π] with a cosine value of − √ c. To evaluate cos−1  − √ _ _ 2 2 . The — — — — 3 angle that satisfies this is cos−1  − √  = _ 2 5π _ . 6 π d. Evalua
ting tan−1(1), we are looking for an angle in the interval  − π _ _  with a tangent value of 1. The correct , 2 2 π _ angle is tan−1(1) = . 4 Try It #2 Evaluate each of the following. a. sin−1(−1) b. tan−1 (−1) c. cos−1 (−1) 1 d. cos−1  _  2 Using a Calculator to evaluate Inverse Trigonometric Functions To evaluate inverse trigonometric functions that do not involve the special angles discussed previously, we will need to use a calculator or other type of technology. Most scientific calculators and calculator-emulating applications have specific keys or buttons for the inverse sine, cosine, and tangent functions. These may be labeled, for example, SIN-1, ARCSIN, or ASIN. In the previous chapter, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trigonometric functions, we can solve for the angles of a right triangle given two sides, and we can use a calculator to find the values to several decimal places. In these examples and exercises, the answers will be interpreted as angles and we will use θ as the independent variable. The value displayed on the calculator may be in degrees or radians, so be sure to set the mode appropriate to the application. Example 3 Evaluating the Inverse Sine on a Calculator Evaluate sin−1(0.97) using a calculator. Solution Because the output of the inverse function is an angle, the calculator will give us a degree value if in degree mode and a radian value if in radian mode. Calculators also use the same domain restrictions on the angles as we are using. In radian mode, sin−1(0.97) ≈ 1.3252. In degree mode, sin−1(0.97) ≈ 75.93°. Note that in calculus and beyond we will use radians in almost all cases. Try It #3 Evaluate cos−1 (−0.4) using a calculator. SECTION 6.3 inverse trigonometric Functions 545 How To… Given two sides of a right triangle like the one shown in Figure 7, find an angle. h p θ a Figure 7 1. If one given side is the hypotenuse of length h and the side of length a adjacent to the desired angle is given, use a __ the equation θ = cos−1   . h 2. If one given side is the hypotenuse of length h and the side of length p opposite to the desired angle is given, use p  the equation θ = sin−1  _ h p a  . 3. If the two legs (the sides adjacent to the right angle) are given, then use the equation θ = tan−1  _ Example 4 Applying the Inverse Cosine to a Right Triangle Solve the triangle in Figure 8 for the angle θ. 12 θ 9 Figure 8 Solution Because we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function. cos θ = 9 __ 12 9 _  θ = cos−1  12 θ ≈ 0.7227 or about 41.4096° Apply definition of the inverse. Evaluate. Try It #4 Solve the triangle in Figure 9 for the angle θ. 10 6 θ Figure 9 546 CHAPTER 6 periodic Functions Finding exact Values of Composite Functions with Inverse Trigonometric Functions There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can usually find exact values for the resulting expressions without resorting to a calculator. Even when the input to the composite function is a variable or an expression, we can often find an expression for the output. To help sort out different cases, let f (x) and g(x) be two different trigonometric functions belonging to the set {sin(x), cos(x), tan(x)} and let f −1(y) and g −1(y) be their inverses. Evaluating Compositions of the Form f (f −1(y )) and f −1(f (x )) For any trigonometric function, f (f −1 (y)) = y for all y in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of f was defined to be identical to the domain of f −1. However, we have to be a little more careful with expressions of the form f −1(f (x)). compositions of a trigonometric function and its inverse sin(sin−1 x) = x for −1 ≤ x ≤ 1 cos(cos−1 x) = x for −1 ≤ x ≤ 1 tan(tan−1 x) = x for −∞ < x < ∞ π π __ __ sin−1(sin x) = x only for − ≤ x ≤ 2 2 cos−1(cos x) = x only for 0 ≤ x ≤ π π π __ __ tan−1(tan x) = x only for − < x < 2 2 Q & A… Is it correct that sin−1(sin x) = x? π π __ __ No. This equation is correct if x belongs to the restricted domain  −  , but sine is defined for all real input , 2 2 values, and for x outside the restricted interval, the equation is not correct because its inverse always returns a value in π π __ __  . The situation is similar for cosine and tangent and their inverses. For example, sin−1  sin   − , 2 2 π 3π __ __   = . 4 4 How To… Given an expression of the form f −1(f(θ)) where f(θ) = sin θ, cos θ, or tan θ, evaluate. 1. If θ is in the restricted domain of f , then f −1(f(θ)) = θ. 2. If not, then find an angle ϕ within the restricted domain of f such that f(ϕ) = f(θ). Then f −1(f(θ)) = ϕ. Example 5 Using Inverse Trigonometric Functions Evaluate the following: π __ a. sin−1  sin    3 b. sin−1  sin  2π ___   3 c. cos−1  cos  2π ___   3 d. cos−1  cos  − π __   3 Solution π π π π a. π __ __ __ __ __  , so sin−1  sin  is in  −   = . , 3 3 3 2 2 π 2π π π b. 2π ___ __ ___ __ __  , so sin−1  sin   = sin   , but sin  is not in  − , 3 3 2 3 2 2π 2π c. 2π ___ ___ ___ is in [0, π], so cos−1  cos    = . 3 3 3 π is not in [0, π], but cos  − π  because cosine is an even function. π __ d. − π __ __ __  = cos  3 π  − π 3 3 3 __ __   = . 3 3 π 2π __ ___   = . 3 3 is in [0, π], so cos−1  cos SECTION 6.3 inverse trigonometric Functions 547 Try It #5 π __   and tan−1  tan  Evaluate tan−1  tan  8 11π ___ 9   . Evaluating Compositions of the Form f −1(g(x )) Now that we can compose a trigonometric function with its inverse, we can explore how to evaluate a composition of a trigonometric function and the inverse of another trigonometric function. We will begin with compositions of the form f−1(g(x)). For special values of x, we can exactly evaluate the inner function and then the outer, inverse function. However, we can find a more general approach by considering the relation between the two acute angles of a right π __ − θ. Consider the sine and cosine of each angle of the right triangle in triangle where one is θ, making the other 2 Figure 10. π – θ 2 a c θ b Figure 10 Right triangle illustrating the cofunction relationships π b − θ  , we have sin−1 (cos θ) = π __ __ __ = sin  Because cos θ = 2 c 2 − θ if 0 ≤ θ ≤ π. If θ is not in this domain, then we need to find another angle that has the same cosine as θ and does belong to the restricted domain; we then subtract π π a π − θ if − π __ __ __ __ __ − θ  , so cos−1 (sin θ) = = cos  . Similarly, sin θ = this angle from 2 2 c 2 2 π __ ≤ θ ≤ . These are just the 2 function-cofunction relationships presented in another way. How To… Given functions of the form sin−1 (cos x) and cos−1 (sin x), evaluate them. π __ 1. If x is in [0, π], then sin−1 (cos x) = − x. 2 2. If x is not in [0, π], then find another angle y in [0, π] such that cos y = cos x. π __ sin−1 (cos x) = − y 2 π π 3. If x is in  − π __ __ __  , then cos−1 (sin x) = − x. , 2 2 2 π  , then find another angle y in  − π π 4. If x is not in  − π __ __ __ __  such that sin y = sin x. , , 2 2 2 2 π __ cos−1 (sin x) = − y 2 Example 6 Evaluating the Composition of an Inverse Sine with a Cosine Evaluate sin−1  cos    a. by direct evaluation. 13π ___ 6 b. by the method described previously. Solution a. Here, we can directly evaluate the inside of the composition. cos  13π ___ 6 π _  = cos  + 2π  6 π _  = cos  6 3 √ ____ 2 = — 548 CHAPTER 6 periodic Functions Now, we can evaluate the inverse function as we did earlier. b. We have x = 13π _ 6 π _ , y = , and 6 — sin−1  3 π √  = _ ____ 2 3 sin−1  cos  13π ___ Try It #6 Evaluate cos−1  sin  − 11π ___ 4   . Evaluating Compositions of the Form f (g −1(x )) To evaluate compositions of the form f (g −1(x)), where f and g are any two of the functions sine, cosine, or tangent and 1 − x2 . When we need to use x is any input in the domain of g −1, we have exact formulas, such as sin(cos−1 x) = √ them, we can derive these formulas by using the trigonometric relations between the angles and sides of a right triangle, together with the use of Pythagorean’s relation between the lengths of the sides. We can use the Pythagorean identity, sin2 x + cos2 x = 1, to solve for one when given the other. We can also use the inverse trigonometric functions to find compositions involving algebraic expressions. — Evaluating the Composition of a Sine with an Inverse Cosine Example 7 4 Find an exact value for sin  cos−1  _   . 5 4  , which means cos θ = 4 __ __ Solution Beginning with the inside, we can say there is some angle such that θ = cos−1  , 5 5 and we are looking for sin θ. We can use the Pythagorean identity to do this. sin2 θ + cos2 θ = 1 Use our known value for cosine. 4 __  sin2 θ +  5 2 = 1 Solve for sine. sin2 θ = 1 − 16 _ 25 ___ 9 _ sin θ = ± √ 25 = ± 3 __ 5 3 4 __ __  is in quadrant I, sin θ must be positive, so the solution is Since θ = cos−1  . See Figure 11. 5 5 3 5 4 θ 4 3 __ __ Figure 11 Right triangle illustrating that if cos θ = , then sin θ = 5 5 We know that the inverse cosine always gives an angle on the interval [0, π], so we know that the sine of that angle must be positive; therefore sin  cos−1  4 __ 5   = sin θ = 3 __ . 5 Try It #7 Evaluate cos  tan−1  5 __ 12   . SECTION 6.3 inverse trigonometric Functions 549 Evaluating the Composition of a Sine with an Inverse Tangent Example 8 7 __   . Find an exact value for sin  tan−1  4 Solution While we could use a similar technique as in Example 6, we will demonstrate a different technique here. 7 __ From the inside, we know there is an angle such that tan θ = . We can envision this as the opposite and adjacent 4 sides on a right triangle, as shown in Figure 12. θ 4 Using the Pythagorean Theorem, we can find the hypotenuse of this triangle. 7 Figure 12 A right triang
le with two sides known 42 + 72 = hypotenuse 2 hypotenuse = √ — 65 Now, we can evaluate the sine of the angle as the opposite side divided by the hypotenuse. This gives us our desired composition. sin θ = 7 _ — 65 √ 7 _   = sin θ sin  tan−1  4 7 _ = — 65 √ = — 65 7 √ ______ 65 Try It #8 7 __   . Evaluate cos  sin−1  9 Finding the Cosine of the Inverse Sine of an Algebraic Expression Example 9 x _ Find a simplified expression for cos  sin−1    for −3 ≤ x ≤ 3. 3 x _ Solution We know there is an angle θ such that sin θ = . 3 sin2 θ + cos2 θ = 1 x _   3 2 + cos2 θ = 1 cos2 θ = 1 − Use the Pythagorean Theorem. Solve for cosine. x2 _ 9 ______ 9 − x2 ______ 9 — 9 − x2 = ± √ _______ 3 π Because we know that the inverse sine must give an angle on the interval  − π __ __  , we can deduce that the cosine of , 2 2 that angle must be positive. cos cos  sin−1  3 — 9 − x2 _______ 3 Try It #9 Find a simplified expression for sin(tan−1 (4x)) for − Access this online resource for additional instruction and practice with inverse trigonometric functions. • evaluate expressions Involving Inverse Trigonometric Functions (http://openstaxcollege.org/l/evalinverstrig) 550 CHAPTER 6 periodic Functions 6.3 SeCTIOn exeRCISeS VeRBAl 1. Why do the functions f (x) = sin−1 x and g(x) = cos−1 x have different ranges? π __ = arcsin(0.5). 3. Explain the meaning of 6 5. Why must the domain of the sine function, sin x, π π __ ___ be restricted to  −  for the inverse sine , 2 2 function to exist? 7. Determine whether the following statement is true or false and explain your answer: arccos(−x) = π − arccos x. AlGeBRAIC For the following exercises, evaluate the expressions. 9. sin−1  − 1 __  2 2 √  ____ 2 — 8. sin−1  2 11. cos−1  − √  ____ 2 14. tan−1 (−1) — 2. Since the functions y = cos x and y = cos−1 x are π ___ inverse functions, why is cos−1  cos  −   not equal 6 π __ to − ? 6 4. Most calculators do not have a key to evaluate sec−1(2). Explain how this can be done using the cosine function or the inverse cosine function. 6. Discuss why this statement is incorrect: arccos (cos x) = x for all x. 1 10. cos−1  __  2 12. tan−1 (1) 15. tan−1  √ — 3  — 3  13. tan−1  − √ 16. tan−1  −1 _  3 √ — For the following exercises, use a calculator to evaluate each expression. Express answers to the nearest hundredth. 17. cos−1 (−0.4) 20. cos−1 (0.8) 18. arcsin(0.23) 21. tan−1 (6) 3 _  19. arccos  5 For the following exercises, find the angle θ in the given right triangle. Round answers to the nearest hundredth. 22. 23. 10 7 12 θ θ 19 For the following exercises, find the exact value, if possible, without a calculator. If it is not possible, explain why. 24. sin−1(cos(π)) 25. tan−1(sin(π)) π _ 27. tan−1  sin    3 30. sin−1  sin  5π _   6 3 _   33. sin  cos−1  5 1 _   36. cos  sin−1  2 28. sin−1  cos  −π _   2 31. tan−1  sin  −5π _ 2   4 _   34. sin  tan−1  3 π _ 26. cos−1  sin    3 4π _   29. tan−1  sin  3 4 _   32. cos  sin−1  5 35. cos  tan−1  12 __   5 SECTION 6.3 section exercises 551 For the following exercises, find the exact value of the expression in terms of x with the help of a reference triangle. 37. tan(sin−1 (x − 1)) 38. sin(cos−1 (1 − x)) 1 x   39. cos  sin−1  _ 40. cos(tan−1 (3x − 1)) 1 __   41. tan  sin−1  x + 2 exTenSIOnS For the following exercise, evaluate the expression without using a calculator. Give the exact value. 3 2 √ √ 42. sin−1  1 __  − cos−1(1)  + sin−1   − cos−1  _ ___ 2 2 2 3 2 √ 1 √  − sin−1  cos−1   + cos−1  __ ___ _  − sin−1(0) 2 2 2 — — — — For the following exercises, find the function if sin t = x ____ . x + 1 43. cos t 44. sec t 45. cot t 46. cos  sin−1  x ____   x + 1 GRAPHICAl 47. tan−1  x _  2x + 1 √ — 48. Graph y = sin−1 x and state the domain and range of 49. Graph y = arccos x and state the domain and range the function. of the function. 50. Graph one cycle of y = tan−1 x and state the domain and range of the function. 52. For what value of x does cos x = cos−1 x? Use a graphing calculator to approximate the answer. ReAl-WORlD APPlICATIOnS 51. For what value of x does sin x = sin−1 x? Use a graphing calculator to approximate the answer. 53. Suppose a 13-foot ladder is leaning against a 54. Suppose you drive 0.6 miles on a road so that the building, reaching to the bottom of a second-floor window 12 feet above the ground. What angle, in radians, does the ladder make with the building? 55. An isosceles triangle has two congruent sides of length 9 inches. The remaining side has a length of 8 inches. Find the angle that a side of 9 inches makes with the 8-inch side. 57. A truss for the roof of a house is constructed from two identical right triangles. Each has a base of 12 feet and height of 4 feet. Find the measure of the acute angle adjacent to the 4-foot side. 3 _ 59. The line y = − x passes through the origin in the 7 x,y-plane. What is the measure of the angle that the line makes with the negative x-axis? 61. A 20-foot ladder leans up against the side of a building so that the foot of the ladder is 10 feet from the base of the building. If specifications call for the ladder’s angle of elevation to be between 35 and 45 degrees, does the placement of this ladder satisfy safety specifications? vertical distance changes from 0 to 150 feet. What is the angle of elevation of the road? 56. Without using a calculator, approximate the value of arctan(10,000). Explain why your answer is reasonable. 3 __ 58. The line y = x passes through the origin in the x,y5 plane. What is the measure of the angle that the line makes with the positive x-axis? 60. What percentage grade should a road have if the angle of elevation of the road is 4 degrees? (The percentage grade is defined as the change in the altitude of the road over a 100-foot horizontal distance. For example a 5% grade means that the road rises 5 feet for every 100 feet of horizontal distance.) 62. Suppose a 15-foot ladder leans against the side of a house so that the angle of elevation of the ladder is 42 degrees. How far is the foot of the ladder from the side of the house? 552 CHAPTER 6 periodic Functions CHAPTeR 6 ReVIeW Key Terms amplitude the vertical height of a function; the constant A appearing in the definition of a sinusoidal function arccosine another name for the inverse cosine; arccos x = cos−1 x arcsine another name for the inverse sine; arcsin x = sin−1 x arctangent another name for the inverse tangent; arctan x = tan−1 x inverse cosine function the function cos−1 x, which is the inverse of the cosine function and the angle that has a cosine equal to a given number inverse sine function the function sin−1 x, which is the inverse of the sine function and the angle that has a sine equal to a given number inverse tangent function the function tan−1 x, which is the inverse of the tangent function and the angle that has a tangent equal to a given number midline the horizontal line y = D, where D appears in the general form of a sinusoidal function periodic function a function f (x) that satisfies f (x + P) = f (x) for a specific constant P and any value of x phase shift the horizontal displacement of the basic sine or cosine function; the constant sinusoidal function any function that can be expressed in the form f (x) = Asin(Bx − C) + D or f (x) = Acos(Bx − C) + D C __ B Key equations Sinusoidal functions f (x) = Asin(Bx − C) + D f (x) = Acos(Bx − C) + D Shifted, compressed, and/or stretched tangent function y = A tan(Bx − C) + D Shifted, compressed, and/or stretched secant function y = A sec(Bx − C) + D Shifted, compressed, and/or stretched cosecant function y = A csc(Bx − C) + D Shifted, compressed, and/or stretched cotangent function y = A cot(Bx − C) + D Key Concepts 6.1 Graphs of the Sine and Cosine Functions • Periodic functions repeat after a given value. The smallest such value is the period. The basic sine and cosine functions have a period of 2π. • The function sin x is odd, so its graph is symmetric about the origin. The function cos x is even, so its graph is symmetric about the y-axis. • The graph of a sinusoidal function has the same general shape as a sine or cosine function. • In the general formula for a sinusoidal function, the period is P = See Example 1. 2π _ ∣B∣ • In the general formula for a sinusoidal function, ∣A∣ represents amplitude. If ∣A∣ > 1, the function is stretched, whereas if ∣A∣ < 1, the function is compressed. See Example 2. • The value in the general formula for a sinusoidal function indicates the phase shift. See Example 3. C __ B • The value D in the general formula for a sinusoidal function indicates the vertical shift from the midline. See Example 4. • Combinations of variations of sinusoidal functions can be detected from an equation. See Example 5. • The equation for a sinusoidal function can be determined from a graph. See Example 6 and Example 7. • A function can be graphed by identifying its amplitude and period. See Example 8 and Example 9. • A function can also be graphed by identifying its amplitude, period, phase shift, and horizontal shift. See Example 10. • Sinusoidal functions can be used to solve real-world problems. See Example 11, Example 12, and Example 13. CHAPTER 6 review 553 6.2 Graphs of the Other Trigonometric Functions • The tangent function has period π. • f (x) = Atan(Bx − C) + D is a tangent with vertical and/or horizontal stretch/compression and shift. See Example 1, Example 2, and Example 3. • The secant and cosecant are both periodic functions with a period of 2π. f (x) = Asec(Bx − C) + D gives a shifted, compressed, and/or stretched secant function graph. See Example 4 and Example 5. • f (x) = Acsc(Bx − C) + D gives a shifted, compressed, and/or stretched cosecant function graph. See Example 6 and Example 7. • The cotangent function has period π and vertical asymptotes at 0, ±π, ±2π, ... • The range of cotangent is (−∞, ∞), and the function is decreasing at each point in its range. • T
he cotangent is zero at ± π __ 2 • f (x) = Acot(Bx − C) + D is a cotangent with vertical and/or horizontal stretch/compression and shift. See Example , ± 3π __ 2 , ... 8 and Example 9. • Real-world scenarios can be solved using graphs of trigonometric functions. See Example 10. 6.3 Inverse Trigonometric Functions • An inverse function is one that “undoes” another function. The domain of an inverse function is the range of the original function and the range of an inverse function is the domain of the original function. • Because the trigonometric functions are not one-to-one on their natural domains, inverse trigonometric functions are defined for restricted domains. • For any trigonometric function f (x), if x = f−1(y), then f (x) = y. However, f (x) = y only implies x = f−1(y) if x is in the restricted domain of f . See Example 1. • Special angles are the outputs of inverse trigonometric functions for special input values; for example, 1 π π __ __ __  . See Example 2. = sin−1  = tan−1(1) and 2 6 4 • A calculator will return an angle within the restricted domain of the original trigonometric function. See Example 3. • Inverse functions allow us to find an angle when given two sides of a right triangle. See Example 4. • In function composition, if the inside function is an inverse trigonometric function, then there are exact expressions; for example, sin(cos−1 (x)) = √ 1 − x2 . See Example 5. — π _ • If the inside function is a trigonometric function, then the only possible combinations are sin−1 (cos x) = − x if and cos−1 (sin x) = ≤ x ≤ − x if − . See Example 6 and Example 7. 2 2 2 • When evaluating the composition of a trigonometric function with an inverse trigonometric function, draw a reference triangle to assist in determining the ratio of sides that represents the output of the trigonometric function. See Example 8. • When evaluating the composition of a trigonometric function with an inverse trigonometric function, you may use trig identities to assist in determining the ratio of sides. See Example 9. 554 CHAPTER 6 periodic Functions CHAPTeR 6 ReVIeW exeRCISeS GRAPHS OF THe SIne AnD COSIne FUnCTIOnS For the following exercises, graph the functions for two periods and determine the amplitude or stretching factor, period, midline equation, and asymptotes. 1. f (x) = −3cos x + 3 2π _ 4. f (x) = −2sin  . f (x) = 6sin  3x − 6 1 _ 2. f (x) = sin x 4 π _ 5. f (x) = 3sin  x −  − 4 4 8. f (x) = −100sin(50x − 20) GRAPHS OF THe OTHeR TRIGOnOMeTRIC FUnCTIOnS π _  3. f (x) = 3cos  x + 6 4π _  + 1  6. f (x) = 2  cos  x − 3 For the following exercises, graph the functions for two periods and determine the amplitude or stretching factor, period, midline equation, and asymptotes. 9. f (x) = tan x − 4 12. f (x) = 0.2cos(0.1x) + 0.3 π _ 10. f (x) = 2tan  x −  6 11. f (x) = −3tan(4x) − 2 For the following exercises, graph two full periods. Identify the period, the phase shift, the amplitude, and asymptotes. 1 _ 13. f (x) = sec x 3 1 _ x  16. f (x) = 8sec  4 14. f (x) = 3cot x 15. f (x) = 4csc(5x) 2 1 _ __ x  csc  17. f (x) = 3 2 18. f (x) = −csc(2x + π) For the following exercises, use this scenario: The population of a city has risen and fallen over a 20-year interval. Its population may be modeled by the following function: y = 12,000 + 8,000sin(0.628x), where the domain is the years since 1980 and the range is the population of the city. 19. What is the largest and smallest population the city 20. Graph the function on the domain of [0, 40]. may have? 21. What are the amplitude, period, and phase shift for 22. Over this domain, when does the population reach the function? 18,000? 13,000? 23. What is the predicted population in 2007? 2010? For the following exercises, suppose a weight is attached to a spring and bobs up and down, exhibiting symmetry. 24. Suppose the graph of the displacement function is shown in Figure 1, where the values on the x-axis represent the time in seconds and the y-axis represents the displacement in inches. Give the equation that models the vertical displacement of the weight on the spring. y 5 4 3 2 1 –1 –2 –3 –4 –5 2 4 68 1 0 x Figure 1 CHAPTER 6 review 555 25. At time = 0, what is the displacement of the weight? 26. At what time does the displacement from the equilibrium point equal zero? 27. What is the time required for the weight to return to its initial height of 5 inches? In other words, what is the period for the displacement function? InVeRSe TRIGOnOMeTRIC FUnCTIOnS For the following exercises, find the exact value without the aid of a calculator. 28. sin−1(1) 1 _  31. cos−1  — 2 √ 34. cos−1  tan  3π _   4 37. tan  cos−1  5 _ 13   29. cos−1  — 3 √  _ 2 32. sin− 35. sin  sec−1    5 38. sin  cos−1  x _   x + 1 30. tan−1(−1) π _ 33. sin−1  cos    6 3 _ 36. cot  sin−1    5 39. Graph f (x) = cos x and f (x) = sec x on the interval [0, 2π) and explain any observations. 40. Graph f (x) = sin x and f (x) = csc x and explain any observations. x _ 41. Graph the function f (x) = − 1 x3 _ 3! x5 _ 5! x7 _ 7! + − on the interval [−1, 1] and compare the graph to the graph of f (x) = sin x on the same interval. Describe any observations. 556 CHAPTER 6 periodic Functions CHAPTeR 6 PRACTICe TeST For the following exercises, sketch the graph of each function for two full periods. Determine the amplitude, the period, and the equation for the midline. 1. f (x) = 0.5sin x 4. f (x) = sin(3x) 1 _ 7. f (x) = 3cos  x − 3 5π _  6 2. f (x) = 5cos x 3. f (x) = 5sin x π _  + 1 5. f (x) = −cos  x + 3 8. f (x) = tan(4x) π _   + 4 6. f (x) = 5sin  3  x − 6 9. f (x) = −2tan  x − 7π _  + 2 6 π _ 12. f (x) = πsec  x  2 10. f (x) = πcos(3x + π) 11. f (x) = 5csc(3x) π _  − 3 13. f (x) = 2csc  x + 4 For the following exercises, determine the amplitude, period, and midline of the graph, and then find a formula for the function. 14. Give in terms of a sine function. y 15. Give in terms of a sine function. y 16. Give in terms of a tangent function. y 6 4 2 6 4 2 12 8 4 –3 –2 –1 1 2 3 x –3 –2 –1 1 2 3 x – 3π 2 –π – π 2 π 2 π 3π 2 x –2 –4 –6 –2 –4 –6 –4 –8 –12 For the following exercises, find the amplitude, period, phase shift, and midline. π _ x + π  − 3 17. y = sin  6 18. y = 8sin  7π _ 6 x + 7π _  + 6 2 19. The outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the temperature is 68°F at midnight and the high and low temperatures during the day are 80°F and 56°F, respectively. Assuming t is the number of hours since midnight, find a function for the temperature, D, in terms of t. 20. Water is pumped into a storage bin and empties according to a periodic rate. The depth of the water is 3 feet at its lowest at 2:00 a.m. and 71 feet at its highest, which occurs every 5 hours. Write a cosine function that models the depth of the water as a function of time, and then graph the function for one period. For the following exercises, find the period and horizontal shift of each function. 21. g(x) = 3tan(6x + 42) 22. n(x) = 4csc  5π _ 3 x − 20π _  3 23. Write the equation for the graph in Figure 1 in terms of the secant function and give the period and phase shift. y 6 4 2 –3 –2 –1 1 2 3 x –2 –4 –6 Figure 1 CHAPTER 6 practice test 557 24. If tan x = 3, find tan(−x). 25. If sec x = 4, find sec(−x). For the following exercises, graph the functions on the specified window and answer the questions. 26. Graph m(x) = sin(2x) + cos(3x) on the viewing window [−10, 10] by [−3, 3]. Approximate the graph’s period. 28. Graph f (x) = observations. sin x _____ x on [−0.5, 0.5] and explain any 27. Graph n(x) = 0.02sin(50πx) on the following domains in x: [0, 1] and [0, 3]. Suppose this function models sound waves. Why would these views look so different? 3 __ For the following exercises, let f (x) = cos(6x). 5 29. What is the largest possible value for f (x)? 31. Where is the function increasing on the interval [0, 2π]? 30. What is the smallest possible value for f (x)? For the following exercises, find and graph one period of the periodic function with the given amplitude, period, and phase shift. π _ 32. Sine curve with amplitude 3, period , and phase 3 π _ , 2  shift (h, k) =  4 π _ 33. Cosine curve with amplitude 2, period , and phase 6 shift (h, k) =  − π _ 4 , 3  For the following exercises, graph the function. Describe the graph and, wherever applicable, any periodic behavior, amplitude, asymptotes, or undefined points. 34. f (x) = 5cos(3x) + 4sin(2x) 35. f (x) = esint — For the following exercises, find the exact value. 3  36. sin−1  √ _ 2 39. cos−1(sin(π)) 37. tan−1  √ — 3  40. cos−1  tan  7π _   4 42. cos−1(−0.4) 43. cos(tan−1(x2)) For the following exercises, suppose sin t = x _ . x + 1 44. tan t 45. csc t 46. Given Figure 2, find the measure of angle θ to three decimal places. Answer in radians. For the following exercises, determine whether the equation is true or false. 47. arcsin  sin  5π _   = 6 5π _ 6 48. arccos  cos  5π _   = 6 5π _ 6 — 3  38. cos−1  − √ _ 2 41. cos(sin−1(1 − 2x)) 12 θ 19 49. The grade of a road is 7%. This means that for every horizontal distance of 100 feet on the road, the vertical rise is 7 feet. Find the angle the road makes with the horizontal in radians. Trigonometric Identities and Equations 7 Figure 1 A sine wave models disturbance. (credit: modification of work by Mikael Altemark, Flickr). CHAPTeR OUTlIne 7.1 Solving Trigonometric equations with Identities 7.2 Sum and Difference Identities 7.3 Double-Angle, Half-Angle, and Reduction Formulas 7.4 Sum-to-Product and Product-to-Sum Formulas 7.5 Solving Trigonometric equations 7.6 Modeling with Trigonometric equations Introduction Math is everywhere, even in places we might not immediately recognize. For example, mathematical relationships describe the transmission of images, light, and sound. The sinusoidal graph in Figure 1 models music playing on a phone, radio, or computer. Such graphs are described using trigono
metric equations and functions. In this chapter, we discuss how to manipulate trigonometric equations algebraically by applying various formulas and trigonometric identities. We will also investigate some of the ways that trigonometric equations are used to model real-life phenomena. 559 560 CHAPTER 7 trigonometric identities and eQuations leARnInG OBjeCTIVeS In this section, you will: • • Verify the fundamental trigonometric identities. Simplify trigonometric expressions using algebra and the identities. 7.1 SOlVInG TRIGOnOMeTRIC eQUATIOnS WITH IDenTITIeS Figure 1 International passports and travel documents In espionage movies, we see international spies with multiple passports, each claiming a different identity. However, we know that each of those passports represents the same person. The trigonometric identities act in a similar manner to multiple passports—there are many ways to represent the same trigonometric expression. Just as a spy will choose an Italian passport when traveling to Italy, we choose the identity that applies to the given scenario when solving a trigonometric equation. In this section, we will begin an examination of the fundamental trigonometric identities, including how we can verify them and how we can use them to simplify trigonometric expressions. Verifying the Fundamental Trigonometric Identities Identities enable us to simplify complicated expressions. They are the basic tools of trigonometry used in solving trigonometric equations, just as factoring, finding common denominators, and using special formulas are the basic tools of solving algebraic equations. In fact, we use algebraic techniques constantly to simplify trigonometric expressions. Basic properties and formulas of algebra, such as the difference of squares formula and the perfect squares formula, will simplify the work involved with trigonometric expressions and equations. We already know that all of the trigonometric functions are related because they all are defined in terms of the unit circle. Consequently, any trigonometric identity can be written in many ways. To verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation. Sometimes we have to factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desired result. In this first section, we will work with the fundamental identities: the Pythagorean identities, the even-odd identities, the reciprocal identities, and the quotient identities. We will begin with the Pythagorean identities (see Table 1), which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identifies, but now we will also use additional identities. sin2 θ + cos2 θ = 1 Pythagorean Identities 1 + cot2 θ = csc2 θ Table 1 1 + tan2 θ = sec2 θ SECTION 7.1 solving trigonometric eQuations with identities 561 The second and third identities can be obtained by manipulating the first. The identity 1 + cot2 θ = csc2 θ is found by rewriting the left side of the equation in terms of sine and cosine. Prove: 1 + cot2 θ = csc2 θ Rewrite the left side. Write both terms with the common denominator. cos2 θ  1 + cot2 θ =  1 + _____ sin2 θ cos2 θ sin2 θ   +  _____ _____ sin2 θ sin2 θ sin2 θ + cos2 θ ___________ sin2 θ =  = = 1 ____ sin2 θ = csc2 θ Similarly, 1 + tan2 θ = sec2 θ can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives Rewrite left side. Write both terms with the common denominator. 2 sin θ 1 + tan2 θ = 1 +   ____ cos θ 2 2 cos θ sin θ   ____ ____ cos θ cos θ cos2 θ + sin2 θ ___________ cos2 θ +  =  = = 1 _____ cos2 θ = sec2 θ The next set of fundamental identities is the set of even-odd identities. The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle and determine whether the identity is odd or even. (See Table 2). tan(−θ) = −tan θ cot(−θ) = −cot θ Even-Odd Identities sin(−θ) = −sin θ csc(−θ) = −csc θ Table 2 cos(−θ) = cos θ sec(−θ) = sec θ Recall that an odd function is one in which f (−x) = −f (x) for all x in the domain of f. The sine function is an odd function because sin(−θ) = −sin θ. The graph of an odd function is symmetric about the origin. For example, consider  is opposite the output of sin  − π π π π __ __ __ __ and −  . Thus, . The output of sin  corresponding inputs of 2 2 2 2 π __  = 1 sin  2 π sin  − π __ __  = −sin   2 2 = −1 and This is shown in Figure 2. y 2 π , 12 f(x) = sin x x π 2π –2π – π , 2 –π –1 –2 Figure 2 Graph of y = sin θ Recall that an even function is one in which f (−x) = f (x) for all x in the domain of f 562 CHAPTER 7 trigonometric identities and eQuations The graph of an even function is symmetric about the y-axis. The cosine function is an even function because π π π __ __ __ cos(−θ) = cos θ. For example, consider corresponding inputs and −  is the same as the . The output of cos  4 4 4 output of cos  − π __  . Thus, 4 π cos  − π __ __  = cos   4 4 ≈ 0.707 See Figure 3. y – π , 4 0.707 2 π , 4 0.707 –2π –π –2 x π 2π f(x) = cos x For all θ in the domain of the sine and cosine functions, respectively, we can state the following: Figure 3 Graph of y = cos θ • Since sin(−θ) = −sin θ, sine is an odd function. • Since, cos(−θ) = cos θ, cosine is an even function. The other even-odd identities follow from the even and odd nature of the sine and cosine functions. For example, consider the tangent identity, tan(−θ) = −tan θ. We can interpret the tangent of a negative angle as = −tan θ. Tangent is therefore an odd function, which means that tan(−θ) = −tan(θ) tan(−θ) = sin(−θ) ______ = cos(−θ) −sin θ ______ cos θ for all θ in the domain of the tangent function. The cotangent identity, cot(−θ) = −cot θ, also follows from the sine and cosine identities. We can interpret the = −cot θ. Cotangent is therefore an odd function, cotangent of a negative angle as cot(−θ) = which means that cot(−θ) = −cot(θ) for all θ in the domain of the cotangent function. cos(−θ) _______ = sin(−θ) cos θ _____ −sin θ The cosecant function is the reciprocal of the sine function, which means that the cosecant of a negative angle will be interpreted as csc(−θ) = = −csc θ. The cosecant function is therefore odd. 1 ______ = sin(−θ) 1 _____ −sin θ Finally, the secant function is the reciprocal of the cosine function, and the secant of a negative angle is interpreted as sec(−θ) = = sec θ. The secant function is therefore even. 1 ______ = cos(−θ) 1 ____ cos θ To sum up, only two of the trigonometric functions, cosine and secant, are even. The other four functions are odd, verifying the even-odd identities. The next set of fundamental identities is the set of reciprocal identities, which, as their name implies, relate trigonometric functions that are reciprocals of each other. See Table 3. Reciprocal Identities 1 ____ csc θ 1 ____ sec θ 1 ____ cot θ csc θ = sec θ = cot θ = 1 ____ sin θ 1 ____ cos θ 1 ____ tan θ sin θ = cos θ = tan θ = The final set of identities is the set of quotient identities, which define relationships among certain trigonometric functions and can be very helpful in verifying other identities. See Table 4. Table 3 Quotient Identities tan θ = sin θ ____ cos θ cot θ = cos θ ____ sin θ Table 4 SECTION 7.1 solving trigonometric eQuations with identities 563 The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions. summarizing trigonometric identities The Pythagorean identities are based on the properties of a right triangle. cos2 θ + sin2 θ = 1 1 + cot2 θ = csc2 θ 1 + tan2 θ = sec2 θ The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle. tan(−θ) = −tan θ cot(−θ) = −cot θ sin(−θ) = −sin θ csc(−θ) = −csc θ cos(−θ) = cos θ sec(−θ) = sec θ The reciprocal identities define reciprocals of the trigonometric functions. sin θ = cos θ = tan θ = csc θ = sec θ = cot θ = 1 ____ csc θ 1 ____ sec θ 1 ____ cot θ 1 ____ sin θ 1 ____ cos θ 1 ____ tan θ The quotient identities define the relationship among the trigonometric functions. tan θ = cot θ = sin θ ____ cos θ cos θ ____ sin θ Example 1 Graphing the Equations of an Identity Graph both sides of the identity cot θ = . In other words, on the graphing calculator, graph y = cot θ and y = 1 ____ tan θ 1 ____ . tan θ Solution See Figure 4. y = cot θ = 1 tan θ y 10 6 2 – 5π 2 – 3π 2 π 2– π 2 3π 2 5π 2 θ –6 –10 Figure 4 564 CHAPTER 7 trigonometric identities and eQuations Analysis We see only one graph because both expressions generate the same image. One is on top of the other. This is a good way to prove any identity. If both expressions give the same graph, then they must be identities. How To… Given a trigonometric identity, verify that it is true. 1. Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build. 2. Look for opportunities to factor expressions, square a binomial, or add fractions. 3. Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions. 4. If these steps do not yield the desired result, try converting all terms to sines and cosines. Example 2 Verifying a Trigonometric Identity Verify tan θ cos θ = sin θ. Solution We will start on the left side, as it is the more complicated side: tan θcos θ =  sin θ ____ cos θ sin θ ____ cos θ  cos θ  cos θ =  = sin θ Analysis This identity was fairly simple to verify, as it only required writing tan θ in terms of sin θ and cos θ. Try It #1 Verify the identity csc θ cos θ tan θ = 1. Example 3 Verifying the Equivalency Using the Even-Odd Identities Verify the foll
owing equivalency using the even-odd identities: (1 + sin x)[1 + sin(−x)] = cos2 x Solution Working on the left side of the equation, we have (1 + sin x)[1 + sin(−x)] = (1 + sin x)(1 − sin x) Example 4 Verifying a Trigonometric Identity Involving sec2 θ = 1 − sin2 x = cos2 x Since sin(−x) = −sin x Difference of squares cos2 x = 1 − sin2 x Verify the identity sec2 θ − 1 ________ sec2 θ = sin2 θ Solution As the left side is more complicated, let’s begin there. sec2 θ − 1 ________ = sec2 θ (tan2 θ + 1) − 1 _____________ sec2 θ sec2 θ = tan2 θ + 1 = tan2 θ _____ sec2 θ 1 ____  = tan2 θ  sec2 θ = tan2 θ(cos2 θ) =  =  = sin2 θ  (cos2 θ)  (cos2 θ) sin2 θ _____ cos2 θ sin2 θ _____ cos2 θ cos2 θ = tan2 θ = 1 ____ sec2 θ sin2 θ _____ cos2 θ SECTION 7.1 solving trigonometric eQuations with identities 565 There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side. sec2 θ _____ sec2 θ = 1 − cos2 θ = sin2 θ sec2 θ − 1 ________ = sec2 θ − 1 ____ sec2 θ In the first method, we used the identity sec2 θ = tan2 θ + 1 and continued to simplify. In the second method, Analysis we split the fraction, putting both terms in the numerator over the common denominator. This problem illustrates that there are multiple ways we can verify an identity. Employing some creativity can sometimes simplify a procedure. As long as the substitutions are correct, the answer will be the same. Try It #2 Show that cot θ ____ csc θ = cos θ. Example 5 Creating and Verifying an Identity Create an identity for the expression 2tan θ sec θ by rewriting strictly in terms of sine. Solution There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression: 2 tan θ sec θ = 2  sin θ   ____ cos θ 1 ____  cos θ Thus, = 2 sin θ _____ cos2 θ = 2 sin θ ________ 1 − sin2 θ 2tan θ sec θ = 2 sin θ ________ 1 − sin2 θ Substitute 1 − sin2 θ for cos2 θ Example 6 Verifying an Identity Using Algebra and Even/Odd Identities Verify the identity: sin2(−θ) − cos2(−θ) _________________ sin(−θ) − cos(−θ) = cos θ − sin θ Solution Let’s start with the left side and simplify: sin2(−θ) − cos2(−θ) _________________ = sin(−θ) − cos(−θ) [sin(−θ)]2 − [cos(−θ)]2 __________________ sin(−θ) − cos(−θ) sin(−x) = −sin x and cos(−x) = cos x = (−sin θ)2 − (cos θ)2 ________________ −sin θ − cos θ = (sin θ)2 − (cos θ)2 −sin θ − cos θ ______________ Difference of squares = (sin θ − cos θ)(sin θ + cos θ) ______________________ −(sin θ + cos θ) = (sin θ − cos θ)(sin θ + cos θ) ______________________ −(sin θ + cos θ) = cos θ − sin θ Try It #3 Verify the identity sin2 θ − 1 _____________ tan θ sin θ − tan θ = sin θ + 1 _______ . tan θ 566 CHAPTER 7 trigonometric identities and eQuations Example 7 Verifying an Identity Involving Cosines and Cotangents Verify the identity: (1 − cos2 x)(1 + cot2 x) = 1. Solution We will work on the left side of the equation (1 − cos2 x)(1 + cot2 x) = (1 − cos2 x)  1 + sin2 x _____ sin2 x = (1 − cos2 x)  cos2 x  _____ sin2 x cos2 x  _____ sin2 x sin2 x + cos2 x  ___________ sin2 x + = (1 − cos2 x)  1 ____  = (sin2 x)  sin2 x = 1 Find the common denominator. Using Algebra to Simplify Trigonometric expressions We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions before solving. Being familiar with the basic properties and formulas of algebra, such as the difference of squares formula, the perfect square formula, or substitution, will simplify the work involved with trigonometric expressions and equations. For example, the equation (sin x + 1)(sin x − 1) = 0 resembles the equation (x + 1)(x − 1) = 0, which uses the factored form of the difference of squares. Using algebra makes finding a solution straightforward and familiar. We can set each factor equal to zero and solve. This is one example of recognizing algebraic patterns in trigonometric expressions or equations. Another example is the difference of squares formula, a2 − b2 = (a − b)(a + b), which is widely used in many areas other than mathematics, such as engineering, architecture, and physics. We can also create our own identities by continually expanding an expression and making the appropriate substitutions. Using algebraic properties and formulas makes many trigonometric equations easier to understand and solve. Example 8 Writing the Trigonometric Expression as an Algebraic Expression Write the following trigonometric expression as an algebraic expression: 2cos2 θ + cos θ − 1. Solution Notice that the pattern displayed has the same form as a standard quadratic expression, ax2 + bx + c. Letting cos θ = x, we can rewrite the expression as follows: 2x2 + x − 1 This expression can be factored as (2x − 1)(x + 1). If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for x. At this point, we would replace x with cos θ and solve for θ. Example 9 Rewriting a Trigonometric Expression Using the Difference of Squares Rewrite the trigonometric expression: 4 cos2 θ − 1. Solution Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares. Thus, 4 cos2 θ − 1 = (2 cos θ)2 − 1 = (2 cos θ − 1)(2 cos θ + 1) Analysis If this expression were written in the form of an equation set equal to zero, we could solve each factor using the zero factor property. We could also use substitution like we did in the previous problem and let cos θ = x, rewrite the expression as 4x2 − 1, and factor (2x − 1)(2x + 1). Then replace x with cos θ and solve for the angle. Try It #4 Rewrite the trigonometric expression: 25 − 9 sin2 θ. SECTION 7.1 solving trigonometric eQuations with identities 567 Example 10 Simplify by Rewriting and Using Substitution Simplify the expression by rewriting and using identities: Solution We can start with the Pythagorean Identity. Now we can simplify by substituting 1 + cot2 θ for csc2 θ. We have 1 + cot2 θ = csc2 θ csc2 θ − cot2 θ csc2 θ − cot2 θ = 1 + cot2 θ − cot2 θ = 1 Try It #5 Use algebraic techniques to verify the identity: cos θ _______ = 1 + sin θ 1 − sin θ _______ cos θ . (Hint: Multiply the numerator and denominator on the left side by 1 − sin θ.) Access these online resources for additional instruction and practice with the fundamental trigonometric identities. • Fundamental Trigonometric Identities (http://openstaxcollege.org/l/funtrigiden) • Verifying Trigonometric Identities (http://openstaxcollege.org/l/verifytrigiden) 568 CHAPTER 7 trigonometric identities and eQuations 7.1 SeCTIOn exeRCISeS VeRBAl 1. We know g(x) = cos x is an even function, and f (x) = sin x and h(x) = tan x are odd functions. What about G(x) = cos2 x, F (x) = sin2 x, and H(x) = tan2 x? Are they even, odd, or neither? Why? 2. Examine the graph of f (x) = sec x on the interval [−π, π]. How can we tell whether the function is even or odd by only observing the graph of f (x) = sec x? 3. After examining the reciprocal identity for sec t, explain why the function is undefined at certain points. 4. All of the Pythagorean identities are related. Describe how to manipulate the equations to get from sin2 t + cos2 t = 1 to the other forms. AlGeBRAIC For the following exercises, use the fundamental identities to fully simplify the expression. 5. sin x cos x sec x 7. tan x sin x + sec x cos2 x 9. cot t + tan t __________ sec(−t) 11. −tan(−x)cot(−x) 6. sin(−x)cos(−x)csc(−x) 8. csc x + cos x cot(−x) 10. 3 sin3 t csc t + cos2 t + 2 cos(−t)cos t 12. −sin(−x)cos x sec x csc x tan x ________________________ cot x + sin2 θ + 1 ____ sec2 θ 13. 1 + tan2 θ ________ csc2 θ 15. 1 − cos2 x ________ tan2 x + 2 sin2 x 14.  tan x ____ csc2 x + tan x ____   sec2 x 1 + tan x ________  − 1 + cot x 1 _____ cos2 x For the following exercises, simplify the first trigonometric expression by writing the simplified form in terms of the second expression. 16. tan x + cot x __________ csc x ; cos x 17. sec x + csc x __________ ; sin x 1 + tan x 18. cos x _______ 1 + sin x + tan x; cos x 19. 1 ________ sin x cos x − cot x; cot x 20. 1 _______ − 1 − cos x cos x _______ 1 + cos x ; csc x 21. (sec x + csc x)(sin x + cos x) − 2 − cot x; tan x 22. 1 __________ csc x − sin x ; sec x and tan x 23. 1 − sin x _______ − 1 + sin x 1 + sin x _______ 1 − sin x ; sec x and tan x 24. tan x; sec x 26. sec x; sin x 28. cot x; csc x 25. sec x; cot x 27. cot x; sin x For the following exercises, verify the identity. 29. cos x − cos3 x = cos x sin2 x 30. cos x(tan x − sec(−x)) = sin x − 1 31. 1 + sin2 x ________ = cos2 x 1 _____ cos2 x + sin2 x _____ cos2 x = 1 + 2 tan2 x 33. cos2 x − tan2 x = 2 − sin2 x − sec2 x 32. (sin x + cos x)2 = 1 + 2 sin x cos x SECTION 7.1 section exercises 569 exTenSIOnS For the following exercises, prove or disprove the identity. 34. 1 _______ − 1 + cos x 1 ___________ 1 − cos( − x) = −2 cot x csc x 35. csc2 x(1 + sin2 x) = cot2 x 36.  sec2(−x) − tan2 x   ______________ tan x 2 + 2 tan x _________ 2 + 2 cot x  − 2 sin2 x = cos 2x 37. tan x ____ sec x sin(−x) = cos2 x 38. sec(−x) __________ tan x + cot x = −sin(−x) 39. 1 + sin x _______ = cos x cos x _________ 1 + sin(−x) For the following exercises, determine whether the identity is true or false. If false, find an appropriate equivalent expression. 40. cos2 θ − sin2 θ ___________ 1 − tan2 θ = sin2 θ 42. sec θ + tan θ __________ cot θ + cos θ = sec2 θ 41. 3 sin2 θ + 4 cos2 θ = 3 + cos2 θ 570 CHAPTER 7 trigonometric identities and eQuations leARnInG OBjeCTIVeS In this section, you will: • • • • • Use sum and difference formulas for cosine. Use sum and difference formulas for sine. Use sum and difference formulas for tangent. Use sum and difference formulas for cofunctions. Use sum and difference formulas to verify identities. 7. 2 SUM AnD DIFFeRenCe IDenTITIeS Figure 1 Mount McKinley, in Denali National Park, Alas
ka, rises 20,237 feet (6,168 m) above sea level. It is the highest peak in North America. (credit: Daniel A. Leifheit, Flickr) How can the height of a mountain be measured? What about the distance from Earth to the sun? Like many seemingly impossible problems, we rely on mathematical formulas to find the answers. The trigonometric identities, commonly used in mathematical proofs, have had real-world applications for centuries, including their use in calculating long distances. The trigonometric identities we will examine in this section can be traced to a Persian astronomer who lived around 950 AD, but the ancient Greeks discovered these same formulas much earlier and stated them in terms of chords. These are special equations or postulates, true for all values input to the equations, and with innumerable applications. In this section, we will learn techniques that will enable us to solve problems such as the ones presented above. The formulas that follow will simplify many trigonometric expressions and equations. Keep in mind that, throughout this section, the term formula is used synonymously with the word identity. Using the Sum and Difference Formulas for Cosine Finding the exact value of the sine, cosine, or tangent of an angle is often easier if we can rewrite the given angle in terms of two angles that have known trigonometric values. We can use the special angles, which we can review in the unit circle shown in Figure 2. , , 1 2 135° 3π 4 120° 2π 3 90° (0, 1) π 2 60° π 3 , 1 2 150° 5π 6 (–1, 0) 180° π , 1 2 , 210° 7π 6 225° 5π 4 , 1 2 , 45° π 4 π 6 30° , 1 2 0°, (1, 0) 330° , 1 2 2π 11π 6 7π 4 315° , ,1 2 3π 2 4π 3 5π 3 240° 270° (0, –1) Figure 2 The Unit Circle 300° ,1 2 SECTION 7.2 sum and diFFerence identities 571 We will begin with the sum and difference formulas for cosine, so that we can find the cosine of a given angle if we can break it up into the sum or difference of two of the special angles. See Table 1. Sum formula for cosine Difference formula for cosine cos(α + β) = cos α cos β − sin α sin β cos(α − β) = cos α cos β + sin α sin β Table 1 First, we will prove the difference formula for cosines. Let’s consider two points on the unit circle. See Figure 3. Point P is at an angle α from the positive x-axis with coordinates (cos α, sin α) and point Q is at an angle of β from the positive x-axis with coordinates (cos β, sin β). Note the measure of angle POQ is α − β. Label two more points: A at an angle of (α − β) from the positive x-axis with coordinates (cos(α − β), sin(α − β)); and point B with coordinates (1, 0). Triangle POQ is a rotation of triangle AOB and thus the distance from P to Q is the same as the distance from A to B. y 2 π , 12 f(x) = sin x x π 2π –2π – π , 2 –π –1 –2 Figure 3 We can find the distance from P to Q using the distance formula. dPQ = √ = √ (cos α − cos β)2 + (sin α − sin β)2 cos2 α − 2 cos α cos β + cos2 β + sin2 α − 2 sin α sin β + sin2 β ———— —— Then we apply the Pythagorean Identity and simplify. = √ = √ = √ ———— —— (cos2 α + sin2 α) + (cos2 β + sin2 β) − 2 cos α cos β − 2 sin α sin β 1 + 1 − 2 cos α cos β − 2 sin α sin β 2 − 2 cos α cos β − 2 sin α sin β —— Similarly, using the distance formula we can find the distance from A to B. —— dAB = √ = √ (cos(α − β) − 1)2 + (sin(α − β) − 0)2 cos2(α − β) − 2 cos(α − β) + 1 + sin2(α − β) ——— Applying the Pythagorean Identity and simplifying we get: = √ = √ = √ ——— —— (cos2(α − β) + sin2(α − β)) − 2 cos(α − β) + 1 1 − 2 cos(α − β) + 1 2 − 2 cos(α − β) — Because the two distances are the same, we set them equal to each other and simplify. √ —— 2 − 2 cos α cos β − 2 sin α sin β = √ 2 − 2 cos α cos β − 2 sin α sin β = 2 − 2 cos(α − β) 2 − 2 cos(α − β) — Finally we subtract 2 from both sides and divide both sides by −2. cos α cos β + sin α sin β = cos(α − β) Thus, we have the difference formula for cosine. We can use similar methods to derive the cosine of the sum of two angles. 572 CHAPTER 7 trigonometric identities and eQuations sum and difference formulas for cosine These formulas can be used to calculate the cosine of sums and differences of angles. cos(α + β) = cos α cos β − sin α sin β cos(α − β) = cos α cos β + sin α sin β How To… Given two angles, find the cosine of the difference between the angles. 1. Write the difference formula for cosine. 2. Substitute the values of the given angles into the formula. 3. Simplify. Finding the Exact Value Using the Formula for the Cosine of the Difference of Two Angles Example 1 Using the formula for the cosine of the difference of two angles, find the exact value of cos  Solution Use the formula for the cosine of the difference of two angles. We have 5π __ 4 π __  . − 6 cos(α − β) = cos α cos β + sin α sin β cos  5π ___ 4 π __  = cos  − 6 — — — π 5π π 5π __ ___ __ ___  + sin   cos   sin   √ __ ____ ____ ____  2 2 2 2 — — 2 6 √ √ ____ ____ 4 4 — — 2 6 − √ − √ __________ 4 = − − = Try It #1 π __ π __ −  . Find the exact value of cos  4 3 Example 2 Finding the Exact Value Using the Formula for the Sum of Two Angles for Cosine Find the exact value of cos(75°). Solution As 75° = 45° + 30°, we can evaluate cos(75°) as cos(45° + 30°). Thus, cos(45° + 30°) = cos(45°)cos(30°) − sin(45°)sin(30°) = = = — — — — 2 3 2  −  √ √ 1 √ __ ____ ____ ____   2 2 2 2 — 2 6 √ √ ____ ____ 4 4 — — 6 − √ 2 √ __________ 4 − Try It #2 Find the exact value of cos(105°). Using the Sum and Difference Formulas for Sine The sum and difference formulas for sine can be derived in the same manner as those for cosine, and they resemble the cosine formulas. SECTION 7.2 sum and diFFerence identities 573 sum and difference formulas for sine These formulas can be used to calculate the sines of sums and differences of angles. sin(α + β) = sin α cos β + cos α sin β sin(α − β) = sin α cos β − cos α sin β How To… Given two angles, find the sine of the difference between the angles. 1. Write the difference formula for sine. 2. Substitute the given angles into the formula. 3. Simplify. Example 3 Using Sum and Difference Identities to Evaluate the Difference of Angles Use the sum and difference identities to evaluate the difference of the angles and show that part a equals part b. a. sin(45° − 30°) b. sin(135° − 120°) Solution a. Let’s begin by writing the formula and substitute the given angles. sin(α − β) = sin α cos β − cos α sin β sin(45° − 30°) = sin(45°)cos(30°) − cos(45°)sin(30°) Next, we need to find the values of the trigonometric expressions. sin(45°) = , cos(30°) = , cos(45°) = — 2 √ ____ 2 — 3 √ ____ 2 — 2 , sin(30°) = 1 √ __ ____ 2 2 Now we can substitute these values into the equation and simplify. 2 3 2 √  −  1 √ √ __ ____ ____ ____   2 2 2 2 — 6 − √ 2 √ _________ 4 sin(45° − 30°) = = — — — — b. Again, we write the formula and substitute the given angles. sin(α − β) = sin α cos β − cos α sin β sin(135° − 120°) = sin(135°)cos(120°) − cos(135°)sin(120°) Next, we find the values of the trigonometric expressions. sin(135°) = — 2 √ , cos(120°) = − 1 __ ____ , cos(135°) = − 2 2 2 √ ____ 2 — , sin(120°) = — 3 √ ____ 2 Now we can substitute these values into the equation and simplify. sin(135° − 120°) = sin(135° − 120°) = — — — — — — — ____ ____ __ ____ 2 2 2 2 2 + √ 6 − √ __________ 4 6 − √ √ _________ ____ ____ __ ____ 2 2 2 2 2 + √ 6 − √ __________ 4 6 − √ √ _________ 4 2 — — — — — — — = = = = 574 CHAPTER 7 trigonometric identities and eQuations Finding the Exact Value of an Expression Involving an Inverse Trigonometric Function Example 4 3 1 __ __ Find the exact value of sin  cos−1  . + sin−1 5 2 3 1 __ __ and β = sin−1 Solution The pattern displayed in this problem is sin(α + β). Let α = cos−1 . Then we can write 5 2 cos α = 1 __ , 0 ≤ α ≤ π 2 π π sin β = 3 __ __ __ ≤ β ≤ , − 2 2 5 We will use the Pythagorean identities to find sin α and cos β. sin α = √ — = √ = √ 1 − cos2 α ______ 1 − 1 __ 4 __ 3 __ 4 — 3 √ _ 2 = cos β = √ — 1 − sin2 β ______ 9 __ 1 − 25 ___ 16 __ 25 = √ = √ = 4 __ 5 Using the sum formula for sine, 3 1 __ __  = sin(α + β) sin  cos−1 + sin−1 5 2 — = = sin α cos β + cos α sin β 3 4 √ 3 + 1 ____ __ __ __ · · 2 5 2 5 — 3 + 3 4 √ ________ 10 = Using the Sum and Difference Formulas for Tangent Finding exact values for the tangent of the sum or difference of two angles is a little more complicated, but again, it is a matter of recognizing the pattern. Finding the sum of two angles formula for tangent involves taking quotient of the sum formulas for sine and cosine and simplifying. Recall, tan x = , cos x ≠ 0. sin x ____ cos x Let’s derive the sum formula for tangent. tan(α + β) = sin(α + β) _ cos(α + β) = sin α cos β + cos α sin β __________________ cos α cos β − sin α sin β = sin α cos β + cos α sin β __________________ cos α cos β cos α cos β − sin α sin β __________________ cos α cos β = sin α cos β ________ + cos α cos β cos α cos β _________ − cos α cos β cos α sin β ________ cos α cos β sin α sin β ________ cos α cos β Divide the numerator and denominator by cos α cos β = = + sin α ____ cos α 1 − sin β ____ cos β sin α sin β ________ cos α cos β tan α + tan β ____________ 1 − tan α tan β We can derive the difference formula for tangent in a similar way. SECTION 7.2 sum and diFFerence identities 575 sum and difference formulas for tangent The sum and difference formulas for tangent are: tan(α + β) = tan α + tan β ___________ 1 − tan α tan β tan(α − β) = tan α − tan β ___________ 1 + tan α tan β How To… Given two angles, find the tangent of the sum of the angles. 1. Write the sum formula for tangent. 2. Substitute the given angles into the formula. 3. Simplify. Example 5 Finding the Exact Value of an Expression Involving Tangent π π __ __ +  . Find the exact value of tan  4 6 Solution Let’s first write the sum formula for tangent and substitute the given angles into the formula. tan(α + β) = tan α + tan β __ 1 − tan α tan β π π __ __  = + tan  4 6 π π __ __  + tan   tan  4 6 ___ π π __ __ 1 −  tan     tan  4 6 Next, we determine the individual tangents within th
e formulas: So we have π __  = tan  6 1 _ — 3 √ π __  = 1 tan  4 π π __ __  = + tan  4 6 1 _ + 1 — 3 √ __ 1 −  1 _  (1 √ — Try It #3 Find the exact value of tan  2π __ 3 π __  . + 4 Finding Multiple Sums and Differences of Angles Example 6 π Given sin α = 3 __ __ , cos . sin(α + β) b. cos(α + β) , π < β < 5 __ 13 c. tan(α + β) 3π __ , find 2 d. tan(α − β) 576 CHAPTER 7 trigonometric identities and eQuations Solution We can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas. π a. To find sin(α + β), we begin with sin α = 3 __ __ and 0 < α < . The side opposite α has length 3, the hypotenuse 2 5 has length 5, and α is in the first quadrant. See Figure 4. Using the Pythagorean Theorem,we can find the length of side a: a2 + 32 = 52 a2 = 16 a = 4 y (4, 3) x (4, 0) 5 α Figure 4 3π __ Since cos β = − 2 quadrant. See Figure 5. Again, using the Pythagorean Theorem, we have and π < β < 5 __ 13 , the side adjacent to β is −5, the hypotenuse is 13, and β is in the third (−5)2 + a2 = 132 25 + a2 = 169 a2 = 144 a = ±12 Since β is in the third quadrant, a = –12. (–5, 0) y x β 13 (–5, –12) Figure 5 The next step is finding the cosine of α and the sine of β. The cosine of α is the adjacent side over the hypotenuse. We can find it from the triangle in Figure 5: cos α = 4 __ . We can also find the sine of β from the triangle in Figure 5, 5 12 __ . Now we are ready to evaluate sin(α + β). as opposite side over the hypotenuse: sin β = − 13 sin(α + β) = sin αcos β + cos αsin β   − 12 4 5 __ __ __   +  13 13 5 48 __ 65 = − 3 __   − =  5 15 __ 65 63 __ 65 = − − SECTION 7.2 sum and diFFerence identities 577 b. We can find cos(α + β) in a similar manner. We substitute the values according to the formula. cos(α + β) = cos α cos β − sin α sin β   − 12 3 __ __   −  13 5 4 __   − =  5 5 __ 13 = − 20 __ 65 + 36 __ 65 = 16 __ 65 and cos α = 4 c. For tan(α + β), if sin α = 3 __ __ , then 5 5 tan α = 3 __ 5 = 3 __ _ 4 4 __ 5 If sin β = − and cos β = − , then 12 __ 13 5 __ 13 Then, tan β = −12 _ 13 _ −5 _ 13 = 12 __ 5 tan(α + β) = tan α + tan β __ 1 − tan α tan β = 12 3 __ __ + 5 4 __ 12 1 − 3 __ __   5 4 = 63 __ 20 _ − 16 __ 20 = − 63 __ 16 d. To find tan(α − β), we have the values we need. We can substitute them in and evaluate. tan(α − β) = tan α − tan β __ 1 + tan α tan β = = 12 3 __ __ − 5 4 __ 1 + 3 12 __ __   5 4 − 33 __ 20 _ 56 __ 20 = − 33 _ 56 Analysis β are angles in the same triangle, which of course, they are not. Also note that A common mistake when addressing problems such as this one is that we may be tempted to think that α and tan(α + β) = sin(α + β) ________ cos(α + β) 578 CHAPTER 7 trigonometric identities and eQuations Using Sum and Difference Formulas for Cofunctions Now that we can find the sine, cosine, and tangent functions for the sums and differences of angles, we can use them to do the same for their cofunctions. You may recall from Right Triangle Trigonometry that, if the sum of two π π __ __ , , those two angles are complements, and the sum of the two acute angles in a right triangle is positive angles is 2 2 so they are also complements. In Figure 6, notice that if one of the acute angles is labeled as θ, then the other acute π __ − θ  . angle must be labeled  2 π __ − θ  : opposite over hypotenuse. Thus, when two angles are complimentary, we can Notice also that sin θ = cos  2 say that the sine of θ equals the cofunction of the complement of θ. Similarly, tangent and cotangent are cofunctions, and secant and cosecant are cofunctions. π – θ 2 θ Figure 6 From these relationships, the cofunction identities are formed. cofunction identities The cofunction identities are summarized in Table 2. π __ − θ  sin θ = cos  2 π __ − θ  sec θ = csc  2 π __ − θ  cos θ = sin  2 π __ − θ  csc θ = sec  2 Table 2 π __ − θ  tan θ = cot  2 π __ − θ  cot θ = tan  2 Notice that the formulas in the table may also be justified algebraically using the sum and difference formulas. For example, using we can write cos(α − β) = cos αcos β + sin αsin β, π π π __ __ __ cos θ + sin − θ  = cos cos  sin θ 2 2 2 = (0)cos θ + (1)sin θ = sin θ Example 7 Finding a Cofunction with the Same Value as the Given Expression π __ Write tan in terms of its cofunction. 9 π __ − θ  . Thus, Solution The cofunction of tan θ = cot  2 π π π __ __ __ −  = cot   tan  9 2 9 = cot  9π ___ 18 − 2π ___  18 = cot  7π ___  18 SECTION 7.2 sum and diFFerence identities 579 Try It #4 π __ in terms of its cofunction. Write sin 7 Using the Sum and Difference Formulas to Verify Identities Verifying an identity means demonstrating that the equation holds for all values of the variable. It helps to be very familiar with the identities or to have a list of them accessible while working the problems. Reviewing the general rules from Solving Trigonometric Equations with Identities may help simplify the process of verifying an identity. How To… Given an identity, verify using sum and difference formulas. 1. Begin with the expression on the side of the equal sign that appears most complex. Rewrite that expression until it matches the other side of the equal sign. Occasionally, we might have to alter both sides, but working on only one side is the most efficient. 2. Look for opportunities to use the sum and difference formulas. 3. Rewrite sums or differences of quotients as single quotients. 4. If the process becomes cumbersome, rewrite the expression in terms of sines and cosines. Example 8 Verifying an Identity Involving Sine Verify the identity sin(α + β) + sin(α − β) = 2 sin α cos β. Solution We see that the left side of the equation includes the sines of the sum and the difference of angles. sin(α + β) = sin α cos β + cos α sin β sin(α − β) = sin α cos β − cos α sin β We can rewrite each using the sum and difference formulas. sin(α + β) + sin(α − β) = sin α cos β + cos α sin β + sin α cos β − cos α sin β We see that the identity is verified. = 2 sin α cos β Example 9 Verifying an Identity Involving Tangent Verify the following identity. sin(α − β) _ cos α cos β = tan α − tan β Solution We can begin by rewriting the numerator on the left side of the equation. sin(α − β) _ = cos α cos β sin α cos β − cos α sin β __ cos α cos β = sin α cos β _ − cos α cos β cos α sin β _ cos α cos β = sin α _ cos α − sin β _ cos β Rewrite using a common denominator. Cancel. We see that the identity is verified. In many cases, verifying tangent identities can successfully be accomplished by writing the tangent in terms of sine and cosine. = tan α − tan β Rewrite in terms of tangent. 580 CHAPTER 7 trigonometric identities and eQuations Try It #5 Verify the identity: tan(π − θ) = −tan θ. Example 10 Using Sum and Difference Formulas to Solve an Application Problem Let L1 and L2 denote two non-vertical intersecting lines, and let θ denote the acute angle between L1 and L2. See Figure 7. Show that tan θ = m2 − m1 _ 1 + m1 m2 where m1 and m2 are the slopes of L1 and L2 respectively. (Hint: Use the fact that tan θ1 = m1 and tan θ2 = m2.) y L1 L2 θ θ1 θ2 x Figure 7 Solution Using the difference formula for tangent, this problem does not seem as daunting as it might. tan θ = tan(θ2 − θ1) = = tan θ2 − tan θ1 __ 1 + tan θ1 tan θ2 m2 − m1 _ 1 + m1m2 Example 11 Investigating a Guy-wire Problem For a climbing wall, a guy-wire R is attached 47 feet high on a vertical pole. Added support is provided by another guy-wire S attached 40 feet above ground on the same pole. If the wires are attached to the ground 50 feet from the pole, find the angle α between the wires. See Figure 8. 47 ft 40 ft R S α β 50 ft Figure 8 Solution Let’s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that tan β = then use difference formula for tangent. 47 __ 50 , and tan( β − α) = 40 __ 50 = 4 __ . We can 5 tan(β − α) = tan β − tan α __ 1 + tan β tan α SECTION 7.2 sum and diFFerence identities 581 Now, substituting the values we know into the formula, we have 4 __ = 5 47 __ − tan α 50 __ 47 ___ 1 + 50 tan α 4  1 + tan α  = 5  47 __ 50 47 __ 50 − tan α  Use the distributive property, and then simplify the functions. 4(1) + 4   tan α = 5  47 __ 50 47 __ 50  − 5tan α 4 + 3.76tan α = 4.7 − 5tan α 5tan α + 3.76tan α = 0.7 8.76tan α = 0.7 tan α ≈ 0.07991 tan−1(0.07991) ≈ .079741 Now we can calculate the angle in degrees. α ≈ 0.079741  180 ___ π  ≈ 4.57° Analysis Occasionally, when an application appears that includes a right triangle, we may think that solving is a matter of applying the Pythagorean Theorem. That may be partially true, but it depends on what the problem is asking and what information is given. Access these online resources for additional instruction and practice with sum and difference identities. • Sum and Difference Identities for Cosine (http://openstaxcollege.org/l/sumdifcos) • Sum and Difference Identities for Sine (http://openstaxcollege.org/l/sumdifsin) • Sum and Difference Identities for Tangent (http://openstaxcollege.org/l/sumdiftan) 582 CHAPTER 7 trigonometric identities and eQuations 7.2 SeCTIOn exeRCISeS VeRBAl 1. Explain the basis for the cofunction identities and when they apply. 3. Explain to someone who has forgotten the even-odd properties of sinusoidal functions how the addition and subtraction formulas can determine this characteristic for f(x) = sin(x) and g(x) = cos(x). (Hint: 0 − x = −x) AlGeBRAIC For the following exercises, find the exact value. 4. cos  7π ___  12 8. tan  − π __  12 5. cos  π __  12 9. tan  19π ___  12 2. Is there only one way to evaluate cos  5π ___  ? Explain 4 how to set up the solution in two different ways, and then compute to make sure they give the same answe
r. 6. sin  5π ___  12 7. sin  11π ___  12 For the following exercises, rewrite in terms of sin x and cos x. 10. sin  x + 11π ___  6 11. sin  x − 3π ___  4 12. cos  x − 5π ___  6 13. cos  x + 2π ___  3 For the following exercises, simplify the given expression. π __ − t  14. csc  2 π __ − θ  15. sec  2 π __ − x  16. cot  2 π __ − x  17. tan  2 18. sin(2x) cos(5x) − sin(5x) cos(2x) 19. 7 3 __ __ x  − tan  tan  x  5 2 7 3 __ __ x  tan  1 + tan  x  5 2 For the following exercises, find the requested information. and cos b = − 1 20. Given that sin a = 2 π __ __ __ , π  , find sin(a + b) and cos(a − b). , with a and b both in the interval  2 4 3 , and cos b = 1 21. Given that sin a = 4 π __ __ __  , find sin(a − b) and cos(a + b). , with a and b both in the interval  0, 2 3 5 For the following exercises, find the exact value of each expression. 1 __   22. sin  cos−1(0) − cos−1  2 1 1 __ __    − cos−1  24. tan  sin−1  2 2 3 2    + sin− 1  23. cos  cos−1  √ √ ____ ____ 2 2 — — SECTION 7.2 section exercises 583 GRAPHICAl For the following exercises, simplify the expression, and then graph both expressions as functions to verify the graphs are identical. π __ − x  25. cos  2 π __ − x  29. tan  4 26. sin(π − x) 30. cos  7π ___ + x  6 π __ + x  27. tan  3 π __ + x  31. sin  4 π __ + x  28. sin  3 32. cos  5π ___ + x  4 For the following exercises, use a graph to determine whether the functions are the same or different. If they are the same, show why. If they are different, replace the second function with one that is identical to the first. (Hint: think 2x = x + x. ) 33. f(x) = sin(4x) − sin(3x)cos x, g(x) = sin x cos(3x) 34. f(x) = cos(4x) + sin x sin(3x), g(x) = −cos x cos(3x) 35. f(x) = sin(3x)cos(6x), g(x) = −sin(3x)cos(6x) 36. f(x) = sin(4x), g(x) = sin(5x)cos x − cos(5x)sin x 37. f(x) = sin(2x), g(x) = 2 sin x cos x 38. f(θ) = cos(2θ), g(θ) = cos2 θ − sin2 θ 39. f(θ) = tan(2θ), g(θ) = tan θ _ 1 + tan2θ 41. f(x) = tan(−x), g(x) = tan x − tan(2x) __ 1 − tan x tan(2x) TeCHnOlOGY 40. f(x) = sin(3x)sin x, g(x) = sin2(2x)cos2 x − cos2(2x)sin2 x For the following exercises, find the exact value algebraically, and then confirm the answer with a calculator to the fourth decimal point. 43. sin(195°) 44. cos(165°) 45. cos(345°) 42. sin(75°) 46. tan(−15°) exTenSIOnS For the following exercises, prove the identities provided. π __  = 47. tan  x + 4 tan x + 1 ________ 1 − tan x 49. cos(a + b) ________ cos a cos b = 1 − tan a tan b 48. tan(a + b) ________ = tan(a − b) sin a cos a + sin b cos b __________________ sin a cos a − sin b cos b 50. cos(x + y)cos(x − y) = cos2 x − sin2 y 51. cos(x + h) − cos x ______________ h = cos x cos h − 1 _ h − sin x sin h _ h For the following exercises, prove or disprove the statements. 52. tan(u + v) = tan u + tan v ___________ 1 − tan u tan v 53. tan(u − v) = tan u − tan v ___________ 1 + tan u tan v 54. tan(x + y) __ = 1 + tan x tan x tan x + tan y __ 1 − tan2 x tan2 y 55. If α, β, and γ are angles in the same triangle, then prove or disprove sin(α + β) = sin γ. 56. If α, β, and γ are angles in the same triangle, then prove or disprove: tan α + tan β + tan γ = tan α tan β tan γ. 584 CHAPTER 7 trigonometric identities and eQuations leARnInG OBjeCTIVeS In this section, you will: • • • • Use double-angle formulas to find exact values. Use double-angle formulas to verify identities. Use reduction formulas to simplify an expression. Use half-angle formulas to find exact values. 7. 3 DOUBle-AnGle, HAlF-AnGle, AnD ReDUCTIOn FORMUlAS Figure 1 Bicycle ramps for advanced riders have a steeper incline than those designed for novices. Bicycle ramps made for competition (see Figure 1) must vary in height depending on the skill level of the competitors. 5 __ For advanced competitors, the angle formed by the ramp and the ground should be θ such that tan θ = . The angle 3 is divided in half for novices. What is the steepness of the ramp for novices ? In this section, we will investigate three additional categories of identities that we can use to answer questions such as this one. Using Double-Angle Formulas to Find exact Values In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where α = β. Deriving the double-angle formula for sine begins with the sum formula, If we let α = β = θ, then we have sin(α + β) = sin α cos β + cos α sin β sin(θ + θ) = sin θ cos θ + cos θ sin θ sin(2θ) = 2sin θ cos θ Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, cos(α + β) = cos α cos β − sin α sin β, and letting α = β = θ, we have cos(θ + θ) = cos θ cos θ − sin θ sin θ cos(2θ) = cos2 θ − sin2 θ Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is: The second interpretation is: cos(2θ) = cos2 θ − sin2 θ = (1 − sin2 θ) − sin2 θ = 1 − 2sin2 θ cos(2θ) = cos2 θ − sin2 θ = cos2 θ − (1 − cos2 θ) = 2 cos2 θ − 1 SECTION 7.3 douBle-angle, halF-angle, and reduction Formulas 585 Similarly, to derive the double-angle formula for tangent, replacing α = β = θ in the sum formula gives tan(α + β) = tan α + tan β ____________ 1 − tan α tan β tan(θ + θ) = tan θ + tan θ ___________ 1 − tan θ tan θ tan(2θ) = 2tan θ ________ 1 − tan2 θ double-angle formulas The double-angle formulas are summarized as follows: sin(2θ) = 2 sin θ cos θ cos(2θ) = cos2 θ − sin2 θ = 1 − 2 sin2 θ = 2 cos2 θ − 1 tan(2θ) = 2 tan θ ________ 1 − tan2 θ How To… Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value. 1. Draw a triangle to reflect the given information. 2. Determine the correct double-angle formula. 3. Substitute values into the formula based on the triangle. 4. Simplify. Example 1 Using a Double-Angle Formula to Find the Exact Value Involving Tangent Given that tan θ = − 3 __ 4 and θ is in quadrant II, find the following: a. sin(2θ) b. cos(2θ) c. tan(2θ) Solution If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given tan θ = − 3 __ , such that θ is in quadrant II. The tangent of an angle is equal to the opposite 4 side over the adjacent side, and because θ is in the second quadrant, the adjacent side is on the x-axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse: (−4)2 + (3)2 = c2 16 + 9 = c2 25 = c2 c = 5 Now we can draw a triangle similar to the one shown in Figure 2. y (–4, 3) (–4, 0) 5 θ Figure 2 x 586 CHAPTER 7 trigonometric identities and eQuations a. Let’s begin by writing the double-angle formula for sine. sin(2θ) = 2 sin θ cos θ We see that we to need to find sin θ and cos θ. Based on Figure 2, we see that the hypotenuse equals 5, so sin θ = 3 __ 5 Thus, . Substitute these values into the equation, and simplify. , and cos θ = − 4 __ 5 4 3 __ __   − sin(2θ) = 2   5 5 = − 24 __ 25 b. Write the double-angle formula for cosine. Again, substitute the values of the sine and cosine into the equation, and simplify. cos(2θ) = cos2 θ − sin2 θ 2 2 3 4 __ __ −   cos(2θ) =  −  5 5 9 16 __ __ − 25 25 7 __ 25 = = c. Write the double-angle formula for tangent. In this formula, we need the tangent, which we were given as tan θ = − 3 __ 4 equation, and simplify. . Substitute this value into the tan(2θ) = 2 tan θ ________ 1 − tan2 θ tan(2θ) = 2 = 3 __  2  − 4 __ 3 __  1 −  − 4 3 __ − 2 _ 9 __ 1 − 16 16 3 __ __   = − 7 2 24 __ 7 = − Try It #1 5 __ , with θ in quadrant I, find cos(2α). Given sin α = 8 Example 2 Using the Double-Angle Formula for Cosine without Exact Values Use the double-angle formula for cosine to write cos(6x) in terms of cos(3x). Solution cos(6x) = cos(2(3x)) = cos2 3x − sin2 3x = 2cos2 3x − 1 Analysis This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function. Using Double-Angle Formulas to Verify Identities Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side. Example 3 Using the Double-Angle Formulas to Establish an Identity Establish the following identity using double-angle formulas: 1 + sin(2θ) = (sin θ + cos θ)2 SECTION 7.3 douBle-angle, halF-angle, and reduction Formulas 587 Solution We will work on the right side of the equal sign and rewrite the expression until it matches the left side. (sin θ + cos θ)2 = sin2 θ + 2 sin θ cos θ + cos2 θ = (sin2 θ + cos2 θ) + 2 sin θ cos θ = 1 + 2 sin θ cos θ = 1 + sin(2θ) Analysis This process is not complicated, as long as we recall the perfect square formula from algebra: (a ± b)2 = a2 ± 2ab + b2 where a = sin θ and b = cos θ. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent. Try It #2 Establish the identity: cos4 θ − sin4 θ = cos(2θ). Example 4 Verifying a Double-Angle Identity for Tangent Verify the identity: tan(2θ) = 2 __________ cot θ − tan θ Solution In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation. tan(2θ) = 2 tan θ ________ 1 − tan2 θ Double-angle formula Multiply by a term that results in desired numerator. = = 1 ____  2 tan θ  tan θ __ 1 ____  (1 − tan2 θ)  tan θ 2 __ tan2 θ 1 _____ ____ tan θ tan θ − Analysis Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been
working backwards to arrive at the equivalency. For example, suppose that we wanted to show = 2 __________ cot θ − tan θ Use reciprocal identity for 1 ____ . tan θ Let’s work on the right side. 2tan θ ________ = 1 − tan2 θ 2 __________ cot θ − tan θ 2 __________ = cot θ − tan θ tan θ  ____ tan θ 2  __ 1 ____ − tan θ tan θ ___ (tan θ) − tan θ(tan θ) 2 tan θ = 1 _ tan θ When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier. = 2 tan θ _ 1 − tan2 θ Try It #3 Verify the identity: cos(2θ)cos θ = cos3 θ − cos θ sin2 θ. 588 CHAPTER 7 trigonometric identities and eQuations Use Reduction Formulas to Simplify an expression The double-angle formulas can be used to derive the reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas. We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with cos(2θ) = 1 − 2 sin2 θ. Solve for sin2 θ: cos(2θ) = 1 − 2 sin2 θ 2 sin2 θ = 1 − cos(2θ) 1 − cos(2θ) _________ 2 Next, we use the formula cos(2θ) = 2 cos2 θ − 1. Solve for cos2 θ: sin2 θ = cos(2θ) = 2 cos2 θ − 1 1 + cos(2θ) = 2 cos2 θ 1 + cos(2θ) _________ 2 = cos2 θ The last reduction formula is derived by writing tangent in terms of sine and cosine: tan2 θ = sin2 θ _____ cos2 θ 1 − cos(2θ) _________ 2 __ 1 + cos(2θ) _________ 2 1 − cos(2θ)   _________ 2 1 − cos(2θ) _________ 1 + cos(2θ) = =  = Substitute the reduction formulas. 2 _________ 1 + cos(2θ)  reduction formulas The reduction formulas are summarized as follows: sin2 θ = 1 − cos(2θ) _________ 2 cos2 θ = 1 + cos(2θ) _________ 2 tan2 θ = 1 − cos(2θ) _________ 1 + cos(2θ) Example 5 Writing an Equivalent Expression Not Containing Powers Greater Than 1 Write an equivalent expression for cos4 x that does not involve any powers of sine or cosine greater than 1. Solution We will apply the reduction formula for cosine twice. 1 + cos(2x) 2  _________ 2 Substitute reduction formula for cos2 x. cos4 x = (cos2 x)2 =  = 1 __ (1 + 2cos(2x) + cos2(2x)) 4 1 + cos2(2x) 1 1 = 1   __________ __ __ __ cos(2x __ __ __ __ + cos(2x) + + cos(4x) 8 2 4 8 1 1 = 3 __ __ __ cos(2x) + + cos(4x) 8 2 8 Analysis The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra. Substitute reduction formula for cos2 x. SECTION 7.3 douBle-angle, halF-angle, and reduction Formulas 589 Example 6 Using the Power-Reducing Formulas to Prove an Identity Use the power-reducing formulas to prove sin3 (2x) =  1 __ sin(2x)  [1 − cos(4x)] 2 Solution We will work on simplifying the left side of the equation: 1 − cos(4x) _________ 2 sin3(2x) = [sin(2x)][sin2(2x)] = sin(2x)  1 __  [1 − cos(4x)] = sin(2x)  2 1 __ = [sin(2x)][1 − cos(4x)] 2  Substitute the power-reduction formula. Analysis Note that in this example, we substituted for sin2(2x). The formula states We let θ = 2x, so 2θ = 4x. 1 − cos(4x) _________ 2 sin2 θ = 1 − cos(2θ) _________ 2 Try It #4 Use the power-reducing formulas to prove that 10 cos4 x = 15 __ 4 5 __ + 5 cos(2x) + cos(4x). 4 Using Half-Angle Formulas to Find exact Values The next set of identities is the set of half-angle formulas, which can be derived from the reduction formulas and we can α __ use when we have an angle that is half the size of a special angle. If we replace θ with , the half-angle formula for sine 2 α __  . Note that the half-angle formulas are preceded by a ± sign. is found by simplifying the equation and solving for sin  2 This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in α __ terminates. which 2 The half-angle formula for sine is derived as follows: sin2 θ = 1 − cos(2θ) _________ 2 α __  = sin2  2 α __ 1 − cos  2 ⋅  2 __ 2 = 1 − cos α ________ 2 _________ 1 − cos α ________ 2 α __ sin  2  = ± √ To derive the half-angle formula for cosine, we have cos2 θ = 1 + cos(2θ) _________ 2 α __  = cos2  2 α __ 1 + cos  2 ⋅  2 2 __ = 1 + cos α ________ 2 _________ 1 + cos α ________ 2 α __ cos  2  = ± √ 590 CHAPTER 7 trigonometric identities and eQuations For the tangent identity, we have tan2 θ = 1 − cos(2θ) _________ 1 + cos(2θ) α __  = tan2  2 =  = ± √ α __ tan  2 α __ 1 − cos  2 ⋅  2 __ α __ 1 + cos  2 ⋅  2 1 − cos α ________ 1 + cos α _________ 1 − cos α _ 1 + cos α half-angle formulas The half-angle formulas are as follows:  = ± √ α __ sin  2  = ± √ α __ cos  2  = ± √ α __ tan  2 _________ 1 − cos α ________ 2 _________ 1 + cos α ________ 2 _________ 1 − cos α ________ 1 + cos α sin α ________ 1 + cos α 1 − cos α ________ sin α = = Example 7 Using a Half-Angle Formula to Find the Exact Value of a Sine Function Find sin(15°) using a half-angle formula. Solution Since 15° = , we use the half-angle formula for sine: 30° ___ 2 sin 30° ___ 2 = √ = √ = √ = √ √ = __________ 1 − cos30° _________ 2 ————— 3 √ _ 1 − 2 _ 2 ————— — — 3 2 − √ _______ 2 _ 2 ________ — 3 2 − √ _______ 4 _______ — 2 − √ 3 _ 2 Analysis Notice that we used only the positive root because sin(15°) is positive. How To… Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle. 1. Draw a triangle to represent the given information. 2. Determine the correct half-angle formula. 3. Substitute values into the formula based on the triangle. 4. Simplify. SECTION 7.3 douBle-angle, halF-angle, and reduction Formulas 591 Example 8 Finding Exact Values Using Half-Angle Identities Given that tan α = and α lies in quadrant III, find the exact value of the following: 8 __ 15 α __ b. cos   2 α __ a. sin   2 α __ c. tan   2 Solution Using the given information, we can draw the triangle shown in Figure 3. Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate sin α = − and cos α = − 8 __ 17 15 __ . 17 y α x (–15, 0) (–15, –8) 17 Figure 3 a. < α __ Before we start, we must remember that, if α is in quadrant III, then 180° < α < 270°, so 2 α __ is in quadrant II, since 90° < α __ α __ < 135°. To find sin , we begin by writing This means that the terminal side of 2 2 2 180° ____ 2 270° ____ . 2 < the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in Figure 3 and simplify. α _ sin 2 _________ 1 − cos α _ 2 ——————— 15 1 −  − _  17 __ 2 = ± √ = ± √ = ± √ = ± √ = ± √ ——— 32 _ 17 _ 2 ______ 32 1 _ _ · 2 17 ___ 16 _ 17 = ± 4 _ — 17 √ = — 17 4 √ _ 17 We choose the positive value of sin α __ because the angle terminates in quadrant II and sine is positive in 2 quadrant II. b. α __ To find cos , we will write the half-angle formula for cosine, substitute the value of the cosine we found from 2 the triangle in Figure 3, and simplify. 592 CHAPTER 7 trigonometric identities and eQuations __________ 1 + cos α ________ 2 ——————————— 15 ___  17 1 +  − __ 2 α __  = ± √ cos  2 = ± √ ——— 2 _ = ± √ 17 _ 2 ______ 1 2 __ __ = ± √ · 17 2 ___ 1 __ = ± √ 17 — 17 √ _ 17 = − α __ We choose the negative value of cos because the angle is in quadrant II because cosine is negative in 2 quadrant II. c. α __ To find tan , we write the half-angle formula for tangent. Again, we substitute the value of the cosine we 2 found from the triangle in Figure 3 and simplify. _________ 1 − cos α _ 1 + cos α ———————— 15 _  1 −  − 17 __ 15 _  1 +  − 17 α _ tan 2 = ± √ = ± √ = ± √ = ± √ = − √ —— 32 _ 17 _ 2 _ 17 ___ 32 _ 2 16 — α __ α __ We choose the negative value of tan lies in quadrant II, and tangent is negative in quadrant II. because 2 2 = −4 Try It #5 Given that sin α = − 4 α __ __ and α lies in quadrant IV, find the exact value of cos   . 2 5 Example 9 Finding the Measurement of a Half Angle Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of θ formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If tan θ = 5 __ for higher-level competition, what is the measurement of the angle for novice 3 competition? SECTION 7.3 douBle-angle, halF-angle, and reduction Formulas 593 Solution Since the angle for novice competition measures half the steepness of the angle for the high-level competition, and tan θ = 5 __ for high-level competition, we can find cos θ from the right triangle and the Pythagorean theorem so 3 that we can use the half-angle identities. See Figure 4. 32 + 52 = 34 c = √ — 34 34 5 θ 3 Figure 4 We see that cos θ = 3 _ — 34 √ = — 34 3 √ _ 34 θ _ . We can use the half-angle formula for tangent: tan 2 = √ _________ 1 − cos θ _ 1 + cos θ . Since tan θ θ _ is in the first quadrant, so is tan . Thus, 2 θ _ tan 2 —————— — 34 3 √ _ 1 − 34 __ 34 3 √ _ 1 + 34 — —————— — 34 34 − 3 √ __ 34 __ 34 + 3 √ __ 34 34 — ___________ — 34 − 3 √ 34 __________ — 34 + 3 √ 34 = √ = √ = √ ≈ 0.57 We can take the inverse tangent to find the angle: tan−1 (0.57) ≈ 29.7°. So the angle of the ramp for novice competition is ≈ 29.7°. Access these online resources for additional instruction and practice with double-angle, half-angle, and reduction formulas. • Double-Angle Identities (http://openstaxcollege.org/l/doubleangiden) • Half-Angle Identities (http://openstaxcollege.org/l/halfangleident) 594 CHAPTER 7 trigonometric identities and eQuations 7.3 SeCTIOn exeRCISeS VeRBAl 1. Explain how to determine the reduction identities 2. Explain how to
determine the double-angle formula from the double-angle identity cos(2x) = cos2 x − sin2 x. for tan(2x) using the double-angle formulas for cos(2x) and sin(2x). 3. We can determine the half-angle formula for tan — — by dividing the formula for 1 − cos x  = ± √ x __ __  2 1 + cos x √ x x __ __  by cos  sin   . Explain how to determine two 2 2 x __ formulas for tan   that do not involve any square 2 roots. 4. For the half-angle formula given in the previous x _ exercise for tan   , explain why dividing by 0 is not a 2 concern. (Hint: examine the values of cos x necessary for the denominator to be 0.) AlGeBRAIC For the following exercises, find the exact values of a) sin(2x), b) cos(2x), and c) tan(2x) without solving for x. 5. If sin x = 1 __ , and x is in quadrant I. 8 6. If cos x = 2 __ , and x is in quadrant I. 3 7. If cos x = − 1 __ , and x is in quadrant III. 2 8. If tan x = −8, and x is in quadrant IV. For the following exercises, find the values of the six trigonometric functions if the conditions provided hold. 9. cos(2θ) = 3 __ and 90° ≤ θ ≤ 180° 5 and 180° ≤ θ ≤ 270° 10. cos(2θ) = 1 _ — 2 √ For the following exercises, simplify to one trigonometric expression. π π __ __   cos  12. 4 sin  8 8 π π __ __   2 cos  11. 2 sin  4 4 For the following exercises, find the exact value using half-angle formulas. 14. cos  − 11π ___  12 π __  13. sin  8 15. sin  11π ___  12 16. cos  7π ___  8 17. tan  5π ___  12 18. tan  − 3π ___  12 19. tan  − 3π ___  8 x x x __ __ __ For the following exercises, find the exact values of a) sin   , and c) tan   , b) cos   without solving for x, when 2 2 2 0 ≤ x ≤ 360°. 4 __ 20. If tan x = − , and x is in quadrant IV. 3 21. If sin x = − , and x is in quadrant III. 12 __ 13 22. If csc x = 7, and x is in quadrant II. 23. If sec x = −4, and x is in quadrant II. SECTION 7.3 section exercises 595 For the following exercises, use Figure 5 to find the requested half and double angles. α 5 θ 12 Figure 5 24. Find sin(2θ), cos(2θ), and tan(2θ). θ θ θ  .  , and tan   , cos  26. Find sin  __ __ __ 2 2 2 25. Find sin(2α), cos(2α), and tan(2α). α __ α __ α __  , and tan   , cos  27. Find sin   . 2 2 2 For the following exercises, simplify each expression. Do not evaluate. 28. cos2(28°) − sin2(28°) 31. cos2(9x) − sin2(9x) 29. 2 cos2(37°) − 1 32. 4 sin(8x) cos(8x) 30. 1 − 2 sin2(17°) 33. 6 sin(5x) cos(5x) For the following exercises, prove the identity given. 34. (sin t − cos t)2 = 1 − sin(2t) 35. sin(2x) = −2 sin(−x) cos(−x) 36. cot x − tan x = 2 cot(2x) 37. sin(2θ) _________ 1 + cos(2θ) tan2 θ = tan3 θ For the following exercises, rewrite the expression with an exponent no higher than 1. 38. cos2(5x) 42. cos2 x sin4 x 39. cos2(6x) 43. cos4 x sin2 x 40. sin4(8x) 44. tan2 x sin2 x 41. sin4(3x) TeCHnOlOGY For the following exercises, reduce the equations to powers of one, and then check the answer graphically. 45. tan4 x 46. sin2(2x) 49. tan4 x cos2 x 50. cos2 x sin(2x) 47. sin2 x cos2 x 51. cos2 (2x)sin x 48. tan2 x sin x x __ 52. tan2   sin x 2 For the following exercises, algebraically find an equivalent function, only in terms of sin x and/or cos x, and then check the answer by graphing both equations. 53. sin(4x) 54. cos(4x) exTenSIOnS For the following exercises, prove the identities. 55. sin(2x) = 2 tan x ________ 1 + tan2 x 57. tan(2x) = 2 sin x cos x _________ 2 cos2 x − 1 56. cos(2α) = 1 − tan2 α ________ 1 + tan2 α 58. (sin2 x − 1)2 = cos(2x) + sin4 x 59. sin(3x) = 3 sin x cos2 x − sin3 x 60. cos(3x) = cos3 x − 3 sin2 x cos x 61. 1 + cos(2t) ___________ = sin(2t) − cos t 2 cos t ________ 2 sin t − 1 62. sin(16x) = 16 sin x cos x cos(2x)cos(4x)cos(8x) 63. cos(16x) = (cos2 (4x) − sin2 (4x) − sin(8x))(cos2 (4x) − sin2 (4x) + sin(8x)) 596 CHAPTER 7 trigonometric identities and eQuations leARnInG OBjeCTIVeS In this section, you will: • • Express products as sums. Express sums as products. 7.4 SUM-TO-PRODUCT AnD PRODUCT-TO-SUM FORMUlAS Figure 1 The UCLA marching band (credit: Eric Chan, Flickr). A band marches down the field creating an amazing sound that bolsters the crowd. That sound travels as a wave that can be interpreted using trigonometric functions. For example, Figure 2 represents a sound wave for the musical note A. In this section, we will investigate trigonometric identities that are the foundation of everyday phenomena such as sound waves. y 1 –1 0.002 0.004 0.006 0.008 0.01 x Figure 2 expressing Products as Sums We have already learned a number of formulas useful for expanding or simplifying trigonometric expressions, but sometimes we may need to express the product of cosine and sine as a sum. We can use the product-to-sum formulas, which express products of trigonometric functions as sums. Let’s investigate the cosine identity first and then the sine identity. Expressing Products as Sums for Cosine We can derive the product-to-sum formula from the sum and difference identities for cosine. If we add the two equations, we get: cos α cos β + sin α sin β = cos(α − β) + cos α cos β − sin α sin β = cos(α + β) 2 cos α cos β = cos(α − β) + cos(α + β) Then, we divide by 2 to isolate the product of cosines: 1 __ [cos(α − β) + cos(α + β)] cos α cos β = 2 SECTION 7.4 sum-to-product and product-to-sum Formulas 597 How To… Given a product of cosines, express as a sum. 1. Write the formula for the product of cosines. 2. Substitute the given angles into the formula. 3. Simplify. Example 1 Write the following product of cosines as a sum: 2 cos  Solution We begin by writing the formula for the product of cosines: Writing the Product as a Sum Using the Product-to-Sum Formula for Cosine 7x __  cos 2 3x __ . 2 We can then substitute the given angles into the formula and simplify. 1 __ cos α cos β = [cos(α − β) + cos(α + β)] 2 2 cos  7x __  cos  2 1 3x __ __   cos   = (2)  2 2 7x __ 2 − 3x __  + cos  2 7x __ 2 + 3x __   2 =  cos  4x __  + cos  2 10x ___ 2   = cos 2x + cos 5x Try It #1 Use the product-to-sum formula to write the product as a sum or difference: cos(2θ)cos(4θ). Expressing the Product of Sine and Cosine as a Sum Next, we will derive the product-to-sum formula for sine and cosine from the sum and difference formulas for sine. If we add the sum and difference identities, we get: sin(α + β) = sin α cos β + cos α sin β + sin(α − β) = sin α cos β − cos α sin β sin(α + β) + sin(α − β) = 2 sin α cos β Then, we divide by 2 to isolate the product of cosine and sine: 1 __ [sin(α +β) + sin(α − β)] sin α cos β = 2 Example 2 Writing the Product as a Sum Containing only Sine or Cosine Express the following product as a sum containing only sine or cosine and no products: sin(4θ)cos(2θ). Solution Write the formula for the product of sine and cosine. Then substitute the given values into the formula and simplify. 1 __ sin α cos β = [sin(α + β) + sin(α − β)] 2 1 __ [sin(4θ + 2θ) + sin(4θ − 2θ)] sin(4θ)cos(2θ) = 2 1 __ [sin(6θ) + sin(2θ)] = 2 Try It #2 Use the product-to-sum formula to write the product as a sum: sin(x + y)cos(x − y). 598 CHAPTER 7 trigonometric identities and eQuations Expressing Products of Sines in Terms of Cosine Expressing the product of sines in terms of cosine is also derived from the sum and difference identities for cosine. In this case, we will first subtract the two cosine formulas: − cos(α − β) = cos α cos β + sin α sin β cos(α + β) = − (cos α cos β − sin α sin β) cos(α − β) − cos(α + β) = 2 sin α sin β Then, we divide by 2 to isolate the product of sines: 1 __ [cos(α − β) − cos(α + β)] sin α sin β = 2 Similarly we could express the product of cosines in terms of sine or derive other product-to-sum formulas. the product-to-sum formulas The product-to-sum formulas are as follows: 1 __ [cos(α − β) + cos(α + β)] cos α cos β = 2 1 __ [cos(α − β) − cos(α + β)] sin α sin β = 2 1 __ [sin(α + β) + sin(α − β)] sin α cos β = 2 1 __ [sin(α + β) − sin(α − β)] cos α sin β = 2 Example 3 Express the Product as a Sum or Difference Write cos(3θ) cos(5θ) as a sum or difference. Solution We have the product of cosines, so we begin by writing the related formula. Then we substitute the given angles and simplify. 1 __ cos α cos β = [cos(α − β) + cos(α + β)] 2 1 __ [cos(3θ − 5θ) + cos(3θ + 5θ)] cos(3θ)cos(5θ) = 2 1 __ [cos(2θ) + cos(8θ)] = 2 Use even-odd identity. Try It #3 Use the product-to-sum formula to evaluate cos 11π ___ 12 cos π __ . 12 expressing Sums as Products Some problems require the reverse of the process we just used. The sum-to-product formulas allow us to express sums of sine or cosine as products. These formulas can be derived from the product-to-sum identities. For example, with a few substitutions, we can derive the sum-to-product identity for sine. Let u + v _____ 2 = α and u − v _____ 2 = β. Then, α + β = + u − v _____ 2 α − β = − u − v _____ 2 u + v _____ 2 2u __ 2 = u = u + v _____ 2 2v __ 2 = v = SECTION 7.4 sum-to-product and product-to-sum Formulas 599 Thus, replacing α and β in the product-to-sum formula with the substitute expressions, we have 1 __ [sin(α + β) + sin(α − β)] sin α cos β = 2 sin  2 sin  u + v _____ 2 u + v _____ 2  cos   cos  u − v _____ 2 u − v _____ 2 1 __ [sin u + sin v]  = 2  = sin u + sin v Substitute for (α + β) and (α − β) The other sum-to-product identities are derived similarly. sum-to-product formulas The sum-to-product formulas are as follows: sin α + sin β = 2 sin  α + β _____ 2  cos  α − β  _____ 2 cos α − cos β = −2 sin  α + β _____ 2  sin  α − β  _____ 2 sin α − sin β = 2 sin  α − β _____ 2  cos  α + β  _____ 2 cos α + cos β = 2 cos  α + β _____ 2  cos  α − β  _____ 2 Example 4 Writing the Difference of Sines as a Product Write the following difference of sines expression as a product: sin(4θ) − sin(2θ). Solution We begin by writing the formula for the difference of sines. sin α − sin β = 2 sin  α − β _____ 2  cos  α + β  _____ 2 Substitute the values into the formula, and simplify. sin(4θ) − sin(2θ) = 2 sin  4θ − 2θ ______ 2  cos  4θ +
2θ  ______ 2 2θ  cos  __ 2 = 2 sin  = 2 sin θ cos(3θ) 6θ  __ 2 Try It #4 Use the sum-to-product formula to write the sum as a product: sin(3θ) + sin(θ). Example 5 Evaluating Using the Sum-to-Product Formula Evaluate cos(15°) − cos(75°). Solution We begin by writing the formula for the difference of cosines. α + β _____ 2 cos α − cos β = −2 sin   sin  α − β  _____ 2 Then we substitute the given angles and simplify. cos(15°) − cos(75°) = −2 sin  15° + 75° ________ 2  sin  15° − 75° ________  2 — = −2 sin(45°)sin(−30°) 2   − = −2  1 √ __ ___  2 2 2 √ ___ 2 = — 600 CHAPTER 7 trigonometric identities and eQuations Example 6 Proving an Identity Prove the identity: cos(4t) − cos(2t) _____________ sin(4t) + sin(2t) = −tan t Solution it matches the right side. We will start with the left side, the more complicated side of the equation, and rewrite the expression until cos(4t) − cos(2t) _____________ sin(4t) + sin(2t) = −2 sin  ____  cos  2 sin  4t + 2t ______ 2 4t + 2t ______ 2 4t − 2t ______  2 4t − 2t ______  2  sin  = −2 sin(3t)sin t ___________ 2 sin(3t)cos t = −2 sin(3t)sin t ___________ 2 sin(3t)cos t = − sin t ____ cos t = −tan t Analysis Recall that verifying trigonometric identities has its own set of rules. The procedures for solving an equation are not the same as the procedures for verifying an identity. When we prove an identity, we pick one side to work on and make substitutions until that side is transformed into the other side. Example 7 Verifying the Identity Using Double-Angle Formulas and Reciprocal Identities Verify the identity csc2 θ − 2 = cos(2θ) ______ . sin2 θ Solution For verifying this equation, we are bringing together several of the identities. We will use the double-angle formula and the reciprocal identities. We will work with the right side of the equation and rewrite it until it matches the left side. cos(2θ) ______ = sin2 θ 1 − 2 sin2 θ _________ sin2 θ = 1 ____ sin2 θ − 2 sin2 θ ______ sin2 θ = csc2 θ − 2 Try It #5 Verify the identity tan θ cot θ − cos2θ = sin2θ. Access these online resources for additional instruction and practice with the product-to-sum and sum-to-product identities. • Sum to Product Identities (http://openstaxcollege.org/l/sumtoprod) • Sum to Product and Product to Sum Identities (http://openstaxcollege.org/l/sumtpptsum) SECTION 7.4 section exercises 601 7.4 SeCTIOn exeRCISeS VeRBAl 1. Starting with the product to sum formula sin α cos β = 1 __ [sin(α + β) + sin(α − β)], explain 2 how to determine the formula for cos α sin β. 3. Explain a situation where we would convert an equation from a sum to a product and give an example. 2. Explain two different methods of calculating cos(195°)cos(105°), one of which uses the product to sum. Which method is easier? 4. Explain a situation where we would convert an equation from a product to a sum, and give an example. AlGeBRAIC For the following exercises, rewrite the product as a sum or difference. 5. 16sin(16x)sin(11x) 6. 20cos(36t)cos(6t) 7. 2sin(5x)cos(3x) 8. 10cos(5x)sin(10x) 9. sin(−x)sin(5x) 10. sin(3x)cos(5x) For the following exercises, rewrite the sum or difference as a product. 11. cos(6t) + cos(4t) 12. sin(3x) + sin(7x) 13. cos(7x) + cos(−7x) 14. sin(3x) − sin(−3x) 15. cos(3x) + cos(9x) 16. sin h − sin(3h) For the following exercises, evaluate the product using a sum or difference of two functions. Evaluate exactly. 17. cos(45°)cos(15°) 18. cos(45°)sin(15°) 19. sin(−345°)sin(−15°) 20. sin(195°)cos(15°) 21. sin(−45°)sin(−15°) For the following exercises, evaluate the product using a sum or difference of two functions. Leave in terms of sine and cosine. 22. cos(23°)sin(17°) 23. 2sin(100°)sin(20°) 24. 2sin(−100°)sin(−20°) 25. sin(213°)cos(8°) 26. 2cos(56°)cos(47°) For the following exercises, rewrite the sum as a product of two functions. Leave in terms of sine and cosine. 27. sin(76°) + sin(14°) 28. cos(58°) − cos(12°) 29. sin(101°) − sin(32°) 30. cos(100°) + cos(200°) 31. sin(−1°) + sin(−2°) For the following exercises, prove the identity. 32. cos(a + b) ________ = cos(a − b) 1 − tan a tan b ___________ 1 + tan a tan b 34. 6cos(8x)sin(2x) ____________ sin(−6x) = −3 sin(10x)csc(6x) + 3 33. 4sin(3x)cos(4x) = 2 sin(7x) − 2 sinx 35. sin x + sin(3x) = 4sin x cos2 x 36. 2(cos3 x − cos x sin2 x)= cos(3x) + cos x 37. 2 tan x cos(3x) = sec x(sin(4x) − sin(2x)) 38. cos(a + b) + cos(a − b) = 2cos a cos b 602 CHAPTER 7 trigonometric identities and eQuations nUMeRIC For the following exercises, rewrite the sum as a product of two functions or the product as a sum of two functions. Give your answer in terms of sines and cosines. Then evaluate the final answer numerically, rounded to four decimal places. 39. cos(58°) + cos(12°) 40. sin(2°) − sin(3°) 41. cos(44°) − cos(22°) 42. cos(176°)sin(9°) 43. sin(−14°)sin(85°) TeCHnOlOGY For the following exercises, algebraically determine whether each of the given expressions is a true identity. If it is not an identity, replace the right-hand side with an expression equivalent to the left side. Verify the results by graphing both expressions on a calculator. 44. 2sin(2x)sin(3x) = cos x − cos(5x) 46. sin(3x) − sin(5x) _____________ cos(3x) + cos(5x) = tan x 45. cos(10θ) + cos(6θ) _______________ cos(6θ) − cos(10θ) = cot(2θ)cot(8θ) 47. 2cos(2x)cos x + sin(2x)sin x = 2 sin x 48. sin(2x) + sin(4x) _____________ sin(2x) − sin(4x) = −tan(3x)cot x For the following exercises, simplify the expression to one term, then graph the original function and your simplified version to verify they are identical. 49. sin(9t) − sin(3t) _____________ cos(9t) + cos(3t) 51. sin(3x) − sin x ____________ sin x 53. sin x cos(15x) − cos x sin(15x) exTenSIOnS 50. 2sin(8x)cos(6x) − sin(2x) 52. cos(5x) + cos(3x) ______________ sin(5x) + sin(3x) For the following exercises, prove the following sum-to-product formulas. 54. sin x − sin y = 2 sin   cos  55. cos x + cos y = 2cos   cos  x − y  _ 2 For the following exercises, prove the identity. 56. sin(6x) + sin(4x) _____________ sin(6x) − sin(4x) = tan (5x)cot x 58. cos(6y) + cos(8y) ______________ sin(6y) − sin(4y) = cot y cos(7y) sec(5y) 60. sin(10x) − sin(2x) ______________ cos(10x) + cos(2x) = tan(4x) 57. cos(3x) + cos x ____________ cos(3x) − cos x = −cot (2x)cot x 59. cos(2y) − cos(4y) ______________ sin(2y) + sin(4y) = tan y 61. cos x − cos(3x) = 4 sin2 x cos x 62. (cos(2x) − cos(4x))2 + (sin(4x) + sin(2x))2 = 4 sin2(3x) π __ − t  = 63. tan  4 1 − tan t _ 1 + tan t SECTION 7.5 solving trigonometric eQuations 603 leARnInG OBjeCTIVeS In this section, you will: • • • • • • • Solve linear trigonometric equations in sine and cosine. Solve equations involving a single trigonometric function. Solve trigonometric equations using a calculator. Solve trigonometric equations that are quadratic in form. Solve trigonometric equations using fundamental identities. Solve trigonometric equations with multiple angles. Solve right triangle problems. 7. 5 SOlVInG TRIGOnOMeTRIC eQUATIOnS Figure 1 Egyptian pyramids standing near a modern city. (credit: Oisin Mulvihill) Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles, which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles. In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids. Solving linear Trigonometric equations in Sine and Cosine Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is 2π. In other words, every 2π units, the y-values repeat. If we need to find all possible solutions, then we must add 2πk, where k is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is 2π: sin θ = sin(θ ± 2kπ) There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections. 604 CHAPTER 7 trigonometric identities and eQuations Solving a Linear Trigonometric Equation Involving the Cosine Function Example 1 Find all possible exact solutions for the equation cos θ = 1 __ . 2 Solution From the unit circle, we know that cos θ = 1 __ 2 π __ θ = , 3 5π ___ 3 These are the solutions in the interval [0, 2π]. All possible solutions are given by where k is an integer. π __ ± 2kπ and θ = θ = 3 5π ___ ± 2kπ 3 Solving a Linear Equation Involving the Sine Function Exampl
e 2 Find all possible exact solutions for the equation sin t = 1 __ . 2 Solution Solving for all possible values of t means that solutions include angles beyond the period of 2π. From Section π __ and t = 7.2 Figure 2, we can see that the solutions are t = 6 5π ___ . But the problem is asking for all possible values that 6 solve the equation. Therefore, the answer is where k is an integer. π __ ± 2πk and t = 6 t = 5π ___ ± 2πk 6 How To… Given a trigonometric equation, solve using algebra. 1. Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity. 2. Substitute the trigonometric expression with a single variable, such as x or u. 3. Solve the equation the same way an algebraic equation would be solved. 4. Substitute the trigonometric expression back in for the variable in the resulting expressions. 5. Solve for the angle. Example 3 Solve the Trigonometric Equation in Linear Form Solve the equation exactly: 2cos θ − 3 = − 5, 0 ≤ θ < 2π. Solution Use algebraic techniques to solve the equation. 2cos θ − 3 = −5 2cos θ = −2 cos θ = −1 θ = π Try It #1 Solve exactly the following linear equation on the interval [0, 2π): 2sin x + 1 = 0. Solving equations Involving a Single Trigonometric Function When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see Section 7.2 Figure 2). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is π, not 2π. π __ Further, the domain of tangent is all real numbers with the exception of odd integer multiples of , unless, of course, 2 a problem places its own restrictions on the domain. SECTION 7.5 solving trigonometric eQuations 605 Example 4 Solving a Problem Involving a Single Trigonometric Function Solve the problem exactly: 2sin2 θ − 1 = 0, 0 ≤ θ < 2π. Solution As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate sin θ. Then we will find the angles. 2sin2 θ − 1 = 0 √ — 2sin2 θ = 1 sin2 θ = 1 __ 2 sin2 θ = ± √ sin θ = ± 1 _ = ± 2 √ 3π ___ , 4 π __ θ = , 4 __ 1 __ 2 5π ___ , 4 — — 2 √ ____ 2 7π ___ 4 Example 5 Solving a Trigonometric Equation Involving Cosecant Solve the following equation exactly: csc θ = −2, 0 ≤ θ < 4π. Solution We want all values of θ for which csc θ = −2 over the interval 0 ≤ θ < 4π. csc θ = −2 = −2 1 ____ sin θ sin θ = − 1 __ 2 θ = 7π ___ , 6 23π 19π 11π ___ ___ ___ , , 6 6 6 Analysis As sin θ = − 1 _ , notice that all four solutions are in the third and fourth quadrants. 2 Example 6 Solving an Equation Involving Tangent π __  = 1, 0 ≤ θ < 2π. Solve the equation exactly: tan  θ − 2 π __ Solution Recall that the tangent function has a period of π. On the interval [0, π), and at the angle of , the tangent has a 4 π π __ __  = 1, then value of 1. However, the angle we want is  θ −  . Thus, if tan  4 2 π π __ __ = θ − 4 2 Over the interval [0, 2π), we have two solutions: θ = 3π ___ ± kπ 4 θ = 3π ___ and θ = 4 3π ___ 4 + π = 7π ___ 4 Try It #2 Find all solutions for tan x = √ — 3 . Example 7 Identify all Solutions to the Equation Involving Tangent Identify all exact solutions to the equation 2(tan x + 3) = 5 + tan x, 0 ≤ x < 2π. 606 CHAPTER 7 trigonometric identities and eQuations Solution We can solve this equation using only algebra. Isolate the expression tan x on the left side of the equals sign. 2(tan x) + 2(3) = 5 + tan x 2tan x + 6 = 5 + tan x 2tan x − tan x = 5 − 6 tan x = −1 There are two angles on the unit circle that have a tangent value of −1: θ = and θ = 3π __ 4 7π __ . 4 Solve Trigonometric equations Using a Calculator Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem. Example 8 Using a Calculator to Solve a Trigonometric Equation Involving Sine Use a calculator to solve the equation sin θ = 0.8, where θ is in radians. Solution Make sure mode is set to radians. To find θ, use the inverse sine function. On most calculators, you will need to push the 2ND button and then the SIN button to bring up the sin−1 function. What is shown on the screen is sin−1( . The calculator is ready for the input within the parentheses. For this problem, we enter sin−1 (0.8), and press ENTER. Thus, to four decimals places, The solution is The angle measurement in degrees is sin−1(0.8) ≈ 0.9273 θ ≈ 0.9273 ± 2πk θ ≈ 53.1° θ ≈ 180° − 53.1° ≈ 126.9° Analysis Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using π − θ. Example 9 Using a Calculator to Solve a Trigonometric Equation Involving Secant Use a calculator to solve the equation sec θ = −4, giving your answer in radians. Solution We can begin with some algebra. sec θ = −4 = −4 1 ____ cos θ cos θ = − 1 __ 4 Check that the MODE is in radians. Now use the inverse cosine function. cos−1  − 1 __  ≈ 1.8235 4 θ ≈ 1.8235 + 2πk π __ ≈ 1.57 and π ≈ 3.14, 1.8235 is between these two numbers, thus θ ≈ 1.8235 is in quadrant II. Since 2 Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure 2. SECTION 7.5 solving trigonometric eQuations 607 y θ ≈ 1.8235 x θʹ ≈ π – 1.8235 ≈ 1.3181 θʹ ≈ π + 1.3181 ≈ 4.4597 Figure 2 So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is θ´≈ π − 1.8235 ≈ 1.3181. The other solution in quadrant III is θ´ ≈ π + 1.3181 ≈ 4.4597. The solutions are θ ≈ 1.8235 ± 2πk and θ ≈ 4.4597 ± 2πk. Try It #3 Solve cos θ = −0.2. Solving Trigonometric equations in Quadratic Form Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as x or u. If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations. Example 10 Solving a Trigonometric Equation in Quadratic Form Solve the equation exactly: cos2 θ + 3 cos θ − 1 = 0, 0 ≤ θ < 2π. Solution We begin by using substitution and replacing cos θ with x. It is not necessary to use substitution, but it may make the problem easier to solve visually. Let cos θ = x. We have The equation cannot be factored, so we will use the quadratic formula x = x2 + 3x − 1 = 0 — b2 − 4ac −b ± √ _____________ 2a —— . (−3)2 − 4(1)(−1) −3 ± √ ______________________ x = Replace x with cos θ, and solve. Thus, Note that only the + sign is used. This is because we get an error when we solve θ = cos−1  calculator, since the domain of the inverse cosine function is [−1, 1]. However, there is a second solution: −3 − √ _________ 2 13  on a — 2 — = −3 ± √ _________ 2 13 cos θ = — 13 −3 ± √ _________ 2 θ = cos−1  — 13 −3 + √  _________ 2 — 13 −3 + √  _________ 2 θ = cos−1  ≈ 1.26 608 CHAPTER 7 trigonometric identities and eQuations This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is θ = 2π − cos−1  ≈ 5.02 — 13 −3 + √  _________ 2 Try It #4 Solve sin2 θ = 2 cos θ + 2, 0 ≤ θ ≤ 2π. [Hint: Make a substitution to express the equation only in terms of cosine.] Example 11 Solving a Trigonometric Equation in Quadratic Form by Factoring Solve the equation exactly: 2 sin2 θ − 5 sin θ + 3 = 0, 0 ≤ θ ≤ 2π. Solution Using grouping, this quadratic can be factored. Either make the real substitution, sin θ = u, or imagine it, as we factor: Now set each factor equal to zero. 2sin2 θ − 5sin θ + 3 = 0 (2sin θ − 3)(sin θ − 1) = 0 2sin θ − 3 = 0 2sin θ = 3 sin θ = 3 __ 2 sin θ − 1 = 0 sin θ = 1 Next solve for θ: sin θ ≠ 3 π __ __ , as the range of the sine function is [−1, 1]. However, sin θ = 1, giving the solution θ = . 2 2 Analysis Make sure to check all solutions on the given domain as some factors have no solution. Try It #5 Solve sin2 θ = 2cos θ + 2, 0 ≤ θ ≤ 2π. [Hint: Make a substitution to express the equation only in terms of cosine.] Example 12 Solving a Trigonometric Equation Using Algebra Solve exactly: 2sin2 θ + sin θ = 0; 0 ≤ θ < 2π Solution This problem should appear familiar as it is similar to a quadratic. Let sin θ = x. The equation becomes 2x2 + x = 0. We begin by factoring: Set each factor equal to zero. 2x2 + x = 0 x(2x + 1) = 0 x = 0 (2x + 1) = 0 1 __ x = − 2 Then, substitute back into the equation the original expression sin θ for x. Thus, sin θ = 0 The solutions within the domain 0 ≤ θ < 2π are θ = 0, π, θ = 0, π sin θ = − 1 __ 2 11π 7π ___ ___ , 6 6 11π ___ . 6 7π ___ , 6 θ = If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero. SECTION 7.5 solving trigonometric eQuations 609 2sin2 θ + sin θ = 0 sin θ(2sin θ + 1) = 0 sin θ = 0 θ = 0, π 2sin θ + 1 = 0 2sin θ = −1 sin θ = − 1 __ 2 θ = 7π ___ , 6 11π ___ 6 Analysis We can see t
he solutions on the graph in Figure 3. On the interval 0 ≤ θ < 2π, the graph crosses the x-axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value. f(θ) = 2 sin2 θ + sin 1 π 6 π 3 π 2 2π 3 5π 6 π 7π 6 4π 3 3π 2 5π 3 11π 6 2π 13π 6 θ Figure 3 We can verify the solutions on the unit circle in Section 7.2 Figure 2 as well. Example 13 Solving a Trigonometric Equation Quadratic in Form Solve the equation quadratic in form exactly: 2sin2 θ − 3sin θ + 1 = 0, 0 ≤ θ < 2π. Solution We can factor using grouping. Solution values of θ can be found on the unit circle: (2sin θ − 1)(sin θ − 1) = 0 2sin θ − 1 = 0 sin θ = 1 __ 2 5π ___ 6 π __ θ = , 6 sin θ = 1 π __ θ = 2 Try It #6 Solve the quadratic equation 2cos2 θ + cos θ = 0. Solving Trigonometric equations Using Fundamental Identities While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation. 610 CHAPTER 7 trigonometric identities and eQuations Example 14 Use Identities to Solve an Equation Use identities to solve exactly the trigonometric equation over the interval 0 ≤ x < 2π. Solution Notice that the left side of the equation is the difference formula for cosine. cos x cos(2x) + sin x sin(2x) = cos x cos(2x) + sin x sin(2x) = — 3 √ ____ 2 — 3 √ ____ 2 — 3 √ ____ 2 — 3 √ ____ 2 — 3 √ ____ 2 cos(x − 2x) = cos(−x) = cos x = Difference formula for cosine Use the negative angle identity. From the unit circle in Figure 2, we see that cos x = — 3 √ ____ 2 π __ when x = , 6 11π ___ . 6 Example 15 Solving the Equation Using a Double-Angle Formula Solve the equation exactly using a double-angle formula: cos(2θ) = cos θ. Solution We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine: cos(2θ) = cos θ 2cos2 θ − 1 = cos θ 2cos2 θ − cos θ − 1 = 0 (2cos θ + 1)(cos θ − 1) = 0 2cos θ + 1 = 0 cos θ = − 1 __ 2 cos θ − 1 = 0 cos θ = 1 So, if cos θ = − 1 __ , then θ = 2 2π __ 3 ± 2πk and θ = 4π __ ± 2πk; if cos θ = 1, then θ = 0 ± 2πk. 3 Example 16 Solving an Equation Using an Identity Solve the equation exactly using an identity: 3cos θ + 3 = 2sin2 θ, 0 ≤ θ < 2π. Solution If we rewrite the right side, we can write the equation in terms of cosine: 3cos θ + 3 = 2sin2 θ 3cos θ + 3 = 2(1 − cos2 θ) 3cos θ + 3 = 2 − 2cos2 θ 2cos2 θ + 3cos θ + 1 = 0 (2cos θ + 1)(cos θ + 1) = 0 2cos θ + 1 = 0 cos θ = − 1 __ 2 2π ___ , 3 cos θ + 1 = 0 θ = 4π ___ 3 cos θ = −1 θ = π Our solutions are θ = 2π ___ , 3 4π ___ 3 , π SECTION 7.5 solving trigonometric eQuations 611 Solving Trigonometric equations with Multiple Angles Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as sin(2x) or cos(3x). When confronted with these equations, recall that y = sin(2x) is a horizontal compression by a factor of 2 of the function y = sin x. On an interval of 2π, we can graph two periods of y = sin(2x), as opposed to one cycle of y = sin x. This compression of the graph leads us to believe there may be twice as many x-intercepts or solutions to sin(2x) = 0 compared to sin x = 0. This information will help us solve the equation. Example 17 Solving a Multiple Angle Trigonometric Equation Solve exactly: cos(2x) = 1 __ on [0, 2π). 2 Solution We can see that this equation is the standard equation with a multiple of an angle. If cos(α) = 1 __ , we know 2 α is in quadrants I and IV. While θ = cos−1 1 __ will only yield solutions in quadrants I and II, we recognize that the 2 solutions to the equation cos θ = 1 __ will be in quadrants I and IV. 2 π 5π π __ __ __ . So, 2x = or 2x = and θ = Therefore, the possible angles are __ __ __   = cos  Does this make sense? Yes, because cos  2  . 2 3 6 5π __ 3 π __ , which means that x = or x = 6 5π __ . 6 Are there any other possible answers? Let us return to our first step. π π __ __ In quadrant I, 2x = , so x = as noted. Let us revolve around the circle again: 6 3 so x = 7π ___ . 6 One more rotation yields π __ + 2π 2x = 3 π __ + = 3 6π ___ 3 = 7π ___ 3 π __ + 4π 2x = 3 π __ + = 3 12π ___ 3 = 13π ___ 3 x = 13π ___ 6 > 2π, so this value for x is larger than 2π, so it is not a solution on [0, 2π). In quadrant IV, 2x = , so x = as noted. Let us revolve around the circle again: 5π __ 3 5π __ 6 2x = + 2π 5π ___ 3 = 5π ___ 3 + 6π ___ 3 so x = 11π ___ . 6 = 11π ___ 3 612 CHAPTER 7 trigonometric identities and eQuations One more rotation yields 2x = + 4π 5π ___ 3 = 5π ___ 3 + 12π ___ 3 = 17π ___ 3 > 2π, so this value for x is larger than 2π, so it is not a solution on [0, 2π). x = 17π ___ 6 π __ Our solutions are x = , 6 5π __ , 6 7π __ 6 , and 11π ___ 6 must go around the unit circle n times. . Note that whenever we solve a problem in the form of sin(nx) = c, we Solving Right Triangle Problems We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the Pythagorean Theorem. We begin with the familiar Pythagorean Theorem, a2 + b2 = c2, and model an equation to fit a situation. Example 18 Using the Pythagorean Theorem to Model an Equation Use the Pythagorean Theorem, and the properties of right triangles to model an equation that fits the problem. One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 69.5 meters above the ground, and the second anchor on the ground is 23 meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)? See Figure 4. 69.5 θ 23 Figure 4 Solution Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem. a2 + b2 = c2 (23)2 + (69.5)2 ≈ 5359 — 5359 ≈ 73.2 m √ The angle of elevation is θ, formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places. tan θ = 69.5 ____ 23 The angle of elevation is approximately 71.7°, and the length of the cable is 73.2 meters. tan−1  69.5 ____ 23  ≈ 1.2522 ≈ 71.69° SECTION 7.5 solving trigonometric eQuations 613 Example 19 Using the Pythagorean Theorem to Model an Abstract Problem OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall. Solution For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder’s length. Equivalently, if the base of the ladder is “a” feet from the wall, the length of the ladder will be 4a feet. See Figure 5. 4a b θ a Figure 5 The side adjacent to θ is a and the hypotenuse is 4a. Thus, cos θ = a __ 4a = 1 __ 4 The elevation of the ladder forms an angle of 75.5° with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem: 1 __  ≈ 75.5° cos−1  4 a2 + b2 = (4a)2 b2 = (4a)2 − a2 b2 = 16a2 − a2 b2 = 15a2 Thus, the ladder touches the wall at a √ 15 15 feet from the ground. b = a √ — — Access these online resources for additional instruction and practice with solving trigonometric equations. • Solving Trigonometric equations I (http://openstaxcollege.org/l/solvetrigeqI) • Solving Trigonometric equations II (http://openstaxcollege.org/l/solvetrigeqII) • Solving Trigonometric equations III (http://openstaxcollege.org/l/solvetrigeqIII) • Solving Trigonometric equations IV (http://openstaxcollege.org/l/solvetrigeqIV) • Solving Trigonometric equations V (http://openstaxcollege.org/l/solvetrigeqV) • Solving Trigonometric equations VI (http://openstaxcollege.org/l/solvetrigeqVI) 614 CHAPTER 7 trigonometric identities and eQuations 7.5 SeCTIOn exeRCISeS VeRBAl 1. Will there always be solutions to trigonometric 2. When solving a trigonometric equation involving function equations? If not, describe an equation that would not have a solution. Explain why or why not. more than one trig function, do we always want to try to rewrite the equation so it is expressed in terms of one trigonometric function? Why or why not? 3. When solving linear trig equations in terms of only sine or cosine, how do we know whether there will be solutions? AlGeBRAIC For the following exercises, find all solutions exactly on the interval 0 ≤ θ < 2π. 4. 2sin θ = − √ 7. 2cos θ = − √ 10. cot . 2sin θ = √ 8. tan θ = −1 — 3 11. 4sin2 x − 2 = 0 For the following exercises, solve exactly on [0, 2π). 13. 2cos θ = √ — 2 16. 2sin θ = − √ — 3 19. 2cos(3θ) = − √ — 2 π __ θ  = √ 22. 2cos  5 — 3 14. 2cos θ = −1 17. 2sin(3θ) = 1 3 20. cos(2θ) = − √ ____ 2 — 6. 2cos θ = 1 9. tan x = 1 12. csc2 x − 4 = 0 15. 2sin θ = −1 18. 2sin(2θ) = √ — 3 21. 2sin(πθ) = 1 For the following exercises, find all exact solutions on [0, 2π). 23. sec(x)sin(x) − 2sin(x) = 0 24. tan(x) − 2sin(x)tan(x) = 0 25. 2cos2 t + cos(t) = 1 27. 2sin(x)cos(x) − sin(x) + 2cos(x) − 1 = 0 29. sec2 x = 1 26. 2tan2(t) = 3sec(t) 28. cos2 θ = 1 __ 2 30. tan2 (x) = −1 + 2tan(−x) 31. 8sin2(x) + 6sin(x) + 1 = 0 32. tan5(x) = tan(x) For the following exercises, solve with the methods shown in this section exactly on the interval [0, 2π). 33. sin
(3x)cos(6x) − cos(3x)sin(6x) = −0.9 34. sin(6x)cos(11x) − cos(6x)sin(11x) = −0.1 35. cos(2x)cos x + sin(2x)sin x = 1 36. 6sin(2t) + 9sin t = 0 37. 9cos(2θ) = 9cos2 θ − 4 39. cos(2t) = sin t 38. sin(2t) = cos t 40. cos(6x) − cos(3x) = 0 For the following exercises, solve exactly on the interval [0, 2π). Use the quadratic formula if the equations do not factor. 41. tan2 x − √ — 3 tan x = 0 42. sin2 x + sin x − 2 = 0 43. sin2 x − 2sin x − 4 = 0 44. 5cos2 x + 3cos x − 1 = 0 45. 3cos2 x − 2cos x − 2 = 0 46. 5sin2 x + 2sin x − 1 = 0 47. tan2 x + 5tan x − 1 = 0 48. cot2 x = −cot x 49. −tan2 x − tan x − 2 = 0 SECTION 7.5 section exercises 615 For the following exercises, find exact solutions on the interval [0, 2π). Look for opportunities to use trigonometric identities. 50. sin2 x − cos2 x − sin x = 0 51. sin2 x + cos2 x = 0 52. sin(2x) − sin x = 0 53. cos(2x) − cos x = 0 54. 2 tan x ________ 2 − sec2 x − sin2 x = cos2 x 55. 1 − cos(2x) = 1 + cos(2x) 56. sec2 x = 7 57. 10sin x cos x = 6cos x 58. −3sin t = 15cos t sin t 59. 4cos2 x − 4 = 15cos x 60. 8sin2 x + 6sin x + 1 = 0 61. 8cos2 θ = 3 − 2cos θ 62. 6cos2 x + 7sin x − 8 = 0 63. 12sin2 t + cos t − 6 = 0 64. tan x = 3sin x 65. cos3 t = cos t GRAPHICAl For the following exercises, algebraically determine all solutions of the trigonometric equation exactly, then verify the results by graphing the equation and finding the zeros. 66. 6sin2 x − 5sin x + 1 = 0 68. 100tan2 x + 20tan x − 3 = 0 67. 8cos2 x − 2cos x − 1 = 0 69. 2cos2 x − cos x + 15 = 0 70. 20sin2 x − 27sin x + 7 = 0 71. 2tan2 x + 7tan x + 6 = 0 72. 130tan2 x + 69tan x − 130 = 0 TeCHnOlOGY For the following exercises, use a calculator to find all solutions to four decimal places. 73. sin x = 0.27 74. sin x = −0.55 75. tan x = −0.34 76. cos x = 0.71 For the following exercises, solve the equations algebraically, and then use a calculator to find the values on the interval [0, 2π). Round to four decimal places. 77. tan2 x + 3tan x − 3 = 0 78. 6tan2 x + 13tan x = −6 79. tan2 x − sec x = 1 80. sin2 x − 2cos2 x = 0 81. 2tan2 x + 9tan x − 6 = 0 82. 4sin2 x + sin(2x)sec x − 3 = 0 exTenSIOnS For the following exercises, find all solutions exactly to the equations on the interval [0, 2π). 83. csc2 x − 3csc x − 4 = 0 84. sin2 x − cos2 x − 1 = 0 85. sin2 x( 1 − sin2 x) + cos2 x( 1 − sin2 x) = 0 86. 3sec2 x + 2 + sin2 x − tan2 x + cos2 x = 0 87. sin2 x − 1 + 2 cos(2x) − cos2 x = 1 88. tan2 x − 1 − sec3 x cos x = 0 89. sin(2x) ______ sec2 x = 0 91. 2cos2 x − sin2 x − cos x − 5 = 0 90. sin(2x) ______ = 0 2 csc2 x 92. 1 ____ sec2 x + 2 + sin2 x + 4cos2 x = 4 616 CHAPTER 7 trigonometric identities and eQuations ReAl-WORlD APPlICATIOnS 93. An airplane has only enough gas to fly to a city 200 miles northeast of its current location. If the pilot knows that the city is 25 miles north, how many degrees north of east should the airplane fly? 95. If a loading ramp is placed next to a truck, at a height of 2 feet, and the ramp is 20 feet long, what angle does the ramp make with the ground? 97. An astronaut is in a launched rocket currently 15 miles in altitude. If a man is standing 2 miles from the launch pad, at what angle is she looking down at him from horizontal? (Hint: this is called the angle of depression.) 99. A man is standing 10 meters away from a 6-meter tall building. Someone at the top of the building is looking down at him. At what angle is the person looking at him? 101. A 90-foot tall building has a shadow that is 2 feet long. What is the angle of elevation of the sun? 103. A spotlight on the ground 3 feet from a 5-foot tall woman casts a 15-foot tall shadow on a wall 6 feet from the woman. At what angle is the light? 94. If a loading ramp is placed next to a truck, at a height of 4 feet, and the ramp is 15 feet long, what angle does the ramp make with the ground? 96. A woman is watching a launched rocket currently 11 miles in altitude. If she is standing 4 miles from the launch pad, at what angle is she looking up from horizontal? 98. A woman is standing 8 meters away from a 10-meter tall building. At what angle is she looking to the top of the building? 100. A 20-foot tall building has a shadow that is 55 feet long. What is the angle of elevation of the sun? 102. A spotlight on the ground 3 meters from a 2-meter tall man casts a 6 meter shadow on a wall 6 meters from the man. At what angle is the light? For the following exercises, find a solution to the word problem algebraically. Then use a calculator to verify the result. Round the answer to the nearest tenth of a degree. 104. A person does a handstand with his feet touching a wall and his hands 1.5 feet away from the wall. If the person is 6 feet tall, what angle do his feet make with the wall? 105. A person does a handstand with her feet touching a wall and her hands 3 feet away from the wall. If the person is 5 feet tall, what angle do her feet make with the wall? 106. A 23-foot ladder is positioned next to a house. If the ladder slips at 7 feet from the house when there is not enough traction, what angle should the ladder make with the ground to avoid slipping? SECTION 7.6 modeling with trigonometric eQuations 617 leARnInG OBjeCTIVeS In this section, you will: • • • • Determine the amplitude and period of sinusoidal functions. Model equations and graph sinusoidal functions. Model periodic behavior. Model harmonic motion functions. 7. 6 MODelInG WITH TRIGOnOMeTRIC eQUATIOnS Figure 1 The hands on a clock are periodic: they repeat positions every twelve hours. (credit: “zoutedrop”/Flickr) Suppose we charted the average daily temperatures in New York City over the course of one year. We would expect to find the lowest temperatures in January and February and highest in July and August. This familiar cycle repeats year after year, and if we were to extend the graph over multiple years, it would resemble a periodic function. Many other natural phenomena are also periodic. For example, the phases of the moon have a period of approximately 28 days, and birds know to fly south at about the same time each year. So how can we model an equation to reflect periodic behavior? First, we must collect and record data. We then find a function that resembles an observed pattern. Finally, we make the necessary alterations to the function to get a model that is dependable. In this section, we will take a deeper look at specific types of periodic behavior and model equations to fit data. Determining the Amplitude and Period of a Sinusoidal Function Any motion that repeats itself in a fixed time period is considered periodic motion and can be modeled by a sinusoidal function. The amplitude of a sinusoidal function is the distance from the midline to the maximum value, or from the midline to the minimum value. The midline is the average value. Sinusoidal functions oscillate above and below the midline, are periodic, and repeat values in set cycles. Recall from Graphs of the Sine and Cosine Functions that the period of the sine function and the cosine function is 2π. In other words, for any value of x, sin(x ± 2πk) = sin x and cos(x ± 2πk) = cos x where k is an integer standard form of sinusoidal equations The general forms of a sinusoidal equation are given as y = Asin(Bt − C) + D or y = Acos(Bt − C) + D where amplitude = ∣ A ∣ , B is related to period such that the period = denotes the horizontal shift, and D represents the vertical shift from the graph’s parent graph. , C is the phase shift such that 2π _ B C _ B Note that the models are sometimes written as y = a sin(ω t ± C) + D or y = a cos(ω t ± C) + D, and period is given as 2π _ ω . The difference between the sine and the cosine graphs is that the sine graph begins with the average value of the function and the cosine graph begins with the maximum or minimum value of the function. 618 CHAPTER 7 trigonometric identities and eQuations Showing How the Properties of a Trigonometric Function Can Transform a Graph Example 1 π _ Show the transformation of the graph of y = sin x into the graph of y = 2sin  4x −  + 2. 2 Solution Consider the series of graphs in Figure 2 and the way each change to the equation changes the image. y y y – 5π 2 2π – – 3π 2 – π – y = sin x π 2 π 3π 2 2π 5π 2 4 3 2 1 π –1 2 –2 –3 –4 (a) – 5π 2 2π – – 3π 1 2 –2 –3 –4 (d) – 5π 2 2π – – 3π 2 –3 –4 (b) y = 2sin x π 2 π 3π 2 2π 5π 2 x – 5π 2 2π – – 3π 2 π – – y = 2sin (4x) π 2 π 3π 2 2π 5π 2 x 4 3 2 1 π 2 –2 –3 –4 (c) π 2 π 3π 2 2π 5π 2 y = 2sin 4x x π 2 – 5π 2 2π – – 3π 2 π – – π 2 π 3π 2 2π 5π 2 x y = 2sin 4x 1 2 –2 –3 –4 (e) Figure 2 (a) The basic graph of y = sinx (b) Changing the amplitude from 1 to 2 generates the graph of y = 2sinx. (c) The period of the sine function changes 2π ___ B . Here we have B = 4, which translates to a period of π _ 2 . The graph completes one full cycle in π _ 2 units. (d) The graph . (e) Finally, the graph is shifted vertically by the value of D. In this case, the graph is shifted up by 2 units. with the value of B, such that period = C __ displays a horizontal shift equal to B π __ = π _ 2 __ , or 4 8 Example 2 Finding the Amplitude and Period of a Function Find the amplitude and period of the following functions and graph one cycle. 1 _ x  a. y = 2sin  4 π _ b. y = −3sin  2x +  2 c. y = cos x + 3 Solution We will solve these problems according to the models. 1 _ x  involves sine, so we use the form a. y = 2sin  4 y = Asin(Bt − C) + D We know that ∣ A ∣ is the amplitude, so the amplitude is 2. Period is , so the period is 2π _ B 2π _ B = 2π _ 1 __ 4 = 8π See the graph in Figure 3. y 2 1 –1 –2 y = 2 sin 4 1 x Amplitude = 2 2π 4π 6π 8π x Period = 8π Figure 3 SECTION 7.6 modeling with trigonometric eQuations 619 π _ b. y = −3sin  2x +  involves sine, so we use the form 2 y = Asin(Bt − C) + D Amplitude is ∣ A ∣ , so the amplitude is ∣ −3 ∣ = 3. Since A is negative, the graph is reflected over the x-axis. Period is , so the period is 2π _ B 2π _ B = 2π _ 2 = π The graph is shifted to the left by π _ π 2 _ _ = units. See
Figure 41 –2 – π 4 y = –3 sin + π 2 2x π 4 π 2 3π 4 x Figure 4 c. y = cos x + 3 involves cosine, so we use the form y = Acos(Bt − C) + D Amplitude is ∣ A ∣ , so the amplitude is 1. The period is 2π. See Figure 5. This is the standard cosine function shifted up three units. y 4 2 1 Midline: y = 3 y = cos x + 3 π 2 π 3π 2 Figure 5 x 2π Try It #1 What are the amplitude and period of the function y = 3cos(3πx)? Finding equations and Graphing Sinusoidal Functions One method of graphing sinusoidal functions is to find five key points. These points will correspond to intervals of 1 _ of the period. The key points will indicate the location of maximum and minimum values. equal length representing 4 If there is no vertical shift, they will also indicate x-intercepts. For example, suppose we want to graph the function y = cos θ. We know that the period is 2π, so we find the interval between key points as follows. 2π _ 4 π _ = 2 π _ Starting with θ = 0, we calculate the first y-value, add the length of the interval to 0, and calculate the second 2 π _ repeatedly until the five key points are determined. The last value should equal the first value, y-value. We then add 2 as the calculations cover one full period. Making a table similar to Table 1, we can see these key points clearly on the graph shown in Figure 6. 620 CHAPTER 7 trigonometric identities and eQuations θ y = cos θ 0 1 π _ 2 0 Table 1 π −1 3π _ 2 0 2π 1 y 1 –1 π 2 π 3π 2 y = cos θ θ 2π Figure 6 Example 3 Graphing Sinusoidal Functions Using Key Points Graph the function y = −4cos(πx) using amplitude, period, and key points. 2π 2π _ _ Solution The amplitude is ∣ −4 ∣ = 4. The period is π = 2. (Recall that we sometimes refer to B as ω.) One ω = cycle of the graph can be drawn over the interval [0, 2]. To find the key points, we divide the period by 4. Make a table 1 _ similar to Table 2, starting with x = 0 and then adding successively to x and calculate y. See the graph in Figure 7. 2 x y = −4 cos(πx) 0 −4 1 _ 2 0 Table 2 1 4 3 _ 2 0 2 −4 y (1, 4) y = –4 cos (πx) 4 2 –2 –4 1 2 3 4 x (2, −4) Figure 7 Try It #2 Graph the function y = 3sin(3x) using the amplitude, period, and five key points. Modeling Periodic Behavior We will now apply these ideas to problems involving periodic behavior. Example 4 Modeling an Equation and Sketching a Sinusoidal Graph to Fit Criteria The average monthly temperatures for a small town in Oregon are given in Table 3. Find a sinusoidal function of the form y = Asin(Bt − C) + D that fits the data (round to the nearest tenth) and sketch the graph. SECTION 7.6 modeling with trigonometric eQuations 621 Month January February March April May June July August September October November December Temperature, ° F 42.5 44.5 48.5 52.5 58 63 68.5 69 64.5 55.5 46.5 43.5 Solution Recall that amplitude is found using the formula Table 3 Thus, the amplitude is A = largest value − smallest value ______________________ 2 ∣ A ∣ = 69 − 42.5 ________ 2 = 13.25 The data covers a period of 12 months, so = 12 which gives B = 2π _ B 2π _ 12 π _ = . 6 The vertical shift is found using the following equation. Thus, the vertical shift is D = highest value + lowest value _____________________ 2 D = 69 + 42.5 ________ 2 = 55.8 π _ So far, we have the equation y = 13.3sin  x − C  + 55.8. 6 To find the horizontal shift, we input the x and y values for the first month and solve for C. π _ (1) − C  + 55.8 42.5 = 13.3sin  6 π _ − C  −13.3 = 13.3sin  6 π π _ _ − C  sin θ = −1 → θ = − −1 = sin  = 2π _ 3 622 CHAPTER 7 trigonometric identities and eQuations π _ We have the equation y = 13.3sin  x − 6 2π _  + 55.8. See the graph in Figure 8. 3 Average Amplitude = 13. 80 70 60 50 40 30 20 10 0 1 2 3 4 8 9 10 11 12 5 7 6 Months Figure 8 Example 5 Describing Periodic Motion The hour hand of the large clock on the wall in Union Station measures 24 inches in length. At noon, the tip of the hour hand is 30 inches from the ceiling. At 3 PM, the tip is 54 inches from the ceiling, and at 6 PM, 78 inches. At 9 PM, it is again 54 inches from the ceiling, and at midnight, the tip of the hour hand returns to its original position 30 inches from the ceiling. Let y equal the distance from the tip of the hour hand to the ceiling x hours after noon. Find the equation that models the motion of the clock and sketch the graph. Solution Begin by making a table of values as shown in Table 4. x Noon 3 PM 6 PM 9 PM Midnight y 30 in 54 in 78 in 54 in 30 in Table 4 Points to plot (0, 30) (3, 54) (6, 78) (9, 54) (12, 30) To model an equation, we first need to find the amplitude. ∣ A ∣ =  78 − 30  ______ 2 The clock’s cycle repeats every 12 hours. Thus, The vertical shift is = 24 B = 2π _ 12 π _ = 6 D = 78 + 30 ______ 2 = 54 There is no horizontal shift, so C = 0. Since the function begins with the minimum value of y when x = 0 (as opposed to the maximum value), we will use the cosine function with the negative value for A. In the form y = A cos(Bx ± C) + D, the equation is See Figure 9. π _ y = −24cos  x  + 54 6 80 70 60 50 40 30 20 10 y (6, 78) (3, 54) (9, 54) (12, 30) (0, 30) 10 2 3 4 5 6 7 8 9 10 11 12 x Figure 9 SECTION 7.6 modeling with trigonometric eQuations 623 Example 6 Determining a Model for Tides The height of the tide in a small beach town is measured along a seawall. Water levels oscillate between 7 feet at low tide and 15 feet at high tide. On a particular day, low tide occurred at 6 AM and high tide occurred at noon. Approximately every 12 hours, the cycle repeats. Find an equation to model the water levels. Solution As the water level varies from 7 ft to 15 ft, we can calculate the amplitude as The cycle repeats every 12 hours; therefore, B is ∣ A ∣ =  (15 − 7)  _______ 2 = 4 2π _ 12 π _ = 6 There is a vertical translation of = 11. Since the value of the function is at a maximum at t = 0, we will use the cosine function, with the positive value for A. (15 + 7) _______ 2 See Figure 10. π _ y = 4cos   t + 11 6 (0, 15) (12, 15) Midline: y = 11 (6, 7) 16 14 12 10 10 11 12 Time Figure 10 Try It #3 The daily temperature in the month of March in a certain city varies from a low of 24°F to a high of 40°F. Find a sinusoidal function to model daily temperature and sketch the graph. Approximate the time when the temperature reaches the freezing point 32°F. Let t = 0 correspond to noon. Example 7 Interpreting the Periodic Behavior Equation The average person’s blood pressure is modeled by the function f(t) = 20 sin(160πt) + 100, where f(t) represents the blood pressure at time t, measured in minutes. Interpret the function in terms of period and frequency. Sketch the graph and find the blood pressure reading. Solution The period is given by 2π ____ 160π 1 __ 80 In a blood pressure function, frequency represents the number of heart beats per minute. Frequency is the reciprocal of period and is given by 2π ___ ω = = ω __ 2π = 160π ____ 2π = 80 See the graph in Figure 11. f(t) = 20sin(160πt) + 100 f(t) 120 100 80 60 1 40 1 20 3 40 t 1 10 Figure 11 The blood pressure reading on the graph is 120  ____ 80 maximum  . _________ minimum 624 CHAPTER 7 trigonometric identities and eQuations 120 _ 80 Analysis Blood pressure of measures the pressure in the arteries when the heart contracts. The bottom number is the minimum or diastolic reading, which measures the pressure in the arteries as the heart relaxes between beats, refilling with blood. Thus, normal blood pressure can be modeled by a periodic function with a maximum of 120 and a minimum of 80. is considered to be normal. The top number is the maximum or systolic reading, which Modeling Harmonic Motion Functions Harmonic motion is a form of periodic motion, but there are factors to consider that differentiate the two types. While general periodic motion applications cycle through their periods with no outside interference, harmonic motion requires a restoring force. Examples of harmonic motion include springs, gravitational force, and magnetic force. Simple Harmonic Motion A type of motion described as simple harmonic motion involves a restoring force but assumes that the motion will continue forever. Imagine a weighted object hanging on a spring, When that object is not disturbed, we say that the object is at rest, or in equilibrium. If the object is pulled down and then released, the force of the spring pulls the object back toward equilibrium and harmonic motion begins. The restoring force is directly proportional to the displacement of the object from its equilibrium point. When t = 0, d = 0. simple harmonic motion We see that simple harmonic motion equations are given in terms of displacement: d = acos(ωt) or d = asin(ωt) where \|a\| is the amplitude, 2π _ ω is the period, and ω _ 2π is the frequency, or the number of cycles per unit of time. Example 8 Finding the Displacement, Period, and Frequency, and Graphing a Function For the given functions, 1. Find the maximum displacement of an object. 2. Find the period or the time required for one vibration. 3. Find the frequency. 4. Sketch the graph. a. y = 5sin(3t) b. y = 6cos(πt) π _ c. y = 5cos  t  2 Solution a. y = 5sin(3t) 1. The maximum displacement is equal to the amplitude, ∣ a ∣ , which is 5. 2. The period is 2π _ ω = 2π _ . 3 3. The frequency is given as ω _ 2π = 3 _ . 2π 4. See Figure 12. The graph indicates the five key points. b. y = 6cos(πt) 2. The period is 1. The maximum displacement is 6. 2π _ ω = ω _ 2π 2π _ π = 2. π 1 _ _ = = . 2π 2 3. The frequency is 4. See Figure 13. y 5 3 1 –1 –3 –5 y 6 4 2 –2 –4 –6 y = 5sin(3t) π 6 π 3 π 2 t 2π 3 Figure 12 y = 6cos(πt) 1 2 3 4 t Figure 13 SECTION 7.6 modeling with trigonometric eQuations 625 π _ c. y = 5cos   t 2 1. The maximum displacement is 5. 2. The period is = 4. 2π _ ω = 2π _ π _ 2 1 _ 3. The frequency is . 4 4. See Figure 14. Damped Harmonic Motion y 5 3 1 –1 –3 –5 π t y = 5cos 2 1 2 3 4 t Figure 14 In reality, a pendulum does not swing back and forth forever, nor does an object on a spring bounce up and down for
ever. Eventually, the pendulum stops swinging and the object stops bouncing and both return to equilibrium. Periodic motion in which an energy-dissipating force, or damping factor, acts is known as damped harmonic motion. Friction is typically the damping factor. In physics, various formulas are used to account for the damping factor on the moving object. Some of these are calculus-based formulas that involve derivatives. For our purposes, we will use formulas for basic damped harmonic motion models. damped harmonic motion In damped harmonic motion, the displacement of an oscillating object from its rest position at time t is given as where c is a damping factor, ∣ a ∣ is the initial displacement and 2π _ ω is the period. f(t) = ae−ctsin(ωt) or f(t) = ae−ctcos(ωt) Example 9 Modeling Damped Harmonic Motion Model the equations that fit the two scenarios and use a graphing utility to graph the functions: Two mass-spring systems exhibit damped harmonic motion at a frequency of 0.5 cycles per second. Both have an initial displacement of 10 cm. The first has a damping factor of 0.5 and the second has a damping factor of 0.1. Solution At time t = 0, the displacement is the maximum of 10 cm, which calls for the cosine function. The cosine function will apply to both models. ω _ 2π We are given the frequency f = of 0.5 cycles per second. Thus, = 0.5 ω _ 2π ω = (0.5)2π The first spring system has a damping factor of c = 0.5. Following the general model for damped harmonic motion, we have f(t) = 10e−0.5t cos(πt) Figure 15 models the motion of the first spring system. = π f(t) 12 8 4 f(t) = 10e–0.5tcos(πt) –15 –9 –3 9 15 t –4 –8 Figure 15 626 CHAPTER 7 trigonometric identities and eQuations The second spring system has a damping factor of c = 0.1 and can be modeled as f (t) = 10e−0.1tcos(πt) Figure 16 models the motion of the second spring system. f(t) f(t) = 10e–0.1tcos (πt) 12 8 4 –12 –8 –4 4 8 12 t –4 –8 –12 Figure 16 Analysis Notice the differing effects of the damping constant. The local maximum and minimum values of the function with the damping factor c = 0.5 decreases much more rapidly than that of the function with c = 0.1. Example 10 Finding a Cosine Function that Models Damped Harmonic Motion Find and graph a function of the form y = ae−ctcos(ωt) that models the information given. a. a = 20, c = 0.05, p = 4 b. a = 2, c = 1.5, f = 3 Solution Substitute the given values into the model. Recall that period is 2π _ ω and frequency is ω _ . 2π π _ a. y = 20e−0.05tcos  t  . See Figure 17. 2 b. y = 2e−1.5tcos(6πt). See Figure 18. y 30 20 10 π t y = 20e–0.05t cos 2 –24 –16 –8 8 16 24 t –10 –20 –30 Figure 17 y 6 4 2 y = 2e–1.5t cos(6πt) –6 –4 –2 2 4 6 t –2 –4 –6 Figure 18 Try It #4 The following equation represents a damped harmonic motion model: f(t) = 5e−6tcos(4t) Find the initial displacement, the damping constant, and the frequency. SECTION 7.6 modeling with trigonometric eQuations 627 Example 11 Finding a Sine Function that Models Damped Harmonic Motion Find and graph a function of the form y = ae−ct sin(ωt) that models the information given. π _ a. a = 7, c = 10, p = 6 b. a = 0.3, c = 0.2, f = 20 Solution Calculate the value of ω and substitute the known values into the model. a. As period is 2π _ ω , we have π 2π _ _ = ω 6 ωπ = 6(2π) ω = 12 The damping factor is given as 10 and the amplitude is 7. Thus, the model is y = 7e−10tsin(12t). See Figure 19. y 3 2 1 –1 y = 7e–10t sin(12t) 1 2 3 4 t Figure 19 20 = ω _ 2π 40π = ω b. As frequency is , we have ω _ 2π The damping factor is given as 0.2 and the amplitude is 0.3. The model is y = 0.3e−0.2tsin(40πt). See Figure 20. y 1 –1 5 10 15 20 25 t –1 Figure 20 Analysis A comparison of the last two examples illustrates how we choose between the sine or cosine functions to model sinusoidal criteria. We see that the cosine function is at the maximum displacement when t = 0, and the sine function π _ is at the equilibrium point when t = 0. For example, consider the equation y = 20e−0.05tcos  t  from Example 10. We 2 can see from the graph that when t = 0, y = 20, which is the initial amplitude. Check this by setting t = 0 in the cosine equation: π _ y = 20e−0.05(0)cos   (0) 2 = 20(1)(1) Using the sine function yields Thus, cosine is the correct function. = 20 π _ y = 20e−0.05(0)sin   (0) 2 = 20(1)(0) = 0 628 CHAPTER 7 trigonometric identities and eQuations Try It #5 Write the equation for damped harmonic motion given a = 10, c = 0.5, and p = 2. Example 12 Modeling the Oscillation of a Spring A spring measuring 10 inches in natural length is compressed by 5 inches and released. It oscillates once every 3 seconds, and its amplitude decreases by 30% every second. Find an equation that models the position of the spring t seconds after being released. Solution The amplitude begins at 5 in. and deceases 30% each second. Because the spring is initially compressed, we will write A as a negative value. We can write the amplitude portion of the function as We put (1 − 0.30)t in the form ect as follows: A(t) = 5(1 − 0.30)t 0.7 = ec c = ln 0.7 c = −0.357 Now let’s address the period. The spring cycles through its positions every 3 seconds, this is the period, and we can use the formula to find omega. 3 = ω = 2π _ ω 2π _ 3 The natural length of 10 inches is the midline. We will use the cosine function, since the spring starts out at its maximum displacement. This portion of the equation is represented as Finally, we put both functions together. Our the model for the position of the spring at t seconds is given as y = cos  2π _ t  + 10 3 See the graph in Figure 21. y = −5e−0.357t cos  2π _ t  + 10 3 y 24 16 8 –8 –24 –16 –8 y = –5e–0.357t cos 3 2πt + 10 8 16 24 t Figure 21 Try It #6 A mass suspended from a spring is raised a distance of 5 cm above its resting position. The mass is released at time 1 _ t = 0 and allowed to oscillate. After second, it is observed that the mass returns to its highest position. Find a 3 function to model this motion relative to its initial resting position. Example 13 Finding the Value of the Damping Constant c According to the Given Criteria A guitar string is plucked and vibrates in damped harmonic motion. The string is pulled and displaced 2 cm from its resting position. After 3 seconds, the displacement of the string measures 1 cm. Find the damping constant. Solution The displacement factor represents the amplitude and is determined by the coefficient ae−ct in the model for damped harmonic motion. The damping constant is included in the term e−ct. It is known that after 3 seconds, the local maximum measures one-half of its original value. Therefore, we have the equation ae−c(t + 3) = 1 _ ae−ct 2 SECTION 7.6 modeling with trigonometric eQuations 629 Use algebra and the laws of exponents to solve for c. Divide out a. Divide out e−ct. Take reciprocals. ae−c(t + 3) = 1 _ ae−ct 2 e−ct · e−3c = 1 _ e−ct 2 e−3c = 1 _ 2 e3c = 2 e3c = 2 3c = ln(2) c = ln(2) _ 3 Then use the laws of logarithms. The damping constant is ln(2) ___ . 3 Bounding Curves in Harmonic Motion Harmonic motion graphs may be enclosed by bounding curves. When a function has a varying amplitude, such that the amplitude rises and falls multiple times within a period, we can determine the bounding curves from part of the function. Example 14 Graphing an Oscillating Cosine Curve Graph the function f(x) = cos(2πx) cos(16πx). Solution The graph produced by this function will be shown in two parts. The first graph will be the exact function f(x) (see Figure 22), and the second graph is the exact function f(x) plus a bounding function (see Figure 23). The graphs look quite different. f(x) 1 –1 f(x) = cos (2πx) cos (16πx) 1 x 2 Figure 22 y 2 1 –1 –2 f(x) = cos (2πx) cos (16πx) y = cos (2πx) 1 y = –cos (2πx) x 2 Figure 23 Analysis The curves y = cos(2πx) and y = −cos(2πx) are bounding curves: they bound the function from above and below, tracing out the high and low points. The harmonic motion graph sits inside the bounding curves. This is an example of a function whose amplitude not only decreases with time, but actually increases and decreases multiple times within a period. Access these online resources for additional instruction and practice with trigonometric applications. • Solving Problems Using Trigonometry (http://openstaxcollege.org/l/solvetrigprob) • Ferris Wheel Trigonometry (http://openstaxcollege.org/l/ferriswheel) • Daily Temperatures and Trigonometry (http://openstaxcollege.org/l/dailytemp) • Simple Harmonic Motion (http://openstaxcollege.org/l/simpleharm) 630 CHAPTER 7 trigonometric identities and eQuations 7.6 SeCTIOn exeRCISeS VeRBAl 1. Explain what types of physical phenomena are best modeled by sinusoidal functions. What are the characteristics necessary? 3. If we want to model cumulative rainfall over the course of a year, would a sinusoidal function be a good model? Why or why not? 2. What information is necessary to construct a trigonometric model of daily temperature? Give examples of two different sets of information that would enable modeling with an equation. 4. Explain the effect of a damping factor on the graphs of harmonic motion functions. AlGeBRAIC For the following exercises, find a possible formula for the trigonometric function represented by the given table of values. 5. x 3 0 y −4 −1 7. x y 0 2 9. x 0 y −2 π _ 4 7 1 4 9 12 15 6 2 −1 −4 −1 18 2 π 3π π _ _ 4 2 2 −3 2 5π _ 4 7 3π _ 2 2 2 10 4 3 4 −2 5 4 6 10 6. x y 8. x y 10. x y 0 5 4 2 1 −3 6 1 8 5 12 10 1 −3 2 1 3 0 1 −3 −7 −3 6 5 4 1 −3 −7 1 0 5 −3 2 5 3 13 5 4 5 −3 6 5 11. x y 12. x −3 −1 − √ — 2 −1 y — 3 − 2 √ −2 −1 0 0 − √ ____ GRAPHICAl For the following exercises, graph the given function, and then find a possible physical process that the equation could model. 13. f (x) = −30cos  xπ _  − 20cos2  6 xπ _  + 80 [0, 12] 6 14. f (x) = −18cos   − 5sin   + 100 on the interval [0, 24] xπ _ 12 xπ _ 12 15. f (x) = 10 − sin  xπ _  + 24tan  6 xπ _ 240  on the interval [0, 80] TeCHnOlOGY For the follow
ing exercise, construct a function modeling behavior and use a calculator to find desired results. 16. A city’s average yearly rainfall is currently 20 inches and varies seasonally by 5 inches. Due to unforeseen circumstances, rainfall appears to be decreasing by 15% each year. How many years from now would we expect rainfall to initially reach 0 inches? Note, the model is invalid once it predicts negative rainfall, so choose the first point at which it goes below 0. SECTION 7.6 section exercises 631 ReAl-WORlD APPlICATIOnS For the following exercises, construct a sinusoidal function with the provided information, and then solve the equation for the requested values. 17. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of 105°F occurs at 5PM and the average temperature for the day is 85°F. Find the temperature, to the nearest degree, at 9AM. 18. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of 84°F occurs at 6PM and the average temperature for the day is 70°F. Find the temperature, to the nearest degree, at 7AM. 19. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the temperature varies between 47°F and 63°F during the day and the average daily temperature first occurs at 10 AM. How many hours after midnight does the temperature first reach 51°F? 20. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the temperature varies between 64°F and 86°F during the day and the average daily temperature first occurs at 12 AM. How many hours after midnight does the temperature first reach 70°F? 21. A Ferris wheel is 20 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o’clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 6 minutes. How much of the ride, in minutes and seconds, is spent higher than 13 meters above the ground? 22. A Ferris wheel is 45 meters in diameter and boarded from a platform that is 1 meter above the ground. The six o’clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. How many minutes of the ride are spent higher than 27 meters above the ground? Round to the nearest second 23. The sea ice area around the North Pole fluctuates between about 6 million square kilometers on September 1 to 14 million square kilometers on March 1. Assuming a sinusoidal fluctuation, when are there less than 9 million square kilometers of sea ice? Give your answer as a range of dates, to the nearest day. 24. The sea ice area around the South Pole fluctuates between about 18 million square kilometers in September to 3 million square kilometers in March. Assuming a sinusoidal fluctuation, when are there more than 15 million square kilometers of sea ice? Give your answer as a range of dates, to the nearest day. 25. During a 90-day monsoon season, daily rainfall can be modeled by sinusoidal functions. If the rainfall fluctuates between a low of 2 inches on day 10 and 12 inches on day 55, during what period is daily rainfall more than 10 inches? 26. During a 90-day monsoon season, daily rainfall can be modeled by sinusoidal functions. A low of 4 inches of rainfall was recorded on day 30, and overall the average daily rainfall was 8 inches. During what period was daily rainfall less than 5 inches? 27. In a certain region, monthly precipitation peaks at 8 inches on June 1 and falls to a low of 1 inch on December 1. Identify the periods when the region is under flood conditions (greater than 7 inches) and drought conditions (less than 2 inches). Give your answer in terms of the nearest day. 28. In a certain region, monthly precipitation peaks at 24 inches in September and falls to a low of 4 inches in March. Identify the periods when the region is under flood conditions (greater than 22 inches) and drought conditions (less than 5 inches). Give your answer in terms of the nearest day. For the following exercises, find the amplitude, period, and frequency of the given function. 29. The displacement h(t) in centimeters of a mass 30. The displacement h(t) in centimeters of a mass suspended by a spring is modeled by the function h(t) = 8sin(6πt),where t is measured in seconds. Find the amplitude, period, and frequency of this displacement. suspended by a spring is modeled by the function h(t) = 11sin(12πt), where t is measured in seconds. Find the amplitude, period, and frequency of this displacement. 632 CHAPTER 7 trigonometric identities and eQuations 31. The displacement h(t) in centimeters of a mass suspended by a spring is modeled by the function π _ h(t) = 4cos  t  , where t is measured in seconds. Find the amplitude, period, and frequency of this displacement. 2 For the following exercises, construct an equation that models the described behavior. 32. The displacement h(t), in centimeters, of a mass suspended by a spring is modeled by the function h(t) = −5 cos(60πt), where t is measured in seconds. Find the amplitude, period, and frequency of this displacement. For the following exercises, construct an equation that models the described behavior. 33. A deer population oscillates 19 above and below average during the year, reaching the lowest value in January. The average population starts at 800 deer and increases by 160 each year. Find a function that models the population, P, in terms of months since January, t. 34. A rabbit population oscillates 15 above and below average during the year, reaching the lowest value in January. The average population starts at 650 rabbits and increases by 110 each year. Find a function that models the population, P, in terms of months since January, t. 35. A muskrat population oscillates 33 above and below average during the year, reaching the lowest value in January. The average population starts at 900 muskrats and increases by 7% each month. Find a function that models the population, P, in terms of months since January, t. 36. A fish population oscillates 40 above and below average during the year, reaching the lowest value in January. The average population starts at 800 fish and increases by 4% each month. Find a function that models the population, P, in terms of months since January, t. 37. A spring attached to the ceiling is pulled 10 cm down from equilibrium and released. The amplitude decreases by 15% each second. The spring oscillates 18 times each second. Find a function that models the distance, D, the end of the spring is from equilibrium in terms of seconds, t, since the spring was released. 38. A spring attached to the ceiling is pulled 7 cm down from equilibrium and released. The amplitude decreases by 11% each second. The spring oscillates 20 times each second. Find a function that models the distance, D, the end of the spring is from equilibrium in terms of seconds, t, since the spring was released. 39. A spring attached to the ceiling is pulled 17 cm down 40. A spring attached to the ceiling is pulled 19 cm down from equilibrium and released. After 3 seconds, the amplitude has decreased to 13 cm. The spring oscillates 14 times each second. Find a function that models the distance, D, the end of the spring is from equilibrium in terms of seconds, t, since the spring was released. from equilibrium and released. After 4 seconds, the amplitude has decreased to 14 cm. The spring oscillates 13 times each second. Find a function that models the distance, D, the end of the spring is from equilibrium in terms of seconds, t, since the spring was released. For the following exercises, create a function modeling the described behavior. Then, calculate the desired result using a calculator. 41. A certain lake currently has an average trout population of 20,000. The population naturally oscillates above and below average by 2,000 every year. This year, the lake was opened to fishermen. If fishermen catch 3,000 fish every year, how long will it take for the lake to have no more trout? 42. Whitefish populations are currently at 500 in a lake. The population naturally oscillates above and below by 25 each year. If humans overfish, taking 4% of the population every year, in how many years will the lake first have fewer than 200 whitefish? 43. A spring attached to a ceiling is pulled down 11 cm from equilibrium and released. After 2 seconds, the amplitude has decreased to 6 cm. The spring oscillates 8 times each second. Find when the spring first comes between −0.1 and 0.1 cm, effectively at rest. 44. A spring attached to a ceiling is pulled down 21 cm from equilibrium and released. After 6 seconds, the amplitude has decreased to 4 cm. The spring oscillates 20 times each second. Find when the spring first comes between −0.1 and 0.1 cm, effectively at rest. SECTION 7.6 section exercises 633 45. Two springs are pulled down from the ceiling and released at the same time. The first spring, which oscillates 8 times per second, was initially pulled down 32 cm from equilibrium, and the amplitude decreases by 50% each second. The second spring, oscillating 18 times per second, was initially pulled down 15 cm from equilibrium and after 4 seconds has an amplitude of 2 cm. Which spring comes to rest first, and at what time? Consider “rest” as an amplitude less than 0.1 cm. 46. Two springs are pulled down from the ceiling and released at the same time. The first spring, which oscillates 14 times per second, was initially pulled down 2 cm from equilibrium, and the amplitude decreases by 8% each second. The second spring, oscillating 22 times per second, was initially pulled down 10 cm from equilibrium and after 3 seconds has an amplitude of 2 cm. Which spring comes to rest first, and at what time? Consider “rest” as an amplitude less than 0.1 cm. exTenSIOnS 47. A plane flies 1 hour at 150 mph at 22° east of nort
h, then continues to fly for 1.5 hours at 120 mph, this time at a bearing of 112° east of north. Find the total distance from the starting point and the direct angle flown north of east. 48. A plane flies 2 hours at 200 mph at a bearing of 60°, then continues to fly for 1.5 hours at the same speed, this time at a bearing of 150°. Find the distance from the starting point and the bearing from the starting point. Hint: bearing is measured counterclockwise from north. π _ For the following exercises, find a function of the form y = abx + csin  x  that fits the given data. 2 49. x y 0 6 1 29 2 96 3 379 50. x y 0 6 1 34 2 150 3 746 51. x y 0 4 1 0 3 2 16 −40 π _ x  + c that fits the given data. For the following exercises, find a function of the form y = abx cos  2 52. x y 53. x y 1 2 1 −11 0 11 634 CHAPTER 7 trigonometric identities and eQuations CHAPTeR 7 ReVIeW Key Terms damped harmonic motion oscillating motion that resembles periodic motion and simple harmonic motion, except that the graph is affected by a damping factor, an energy dissipating influence on the motion, such as friction double-angle formulas identities derived from the sum formulas for sine, cosine, and tangent in which the angles are equal even-odd identities set of equations involving trigonometric functions such that if f (−x) = −f (x), the identity is odd, and if f (−x) = f (x), the identity is even half-angle formulas identities derived from the reduction formulas and used to determine half-angle values of trigonometric functions product-to-sum formula a trigonometric identity that allows the writing of a product of trigonometric functions as a sum or difference of trigonometric functions Pythagorean identities set of equations involving trigonometric functions based on the right triangle properties quotient identities pair of identities based on the fact that tangent is the ratio of sine and cosine, and cotangent is the ratio of cosine and sine reciprocal identities set of equations involving the reciprocals of basic trigonometric definitions reduction formulas identities derived from the double-angle formulas and used to reduce the power of a trigonometric function simple harmonic motion a repetitive motion that can be modeled by periodic sinusoidal oscillation sum-to-product formula a trigonometric identity that allows, by using substitution, the writing of a sum of trigonometric functions as a product of trigonometric functions Key equations Pythagorean identities Even-odd identities Reciprocal identities sin2 θ + cos2 θ = 1 1 + cot2 θ = csc2 θ 1 + tan2 θ = sec2 θ tan(−θ) = −tan θ cot(−θ) = −cot θ sin(−θ) = −sin θ csc(−θ) = −csc θ cos(−θ) = cos θ sec(−θ) = sec θ sin θ = 1 _ csc θ cos θ = 1 _ sec θ tan θ = 1 _ cot θ csc θ = 1 _ sin θ sec θ = 1 _ cos θ cot θ = 1 _ tan θ CHAPTER 7 review 635 Quotient identities tan θ = sin θ _ cos θ cot θ = cos θ _ sin θ Sum Formula for Cosine cos(α + β) = cos α cos β − sin α sin β Difference Formula for Cosine cos(α − β) = cos α cos β + sin α sin β Sum Formula for Sine sin(α + β)= sin α cos β + cos α sin β Difference Formula for Sine sin(α − β) = sin α cos β − cos α sin β Sum Formula for Tangent Difference Formula for Tangent Cofunction identities Double-angle formulas Reduction formulas tan(α + β) = tan α + tan β ___________ 1 − tan α tan β tan(α − β) = tan α − tan β ___________ 1 + tan α tan β π _ sin θ = cos  − θ  2 π _ − θ  cos θ = sin  2 π _ − θ  tan θ = cot  2 π _ − θ  cot θ = tan  2 π _ − θ  sec θ = csc  2 π _ − θ  csc θ = sec  2 sin(2θ) = 2sin θ cos θ cos(2θ) = cos2 θ − sin2 θ = 1 − 2sin2 θ = 2cos2 θ − 1 tan(2θ) = 2tan θ ________ 1 − tan2 θ sin2θ = 1 − cos(2θ) _________ 2 cos2θ = 1 + cos(2θ) _________ 2 tan2θ = 1 − cos(2θ) _________ 1 + cos(2θ) 636 CHAPTER 7 trigonometric identities and eQuations Half-angle formulas Product-to-sum Formulas Sum-to-product Formulas α _ sin 2 α _ cos 2 α _ tan 2 = ± √ = ± √ = ± √ _________ 1 − cos α _______ 2 _________ 1 + cos α _______ 2 _________ 1 − cos α _______ 1 + cos α = = sin α _______ 1 + cos α 1 − cos α _______ sin α 1 __ [cos(α − β) + cos(α + β)] cos α cos β = 2 1 __ [sin(α + β) + sin(α − β)] sin α cos β = 2 1 __ [cos(α − β) − cos(α + β)] sin α sin β = 2 1 __ [sin(α + β) − sin(α − β)] cos α sin β = 2 sin α + sin β = 2 sin  sin α − sin β = 2 sin  cos α − cos β = −2 sin  cos α + cos β = 2cos   cos   cos  α + β _____ 2 α − β _____ 2 α + β _____ 2 α + β _____ 2  cos  α − β  _____ 2 α + β  _____ 2 α − β  _____ 2 α − β  _____ 2  sin  Standard form of sinusoidal equation y = A sin(Bt − C) + D or y = A cos(Bt − C) + D Simple harmonic motion Damped harmonic motion Key Concepts d = a cos(ωt) or d = a sin(ωt) f (t) = ae−ct sin(ωt) or f (t) = ae−ct cos(ωt) 7.1 Solving Trigonometric Equations with Identities • There are multiple ways to represent a trigonometric expression. Verifying the identities illustrates how expressions can be rewritten to simplify a problem. • Graphing both sides of an identity will verify it. See Example 1. • Simplifying one side of the equation to equal the other side is another method for verifying an identity. See Example 2 and Example 3. • The approach to verifying an identity depends on the nature of the identity. It is often useful to begin on the more complex side of the equation. See Example 4. • We can create an identity by simplifying an expression and then verifying it. See Example 5. • Verifying an identity may involve algebra with the fundamental identities. See Example 6 and Example 7. CHAPTER 7 review 637 • Algebraic techniques can be used to simplify trigonometric expressions. We use algebraic techniques throughout this text, as they consist of the fundamental rules of mathematics. See Example 8, Example 9, and Example 10. 7.2 Sum and Difference Identities • The sum formula for cosines states that the cosine of the sum of two angles equals the product of the cosines of the angles minus the product of the sines of the angles. The difference formula for cosines states that the cosine of the difference of two angles equals the product of the cosines of the angles plus the product of the sines of the angles. • The sum and difference formulas can be used to find the exact values of the sine, cosine, or tangent of an angle. See Example 1 and Example 2. • The sum formula for sines states that the sine of the sum of two angles equals the product of the sine of the first angle and cosine of the second angle plus the product of the cosine of the first angle and the sine of the second angle. The difference formula for sines states that the sine of the difference of two angles equals the product of the sine of the first angle and cosine of the second angle minus the product of the cosine of the first angle and the sine of the second angle. See Example 3. • The sum and difference formulas for sine and cosine can also be used for inverse trigonometric functions. See Example 4. • The sum formula for tangent states that the tangent of the sum of two angles equals the sum of the tangents of the angles divided by 1 minus the product of the tangents of the angles. The difference formula for tangent states that the tangent of the difference of two angles equals the difference of the tangents of the angles divided by 1 plus the product of the tangents of the angles. See Example 5. • The Pythagorean Theorem along with the sum and difference formulas can be used to find multiple sums and differences of angles. See Example 6. • The cofunction identities apply to complementary angles and pairs of reciprocal functions. See Example 7. • Sum and difference formulas are useful in verifying identities. See Example 8 and Example 9. • Application problems are often easier to solve by using sum and difference formulas. See Example 10 and Example 11. 7.3 Double-Angle, Half-Angle, and Reduction Formulas • Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent. See Example 1, Example 2, Example 3, and Example 4. • Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term. See Example 5 and Example 6. • Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not. See Example 7, Example 8, and Example 9. 7.4 Sum-to-Product and Product-to-Sum Formulas • From the sum and difference identities, we can derive the product-to-sum formulas and the sum-to-product formulas for sine and cosine. • We can use the product-to-sum formulas to rewrite products of sines, products of cosines, and products of sine and cosine as sums or differences of sines and cosines. See Example 1, Example 2, and Example 3. • We can also derive the sum-to-product identities from the product-to-sum identities using substitution. • We can use the sum-to-product formulas to rewrite sum or difference of sines, cosines, or products sine and cosine as products of sines and cosines. See Example 4. 638 CHAPTER 7 trigonometric identities and eQuations • Trigonometric expressions are often simpler to evaluate using the formulas. See Example 5. • The identities can be verified using other formulas or by converting the expressions to sines and cosines. To verify an identity, we choose the more complicated side of the equals sign and rewrite it until it is transformed into the other side. See Example 6 and Example 7. 7.5 Solving Trigonometric Equations • When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution. See Example 1, Example 2, and Example 3. • Equations involving a single trigonometric function can be solved or verified using the unit circle. See Example 4, Example 5, and Example 6, and Example 7. • We can also solve trigonometric equations using a graphing calculator. See Exampl
e 8 and Example 9. • Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc. See Example 10, Example 11, Example 12, and Example 13. • We can also use the identities to solve trigonometric equation. See Example 14, Example 15, and Example 16. • We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval. See Example 17. • Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions. See Example 18. 7.6 Modeling with Trigonometric Equations • Sinusoidal functions are represented by the sine and cosine graphs. In standard form, we can find the amplitude, period, and horizontal and vertical shifts. See Example 1 and Example 2. • Use key points to graph a sinusoidal function. The five key points include the minimum and maximum values and the midline values. See Example 3. • Periodic functions can model events that reoccur in set cycles, like the phases of the moon, the hands on a clock, and the seasons in a year. See Example 4, Example 5, Example 6 and Example 7. • Harmonic motion functions are modeled from given data. Similar to periodic motion applications, harmonic motion requires a restoring force. Examples include gravitational force and spring motion activated by weight. See Example 8. • Damped harmonic motion is a form of periodic behavior affected by a damping factor. Energy dissipating factors, like friction, cause the displacement of the object to shrink. See Example 9, Example 10, Example 11, Example 12, and Example 13. • Bounding curves delineate the graph of harmonic motion with variable maximum and minimum values. See Example 14. CHAPTER 7 review 639 CHAPTeR 7 ReVIeW exeRCISeS SOlVInG TRIGOnOMeTRIC eQUATIOnS WITH IDenTITIeS For the following exercises, find all solutions exactly that exist on the interval [0, 2π). 1. csc2 t = 3 4. tan x sin x + sin(−x) = 0 2. cos2 x = 1 __ 4 5. 9sin ω − 2 = 4 sin2 ω 3. 2sin θ = −1 6. 1 − 2tan(ω) = tan2(ω) For the following exercises, use basic identities to simplify the expression. 7. sec x cos x + cos x − 1 ____ sec x 8. sin3 x + cos2 x sin x For the following exercises, determine if the given identities are equivalent. 9. sin2 x + sec2 x − 1 = (1 − cos2 x)(1 + cos2 x) __ cos2 x 10. tan3 x csc2 x cot2 x cos x sin x = 1 SUM AnD DIFFeRenCe IDenTITIeS For the following exercises, find the exact value. 11. tan  7π ___  12 12. cos  25π ___  12 13. sin(70°)cos(25°) − cos(70°)sin(25°) 14. cos(83°)cos(23°) + sin(83°)sin(23°) For the following exercises, prove the identity. 15. cos(4x) − cos(3x)cosx = sin2 x − 4cos2 x sin2 x 16. cos(3x) − cos3 x = − cos x sin2 x − sin x sin(2x) For the following exercise, simplify the expression. 1 1 __ __ x  + tan  x  tan  8 2 ___ 1 1 __ __ x  x  tan  1 − tan  2 8 17. For the following exercises, find the exact value. 1 __   18. cos  sin−1 (0) − cos−1  2 1 __   19. tan  sin−1 (0) + sin−1  2 DOUBle-AnGle, HAlF-AnGle, AnD ReDUCTIOn FORMUlAS For the following exercises, find the exact value. 20. Find sin(2θ), cos(2θ), and tan(2θ) given π cos θ = − 1 __ __ , π  and θ is in the interval  3 2 21. Find sin(2θ), cos(2θ), and tan(2θ) given π sec θ = − 5 __ __ , π  and θ is in the interval  2 3 22. sin  7π ___  8 23. sec  3π ___  8 640 CHAPTER 7 trigonometric identities and eQuations For the following exercises, use Figure 1 to find the desired quantities. α 25 β 24 Figure 1 24. sin(2β), cos(2β), tan(2β), sin(2α), cos(2α), and tan(2α) α α β β β  , sin   , tan   , cos  25. sin  __ __ __ __ __  , cos   , 2 2 2 2 2 α __ and tan   2 For the following exercises, prove the identity. 26. 2cos(2x) _______ sin(2x) = cot x − tan x 27. cot x cos(2x) = − sin(2x) + cot x For the following exercises, rewrite the expression with no powers. 28. cos2 x sin4(2x) 29. tan2 x sin3 x SUM-TO-PRODUCT AnD PRODUCT-TO-SUM FORMUlAS For the following exercises, evaluate the product for the given expression using a sum or difference of two functions. Write the exact answer. π π _ __   sin  30. cos  3 4 π π __ __   cos  32. 2cos  3 5 31. 2sin  2π __  sin  3 5π __  6 For the following exercises, evaluate the sum by using a product formula. Write the exact answer. 33. sin  π __ 12  − sin  7π __  12 34. cos  5π __ 12  + cos  7π __  12 For the following exercises, change the functions from a product to a sum or a sum to a product. 35. sin(9x)cos(3x) 37. sin(11x) + sin(2x) 36. cos(7x)cos(12x) 38. cos(6x) + cos(5x) SOlVInG TRIGOnOMeTRIC eQUATIOnS For the following exercises, find all exact solutions on the interval [0, 2π). 39. tan x + 1 = 0 40. 2sin(2x) + √ — 2 = 0 For the following exercises, find all exact solutions on the interval [0, 2π). 41. 2sin2 x − sin x = 0 43. 2sin2 x + 5 sin x + 3 = 0 42. cos2 x − cos x − 1 = 0 44. cos x − 5sin(2x) = 0 45. 1 ____ sec2 x + 2 + sin2 x + 4cos2 x = 0 CHAPTER 7 review 641 For the following exercises, simplify the equation algebraically as much as possible. Then use a calculator to find the solutions on the interval [0, 2π). Round to four decimal places. 46. √ — 3 cot2 x + cot x = 1 47. csc2 x − 3csc x − 4 = 0 For the following exercises, graph each side of the equation to find the zeroes on the interval [0, 2π). 48. 20cos2 x + 21cos x + 1 = 0 49. sec2 x − 2sec x = 15 MODelInG WITH TRIGOnOMeTRIC eQUATIOnS For the following exercises, graph the points and find a possible formula for the trigonometric values in the given table. 50. x y 52. x y 0 1 1 6 2 11 −2 3 4 1 5 6 51. x y 0 −2 −2 −5 −1 3 −1 − 2 √ — 2 53. A man with his eye level 6 feet above the ground is standing 3 feet away from the base of a 15-foot vertical ladder. If he looks to the top of the ladder, at what angle above horizontal is he looking? 54. Using the ladder from the previous exercise, if a 6-foot-tall construction worker standing at the top of the ladder looks down at the feet of the man standing at the bottom, what angle from the horizontal is he looking? For the following exercises, construct functions that model the described behavior. 55. A population of lemmings varies with a yearly low of 500 in March. If the average yearly population of lemmings is 950, write a function that models the population with respect to t, the month. 56. Daily temperatures in the desert can be very extreme. If the temperature varies from 90°F to 30°F and the average daily temperature first occurs at 10 AM, write a function modeling this behavior. For the following exercises, find the amplitude, frequency, and period of the given equations. 57. y = 3cos(xπ) 58. y = −2sin(16xπ) For the following exercises, model the described behavior and find requested values. 59. An invasive species of carp is introduced to Lake Freshwater. Initially there are 100 carp in the lake and the population varies by 20 fish seasonally. If by year 5, there are 625 carp, find a function modeling the population of carp with respect to t, the number of years from now. 60. The native fish population of Lake Freshwater averages 2500 fish, varying by 100 fish seasonally. Due to competition for resources from the invasive carp, the native fish population is expected to decrease by 5% each year. Find a function modeling the population of native fish with respect to t, the number of years from now. Also determine how many years it will take for the carp to overtake the native fish population. 642 CHAPTER 7 trigonometric identities and eQuations CHAPTeR 7 PRACTICe TeST For the following exercises, simplify the given expression. 1. cos(−x)sin x cot x + sin2 x 2. sin(−x)cos(−2x)−sin(−x)cos(−2x) For the following exercises, find the exact value. 3. cos  7π __  12 4. tan  3π ___  8 — 2  + tan−1 √ 5. tan  sin−1  √ ____ 2 — 3  π π __ __   sin  6. 2sin  6 4 For the following exercises, find all exact solutions to the equation on [0, 2π). 7. cos2 x − sin2 x − 1 = 0 9. cos(2x) + sin2 x = 0 11. Rewrite the expression as a product instead of a sum: cos(2x) + cos(−8x). 13. Find the solutions of sec2 x − 2sec x = 15 on the interval [0, 2π) algebraically; then graph both sides of the equation to determine the answer. θ θ θ  given  , and tan   , cos  15. Find sin  __ __ __ 2 2 2 7 __ cos θ = 25 and θ is in quadrant IV. For the following exercises, prove the identity. 8. cos2 x = cos x 4sin2 x + 2sin x − 3 = 0 10. 2sin2 x − sin x = 0 12. Find all solutions of tan(x) − √ — 3 = 0. 14. Find sin(2θ), cos(2θ), and tan(2θ) given cot θ = − 3 π __ __ , π  . and θ is on the interval  2 4 16. Rewrite the expression sin4 x with no powers greater than 1. 17. tan3 x − tan x sec2 x = tan(−x) 18. sin(3x) − cos x sin(2x) = cos2 x sin x − sin3 x 19. sin(2x) ______ − sin x cos(2x) ______ cos x = sec x 20. Plot the points and find a function of the form y = Acos(Bx + C) + D that fits the given data. x y 0 −2 1 2 2 −2 3 2 4 −2 5 2 21. The displacement h(t) in centimeters of a mass 22. A woman is standing 300 feet away from a 2,000- suspended by a spring is modeled by the function 1 __ h(t) = sin(120πt), where t is measured in seconds. 4 Find the amplitude, period, and frequency of this displacement. foot building. If she looks to the top of the building, at what angle above horizontal is she looking? A bored worker looks down at her from the 15th floor (1500 feet above her). At what angle is he looking down at her? Round to the nearest tenth of a degree. 23. Two frequencies of sound are played on an instrument governed by the equation n(t) = 8 cos(20πt)cos(1,000πt). What are the period and frequency of the “fast” and “slow” oscillations? What is the amplitude? 24. The average monthly snowfall in a small village in the Himalayas is 6 inches, with the low of 1 inch occurring in July. Construct a function
that models this behavior. During what period is there more than 10 inches of snowfall? 25. A spring attached to a ceiling is pulled down 20 cm. After 3 seconds, wherein it completes 6 full periods, the amplitude is only 15 cm. Find the function modeling the position of the spring t seconds after being released. At what time will the spring come to rest? In this case, use 1 cm amplitude as rest. 26. Water levels near a glacier currently average 9 feet, varying seasonally by 2 inches above and below the average and reaching their highest point in January. Due to global warming, the glacier has begun melting faster than normal. Every year, the water levels rise by a steady 3 inches. Find a function modeling the depth of the water t months from now. If the docks are 2 feet above current water levels, at what point will the water first rise above the docks? 8 Further Applications of Trigonometry Figure 1 General Sherman, the world’s largest living tree. (credit: Mike Baird, Flickr) CHAPTeR OUTlIne 8.1 non-right Triangles: law of Sines 8.2 non-right Triangles: law of Cosines 8.3 Polar Coordinates 8.4 Polar Coordinates: Graphs 8.5 Polar Form of Complex numbers 8.6 Parametric equations 8.7 Parametric equations: Graphs 8.8 Vectors Introduction The world’s largest tree by volume, named General Sherman, stands 274.9 feet tall and resides in Northern California.[27] Just how do scientists know its true height? A common way to measure the height involves determining the angle of elevation, which is formed by the tree and the ground at a point some distance away from the base of the tree. This method is much more practical than climbing the tree and dropping a very long tape measure. In this chapter, we will explore applications of trigonometry that will enable us to solve many different kinds of problems, including finding the height of a tree. We extend topics we introduced in Trigonometric Functions and investigate applications more deeply and meaningfully. 27 Source: National Park Service. "The General Sherman Tree." http://www.nps.gov/seki/naturescience/sherman.htm. Accessed April 25, 2014. 643 644 CHAPTER 8 Further applications oF trigonometry leARnInG OBjeCTIVeS In this section, you will: • Use the Law of Sines to solve oblique triangles. • Find the area of an oblique triangle using the sine function. • Solve applied problems using the Law of Sines. 8.1 nOn-RIGHT TRIAnGleS: lAW OF SIneS Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? We see in Figure 1 that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving non-right triangles. 15° 35° 20 miles Figure 1 Using the law of Sines to Solve Oblique Triangles In any triangle, we can draw an altitude, a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles. Any triangle that is not a right triangle is an oblique triangle. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations: 1. ASA (angle-side-angle) We know the measurements of two angles and the included side. See Figure 2. β α γ Figure 2 2. AAS (angle-angle-side) We know the measurements of two angles and a side that is not between the known angles. See Figure 3. β α γ Figure 3 3. SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides. See Figure 4. β α γ Figure 4 SECTION 8.1 non-right triangles: law oF sines 645 Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let’s see how this statement is derived by considering the triangle shown in Figure 5. γ b h a α c Figure 5 β h h _ _ and sin β = Using the right triangle relationships, we know that sin α = a . Solving both equations for h gives two b different expressions for h. We then set the expressions equal to each other. h = bsin α and h = asin β bsin α = asin β Similarly, we can compare the other ratios. 1 _  ab 1 _  (bsin α) = (asin β)  ab = sin α _ a sin β _ b 1 _  Multiply both sides by . ab sin α _ a = sin γ _ c and sin β _ b = sin γ _ c Collectively, these relationships are called the Law of Sines. sin α _ a = sin β _ b = sin γ _ c Note the standard way of labeling triangles: angle α (alpha) is opposite side a; angle β (beta) is opposite side b; and angle γ (gamma) is opposite side c. See Figure 6. While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified. β c a α b Figure 6 γ Law of Sines Given a triangle with angles and opposite sides labeled as in Figure 6, the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines is based on proportions and is presented symbolically two ways. = = sin α _ a a _ sin α = sin β _ b b _ sin β = sin γ _ c c _ sin γ To solve an oblique triangle, use any pair of applicable ratios. 646 CHAPTER 8 Further applications oF trigonometry Example 1 Solving for Two Unknown Sides and Angle of an AAS Triangle Solve the triangle shown in Figure 7 to the nearest tenth. β c 50° α 10 b Figure 7 30° γ Solution The three angles must add up to 180 degrees. From this, we can determine that β = 180° − 50° − 30° = 100° To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle α = 50° and its corresponding side a = 10. We can use the following proportion from the Law of Sines to find the length of c. = sin(30°) _ c sin(50°) _ 10 sin(50°) _ c 10 = sin(30°) c = sin(30°) c ≈ 6.5 10 _ sin(50°) Multiply both sides by c. Multiply by the reciprocal to isolate c. Similarly, to solve for b, we set up another proportion. sin(50°) _ 10 = sin(100°) _ b bsin(50°) = 10sin(100°) Multiply both sides by b. b = 10sin(100°) _ sin(50°) b ≈ 12.9 Multiply by the reciprocal to isolate b. Therefore, the complete set of angles and sides is α = 50° β = 100° γ = 30° a = 10 b ≈ 12.9 c ≈ 6.5 Try It #1 Solve the triangle shown in Figure 8 to the nearest tenth. β c a 43° 98° α 22 Figure 8 Using The law of Sines to Solve SSA Triangles We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution. SECTION 8.1 non-right triangles: law oF sines 647 possible outcomes for SSA triangles Oblique triangles in the category SSA may have four different outcomes. Figure 9 illustrates the solutions with the known sides a and b and known angle α. No triangle, a < h γ γ Right triangle, a = h Two triangles, a > h, a < b One trianglea) β α (b) β α (c) Figure 9 β α (d) β Example 2 Solving an Oblique SSA Triangle Solve the triangle in Figure 10 for the missing side and find the missing angle measures to the nearest tenth. γ 8 6 35° α β Figure 10 Solution Use the Law of Sines to find angle β and angle γ, and then side c. Solving for β, we have the proportion = = sin β _ b sin β _ 8 sin α _ a sin(35°) _ 6 8sin(35°) _ 6 0.7648 ≈ sin β sin−1(0.7648) ≈ 49.9° β ≈ 49.9° = sin β However, in the diagram, angle β appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of β? Let’s investigate further. Dropping a perpendicular from γ and viewing the triangle from a right angle perspective, we have Figure 11. It appears that there may be a second triangle that will fit the given criteria. γ' 8 6 6 α' 35° β Figure 11 β' φ The angle supplementary to β is approximately equal to 49.9°, which means that β = 180° − 49.9° = 130.1°. (Remember that the sine function is positive in both the first and second quadrants.) Solving for γ, we have γ = 180° − 35° − 130.1° ≈ 14.9° 648 CHAPTER 8 Further applications oF trigonometry We can then use these measurements to solve the other triangle. Since γ' is supplementary to α and β, we have γ' = 180° − 35° − 49.9° ≈ 95.1° Now we need to find c and c'. We have Finally, c _ = sin(14.9°) 6 _ sin(35°) c = 6sin(14.9°) _ sin(35°) ≈ 2.7 c' _ = sin(95.1°) 6 _ sin(35°) c' = 6sin(95.1°) _ sin(35°) ≈ 10.4 To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in Figure 12. γ 14.9° a = 6 130.1° β 35° α' Figure 12 γ' 95.1° a' = 6 49.9° β' b' = 8 c' ≈ 10.4 (b) b = 8 35° α c ≈ 2.7 (a) However, we were looking for the values for the triangle with an obtuse angle β. We can see them in the first triangle (a) in Figure 12. Try It #2 Given α = 80°, a = 120, and b = 121, find the missing side and angles. If there is more than one possible solution, show both. Example 3 Solving for the Unknown Sides and Angles of a SSA Triangle In the triangle shown in Figure 13, solve for the unknown side and angles. Round your an
swers to the nearest tenth. 12 β a α 85° 9 Figure 13 Solution In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle γ = 85°, and its corresponding side c = 12, and we know side b = 9. We will use this proportion to solve for β. sin(85°) _ 12 = sin β _ 9 9sin(85°) _ 12 = sin β Isolate the unknown. SECTION 8.1 non-right triangles: law oF sines 649 To find β, apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for β. It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions. 9sin(85°)  β = sin−1  _ 12 β ≈ sin−1(0.7471) β ≈ 48.3° In this case, if we subtract β from 180°, we find that there may be a second possible solution. Thus, β = 180° − 48.3° ≈ 131.7°. To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives α = 180° − 85° − 131.7° ≈ − 36.7°, which is impossible, and so β ≈ 48.3°. To find the remaining missing values, we calculate α = 180° − 85° − 48.3° ≈ 46.7°. Now, only side a is needed. Use the Law of Sines to solve for a by one of the proportions. sin(85°) _ 12 sin(85°) _ 12 a = sin(46.7°) _ a = sin(46.7°) a = 12sin(46.7°) __ sin(85°) ≈ 8.8 The complete set of solutions for the given triangle is α ≈ 46.7° β ≈ 48.3° γ = 85° a ≈ 8.8 b = 9 c = 12 Try It #3 Given α = 80°, a = 100, b = 10, find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth. Example 4 Finding the Triangles That Meet the Given Criteria Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10. Solution Using the given information, we can solve for the angle opposite the side of length 10. See Figure 14. = sin α _ 10 sin(50°) _ 4 10sin(50°) _________ 4 sin α ≈ 1.915 sin α = α 4 50° 10 Figure 14 We can stop here without finding the value of α. Because the range of the sine function is [−1, 1], it is impossible for the sine value to be 1.915. In fact, inputting sin−1 (1.915) in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions. Try It #4 Determine the number of triangles possible given a = 31, b = 26, β = 48°. 650 CHAPTER 8 Further applications oF trigonometry Finding the Area of an Oblique Triangle Using the Sine Function Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the 1 _ area of an oblique triangle. Recall that the area formula for a triangle is given as Area = bh, where b is base and h is 2 property sin α = height. For oblique triangles, we must find h before we can use the area formula. Observing the two triangles in Figure 15, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric h _ to write an equation for area in oblique triangles. In the acute triangle, we have sin α = c or csin α = h. However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base b to form a right triangle. The angle used in calculation is α', or 180 − α. opposite _ hypotenuse ' α γ Figure 15 a b γ Thus, Similarly, 1 1 _ _ Area = (base)(height) = b(csin α) 2 2 1 1 _ _ a(bsin γ) = Area = a(csin β) 2 2 area of an oblique triangle The formula for the area of an oblique triangle is given by 1 _ Area = bcsin α 2 1 _ = acsin β 2 1 _ = absin γ 2 This is equivalent to one-half of the product of two sides and the sine of their included angle. Example 5 Finding the Area of an Oblique Triangle Find the area of a triangle with sides a = 90, b = 52, and angle γ = 102°. Round the area to the nearest integer. Solution Using the formula, we have 1 _ Area = absin γ 2 1 _ Area = (90)(52)sin(102°) 2 Area ≈ 2289 square units Try It #5 Find the area of the triangle given β = 42°, a = 7.2 ft, c = 3.4 ft. Round the area to the nearest tenth. Solving Applied Problems Using the law of Sines The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion. SECTION 8.1 non-right triangles: law oF sines 651 Example 6 Finding an Altitude Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 16. Round the altitude to the nearest tenth of a mile. a 15° 20 miles Figure 16 35° Solution To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side a, and then use right triangle relationships to find the height of the aircraft, h. Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180° − 15° − 35° = 130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship. sin(130°) _ = 20 sin(35°) _ a asin(130°) = 20sin(35°) a = 20sin(35°) _ sin(130°) a ≈ 14.98 The distance from one station to the aircraft is about 14.98 miles. Now that we know a, we can use right triangle relationships to solve for h. sin(15°) = opposite _ hypotenuse h _ sin(15°) = a h _ 14.98 sin(15°) = h = 14.98sin(15°) h ≈ 3.88 The aircraft is at an altitude of approximately 3.9 miles. Try It #6 The diagram shown in Figure 17 represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70°, the angle of elevation from the northern end zone, point B, is 62°, and the distance between the viewing points of the two end zones is 145 yards. C 70° A 145 yards Figure 17 62° B Access the following online resources for additional instruction and practice with trigonometric applications. • law of Sines: The Basics (http://openstaxcollege.org/l/sinesbasic) • law of Sines: The Ambiguous Case (http://openstaxcollege.org/l/sinesambiguous) 652 CHAPTER 8 Further applications oF trigonometry 8.1 SeCTIOn exeRCISeS VeRBAl 1. Describe the altitude of a triangle. 3. When can you use the Law of Sines to find a missing angle? 2. Compare right triangles and oblique triangles. 4. In the Law of Sines, what is the relationship between the angle in the numerator and the side in the denominator? 5. What type of triangle results in an ambiguous case? AlGeBRAIC For the following exercises, assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve each triangle, if possible. Round each answer to the nearest tenth. 6. α = 43°, γ = 69°, a = 20 9. a = 4, α = 60°, β = 100° 7. α = 35°, γ = 73°, c = 20 10. b = 10, β = 95°, γ = 30° 8. α = 60°, β = 60°, γ = 60° For the following exercises, use the Law of Sines to solve for the missing side for each oblique triangle. Round each answer to the nearest hundredth. Assume that angle A is opposite side a, angle B is opposite side b, and angle C is opposite side c. 11. Find side b when A = 37°, B = 49°, c = 5. 13. Find side c when B = 37°, C = 21, b = 23. 12. Find side a when A = 132°, C = 23°, b = 10. For the following exercises, assume α is opposite side a, β is opposite side b, and γ is opposite side c. Determine whether there is no triangle, one triangle, or two triangles. Then solve each triangle, if possible. Round each answer to the nearest tenth. 14. α = 119°, a = 14, b = 26 17. a = 12, c = 17, α = 35° 20. a = 7, b = 3, β = 24° 23. β = 119°, b = 8.2, a = 11.3 15. γ = 113°, b = 10, c = 32 18. a = 20.5, b = 35.0, β = 25° 21. b = 13, c = 5, γ = 10° 16. b = 3.5, c = 5.3, γ = 80° 19. a = 7, c = 9, α = 43° 22. a = 2.3, c = 1.8, γ = 28° For the following exercises, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. Round each answer to the nearest tenth. 24. Find angle A when a = 24, b = 5, B = 22°. 26. Find angle B when A = 12°, a = 2, b = 9. 25. Find angle A when a = 13, b = 6, B = 20°. For the following exercises, find the area of the triangle with the given measurements. Round each answer to the nearest tenth. 27. a = 5, c = 6, β = 35° 30. a = 7.2, b = 4.5, γ = 43° GRAPHICAl 28. b = 11, c = 8, α = 28° 29. a = 32, b = 24, γ = 75° For the following exercises, find the length of side x. Round to the nearest tenth. 31. 32. 33. 10 x 70° 50° 25° 6 120° x 45° 15 x 75° SECTION 8.1 section exercises 653 34. 35. 18 x 40° 110° 14 50° x 42° 36. 8.6 111° 22° x For the following exercises, find the measure of angle x, if possible. Round to the nearest tenth. 98° 5 38. 37° 8 39. 5 x 13 22° 37. 40. x 5.7 x 10 5.3 59° x 11 41. Notice that x is an obtuse angle. 42. 65° 12 x 10 21 24 x 55° For the following exercises, solve the triangle. Round each answer to the nearest tenth. 43. A 24.1 32.6 93° B C For the following exercises, find the area of each triangle. Round each answer to the nearest tenth. 44. 45. 18 46. 16 30° 10 25° 15 4.5 51° 2.9 47. 58° 9 11 51° 48. 49. 25 18 40° 30° 3.5 115° 30 50 654 CHAPTER 8 Further applications oF trigonometry exTenSIOnS 50. Find the radius of the circle in Figure 18. Round to 51. Find the diameter of the circle in Figure 19. Round the nearest tenth. to the nearest tenth. 145° 3 8.3 110° Figure 18 Figure 19 52. Find m ∠ADC in Figure 20. Round to the nearest 53. Find AD in Figure 21. Round to the nearest tenth. tenth. A 10 9 8 60° B C D Figure 20 A 13 Figure 21 12 53° B 44° D C 54. Solve both triangles in Figure 22. Round each 55. Find AB in the parallelogram shown in Figure 23. answer to the nearest tenth. A 48° 4.2 B 46° E Figure 22 48° C 2 D B A 130° 10 12 130° D C Figure 23 56. Solve the triangle in Figure 24. (Hint: Draw a 57. Solve the triangle in Figure 25. (Hint: Draw a perpendicular from H to JK). Round each answer to the nearest tenth. perpendicular from N to LM). Round each answer to the nearest tenth. H 7 10 Figure 24 20° J K L 5 74° M 4.6 Figure 25 N SECTION 8.1 section exer
cises 655 58. In Figure 26, ABCD is not a parallelogram. ∠m is obtuse. Solve both triangles. Round each answer to the nearest tenth. A m x 35° n 29 45 h k D B y 40 Figure 26 65° C ReAl-WORlD APPlICATIOnS 59. A pole leans away from the sun at an angle of 7° to the vertical, as shown in Figure 27. When the elevation of the sun is 55°, the pole casts a shadow 42 feet long on the level ground. How long is the pole? Round the answer to the nearest tenth. 60. To determine how far a boat is from shore, two radar stations 500 feet apart find the angles out to the boat, as shown in Figure 28. Determine the distance of the boat from station A and the distance of the boat from shore. Round your answers to the nearest whole foot. Figure 27 70° 60° A B Figure 28 61. Figure 29 shows a satellite orbiting Earth. The satellite passes directly over two tracking stations A and B, which are 69 miles apart. When the satellite is on one side of the two stations, the angles of elevation at A and B are measured to be 86.2° and 83.9°, respectively. How far is the satellite from station A and how high is the satellite above the ground? Round answers to the nearest whole mile. 62. A communications tower is located at the top of a steep hill, as shown in Figure 30. The angle of inclination of the hill is 67°. A guy wire is to be attached to the top of the tower and to the ground, 165 meters downhill from the base of the tower. The angle formed by the guy wire and the hill is 16°. Find the length of the cable required for the guy wire to the nearest whole meter. 83.9° A 86.2° B Figure 29 16° 165m 67° Figure 30 656 CHAPTER 8 Further applications oF trigonometry 63. The roof of a house is at a 20° angle. An 8-foot solar panel is to be mounted on the roof and should be angled 38° relative to the horizontal for optimal results. (See Figure 31). How long does the vertical support holding up the back of the panel need to be? Round to the nearest tenth. 64. Similar to an angle of elevation, an angle of depression is the acute angle formed by a horizontal line and an observer’s line of sight to an object below the horizontal. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 6.6 km apart, to be 37° and 44°, as shown in Figure 32. Find the distance of the plane from point A to the nearest tenth of a kilometer. 8 ft 38° 20° Figure 31 65. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 4.3 km apart, to be 32° and 56°, as shown in Figure 33. Find the distance of the plane from point A to the nearest tenth of a kilometer. 32° 56° A Figure 33 B 37° 44° A B Figure 32 66. In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 39°. They then move 300 feet closer to the building and find the angle of elevation to be 50°. Assuming that the street is level, estimate the height of the building to the nearest foot. 67. In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 35°. They then move 250 feet closer to the building and find the angle of elevation to be 53°. Assuming that the street is level, estimate the height of the building to the nearest foot. 1 _ miles apart spot 69. A man and a woman standing 3 2 a hot air balloon at the same time. If the angle of elevation from the man to the balloon is 27°, and the angle of elevation from the woman to the balloon is 41°, find the altitude of the balloon to the nearest foot. 68. Points A and B are on opposite sides of a lake. Point C is 97 meters from A. The measure of angle BAC is determined to be 101°, and the measure of angle ACB is determined to be 53°. What is the distance from A to B, rounded to the nearest whole meter? 70. Two search teams spot a stranded climber on a mountain. The first search team is 0.5 miles from the second search team, and both teams are at an altitude of 1 mile. The angle of elevation from the first search team to the stranded climber is 15°. The angle of elevation from the second search team to the climber is 22°. What is the altitude of the climber? Round to the nearest tenth of a mile. SECTION 8.1 section exercises 657 71. A street light is mounted on a pole. A 6-foot-tall man is standing on the street a short distance from the pole, casting a shadow. The angle of elevation from the tip of the man’s shadow to the top of his head of 28°. A 6-foot-tall woman is standing on the same street on the opposite side of the pole from the man. The angle of elevation from the tip of her shadow to the top of her head is 28°. If the man and woman are 20 feet apart, how far is the street light from the tip of the shadow of each person? Round the distance to the nearest tenth of a foot. 73. Two streets meet at an 80° angle. At the corner, a park is being built in the shape of a triangle. Find the area of the park if, along one road, the park measures 180 feet, and along the other road, the park measures 215 feet. 72. Three cities, A, B, and C, are located so that city A is due east of city B. If city C is located 35° west of north from city B and is 100 miles from city A and 70 miles from city B, how far is city A from city B? Round the distance to the nearest tenth of a mile. 74. Brian’s house is on a corner lot. Find the area of the front yard if the edges measure 40 and 56 feet, as shown in Figure 34. House 56 ft 135° 40 ft Figure 34 75. The Bermuda triangle is a region of the Atlantic 76. A yield sign measures 30 inches on all three sides. What is the area of the sign? Ocean that connects Bermuda, Florida, and Puerto Rico. Find the area of the Bermuda triangle if the distance from Florida to Bermuda is 1030 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the angle created by the two distances is 62°. 77. Naomi bought a modern dining table whose top is in the shape of a triangle. Find the area of the table top if two of the sides measure 4 feet and 4.5 feet, and the smaller angles measure 32° and 42°, as shown in Figure 35. 4.5 feet 32° Figure 35 4 feet 42° 658 CHAPTER 8 Further applications oF trigonometry leARnInG OBjeCTIVeS In this section, you will: • Use the Law of Cosines to solve oblique triangles. • Solve applied problems using the Law of Cosines. Use Heron’s formula to find the area of a triangle. • 8.2 nOn-RIGHT TRIAnGleS: lAW OF COSIneS Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles as shown in Figure 1. How far from port is the boat? 8 mi 20° 10 mi Port Figure 1 Unfortunately, while the Law of Sines enables us to address many non-right triangle cases, it does not help us with triangles where the known angle is between two known sides, a SAS (side-angle-side) triangle, or when all three sides are known, but no angles are known, a SSS (side-side-side) triangle. In this section, we will investigate another tool for solving oblique triangles described by these last two cases. Using the law of Cosines to Solve Oblique Triangles The tool we need to solve the problem of the boat’s distance from the port is the Law of Cosines, which defines the relationship among angle measurements and side lengths in oblique triangles. Three formulas make up the Law of Cosines. At first glance, the formulas may appear complicated because they include many variables. However, once the pattern is understood, the Law of Cosines is easier to work with than most formulas at this mathematical level. Understanding how the Law of Cosines is derived will be helpful in using the formulas. The derivation begins with the Generalized Pythagorean Theorem, which is an extension of the Pythagorean Theorem to non-right triangles. Here is how it works: An arbitrary non-right triangle ABC is placed in the coordinate plane with vertex A at the origin, side c drawn along the x-axis, and vertex C located at some point (x, y) in the plane, as illustrated in Figure 2. Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted. y C (b cosθ, b sinθ) A (0, 0 Figure 2 SECTION 8.2 non-right triangles: law oF cosines 659 We can drop a perpendicular from C to the x-axis (this is the altitude or height). Recalling the basic trigonometric identities, we know that cos θ = x(adjacent) __ b(hypotenuse) and sin θ = y(opposite) __ b(hypotenuse) In terms of θ, x = bcos θ and y = bsin θ. The (x, y) point located at C has coordinates (bcos θ, bsin θ). Using the side (x − c) as one leg of a right triangle and y as the second leg, we can find the length of hypotenuse a using the Pythagorean Theorem. Thus, a2 = (x − c)2 + y2 = (bcos θ − c)2 + (bsin θ)2 Substitute (bcos θ) for x and (bsin θ) for y. = (b2 cos2 θ − 2bccos θ + c2) + b2 sin2 θ Expand the perfect square. = b2 cos2 θ + b2 sin2 θ + c2 − 2bccos θ Group terms noting that cos2 θ + sin2 θ = 1. = b2(cos2 θ + sin2 θ) + c2 − 2bccos θ Factor out b2. a2 = b2 + c2 − 2bccos θ The formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion. Keep in mind that it is always helpful to sketch the triangle when solving for angles or sides. In a real-world scenario, try to draw a diagram of the situation. As more information emerges, the diagram may have to be altered. Make those alterations to the diagram and, in the end, the problem will be easier to solve. Law of Cosines The Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle. For triangles labeled as in Figure 3, with angles α, β, and γ, and opposite corresponding sides a, b, and c, respectively, the
Law of Cosines is given as three equations. a2 = b2 + c2 − 2bc cos α b2 = a2 + c2 − 2ac cos β c2 = a2 + b2 − 2ab cos γ β a To solve for a missing side measurement, the corresponding opposite angle measure is needed. When solving for an angle, the corresponding opposite side measure is needed. We can use another version of the Law of Cosines to solve for an angle. c α b Figure 3 γ cos α = b2 + c2 − a2 _ 2bc cos β = a2 + c2 − b2 _ 2ac cos γ = a2 + b2 − c2 _ 2ab How To… Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle. 1. Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles. 2. Apply the Law of Cosines to find the length of the unknown side or angle. 3. Apply the Law of Sines or Cosines to find the measure of a second angle. 4. Compute the measure of the remaining angle. 660 CHAPTER 8 Further applications oF trigonometry Example 1 Finding the Unknown Side and Angles of a SAS Triangle Find the unknown side and angles of the triangle in Figure 4. γ b a 5 10 α c 5 12 Figure 4 30° β Solution First, make note of what is given: two sides and the angle between them. This arrangement is classified as SAS and supplies the data needed to apply the Law of Cosines. Each one of the three laws of cosines begins with the square of an unknown side opposite a known angle. For this example, the first side to solve for is side b, as we know the measurement of the opposite angle β. b2 = a2 + c2−2accos β b2 = 102 + 122 − 2(10)(12)cos(30°) — 3 √  b2 = 100 + 144 − 240  _ 2 b2 = 244 − 120 √ b = √ b ≈ 6.013 — — 3 244 − 120 √ 3 — Substitute the measurements for the known quantities. Evaluate the cosine and begin to simplify. Use the square root property. Because we are solving for a length, we use only the positive square root. Now that we know the length b, we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle α, we have sin β _ b = sin α _ a sin α _ 10 = sin(30°) _ 6.013 sin α = 10sin(30°) _ 6.013 10sin(30°)  _ 6.013) α = sin−1  α ≈ 56.3° Multiply both sides of the equation by 10. Find the inverse sine of 10sin(30°) _ 6.013 . The other possibility for α would be α = 180° − 56.3° ≈ 123.7°. In the original diagram, α is adjacent to the longest side, so α is an acute angle and, therefore, 123.7° does not make sense. Notice that if we choose to apply the Law of Cosines, we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between 0° and 180°. Proceeding with α ≈ 56.3°, we can then find the third angle of the triangle. The complete set of angles and sides is γ = 180° − 30° − 56.3° ≈ 93.7° α ≈ 56.3° β = 30° γ ≈ 93.7° a = 10 b ≈ 6.013 c = 12 Try It #1 Find the missing side and angles of the given triangle: α = 30°, b = 12, c = 24. SECTION 8.2 non-right triangles: law oF cosines 661 Example 2 Solving for an Angle of a SSS Triangle Find the angle α for the given triangle if side a = 20, side b = 25, and side c = 18. Solution For this example, we have no angles. We can solve for any angle using the Law of Cosines. To solve for angle α, we have a2 = b2 + c2 −2bccos α 202 = 252 + 182−2(25)(18)cos α Substitute the appropriate measurements. 400 = 625 + 324 − 900cos α Simplify in each step. 400 = 949 − 900cos α −549 = −900cos α −549 _ −900 = cos α 0.61 ≈ cos α cos−1(0.61) ≈ α α ≈ 52.4° Isolate cos α. Find the inverse cosine. See Figure 5. γ b = 25 α 52.4° a = 20 c = 18 β Figure 5 Analysis Because the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases using this method. Try It #2 Given a = 5, b = 7, and c = 10, find the missing angles. Solving Applied Problems Using the law of Cosines Just as the Law of Sines provided the appropriate equations to solve a number of applications, the Law of Cosines is applicable to situations in which the given data fits the cosine models. We may see these in the fields of navigation, surveying, astronomy, and geometry, just to name a few. Example 3 Using the Law of Cosines to Solve a Communication Problem On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is accomplished through a process called triangulation, which works by using the distances from two known points. Suppose there are two cell phone towers within range of a cell phone. The two towers are located 6,000 feet apart along a straight highway, running east to west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is 5,050 feet from the first tower and 2,420 feet from the second tower. Determine the position of the cell phone north and east of the first tower, and determine how far it is from the highway. 662 CHAPTER 8 Further applications oF trigonometry Solution For simplicity, we start by drawing a diagram similar to Figure 6 and labeling our given information. 2,420 ft. 5,050 ft. θ 6,000 ft. Figure 6 Using the Law of Cosines, we can solve for the angle θ. Remember that the Law of Cosines uses the square of one side to find the cosine of the opposite angle. For this example, let a = 2420, b = 5050, and c = 6000. Thus, θ corresponds to the opposite side a = 2420. a2 = b2 + c2 − 2bccos θ (2420)2 = (5050)2 + (6000)2 − 2(5050)(6000)cos θ (2420)2 − (5050)2 − (6000)2 = − 2(5050)(6000)cos θ (2420)2 − (5050)2 − (6000)2 ___ −2(5050)(6000) = cos θ cos θ ≈ 0.9183 θ ≈ cos−1(0.9183) θ ≈ 23.3° To answer the questions about the phone’s position north and east of the tower, and the distance to the highway, drop a perpendicular from the position of the cell phone, as in Figure 7. This forms two right triangles, although we only need the right triangle that includes the first tower for this problem. y 5,050 ft. 23.3° x Figure 7 Using the angle θ = 23.3° and the basic trigonometric identities, we can find the solutions. Thus cos(23.3°) = x _ 5050 x = 5050 cos(23.3°) x ≈ 4638.15 feet sin(23.3°) = y _ 5050 y = 5050sin(23.3°) y ≈ 1997.5 feet The cell phone is approximately 4,638 feet east and 1,998 feet north of the first tower, and 1,998 feet from the highway. Example 4 Calculating Distance Traveled Using a SAS Triangle Returning to our problem at the beginning of this section, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat? The diagram is repeated here in Figure 8. SECTION 8.2 non-right triangles: law oF cosines 663 8 mi 20° 10 mi Port Figure 8 Solution The boat turned 20 degrees, so the obtuse angle of the non-right triangle is the supplemental angle, 180° − 20° = 160°. With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle—the distance of the boat to the port. x2 = 82 + 102 − 2(8)(10)cos(160°) x2 = 314.35 x = √ — 314.35 x ≈ 17.7 miles The boat is about 17.7 miles from port. Using Heron’s Formula to Find the Area of a Triangle We already learned how to find the area of an oblique triangle when we know two sides and an angle. We also know the formula to find the area of a triangle using the base and the height. When we know the three sides, however, we can use Heron’s formula instead of finding the height. Heron of Alexandria was a geometer who lived during the first century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known. Heron’s formula Heron’s formula finds the area of oblique triangles in which sides a, b, and c are known. Area = √ —— s(s − a)(s − b)(s − c) where s = (a + b + c) _ 2 is one half of the perimeter of the triangle, sometimes called the semi-perimeter. Example 5 Using Heron’s Formula to Find the Area of a Given Triangle Find the area of the triangle in Figure 9 using Heron’s formula. c = 7 A B b = 15 Figure 9 a = 10 C Solution First, we calculate s. s = s = (a + b + c) _ 2 (10 + 15 + 7) __ 2 = 16 664 CHAPTER 8 Further applications oF trigonometry Then we apply the formula. Area = √ Area = √ —— s(s − a)(s − b)(s − c) 16(16 − 10)(16 − 15)(16 − 7) —— The area is approximately 29.4 square units. Area ≈ 29.4 Try It #3 Use Heron’s formula to find the area of a triangle with sides of lengths a = 29.7 ft, b = 42.3 ft, and c = 38.4 ft. Example 6 Applying Heron’s Formula to a Real-World Problem A Chicago city developer wants to construct a building consisting of artist’s lofts on a triangular lot bordered by Rush Street, Wabash Avenue, and Pearson Street. The frontage along Rush Street is approximately 62.4 meters, along Wabash Avenue it is approximately 43.5 meters, and along Pearson Street it is approximately 34.1 meters. How many square meters are available to the developer? See Figure 10 for a view of the city property ) Figure 10 Solution Find the measurement for s, which is one-half of the perimeter. 62.4 + 43.5 + 34.1 __ 2 s = s = 70 m Apply Heron’s formula. Area = √ Area = √ —— 70(70 − 62.4)(70 − 43.5)(70 − 34.1) 506,118.2 — The developer has about 711.4 square meters. Area ≈ 711.4 Try It #4 Find the area of a triangle given a = 4.38 ft , b = 3.79 ft, and c = 5.22 ft. Access these online resources for additional instruction and practice with the Law of Cosines. • law of Cosines (http://openstaxcollege.org/l/lawcosines) • law of Cosines: Applications (http://openstaxcollege.org/l/cosineapp) • law of Cosines: Applications 2 (http://openstaxcollege.org/l/cosineapp2) SECTION 8.2 section exercises 665 8.2 SeCTIOn exeRCISeS VeRBAl 1. If you are looking for a missing side of a triangle, what do you need to know when using the Law of Cosines? 2. If you are looking for a missing angle of a triangle, what do you need to know when using the Law of Cosines? 3. Explain what s represents in Heron’s formula. 4. Explain the relationship between the Pythagorean Theorem and the Law of Cosines. 5. When must you use the Law of Cosines instead of the Pythagorean Theorem? AlGeBRAIC For the following e
xercises, assume α is opposite side a, β is opposite side b, and γ is opposite side c. If possible, solve each triangle for the unknown side. Round to the nearest tenth. 6. γ = 41.2°, a = 2.49, b = 3.13 8. β = 58.7°, a = 10.6, c = 15.7 10. α = 119°, a = 26, b = 14 12. β = 67°, a = 49, b = 38 14. α = 36.6°, a = 186.2, b = 242.2 7. α = 120°, b = 6, c =7 9. γ = 115°, a = 18, b = 23 11. γ = 113°, b = 10, c = 32 13. α = 43.1°, a = 184.2, b = 242.8 15. β = 50°, a = 105, b = 45 For the following exercises, use the Law of Cosines to solve for the missing angle of the oblique triangle. Round to the nearest tenth. 16. a = 42, b = 19, c = 30; find angle A. 18. a = 16, b = 31, c = 20; find angle B. 20. a = 108, b = 132, c = 160; find angle C. 17. a = 14, b = 13, c = 20; find angle C. 19. a = 13, b = 22, c = 28; find angle A. For the following exercises, solve the triangle. Round to the nearest tenth. 21. A = 35°, b = 8, c = 11 23. C = 121°, a = 21, b = 37 25. a = 3.1, b = 3.5, c = 5 22. B = 88°, a = 4.4, c = 5.2 24. a = 13, b = 11, c = 15 26. a = 51, b = 25, c = 29 For the following exercises, use Heron’s formula to find the area of the triangle. Round to the nearest hundredth. 27. Find the area of a triangle with sides of length 18 in, 28. Find the area of a triangle with sides of length 20 cm, 21 in, and 32 in. Round to the nearest tenth. 26 cm, and 37 cm. Round to the nearest tenth. 1 1 1 _ _ _ m, c = m, b = 29. a = m 4 3 2 31. a = 1.6 yd, b = 2.6 yd, c = 4.1 yd GRAPHICAl 30. a = 12.4 ft, b = 13.7 ft, c = 20.2 ft For the following exercises, find the length of side x. Round to the nearest tenth. 32. x 33. 4.5 6.5 5 72° 42º 3.4 x 34. 12 A 40° x 15 B 666 CHAPTER 8 Further applications oF trigonometry 35. 30 36. 225 37. 65° 23 x 50° x 305 1 5 123° x 1 3 For the following exercises, find the measurement of angle A. 38. 1.5 A 2.5 2.3 39. 115 125 40. 4.3 A 6.8 8.2 100 A 41. 40.6 38.7 A 23.3 42. Find the measure of each angle in the triangle shown in Figure 11. Round to the nearest tenth. C 12 A 10 7 B Figure 11 For the following exercises, solve for the unknown side. Round to the nearest tenth. 44. 60° 20 28 16 30° 10 45. 22° 13 20 43. 46. 9 88° 5 SECTION 8.2 section exercises 667 For the following exercises, find the area of the triangle. Round to the nearest hundredth. 47. 8 12 17 48. 49. 2.6 50 22 36 1.9 4.3 50. 12.5 51. 8.9 16.2 exTenSIOnS 52. A parallelogram has sides of length 16 units and 10 units. The shorter diagonal is 12 units. Find the measure of the longer diagonal. 53. The sides of a parallelogram are 11 feet and 17 feet. The longer diagonal is 22 feet. Find the length of the shorter diagonal. 54. The sides of a parallelogram are 28 centimeters and 40 centimeters. The measure of the larger angle is 100°. Find the length of the shorter diagonal. 55. A regular octagon is inscribed in a circle with a radius of 8 inches. (See Figure 12.) Find the perimeter of the octagon. 56. A regular pentagon is inscribed in a circle of radius 12 cm. (See Figure 13.) Find the perimeter of the pentagon. Round to the nearest tenth of a centimeter. Figure 12 Figure 13 For the following exercises, suppose that x2 = 25 + 36 − 60 cos(52) represents the relationship of three sides of a triangle and the cosine of an angle. 57. Draw the triangle. 58. Find the length of the third side. 668 CHAPTER 8 Further applications oF trigonometry For the following exercises, find the area of the triangle. 59. 5.3 22° 60. 3.4 61. 8 80° 6 18° 18.8 12.8 ReAl-WORlD APPlICATIOnS 62. A surveyor has taken the measurements shown in Figure 14. Find the distance across the lake. Round answers to the nearest tenth. 63. A satellite calculates the distances and angle shown in Figure 15 (not to scale). Find the distance between the two cities. Round answers to the nearest tenth. 800 ft 900 ft 70° Figure 14 370 km 2.1° 350 km Figure 15 64. An airplane flies 220 miles with a heading of 40°, and then flies 180 miles with a heading of 170°. How far is the plane from its starting point, and at what heading? Round answers to the nearest tenth. 65. A 113-foot tower is located on a hill that is inclined 34° to the horizontal, as shown in Figure 16. A guywire is to be attached to the top of the tower and anchored at a point 98 feet uphill from the base of the tower. Find the length of wire needed. 113 ft 98 ft 34° Figure 16 SECTION 8.2 section exercises 669 66. Two ships left a port at the same time. One ship 67. The graph in Figure 17 represents two boats traveled at a speed of 18 miles per hour at a heading of 320°. The other ship traveled at a speed of 22 miles per hour at a heading of 194°. Find the distance between the two ships after 10 hours of travel. departing at the same time from the same dock. The first boat is traveling at 18 miles per hour at a heading of 327° and the second boat is traveling at 4 miles per hour at a heading of 60°. Find the distance between the two boats after 2 hours. 4 mph 18 mph Figure 17 68. A triangular swimming pool measures 40 feet on one side and 65 feet on another side. These sides form an angle that measures 50°. How long is the third side (to the nearest tenth)? 69. A pilot flies in a straight path for 1 hour 30 min. She then makes a course correction, heading 10° to the right of her original course, and flies 2 hours in the new direction. If she maintains a constant speed of 680 miles per hour, how far is she from her starting position? 70. Los Angeles is 1,744 miles from Chicago, Chicago is 714 miles from New York, and New York is 2,451 miles from Los Angeles. Draw a triangle connecting these three cities, and find the angles in the triangle. 71. Philadelphia is 140 miles from Washington, D.C., Washington, D.C. is 442 miles from Boston, and Boston is 315 miles from Philadelphia. Draw a triangle connecting these three cities and find the angles in the triangle. 72. Two planes leave the same airport at the same time. One flies at 20° east of north at 500 miles per hour. The second flies at 30° east of south at 600 miles per hour. How far apart are the planes after 2 hours? 73. Two airplanes take off in different directions. One travels 300 mph due west and the other travels 25° north of west at 420 mph. After 90 minutes, how far apart are they, assuming they are flying at the same altitude? 74. A parallelogram has sides of length 15.4 units and 9.8 units. Its area is 72.9 square units. Find the measure of the longer diagonal. 75. The four sequential sides of a quadrilateral have lengths 4.5 cm, 7.9 cm, 9.4 cm, and 12.9 cm. The angle between the two smallest sides is 117°. What is the area of this quadrilateral? 76. The four sequential sides of a quadrilateral have lengths 5.7 cm, 7.2 cm, 9.4 cm, and 12.8 cm. The angle between the two smallest sides is 106°. What is the area of this quadrilateral? 77. Find the area of a triangular piece of land that measures 30 feet on one side and 42 feet on another; the included angle measures 132°. Round to the nearest whole square foot. 78. Find the area of a triangular piece of land that measures 110 feet on one side and 250 feet on another; the included angle measures 85°. Round to the nearest whole square foot. 670 CHAPTER 8 Further applications oF trigonometry leARnInG OBjeCTIVeS In this section, you will: • Plot points using polar coordinates. • Convert from polar coordinates to rectangular coordinates. • Convert from rectangular coordinates to polar coordinates. • Transform equations between polar and rectangular forms. • Identify and graph polar equations by converting to rectangular equations. 8.3 POlAR COORDInATeS Over 12 kilometers from port, a sailboat encounters rough weather and is blown off course by a 16-knot wind (see Figure 1). How can the sailor indicate his location to the Coast Guard? In this section, we will investigate a method of representing location that is different from a standard coordinate grid. 16-knot wind 90° 120° 60° 0 150° 180° 210° Port 30° 6 9 12 0°/360° 330° 300° 240° 270° Figure 1 Plotting Points Using Polar Coordinates When we think about plotting points in the plane, we usually think of rectangular coordinates (x, y) in the Cartesian coordinate plane. However, there are other ways of writing a coordinate pair and other types of grid systems. In this section, we introduce to polar coordinates, which are points labeled (r, θ) and plotted on a polar grid. The polar grid is represented as a series of concentric circles radiating out from the pole, or the origin of the coordinate plane. The polar grid is scaled as the unit circle with the positive x-axis now viewed as the polar axis and the origin as the pole. The first coordinate r is the radius or length of the directed line segment from the pole. The angle θ, measured in radians, indicates the direction of r. We move counterclockwise from the polar axis by an angle of θ, and measure a directed line segment the length of r in the direction of θ. Even though we measure θ first and then r, the polar point is written with π π _ _  , we would move the r-coordinate first. For example, to plot the point  2, units in the counterclockwise direction 4 4 and then a length of 2 from the pole. This point is plotted on the grid in Figure 2. 4 3 2 1 –4 –3 –2 –1 1 2 3 4 –1 –2 –3 –4 Polar Grid Figure 2 SECTION 8.3 polar coordinates 671 Example 1 Plotting a Point on the Polar Grid π _  on the polar grid. Plot the point  3, 2 π _ Solution The angle is found by sweeping in a counterclockwise direction 90° from the polar axis. The point is located 2 π _ direction, as shown in Figure 3. at a length of 3 units from the pole in the 2 4 3 2 1 –4 –3 –2 –1 1 2 3 4 –1 –2 –3 –4 Figure 3 Try It #1 π _  in the polar grid. Plot the point  2, 3 Example 2 Plotting a Point in the Polar Coordinate System with a Negative Component π _ Plot the point  −2,  on the polar grid. 6 π _ is located in the first quadrant. However, r = −2. We can approach plotting a point with a Solution We know that 6 negative r in two ways: π π _ _  by movin
g 1. Plot the point  2, in the counterclockwise direction and extending a directed line segment 2 units 6 6 into the first quadrant. Then retrace the directed line segment back through the pole, and continue 2 units into the third quadrant; π _ 2. Move in the counterclockwise direction, and draw the directed line segment from the pole 2 units in the negative 6 direction, into the third quadrant. π _  shown in Figure 4(b). See Figure 4(a). Compare this to the graph of the polar coordinate  2, 6 4 3 2 1 – – 44 –4 –3 –2 –1 1 2 3 4 –1 –2 –3 –4 (b) (a) Figure 4 Try It #2 π _  and  2, Plot the points  3, − 6 9π _  on the same polar grid. 4 672 CHAPTER 8 Further applications oF trigonometry Converting from Polar Coordinates to Rectangular Coordinates When given a set of polar coordinates, we may need to convert them to rectangular coordinates. To do so, we can recall the relationships that exist among the variables x, y, r, and θ. x _ cos θ = r → x = rcos θ y _ sin θ = r → y = rsin θ Dropping a perpendicular from the point in the plane to the x-axis forms a right triangle, as illustrated in Figure 5. An easy way to remember the equations above is to think of cos θ as the adjacent side over the hypotenuse and sin θ as the opposite side over the hypotenuse. y (x, y) or (r, θ) y r θ x x Figure 5 converting from polar coordinates to rectangular coordinates To convert polar coordinates (r, θ) to rectangular coordinates (x, y), let x _ r → x = rcos θ cos θ = y _ r → y = rsin θ sin θ = How To… Given polar coordinates, convert to rectangular coordinates. 1. Given the polar coordinate (r, θ), write x = rcos θ and y = rsin θ. 2. Evaluate cos θ and sin θ. 3. Multiply cos θ by r to find the x-coordinate of the rectangular form. 4. Multiply sin θ by r to find the y-coordinate of the rectangular form. Example 3 Writing Polar Coordinates as Rectangular Coordinates π _  as rectangular coordinates. Write the polar coordinates  3, 2 Solution Use the equivalent relationships. x = rcos θ π _ = 0 x = 3cos 2 y = rsin θ π _ = 3 y = 3sin 2 The rectangular coordinates are (0, 3). See Figure 6. SECTION 8.3 polar coordinates 673 4 3 2 1 –4 –3 –2 –1 1 2 3 4 –5 –4 –3 –2 –1 –2 –3 –4 y 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 (0, 3) 21 3 4 5 x Polar Grid Coordinate Grid Figure 6 Example 4 Writing Polar Coordinates as Rectangular Coordinates Write the polar coordinates (−2, 0) as rectangular coordinates. Solution See Figure 7. Writing the polar coordinates as rectangular, we have x = rcos θ x = −2cos(0) = −2 y = rsin θ y = −2sin(0) = 0 The rectangular coordinates are also (−2, 0). y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 5 –4 –3 –2 Figure 7 Try It #3 Write the polar coordinates  −1, 2π _  as rectangular coordinates. 3 Converting from Rectangular Coordinates to Polar Coordinates To convert rectangular coordinates to polar coordinates, we will use two other familiar relationships. With this conversion, however, we need to be aware that a set of rectangular coordinates will yield more than one polar point. 674 CHAPTER 8 Further applications oF trigonometry converting from rectangular coordinates to polar coordinates Converting from rectangular coordinates to polar coordinates requires the use of one or more of the relationships illustrated in Figure 8. y (x, y), (r, θ) x _ cos θ = r or x = rcos θ y _ sin θ = r or y = rsin tan θ = x y x r x θ Figure 8 Example 5 Writing Rectangular Coordinates as Polar Coordinates Convert the rectangular coordinates (3, 3) to polar coordinates. y _ Solution We see that the original point (3, 3) is in the first quadrant. To find θ, use the formula tan θ = x . This gives 3 _ tan θ = 3 tan θ = 1 θ = tan−1(1) To find r, we substitute the values for x and y into the formula r = √ in the first quadrant. Thus — π _ x 2 + y 2 . We know that r must be positive, as is 18 = 3 √ — — 2 So, r = 3 √ — π _ , giving us the polar point  3 √ 2 and θ = 4 — π _  . See Figure 95 –4 –3 –2 –1–1 –2 –3 –4 –5 1 2 3 54 –5 –4 –3 –2 Figure 9 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (3, 3) 21 3 4 5 x — Analysis There are other sets of polar coordinates that will be the same as our first solution. For example, the points  −3 √ π _ and  3 √  . The point  −3 √ 2 , 4 π _ . The radius is expressed as −3 √ further counterclockwise by π, which is directly opposite 4 7π _  and will coincide with the original solution of  3 √ 4 5π  ___ 2 , 4 5π _  indicates a move 4 is 2 . However, the angle 5π _ 4 2 , − 2 , — — — — located in the third quadrant and, as r is negative, we extend the directed line segment in the opposite direction, into the π 7π _ _ first quadrant. This is the same point as  3 √ , from . 4 4 2 , is the same. The radius, 3 √ 7π _  is a move further clockwise by − 4 π __  . The point  3 √ 2 , 4 2 , − — — — SECTION 8.3 polar coordinates 675 Transforming equations between Polar and Rectangular Forms We can now convert coordinates between polar and rectangular form. Converting equations can be more difficult, but it can be beneficial to be able to convert between the two forms. Since there are a number of polar equations that cannot be expressed clearly in Cartesian form, and vice versa, we can use the same procedures we used to convert points between the coordinate systems. We can then use a graphing calculator to graph either the rectangular form or the polar form of the equation. How To… Given an equation in polar form, graph it using a graphing calculator. 1. Change the MODE to POL, representing polar form. 2. Press the Y= button to bring up a screen allowing the input of six equations: r1, r2, ... , r6. 3. Enter the polar equation, set equal to r. 4. Press GRAPH. Example 6 Writing a Cartesian Equation in Polar Form Write the Cartesian equation x 2 + y 2 = 9 in polar form. Solution The goal is to eliminate x and y from the equation and introduce r and θ. Ideally, we would write the equation r as a function of θ. To obtain the polar form, we will use the relationships between (x, y) and (r, θ). Since x = rcos θ and y = rsin θ, we can substitute and solve for r. (rcos θ) 2 + (rsin θ) 2 = 9 r 2 cos 2 θ + r 2 sin 2 θ = 9 r 2 (cos 2 θ + sin 2 θ) = 9 r 2 (1) = 9 Substitute cos 2 θ + sin 2 θ = 1. Thus, x 2 + y 2 = 9, r = 3, and r = −3 should generate the same graph. See Figure 10. r = ± 3 Use the square root property. –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (a) 21 3 4 5 x 1 2 3 4 (b) To graph a circle in rectangular form, we must first solve for y. Figure 10 (a) Cartesian form x 2 + y 2 = 9 (b) Polar form Note that this is two separate functions, since a circle fails the vertical line test. Therefore, we need to enter the positive — and negative square roots into the calculator separately, as two equations in the form Y1 = √ Press GRAPH. 9− x 2 and Y2 = − √ 9− x 2 . — 676 CHAPTER 8 Further applications oF trigonometry Example 7 Rewriting a Cartesian Equation as a Polar Equation Rewrite the Cartesian equation x 2 + y 2 = 6y as a polar equation. Solution This equation appears similar to the previous example, but it requires different steps to convert the equation. We can still follow the same procedures we have already learned and make the following substitutions: r 2 = 6y Use x 2 + y 2 = r 2. r 2 = 6rsin θ Substitute y = rsin θ. r 2−6rsin θ = 0 Set equal to 0. r (r − 6sin θ) = 0 Factor and solve. r = 0 We reject r = 0, as it only represents one point, (0, 0). or r = 6sin θ Therefore, the equations x 2 + y 2 = 6y and r = 6sin θ should give us the same graph. See Figure 11. y 6 5 4 3 2 1 –6 –5 –4 –3 –2 –1 21 3 4 5 6 x –6 –5 –4 –3 –2 –1 321 4 5 6 (a) (b) Figure 11 (a) Cartesian form x 2 + y 2 = 6y (b) polar form r = 6sin θ The Cartesian or rectangular equation is plotted on the rectangular grid, and the polar equation is plotted on the polar grid. Clearly, the graphs are identical. Example 8 Rewriting a Cartesian Equation in Polar Form Rewrite the Cartesian equation y = 3x + 2 as a polar equation. Solution We will use the relationships x = rcos θ and y = rsin θ. y = 3x + 2 rsin θ = 3rcos θ + 2 rsin θ − 3rcos θ = 2 r(sin θ − 3cos θ) = 2 r = 2 __ sin θ − 3cos θ Isolate r. Solve for r. Try It #4 Rewrite the Cartesian equation y 2 = 3 − x 2 in polar form. Identify and Graph Polar equations by Converting to Rectangular equations We have learned how to convert rectangular coordinates to polar coordinates, and we have seen that the points are indeed the same. We have also transformed polar equations to rectangular equations and vice versa. Now we will demonstrate that their graphs, while drawn on different grids, are identical. SECTION 8.3 polar coordinates 677 Example 9 Graphing a Polar Equation by Converting to a Rectangular Equation Covert the polar equation r = 2sec θ to a rectangular equation, and draw its corresponding graph. Solution The conversion is r = 2sec θ r = 2 _ cos θ rcos θ = 2 Notice that the equation r = 2sec θ drawn on the polar grid is clearly the same as the vertical line x = 2 drawn on the rectangular grid (see Figure 12). Just as x = c is the standard form for a vertical line in rectangular form, r = csec θ is the standard form for a vertical line in polar form. x = 2 r = 2 sec 4 –3 –2 –1 1 2 3 4 –5 –4 –3 –2 –1 –2 –3 –4 (a) 21 1 –1 –2 –3 –4 –5 (b) A similar discussion would demonstrate that the graph of the function r = 2csc θ will be the horizontal line y = 2. In fact, r = ccsc θ is the standard form for a horizontal line in polar form, corresponding to the rectangular form y = c. Figure 12 (a) Polar grid (b) Rectangular coordinate system Example 10 Rewriting a Polar Equation in Cartesian Form Rewrite the polar equation r = as a Cartesian equation. 3 _ 1 − 2cos θ Solution The goal is to eliminate θ and r, and introduce x and y. We clear the fraction, and then use substitution. In order to replace r with x and y, we must use the expression x 2 + y 2 = r 2. r = 3 _ 1 − 2cos θ r(1 − 2cos θ2x = 3 x _ Use cos θ = r to eliminate θ. r = 3 + 2x Isolate r. r 2 = (3 + 2x) 2 Square both sides. x 2 + y 2 = (3 + 2x) 2 Use x 2 + y 2 = r 2. The Cartesian e
quation is x 2 + y 2 = (3 + 2x) 2. However, to graph it, especially using a graphing calculator or computer program, we want to isolate y. x 2 + y 2 = (3 + 2x) 2 y 2 = (3 + 2x3 + 2x) 2 − x 2 When our entire equation has been changed from r and θ to x and y, we can stop, unless asked to solve for y or simplify. See Figure 13. 678 CHAPTER 8 Further applications oF trigonometry y x2 + y2 = (3 + 2x)2 –5 –4 –3 –2 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 21 3 4 5 x –5 r = 3 1 – 2 cosθ 1 2 3 54 5 4 3 2 1 –4 –3 –2 –1–1 –2 –3 –4 –5 Coordinate Grid Polar Grid Figure 13 The “hour-glass” shape of the graph is called a hyperbola. Hyperbolas have many interesting geometric features and applications, which we will investigate further in Analytic Geometry. Analysis In this example, the right side of the equation can be expanded and the equation simplified further, as shown above. However, the equation cannot be written as a single function in Cartesian form. We may wish to write the rectangular equation in the hyperbola’s standard form. To do this, we can start with the initial equation. x 2 + y 2 = (3 + 2x) 2 x 2 + y 2 − (3 + 2x9 + 12x + 4x 2 − 12x −4x 2 = 0 −3x 2 − 12x + y 2 = 9 3x 2 + 12x − y 2 = −9 Multiply through by −1. 3(x 2 + 4x + ) − y 2 = −9 + 12 Organize terms to complete the square for x. 3(x 2 + 4x + 4) − y 2 = −9 + 12 3(x + 2)2 − y 2 = 3 (x + 2)2 − y 2 _ = 1 3 Try It #5 Rewrite the polar equation r = 2sin θ in Cartesian form. Example 11 Rewriting a Polar Equation in Cartesian Form Rewrite the polar equation r = sin(2θ) in Cartesian form. Solution r = sin(2θ) r = 2sin θcos θ y x   r = 2  __ __  r r 2xy _ r = r 2 r3 = 2xy 3 = 2xy — x 2 + y 2   √ Use the double angle identity for sine. y x __ r . __ r and sin θ = Use cos θ = Simplify. Multiply both sides by r 2. As x 2 + y 2 = r 2 . This equation can also be written as 3 2 _ _ = 2xy or x 2 + y 2 = (2xy) (x 2 + y 2) . 3 2 Access these online resources for additional instruction and practice with polar coordinates. Introduction to Polar Coordinates (http://openstaxcollege.org/l/intropolar) • • Comparing Polar and Rectangular Coordinates (http://openstaxcollege.org/l/polarrect) SECTION 8.3 section exercises 679 8.3 SeCTIOn exeRCISeS VeRBAl 1. How are polar coordinates different from 2. How are the polar axes different from the x- and rectangular coordinates? 3. Explain how polar coordinates are graphed. π π _ _  and  3, − 5. Explain why the points  −3,  are 2 2 the same. y-axes of the Cartesian plane? π π _ _  and  −3,  related? 4. How are the points  3, 2 2 AlGeBRAIC For the following exercises, convert the given polar coordinates to Cartesian coordinates with r > 0 and 0 ≤ θ ≤2π. Remember to consider the quadrant in which the given point is located when determining θ for the point. 6.  7, 7π _  6 7. (5, π) π _ 8.  6, −  4 π _ 9.  −3,  6 10.  4, 7π _  4 For the following exercises, convert the given Cartesian coordinates to polar coordinates with r > 0, 0 ≤ θ <2π. Remember to consider the quadrant in which the given point is located. 11. (4, 2) 12. (−4, 6) 13. (3, −5) 14. (−10, −13) 15. (8, 8) For the following exercises, convert the given Cartesian equation to a polar equation. 16. x = 3 20. x 2 + y 2 = 4y 24. x 2 + y 2 = 9 17. y = 4 21. x 2 + y 2 = 3x 25. x 2 = 9y 18. y = 4x 2 22. x 2 − y 2 = x 26. y 2 = 9x 19. y = 2x4 23. x 2 − y 2 = 3y 27. 9xy = 1 For the following exercises, convert the given polar equation to a Cartesian equation. Write in the standard form of a conic if possible, and identify the conic section represented. 28. r = 3sin θ 29. r = 4cos θ 32. r = 2sec θ 33. r = 3csc θ 36. r = 4 37. r 2 = 4 30. r = 4 __ sin θ + 7cos θ 31. r = 6 __ cos θ + 3sin θ 34. r = √ 38. r = — rcos θ + 2 1 __ 4cos θ − 3sin θ GRAPHICAl For the following exercises, find the polar coordinates of the point. 40. π 2 3π 2 41. 0 π π 2 3π 2 42. 0 π 35. r 2 = 4sec θ csc θ 39. r = 3 __ cos θ − 5sin θ π 2 3π 2 0 680 CHAPTER 8 Further applications oF trigonometry 43. π π 2 3π 2 44. 0 π π 2 3π 2 0 For the following exercises, plot the points. π _ 45.  −2,  3 −5π _  4 50.  4, π _ 46.  −1, −  2 51.  3, 5π _  6 47.  3.5, 7π _  4 52.  −1.5, 7π _  6 π _ 48.  −4,  3 π _ 53.  −2,  4 π _  49.  5, 2 54.  1, 3π _  2 For the following exercises, convert the equation from rectangular to polar form and graph on the polar axis. 55. 5x − y = 6 59. x = 2 56. 2x + 7y = −3 60. x 2 + y 2 = 5y 57. x 2 +(y − 1) 2 = 1 61. x 2 + y 2 = 3x 58. (x + 2) 2 +(y + 3) 2 = 13 For the following exercises, convert the equation from polar to rectangular form and graph on the rectangular plane. 62. r = 6 66. r = sec θ TeCHnOlOGY 63. r = − 4 67. r = −10sin θ 64. θ = − 2π _ 3 68. r = 3cos θ π _ 65. θ = 4 69. Use a graphing calculator to find the rectangular π _ coordinates of  2, −  . Round to the nearest 5 thousandth. 70. Use a graphing calculator to find the rectangular 3π _  . Round to the nearest 7 coordinates of  −3, thousandth. 71. Use a graphing calculator to find the polar 72. Use a graphing calculator to find the polar coordinates of (−7, 8) in degrees. Round to the nearest thousandth. coordinates of (3, −4) in degrees. Round to the nearest hundredth. 73. Use a graphing calculator to find the polar coordinates of (−2, 0) in radians. Round to the nearest hundredth. exTenSIOnS 74. Describe the graph of r = asec θ; a > 0. 76. Describe the graph of r = acsc θ; a > 0. 78. What polar equations will give an oblique line? For the following exercises, graph the polar inequality. 79 81 83. 0 ≤ θ ≤ 3 75. Describe the graph of r = asec θ; a < 0. 77. Describe the graph of r = acsc θ; a < 0. π _ 803 823 < r < 2 < θ ≤ 84. − 3 6 SECTION 8.4 polar coordinates: graphs 681 leARnInG OBjeCTIVeS In this section, you will: • Test polar equations for symmetry. • Graph polar equations by plotting points. 8.4 POlAR COORDInATeS: GRAPHS The planets move through space in elliptical, periodic orbits about the sun, as shown in Figure 1. They are in constant motion, so fixing an exact position of any planet is valid only for a moment. In other words, we can fix only a planet’s instantaneous position. This is one application of polar coordinates, represented as (r, θ). We interpret r as the distance from the sun and θ as the planet’s angular bearing, or its direction from a fixed point on the sun. In this section, we will focus on the polar system and the graphs that are generated directly from polar coordinates. Venus Mercury Earth Mars Figure 1 Planets follow elliptical paths as they orbit around the Sun. (credit: modification of work by nASA/jPl-Caltech) Testing Polar equations for Symmetry Just as a rectangular equation such as y = x2 describes the relationship between x and y on a Cartesian grid, a polar equation describes a relationship between r and θ on a polar grid. Recall that the coordinate pair (r, θ) indicates that we move counterclockwise from the polar axis (positive x-axis) by an angle of θ, and extend a ray from the pole (origin) r units in the direction of θ. All points that satisfy the polar equation are on the graph. Symmetry is a property that helps us recognize and plot the graph of any equation. If an equation has a graph that is symmetric with respect to an axis, it means that if we folded the graph in half over that axis, the portion of the graph on one side would coincide with the portion on the other side. By performing three tests, we will see how to apply the properties of symmetry to polar equations. Further, we will use symmetry (in addition to plotting key points, zeros, and maximums of r) to determine the graph of a polar equation. π _ In the first test, we consider symmetry with respect to the line θ = (y-axis). We replace (r, θ) with (−r, −θ) to 2 determine if the new equation is equivalent to the original equation. For example, suppose we are given the equation r = 2sin θ; r = 2sin θ −r = 2sin(−θ) −r = −2sin θ r = 2sin θ Replace (r, θ) with (−r, −θ). Identity: sin(−θ)= −sin θ. Multiply both sides by−1. π _ This equation exhibits symmetry with respect to the line θ = . 2 In the second test, we consider symmetry with respect to the polar axis (x -axis). We replace (r, θ) with (r, −θ) or (−r, π − θ) to determine equivalency between the tested equation and the original. For example, suppose we are given the equation r = 1 − 2cos θ. r = 1 − 2cos θ r = 1 − 2cos(−θ) r = 1 − 2cos θ Replace (r, θ) with (r, −θ). Even/Odd identity 682 CHAPTER 8 Further applications oF trigonometry The graph of this equation exhibits symmetry with respect to the polar axis. In the third test, we consider symmetry with respect to the pole (origin). We replace (r, θ) with (−r, θ) to determine if the tested equation is equivalent to the original equation. For example, suppose we are given the equation r = 2sin(3θ). r = 2sin(3θ) −r = 2sin(3θ) The equation has failed the symmetry test, but that does not mean that it is not symmetric with respect to the pole. Passing one or more of the symmetry tests verifies that symmetry will be exhibited in a graph. However, failing the π _ symmetry tests does not necessarily indicate that a graph will not be symmetric about the line θ = , the polar axis, 2 or the pole. In these instances, we can confirm that symmetry exists by plotting reflecting points across the apparent axis of symmetry or the pole. Testing for symmetry is a technique that simplifies the graphing of polar equations, but its application is not perfect. symmetry tests A polar equation describes a curve on the polar grid. The graph of a polar equation can be evaluated for three types of symmetry, as shown in Figure 2. θ θ θ θ (a) (b) θ θ (c) Figure 2 (a) A graph is symmetric with respect to the line θ = π _ (y-axis) if replacing (r, θ ) with (−r, −θ ) yields an equivalent equation. 2 (b) A graph is symmetric with respect to the polar axis (x-axis) if replacing (r, θ) with (r, −θ ) or (−r, π−θ ) yields an equivalent equation. (c) A graph is symmetric with respect to the pole (origin) if replacing (r, θ ) with
(−r, θ ) yields an equivalent equation. How To… Given a polar equation, test for symmetry. π _ symmetry; (r,− θ) for polar 1. Substitute the appropriate combination of components for (r, θ): (−r,− θ) for θ = 2 axis symmetry; and (−r, θ) for symmetry with respect to the pole. 2. If the resulting equations are equivalent in one or more of the tests, the graph produces the expected symmetry. Example 1 Testing a Polar Equation for Symmetry Test the equation r = 2sin θ for symmetry. Solution Test for each of the three types of symmetry. 1) Replacing (r, θ) with (−r, −θ) yields the same result. Thus, the graph is symmetric with respect to the line θ = π _ . 2 −r = 2sin(−θ) −r = −2sin θ r = 2sin θ Even-odd identity Multiply by −1 2) Replacing θ with −θ does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis. Passed Failed r = 2sin(−θ) r = −2sin θ r = −2sin θ ≠ 2sin θ Even-odd identity 3) Replacing r with −r changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole. −r = 2sin θ r = −2sin θ ≠ 2sin θ Failed Table 1 SECTION 8.4 polar coordinates: graphs 683 Analysis Using a graphing calculator, we can see that the equation r = 2sin θ is a circle centered at (0, 1) with radius π π _ r = 1 and is indeed symmetric to the line θ = . We can also see that the graph is not symmetric with the polar axis or the 2 pole. See Figure 3. 4 3 2 1 –4 –3 –2 –1 1 2 3 4 –1 –2 –3 –4 Figure 3 Try It #1 Test the equation for symmetry: r = − 2cos θ. Graphing Polar equations by Plotting Points To graph in the rectangular coordinate system we construct a table of x and y values. To graph in the polar coordinate system we construct a table of θ and r values. We enter values of θ into a polar equation and calculate r. However, using the properties of symmetry and finding key values of θ and r means fewer calculations will be needed. Finding Zeros and Maxima To find the zeros of a polar equation, we solve for the values of θ that result in r = 0. Recall that, to find the zeros of polynomial functions, we set the equation equal to zero and then solve for x. We use the same process for polar equations. Set r = 0, and solve for θ. For many of the forms we will encounter, the maximum value of a polar equation is found by substituting those values of θ into the equation that result in the maximum value of the trigonometric functions. Consider r = 5cos θ; the maximum distance between the curve and the pole is 5 units. The maximum value of the cosine function is 1 when θ = 0, so our polar equation is 5cos θ, and the value θ = 0 will yield the maximum \| r \| . π _ Similarly, the maximum value of the sine function is 1 when θ = , and if our polar equation is r = 5sin θ, the value 2 π _ will yield the maximum \| r \| . We may find additional information by calculating values of r when θ = 0. These points θ = 2 would be polar axis intercepts, which may be helpful in drawing the graph and identifying the curve of a polar equation. Example 2 Finding Zeros and Maximum Values for a Polar Equation Using the equation in Example 1, find the zeros and maximum \| r \| and, if necessary, the polar axis intercepts of r = 2sin θ. Solution To find the zeros, set r equal to zero and solve for θ. 2sin θ = 0 sin θ = 0 θ = sin−1 0 θ = nπ where n is an integer Substitute any one of the θ values into the equation. We will use 0. r = 2sin(0) r =0 The points (0, 0) and (0, ± nπ) are the zeros of the equation. They all coincide, so only one point is visible on the graph. This point is also the only polar axis intercept. 684 CHAPTER 8 Further applications oF trigonometry To find the maximum value of the equation, look at the maximum value of the trigonometric function sin θ, which π π π _ _ _  = 1. Substitute ± 2kπ resulting in sin  occurs when θ = for θ. 2 2 2 π _ r = 2sin   2 r = 2(1) r = 2 π _ Analysis The point  2,  will be the maximum value on the graph. Let’s plot a few more points to verify the graph of a 2 circle. See Table 2 and Figure 4 2π _ 3 5π _ 6 π r = 2sin θ r = 2sin(0) = 0 π _ r = 2sin   = 1 6 π _ r = 2sin   ≈ 1.73 3 π _ r = 2sin   = 2 2 r = 2sin  2π _  ≈ 1.73 3 r = 2sin  5π _  = 1 6 r = 2sin(π) = 0 r 0 1 1.73 2 1.73 1 0 –4 4 4 4 4 –3 3 3 3 3 3 Table 2 4444 4 2 2 333––– –3 444––– –4 Figure Try It #2 Without converting to Cartesian coordinates, test the given equation for symmetry and find the zeros and maximum values of \| r \| : r = 3cos θ. Investigating Circles Now we have seen the equation of a circle in the polar coordinate system. In the last two examples, the same equation was used to illustrate the properties of symmetry and demonstrate how to find the zeros, maximum values, and plotted points that produced the graphs. However, the circle is only one of many shapes in the set of polar curves. There are five classic polar curves: cardioids, limaçons, lemniscates, rose curves, and Archimedes’ spirals. We will briefly touch on the polar formulas for the circle before moving on to the classic curves and their variations. SECTION 8.4 polar coordinates: graphs 685 formulas for the equation of a circle Some of the formulas that produce the graph of a circle in polar coordinates are given by r = acos θ and r = asin θ, where a is the diameter of the circle or the distance from the pole to the farthest point on the circumference. The a a _ _ , or one-half the diameter. For r = acos θ, the center is  , 0  . For r = asin θ, the center is  , π  . 2 2 radius is \| a \| _ 2 Figure 5 shows the graphs of these four circles. r = acos θ, a > 0 (a) r = acos θ, a < 0 (b) r = asin θ, a > 0 (c) r = asin θ, a < 0 (d) Figure 5 Example 3 Sketching the Graph of a Polar Equation for a Circle Sketch the graph of r = 4cos θ. Solution First, testing the equation for symmetry, we find that the graph is symmetric about the polar axis. Next, we π _ find the zeros and maximum \| r \| for r = 4cos θ. First, set r = 0, and solve for θ . Thus, a zero occurs at θ = ± kπ. 2 π _ A key point to plot is  0,  2 To find the maximum value of r, note that the maximum value of the cosine function is 1 when θ = 0 ± 2kπ. Substitute θ = 0 into the equation: r = 4cos θ r = 4cos(0) r = 4(1) = 4 The maximum value of the equation is 4. A key point to plot is (4, 0). As r = 4cos θ is symmetric with respect to the polar axis, we only need to calculate r-values for θ over the interval [0, π]. Points in the upper quadrant can then be reflected to the lower quadrant. Make a table of values similar to Table 3. The graph is shown in Figure 6. θ r 0 4 π _ 6 3.46 π _ 4 2.83 π _ 3 2 π _ 2 0 2π _ 3 −2 3π _ 4 5π _ 6 −2.83 −3.46 π 4 Table 3 –4 –3 –2 –1 4 3 2 1 –1 –2 –3 –4 1 2 3 4(4, 0) Figure 6 686 CHAPTER 8 Further applications oF trigonometry Investigating Cardioids While translating from polar coordinates to Cartesian coordinates may seem simpler in some instances, graphing the classic curves is actually less complicated in the polar system. The next curve is called a cardioid, as it resembles a heart. This shape is often included with the family of curves called limaçons, but here we will discuss the cardioid on its own. formulas for a cardioid The formulas that produce the graphs of a cardioid are given by r = a ± bcos θ and r = a ± bsin θ where a > 0, b > 0, and a __ = 1. The cardioid graph passes through the pole, as we can see in Figure 7. b r = a + bcosθ (a) r = a − bcosθ (b) r = a + bsinθ (c) r = a − bsinθ (d) Figure 7 How To… Given the polar equation of a cardioid, sketch its graph. 1. Check equation for the three types of symmetry. 2. Find the zeros. Set r = 0. 3. Find the maximum value of the equation according to the maximum value of the trigonometric expression. 4. Make a table of values for r and θ. 5. Plot the points and sketch the graph. Example 4 Sketching the Graph of a Cardioid Sketch the graph of r = 2 + 2cos θ. Solution First, testing the equation for symmetry, we find that the graph of this equation will be symmetric about the polar axis. Next, we find the zeros and maximums. Setting r = 0, we have θ = π +2kπ. The zero of the equation is located at (0, π). The graph passes through this point. The maximum value of r = 2 + 2cos θ occurs when cos θ is a maximum, which is when cos θ =1 or when θ = 0. Substitute θ =0 into the equation, and solve for r. r = 2 + 2cos(0) r = 2 + 2(1) = 4 The point (4, 0) is the maximum value on the graph. We found that the polar equation is symmetric with respect to the polar axis, but as it extends to all four quadrants, we need to plot values over the interval [0, π]. The upper portion of the graph is then reflected over the polar axis. Next, we make a table of values, as in Table 4, and then we plot the points and draw the graph. See Figure 8. θ r 0 4 π _ 4 3.41 Table 4 π _ 2 2 2π _ 3 1 π 0 SECTION 8.4 polar coordinates: graphs 687 –4 –3 –2 –1 4 3 2 1 –1 –2 –3 –4 (4, 0) ) 2 3 4 Investigating Limaçons Figure 8 The word limaçon is Old French for “snail,” a name that describes the shape of the graph. As mentioned earlier, the cardioid is a member of the limaçon family, and we can see the similarities in the graphs. The other images in this category include the one-loop limaçon and the two-loop (or inner-loop) limaçon. One-loop limaçons are sometimes a a __ __ ≥ 2. < 2 and convex limaçons when referred to as dimpled limaçons when 1 < b b formulas for one-loop limaçons The formulas that produce the graph of a dimpled one-loop limaçon are given by r = a ± bcos θ and r = a ± bsin θ where a > 0, b > 0, and 1 < a __ < 2. All four graphs are shown in Figure 9. b r = a + bcosθ (a) r = a − bcosθ (b) r = a + bsinθ (c) r = a − bsinθ (d) Figure 9 Dimpled limaçons How To… Given a polar equation for a one-loop limaçon, sketch the graph. 1. Test the equation for symmetry. Remember that failing a symmetry test does not mean that the shape will not exhibit symmetry. Often the symmetry may reveal itself when the points are plotted. 2. Find the zeros. 3. Find the
maximum values according to the trigonometric expression. 4. Make a table. 5. Plot the points and sketch the graph. Example 5 Sketching the Graph of a One-Loop Limaçon Graph the equation r = 4 − 3sin θ. Solution First, testing the equation for symmetry, we find that it fails all three symmetry tests, meaning that the graph may or may not exhibit symmetry, so we cannot use the symmetry to help us graph it. However, this equation has a π __ graph that clearly displays symmetry with respect to the line θ = 2 calculator will immediately illustrate the graph’s reflective quality. , yet it fails all the three symmetry tests. A graphing Next, we find the zeros and maximum, and plot the reflecting points to verify any symmetry. Setting r =0 results in θ being undefined. What does this mean? How could θ be undefined? The angle θ is undefined for any value of sin θ > 1. Therefore, θ is undefined because there is no value of θ for which sin θ > 1. Consequently, the graph does not pass 688 CHAPTER 8 Further applications oF trigonometry through the pole. Perhaps the graph does cross the polar axis, but not at the pole. We can investigate other intercepts by calculating r when θ = 0. r(0) = 4 − 3sin(0) r = 4−3 ⋅ 0 = 4 So, there is at least one polar axis intercept at (4, 0). π π _ _ , we will substitute θ = Next, as the maximum value of the sine function is 1 when θ = into the equation and solve 2 2 for r. Thus, r = 1. Make a table of the coordinates similar to Table 5. θ r 0 4 π _ 6 2.5 π _ 3 1.4 π _ 2 1 2π _ 3 1.4 5π _ 6 2.5 π 4 7π _ 6 5.5 4π _ 3 6.6 3π _ 2 7 5π _ 3 6.6 11π _ 6 5.5 2π 4 The graph is shown in Figure 10. Table 5 1 ) –4 Figure 10 Analysis This is an example of a curve for which making a table of values is critical to producing an accurate graph. The symmetry tests fail; the zero is undefined. While it may be apparent that an equation involving sin θ is likely symmetric π __ with respect to the line θ = , evaluating more points helps to verify that the graph is correct. 2 Try It #3 Sketch the graph of r = 3 − 2cos θ. Another type of limaçon, the inner-loop limaçon, is named for the loop formed inside the general limaçon shape. It was discovered by the German artist Albrecht Dürer(1471-1528), who revealed a method for drawing the inner-loop limaçon in his 1525 book Underweysung der Messing. A century later, the father of mathematician Blaise Pascal, Étienne Pascal(1588-1651), rediscovered it. formulas for inner-loop limaçons The formulas that generate the inner-loop limaçons are given by r = a ± bcos θ and r = a ± bsin θ where a > 0, b > 0, and a < b. The graph of the inner-loop limaçon passes through the pole twice: once for the outer loop, and once for the inner loop. See Figure 11 for the graphs. r = a + bcosθ, a < b (a) r = a − bcosθ, a < b (b) r = a + bsinθ, a < b (c) r = a − bsinθ, a < b (d) Figure 11 SECTION 8.4 polar coordinates: graphs 689 Example 6 Sketching the Graph of an Inner-Loop Limaçon Sketch the graph of r = 2 + 5cos θ. Solution Testing for symmetry, we find that the graph of the equation is symmetric about the polar axis. Next, finding the zeros reveals that when r = 0, θ = 1.98. The maximum \| r \| is found when cos θ =1 or when θ = 0. Thus, the maximum is found at the point (7, 0). Even though we have found symmetry, the zero, and the maximum, plotting more points will help to define the shape, and then a pattern will emerge. See Table 6. θ r 0 7 π __ 6 6.3 π __ 3 4.5 π __ 2 2 2π ___ 3 5π ___ 6 −0.5 −2.3 π −3 7π ___ 6 4π ___ 3 −2.3 −0.5 3π ___ 2 2 5π ___ 3 4.5 11π ____ 6 6.3 2π 7 Table 6 As expected, the values begin to repeat after θ = π. The graph is shown in Figure 12. (−3, π) (7, 0) Investigating Lemniscates Figure 12 Inner-loop limaçon The lemniscate is a polar curve resembling the infinity symbol ∞ or a figure 8. Centered at the pole, a lemniscate is symmetrical by definition. formulas for lemniscates The formulas that generate the graph of a lemniscate are given by r2 = a2 cos 2θ and r2 = a2 sin 2θ where a ≠ 0. The formula r2 = a2 sin 2θ is symmetric with respect to the pole. The formula r2 = a2 cos 2θ is symmetric with respect to the pole, the line θ = π __ 2 , and the polar axis. See Figure 13 for the graphs. r 2 = a 2cos(2θ) (a) r 2 = −a 2cos(2θ) (b) r 2 = a 2sin(2θ) (c) r 2 = −a 2sin(2θ) (d) Figure 13 Example 7 Sketching the Graph of a Lemniscate Sketch the graph of r2 = 4cos 2θ. π _ Solution The equation exhibits symmetry with respect to the line θ = , the polar axis, and the pole. 2 Let’s find the zeros. It should be routine by now, but we will approach this equation a little differently by making the substitution u = 2θ. 690 CHAPTER 8 Further applications oF trigonometry 0 = 4cos 2θ 0 = 4cos u 0 = cos u π _ cos− 2θ = 2 π _ θ = 4 Substitute 2θ back in for u. π _  is a zero of the equation. So, the point  0, 4 Now let’s find the maximum value. Since the maximum of cos u = 1 when u = 0, the maximum cos 2θ = 1 when 2θ = 0. Thus, r2 = 4cos(0) r2 = 4(1 We have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line θ = , and the polar axis, 2 we only need to plot points in the first quadrant. Make a table similar to Table 7 √ Table Plot the points on the graph, such as the one shown in Figure 14. 4 3 2 1 (2, 0) –4 –3 –2 –1 1 2 3 4 –1 –2 –3 –4 Figure 14 lemniscate Analysis Making a substitution such as u = 2θ is a common practice in mathematics because it can make calculations simpler. However, we must not forget to replace the substitution term with the original term at the end, and then solve for the unknown. Some of the points on this graph may not show up using the Trace function on the TI-84 graphing calculator, and the calculator table may show an error for these same points of r. This is because there are no real square roots for 4cos(2θ) are complex numbers because there is a negative these values of θ. In other words, the corresponding r-values of √ number under the radical. — Investigating Rose Curves The next type of polar equation produces a petal-like shape called a rose curve. Although the graphs look complex, a simple polar equation generates the pattern. SECTION 8.4 polar coordinates: graphs 691 rose curves The formulas that generate the graph of a rose curve are given by r = acos nθ and r = asin nθ where a ≠ 0. If n is even, the curve has 2n petals. If n is odd, the curve has n petals. See Figure 15. r = acos(nθ), n even r = asin(nθ), n odd (a) (b) Figure 15 Example 8 Sketching the Graph of a Rose Curve (n Even) Sketch the graph of r = 2cos 4θ. Solution Testing for symmetry, we find again that the symmetry tests do not tell the whole story. The graph is not only π _ symmetric with respect to the polar axis, but also with respect to the line θ = and the pole. 2 Now we will find the zeros. First make the substitution u = 4θ. 0 = 2cos 4θ 0 = cos 4θ 0 = cos u cos− 4θ = The zero is θ =  is on the curve. . The point  0, 8 8 Next, we find the maximum \| r \| . We know that the maximum value of cos u = 1 when θ = 0. Thus, r = 2cos(4 ⋅ 0) r = 2cos(0) r = 2(1) = 2 The point (2, 0) is on the curve. The graph of the rose curve has unique properties, which are revealed in Table 82 3π _ 8 0 π _ 2 2 5π _ 8 0 3π _ 4 −2 Table 8 π __ , it makes sense to divide values in the table by 8 π __ As r = 0 when θ = units. A definite pattern emerges. Look at the 8 range of r-values: 2, 0, −2, 0, 2, 0, −2, and so on. This represents the development of the curve one petal at a time. Starting at r = 0, each petal extends out a distance of r = 2, and then turns back to zero 2n times for a total of eight petals. See the graph in Figure 16. 692 CHAPTER 8 Further applications oF trigonometry –2 –1 2 11 1 –2 n = 4 444444 1 a 2 Figure 16 Rose curve, n even Analysis When these curves are drawn, it is best to plot the points in order, as in the Table 8. This allows us to see how the graph hits a maximum (the tip of a petal), loops back crossing the pole, hits the opposite maximum, and loops back to the pole. The action is continuous until all the petals are drawn. Try It #4 Sketch the graph of r = 4sin(2θ). Example 9 Sketching the Graph of a Rose Curve ( n Odd) Sketch the graph of r = 2sin(5θ). π _ Solution The graph of the equation shows symmetry with respect to the line θ = . Next, find the zeros and maximum. 2 We will want to make the substitution u = 5θ. 0 = 2sin(5θ) 0 = sin u sin −1 0 = 0 u = 0 5θ = 0 θ = 0 The maximum value is calculated at the angle where sin θ is a maximum. Therefore, π _ r = 2sin  5 ⋅  2 r = 2(1) = 2 Thus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for n odd yields the same number of petals as n, there will be five petals on the graph. See Figure 17. –4 –3 –2 –1 4 3 – –2 –3 –4 a 1 2 3 4 Create a table of values similar to Table 9. Figure 17 Rose curve, n odd 1.73 Table 9 π _ 2 2 2π _ 3 −1.73 5π _ 6 1 π 0 SECTION 8.4 polar coordinates: graphs 693 Try It #5 Sketch the graph of r = 3cos(3θ). Investigating the Archimedes’ Spiral The final polar equation we will discuss is the Archimedes’ spiral, named for its discoverer, the Greek mathematician Archimedes (c. 287 BCE–c. 212 BCE), who is credited with numerous discoveries in the fields of geometry and mechanics. Archimedes’ spiral The formula that generates the graph of the Archimedes’ spiral is given by r = θ for θ ≥ 0. As θ increases, r increases at a constant rate in an ever-widening, never-ending, spiraling path. See Figure 18. r = θ, [0, 2π] (a) r = θ, [0, 4π] (b) Figure 18 How To… Given an Archimedes’ spiral over [0, 2π], sketch the graph. 1. Make a table of values for r and θ over the given domain. 2. Plot the points and sketch the graph. Example 10 Sketching the Graph of an Archimedes’ Spiral Sketch the graph of r = θ over [0, 2π]. Solution As r is equal to θ, the plot of the Archimedes’ spiral begins at the pole at the point (0, 0). While the graph hints of symmetry, there is no formal symmetry with regard to passing the symmetry tests. Furt
her, there is no maximum value, unless the domain is restricted. Create a table such as Table 10. θ π _ 4 0.785 π _ 2 1.57 π 2π 3π _ 2 4.71 7π _ 4 5.50 r 3.14 Table 10 Notice that the r-values are just the decimal form of the angle measured in radians. We can see them on a graph in Figure 19. 6.28 (π, π) –5 –4 –3 –2 –7 –6 7 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –7 (2π, 2π) 1 2 3 4 5 7 Figure 19 Archimedes’ spiral 694 CHAPTER 8 Further applications oF trigonometry Analysis The domain of this polar curve is [0, 2π]. In general, however, the domain of this function is (−∞, ∞). Graphing the equation of the Archimedes’ spiral is rather simple, although the image makes it seem like it would be complex. Try It #6 Sketch the graph of r = −θ over the interval [0, 4π]. Summary of Curves We have explored a number of seemingly complex polar curves in this section. Figure 20 and Figure 21 summarize the graphs and equations for each of these curves. Circle Cardioid One-Loop Limaçon Inner-Loop Limaçon r = asin θ r = acos θ r = a ± bcos θ r = a ± bsin θ a > 0, b > 0, a/b = 1 r = a ± bcos θ r = a ± bsin θ a > 0, b > 0, 1 < a/b < 2 r = a ± bcos θ r = a ± bsin θ a > 0, b > 0, a < b (a) (b) (c) (d) Figure 20 Lemniscate Rose Curve (n even) Rose Curve (n odd) Archimedes’ Spiral r2 = a2cos 2θ r2 = a2sin 2θ a ≠ 0 (a) r = acos nθ r = asin nθ n even, 2n petals (b) r = acos nθ r = asin nθ n odd, n petals (c) r = θ θ ≥ 0 (d) Figure 21 Access these online resources for additional instruction and practice with graphs of polar coordinates. • Graphing Polar equations Part 1 (http://openstaxcollege.org/l/polargraph1) • Graphing Polar equations Part 2 (http://openstaxcollege.org/l/polargraph2) • Animation: The Graphs of Polar equations (http://openstaxcollege.org/l/polaranim) • Graphing Polar equations on the TI-84 (http://openstaxcollege.org/l/polarTI84) SECTION 8.4 section exercises 695 8.4 SeCTIOn exeRCISeS VeRBAl 1. Describe the three types of symmetry in polar 2. Which of the three types of symmetries for polar graphs, and compare them to the symmetry of the Cartesian plane. graphs correspond to the symmetries with respect to the x-axis, y-axis, and origin? 3. What are the steps to follow when graphing polar 4. Describe the shapes of the graphs of cardioids, equations? limaçons, and lemniscates. 5. What part of the equation determines the shape of the graph of a polar equation? GRAPHICAl For the following exercises, test the equation for symmetry. 6. r = 5cos 3θ 7. r = 3 − 3cos θ 8. r = 3 + 2sin θ 9. r = 3sin 2θ 10. r = 4 11. r = 2θ 14. r = 3 √ — 1−cos2θ 15. r = √ — 5sin 2θ θ _ 12. r = 4cos 2 2 _ 13. r = θ For the following exercises, graph the polar equation. Identify the name of the shape. 16. r = 3cos θ 17. r = 4sin θ 18. r = 2 + 2cos θ 19. r = 2 − 2cos θ 20. r = 5 − 5sin θ 21. r = 3 + 3sin θ 22. r = 3 + 2sin θ 23. r = 7 + 4sin θ 24. r = 4 + 3cos θ 25. r = 5 + 4cos θ 26. r = 10 + 9cos θ 27. r = 1 + 3sin θ 28. r = 2 + 5sin θ 29. r = 5 + 7sin θ 30. r = 2 + 4cos θ 31. r = 5 + 6cos θ 32. r 2 = 36cos(2θ) 33. r 2 = 10cos(2θ) 34. r 2 = 4sin(2θ) 35. r 2 = 10sin(2θ) 36. r = 3sin(2θ) 40. r = 4sin(5θ) TeCHnOlOGY 37. r = 3cos(2θ) 38. r = 5sin(3θ) 39. r = 4sin(4θ) 41. r = −θ 42. r = 2θ 43. r = − 3θ For the following exercises, use a graphing calculator to sketch the graph of the polar equation. 1 _ 44. r = θ 47. r = 2 √ — 1 − sin2 θ , a hippopede 50. r = θ2 53. r = θcos θ 45. r = 1 _ — θ √ 48. r = 5 + cos(4θ) 51. r = θ + 1 46. r = 2sin θ tan θ, a cissoid 49. r = 2 − sin(2θ) 52. r = θsin θ For the following exercises, use a graphing utility to graph each pair of polar equations on a domain of [0, 4π] and then explain the differences shown in the graphs. 54. r = θ, r = −θ 55. r = θ, r = θ + sin θ 56. r = sin θ + θ, r = sin θ − θ θ θ   , r = θsin  57. r = 2sin  _ _ 2 2 58. r = sin(cos(3θ)) r = sin(3θ) 696 CHAPTER 8 Further applications oF trigonometry 59. On a graphing utility, graph r = sin  16 _ θ  on 5 [0, 4π], [0, 8π], [0, 12π], and [0, 16π]. Describe the effect of increasing the width of the domain. 60. On a graphing utility, graph and sketch 5 _ θ   r = sin θ +  sin  2 3 on [0, 4π]. 61. On a graphing utility, graph each polar equation. 62. On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs. Explain the similarities and differences you observe in the graphs. r1 = 3sin(3θ) r2 = 2sin(3θ) r3 = sin(3θ) r1 = 3 + 3cos θ r2 = 2 + 2cos θ r3 = 1 + cos θ 63. On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs. r1 = 3θ r2 = 2θ r3 = θ exTenSIOnS For the following exercises, draw each polar equation on the same set of polar axes, and find the points of intersection. 64. r1 = 3 + 2sin θ, r2 = 2 66. r1 = 1 + sin θ, r2 = 3sin θ 68. r1 = cos(2θ), r2 = sin(2θ) 70. r1 = √ — 3 , r2 = 2sin(θ) 72. r1 = 1 + cos θ, r2 = 1 − sin θ 65. r1 = 6 − 4cos θ, r2 = 4 67. r1 = 1 + cos θ, r2 = 3cos θ 69. r1 = sin2 (2θ), r2 = 1 − cos(4θ) 71. r1 2 = sin θ, r2 2 = cos θ SECTION 8.5 polar Form oF complex numBers 697 leARnInG OBjeCTIVeS In this section, you will: • Plot complex numbers in the complex plane. • Find the absolute value of a complex number. • Write complex numbers in polar form. • Convert a complex number from polar to rectangular form. • Find products of complex numbers in polar form. • Find quotients of complex numbers in polar form. • Find powers of complex numbers in polar form. • Find roots of complex numbers in polar form. 8.5 POlAR FORM OF COMPlex nUMBeRS “God made the integers; all else is the work of man.” This rather famous quote by nineteenth-century German mathematician Leopold Kronecker sets the stage for this section on the polar form of a complex number. Complex numbers were invented by people and represent over a thousand years of continuous investigation and struggle by mathematicians such as Pythagoras, Descartes, De Moivre, Euler, Gauss, and others. Complex numbers answered questions that for centuries had puzzled the greatest minds in science. We first encountered complex numbers in Complex Numbers. In this section, we will focus on the mechanics of working with complex numbers: translation of complex numbers from polar form to rectangular form and vice versa, interpretation of complex numbers in the scheme of applications, and application of De Moivre’s Theorem. Plotting Complex numbers in the Complex Plane Plotting a complex number a + bi is similar to plotting a real number, except that the horizontal axis represents the real part of the number, a, and the vertical axis represents the imaginary part of the number, bi. How To… Given a complex number a + bi, plot it in the complex plane. 1. Label the horizontal axis as the real axis and the vertical axis as the imaginary axis. 2. Plot the point in the complex plane by moving a units in the horizontal direction and b units in the vertical direction. Example 1 Plotting a Complex Number in the Complex Plane Plot the complex number 2 − 3i in the complex plane. Solution From the origin, move two units in the positive horizontal direction and three units in the negative vertical direction. See Figure 1. Imaginary –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 Real 21 3 4 5 2 – 3i Figure 1 698 CHAPTER 8 Further applications oF trigonometry Try It #1 Plot the point 1 + 5i in the complex plane. Finding the Absolute Value of a Complex number The first step toward working with a complex number in polar form is to find the absolute value. The absolute value of a complex number is the same as its magnitude, or ∣ z ∣ . It measures the distance from the origin to a point in the plane. For example, the graph of z = 2 + 4i, in Figure 2, shows ∣ z ∣ . Imaginary (2 + 4i) \|z\| = 20 21 3 4 5 Real –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 Figure 2 absolute value of a complex number Given z = x + yi, a complex number, the absolute value of z is defined as ∣ z ∣ = √ It is the distance from the origin to the point (x, y). — x2 + y2 Notice that the absolute value of a real number gives the distance of the number from 0, while the absolute value of a complex number gives the distance of the number from the origin, (0, 0). Example 2 Finding the Absolute Value of a Complex Number with a Radical Find the absolute value of z = √ — 5 − i. Solution Using the formula, we have See Figure 3. — x2 + y2 — — √ 5 2 + (−1) Imaginary –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 Real 21 3 4 5 5 Figure 3 SECTION 8.5 polar Form oF complex numBers 699 Try It #2 Find the absolute value of the complex number z = 12 − 5i. Example 3 Finding the Absolute Value of a Complex Number Given z = 3 − 4i, find ∣ z ∣ . Solution Using the formula, we have — — x2+ y2 (3)2 + (−4)2 9 + 16 25 — — ∣ The absolute value of z is 5. See Figure 4. Imaginary –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 Real 21 5 4 3 \|z\| = 5 (3 − 4i) Figure 4 Try It #3 Given z =1 − 7i, find ∣ z ∣ . Writing Complex numbers in Polar Form The polar form of a complex number expresses a number in terms of an angle θ and its distance from the origin r. Given a complex number in rectangular form expressed as z = x + yi, we use the same conversion formulas as we do to write the number in trigonometric form: x = rcos θ y = rsin θ r = √ — x2+ y2 We review these relationships in Figure 5. Imaginary x + yi y Real Figure 5 r θ x 7 00 CHAPTER 8 Further applications oF trigonometry We use the term modulus to represent the absolute value of a complex number, or the distance from the origin to the point (x, y). The modulus, then, is the same as r, the radius in polar form. We use θ to indicate the angle of direction (just as with polar coordinates). Substituting, we have z = x + yi z = rcos θ + (rsin θ)i z = r(cos θ + isin θ) polar form of a complex number Writing a complex number in polar form involves the following conversion formulas: Making a direct substitution, we have x = rcos θ y = rsin θ r = √ — x2+ y2 z = x + yi z = (rcos θ) + i(rsin θ) z = r(cos θ + isin θ) where r is the modulus and θ
is the argument. We often use the abbreviation rcis θ to represent r(cos θ + isin θ). Example 4 Expressing a Complex Number Using Polar Coordinates Express the complex number 4i using polar coordinates. Solution On the complex plane, the number z = 4i is the same as z = 0 + 4i. Writing it in polar form, we have to calculate r first. r = √ r = √ r = √ — — x2+ y2 02 + 42 16 — r = 4 π _ . In polar coordinates, the complex number z = 0 + 4i can be Next, we look at x. If x = rcos θ, and x = 0, then θ = 2 π π π _ _ _  + isin  written as z = 4  cos   . See Figure 6.   or 4cis  2 2 2 Imaginary z = 4i π 2 21 3 4 5 Real –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 Figure 6 Try It #4 Express z = 3i as rcis θ in polar form. SECTION 8.5 polar Form oF complex numBers 701 Example 5 Finding the Polar Form of a Complex Number Find the polar form of −4 + 4i. Solution First, find the value of r √ — — x2+ y2 (−4)2 + (42) 32 2 — — Find the angle θ using the formula: Thus, the solution is 4 √ — 2 cis  3π _  . 4 — cos θ = x _ cos θ = r −4 ______ 2 4 √ cos θ = − 1 _____ 2 √ 3π θ = cos−1  − 1 _____ _  = 4 2 √ — — Try It #5 _ 3 + i in polar form. Write z = √ Converting a Complex number from Polar to Rectangular Form Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. In other words, given z = r(cos θ + isin θ), first evaluate the trigonometric functions cos θ and sin θ. Then, multiply through by r. Example 6 Converting from Polar to Rectangular Form Convert the polar form of the given complex number to rectangular form: Solution We begin by evaluating the trigonometric expressions. π π _ _ z = 12  cos   + isin    6 6 — π _  = cos  6 3 √ _ 2 π 1 _ _  = and sin  2 6 After substitution, the complex number is We apply the distributive property: — 3 z = 12  √ = 12  √ _ 2 i  1 _ + 2 The rectangular form of the given point in complex form is 6 √ — 3 + 6i. = 6 √ — 3 + 6i = (1212) i 2 7 02 CHAPTER 8 Further applications oF trigonometry Example 7 Finding the Rectangular Form of a Complex Number Find the rectangular form of the complex number given r = 13 and tan θ = y _ , and tan θ = x , we first determine r = √ Solution If tan θ = — 5 _ . 12 x2+ y2 = √ 5 _ 12 — x _ 122 + 52 = 13. We then find cos θ = r y _ and sin θ = r . z = 13(cos θ + isin θ) + 5 _ i  13 12 _ = 13  13 = 12 + 5i The rectangular form of the given number in complex form is 12 + 5i. Try It #6 Convert the complex number to rectangular form: z = 4  cos 11π _ 6 + isin 11π _  6 Finding Products of Complex numbers in Polar Form Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. For the rest of this section, we will work with formulas developed by French mathematician Abraham De Moivre (1667–1754). These formulas have made working with products, quotients, powers, and roots of complex numbers much simpler than they appear. The rules are based on multiplying the moduli and adding the arguments. products of complex numbers in polar form If z1 = r1(cos θ1 + isin θ1) and z2 = r2(cos θ2 + isin θ2), then the product of these numbers is given as: Notice that the product calls for multiplying the moduli and adding the angles. z1z2 = r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)] z1z2 = r1r2cis(θ1 + θ2) Example 8 Finding the Product of Two Complex Numbers in Polar Form Find the product of z1z2, given z1 = 4(cos(80°) + isin(80°)) and z2 = 2(cos(145°) + isin(145°)). Solution Follow the formula z1z2 = 4 ⋅ 2[cos(80° + 145°) + isin(80° + 145°)] z1z2 = 8[cos(225°) + isin(225°)] z1z2 = 8  cos  5π _  + isin  4 5π  − z1z2 = z1z2 = − 4 √ 2 − 4i √ — Finding Quotients of Complex numbers in Polar Form The quotient of two complex numbers in polar form is the quotient of the two moduli and the difference of the two arguments. SECTION 8.5 polar Form oF complex numBers 703 quotients of complex numbers in polar form If z1 = r1(cos θ1 + isin θ1) and z2 = r2(cos θ2 + isin θ2), then the quotient of these numbers is z1 __ z2 z1 __ z2 r1 __ r2 [cos(θ1 −θ2) + isin(θ1 − θ2)], z2 ≠ 0 r1 __ r2 cis(θ1 − θ2), z2 ≠ 0 = = Notice that the moduli are divided, and the angles are subtracted. How To… Given two complex numbers in polar form, find the quotient. . 1. Divide r1 __ r2 2. Find θ1 − θ2. 3. Substitute the results into the formula: z = r(cos θ + isin θ). Replace r with r1 __ r2 , and replace θ with θ1 − θ2. 4. Calculate the new trigonometric expressions and multiply through by r. Example 9 Finding the Quotient of Two Complex Numbers Find the quotient of z1 = 2(cos(213°) + isin(213°)) and z2 = 4(cos(33°) + isin(33°)). Solution Using the formula, we have 2 __ [cos(213° − 33°) + isin(213° − 33°)] = 4 1 __ [cos(180°) + isin(180°)] = 2 z1 __ z2 z1 __ z2 z1 __ z2 z1 __ z2 z1 __ z2 1 __ [ − 1 + 0i] = 2 1 __ + 0i = − 2 1 __ = − 2 Try It #7 Find the product and the quotient of z1 = 2 √ — 3 (cos(150°) + isin(150°)) and z2 = 2(cos(30°) + isin(30°)). Finding Powers of Complex numbers in Polar Form Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem. It states that, for a positive integer n, zn is found by raising the modulus to the nth power and multiplying the argument by n. It is the standard method used in modern mathematics. De Moivre’s Theorem If z = r(cos θ + isin θ) is a complex number, then zn = r n[cos(nθ) + isin(nθ)] zn = r n cis(nθ) where n is a positive integer. 7 04 CHAPTER 8 Further applications oF trigonometry Example 10 Evaluating an Expression Using De Moivre’s Theorem Evaluate the expression (1 + i)5 using De Moivre’s Theorem. Solution Since De Moivre’s Theorem applies to complex numbers written in polar form, we must first write (1 + i) in polar form. Let us find r. r = √ — x2 + y2 r = √ — (1)2 + (1)2 r = √ — 2 y _ Then we find θ. Using the formula tan θ = x gives 1 _ tan θ = 1 tan θ = 1 π _ θ = 4 Use De Moivre’s Theorem to evaluate the expression. (a + bi)n = rn[cos(nθ) + isin(nθ)] (1 + i)5  cos  5 ⋅  + isin  5 ⋅   4 4 (1 + i)5 = 4 √ — 2  cos  5π _  + isin  4 5π _   4 — — — (1 + i)1 + i)5 = − 4 − 4i Finding Roots of Complex numbers in Polar Form To find the nth root of a complex number in polar form, we use the nth Root Theorem or De Moivre’s Theorem and raise the complex number to a power with a rational exponent. There are several ways to represent a formula for finding nth roots of complex numbers in polar form. the nth root theorem To find the nth root of a complex number in polar form, use the formula given as 1 1 __ n = r z θ __ n  cos  _ n + θ 2kπ n  + isin  _ _ n + 2kπ n   _ where k = 0, 1, 2, 3, . . . , n − 1. We add θ 2kπ _ _ n to n in order to obtain the periodic roots. SECTION 8.5 polar Form oF complex numBers 705 Example 11 Finding the nth Root of a Complex Number Evaluate the cube roots of z = 8  cos  Solution We have 2π  + isin  ___ 3 2π   . ___ 3 1 1 __ __ = 8 z 3 3  cos  2π ___ 3 _ 3 +  + isin  2kπ ____ 3 2π ___ 3 _ 3 +   2kπ ____ 3 1 __ = 2  cos  z 3 2π ___ + 9 2kπ ____ 3  + isin  2π ___ 9 + 2kπ ____ 3   There will be three roots: k = 0, 1, 2. When k = 0, we have 1 __ = 2  cos  z 3 2π  + isin  ___ 9 2π   ___ 9 When k = 1, we have When k = 2, we have 1 __ = 2  cos  z 3 2π ___ 9 + 6π  + isin  ___ 9 2π ___ 9 + 6π   Add ___ 9 2(1)π _ 3 to each angle. 1 __ = 2  cos  z 3 8π  + isin  ___ 9 8π   ___ 9 1 __ = 2  cos  z 3 2π ___ + 9 12π ____ 9  + isin  2π ___ 9 + 12π ____ 9   Add 2(2)π _ 3 to each angle. 1 __ = 2  cos  z 3 14π ____ 9  + isin  14π   ____ 9 Remember to find the common denominator to simplify fractions in situations like this one. For k = 1, the angle simplification is 2π ___ 3 _ 3 + 2(1)π _____ 3 = 2π 1  +  __ ___ 3 3 3 2(1)π   __ _____ 3 3 = 2π ___ 9 + 6π ___ 9 = 8π ___ 9 Try It #8 Find the four fourth roots of 16(cos(120°) + isin(120°)). Access these online resources for additional instruction and practice with polar forms of complex numbers. • The Product and Quotient of Complex numbers in Trigonometric Form (http://openstaxcollege.org/l/prodquocomplex) • De Moivre’s Theorem (http://openstaxcollege.org/l/demoivre) 7 06 CHAPTER 8 Further applications oF trigonometry 8.5 SeCTIOn exeRCISeS VeRBAl 1. A complex number is a + bi. Explain each part. 2. What does the absolute value of a complex number 3. How is a complex number converted to polar form? 5. What is De Moivre’s Theorem and what is it used for? AlGeBRAIC represent? 4. How do we find the product of two complex numbers? For the following exercises, find the absolute value of the given complex number. 6. 5 + 3i _ 2 − 6i 9. √ 7. −7 + i 10. 2i For the following exercises, write the complex number in polar form. 12. 2 + 2i 15. √ — 3 + i 13. 8 − 4i 16. 3i 8. −3 − 3i 11. 2.2 − 3.1i 1 1 __ __ − 14. − i 2 2 For the following exercises, convert the complex number from polar to rectangular form. π _ 17. z = 7cis   6 20. z = 7cis(25°) π _ 18. z = 2cis   3 21. z = 3cis(240°) 19. z = 4cis  7π _  6 22. z = √ — 2 cis(100°) For the following exercises, find z1 z2 in polar form. 23. z1 = 2 √ — 3 cis(116°); z2 = 2cis(82°) 1 _ 25. z1 = 3cis(120°); z2 = cis(60°) 4 27. z1 = √ — 5 cis  5π _  ; z2 = √ 8 — 15 cis  π _  12 For the following exercises, find z1 _ in polar form. z2 29. z1 = 21cis(135°); z2 = 3cis(65°) 31. z1 = 15cis(120°); z2 = 3cis(40°) 33. z1 = 5 √ — 2 cis(π); z2 = √ — 2 cis  2π _  3 24. z1 = √ — 2 cis(205°); z2 = 2 √ — 2 cis(118°) π π _ _  ; z2 = 5cis  26. z1 = 3cis   4 6 π π _ _  ; z2 = 2cis  28. z1 = 4cis   4 2 30. z1 = √ — 2 cis(90°); z2 = 2cis(60°) π π _ _  ; z2 = 2cis  32. z1 = 6cis   4 3 34. z1 = 2cis  π 3π _ _  ; z2 = 3cis   5 4 For the following exercises, find the powers of each complex number in polar form. 35. Find z3 when z = 5cis(45°). 37. Find z2 when z = 3cis(120°). 39. Find z4 when z = cis  3π _  . 16 36. Find z4 when z = 2cis(70°). π _ 38. Find z2 when z = 4cis   . 4 40. Find z3 when z = 3cis  5π _  . 3 SECTION 8.5
section exercises 707 For the following exercises, evaluate each root. 41. Evaluate the cube root of z when z = 27cis(240°). 43. Evaluate the cube root of z when z = 32cis  7π _  . 4 45. Evaluate the cube root of z when z = 8cis  2π _  . 3 42. Evaluate the square root of z when z = 16cis(100°). 44. Evaluate the square root of z when z = 32cis(π). GRAPHICAl For the following exercises, plot the complex number in the complex plane. 46. 2 + 4i 47. −3 − 3i 49. −1 − 5i 52. −4 55. 1 − 4i TeCHnOlOGY 50. 3 + 2i 53. 6 − 2i 48. 5 − 4i 51. 2i 54. −2 + i For the following exercises, find all answers rounded to the nearest hundredth. 56. Use the rectangular to polar feature on the graphing 57. Use the rectangular to polar feature on the graphing calculator to change 5 + 5i to polar form. calculator to change 3 − 2i to polar form. 58. Use the rectangular to polar feature on the graphing calculator to change −3 − 8i to polar form. 59. Use the polar to rectangular feature on the graphing calculator to change 4cis(120°) to rectangular form. 60. Use the polar to rectangular feature on the graphing calculator to change 2cis(45°) to rectangular form. 61. Use the polar to rectangular feature on the graphing calculator to change 5cis(210°) to rectangular form. 7 08 CHAPTER 8 Further applications oF trigonometry leARnInG OBjeCTIVeS In this section, you will: • Parameterize a curve. • Eliminate the parameter. • Find a rectangular equation for a curve defined parametrically. • Find parametric equations for curves defined by rectangular equations. 8.6 PARAMeTRIC eQUATIOnS Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in Figure 1. At any moment, the moon is located at a particular spot relative to the planet. But how do we write and solve the equation for the position of the moon when the distance from the planet, the speed of the moon’s orbit around the planet, and the speed of rotation around the sun are all unknowns? We can solve only for one variable at a time. Figure 1 In this section, we will consider sets of equations given by x(t) and y(t) where t is the independent variable of time. We can use these parametric equations in a number of applications when we are looking for not only a particular position but also the direction of the movement. As we trace out successive values of t, the orientation of the curve becomes clear. This is one of the primary advantages of using parametric equations: we are able to trace the movement of an object along a path according to time. We begin this section with a look at the basic components of parametric equations and what it means to parameterize a curve. Then we will learn how to eliminate the parameter, translate the equations of a curve defined parametrically into rectangular equations, and find the parametric equations for curves defined by rectangular equations. Parameterizing a Curve When an object moves along a curve—or curvilinear path—in a given direction and in a given amount of time, the position of the object in the plane is given by the x-coordinate and the y-coordinate. However, both x and y vary over time and so are functions of time. For this reason, we add another variable, the parameter, upon which both x and y are dependent functions. In the example in the section opener, the parameter is time, t. The x position of the moon at time, t, is represented as the function x(t), and the y position of the moon at time, t, is represented as the function y(t). Together, x(t) and y(t) are called parametric equations, and generate an ordered pair (x(t), y(t)). Parametric equations primarily describe motion and direction. When we parameterize a curve, we are translating a single equation in two variables, such as x and y, into an equivalent pair of equations in three variables, x, y, and t. One of the reasons we parameterize a curve is because the parametric equations yield more information: specifically, the direction of the object’s motion over time. When we graph parametric equations, we can observe the individual behaviors of x and of y. There are a number of shapes that cannot be represented in the form y = f (x), meaning that they are not functions. For example, consider r2 − x2 and the graph of a circle, given as r 2 = x 2 + y 2. Solving for y gives y = ± √ r2 − x2 . If we graph y1 and y2 together, the graph will not pass the vertical line test, as shown in Figure 2. Thus, y2 = − √ the equation for the graph of a circle is not a function. r2 − x2 , or two equations: y1 = √ — — — SECTION 8.6 parametric eQuations 709 y Vertical line test on circle r2 = x2 + y2 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 Figure 2 However, if we were to graph each equation on its own, each one would pass the vertical line test and therefore would represent a function. In some instances, the concept of breaking up the equation for a circle into two functions is similar to the concept of creating parametric equations, as we use two functions to produce a non-function. This will become clearer as we move forward. parametric equations Suppose t is a number on an interval, I. The set of ordered pairs, (x(t), y(t)), where x = f (t) and y = g(t), forms a plane curve based on the parameter t. The equations x = f (t) and y = g(t) are the parametric equations. Example 1 Parameterizing a Curve Parameterize the curve y = x2 − 1 letting x(t) = t. Graph both equations. Solution If x(t) = t, then to find y(t) we replace the variable x with the expression given in x(t). In other words, y(t) = t 2 − 1. Make a table of values similar to Table 1, and sketch the graph. t −4 −3 −2 −1 0 1 2 3 4 x(t) −4 −3 −2 −1 0 1 2 3 4 y(t) y(−4) = (−4)2 − 1 = 15 y(−3) = (−3)2 − 1 = 8 y(−2) = (−2)2 − 1 = 3 y(−1) = (−1)2 − 1 = 0 y(0) = (0)2 − 1 = − 1 y(1) = (1)2 − 1 = 0 y(2) = (2)2 − 1 = 3 y(3) = (3)2 − 1 = 8 y(4) = (4)2 − 1 = 15 Table 1 See the graphs in Figure 3. It may be helpful to use the TRACE feature of a graphing calculator to see how the points are generated as t increases. –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (a) 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (b) 21 3 4 5 x Figure 3 (a) Parametric y(t ) = t 2 − 1 (b) Rectangular y = x 2 − 1 7 10 CHAPTER 8 Further applications oF trigonometry Analysis The arrows indicate the direction in which the curve is generated. Notice the curve is identical to the curve of y = x2 − 1. Try It #1 Construct a table of values and plot the parametric equations: x(t) = t − 3, y(t) = 2t + 4; − 1 ≤ t ≤ 2. Example 2 Finding a Pair of Parametric Equations Find a pair of parametric equations that models the graph of y = 1 − x2, using the parameter x(t) = t. Plot some points and sketch the graph. Solution If x(t) = t and we substitute t for x into the y equation, then y(t) = 1 − t2. Our pair of parametric equations is x(t) = t y(t) = 1 − t2 To graph the equations, first we construct a table of values like that in Table 2. We can choose values around t = 0, from t = − 3 to t = 3. The values in the x(t) column will be the same as those in the t column because x(t) = t. Calculate values for the column y(t). t −3 −2 −1 0 1 2 3 x(t) = t −3 −2 −1 0 1 2 3 Table 2 y(t) = 1 − t2 y(−3) = 1 − (−3)2 = − 8 y(−2) = 1 − (−2)2 = − 3 y(−1) = 1 − (−1)2 = 0 y(0) = 1 − 0 = 1 y(1) = 1 − (1)2 = 0 y(2) = 1 − (2)2 = − 3 y(3) = 1 − (3)2 = − 8 The graph of y = 1 − t2 is a parabola facing downward, as shown in Figure 4. We have mapped the curve over the interval [−3, 3], shown as a solid line with arrows indicating the orientation of the curve according to t. Orientation refers to the path traced along the curve in terms of increasing values of t. As this parabola is symmetric with respect to the line x = 0, the values of x are reflected across the y-axis. 321 4 5 x –5 –4 –3 –2 y 2 1 –1 –1 –2 –3 –4 –5 –6 –7 –8 –9 –10 Figure 4 Try It #2 Parameterize the curve given by x = y3 − 2y. SECTION 8.6 parametric eQuations 711 Example 3 Finding Parametric Equations That Model Given Criteria An object travels at a steady rate along a straight path (−5, 3) to (3, −1) in the same plane in four seconds. The coordinates are measured in meters. Find parametric equations for the position of the object. Solution The parametric equations are simple linear expressions, but we need to view this problem in a step-by-step fashion. The x-value of the object starts at −5 meters and goes to 3 meters. This means the distance x has changed by , or 2 m/s. We can write the x-coordinate as a linear function with respect 8 meters in 4 seconds, which is a rate of 8m _ 4s to time as x(t) = 2t − 5. In the linear function template y = mx + b, 2t = mx and −5 = b. Similarly, the y-value of the −4m _ object starts at 3 and goes to −1, which is a change in the distance y of −4 meters in 4 seconds, which is a rate of , 4s or −1 m/s. We can also write the y-coordinate as the linear function y(t) = −t + 3. Together, these are the parametric equations for the position of the object, where x and y are expressed in meters and t represents time: Using these equations, we can build a table of values for t, x, and y (see Table 3). In this example, we limited values of t to non-negative numbers. In general, any value of t can be used. x(t) = 2t − 5 y(t) = −(t) = 2t − 5 x = 2(0) − 5 = −5 x = 2(1) − 5 = −3 x = 2(2) − 5 = −1 x = 2(3) − 5 = 1 x = 2(4) − 5 = 3 Table 3 y(t) = −t + 3 y = −(0) + 3 = 3 y = −(1) + 3 = 2 y = −(2) + 3 = 1 y = −(3) + 3 = 0 y = −(4) + 3 = − 1 From this table, we can create three graphs, as shown in Figure 5. –6 –5 –4 –3 –2 x 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 – 6 (a) 21 3 4 5 6 t –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 (b) y 5 4 3 2 1 t =1 21 3 4 5 6 t –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 (c) t =3 21 3 4 5 x Figure 5 (a) A graph of x vs. t, representing the horizontal position over time. (b) A graph of y vs. t, representing the vertical position over time. (c) A graph of y vs. x, representing the position of the object in the plane at time t. Analysis Again, we see that, in F
igure 5(c), when the parameter represents time, we can indicate the movement of the object along the path with arrows. eliminating the Parameter In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as x and y. Eliminating the parameter is a method that may make graphing some curves easier. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate the orientation of the curve as well. There are various methods for eliminating the parameter t from a set of parametric equations; not every method works for every type of equation. Here we will review the methods for the most common types of equations. 7 12 CHAPTER 8 Further applications oF trigonometry Eliminating the Parameter from Polynomial, Exponential, and Logarithmic Equations For polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve for t. We substitute the resulting expression for t into the second equation. This gives one equation in x and y. Example 4 Eliminating the Parameter in Polynomials Given x(t) = t2 + 1 and y(t) = 2 + t, eliminate the parameter, and write the parametric equations as a Cartesian equation. Solution We will begin with the equation for y because the linear equation is easier to solve for t. Next, substitute y − 2 for t in x(t). = t2 + 1 x = (y − 2)2 + 1 Substitute the expression for t into x. x = y2 − 4y + 4 + 1 x = y2 − 4y + 5 x = y2 − 4y + 5 The Cartesian form is x = y2 − 4y + 5. Analysis This is an equation for a parabola in which, in rectangular terms, x is dependent on y. From the curve’s vertex at (1, 2), the graph sweeps out to the right. See Figure 6. In this section, we consider sets of equations given by the functions x(t) and y(t), where t is the independent variable of time. Notice, both x and y are functions of time; so in general y is not a function of x. y 6 5 4 3 2 1 –2 –1 –1 –2 –3 –4 –5 21 3 4 5 6 7 8 9 10 11 12 13 x Figure 6 Try It #3 Given the equations below, eliminate the parameter and write as a rectangular equation for y as a function of x. x(t) = 2t2 + 6 y(t) = 5 − t Example 5 Eliminating the Parameter in Exponential Equations Eliminate the parameter and write as a Cartesian equation: x(t) = e−t and y(t) = 3et, t > 0. Solution Isolate et. x = e−t et = 1 _ x SECTION 8.6 parametric eQuations 713 Substitute the expression into y(t). 3 _ The Cartesian form is y = x y = 3et Analysis The graph of the parametric equation is shown in Figure 7(a). The domain is restricted to t > 0. The Cartesian equation, y = 3x is shown in Figure 7(b) and has only one restriction on the domain, x ≠ 0. –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 (a) x(t) = e−t y(t) = 3et 21 3 4 5 6 x –6 –5 –4 –3 –2 Figure 7 y = 3 x 21 1 –1 –2 –3 –4 –5 –6 (b) Example 6 Eliminating the Parameter in Logarithmic Equations — Eliminate the parameter and write as a Cartesian equation: x(t) = √ t + 2 and y(t) = log(t). Solution Solve the first equation for tx − 2)2 = t Square both sides. Then, substitute the expression for t into the y equation. y = log(t) y = log(x − 2)2 The Cartesian form is y = log(x − 2)2. Analysis To be sure that the parametric equations are equivalent to the Cartesian equation, check the domains. The t + 2 to t > 0; we restrict the domain on x to x > 2. The domain for parametric equations restrict the domain on x = √ the parametric equation y = log(t) is restricted to t > 0; we limit the domain on y = log(x − 2)2 to x > 2. — Try It #4 Eliminate the parameter and write as a rectangular equation. x(t) = t2 y(t) = ln(t) t > 0 Eliminating the Parameter from Trigonometric Equations Eliminating the parameter from trigonometric equations is a straightforward substitution. We can use a few of the familiar trigonometric identities and the Pythagorean Theorem. 7 14 CHAPTER 8 Further applications oF trigonometry First, we use the identities: Solving for cos t and sin t, we have Then, use the Pythagorean Theorem: Substituting gives x(t) = acos t y(t) = bsin t x _ a = cos t y _ a = sin t cos2 t + sin2 t = 1 x _ cos2 t + sin2 Example 7 Eliminating the Parameter from a Pair of Trigonometric Parametric Equations Eliminate the parameter from the given pair of trigonometric equations where 0 ≤ t ≤ 2π and sketch the graph. Solution Solving for cos t and sin t, we have x(t) = 4cos t y(t) = 3sin t x = 4cos t x _ = cos t 4 y = 3sin t y _ = sin t 3 Next, use the Pythagorean identity and make the substitutions. 2 2 x _   4 cos2 t + sin2 t = 1 y  +  _ 3 y2 x2 _ _ 9 16 = 1 = 1 + The graph for the equation is shown in Figure 8. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 – 21 3 4 Figure 8 Analysis Applying the general equations for conic sections (introduced in Analytic Geometry, we can identify x2 π _ _ = 1 as an ellipse centered at (0, 0). Notice that when t = 0 the coordinates are (4, 0), and when t = the 2 16 coordinates are (0, 3). This shows the orientation of the curve with increasing values of t. y2 _ 9 + SECTION 8.6 parametric eQuations 715 Try It #5 Eliminate the parameter from the given pair of parametric equations and write as a Cartesian equation: x(t) = 2cos t and y(t) = 3sin t. Finding Cartesian equations from Curves Defined Parametrically When we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially “eliminating the parameter.” However, there are various methods we can use to rewrite a set of parametric equations as a Cartesian equation. The simplest method is to set one equation equal to the parameter, such as x(t) = t. In this case, y(t) can be any expression. For example, consider the following pair of equations. x(t) = t y(t) = t 2 − 3 Rewriting this set of parametric equations is a matter of substituting x for t. Thus, the Cartesian equation is y = x2 − 3. Example 8 Finding a Cartesian Equation Using Alternate Methods Use two different methods to find the Cartesian equation equivalent to the given set of parametric equations. x(t) = 3t − 2 y(t) = t + 1 Solution Method 1. First, let’s solve the x equation for t. Then we can substitute the result into the y equation. x = 3t − 2 x + 2 = 3t x + 2 _ 3 = t Now substitute the expression for t into the y equation Method 2. Solve the y equation for t and substitute this expression in the x equation. Make the substitution and then solve for y(y − 1) − 2 x = 3y − 3 − 2 x = 3y − 5 x + 5 = 3y 16 CHAPTER 8 Further applications oF trigonometry Try It #6 Write the given parametric equations as a Cartesian equation: x(t) = t3 and y(t) = t 6. Finding Parametric equations for Curves Defined by Rectangular equations Although we have just shown that there is only one way to interpret a set of parametric equations as a rectangular equation, there are multiple ways to interpret a rectangular equation as a set of parametric equations. Any strategy we may use to find the parametric equations is valid if it produces equivalency. In other words, if we choose an expression to represent x, and then substitute it into the y equation, and it produces the same graph over the same domain as the rectangular equation, then the set of parametric equations is valid. If the domain becomes restricted in the set of parametric equations, and the function does not allow the same values for x as the domain of the rectangular equation, then the graphs will be different. Example 9 Finding a Set of Parametric Equations for Curves Defined by Rectangular Equations Find a set of equivalent parametric equations for y = (x + 3)2 + 1. Solution An obvious choice would be to let x(t) = t. Then y(t) = (t + 3)2 + 1. But let’s try something more interesting. What if we let x = t + 3? Then we have The set of parametric equations is See Figure 91 –1 –2 –3 –4 –5 (a) –6 –5 –4 –3 –2 y = (x + 3)2 + 1 y = ((t + 3) + 3)2 + 1 y = (t + 6)2 + 1 x(t) = t + 3 y(t) = (t + 6)2 + 1 Parametric x = t + 3 y = (t + 6)2 + 1 21 3 4 5 6 x – –56 –4 –3 –2 Figure 9 Rectangular y = (x + 3)2 + 1 21 1 –1 –2 –3 –4 (b) Access these online resources for additional instruction and practice with parametric equations. Introduction to Parametric equations (http://openstaxcollege.org/l/introparametric) • • Converting Parametric equations to Rectangular Form (http://openstaxcollege.org/l/convertpara) SECTION 8.6 section exercises 717 8.6 SeCTIOn exeRCISeS VeRBAl 1. What is a system of parametric equations? 2. Some examples of a third parameter are time, length, speed, and scale. Explain when time is used as a parameter. 3. Explain how to eliminate a parameter given a set of 4. What is a benefit of writing a system of parametric parametric equations. equations as a Cartesian equation? 5. What is a benefit of using parametric equations? 6. Why are there many sets of parametric equations to represent on Cartesian function? AlGeBRAIC For the following exercises, eliminate the parameter t to rewrite the parametric equation as a Cartesian equation. x(t) = 5 − t y(t) = 8 − 2t x(t) = 2e t y(t) = 1 − 5t x(t) = t 3 − t y(t) = 2t x(t) = 4 cos t y(t) = 5 sin t x(t) = t − 1 y(t) = t 2 7. 11. 15. 19. 23. { { { { { x(t) = 6 − 3t y(t) = 10 − t x(t) = e –2t y(t) = 2e −t x(t) = t − t 4 y(t) = t + 2 x(t) = 3 sin t y(t) = 6 cos t x(t) = −t y(t) = t 3 + 1 8. 12. 16. 20. 24. { { { { { x(t) = 2t + 1 — y(t) = 3 √ t x(t) = 4 log (t) y(t) = 3 + 2t x(t) = e 2t y(t) = e 6t x(t) = 2 cos2 t y(t) = −sin t x(t) = 2t − 1 y(t) = t3 − 2 9. 13. 17. 21. 25. { { { { { 10. 14. 18. 22. { { { { x(t) = 3t − 1 y(t) = 2t 2 x(t) = log (2t) y(t) = √ — t − 1 x(t) = t 5 y(t) = t10 x(t) = cos t + 4 y(t) = 2 sin2 t For the following exercises, rewrite the parametric equation as a Cartesian equation by building an x-y table. x(t) = 2t − 1 y(t) = t + 4 26. { x(t) = 4 − t y(t) = 3t + 2 27. { x(t) = 2t − 1 y(t) = 5t 28. { x(t) = 4t − 1 y(t) = 4t + 2 29. { For the following exercises, parameterize (write parametric equations
for) each Cartesian equation by setting x(t) = t or by setting y(t) = t. 30. y(x) = 3x2 + 3 31. y(x) = 2 sin x + 1 32. x(y) = 3 log (y) + y 33. x(y) = √ — y + 2y For the following exercises, parameterize (write parametric equations for) each Cartesian equation by using x(t) = a cos t and y(t) = b sin t. Identify the curve. 34 35. x 2 _ 16 + y 2 _ 36 = 1 36. x 2 + y 2 = 16 37. x 2 + y 2 = 10 38. Parameterize the line from (3, 0) to (−2, −5) so that the line is at (3, 0) at t = 0, and at (−2, −5) at t = 1. 40. Parameterize the line from (−1, 5) to (2, 3) so that the line is at (−1, 5) at t = 0, and at (2, 3) at t = 1. 39. Parameterize the line from (−1, 0) to (3, −2) so that the line is at (−1, 0) at t = 0, and at (3, −2) at t = 1. 41. Parameterize the line from (4, 1) to (6, −2) so that the line is at (4, 1) at t = 0, and at (6, −2) at t = 1. 7 18 CHAPTER 8 Further applications oF trigonometry TeCHnOlOGY For the following exercises, use the table feature in the graphing calculator to determine whether the graphs intersect. 42. { x1(t) = 3t y1(t) = 2t − 1 and { x2(t) = t + 3 y2(t) = 4t − 4 43. { x1(t) = t 2 y1(t) = 2t − 1 and x2(t) = −t + 6 y2(t) = t + 1 { For the following exercises, use a graphing calculator to complete the table of values for each set of parametric equations. 44. x1(t) = 3t 2 − 3t + 7 y1(t) = 2t + 3 { 45. x1(t) = t 2 − 4 y1(t) = 2t 2 − 1 { 46. x1(t) = t 4 y1(t1 0 1 exTenSIOnS 1 0 1 2 47. Find two different sets of parametric equations for 48. Find two different sets of parametric equations for y = (x + 1)2. y = 3x − 2. 49. Find two different sets of parametric equations for y = x 2 − 4x + 4. SECTION 8.7 parametric eQuations: graphs 719 leARnInG OBjeCTIVeS In this section, you will: • Graph plane curves described by parametric equations by plotting points. • Graph parametric equations. 8.7 PARAMeTRIC eQUATIOnS: GRAPHS It is the bottom of the ninth inning, with two outs and two men on base. The home team is losing by two runs. The batter swings and hits the baseball at 140 feet per second and at an angle of approximately 45° to the horizontal. How far will the ball travel? Will it clear the fence for a game-winning home run? The outcome may depend partly on other factors (for example, the wind), but mathematicians can model the path of a projectile and predict approximately how far it will travel using parametric equations. In this section, we’ll discuss parametric equations and some common applications, such as projectile motion problems. Figure 1 Parametric equations can model the path of a projectile. (credit: Paul Kreher, Flickr) Graphing Parametric equations by Plotting Points In lieu of a graphing calculator or a computer graphing program, plotting points to represent the graph of an equation is the standard method. As long as we are careful in calculating the values, point-plotting is highly dependable. How To… Given a pair of parametric equations, sketch a graph by plotting points. 1. Construct a table with three columns: t, x(t), and y(t). 2. Evaluate x and y for values of t over the interval for which the functions are defined. 3. Plot the resulting pairs (x, y). Example 1 Sketching the Graph of a Pair of Parametric Equations by Plotting Points Sketch the graph of the parametric equations x(t) = t 2 + 1, y(t) = 2 + t. Solution Construct a table of values for t, x(t), and y(t), as in Table 1, and plot the points in a plane. 7 20 CHAPTER 8 Further applications oF trigonometry t −5 −4 −3 −2 −1 0 1 2 3 4 5 x(t) = t 2 + 1 26 17 10 5 2 1 2 5 10 17 26 Table 1 y(t) = 2 + t −3 −2 −1 0 1 2 3 4 5 6 7 The graph is a parabola with vertex at the point (1, 2), opening to the right. See Figure 2, (10, 5) t = 2, (5, 4) t = 1, (2, 3) t = 0, (1, 2) t = −1, (2, 1) t = −2, (5, 0 10 11 t = −3, (10,−1) Figure 2 Analysis As values for t progress in a positive direction from 0 to 5, the plotted points trace out the top half of the parabola. As values of t become negative, they trace out the lower half of the parabola. There are no restrictions on the domain. The arrows indicate direction according to increasing values of t. The graph does not represent a function, as it will fail the vertical line test. The graph is drawn in two parts: the positive values for t, and the negative values for t. Try It #1 Sketch the graph of the parametric equations x = √ — t , y = 2t + 3, 0 ≤ t ≤ 3. Example 2 Sketching the Graph of Trigonometric Parametric Equations Construct a table of values for the given parametric equations and sketch the graph: x = 2cos t y = 4sin t Solution Construct a table like that in Table 2 using angle measure in radians as inputs for t, and evaluating x and y. Using angles with known sine and cosine values for t makes calculations easier. SECTION 8.7 parametric eQuations: graphs 721 x = 2cos t x = 2cos(0) = 2 π __  = √ x = 2cos  6 — 3 π __  = 1 x = 2cos  3 π __  = 0 x = 2cos  2 y = 4sin t y = 4sin(0) = 0 π __  = 2 y = 4sin  6 __ π __  = 2 √ y = 4sin  3 3 π __  = 4 y = 4sin  2 x = 2cos  2π  = −1 ___ 3 y = 4sin  2π  = −2 √ ___ 3 — 3 x = 2cos  5π  = − √ ___ 6 — 3 x = 2cos(π) = −2 y = 4sin  5π  = 2 ___ 6 y = 4sin(π) = 0 x = 2cos  7π  = − √ ___ 6 — 3 y = 4sin  7π  = −2 ___ 6 x = 2cos  4π  = −1 ___ 3 y = 4sin  4π  = −2 √ ___ 3 — 3 x = 2cos  3π  = 0 ___ 2 x = 2cos  5π  = 1 ___ 3 y = 4sin  3π  = −4 ___ 2 y = 4sin  5π  = −2 √ ___ 3 — 3 x = 2cos  11π ____ 6  = √ — 3 y = 4sin  11π ____ 6  = −2 x = 2cos(2π) = 2 y = 4sin(2π) = 0 Table 2 t 0 π __ 6 π __ 3 π __ 2 2π ___ 3 5π ___ 6 π 7π ___ 6 4π ___ 3 3π ___ 2 5π ___ 3 11π ____ 6 2π Figure 3 shows the graph. (– 3, 2) t = 5π 6 –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (0, 4) t = π 2 (2, 0) t = 0 4 3 5 21 ( 3, –2) t = x 11π 6 Figure 3 By the symmetry shown in the values of x and y, we see that the parametric equations represent an ellipse. The ellipse is mapped in a counterclockwise direction as shown by the arrows indicating increasing t values. Analysis We have seen that parametric equations can be graphed by plotting points. However, a graphing calculator will save some time and reveal nuances in a graph that may be too tedious to discover using only hand calculations. 7 22 CHAPTER 8 Further applications oF trigonometry Make sure to change the mode on the calculator to parametric (PAR). To confirm, the Y= window should show instead of Y1= . X1T = Y1T = Try It #2 Graph the parametric equations: x = 5cos t, y = 3sin t. Example 3 Graphing Parametric Equations and Rectangular Form Together Graph the parametric equations x = 5cos t and y = 2sin t. First, construct the graph using data points generated from the parametric form. Then graph the rectangular form of the equation. Compare the two graphs. Solution Construct a table of values like that in Table 3. t 0 1 2 3 4 5 −1 −2 −3 −4 −5 x = 5cos t x = 5cos(0) = 5 x = 5cos(1) ≈ 2.7 x = 5cos(2) ≈ −2.1 x = 5cos(3) ≈ −4.95 x = 5cos(4) ≈ −3.3 x = 5cos(5) ≈ 1.4 x = 5cos(−1) ≈ 2.7 x = 5cos(−2) ≈ −2.1 x = 5cos(−3) ≈ −4.95 x = 5cos(−4) ≈ −3.3 x = 5cos(−5) ≈ 1.4 Table 3 y = 2sin t y = 2sin(0) = 0 y = 2sin(1) ≈ 1.7 y = 2sin(2) ≈ 1.8 y = 2sin(3) ≈ 0.28 y = 2sin(4) ≈ −1.5 y = 2sin(5) ≈ −1.9 y = 2sin(−1) ≈ −1.7 y = 2sin(−2) ≈ −1.8 y = 2sin(−3) ≈ −0.28 y = 2sin(−4) ≈ 1.5 y = 2sin(−5) ≈ 1.9 Plot the (x, y) values from the table. See Figure 4. Parametric –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 Rectangular y 5 4 3 2 1 21 3 4 5 x –5 –4 –3 –2 Figure 4 –1 –1 –2 –3 –4 –5 21 3 4 5 x Next, translate the parametric equations to rectangular form. To do this, we solve for t in either x(t) or y(t), and then substitute the expression for t in the other equation. The result will be a function y(x) if solving for t as a function of x, or x(y) if solving for t as a function of y. x = 5cos t x _ = cos t 5 y = 2sin t y _ = sin t 2 Solve for cos t. Solve for sin t. SECTION 8.7 parametric eQuations: graphs 723 Then, use the Pythagorean Theorem. cos2 t + sin2 25 + 2 = 1 = 1 In Figure 5, the data from the parametric equations and the rectangular equation are plotted together. The Analysis parametric equations are plotted in blue; the graph for the rectangular equation is drawn on top of the parametric in a dashed style colored red. Clearly, both forms produce the same graph. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x Figure 5 Example 4 Graphing Parametric Equations and Rectangular Equations on the Coordinate System Graph the parametric equations x = t + 1 and y = √ coordinate system. — t , t ≥ 0, and the rectangular equivalent y = √ x − 1 on the same — Solution Construct a table of values for the parametric equations, as we did in the previous example, and graph y = √ t , t ≥ 0 on the same grid, as in Figure 6. — y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –3 –2 21 3 4 5 6 7 8 9 x Figure 6 Analysis With the domain on t restricted, we only plot positive values of t. The parametric data is graphed in blue and the graph of the rectangular equation is dashed in red. Once again, we see that the two forms overlap. Try It #3 Sketch the graph of the parametric equations x = 2cos θ and y = 4sin θ, along with the rectangular equation on the same grid. 7 24 CHAPTER 8 Further applications oF trigonometry Applications of Parametric equations Many of the advantages of parametric equations become obvious when applied to solving real-world problems. Although rectangular equations in x and y give an overall picture of an object's path, they do not reveal the position of an object at a specific time. Parametric equations, however, illustrate how the values of x and y change depending on t, as the location of a moving object at a particular time. A common application of parametric equations is solving problems involving projectile motion. In this type of motion, an object is propelled forward in an upward direction forming an angle of θ to the horizontal, with an initial speed of v0, and at a height h above the horizontal. The path of an object propelled at an inclination of θ to the horizontal, with initial speed v0, and at a height h ab
ove the horizontal, is given by x = (v0cos θ )t 1 __ y = − gt 2 + (v0sin θ )t + h 2 where g accounts for the effects of gravity and h is the initial height of the object. Depending on the units involved in the problem, use g = 32 ft/s2 or g = 9.8 m/s2. The equation for x gives horizontal distance, and the equation for y gives the vertical distance. How To… Given a projectile motion problem, use parametric equations to solve. 1. The horizontal distance is given by x = (v0 cos θ)t. Substitute the initial speed of the object for v0. 2. The expression cos θ indicates the angle at which the object is propelled. Substitute that angle in degrees for cos θ. 1 1 __ __ gt2 + (v0 sin θ)t + h. The term − 3. The vertical distance is given by the formula y = − gt2 represents the effect of 2 2 gravity. Depending on units involved, use g = 32 ft/s2 or g = 9.8 m/s2. Again, substitute the initial speed for v0, and the height at which the object was propelled for h. 4. Proceed by calculating each term to solve for t. Example 5 Finding the Parametric Equations to Describe the Motion of a Baseball Solve the problem presented at the beginning of this section. Does the batter hit the game-winning home run? Assume that the ball is hit with an initial velocity of 140 feet per second at an angle of 45° to the horizontal, making contact 3 feet above the ground. a. Find the parametric equations to model the path of the baseball. b. Where is the ball after 2 seconds? c. How long is the ball in the air? d. Is it a home run? Solution a. Use the formulas to set up the equations. The horizontal position is found using the parametric equation for x. Thus, x = (v0 cos θ)t x = (140cos(45°))t The vertical position is found using the parametric equation for y. Thus, y = − 16t2 + (v0 sin θ)t + h y = − 16t2 + (140sin(45°))t + 3 b. Substitute 2 into the equations to find the horizontal and vertical positions of the ball. x = (140cos(45°))(2) x = 198 feet y = −16(2)2 + (140sin(45°))(2) + 3 y = 137 feet After 2 seconds, the ball is 198 feet away from the batter’s box and 137 feet above the ground. SECTION 8.7 parametric eQuations: graphs 725 c. To calculate how long the ball is in the air, we have to find out when it will hit ground, or when y = 0. Thus, y = −16t 2 +(140sin(45°))t + 3 y = 0 t = 6.2173 Set y (t) = 0 and solve the quadratic. When t = 6.2173 seconds, the ball has hit the ground. (The quadratic equation can be solved in various ways, but this problem was solved using a computer math program.) d. We cannot confirm that the hit was a home run without considering the size of the outfield, which varies from field to field. However, for simplicity’s sake, let’s assume that the outfield wall is 400 feet from home plate in the deepest part of the park. Let’s also assume that the wall is 10 feet high. In order to determine whether the ball clears the wall, we need to calculate how high the ball is when x = 400 feet. So we will set x = 400, solve for t, and input t into y. x = (140cos(45°))t 400 = (140cos(45°))t t = 4.04 y = − 16(4.04)2 + (140sin(45°))(4.04) + 3 y = 141.8 The ball is 141.8 feet in the air when it soars out of the ballpark. It was indeed a home run. See Figure 7. 300 ) t f ( t h g i e H 200 100 Trajectory of ball Position of hitter Outfield wall 0 100 200 400 500 600 300 Distance (ft) Figure 7 Access the following online resource for additional instruction and practice with graphs of parametric equations. • Graphing Parametric equations on the TI-84 (http://openstaxcollege.org/l/graphpara84) 7 26 CHAPTER 8 Further applications oF trigonometry 8.7 SeCTIOn exeRCISeS VeRBAl 1. What are two methods used to graph parametric 2. What is one difference in point-plotting parametric equations? equations compared to Cartesian equations? 3. Why are some graphs drawn with arrows? 4. Name a few common types of graphs of parametric equations. 5. Why are parametric graphs important in understanding projectile motion? GRAPHICAl For the following exercises, graph each set of parametric equations by making a table of values. Include the orientation on the graph. x(t) = t y(t) = t 2 − 1 x(t) = t − 1 y(t) = t 2 7. 6. x y { { −3 −2 −1 0 1 2 t x y t −3 −2 −1 0 1 2 3 x(t) = 2 + t y(t) = 3 − 2t 8(t) = −2 − 2t y(t) = 3 + t 9(t) = t3 y(t) = t + 2 10. { x(t) = t2 y(t) = t + 3 11 For the following exercises, sketch the curve and include the orientation. x(t) = t y(t) = √ — t x(t) = 4sin t y(t) = 2cos t x(t) = sec t y(t) = tan t 12. 16. 20. { { { 13. x(t) = − √ y(t) = t — t 17. 21. x(t) = 2sin t y(t) = 4cos t x(t) = sec t y(t) = tan2 t { { { x(t) = −t + 2 y(t) = 5 − \| t \| x(t) = 3cos2 t y(t) = −3sin2 t 15. 19. { { x(t) = 5 − \| t \| y(t) = t + 2 x(t) = 3cos2 t y(t) = −3sin t x(t) = 1 ___ e2t y(t) = e−t 14. 18. 22. { { { For the following exercises, graph the equation and include the orientation. Then, write the Cartesian equation. x(t) = t − 1 y(t) = −t2 23. { x(t) = t3 y(t) = t + 3 24. { x(t) = 2cos t y(t) = − sin t 25. { SECTION 8.7 section exercises 727 x(t) = 7cos t y(t) = 7sin t 26. { x(t) = e 2t y(t) = −e t 27. { For the following exercises, graph the equation and include the orientation. 28. x = t2, y = 3t, 0 ≤ t ≤ 5 29. x = 2t, y = t2, −5 ≤ t ≤ 5 _ 31. x(t) = −t, y(t) = √ t , t ≥ 5 25 − t2 , 0 < t ≤ 5 π __ 32. x(t) = −2cos t, y = 6sin t 0 ≤ t ≤ π 33. x(t) = −sec t, y = tan t, − 2 30. x = t, y = √ π __ < t < 2 — For the following exercises, use the parametric equations for integers a and b: x(t) = acos((a + b)t) y(t) = acos((a − b)t) 34. Graph on the domain [−π, 0], where a = 2 and 35. Graph on the domain [−π, 0], where a = 3 and b = 1, and include the orientation. b = 2, and include the orientation. 36. Graph on the domain [−π, 0], where a = 4 and 37. Graph on the domain [−π, 0], where a = 5 and b = 4, b = 3, and include the orientation. and include the orientation. 38. If a is 1 more than b, describe the effect the values of a and b have on the graph of the parametric equations. 39. Describe the graph if a = 100 and b = 99. 40. What happens if b is 1 more than a? Describe the 41. If the parametric equations x(t) = t 2 and graph. y(t) = 6 − 3t have the graph of a horizontal parabola opening to the right, what would change the direction of the curve? For the following exercises, describe the graph of the set of parametric equations. 42. x(t) = − t2 and y(t) is linear 44. y(t) = − t2 and x(t) is linear 43. y(t) = t2 and x(t) is linear 45. Write the parametric equations of a circle with center (0, 0), radius 5, and a counterclockwise orientation. 46. Write the parametric equations of an ellipse with center (0, 0), major axis of length 10, minor axis of length 6, and a counterclockwise orientation. For the following exercises, use a graphing utility to graph on the window [−3, 3] by [−3, 3] on the domain [0, 2π) for the following values of a and b , and include the orientation. x(t) = sin(at) y(t) = sin(bt) { 47. a = 1, b = 2 50. a = 5, b = 5 48. a = 2, b = 1 51. a = 2, b = 5 49. a = 3, b = 3 52. a = 5, b = 2 TeCHnOlOGY For the following exercises, look at the graphs that were created by parametric equations of the form Use the parametric mode on the graphing calculator to find the values of a, b, c, and d to achieve each graph. { x(t) = acos(bt) y(t) = csin(dt) 53. –6 –5 –4 –3 –1–1 –2 –3 –4 –5 –6 54. 21 3 4 5 6 x –6 –5 –4 –3 –1–1 –2 –3 –4 –5 –6 55. 21 3 4 5 6 x –6 –5 –4 –3 –1–1 –2 –3 –4 –5 –6 56. 21 3 4 5 6 x –6 –5 –4 –3 –1–1 –2 –3 –4 –5 –6 21 3 4 5 6 x 7 28 CHAPTER 8 Further applications oF trigonometry For the following exercises, use a graphing utility to graph the given parametric equations. x(t) = cos t − 1 y(t) = sin t + t a. { x(t) = cos t + t y(t) = sin t − 1 b. { x(t) = t − sin t y(t) = cos t − 1 c. { 57. Graph all three sets of parametric equations on the 58. Graph all three sets of parametric equations on the domain [0, 2π]. domain [0, 4π]. 59. Graph all three sets of parametric equations on the domain [−4π, 6π]. 60. The graph of each set of parametric equations appears to “creep” along one of the axes. What controls which axis the graph creeps along? 61. Explain the effect on the graph of the parametric 62. Explain the effect on the graph of the parametric equation when we switched sin t and cos t. equation when we changed the domain. exTenSIOnS 63. An object is thrown in the air with vertical velocity of 20 ft/s and horizontal velocity of 15 ft/s. The object’s height can be described by the equation y(t) = − 16t2 + 20t, while the object moves horizontally with constant velocity 15 ft/s. Write parametric equations for the object’s position, and then eliminate time to write height as a function of horizontal position. 64. A skateboarder riding on a level surface at a constant speed of 9 ft/s throws a ball in the air, the height of which can be described by the equation y(t) = − 16t2 + 10t + 5.Write parametric equations for the ball’s position, and then eliminate time to write height as a function of horizontal position. For the following exercises, use this scenario: A dart is thrown upward with an initial velocity of 65 ft/s at an angle of elevation of 52°. Consider the position of the dart at any time t. Neglect air resistance. 65. Find parametric equations that model the problem 66. Find all possible values of x that represent the situation. situation. 67. When will the dart hit the ground? 68. Find the maximum height of the dart. 69. At what time will the dart reach maximum height? For the following exercises, look at the graphs of each of the four parametric equations. Although they look unusual and beautiful, they are so common that they have names, as indicated in each exercise. Use a graphing utility to graph each on the indicated domain. 70. An epicycloid: { x(t) = 14cos t − cos(14t) y(t) = 14sin t + sin(14t) 71. An hypocycloid: { x(t) = 6sin t + 2sin(6t) y(t) = 6cos t − 2cos(6t) on the domain [0, 2π]. on the domain [0, 2π]. 72. An hypotrochoid: { x(t) = 2sin t + 5cos(6t) y(t) = 5cos t − 2sin(6t) on the domain [0, 2π]. 73. A rose: { x(t)
= 5sin(2t) sin t y(t) = 5sin(2t) cos t on the domain [0, 2π]. SECTION 8.8 vectors 729 leARnInG OBjeCTIVeS In this section, you will: • View vectors geometrically. • Find magnitude and direction. • Perform vector addition and scalar multiplication. • Find the component form of a vector. • Find the unit vector in the direction of v. • Perform operations with vectors in terms of i and j. • Find the dot product of two vectors. 8.8 VeCTORS An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour, as shown in Figure 1. What are the ground speed and actual bearing of the plane? N A O 140˚ α 200 X C 16.2 B Figure 1 Ground speed refers to the speed of a plane relative to the ground. Airspeed refers to the speed a plane can travel relative to its surrounding air mass. These two quantities are not the same because of the effect of wind. In an earlier section, we used triangles to solve a similar problem involving the movement of boats. Later in this section, we will find the airplane’s groundspeed and bearing, while investigating another approach to problems of this type. First, however, let’s examine the basics of vectors. A Geometric View of Vectors A vector is a specific quantity drawn as a line segment with an arrowhead at one end. It has an initial point, where it begins, and a terminal point, where it ends. A vector is defined by its magnitude, or the length of the line, and its direction, indicated by an arrowhead at the terminal point. Thus, a vector is a directed line segment. There are various symbols that distinguish vectors from other quantities: → →→ • Lower case, boldfaced type, with or without an arrow on top such as v, u, w, v, u, w. → • Given initial point P and terminal point Q, a vector can be represented as PQ. The arrowhead on top is what indicates that it is not just a line, but a directed line segment. • Given an initial point of (0, 0) and terminal point (a, b), a vector may be represented as 〈a, b〉. This last symbol 〈a, b〉 has special significance. It is called the standard position. The position vector has an initial point (0, 0) and a terminal point 〈a, b〉. To change any vector into the position vector, we think about the change in the x-coordinates and the change in the y-coordinates. Thus, if the initial point of a vector CD is C(x1, y1) and the terminal point is D(x2, y2), then the position vector is found by calculating → → AB = 〈x2 − x1, y2 − y1 〉 = 〈a, b〉 7 30 CHAPTER 8 Further applications oF trigonometry → In Figure 2, we see the original vector CD and the position vector AB. → (x1, y1) C (x2, y2) D B (a, b) 21 1–1 –2 –3 –4 –5 –4 –3 –2 Figure 2 properties of vectors A vector is a directed line segment with an initial point and a terminal point. Vectors are identified by magnitude, or the length of the line, and direction, represented by the arrowhead pointing toward the terminal point. The position vector has an initial point at (0, 0) and is identified by its terminal point 〈a, b〉. Example 1 Find the Position Vector Consider the vector whose initial point is P(2, 3) and terminal point is Q(6, 4). Find the position vector. Solution The position vector is found by subtracting one x-coordinate from the other x-coordinate, and one y-coordinate from the other y-coordinate. Thus v = 〈6 −2, 4 −3〉 = 〈4, 1〉 The position vector begins at (0, 0) and terminates at (4, 1). The graphs of both vectors are shown in Figure 3. y 5 4 3 2 1 0 –1–1 –2 –3 –4 –5 –4 –3 –2 Q (6, 4) P (2, 3) 〈4, 1〉 21 3 4 5 6 7 8 x We see that the position vector is 〈4, 1〉. Figure 3 Example 2 Drawing a Vector with the Given Criteria and Its Equivalent Position Vector Find the position vector given that vector v has an initial point at (−3, 2) and a terminal point at (4, 5), then graph both vectors in the same plane. Solution The position vector is found using the following calculation: Thus, the position vector begins at (0, 0) and terminates at (7, 3). See Figure 4. v = 〈4 − ( − 3), 5 − 2〉 = 〈7, 3〉 SECTION 8.8 vectors 731 y (−3, 2) –4 –3 5 4 3 2 1 (0, 0) –1–1 –2 –2 –3 –4 –5 (4, 5) (7, 3) Position vector x 21 3 4 5 6 7 8 Figure 4 Try It #1 Draw a vector v that connects from the origin to the point (3, 5). Finding Magnitude and Direction To work with a vector, we need to be able to find its magnitude and its direction. We find its magnitude using the Pythagorean Theorem or the distance formula, and we find its direction using the inverse tangent function. magnitude and direction of a vector Given a position vector v = 〈a, b〉, the magnitude is found by \| v \| = √ formed with the x-axis, or with the y-axis, depending on the application. For a position vector, the direction is found by tan θ =  b __ a  ⇒ θ = tan −1  b __ a  , as illustrated in Figure 5. a 2 + b 2 . The direction is equal to the angle — a, b〉 3 4 5 x \|v\| θ 2 1 Figure 5 Two vectors v and u are considered equal if they have the same magnitude and the same direction. Additionally, if both vectors have the same position vector, they are equal. Example 3 Finding the Magnitude and Direction of a Vector Find the magnitude and direction of the vector with initial point P(−8, 1) and terminal point Q(−2, −5). Draw the vector. Solution First, find the position vector. u = 〈−2, − (−8), −5−1〉 = 〈6, − 6〉 We use the Pythagorean Theorem to find the magnitude. \| u \| = √ — — (6) 2 + (−6) 2 72 2 — = √ = 6 √ 7 32 CHAPTER 8 Further applications oF trigonometry The direction is given as tan θ = −6 ___ = −1 ⇒ θ = tan 6 = −45° −1 (−1) However, the angle terminates in the fourth quadrant, so we add 360° to obtain a positive angle. Thus, −45° + 360° = 315°. See Figure 6. y 6 5 4 3 2 1 315° –6 –5 –4 –3 –2 –1–1 –2 –3 –4 –5 –6 21 3 4 5 6 x Example 4 Showing That Two Vectors Are Equal Figure 6 Show that vector v with initial point at (5, −3) and terminal point at (−1, 2) is equal to vector u with initial point at (−1, −3) and terminal point at (−7, 2). Draw the position vector on the same grid as v and u. Next, find the magnitude and direction of each vector. Solution As shown in Figure 7, draw the vector v starting at initial (5, −3) and terminal point (−1, 2). Draw the vector u with initial point (−1, −3) and terminal point (−7, 2). Find the standard position for each. Next, find and sketch the position vector for v and u. We have v = 〈−1 − 5, 2 − ( − 3)〉 = 〈−6, 5〉 u = 〈−7 − (−1), 2 − (−3)〉 = 〈−6, 5〉 Since the position vectors are the same, v and u are the same. An alternative way to check for vector equality is to show that the magnitude and direction are the same for both vectors. To show that the magnitudes are equal, use the Pythagorean Theorem \| = √ = √ = √ = √ — —— — — (−1 − 5) 2 + (2 − (−3)) 2 (−6)2 + (5)2 36 + 25 61 (−7 − (−1))2 + (2 − (−3))2 (−6)2 + (5)2 36 + 25 61 —— — — — As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives 5 5 _ _  ⇒ θ = tan−1  − tan θ = − 6 6 = −39.8° However, we can see that the position vector terminates in the second quadrant, so we add 180°. Thus, the direction is −39.8° + 180° = 140.2°. SECTION 8.8 vectors 733 Position vector –7 –6 –5 –4 –3 –2 u y 5 4 3 2 1 –1–1 –2 –3 –4 –5 140.2° 21 3 4 5 x v Figure 7 Performing Vector Addition and Scalar Multiplication Now that we understand the properties of vectors, we can perform operations involving them. While it is convenient to think of the vector u = 〈x, y〉 as an arrow or directed line segment from the origin to the point (x, y), vectors can be situated anywhere in the plane. The sum of two vectors u and v, or vector addition, produces a third vector u+ v, the resultant vector. To find u + v, we first draw the vector u, and from the terminal end of u, we drawn the vector v. In other words, we have the initial point of v meet the terminal end of u. This position corresponds to the notion that we move along the first vector and then, from its terminal point, we move along the second vector. The sum u + v is the resultant vector because it results from addition or subtraction of two vectors. The resultant vector travels directly from the beginning of u to the end of v in a straight path, as shown in Figure 8. − Figure 8 Vector subtraction is similar to vector addition. To find u − v, view it as u + (−v). Adding −v is reversing direction of v and adding it to the end of u. The new vector begins at the start of u and stops at the end point of −v. See Figure 9 for a visual that compares vector addition and vector subtraction using parallelogramsv u u Figure 9 Example 5 Adding and Subtracting Vectors Given u = 〈3, − 2〉 and v = 〈−1, 4〉, find two new vectors u + v, and u − v. Solution To find the sum of two vectors, we add the components. Thus, u + v = 〈3, − 2〉 + 〈−1, 4〉 = 〈3 + ( − 1), − 2 + 4〉 = 〈2, 2〉 See Figure 10(a). To find the difference of two vectors, add the negative components of v to u. Thus, u + (−v) = 〈3, − 2〉 + 〈1, −4〉 = 〈3 + 1, − 2 + (−4)〉 = 〈4, − 6〉 7 34 CHAPTER 8 Further applications oF trigonometry See Figure 10(b). 5 –4 –3 –2 –5 –4 –3 –2 21 u –1–1 –2 –3 –4 (a) y 2 1 x 321 u 4 5 − v u − v –1–1 –2 –3 –4 –5 –6 (b) Figure 10 (a) Sum of two vectors (b) Difference of two vectors Multiplying By a Scalar While adding and subtracting vectors gives us a new vector with a different magnitude and direction, the process of multiplying a vector by a scalar, a constant, changes only the magnitude of the vector or the length of the line. Scalar multiplication has no effect on the direction unless the scalar is negative, in which case the direction of the resulting vector is opposite the direction of the original vector. scalar multiplication Scalar multiplication involves the product of a vector and a scalar. Each component of the vector is multiplied by the scalar. Thus, to multiply v = 〈a, b〉 by k, we have kv = 〈ka, kb〉 Only the magnitude changes, unless k is negative, and then the vector reverses direction. Example 6 Performing Scalar Multipl
ication 1 _ v, and −v. Given vector v = 〈3, 1〉 , find 3v, 2 Solution See Figure 11 for a geometric interpretation. If v = 〈3, 1〉, then 3v = 〈3 ⋅ 3, 3 ⋅ 1〉 = 〈9, 3〉 ⋅ 1v = 〈−3, −1〉 3v v –v v1 2 Figure 11 1 _ v is half the length of v, and −v is the same length of v, Analysis Notice that the vector 3v is three times the length of v, 2 but in the opposite direction. SECTION 8.8 vectors 735 Try It #2 Find the scalar multiple 3u given u = 〈5, 4〉 . Example 7 Using Vector Addition and Scalar Multiplication to Find a New Vector Given u = 〈3, − 2〉 and v = 〈−1, 4〉, find a new vector w = 3u + 2v. Solution First, we must multiply each vector by the scalar. 3u = 3 〈3, − 2〉 = 〈9, − 6〉 2v = 2 〈−1, 4〉 = 〈−2, 8〉 w = 3u + 2v = 〈9, − 6〉 + 〈−2, 8〉 = 〈9 − 2, − 6 + 8〉 = 〈7, 2〉 Then, add the two together. So, w = 〈7, 2〉. Finding Component Form In some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the x direction, and the vertical component is the y direction. For example, we can see in the graph in Figure 12 that the position vector 〈2, 3〉 comes from adding the vectors v1 and v2. We have v1 with initial point (0, 0) and terminal point (2, 0). We also have v2 with initial point (0, 0) and terminal point (0, 3). v1 = 〈2 − 0, 0 − 0〉 = 〈2, 0〉 Therefore, the position vector is v2 = 〈0 − 0, 3 − 0〉 = 〈0, 3〉 v = 〈2 + 0, 3 + 0〉 = 〈2, 3〉 Using the Pythagorean Theorem, the magnitude of v1 is 2, and the magnitude of v2 is 3. To find the magnitude of v, use the formula with the position vector. __________ \| v \| = √ = √ ∣ v1 ∣ 2 + ∣ v2 ∣ 2 22 + 32 — The magnitude of v is √ — y _ 13 . To find the direction, we use the tangent function tan θ = x . = √ — 13 tan θ = v2 _ v1 3 _ tan θ = 2 3 _  = 56.3° θ = tan−1  2 7 36 CHAPTER 8 Further applications oF trigonometry y 4 3 2 v2 1 –1 –1 \|v\| 56.3° v1 1 (2, 0) 2 3 4 x Figure 12 Thus, the magnitude of v is √ — 13 and the direction is 56.3° off the horizontal. Example 8 Finding the Components of the Vector Find the components of the vector v with initial point (3, 2) and terminal point (7, 4). Solution First find the standard position. See the illustration in Figure 13. v = 〈7 − 3, 4 − 2〉 = 〈4, 2〉 y 4 3 2 1 v2 = 〈0, 2〉 v1 = 〈4, 0〉 –1 1 2 3 4 x –1 Figure 13 The horizontal component is v1 = 〈4, 0〉 and the vertical component is v2 = 〈0, 2〉. Finding the Unit Vector in the Direction of v In addition to finding a vector’s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a unit vector. We can then preserve the direction of the original vector while simplifying calculations. Unit vectors are defined in terms of components. The horizontal unit vector is written as i = 〈1, 0〉 and is directed along the positive horizontal axis. The vertical unit vector is written as j = 〈0, 1〉 and is directed along the positive vertical axis. See Figure 14. y 4 3 2 1 –1 –1 j = 〈0, 1〉 i = 〈1, 0〉 1 2 x 3 4 Figure 14 SECTION 8.8 vectors 737 the unit vectors If v is a nonzero vector, then is a unit vector in the direction of v. Any vector divided by its magnitude is a v _ ∣ v ∣ unit vector. Notice that magnitude is always a scalar, and dividing by a scalar is the same as multiplying by the reciprocal of the scalar. Example 9 Finding the Unit Vector in the Direction of v Find a unit vector in the same direction as v = 〈−5, 12〉. Solution First, we will find the magnitude. \| v \| = √ = √ = √ = 13 — — (−5)2 + (12)2 25 + 144 169 — Then we divide each component by \| v \| , which gives a unit vector in the same direction as v: or, in component form See Figure 15. = − v _ \| v \| = 〈− v _ \| v \| 5 _ 13 i + 12 _ j 13 5 _ , 13 〉 12 _ 13 y 14 13 12 11 10 9 8 7 6 5 4 3 2 1 〈−5, 12〉 5 12 ,− 13 13 –8 –7 –6 –5 –4 –3 –2 0 –1–1 21 Figure 15 x 5 _ Verify that the magnitude of the unit vector equals 1. The magnitude of − 13 _________ 25 ____ + 144 ___ 169 169 ____ 169 _ 169  − 5 ___ √  13 +  12 ___  13 __________________ = √ 2 2 = √ = 1 i + j is given as 12 _ 13 The vector u = i + j is the unit vector in the same direction as v = 〈−5, 12〉. 5 ___ 13 12 ___ 13 Performing Operations with Vectors in Terms of i and j So far, we have investigated the basics of vectors: magnitude and direction, vector addition and subtraction, scalar multiplication, the components of vectors, and the representation of vectors geometrically. Now that we are familiar with the general strategies used in working with vectors, we will represent vectors in rectangular coordinates in terms of i and j. 7 38 CHAPTER 8 Further applications oF trigonometry vectors in the rectangular plane Given a vector v with initial point P = (x1, y1) and terminal point Q = (x2, y2), v is written as v = (x2 − x1)i + (y1 − y2) j The position vector from (0, 0) to (a, b), where (x2 − x1) = a and (y2 − y1) = b, is written as v = ai + bj. This vector sum is called a linear combination of the vectors i and j. The magnitude of v = ai + bj is given as \| v \| = √ — a2 + b2 . See Figure 16. v = ai + bj bj ai Figure 16 Example 10 Writing a Vector in Terms of i and j Given a vector v with initial point P = (2, −6) and terminal point Q = (−6, 6), write the vector in terms of i and j. Solution Begin by writing the general form of the vector. Then replace the coordinates with the given values. v = (x2 − x1)i + (y2 − y1) j = ( −6 − 2)i + (6 − ( − 6)) j = − 8i + 12 j Example 11 Writing a Vector in Terms of i and j Using Initial and Terminal Points Given initial point P1 = (−1, 3) and terminal point P2 = (2, 7), write the vector v in terms of i and j. Solution Begin by writing the general form of the vector. Then replace the coordinates with the given values. v = (x2 − x1)i + (y2 − y1) j v = (2 − ( − 1))i + (7 − 3) j = 3i + 4 j Try It #3 Write the vector u with initial point P = (−1, 6) and terminal point Q = (7, − 5) in terms of i and j. Performing Operations on Vectors in Terms of i and j When vectors are written in terms of i and j, we can carry out addition, subtraction, and scalar multiplication by performing operations on corresponding components. adding and subtracting vectors in rectangular coordinates Given v = ai + bj and u = ci + dj, then v + u = (a + c)i + (b + d)j v − u = (a − c)i + (b − d)j SECTION 8.8 vectors 739 Example 12 Finding the Sum of the Vectors Find the sum of v1 = 2i − 3j and v2 = 4i + 5j. Solution According to the formula, we have v1 + v2 = (2 + 4)i + ( − 3 + 5) j = 6i + 2 j Calculating the Component Form of a Vector: Direction We have seen how to draw vectors according to their initial and terminal points and how to find the position vector. We have also examined notation for vectors drawn specifically in the Cartesian coordinate plane using i and j. For any of these vectors, we can calculate the magnitude. Now, we want to combine the key points, and look further at the ideas of magnitude and direction. Calculating direction follows the same straightforward process we used for polar coordinates. We find the direction of the vector by finding the angle to the horizontal. We do this by using the basic trigonometric identities, but with \|v\| replacing r. vector components in terms of magnitude and direction Given a position vector v = 〈x, y〉 and a direction angle θ, cos θ = and x _ \| v \| sin θ = y _ \| v \| Thus, v = xi + yj = \|v\|cos θi + \|v\|sin θj, and magnitude is expressed as \|v\| = √ — x2 + y2 . x = \| v \| cos θ y = \| v \| sin θ Example 13 Writing a Vector in Terms of Magnitude and Direction Write a vector with length 7 at an angle of 135° to the positive x-axis in terms of magnitude and direction. Solution Using the conversion formulas x = \|v\| cos θi and y =\|v\| sin θ j, we find that x = 7cos(135°) = 7sin(135°) j This vector can be written as v = 7cos(135°)i + 7sin(135°) j or simplified as = — Try It #4 A vector travels from the origin to the point (3, 5). Write the vector in terms of magnitude and direction. Finding the Dot Product of Two Vectors As we discussed earlier in the section, scalar multiplication involves multiplying a vector by a scalar, and the result is a vector. As we have seen, multiplying a vector by a number is called scalar multiplication. If we multiply a vector by a vector, there are two possibilities: the dot product and the cross product. We will only examine the dot product here; you may encounter the cross product in more advanced mathematics courses. The dot product of two vectors involves multiplying two vectors together, and the result is a scalar. 7 40 CHAPTER 8 Further applications oF trigonometry dot product The dot product of two vectors v = 〈a, b〉 and u = 〈c, d〉 is the sum of the product of the horizontal components and the product of the vertical components. To find the angle between the two vectors, use the formula below. v ⋅ u = ac + bd cos θ = v ___ \| v \| ⋅ u ___ \| u \| Example 14 Finding the Dot Product of Two Vectors Find the dot product of v = 〈5, 12〉 and u = 〈−3, 4〉. Solution Using the formula, we have v ⋅ u = 〈5, 12〉 ⋅ 〈−3, 4〉 = 5 ⋅ ( −3) + 12 ⋅ 4 = −15 + 48 = 33 Example 15 Finding the Dot Product of Two Vectors and the Angle between Them Find the dot product of v1 = 5i + 2j and v2 = 3i + 7j. Then, find the angle between the two vectors. Solution Finding the dot product, we multiply corresponding components. v1 ⋅ v2 = 〈5, 2〉 ⋅ 〈3, 7〉 = 5 ⋅ 3 + 2 ⋅ 7 = 15 + 14 = 29 To find the angle between them, we use the formula cos \| ______ — √ 29 〉 ⋅ 〈 3 ______ 58 √ 〉 + 7 ______ 58 √ — — ⋅ ______ 29 √ — = 5 ______ 29 √ ⋅ 3 ______ 58 √ + 2 ______ 29 √ ⋅ 7 ______ 58 √ — — — — = 15 ________ + 1682 √ — 14 ________ = 1682 √ — 29 ________ 1682 √ — = 0.707107 cos−1(0.707107) = 45° See Figure 171–1 –2 –3 –4 –6 –5 –4 –3 –2 45° 21 3 4 5 6 x Figure 17 SECTION 8.8 vectors 741 Example 16 Finding the Angle between Two Vectors Find the angle between u = 〈−3, 4〉 and v = 〈5, 12〉. Solution Using the formula, we have ⋅ θ = cos−3i + 4j _ ⋅ 5 v _
 \| v \| 5i + 12j _ 13 13 4 _  +  ⋅ 5 12 _  13 = − 15 _ 65 + 48 _ 65 = 33 _ 65 θ = cos−1  33 _  65 See Figure 18. = 59.5° y 12 11 10 9 8 7 6 5 4 3 2 1 59.5° –6 –5 –4 –3 –2 –1–1 21 3 4 5 6 x Figure 18 Example 17 Finding Ground Speed and Bearing Using Vectors We now have the tools to solve the problem we introduced in the opening of the section. An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour. What are the ground speed and actual bearing of the plane? See Figure 19. N A O 140˚ α 200 X C 16.2 B Figure 19 Solution The ground speed is represented by x in the diagram, and we need to find the angle α in order to calculate the adjusted bearing, which will be 140° + α . 7 42 CHAPTER 8 Further applications oF trigonometry Notice in Figure 19, that angle BCO must be equal to angle AOC by the rule of alternating interior angles, so angle BCO is 140°. We can find x by the Law of Cosines: x2 = (16.2)2 + (200)2 − 2(16.2)(200)cos(140°) x2 = 45,226.41 — x = √ 45, 226.41 The ground speed is approximately 213 miles per hour. Now we can calculate the bearing using the Law of Sines. x = 212.7 sin α _ 16.2 = sin(140°) _ 212.7 sin α = 16.2sin(140°) __ 212.7 = 0.04896 sin−1(0.04896) = 2.8° Therefore, the plane has a SE bearing of 140° + 2.8° = 142.8°. The ground speed is 212.7 miles per hour. Access these online resources for additional instruction and practice with vectors. • Introduction to Vectors (http://openstaxcollege.org/l/introvectors) • Vector Operations (http://openstaxcollege.org/l/vectoroperation) • The Unit Vector (http://openstaxcollege.org/l/unitvector) SECTION 8.8 section exercises 743 8.8 SeCTIOn exeRCISeS VeRBAl 1. What are the characteristics of the letters that are 2. How is a vector more specific than a line segment? commonly used to represent vectors? 3. What are i and j, and what do they represent? 5. When a unit vector is expressed as 〈a, b〉, which letter is the coefficient of the i and which the j? AlGeBRAIC 4. What is component form? 6. Given a vector with initial point (5, 2) and terminal point (−1, − 3), find an equivalent vector whose initial point is (0, 0). Write the vector in component form 〈a, b〉. 7. Given a vector with initial point (−4, 2) and terminal point (3, − 3), find an equivalent vector whose initial point is (0, 0). Write the vector in component form 〈a, b〉. 8. Given a vector with initial point (7, − 1) and terminal point (−1, − 7), find an equivalent vector whose initial point is (0, 0). Write the vector in component form 〈a, b〉. For the following exercises, determine whether the two vectors u and v are equal, where u has an initial point P1 and a terminal point P2 and v has an initial point P3 and a terminal point P4. 9. P1 = (5, 1), P2 = (3, − 2), P3 = (−1, 3), and P4 = (9, − 4) 11. P1 = (−1, − 1), P2 = (−4, 5), P3 = (−10, 6), and P4 = (−13, 12) 13. P1 = (8, 3), P2 = (6, 5), P3 = (11, 8), and P4 = (9, 10) 15. Given initial point P1 = (6, 0) and terminal point P2 = (−1, − 3), write the vector v in terms of i and j. 10. P1 = (2, −3), P2 = (5, 1), P3 = (6, − 1), and P4 = (9, 3) 12. P1 = (3, 7), P2 = (2, 1), P3 = (1, 2), and P4 = (−1, − 4) 14. Given initial point P1 = (−3, 1) and terminal point P2 = (5, 2), write the vector v in terms of i and j. For the following exercises, use the vectors u = i + 5j, v = −2i − 3j, and w = 4i − j. 16. Find u + (v − w) 17. Find 4v + 2u For the following exercises, use the given vectors to compute u + v, u − v, and 2u − 3v. 18. u = 〈2, − 3〉 , v = 〈1, 5〉 20. Let v = −4i + 3j. Find a vector that is half the length and points in the same direction as v. 19. u = 〈−3, 4〉 , v = 〈−2, 1〉 21. Let v = 5i + 2j. Find a vector that is twice the length and points in the opposite direction as v. For the following exercises, find a unit vector in the same direction as the given vector. 22. a = 3i + 4j 26. u = 100i + 200j 23. b = −2i + 5j 27. u = −14i + 2j 24. c = 10i − j 5 1 __ __ i + 25. d = − j 2 3 For the following exercises, find the magnitude and direction of the vector, 0 ≤ θ < 2π. 28. 〈0, 4〉 29. 〈6, 5〉 30. 〈2, −5〉 31. 〈−4, −6〉 32. Given u = 3i − 4j and v = −2i + 3j, calculate u ⋅ v. 33. Given u = −i − j and v = i + 5j, calculate u ⋅ v. 34. Given u = 〈−2, 4〉 and v = 〈−3, 1〉, calculate u ⋅ v. 35. Given u = 〈−1, 6〉 and v = 〈6, − 1〉, calculate u ⋅ v. 7 44 CHAPTER 8 Further applications oF trigonometry GRAPHICAl 1 __ v. For the following exercises, given v, draw v, 3v and 2 36. 〈2, −1〉 37. 〈−1, 4〉 38. 〈−3, −2〉 For the following exercises, use the vectors shown to sketch u + v, u − v, and 2u. 39. 40. 41. v u u v u v For the following exercises, use the vectors shown to sketch 2u + v. 42. 43. u v v u For the following exercises, use the vectors shown to sketch u − 3v. 44. u v 45. u v For the following exercises, write the vector shown in component form. 46. 47. SECTION 8.8 section exercises 745 48. Given initial point P 1 = (2, 1) and terminal point P 2 = (−1, 2), write the vector v in terms of i and j, then draw the vector on the graph. 49. Given initial point P 1 = (4, − 1) and terminal point P 2 = (−3, 2), write the vector v in terms of i and j. Draw the points and the vector on the graph. 50. Given initial point P 1 = (3, 3) and terminal point P 2 = (−3, 3), write the vector v in terms of i and j. Draw the points and the vector on the graph. exTenSIOnS For the following exercises, use the given magnitude and direction in standard position, write the vector in component form. 51. \| v \| = 6, θ = 45 ° 52. \| v \| = 8, θ = 220° 53. \| v \| = 2, θ = 300° 54. \| v \| = 5, θ = 135° 55. A 60-pound box is resting on a ramp that is inclined 56. A 25-pound box is resting on a ramp that is inclined 12°. Rounding to the nearest tenth, a. Find the magnitude of the normal (perpendicular) 8°. Rounding to the nearest tenth, a. Find the magnitude of the normal (perpendicular) component of the force. component of the force. b. Find the magnitude of the component of the force b. Find the magnitude of the component of the force that is parallel to the ramp. that is parallel to the ramp. 57. Find the magnitude of the horizontal and vertical components of a vector with magnitude 8 pounds pointed in a direction of 27° above the horizontal. Round to the nearest hundredth. 58. Find the magnitude of the horizontal and vertical components of the vector with magnitude 4 pounds pointed in a direction of 127° above the horizontal. Round to the nearest hundredth. 59. Find the magnitude of the horizontal and vertical components of a vector with magnitude 5 pounds pointed in a direction of 55° above the horizontal. Round to the nearest hundredth. 60. Find the magnitude of the horizontal and vertical components of the vector with magnitude 1 pound pointed in a direction of 8° above the horizontal. Round to the nearest hundredth. ReAl-WORlD APPlICATIOnS 61. A woman leaves home and walks 3 miles west, then 2 miles southwest. How far from home is she, and in what direction must she walk to head directly home? 62. A boat leaves the marina and sails 6 miles north, then 2 miles northeast. How far from the marina is the boat, and in what direction must it sail to head directly back to the marina? 63. A man starts walking from home and walks 4 64. A woman starts walking from home and walks 4 miles east, 2 miles southeast, 5 miles south, 4 miles southwest, and 2 miles east. How far has he walked? If he walked straight home, how far would he have to walk? miles east, 7 miles southeast, 6 miles south, 5 miles southwest, and 3 miles east. How far has she walked? If she walked straight home, how far would she have to walk? 65. A man starts walking from home and walks 3 miles at 20° north of west, then 5 miles at 10° west of south, then 4 miles at 15° north of east. If he walked straight home, how far would he have to the walk, and in what direction? 66. A woman starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If she walked straight home, how far would she have to walk, and in what direction? 7 46 CHAPTER 8 Further applications oF trigonometry 67. An airplane is heading north at an airspeed of 68. An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 500 km/hr, but there is a wind blowing from the northwest at 50 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 69. An airplane needs to head due north, but there is a wind blowing from the southwest at 60 km/hr. The plane flies with an airspeed of 550 km/hr. To end up flying due north, how many degrees west of north will the pilot need to fly the plane? 70. An airplane needs to head due north, but there is a wind blowing from the northwest at 80 km/hr. The plane flies with an airspeed of 500 km/hr. To end up flying due north, how many degrees west of north will the pilot need to fly the plane? 71. As part of a video game, the point (5, 7) is rotated 72. As part of a video game, the point (7, 3) is rotated counterclockwise about the origin through an angle of 35°. Find the new coordinates of this point. counterclockwise about the origin through an angle of 40°. Find the new coordinates of this point. 73. Two children are throwing a ball back and forth straight across the back seat of a car. The ball is being thrown 10 mph relative to the car, and the car is traveling 25 mph down the road. If one child doesn't catch the ball, and it flies out the window, in what direction does the ball fly (ignoring wind resistance)? 74. Two children are throwing a ball back and forth straight across the back seat of a car. The ball is being thrown 8 mph relative to the car, and the car is traveling 45 mph down the road. If one child doesn't catch the bal
l, and it flies out the window, in what direction does the ball fly (ignoring wind resistance)? 75. A 50-pound object rests on a ramp that is inclined 19°. Find the magnitude of the components of the force parallel to and perpendicular to (normal) the ramp to the nearest tenth of a pound. 76. Suppose a body has a force of 10 pounds acting on it to the right, 25 pounds acting on it upward, and 5 pounds acting on it 45° from the horizontal. What single force is the resultant force acting on the body? 77. Suppose a body has a force of 10 pounds acting on it to the right, 25 pounds acting on it −135° from the horizontal, and 5 pounds acting on it directed 150° from the horizontal. What single force is the resultant force acting on the body? 78. The condition of equilibrium is when the sum of the forces acting on a body is the zero vector. Suppose a body has a force of 2 pounds acting on it to the right, 5 pounds acting on it upward, and 3 pounds acting on it 45° from the horizontal. What single force is needed to produce a state of equilibrium on the body? 79. Suppose a body has a force of 3 pounds acting on it to the left, 4 pounds acting on it upward, and 2 pounds acting on it 30° from the horizontal. What single force is needed to produce a state of equilibrium on the body? Draw the vector. CHAPTER 8 review 747 CHAPTeR 8 ReVIeW Key Terms altitude a perpendicular line from one vertex of a triangle to the opposite side, or in the case of an obtuse triangle, to the line containing the opposite side, forming two right triangles ambiguous case a scenario in which more than one triangle is a valid solution for a given oblique SSA triangle Archimedes’ spiral a polar curve given by r = θ. When multiplied by a constant, the equation appears as r = aθ. As r = θ, the curve continues to widen in a spiral path over the domain. argument the angle associated with a complex number; the angle between the line from the origin to the point and the positive real axis cardioid a member of the limaçon family of curves, named for its resemblance to a heart; its equation is given as a _ = 1 r = a ± bcos θ and r = a ± bsin θ, where b a _ ≥ 2 convex limaçon a type of one-loop limaçon represented by r = a ± bcos θ and r = a ± bsin θ such that b De Moivre’s Theorem formula used to find the nth power or nth roots of a complex number; states that, for a positive integer n, z n is found by raising the modulus to the nth power and multiplying the angles by n a _ < 2 dimpled limaçon a type of one-loop limaçon represented by r = a ± bcos θ and r = a ± bsin θ such that 1 < b dot product given two vectors, the sum of the product of the horizontal components and the product of the vertical components Generalized Pythagorean Theorem an extension of the Law of Cosines; relates the sides of an oblique triangle and is used for SAS and SSS triangles initial point the origin of a vector inner-loop limaçon a polar curve similar to the cardioid, but with an inner loop; passes through the pole twice; represented by r = a ± bcos θ and r = a ± bsin θ where a < b Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle Law of Sines states that the ratio of the measurement of one angle of a triangle to the length of its opposite side is equal to the remaining two ratios of angle measure to opposite side; any pair of proportions may be used to solve for a missing angle or side lemniscate a polar curve resembling a figure 8 and given by the equation r 2 = a2 cos 2θ and r 2 = a2 sin 2θ, a ≠ 0 magnitude the length of a vector; may represent a quantity such as speed, and is calculated using the Pythagorean Theorem modulus the absolute value of a complex number, or the distance from the origin to the point (x, y); also called the amplitude oblique triangle any triangle that is not a right triangle a _ > 1; one-loop limaçon a polar curve represented by r = a ± bcos θ and r = a ± bsin θ such that a > 0, b > 0, and b may be dimpled or convex; does not pass through the pole parameter a variable, often representing time, upon which x and y are both dependent polar axis on the polar grid, the equivalent of the positive x-axis on the rectangular grid polar coordinates on the polar grid, the coordinates of a point labeled (r, θ), where θ indicates the angle of rotation from the polar axis and r represents the radius, or the distance of the point from the pole in the direction of θ polar equation an equation describing a curve on the polar grid polar form of a complex number a complex number expressed in terms of an angle θ and its distance from the origin r; can be found by using conversion formulas x = rcos θ, y = rsin θ, and r = √ — x2 + y2 pole the origin of the polar grid 7 48 CHAPTER 8 Further applications oF trigonometry resultant a vector that results from addition or subtraction of two vectors, or from scalar multiplication rose curve a polar equation resembling a flower, given by the equations r = acos nθ and r = asin nθ; when n is even there are 2n petals, and the curve is highly symmetrical; when n is odd there are n petals. scalar a quantity associated with magnitude but not direction; a constant scalar multiplication the product of a constant and each component of a vector standard position the placement of a vector with the initial point at (0, 0) and the terminal point (a, b), represented by the change in the x-coordinates and the change in the y-coordinates of the original vector terminal point the end point of a vector, usually represented by an arrow indicating its direction unit vector a vector that begins at the origin and has magnitude of 1; the horizontal unit vector runs along the x-axis and is defined as v1 = 〈1, 0〉 the vertical unit vector runs along the y-axis and is defined as v2 = 〈0, 1〉. vector a quantity associated with both magnitude and direction, represented as a directed line segment with a starting point (initial point) and an end point (terminal point) vector addition the sum of two vectors, found by adding corresponding components Key equations Law of Sines Area for oblique triangles Law of Cosines = = sin α _____ a a _ sin α = sin β ____ b b _ sin β = sin γ ____ c c _ sin γ 1 _ Area = bcsin α 2 1 _ = acsin β 2 1 _ = absin γ 2 a2 = b2 + c2 − 2bccos α b2 = a2 + c2 − 2accos β c2 = a2 + b2 − 2abcos γ Heron’s formula Area = √ —— s(s − a)(s − b)(s − c) where s = (a + b + c) _ 2 Conversion formulas Key Concepts 8.1 Non-right Triangles: Law of Sines x _ r → x = rcos θ cos θ = y _ r → y = rsin θ sin θ = r2 = x2 + y2 y _ tan θ = x • The Law of Sines can be used to solve oblique triangles, which are non-right triangles. • According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side. CHAPTER 8 review 749 • There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution. See Example 1. • The ambiguous case arises when an oblique triangle can have different outcomes. • There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution. See Example 2 and Example 3. • The Law of Sines can be used to solve triangles with given criteria. See Example 4. • The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. See Example 5. • There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation. See Example 6. 8.2 Non-right Triangles: Law of Cosines • The Law of Cosines defines the relationship among angle measurements and lengths of sides in oblique triangles. • The Generalized Pythagorean Theorem is the Law of Cosines for two cases of oblique triangles: SAS and SSS. Dropping an imaginary perpendicular splits the oblique triangle into two right triangles or forms one right triangle, which allows sides to be related and measurements to be calculated. See Example 1 and Example 2. • The Law of Cosines is useful for many types of applied problems. The first step in solving such problems is generally to draw a sketch of the problem presented. If the information given fits one of the three models (the three equations), then apply the Law of Cosines to find a solution. See Example 3 and Example 4. • Heron’s formula allows the calculation of area in oblique triangles. All three sides must be known to apply Heron’s formula. See Example 5 and See Example 6. 8.3 Polar Coordinates • The polar grid is represented as a series of concentric circles radiating out from the pole, or origin. • To plot a point in the form (r, θ), θ > 0, move in a counterclockwise direction from the polar axis by an angle of θ, and then extend a directed line segment from the pole the length of r in the direction of θ. If θ is negative, move in a clockwise direction, and extend a directed line segment the length of r in the direction of θ. See Example 1. • If r is negative, extend the directed line segment in the opposite direction of θ. See Example 2. • To convert from polar coordinates to rectangular coordinates, use the formulas x = rcos θ and y = rsin θ. See Example 3 and Example 4. x _ • To convert from rectangular coordinates to polar coordinates, use one or more of the formulas: cos , and r = √ r , tan θ = sin θ = x2 + y2 . See Example 5. • Transforming equations between polar and rectangular forms means making the appropriate substitutions based on the available formulas, together with algebraic manipulations. See Example 6, Example 7, and Example 8. • Using the appropriate substitutions makes it possible to rewrite a polar equation as a rectangular equation, and then graph it in the rectangular plane. See Example 9, Example 10, a
nd Example 11. 8.4 Polar Coordinates: Graphs π _ • It is easier to graph polar equations if we can test the equations for symmetry with respect to the line θ = , the polar 2 axis, or the pole. • There are three symmetry tests that indicate whether the graph of a polar equation will exhibit symmetry. If an equation fails a symmetry test, the graph may or may not exhibit symmetry. See Example 1. • Polar equations may be graphed by making a table of values for θ and r. • The maximum value of a polar equation is found by substituting the value θ that leads to the maximum value of the trigonometric expression. • The zeros of a polar equation are found by setting r = 0 and solving for θ. See Example 2. • Some formulas that produce the graph of a circle in polar coordinates are given by r = acos θ and r = asin θ. See Example 3. • The formulas that produce the graphs of a cardioid are given by r = a ± bcos θ and r = a ± bsin θ, for a > 0, a _ = 1. See Example 4. b > 0, and b 7 50 CHAPTER 8 Further applications oF trigonometry • The formulas that produce the graphs of a one-loop limaçon are given by r = a ± bcos θ and r = a ± bsin θ for a _ 1 < < 2. See Example 5. b • The formulas that produce the graphs of an inner-loop limaçon are given by r = a ± bcos θ and r = a ± bsin θ for a > 0, b > 0, and a < b. See Example 6. • The formulas that produce the graphs of a lemniscates are given by r2 = a2 cos 2θ and r2 = a2 sin 2θ, where a ≠ 0. See Example 7. • The formulas that produce the graphs of rose curves are given by r = acos nθ and r = asin nθ, where a ≠ 0; if n is even, there are 2n petals, and if n is odd, there are n petals. See Example 8 and Example 9. • The formula that produces the graph of an Archimedes’ spiral is given by r = θ, θ ≥ 0. See Example 10. 8.5 Polar Form of Complex Numbers • Complex numbers in the form a + bi are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Label the x-axis as the real axis and the y-axis as the imaginary axis. See Example 1. • The absolute value of a complex number is the same as its magnitude. It is the distance from the origin to the point: — ∣ z ∣ = √ a2 + b2 . See Example 2 and Example 3. • To write complex numbers in polar form, we use the formulas x = rcos θ, y = rsin θ, and r = √ — x2 + y2 . Then, z = r(cos θ + isin θ). See Example 4 and Example 5. • To convert from polar form to rectangular form, first evaluate the trigonometric functions. Then, multiply through by r. See Example 6 and Example 7. • To find the product of two complex numbers, multiply the two moduli and add the two angles. Evaluate the trigonometric functions, and multiply using the distributive property. See Example 8. • To find the quotient of two complex numbers in polar form, find the quotient of the two moduli and the difference of the two angles. See Example 9. • To find the power of a complex number zn, raise r to the power n, and multiply θ by n. See Example 10. • Finding the roots of a complex number is the same as raising a complex number to a power, but using a rational exponent. See Example 11. 8.6 Parametric Equations • Parameterizing a curve involves translating a rectangular equation in two variables, x and y, into two equations in three variables, x, y, and t. Often, more information is obtained from a set of parametric equations. See Example 1, Example 2, and Example 3. • Sometimes equations are simpler to graph when written in rectangular form. By eliminating t, an equation in x and y is the result. • To eliminate t, solve one of the equations for t, and substitute the expression into the second equation. See Example 4, Example 5, Example 6, and Example 7. • Finding the rectangular equation for a curve defined parametrically is basically the same as eliminating the parameter. Solve for t in one of the equations, and substitute the expression into the second equation. See Example 8. • There are an infinite number of ways to choose a set of parametric equations for a curve defined as a rectangular equation. • Find an expression for x such that the domain of the set of parametric equations remains the same as the original rectangular equation. See Example 9. CHAPTER 8 review 751 8.7 Parametric Equations: Graphs • When there is a third variable, a third parameter on which x and y depend, parametric equations can be used. • To graph parametric equations by plotting points, make a table with three columns labeled t, x(t), and y(t). Choose values for t in increasing order. Plot the last two columns for x and y. See Example 1 and Example 2. • When graphing a parametric curve by plotting points, note the associated t-values and show arrows on the graph indicating the orientation of the curve. See Example 3 and Example 4. • Parametric equations allow the direction or the orientation of the curve to be shown on the graph. Equations that are not functions can be graphed and used in many applications involving motion. See Example 5. • Projectile motion depends on two parametric equations: x = (v0 cos θ)t and y = − 16t2 + (v0 sin θ)t + h. Initial velocity is symbolized as v0. θ represents the initial angle of the object when thrown, and h represents the height at which the object is propelled. 8.8 Vectors • The position vector has its initial point at the origin. See Example 1. • If the position vector is the same for two vectors, they are equal. See Example 2. Vectors are defined by their magnitude and direction. See Example 3. • If two vectors have the same magnitude and direction, they are equal. See Example 4. • Vector addition and subtraction result in a new vector found by adding or subtracting corresponding elements. See Example 5. • Scalar multiplication is multiplying a vector by a constant. Only the magnitude changes; the direction stays the same. See Example 6 and Example 7. • Vectors are comprised of two components: the horizontal component along the positive x-axis, and the vertical component along the positive y-axis. See Example 8. • The unit vector in the same direction of any nonzero vector is found by dividing the vector by its magnitude. • The magnitude of a vector in the rectangular coordinate system is ∣ v ∣ = √ • In the rectangular coordinate system, unit vectors may be represented in terms of i and j where i represents the horizontal component and j represents the vertical component. Then, v = ai + bj is a scalar multiple of v by real numbers a and b. See Example 10 and Example 11. a2 + b2 . See Example 9. — • Adding and subtracting vectors in terms of i and j consists of adding or subtracting corresponding coefficients of i and corresponding coefficients of j. See Example 12. • A vector v = ai + bj is written in terms of magnitude and direction as v = ∣ v ∣ cos θi + ∣ v ∣ sin θj. See Example 13. • The dot product of two vectors is the product of the i terms plus the product of the j terms. See Example 14. • We can use the dot product to find the angle between two vectors. Example 15 and Example 16. • Dot products are useful for many types of physics applications. See Example 17. 7 52 CHAPTER 8 Further applications oF trigonometry CHAPTeR 8 ReVIeW exeRCISeS nOn-RIGHT TRIAnGleS: lAW OF SIneS For the following exercises, assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve each triangle, if possible. Round each answer to the nearest tenth. 1. β = 50°, a = 105, b = 45 3. Solve the triangle. 2. α = 43.1°, a = 184.2, b = 242.8 4. Find the area of the triangle. c 36° 16 24° a C 75° 8 11 5. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 2.1 km apart, to be 25° and 49°, as shown in Figure 1. Find the distance of the plane from point A and the elevation of the plane. 25° 49° A Figure 1 B nOn-RIGHT TRIAnGleS: lAW OF COSIneS 6. Solve the triangle, rounding to the nearest tenth, assuming α is opposite side a, β is opposite side b, and γ is opposite side c: a = 4, b = 6, c = 8. 7. Solve the triangle in Figure 2, rounding to the nearest tenth. 8. Find the area of a triangle with sides of length 8.3, 6.6, and 9.1. 9. To find the distance between two cities, a satellite calculates the distances and angle shown in Figure 3 (not to scale). Find the distance between the cities. Round answers to the nearest tenth. 13 B 54° a Figure 2 15 C 250 km 1.8° 210 km Figure 3 CHAPTER 8 review 753 POlAR COORDInATeS π __ 10. Plot the point with polar coordinates  3,  . 6 3π  to rectangular coordinates. 12. Convert  6, − ___ 4 14. Convert (7, −2) to polar coordinates. 2π 11. Plot the point with polar coordinates  5, −  ___ 3 3π  to rectangular coordinates. 13. Convert  −2, ___ 2 15. Convert (−9, −4) to polar coordinates. For the following exercises, convert the given Cartesian equation to a polar equation. 16. x = − 2 17. x2 + y2 = 64 18. x2 + y2 = − 2y For the following exercises, convert the given polar equation to a Cartesian equation. 19. r = 7cos θ 20. r = −2 ____________ 4cos θ + sin θ For the following exercises, convert to rectangular form and graph. 21. θ = 22. r = 5sec θ 3π ___ 4 POlAR COORDInATeS: GRAPHS For the following exercises, test each equation for symmetry. 23. r = 4 + 4sin θ 25. Sketch a graph of the polar equation r = 1 − 5sin θ. 24. r = 7 26. Sketch a graph of the polar equation r = 5sin(7θ). Label the axis intercepts. 27. Sketch a graph of the polar equation r = 3 − 3cos θ POlAR FORM OF COMPlex nUMBeRS For the following exercises, find the absolute value of each complex number. 28. −2 + 6i 29. 4 − 3i Write the complex number in polar form. 30. 5 + 9i — 1 __ − 31. 2 3 √ ____ i 2 For the following exercises, convert the complex number from polar to rectangular form. 5π  32. z = 5cis  ___ 6 For the following exercises, find the product z1z2 in polar form. 34. z1 = 2cis(89°), z2 = 5cis(23°) 33. z = 3cis(40°) π π __ __   , z2 = 6cis  35. z1 = 10cis  3 6 For the following exercises, find the quotient 36. z1 = 12cis(55°), z2 = 3cis(18°) z1 _
in polar form. z2 π 5π  , z2 = 9cis  37. z1 = 27cis  __ ___  3 3 For the following exercises, find the powers of each complex number in polar form. 38. Find z4 when z = 2cis(70°) For the following exercises, evaluate each root. 40. Evaluate the cube root of z when z = 64cis(210°). 3π  39. Find z2 when z = 5cis  ___ 4 3π  . 41. Evaluate the square root of z when z = 25cis  ___ 2 For the following exercises, plot the complex number in the complex plane. 42. 6 − 2i 43. −1 + 3i PARAMeTRIC eQUATIOnS For the following exercises, eliminate the parameter t to rewrite the parametric equation as a Cartesian equation. 44.  x(t) = 3t − 1 y(t) = √ t — 45.  x(t) = − cos t y(t) = 2sin2 t 7 54 CHAPTER 8 Further applications oF trigonometry 46. Parameterize (write a parametric equation for) each Cartesian equation by using x(t) = acos t and y2 ___ y(t) = bsin t for 16 x2 ___ 25 = 1. + 47. Parameterize the line from (−2, 3) to (4, 7) so that the line is at (−2, 3) at t = 0 and (4, 7) at t = 1. PARAMeTRIC eQUATIOnS: GRAPHS For the following exercises, make a table of values for each set of parametric equations, graph the equations, and include an orientation; then write the Cartesian equation. 48.  x(t) = 3t2 y(t) = 2t − 1 49.  x(t) = e t y(t) = −2e 5t 50.  x(t) = 3cos t y(t) = 2sin t 51. A ball is launched with an initial velocity of 80 feet per second at an angle of 40° to the horizontal. The ball is released at a height of 4 feet above the ground. a. Find the parametric equations to model the path of the ball. b. Where is the ball after 3 seconds? c. How long is the ball in the air? VeCTORS For the following exercises, determine whether the two vectors, u and v, are equal, where u has an initial point P1 and a terminal point P2, and v has an initial point P3 and a terminal point P4. 52. P1 = (−1, 4), P2 = (3, 1), P3 = (5, 5) and 53. P1 = (6, 11), P2 = (−2, 8), P3 = (0, − 1) and P4 = (9, 2) P4 = (−8, 2) For the following exercises, use the vectors u = 2i − j, v = 4i − 3j, and w = −2i + 5j to evaluate the expression. 54. u − v 55. 2v − u + w For the following exercises, find a unit vector in the same direction as the given vector. 56. a = 8i − 6j 57. b = −3i − j For the following exercises, find the magnitude and direction of the vector. 58. 〈6, −2〉 59. 〈−3, −3〉 For the following exercises, calculate u ⋅ v. 60. u = −2i + j and v = 3i + 7j 1 _ 62. Given v = 〈−3, 4〉 draw v, 2v, and v. 2 61. u = i + 4j and v = 4i + 3j 63. Given the vectors shown in Figure 4, sketch u + v, u − v and 3v. u v Figure 4 64. Given initial point P1 = (3, 2) and terminal point P2 = (−5, −1), write the vector v in terms of i and j. Draw the points and the vector on the graph. CHAPTER 8 practice test 755 CHAPTeR 8 PRACTICe TeST 1. Assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve the triangle, if possible, and round each answer to the nearest tenth, given β = 68°, b = 21, c = 16. 3. A pilot flies in a straight path for 2 hours. He then makes a course correction, heading 15° to the right of his original course, and flies 1 hour in the new direction. If he maintains a constant speed of 575 miles per hour, how far is he from his starting position? 5. Convert  2, π __ 3  to rectangular coordinates. 2. Find the area of the triangle in Figure 1. Round each answer to the nearest tenth. 6.25 5 60° 7 Figure 1 4. Convert (2, 2) to polar coordinates, and then plot the point. 6. Convert the polar equation to a Cartesian equation: x2 + y2 = 5y. 7. Convert to rectangular form and graph: r = − 3csc θ. 9. Graph r = 3 + 3cos θ. 8. Test the equation for symmetry: r = − 4sin (2θ). 10. Graph r = 3 − 5sin θ. 11. Find the absolute value of the complex number 12. Write the complex number in polar form: 4 + i. 5 − 9i. 13. Convert the complex number from polar to rectangular form: z = 5cis  2π ___  . 3 Given z1 = 8cis(36°) and z2 = 2cis(15°), evaluate each expression. 14. z1 z2 15. z1 _ z2 18. Plot the complex number −5 − i in the complex plane. 20. Parameterize (write a parametric equation for) the following Cartesian equation by using x(t) = acos t y2 and y(t) = bsin t: x2 ___ ___ 100 36 = 1. + 16. (z2)3 17. √ — z1 19. Eliminate the parameter t to rewrite the following parametric equations as a Cartesian equation: x(t) = t + 1 y(t) = 2t2  21. Graph the set of parametric equations and find the Cartesian equation:  x(t) = −2sin t y(t) = 5cos t 22. A ball is launched with an initial velocity of 95 feet per second at an angle of 52° to the horizontal. The ball is released at a height of 3.5 feet above the ground. a. Find the parametric equations to model the path of the ball. b. Where is the ball after 2 seconds? c. How long is the ball in the air? For the following exercises, use the vectors u = i −3j and v = 2i + 3j. 23. Find 2u − 3v. 24. Calculate u ∙ v. 25. Find a unit vector in the same direction as v. 26. Given vector v has an initial point P1 = (2, 2) and terminal point P2 = (−1, 0), write the vector v in terms of i and j. On the graph, draw v, and − v. 9 Systems of Equations and Inequalities Figure 1 enigma machines like this one, once owned by Italian dictator Benito Mussolini, were used by government and military officials for enciphering and deciphering top-secret communications during World War II. (credit: Dave Addey, Flickr) CHAPTeR OUTlIne 9.1 Systems of linear equations: Two Variables 9.2 Systems of linear equations: Three Variables 9.3 Systems of nonlinear equations and Inequalities: Two Variables 9.4 Partial Fractions 9.5 Matrices and Matrix Operations 9.6 Solving Systems with Gaussian elimination 9.7 Solving Systems with Inverses 9.8 Solving Systems with Cramer's Rule Introduction By 1943, it was obvious to the Nazi regime that defeat was imminent unless it could build a weapon with unlimited destructive power, one that had never been seen before in the history of the world. In September, Adolf Hitler ordered German scientists to begin building an atomic bomb. Rumors and whispers began to spread from across the ocean. Refugees and diplomats told of the experiments happening in Norway. However, Franklin D. Roosevelt wasn’t sold, and even doubted British Prime Minister Winston Churchill’s warning. Roosevelt wanted undeniable proof. Fortunately, he soon received the proof he wanted when a group of mathematicians cracked the “Enigma” code, proving beyond a doubt that Hitler was building an atomic bomb. The next day, Roosevelt gave the order that the United States begin work on the same. The Enigma is perhaps the most famous cryptographic device ever known. It stands as an example of the pivotal role cryptography has played in society. Now, technology has moved cryptanalysis to the digital world. Many ciphers are designed using invertible matrices as the method of message transference, as finding the inverse of a matrix is generally part of the process of decoding. In addition to knowing the matrix and its inverse, the receiver must also know the key that, when used with the matrix inverse, will allow the message to be read. In this chapter, we will investigate matrices and their inverses, and various ways to use matrices to solve systems of equations. First, however, we will study systems of equations on their own: linear and nonlinear, and then partial fractions. We will not be breaking any secret codes here, but we will lay the foundation for future courses. 757 7 58 CHAPTER 9 systems oF eQuations and ineQualities leARnInG OBjeCTIVeS In this section, you will: • Solve systems of equations by graphing. • Solve systems of equations by substitution. • Solve systems of equations by addition. • • Express the solution of a system of dependent equations containing two variables. Identify inconsistent systems of equations containing two variables. 9.1 SYSTeMS OF lIneAR eQUATIOnS: TWO VARIABleS Figure 1 (credit: Thomas Sørenes) A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible? In this section, we will consider linear equations with two variables to answer these and similar questions. Introduction to Systems of equations In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution. In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables. 2x + y = 15 3x − y = 5 The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists. 2(4) + (7) = 15 True 3(4) − (7) = 5 True In addition to considering the number of equations and variables, we can categorize system
s of linear equations by the number of solutions. A consistent system of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have SECTION 9.1 systems oF linear eQuations: two variaBles 759 different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y-intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions. Another type of system of linear equations is an inconsistent system, which is one in which the equations represent two parallel lines. The lines have the same slope and different y-intercepts. There are no points common to both lines; hence, there is no solution to the system. types of linear systems There are three types of systems of linear equations in two variables, and three types of solutions. • An independent system has exactly one solution pair (x, y). The point where the two lines intersect is the only solution. • An inconsistent system has no solution. Notice that the two lines are parallel and will never intersect. • A dependent system has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations. Figure 2 compares graphical representations of each type of system.   7 –, 11   5 5   –6 –4 –2 y 5 4 3 2 1 0 –1 –2 –3 –4 –5 y 5 4 3 2 1 0 –1 –2 –3 2 4 6 x –6 –4 –2 (–1, –2) 2 4 6 x –6 –4 –2 Figure 2 –4 –5 (1, 21 –2 –3 –4 –5 How To… Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution. Independent System Inconsistent System Dependent System 1. Substitute the ordered pair into each equation in the system. 2. Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution. Example 1 Determining Whether an Ordered Pair Is a Solution to a System of Equations Determine whether the ordered pair (5, 1) is a solution to the given system of equations. Solution Substitute the ordered pair (5, 1) into both equations. x + 3y = 8 2x − 9 = y (5) + 3(1) = 8 8 = 8 2(5) − 9 = (1) 1 = 1 True True The ordered pair (5, 1) satisfies both equations, so it is the solution to the system. Analysis We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines. See Figure 3. 7 60 CHAPTER 9 systems oF eQuations and ineQualities y 5 4 3 2 1 –2 –1 –1 –2 –3 –4 –5 x +3y = 8 4 21 3 (5, 1) 5 6 7 8 9 10 x 2x − 9 = y Figure 3 Try It #1 Determine whether the ordered pair (8, 5) is a solution to the following system. 5x − 4y = 20 2x + 1 = 3y Solving Systems of equations by Graphing There are multiple methods of solving systems of linear equations. For a system of linear equations in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes. Example 2 Solving a System of Equations in Two Variables by Graphing Solve the following system of equations by graphing. Identify the type of system. Solution Solve the first equation for y. Solve the second equation for y. 2x + y = −8 x − y = −1 2x + y = −8 y = − 2x −8 x − y = −1 y = x + 1 Graph both equations on the same set of axes as in Figure 4. y 10 8 6 4 2 –1 –2 –4 –6 –8 –10 y = −2x −8 –5 –4 –3 –2 (−3, −2) y = x + 1 21 3 4 5 x Figure 4 SECTION 9.1 systems oF linear eQuations: two variaBles 761 The lines appear to intersect at the point (−3, −2). We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations. 2(−3) + (−2) = −8 −8 = −8 True (−3) − (−2) = −1 −1 = −1 True The solution to the system is the ordered pair (−3, −2), so the system is independent. Try It #2 Solve the following system of equations by graphing. 2x − 5y = −25 −4x + 5y = 35 Q & A… Can graphing be used if the system is inconsistent or dependent? Yes, in both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system. Solving Systems of equations by Substitution Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical. How To… Given a system of two equations in two variables, solve using the substitution method. 1. Solve one of the two equations for one of the variables in terms of the other. 2. Substitute the expression for this variable into the second equation, then solve for the remaining variable. 3. Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair. 4. Check the solution in both equations. Example 3 Solving a System of Equations in Two Variables by Substitution Solve the following system of equations by substitution. Solution First, we will solve the first equation for y. −x + y = −5 2x − 5y = 1 −x + y = −5 y = x − 5 7 62 CHAPTER 9 systems oF eQuations and ineQualities Now we can substitute the expression x − 5 for y in the second equation. 2x − 5y = 1 2x − 5(x − 5) = 1 2x − 5x + 25 = 1 −3x = −24 x = 8 Now, we substitute x = 8 into the first equation and solve for y. −(8) + y = −5 y = 3 Our solution is (8, 3). Check the solution by substituting (8, 3) into both equations. −x + y = − 5 −(8) + (3) = − 5 True 2x − 5y = 1 2(8) − 5(3) = 1 True Try It #3 Solve the following system of equations by substitution. x = y + 3 4 = 3x − 2y Q & A… Can the substitution method be used to solve any linear system in two variables? Yes, but the method works best if one of the equations contains a coefficient of 1 or −1 so that we do not have to deal with fractions. Solving Systems of equations in Two Variables by the Addition Method A third method of solving systems of linear equations is the addition method. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition. How To… Given a system of equations, solve using the addition method. 1. Write both equations with x- and y-variables on the left side of the equal sign and constants on the right. 2. Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable. 3. Solve the resulting equation for the remaining variable. 4. Substitute that value into one of the original equations and solve for the second variable. 5. Check the solution by substituting the values into the other equation. SECTION 9.1 systems oF linear eQuations: two variaBles 763 Example 4 Solving a System by the Addition Method Solve the given system of equations by addition. x + 2y = −1 −x + y = 3 Solution Both equations are already set equal to a constant. Notice that the coefficient of x in the second equation, −1, is the opposite of the coefficient of x in the first equation, 1. We can add the two equations to eliminate x without needing to multiply by a constant. x + 2y = − 1 −x + y = 3 3y = 2 Now that we have eliminated x, we can solve the resulting equation for y. 3y = 2 2 _ y = 3 Then, we substitute this value for y into one of the original equations and solve for x. 2 The solution to this system is  − 7  __ __ , 3 3 Check the solution in the first equation. −x + y = 3 2 __ = 3 −x + 3 2 _ −x = 3 − 3 7 _ − + 2y = −1 __ __ 3 3 4 − 7 __ __ = −1 + 3 3 − 3 __ = −1 3 −1 = −1 True Analysis We gain an important perspective on systems of equations by looking at the graphical representation. See Figure 5 to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution. x + 2y = −1 – ,7 3 2 3 –6 –5 –4 –3 –2 −1 –1 –2 –3 –4 –5 21 3 4 5 6 x Figure 5 7 64 CHAPTER 9 systems oF eQuations and ineQualities Example 5 Using the Addition Method When Multiplication of One Equation Is Required Solve the given system of equations by the addition method. 3x + 5y = −11 x − 2y = 11 Solution Adding these equations as presented will not eliminate a variable. However, we see that the first equation has 3x in it and the second equation has x. So if we multiply the second equation by −3, the x-terms will add to zero. x − 2y = 11 −3(x − 2y) = −3(11) −3x + 6y = −33 Multiply both sides by −3. Use the distributive property. Now, let’s add them. 3x + 5y = −11 −3x + 6y = −33 11y = −44 y = −4 For the last st
ep, we substitute y = −4 into one of the original equations and solve for x. 3x + 5y = − 11 3x + 5( − 4) = − 11 3x − 20 = − 11 3x = 9 x = 3 Our solution is the ordered pair (3, −4). See Figure 6. Check the solution in the original second equation. x − 2y = 11 (3) − 2( − 4) = 3 + 8 11 = 11 True y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 21 3 4 5 6 x (3, –4) x − 2y = 11 3x + 5y = −11 Figure 6 2x − 7y = 2 3x + y = −20 Try It #4 Solve the system of equations by addition. Example 6 Using the Addition Method When Multiplication of Both Equations Is Required Solve the given system of equations in two variables by addition. 2x + 3y = −16 5x − 10y = 30 SECTION 9.1 systems oF linear eQuations: two variaBles 765 Solution One equation has 2x and the other has 5x. The least common multiple is 10x so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate x by multiplying the first equation by −5 and the second equation by 2. Then, we add the two equations together. − 5(2x + 3y) = −5(−16) − 10x − 15y = 80 2(5x − 10y) = 2(30) 10x − 20y = 60 −10x − 15y = 80 10x − 20y = 60 −35y = 140 y = −4 Substitute y = −4 into the original first equation. 2x + 3(−4) = −16 2x − 12 = −16 2x = −4 x = −2 The solution is (−2, −4). Check it in the other equation. See Figure 7. 5x − 10y = 30 5(−2) − 10(−4) = 30 −10 + 40 = 30 30 = 30 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –7 2x + 3y = −16 –6 –5 –4 –3 –2 (−2, −4) 5x − 10y = 30 21 3 4 5 6 x Example 7 Using the Addition Method in Systems of Equations Containing Fractions Solve the given system of equations in two variables by addition. Figure 7 x y __ __ = 3 + 6 3 y x __ __ = 1 − 4 2 Solution First clear each equation of fractions by multiplying both sides of the equation by the least common denominator. y x __ __  = 6(3) + 6  6 3 2x + y = 18 y x __ __  = 4(1) − 4  4 2 2x − y = 4 7 66 CHAPTER 9 systems oF eQuations and ineQualities Now multiply the second equation by −1 so that we can eliminate the x-variable. Add the two equations to eliminate the x-variable and solve the resulting equation. −1(2x − y) = −1(4) −2x + y = −4 Substitute y = 7 into the first equation. 2x + y = 18 −2x + y = −4 2y = 14 y = 7 2x + (7) = 18 2x = 11 11 ___ 2 = 5.5 x = The solution is  11 , 7  . Check it in the other equation. ___ 2 x y __ __ = 1 − 2 4 11 _ 7 2 ____ __ − = 1 2 4 7 11 __ ___ = 1 − 4 4 4 __ = 1 4 Try It #5 Solve the system of equations by addition. 2x + 3y = 8 3x + 5y = 10 Identifying Inconsistent Systems of equations Containing Two Variables Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different y-intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as 12 = 0. Example 8 Solving an Inconsistent System of Equations Solve the following system of equations. x = 9 − 2y x + 2y = 13 Solution We can approach this problem in two ways. Because one equation is already solved for x, the most obvious step is to use substitution. x + 2y = 13 (9 − 2y) + 2y = 13 9 + 0y = 13 9 = 13 Clearly, this statement is a contradiction because 9 ≠ 13. Therefore, the system has no solution. SECTION 9.1 systems oF linear eQuations: two variaBles 767 The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows. x = 9 − 2y 2y = − x + 9 y = − 1 9 __ __ x + 2 2 We then convert the second equation expressed to slope-intercept form. x + 2y = 13 2y = − x + 13 y = − 1 __ x + 2 13 ___ 2 Comparing the equations, we see that they have the same slope but different y-intercepts. Therefore, the lines are parallel and do not intersect. 9 y = − 1 __ __ x + 2 2 13 y = − 1 ___ __ x + 2 2 Analysis Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown in Figure 812 –10 –8 –6 –4 y 12 10 8 6 4 2 –2 –2 –4 –6 –8 –10 –12 1 y = − x + 2 x 13 2 42 6 8 10 12 Try It #6 Solve the following system of equations in two variables. Figure 8 2y − 2x = 2 2y − 2x = 6 expressing the Solution of a System of Dependent equations Containing Two Variables Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as 0 = 0. Example 9 Finding a Solution to a Dependent System of Linear Equations Find a solution to the system of equations using the addition method. x + 3y = 2 3x + 9y = 6 Solution With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating x. If we multiply both sides of the first equation by −3, then we will be able to eliminate the x-variable. 7 68 CHAPTER 9 systems oF eQuations and ineQualities Now add the equations. x + 3y = 2 (−3)(x + 3y) = (−3)(2) −3x − 9y = − 6 −3x − 9y = −6 + 3x + 9y = 6 0 = 0 We can see that there will be an infinite number of solutions that satisfy both equations. Analysis before adding. Let’s look at what happens when we convert the system to slope-intercept form. If we rewrote both equations in the slope-intercept form, we might know what the solution would look like x + 3y = 2 3y = − x + 2 1 2 __ __ y = − x + 3 3 3x + 9y = 6 9y = −3x + 6 3 6 __ __ y = − x + 9 9 2 1 __ __ x + y = − 3 3 1 2  . See Figure 9. Notice the results are the same. The general solution to the system is  x, − _ _ x + 3 3 x + 3y = 2 –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 x 5 3x + 9y = 6 Figure 9 Try It #7 Solve the following system of equations in two variables. y − 2x = 5 −3y + 6x = −15 Using Systems of equations to Investigate Profits Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation R = xp, where x = quantity and p = price. The revenue function is shown in orange in Figure 10. The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in Figure 10. The x-axis represents quantity in hundreds of units. The y-axis represents either cost or revenue in hundreds of dollars. SECTION 9.1 systems oF linear eQuations: two variaBles 769 ) 70 60 50 40 30 20 10 0 Profit (7, 33) Cost Break-even Revenue –10 0 5 10 Quantity (in hundreds of units) 15 20 Figure 10 The point at which the two lines intersect is called the break-even point. We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money. The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as P(x) = R(x) − C(x). Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses. Example 10 Finding the Break-Even Point and the Profit Function Using Substitution Given the cost function C(x) = 0.85x + 35,000 and the revenue function R(x) = 1.55x, find the break-even point and the profit function. Solution Write the system of equations using y to replace function notation. Substitute the expression 0.85x + 35,000 from the first equation into the second equation and solve for x. y = 0.85x + 35,000 y = 1.55x 0.85x + 35,000 = 1.55x 35,000 = 0.7x 50,000 = x Then, we substitute x = 50,000 into either the cost function or the revenue function. The break-even point is (50,000, 77,500). The profit function is found using the formula P(x) = R(x) − C(x). 1.55(50,000) = 77,500 P(x) = 1.55x − (0.85x + 35, 000) = 0.7x − 35, 000 The profit function is P(x) = 0.7x − 35,000. Analysis The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units. See Figure 11. We see from the graph in Figure 12 that the profit function has a negative value until x = 50,000, when the graph crosses the x-axis. Then, the graph emerges into positive y-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss. 7 70 CHAPTER 9 systems oF eQuations and ineQualities 100,000 s r a l l o D 80,000 60,000 40,000 20,000 0 0 Profit Break-even point (50,000, 77,500) Cost C(x) = 0.85x + 35,000 Revenue R(x) = 1.55x 40,000 20,000 Quantity Figure 11 60,000 80,000 100,000 60,000 50,000 40,000 30,000 20,000 10,000 0 –10,000 –20,000 –30,000 –40,000 –20,000 –10,000 Profit Profit P(x) = 0.7x − 35,000 Break-even point (50,000, 0) 0 10,000 20,000 30,000 60,000 50,000 40,000 Quantity 70,000 80,000 90,000 100,000 110,000 120,000 Figure 12 Example 11 Writing and Solving a System of Equations in Two Variables The cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,0
00. How many children and how many adults bought tickets? Solution Let c = the number of children and a = the number of adults in attendance. The total number of people is 2,000. We can use this to write an equation for the number of people at the circus that day. c + a = 2,000 The revenue from all children can be found by multiplying $25.00 by the number of children, 25c. The revenue from all adults can be found by multiplying $50.00 by the number of adults, 50a. The total revenue is $70,000. We can use this to write an equation for the revenue. 25c + 50a = 70,000 We now have a system of linear equations in two variables. c + a = 2,000 25c + 50a = 70,000 In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either c or a. We will solve for a. c + a = 2,000 Substitute the expression 2,000 − c in the second equation for a and solve for c. a = 2,000 − c 25c + 50(2,000 − c) = 70,000 25c + 100,000 − 50c = 70,000 − 25c = −30,000 c = 1,200 Substitute c = 1,200 into the first equation to solve for a. 1,200 + a = 2,000 a = 800 We find that 1,200 children and 800 adults bought tickets to the circus that day. Try It #8 Meal tickets at the circus cost $4.00 for children and $12.00 for adults. If 1,650 meal tickets were bought for a total of $14,200, how many children and how many adults bought meal tickets? Access these online resources for additional instruction and practice with systems of linear equations. • Solving Systems of equations Using Substitution (http://openstaxcollege.org/l/syssubst) • Solving Systems of equations Using elimination (http://openstaxcollege.org/l/syselim) • Applications of Systems of equations (http://openstaxcollege.org/l/sysapp) SECTION 9.1 section exercises 771 9.1 SeCTIOn exeRCISeS VeRBAl 1. Can a system of linear equations have exactly two solutions? Explain why or why not. 3. If you are solving a break-even analysis and get a negative break-even point, explain what this signifies for the company? 5. Given a system of equations, explain at least two different methods of solving that system. AlGeBRAIC 2. If you are performing a break-even analysis for a business and their cost and revenue equations are dependent, explain what this means for the company’s profit margins. 4. If you are solving a break-even analysis and there is no break-even point, explain what this means for the company. How should they ensure there is a break-even point? For the following exercises, determine whether the given ordered pair is a solution to the system of equations. 6. 5x − y = 4 x + 6y = 2 and (4, 0) 9. −2x + 5y = 7 7. −3x − 5y = 13 8. 3x + 7y = 1 − x + 4y = 10 and (−6, 1) 2x + 4y = 0 and (2, 3) 10. x + 8y = 43 2x + 9y = 7 and (−1, 1) 3x − 2y = −1 and (3, 5) For the following exercises, solve each system by substitution. 11. x + 3y = 5 2x + 3y = 4 15. −2x + 3y = 1.2 −3x − 6y = 1.8 1 1 __ __ y = 16 x + 19. 3 2 1 1 __ __ y = 9 x + 4 6 12. 3x − 2y = 18 5x + 10y = −10 16. x − 0.2y = 1 −10x + 2y = 5 3 1 __ __ y = 11 x + 20. − 2 4 1 1 __ __ y = 3 x + − 3 8 13. 4x + 2y = −10 3x + 9y = 0 14. 2x + 4y = −3.8 9x − 5y = 1.3 17. 3 x + 5y = 9 30x + 50y = −90 18. −3x + y = 2 12x − 4y = −8 For the following exercises, solve each system by addition. 21. −2x + 5y = −42 7x + 2y = 30 22. 6x − 5y = −34 2x + 6y = 4 23. 5x − y = −2.6 −4x − 6y = 1.4 24. 7x − 2y = 3 4x + 5y = 3.25 25. −x + 2y = −1 5x − 10y = 6 26. 7x + 6y = 2 −28x − 24y = −8 29. −0.2x + 0.4y = 0.6 x − 2y = −3 30. −0.1x + 0.2y = 0.6 5x − 10y = 1 5 x + 1 __ __ y = 0 27. 4 6 1 1 __ __ y = − x − 2 8 43 ___ 120 2 1 1 __ __ __ y = x + 28. 9 3 9 1 4 1 __ __ __ y = − x + − 3 5 2 For the following exercises, solve each system by any method. 31. 5x + 9y = 16 x + 2y = 4 32. 6x − 8y = −0.6 3x + 2y = 0.9 33. 5x − 2y = 2.25 7x − 4y = 3 34. x − y = − 5 ___ 12 55 ___ 12 5 __ −6x + y = 2 55 ___ 2 7 72 CHAPTER 9 systems oF eQuations and ineQualities 7 __ 35. 7x − 4y = 6 1 __ 2x + 4y = 3 36. 3x + 6y = 11 2x + 4y = 9 39. 2.2x + 1.3y = −0.1 4.2x + 4.2y = 2.1 40. 0.1x + 0.2y = 2 0.35x − 0.3y = 0 GRAPHICAl 37. 7 __ 3 21 ___ x + 6 1 __ y = 2 x − 6 3 ___ 12 − y = −3 1 1 1 __ __ __ y = x + 38. 3 3 2 1 1 3 __ __ __ y = − x + 8 4 2 For the following exercises, graph the system of equations and state whether the system is consistent, inconsistent, or dependent and whether the system has one solution, no solution, or infinite solutions. 41. 3x − y = 0.6 x − 2y = 1.3 44. 3x − 5y = 7 x − 2y = 3 TeCHnOlOGY 42. −x + 2y = 4 2x − 4y = 1 45. 3x − 2y = 5 −9x + 6y = −15 43. x + 2y = 7 2x + 6y = 12 For the following exercises, use the intersect function on a graphing device to solve each system. Round all answers to the nearest hundredth. 46. 0.1x + 0.2y = 0.3 −0.3x + 0.5y = 1 49. 0.15x + 0.27y = 0.39 −0.34x + 0.56y = 1.8 47. −0.01x + 0.12y = 0.62 0.15x + 0.20y = 0.52 50. −0.71x + 0.92y = 0.13 0.83x + 0.05y = 2.1 48. 0.5x + 0.3y = 4 0.25x − 0.9y = 0.46 exTenSIOnS For the following exercises, solve each system in terms of A, B, C, D, E, and F where A – F are nonzero numbers. Note that A ≠ B and AE ≠ BD. 51 52. x + Ay = 1 x + By = 1 53. Ax + y = 0 Bx + y = 1 54. Ax + By = C x + y = 1 55. Ax + By = C Dx + Ey = F ReAl-WORlD APPlICATIOnS For the following exercises, solve for the desired quantity. 56. A stuffed animal business has a total cost of 57. A fast-food restaurant has a cost of production production C = 12x + 30 and a revenue function R = 20x. Find the break-even point. C(x) = 11x + 120 and a revenue function R(x) = 5x. When does the company start to turn a profit? 58. A cell phone factory has a cost of production C(x) = 150x + 10,000 and a revenue function R(x) = 200x. What is the break-even point? 59. A musician charges C(x) = 64x + 20,000, where x is the total number of attendees at the concert. The venue charges $80 per ticket. After how many people buy tickets does the venue break even, and what is the value of the total tickets sold at that point? 60. A guitar factory has a cost of production C(x) = 75x + 50,000. If the company needs to break even after 150 units sold, at what price should they sell each guitar? Round up to the nearest dollar, and write the revenue function. SECTION 9.1 section exercises 773 For the following exercises, use a system of linear equations with two variables and two equations to solve. 61. Find two numbers whose sum is 28 and difference is 13. 63. The startup cost for a restaurant is $120,000, and each meal costs $10 for the restaurant to make. If each meal is then sold for $15, after how many meals does the restaurant break even? 62. A number is 9 more than another number. Twice the sum of the two numbers is 10. Find the two numbers. 64. A moving company charges a flat rate of $150, and an additional $5 for each box. If a taxi service would charge $20 for each box, how many boxes would you need for it to be cheaper to use the moving company, and what would be the total cost? 65. A total of 1,595 first- and second-year college students gathered at a pep rally. The number of freshmen exceeded the number of sophomores by 15. How many freshmen and sophomores were in attendance? 66. 276 students enrolled in a freshman-level chemistry class. By the end of the semester, 5 times the number of students passed as failed. Find the number of students who passed, and the number of students who failed. 67. There were 130 faculty at a conference. If there were 18 more women than men attending, how many of each gender attended the conference? 68. A jeep and BMW enter a highway running east- west at the same exit heading in opposite directions. The jeep entered the highway 30 minutes before the BMW did, and traveled 7 mph slower than the BMW. After 2 hours from the time the BMW entered the highway, the cars were 306.5 miles apart. Find the speed of each car, assuming they were driven on cruise control. 69. If a scientist mixed 10% saline solution with 60% saline solution to get 25 gallons of 40% saline solution, how many gallons of 10% and 60% solutions were mixed? 70. An investor earned triple the profits of what she earned last year. If she made $500,000.48 total for both years, how much did she earn in profits each year? 71. An investor who dabbles in real estate invested 1.1 million dollars into two land investments. On the first investment, Swan Peak, her return was a 110% increase on the money she invested. On the second investment, Riverside Community, she earned 50% over what she invested. If she earned $1 million in profits, how much did she invest in each of the land deals? 73. If an investor invests $23,000 into two bonds, one that pays 4% in simple interest, and the other paying 2% simple interest, and the investor earns $710.00 annual interest, how much was invested in each account? 72. If an investor invests a total of $25,000 into two bonds, one that pays 3% simple interest, and the 7 __ % interest, and the investor other that pays 2 8 earns $737.50 annual interest, how much was invested in each account? 74. CDs cost $5.96 more than DVDs at All Bets Are Off Electronics. How much would 6 CDs and 2 DVDs cost if 5 CDs and 2 DVDs cost $127.73? 75. A store clerk sold 60 pairs of sneakers. The high-tops sold for $98.99 and the low-tops sold for $129.99. If the receipts for the two types of sales totaled $6,404.40, how many of each type of sneaker were sold? 76. A concert manager counted 350 ticket receipts the day after a concert. The price for a student ticket was $12.50, and the price for an adult ticket was $16.00. The register confirms that $5,075 was taken in. How many student tickets and adult tickets were sold? 77. Admission into an amusement park for 4 children and 2 adults is $116.90. For 6 children and 3 adults, the admission is $175.35. Assuming a different price for children and adults, what is the price of the child’s ticket and the price of the adult ticket? 7 74 CHAPTER 9 systems oF eQuations and ineQualities leARnInG OBjeCTIVeS In this section, you will: • Solve systems of three equations in three vari
ables. • • Express the solution of a system of dependent equations containing three variables. Identify inconsistent systems of equations containing three variables. 9.2 SYSTeMS OF lIneAR eQUATIOnS: THRee VARIABleS Figure 1 (credit: “elembis,” Wikimedia Commons) John received an inheritance of $12,000 that he divided into three parts and invested in three ways: in a money-market fund paying 3% annual interest; in municipal bonds paying 4% annual interest; and in mutual funds paying 7% annual interest. John invested $4,000 more in municipal funds than in municipal bonds. He earned $670 in interest the first year. How much did John invest in each type of fund? Understanding the correct approach to setting up problems such as this one makes finding a solution a matter of following a pattern. We will solve this and similar problems involving three equations and three variables in this section. Doing so uses similar techniques as those used to solve systems of two equations in two variables. However, finding solutions to systems of three equations requires a bit more organization and a touch of visual gymnastics. Solving Systems of Three equations in Three Variables In order to solve systems of equations in three variables, known as three-by-three systems, the primary tool we will be using is called Gaussian elimination, named after the prolific German mathematician Karl Friedrich Gauss. While there is no definitive order in which operations are to be performed, there are specific guidelines as to what type of moves can be made. We may number the equations to keep track of the steps we apply. The goal is to eliminate one variable at a time to achieve upper triangular form, the ideal form for a three-by-three system because it allows for straightforward back-substitution to find a solution (x, y, z), which we call an ordered triple. A system in upper triangular form looks like the following: Ax + By + Cz = D Ey + Fz = G Hz = K The third equation can be solved for z, and then we back-substitute to find y and x. To write the system in upper triangular form, we can perform the following operations: 1. Interchange the order of any two equations. 2. Multiply both sides of an equation by a nonzero constant. 3. Add a nonzero multiple of one equation to another equation. SECTION 9.2 systems oF linear eQuations: three variaBles 775 The solution set to a three-by-three system is an ordered triple {(x, y, z)}. Graphically, the ordered triple defines the point that is the intersection of three planes in space. You can visualize such an intersection by imagining any corner in a rectangular room. A corner is defined by three planes: two adjoining walls and the floor (or ceiling). Any point where two walls and the floor meet represents the intersection of three planes. number of possible solutions Figure 2 and Figure 3 illustrate possible solution scenarios for three-by-three systems. • Systems that have a single solution are those which, after elimination, result in a solution set consisting of an ordered triple {(x, y, z)}. Graphically, the ordered triple defines a point that is the intersection of three planes in space. • Systems that have an infinite number of solutions are those which, after elimination, result in an expression that is always true, such as 0 = 0. Graphically, an infinite number of solutions represents a line or coincident plane that serves as the intersection of three planes in space. • Systems that have no solution are those that, after elimination, result in a statement that is a contradiction, such as 3 = 0. Graphically, a system with no solution is represented by three planes with no point in common. Figure 2 ( a)Three planes intersect at a single point, representing a three-by-three system with a single solution. ( b) Three planes intersect in a line, representing a three-by-three system with infinite solutions. (a) (b) (a) (b) (c) Figure 3 All three figures represent three-by-three systems with no solution. ( a) The three planes intersect with each other, but not at a common point. (b) Two of the planes are parallel and intersect with the third plane, but not with each other. ( c) All three planes are parallel, so there is no point of intersection. Example 1 Determining Whether an Ordered Triple Is a Solution to a System Determine whether the ordered triple (3, −2, 1) is a solution to the system. x + y + z = 2 6x − 4y + 5z = 31 5x + 2y + 2z = 13 Solution We will check each equation by substituting in the values of the ordered triple for x, y, and z. x + y + z = 2 (3) + (−2) + (1) = 2 True 6x − 4y + 5z = 31 6(3) − 4(−2) + 5(1) = 31 18 + 8 + 5 = 31 True 7 76 CHAPTER 9 systems oF eQuations and ineQualities 5x + 2y + 2z = 13 5(3) + 2(−2) + 2(1) = 13 15 − 4 + 2 = 13 True The ordered triple (3, −2, 1) is indeed a solution to the system. How To… Given a linear system of three equations, solve for three unknowns. 1. Pick any pair of equations and solve for one variable. 2. Pick another pair of equations and solve for the same variable. 3. You have created a system of two equations in two unknowns. Solve the resulting two-by-two system. 4. Back-substitute known variables into any one of the original equations and solve for the missing variable. Example 2 Solving a System of Three Equations in Three Variables by Elimination Find a solution to the following system: x − 2y + 3z = 9 −x + 3y − z = −6 2x − 5y + 5z = 17 (1) (2) (3) Solution There will always be several choices as to where to begin, but the most obvious first step here is to eliminate x by adding equations (1) and (2). x − 2y + 3z = 9 −x + 3y − z = −6 y + 2z = 3 (1) (2) (3) The second step is multiplying equation (1) by −2 and adding the result to equation (3). These two steps will eliminate the variable x. −2x + 4y − 6z = −18 2x − 5y + 5z = 17 − y − z = −1 (1) multiplied by − 2 (3) (5) In equations (4) and (5), we have created a new two-by-two system. We can solve for z by adding the two equations. y + 2z = 3 −4) (5) (6) Choosing one equation from each new system, we obtain the upper triangular form: x − 2y + 3z = 9 y + 2z = 3 z = 2 (1) (4) (6) Next, we back-substitute z = 2 into equation (4) and solve for y. y + 2(21 Finally, we can back-substitute z = 2 and y = −1 into equation (1). This will yield the solution for x. x − 2(−1) + 3(2 SECTION 9.2 systems oF linear eQuations: three variaBles 777 The solution is the ordered triple (1, −1, 2). See Figure 4. (1, −1, 22 Figure 4 Example 3 Solving a Real-World Problem Using a System of Three Equations in Three Variables In the problem posed at the beginning of the section, John invested his inheritance of $12,000 in three different funds: part in a money-market fund paying 3% interest annually; part in municipal bonds paying 4% annually; and the rest in mutual funds paying 7% annually. John invested $4,000 more in mutual funds than he invested in municipal bonds. The total interest earned in one year was $670. How much did he invest in each type of fund? Solution To solve this problem, we use all of the information given and set up three equations. First, we assign a variable to each of the three investment amounts: x = amount invested in money-market fund y = amount invested in municipal bonds z = amount invested in mutual funds The first equation indicates that the sum of the three principal amounts is $12,000. x + y + z = 12,000 We form the second equation according to the information that John invested $4,000 more in mutual funds than he invested in municipal bonds. z = y + 4,000 The third equation shows that the total amount of interest earned from each fund equals $670. Then, we write the three equations as a system. 0.03x + 0.04y + 0.07z = 670 x + y + z = 12,000 − y + z = 4,000 0.03x + 0.04y + 0.07z = 670 To make the calculations simpler, we can multiply the third equation by 100. Thus, x + y + z = 12,000 − y + z = 4,000 3x + 4y + 7z = 67,000 (1) (2) (3) Step 1. Interchange equation (2) and equation (3) so that the two equations with three variables will line up. x + y + z = 12,000 3x + 4y + 7z = 67,000 − y + z = 4,000 Step 2. Multiply equation (1) by −3 and add to equation (2). Write the result as row 2. x + y + z = 12,000 y + 4z = 31,000 − y + z = 4,000 7 78 CHAPTER 9 systems oF eQuations and ineQualities Step 3. Add equation (2) to equation (3) and write the result as equation (3). x + y + z = 12,000 y + 4z = 31,000 5z = 35,000 Step 4. Solve for z in equation (3). Back-substitute that value in equation (2) and solve for y. Then, back-substitute the values for z and y into equation (1) and solve for x. 5z = 35,000 z = 7,000 y + 4(7,000) = 31,000 y = 3,000 x + 3,000 + 7,000 = 12,000 x = 2,000 John invested $2,000 in a money-market fund, $3,000 in municipal bonds, and $7,000 in mutual funds. Try It #1 Solve the system of equations in three variables. 2x + y − 2z = −1 3x − 3y − z = 5 x − 2y + 3z = 6 Identifying Inconsistent Systems of equations Containing Three Variables Just as with systems of equations in two variables, we may come across an inconsistent system of equations in three variables, which means that it does not have a solution that satisfies all three equations. The equations could represent three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. The process of elimination will result in a false statement, such as 3 = 7 or some other contradiction. Example 4 Solving an Inconsistent System of Three Equations in Three Variables Solve the following system. x − 3y + z = 4 −x + 2y − 5z = 3 5x − 13y + 13z = 8 (1) (2) (3) Solution Looking at the coefficients of x, we can see that we can eliminate x by adding equation (1) to equation (2). x − 3y + z = 4 −x + 2y − 5z = 3 −y − 4z = 7 (1) (2) (4) Next, we multiply equation (1) by −5 and add it to equation (3). −5x + 15y − 5z = −20 5x − 13y + 13z = 8 (1) multiplied by −5 (3) 2y + 8z = −12 (5) Then, we multiply equation (4) by 2 and add it to equation (5). −2y −
8z = 14 (4) multiplied by 2 2y + 8z = −12 0 = 2 (5) SECTION 9.2 systems oF linear eQuations: three variaBles 779 The final equation 0 = 2 is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution. Analysis In this system, each plane intersects the other two, but not at the same location. Therefore, the system is inconsistent. Try It #2 Solve the system of three equations in three variables. x + y + z = 2 y − 3z = 1 2x + y + 5z = 0 expressing the Solution of a System of Dependent equations Containing Three Variables We know from working with systems of equations in two variables that a dependent system of equations has an infinite number of solutions. The same is true for dependent systems of equations in three variables. An infinite number of solutions can result from several situations. The three planes could be the same, so that a solution to one equation will be the solution to the other two equations. All three equations could be different but they intersect on a line, which has infinite solutions. Or two of the equations could be the same and intersect the third on a line. Example 5 Finding the Solution to a Dependent System of Equations Find the solution to the given system of three equations in three variables. 2x + y − 3z = 0 4x + 2y − 6z = 0 x − y + z = 0 (1) (2) (3) Solution First, we can multiply equation (1) by −2 and add it to equation (2). −4x − 2y + 6z = 0 equation (1) multiplied by −2 4x + 2y − 6z = 0 (2) 0 = 0 We do not need to proceed any further. The result we get is an identity, 0 = 0, which tells us that this system has an infinite number of solutions. There are other ways to begin to solve this system, such as multiplying equation (3) by −2, and adding it to equation (1). We then perform the same steps as above and find the same result, 0 = 0. When a system is dependent, we can find general expressions for the solutions. Adding equations (1) and (3), we have We then solve the resulting equation for z. 2x + y − 3z = 0 x − y + z = 0 3x − 2z = 0 3x − 2z = 0 3 __ z = x 2 We back-substitute the expression for z into one of the equations and solve for y. 7 80 CHAPTER 9 systems oF eQuations and ineQualities 3 x  = 0 2x + y − 3  __ 2 9 __ x = 0 2x + y − 2 9 __ x − 2x y = 2 5 __ y = x 2 3 5 x  . In this solution, x can be any real number. The values of y and z are dependent So the general solution is  x, __ __ x, 2 2 on the value selected for x. Analysis As shown in Figure 5, two of the planes are the same and they intersect the third plane on a line. The solution set is infinite, as all points along the intersection line will satisfy all three equations. x − y + z = 0 −4x − 2y + 6z = 0 4x + 2y − 6z = 0 Figure 5 Q & A… Does the generic solution to a dependent system always have to be written in terms of x? No, you can write the generic solution in terms of any of the variables, but it is common to write it in terms of x and if needed x and y. Try It #3 Solve the following system. x + y + z = 7 3x − 2y − z = 4 x + 6y + 5z = 24 Access these online resources for additional instruction and practice with systems of equations in three variables. • ex 1: System of Three equations with Three Unknowns Using elimination (http://openstaxcollege.org/l/systhree) • ex. 2: System of Three equations with Three Unknowns Using elimination (http://openstaxcollege.org/l/systhelim) SECTION 9.2 section exercises 781 9.2 SeCTIOn exeRCISeS VeRBAl 1. Can a linear system of three equations have exactly 2. If a given ordered triple solves the system of equations, is that solution unique? If so, explain why. If not, give an example where it is not unique. 4. Using the method of addition, is there only one way to solve the system? two solutions? Explain why or why not 3. If a given ordered triple does not solve the system of equations, is there no solution? If so, explain why. If not, give an example. 5. Can you explain whether there can be only one method to solve a linear system of equations? If yes, give an example of such a system of equations. If not, explain why not. AlGeBRAIC For the following exercises, determine whether the ordered triple given is the solution to the system of equations. 6. 2x − 6y + 6z = −12 x + 4y + 5z = −1 and (0, 1, −1) −x + 2y + 3z = −1 8. 6x − 7y + z = 2 −x − y + 3z = 4 and (4, 2, −6) 2x + y − z = 1 10. −x − y + 2z = 3 5x + 8y − 3z = 4 and (4, 1, −7) −x + 3y − 5z = −5 7. 6x − y + 3z = 6 3x + 5y + 2z = 0 and (3, −3, −5) x + y = 0 9 and (4, 4, −1) x − y + z = −1 For the following exercises, solve each system by substitution. 11. 3x − 4y + 2z = −15 2x + 4y + z = 16 2x + 3y + 5z = 20 14. 4x − 3y + 5z = 31 −x + 2y + 4z = 20 x + 5y − 2z = −29 12. 5x − 2y + 3z = 20 2x − 4y − 3z = −9 x + 6y − 8z = 21 15. 5x − 2y + 3z = 4 −4x + 6y − 7z = −1 3x + 2y − z = 4 13. 5x + 2y + 4z = 9 −3x + 2y + z = 10 4x − 3y + 5z = −3 16. 4x + 6y + 9z = 0 −5x + 2y − 6z = 3 7x − 4y + 3z = −3 For the following exercises, solve each system by Gaussian elimination. 17. 2x − y + 3z = 17 −5x + 4y − 2z = −46 2y + 5z = −7 20. 4x + 6y − 2z = 8 6x + 9y − 3z = 12 −2x − 3y + z = −4 23. x + y + z = 14 2y + 3z = −14 −16y − 24z = −112 18. 5x − 6y + 3z = 50 − x + 4y = 10 2x − z = 10 21. 2x + 3y − 4z = 5 −3x + 2y + z = 11 −x + 5y + 3z = 4 19. 2x + 3y − 6z = 1 −4x − 6y + 12z = −2 x + 2y + 5z = 10 22. 10x + 2y − 14z = 8 −x − 2y − 4z = −1 −12x − 6y + 6z = −12 24. 5x − 3y + 4z = −1 −4x + 2y − 3z = 0 −x + 5y + 7z = −11 25. x + y + z = 0 2x − y + 3z = 0 x − z = 0 7 82 CHAPTER 9 systems oF eQuations and ineQualities 26. 3x + 2y − 5z = 6 5x − 4y + 3z = −12 4x + 5y−2z = 15 27. x + y + z = 0 2x − y + 3z = 0 x − z = 1 29. 6x − 5y + 6z = 38 3 1 1 __ __ __ __ y − z = −74 −4x − 2 3 1 1 __ __ __ z = 0 y + x − 32. 4 4 2 1 2 1 __ ___ __ z = −2 x − y + 5 4 10 1 1 1 __ __ __ __ 35. − 1 __ 4 4 x − 5 __ − 1 __ 2 3 − 1 __ x − 1 __ 3 3 y + 5 __ 2 y + 5 __ 4 y + 1 __ 3 z = −5 z = 55 ___ 12 z = 5 __ 3 2 1 1 13 __ __ __ ___ z = − y + x − 30. 5 5 2 10 1 2 7 1 __ __ __ ___ 20 5 1 3 1 __ __ __ __ __ y + 1 __ 33. 4 __ z = 1 5 8 2 x − 3 __ y + 1 __ − 4 __ 5 4 3 x − 7 __ y + 1 __ − 2 __ 5 2 8 z = −5 z = −8 y + 1 ___ x + 1 ___ 36. 1 ___ 80 40 60 − 1 __ y − 1 __ x − 1 __ 4 3 2 y + 3 ___ x + 3 ___ 3 __ 12 8 16 z = 1 ___ 100 z = − 1 __ 5 z = 3 ___ 20 1 1 __ __ y − z = − 28. 3x − 2 2 4x + z = 3 3 5 __ __ − __ __ __ __ z = y − x − 31. − 4 2 3 4 1 1 1 __ __ __ __ __ __ __ __ y + 1 __ x − 1 __ 34. − 1 __ 8 3 3 6 z = − 23 ___ x − 7 __ y + 1 __ − 2 __ 3 3 3 8 y + 5 __ x − 5 __ − 1 __ z = 0 8 3 6 37. 0.1x − 0.2y + 0.3z = 2 0.5x − 0.1y + 0.4z = 8 0.7x − 0.2y + 0.3z = 8 38. 0.2x + 0.1y − 0.3z = 0.2 0.8x + 0.4y − 1.2z = 0.1 1.6x + 0.8y − 2.4z = 0.2 39. 1.1x + 0.7y − 3.1z = −1.79 2.1x + 0.5y − 1.6z = −0.13 0.5x + 0.4y − 0.5z = −0.07 40. 0.5x − 0.5y + 0.5z = 10 0.2x − 0.2y + 0.2z = 4 0.1x − 0.1y + 0.1z = 2 41. 0.1x + 0.2y + 0.3z = 0.37 0.1x − 0.2y − 0.3z = −0.27 0.5x − 0.1y − 0.3z = −0.03 42. 0.5x − 0.5y − 0.3z = 0.13 0.4x − 0.1y − 0.3z = 0.11 0.2x − 0.8y − 0.9z = −0.32 43. 0.5x + 0.2y − 0.3z = 1 0.4x − 0.6y + 0.7z = 0.8 0.3x − 0.1y − 0.9z = 0.6 44. 0.3x + 0.3y + 0.5z = 0.6 0.4x + 0.4y + 0.4z = 1.8 0.4x + 0.2y + 0.1z = 1.6 45. 0.8x + 0.8y + 0.8z = 2.4 0.3x − 0.5y + 0.2z = 0 0.1x + 0.2y + 0.3z = 0.6 exTenSIOnS For the following exercises, solve the system for x, y, and z. 46. x + y + z = 3 47. 5x − 3y − z + 1 _____ 2 = 1 __ 2 + 2z = −3 6x + y − 9 _____ 2 x + 8 _____ 2 − 4y + z = 4 48. x + 4 _____ 7 x − 2 _____ 4 x + 6 _____ 3 − + − y − 1 _____ 6 y + 1 _____ 8 y + 2 _____ 3 + − + z + 2 _____ 3 z + 8 _____ 12 z + 4 _____ _____ 2 x − 2 _____ 3 + + y − 3 _____ 2 y + 4 _____ 3 + z + 1 _____ 2 + z − 3 _____ 3 = 0 = 2 __ 3 + + 49. x − 3 _____ 6 x + 2 _____ 4 x + 6 _____ 2 y + 2 _____ 2 y − 5 _____ 2 y − 3 _____ 2 − − z − 3 _____ 3 z + 4 _____ 50. x − 1 ____ 3 + y + 3 _____ 4 + z + 2 _____ 6 = 1 4x + 3y − 2z = 11 0.02x + 0.015y − 0.01z = 0.065 SECTION 9.2 section exercises 783 ReAl-WORlD APPlICATIOnS 51. Three even numbers sum up to 108. The smaller 3 _ the is half the larger and the middle number is 4 larger. What are the three numbers? 52. Three numbers sum up to 147. The smallest number is half the middle number, which is half the largest number. What are the three numbers? 53. At a family reunion, there were only blood relatives, consisting of children, parents, and grandparents, in attendance. There were 400 people total. There were twice as many parents as grandparents, and 50 more children than parents. How many children, parents, and grandparents were in attendance? 55. Your roommate, Sarah, offered to buy groceries for you and your other roommate. The total bill was $82. She forgot to save the individual receipts but remembered that your groceries were $0.05 cheaper than half of her groceries, and that your other roommate’s groceries were $2.10 more than your groceries. How much was each of your share of the groceries? 57. Three coworkers work for the same employer. Their jobs are warehouse manager, office manager, and truck driver. The sum of the annual salaries of the warehouse manager and office manager is $82,000. The office manager makes $4,000 more than the truck driver annually. The annual salaries of the warehouse manager and the truck driver total $78,000. What is the annual salary of each of the co-workers? 59. A local band sells out for their concert. They sell all 1,175 tickets for a total purse of $28,112.50. The tickets were priced at $20 for student tickets, $22.50 for children, and $29 for adult tickets. If the band sold twice as many adult as children tickets, how many of each type was sold? 61. Last year, at Haven’s Pond Car Dealership, for a particular model of BMW, Jeep, and Toyota, one could purchase all three cars for a total of $140,000. This year, due to inflation, the same cars would cost $151,830. The cost of the BMW increased by 8%, the Jeep by 5%, and the Toyota by 12%. If the price of last year’s Jeep was $7,000 less than the price of last year’s BMW, what was the price of each of the three cars last y
ear? 54. An animal shelter has a total of 350 animals comprised of cats, dogs, and rabbits. If the number of rabbits is 5 less than one-half the number of cats, and there are 20 more cats than dogs, how many of each animal are at the shelter? 56. Your roommate, John, offered to buy household supplies for you and your other roommate. You live near the border of three states, each of which has a different sales tax. The total amount of money spent was $100.75. Your supplies were bought with 5% tax, John’s with 8% tax, and your third roommate’s with 9% sales tax. The total amount of money spent without taxes is $93.50. If your supplies before tax were $1 more than half of what your third roommate’s supplies were before tax, how much did each of you spend? Give your answer both with and without taxes. 58. At a carnival, $2,914.25 in receipts were taken at the end of the day. The cost of a child’s ticket was $20.50, an adult ticket was $29.75, and a senior citizen ticket was $15.25. There were twice as many senior citizens as adults in attendance, and 20 more children than senior citizens. How many children, adult, and senior citizen tickets were sold? 60. In a bag, a child has 325 coins worth $19.50. There were three types of coins: pennies, nickels, and dimes. If the bag contained the same number of nickels as dimes, how many of each type of coin was in the bag? 62. A recent college graduate took advantage of his business education and invested in three investments immediately after graduating. He invested $80,500 into three accounts, one that paid 4% simple interest, one that paid 4% simple interest, one that paid 3 1 __ % simple interest, and one that 8 paid 2 1 __ 2 at the end of one year. If the amount of the money invested in the second account was four times the amount invested in the third account, how much was invested in each account? % simple interest. He earned $2,670 interest 7 84 CHAPTER 9 systems oF eQuations and ineQualities 63. You inherit one million dollars. You invest it all in three accounts for one year. The first account pays 3% compounded annually, the second account pays 4% compounded annually, and the third account pays 2% compounded annually. After one year, you earn $34,000 in interest. If you invest four times the money into the account that pays 3% compared to 2%, how much did you invest in each account? 64. You inherit one hundred thousand dollars. You invest it all in three accounts for one year. The first account pays 4% compounded annually, the second account pays 3% compounded annually, and the third account pays 2% compounded annually. After one year, you earn $3,650 in interest. If you invest five times the money in the account that pays 4% compared to 3%, how much did you invest in each account? 65. The top three countries in oil consumption in a certain year are as follows: the United States, Japan, and China. In millions of barrels per day, the three top countries consumed 39.8% of the world’s consumed oil. The United States consumed 0.7% more than four times China’s consumption. The United States consumed 5% more than triple Japan’s consumption. What percent of the world oil consumption did the United States, Japan, and China consume?[28] 66. The top three countries in oil production in the same year are Saudi Arabia, the United States, and Russia. In millions of barrels per day, the top three countries produced 31.4% of the world’s produced oil. Saudi Arabia and the United States combined for 22.1% of the world’s production, and Saudi Arabia produced 2% more oil than Russia. What percent of the world oil production did Saudi Arabia, the United States, and Russia produce?[29] 67. The top three sources of oil imports for the United States in the same year were Saudi Arabia, Mexico, and Canada. The three top countries accounted for 47% of oil imports. The United States imported 1.8% more from Saudi Arabia than they did from Mexico, and 1.7% more from Saudi Arabia than they did from Canada. What percent of the United States oil imports were from these three countries?[30] 68. The top three oil producers in the United States in a certain year are the Gulf of Mexico, Texas, and Alaska. The three regions were responsible for 64% of the United States oil production. The Gulf of Mexico and Texas combined for 47% of oil production. Texas produced 3% more than Alaska. What percent of United States oil production came from these regions?[31] 69. At one time, in the United States, 398 species of 70. Meat consumption in the United States can be animals were on the endangered species list. The top groups were mammals, birds, and fish, which comprised 55% of the endangered species. Birds accounted for 0.7% more than fish, and fish accounted for 1.5% more than mammals. What percent of the endangered species came from mammals, birds, and fish? broken into three categories: red meat, poultry, and fish. If fish makes up 4% less than one-quarter of poultry consumption, and red meat consumption is 18.2% higher than poultry consumption, what are the percentages of meat consumption?[32] 28 “Oil reserves, production and consumption in 2001,” accessed April 6, 2014, http://scaruffi.com/politics/oil.html. 29 “Oil reserves, production and consumption in 2001,” accessed April 6, 2014, http://scaruffi.com/politics/oil.html. 30 “Oil reserves, production and consumption in 2001,” accessed April 6, 2014, http://scaruffi.com/politics/oil.html. 31 “USA: The coming global oil crisis,” accessed April 6, 2014, http://www.oilcrisis.com/us/. 32 “The United States Meat Industry at a Glance,” accessed April 6, 2014, http://www.meatami.com/ht/d/sp/i/47465/pid/47465. SECTION 9.3 systems oF nonlinear eQuations and ineQualities: two variaBles 785 leARnInG OBjeCTIVeS In this section, you will: • Solve a system of nonlinear equations using substitution. • Solve a system of nonlinear equations using elimination. • Graph a nonlinear inequality. • Graph a system of nonlinear inequalities. 9.3 SYSTeMS OF nOnlIneAR eQUATIOnS AnD IneQUAlITIeS: TWO VARIABleS Halley’s Comet (Figure 1) orbits the sun about once every 75 years. Its path can be considered to be a very elongated ellipse. Other comets follow similar paths in space. These orbital paths can be studied using systems of equations. These systems, however, are different from the ones we considered in the previous section because the equations are not linear. In this section, we will consider the intersection of a parabola and a line, a circle and a line, and a circle and an ellipse. The methods for solving systems of nonlinear equations are similar to those for linear equations. Figure 1 Halley’s Comet (credit: "nASA Blueshift"/Flickr) Solving a System of nonlinear equations Using Substitution A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear. Recall that a linear equation can take the form Ax + By + C = 0. Any equation that cannot be written in this form in nonlinear. The substitution method we used for linear systems is the same method we will use for nonlinear systems. We solve one equation for one variable and then substitute the result into the second equation to solve for another variable, and so on. There is, however, a variation in the possible outcomes. Intersection of a Parabola and a Line There are three possible types of solutions for a system of nonlinear equations involving a parabola and a line. possible types of solutions for points of intersection of a parabola and a line Figure 2 illustrates possible solution sets for a system of equations involving a parabola and a line. • No solution. The line will never intersect the parabola. • One solution. The line is tangent to the parabola and intersects the parabola at exactly one point. • Two solutions. The line crosses on the inside of the parabola and intersects the parabola at two points. No solutions y 42 6 8 x –8 –6 –4 6 4 2 –2 –2 –4 –6 –8 (a) One solutions Two solutions 321 4 5 x –2 y 1 –1 –1 –2 –3 –4 –5 –6 (b) Figure 2 y 6 5 4 3 2 1 –3 –2 –1 –1 21 3 4 (c) x 7 86 CHAPTER 9 systems oF eQuations and ineQualities How To… Given a system of equations containing a line and a parabola, find the solution. 1. Solve the linear equation for one of the variables. 2. Substitute the expression obtained in step one into the parabola equation. 3. Solve for the remaining variable. 4. Check your solutions in both equations. Example 1 Solving a System of Nonlinear Equations Representing a Parabola and a Line Solve the system of equations. x − y = −1 y = x 2 + 1 Solution Solve the first equation for x and then substitute the resulting expression into the second equation. Expand the equation and set it equal to zero. x − y = −y − 1) 2 + 1 Solve for x. Substitute expression for x. y = (y − 1) 2 = ( y 2 − 2y + 1) + 1 = y 2 − 2y + 2 0 = y 2 −3y + 2 = (y − 2)(y − 1) Solving for y gives y = 2 and y = 1. Next, substitute each value for y into the first equation to solve for x. Always substitute the value into the linear equation to check for extraneous solutions. x − y = −1 x − (2) = −1 x = 1 x − (1) = −1 x = 0 The solutions are (1, 2) and (0, 1), which can be verified by substituting these (x, y) values into both of the original equations. See Figure 31 (1, 2) (0, 1) 1 2 3 4 5 6 x –6 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 –6 Figure 3 SECTION 9.3 systems oF nonlinear eQuations and ineQualities: two variaBles 787 Q & A… Could we have substituted values for y into the second equation to solve for x in Example 1? Yes, but because x is squared in the second equation this could give us extraneous solutions for x. For y = 1 This gives us the same value as in the solution. For Notice that −1 is an extraneous solution. Try It #1 Solve the given system of equations by substitution. 3x − y = −2 2 x 2 − y = 0 Intersection of a Circle and a Line Just as with a parabola and a line, there are three possible outcomes when solving a system of equations represen
ting a circle and a line. possible types of solutions for the points of intersection of a circle and a line Figure 4 illustrates possible solution sets for a system of equations involving a circle and a line. • No solution. The line does not intersect the circle. • One solution. The line is tangent to the circle and intersects the circle at exactly one point. • Two solutions. The line crosses the circle and intersects it at two points. No solutions One solution Two solutions Figure 4 How To… Given a system of equations containing a line and a circle, find the solution. 1. Solve the linear equation for one of the variables. 2. Substitute the expression obtained in step one into the equation for the circle. 3. Solve for the remaining variable. 4. Check your solutions in both equations. 7 88 CHAPTER 9 systems oF eQuations and ineQualities Example 2 Finding the Intersection of a Circle and a Line by Substitution Find the intersection of the given circle and the given line by substitution. x 2 + y 2 = 5 y = 3x − 5 Solution One of the equations has already been solved for y. We will substitute y = 3x − 5 into the equation for the circle. Now, we factor and solve for x. x 2 + (3x − 530x + 25 = 5 10 x 2 −30x + 20 = 0 10( x 2 − 3x + 2) = 0 10(x − 2)(x − 1) = 0 x = 2 x = 1 Substitute the two x-values into the original linear equation to solve for y. y = 3(2)−5 = 1 y = 3(1)−5 = −2 The line intersects the circle at (2, 1) and (1, −2), which can be verified by substituting these (x, y) values into both of the original equations. See Figure 5. y 5 4 3 2 1 x2 + y2 = 5 –5 –4 –3 –2 0 –1 –1 –2 –3 –4 –5 y = 3x − 5 (2, 1) 21 3 4 5 x (1, –2) Figure 5 x 2 + y 2 = 10 x − 3y = −10 Try It #2 Solve the system of nonlinear equations. Solving a System of nonlinear equations Using elimination We have seen that substitution is often the preferred method when a system of equations includes a linear equation and a nonlinear equation. However, when both equations in the system have like variables of the second degree, solving them using elimination by addition is often easier than substitution. Generally, elimination is a far simpler method when the system involves only two equations in two variables (a two-by-two system), rather than a three-by-three system, as there are fewer steps. As an example, we will investigate the possible types of solutions when solving a system of equations representing a circle and an ellipse. SECTION 9.3 systems oF nonlinear eQuations and ineQualities: two variaBles 789 possible types of solutions for the points of intersection of a circle and an ellipse Figure 6 illustrates possible solution sets for a system of equations involving a circle and an ellipse. • No solution. The circle and ellipse do not intersect. One shape is inside the other or the circle and the ellipse are a distance away from the other. • One solution. The circle and ellipse are tangent to each other, and intersect at exactly one point. • Two solutions. The circle and the ellipse intersect at two points. • Three solutions. The circle and the ellipse intersect at three points. • Four solutions. The circle and the ellipse intersect at four points. No solution One solution Two solutions Three solutions Four solutions Figure 6 Example 3 Solving a System of Nonlinear Equations Representing a Circle and an Ellipse Solve the system of nonlinear equations. x 2 + y 2 = 26 3 x 2 + 25 y 2 = 100 (1) (2) Solution Let’s begin by multiplying equation (1) by −3, and adding it to equation (2). (−3)( x 2 + y 2 ) = (−3)(26) −3 x 2 − 3 y 2 = − 78 3 x 2 + 25 y 2 = 100 22 y 2 = 22 After we add the two equations together, we solve for y Substitute y = ± 1 into one of the equations and solve for x. x 2 + (1) 2 = 26 x 2 + 1 = 26 x 2 = 25 x = ± √ — 25 = ± 5 x 2 + (−1) 2 = 26 x 2 + 1 = 26 x 2 = 25 = ± 5 There are four solutions: (5, 1), (−5, 1), (5, −1), and (−5, −1). See Figure 7. (−5, 1) –5 –6 (−5, −1) –4 –3 –1 –1 –2 –3 –4 –5 –6 21 3 4 (5, 1) x 5 6 (5, −1) Figure 7 7 90 CHAPTER 9 systems oF eQuations and ineQualities Try It #3 Find the solution set for the given system of nonlinear equations. 4 x 2 + y 2 = 13 x 2 + y 2 = 10 Graphing a nonlinear Inequality All of the equations in the systems that we have encountered so far have involved equalities, but we may also encounter systems that involve inequalities. We have already learned to graph linear inequalities by graphing the corresponding equation, and then shading the region represented by the inequality symbol. Now, we will follow similar steps to graph a nonlinear inequality so that we can learn to solve systems of nonlinear inequalities. A nonlinear inequality is an inequality containing a nonlinear expression. Graphing a nonlinear inequality is much like graphing a linear inequality. Recall that when the inequality is greater than, y > a, or less than, y < a, the graph is drawn with a dashed line. When the inequality is greater than or equal to, y ≥ a, or less than or equal to, y ≤ a, the graph is drawn with a solid line. The graphs will create regions in the plane, and we will test each region for a solution. If one point in the region works, the whole region works. That is the region we shade. See Figure 8. y > x2 − 4 21 3 4 5 6 x –6 –5 –4 –3 –2 –6 –5 –4 –3 –2 y 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 (a) y ≥ x2 − 4 21 3 4 5 6 x –6 –5 –4 –3 –2 y 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 (b) y 5 4 3 2 1 0 –1 –1 –2 –3 –4 –5 (c) y < x2 − 4 21 3 4 5 6 x –6 –5 –4 –3 –2 y ≤ x2 − 4 21 1 –1 –2 –3 –4 –5 (d) Figure 8 (a) an example of y > a; (b) an example of y ≥ a; (c) an example of y < a; (d) an example of y ≤ a How To… Given an inequality bounded by a parabola, sketch a graph. 1. Graph the parabola as if it were an equation. This is the boundary for the region that is the solution set. 2. If the boundary is included in the region (the operator is ≤ or ≥), the parabola is graphed as a solid line. 3. If the boundary is not included in the region (the operator is < or >), the parabola is graphed as a dashed line. 4. Test a point in one of the regions to determine whether it satisfies the inequality statement. If the statement is true, the solution set is the region including the point. If the statement is false, the solution set is the region on the other side of the boundary line. 5. Shade the region representing the solution set. Graphing an Inequality for a Parabola Example 4 Graph the inequality y > x 2 + 1. Solution First, graph the corresponding equation y = x 2 + 1. Since y > x 2 + 1 has a greater than symbol, we draw the graph with a dashed line. Then we choose points to test both inside and outside the parabola. Let’s test the points (0, 2) and (2, 0). One point is clearly inside the parabola and the other point is clearly outside0) 2 + 1 2 > 1 True 0 > (2) 2 + 1 0 > 5 False SECTION 9.3 systems oF nonlinear eQuations and ineQualities: two variaBles 791 The graph is shown in Figure 9. We can see that the solution set consists of all points inside the parabola, but not on the graph itself. y 6 5 4 3 2 1 –6 –5 –4 –3 –2 –1 –1 (0, 2) (2, 0) 4 3 21 x 5 6 Figure 9 Graphing a System of nonlinear Inequalities Now that we have learned to graph nonlinear inequalities, we can learn how to graph systems of nonlinear inequalities. A system of nonlinear inequalities is a system of two or more inequalities in two or more variables containing at least one inequality that is not linear. Graphing a system of nonlinear inequalities is similar to graphing a system of linear inequalities. The difference is that our graph may result in more shaded regions that represent a solution than we find in a system of linear inequalities. The solution to a nonlinear system of inequalities is the region of the graph where the shaded regions of the graph of each inequality overlap, or where the regions intersect, called the feasible region. How To… Given a system of nonlinear inequalities, sketch a graph. 1. Find the intersection points by solving the corresponding system of nonlinear equations. 2. Graph the nonlinear equations. 3. Find the shaded regions of each inequality. 4. Identify the feasible region as the intersection of the shaded regions of each inequality or the set of points common to each inequality. Example 5 Graphing a System of Inequalities Graph the given system of inequalities ≤ 12 Solution These two equations are clearly parabolas. We can find the points of intersection by the elimination process: Add both equations and the variable y will be eliminated. Then we solve for x = 12 3 x 2 = 12 x 2 = 4 x = ± 2 Substitute the x-values into one of the equations and solve for y. x 2 − y = 0 (2 (−2 92 CHAPTER 9 systems oF eQuations and ineQualities The two points of intersection are (2, 4) and (−2, 4). Notice that the equations can be rewritten as follows ≤ 12 y ≤ − 2 x 2 + 12 Graph each inequality. See Figure 10. The feasible region is the region between the two equations bounded by 2 x 2 + y ≤ 12 on the top and x 2 − y ≤ 0 on the bottom. y 14 12 10 8 6 4 2 (–2, 4) (2, 4) –– 12 10 –8 –6 –4 –2 2– 42 6 8 10 12 Figure 10 Try It #4 Graph the given system of inequalities. Access these online resources for additional instruction and practice with nonlinear equations. • Solve a System of nonlinear equations Using Substitution (http://openstaxcollege.org/l/nonlinsub) • Solve a System of nonlinear equations Using elimination (http://openstaxcollege.org/l/nonlinelim) SECTION 9.3 section exercises 793 9.3 SeCTIOn exeRCISeS VeRBAl 1. Explain whether a system of two nonlinear equations can have exactly two solutions. What about exactly three? If not, explain why not. If so, give an example of such a system, in graph form, and explain why your choice gives two or three answers. 3. When you graph a system of inequalities, will there always be a feasible region? If so, explain why. If not, give an example of a graph of inequalities that does not have a feasible region. Why does it not have a feasible region? 5. If you perform your break-even analysis and there is more tha
n one solution, explain how you would determine which x-values are profit and which are not. AlGeBRAIC 2. When graphing an inequality, explain why we only need to test one point to determine whether an entire region is the solution? 4. If you graph a revenue and cost function, explain how to determine in what regions there is profit. For the following exercises, solve the system of nonlinear equations using substitution. 6 10 For the following exercises, solve the system of nonlinear equations using elimination. 11. 4 x 2 − 9 y 2 = 36 4 x 2 + 9 y 2 = 36 14 12. x 2 + y 2 = 25 x 2 − y 2 = 1 134 y 2 = 25x − 10 15. x 2 + y 2 + = 2500 y = 2 x 2 1 _ 16 For the following exercises, use any method to solve the system of nonlinear equations. 16. −2 x 2 + y = −5 6x − y = 9 19 22 17. − x 2 + y = 2 −x + y = 2 20 23 For the following exercises, use any method to solve the nonlinear system. 24 27 30. x 2 + y 2 = 25 x 2 − y 2 = 36 25 28. − x 2 + y = 2 −4x + y = −1 31 18. x 2 + y 2 = 1 y = 20 x 2 −1 21. 9 x 2 + 25 y 2 = 225 (x − 6) 2 + y 2 = 1 26 29. − x 2 + y = 2 2y = − x 32. 16 x 2 − 9 y 2 + 144 = 0 y 2 + x 2 = 16 7 94 CHAPTER 9 systems oF eQuations and ineQualities 33. 3 x 2 − y 2 = 12 (x − 1) 2 + y 2 = 1 34. 3 x 2 − y 2 = 12 (x − 1) 2 + y 2 = 4 36. x 2 − y 2 − 6x − 4y − 11 = 0 − x 2 + y 2 = 5 37. x 2 + y 2 − 6y = 7 x 2 + y = 1 35. 3 x 2 − y 2 = 12 x 2 + y 2 = 16 38. x 2 + y 2 = 6 xy = 1 GRAPHICAl For the following exercises, graph the inequality. 39. x 2 + y < 9 40. x 2 + y 2 < 4 For the following exercises, graph the system of inequalities. Label all points of intersection. 41. x 2 + y < 1 y > 2x 44. x 2 − y 2 > −4 x 2 + y 2 < 12 42. x 2 + y < −5 y > 5x + 10 45. x 2 + 3 y 2 > 16 3 x 2 − y 2 < 1 exTenSIOnS For the following exercises, graph the inequality. 46. y ≥ e x y ≤ ln(x) + 5 47. y ≤ −log(x) y ≤ e x 43. x 2 + y 2 < 25 3 x 2 − y 2 > 12 For the following exercises, find the solutions to the nonlinear equations with two variables. 48. = 24 + 1 __ y 2 + 4 = 0 4 __ x 2 − 2 __ y 2 5 __ x 2 49. 6 __ x 2 1 __ x 2 − 1 __ y 2 − 6 __ y 2 = 8 = 1 __ 8 50. x 2 − xy + y 2 −2 = 0 x + 3y = 4 51. x 2 − xy − 2 y 2 − 52. x 2 + 4xy − TeCHnOlOGY For the following exercises, solve the system of inequalities. Use a calculator to graph the system to confirm the answer. 53. xy < 1 y > √ — x 54. x 2 + y < 3 y > 2x ReAl-WORlD APPlICATIOnS For the following exercises, construct a system of nonlinear equations to describe the given behavior, then solve for the requested solutions. 55. Two numbers add up to 300. One number is twice the square of the other number. What are the numbers? 56. The squares of two numbers add to 360. The second number is half the value of the first number squared. What are the numbers? 57. A laptop company has discovered their cost and 58. A cell phone company has the following cost and revenue functions for each day: C(x) = 3 x 2 − 10x + 200 and R(x) = −2 x 2 + 100x + 50. If they want to make a profit, what is the range of laptops per day that they should produce? Round to the nearest number which would generate profit. revenue functions: C(x) = 8 x 2 − 600x + 21,500 and R(x) = −3 x 2 + 480x. What is the range of cell phones they should produce each day so there is profit? round to the nearest number that generates profit. SECTION 9.4 partial Fractions 795 leARnInG OBjeCTIVeS In this section, you will: • Decompose P (x )Q (x ), where Q (x ) has only nonrepeated linear factors. • Decompose P (x )Q (x ), where Q (x ) has repeated linear factors. • Decompose P (x )Q (x ), where Q (x ) has a nonrepeated irreducible quadratic factor. • Decompose P (x )Q (x ), where Q (x ) has a repeated irreducible quadratic factor. 9.4 PARTIAl FRACTIOnS Earlier in this chapter, we studied systems of two equations in two variables, systems of three equations in three variables, and nonlinear systems. Here we introduce another way that systems of equations can be utilized—the decomposition of rational expressions. Fractions can be complicated; adding a variable in the denominator makes them even more so. The methods studied in this section will help simplify the concept of a rational expression. Decomposing Where Q(x ) Has Only nonrepeated linear Factors P(x ) ____ Q(x ) Recall the algebra regarding adding and subtracting rational expressions. These operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression. In this section, we will look at partial fraction decomposition, which is the undoing of the procedure to add or subtract rational expressions. In other words, it is a return from the single simplified rational expression to the original expressions, called the partial fractions. For example, suppose we add the following fractions: We would first need to find a common denominator, (x + 2)(x − 3). Next, we would write each expression with this common denominator and find the sum of the terms. 2 _____ x − 3 + −1 _____ x + 2 x + 2 2 _____ _____  x + 2 x − 3  + −1 _____ x + 2  x − 3 _____ x − 3  = 2x + 4 − x + 3 ____________ = (x + 2)(x − 3) x + 7 ________ x 2 − x − 6 Partial fraction decomposition is the reverse of this procedure. We would start with the solution and rewrite (decompose) it as the sum of two fractions. x + 7 _______ x 2 − x − 6 Simplified sum = 2 ____ x − 3 + −1 _____ x + 2 Partial fraction decomposition We will investigate rational expressions with linear factors and quadratic factors in the denominator where the degree of the numerator is less than the degree of the denominator. Regardless of the type of expression we are decomposing, the first and most important thing to do is factor the denominator. When the denominator of the simplified expression contains distinct linear factors, it is likely that each of the original rational expressions, which were added or subtracted, had one of the linear factors as the denominator. In other words, using the example above, the factors of x 2 − x − 6 are (x − 3)(x + 2), the denominators of the decomposed rational expression. So we will rewrite the simplified form as the sum of individual fractions and use a variable for each numerator. Then, we will solve for each numerator using one of several methods available for partial fraction decomposition. 7 96 CHAPTER 9 systems oF eQuations and ineQualities partial fraction decomposition of : Q(x) has nonrepeated linear factors P(x) ____ Q(x) The partial fraction decomposition of P(x) is less than the degree of Q(x) is P(x) ____ Q(x) when Q(x) has nonrepeated linear factors and the degree of P(x) ____ Q(x) = A 1 __ +  a 1 x + b 1  A 2 __ +  a 2 x + b 2  A 3 __  __ .  a n x + b n  How To… Given a rational expression with distinct linear factors in the denominator, decompose it. 1. Use a variable for the original numerators, usually A, B, or C, depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use A n for each numerator P(x) ____ Q(x) = A 1 __ +  a 1 x + b 1  A 2 __  __ .  a n x + b n  2. Multiply both sides of the equation by the common denominator to eliminate fractions. 3. Expand the right side of the equation and collect like terms. 4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators. Example 1 Decomposing a Rational Function with Distinct Linear Factors Decompose the given rational expression with distinct linear factors. 3x ___________ (x + 2)(x − 1) Solution We will separate the denominator factors and give each numerator a symbolic label, like A, B, or C. = A ______ (x + 2) 3x ___________ (x + 2)(x − 1) + B _____ (x − 1) Multiply both sides of the equation by the common denominator to eliminate the fractions: (x + 2)(x − 1)  3x __________ (x + 2)(x − 1)  =  (x + 2) (x − 1)  A ______  (x + 2)  + (x + 2)  (x − 1)  B _____   (x − 1) The resulting equation is Expand the right side of the equation and collect like terms. 3x = A(x − 1) + B(x + 2) 3x = Ax − A + Bx + 2B 3x = (A + B)x − A + 2B Set up a system of equations associating corresponding coefficients. Add the two equations and solve for B. 3 = A + B 0 = −A + 2B 3 = A + B 0 = −A + 2B 3 = 0 + 3B 1 = B Substitute B = 1 into one of the original equations in the system. 3 = A + 1 2 = A Thus, the partial fraction decomposition is 3x __________ (x + 2)(x − 1) = 2 ______ (x + 2) + 1 _____ (x − 1) SECTION 9.4 partial Fractions 797 Another method to use to solve for A or B is by considering the equation that resulted from eliminating the fractions and substituting a value for x that will make either the A- or B-term equal 0. If we let x = 1, the A-term becomes 0 and we can simply solve for B. 3x = A(x − 1) + B(x + 2) 3(1) = A[(1) − 1] + B[(1) + 2] 3 = 0 + 3B 1 = B Next, either substitute B = 1 into the equation and solve for A, or make the B-term 0 by substituting x = −2 into the equation. 3x = A(x − 1) + B(x + 2) 3(−2) = A[(−2) − 1] + B[(−2) + 2] −6 = −3A + 0 −6 ___ −3 = A 2 = A We obtain the same values for A and B using either method, so the decompositions are the same using either method. 3x ___________ (x + 2)(x − 1) = 2 ______ (x + 2) + 1 _____ (x − 1) Although this method is not seen very often in textbooks, we present it here as an alternative that may make some partial fraction decompositions easier. It is known as the Heaviside method, named after Charles Heaviside, a pioneer in the study of electronics. Try It #1 Find the partial fraction decomposition of the following expression. x ____________ (x − 3)(x − 2) Decomposing Where Q(x ) Has Repeated linear Factors P(x ) ____ Q(x ) Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers. partial fraction decomposition of : Q(x) has repeated linear factors P(x) ____ Q(x) The partial fraction
decomposition of when Q(x) has repeated linear factor occurring n times and the degree of P(x) ____ Q(x) P(x) is less than the degree of Q(x), is A 1 _______ + (ax + b) P(x) ____ Q(x) = A 2 ________ + (ax + b) 2 A 3 ________ (ax + b) 3 + … + A n ________ (ax + b) n Write the denominator powers in increasing order. How To… Given a rational expression with repeated linear factors, decompose it. 1. Use a variable like A, B, or C for the numerators and account for increasing powers of the denominators. P(x) ____ Q(x) = A 1 _______ + (ax + b) A 2 ________ (ax + b) 2 + … + A n ________ (ax + b) n 2. Multiply both sides of the equation by the common denominator to eliminate fractions. 3. Expand the right side of the equation and collect like terms. 4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators. 7 98 CHAPTER 9 systems oF eQuations and ineQualities Example 2 Decomposing with Repeated Linear Factors Decompose the given rational expression with repeated linear factors. − x 2 + 2x + 4 ___________ x 3 − 4 x 2 + 4x Solution The denominator factors are x (x − 2) 2 . To allow for the repeated factor of (x − 2), the decomposition will include three denominators: x, (x − 2), and (x − 2) 2 . Thus, Next, we multiply both sides by the common denominator. − x 2 + 2x + 4 __________ x 3 − 4 x 2 + 4x = A __ x + B ______ (x − 2) + C _______ (x − 2) 2 x (x − 2) 2  − x 2 + 2x + 4  =  A __ x + B ______ __________ _______ (x − 2) (x − 2) 2 x (x − 2) 2 − x 2 + 2x + 4 = A (x − 2) 2 + Bx(x−2) + Cx + C  x (x − 2) 2 On the right side of the equation, we expand and collect like terms. − x 2 + 2x + 4 = A( x 2 − 4x + 4) + B( x 2 − 2x) + Cx = A x 2 − 4Ax + 4A + B x 2 − 2Bx + Cx = (A + B) x 2 + ( −4A − 2B + C)x + 4A Next, we compare the coefficients of both sides. This will give the system of equations in three variables: − x 2 + 2x + 4 = (A + B) x 2 + (−4A − 2B + C)x + 4A (1) A + B = −1 −4A − 2B + C = 2 4A = 4 (2) (3) Solving for A, we have Substitute A = 1 into equation (1). 4A = 4 A = 1 A + B = −1 (1) + B = −1 B = −2 Then, to solve for C, substitute the values for A and B into equation (2). −4A − 2B + C = 2 −4(1) − 2(−2) + C = 2 − Thus, − x 2 + 2x + 4 __________ x 3 − 4 x 2 + 4x = 1 __ x − 2 ______ (x − 2) + 2 _______ (x − 2) 2 Try It #2 Find the partial fraction decomposition of the expression with repeated linear factors. 6x − 11 _______ (x − 1) 2 SECTION 9.4 partial Fractions 799 Decomposing Where Q(x ) Has a nonrepeated Irreducible Quadratic Factor P(x ) ____ Q(x ) So far, we have performed partial fraction decomposition with expressions that have had linear factors in the denominator, and we applied numerators A, B, or C representing constants. Now we will look at an example where one of the factors in the denominator is a quadratic expression that does not factor. This is referred to as an irreducible quadratic factor. In cases like this, we use a linear numerator such as Ax + B, Bx + C, etc. decomposition of Q(x) has a nonrepeated irreducible quadratic factor P(x) _____ Q(x): The partial fraction decomposition of such that Q(x) has a nonrepeated irreducible quadratic factor and the P(x) ____ Q(x) degree of P(x) is less than the degree of Q(x), is written as P(x) ____ Q(x) = A 1 x + B 1 ______________ ______________ ( ______________ ( ) The decomposition may contain more rational expressions if there are linear factors. Each linear factor will have a different constant numerator: A, B, C, and so on. How To… Given a rational expression where the factors of the denominator are distinct, irreducible quadratic factors, decompose it. 1. Use variables such as A, B, or C for the constant numerators over linear factors, and linear expressions such as A 1 P(x) ____ Q(x , etc., for the numerators of each quadratic factor in the denominator. A 1 x + B 1 _____________ ( . Multiply both sides of the equation by the common denominator to eliminate fractions. 3. Expand the right side of the equation and collect like terms. 4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of A n x + B n _____________ ( _____________ ( ______ ax + b + … + + + equations to solve for the numerators. Example 3 Decomposing When Q(x) Contains a Nonrepeated Irreducible Quadratic Factor P(x) ____ Q(x) Find a partial fraction decomposition of the given expression. 8 x 2 + 12x − 20 ________________ (x + 3)( x 2 + x + 2) Solution We have one linear factor and one irreducible quadratic factor in the denominator, so one numerator will be a constant and the other numerator will be a linear expression. Thus, 8 x 2 + 12x−20 ________________ (x + 3)( x 2 + x + 2) = A ______ (x + 3) + Bx + C __________ ( x 2 + x + 2) We follow the same steps as in previous problems. First, clear the fractions by multiplying both sides of the equation by the common denominator. (x + 3)( x 2 + x + 2)  8 x 2 + 12x − 20 ________________ (x + 3)( x 2 + x + 2)  =  A ______ (x + 3) + Bx + C __________ ( x 2 + x + 2)  (x + 3)( x 2 + x + 2) 8 x 2 + 12x − 20 = A( x 2 + x + 2) + (Bx + C)(x + 3) 800 CHAPTER 9 systems oF eQuations and ineQualities Notice we could easily solve for A by choosing a value for x that will make the Bx + C term equal 0. Let x = −3 and substitute it into the equation. 8 x 2 + 12x − 20 = A( x 2 + x + 2) + (Bx + C)(x + 3) 8 (−3) 2 + 12(−3) − 20 = A( (−3) 2 + (−3) + 2) + (B(−3) + C)((−3) + 3) 16 = 8A A = 2 Now that we know the value of A, substitute it back into the equation. Then expand the right side and collect like terms. 8 x 2 + 12x − 20 = 2( x 2 + x + 2) + (Bx + C)(x + 3) 8 x 2 + 12x − 20 = 2 x 2 + 2x + 4 + B x 2 + 3B + Cx + 3C 8 x 2 + 12x − 20 = (2 + B) x 2 + (2 + 3B + C)x + (4 + 3C) Setting the coefficients of terms on the right side equal to the coefficients of terms on the left side gives the system of equations. 2 + B = 8 2 + 3B + C = 12 4 + 3C = −20 (1) (2) (3) Solve for B using equation (1) and solve for C using equation (3). 2 + B = 8 (1) B = 6 4 + 3C = −20 (3) 3C = −24 C = −8 Thus, the partial fraction decomposition of the expression is 8 x 2 + 12x − 20 _______________ = (x + 3)( x 2 + x + 2) 2 ______ (x + 3) + 6x − 8 __________ ( x 2 + x + 2) Q & A… Could we have just set up a system of equations to solve Example 3? Yes, we could have solved it by setting up a system of equations without solving for A first. The expansion on the right would be: 8 x 2 + 12x − 20 = A x 2 + Ax + 2A + B x 2 + 3B + Cx + 3C 8 x 2 + 12x − 20 = (A + B) x 2 + (A + 3B + C)x + (2A + 3C) So the system of equations would be: A + B = 8 A + 3B + C = 12 2A + 3C = −20 Try It #3 Find the partial fraction decomposition of the expression with a nonrepeating irreducible quadratic factor. 5 x 2 − 6x + 7 ____________ (x − 1)( x 2 + 1) SECTION 9.4 partial Fractions 801 Decomposing Where Q(x ) Has a Repeated Irreducible Quadratic Factor P(x ) ____ Q(x ) Now that we can decompose a simplified rational expression with an irreducible quadratic factor, we will learn how to do partial fraction decomposition when the simplified rational expression has repeated irreducible quadratic factors. The decomposition will consist of partial fractions with linear numerators over each irreducible quadratic factor represented in increasing powers. decomposition of : when Q(x) has a repeated irreducible quadratic factor P(x) ____ Q(x) P(x) ____ Q(x) of P(x) is less than the degree of Q(x), is The partial fraction decomposition of , when Q(x) has a repeated irreducible quadratic factor and the degree P(x) _____________ n (a x 2 + bx + c) = A 1 x + B 1 ____________ (a x 2 + bx + c) + A 2 x + B 2 _____________ (a x 2 + bx + c) 2 + A 3 x + B 3 _____________ (a x 2 + bx + c _____________ n (a x 2 + bx + c) Write the denominators in increasing powers. How To… Given a rational expression that has a repeated irreducible factor, decompose it. 1. Use variables like A, B, or C for the constant numerators over linear factors, and linear expressions such as , etc., for the numerators of each quadratic factor in the denominator written in increasing powers, such as P(x) ____ Q(x) = A ______ ax + b + A 1 x + B 1 ____________ (a x 2 + bx + c) + A 2 x + B 2 ____________ (a x 2 + bx + c ____________ n (a x 2 + bx + c) 2. Multiply both sides of the equation by the common denominator to eliminate fractions. 3. Expand the right side of the equation and collect like terms. 4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators. Example 4 Decomposing a Rational Function with a Repeated Irreducible Quadratic Factor in the Denominator Decompose the given expression that has a repeated irreducible factor in the denominator ________________ x ( x 2 + 1) 2 Solution The factors of the denominator are x, ( x 2 + 1), and ( x 2 + 1) 2 . Recall that, when a factor in the denominator is a quadratic that includes at least two terms, the numerator must be of the linear form Ax + B. So, let’s begin the decomposition ________________ x ( x 2 + 1) 2 = A __ x + Bx + C _______ ( x 2 + 1) + Dx + E _______ ( x 2 + 1) 2 We eliminate the denominators by multiplying each term by x ( x 2 + 1) 2 . Thus) 2 + (Bx + C)(x)( x 2 + 1) + (Dx + E)(x) Expand the right side. Now we will collect like terms( x 4 + 2 x 2 + 1 + Cx + D x 2 + Ex = A x 4 + 2A + Cx + D x 2 + Ex A + B) x 4 + (C) x 3 + (2A + B + D) x 2 + (C + E)x + A 802 CHAPTER 9 systems oF eQuations and ineQualities Set up the system of equations matching corresponding coefficients on each side of the equal sign. A + B = 1 C = 1 2A + 1 A = 1 We can use substitution from this point. Substitute A = 1 into the first equation. Substitute A = 1 and B = 0 into the third equation. 1 + B = 1 B = 0 Substitute C = 1 into the fourth equation. 2(11 1 + E = −1 E = −2 Now we have solved for all of the unknowns on the right side of the equal sign. We have
A = 1, B = 0, C = 1, D = −1, and E = −2. We can write the decomposition as follows ________________ x ( x 2 + 1) 2 = 1 __ x + 1 _______ ( x 2 + 1) − x + 2 _______ ( x 2 + 1) 2 Try It #4 Find the partial fraction decomposition of the expression with a repeated irreducible quadratic factor. x 3 − 4 x 2 + 9x−5 ______________ ( x 2 −2x + 3) 2 Access these online resources for additional instruction and practice with partial fractions. • Partial Fraction Decomposition (http://openstaxcollege.org/l/partdecomp) • Partial Fraction Decomposition With Repeated linear Factors (http://openstaxcollege.org/l/partdecomprlf) • Partial Fraction Decomposition With linear and Quadratic Factors (http://openstaxcollege.org/l/partdecomlqu) SECTION 9.4 section exercises 803 9.4 SeCTIOn exeRCISeS VeRBAl 1. Can any quotient of polynomials be decomposed into at least two partial fractions? If so, explain why, and if not, give an example of such a fraction 2. Can you explain why a partial fraction decomposition is unique? (Hint: Think about it as a system of equations.) 3. Can you explain how to verify a partial fraction decomposition graphically? 4. You are unsure if you correctly decomposed the partial fraction correctly. Explain how you could double check your answer. 5. Once you have a system of equations generated by the partial fraction decomposition, can you explain another method to solve it? For example if you had 7x + 13 ___________ 3 x 2 + 8x + 15 = A _____ x + 1 simplify to 7x + 13 = A(3x + 5) + B(x + 1). Explain how you could intelligently choose an x-value that will eliminate either A or B and solve for A and B. + B ______ 3x + 5 , we eventually AlGeBRAIC For the following exercises, find the decomposition of the partial fraction for the nonrepeating linear factors. 6. 5x + 16 ___________ x 2 + 10x + 24 9. 10x + 47 __________ x 2 + 7x + 10 12. x + 1 __________ x 2 + 7x + 10 15. 6x _____ x 2 − 4 18. 4x + 3 __________ x 2 + 8x + 15 7. 3x − 79 __________ x 2 − 5x − 24 10. x ____________ 6 x 2 + 25x + 25 13. 5x _ x 2 − 9 16. 2x − 3 _________ x 2 − 6x + 5 19. 3x − 1 _________ x 2 −5x + 6 8. −x − 24 __________ x 2 − 2x − 24 11. 32x − 11 ____________ 20 x 2 − 13x + 2 14. 10x ______ x 2 − 25 17. 4x − 1 _ x 2 − x − 6 For the following exercises, find the decomposition of the partial fraction for the repeating linear factors. 20. −5x − 19 ________ (x + 4) 2 23. −24x − 27 _________ (4x + 5) 2 26. 5x + 14 ____________ 2 x 2 + 12x + 18 21. x _______ (x − 2) 2 24. −24x − 27 _________ (6x − 7) 2 22. 7x + 14 _______ (x + 3) 2 25. 5 − x _______ (x − 7) 2 27. 5 x 2 + 20x + 8 ___________ 2x (x + 1) 2 28. 4 x 2 + 55x + 25 ____________ 5x (3x + 5) 2 29. 54 x 3 + 127 x 2 + 80x + 16 ____________________ 2 x 2 (3x + 2) 2 30. x 3 − 5 x 2 + 12x + 144 _________________ x 2 ( x 2 + 12x + 36) 804 CHAPTER 9 systems oF eQuations and ineQualities For the following exercises, find the decomposition of the partial fraction for the irreducible non repeating quadratic factor. 31. 4 x 2 + 6x + 11 _______________ (x + 2)( x 2 + x + 3) 34. x 2 + 3x + 1 ________________ (x + 1)( x 2 + 5x − 2) 37. 4 x 2 + 5x + 3 __________ x 3 − 1 40. x 2 + 2x + 40 __________ x 3 − 125 43. −2 x 3 − 30 x 2 + 36x + 216 ____________________ x 4 + 216x 32. 4 x 2 + 9x + 23 _________________ (x − 1)( x 2 + 6x + 11) 33. −2 x 2 + 10x + 4 ________________ (x − 1)( x 2 + 3x + 8) 35. 4 x 2 + 17x − 1 ________________ (x + 3)( x 2 + 6x + 1) 36. 4 x 2 ________________ (x + 5)( x 2 + 7x − 5) 38. −5 x 2 + 18x − 4 ____________ x 3 + 8 41. 4 x 2 + 4x + 12 ___________ 8 x 3 − 27 39. 3 x 2 − 7x + 33 ___________ x 3 + 27 42. −50 x 2 + 5x − 3 ____________ 125 x 3 − 1 For the following exercises, find the decomposition of the partial fraction for the irreducible repeating quadratic factor. 44. 3 x 3 + 2 x 2 + 14x + 15 _________________ ( x 2 + 4) 2 47. x 2 + 5x + 5 _________ (x + 2) 2 50. 2 x 3 + 11x + 7x + 70 ________________ (2 x 2 + x + 14) 2 53. 2x − 9 _______ ( x 2 − x) 2 exTenSIOnS 45. 48. x 3 + 6 x 2 + 5x + 9 ______________ ( x 2 + 1) 2 x 3 + 2 x 2 + 4x ___________ ( x 2 + 2x + 9) 2 46. 49 ___________ ( x 2 − 3) 2 x 2 + 25 ____________ ( x 2 + 3x + 25) 2 51. 5x + 2 ________ x( x 2 + 4) 2 54. 5 x 3 − 2x + 1 __________ ( x 2 + 2x) 2 52 + 6x + 36 ___________________ x( x 2 + 6) 2 For the following exercises, find the partial fraction expansion. 55. x 2 + 4 _______ (x + 1) 3 56. x 3 − 4 x 2 + 5x + 4 ______________ (x − 2) 3 For the following exercises, perform the operation and then find the partial fraction decomposition. 57. 7 _____ x + 8 + 5 _____ x − 2 − x − 1 __________ x 2 −6x − 16 58. 1 _____ x − 4 − 3 _____ x + 6 − 2x + 7 __________ x 2 + 2x − 24 59. 2x ______ x 2 − 16 − 1−2x _________ x 2 + 6x + 8 − x − 5 ______ x 2 − 4x SECTION 9.5 matrices and matrix operations 805 leARnInG OBjeCTIVeS In this section, you will: • Find the sum and difference of two matrices. • Find scalar multiples of a matrix. • Find the product of two matrices. 9.5 MATRICeS AnD MATRIx OPeRATIOnS Two club soccer teams, the Wildcats and the Mud Cats, are hoping to obtain new equipment for an upcoming season. Table 1 shows the needs of both teams. Figure 1 (credit: “SD Dirk,” Flickr) Goals Balls Jerseys Wildcats 6 30 14 Table 1 Mud Cats 10 24 20 A goal costs $300; a ball costs $10; and a jersey costs $30. How can we find the total cost for the equipment needed for each team? In this section, we discover a method in which the data in the soccer equipment table can be displayed and used for calculating other information. Then, we will be able to calculate the cost of the equipment. Finding the Sum and Difference of Two Matrices To solve a problem like the one described for the soccer teams, we can use a matrix, which is a rectangular array of numbers. A row in a matrix is a set of numbers that are aligned horizontally. A column in a matrix is a set of numbers that are aligned vertically. Each number is an entry, sometimes called an element, of the matrix. Matrices (plural) are enclosed in [ ] or ( ), and are usually named with capital letters. For example, three matrices named A, B, and C are shown below1 3 0 2 3 1  806 CHAPTER 9 systems oF eQuations and ineQualities Describing Matrices A matrix is often referred to by its size or dimensions: m × n indicating m rows and n columns. Matrix entries are defined first by row and then by column. For example, to locate the entry in matrix A identified as a ij , we look for the entry in row i, column j. In matrix A, shown below, the entry in row 2, column 3 is a 23 . A =  a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33  A square matrix is a matrix with dimensions n × n, meaning that it has the same number of rows as columns. The 3 × 3 matrix above is an example of a square matrix. A row matrix is a matrix consisting of one row with dimensions 1 × n. [ a 11 a 12 a 13 ] A column matrix is a matrix consisting of one column with dimensions m × 1. a 11   a 12 a 13 A matrix may be used to represent a system of equations. In these cases, the numbers represent the coefficients of the variables in the system. Matrices often make solving systems of equations easier because they are not encumbered with variables. We will investigate this idea further in the next section, but first we will look at basic matrix operations. matrices A matrix is a rectangular array of numbers that is usually named by a capital letter: A, B, C, and so on. Each entry in a matrix is referred to as a ij , such that i represents the row and j represents the column. Matrices are often referred to by their dimensions: m × n indicating m rows and n columns. Example 1 Given matrix A: Finding the Dimensions of the Given Matrix and Locating Entries a. What are the dimensions of matrix A? b. What are the entries at a 31 and a 22 ? 2  Solution a. The dimensions are 3 × 3 because there are three rows and three columns. b. Entry a 31 is the number at row 3, column 1, which is 3. The entry a 22 is the number at row 2, column 2, which is 4. Remember, the row comes first, then the column. Adding and Subtracting Matrices We use matrices to list data or to represent systems. Because the entries are numbers, we can perform operations on matrices. We add or subtract matrices by adding or subtracting corresponding entries. In order to do this, the entries must correspond. Therefore, addition and subtraction of matrices is only possible when the matrices have the same dimensions. We can add or subtract a 3 × 3 matrix and another 3 × 3 matrix, but we cannot add or subtract a 2 × 3 matrix and a 3 × 3 matrix because some entries in one matrix will not have a corresponding entry in the other matrix. SECTION 9.5 matrices and matrix operations 807 adding and subtracting matrices Given matrices A and B of like dimensions, addition and subtraction of A and B will produce matrix C or matrix D of the same dimension. A + B = C such that a ij + b ij = c ij A − B = D such that a ij − b ij = d ij Matrix addition is commutative. It is also associative. A + B = B + A (A + B) + C = A + (B + C) Example 2 Finding the Sum of Matrices Find the sum of A and B, given Solution Add corresponding entries. A =  a b c d  and  Example 3 Adding Matrix A and Matrix B Find the sum of A and B. A =  4 1 3 2  and B =  5 9  0 7 Solution Add corresponding entries. Add the entry in row 1, column 1, a 11 , of matrix A to the entry in row 1, column 1, b 11 , of B. Continue the pattern until all entries have been added 10  9 Example 4 Finding the Difference of Two Matrices Find the difference of A and B. A =  −2 3 0 1  and B =  8 1 5 4  Solution We subtract the corresponding entries of each matrix. A − B =  −10 2  −5 −3 808 CHAPTER 9 systems oF eQuations and ineQualities Example 5 Finding the Sum and Difference of Two 3 x 3 Matrices Given A and B : a. Find the sum. b. Find the difference. A =  2 −10 −2 14 12 10 4 −2 2  and B =  10 −2 6 0 −12 −4 −5 2 −2  Solution a. Add the corresponding entries. 10 −2 6 0 −12 −4 −5 2 −2  A + B =  =  =  10 2  +  2 −10
−2 12 14 −2 4 2 + 6 −10 + 10 −2 − 2 14 + 0 12 − 12 4 − 5 −2 + 2 8 0 −4 14 0 −6 −1 0 −0  10 − 4 2 − 2  b. Subtract the corresponding entries. 10 −2 6 0 −12 −4 −5 2 −2  A − B =  =  =  10 2  −  2 −10 −2 12 14 −2 4 2 − 6 −10 − 10 −2 + 2 12 + 12 14 − 0 −2 − 2 4 + 5 −4 −20 0 14 9 24 14 −4  4 10 + 4 2 + 2  Try It #1 Add matrix A and matrix B. A =  2 1 1 6 0  and B =  3 1 −4 −3 −2  5 3 Finding Scalar Multiples of a Matrix Besides adding and subtracting whole matrices, there are many situations in which we need to multiply a matrix by a constant called a scalar. Recall that a scalar is a real number quantity that has magnitude, but not direction. For example, time, temperature, and distance are scalar quantities. The process of scalar multiplication involves multiplying each entry in a matrix by a scalar. A scalar multiple is any entry of a matrix that results from scalar multiplication. Consider a real-world scenario in which a university needs to add to its inventory of computers, computer tables, and chairs in two of the campus labs due to increased enrollment. They estimate that 15% more equipment is needed in both labs. The school’s current inventory is displayed in Table 2. Lab A Lab B Computers Computer Tables Chairs 15 16 16 Table 2 27 34 34 SECTION 9.5 matrices and matrix operations 809 Converting the data to a matrix, we have C 2013 =  15 27 16 34 16 34  To calculate how much computer equipment will be needed, we multiply all entries in matrix C by 0.15. We must round up to the next integer, so the amount of new equipment needed is (0.15) C 2013 =  (0.15)15 (0.15)16 (0.15)16 (0.15)27 (0.15)34 (0.15)34  =  2.25 4.05 2.4 5.1 2.4 5.1  Adding the two matrices as shown below, we see the new inventory amounts. 3 5 3 6 3 6   This means 15 27 16 34 16 34  +  3 5 3 6 3 6   =  18 32 19 40 19 40  C 2014 =  18 32 19 40 19 40  Thus, Lab A will have 18 computers, 19 computer tables, and 19 chairs; Lab B will have 32 computers, 40 computer tables, and 40 chairs. scalar multiplication Scalar multiplication involves finding the product of a constant by each entry in the matrix. Given the scalar multiple cA is A =  a 11 a 21 a 12  a 22 cA = c  =  a 11 a 21 ca 11 ca 21 a 12  a 22 ca 12  ca 22 Scalar multiplication is distributive. For the matrices A, B, and C with scalars a and b, a(A + B) = aA + aB (a + b)A = aA + bA Example 6 Multiplying the Matrix by a Scalar Multiply matrix A by the scalar 3. Solution Multiply each entry in A by the scalar 3. A =  8 1  5 4 3A = 24 3  15 12 =  =  810 CHAPTER 9 systems oF eQuations and ineQualities Try It #2 Given matrix B, find −2B where A =  4 1 3 2  Example 7 Finding the Sum of Scalar Multiples Find the sum 3A + 2B. Solution First, find 3A, then 2B. A =  1 −2 0 0 −1 2 3 −6 4  and B =  −1 2 1 0 −3 2 1 −4 0  3A =  =  2B =  =   3 ⋅ 1 3(−2) 3 ⋅ 0 3 ⋅ 0 3(−1(−6) 3 −6 0 0 −3 6 12 9 −18 2(−12 4 2 2(−3) 2 ⋅ 2 2(−46 4 0 2 −8  Now, add 3A + 2B. 3A + 2B =  =  =  −2 4 2 0 −6 4 0 2 −8   +  3 −6 0 0 −3 6 12 9 −18 3 − 2 −3 − 6 6 + 4 12 + 0 9 + 2 −18 − 8  1 −2 2 0 −9 10 12 11 −26  Finding the Product of Two Matrices In addition to multiplying a matrix by a scalar, we can multiply two matrices. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. If A is an m × r matrix and B is an r × n matrix, then the product matrix AB is an m × n matrix. For example, the product AB is possible because the number of columns in A is the same as the number of rows in B. If the inner dimensions do not match, the product is not defined same We multiply entries of A with entries of B according to a specific pattern as outlined below. The process of matrix multiplication becomes clearer when working a problem with real numbers. To obtain the entries in row i of AB, we multiply the entries in row i of A by column j in B and add. For example, given matrices A and B, where the dimensions of A are 2 × 3 and the dimensions of B are 3 × 3, the product of AB will be a 2 × 3 matrix. A =  a 11 a 21 a 12 a 22 a 13 a 23  and B =  b 11 b 21 b 31 b 12 b 22 b 32 b 13 b 23 b 33  SECTION 9.5 matrices and matrix operations 811 Multiply and add as follows to obtain the first entry of the product matrix AB. 1. To obtain the entry in row 1, column 1 of AB, multiply the first row in A by the first column in B, and add. [ a 11 a 12 a 13 ]  b 11 b 21 b 31  = a 11 ⋅ b 11 + a 12 ⋅ b 21 + a 13 ⋅ b 31 2. To obtain the entry in row 1, column 2 of AB, multiply the first row of A by the second column in B, and add. [ a 11 a 12 a 13 ]  b 12 b 22 b 32  = a 11 ⋅ b 12 + a 12 ⋅ b 22 + a 13 ⋅ b 32 3. To obtain the entry in row 1, column 3 of AB, multiply the first row of A by the third column in B, and add. [ a 11 a 12 a 13 ]  b 13 b 23 b 33  = a 11 ⋅ b 13 + a 12 ⋅ b 23 + a 13 ⋅ b 33 We proceed the same way to obtain the second row of AB. In other words, row 2 of A times column 1 of B; row 2 of A times column 2 of B; row 2 of A times column 3 of B. When complete, the product matrix will be AB =  a 11 ⋅ b 11 + a 12 ⋅ b 21 + a 13 ⋅ b 31 a 21 ⋅ b 11 + a 22 ⋅ b 21 + a 23 ⋅ b 31 a 11 ⋅ b 12 + a 12 ⋅ b 22 + a 13 ⋅ b 32 a 21 ⋅ b 12 + a 22 ⋅ b 22 + a 23 ⋅ b 32 a 11 ⋅ b 13 + a 12 ⋅ b 23 + a 13 ⋅ b 33  a 21 ⋅ b 13 + a 22 ⋅ b 23 + a 23 ⋅ b 33 properties of matrix multiplication For the matrices A, B, and C the following properties hold. • Matrix multiplication is associative: • Matrix multiplication is distributive: (AB)C = A(BC). C(A + B) = CA + CB, (A + B)C = AC + BC. Note that matrix multiplication is not commutative. Example 8 Multiplying Two Matrices Multiply matrix A and matrix B.  and B =  Solution First, we check the dimensions of the matrices. Matrix A has dimensions 2 × 2 and matrix B has dimensions 2 × 2. The inner dimensions are the same so we can perform the multiplication. The product will have the dimensions 2 × 2 We perform the operations outlined previously. AB =  =  =  (5) + 2(7) 1(6) + 2(8)  3(5) + 4(7) 3(6) + 4(8) 19 22  43 50 Example 9 Multiplying Two Matrices Given A and B : a. Find AB. b. Find BA. A =  −1 2 3 4 0 5  and B =  5 −1 −4 0 2 3  812 CHAPTER 9 systems oF eQuations and ineQualities Solution a. As the dimensions of A are 2 × 3 and the dimensions of B are 3 × 2, these matrices can be multiplied together because the number of columns in A matches the number of rows in B. The resulting product will be a 2 × 2 matrix, the number of rows in A by the number of columns in B. AB =  −1 2 3 4 0 5 5 −1 −1(5) + 2(−4) + 3(2) −1(−1) + 2(0) + 3(3) 4(−1) + 0(0) + 5(3) 4(5) + 0(−4) + 5(2)  =  −7 10  30 11 b. The dimensions of B are 3 × 2 and the dimensions of A are 2 × 3. The inner dimensions match so the product is defined and will be a 3 × 3 matrix. BA =  =  =  5 −1 −4 0 2 3   −1 2 3 4 0 5  5(−1) + −1(4) 5(2) + −1(0) 5(3) + −1(5) −4(−1) + 0(4) −4(2) + 0(0) −4(3) + 0(5) 2(−1) + 3(4) 2(2) + 3(0) 2(3) + 3(5) −9 10 10  4 −8 −12 10 4 21  Analysis Notice that the products AB and BA are not equal. AB =  −7 10 30 11  ≠  −9 10 10 4 −8 −12 10 4 21  = BA This illustrates the fact that matrix multiplication is not commutative. Q & A… Is it possible for AB to be defined but not BA? Yes, consider a matrix A with dimension 3 × 4 and matrix B with dimension 4 × 2. For the product AB the inner dimensions are 4 and the product is defined, but for the product BA the inner dimensions are 2 and 3 so the product is undefined. Example 10 Using Matrices in Real-World Problems Let’s return to the problem presented at the opening of this section. We have Table 3, representing the equipment needs of two soccer teams. Goals Balls Jerseys Wildcats 6 30 14 Table 3 Mud Cats 10 24 20 We are also given the prices of the equipment, as shown in Table 4. Goals Balls Jerseys Table 4 $300 $10 $30 SECTION 9.5 matrices and matrix operations 813 We will convert the data to matrices. Thus, the equipment need matrix is written as E =  6 10 30 24 14 20  The cost matrix is written as C = [300 10 30] We perform matrix multiplication to obtain costs for the equipment. CE = [300 10 30]  6 10 30 24 14 20  = [300(6) + 10(30) + 30(14) 300(10) + 10(24) + 30(20)] = [2,520 3,840] The total cost for equipment for the Wildcats is $2,520, and the total cost for equipment for the Mud Cats is $3,840. How To… Given a matrix operation, evaluate using a calculator. 1. Save each matrix as a matrix variable [A], [B], [C], ... 2. Enter the operation into the calculator, calling up each matrix variable as needed. 3. If the operation is defined, the calculator will present the solution matrix; if the operation is undefined, it will display an error message. Example 11 Using a Calculator to Perform Matrix Operations Find AB − C given A =  −15 25 32 41 −7 −28 10 34 −2  , B =  45 21 −37 −24 52 19 6 −48 −31  , and C =  −100 −89 −98 25 −56 74 −67 42 −75  . Solution On the matrix page of the calculator, we enter matrix A above as the matrix variable [A], matrix B above as the matrix variable [B], and matrix C above as the matrix variable [C]. On the home screen of the calculator, we type in the problem and call up each matrix variable as needed. The calculator gives us the following matrix. [A]×[B] − [C] −983 −462 136 1,820 1,897 −856 −311 2,032 413   Access these online resources for additional instruction and practice with matrices and matrix operations. • Dimensions of a Matrix (http://openstaxcollege.org/l/matrixdimen) • Matrix Addition and Subtraction (http://openstaxcollege.org/l/matrixaddsub) • Matrix Operations (http://openstaxcollege.org/l/matrixoper) • Matrix Multiplication (http://openstaxcollege.org/l/matrixmult) 814 CHAPTER 9 systems oF eQuations and ineQualities 9.5 SeCTIOn exeRCISeS VeRBAl 1. Can we add any two matrices together? If so, explain why; if not, explain why not and give an example of two matrices that cannot be added together. 3. Can both the pr
oducts AB and BA be defined? If so, explain how; if not, explain why. 5. Does matrix multiplication commute? That is, does AB = BA? If so, prove why it does. If not, explain why it does not. AlGeBRAIC 2. Can we multiply any column matrix by any row matrix? Explain why or why not. 4. Can any two matrices of the same size be multiplied? If so, explain why, and if not, explain why not and give an example of two matrices of the same size that cannot be multiplied together. For the following exercises, use the matrices below and perform the matrix addition or subtraction. Indicate if the operation is undefined 14 22 6  , C =  1 5 8 92 12 6  , D =  10 14 7 2 5 61  , E =  6 12 14 5  , F =  0 9 78 17 15 4  6. A + B 7. C + D 8. A + C 9. B − E 10. C + F 11. D − B For the following exercises, use the matrices below to perform scalar multiplication. A =  4 6 13 12  , B =  3 9 21 12 0 64  , C =  16 3 7 18 90 5 3 29  , D =  18 12 13 8 14 6 7 4 21  12. 5A 13. 3B 14. −2B 15. −4C 1 __ 16. C 2 17. 100D For the following exercises, use the matrices below to perform matrix multiplication. A =  −8 0 12  , C =  4 10 −3 12 9 3 1 0 8 −10  18. AB 19. BC 20. CA 21. BD 22. DC 23. CB For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. A =  2 −5 6 7  , B =  −9 6 −8 7 −6 −5 1 0 9  24. A + B − C 25. 4A + 5D 26. 2C + B 27. 3D + 4E 28. C − 0.5D 29. 100D − 10E SECTION 9.5 section exercises 815 For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. (Hint: A 2 = A ⋅ A) A =  −10 20 5 25  , B =  40 10 −20 30  , C =  −1 0 0 −1 1 0  30. AB 36. C 2 31. BA 37. B 2 A 2 32. CA 38. A 2 B 2 33. BC 34. A 2 35. B 2 39. (AB) 2 40. (BA) 2 For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. (Hint: A 2 = A ⋅ A2 3 4 −1 1 −5  , C =  0.5 0.1 1 0.2 −0.5 0.3  , D =  1 0 −1 −6 7 5 4 2 1  42. BA 43. BD 44. DC 45. D 2 46. A 2 48. (AB)C 49. A(BC) 41. AB 47. D 3 TeCHnOlOGY For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. Use a calculator to verify your solution. A =  −2 0 9 1 8 −3 0.5 4 5  , B =  0.5 3 0 −  50. AB 51. BA 52. CA 53. BC 54. ABC exTenSIOnS For the following exercises, use the matrix below to perform the indicated operation on the given matrix 55. B2 56. B3 57. B4 58. B5 59. Using the above questions, find a formula for Bn. Test the formula for B201 and B202, using a calculator. 816 CHAPTER 9 systems oF eQuations and ineQualities leARnInG OBjeCTIVeS In this section, you will: • Write the augmented matrix of a system of equations. • Write the system of equations from an augmented matrix. • Perform row operations on a matrix. • Solve a system of linear equations using matrices. 9.6 SOlVInG SYSTeMS WITH GAUSSIAn elIMInATIOn Figure 1 German mathematician Carl Friedrich Gauss (1777–1855). Carl Friedrich Gauss lived during the late 18th century and early 19th century, but he is still considered one of the most prolific mathematicians in history. His contributions to the science of mathematics and physics span fields such as algebra, number theory, analysis, differential geometry, astronomy, and optics, among others. His discoveries regarding matrix theory changed the way mathematicians have worked for the last two centuries. We first encountered Gaussian elimination in Systems of Linear Equations: Two Variables. In this section, we will revisit this technique for solving systems, this time using matrices. Writing the Augmented Matrix of a System of equations A matrix can serve as a device for representing and solving a system of equations. To express a system in matrix form, we extract the coefficients of the variables and the constants, and these become the entries of the matrix. We use a vertical line to separate the coefficient entries from the constants, essentially replacing the equal signs. When a system is written in this form, we call it an augmented matrix. For example, consider the following 2 × 2 system of equations. We can write this system as an augmented matrix: 3x + 4y = 7 4x − 2y = 5 3 4  4 −2 7  5 ∣ We can also write a matrix containing just the coefficients. This is called the coefficient matrix. 3 4   4 −2 SECTION 9.6 solving systems with gaussian elimination 817 A three-by-three system of equations such as has a coefficient matrix and is represented by the augmented matrix 3x − 2x − 3z = 2 3 −1 −1 1 1 0 2 0 −3   3 −1 −1 0 5 1 1 0 2 0 −3 2   ∣ Notice that the matrix is written so that the variables line up in their own columns: x-terms go in the first column, y-terms in the second column, and z-terms in the third column. It is very important that each equation is written in standard form ax + by + cz = d so that the variables line up. When there is a missing variable term in an equation, the coefficient is 0. How To… Given a system of equations, write an augmented matrix. 1. Write the coefficients of the x-terms as the numbers down the first column. 2. Write the coefficients of the y-terms as the numbers down the second column. 3. If there are z-terms, write the coefficients as the numbers down the third column. 4. Draw a vertical line and write the constants to the right of the line. Example 1 Writing the Augmented Matrix for a System of Equations Write the augmented matrix for the given system of equations. x + 2y − z = 3 2x − y + 2z = 6 x − 3y + 3z = 4 Solution The augmented matrix displays the coefficients of the variables, and an additional column for the constants. 1 2 −1 3 2 −1 2 6 1 −3 3 4   ∣ Try It #1 Write the augmented matrix of the given system of equations. 4x − 3y = 11 3x + 2y = 4 Writing a System of equations from an Augmented Matrix We can use augmented matrices to help us solve systems of equations because they simplify operations when the systems are not encumbered by the variables. However, it is important to understand how to move back and forth between formats in order to make finding solutions smoother and more intuitive. Here, we will use the information in an augmented matrix to write the system of equations in standard form. 818 CHAPTER 9 systems oF eQuations and ineQualities Example 2 Writing a System of Equations from an Augmented Matrix Form Find the system of equations from the augmented matrix. Solution When the columns represent the variables x, y, and z, 1 −3 −5 2 −5 −4 −3 5 4  −2 5 6  ∣ 1 −3 −5 2 −5 −4 −3 5 4 −2 5 6  →  x − 3y − 5z = − 2 2x − 5y − 4z = 5 −3x + 5y + 4z = 6 ∣ Try It #2 Write the system of equations from the augmented matrix. 1 −1 1 2 −1 3 0 1 1 5 1 −9   ∣ Performing Row Operations on a Matrix Now that we can write systems of equations in augmented matrix form, we will examine the various row operations that can be performed on a matrix, such as addition, multiplication by a constant, and interchanging rows. Performing row operations on a matrix is the method we use for solving a system of equations. In order to solve the system of equations, we want to convert the matrix to row-echelon form, in which there are ones down the main diagonal from the upper left corner to the lower right corner, and zeros in every position below the main diagonal as shown. Row-echelon form 1 a b 0 1 d 0 0 1   We use row operations corresponding to equation operations to obtain a new matrix that is row-equivalent in a simpler form. Here are the guidelines to obtaining row-echelon form. 1. In any nonzero row, the first nonzero number is a 1. It is called a leading 1. 2. Any all-zero rows are placed at the bottom on the matrix. 3. Any leading 1 is below and to the right of a previous leading 1. 4. Any column containing a leading 1 has zeros in all other positions in the column. To solve a system of equations we can perform the following row operations to convert the coefficient matrix to rowechelon form and do back-substitution to find the solution. 1. Interchange rows. (Notation: R i ↔ R j ) 2. Multiply a row by a constant. (Notation: c R i ) 3. Add the product of a row multiplied by a constant to another row. (Notation: R i + c R j ) Each of the row operations corresponds to the operations we have already learned to solve systems of equations in three variables. With these operations, there are some key moves that will quickly achieve the goal of writing a matrix in row-echelon form. To obtain a matrix in row-echelon form for finding solutions, we use Gaussian elimination, a method that uses row operations to obtain a 1 as the first entry so that row 1 can be used to convert the remaining rows. SECTION 9.6 solving systems with gaussian elimination 819 Gaussian elimination The Gaussian elimination method refers to a strategy used to obtain the row-echelon form of a matrix. The goal is to write matrix A with the number 1 as the entry down the main diagonal and have all zeros below. A =  a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33  After Gaussian elimination A =  b 13 b 23 b 12 1 0 1 0 0 1  The first step of the Gaussian strategy includes obtaining a 1 as the first entry, so that row 1 may be used to alter the rows below. How To… Given an augmented matrix, perform row operations to achieve row-echelon form. 1. The first equation should have a leading coefficient of 1. Interchange rows or multiply by a constant, if necessary. 2. Use row operations to obtain zeros down the first column below the first entry of 1. 3. Use row operations to obtain a 1 in row 2, column 2. 4. Use row operations to obtain zeros down column 2, below the entry of 1. 5. Use row operations to obtain a 1 in row 3, column 3. 6. Continue this process for all rows until there is a 1 in every entry down the main diagonal and there are only zeros below. 7
. If any rows contain all zeros, place them at the bottom. Example 3 Solving a 2 × 2 System by Gaussian Elimination Solve the given system by Gaussian elimination. Solution First, we write this as an augmented matrix. 2x + 3y = 6 x − y = 1 __ 2 2 3  1 −1 2 3 1 _  2 6 ∣ ∣ ∣ ∣ We want a 1 in row 1, column 1. This can be accomplished by interchanging row 1 and row 2. We now have a 1 as the first entry in row 1, column 1. Now let’s obtain a 0 in row 2, column 1. This can be accomplished by multiplying row 1 by −2, and then adding the result to row 2. −1 1 _  2 5 0 5 We only have one more step, to multiply row 2 by 1 __ . 5 R 2 = R 2 →  1 __ 5 Use back-substitution. The second row of the matrix represents y = 1. Back-substitute y = 1 into the first equation. x − (1) = 1 __ 2 x = 3 __ 2 1 _  2 1 1 −1 0 1 The solution is the point  3 __ 2 , 1  . 820 CHAPTER 9 systems oF eQuations and ineQualities Try It #3 Solve the given system by Gaussian elimination. 4x + 3y = 11 x − 3y = −1 Example 4 Using Gaussian Elimination to Solve a System of Equations Use Gaussian elimination to solve the given 2 × 2 system of equations. 2x + y = 1 4x + 2y = 6 Solution Write the system as an augmented matrix. 2 1  4 2 1 6 ∣  Obtain a 1 in row 1, column 1. This can be accomplished by multiplying the first row by 1 __ . 2 Next, we want a 0 in row 2, column 1. Multiply row 1 by −4 and add row 1 to row 2. R 1 = R 1 →  1 __ The second row represents the equation 0 = 4. Therefore, the system is inconsistent and has no solution. Example 5 Solving a Dependent System Solve the system of equations. 3x + 4y = 12 6x + 8y = 24 Solution Perform row operations on the augmented matrix to try and achieve row-echelon form →  12  24 − 1 __ 2 0 0 6 8 0  24 24  0 The matrix ends up with all zeros in the last row: 0y = 0. Thus, there are an infinite number of solutions and the system is classified as dependent. To find the generic solution, return to one of the original equations and solve for y. 3x + 4y = 12 4y = 12 − 3x y = 3 − 3 __ x 4 So the solution to this system is  x, 3 − 3 __ 4 x  . Example 6 Performing Row Operations on a 3 × 3 Augmented Matrix to Obtain Row-Echelon Form Perform row operations on the given matrix to obtain row-echelon form. Solution The first row already has a 1 in row 1, column 1. The next step is to multiply row 1 by −2 and add it to row 2. 1 −3 4 3 2 −5 6 6 6 −3 3 4   ∣ SECTION 9.6 solving systems with gaussian elimination 821 Then replace row 2 with the result. Next, obtain a zero in row 3, column 1. Next, obtain a zero in row 3, column 2. The last step is to obtain a 1 in row 3, column 3. −3 4 3 0 1 −2 0 6 −3 4 3 0 1 −2 0 0 −6 16 15  3 4 3 0 1 −2 0 15 0 0 4 1 __ 3 4 3 −6 0 1 −2 21 __ 0 0 1 2 ∣ ∣ ∣ ∣   Try It #4 Write the system of equations in row-echelon form. x − 2y + 3z = 9 −x + 3y = − 4 2x − 5y + 5z = 17 Solving a System of linear equations Using Matrices We have seen how to write a system of equations with an augmented matrix, and then how to use row operations and back-substitution to obtain row-echelon form. Now, we will take row-echelon form a step farther to solve a 3 by 3 system of linear equations. The general idea is to eliminate all but one variable using row operations and then backsubstitute to solve for the other variables. Example 7 Solving a System of Linear Equations Using Matrices Solve the system of linear equations using matrices. x − y + z = 8 2x + 3y − z = −2 3x − 2y − 9z = 9 Solution First, we write the augmented matrix. 1 −1 1 8  2 3 −1 −2 3 −2 −9 9  ∣ −3 R 1 + R 3 = R 3 →  Next, we perform row operations to obtain row-echelon form. −1 1 0 5 −3 3 −2 −9 8  −18 9 ∣ Interchange R 2 and R 3 →  1 −1 1 8 0 1 −12 −15 0 5 −3 −18  The easiest way to obtain a 1 in row 2 of column 1 is to interchange R 2 and R 3 . 1 −1 1 0 5 −3 0 1 −12 8  −18 −15 ∣ 822 Then CHAPTER 9 systems oF eQuations and ineQualities 1 −1 1 0 1 −12 0 0 57 The last matrix represents the equivalent system. −15 57 − 1 ___ 57 R 3 = R 3 →  1 −1 1 0 1 −12 0 0 1 8  −15 − 12z = −15 z = 1 Using back-substitution, we obtain the solution as (4, −3, 1). Example 8 Solving a Dependent System of Linear Equations Using Matrices Solve the following system of linear equations using matrices. −x − 2y + z = −1 2x + 3y = 2 y − 2z = 0 Solution Write the augmented matrix. −1 −2 1 2 3 0 0 1 −2  −1 2 0  ∣ First, multiply row 1 by −1 to get a 1 in row 1, column 1. Then, perform row operations to obtain row-echelon form. − R 1 →  −1 −2 1 2 3 0 0 1 −2 −1 1 0 1 −2 0 2 3 0 2 −1 1 0 1 −2 0 0 −1 2 0 1 −2 1 0 0 0 0 ∣ ∣  ∣ ∣  The last matrix represents the following system. x + 2y − z = 1 y − 2z = 0 0 = 0 We see by the identity 0 = 0 that this is a dependent system with an infinite number of solutions. We then find the generic solution. By solving the second equation for y and substituting it into the first equation we can solve for z in terms of x. x + 2y − z = 1 y = 2z x + 2(2z) − z = 1 x + 3z = 1 z = 1 − x _____ 3 SECTION 9.6 solving systems with gaussian elimination 823 Now we substitute the expression for z into the second equation to solve for y in terms of x. y − 2z = 0 The generic solution is  x, 2 − 2x ______ 3 , 1 − x _____  . 3 Try It #5 Solve the system using matrices. y − 2  1 − x _____ 3 z = 1 − x _____ 3  = 0 y = 2 − 2x ______ 3 x + 4y − z = 4 2x + 5y + 8z = 15 x + 3y − 3z = 1 Q & A… Can any system of linear equations be solved by Gaussian elimination? Yes, a system of linear equations of any size can be solved by Gaussian elimination. How To… Given a system of equations, solve with matrices using a calculator. 1. Save the augmented matrix as a matrix variable [A], [B], [C], … . 2. Use the ref( function in the calculator, calling up each matrix variable as needed. Example 9 Solving Systems of Equations with Matrices Using a Calculator Solve the system of equations. 5x + 3y + 9z = −1 −2x + 3y − z = −2 −x − 4y + 5z = 1 Solution Write the augmented matrix for the system of equations. On the matrix page of the calculator, enter the augmented matrix above as the matrix variable [A]. 5 3 9 −2 3 −1 −1 −4 5  −1  −2 −1 ∣ [A] =  5 3 9 −1 −2 3 −1 −2 −1 −4 5 1  Use the ref( function in the calculator, calling up the matrix variable [A]. ref([A]) Evaluate 13 __ 21 − 4 _ 7 − 24 ___ 187 + 13 __ 21 z = − 4 _ 7 z = − 24 ___ 187 Using back-substitution, the solution is  61 ___ 187 , − 92 ___ 187 , − 24 ___  . 187 824 CHAPTER 9 systems oF eQuations and ineQualities Example 10 Applying 2 × 2 Matrices to Finance Carolyn invests a total of $12,000 in two municipal bonds, one paying 10.5% interest and the other paying 12% interest. The annual interest earned on the two investments last year was $1,335. How much was invested at each rate? Solution We have a system of two equations in two variables. Let x = the amount invested at 10.5% interest, and y = the amount invested at 12% interest. x + y = 12,000 0.105x + 0.12y = 1,335 As a matrix, we have Multiply row 1 by −0.105 and add the result to row 2. 1 1  0 0.015 Then, 1 1  0.105 0.12 12,000 1,335  ∣ ∣ 12,000  75 0.015y = 75 y = 5,000 So 12,000 − 5,000 = 7,000. Thus, $5,000 was invested at 12% interest and $7,000 at 10.5% interest. Example 11 Applying 3 × 3 Matrices to Finance Ava invests a total of $10,000 in three accounts, one paying 5% interest, another paying 8% interest, and the third paying 9% interest. The annual interest earned on the three investments last year was $770. The amount invested at 9% was twice the amount invested at 5%. How much was invested at each rate? Solution We have a system of three equations in three variables. Let x be the amount invested at 5% interest, let y be the amount invested at 8% interest, and let z be the amount invested at 9% interest. Thus, x + y + z = 10,000 0.05x + 0.08y + 0.09z = 770 2x − z = 0 As a matrix, we have 1 1 1 0.05 0.08 0.09 2 0 −1  10, 000 770  0 ∣ Now, we perform Gaussian elimination to achieve row-echelon form. ∣ ∣ −0.05 .03 0.04 2 0 −1 10,000 270  0 1 ____ 0.03 −.03 0.04 0 −2 −3 10,000 270  −20,000 2 −3 10,000 9,000  −20,000 10,000 9,000 −2,000  ∣ ∣ The third row tells us − 1 __ 3 z = −2,000; thus z = 6,000. SECTION 9.6 solving systems with gaussian elimination 825 The second row tells us y + 4 __ 3 z = 9,000. Substituting z = 6,000, we get y + 4 __ (6,000) = 9,000 3 y + 8,000 = 9,000 y = 1,000 The first row tells us x + y + z = 10, 000. Substituting y = 1,000 and z = 6,000, we get x + 1,000 + 6,000 = 10,000 x = 3,000 The answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest. Try It #6 A small shoe company took out a loan of $1,500,000 to expand their inventory. Part of the money was borrowed at 7%, part was borrowed at 8%, and part was borrowed at 10%. The amount borrowed at 10% was four times the amount borrowed at 7%, and the annual interest on all three loans was $130,500. Use matrices to find the amount borrowed at each rate. Access these online resources for additional instruction and practice with solving systems of linear equations using Gaussian elimination. • Solve a System of Two equations Using an Augmented Matrix (http://openstaxcollege.org/l/system2augmat) • Solve a System of Three equations Using an Augmented Matrix (http://openstaxcollege.org/l/system3augmat) • Augmented Matrices on the Calculator (http://openstaxcollege.org/l/augmatcalc) 826 CHAPTER 9 systems oF eQuations and ineQualities 9.6 SeCTIOn exeRCISeS VeRBAl 1. Can any system of linear equations be written as an 2. Can any matrix be written as a system of linear augmented matrix? Explain why or why not. Explain how to write that augmented matrix. equations? Explain why or why not. Explain how to write that system of equations. 3. Is there only one correct method of using row operations on a matrix? Try to explain two different row operations possible to solve the augmented matrix  9 3 1 −2 0  . 6 ∣ 4. Can a matrix whose entry is 0 on the diagonal be solved? Explain why or why not. What would you do t
o remedy the situation? 5. Can a matrix that has 0 entries for an entire row have one solution? Explain why or why not. AlGeBRAIC For the following exercises, write the augmented matrix for the linear system. 6. 8x − 37y = 8 2x + 12y = 3 9. x + 5y + 8z = 19 12x + 3y = 4 3x + 4y + 9z = −7 7. 16y = 4 9x − y = 2 10. 6x + 12y + 16z = 4 19x − 5y + 3z = −9 x + 2y = −8 For the following exercises, write the linear system from the augmented matrix. 8. 3x + 2y + 10z = 3 −6x + 2y + 5z = 13 4x + z = 18 11.  −2 5 6 −18 5 26 ∣  12.  3 4 10 17 10  439 ∣ 14.  43 8 29 1 −1 7 5 38 10 0 0 3  ∣ 15.  4 5 −2 0 1 58 8 7 −3 12 2 −5  ∣ For the following exercises, solve the system by Gaussian elimination. 13.  3 2 0 3  −1 −9 4 −1 8 8 5 7 ∣ 1 0 16.  0 0 20.  −2 0 0 2 1  −1 ∣ 3  0 ∣ 1 0 17.  1 0 1  2 ∣ 21. 2x − 3y = − 9 5x + 4y = 58 1 2 18.  4 5 3  6 ∣ 22. 6x + 2y = −4 3x + 4y = −17 −1 2 19.  4 −5 −3 6 ∣  23. 2x + 3y = 12 4x + y = 14 24. −4x − 3y = −2 3x − 5y = −13 25. −5x + 8y = 3 10x + 6y = 5 26. 3x + 4y = 12 −6x − 8y = −24 27. −60x + 45y = 12 20x − 15y = −4 28. 11x + 10y = 43 15x + 20y = 65 29. 2x − y = 2 3x + 2y = 17 30. −1.06x−2.25y = 5.51 −5.03x − 1.08y = 5.40 32. 1 __ 4 1 __ 2 x − 2 __ y = −1 3 x + 1 __ y = 3 3 33 31 45 87   ∣ 34 50 20 −90   ∣ x − 3 __ 31. 3 __ 5 4 1 x + 2 __ __ 3 4 y = 4 y = 1 35   ∣ 36. −0.1 0.3 −0.1 −0.4 0.2 0.1 0.6 0.1 0.7 0.2 0.8 −0.8   ∣ 39. 2x + 3y + 2z = 1 −4x − 6y − 4z = −2 10x + 15y + 10z = 5 42y − z = −3 x + 1 __ 45. − 1 __ 2 2 1 __ x − 1 __ 2 2 x + 1 __ 1 __ 5 4 y + 1 __ 7 y + 1 __ 4 y + 1 __ 3 z = − 53 ___ 14 z = 3 z = 23 ___ 15 SECTION 9.6 section exercises 827 38. x + y − 4z = −4 5x − 3y − 2z = 0 2x + 6y + 7z = 30 41. x + 2y − z = 1 −x − 2y + 2z = −2 3x + 6y − 3z = 3 x − 2 __ 44. 1 __ z = − 1 __ 4 3 2 x + 1 __ y = 4 __ 1 __ 7 5 3 z = 2 __ y − 1 __ 1 __ 5 9 3 37. −2x + 3y − 2z = 3 4x + 2y − z = 9 4x − 8y + 2z = −6 40. x + 2y − z = 1 −x − 2y + 2z = −2 3x + 6y − 3z = 5 43. x + y + z = 100 x + 2z = 125 −y + 2z = 25 46. − 1 __ y + 1 __ x − 1 __ 4 2 3 1 __ y − 1 __ x + 1 __ 7 5 6 − 1 __ y + 1 ___ x + 1 __ 10 9 8 z = − 29 ___ 6 z = 431 ___ 210 z = − 49 ___ 45 exTenSIOnS For the following exercises, use Gaussian elimination to solve the system. 47. x − 1 _____ 7 + y − 2 _____ 8 + z − 3 _____ _____ 3 + 2y + z−3 ____ 3 = 5 48. x − 1 _____ 4 x + 5 _____ 2 + 3z = −1 − y + 1 _____ 4 y + 7 _____ 4 x + y − z−2 ____ 2 + − z = 4 = 1 − 49. x − 3 _____ 4 x + 5 _____ 2 y − 1 _____ 3 y + 5 _____ 2 + + + 2z = −1 z + 5 _____ 2z = 3 51. 50. x − 3 _____ 10 x + 5 _____ 4 x − 1 _____ 4 + − y + 3 _____ 2 y − 1 _____ 8 y + 4 _____ 2 + + z = 3 __ 2 + 3z = 3 __ 2 − x − 3 _____ 4 x + 5 _____ 2 y − 1 _____ 3 y + 5 _____ 2 + + + 2z = −1 z + 5 _____ ReAl-WORlD APPlICATIOnS For the following exercises, set up the augmented matrix that describes the situation, and solve for the desired solution. 52. Every day, a cupcake store sells 5,000 cupcakes in 53. At a competing cupcake store, $4,520 worth of chocolate and vanilla flavors. If the chocolate flavor is 3 times as popular as the vanilla flavor, how many of each cupcake sell per day? 54. You invested $10,000 into two accounts: one that has simple 3% interest, the other with 2.5% interest. If your total interest payment after one year was $283.50, how much was in each account after the year passed? cupcakes are sold daily. The chocolate cupcakes cost $2.25 and the red velvet cupcakes cost $1.75. If the total number of cupcakes sold per day is 2,200, how many of each flavor are sold each day? 55. You invested $2,300 into account 1, and $2,700 into account 2. If the total amount of interest after one year is $254, and account 2 has 1.5 times the interest rate of account 1, what are the interest rates? Assume simple interest rates. 828 CHAPTER 9 systems oF eQuations and ineQualities 56. Bikes’R’Us manufactures bikes, which sell for $250. It costs the manufacturer $180 per bike, plus a startup fee of $3,500. After how many bikes sold will the manufacturer break even? 57. A major appliance store is considering purchasing vacuums from a small manufacturer. The store would be able to purchase the vacuums for $86 each, with a delivery fee of $9,200, regardless of how many vacuums are sold. If the store needs to start seeing a profit after 230 units are sold, how much should they charge for the vacuums? 58. The three most popular ice cream flavors are 59. At an ice cream shop, three flavors are increasing chocolate, strawberry, and vanilla, comprising 83% of the flavors sold at an ice cream shop. If vanilla sells 1% more than twice strawberry, and chocolate sells 11% more than vanilla, how much of the total ice cream consumption are the vanilla, chocolate, and strawberry flavors? in demand. Last year, banana, pumpkin, and rocky road ice cream made up 12% of total ice cream sales. This year, the same three ice creams made up 16.9% of ice cream sales. The rocky road sales doubled, the banana sales increased by 50%, and the pumpkin sales increased by 20%. If the rocky road ice cream had one less percent of sales than the banana ice cream, find out the percentage of ice cream sales each individual ice cream made last year. 60. A bag of mixed nuts contains cashews, pistachios, and almonds. There are 1,000 total nuts in the bag, and there are 100 less almonds than pistachios. The cashews weigh 3 g, pistachios weigh 4 g, and almonds weigh 5 g. If the bag weighs 3.7 kg, find out how many of each type of nut is in the bag. 61. A bag of mixed nuts contains cashews, pistachios, and almonds. Originally there were 900 nuts in the bag. 30% of the almonds, 20% of the cashews, and 10% of the pistachios were eaten, and now there are 770 nuts left in the bag. Originally, there were 100 more cashews than almonds. Figure out how many of each type of nut was in the bag to begin with. SECTION 9.7 solving systems with inverses 829 leARnInG OBjeCTIVeS In this section, you will: • Find the inverse of a matrix. • Solve a system of linear equations using an inverse matrix. 9.7 SOlVInG SYSTeMS WITH InVeRSeS Nancy plans to invest $10,500 into two different bonds to spread out her risk. The first bond has an annual return of 10%, and the second bond has an annual return of 6%. In order to receive an 8.5% return from the two bonds, how much should Nancy invest in each bond? What is the best method to solve this problem? There are several ways we can solve this problem. As we have seen in previous sections, systems of equations and matrices are useful in solving real-world problems involving finance. After studying this section, we will have the tools to solve the bond problem using the inverse of a matrix. Finding the Inverse of a Matrix We know that the multiplicative inverse of a real number a is a −1 , and a a −1 = a −1 a =  1 __ a  a = 1. For example, 2 −1 = 1 __ 2 and  1 __  2 = 1. The multiplicative inverse of a matrix is similar in concept, except that the product of matrix A and its 2 inverse A −1 equals the identity matrix. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by I n where n represents the dimension of the matrix. The following equations are the identity matrices for a 2 × 2 matrix and a 3 × 3 matrix, respectively  The identity matrix acts as a 1 in matrix algebra. For example, AI = IA = A. A matrix that has a multiplicative inverse has the properties AA −1 = I A −1 A = I A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility, AA −1 = A −1 A = I, is a requirement. Not all square matrices have an inverse, but if A is invertible, then A −1 is unique. We will look at two methods for finding the inverse of a 2 × 2 matrix and a third method that can be used on both 2 × 2 and 3 × 3 matrices. the identity matrix and multiplicative inverse The identity matrix, I n , is a square matrix containing ones down the main diagonal and zeros everywhere else  If A is an n × n matrix and B is an n × n matrix such that AB = BA = I n , then B = A −1 , the multiplicative inverse of a matrix A. 830 CHAPTER 9 systems oF eQuations and ineQualities Example 1 Showing That the Identity Matrix Acts as a 1 Given matrix A, show that AI = IA = A. A =  3 4  −2 5 Solution Use matrix multiplication to show that the product of A and the identity is equal to the product of the identity and A. AI =  3 4 −2 5  1 0  0 1  =  AI =  1 0 0 1  3 4  −2 ⋅ 1 + 5 ⋅ 0 − ⋅ (−2 ⋅ (−22 5  =  3 4 −2 5  How To… Given two matrices, show that one is the multiplicative inverse of the other. 1. Given matrix A of order n × n and matrix B of order n × n multiply AB. 2. If AB = I, then find the product BA. If BA = I, then B = A −1 and A = B −1 . Example 2 Showing That Matrix A Is the Multiplicative Inverse of Matrix B Show that the given matrices are multiplicative inverses of each other. 1 5 −2 −9 −9 −5 2 1  , B =  A =   Solution Multiply AB and BA. If both products equal the identity, then the two matrices are inverses of each other. AB =  1 5 −2 −9   −9 −5 2 1  =  1(−9) + 5(2) 1(−5) + 5(1)  −2(−9) − 9(2) −2(−5) − 9(1) =  1 0  0 1 −9 −5 2 1 BA =    1 5 −2 −9  =  −9(1) − 5(−2) −9(5) − 5(−9)  2(1) + 1(−2) 2(−5) + 1(−9) =  1 0 0 1  A and B are inverses of each other. Try It #1 Show that the following two matrices are inverses of each other. A =  1 4 −1 −3  , B =  −3 −4 1 1  Finding the Multiplicative Inverse Using Matrix Multiplication We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication. SECTION 9.7 solving systems with inverses 831 Example 3 Finding the Multiplicative Inverse Using Matrix Multiplication Use matrix multiplication to find the inverse of the given matrix. A =  Solution For
this method, we multiply A by a matrix containing unknown constants and set it equal to the identity. 1 −2  2 −3  Find the product of the two matrices on the left side of the equal sign.  =  1 −2   2 −  =  Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0.  1a − 2c 1b − 2d 2a − 3c 2b − 3d 1 −2   2 −3 a b  c d Using row operations, multiply and add as follows: (−2) R 1 + R 2 → R 2 . Add the equations, and solve for c. 1a − 2c = 1 2a − 3c = 0 R 1 R 2 Back-substitute to solve for a. 1a − 2c = 1 0 + 1c = −2 c = −2 a − 2(−23 Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity. Using row operations, multiply and add as follows: (−2) R 1 + R 2 = R 2 . Add the two equations and solve for d. 1b − 2d = 0 2b − 3d = 1 R 1 R 2 Once more, back-substitute and solve for b. 1b − 2d = 0 0 + 1d = 1 d = 1 b − 2(11 =  −3 2  −2 1 Finding the Multiplicative Inverse by Augmenting with the Identity Another way to find the multiplicative inverse is by augmenting with the identity. When matrix A is transformed into I, the augmented matrix I transforms into A −1 . For example, given augment A with the identity A =  2 1  5 3 832 CHAPTER 9 systems oF eQuations and ineQualities  Perform row operations with the goal of turning A into the identity. Switch row 1 and row 2. 2. Multiply row 2 by −2 and add to row 1. 3. Multiply row 1 by −2 and add to row 2. 4. Add row 2 to row 1. 5. Multiply row 2 by −1. The matrix we have found is A −1 −2 1  5 −2 1 0  0 −1 3 −1 5 −2  1 0  0 1 3 −1  −5 2 ∣ A −1 =  3 −1  −5 2 Finding the Multiplicative Inverse of 2 × 2 Matrices Using a Formula When we need to find the multiplicative inverse of a 2 × 2 matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity. If A is a 2 × 2 matrix, such as the multiplicative inverse of A is given by the formula 1 = d −b 1  _______ ad − bc −c a  where ad − bc ≠ 0. If ad − bc = 0, then A has no inverse. Example 4 Using the Formula to Find the Multiplicative Inverse of Matrix A Use the formula to find the multiplicative inverse of Solution Using the formula, we have A =  1 −2 2 −3  A −1 = −3 2 1   ________________ (1)(−3) − (−2)(2) −2 1 = −3 2 1   _______ −3 + 4 −2 1 −3 2  −2 1 =  Analysis We can check that our formula works by using one of the other methods to calculate the inverse. Let’s augment A with the identity. 1 0  0 1 Perform row operations with the goal of turning A into the identity. 1 −2  2 −3 ∣ SECTION 9.7 solving systems with inverses 833 1. Multiply row 1 by −2 and add to row 2. 2. Multiply row 1 by 2 and add to row 1. So, we have verified our original solution. 1 −2  0 1 1 0 −2 1  1 0  0 1 −3 2 −2 1  ∣ ∣ A −1 =  −3 2  −2 1 Try It #2 Use the formula to find the inverse of matrix A. Verify your answer by augmenting with the identity matrix. A −1 =  1 −1 2 3  Example 5 Finding the Inverse of the Matrix, If It Exists Find the inverse, if it exists, of the given matrix. A =  3 6 1 2  Solution We will use the method of augmenting with the identity. Switch row 1 and row 2. 2. Multiply row 1 by −3 and add it to row 2. There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse. 1 0 −3 1  Finding the Multiplicative Inverse of 3 × 3 Matrices Unfortunately, we do not have a formula similar to the one for a 2 × 2 matrix to find the inverse of a 3 × 3 matrix. Instead, we will augment the original matrix with the identity matrix and use row operations to obtain the inverse. Given a 3 × 3 matrix augment A with the identity matrix  To begin, we write the augmented matrix with the identity on the right and A on the left. Performing elementary row operations so that the identity matrix appears on the left, we will obtain the inverse matrix on the right. We will find the inverse of this matrix in the next example. How To… Given a 3 × 3 matrix, find the inverse 1. Write the original matrix augmented with the identity matrix on the right. 2. Use elementary row operations so that the identity appears on the left. 3. What is obtained on the right is the inverse of the original matrix. 4. Use matrix multiplication to show that A A −1 = I and A −1 A = I. 834 CHAPTER 9 systems oF eQuations and ineQualities Example 6 Finding the Inverse of a 3 × 3 Matrix Given the 3 × 3 matrix A, find the inverse  Solution Augment A with the identity matrix, and then begin row operations until the identity matrix replaces A. The matrix on the right will be the inverse of A   ∣ Interchange R 2 and 1 0 1 −1 1 0 −1 0 1 1 0 0 −1 1 0 −1 0 1 3 −2 0 −1 1 0 −1 0 1 6 −2 −3   Thus, Analysis To prove that B = A −1 , let’s multiply the two matrices together to see if the product equals the identity, if A A −1 = I and A −1 A = I. A −1 = B =  −1 1 0 −1 0 1 6 −2 −1 1 0 −1 0 1 6 −2 −1 1 0 −1 0 1 6 −2 −1 =  =  =  A −(−1) + 3(−1) + 1(6) 2(1) + 3(0) + 1(−2) 2(0) + 3(1) + 1(−3) 3(−1) + 3(−1) + 1(6) 3(1) + 3(0) + 1(−2) 3(0) + 3(1) + 1(−3) 2(−1) + 4(−1) + 1(6) 2(1) + 4(0) + 1(−2) 2(0) + 4(1) + 1(−3)  −1(2) + 1(3) + 0(2) −1(2) + 0(3) + 1(2) −1(3) + 1(3) + 0(4) −1(3) + 0(3) + 1(4) 6(2) + −2(3) + −3(2) 6(3) + −2(3) + −3(4) 6(1) + −2(1) + −3(1) −1(1) + 1(1) + 0(1) −1(1) + 0(1) + 1(1)  SECTION 9.7 solving systems with inverses 835 Try It #3 Find the inverse of the 3 × 3 matrix. A =  2 −17 11 −1 11 −7 0 3 −2  Solving a System of linear equations Using the Inverse of a Matrix Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: X is the matrix representing the variables of the system, and B is the matrix representing the constants. Using matrix multiplication, we may define a system of equations with the same number of equations as variables as AX = B To solve a system of linear equations using an inverse matrix, let A be the coefficient matrix, let X be the variable matrix, and let B be the constant matrix. Thus, we want to solve a system AX = B. For example, look at the following system of equations. From this system, the coefficient matrix is The variable matrix is And the constant matrix is Then AX = B looks like Recall the discussion earlier in this section regarding multiplying a real number by its inverse, ( 2 −1 ) 2 =  1 __  2 = 1. To 2 solve a single linear equation ax = b for x, we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of a. Thus, ax = b  1 __ a  ax =  1 __ a  b ( a −1 )ax = ( a −1 )b [( a −1 )a]x = ( a −1 )b 1x = ( a −1 )b x = ( a −1 )b The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same—to isolate the variable. We will investigate this idea in detail, but it is helpful to begin with a 2 × 2 system and then move on to a 3 × 3 system. 836 CHAPTER 9 systems oF eQuations and ineQualities solving a system of equations using the inverse of a matrix Given a system of equations, write the coefficient matrix A, the variable matrix X, and the constant matrix B. Then Multiply both sides by the inverse of A to obtain the solution. AX = B ( A −1 )AX = ( A −1 )B [( A −1 )A]X = ( A −1 )B IX = ( A −1 )B X = ( A −1 )B Q & A… If the coefficient matrix does not have an inverse, does that mean the system has no solution? No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions. Example 7 Solving a 2 × 2 System Using the Inverse of a Matrix Solve the given system of equations using the inverse of a matrix. 3x + 8y = 5 4x + 11y = 7 Solution Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix. Then 11 7 y  First, we need to calculate A −1 . Using the formula to calculate the inverse of a 2 by 2 matrix, we have:  8 3  4 11 1 = d −b 1   _______ ad − bc −c a 11 −8  −4 3 1  ___________ 3(11) − 8(4) = = 1 __  1 11 −8  −4 3 So, 11 −8  −4 3 Now we are ready to solve. Multiply both sides of the equation by A −1 . A −1 =  ( A −1 )AX = ( A −1 )B 11 −8   −4 3 3 8  4 11  1 0  0 1  The solution is (−1, 1).  11 −8 5   −4 3 7 11(5) + (−8)7  −4(5) + 3(7) −1  1 SECTION 9.7 solving systems with inverses 837 Q & A… Can we solve for X by finding the product B A −1 ? No, recall that matrix multiplication is not commutative, so A −1 B ≠ B A −1 . Consider our steps for solving the matrix equation. ( A −1 )AX = ( A −1 )B [( A −1 )A]X = ( A −1 )B IX = ( A −1 )B X = ( A −1 )B Notice in the first step we multiplied both sides of the equation by A −1 , but the A −1 was to the left of A on the left side and to the left of B on the right side. Because matrix multiplication is not commutative, order matters. Example 8 Solving a 3 × 3 System Using the Inverse of a Matrix Solve the following system using the inverse of a matrix. 5x + 15y + 56z = 35 −4x − 11y − 41z = −26 −x − 3y − 11z = −7 Solution Write the equation AX = B. First, we will find the inverse of A by augmenting with the identity. 1 0 15 −4 −11 −41 0 1 0 −3 −1 0 0 −11  56 0 1 5 Multiply row 1 by 1 __ . 5 5 15 56 −4 −11 −41 −1 −3 −11   35  −26 −7 1 1 _ 5 56 __ 5 0 3 0 1 0 −4 −11 −41 −3 −1 0 0 −11 0 1 x  =   y z ∣ ∣ ∣   56 __ 5 1 _ 5 0 0 19 __ 5 4 _ 5 1 0 −11 1 0 0 1 0 −1 3 1 −3 3 1  0 1 0 0 56 __ 5 1 _ 5 0 0 19 __ __ 5 19 __ 5 1 _ 5 − 11 _ 5 4 _ 5 1 _ 5 − __ 5 19 __ 5 1 − 11 _ 5 4 _ 5 1 −3 0 1 0 5 0   ∣ ∣ ∣      Multiply row 1 by 4 and add to row 2. Add row 1 to row 3. Multiply row 2 by −3 and add to row 1. Multiply row 3 by 5. 838 CHAPT
ER 9 systems oF eQuations and ineQualities Multiply row 3 by 1 __ 5 and add to row 1. Multiply row 3 by − 19 ___ 5 and add to row 2. So, 0 19 __ 5 1 ∣ ∣ −2 −2 4 _ 5 1 −3 1 1 −19 5 0  A −1 =  −2 −3 1 −3 1 −19 1 0 5  Multiply both sides of the equation by A −1 . We want A −1 AX = A −1 B: −2 −3 1 −3 1 −19 1 0 5    5 15 56 −4 −11 −41 −1 −3 −11 x    =  y z −2 −3 1 −3 1 −19 1 0 5 35    −26 −7 Thus, The solution is (1, 2, 0). A −1 B =  −70 + 78−7 −105 − 26 + 133 35 + 0 − 35  =  1 2 0  Try It #4 Solve the system using the inverse of the coefficient matrix. 2x − 17y + 11z = 0 −x + 11y − 7z = 8 3y − 2z = −2 How To… Given a system of equations, solve with matrix inverses using a calculator. 1. Save the coefficient matrix and the constant matrix as matrix variables [A] and [B]. 2. Enter the multiplication into the calculator, calling up each matrix variable as needed. 3. If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message. Example 9 Using a Calculator to Solve a System of Equations with Matrix Inverses Solve the system of equations with matrix inverses using a calculator 2x + 3y + z = 32 3x + 3y + z = −27 2x + 4y + z = −2 Solution On the matrix page of the calculator, enter the coefficient matrix as the matrix variable [A], and enter the constant matrix as the matrix variable [B]. [AB] =  32  −27 −2 SECTION 9.7 solving systems with inverses 839 On the home screen of the calculator, type in the multiplication to solve for X, calling up each matrix variable as needed. Evaluate the expression. [A] −1 × [B] −59   −34 252 Access these online resources for additional instruction and practice with solving systems with inverses. • The Identity Matrix (http://openstaxcollege.org/l/identmatrix) • Determining Inverse Matrices (http://openstaxcollege.org/l/inversematrix) • Using a Matrix equation to Solve a System of equations (http://openstaxcollege.org/l/matrixsystem) 840 CHAPTER 9 systems oF eQuations and ineQualities 9.7 SeCTIOn exeRCISeS VeRBAl 1. In a previous section, we showed that matrix 2. Does every 2 × 2 matrix have an inverse? Explain multiplication is not commutative, that is, AB ≠ BA in most cases. Can you explain why matrix multiplication is commutative for matrix inverses, that is, A −1 A = A A −1 ? why or why not. Explain what condition is necessary for an inverse to exist. 3. Can you explain whether a 2 × 2 matrix with an 4. Can a matrix with an entire column of zeros have an entire row of zeros can have an inverse? inverse? Explain why or why not. 5. Can a matrix with zeros on the diagonal have an inverse? If so, find an example. If not, prove why not. For simplicity, assume a 2 × 2 matrix. AlGeBRAIC In the following exercises, show that matrix A is the inverse of matrix B. 6. A =  1 0 − __ 35 9. A =  −1 −2 −1 −6 −4  10 __ 2 2 1 −1 0 1 1 0 −1 1  12 12   , B = 1 ___ 36 −6 84 −6 7 −26 1 −1 −22 5  11 __ 4 6 0 −2 17 −3 −5 −12 2 4  For the following exercises, find the multiplicative inverse of each matrix, if it exists. 13.  3 −2 1 9  17.  1 1  2 2 14.  −2 2 3 1  18.  0 1 1 0  15.  −3 7 9 2  16.  −4 −3  −5 8 19.  0.5 1.5  1 −0.5 20.  1 0 6 −2 1 7 3 0 2  21.  0 1 −3 4 1 0 1 0 5  22.  1 2 −1 −3 4 1 −2 −4 −5  23.  1 9 −3 2 5 6 4 −2 7  25 26  24.  1 −2 3 −4 8 −12 1 4 2  SECTION 9.7 section exercises 841 For the following exercises, solve the system using the inverse of a 2 × 2 matrix. 27. 5x − 6y = −61 4x + 3y = −2 28. 8x + 4y = −100 3x − 4y = 1 29. 3x − 2y = 6 −x + 5y = −2 30. 5x−4y = −5 4x + y = 2.3 31. −3x − 4y = 9 12x + 4y = −6 32. −2x + 3y = 3 ___ 10 − x + 5y = 1 __ 2 33. 8 __ y = 2 __ x − 4 __ 5 5 5 y = 7 ___ − 8 __ x + 1 __ 5 5 10 34. 1 __ 2 1 __ 2 y = − 1 __ x + 1 __ 5 4 y = − 9 __ x − 3 __ 5 4 For the following exercises, solve a system using the inverse of a 3 × 3 matrix. 35. 3x − 2y + 5z = 21 5x + 4y = 37 x − 2y − 5z = 5 36. 4x + 4y + 4z = 40 2x − 3y + 4z = −12 −x + 3y + 4z = 9 37. 6x − 5y − z = 31 −x + 2y + z = −6 3x + 3y + 2z = 13 38. 6x − 5y + 2z = −4 2x + 5y − z = 12 2x + 5y + z = 12 39. 4x − 2y + 3z = −12 2x + 2y − 9z = 33 6y − 4z = 1 z = 31 ___ 41. 1 __ y + 1 __ x − 1 __ 5 5 100 2 z = 7 ___ − 3 __ y + 1 __ x − 1 __ 4 4 40 2 4 y + 3 __ x − 1 __ __ z = 14 − 2 2 5 TeCHnOlOGY y + 4z = − 41 ___ 2 x − 1 __ 40. 1 ___ 5 10 1 __ x − 20y + 2 __ z = −101 5 5 x + 4y − 3 ___ 3 ___ 10 10 z = 23 42. 0.1x + 0.2y + 0.3z = −1.4 0.1x − 0.2y + 0.3z = 0.6 0.4y + 0.9z = −2 For the following exercises, use a calculator to solve the system of equations with matrix inverses. 43. 2x − y = −3 −x + 2y = 2.3 45. 12.3x − 2y − 2.5z = 2 36.9x + 7y − 7.5z = −7 8y − 5z = −10 exTenSIOnS x − 3 __ y = − 43 ___ 44. − 1 __ 20 2 2 y = 31 ___ 5 __ x + 11 ___ 5 4 2 46. 0.5x − 3y + 6z = −0.8 0.7x − 2y = −0.06 0.5x + 4y + 5z = 0 For the following exercises, find the inverse of the given matrix. 47. 51   48.  −1 0 2 −3 1 0  492 3 −5 0 1 1   50   842 CHAPTER 9 systems oF eQuations and ineQualities ReAl-WORlD APPlICATIOnS For the following exercises, write a system of equations that represents the situation. Then, solve the system using the inverse of a matrix. 52. 2,400 tickets were sold for a basketball game. If the prices for floor 1 and floor 2 were different, and the total amount of money brought in is $64,000, how much was the price of each ticket? 53. In the previous exercise, if you were told there were 400 more tickets sold for floor 2 than floor 1, how much was the price of each ticket? 54. A food drive collected two different types of canned goods, green beans and kidney beans. The total number of collected cans was 350 and the total weight of all donated food was 348 lb, 12 oz. If the green bean cans weigh 2 oz less than the kidney bean cans, how many of each can was donated? 55. Students were asked to bring their favorite fruit to class. 95% of the fruits consisted of banana, apple, and oranges. If oranges were twice as popular as bananas, and apples were 5% less popular than bananas, what are the percentages of each individual fruit? 56. A sorority held a bake sale to raise money and sold brownies and chocolate chip cookies. They priced the brownies at $1 and the chocolate chip cookies at $0.75. They raised $700 and sold 850 items. How many brownies and how many cookies were sold? 58. Anna, Ashley, and Andrea weigh a combined 370 lb. If Andrea weighs 20 lb more than Ashley, and Anna weighs 1.5 times as much as Ashley, how much does each girl weigh? 60. A farmer constructed a chicken coop out of chicken wire, wood, and plywood. The chicken wire cost $2 per square foot, the wood $10 per square foot, and the plywood $5 per square foot. The farmer spent a total of $51, and the total amount of materials used was 14 ft 2 . He used 3 ft 2 more chicken wire than plywood. How much of each material in did the farmer use? 57. A clothing store needs to order new inventory. It has three different types of hats for sale: straw hats, beanies, and cowboy hats. The straw hat is priced at $13.99, the beanie at $7.99, and the cowboy hat at $14.49. If 100 hats were sold this past quarter, $1,119 was taken in by sales, and the amount of beanies sold was 10 more than cowboy hats, how many of each should the clothing store order to replace those already sold? 59. Three roommates shared a package of 12 ice cream bars, but no one remembers who ate how many. If Tom ate twice as many ice cream bars as Joe, and Albert ate three less than Tom, how many ice cream bars did each roommate eat? 61. Jay has lemon, orange, and pomegranate trees in his backyard. An orange weighs 8 oz, a lemon 5 oz, and a pomegranate 11 oz. Jay picked 142 pieces of fruit weighing a total of 70 lb, 10 oz. He picked 15.5 times more oranges than pomegranates. How many of each fruit did Jay pick? SECTION 9.8 solving systems with cramer's rule 843 leARnInG OBjeCTIVeS In this section, you will: • Evaluate 2 × 2 determinants. • Use Cramer’s Rule to solve a system of equations in two variables. • Evaluate 3 × 3 determinants. • Use Cramer’s Rule to solve a system of three equations in three variables. • Know the properties of determinants. 9.8 SOlVInG SYSTeMS WITH CRAMeR'S RUle We have learned how to solve systems of equations in two variables and three variables, and by multiple methods: substitution, addition, Gaussian elimination, using the inverse of a matrix, and graphing. Some of these methods are easier to apply than others and are more appropriate in certain situations. In this section, we will study two more strategies for solving systems of equations. evaluating the Determinant of a 2 × 2 Matrix A determinant is a real number that can be very useful in mathematics because it has multiple applications, such as calculating area, volume, and other quantities. Here, we will use determinants to reveal whether a matrix is invertible by using the entries of a square matrix to determine whether there is a solution to the system of equations. Perhaps one of the more interesting applications, however, is their use in cryptography. Secure signals or messages are sometimes sent encoded in a matrix. The data can only be decrypted with an invertible matrix and the determinant. For our purposes, we focus on the determinant as an indication of the invertibility of the matrix. Calculating the determinant of a matrix involves following the specific patterns that are outlined in this section. find the determinant of a 2 × 2 matrix The determinant of a 2 × 2 matrix, given is defined as A =  det(A  ∣ = ad − cb Notice the change in notation. There are several ways to indicate the determinant, including det(A) and replacing the brackets in a matrix with straight lines, ∣ A ∣ . Example 1 Finding the Determinant of a 2 × 2 Matrix Find the determinant of the given matrix. Solution A =  5 2 −6 3 det(A) = ∣ 5 2 −6 3  ∣ = 5(3) − (−6)(2) = 27 Using Cramer’s Rule to Solve a System of Two equations in Two Variables We will now int
roduce a final method for solving systems of equations that uses determinants. Known as Cramer’s Rule, this technique dates back to the middle of the 18th century and is named for its innovator, the Swiss mathematician Gabriel Cramer (1704-1752), who introduced it in 1750 in Introduction à l'Analyse des lignes Courbes algébriques. Cramer’s Rule is a viable and efficient method for finding solutions to systems with an arbitrary number of unknowns, provided that we have the same number of equations as unknowns. 844 CHAPTER 9 systems oF eQuations and ineQualities Cramer’s Rule will give us the unique solution to a system of equations, if it exists. However, if the system has no solution or an infinite number of solutions, this will be indicated by a determinant of zero. To find out if the system is inconsistent or dependent, another method, such as elimination, will have to be used. To understand Cramer’s Rule, let’s look closely at how we solve systems of linear equations using basic row operations. Consider a system of two equations in two variables1) (2) We eliminate one variable using row operations and solve for the other. Say that we wish to solve for x. If equation (2) is multiplied by the opposite of the coefficient of y in equation (1), equation (1) is multiplied by the coefficient of y in equation (2), and we add the two equations, the variable y will be eliminated. Now, solve for x Multiply R 1 by Multiply R 2 by − Similarly, to solve for y, we will eliminate x. Solving for y gives Multiply R 1 by Multiply R 2 by − Notice that the denominator for both x and y is the determinant of the coefficient matrix. We can use these formulas to solve for x and y, but Cramer’s Rule also introduces new notation: • D : determinant of the coefficient matrix • D x : determinant of the numerator in the solution of x • D y : determinant of the numerator in the solution of The key to Cramer’s Rule is replacing the variable column of interest with the constant column and calculating the determinants. We can then express x and y as a quotient of two determinants. SECTION 9.8 solving systems with cramer's rule 845 Cramer’s Rule for 2 × 2 systems Cramer’s Rule is a method that uses determinants to solve systems of equations that have the same number of equations as variables. Consider a system of two linear equations in two variables The solution using Cramer’s Rule is given as ; y = = , D ≠ 0 If we are solving for x, the x column is replaced with the constant column. If we are solving for y, the y column is replaced with the constant column. Example 2 Using Cramer’s Rule to Solve a 2 × 2 System Solve the following 2 × 2 system using Cramer’s Rule. 12x + 3y = 15 2x − 3y = 13 Solution Solve for x. x = = D x _ D −45 − 39 ________ = −36 − 6 −84 ____ −42 = 2 y = = D y _ D = 156 − 30 ________ −36 − 6 = − 126 ___ 42 = −3 15 3 13 −3 _ = 12 3 2 −3 ∣ ∣ 12 15 2 13 _ 12 3 2 −3 ∣ ∣ ∣ ∣ ∣ ∣ Solve for y. The solution is (2, −3). Try It #1 Use Cramer’s Rule to solve the 2 × 2 system of equations. x + 2y = −11 −2x + y = −13 evaluating the Determinant of a 3 × 3 Matrix Finding the determinant of a 2 × 2 matrix is straightforward, but finding the determinant of a 3 × 3 matrix is more complicated. One method is to augment the 3 × 3 matrix with a repetition of the first two columns, giving a 3 × 5 matrix. Then we calculate the sum of the products of entries down each of the three diagonals (upper left to lower right), and subtract the products of entries up each of the three diagonals (lower left to upper right). This is more easily understood with a visual and an example. Find the determinant of the 3 × 3 matrix  846 CHAPTER 9 systems oF eQuations and ineQualities 1. Augment A with the first two columns. det(A1 1 4 0 1 0 2 3 −1 4 2 ∣  ∣ 2. From upper left to lower right: Multiply the entries down the first diagonal. Add the result to the product of entries down the second diagonal. Add this result to the product of the entries down the third diagonal. 3. From lower left to upper right: Subtract the product of entries up the first diagonal. From this result subtract the product of entries up the second diagonal. From this result, subtract the product of entries up the third diagonal. det(A) = ∣ a 1 a 2 a 3 The algebra is as follows Example 3 Finding the Determinant of a 3 × 3 Matrix Find the determinant of the 3 × 3 matrix given Solution Augment the matrix with the first two columns and then follow the formula. Thus, = 0(−1)(1) + 2(1)(4) + 1(3)(0) − 4(−1)(1) − 0(1)(0) − 1(3)(2 Try It #2 Find the determinant of the 3 × 3 matrix. det(A) = ∣ 1 −3 7 1 1 1 1 −2 3 ∣ Q & A… Can we use the same method to find the determinant of a larger matrix? No, this method only works for 2 × 2 and 3 × 3 matrices. For larger matrices it is best to use a graphing utility or computer software. Using Cramer’s Rule to Solve a System of Three equations in Three Variables Now that we can find the determinant of a 3 × 3 matrix, we can apply Cramer’s Rule to solve a system of three equations in three variables. Cramer’s Rule is straightforward, following a pattern consistent with Cramer’s Rule for 2 × 2 matrices. As the order of the matrix increases to 3 × 3, however, there are many more calculations required. When we calculate the determinant to be zero, Cramer’s Rule gives no indication as to whether the system has no solution or an infinite number of solutions. To find out, we have to perform elimination on the system. SECTION 9.8 solving systems with cramer's rule 847 Consider a 3 × 3 system of equations where ∣ If we are writing the determinant D x , we replace the x column with the constant column. If we are writing the determinant D y , we replace the y column with the constant column. If we are writing the determinant D z , we replace the z column with the constant column. Always check the answer. Example 4 Solving a 3 × 3 System Using Cramer’s Rule Find the solution to the given 3 × 3 system using Cramer’s Rule. x + y − z = 6 3x − 2y + z = −5 x + 3y − 2z = 14 Solution Use Cramer’s Rule. D = ∣ 1 1 −1 3 −2 1 1 3 −1 −5 −2 1 14 3 −1 3 −5 1 1 14 −2 −5 1 3 14 ∣ Then3 ___ −3 _ D D y −9 ___ −3 _ D D z = 6 ___ − The solution is (1, 3, −2). Try It #3 Use Cramer’s Rule to solve the 3 × 3 matrix. x − 3y + 7z = 13 x + y + z = 1 x − 2y + 3z = 4 Example 5 Using Cramer’s Rule to Solve an Inconsistent System Solve the system of equations using Cramer’s Rule. (2) Solution We begin by finding the determinants D, D x , and D y . 6x − 4y = 0 3x − 2y = 4 (1) D = ∣ 3 −2 6 −4 ∣ = 3(−4) − 6(−2) = 0 848 CHAPTER 9 systems oF eQuations and ineQualities We know that a determinant of zero means that either the system has no solution or it has an infinite number of solutions. To see which one, we use the process of elimination. Our goal is to eliminate one of the variables. 1. Multiply equation (1) by −2. 2. Add the result to equation (2). −6x + 4y = −8 6x − 4y = 0 0 = −8 We obtain the equation 0 = −8, which is false. Therefore, the system has no solution. Graphing the system reveals two parallel lines. See Figure 1. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 – 21 3 4 5 Figure 1 Example 6 Use Cramer’s Rule to Solve a Dependent System Solve the system with an infinite number of solutions. x − 2y + 3z = 0 3x + y − 2z = 0 2x − 4y + 6z = 0 (1) (2) (3) Solution Let’s find the determinant first. Set up a matrix augmented by the first two columns. Then, ∣ 1 −2 3 3 1 −2 2 −4 6 1 −2 3 1 2 −4 ∣ ∣ 1(1)(6) + (−2)(−2)(2) + 3(3)(−4) − 2(1)(3) − (−4)(−2)(1) − 6(3)(−2) = 0 As the determinant equals zero, there is either no solution or an infinite number of solutions. We have to perform elimination to find out. 1. Multiply equation (1) by −2 and add the result to equation (3): −2x + 4y − 6x = 0 2x − 4y + 6z = 0 0 = 0 2. Obtaining an answer of 0 = 0, a statement that is always true, means that the system has an infinite number of solutions. Graphing the system, we can see that two of the planes are the same and they both intersect the third plane on a line. See Figure 2. SECTION 9.8 solving systems with cramer's rule 849 x − 2y + 3z = 0 2x − 4y + 6z = 0 3x + y + 2z = 0 Figure 2 Understanding Properties of Determinants There are many properties of determinants. Listed here are some properties that may be helpful in calculating the determinant of a matrix. properties of determinants 1. If the matrix is in upper triangular form, the determinant equals the product of entries down the main diagonal. 2. When two rows are interchanged, the determinant changes sign. 3. If either two rows or two columns are identical, the determinant equals zero. 4. If a matrix contains either a row of zeros or a column of zeros, the determinant equals zero. 5. The determinant of an inverse matrix A −1 is the reciprocal of the determinant of the matrix A. 6. If any row or column is multiplied by a constant, the determinant is multiplied by the same factor. Example 7 Illustrating Properties of Determinants Illustrate each of the properties of determinants. Solution Property 1 states that if the matrix is in upper triangular form, the determinant is the product of the entries down the main diagonal1  Augment A with the first two columns1 1 2 0 2 0 0  ∣ Then det(A) = 1(2)(−1) + 2(1)(0) + 3(0)(0) − 0(2)(3) − 0(1)(1) + 1(0)(2) = −2 Property 2 states that interchanging rows changes the sign. Given A =  B =  −1 5 4 −3 4 −3 −1 5  , det(A) = (−1)(−3) − (4)(5) = 3 − 20 = −17  , det(B) = (4)(5) − (−1)(−3) = 20 − 3 = 17 850 CHAPTER 9 systems oF eQuations and ineQualities Property 3 states that if two rows or two columns are identical, the determinant equals zero1 2  −1 2 2 ∣ det(A) = 1(2)(2) + 2(2)(−1) + 2(2)(2) + 1(2)(2) − 2(2)(1) − 2(2)(2 Property 4 states that if a row or column equals zero, the determinant equals zero. Thus, 1 2 0 0  , det(A) = 1(0) − 2(0) = 0 A =  Property 5 states that the determinant of an inverse matrix A −1 is the reciprocal of the determinant A. Thus, A =  1 2 3 4  , det(A) = 1(4) − 3(2) = −2 A −1 =  −2 1 − 1 _ 2
3 _ 2  , det( A −1 ) = −2  − 1 __ 2  −  3 __ 2  (1) = − 1 __ 2 Property 6 states that if any row or column of a matrix is multiplied by a constant, the determinant is multiplied by the same factor. Thus, A =  1 2 3 4  , det(A) = 1(4) − 2(3) = −2 B =  2(1) 3 2(2) 4  , det(B) = 2(4) − 3(4) = −4 Example 8 Using Cramer’s Rule and Determinant Properties to Solve a System Find the solution to the given 3 × 3 system. 2x + 4y + 4z = 2 3x + 7y + 7z = −5 x + 2y + 2z = 4 (1) (2) (3) Solution Using Cramer’s Rule, we have ∣ Notice that the second and third columns are identical. According to Property 3, the determinant will be zero, so there is either no solution or an infinite number of solutions. We have to perform elimination to find out. 1. Multiply equation (3) by −2 and add the result to equation (1). −2x − 4y − 4x = −8 2x + 4y + 4z = 2 0 = −6 Obtaining a statement that is a contradiction means that the system has no solution. Access these online resources for additional instruction and practice with Cramer’s Rule. • Solve a System of Two equations Using Cramer's Rule (http://openstaxcollege.org/l/system2cramer) • Solve a Systems of Three equations using Cramer's Rule (http://openstaxcollege.org/l/system3cramer) SECTION 9.8 section exercises 851 9.8 SeCTIOn exeRCISeS VeRBAl 1. Explain why we can always evaluate the determinant 2. Examining Cramer’s Rule, explain why there is no of a square matrix. 3. Explain what it means in terms of an inverse for a matrix to have a 0 determinant. AlGeBRAIC For the following exercises, find the determinant. unique solution to the system when the determinant of your matrix is 0. For simplicity, use a 2 × 2 matrix. 4. The determinant of 2 × 2 matrix A is 3. If you switch the rows and multiply the first row by 6 and the second row by 2, explain how to find the determinant and provide the answer4 ∣ −2 3.1 −3 4,000 ∣ . ∣ 9. ∣ 13. ∣ 17. ∣ 21. ∣ 5 1 −1 2 3 1 3 −6 −3 ∣ −1 2 3 −4 10 20 0 −10 ∣ ∣ −1.1 0.6 7.2 −0.5 2 −3 1 3 −4 1 −5 6 1 ∣ ∣ 6. ∣ 10. ∣ 14. ∣ 18. ∣ 22. ∣ 2 −5 −1 6 10 0.2 5 0.1 ∣ ∣ −1 0 0 0 1 0 0 0 −3 ∣ −2 1 4 −4 2 −8 2 −8 −3 ∣ 7. ∣ 11. ∣ 15. ∣ 19. ∣ 23. ∣ 1.1 −4 4.1 2 0 −0.4 −1 0 2.5 ∣ 2 1.1 −9.3 −1.6 3.1 3 −8 0 2 ∣ 8. ∣ 12. ∣ 16. ∣ 20. ∣ 24. ∣ −8 4 −1 5 6 −3 4 8 ∣ ∣ −1 4 0 0 2 3 0 0 −3 6 −1 2 −4 −3 5 1 9 − For the following exercises, solve the system of linear equations using Cramer’s Rule. 25. 2x − 3y = −1 4x + 5y = 9 29. 4x + 3y = 23 2x − y = −1 26. 5x − 4y = 2 −4x + 7y = 6 27. 6x − 3y = 2 −8x + 9y = −1 28. 2x + 6y = 12 5x − 2y = 13 30. 10x − 6y = 2 −5x + 8y = −1 31. 4x − 3y = −3 2x + 6y = −4 32. 4x − 5y = 7 −3x + 9y = 0 33. 4x + 10y = 180 −3x − 5y = −105 34. 8x − 2y = −3 −4x + 6y = 4 For the following exercises, solve the system of linear equations using Cramer’s Rule. 35. x + 2y − 4z = − 1 7x + 3y + 5z = 26 −2x − 6y + 7z = − 6 36. −5x + 2y − 4z = − 47 4x − 3y − z = − 94 3x − 3y + 2z = 94 37. 4x + 5y − z = −7 −2x − 9y + 2z = 8 5y + 7z = 21 38. 4x − 3y + 4z = 10 5x − 2z = − 2 3x + 2y − 5z = − 9 39. 4x − 2y + 3z = 6 40. 5x + 2y − z = 1 − 6x + y = − 2 2x + 7y + 8z = 24 −7x − 8y + 3z = 1.5 6x − 12y + z = 7 41. 13x − 17y + 16z = 73 −11x + 15y + 17z = 61 46x + 10y − 30z = − 18 42. −4x − 3y − 8z = − 7 2x − 9y + 5z = 0.5 5x − 6y − 5z = − 2 852 CHAPTER 9 systems oF eQuations and ineQualities 43. 4x − 6y + 8z = 10 −2x + 3y − 4z = − 5 x + y + z = 1 44. 4x − 6y + 8z = 10 −2x + 3y − 4z = − 5 12x + 18y − 24z = − 30 TeCHnOlOGY For the following exercises, use the determinant function on a graphing utility. 45 ∣ ∣ 469 1 3 −1 −2 0 −2 1 1 ∣ 47. 1 _ 2 100 ,000 0 0 0 2 48 ∣ ∣ ∣ ∣ ReAl-WORlD APPlICATIOnS For the following exercises, create a system of linear equations to describe the behavior. Then, calculate the determinant. Will there be a unique solution? If so, find the unique solution. 49. Two numbers add up to 56. One number is 20 less 50. Two numbers add up to 104. If you add two times than the other. the first number plus two times the second number, your total is 208 51. Three numbers add up to 106. The first number is 3 less than the second number. The third number is 4 more than the first number. 52. Three numbers add to 216. The sum of the first two numbers is 112. The third number is 8 less than the first two numbers combined. For the following exercises, create a system of linear equations to describe the behavior. Then, solve the system for all solutions using Cramer’s Rule. 53. You invest $10,000 into two accounts, which receive 8% interest and 5% interest. At the end of a year, you had $10,710 in your combined accounts. How much was invested in each account? 55. A movie theater needs to know how many adult tickets and children tickets were sold out of the 1,200 total tickets. If children’s tickets are $5.95, adult tickets are $11.15, and the total amount of revenue was $12,756, how many children’s tickets and adult tickets were sold? 57. You decide to paint your kitchen green. You create the color of paint by mixing yellow and blue paints. You cannot remember how many gallons of each color went into your mix, but you know there were 10 gal total. Additionally, you kept your receipt, and know the total amount spent was $29.50. If each gallon of yellow costs $2.59, and each gallon of blue costs $3.19, how many gallons of each color go into your green mix? 54. You invest $80,000 into two accounts, $22,000 in one account, and $58,000 in the other account. At the end of one year, assuming simple interest, you have earned $2,470 in interest. The second account receives half a percent less than twice the interest on the first account. What are the interest rates for your accounts? 56. A concert venue sells single tickets for $40 each and couple’s tickets for $65. If the total revenue was $18,090 and the 321 tickets were sold, how many single tickets and how many couple’s tickets were sold? 58. You sold two types of scarves at a farmers’ market and would like to know which one was more popular. The total number of scarves sold was 56, the yellow scarf cost $10, and the purple scarf cost $11. If you had total revenue of $583, how many yellow scarves and how many purple scarves were sold? SECTION 9.8 section exercises 853 59. Your garden produced two types of tomatoes, one green and one red. The red weigh 10 oz, and the green weigh 4 oz. You have 30 tomatoes, and a total weight of 13 lb, 14 oz. How many of each type of tomato do you have? 60. At a market, the three most popular vegetables make up 53% of vegetable sales. Corn has 4% higher sales than broccoli, which has 5% more sales than onions. What percentage does each vegetable have in the market share? 61. At the same market, the three most popular fruits 62. Three bands performed at a concert venue. The make up 37% of the total fruit sold. Strawberries sell twice as much as oranges, and kiwis sell one more percentage point than oranges. For each fruit, find the percentage of total fruit sold. first band charged $15 per ticket, the second band charged $45 per ticket, and the final band charged $22 per ticket. There were 510 tickets sold, for a total of $12,700. If the first band had 40 more audience members than the second band, how many tickets were sold for each band? 63. A movie theatre sold tickets to three movies. The 64. Men aged 20–29, 30–39, and 40–49 made up 78% tickets to the first movie were $5, the tickets to the second movie were $11, and the third movie was $12. 100 tickets were sold to the first movie. The total number of tickets sold was 642, for a total revenue of $6,774. How many tickets for each movie were sold? of the population at a prison last year. This year, the same age groups made up 82.08% of the population. The 20–29 age group increased by 20%, the 30–39 age group increased by 2%, and the 40–49 age group decreased to 3 __ of their previous population. 4 Originally, the 30–39 age group had 2% more prisoners than the 20–29 age group. Determine the prison population percentage for each age group last year. 65. At a women’s prison down the road, the total number of inmates aged 20–49 totaled 5,525. This year, the 20–29 age group increased by 10%, the 30–39 age group decreased by 20%, and the 40–49 age group doubled. There are now 6,040 prisoners. Originally, there were 500 more in the 30–39 age group than the 20–29 age group. Determine the prison population for each age group last year. For the following exercises, use this scenario: A health-conscious company decides to make a trail mix out of almonds, dried cranberries, and chocolate-covered cashews. The nutritional information for these items is shown in Table 1. Almonds (10) Cranberries (10) Cashews (10) Fat (g) 6 0.02 7 Protein (g) 2 0 3.5 Carbohydrates (g) 3 8 5.5 Table 1 67. For the “hiking” mix, there are 1,000 pieces in the mix, containing 390.8 g of fat, and 165 g of protein. If there is the same amount of almonds as cashews, how many of each item is in the trail mix? 66. For the special “low-carb” trail mix, there are 1,000 pieces of mix. The total number of carbohydrates is 425 g, and the total amount of fat is 570.2 g. If there are 200 more pieces of cashews than cranberries, how many of each item is in the trail mix? 68. For the “energy-booster” mix, there are 1,000 pieces in the mix, containing 145 g of protein and 625 g of carbohydrates. If the number of almonds and cashews summed together is equivalent to the amount of cranberries, how many of each item is in the trail mix? 854 CHAPTER 9 systems oF eQuations and ineQualities CHAPTeR 9 ReVIeW Key Terms addition method an algebraic technique used to solve systems of linear equations in which the equations are added in a way that eliminates one variable, allowing the resulting equation to be solved for the remaining variable; substitution is then used to solve for the first variable augmented matrix a coefficient matrix adjoined with the constant column separated by a vertical line within the matrix brackets break-even point the point at which a cost function intersects a revenue functio
n; where profit is zero coefficient matrix a matrix that contains only the coefficients from a system of equations column a set of numbers aligned vertically in a matrix consistent system a system for which there is a single solution to all equations in the system and it is an independent system, or if there are an infinite number of solutions and it is a dependent system cost function the function used to calculate the costs of doing business; it usually has two parts, fixed costs and variable costs Cramer’s Rule a method for solving systems of equations that have the same number of equations as variables using determinants dependent system a system of linear equations in which the two equations represent the same line; there are an infinite number of solutions to a dependent system determinant a number calculated using the entries of a square matrix that determines such information as whether there is a solution to a system of equations entry an element, coefficient, or constant in a matrix feasible region the solution to a system of nonlinear inequalities that is the region of the graph where the shaded regions of each inequality intersect Gaussian elimination using elementary row operations to obtain a matrix in row-echelon form identity matrix a square matrix containing ones down the main diagonal and zeros everywhere else; it acts as a 1 in matrix algebra inconsistent system a system of linear equations with no common solution because they represent parallel lines, which have no point or line in common independent system a system of linear equations with exactly one solution pair (x, y) main diagonal entries from the upper left corner diagonally to the lower right corner of a square matrix matrix a rectangular array of numbers multiplicative inverse of a matrix a matrix that, when multiplied by the original, equals the identity matrix nonlinear inequality an inequality containing a nonlinear expression partial fraction decomposition the process of returning a simplified rational expression to its original form, a sum or difference of simpler rational expressions partial fractions the individual fractions that make up the sum or difference of a rational expression before combining them into a simplified rational expression profit function the profit function is written as P(x) = R(x) − C(x), revenue minus cost revenue function the function that is used to calculate revenue, simply written as R = xp, where x = quantity and p = price row a set of numbers aligned horizontally in a matrix row operations adding one row to another row, multiplying a row by a constant, interchanging rows, and so on, with the goal of achieving row-echelon form row-echelon form after performing row operations, the matrix form that contains ones down the main diagonal and zeros at every space below the diagonal row-equivalent two matrices A and B are row-equivalent if one can be obtained from the other by performing basic row operations scalar multiple an entry of a matrix that has been multiplied by a scalar solution set the set of all ordered pairs or triples that satisfy all equations in a system of equations substitution method an algebraic technique used to solve systems of linear equations in which one of the two equations is solved for one variable and then substituted into the second equation to solve for the second variable CHAPTER 9 review 855 system of linear equations a set of two or more equations in two or more variables that must be considered simultaneously. system of nonlinear equations a system of equations containing at least one equation that is of degree larger than one system of nonlinear inequalities a system of two or more inequalities in two or more variables containing at least one inequality that is not linear Key equations Identity matrix for a 2 × 2 matrix Identity matrix for a 3 × 3 matrix Multiplicative inverse of a 2 × 2 matrix Key Concepts 9.1 Systems of Linear Equations: Two Variables 1 = d −b 1  _______ ad − bc −c a  , where ad − bc ≠ 0 • A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously. • The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. See Example 1. • Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution. • One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes. See Example 2. • Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation. See Example 3. • A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables. See Example 4. • It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together. See Example 5, Example 6, and Example 7. • Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect. See Example 8. • The solution to a system of dependent equations will always be true because both equations describe the same line. See Example 9. • Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit. See Example 10 and Example 11. 9.2 Systems of Linear Equations: Three Variables • A solution set is an ordered triple {(x, y, z)} that represents the intersection of three planes in space. See Example 1. • A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. The steps include interchanging the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a nonzero multiple of one equation to another equation. See Example 2. • Systems of three equations in three variables are useful for solving many different types of real-world problems. See Example 3. • A system of equations in three variables is inconsistent if no solution exists. After performing elimination operations, the result is a contradiction. See Example 4. • Systems of equations in three variables that are inconsistent could result from three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. 856 CHAPTER 9 systems oF eQuations and ineQualities • A system of equations in three variables is dependent if it has an infinite number of solutions. After performing elimination operations, the result is an identity. See Example 5. • Systems of equations in three variables that are dependent could result from three identical planes, three planes intersecting at a line, or two identical planes that intersect the third on a line. 9.3 Systems of Nonlinear Equations and Inequalities: Two Variables • There are three possible types of solutions to a system of equations representing a line and a parabola: (1) no solution, the line does not intersect the parabola; (2) one solution, the line is tangent to the parabola; and (3) two solutions, the line intersects the parabola in two points. See Example 1. • There are three possible types of solutions to a system of equations representing a circle and a line: (1) no solution, the line does not intersect the circle; (2) one solution, the line is tangent to the parabola; (3) two solutions, the line intersects the circle in two points. See Example 2. • There are five possible types of solutions to the system of nonlinear equations representing an ellipse and a circle: (1) no solution, the circle and the ellipse do not intersect; (2) one solution, the circle and the ellipse are tangent to each other; (3) two solutions, the circle and the ellipse intersect in two points; (4) three solutions, the circle and ellipse intersect in three places; (5) four solutions, the circle and the ellipse intersect in four points. See Example 3. • An inequality is graphed in much the same way as an equation, except for > or <, we draw a dashed line and shade the region containing the solution set. See Example 4. • Inequalities are solved the same way as equalities, but solutions to systems of inequalities must satisfy both inequalities. See Example 5. 9.4 Partial Fractions P(x) ____ Q(x) • Decompose by writing the partial fractions as A ________ + a 1 x + b 1 the right side, collecting like terms, and setting corresponding coefficients equal to each other, then setting up and solving a system of equations. See Example 1. . Solve by clearing the fractions, expanding B ________ a 2 x + b 2 • The decomposition of with repeated linear factors must account for the factors of the denominator in P(x) ____ Q(x) increasing powers. See Example 2. with a nonrepeated irreducible quadratic factor needs a linear numerator over the • The decomposition of P(x) ____ Q(x) quadratic factor, as in A __ x + . See Example 3. Bx + C ____________ (a x 2 + bx + c) P(x) ____ Q(x) • In the decomposition of , where Q(x) has a repeated irreducible quadratic factor, when the irreducible quadratic factors are repeated, powers of the denominator factors must be represented in increasing powers as Ax + B __ + (a x 2 + bx + c) A 2 x + B 2 __ (a x 2 + bx + c __ n (a x 2 + bx + c) . See Example 4. 9.5 Matrices and Matrix Operations • A matrix is a rectangular array of numbers. Entries are arranged in rows and columns. • The dimensions of a matrix refer to the number of rows and the number of columns. A 3 × 2 matrix has three rows and two columns. See Example 1. • We add and subtract matrices of
equal dimensions by adding and subtracting corresponding entries of each matrix. See Example 2, Example 3, Example 4, and Example 5. • Scalar multiplication involves multiplying each entry in a matrix by a constant. See Example 6. • Scalar multiplication is often required before addition or subtraction can occur. See Example 7. • Multiplying matrices is possible when inner dimensions are the same—the number of columns in the first matrix must match the number of rows in the second. • The product of two matrices, A and B, is obtained by multiplying each entry in row 1 of A by each entry in column 1 of B; then multiply each entry of row 1 of A by each entry in columns 2 of B, and so on. See Example 8 and Example 9. CHAPTER 9 review 857 • Many real-world problems can often be solved using matrices. See Example 10. • We can use a calculator to perform matrix operations after saving each matrix as a matrix variable. See Example 11. 9.6 Solving Systems with Gaussian Elimination • An augmented matrix is one that contains the coefficients and constants of a system of equations. See Example 1. • A matrix augmented with the constant column can be represented as the original system of equations. See Example 2. • Row operations include multiplying a row by a constant, adding one row to another row, and interchanging rows. • We can use Gaussian elimination to solve a system of equations. See Example 3, Example 4, and Example 5. • Row operations are performed on matrices to obtain row-echelon form. See Example 6. • To solve a system of equations, write it in augmented matrix form. Perform row operations to obtain row-echelon form. Back-substitute to find the solutions. See Example 7 and Example 8. • A calculator can be used to solve systems of equations using matrices. See Example 9. • Many real-world problems can be solved using augmented matrices. See Example 10 and Example 11. 9.7 Solving Systems with Inverses • An identity matrix has the property AI = IA = A. See Example 1. • An invertible matrix has the property A A −1 = A −1 A = I. See Example 2. • Use matrix multiplication and the identity to find the inverse of a 2 × 2 matrix. See Example 3. • The multiplicative inverse can be found using a formula. See Example 4. • Another method of finding the inverse is by augmenting with the identity. See Example 5. • We can augment a 3 × 3 matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse. See Example 6. • Write the system of equations as AX = B, and multiply both sides by the inverse of A: A −1 AX = A −1 B. See Example 7 and Example 8. • We can also use a calculator to solve a system of equations with matrix inverses. See Example 9. 9.8 Solving Systems with Cramer's Rule  is ad − bc. See Example 1. • The determinant for  • Cramer’s Rule replaces a variable column with the constant column. Solutions are = . See Example 2. • To find the determinant of a 3 × 3 matrix, augment with the first two columns. Add the three diagonal entries (upper left to lower right) and subtract the three diagonal entries (lower left to upper right). See Example 3. • To solve a system of three equations in three variables using Cramer’s Rule, replace a variable column with the constant D z _ D • Cramer’s Rule is also useful for finding the solution of a system of equations with no solution or infinite solutions. See column for each desired solution: x = . See Example 4 = Example 5 and Example 6. • Certain properties of determinants are useful for solving problems. For example: ○ If the matrix is in upper triangular form, the determinant equals the product of entries down the main diagonal. ○ When two rows are interchanged, the determinant changes sign. ○ If either two rows or two columns are identical, the determinant equals zero. ○ If a matrix contains either a row of zeros or a column of zeros, the determinant equals zero. ○ The determinant of an inverse matrix A −1 is the reciprocal of the determinant of the matrix A. ○ If any row or column is multiplied by a constant, the determinant is multiplied by the same factor. See Example 7 and Example 8. 858 CHAPTER 9 systems oF eQuations and ineQualities CHAPTeR 9 ReVIeW exeRCISeS SYSTeMS OF lIneAR eQUATIOnS: TWO VARIABleS For the following exercises, determine whether the ordered pair is a solution to the system of equations. 1. 3x − y = 4 x + 4y = − 3 and ( − 1, 1) 2. 6x − 2y = 24 −3x + 3y = 18 and (9, 15) For the following exercises, use substitution to solve the system of equations. 3. 10x + 5y = −5 3x − 2y = −12 4. 4 __ 7 5 __ 6 y = 43 ___ x + 1 __ 5 70 y = − 2 __ x − 1 __ 3 3 5. 5x + 6y = 14 4x + 8y = 8 For the following exercises, use addition to solve the system of equations. 6. 3x + 2y = −7 2x + 4y = 6 7. 3x + 4y = 2 9x + 12y = 3 8. 8x + 4y = 2 6x − 5y = 0.7 For the following exercises, write a system of equations to solve each problem. Solve the system of equations. 9. A factory has a cost of production C(x) = 150x + 15,000 and a revenue function R(x) = 200x. What is the break-even point? 10. A performer charges C(x) = 50x + 10,000, where x is the total number of attendees at a show. The venue charges $75 per ticket. After how many people buy tickets does the venue break even, and what is the value of the total tickets sold at that point? SYSTeMS OF lIneAR eQUATIOnS: THRee VARIABleS For the following exercises, solve the system of three equations using substitution or addition. 11. 0.5x − 0.5y = 10 − 0.2y + 0.2x = 4 0.1x + 0.1z = 2 14. 2x − 3y + z = −1 x + y + z = −4 4x + 2y − 3z = 33 17. 2x − 3y + z = 0 2x + 4y − 3z = 0 6x − 2y − z = 0 12. 5x + 3y − z = 5 3x − 2y + 4z = 13 4x + 3y + 5z = 22 15. 3x + 2y − z = −10 x − y + 2z = 7 −x + 3y + z = −2 18. 6x − 4y − 2z = 2 3x + 2y − 5z = 4 6y − 7z = 5 13. x + y + z = 1 2x + 2y + 2z = 1 3x + 3y = 2 16. 3x + 4z = −11 x − 2y = 5 4y − z = −10 For the following exercises, write a system of equations to solve each problem. Solve the system of equations. 19. Three odd numbers sum up to 61. The smaller is 20. A local theatre sells out for their show. They sell all one-third the larger and the middle number is 16 less than the larger. What are the three numbers? 500 tickets for a total purse of $8,070.00. The tickets were priced at $15 for students, $12 for children, and $18 for adults. If the band sold three times as many adult tickets as children’s tickets, how many of each type was sold? CHAPTER 9 review 859 SYSTeMS OF nOnlIneAR eQUATIOnS AnD IneQUAlITIeS: TWO VARIABleS For the following exercises, solve the system of nonlinear equations. 21. y = x 2 − 7 y = 5x − 13 22. y = x 2 − 4 y = 5x + 10 23. x 2 + y 2 = 16 24. x 2 + y 2 = 25 25 For the following exercises, graph the inequality. 26. y > x 2 − 1 27. 1 __ 4 x 2 + y 2 < 4 For the following exercises, graph the system of inequalities. 28. x 2 + y 2 + 2x < 3 y > − x 2 − 3 29. x 2 − 2x + y 2 − 4x < 4 y < − x + 4 30 PARTIAl FRACTIOnS For the following exercises, decompose into partial fractions. 31. −2x + 6 __________ x 2 + 3x + 2 34. x − 18 ____________ x 2 − 12x + 36 32. 10x + 2 ___________ 4 x 2 + 4x + 1 35. − x 2 + 36x + 70 _____________ x 3 − 125 33. 7x + 20 ____________ x 2 + 10x + 25 36. −5 x 2 + 6x − 2 ____________ x 3 + 27 37. x 3 − 4 x 2 + 3x + 11 ________________ ( x 2 − 2) 2 38. 4 x 4 − 2 x 3 + 22 x 2 − 6x + 48 _______________________ x( x 2 + 4) 2 MATRICeS AnD MATRIx OPeRATIOnS For the following exercises, perform the requested operations on the given matrices. A =  4 −3 11 −2 4  , C =  6 7 11 −2 14 0  , D =  1 −4 9 10 5 −14 3 2 −1 3 0 1 9  39. −4A 45. CB 40. 10D − 6E 46. DE 41. B + C 47. ED 42. AB 48. EC 43. BA 49. CE 44. BC 50. A 3 SOlVInG SYSTeMS WITH GAUSSIAn elIMInATIOn For the following exercises, write the system of linear equations from the augmented matrix. Indicate whether there will be a unique solution. 51.  1 0 −3 7  −5 0 1 2 0 0 0 0 ∣ 52.  −9 1 0 5 0 1 −2 4 3 0 0 0  ∣ For the following exercises, write the augmented matrix from the system of linear equations. 53. −2x + 2y + z = 7 2x − 8y + 5z = 0 19x − 10y + 22z = 3 54. 4x + 2y − 3z = 14 −12x + 3y + z = 100 9x − 6y + 2z = 31 55. x + 3z = 12 −x + 4y = 0 y + 2z = − 7 860 CHAPTER 9 systems oF eQuations and ineQualities For the following exercises, solve the system of linear equations using Gaussian elimination. 56. 3x − 4y = − 7 −6x + 8y = 14 57. 3x − 4y = 1 −6x + 8y = 6 58. −1.1x − 2.3y = 6.2 −5.2x − 4.1y = 4.3 59. 2x + 3y + 2z = 1 −4x − 6y − 4z = − 2 10x + 15y + 10z = 0 60. −x + 2y − 4z = 8 3y + 8z = − 4 −7x + y + 2z = 1 SOlVInG SYSTeMS WITH InVeRSeS For the following exercises, find the inverse of the matrix. 61.  −0.2 1.4  1.2 −0.4 62.  1 __ 2 − 1 __ 4 − 1 __ 2 3 __ 4  63.  12 9 −6 −1 3 2 −4 −3 2  64  For the following exercises, find the solutions by computing the inverse of the matrix. 65. 0.3x − 0.1y = −10 −0.1x + 0.3y = 14 66. 0.4x − 0.2y = −0.6 −0.1x + 0.05y = 0.3 67. 4x + 3y − 3z = −4.3 5x − 4y − z = −6.1 x + z = −0.7 68. −2x − 3y + 2z = 3 −x + 2y + 4z = −5 −2y + 5z = −3 For the following exercises, write a system of equations to solve each problem. Solve the system of equations. 69. Students were asked to bring their favorite fruit to class. 90% of the fruits consisted of banana, apple, and oranges. If oranges were half as popular as bananas and apples were 5% more popular than bananas, what are the percentages of each individual fruit? 70. A sorority held a bake sale to raise money and sold brownies and chocolate chip cookies. They priced the brownies at $2 and the chocolate chip cookies at $1. They raised $250 and sold 175 items. How many brownies and how many cookies were sold? SOlVInG SYSTeMS WITH CRAMeR'S RUle For the following exercises, find the determinant. 71.  100 0  0 0 72.  0.2 −0.6 0.7 −1.1  73.  −1 4 3 0 2 3 0 0 −3  √ 74 — For the following exercises, use Cramer’s Rule to solve the linear systems of equations. 75. 4x − 2y = 23 −5x − 10y = −35 76. 0.2x − 0.1y = 0 −0.3x + 0.3y = 2.5 7
7. −0.5x + 0.1y = 0.3 −0.25x + 0.05y = 0.15 78. x + 6y + 3z = 4 2x + y + 2z = 3 3x − 2y + z = 0 79. 4x − 3y + 5z = − 5 __ 2 7x − 9y − 3z = 3 __ 2 x − 5y − 5z = 5 __ 2 80. 3 ___ 10 1 ___ 10 2 __ 5 x − 1 __ 5 x − 1 ___ 10 x − 1 __ 2 y − 3 ___ 10 y − 1 __ 2 y − 3 __ 5 z = − 1 ___ 50 z = − 9 ___ 50 z = − 1 __ 5 CHAPTER 9 practice test 861 CHAPTeR 9 PRACTICe TeST Is the following ordered pair a solution to the system of equations? 1. −5x − y = 12 x + 4y = 9 with ( − 3, 3) For the following exercises, solve the systems of linear and nonlinear equations using substitution or elimination. Indicate if no solution exists. y = 4 x − 1 __ 3 2. 1 __ 2 3 __ x − y = 0 2 3. − 1 __ 2 x − 4y = 4 2x + 16y = 2 4. 5x − y = 1 −10x + 2y = − 2 6. x + z = 20 x + y + z = 20 x + 2y + z = 10 7. 5x − 4y − 3z = 0 2x + y + 2z = 0 x − 6y − 7z = 0 8. y = x 2 + 2x − 3 y = x − 1 For the following exercises, graph the following inequalities. 10. y < x 2 + 9 11. 4x − 6y − 2z = 1 ___ 10 x − 7y + 5z = − 1 __ 4 3x + 6y − 9z = 6 __ 5 9. y 2 + x 2 = 25 y 2 − 2 x 2 = 1 For the following exercises, write the partial fraction decomposition. 12. −8x − 30 ____________ x 2 + 10x + 25 13. 13x + 2 ________ (3x + 1) 2 14. x 4 − x 3 + 2x − 1 ______________ x ( x 2 + 1) 2 For the following exercises, perform the given matrix operations. 15. 5  4 9 −2 3 −6 12  + 1 __   2 4 −8 18. det ∣ 0 0 400 4,000 ∣ 16.  1 4 −7 −2 9 5 12 0 −4   3 −4 1 3 5 10  17.  1 __ 2 1 __ 4 1 __ 2 − 1 __ 2 0 19. det ∣ − 1 __ 2 0 1 __ 2 0 1 __ 2 0 ∣ −1 1 __ 3 1 __ 5  20. If det(A) = −6, what would be the determinant if you switched rows 1 and 3, multiplied the second row by 12, and took the inverse? 21. Rewrite the system of linear equations as an augmented matrix. 14x − 2y + 13z = 140 −2x + 3y − 6z = −1 x − 5y + 12z = 11 22. Rewrite the augmented matrix as a system of linear equations. 1 0 3 −2 4 9 −6 1 2  12  −5 8 ∣ 862 CHAPTER 9 systems oF eQuations and ineQualities For the following exercises, use Gaussian elimination to solve the systems of equations. 23. x − 6y = 4 2x − 12y = 0 24. 2x + y + z = −3 x − 2y + 3z = 6 x − y − z = 6 For the following exercises, use the inverse of a matrix to solve the systems of equations. 25. 4x − 5y = −50 −x + 2y = 80 z = −49 26. 1 ___ 100 3 ___ 100 9 ___ 100 x − 3 ___ 100 x − 7 ___ 100 x − 9 ___ 100 y + 1 ___ 20 y − 1 ___ 100 y − 9 ___ 100 z = 13 z = 99 For the following exercises, use Cramer’s Rule to solve the systems of equations. 27. 200x − 300y = 2 400x + 715y = 4 28. 0.1x + 0.1y − 0.1z = −1.2 0.1x − 0.2y + 0.4z = −1.2 0.5x − 0.3y + 0.8z = −5.9 For the following exercises, solve using a system of linear equations. 29. A factory producing cell phones has the following 30. A small fair charges $1.50 for students, $1 for cost and revenue functions: C(x) = x 2 + 75x + 2,688 and R(x) = x 2 + 160x. What is the range of cell phones they should produce each day so there is profit? Round to the nearest number that generates profit. children, and $2 for adults. In one day, three times as many children as adults attended. A total of 800 tickets were sold for a total revenue of $1,050. How many of each type of ticket was sold? Analytic Geometry 10 Figure 1 (a) Greek philosopher Aristotle (384–322 BCe) (b) German mathematician and astronomer johannes Kepler (1571–1630) a b CHAPTeR OUTlIne 10.1 The ellipse 10.2 The Hyperbola 10.3 The Parabola 10.4 Rotation of Axes 10.5 Conic Sections in Polar Coordinates Introduction The Greek mathematician Menaechmus (c. 380–c. 320 BCE) is generally credited with discovering the shapes formed by the intersection of a plane and a right circular cone. Depending on how he tilted the plane when it intersected the cone, he formed different shapes at the intersection—beautiful shapes with near-perfect symmetry. It was also said that Aristotle may have had an intuitive understanding of these shapes, as he observed the orbit of the planet to be circular. He presumed that the planets moved in circular orbits around Earth, and for nearly 2,000 years this was the commonly held belief. It was not until the Renaissance movement that Johannes Kepler noticed that the orbits of the planet were not circular in nature. His published law of planetary motion in the 1600s changed our view of the solar system forever. He claimed that the sun was at one end of the orbits, and the planets revolved around the sun in an oval-shaped path. In this chapter, we will investigate the two-dimensional figures that are formed when a right circular cone is intersected by a plane. We will begin by studying each of three figures created in this manner. We will develop defining equations for each figure and then learn how to use these equations to solve a variety of problems. 863 864 CHAPTER 10 analytic geometry leARnInG OBjeCTIVeS In this section, you will: • Write equations of ellipses in standard form. • Graph ellipses centered at the origin. • Graph ellipses not centered at the origin. • Solve applied problems involving ellipses. 10.1 THe ellIPSe Figure 1 The national Statuary Hall in Washington, D.C. (credit: Greg Palmer, Flickr) Can you imagine standing at one end of a large room and still being able to hear a whisper from a person standing at the other end? The National Statuary Hall in Washington, D.C., shown in Figure 1, is such a room.[33] It is an ovalshaped room called a whispering chamber because the shape makes it possible for sound to travel along the walls. In this section, we will investigate the shape of this room and its real-world applications, including how far apart two people in Statuary Hall can stand and still hear each other whisper. Writing equations of ellipses in Standard Form A conic section, or conic, is a shape resulting from intersecting a right circular cone with a plane. The angle at which the plane intersects the cone determines the shape, as shown in Figure 2. Ellipse Hyperbola Figure 2 Parabola Conic sections can also be described by a set of points in the coordinate plane. Later in this chapter, we will see that the graph of any quadratic equation in two variables is a conic section. The signs of the equations and the coefficients of the variable terms determine the shape. This section focuses on the four variations of the standard form of the 33. Architect of the Capitol. http://www.aoc.gov. Accessed April 15, 2014. SECTION 10.1 the ellipse 865 equation for the ellipse. An ellipse is the set of all points (x, y) in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci). We can draw an ellipse using a piece of cardboard, two thumbtacks, a pencil, and string. Place the thumbtacks in the cardboard to form the foci of the ellipse. Cut a piece of string longer than the distance between the two thumbtacks (the length of the string represents the constant in the definition). Tack each end of the string to the cardboard, and trace a curve with a pencil held taut against the string. The result is an ellipse. See Figure 3. Foci Figure 3 Every ellipse has two axes of symmetry. The longer axis is called the major axis, and the shorter axis is called the minor axis. Each endpoint of the major axis is the vertex of the ellipse (plural: vertices), and each endpoint of the minor axis is a co-vertex of the ellipse. The center of an ellipse is the midpoint of both the major and minor axes. The axes are perpendicular at the center. The foci always lie on the major axis, and the sum of the distances from the foci to any point on the ellipse (the constant sum) is greater than the distance between the foci. See Figure 4. y Co-vertex Minor Axis Vertex Focus Major Axis Center Focus Vertex x Co-vertex Figure 4 In this section, we restrict ellipses to those that are positioned vertically or horizontally in the coordinate plane. That is, the axes will either lie on or be parallel to the x- and y-axes. Later in the chapter, we will see ellipses that are rotated in the coordinate plane. To work with horizontal and vertical ellipses in the coordinate plane, we consider two cases: those that are centered at the origin and those that are centered at a point other than the origin. First we will learn to derive the equations of ellipses, and then we will learn how to write the equations of ellipses in standard form. Later we will use what we learn to draw the graphs. 866 CHAPTER 10 analytic geometry Deriving the Equation of an Ellipse Centered at the Origin To derive the equation of an ellipse centered at the origin, we begin with the foci (−c, 0) and (c, 0). The ellipse is the set of all points (x, y) such that the sum of the distances from (x, y) to the foci is constant, as shown in Figure 5. (x, y) d1 y d2 (–c, 0) (c, 0) (a, 0) x (–a, 0) Figure 5 If (a, 0) is a vertex of the ellipse, the distance from (−c, 0) to (a, 0) is a − ( −c) = a + c. The distance from (c, 0) to (a, 0) is a − c . The sum of the distances from the foci to the vertex is (a + c) + (a − c) = 2a If (x, y) is a point on the ellipse, then we can define the following variables: d1 = the distance from (−c, 0) to (x, y) d2 = the distance from (c, 0) to (x, y) By the definition of an ellipse, d1 + d2 is constant for any point (x, y) on the ellipse. We know that the sum of these distances is 2a for the vertex (a, 0). It follows that d1 + d2 = 2a for any point on the ellipse. We will begin the derivation by applying the distance formula. The rest of the derivation is algebraic. d1 + d2 = √ —— (x − ( − c))2 + (y − 0)2 + √ — (x − c)2 + (y − 0)2 = 2a Distance formula — (x + c)2 + y 2 + √ — (x − c)2 + y 2 = 2a √ — (x + c)2 + y 2 = 2a − √ — (x − c)2 + y 2 √ (x + c)2 + y 2 = [2a − √ x 2 + 2cx + c2 + y 2 = 4a2 − 4a √ — (x − c)2 + y 2 ] — 2 x 2 + 2cx + c2 + y 2 = 4a2 − 4a √ (x − c)2 + y 2 + (x − c)2 + y 2 (x − c)2 + y 2 + x 2 − 2cx + c2 + y 2 — 2cx = 4a2 − 4a √ (x − c)2 + y 2 − 2cx — 4cx − 4a2 = −4a √ — — (x − c)2 + y 2 (x − c)2 + y 2 2 (x − c)2 + y 2  cx − a2 = − a √ = a2  √ c2 x 2 − 2a2 cx + a4 = a2  x 2 − 2cx + c
2 + y 2  2  cx − a2  — c2 x 2 − 2a2 cx + a4 = a2 x 2 − 2a2 cx + a2 c2 + a2 y 2 a2 x 2 − c2 x 2 + a2 y 2 = a4 − a2 c2 x 2(a2 − c2) + a2 y 2 = a2(a2 − c2) x 2 b2 + a2 y 2 = a2 b2 a2b2 ____ a2b2 x 2b2 a2y 2 ____ ____ a2b2 + a2b2 = y 2 x 2 __ __ b2 = 1 a2 + Simplify expressions. Move radical to opposite side. Square both sides. Expand the squares. Expand remaining squares. Combine like terms. Isolate the radical. Divide by 4. Square both sides. Expand the squares. Distribute a2. Rewrite. Factor common terms. Set b2 = a2 − c2. Divide both sides by a2b2. Simplify. Thus, the standard equation of an ellipse is y 2 x 2 _ __ b2 = 1. This equation defines an ellipse centered at the origin. a2 + If a > b, the ellipse is stretched further in the horizontal direction, and if b > a, the ellipse is stretched further in the vertical direction. SECTION 10.1 the ellipse 867 Writing Equations of Ellipses Centered at the Origin in Standard Form Standard forms of equations tell us about key features of graphs. Take a moment to recall some of the standard forms of equations we’ve worked with in the past: linear, quadratic, cubic, exponential, logarithmic, and so on. By learning to interpret standard forms of equations, we are bridging the relationship between algebraic and geometric representations of mathematical phenomena. The key features of the ellipse are its center, vertices, co-vertices, foci, and lengths and positions of the major and minor axes. Just as with other equations, we can identify all of these features just by looking at the standard form of the equation. There are four variations of the standard form of the ellipse. These variations are categorized first by the location of the center (the origin or not the origin), and then by the position (horizontal or vertical). Each is presented along with a description of how the parts of the equation relate to the graph. Interpreting these parts allows us to form a mental picture of the ellipse. standard forms of the equation of an ellipse with center (0, 0) The standard form of the equation of an ellipse with center (0, 0) and major axis on the x-axis is where y 2 x 2 _ _ b2 = 1 a2 + • a > b • the length of the major axis is 2a • the coordinates of the vertices are (±a, 0) • the length of the minor axis is 2b • the coordinates of the co-vertices are (0, ±b) • the coordinates of the foci are (±c, 0) , where c2 = a2 − b2. See Figure 6a. The standard form of the equation of an ellipse with center (0, 0) and major axis on the y-axis is where y 2 x 2 _ _ a2 = 1 b2 + • a > b • the length of the major axis is 2a • the coordinates of the vertices are (0, ± a) • the length of the minor axis is 2b • the coordinates of the co-vertices are (±b, 0) • the coordinates of the foci are (0, ± c) , where c2 = a2 − b2. See Figure 6b. Note that the vertices, co-vertices, and foci are related by the equation c2 = a2 − b2. When we are given the coordinates of the foci and vertices of an ellipse, we can use this relationship to find the equation of the ellipse in standard form. y (0, b) Minor Axis (−a, 0) (−c, 0) Major Axis (0, 0) (c, 0) (a, 0) x (−b, 0) Minor Axis (0, −b) (a) Figure 6 (a) Horizontal ellipse with center (0, 0) (b) Vertical ellipse with center (0, 0) y (0, a) (0, c) Major Axis (0, 0) (0, −c) (0, −a) (b) (b, 0) x 868 CHAPTER 10 analytic geometry How To… Given the vertices and foci of an ellipse centered at the origin, write its equation in standard form. 1. Determine whether the major axis lies on the x- or y-axis. a. If the given coordinates of the vertices and foci have the form (±a, 0) and ( ±c, 0) respectively, then the major axis is the x-axis. Use the standard form axis is the y-axis. Use the standard form y 2 x 2 _ _ b2 = 1. a2 + y 2 x 2 _ _ a2 = 1. b2 + b. If the given coordinates of the vertices and foci have the form (0, ±a) and ( ±c, 0), respectively, then the major 2. Use the equation c2 = a2 − b2, along with the given coordinates of the vertices and foci, to solve for b2. 3. Substitute the values for a2 and b2 into the standard form of the equation determined in Step 1. Example 1 Writing the Equation of an Ellipse Centered at the Origin in Standard Form What is the standard form equation of the ellipse that has vertices (±8, 0) and foci (±5, 0)? Solution The foci are on the x-axis, so the major axis is the x-axis. Thus, the equation will have the form y 2 x 2 _ __ b2 = 1 a2 + The vertices are (±8, 0), so a = 8 and a2 = 64. The foci are (±5, 0), so c = 5 and c2 = 25. We know that the vertices and foci are related by the equation c 2 = a 2 − b 2. Solving for b 2, we have: c 2 = a 2 − b2 25 = 64 − b2 b 2 = 39 Substitute for c 2 and a 2. Solve for b2. Now we need only substitute a2 = 64 and b2 = 39 into the standard form of the equation. The equation of the ellipse is + = 1. x 2 _ 64 y 2 _ 39 Try It #1 What is the standard form equation of the ellipse that has vertices (0, ± 4) and foci (0, ± √ — 15 )? Q & A… Can we write the equation of an ellipse centered at the origin given coordinates of just one focus and vertex? Yes. Ellipses are symmetrical, so the coordinates of the vertices of an ellipse centered around the origin will always have the form (±a, 0) or (0, ± a). Similarly, the coordinates of the foci will always have the form (±c, 0) or (0, ± c). Knowing this, we can use a and c from the given points, along with the equation c2 = a2 − b2, to find b2. Writing Equations of Ellipses Not Centered at the Origin Like the graphs of other equations, the graph of an ellipse can be translated. If an ellipse is translated h units horizontally and k units vertically, the center of the ellipse will be (h, k). This translation results in the standard form of the equation we saw previously, with x replaced by (x − h) and y replaced by (y − k). SECTION 10.1 the ellipse 869 standard forms of the equation of an ellipse with center (h, k) The standard form of the equation of an ellipse with center (h, k) and major axis parallel to the x-axis is where (x − h)2 _ a2 + (y − k)2 _ b2 = 1 • a > b • the length of the major axis is 2a • the coordinates of the vertices are (h ± a, k) • the length of the minor axis is 2b • the coordinates of the co-vertices are (h, k ± b) • the coordinates of the foci are (h ± c, k), where c2 = a2 − b2. See Figure 7a. The standard form of the equation of an ellipse with center (h, k) and major axis parallel to the y-axis is where (x − h)2 _ b2 + (y − k)2 _ a2 = 1 • a > b • the length of the major axis is 2a • the coordinates of the vertices are (h, k ± a) • the length of the minor axis is 2b • the coordinates of the co-vertices are (h ± b, k) • the coordinates of the foci are (h, k ± c), where c2 = a2 − b2. See Figure 7b. Just as with ellipses centered at the origin, ellipses that are centered at a point (h, k) have vertices, co-vertices, and foci that are related by the equation c2 = a2 − b2. We can use this relationship along with the midpoint and distance formulas to find the equation of the ellipse in standard form when the vertices and foci are given. y (h, k + b) y (h, k + a) (h − a, k) Minor Axis (h + c1, k) (h − c1, k) (h, k) Major Axis (h, k − b) (a) (h + a, k) x (h − b, k) Major Axis (h, k) Minor Axis (h, k + c) (h + b, k) x (h, k − c) (h, k − a) (b) Figure 7 (a) Horizontal ellipse with center (h, k) (b) Vertical ellipse with center (h, k) How To… Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form. 1. Determine whether the major axis is parallel to the x- or y-axis. a. If the y-coordinates of the given vertices and foci are the same, then the major axis is parallel to the x-axis. Use the standard form (x − h)2 _ a2 + (y − k)2 _ b2 = 1. b. If the x-coordinates of the given vertices and foci are the same, then the major axis is parallel to the y-axis. Use the standard form (x − h)2 _ b2 + (y − k)2 _ a2 = 1. 870 CHAPTER 10 analytic geometry 2. Identify the center of the ellipse (h, k) using the midpoint formula and the given coordinates for the vertices. 3. Find a2 by solving for the length of the major axis, 2a, which is the distance between the given vertices. 4. Find c2 using h and k, found in Step 2, along with the given coordinates for the foci. 5. Solve for b2 using the equation c2 = a2 − b2. 6. Substitute the values for h, k, a2, and b2 into the standard form of the equation determined in Step 1. Example 2 Writing the Equation of an Ellipse Centered at a Point Other Than the Origin What is the standard form equation of the ellipse that has vertices (−2, −8) and (−2, 2) and foci (−2, −7) and (−2, 1)? Solution The x-coordinates of the vertices and foci are the same, so the major axis is parallel to the y-axis. Thus, the equation of the ellipse will have the form (y − k)2 _ a2 First, we identify the center, (h, k). The center is halfway between the vertices, (−2, − 8) and (−2, 2). (x − h)2 _ b2 = 1 + Applying the midpoint formula, we have: (h, k) =  −2 + (−2) _________ , 2 −8 + 2  _______ 2 Next, we find a2. The length of the major axis, 2a, is bounded by the vertices. We solve for a by finding the distance between the y-coordinates of the vertices. = (−2, −3) 2a = 2 − (−8) 2a = 10 a = 5 So a2 = 25. Now we find c2. The foci are given by (h, k ± c). So, (h, k − c) = (−2, −7) and (h, k + c) = (−2, 1). We substitute k = −3 using either of these points to solve for c. k + c = 1 −3 + c = 1 c = 4 So c2 = 16. Next, we solve for b2 using the equation c2 = a2 − b2. c2 = a2 − b2 16 = 25 − b2 b2 = 9 Finally, we substitute the values found for h, k, a2, and b2 into the standard form equation for an ellipse: (y + 3)2 _ 25 (x + 2)2 _ 9 = 1 + Try It #2 What is the standard form equation of the ellipse that has vertices (−3, 3) and (5, 3) and foci (1 − 2 √ (1 + 2 √ 3 , 3)? — — 3 , 3) and Graphing ellipses Centered at the Origin Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph a2 = 1, b2 + b2 = 1, a > b for horizontal ellipses and a2 + e
llipses centered at the origin, we use the standard form a > b for vertical ellipses. SECTION 10.1 the ellipse 871 How To… Given the standard form of an equation for an ellipse centered at (0, 0), sketch the graph. 1. Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci. a. If the equation is in the form y 2 x 2 _ _ b2 = 1, where a > b, then a2 + • the major axis is the x-axis • the coordinates of the vertices are (±a, 0) • the coordinates of the co-vertices are (0, ± b) • the coordinates of the foci are (±c, 0) y 2 x 2 _ _ a2 = 1, where a > b, then b2 + b. If the equation is in the form • the major axis is the y-axis • the coordinates of the vertices are (0, ± a) • the coordinates of the co-vertices are (±b, 0) • the coordinates of the foci are (0, ± c) 2. Solve for c using the equation c2 = a2 − b2. 3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse. Example 3 Graphing an Ellipse Centered at the Origin x 2 _ 9 + y 2 _ 25 Graph the ellipse given by the equation, = 1. Identify and label the center, vertices, co-vertices, and foci. Solution First, we determine the position of the major axis. Because 25 > 9, the major axis is on the y-axis. Therefore, where b 2 = 9 and a2 = 25. It follows that: b 2 + the equation is in the form • the center of the ellipse is (0, 0) • the coordinates of the vertices are (0, ± a) = (0, ± √ • the coordinates of the co-vertices are (±b, 0) = (± √ • the coordinates of the foci are (0, ± c), where c 2 = a 2 − b 2 Solving for c, we have: 25 ) = (0, ± 5) 9 , 0) = (±3, 0) — — — — a2 − b2 25 − 9 16 — Therefore, the coordinates of the foci are (0, ± 4). Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See Figure 8. y (0, 5) (0, 4) (–3, 0) (0, 0) (3, 0) x (0, –4) (0, –5) Figure 8 872 CHAPTER 10 analytic geometry Try It #3 Graph the ellipse given by the equation x 2 __ 36 y 2 _ 4 + = 1. Identify and label the center, vertices, co-vertices, and foci. Example 4 Graphing an Ellipse Centered at the Origin from an Equation Not in Standard Form Graph the ellipse given by the equation 4x 2 + 25y 2 = 100. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci. Solution First, use algebra to rewrite the equation in standard form. + 4x 2 + 25y 2 = 100 25y 2 4x 2 100 ____ _ ___ 100 100 100 y 2 _ 4 x 2 _ 25 = 1 = + Next, we determine the position of the major axis. Because 25 > 4, the major axis is on the x-axis. Therefore, the equation is in the form y 2 x 2 _ _ a2 + b2 = 1, where a2 = 25 and b2 = 4. It follows that: • the center of the ellipse is (0, 0) • the coordinates of the vertices are (±a, 0) = (± √ • the coordinates of the co-vertices are (0, ± b) = (0, ± √ • the coordinates of the foci are (±c, 0), where c 2 = a 2 − b 2. Solving for c, we have: 25 , 0) = (±5, 0) 4 ) = (0, ± 2) — — c = ± √ = ± √ = ± √ — — a2 − b2 25 − 4 21 — Therefore the coordinates of the foci are (± √ Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See Figure 9. 21 , 0). — (–5, 0) y (0, 2) (0, 0) (0, −2) Figure 9 (5, 0) x Try It #4 Graph the ellipse given by the equation 49x 2 + 16y 2 = 784. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci. Graphing ellipses not Centered at the Origin When an ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point, (h, k), we use the standard forms (y − k)2 _ a2 (x − h)2 _ b 2 horizontal ellipses and + = 1, a > b for vertical ellipses. From these standard equations, we can (x − h)2 _ a 2 + (y − k)2 _ b 2 = 1, a > b for easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes. SECTION 10.1 the ellipse 873 How To… Given the standard form of an equation for an ellipse centered at (h, k), sketch the graph. 1. Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci. a. If the equation is in the form = 1, where a > b, then (x − h)2 _ a2 + (y − k)2 _ b2 • the center is (h, k) • the major axis is parallel to the x-axis • the coordinates of the vertices are (h ± a, k) • the coordinates of the co-vertices are (h, k ± b) • the coordinates of the foci are (h ± c, k) b. If the equation is in the form = 1, where a > b, then (x − h)2 _ b2 + (y − k)2 _ a2 • the center is (h, k) • the major axis is parallel to the y-axis • the coordinates of the vertices are (h, k ± a) • the coordinates of the co-vertices are (h ± b, k) • the coordinates of the foci are (h, k ± c) 2. Solve for c using the equation c 2 = a2 − b2. 3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse. Example 5 Graphing an Ellipse Centered at (h, k) Graph the ellipse given by the equation, and foci. (x + 2)2 _ 4 + (y − 5)2 _ 9 = 1. Identify and label the center, vertices, co-vertices, Solution First, we determine the position of the major axis. Because 9 > 4, the major axis is parallel to the y-axis. Therefore, the equation is in the form = 1, where b2 = 4 and a2 = 9. It follows that: (x − h)2 _ b2 + (y − k)2 _ a2 • the center of the ellipse is (h, k) = (−2, 5) • the coordinates of the vertices are (h, k ± a) = (−2, 5 ± √ • the coordinates of the co-vertices are (h ± b, k) = (−2 ± √ • the coordinates of the foci are (h, k ± c), where c 2 = a 2 − b 2. Solving for c, we have: 9 ) = (−2, 5 ± 3), or (−2, 2) and (−2, 8) 4 , 5) = (−2 ± 2, 5), or (−4, 5) and (0, 5) — — c = ± √ = ± √ = ± √ — — a2 − b2 9 − 4 5 — Therefore, the coordinates of the foci are (−2, 5 − √ — 5 ) and (−2, 5 + √ — 5 ). Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. y (−2, 8) (−2, 5 + √5) (−4, 5) (0, 5) (−2, 5) (−2, 5 −√5) (−2, 2) Figure 10 x 874 CHAPTER 10 analytic geometry Try It #5 Graph the ellipse given by the equation and foci. (x − 4)2 _ 36 + (y − 2)2 _ 20 = 1. Identify and label the center, vertices, co-vertices, How To… Given the general form of an equation for an ellipse centered at (h, k), express the equation in standard form. 1. Recognize that an ellipse described by an equation in the form ax 2 + by 2 + cx + dy + e = 0 is in general form. 2. Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation. 3. Factor out the coefficients of the x 2 and y 2 terms in preparation for completing the square. 4. Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two binomials squared set equal to a constant, m1 (x − h)2 + m2(y − k)2 = m3, where m1, m2, and m3 are constants. 5. Divide both sides of the equation by the constant term to express the equation in standard form. Example 6 Graphing an Ellipse Centered at ( h, k) by First Writing It in Standard Form Graph the ellipse given by the equation 4x 2 + 9y 2 − 40x + 36y + 100 = 0. Identify and label the center, vertices, co-vertices, and foci. Solution We must begin by rewriting the equation in standard form. 4x 2 + 9y 2 − 40x + 36y + 100 = 0 Group terms that contain the same variable, and move the constant to the opposite side of the equation. (4x 2 − 40x) + (9y 2 + 36y) = −100 Factor out the coefficients of the squared terms. 4(x 2 − 10x)+ 9(y 2 + 4y) = −100 Complete the square twice. Remember to balance the equation by adding the same constants to each side. 4(x 2 − 10x + 25)+ 9(y 2 + 4y + 4) = −100 + 100 + 36 Rewrite as perfect squares. Divide both sides by the constant term to place the equation in standard form. 4(x − 5)2 + 9(y + 2)2 = 36 Now that the equation is in standard form, we can determine the position of the major axis. Because 9 > 4, the major (x − 5)2 _ 9 + (y + 2)2 _ 4 = 1 axis is parallel to the x-axis. Therefore, the equation is in the form = 1, where a2 = 9 and b2 = 4. (x − h)2 _ a2 + (y − k)2 _ b2 It follows that: • the center of the ellipse is (h, k) = (5, −2) • the coordinates of the vertices are (h ± a, k) = (5 ± √ • the coordinates of the co-vertices are (h, k ± b) = (5, −2 ± √ • the coordinates of the foci are (h ± c, k), where c2 = a2 − b2. Solving for c, we have: 9 , −2) = (5 ± 3, −2), or (2, −2) and (8, −2) 4 )= (5, −2 ± 2), or (5, −4) and (5, 0) — — = ± √ 5 , −2) and (5 + √ Therefore, the coordinates of the foci are (5 − √ Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse as shown in Figure 11. — — c = ± √ = ± √ — — a2 − b2 9 − 4 5 5 , −2). — SECTION 10.1 the ellipse 875 –1 0 –1 –2 –3 –4 1 (5 −√5, −2) 1 2 3 4 (5 +√5, −2) 6 7 8 9 x (5, 0) 5 (2, −2) (5, −2) (8, −2) (5, −4) Figure 11 Try It #6 Express the equation of the ellipse given in standard form. Identify the center, vertices, co-vertices, and foci of the ellipse. 4x 2 + y 2 − 24x + 2y + 21 = 0 Solving Applied Problems Involving ellipses Many real-world situations can be represented by ellipses, including orbits of planets, satellites, moons and comets, and shapes of boat keels, rudders, and some airplane wings. A medical device called a lithotripter uses elliptical reflectors to break up kidney stones by generating sound waves. Some buildings, called whispering chambers, are designed with elliptical domes so that a person whispering at one focus can easily be heard by someone standing at the other focus. This occurs because of the acoustic properties of an ellipse. When a sound wave originates at one focus of a whispering chamber, the sound wave will be reflected off the elliptical dome and back to the other focus. See Figure 12. In the whisper chamber at the Museum of Science and Industry in Chicago, two people standing at the foci—about 43
feet apart—can hear each other whisper. Figure 12 Sound waves are reflected between foci in an elliptical room, called a whispering chamber. Example 7 Locating the Foci of a Whispering Chamber The Statuary Hall in the Capitol Building in Washington, D.C. is a whispering chamber. Its dimensions are 46 feet wide by 96 feet long as shown in Figure 13. a. What is the standard form of the equation of the ellipse representing the outline of the room? Hint: assume a horizontal ellipse, and let the center of the room be the point (0, 0). b. If two senators standing at the foci of this room can hear each other whisper, how far apart are the senators? Round to the nearest foot. 46 feet 96 feet Figure 13 876 Solution CHAPTER 10 analytic geometry a. We are assuming a horizontal ellipse with center (0, 0), so we need to find an equation of the form y 2 x 2 _ _ b2 = 1, where a > b. We know that the length of the major axis, 2a, is longer than the length of the a2 + minor axis, 2b. So the length of the room, 96, is represented by the major axis, and the width of the room, 46, is represented by the minor axis. • Solving for a, we have 2a = 96, so a = 48, and a2 = 2304. • Solving for b, we have 2b = 46, so b = 23, and b2 = 529. Therefore, the equation of the ellipse is y 2 _ 529 b. To find the distance between the senators, we must find the distance between the foci, (±c, 0), where x 2 _ 2304 = 1. + c2 = a2 − b2. Solving for c, we have: c2 = a2 − b2 c2 = 2304 − 529 Substitute using the values found in part (a). — c = ± √ c = ± √ 2304 − 529 1775 — Subtract. Take the square root of both sides. The points (±42, 0) represent the foci. Thus, the distance between the senators is 2(42) = 84 feet. c ≈ ± 42 Round to the nearest foot. Try It #7 Suppose a whispering chamber is 480 feet long and 320 feet wide. a. What is the standard form of the equation of the ellipse representing the room? Hint: assume a horizontal ellipse, and let the center of the room be the point (0, 0). b. If two people are standing at the foci of this room and can hear each other whisper, how far apart are the people? Round to the nearest foot. Access these online resources for additional instruction and practice with ellipses. • Conic Sections: The ellipse (http://openstaxcollege.org/l/conicellipse) • Graph an ellipse with Center at the Origin (http://openstaxcollege.org/l/grphellorigin) • Graph an ellipse with Center not at the Origin (http://openstaxcollege.org/l/grphellnot) SECTION 10.1 section exercises 877 10.1 SeCTIOn exeRCISeS VeRBAl 1. Define an ellipse in terms of its foci. 3. What special case of the ellipse do we have when the 2. Where must the foci of an ellipse lie? 4. For the special case mentioned in the previous major and minor axis are of the same length? question, what would be true about the foci of that ellipse? 5. What can be said about the symmetry of the graph of an ellipse with center at the origin and foci along the y-axis? AlGeBRAIC For the following exercises, determine whether the given equations represent ellipses. If yes, write in standard form. 6. 2x 2 + y = 4 9. 4x 2 + 9y 2 = 1 7. 4x 2 + 9y 2 = 36 8. 4x 2 − y 2 = 4 10. 4x 2 − 8x + 9y 2 − 72y + 112 = 0 For the following exercises, write the equation of an ellipse in standard form, and identify the end points of the major and minor axes as well as the foci. 11 49 14. 4x 2 + 16y 2 = 1 17. (x + 5)2 _ 4 + (y − 7)2 _ 9 = 1 20. 9x 2 − 54x + 9y 2 − 54y + 81 = 0 22. 4x 2 + 24x + 16y 2 − 128y + 228 = 0 24. x 2 + 2x + 100y 2 − 1000y + 2401 = 0 26. 9x 2 + 72x + 16y 2 + 16y + 4 = 0 12. x 2 _ 100 + y 2 _ 64 = 1 15. (x − 2)2 _______ 49 + (y − 4)2 _______ 25 = 1 13. x 2 + 9y 2 = 1 16. (x − 2)2 _ 81 + (y + 1)2 _ 16 = 1 18. (x − 7)2 _______ 49 + = 1 (y − 7)2 _______ 49 21. 4x 2 − 24x + 36y 2 − 360y + 864 = 0 23. 4x 2 + 40x + 25y 2 − 100y + 100 = 0 25. 4x 2 + 24x + 25y 2 + 200y + 336 = 0 19. 4x 2 − 8x + 9y 2 − 72y + 112 = 0 For the following exercises, find the foci for the given ellipses. 27. (x + 3)2 _ 25 + (y + 1)2 _ 36 = 1 28. (x + 1)2 _ 100 + (y − 2)2 _ 4 = 1 29. x 2 + y 2 = 1 30. x 2 + 4y 2 + 4x + 8y = 1 31. 10x 2 + y 2 + 200x = 0 GRAPHICAl For the following exercises, graph the given ellipses, noting center, vertices, and foci. 32. x 2 _ 25 + y 2 _ 36 = 1 35. 81x 2 + 49y 2 = 1 38. + x 2 _ 2 (y + 1)2 _ 5 = 1 33. 36 16 (x − 2)2 _ 64 + 34. 4x 2 + 9y 2 = 1 = 1 (y − 4)2 _ 16 39. 4x 2 − 8x + 16y 2 − 32y − 44 = 0 (x + 3)2 _ 9 + 37. (y − 3)2 _ 9 = 1 40. x 2 − 8x + 25y 2 − 100y + 91 = 0 42. 64x 2 + 128x + 9y 2 − 72y − 368 = 0 44. 100x 2 + 1000x + y 2 − 10y + 2425 = 0 41. x 2 + 8x + 4y 2 − 40y + 112 = 0 43. 16x 2 + 64x + 4y 2 − 8y + 4 = 0 45. 4x 2 + 16x + 4y 2 + 16y + 16 = 0 For the following exercises, use the given information about the graph of each ellipse to determine its equation. 46. Center at the origin, symmetric with respect to the x- and y-axes, focus at (4, 0), and point on graph (0, 3). 47. Center at the origin, symmetric with respect to the x- and y-axes, focus at (0, −2), and point on graph (5, 0). 878 CHAPTER 10 analytic geometry 48. Center at the origin, symmetric with respect to the x- and y-axes, focus at (3, 0), and major axis is twice as long as minor axis. 49. Center (4, 2); vertex (9, 2); one focus: (4 + 2 √ 6 , 2). — 50. Center (3, 5); vertex (3, 11); one focus: (3, 5 + 4 √ For the following exercises, given the graph of the ellipse, determine its equation. 2 ) — 51. Center (−3, 4); vertex (1, 4); one focus: (−3 + 2 √ — 3 , 4) y –8 –6 –4 –2 0 –2 –4 –6 –5 –6 –4 –2 52. 55 53. y 6 4 2 –10 –8 –6 –4 –2 0 –2 2 4 6 8 10 x 56. –4 –6 y 5 4 3 2 1 54. –6 –4 –2 y –8 –6 –4 –2 0 –2 –4 –6 –5 x 2 4 6 –5 –4 –3 –2 –1 x 1 0 –1 –3 –2 –1 0 –1 1 2 3 4 5 x exTenSIOnS For the following exercises, find the area of the ellipse. The area of an ellipse is given by the formula Area = a ⋅ b ⋅ π. 57. (x − 3)2 _ 9 + (y − 3)2 _ 16 = 1 58. (x + 6)2 _ 16 + (y − 6)2 _ 36 = 1 59. (x + 1)2 _ 4 + (y − 2)2 _ 5 = 1 60. 4x 2 − 8x + 9y 2 − 72y + 112 = 0 61. 9x 2 − 54x + 9y 2 − 54y + 81 = 0 ReAl-WORlD APPlICATIOnS 62. Find the equation of the ellipse that will just fit 63. Find the equation of the ellipse that will just fit inside a box that is 8 units wide and 4 units high. 64. An arch has the shape of a semi-ellipse (the top half of an ellipse). The arch has a height of 8 feet and a span of 20 feet. Find an equation for the ellipse, and use that to find the height to the nearest 0.01 foot of the arch at a distance of 4 feet from the center. 66. A bridge is to be built in the shape of a semi- elliptical arch and is to have a span of 120 feet. The height of the arch at a distance of 40 feet from the center is to be 8 feet. Find the height of the arch at its center. 68. A person is standing 8 feet from the nearest wall in a whispering gallery. If that person is at one focus, and the other focus is 80 feet away, what is the length and height at the center of the gallery? inside a box that is four times as wide as it is high. Express in terms of h, the height. 65. An arch has the shape of a semi-ellipse. The arch has a height of 12 feet and a span of 40 feet. Find an equation for the ellipse, and use that to find the distance from the center to a point at which the height is 6 feet. Round to the nearest hundredth. 67. A person in a whispering gallery standing at one focus of the ellipse can whisper and be heard by a person standing at the other focus because all the sound waves that reach the ceiling are reflected to the other person. If a whispering gallery has a length of 120 feet, and the foci are located 30 feet from the center, find the height of the ceiling at the center. SECTION 10.2 the hyperBola 879 leARnInG OBjeCTIVeS In this section, you will: • Locate a hyperbola’s vertices and foci. • Write equations of hyperbolas in standard form. • Graph hyperbolas centered at the origin. • Graph hyperbolas not centered at the origin. • Solve applied problems involving hyperbolas. 10. 2 THe HYPeRBOlA What do paths of comets, supersonic booms, ancient Grecian pillars, and natural draft cooling towers have in common? They can all be modeled by the same type of conic. For instance, when something moves faster than the speed of sound, a shock wave in the form of a cone is created. A portion of a conic is formed when the wave intersects the ground, resulting in a sonic boom. See Figure 1. Wake created from shock wave Portion of a hyperbola Figure 1 A shock wave intersecting the ground forms a portion of a conic and results in a sonic boom. Most people are familiar with the sonic boom created by supersonic aircraft, but humans were breaking the sound barrier long before the first supersonic flight. The crack of a whip occurs because the tip is exceeding the speed of sound. The bullets shot from many firearms also break the sound barrier, although the bang of the gun usually supersedes the sound of the sonic boom. locating the Vertices and Foci of a Hyperbola In analytic geometry, a hyperbola is a conic section formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. This intersection produces two separate unbounded curves that are mirror images of each other. See Figure 2. Like the ellipse, the hyperbola can also be defined as a set of points in the coordinate plane. A hyperbola is the set of all points (x, y) in a plane such that the difference of the distances between (x, y) and the foci is a positive constant. Figure 2 A hyperbola 880 CHAPTER 10 analytic geometry Notice that the definition of a hyperbola is very similar to that of an ellipse. The distinction is that the hyperbola is defined in terms of the difference of two distances, whereas the ellipse is defined in terms of the sum of two distances. As with the ellipse, every hyperbola has two axes of symmetry. The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line that contains the transverse axis. The conjugate axis is pe
rpendicular to the transverse axis and has the co-vertices as its endpoints. The center of a hyperbola is the midpoint of both the transverse and conjugate axes, where they intersect. Every hyperbola also has two asymptotes that pass through its center. As a hyperbola recedes from the center, its branches approach these asymptotes. The central rectangle of the hyperbola is centered at the origin with sides that pass through each vertex and co-vertex; it is a useful tool for graphing the hyperbola and its asymptotes. To sketch the asymptotes of the hyperbola, simply sketch and extend the diagonals of the central rectangle. See Figure 3. y Conjugate axis Transverse axis Co-vertex Focus Vertex Vertex x Focus Co-vertex Center Asymptote Asymptote Figure 3 Key features of the hyperbola In this section, we will limit our discussion to hyperbolas that are positioned vertically or horizontally in the coordinate plane; the axes will either lie on or be parallel to the x- and y-axes. We will consider two cases: those that are centered at the origin, and those that are centered at a point other than the origin. Deriving the Equation of an Ellipse Centered at the Origin Let (−c, 0) and (c, 0) be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points (x, y) such that the difference of the distances from (x, y) to the foci is constant. See Figure 4. y (x, y) d2 d1 (c, 0) x (–c, 0) (–a, 0) (a, 0) Figure 4 If (a, 0) is a vertex of the hyperbola, the distance from (−c, 0) to (a, 0) is a − (−c) = a + c. The distance from (c, 0) to (a, 0) is c − a. The sum of the distances from the foci to the vertex is (a + c) − (c − a) = 2a SECTION 10.2 the hyperBola 881 If (x, y) is a point on the hyperbola, we can define the following variables: d2 = the distance from (−c, 0) to (x, y) d1 = the distance from (c, 0) to (x, y) By definition of a hyperbola, d2 − d1 is constant for any point (x, y) on the hyperbola. We know that the difference of these distances is 2a for the vertex (a, 0). It follows that d2 − d1 = 2a for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula. The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses. — —— (x − ( − c))2 + (y − 0)2 − √ (x − c)2 + (y − 0)2 = 2a Distance formula — (x + c)2 + y 2 − √ √ d2 − d1 = √ (x − c)2 + y 2 = 2a — — (x + c)2 + y 2 = 2a + √ — (x − c)2 + y 2 √ (x + c)2 + y 2 = (2a + √ x 2 + 2cx + c2 + y 2 = 4a2 + 4a √ — (x − c)2 + y 2 )2 — (x − c)2 + y 2 + (x − c)2 + y 2 x 2 + 2cx + c2 + y 2 = 4a2 + 4a √ — (x − c)2 + y 2 + x 2 − 2cx + c2 + y 2 2cx = 4a2 + 4a √ — (x − c)2 + y 2 − 2cx 4cx − 4a2 = 4a √ — (x − c)2 + y 2 — (x − c)2 + y 2 cx − a2 = a √ (cx − a2) 2 = a2  √ (x − c)2 + y 2  c2 x 2 − 2a2 cx + a4 = a2(x 2 − 2cx + c2 + y 2) — 2 c2 x 2 − 2a2 cx + a4 = a2 x 2 − 2a2 cx + a2 c2 + a2 y 2 a4 + c2 x 2 = a2 x 2 + a2 c2 + a2 y 2 c2 x 2 − a2 x 2 − a2 y 2 = a2 c2 − a4 x 2 (c2 − a2) − a2 y 2 = a2 (c2 − a2) x 2 b2 − a2 y 2 = a2 b2 a2b2 ____ a2b2 x 2b2 a2y 2 ____ ____ a2b2 − a2b2 = y 2 x 2 _ _ b2 = 1 a2 − Simplify expressions. Move radical to opposite side. Square both sides. Expand the squares. Expand remaining square. Combine like terms. Isolate the radical. Divide by 4. Square both sides. Expand the squares. Distribute a2. Combine like terms. Rearrange terms. Factor common terms Set b2 = c2 − a2. Divide both sides by a2b2. This equation defines a hyperbola centered at the origin with vertices (±a, 0) and co-vertices (0 ± b). standard forms of the equation of a hyperbola with center (0, 0) The standard form of the equation of a hyperbola with center (0, 0) and major axis on the x-axis is y 2 x 2 _ _ b2 = 1 a2 − where • the length of the transverse axis is 2a • the coordinates of the vertices are (±a, 0) • the length of the conjugate axis is 2b • the coordinates of the co-vertices are (0, ±b) • the distance between the foci is 2c, where c2 = a2 + b2 • the coordinates of the foci are (±c, 0) b __ a x • the equations of the asymptotes are y = ± 882 CHAPTER 10 analytic geometry See Figure 5a. The standard form of the equation of a hyperbola with center (0, 0) and transverse axis on the y-axis is y 2 x 2 _ _ b2 = 1 a2 − where • the length of the transverse axis is 2a • the coordinates of the vertices are (0, ± a) • the length of the conjugate axis is 2b • the coordinates of the co-vertices are (±b, 0) • the distance between the foci is 2c, where c2 = a2 + b2 • the coordinates of the foci are (0, ± c) a _ • the equations of the asymptotes are y = ± x b See Figure 5b. Note that the vertices, co-vertices, and foci are related by the equation c2 = a2 + b2. When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci0, b) y = − a x b (−c, 0) (−a, 0) (c, 0) (a, 0) (0, 0) x (−b, 0) (0, −b) (a) y = a x b x (b, 0) y (0, c ) (0, a) (0, 0) (0, −a) (0, −c) (b) Figure 5 (a) Horizontal hyperbola with center (0, 0) (b) Vertical hyperbola with center (0, 0) How To… Given the equation of a hyperbola in standard form, locate its vertices and foci. 1. Determine whether the transverse axis lies on the x- or y-axis. Notice that a2 is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices. a. If the equation has the form y 2 x 2 _ _ b2 = 1, then the transverse axis lies on the x-axis. The vertices are located at a2 − ( ± a, 0), and the foci are located at (± c, 0). b. If the equation has the form y 2 x 2 _ _ b2 = 1, then the transverse axis lies on the y-axis. The vertices are located at a2 − (0, ± a), and the foci are located at (0, ± c). 2. Solve for a using the equation a = √ 3. Solve for c using the equation c = √ — a2 . a2 + b2. — SECTION 10.2 the hyperBola 883 Example 1 Locating a Hyperbola’s Vertices and Foci y 2 _ 49 − = 1. x 2 _ 32 Identify the vertices and foci of the hyperbola with equation Solution The equation has the form y 2 x 2 _ _ b2 = 1, so the transverse axis lies on the y-axis. The hyperbola is centered a2 − at the origin, so the vertices serve as the y-intercepts of the graph. To find the vertices, set x = 0, and solve for y. − − x 2 _ 32 02 _ 32 1 = 1 = y 2 _ 49 y 2 _ 49 y 2 _ 49 y 2 = 49 1 = The foci are located at (0, ± c). Solving for c, y = ± √ — 49 = ± 7 c = √ — a2 + b2 = √ — 49 + 32 = √ — 81 = 9 Therefore, the vertices are located at (0, ± 7), and the foci are located at (0, 9). Try It #1 Identify the vertices and foci of the hyperbola with equation x 2 _ 9 − y 2 _ 25 = 1. Writing equations of Hyperbolas in Standard Form Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin. Hyperbolas Centered at the Origin Reviewing the standard forms given for hyperbolas centered at (0, 0), we see that the vertices, co-vertices, and foci are related by the equation c2 = a2 + b2. Note that this equation can also be rewritten as b2 = c2 − a2. This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices. How To… Given the vertices and foci of a hyperbola centered at (0, 0), write its equation in standard form. 1. Determine whether the transverse axis lies on the x- or y-axis. transverse axis is the x-axis. Use the standard form a. If the given coordinates of the vertices and foci have the form (± a, 0) and (± c, 0), respectively, then the y 2 x 2 _ _ b2 = 1. a2 − b. If the given coordinates of the vertices and foci have the form (0, ± a) and (0, ± c), respectively, then the y 2 x 2 _ _ b2 = 1. a2 − transverse axis is the y-axis. Use the standard form 2. Find b2 using the equation b2 = c2 − a2. 3. Substitute the values for a2 and b2 into the standard form of the equation determined in Step 1. 884 CHAPTER 10 analytic geometry Example 2 What is the standard form equation of the hyperbola that has vertices (±6, 0) and foci (±2 √ Finding the Equation of a Hyperbola Centered at (0, 0) Given its Foci and Vertices 10 , 0)? — Solution The vertices and foci are on the x-axis. Thus, the equation for the hyperbola will have the form The vertices are (±6, 0), so a = 6 and a2 = 36. The foci are (±2 √ Solving for b2, we have 10 , 0), so c = 2 √ — — 10 and c2 = 40. b2 = c2 − a2 b2 = 40 − 36 b2 = 4 Substitute for c2 and a2. Subtract. y 2 x 2 _ _ b2 = 1. a2 − Finally, we substitute a2 = 36 and b2 = 4 into the standard form of the equation, hyperbola is − = 1, as shown in Figure 6. x 2 _ 36 b2 = 1. The equation of the a2 − –10 –8 –6 –4 –2 y 6 4 2 0 –2 –4 –6 Figure 6 2 4 6 8 10 x Try It #2 What is the standard form equation of the hyperbola that has vertices (0, ± 2) and foci (0, ± 2 √ — 5 )? Hyperbolas Not Centered at the Origin Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated h units horizontally and k units vertically, the center of the hyperbola will be (h, k). This translation results in the standard form of the equation we saw previously, with x replaced by (x − h) and y replaced by (y − k). standard forms of the equation of a hyperbola with center (h, k) The standard form of the equation of a hyperbola with center (h, k) and transverse axis parallel to the x-axis is where (x − h)2 _ a2 − (y − k)2 _ b2 = 1 • the length of the transverse axis is 2a • the coordinates of the vertices are (h ± a, k) • the length of the conjugate axis is 2b • the coordinates of the co-verti
ces are (h, k ± b) • the distance between the foci is 2c, where c 2 = a 2 + b 2 • the coordinates of the foci are (h ± c, k) SECTION 10.2 the hyperBola 885 The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle b __ is 2a and its width is 2b. The slopes of the diagonals are ± , and each diagonal passes through the center (h, k). a b __ Using the point-slope formula, it is simple to show that the equations of the asymptotes are y = ± (x − h) + k. a See Figure 7a. The standard form of the equation of a hyperbola with center (h, k) and transverse axis parallel to the y-axis is where (y − k)2 _ a2 − (x − h)2 _ b2 = 1 • the length of the transverse axis is 2a • the coordinates of the vertices are (h, k ± a) • the length of the conjugate axis is 2b • the coordinates of the co-vertices are (h ± b, k) • the distance between the foci is 2c, where c 2 = a 2 + b 2 • the coordinates of the foci are (h, k ± c) a __ (x − h) + k. See Figure 7b. Using the reasoning above, the equations of the asymptotes are x − h) + k a (h, k + b) y (h − c, k) (h − a, k) (h − b, k) (h, k) (h + c, k) (h + a, k) x (h, k − b) y = b (x − h) + k a (a) (h, k + c) (h, k + a) (h, k) (h, k − a) (h, k − c) (b) Figure 7 (a) Horizontal hyperbola with center (h, k) (b) Vertical hyperbola with center (h, k) y = a (x − h) + k b (h + b, k) y = − a (x − h) + k b x Like hyperbolas centered at the origin, hyperbolas centered at a point (h, k) have vertices, co-vertices, and foci that are related by the equation c2 = a2 + b2. We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given. How To… Given the vertices and foci of a hyperbola centered at (h, k), write its equation in standard form. 1. Determine whether the transverse axis is parallel to the x- or y-axis. a. If the y-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the x-axis. Use the standard form (x − h)2 _ a2 − (y − k)2 _ b2 = 1. b. If the x-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the y-axis. Use the standard form (y − k)2 _ a2 − (x − h)2 _ b2 = 1. 2. Identify the center of the hyperbola, (h, k), using the midpoint formula and the given coordinates for the vertices. 886 CHAPTER 10 analytic geometry 3. Find a2 by solving for the length of the transverse axis, 2a, which is the distance between the given vertices. 4. Find c2 using h and k found in Step 2 along with the given coordinates for the foci. 5. Solve for b2 using the equation b2 = c2 − a2. 6. Substitute the values for h, k, a2, and b2 into the standard form of the equation determined in Step 1. Example 3 Finding the Equation of a Hyperbola Centered at (h, k) Given its Foci and Vertices What is the standard form equation of the hyperbola that has vertices at (0, −2) and (6, −2) and foci at (−2, −2) and (8, −2)? Solution The y-coordinates of the vertices and foci are the same, so the transverse axis is parallel to the x-axis. Thus, the equation of the hyperbola will have the form (x − h)2 _ a2 − (y − k)2 _ b2 = 1 First, we identify the center, (h, k). The center is halfway between the vertices (0, −2) and (6, −2). Applying the midpoint formula, we have Next, we find a2. The length of the transverse axis, 2a, is bounded by the vertices. So, we can find a2 by finding the distance between the x-coordinates of the vertices. (h, k) =  0 + 6 −2 + (−2) _________ _____ , 2 2  = (3, −2) 2a = \| 0 − 6 \| 2a = 6 a = 3 a2 = 9 Now we need to find c2. The coordinates of the foci are (h ± c, k). So (h − c, k) = (−2, −2) and (h + c, k) = (8, −2). We can use the x-coordinate from either of these points to solve for c. Using the point (8, −2), and substituting h = 3 c2 = 25 Next, solve for b2 using the equation b2 = c2 − a2 : b2 = c2 − a2 = 25 − 9 = 16 Finally, substitute the values found for h, k, a2, and b2 into the standard form of the equation. (x − 3)2 _ 9 − (y + 2)2 _ 16 = 1 Try It #3 What is the standard form equation of the hyperbola that has vertices (1, −2) and (1, 8) and foci (1, −10) and (1, 16)? Graphing Hyperbolas Centered at the Origin When we have an equation in standard form for a hyperbola centered at the origin, we can interpret its parts to identify the key features of its graph: the center, vertices, co-vertices, asymptotes, foci, and lengths and positions of the y 2 x 2 _ _ b2 = 1 for a2 − transverse and conjugate axes. To graph hyperbolas centered at the origin, we use the standard form y 2 x 2 _ _ b2 = 1 for vertical hyperbolas. a2 − horizontal hyperbolas and the standard form SECTION 10.2 the hyperBola 887 How To… Given a standard form equation for a hyperbola centered at (0, 0), sketch the graph. 1. Determine which of the standard forms applies to the given equation. 2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the a. If the equation is in the form vertices, co-vertices, and foci; and the equations for the asymptotes. y 2 x 2 _ _ b2 = 1, then a2 − • the transverse axis is on the x-axis • the coordinates of the vertices are (±a, 0) • the coordinates of the co-vertices are (0, ± b) • the coordinates of the foci are (±c, 0) b _ • the equations of the asymptotes are b2 = 1, then a2 − • the transverse axis is on the y-axis • the coordinates of the vertices are (0, ± a) • the coordinates of the co-vertices are (±b, 0) • the coordinates of the foci are (0, ± c) a __ • the equations of the asymptotes are y = ± x b b. If the equation is in the form 3. Solve for the coordinates of the foci using the equation c = ± √ — a2 + b2 . 4. Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola. Example 4 Graphing a Hyperbola Centered at (0, 0) Given an Equation in Standard Form Graph the hyperbola given by the equation asymptotes. y 2 _ 64 x 2 _ 36 − = 1. Identify and label the vertices, co-vertices, foci, and Solution The standard form that applies to the given equation is y-axis y 2 x 2 _ _ b2 = 1. Thus, the transverse axis is on the a2 − The coordinates of the vertices are (0, ± a) = (0, ± √ — 64 ) = (0, ± 8) The coordinates of the co-vertices are (±b, 0) = (± √ — 36 , 0) = (±6, 0) The coordinates of the foci are (0, ± c), where c = ± √ — a2 + b2 . Solving for c, we have c = ± √ — a2 + b2 = ± √ — 64 + 36 = ± √ — 100 = ± 10 Therefore, the coordinates of the foci are (0, ± 10) a 8 4 __ __ __ The equations of the asymptotes are Plot and label the vertices and co-vertices, and then sketch the central rectangle. Sides of the rectangle are parallel to the axes and pass through the vertices and co-vertices. Sketch and extend the diagonals of the central rectangle to show the asymptotes. The central rectangle and asymptotes provide the framework needed to sketch an accurate graph of the hyperbola. Label the foci and asymptotes, and draw a smooth curve to form the hyperbola, as shown in Figure 8. 888 CHAPTER 10 analytic geometry y (0, 10) (0, 8) y = − 4 x 3 (−6, 8) y = 4 x 3 y (0, 8) (–6, 0) (0, 0) (6, 0) x (−6, 0) (0, 0) (6, 8) (6, 0) x (0, –8) (0, – 10) (0, −8) (−6, −8) (6, −8) Figure 8 Try It #4 Graph the hyperbola given by the equation asymptotes. x 2 _ 144 − y 2 _ 81 = 1. Identify and label the vertices, co-vertices, foci, and Graphing Hyperbolas not Centered at the Origin Graphing hyperbolas centered at a point (h, k) other than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard forms (y − k)2 _ a2 (x − h)2 _ b2 − = 1 for vertical hyperbolas. From these standard form equations we can easily calculate and plot (x − h)2 _ a2 − (y − k)2 _ b2 = 1 for horizontal hyperbolas, and key features of the graph: the coordinates of its center, vertices, co-vertices, and foci; the equations of its asymptotes; and the positions of the transverse and conjugate axes. How To… Given a general form for a hyperbola centered at (h, k), sketch the graph. 1. Determine which of the standard forms applies to the given equation. Convert the general form to that standard form. 2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the center, vertices, co-vertices, foci; and equations for the asymptotes. a. If the equation is in the form • the transverse axis is parallel to the x-axis (x − h)2 _ a2 − (y − k)2 _ b2 = 1, then • the center is (h, k) • the coordinates of the vertices are (h ± a, k) • the coordinates of the co-vertices are (h, k ± b) • the coordinates of the foci are (h ± c, k) b __ (x − h) + k • the equations of the asymptotes are y = ± a SECTION 10.2 the hyperBola 889 b. If the equation is in the form (y − k)2 _______ a2 − (x − h)2 _______ b2 = 1, then • the transverse axis is parallel to the y-axis • the center is (h, k) • the coordinates of the vertices are (h, k ± a) • the coordinates of the co-vertices are (h ± b, k) • the coordinates of the foci are (h, k ± c) • the equations of the asymptotes are y = ± 3. Solve for the coordinates of the foci using the equation c = ± √ — a2 + b2 . a _ (x − h) + k b 4. Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw a smooth curve to form the hyperbola. Example 5 Graphing a Hyperbola Centered at (h, k) Given an Equation in General Form Graph the hyperbola given by the equation 9x 2 − 4y 2 − 36x − 40y − 388 = 0. Identify and label the center, vertices, co-vertices, foci, and asymptotes. Solution Start by expressing the equation in standard form. Group terms that contain the same variable, and move the constant to the opposite side of the equation. Factor the leading coefficient of each expression. (9x 2 − 36x) − (4y 2 + 40y) = 388 9(x 2 − 4x) − 4(y 2 + 10y) = 388 Complete the square twice. Remember to balance the equation by adding the same constants to each side. 9(x 2 − 4x + 4) −4(y 2 +
10y + 25) = 388 + 36 − 100 Rewrite as perfect squares. 9(x − 2)2 − 4(y + 5)2 = 324 Divide both sides by the constant term to place the equation in standard form. (x − 2)2 _ 36 − (y + 5)2 _ 81 = 1 The standard form that applies to the given equation is = 1, where a2 = 36 and b2 = 81, or a = 6 (x − h)2 _ a2 − (y − k)2 _ b2 and b = 9. Thus, the transverse axis is parallel to the x-axis. It follows that: • the center of the ellipse is (h, k) = (2, −5) • the coordinates of the vertices are (h ± a, k) = (2 ± 6, −5), or (−4, −5) and (8, −5) • the coordinates of the co-vertices are (h, k ± b) = (2, − 5 ± 9), or (2, − 14) and (2, 4) a2 + b 2 . Solving for c, we have • the coordinates of the foci are (h ± c, k), where c = ± √ — c = ± √ — 36 + 81 = ± √ — 117 = ± 3 √ — 13 Therefore, the coordinates of the foci are (2 − 3 √ — 13 , −5) and (2 + 3 √ — 13 , −5). b 3 __ __ The equations of the asymptotes are y = ± (x − 2) − 5. (x − h) + k = ± 2 a Next, we plot and label the center, vertices, co-vertices, foci, and asymptotes and draw smooth curves to form the hyperbola, as shown in Figure 9. 890 CHAPTER 10 analytic geometry y (2, 4) x (−4, −5) (8, −5) (2, −14) (2, −5) Figure 9 Try It #5 Graph the hyperbola given by the standard form of an equation vertices, co-vertices, foci, and asymptotes. (y + 4)2 _ − 100 (x − 3)2 _ 64 = 1. Identify and label the center, Solving Applied Problems Involving Hyperbolas As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. The design efficiency of hyperbolic cooling towers is particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are often touted for their ability to generate power efficiently. Because of their hyperbolic form, these structures are able to withstand extreme winds while requiring less material than any other forms of their size and strength. See Figure 10. For example, a 500-foot tower can be made of a reinforced concrete shell only 6 or 8 inches wide! Figure 10 Cooling towers at the Drax power station in north Yorkshire, United Kingdom (credit: les Haines, Flickr) The first hyperbolic towers were designed in 1914 and were 35 meters high. Today, the tallest cooling towers are in France, standing a remarkable 170 meters tall. In Example 6 we will use the design layout of a cooling tower to find a hyperbolic equation that models its sides. SECTION 10.2 the hyperBola 891 Example 6 Solving Applied Problems Involving Hyperbolas The design layout of a cooling tower is shown in Figure 11. The tower stands 179.6 meters tall. The diameter of the top is 72 meters. At their closest, the sides of the tower are 60 meters apart. 72 m 60 m 79.6 m 179.6 m Figure 11 Project design for a natural draft cooling tower Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola— indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places. Solution We are assuming the center of the tower is at the origin, so we can use the standard form of a horizontal y 2 x 2 _ _ b2 = 1, where the branches of the hyperbola form the sides of the cooling tower. a2 − hyperbola centered at the origin: We must find the values of a2 and b2 to complete the model. First, we find a2. Recall that the length of the transverse axis of a hyperbola is 2a. This length is represented by the distance where the sides are closest, which is given as 65.3 meters. So, 2a = 60. Therefore, a = 30 and a2 = 900. To solve for b2, we need to substitute for x and y in our equation using a known point. To do this, we can use the dimensions of the tower to find some point (x, y) that lies on the hyperbola. We will use the top right corner of the tower to represent that point. Since the y-axis bisects the tower, our x-value can be represented by the radius of the top, or 36 meters. The y-value is represented by the distance from the origin to the top, which is given as 79.6 meters. Therefore, y 2 x 2 _ _ b2 = 1 a2 − Standard form of horizontal hyperbola. b2 = y 2 _ x 2 _ a2 − 1 = (79.6)2 _ (36)2 ____ − 1 900 Isolate b2 Substitute for a2, x, and y The sides of the tower can be modeled by the hyperbolic equation ≈ 14400.3636 Round to four decimal places x 2 _ 900 − y 2 _ 14400.3636 = 1, or x 2 _ 302 − y 2 _ 120.00152 = 1 892 CHAPTER 10 analytic geometry Try It #6 A design for a cooling tower project is shown in Figure 12. Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places. 60 m 40 m 67.082 m 167.082 m Figure 12 Access these online resources for additional instruction and practice with hyperbolas. • Conic Sections: The Hyperbola Part 1 of 2 (http://openstaxcollege.org/l/hyperbola1) • Conic Sections: The Hyperbola Part 2 of 2 (http://openstaxcollege.org/l/hyperbola2) • Graph a Hyperbola with Center at Origin (http://openstaxcollege.org/l/hyperbolaorigin) • Graph a Hyperbola with Center not at Origin (http://openstaxcollege.org/l/hbnotorigin) SECTION 10.2 section exercises 893 10.2 SeCTIOn exeRCISeS VeRBAl 1. Define a hyperbola in terms of its foci. 2. What can we conclude about a hyperbola if its asymptotes intersect at the origin? 3. What must be true of the foci of a hyperbola? 4. If the transverse axis of a hyperbola is vertical, what do we know about the graph? 5. Where must the center of hyperbola be relative to its foci? AlGeBRAIC For the following exercises, determine whether the following equations represent hyperbolas. If so, write in standard form. 6. 3y 2 + 2x = 6 8. 5y 2 + 4x 2 = 6x 10. −9x 2 + 18x + y 2 + 4y − 14 = 0 7. − x 2 ___ 36 y 2 __ = 1 9 9. 25x 2 − 16y 2 = 400 For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes. 11. 13 ___ ___ 36 25 y 2 x 2 __ ___ − 4 81 (x − 1)2 _______ 9 (x − 2)2 _______ 49 15. 17. − − = 1 = 1 (y − 2)2 _______ 16 (y + 7)2 _______ 49 19. −9x 2 − 54x + 9y 2 − 54y + 81 = 0 21. −4x 2 + 24x + 16y 2 − 128y + 156 = 0 23. x 2 + 2x − 100y 2 − 1000y + 2401 = 0 25. 4x 2 + 24x − 25y 2 + 200y − 464 = 0 12. x 2 ___ 100 − y 2 __ = 1 9 14. 9y 2 − 4x 2 = 1 16. (y − 6)2 _______ 36 − (x + 1)2 _______ 16 = 1 18. 4x 2 − 8x − 9y 2 − 72y + 112 = 0 20. 4x 2 − 24x − 36y 2 − 360y + 864 = 0 22. −4x 2 + 40x + 25y 2 − 100y + 100 = 0 24. −9x 2 + 72x + 16y 2 + 16y + 4 = 0 For the following exercises, find the equations of the asymptotes for each hyperbola. (y + 4)2 _______ 22 (x − 3)2 _______ 52 − 26. 27. = 1 x 2 y 2 __ __ 32 = 1 32 − (y − 3)2 _______ 32 28. − (x + 5)2 _______ 62 30. 16y 2 + 96y − 4x 2 + 16x + 112 = 0 = 1 29. 9x 2 − 18x − 16y 2 + 32y − 151 = 0 GRAPHICAl For the following exercises, sketch a graph of the hyperbola, labeling vertices and foci. = 1 = 1 31. 33. 35. 37. − − x 2 y 2 ___ ___ 49 16 x 2 y 2 ___ __ 9 25 (y + 5)2 _______ 9 (y − 3)2 _______ 9 − − (x − 4)2 _______ 25 (x − 3)2 _______ 9 = 1 = 1 32. − = 1 x 2 ___ 64 y 2 __ 4 34. 81x 2 − 9y 2 = 1 36. (x − 2)2 _______ 8 − (y + 3)2 _______ 27 = 1 38. −4x 2 − 8x + 16y 2 − 32y − 52 = 0 894 CHAPTER 10 analytic geometry 39. x 2 − 8x − 25y 2 − 100y − 109 = 0 40. −x 2 + 8x + 4y 2 − 40y + 88 = 0 41. 64x 2 + 128x − 9y 2 − 72y − 656 = 0 42. 16x 2 + 64x − 4y 2 − 8y − 4 = 0 43. −100x 2 + 1000x + y 2 − 10y − 2575 = 0 44. 4x 2 + 16x − 4y 2 + 16y + 16 = 0 For the following exercises, given information about the graph of the hyperbola, find its equation. 45. Vertices at (3, 0) and (−3, 0) and one focus at (5, 0). 46. Vertices at (0, 6) and (0, −6) and one focus at (0, −8). 47. Vertices at (1, 1) and (11, 1) and one focus at (12, 1). 48. Center: (0, 0); vertex: (0, −13); one focus: (0, √ — 313 ). 49. Center: (4, 2); vertex: (9, 2); one focus: (4 + √ — 26 , 2). 50. Center: (3, 5); vertex: (3, 11); one focus: (3, 5 + 2 √ — 10 ). For the following exercises, given the graph of the hyperbola, find its equation. 52. y Foci Vertices Center (1, 1) Foci x 54. y Foci Center (3, 1) Foci Vertices 51. 53. 55. y 10 8 6 4 2 0 –2 –2 –4 –6 –8 –10 –10 –8 –6 –4 42 6 8 10 x y Foci (−1, 3) Center (−1, 0) (−1, −3) Foci y (–3, –3) Center Vertices Vertices Foci (–8, –3) x x Foci (2, –3) SECTION 10.2 section exercises 895 exTenSIOnS For the following exercises, express the equation for the hyperbola as two functions, with y as a function of x. Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. 56 57 58. (x − 2)2 _ 16 − (y + 3)2 _ 25 = 1 59. −4x 2 − 16x + y 2 − 2y − 19 = 0 60. 4x 2 − 24x − y 2 − 4y + 16 = 0 ReAl-WORlD APPlICATIOnS For the following exercises, a hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph. 61. The hedge will follow the asymptotes y = x and y = −x, and its closest distance to the center fountain is 5 yards. 62. The hedge will follow the asymptotes y = 2x and y = −2x, and its closest distance to the center fountain is 6 yards. 1 __ 63. The hedge will follow the asymptotes y = x and 2 1 __ y = − x, and its closest distance to the center 2 2 __ 64. The hedge will follow the asymptotes y = x and 3 2 __ y = − x, and its closest distance to the center 3 fountain is 10 yards. fountain is 12 yards. 3 __ 65. The hedge will follow the asymptotes y = x and 4 3 __ y = − x, and its closest distance to the center 4 fountain is 20 yards. For the following exercises, assume an object enters our solar system and we want to graph its path on a coordinate system with the sun at the origin and the x-axis as the axis of symmetry for the object's path. Give the equation of the flight path of each object using the
given information. 66. The object enters along a path approximated by the line y = x − 2 and passes within 1 au (astronomical unit) of the sun at its closest approach, so that the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = −x + 2. 68. The object enters along a path approximated by the line y = 0.5x + 2 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = −0.5x − 2. 67. The object enters along a path approximated by the line y = 2x − 2 and passes within 0.5 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = −2x + 2. 69. The object enters along a path approximated by the 1 __ x − 1 and passes within 1 au of the sun line y = 3 at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a 1 __ path approximated by the line y = − x + 1. 3 70. The object enters along a path approximated by the line y = 3x − 9 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = −3x + 9. 896 CHAPTER 10 analytic geometry leARnInG OBjeCTIVeS In this section, you will: • Graph parabolas with vertices at the origin. • Write equations of parabolas in standard form. • Graph parabolas with vertices not at the origin. • Solve applied problems involving parabolas. 10. 3 THe PARABOlA Figure 1 The Olympic torch concludes its journey around the world when it is used to light the Olympic cauldron during the opening ceremony. (credit: Ken Hackman, U.S. Air Force) Did you know that the Olympic torch is lit several months before the start of the games? The ceremonial method for lighting the flame is the same as in ancient times. The ceremony takes place at the Temple of Hera in Olympia, Greece, and is rooted in Greek mythology, paying tribute to Prometheus, who stole fire from Zeus to give to all humans. One of eleven acting priestesses places the torch at the focus of a parabolic mirror (see Figure 1), which focuses light rays from the sun to ignite the flame. Parabolic mirrors (or reflectors) are able to capture energy and focus it to a single point. The advantages of this property are evidenced by the vast list of parabolic objects we use every day: satellite dishes, suspension bridges, telescopes, microphones, spotlights, and car headlights, to name a few. Parabolic reflectors are also used in alternative energy devices, such as solar cookers and water heaters, because they are inexpensive to manufacture and need little maintenance. In this section we will explore the parabola and its uses, including low-cost, energy-efficient solar designs. Graphing Parabolas with Vertices at the Origin In The Ellipse, we saw that an ellipse is formed when a plane cuts through a right circular cone. If the plane is parallel to the edge of the cone, an unbounded curve is formed. This curve is a parabola. See Figure 2. Figure 2 Parabola SECTION 10.3 the paraBola 897 Like the ellipse and hyperbola, the parabola can also be defined by a set of points in the coordinate plane. A parabola is the set of all points (x, y) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix. In Quadratic Functions, we learned about a parabola’s vertex and axis of symmetry. Now we extend the discussion to include other key features of the parabola. See Figure 3. Notice that the axis of symmetry passes through the focus and vertex and is perpendicular to the directrix. The vertex is the midpoint between the directrix and the focus. The line segment that passes through the focus and is parallel to the directrix is called the latus rectum. The endpoints of the latus rectum lie on the curve. By definition, the distance d from the focus to any point P on the parabola is equal to the distance from P to the directrix. y Latus rectum Axis of symmetry Focus Vertex Directrix x Figure 3 Key features of the parabola To work with parabolas in the coordinate plane, we consider two cases: those with a vertex at the origin and those with a vertex at a point other than the origin. We begin with the former. y y = −p d (x, y) d x (x, −p) (0, p) (0, 0) Figure 4 Let (x, y) be a point on the parabola with vertex (0, 0), focus (0, p), and directrix y = −p as shown in Figure 4. The distance d from point (x, y) to point (x, −p) on the directrix is the difference of the y-values: d = y + p. The distance from the focus (0, p) to the point (x, y) is also equal to d and can be expressed using the distance formula. d = √ = √ — (x − 0)2 + (y − p)2 x 2 + (y − p)2 — Set the two expressions for d equal to each other and solve for y to derive the equation of the parabola. We do this because the distance from (x, y) to (0, p) equals the distance from (x, y) to (x, −p). — x 2 + (y − p)2 = y + p √ We then square both sides of the equation, expand the squared terms, and simplify by combining like terms. x 2 + (y − p)2 = (y + p)2 x 2 + y 2 − 2py + p2 = y 2 + 2py + p2 x 2 − 2py = 2py x 2 = 4py The equations of parabolas with vertex (0, 0) are y 2 = 4px when the x-axis is the axis of symmetry and x 2 = 4py when the y-axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features. 898 CHAPTER 10 analytic geometry standard forms of parabolas with vertex (0, 0) Table 1 and Figure 5 summarize the standard features of parabolas with a vertex at the origin. Focus (p, 0) (0, p) Table 1 x Axis of Symmetry Equation x-axis y-axis y 2 = 4px x 2 = 4py y y 2 = 4px p > 0 (p, \|2p\|) (p, 0) (p, −\|2p\|) (a) y x2 = 4py p > 0 (0, 0) x = −p Directrix Endpoints of Latus Rectum x = −p y = −p (p, ± 2p) (± 2p, p) y 2 = 4px p < 0 y (p, \|2p\|) (p, 0) (p, −\|2p\|) x (0, 0) (b) y (0, 0) x = −p y = −p x (−\|2p\|, p) (0, p) (\|2p\|, p) x y = −p (0, 0) (c) (−\|2p\|, p) (0, p) (\|2p\|, p) x2 = 4py p < 0 (d) Figure 5 (a) When p > 0 and the axis of symmetry is the x-axis, the parabola opens right. (b) When p < 0 and the axis of symmetry is the x-axis, the parabola opens left. (c) When p < 0 and the axis of symmetry is the y-axis, the parabola opens up. (d) When p < 0 and the axis of symmetry is the y-axis, the parabola opens down. The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and latus rectum. See Figure 5. When given a standard equation for a parabola centered at the origin, we can easily identify the key features to graph the parabola. A line is said to be tangent to a curve if it intersects the curve at exactly one point. If we sketch lines tangent to the parabola at the endpoints of the latus rectum, these lines intersect on the axis of symmetry, as shown in Figure 6. y y 2 = 24x (6, 12) (0, 0) (6, 0) x (6, −12) x = −6 Figure 6 SECTION 10.3 the paraBola 899 How To… Given a standard form equation for a parabola centered at (0, 0), sketch the graph. 1. Determine which of the standard forms applies to the given equation: y 2 = 4px or x 2 = 4py. 2. Use the standard form identified in Step 1 to determine the axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum. a. If the equation is in the form y 2 = 4px, then • the axis of symmetry is the x-axis, y = 0 • set 4p equal to the coefficient of x in the given equation to solve for p. If p > 0, the parabola opens right. If p < 0, the parabola opens left. • use p to find the coordinates of the focus, (p, 0) • use p to find the equation of the directrix, x = − p • use p to find the endpoints of the latus rectum, (p, ± 2p). Alternately, substitute x = p into the original equation. b. If the equation is in the form x 2 = 4py, then • the axis of symmetry is the y-axis, x = 0 • set 4p equal to the coefficient of y in the given equation to solve for p. If p > 0, the parabola opens up. If p < 0, the parabola opens down. • use p to find the coordinates of the focus, (0, p) • use p to find equation of the directrix, y = − p • use p to find the endpoints of the latus rectum, (±2p, p) 3. Plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. Example 1 Graphing a Parabola with Vertex (0, 0) and the x-axis as the Axis of Symmetry Graph y 2 = 24x. Identify and label the focus, directrix, and endpoints of the latus rectum. Solution The standard form that applies to the given equation is y 2 = 4px. Thus, the axis of symmetry is the x-axis. It follows that: • 24 = 4p, so p = 6. Since p > 0, the parabola opens right • the coordinates of the focus are (p, 0) = (6, 0) • the equation of the directrix is x = −p = − 6 • the endpoints of the latus rectum have the same x-coordinate at the focus. To find the endpoints, substitute x = 6 into the original equation: (6, ± 12) Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See Figure 7. x = −6 (0, 0) y 20 16 12 8 4 –10 –8 –6 –4 –2 (6, 12) (6, 0) 4 6 8 10 x 2 0 –4 –8 –12 –16 –20 Figure 7 (6, –12) 900 CHAPTER 10 analytic geometry Try It #1 Graph y 2 = −16x. Identify and label the focus, directrix, and endpoints of the latus rectum. Example 2 Graphing a Parabola with Vertex (0, 0) and the y-axis as the Axis of Symmetry Graph x 2 = −6y. Identify and label the focus, directrix, and endpoints of the latus rectum. Solution The standard form that applies to the given equation is x 2 = 4py. Thus, the axis of symmetry is the y-axis. It follows that: 3 __ • −6 = 4p, so p = − Since p < 0, the parabola opens down. 2 3 • the coordinates of the focus are (0, p) =  0, −  __ 2 3 __ • the equation of the directrix is y = − p = 2 3 3  into the original equation,  ±3, − __ __ • the endpoints of the latus rectum can be found by substituting y = 2 2 Next we plot the focus, directrix, and la
tus rectum, and draw a smooth curve to form the parabola. −3, − 3 2 y (0, 0) 0, − 3 2 Figure 8 y = 3 2 x 3, − 3 2 x2 = −6y Try It #2 Graph x 2 = 8y. Identify and label the focus, directrix, and endpoints of the latus rectum. Writing equations of Parabolas in Standard Form In the previous examples, we used the standard form equation of a parabola to calculate the locations of its key features. We can also use the calculations in reverse to write an equation for a parabola when given its key features. How To… Given its focus and directrix, write the equation for a parabola in standard form. 1. Determine whether the axis of symmetry is the x- or y-axis. a. If the given coordinates of the focus have the form (p, 0), then the axis of symmetry is the x-axis. Use the standard form y 2 = 4px. b. If the given coordinates of the focus have the form (0, p), then the axis of symmetry is the y-axis. Use the standard form x 2 = 4py. 2. Multiply 4p. 3. Substitute the value from Step 2 into the equation determined in Step 1. Example 3 Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix 1 1 What is the equation for the parabola with focus  − , 0  and directrix x = __ __ ? 2 2 Solution The focus has the form (p, 0), so the equation will have the form y 2 = 4px. 1 _ ) = −2. • Multiplying 4p, we have 4p = 4(− 2 • Substituting for 4p, we have y 2 = 4px = −2x. Therefore, the equation for the parabola is y 2 = −2x. SECTION 10.3 the paraBola 901 Try It #3 7 7  and directrix y = − What is the equation for the parabola with focus  0, __ __ ? 2 2 Graphing Parabolas with Vertices not at the Origin Like other graphs we’ve worked with, the graph of a parabola can be translated. If a parabola is translated h units horizontally and k units vertically, the vertex will be (h, k). This translation results in the standard form of the equation we saw previously with x replaced by (x − h) and y replaced by (y − k). To graph parabolas with a vertex (h, k) other than the origin, we use the standard form (y − k)2 = 4p(x − h) for parabolas that have an axis of symmetry parallel to the x-axis, and (x − h)2 = 4p(y − k) for parabolas that have an axis of symmetry parallel to the y-axis. These standard forms are given below, along with their general graphs and key features. standard forms of parabolas with vertex (h, k) Table 2 and Figure 9 summarize the standard features of parabolas with a vertex at a point (h, k). Axis of Symmetry Equation Focus Directrix Endpoints of Latus Rectum y = k x = h (y − k)2 = 4p(x − h) (h + p, k) (x − h)2 = 4p(y − k) (h, k + p) x = h −p y = k −p (h + p, k ± 2p) (h ± 2p, k + p) Table 2 y (y − k)2 = 4p(x − h) p > 0 y (y − k)2 = 4p(x − h) p < 0 (h + p, k + 2p) y = k (h + p, k) (h + p, k − 2p) (h, k) x = h − p (a) x (h + p, k + \|2p\|) (h + p, k) (h + p, k − \|2p\|) y = k x (h, k) x = h − p (b) y (x − h)2 = 4p(y − h) p > 0 y (x − h)2 = 4p(y − k) p < 0 x = h (h, k + ph, k) (h − 2p, k + p) (h + 2p, k + p) y = k − p (h, k) (h − \|2p\|, k + p) (h + \|2p\|, k + p) (h, k + p) x x (c) (d) Figure 9 (a) When p > 0, the parabola opens right. (b) When p < 0, the parabola opens left. (c) When p > 0, the parabola opens up. (d) When p < 0, the parabola opens down. 902 CHAPTER 10 analytic geometry How To… Given a standard form equation for a parabola centered at (h, k), sketch the graph. 1. Determine which of the standard forms applies to the given equation: (y − k)2 = 4p(x − h) or (x − h)2 = 4p(y − k). 2. Use the standard form identified in Step 1 to determine the vertex, axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum. a. If the equation is in the form (y − k)2 = 4p(x − h), then: • use the given equation to identify h and k for the vertex, (h, k) • use the value of k to determine the axis of symmetry, y = k • set 4p equal to the coefficient of (x − h) in the given equation to solve for p. If p > 0, the parabola opens right. If p < 0, the parabola opens left. • use h, k, and p to find the coordinates of the focus, (h + p, k) • use h and p to find the equation of the directrix, x = h − p • use h, k, and p to find the endpoints of the latus rectum, (h + p, k ± 2p) b. If the equation is in the form (x − h)2 = 4p(y − k), then: • use the given equation to identify h and k for the vertex, (h, k) • use the value of h to determine the axis of symmetry, x = h • set 4p equal to the coefficient of (y − k) in the given equation to solve for p. If p > 0, the parabola opens up. If p < 0, the parabola opens down. • use h, k, and p to find the coordinates of the focus, (h, k + p) • use k and p to find the equation of the directrix, y = k − p • use h, k, and p to find the endpoints of the latus rectum, (h ± 2p, k + p) 3. Plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. Example 4 Graphing a Parabola with Vertex ( h, k) and Axis of Symmetry Parallel to the x-axis Graph (y − 1)2 = −16(x + 3). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum. Solution The standard form that applies to the given equation is (y − k)2 = 4p(x − h). Thus, the axis of symmetry is parallel to the x-axis. It follows that: • the vertex is (h, k) = (−3, 1) • the axis of symmetry is y = k = 1 • −16 = 4p, so p = −4. Since p < 0, the parabola opens left. • the coordinates of the focus are (h + p, k) = (−3 + (−4), 1) = (−7, 1) • the equation of the directrix is x = h − p = −3 − (−4) = 1 • the endpoints of the latus rectum are (h + p, k ± 2p) = (−3 + (−4), 1 ± 2(−4)), or (−7, −7) and (−7, 9) Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See Figure 10. y (y − 1)2 = −16(x + 3) (−7 , 9) (−7 , 1) (−3, 1) (−7 , −7 ) Figure 10 y = 1 x x = 1 SECTION 10.3 the paraBola 903 Try It #4 Graph (y + 1)2 = 4(x − 8). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum. Example 5 Graphing a Parabola from an Equation Given in General Form Graph x 2 − 8x − 28y − 208 = 0. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum. Solution Start by writing the equation of the parabola in standard form. The standard form that applies to the given equation is (x − h)2 = 4p (y − k). Thus, the axis of symmetry is parallel to the y-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable x in order to complete the square. x 2 − 8x − 28y − 208 = 0 x 2 − 8x = 28y + 208 x 2 − 8x + 16 = 28y + 208 + 16 (x − 4)2 = 28y + 224 (x − 4)2 = 28(y + 8) (x − 4)2 = 4 ⋅ 7 ⋅ (y + 8) It follows that: • the vertex is (h, k) = (4, −8) • the axis of symmetry is x = h = 4 • since p = 7, p > 0 and so the parabola opens up • the coordinates of the focus are (h, k + p) = (4, −8 + 7) = (4, −1) • the equation of the directrix is y = k − p = −8 − 7 = −15 • the endpoints of the latus rectum are (h ± 2p, k + p) = (4 ± 2(7), −8 + 7), or (−10, −1) and (18, −1) Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See Figure 11. (−10, −1) y (x − 4)2 = 28(y + 8) x (18, −1) y = −15 (4, −1) (4, −8) x = 4 Figure 11 Try It #5 Graph (x + 2)2 = −20 (y − 3). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum. Solving Applied Problems Involving Parabolas As we mentioned at the beginning of the section, parabolas are used to design many objects we use every day, such as telescopes, suspension bridges, microphones, and radar equipment. Parabolic mirrors, such as the one used to light the Olympic torch, have a very unique reflecting property. When rays of light parallel to the parabola’s axis of symmetry are directed toward any surface of the mirror, the light is reflected directly to the focus. See Figure 12. This is why the Olympic torch is ignited when it is held at the focus of the parabolic mirror. 904 CHAPTER 10 analytic geometry Parallel rays of sunlight Focus Parabolic reflector Figure 12 Parabolic mirrors have the ability to focus the sun’s energy to a single point, raising the temperature hundreds of degrees in a matter of seconds. Thus, parabolic mirrors are featured in many low-cost, energy efficient solar products, such as solar cookers, solar heaters, and even travel-sized fire starters. Example 6 Solving Applied Problems Involving Parabolas A cross-section of a design for a travel-sized solar fire starter is shown in Figure 13. The sun’s rays reflect off the parabolic mirror toward an object attached to the igniter. Because the igniter is located at the focus of the parabola, the reflected rays cause the object to burn in just seconds. a. Find the equation of the parabola that models the fire starter. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane. b. Use the equation found in part ( a) to find the depth of the fire starter. Igniter 1.7 in Depth 4.5 in Figure 13 Cross-section of a travel-sized solar fire starter Solution a. The vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form x 2 = 4py, where p > 0. The igniter, which is the focus, is 1.7 inches above the vertex of the dish. Thus we have p = 1.7. x 2 = 4py Standard form of upward-facing parabola with vertex (0, 0) x 2 = 4(1.7)y Substitute 1.7 for p. x 2 = 6.8y Multiply. b. The dish extends = 2.25 inches on either side of the origin. We can substitute 2.25 for x in the equation 4.5 ___ 2 from part ( a) to find the depth of the dish. x 2 = 6.8y (2.25)2 = 6.8y Equation found in part ( a). Substitute 2.25 for x. y ≈ 0.74 Solve for y. The dish is about 0.74 inches deep. SECTION 10.3 the paraBola 905 Try It #6 Balcony-sized solar cookers have been designed for families living in India. The top of a dish has a diameter of 1,600 mm. The sun’s rays reflect off the parabolic mirror toward the “
cooker,” which is placed 320 mm from the base. a. Find an equation that models a cross-section of the solar cooker. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane, and that the parabola opens to the right (i.e., has the x-axis as its axis of symmetry). b. Use the equation found in part (a) to find the depth of the cooker. Access these online resources for additional instruction and practice with parabolas. • Conic Sections: The Parabola Part 1 of 2 (http://openstaxcollege.org/l/parabola1) • Conic Sections: The Parabola Part 2 of 2 (http://openstaxcollege.org/l/parabola2) • Parabola with Vertical Axis (http://openstaxcollege.org/l/parabolavertcal) • Parabola with Horizontal Axis (http://openstaxcollege.org/l/parabolahoriz) 906 CHAPTER 10 analytic geometry 10.3 SeCTIOn exeRCISeS VeRBAl 1. Define a parabola in terms of its focus and directrix. 3. If the equation of a parabola is written in standard form and p is negative and the directrix is a horizontal line, then what can we conclude about its graph? 5. As the graph of a parabola becomes wider, what will happen to the distance between the focus and directrix? AlGeBRAIC 2. If the equation of a parabola is written in standard form and p is positive and the directrix is a vertical line, then what can we conclude about its graph? 4. What is the effect on the graph of a parabola if its equation in standard form has increasing values of p? For the following exercises, determine whether the given equation is a parabola. If so, rewrite the equation in standard form. 6. y 2 = 4 − x 2 9. (y − 3)2 = 8(x − 2) 7. y = 4x 2 10. y 2 + 12x − 6y − 51 = 0 8. 3x 2 − 6y 2 = 12 For the following exercises, rewrite the given equation in standard form, and then determine the vertex (V), focus (F), and directrix (d) of the parabola. 11. x = 8y 2 1 __ 12. y = x 2 4 13. y = −4x 2 1 __ 14. x = y 2 8 15. x = 36y 2 16. x = 1 ___ 36 y 2 17. (x − 1)2 = 4(y − 1) 20. (x + 1)2 = 2(y + 4) 23. y 2 + 12x − 6y + 21 = 0 26. y 2 − 24x + 4y − 68 = 0 29. 3y 2 − 4x − 6y + 23 = 0 GRAPHICAl 4 __ (x + 4) 18. (y − 2)2 = 5 21. (x + 4)2 = 24(y + 1) 24. x 2 − 4x − 24y + 28 = 0 27. x 2 − 4x + 2y − 6 = 0 30. x 2 + 4x + 8y − 4 = 0 19. (y − 4)2 = 2(x + 3) 22. (y + 4)2 = 16(x + 4) 25. 5x 2 − 50x − 4y + 113 = 0 28. y 2 − 6y + 12x − 3 = 0 For the following exercises, graph the parabola, labeling the focus and the directrix. 1 __ 31. x = y 2 8 32. y = 36x 2 33. y = 1 ___ 36 x 2 34. y = −9x 2 37. −6(y + 5)2 = 4(x − 4) 4 __ (x + 2) 35. (y − 2)2 = − 3 38. y 2 − 6y − 8x + 1 = 0 36. −5(x + 5)2 = 4(y + 5) 39. x 2 + 8x + 4y + 20 = 0 40. 3x 2 + 30x − 4y + 95 = 0 41. y 2 − 8x + 10y + 9 = 0 42. x 2 + 4x + 2y + 2 = 0 43. y 2 + 2y − 12x + 61 = 0 44. −2x 2 + 8x − 4y − 24 = 0 For the following exercises, find the equation of the parabola given information about its graph. 45. Vertex is (0, 0); directrix is y = 4, focus is (0, −4). 46. Vertex is (0, 0); directrix is x = 4, focus is (−4, 0). 47. Vertex is (2, 2); directrix is x = 2 − √ — (2 + √ 2 , 2). — 2 , focus is 1 7 , 3  . , focus is  − __ __ 48. Vertex is (−2, 3); directrix is x = − 2 2 49. Vertex is ( √ 3 ). is (0 ); directrix is x = 2 √ — 2 , focus 50. Vertex is (1, 2); directrix is y = 1 11  . , focus is  1, __ ___ 3 3 SECTION 10.3 section exercises 907 For the following exercises, determine the equation for the parabola from its graph. 52. y Focus (−1, 2) Vertex (3, 2) Axis of symmetry 54. y Vertex (−3, 5) Focus −3, 319 64 Axis of symmetry x x 51. 53. Axis of symmetry Focus 1 4 0, y y Vertex (0, 0) Vertex (−2, 2) Axis of symmetry Focus − 31 , 2 16 x x 55. y Vertex Focus Axis of symmetry x 908 CHAPTER 10 analytic geometry exTenSIOnS For the following exercises, the vertex and endpoints of the latus rectum of a parabola are given. Find the equation. 56. V(0, 0), Endpoints (2, 1), (−2, 1) 57. V(0, 0), Endpoints (−2, 4), (−2, −4) 58. V(1, 2), Endpoints (−5, 5), (7, 5) 59. V(−3, −1), Endpoints (0, 5), (0, −7) 7 7   ,  3, − 60. V(4, −3), Endpoints  5, − __ __ 2 2 ReAl-WORlD APPlICATIOnS 61. The mirror in an automobile headlight has a parabolic cross-section with the light bulb at the focus. On a schematic, the equation of the parabola is given as x 2 = 4y. At what coordinates should you place the light bulb? 62. If we want to construct the mirror from the previous exercise such that the focus is located at (0, 0.25), what should the equation of the parabola be? 63. A satellite dish is shaped like a paraboloid of 64. Consider the satellite dish from the previous revolution. This means that it can be formed by rotating a parabola around its axis of symmetry. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver be placed? exercise. If the dish is 8 feet across at the opening and 2 feet deep, where should we place the receiver? 65. A searchlight is shaped like a paraboloid of revolution. A light source is located 1 foot from the base along the axis of symmetry. If the opening of the searchlight is 3 feet across, find the depth. 66. If the searchlight from the previous exercise has the light source located 6 inches from the base along the axis of symmetry and the opening is 4 feet, find the depth. 67. An arch is in the shape of a parabola. It has a span of 100 feet and a maximum height of 20 feet. Find the equation of the parabola, and determine the height of the arch 40 feet from the center. 68. If the arch from the previous exercise has a span of 160 feet and a maximum height of 40 feet, find the equation of the parabola, and determine the distance from the center at which the height is 20 feet. 69. An object is projected so as to follow a parabolic path given by y = −x 2 + 96x, where x is the horizontal distance traveled in feet and y is the height. Determine the maximum height the object reaches. 70. For the object from the previous exercise, assume the path followed is given by y = −0.5x 2 + 80x. Determine how far along the horizontal the object traveled to reach maximum height. SECTION 10.4 rotation oF axis 909 leARnInG OBjeCTIVeS In this section, you will: • Identify nondegenerate conic sections given their general form equations. • Use rotation of axes formulas. • Write equations of rotated conics in standard form. • Identify conics without rotating axes. 10. 4 ROTATIOn OF AxIS As we have seen, conic sections are formed when a plane intersects two right circular cones aligned tip to tip and extending infinitely far in opposite directions, which we also call a cone. The way in which we slice the cone will determine the type of conic section formed at the intersection. A circle is formed by slicing a cone with a plane perpendicular to the axis of symmetry of the cone. An ellipse is formed by slicing a single cone with a slanted plane not perpendicular to the axis of symmetry. A parabola is formed by slicing the plane through the top or bottom of the double-cone, whereas a hyperbola is formed when the plane slices both the top and bottom of the cone. See Figure 1. Diagonal Slice Horizontal Slice Deep Vertical Slice Vertical Slice Ellipse Circle Hyperbola Parabola Figure 1 The nondegenerate conic sections Ellipses, circles, hyperbolas, and parabolas are sometimes called the nondegenerate conic sections, in contrast to the degenerate conic sections, which are shown in Figure 2. A degenerate conic results when a plane intersects the double cone and passes through the apex. Depending on the angle of the plane, three types of degenerate conic sections are possible: a point, a line, or two intersecting lines. Intersecting Lines Single Line Single Point Figure 2 Degenerate conic sections 910 CHAPTER 10 analytic geometry Identifying nondegenerate Conics in General Form In previous sections of this chapter, we have focused on the standard form equations for nondegenerate conic sections. In this section, we will shift our focus to the general form equation, which can be used for any conic. The general form is set equal to zero, and the terms and coefficients are given in a particular order, as shown below. Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 where A, B, and C are not all zero. We can use the values of the coefficients to identify which type conic is represented by a given equation. You may notice that the general form equation has an xy term that we have not seen in any of the standard form equations. As we will discuss later, the xy term rotates the conic whenever B is not equal to zero. Conic Sections Example ellipse circle hyperbola parabola one line intersecting lines parallel lines a point no graph 4x 2 + 9y 2 = 1 4x 2 + 4y 2 = 1 4x 2 − 9y 2 = 1 4x 2 = 9y or 4y 2 = 9x 4x + 9y = 1 (x − 4) (y + 4)= 0 (x − 4)(x − 9) = 0 4x 2 + 4y 2 = 0 4x 2 + 4y 2 = − 1 Table 1 general form of conic sections A conic section has the general form where A, B, and C are not all zero. Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 Table 2 summarizes the different conic sections where B = 0, and A and C are nonzero real numbers. This indicates that the conic has not been rotated. Conic Sections Example ellipse circle hyperbola parabola Ax 2 + Cy 2 + Dx + Ey + F = 0, A ≠ C and AC > 0 Ax 2 + Cy 2 + Dx + Ey + F = 0, A = C Ax 2 − Cy 2 + Dx + Ey + F = 0 or − Ax 2 + Cy 2 + Dx + Ey + F = 0, where A and C are positive Ax 2 + Dx + Ey + F = 0 or Cy 2 + Dx + Ey + F = 0 Table 2 How To… Given the equation of a conic, identify the type of conic. 1. Rewrite the equation in the general form, Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. 2. Identify the values of A and C from the general form. a. If A and C are nonzero, have the same sign, and are not equal to each other, then the graph may be an ellipse. b. If A and C are equal and nonzero and have the same sign, then the graph may be a circle. SECTION 10.4 rotation oF axis 911 c. If A and C are nonzero and have opposite signs, then the graph may be a hyperbola. d. If either A or C is zero, then the graph may be a parabola. If B = 0, the conic section will ha
ve a vertical and/or horizontal axes. If B does not equal 0, as shown below, the conic section is rotated. Notice the phrase “may be” in the definitions. That is because the equation may not represent a conic section at all, depending on the values of A, B, C, D, E, and F. For example, the degenerate case of a circle or an ellipse is a point: Ax 2 +By 2=0, when A and B have the same sign. The degenerate case of a hyperbola is two intersecting straight lines: Ax 2 +By 2=0, when A and B have opposite signs. On the other hand, the equation Ax 2 +By 2+1=0, when A and B are positive does not represent a graph at all, since there are no real ordered pairs which satisfy it. Example 1 Identifying a Conic from Its General Form Identify the graph of each of the following nondegenerate conic sections. a. 4x 2 − 9y 2 + 36x + 36y − 125 = 0 b. 9y 2 + 16x + 36y − 10 = 0 c. 3x 2 + 3y 2 − 2x − 6y − 4 = 0 d. −25x 2 − 4y 2 + 100x + 16y + 20 = 0 Solution a. Rewriting the general form, we have Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 4x 2 + 0xy + (−9)y 2 + 36x + 36y + (−125) = 0 A = 4 and C = −9, so we observe that A and C have opposite signs. The graph of this equation is a hyperbola. b. Rewriting the general form, we have Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 0x 2 + 0xy + 9y 2 + 16x + 36y + (−10) = 0 A = 0 and C = 9. We can determine that the equation is a parabola, since A is zero. c. Rewriting the general form, we have Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 3x 2 + 0xy + 3y 2 + (−2)x + (−6)y + (−4) = 0 A = 3 and C = 3. Because A = C, the graph of this equation is a circle. d. Rewriting the general form, we have Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 (−25)x 2 + 0xy + (−4)y 2 + 100x + 16y + 20 = 0 A = −25 and C = −4. Because AC > 0 and A ≠ C, the graph of this equation is an ellipse. Try It #1 Identify the graph of each of the following nondegenerate conic sections. a. 16y 2 − x 2 + x − 4y − 9 = 0 b. 16x 2 + 4y 2 + 16x + 49y − 81 = 0 912 CHAPTER 10 analytic geometry Finding a New Representation of the Given Equation after Rotating through a Given Angle Until now, we have looked at equations of conic sections without an xy term, which aligns the graphs with the x- and y- axes. When we add an xy term, we are rotating the conic about the origin. If the x- and y-axes are rotated through an angle, say θ, then every point on the plane may be thought of as having two representations: (x, y) on the Cartesian plane with the original x-axis and y-axis, and (x′, y′) on the new plane defined by the new, rotated axes, called the x′-axis and y′-axis. See Figure 3. y y’ x’ θ x Figure 3 The graph of the rotated ellipse x 2 + y 2 − xy − 15 = 0 We will find the relationships between x and y on the Cartesian plane with x′ and y′ on the new rotated plane. See Figure 4. y θ y' cos θ −sin θ x' sin θ θ cos θ x Figure 4 The Cartesian plane with x- and y-axes and the resulting x′− and y′−axes formed by a rotation by an angle θ. The original coordinate x- and y-axes have unit vectors i and j. The rotated coordinate axes have unit vectors i′ and j′. The angle θ is known as the angle of rotation. See Figure 5. We may write the new unit vectors in terms of the original ones. i′ = cos θi + sin θ j j′ = −sin θi + cos θ j y j y' j' cos θ −sin θ x' sin θ x i' i θ cos θ Figure 5 Relationship between the old and new coordinate planes. SECTION 10.4 rotation oF axis 913 Consider a vector u in the new coordinate plane. It may be represented in terms of its coordinate axes. u = x′ i′ + y′ j′ u = x′(i cos θ + j sin θ) + y′( − i sin θ + j cos θ) u = ix ' cos θ + jx ' sin θ − iy' sin θ + jy' cos θ Substitute. Distribute. u = ix ' cos θ − iy' sin θ + jx ' sin θ + jy' cos θ Apply commutative property. u = (x ' cos θ − y' sin θ)i + (x' sin θ + y' cos θ) j Factor by grouping. Because u = x′ i′ + y′ j′, we have representations of x and y in terms of the new coordinate system. x = x′ cos θ − y′ sin θ and y = x′ sin θ + y′ cos θ equations of rotation If a point (x, y) on the Cartesian plane is represented on a new coordinate plane where the axes of rotation are formed by rotating an angle θ from the positive x-axis, then the coordinates of the point with respect to the new axes are (x′, y′). We can use the following equations of rotation to define the relationship between (x, y) and (x′, y′): x = x′ cos θ − y′ sin θ and y = x′ sin θ + y′ cos θ How To… Given the equation of a conic, find a new representation after rotating through an angle. 1. Find x and y where x = x′ cos θ − y′ sin θ and y = x′ sin θ + y′ cos θ. 2. Substitute the expression for x and y into in the given equation, then simplify. 3. Write the equations with x′ and y′ in standard form. Example 2 Finding a New Representation of an Equation after Rotating through a Given Angle Find a new representation of the equation 2x 2 − xy + 2y 2 − 30 = 0 after rotating through an angle of θ = 45°. Solution Find x and y, where x = x′ cos θ − y′ sin θ and y = x′ sin θ + y′ cos θ. Because θ = 45°, and x = x′ cos(45°) − y′ sin(45°)  − y′  1 _ — 2 √ x′ − y′ sin(45°) + y′ cos(45°)  + y′  1 _ — 2 √ x′ + y′ _ 2 √ — y = Substitute x = x′ cosθ − y′ sinθ and y = x′ sin θ + y′ cos θ into 2x 2 − xy + 2y 2 − 30 = 0. 2  2 x′ − y′ _ 2 √  — −  x′ − y′ _ 2 √   x′ + y′ + y′ _ − 30 = 0 2 √  — 914 Simplify. CHAPTER 10 analytic geometry / / 2 (x′ − y′)(x′ − y′) __ − / 2 (x′ + y′)(x′ + y′) __ / 2 − 30 = 0 + / 2 (x′ − y′)(x′ + y′) __ 2 (x′ 2 − y′ 2) _ 2 x′ 2  −2x′ y′ + y′ 2 − + x′ 2  + 2x′ y′ + y′ 2 − 30 = 0 2x′ 2 + 2y′ 2 − (x′ 2 − y′ 2) _ 2 = 30 FOIL method Combine like terms. Combine like terms. 2  2x′ 2 + 2y′ 2 − (x′ 2 − y′ 2) _ 2  = 2(30) Multiply both sides by 2. 4x′ 2 + 4y′ 2 − (x′ 2 − y′ 2) = 60 Simplify. 4x′ 2 + 4y′ 2 − x′ 2 + y′ 2 = 60 Distribute. 3x′ 2 _ 60 + 5y′ 2 _ 60 = 60 _ 60 Set equal to 1. Write the equations with x′ and y′ in the standard form. This equation is an ellipse. Figure 6 shows the graph. x′ 2 _ 20 + y′ 2 _ 12 = 1 x’ θ = 45° 2 3 1 4 x y’ –4 –3 –2 –1 4 3 2 1 0 –1 –2 –3 –4 Figure 6 Writing equations of Rotated Conics in Standard Form Now that we can find the standard form of a conic when we are given an angle of rotation, we will learn how to transform the equation of a conic given in the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. into standard form by rotating the axes. To do so, we will rewrite the general form as an equation in the x′ and y′ coordinate system without the x′y′ term, by rotating the axes by a measure of θ that satisfies We have learned already that any conic may be represented by the second degree equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. cot(2θ) = A − C ______ B where A, B, and C are not all zero. However, if B ≠ 0, then we have an xy term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle θ where cot(2θ) = A − C ______ . B • If cot(2θ) > 0, then 2θ is in the first quadrant, and θ is between (0°, 45°). • If cot(2θ) < 0, then 2θ is in the second quadrant, and θ is between (45°, 90°). • If A = C, then θ = 45°. How To… Given an equation for a conic in the x′ y′ system, rewrite the equation without the x′ y′ term in terms of x′ and y′, where the x′ and y′ axes are rotations of the standard axes by θ degrees. SECTION 10.4 rotation oF axis 915 1. Find cot(2θ). 2. Find sin θ and cos θ. 3. Substitute sin θ and cos θ into x = x′ cos θ − y′ sin θ and y = x′ sin θ + y′ cos θ. 4. Substitute the expression for x and y into in the given equation, and then simplify. 5. Write the equations with x′ and y′ in the standard form with respect to the rotated axes. Example 3 Rewriting an Equation with respect to the x´ and y´ axes without the x´y´ Term Rewrite the equation 8x 2 − 12xy + 17y 2 = 20 in the x′ y′ system without an x′ y′ term. Solution First, we find cot(2θ). See Figure 7. 8x 2 − 12xy + 17y 2 = 20 ⇒ A = 8, B = − 12 and C = 17 cot(2θ) = cot(2θ) = = A − C ______ B −9 ____ −12 3 __ = 4 8 − 17 ______ −12 y h 2θ 3 Figure 7 4 x 3 __ = cot(2θ) = 4 adjacent _ opposite 32 + 42 = h2 9 + 16 = h2 25 = h2 h = 5 ———— ————— So the hypotenuse is Next, we find sin θ and cos θ. ___________ 1 − cos(2θ) __________ 2 = √ 3 __ 1 − 5 _ 2 = √ 3 5 __ __ − 5 5 _ 2 = √ _________ 5 − 3 1 = √ __ _____ ⋅ 5 2 ___ 2 = √ __ 10 __ 1 __ 5 ———— ————— = √ 3 __ 1 + 5 _ 2 = √ 3 5 __ __ + 5 5 _ 2 = √ _________ 5 + 3 1 = √ __ _____ ⋅ 5 2 ___ 8 = √ _ 10 __ 4 __ 5 sin θ = √ sin θ = cos θ = √ cos √ ___________ 1 + cos(2θ) __________ 2 Substitute the values of sin θ and cos θ into x = x′ cos θ − y′ sin θ and y = x′ sin θ + y′ cos θ. x = x′ cos θ − y′ sin θ and 1 2 _ _   − y′  x = x′  — — 5 5 √ √ 2x′ − y′ sin θ + y′ cos θ 916 CHAPTER 10 analytic geometry Substitute the expressions for x and y into in the given equation, and then simplify. 2 1 _ _   + y′  y = x′  — — 5 5 √ √ x′ + 2y′ _ 5 y = √ — 2 2x′ − y′  8  _ 5 √ — − 12  2x′ − y′   _ 5 √ x′ + 2y′ _ 5 √ — —  + 17  2 x′ + 2y′  _ = 20 5 √ — 8  (2x′ − y′)(2x′ − y′) __ 5  = 20 8 (4x′ 2 − 4x′ y′ + y′ 2) − 12(2x′ 2 + 3x′ y′ − 2y′ 2) + 17(x′ 2 + 4x′ y′ + 4y′ 2) = 100 (2x′ − y′)(x′ + 2y′) __ 5 (x′ + 2y′)(x′ + 2y′) __ 5  + 17   − 12  32x′ 2 − 32x′ y′ + 8y′ 2 − 24x′ 2 − 36x′ y′ + 24y′ 2 + 17x′ 2 + 68x′ y′ + 68y′ 2 = 100 25x′ 2 + 100y′ 2 = 100 100 ___ 100 Write the equations with x′ and y′ in the standard form with respect to the new coordinate system. 100 ___ 100 y′ 2 = x′ 2 + 25 ____ 100 Figure 8 shows the graph of the ellipse. x′ 2 _ 4 + y3 –2 –1 1 2 3 x –1 –2 Figure 8 Try It #2 Rewrite the 13x 2 − 6 √ — 3 xy + 7y 2 = 16 in the x′ y′ system without the x′ y′ term. Example 4 Graphing an Equation That Has No x´y´ Terms Graph the following equation relative to the x′ y′ system: x 2 + 12xy − 4y 2 = 30 Solution First, we find cot(2θ). x 2 + 12xy − 4y 2 = 20 ⇒ A = 1, B = 12, and C = −4 cot(2θ) = A − C ______ B 1 − (−4) ________ 12 cot(2θ) = cot(2θ) = 5 ___ 12 SECTION 10.4 rotation oF axis 917 Because cot(2θ) = , we can draw a reference triangle as in Figure 9. 5 __ 12 y 12 cot(2θ) = 5 12 x 2θ 5 Figure 9 cot(2θ) = = 5 ___ 12 adja
cent _ opposite Thus, the hypotenuse is 52 + 122 = h2 25 + 144 = h2 169 = h2 h = 13 Next, we find sin θ and cos θ. We will use half-angle identities. ———— ————— sin θ = √ ___________ 1 − cos(2θ) __________ 2 cos θ = √ ___________ 1 + cos(2θ) __________ 2 = √ = √ 5 __ 1 − 13 _ 2 ———— 5 __ 1 + 13 _ 2 = √ = √ 5 _ 13 13 _ − 13 _ 2 ————— = √ ______ 8 1 __ _____ = ⋅ 2 13 2 _ — √ 13 5 _ 13 13 _ + 13 _ 2 = √ ______ 18 1 __ ___ = ⋅ 2 13 3 _ — √ 13 Now we find x and y. x = x′ cos θ − y′ sin θ and  − y′   2 _ — 13 √ x = x′  3 _ — 13 √ 3x′ − 2y′ _ 13 √ — x = y = x′ sin θ + y′ cos θ  + y′   3 _ — 13 √ y = x′  2 _ — 13 √ 2x′ + 3y′ _ 13 √ — y = Now we substitute x = and y = into x 2 + 12xy − 4y 2 = 30. — 3x′ − 2y′ _ 13 √ 2 3x′ − 2y′ _ 13  √ —  — 2x′ + 3y′ _ 13 √ 3x′ − 2y′ _ 13   √ — + 12  2x′ + 3y′ _ 13  − 4  √ — 2 2x′ + 3y′ _ = 30 13  √ — 918 CHAPTER 10 analytic geometry 1  ___ 13  [(3x′ − 2y′ )2 + 12(3x′ − 2y′ )(2x′ + 3y′ ) − 4 (2x′ + 3y′ )2] = 30 1  [9x′ 2 − 12x′ y′ + 4y′ 2 + 12 (6x′ 2 + 5x′ y′ − 6y′ 2) − 4 (4x′ 2 + 12x′ y′ + 9y′ 2)] = 30  ___ 13 1  [9x′ 2 − 12x′ y′ + 4y′ 2 + 72x′ 2 + 60x′ y′ − 72y′ 2 − 16x′ 2 − 48x′ y′ − 36y′ 2] = 30  ___ 13  [65x′ 2 − 104y′ 2] = 30 1  ___ 13 Factor. Multiply. Distribute. Combine like terms. Figure 10 shows the graph of the hyperbola x′ 2 _ 6 − 4y′ 2 _ 15 = 1. 65x′ 2 − 104y′ 2 = 390 Multiply. x′ 2 _ − 6 4y′ 2 _ 15 = 1 Divide by 390. x’ 1 2 3 4 5 x y’ –5 –4 –3 –2 –1 y 5 4 3 2 1 0 –1 –2 –3 –4 –5 Identifying Conics without Rotating Axes Figure 10 Now we have come full circle. How do we identify the type of conic described by an equation? What happens when the axes are rotated? Recall, the general form of a conic is If we apply the rotation formulas to this equation we get the form A′ x′ 2 + B′x′y′ + C′y′ 2 + D′x′ + E′y′ + F′ = 0 Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 It may be shown that B2 − 4AC = B′ 2 − 4A′ C′. The expression does not vary after rotation, so we call the expression invariant. The discriminant, B2 − 4AC, is invariant and remains unchanged after rotation. Because the discriminant remains unchanged, observing the discriminant enables us to identify the conic section. using the discriminant to identify a conic If the equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 is transformed by rotating axes into the equation A′x′ 2 + B′x′y′ + C′y′ 2 + D′x′ + E′y′ + F′ = 0, then B2 − 4AC = B′ 2 − 4A′C′. The equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 is an ellipse, a parabola, or a hyperbola, or a degenerate case of one of these. If the discriminant, B2 − 4AC, is • < 0, the conic section is an ellipse • = 0, the conic section is a parabola • > 0, the conic section is a hyperbola SECTION 10.4 rotation oF axis 919 Example 5 Identifying the Conic without Rotating Axes Identify the conic for each of the following without rotating axes. a. 5x 2 + 2 √ — b. 5x 2 + 2 √ 3 xy + 2y 2 − 5 = 0 3 xy + 12y 2 − 5 = 0 — Solution a. Let’s begin by determining A, B, and C xy + 2 C { y 2 − 5 = 0 Now, we find the discriminant. — B2 − 4AC =  2 √ 2 3  − 4(5)(2) = 4(3) − 40 = 12 − 40 = − 28 < 0 3 xy + 2y 2 − 5 = 0 represents an ellipse. — Therefore, 5x 2 + 2 √ b. Again, let’s begin by determining A, B, and . — 3 x 2 + 2 √ B Now, we find the discriminant. 5 A { xy + 12 C { y 2 − 5 = 0 B2 − 4AC =  2 √ — 2 3  − 4(5)(12) = 4(3) − 240 = 12 − 240 = − 228 < 0 3 xy + 12y 2 − 5 = 0 represents an ellipse. — Therefore, 5x 2 + 2 √ Try It #3 Identify the conic for each of the following without rotating axes. a. x 2 − 9xy + 3y 2 − 12 = 0 b. 10x 2 − 9xy + 4y 2 − 4 = 0 Access this online resource for additional instruction and practice with conic sections and rotation of axes. • Introduction to Conic Sections (http://openstaxcollege.org/l/introconic) 920 CHAPTER 10 analytic geometry 10.4 SeCTIOn exeRCISeS VeRBAl 1. What effect does the xy term have on the graph of a conic section? 3. If the equation of a conic section is written in the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0, and B2 − 4AC > 0, what can we conclude? 2. If the equation of a conic section is written in the form Ax 2 + By 2 + Cx + Dy + E = 0 and AB = 0, what can we conclude? 4. Given the equation ax 2 + 4x + 3y 2 − 12 = 0, what can we conclude if a > 0? 5. For the equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0, the value of θ that satisfies cot(2θ) = gives us A − C ______ B what information? AlGeBRAIC For the following exercises, determine which conic section is represented based on the given equation. 6. 9x 2 + 4y 2 + 72x + 36y − 500 = 0 8. 2x 2 − 2y 2 + 4x − 6y − 2 = 0 10. 4y 2 − 5x + 9y + 1 = 0 7. x 2 − 10x + 4y − 10 = 0 9. 4x 2 − y 2 + 8x − 1 = 0 11. 2x 2 + 3y 2 − 8x − 12y + 2 = 0 12. 4x 2 + 9xy + 4y 2 − 36y − 125 = 0 13. 3x 2 + 6xy + 3y 2 − 36y − 125 = 0 14. −3x 2 + 3 √ — 3 xy − 4y 2 + 9 = 0 16. −x 2 + 4 √ — 2 xy + 2y 2 − 2y + 1 = 0 15. 2x 2 + 4 √ — 3 xy + 6y 2 − 6x − 3 = 0 17. 8x 2 + 4 √ — 2 xy + 4y 2 − 10x + 1 = 0 For the following exercises, find a new representation of the given equation after rotating through the given angle. 18. 3x 2 + xy + 3y 2 − 5 = 0, θ = 45° 20. 2x 2 + 8xy − 1 = 0, θ = 30° 22. 4x 2 + √ 2 xy + 4y 2 + y + 2 = 0, θ = 45° — 19. 4x 2 − xy + 4y 2 − 2 = 0, θ = 45° 21. −2x 2 + 8xy + 1 = 0, θ = 45° For the following exercises, determine the angle θ that will eliminate the xy term and write the corresponding equation without the xy term. 23. x 2 + 3 √ — 3 xy + 4y 2 + y − 2 = 0 — 25. 9x 2 − 3 √ 27. 16x 2 + 24xy + 9y 2 + 6x − 6y + 2 = 0 3 xy + 6y 2 + 4y − 3 = 0 29. x 2 + 4xy + y 2 − 2x + 1 = 0 GRAPHICAl 24. 4x 2 + 2 √ — 3 xy + 6y 2 + y − 2 = 0 — 3 xy − 2y 2 − x = 0 26. −3x 2 − √ 28. x 2 + 4xy + 4y 2 + 3x − 2 = 0 30. 4x 2 − 2 √ — 3 xy + 6y 2 − 1 = 0 For the following exercises, rotate through the given angle based on the given equation. Give the new equation and graph the original and rotated equation. 31. y = − x 2, θ = − 45° 32. x = y 2, θ = 45° 33, θ = 45° 1 y 2 _ 16 34. + x 2 _ = 1, θ = 45° 9 37. x = (y − 1)2, θ = 30° 35. y 2 − x 2 = 1, θ = 45° y 2 _ = 1, θ = 30° 4 x 2 _ 9 + 38. 36. y = x 2 _ , θ = 30° 2 SECTION 10.4 section exercises 921 For the following exercises, graph the equation relative to the x′ y′ system in which the equation has no x′ y′ term. 39. xy = 9 41. x 2 − 10xy + y 2 − 24 = 0 43. 6x 2 + 2 √ — 3 xy + 4y 2 − 21 = 0 45. 21x 2 + 2 √ 3 xy + 19y 2 − 18 = 0 — 47. 16x 2 + 24xy + 9y 2 − 60x + 80y = 0 49. 4x 2 − 4xy + y 2 − 8 √ — 5 x − 16 √ — 5 y = 0 40. x 2 + 10xy + y 2 − 6 = 0 42. 4x 2 − 3 √ — 3 xy + y 2 − 22 = 0 44. 11x 2 + 10 √ — 3 xy + y 2 − 64 = 0 46. 16x 2 + 24xy + 9y 2 − 130x + 90y = 0 48. 13x 2 − 6 √ — 3 xy + 7y 2 − 16 = 0 For the following exercises, determine the angle of rotation in order to eliminate the xy term. Then graph the new set of axes. 50. 6x 2 − 5 √ — 3 xy + y 2 + 10x − 12y = 0 52. 6x 2 − 8 √ — 3 xy + 14y 2 + 10x − 3y = 0 51. 6x 2 − 5xy + 6y 2 + 20x − y = 0 53. 4x 2 + 6 √ — 3 xy + 10y 2 + 20x − 40y = 0 54. 8x 2 + 3xy + 4y 2 + 2x − 4 = 0 55. 16x 2 + 24xy + 9y 2 + 20x − 44y = 0 For the following exercises, determine the value of k based on the given equation. 56. Given 4x 2 + kxy + 16y 2 + 8x + 24y − 48 = 0, 57. Given 2x 2 + kxy + 12y 2 + 10x − 16y + 28 = 0, find k for the graph to be a parabola. find k for the graph to be an ellipse. 58. Given 3x 2 + kxy + 4y 2 − 6x + 20y + 128 = 0, 59. Given kx 2 + 8xy + 8y 2 − 12x + 16y + 18 = 0, find k for the graph to be a hyperbola. find k for the graph to be a parabola. 60. Given 6x 2 + 12xy + ky 2 + 16x + 10y + 4 = 0, find k for the graph to be an ellipse. 922 CHAPTER 10 analytic geometry leARnInG OBjeCTIVeS In this section, you will: • Identify a conic in polar form. • Graph the polar equations of conics. • Define conics in terms of a focus and a directrix. 10. 5 COnIC SeCTIOnS In POlAR COORDInATeS Figure 1 Planets orbiting the sun follow elliptical paths. (credit: nASA Blueshift, Flickr) Most of us are familiar with orbital motion, such as the motion of a planet around the sun or an electron around an atomic nucleus. Within the planetary system, orbits of planets, asteroids, and comets around a larger celestial body are often elliptical. Comets, however, may take on a parabolic or hyperbolic orbit instead. And, in reality, the characteristics of the planets’ orbits may vary over time. Each orbit is tied to the location of the celestial body being orbited and the distance and direction of the planet or other object from that body. As a result, we tend to use polar coordinates to represent these orbits. In an elliptical orbit, the periapsis is the point at which the two objects are closest, and the apoapsis is the point at which they are farthest apart. Generally, the velocity of the orbiting body tends to increase as it approaches the periapsis and decrease as it approaches the apoapsis. Some objects reach an escape velocity, which results in an infinite orbit. These bodies exhibit either a parabolic or a hyperbolic orbit about a body; the orbiting body breaks free of the celestial body’s gravitational pull and fires off into space. Each of these orbits can be modeled by a conic section in the polar coordinate system. Identifying a Conic in Polar Form Any conic may be determined by three characteristics: a single focus, a fixed line called the directrix, and the ratio of the distances of each to a point on the graph. Consider the parabola x = 2 + y 2 shown in Figure 2. P(r, θ) D r θ Polar axis Directrix F, Focus @ pole x = 2 + y2 Figure 2 SECTION 10.5 conic sections in polar coordinates 923 In The Parabola, we learned how a parabola is defined by the focus (a fixed point) and the directrix (a fixed line). In this section, we will learn how to define any conic in the polar coordinate system in terms of a fixed point, the focus P(r, θ) at the pole, and a line, the directrix, which is perpendicular to the polar axis. If F is a fixed point, the focus, and D is a fixed line, the directrix, then we can let e be a fixed positive number, called the eccentricity, which we can define as the ratio of the distances from a point on the graph to the focus and the
point on the graph to the directrix. Then the set of all points P such that e = is a conic. In other words, we can define PF ___ PD a conic as the set of all points P with the property that the ratio of the distance from P to F to the distance from P to D is equal to the constant e. For a conic with eccentricity e, • if 0 ≤ e < 1, the conic is an ellipse • if e = 1, the conic is a parabola • if e > 1, the conic is an hyperbola With this definition, we may now define a conic in terms of the directrix, x = ± p, the eccentricity e, and the angle θ. Thus, each conic may be written as a polar equation, an equation written in terms of r and θ. the polar equation for a conic For a conic with a focus at the origin, if the directrix is x = ± p, where p is a positive real number, and the eccentricity is a positive real number e, the conic has a polar equation r = ep _ 1 ± e cos θ For a conic with a focus at the origin, if the directrix is y = ± p, where p is a positive real number, and the eccentricity is a positive real number e, the conic has a polar equation r = ep _ 1 ± e sin θ How To… Given the polar equation for a conic, identify the type of conic, the directrix, and the eccentricity. 1. Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form. 2. Identify the eccentricity e as the coefficient of the trigonometric function in the denominator. 3. Compare e with 1 to determine the shape of the conic. 4. Determine the directrix as x = p if cosine is in the denominator and y = p if sine is in the denominator. Set ep equal to the numerator in standard form to solve for x or y. Example 1 Identifying a Conic Given the Polar Form For each of the following equations, identify the conic with focus at the origin, the directrix, and the eccentricity. a. r = 6 _________ 3 + 2 sin θ b. r = 12 _________ 4 + 5 cos θ c. r = 7 _________ 2 − 2 sin θ Solution For each of the three conics, we will rewrite the equation in standard form. Standard form has a 1 as the constant in the denominator. Therefore, in all three parts, the first step will be to multiply the numerator and 1 __ , where c is that constant. denominator by the reciprocal of the constant of the original equation, c 1 __ . a. Multiply the numerator and denominator by 3 r = 1   __ 3 6 _ _________ = ⋅ 3 + 2sin θ 1   __ 3 1  6  __ 3 __ = 1 1  sin θ  + 2  3  __ __ 3 3 2 __ 2 _ 1 + sin θ 3 924 CHAPTER 10 analytic geometry 2 __ Because sin θ is in the denominator, the directrix is y = p. Comparing to standard form, note that e = . 3 Therefore, from the numerator, 2 = ep 2 __ =   __ __ __ p 3 2 2 3 = p 2 __ and the directrix is y = 3. Since e < 1, the conic is an ellipse. The eccentricity is e = 3 1 __ . b. Multiply the numerator and denominator by 4 r = r = 1   _ 12 4 _ _ ⋅ 4 + 5 cos θ 1   _ 4 1 _  12  4 __ 1 1 _ _  cos __ Because cos θ is in the denominator, the directrix is x = p. Comparing to standard form + cos θ 4 Therefore, from the numerator, 3 = ep 5 __ =   __ __ __ p 4 5 5 12 ___ = p 5 5 __ and the directrix is x = Since e > 1, the conic is a hyperbola. The eccentricity is e = 4 12 ___ 5 = 2.4. 1 __ . c. Multiply the numerator and denominator by sin __ r = 1 1  sin − sin θ r = Because sine is in the denominator, the directrix is y = −p. Comparing to standard form, e = 1. Therefore, from the numerator, 7 __ = ep 2 7 __ = (1)p 2 7 __ = p 2 7 __ = −3.5. Because e = 1, the conic is a parabola. The eccentricity is e = 1 and the directrix is y = − 2 SECTION 10.5 conic sections in polar coordinates 925 Try It #1 Identify the conic with focus at the origin, the directrix, and the eccentricity for r = 2 ________ . 3 − cos θ Graphing the Polar equations of Conics When graphing in Cartesian coordinates, each conic section has a unique equation. This is not the case when graphing in polar coordinates. We must use the eccentricity of a conic section to determine which type of curve to graph, and then determine its specific characteristics. The first step is to rewrite the conic in standard form as we have done in the previous example. In other words, we need to rewrite the equation so that the denominator begins with 1. This enables us to determine e and, therefore, the shape of the curve. The next step is to substitute values for θ and solve for r to plot π __ , π, and a few key points. Setting θ equal to 0, 2 3π ___ provides the vertices so we can create a rough sketch of the graph. 2 Example 2 Graphing a Parabola in Polar Form Graph r = 5 _________ 3 + 3 cos θ . Solution First, we rewrite the conic in standard form by multiplying the numerator and denominator by the 1 __ . reciprocal of 3, which is 3 1  5  _ 3 __ 1 1  cos cos θ 5 __ 3 _ 1 + cos θ Because e = 1, we will graph a parabola with a focus at the origin. The function has a cos θ, and there is an addition sign in the denominator, so the directrix is x = p. 5 __ = ep 3 5 __ = (1)p 3 5 __ = p 3 5 __ The directrix is x = . 3 Plotting a few key points as in Table 1 will enable us to see the vertices. See Figure 3. θ A 0 r = 5 _________ 3 + 3 cos θ 5 __ ≈ 0.83 6 B π __ 2 5 __ ≈ 1.67 3 Table 1 C π undefined D 3π ___ 2 5 __ ≈ 1.67 Directrix Figure 3 926 CHAPTER 10 analytic geometry Analysis We can check our result with a graphing utility. See Figure 4. Example 3 Graphing a Hyperbola in Polar Form 2 3 4 5 Figure 4 Graph r = 8 _________ 2 − 3 sin θ . Solution First, we rewrite the conic in standard form by multiplying the numerator and denominator by the 1 __ . reciprocal of 2, which is sin θ 1  8  _ 2 __ 1 1  sin − cos θ 2 3 __ Because e = , e > 1, so we will graph a hyperbola with a focus at the origin. The function has a sin θ term and there 2 is a subtraction sign in the denominator, so the directrix is y = −p. 4 = ep 3 4 =   p __ 2 2  = p 4  __ 3 8 __ = p 3 8 __ The directrix is y = − . 3 Plotting a few key points as in Table 2 will enable us to see the vertices. See Figure 5. θ r = 8 _________ 2 − 3 sin θ A 0 4 B π __ 2 −8 Table 2 C π 4 D 3π ___ 2 8 __ = 1. 10 C D B Figure 5 SECTION 10.5 conic sections in polar coordinates 927 Example 4 Graphing an Ellipse in Polar Form Graph r = 10 __________ 5 − 4 cos θ. Solution First, we rewrite the conic in standard form by multiplying the numerator and denominator by the 1 __ . reciprocal of 5, which is 5 r = 10 _________ 5 − 4 cos θ = 1 _  10  5 __ 1 1 _ _  cos __ 1 − sin θ 5 4 __ , e < 1, so we will graph an ellipse with a focus at the origin. The function has a cos θ, and there is a Because e = 5 subtraction sign in the denominator, so the directrix is x = −p. 2 = ep 4  p 2 =  __ 5 5  = p 2  __ 4 5 __ = p 2 5 __ The directrix is x = − . 2 Plotting a few key points as in Table 3 will enable us to see the vertices. See Figure 6. θ r = 10 _________ 5 − 4 cos θ A 0 10 B π __ 2 2 Table 3 C π 10 ___ 9 ≈ 1.1 D 3π ___ 2 2 B C D 2 4 68 1 A 0 12 r x = − 5 2 Directrix Figure 6 928 CHAPTER 10 analytic geometry Analysis We can check our result with a graphing utility. See Figure 7. Figure 7 r = of [−3, 12, 1] by [ −4, 4, 1], θ min = 0 and θ max = 2π. graphed on a viewing window 10 _ 5 − 4 cos θ Try It #2 Graph r = 2 _________ 4 − cos θ. Defining Conics in Terms of a Focus and a Directrix So far we have been using polar equations of conics to describe and graph the curve. Now we will work in reverse; we will use information about the origin, eccentricity, and directrix to determine the polar equation. How To… Given the focus, eccentricity, and directrix of a conic, determine the polar equation. 1. Determine whether the directrix is horizontal or vertical. If the directrix is given in terms of y, we use the general polar form in terms of sine. If the directrix is given in terms of x, we use the general polar form in terms of cosine. 2. Determine the sign in the denominator. If p < 0, use subtraction. If p > 0, use addition. 3. Write the coefficient of the trigonometric function as the given eccentricity. 4. Write the absolute value of p in the numerator, and simplify the equation. Example 5 Finding the Polar Form of a Vertical Conic Given a Focus at the Origin and the Eccentricity and Directrix Find the polar form of the conic given a focus at the origin, e = 3 and directrix y = − 2. Solution The directrix is y = −p, so we know the trigonometric function in the denominator is sine. Because y = −2, −2 < 0, so we know there is a subtraction sign in the denominator. We use the standard form of and e = 3 and \| −2 \| = 2 = p. Therefore, r = ep ________ 1 − esin θ r = (3)(2) _________ 1 − 3 sin θ r = 6 _________ 1 − 3 sin θ Example 6 Finding the Polar Form of a Horizontal Conic Given a Focus at the Origin and the Eccentricity and Directrix 3 _ , and directrix x = 4. Find the polar form of a conic given a focus at the origin, e = 5 Solution Because the directrix is x = p, we know the function in the denominator is cosine. Because x = 4, 4 > 0, so we know there is an addition sign in the denominator. We use the standard form of 3 __ and \|4\| = 4 = p. and e = 5 r = ep _________ 1 + e cos θ SECTION 10.5 conic sections in polar coordinates 929 Therefore, r = 3  (4)  _ 5 ___________ 3 _ 1 + cos θ 5 r = 12 _ 5 ___________ 3 _ 1 + cos θ 5 r = 12 _ 5 _______________ 3 5  + 1  _ _ cos θ 5 5 r = 12 _ 5 ____________ 3 5 _ _ + cos θ 5 5 r = 12 _ 5 ⋅ 5 _________ 5 + 3 cos θ r = 12 _________ 5 + 3 cos θ Try It #3 Find the polar form of the conic given a focus at the origin, e = 1, and directrix x = −1. Example 7 Converting a Conic in Polar Form to Rectangular Form Convert the conic r = to rectangular form. 1 _________ 5 − 5sin θ Solution We will rearrange the formula to use the identities r = √ — x 2 + y2 , x = r cos θ, and y = r sin θ. r = 1 _________ 5 − 5sin θ 1 _________ 5 − 5sin θ r ⋅ (5 − 5 sin θ) = ⋅ (5 − 5 sin θ) 5r − 5r sin θ = 1 5r = 1 + 5r sin θ 25r 2 = (1 + 5r sin θ)2 25(x 2 + y 2) = (1 + 5y)2 Eliminate the fraction. Distribute. Isolate 5r. Square both sides. Substitute r = √ — x 2 + y 2
and y = r sin θ. 25x 2 + 25y 2 = 1 + 10y + 25y 2 Distribute and use FOIL. 25x 2 − 10y = 1 Rearrange terms and set equal to 1. Try It #4 Convert the conic r = 2 __________ 1 + 2cos θ to rectangular form. Access these online resources for additional instruction and practice with conics in polar coordinates. • Polar equations of Conic Sections (http://openstaxcollege.org/l/determineconic) • Graphing Polar equations of Conics – 1 (http://openstaxcollege.org/l/graphconic1) • Graphing Polar equations of Conics – 2 (http://openstaxcollege.org/l/graphconic2) 930 CHAPTER 10 analytic geometry 10.5 SeCTIOn exeRCISeS VeRBAl 1. Explain how eccentricity determines which conic 2. If a conic section is written as a polar equation, what section is given. must be true of the denominator? 3. If a conic section is written as a polar equation, and the denominator involves sin θ, what conclusion can be drawn about the directrix? 4. If the directrix of a conic section is perpendicular to the polar axis, what do we know about the equation of the graph? 5. What do we know about the focus/foci of a conic section if it is written as a polar equation? AlGeBRAIC For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity. 6. r = 10. r = 6 __________ 1 − 2 cos θ 16 __________ 4 + 3 cos θ 7. r = 11. r = 3 __________ 4 − 4 sin θ 3 ____________ 10 + 10 cos θ 8. r = 12. r = 8 __________ 4 − 3 cos θ 2 ________ 1 − cos θ 9. r = 13. r = 5 __________ 1 + 2 sin θ 4 __________ 7 + 2 cos θ 14. r(1 − cos θ) = 3 15. r(3 + 5sin θ) = 11 16. r(4 − 5sin θ) = 1 17. r(7 + 8cos θ) = 7 For the following exercises, convert the polar equation of a conic section to a rectangular equation. 18. r = 4 __________ 1 + 3 sin θ 22. r = 4 __________ 2 + 2 sin θ 19. r = 2 __________ 5 − 3 sin θ 23. r = 3 __________ 8 − 8 cos θ 20. r = 8 __________ 3 − 2 cos θ 24. r = 2 __________ 6 + 7 cos θ 26. r(5 + 2 cos θ) = 6 27. r(2 − cos θ) = 1 28. r(2.5 − 2.5 sin θ) = 5 30. r = 6csc θ __________ 3 + 2 csc θ 21. r = 3 __________ 2 + 5 cos θ 25. r = 29. r = 5 ___________ 5 − 11 sin θ 6sec θ ___________ −2 + 3 sec θ For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci. 31. r = 35. r = 5 ________ 2 + cos θ 8 __________ 4 − 5 cos θ 32. r = 36. r = 2 __________ 3 + 3 sin θ 3 __________ 4 − 4 cos θ 33. r = 37. r = 10 __________ 5 − 4 sin θ 2 _ 1 − sin θ 34. r = 38. r = 3 __________ 1 + 2 cos θ 6 __________ 3 + 2 sin θ 39. r(1 + cos θ) = 5 40. r(3 − 4sin θ) = 9 41. r(3 − 2sin θ) = 6 42. r(6 − 4cos θ) = 5 For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix. 1 __ 43. Directrix: x = 4; e = 5 1 __ 46. Directrix: y = − 2; e = 2 7 1 __ __ 49. Directrix __ 52. Directrix: x = −2; e = 3 1 __ 55. Directrix: x = −3; e = 3 44. Directrix: x = − 4; e = 5 45. Directrix: y = 2; e = 2 47. Directrix: x = 1; e = 1 48. Directrix: x = −1; e = 1 7 2 __ __ ; e = 50. Directrix: y = 5 2 3 __ 53. Directrix: x = −5; e = 4 3 __ 51. Directrix: y = 4; e = 2 54. Directrix: y = 2; e = 2.5 exTenSIOnS Recall from Rotation of Axes that equations of conics with an xy term have rotated graphs. For the following exercises, express each equation in polar form with r as a function of θ. 57. x 2 + xy + y 2 = 4 56. xy = 2 58. 2x 2 + 4xy + 2y 2 = 9 59. 16x 2 + 24xy + 9y 2 = 4 60. 2xy + y = 1 CHAPTER 10 review 931 CHAPTeR 10 ReVIeW Key Terms angle of rotation an acute angle formed by a set of axes rotated from the Cartesian plane where, if cot(2θ) > 0, then θ is between (0°, 45°); if cot(2θ) < 0, then θ is between (45°, 90°); and if cot(2θ) = 0, then θ = 45° center of a hyperbola the midpoint of both the transverse and conjugate axes of a hyperbola center of an ellipse the midpoint of both the major and minor axes conic section any shape resulting from the intersection of a right circular cone with a plane conjugate axis the axis of a hyperbola that is perpendicular to the transverse axis and has the co-vertices as its endpoints degenerate conic sections any of the possible shapes formed when a plane intersects a double cone through the apex. Types of degenerate conic sections include a point, a line, and intersecting lines. directrix a line perpendicular to the axis of symmetry of a parabola; a line such that the ratio of the distance between the points on the conic and the focus to the distance to the directrix is constant eccentricity the ratio of the distances from a point P on the graph to the focus F and to the directrix D represented by e = PF ___ PD , where e is a positive real number ellipse the set of all points (x, y) in a plane such that the sum of their distances from two fixed points is a constant foci plural of focus focus (of a parabola) a fixed point in the interior of a parabola that lies on the axis of symmetry focus (of an ellipse) one of the two fixed points on the major axis of an ellipse such that the sum of the distances from these points to any point (x, y) on the ellipse is a constant hyperbola the set of all points (x, y) in a plane such that the difference of the distances between (x, y) and the foci is a positive constant latus rectum the line segment that passes through the focus of a parabola parallel to the directrix, with endpoints on the parabola major axis the longer of the two axes of an ellipse minor axis the shorter of the two axes of an ellipse nondegenerate conic section a shape formed by the intersection of a plane with a double right cone such that the plane does not pass through the apex; nondegenerate conics include circles, ellipses, hyperbolas, and parabolas parabola the set of all points (x, y) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix polar equation an equation of a curve in polar coordinates r and θ transverse axis the axis of a hyperbola that includes the foci and has the vertices as its endpoints Key equations Horizontal ellipse, center at origin Vertical ellipse, center at origin Horizontal ellipse, center (h, k) Vertical ellipse, center (h, k) Hyperbola, center at origin, transverse axis on x-axis Hyperbola, center at origin, transverse axis on y-axis = 1 __ __ b2 a2 y 2 x 2 __ __ b2 a2 (x − h)2 _______ a2 (x − h)2 _______ b2 = 1, a > b + + (y − k)2 _______ b2 (y − k)2 _______ a2 = 1, a > b = 1, a > b − = 1 x 2 __ a2 y 2 __ a2 y 2 __ b2 x 2 __ b2 − = 1 932 CHAPTER 10 analytic geometry Hyperbola, center at (h, k), transverse axis parallel to x-axis Hyperbola, center at (h, k), transverse axis parallel to y-axis (x − h)2 _______ a2 (y − k)2 _______ a2 − − (y − k)2 _______ b2 (x − h)2 _______ b2 = 1 = 1 Parabola, vertex at origin, axis of symmetry on x-axis Parabola, vertex at origin, axis of symmetry on y-axis y 2 = 4px x 2 = 4py Parabola, vertex at (h, k), axis of symmetry on x-axis (y − k)2 = 4p(x − h) Parabola, vertex at (h, k), axis of symmetry on y-axis (x − h)2 = 4p(y − k) General Form equation of a conic section Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 Rotation of a conic section Angle of rotation Key Concepts 10.1 The Ellipse x = x' cos θ − y' sin θ y = x' sin θ + y' cos θ θ, where cot(2θ) = A − C ______ B • An ellipse is the set of all points (x, y) in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci). • When given the coordinates of the foci and vertices of an ellipse, we can write the equation of the ellipse in standard form. See Example 1 and Example 2. • When given an equation for an ellipse centered at the origin in standard form, we can identify its vertices, co-vertices, foci, and the lengths and positions of the major and minor axes in order to graph the ellipse. See Example 3 and Example 4. • When given the equation for an ellipse centered at some point other than the origin, we can identify its key features and graph the ellipse. See Example 5 and Example 6. • Real-world situations can be modeled using the standard equations of ellipses and then evaluated to find key features, such as lengths of axes and distance between foci. See Example 7. 10.2 The Hyperbola • A hyperbola is the set of all points (x, y) in a plane such that the difference of the distances between (x, y) and the foci is a positive constant. • The standard form of a hyperbola can be used to locate its vertices and foci. See Example 1. • When given the coordinates of the foci and vertices of a hyperbola, we can write the equation of the hyperbola in standard form. See Example 2 and Example 3. • When given an equation for a hyperbola, we can identify its vertices, co-vertices, foci, asymptotes, and lengths and positions of the transverse and conjugate axes in order to graph the hyperbola. See Example 4 and Example 5. • Real-world situations can be modeled using the standard equations of hyperbolas. For instance, given the dimensions of a natural draft cooling tower, we can find a hyperbolic equation that models its sides. See Example 6. CHAPTER 10 review 933 10.3 The Parabola • A parabola is the set of all points (x, y) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix. • The standard form of a parabola with vertex (0, 0) and the x-axis as its axis of symmetry can be used to graph the parabola. If p > 0, the parabola opens right. If p < 0, the parabola opens left. See Example 1. • The standard form of a parabola with vertex (0, 0) and the y-axis as its axis of symmetry can be used to graph the parabola. If p > 0, the parabola opens up. If p < 0, the parabola opens down. See Example 2. • When given the focus and directrix of a parabola, we can write its equation in standard form. See Example 3. • The standard form of a parabola with vertex (h, k) and axi
s of symmetry parallel to the x-axis can be used to graph the parabola. If p > 0, the parabola opens right. If p < 0, the parabola opens left. See Example 4. • The standard form of a parabola with vertex (h, k) and axis of symmetry parallel to the y-axis can be used to graph the parabola. If p > 0, the parabola opens up. If p < 0, the parabola opens down. See Example 5. • Real-world situations can be modeled using the standard equations of parabolas. For instance, given the diameter and focus of a cross-section of a parabolic reflector, we can find an equation that models its sides. See Example 6. 10.4 Rotation of Axes • Four basic shapes can result from the intersection of a plane with a pair of right circular cones connected tail to tail. They include an ellipse, a circle, a hyperbola, and a parabola. • A nondegenerate conic section has the general form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 where A, B and C are not all zero. The values of A, B, and C determine the type of conic. See Example 1. • Equations of conic sections with an xy term have been rotated about the origin. See Example 2. • The general form can be transformed into an equation in the x' and y' coordinate system without the x' y' term. See Example 3 and Example 4. • An expression is described as invariant if it remains unchanged after rotating. Because the discriminant is invariant, observing it enables us to identify the conic section. See Example 5. 10.5 Conic Sections in Polar Coordinates • Any conic may be determined by a single focus, the corresponding eccentricity, and the directrix. We can also define a conic in terms of a fixed point, the focus P(r, θ) at the pole, and a line, the directrix, which is perpendicular to the polar axis. • A conic is the set of all points e = , where eccentricity e is a positive real number. Each conic may be written PF ___ PD in terms of its polar equation. See Example 1. • The polar equations of conics can be graphed. See Example 2, Example 3, and Example 4. • Conics can be defined in terms of a focus, a directrix, and eccentricity. See Example 5 and Example 6. • We can use the identities cos θ, and y = r sin θ to convert the equation for a conic from polar — to rectangular form. See Example 7. 934 CHAPTER 10 analytic geometry CHAPTeR 10 ReVIeW exeRCISeS THe ellIPSe For the following exercises, write the equation of the ellipse in standard form. Then identify the center, vertices, and foci. 1. x 2 ___ 25 + y 2 ___ 64 = 1 2. (x − 2)2 _______ 100 + (y + 3)2 _______ 36 = 1 3. 9x 2 + y 2 + 54x − 4y + 76 = 0 4. 9x 2 + 36y 2 − 36x + 72y + 36 = 0 For the following exercises, graph the ellipse, noting center, vertices, and foci. 5. x 2 ___ 36 y 2 __ = 1 + 9 6. (x − 4)2 _______ 25 + (y + 3)2 _______ 49 = 1 7. 4x 2 + y 2 + 16x + 4y − 44 = 0 8. 2x 2 + 3y 2 − 20x + 12y + 38 = 0 For the following exercises, use the given information to find the equation for the ellipse. 9. Center at (0, 0), focus at (3, 0), vertex at (−5, 0) 10. Center at (2, −2), vertex at (7, −2), focus at (4, −2) 11. A whispering gallery is to be constructed such that the foci are located 35 feet from the center. If the length of the gallery is to be 100 feet, what should the height of the ceiling be? THe HYPeRBOlA For the following exercises, write the equation of the hyperbola in standard form. Then give the center, vertices, and foci. 12. x 2 ___ 81 y 2 __ = 1 − 9 13. (y + 1)2 _______ 16 − (x − 4)2 _______ 36 = 1 14. 9y 2 − 4x 2 + 54y − 16x + 29 = 0 15. 3x 2 − y 2 − 12x − 6y − 9 = 0 For the following exercises, graph the hyperbola, labeling vertices and foci. x 2 __ − 16. 9 y 2 ___ 16 = 1 17. (y − 1)2 _______ 49 − (x + 1)2 _______ 4 = 1 18. x 2 − 4y 2 + 6x + 32y − 91 = 0 19. 2y 2 − x 2 − 12y − 6 = 0 For the following exercises, find the equation of the hyperbola. 20. Center at (0, 0), vertex at (0, 4), focus at (0, −6) 21. Foci at (3, 7) and (7, 7), vertex at (6, 7) THe PARABOlA For the following exercises, write the equation of the parabola in standard form. Then give the vertex, focus, and directrix. 22. y 2 = 12x 1 __ (y − 1) 23. (x + 2)2 = 2 24. y 2 − 6y − 6x − 3 = 0 25. x 2 + 10x − y + 23 = 0 For the following exercises, graph the parabola, labeling vertex, focus, and directrix. 26. x 2 + 4y = 0 1 __ (x + 3) 27. (y − 1)2 = 2 28. x 2 − 8x − 10y + 46 = 0 29. 2y 2 + 12y + 6x + 15 = 0 CHAPTER 10 review 935 For the following exercises, write the equation of the parabola using the given information. 30. Focus at (−4, 0); directrix is x = 4 9 7  ; directrix is y = 31. Focus at  2, _ _ 8 8 32. A cable TV receiving dish is the shape of a paraboloid of revolution. Find the location of the receiver, which is placed at the focus, if the dish is 5 feet across at its opening and 1.5 feet deep. ROTATIOn OF AxeS For the following exercises, determine which of the conic sections is represented. 33. 16x 2 + 24xy + 9y 2 + 24x − 60y − 60 = 0 34. 4x 2 + 14xy + 5y 2+ 18x − 6y + 30 = 0 35. 4x 2 + xy + 2y 2 + 8x − 26y + 9 = 0 For the following exercises, determine the angle θ that will eliminate the xy term, and write the corresponding equation without the xy term. 36. x 2 + 4xy − 2y 2 − 6 = 0 37. x 2 − xy + y 2 − 6 = 0 For the following exercises, graph the equation relative to the x'y' system in which the equation has no x'y' term. 38. 9x 2 − 24xy + 16y 2 − 80x − 60y + 100 = 0 39. x 2 − xy + y 2 − 2 = 0 40. 6x 2 + 24xy − y 2 − 12x + 26y + 11 = 0 COnIC SeCTIOnS In POlAR COORDInATeS For the following exercises, given the polar equation of the conic with focus at the origin, identify the eccentricity and directrix. 41. r = 10 _________ 1 − 5 cos θ 42. r = 6 _________ 3 + 2 cos θ 43. r = 1 _________ 4 + 3 sin θ 44. r = 3 _________ 5 − 5 sin θ For the following exercises, graph the conic given in polar form. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse or a hyperbola, label the vertices and foci. 45. r = 3 ________ 1 − sin θ 46. r = 8 _________ 4 + 3 sin θ 47. r = 10 _________ 4 + 5 cos θ 48. r = 9 _________ 3 − 6 cos θ For the following exercises, given information about the graph of a conic with focus at the origin, find the equation in polar form. 49. Directrix is x = 3 and eccentricity e = 1 50. Directrix is y = −2 and eccentricity e = 4 936 CHAPTER 10 analytic geometry CHAPTeR 10 PRACTICe TeST For the following exercises, write the equation in standard form and state the center, vertices, and foci. y 2 x 2 __ __ + 1. 4 9 = 1 2. 9y 2 + 16x 2 − 36y + 32x − 92 = 0 For the following exercises, sketch the graph, identifying the center, vertices, and foci. 3. (x − 3)2 _______ 64 + (y − 2)2 _______ 36 = 1 4. 2x 2 + y 2 + 8x − 6y − 7 = 0 5. Write the standard form equation of an ellipse with a center at (1, 2), vertex at (7, 2), and focus at (4, 2). 6. A whispering gallery is to be constructed with a length of 150 feet. If the foci are to be located 20 feet away from the wall, how high should the ceiling be? For the following exercises, write the equation of the hyperbola in standard form, and give the center, vertices, foci, and asymptotes. 7. − = 1 x 2 ___ 49 y 2 ___ 81 8. 16y 2 − 9x 2 + 128y + 112 = 0 For the following exercises, graph the hyperbola, noting its center, vertices, and foci. State the equations of the asymptotes. (x − 3)2 _______ 25 10. y 2 − x 2 + 4y − 4x − 18 = 0 (y + 3)2 _______ 1 = 1 − 9. 11. Write the standard form equation of a hyperbola with foci at (1, 0) and (1, 6), and a vertex at (1, 2). For the following exercises, write the equation of the parabola in standard form, and give the vertex, focus, and equation of the directrix. 12. y 2 + 10x = 0 13. 3x 2 − 12x − y + 11 = 0 For the following exercises, graph the parabola, labeling the vertex, focus, and directrix. 14. (x − 1)2 = −4(y + 3) 16. Write the equation of a parabola with a focus at 15. y 2 + 8x − 8y + 40 = 0 17. A searchlight is shaped like a paraboloid of (2, 3) and directrix y = −1. revolution. If the light source is located 1.5 feet from the base along the axis of symmetry, and the depth of the searchlight is 3 feet, what should the width of the opening be? For the following exercises, determine which conic section is represented by the given equation, and then determine the angle θ that will eliminate the xy term. 18. 3x 2 − 2xy + 3y 2 = 4 19. x 2 + 4xy + 4y 2 + 6x − 8y = 0 For the following exercises, rewrite in the x'y' system without the x'y' term, and graph the rotated graph. 20. 11x 2 + 10 √ — 3 xy + y 2 = 4 21. 16x 2 + 24xy + 9y 2 − 125x = 0 For the following exercises, identify the conic with focus at the origin, and then give the directrix and eccentricity. 22. r = 3 ________ 2 − sin θ 23. r = 5 _________ 4 + 6 cos θ For the following exercises, graph the given conic section. If it is a parabola, label vertex, focus, and directrix. If it is an ellipse or a hyperbola, label vertices and foci. 24. r = 12 _________ 4 − 8 sin θ 25. r = 2 _________ 4 + 4 sin θ 26. Find a polar equation of the conic with focus at the origin, eccentricity of e = 2, and directrix: x = 3. 11 Sequences, Probability and Counting Theory Figure 1 (credit: Robert S. Donovan, Flickr.) CHAPTeR OUTlIne 11.1 Sequences and Their notations 11.2 Arithmetic Sequences 11.3 Geometric Sequences 11.4 Series and Their notations 11.5 Counting Principles 11.6 Binomial Theorem 11.7 Probability Introduction A lottery winner has some big decisions to make regarding what to do with the winnings. Buy a villa in Saint Barthélemy? A luxury convertible? A cruise around the world? The likelihood of winning the lottery is slim, but we all love to fantasize about what we could buy with the winnings. One of the first things a lottery winner has to decide is whether to take the winnings in the form of a lump sum or as a series of regular payments, called an annuity, over the next 30 years or so. This decision is often based on many factors, such as tax implications, interest rates, and investment strategies. There are also personal reasons to consider when making the choice, and one can make many
arguments for either decision. However, most lottery winners opt for the lump sum. In this chapter, we will explore the mathematics behind situations such as these. We will take an in-depth look at annuities. We will also look at the branch of mathematics that would allow us to calculate the number of ways to choose lottery numbers and the probability of winning. 937 938 CHAPTER 11 seQuences, proBaBility and counting theory leARnInG OBjeCTIVeS In this section, you will: • Write the terms of a sequence defined by an explicit formula. • Write the terms of a sequence defined by a recursive formula. • Use factorial notation. 11.1 SeQUenCeS AnD THeIR nOTATIOnS A video game company launches an exciting new advertising campaign. They predict the number of online visits to their website, or hits, will double each day. The model they are using shows 2 hits the first day, 4 hits the second day, 8 hits the third day, and so on. See Table 1. Day Hits 1 2 2 4 Table 1 3 8 4 16 5 32 … … If their model continues, how many hits will there be at the end of the month? To answer this question, we’ll first need to know how to determine a list of numbers written in a specific order. In this section, we will explore these kinds of ordered lists. Writing the Terms of a Sequence Defined by an explicit Formula One way to describe an ordered list of numbers is as a sequence. A sequence is a function whose domain is a subset of the counting numbers. The sequence established by the number of hits on the website is {2, 4, 8, 16, 32, …}. The ellipsis (…) indicates that the sequence continues indefinitely. Each number in the sequence is called a term. The first five terms of this sequence are 2, 4, 8, 16, and 32. Listing all of the terms for a sequence can be cumbersome. For example, finding the number of hits on the website at the end of the month would require listing out as many as 31 terms. A more efficient way to determine a specific term is by writing a formula to define the sequence. One type of formula is an explicit formula, which defines the terms of a sequence using their position in the sequence. Explicit formulas are helpful if we want to find a specific term of a sequence without finding all of the previous terms. We can use the formula to find the nth term of the sequence, where n is any positive number. In our example, each number in the sequence is double the previous number, so we can use powers of 2 to write a formula for the nth term. {2, 4, 8, 16, 22, …, ?, …} ↓ ↓ ↓ ↓ ↓ {21, 22, 23, 24, 25, …, 2n, …} ↓ The first term of the sequence is 21 = 2, the second term is 22 = 4, the third term is 23 = 8, and so on. The nth term of the sequence can be found by raising 2 to the nth power. An explicit formula for a sequence is named by a lower case letter a, b, c... with the subscript n. The explicit formula for this sequence is an = 2n . Now that we have a formula for the nth term of the sequence, we can answer the question posed at the beginning of this section. We were asked to find the number of hits at the end of the month, which we will take to be 31 days. To find the number of hits on the last day of the month, we need to find the 31st term of the sequence. We will substitute 31 for n in the formula. a31 = 231 = 2,147,483,648 SECTION 11.1 seQuences and their notations 939 If the doubling trend continues, the company will get 2,147,483,648 hits on the last day of the month. That is over 2.1 billion hits! The huge number is probably a little unrealistic because it does not take consumer interest and competition into account. It does, however, give the company a starting point from which to consider business decisions. Another way to represent the sequence is by using a table. The first five terms of the sequence and the nth term of the sequence are shown in Table 2. n nth term of the sequence, an 1 2 2 4 Table 2 3 8 4 16 5 32 n 2n Graphing provides a visual representation of the sequence as a set of distinct points. We can see from the graph in Figure 1 that the number of hits is rising at an exponential rate. This particular sequence forms an exponential function. an 36 32 28 24 20 16 12 8 4 0 (5, 32) (4, 16) (3, 8) (2, 4) (1, 2) 1 2 3 4 5 6 n Figure 1 Lastly, we can write this particular sequence as {2, 4, 8, 16, 32, …, 2n, …} A sequence that continues indefinitely is called an infinite sequence. The domain of an infinite sequence is the set of counting numbers. If we consider only the first 10 terms of the sequence, we could write This sequence is called a finite sequence because it does not continue indefinitely. {2, 4, 8, 16, 32, …, 2n, … , 1024}. sequence A sequence is a function whose domain is the set of positive integers. A finite sequence is a sequence whose domain consists of only the first n positive integers. The numbers in a sequence are called terms. The variable a with a number subscript is used to represent the terms in a sequence and to indicate the position of the term in the sequence. a1, a2, a3, … , an, … We call a1 the first term of the sequence, a2 the second term of the sequence, a3 the third term of the sequence, and so on. The term an is called the nth term of the sequence, or the general term of the sequence. An explicit formula defines the nth term of a sequence using the position of the term. A sequence that continues indefinitely is an infinite sequence. Q & A… Does a sequence always have to begin with a1? No. In certain problems, it may be useful to define the initial term as a0 instead of a1. In these problems, the domain of the function includes 0. 940 CHAPTER 11 seQuences, proBaBility and counting theory How To… Given an explicit formula, write the first n terms of a sequence. 1. Substitute each value of n into the formula. Begin with n = 1 to find the first term, a1. 2. To find the second term, a2, use n = 2. 3. Continue in the same manner until you have identified all n terms. Example 1 Writing the Terms of a Sequence Defined by an Explicit Formula Write the first five terms of the sequence defined by the explicit formula an = −3n + 8. Solution Substitute n = 1 into the formula. Repeat with values 2 through 5 for n. n = 1 a1 = −3(1) + 8 = 5 n = 2 a2 = −3(2) + 8 = 2 n = 3 a3 = −3(3) + 8 = −1 n = 4 a4 = −3(4) + 8 = −4 n = 5 a5 = −3(5) + 8 = −7 The first five terms are {5, 2, −1, −4, −7}. Analysis The sequence values can be listed in a table. A table, such as Table 3, is a convenient way to input the function into a graphing utility. n an 2 2 3 −1 4 −4 5 −7 1 5 Table 3 A graph can be made from this table of values. From the graph in Figure 2, we can see that this sequence represents a linear function, but notice the graph is not continuous because the domain is over the positive integers only. an 6 5 4 3 2 1 (1, 5) (2, 2) –1 0 –1 1 2 3 5 4 (3, –1) n 6 –2 –3 –4 –5 –6 –7 –8 (4, –4) (5, –7) Figure 2 Try It #1 Write the first five terms of the sequence defined by the explicit formula tn = 5n − 4. SECTION 11.1 seQuences and their notations 941 Investigating Alternating Sequences Sometimes sequences have terms that are alternate. In fact, the terms may actually alternate in sign. The steps to finding terms of the sequence are the same as if the signs did not alternate. However, the resulting terms will not show increase or decrease as n increases. Let’s take a look at the following sequence. {2, −4, 6, −8} Notice the first term is greater than the second term, the second term is less than the third term, and the third term is greater than the fourth term. This trend continues forever. Do not rearrange the terms in numerical order to interpret the sequence. How To… Given an explicit formula with alternating terms, write the first n terms of a sequence. 1. Substitute each value of n into the formula. Begin with n = 1 to find the first term, a1. The sign of the term is given by the (−1)n in the explicit formula. 2. To find the second term, a2, use n = 2. 3. Continue in the same manner until you have identified all n terms. Example 2 Writing the Terms of an Alternating Sequence Defined by an Explicit Formula Write the first five terms of the sequence. an = (−1)n n 2 _______ n + 1 Solution Substitute n = 1, n = 2, and so on in the formula. n = 1 a1 = n = 2 a2 = n = 3 a3 = n = 4 a4 = 1 __ = − 2 (−1)1 12 _______ 1 + 1 (−1)2 22 _______ 2 + 1 (−1)3 32 _______ 3 + 1 (−1)4 42 _______ 4 + 1 (−1)5 52 _______ 5 + 1 4 __ = 3 9 __ = − 4 = 16 __ 5 = − 25 ___ 6 n = 5 a5 = 25  ___ 6 9 4 1 The first five terms are  − __ __ __ , − , , 4 3 2 16 ___ 5 , − Analysis The graph of this function, shown in Figure 3, looks different from the ones we have seen previously in this section because the terms of the sequence alternate between positive and negative values. an 4 3 2 1 1 4, 3 5 1 2, 1 3 –1 –2 –3 –4 –5 1 1, − 2 3, −2 1 4 5, −4 1 6 Figure 3 942 CHAPTER 11 seQuences, proBaBility and counting theory Q & A… In Example 2, does the (−1) to the power of n account for the oscillations of signs? Yes, the power might be n, n + 1, n − 1, and so on, but any odd powers will result in a negative term, and any even power will result in a positive term. Try It #2 Write the first five terms of the sequence. an = 4n _____ (−2)n Investigating Piecewise Explicit Formulas We’ve learned that sequences are functions whose domain is over the positive integers. This is true for other types of functions, including some piecewise functions. Recall that a piecewise function is a function defined by multiple subsections. A different formula might represent each individual subsection. How To… Given an explicit formula for a piecewise function, write the first n terms of a sequence 1. Identify the formula to which n = 1 applies. 2. To find the first term, a1, use n = 1 in the appropriate formula. 3. Identify the formula to which n = 2 applies. 4. To find the second term, a2, use n = 2 in the appropriate formula. 5. Continue in the same manner until you have identified all n terms. Example 3 Writing the Terms of a Sequence Defined by a Piecewise Expli
cit Formula Write the first six terms of the sequence. Solution Substitute n = 1, n = 2, and so on in the appropriate formula. Use n2 when n is not a multiple of 3. Use n _ 3 when n is a multiple of 3. an = { n2 if n is not divisible by 3 n __ if n is divisible by 3 3 a1 = 12 = 1 a2 = 22 = 4 3 __ a3 = = 1 3 a4 = 42 = 16 a5 = 52 = 25 6 __ = 2 a6 = 3 1 is not a multiple of 3. Use n2. 2 is not a multiple of 3. Use n2. n __ . 3 is a multiple of 3. Use 3 4 is not a multiple of 3. Use n2. 5 is not a multiple of 3. Use n2. n __ . 6 is a multiple of 3. Use 3 The first six terms are {1, 4, 1, 16, 25, 2}. Analysis Every third point on the graph shown in Figure 4 stands out from the two nearby points. This occurs because the sequence was defined by a piecewise function. an 28 24 20 16 12 8 4 0 1 2 34 5 6 7 n Figure 4 SECTION 11.1 seQuences and their notations 943 Try It #3 Write the first six terms of the sequence. an = { 2n3 if n is odd 5n ___ if n is even 2 Finding an Explicit Formula Thus far, we have been given the explicit formula and asked to find a number of terms of the sequence. Sometimes, the explicit formula for the nth term of a sequence is not given. Instead, we are given several terms from the sequence. When this happens, we can work in reverse to find an explicit formula from the first few terms of a sequence. The key to finding an explicit formula is to look for a pattern in the terms. Keep in mind that the pattern may involve alternating terms, formulas for numerators, formulas for denominators, exponents, or bases. How To… Given the first few terms of a sequence, find an explicit formula for the sequence. 1. Look for a pattern among the terms. 2. If the terms are fractions, look for a separate pattern among the numerators and denominators. 3. Look for a pattern among the signs of the terms. 4. Write a formula for an in terms of n. Test your formula for n = 1, n = 2, and n = 3. Example 4 Writing an Explicit Formula for the nth Term of a Sequence Write an explicit formula for the nth term of each sequence. 2 a.  − ___ , 11 3 ___ 13 , − 4 ___ , 15 5 ___ 17 , − 6 ___ 19 , …  b.  − 2 ___ 25 , − 2 ___ 125 , − 2 ___ 625 , − 2 _____ 3,125 , − 2 ______ 15,625 , …  c. {e 4, e 5, e 6, e 7, e 8, … } Solution Look for the pattern in each sequence. a. The terms alternate between positive and negative. We can use (−1)n to make the terms alternate. The numerator can be represented by n + 1. The denominator can be represented by 2n + 9. b. The terms are all negative. an = (−1)n (n + 1) ___________ 2n + 9  − 2 ___ 25 , − 2 ___ 125 , − 2 ___ 625 , − 2 _____ 3,125 , − 2 ______ 15,625 , …  The numerator is 5n  __ __ __ __ __ __ __ 57 , … − 56 , − 55 , − 54 , − 53 , − 52 , − The denominators are increasing powers of 5 So we know that the fraction is negative, the numerator is 2, and the denominator can be represented by 5 n + 1. c. The terms are powers of e. For n = 1, the first term is e 4 so the exponent must be n + 3. an = e n + 3 an = − 2 ______ 5 n + 1 Try It #4 Write an explicit formula for the nth term of the sequence. {9, −81, 729, −6,561, 59,049, …} 944 CHAPTER 11 seQuences, proBaBility and counting theory Try It #5 Write an explicit formula for the nth term of the sequence. 9 3  − __ __ , − , − 8 4 27 ___ 12 , − , − 81 ___ 16 243 ___ 20 , ...  Try It #6 Write an explicit formula for the nth term of the sequence. 1  __ e2 , 1 _ e , 1, e, e 2, ...  Writing the Terms of a Sequence Defined by a Recursive Formula Sequences occur naturally in the growth patterns of nautilus shells, pinecones, tree branches, and many other natural structures. We may see the sequence in the leaf or branch arrangement, the number of petals of a flower, or the pattern of the chambers in a nautilus shell. Their growth follows the Fibonacci sequence, a famous sequence in which each term can be found by adding the preceding two terms. The numbers in the sequence are 1, 1, 2, 3, 5, 8, 13, 21, 34,…. Other examples from the natural world that exhibit the Fibonacci sequence are the Calla Lily, which has just one petal, the Black-Eyed Susan with 13 petals, and different varieties of daisies that may have 21 or 34 petals. Each term of the Fibonacci sequence depends on the terms that come before it. The Fibonacci sequence cannot easily be written using an explicit formula. Instead, we describe the sequence using a recursive formula, a formula that defines the terms of a sequence using previous terms. A recursive formula always has two parts: the value of an initial term (or terms), and an equation defining an in terms of preceding terms. For example, suppose we know the following: We can find the subsequent terms of the sequence using the first term. a1 = 3 an = 2an − 1 − 1, for n ≥ 2 a1 = 3 a2 = 2a1 − 1 = 2(3) − 1 = 5 a3 = 2a2 − 1 = 2(5) − 1 = 9 a4 = 2a3 − 1 = 2(9) − 1 = 17 So the first four terms of the sequence are {3, 5, 9, 17} . The recursive formula for the Fibonacci sequence states the first two terms and defines each successive term as the sum of the preceding two terms. a1 = 1 a2 = 1 an = an − 1 + an − 2 for n ≥ 3 To find the tenth term of the sequence, for example, we would need to add the eighth and ninth terms. We were told previously that the eighth and ninth terms are 21 and 34, so a10 = a9 + a8 = 34 + 21 = 55 recursive formula A recursive formula is a formula that defines each term of a sequence using preceding term(s ). Recursive formulas must always state the initial term, or terms, of the sequence. SECTION 11.1 seQuences and their notations 945 Q & A… Must the first two terms always be given in a recursive formula? No. The Fibonacci sequence defines each term using the two preceding terms, but many recursive formulas define each term using only one preceding term. These sequences need only the first term to be defined. How To… Given a recursive formula with only the first term provided, write the first n terms of a sequence. 1. Identify the initial term, a1, which is given as part of the formula. This is the first term. 2. To find the second term, a2, substitute the initial term into the formula for an − 1. Solve. 3. To find the third term, a3, substitute the second term into the formula. Solve. 4. Repeat until you have solved for the nth term. Example 5 Writing the Terms of a Sequence Defined by a Recursive Formula Write the first five terms of the sequence defined by the recursive formula. a1 = 9 an = 3an − 1 − 20, for n ≥ 2 Solution The first term is given in the formula. For each subsequent term, we replace an − 1 with the value of the preceding term a1 = 9 a2 = 3a1 − 20 = 3(9) − 20 = 27 − 20 = 7 a3 = 3a2 − 20 = 3(7) − 20 = 21 − 20 = 1 a4 = 3a3 − 20 = 3(1) − 20 = 3 − 20 = −17 a5 = 3a4 − 20 = 3( −17) − 20 = −51 − 20 = −71 The first five terms are {9, 7, 1, – 17, – 71}. See Figure 5. an 20 (1, 9) (2, 7) (3, 1) −1 0 −20 −40 −60 −80 1 2 3 45 6 (4, −17) n (5, −71) Figure 5 Try It #7 Write the first five terms of the sequence defined by the recursive formula. a1 = 2 an = 2an − 1 + 1, for n ≥ 2 How To… Given a recursive formula with two initial terms, write the first n terms of a sequence. 1. Identify the initial term, a1, which is given as part of the formula. 2. Identify the second term, a2, which is given as part of the formula. 3. To find the third term, substitute the initial term and the second term into the formula. Evaluate. 4. Repeat until you have evaluated the nth term. 946 CHAPTER 11 seQuences, proBaBility and counting theory Example 6 Writing the Terms of a Sequence Defined by a Recursive Formula Write the first six terms of the sequence defined by the recursive formula. a1 = 1 a2 = 2 an = 3an − 1 + 4an − 2, for n ≥ 3 Solution The first two terms are given. For each subsequent term, we replace an − 1 and an − 2 with the values of the two preceding terms a3 = 3a2 + 4a1 = 3(2) + 4(1) = 10 a4 = 3a3 + 4a2 = 3(10) + 4(2) = 38 a5 = 3a4 + 4a3 = 3(38) + 4(10) = 154 a6 = 3a5 + 4a4 = 3(154) + 4(38) = 614 The first six terms are {1, 2, 10, 38, 154, 614}. See Figure 6. an 700 600 500 400 300 200 100 (2, 2) (1, 1) (3, 10) 0 1 2 4 3 Figure 6 (6, 614) (5, 154) (4, 38) 5 n 6 7 Try It #8 Write the first 8 terms of the sequence defined by the recursive formula. a1 = 0 a2 = 1 a3 = 1 an = an − 1 _____ an − 2 + an − 3, for n ≥ 4 Using Factorial notation The formulas for some sequences include products of consecutive positive integers. n factorial, written as n!, is the product of the positive integers from 1 to n. For example, 4! = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24 5 = 120 An example of formula containing a factorial is an = (n + 1)!. The sixth term of the sequence can be found by substituting 6 for n. The factorial of any whole number n is n(n − 1)! We can therefore also think of 5! as 5 ⋅ 4!. a6 = (6 + 1)! = 7,040 SECTION 11.1 seQuences and their notations 947 n factorial n factorial is a mathematical operation that can be defined using a recursive formula. The factorial of n, denoted n!, is defined for a positive integer n as: 0! = 1 1! = 1 n ! = n(n − 1)(n − 2) ⋯ (2)(1), for n ≥ 2 The special case 0! is defined as 0! = 1. Q & A… Can factorials always be found using a calculator? No. Factorials get large very quickly—faster than even exponential functions! When the output gets too large for the calculator, it will not be able to calculate the factorial. Example 7 Writing the Terms of a Sequence Using Factorials Write the first five terms of the sequence defined by the explicit formula an = Solution Substitute n = 1, n = 2, and so on in the formula. 5n _______ (n + 2)! . 5 __ = 6 5 ______ 3 · 2 · 1 10 _________ = 4 · 3 · 2 · 1 5 ___ 12 = = a1 = = = 5(1) _______ (1 + 2)! 5(2) _______ (2 + 2)! 5(3) _______ (3 + 2)! 5(4) _______ (4 + 2)! 5(5) _______ (5 + 2)! 5 __ 3! 10 ___ 4! 15 ___ 5! 20 ___ 6! 25 ___ 7! a2 = a3 = a4 = a5 = = = 1 15 __ ___________ = = = 20 _____________ ___ 36 = = 25 _______________ = _____ 1,008 5 1 __ ___ , , 8 12 5 The first five terms are  __ , 6 Analysis Figure 7 shows the graph of the sequence. Notice that, since factorial
s grow very quickly, the presence of the factorial term in the denominator results in the denominator becoming much larger than the numerator as n increases. This means the quotient gets smaller and, as the plot of the terms shows, the terms are decreasing and nearing zero. 5  . _____ 1,008 1 ___ , 36 1, 5 6 an , 5 12 3, 1 8 4, 1 36 5, 5 1008 0 1 2 4 3 Figure 7 n 5 6 Try It #9 Write the first five terms of the sequence defined by the explicit formula an = (n + 1)! _______ . 2n Access this online resource for additional instruction and practice with sequences. • Finding Terms in a Sequence (http://openstaxcollege.org/l/findingterms) 948 CHAPTER 11 seQuences, proBaBility and counting theory 11.1 SeCTIOn exeRCISeS VeRBAl 1. Discuss the meaning of a sequence. If a finite 2. Describe three ways that a sequence can be defined. sequence is defined by a formula, what is its domain? What about an infinite sequence? 3. Is the ordered set of even numbers an infinite sequence? What about the ordered set of odd numbers? Explain why or why not. 5. What is a factorial, and how is it denoted? Use an example to illustrate how factorial notation can be beneficial. 4. What happens to the terms an of a sequence when there is a negative factor in the formula that is raised to a power that includes n? What is the term used to describe this phenomenon? AlGeBRAIC For the following exercises, write the first four terms of the sequence. 6. an = 2n − 2 2n + 1 ______ n3 10. an = 14. an = (−10)n + 1 7. an = − 16 _____ n + 1 11. an = 1.25 ⋅ (−4)n − 1 15. an = −  4 ⋅ (−5)n − 1  __________ 5 8. an = −(−5)n − 1 12. an = −4 ⋅ (−6)n − 1 For the following exercises, write the first eight terms of the piecewise sequence. 9. an = 13. an = 2n __ n3 n2 ______ 2n + 1 (3)n − 1 if n is odd 16. an = { (−2)n − 2 if n is even 18. an = { (2n + 1)2 if n is divisible by 4 20. an = { 4(n2 − 2) if n ≤ 3 or n > 6 2 _ n if n is not divisible by 4 n2 − 2 _ 4 if 3 < n ≤ 6 if n ≤ 5 n2 _ 2n + 1 n2 − 5 if n > 5 17. an = { 19. an = { −0.6 ⋅ 5n − 1 if n is prime or 1 2.5 ⋅ (−2)n − 1 if n is composite For the following exercises, write an explicit formula for each sequence. 21. 4, 7, 12, 19, 28, … 22. −4, 2, − 10, 14, − 34, … 4 __ 23. 1, 1, , 2, 3 16 ___ , … 5 24. 0, 1 − e1 _____ 1 + e2 , 1 − e2 _____ 1 + e3 , 1 − e4 1 − e3 ______ ______ , 1 + e5 1 + e4 , … 1 1 1 __ __ __ , − 25. 1, − , , 8 4 2 1 ___ 16 , … For the following exercises, write the first five terms of the sequence. 26. a1 = 9, an = an − 1 + n 27. a1 = 3, an = (−3)an − 1 (−3)n − 1 _ an − 1 − 2 29. a1 = −1, an = 28. a1 = −4, an = an − 1 + 2n _ an − 1 − 1 1 30. a1 = −30, an = (2 + an − 1)   __ 2 n For the following exercises, write the first eight terms of the sequence. 31. a1 = 1 ___ 24 , a2 = 1, an = (2an − 2)(3an − 1) 2(an −1 + 2) __ an − 2 33. a1 = 2, a2 = 10, an = 32. a1 = −1, a2 = 5, an = an − 2(3 − an −1) SECTION 11.1 section exercises 949 For the following exercises, write a recursive formula for each sequence. 34. −2.5, − 5, − 10, − 20, − 40, … 35. −8, − 6, − 3, 1, 6, … 36. 2, 4, 12, 48, 240, … 37. 35, 38, 41, 44, 47, … 3 __ 38. 15, 3, , 5 3 ___ , 25 3 ___ 125 , … For the following exercises, evaluate the factorial. 39. 6! 40.  12  ! ___ 6 41. 12! ___ 6! 42. 100! ____ 99! For the following exercises, write the first four terms of the sequence. 45. an = 44. an = 43. an = 3 ⋅ n ! _____ 4 ⋅ n ! n! __ n2 n ! _________ n2 − n − 1 46. an = 100 ⋅ n ________ n(n − 1)! GRAPHICAl For the following exercises, graph the first five terms of the indicated sequence 47. an = (−1)n _____ n + n 48. an = { if n in even 4 + n _ 2n 3 + n if n is odd 49. a1 = 2, an = (−an − 1 + 1)2 50. an = 1, an = an − 1 + 8 51. an = (n + 1)! _______ (n − 1)! For the following exercises, write an explicit formula for the sequence using the first five points shown on the graph. 52. an 15 13 11 9 7 5 3 0 (5, 13) (4, 11) (3, 9) (2, 7) (1, 5) 1 2 3 4 5 6 7 n 53. an 54. an 5, 8) (4, 4) (3, 2) (2, 1) (1, 0.5) 1 2 3 4 5 6 7 n 18 15 12 9 6 3 0 (1, 12) (2, 9) (3, 6) (4, 3) (5, 0) 1 2 3 4 5 6 7 n For the following exercises, write a recursive formula for the sequence using the first five points shown on the graph. 55. an 56. an 22 20 16 12 8 4 0 (5, 21) (4, 13) (3, 9) (2, 7) (1, 6) 1 2 3 4 5 n 16 12 8 4 0 (1, 16) (2, 8) (3, 4) (4, 2) 1 2 3 4 5 (5, 1) n 950 CHAPTER 11 seQuences, proBaBility and counting theory TeCHnOlOGY Follow these steps to evaluate a sequence defined recursively using a graphing calculator: • On the home screen, key in the value for the initial term a1 and press [ENTER]. • Enter the recursive formula by keying in all numerical values given in the formula, along with the key strokes [2ND] ANS for the previous term an − 1. Press [ENTER]. • Continue pressing [ENTER] to calculate the values for each successive term. For the following exercises, use the steps above to find the indicated term or terms for the sequence. 57. Find the first five terms of the sequence a1 = 58. Find the 15th term of the sequence 87 ___ , 111 . Use the >Frac feature to give a1 = 625, an = 0.8an − 1 + 18. 12 4 ___ __ an − 1 + an = 3 37 fractional results. 59. Find the first five terms of the sequence a1 = 2, an = 2[(an − 1) − 1] + 1. 61. Find the tenth term of the sequence a1 = 2, an = nan − 1 60. Find the first ten terms of the sequence (an − 1 + 1)! _ an − 1! a1 = 8, an = . Follow these steps to evaluate a finite sequence defined by an explicit formula. Using a TI-84, do the following. • In the home screen, press [2ND] LIST. • Scroll over to OPS and choose “seq(” from the dropdown list. Press [ENTER]. • In the line headed “Expr:” type in the explicit formula, using the [X,T, θ, n] button for n • In the line headed “Variable:” type in the variable used on the previous step. • In the line headed “start:” key in the value of n that begins the sequence. • In the line headed “end:” key in the value of n that ends the sequence. • Press [ENTER] 3 times to return to the home screen. You will see the sequence syntax on the screen. Press [ENTER] to see the list of terms for the finite sequence defined. Use the right arrow key to scroll through the list of terms. Using a TI-83, do the following. • In the home screen, press [2ND] LIST. • Scroll over to OPS and choose “seq(” from the dropdown list. Press [ENTER]. • Enter the items in the order “Expr”, “Variable”, “start”, “end” separated by commas. See the instructions above for the description of each item. • Press [ENTER] to see the list of terms for the finite sequence defined. Use the right arrow key to scroll through the list of terms. For the following exercises, use the steps above to find the indicated terms for the sequence. Round to the nearest thousandth when necessary. 62. List the first five terms of the sequence. an = − 5 28 _ ___ n + 9 3 64. List the first five terms of the sequence. an = 15n ⋅ (−2)n − 1 ____________ 47 63. List the first six terms of the sequence. n3 − 3.5n2 + 4.1n − 1.5 ___________________ 2.4n 65. List the first four terms of the sequence. an = an = 5.7n + 0.275(n − 1)! 66. List the first six terms of the sequence an = n! __ . n exTenSIOnS 67. Consider the sequence defined by an = −6 − 8n. Is an = −421 a term in the sequence? Verify the result. 69. Find a recursive formula for the sequence 1, 0, −1, −1, 0, 1, 1, 0, −1, −1, 0, 1, 1, ... (Hint: find a pattern for an based on the first two terms.) 71. Prove the conjecture made in the preceding exercise. 68. What term in the sequence an = the value 41? Verify the result. n2 + 4n + 4 __________ 2(n + 2) has 70. Calculate the first eight terms of the sequences and bn = n3 + 3n2 + 2n, and then an = make a conjecture about the relationship between these two sequences. (n + 2)! _______ (n − 1)! SECTION 11.2 arithmetic seQuences 951 leARnInG OBjeCTIVeS In this section, you will: • Find the common difference for an arithmetic sequence. • Write terms of an arithmetic sequence. • Use a recursive formula for an arithmetic sequence. • Use an explicit formula for an arithmetic sequence. 11. 2 ARITHMeTIC SeQUenCeS Companies often make large purchases, such as computers and vehicles, for business use. The book-value of these supplies decreases each year for tax purposes. This decrease in value is called depreciation. One method of calculating depreciation is straight-line depreciation, in which the value of the asset decreases by the same amount each year. As an example, consider a woman who starts a small contracting business. She purchases a new truck for $25,000. After five years, she estimates that she will be able to sell the truck for $8,000. The loss in value of the truck will therefore be $17,000, which is $3,400 per year for five years. The truck will be worth $21,600 after the first year; $18,200 after two years; $14,800 after three years; $11,400 after four years; and $8,000 at the end of five years. In this section, we will consider specific kinds of sequences that will allow us to calculate depreciation, such as the truck’s value. Finding Common Differences The values of the truck in the example are said to form an arithmetic sequence because they change by a constant amount each year. Each term increases or decreases by the same constant value called the common difference of the sequence. For this sequence, the common difference is −3,400. −3,400 −3,400 −3,400 −3,400 −3,400 {25000, 21600, 18200, 14800, 11400, 8000} The sequence below is another example of an arithmetic sequence. In this case, the constant difference is 3. You can choose any term of the sequence, and add 3 to find the subsequent term. +3 +3 +3 +3 {3, 6, 9, 12, 15, ...} arithmetic sequence An arithmetic sequence is a sequence that has the property that the difference between any two consecutive terms is a constant. This constant is called the common difference. If a1 is the first term of an arithmetic sequence and d is the common difference, the sequence will be: {an} = {a1, a1 + d, a1 + 2d, a1 + 3d,...} Example 1 Finding Common Differences Is each sequence arithmet
ic? If so, find the common difference. a. {1, 2, 4, 8, 16, ... } b. { −3, 1, 5, 9, 13, ... } Solution Subtract each term from the subsequent term to determine whether a common difference exists. a. The sequence is not arithmetic because there is no common difference 16 − 8 = 8 b. The sequence is arithmetic because there is a common difference. The common difference is 4. 1 − (−3 13 − 9 = 4 952 CHAPTER 11 seQuences, proBaBility and counting theory Analysis The graph of each of these sequences is shown in Figure 1. We can see from the graphs that, although both sequences show growth, a is not linear whereas b is linear. Arithmetic sequences have a constant rate of change so their graphs will always be points on a line. an 20 16 12 8 4 0 −4 an 20 16 12 8 4 1 2 3 4 56 (a) n 0 −4 Figure 1 1 2 3 4 5 6 n (b) Q & A… If we are told that a sequence is arithmetic, do we have to subtract every term from the following term to find the common difference? No. If we know that the sequence is arithmetic, we can choose any one term in the sequence, and subtract it from the subsequent term to find the common difference. Try It #1 Is the given sequence arithmetic? If so, find the common difference. {18, 16, 14, 12, 10, … } Try It #2 Is the given sequence arithmetic? If so, find the common difference. {1, 3, 6, 10, 15, … } Writing Terms of Arithmetic Sequences Now that we can recognize an arithmetic sequence, we will find the terms if we are given the first term and the common difference. The terms can be found by beginning with the first term and adding the common difference repeatedly. In addition, any term can also be found by plugging in the values of n and d into formula below. an = a1 + (n − 1)d How To… Given the first term and the common difference of an arithmetic sequence, find the first several terms. 1. Add the common difference to the first term to find the second term. 2. Add the common difference to the second term to find the third term. 3. Continue until all of the desired terms are identified. 4. Write the terms separated by commas within brackets. Example 2 Writing Terms of Arithmetic Sequences Write the first five terms of the arithmetic sequence with a1 = 17 and d = −3. Solution Adding −3 is the same as subtracting 3. Beginning with the first term, subtract 3 from each term to find the next term. The first five terms are {17, 14, 11, 8, 5} SECTION 11.2 arithmetic seQuences 953 Analysis As expected, the graph of the sequence consists of points on a line as shown in Figure 2. an 20 16 12 Figure 2 Try It #3 List the first five terms of the arithmetic sequence with a1 = 1 and d = 5. How To… Given any first term and any other term in an arithmetic sequence, find a given term. 1. Substitute the values given for a1, an, n into the formula an = a1 + (n − 1)d to solve for d. 2. Find a given term by substituting the appropriate values for a1, n, and d into the formula an = a1 + (n − 1)d. Example 3 Writing Terms of Arithmetic Sequences Given a1 = 8 and a4 = 14, find a5. Solution The sequence can be written in terms of the initial term 8 and the common difference d. We know the fourth term equals 14; we know the fourth term has the form a1 + 3d = 8 + 3d. We can find the common difference d. {8, 8 + d, 8 + 2d, 8 + 3d} an = a1 + (n − 1)d a4 = a1 + 3d a4 = 8 + 3d 14 = 8 + 3d d = 2 Write the fourth term of the sequence in terms of a1 and d. Substitute 14 for a4. Solve for the common difference. Find the fifth term by adding the common difference to the fourth term. a5 = a4 + 2 = 16 Analysis Notice that the common difference is added to the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by adding the common difference to the first term nine times or by using the equation an = a1 + (n − 1)d. Try It #4 Given a3 = 7 and a5 = 17, find a2. Using Recursive Formulas for Arithmetic Sequences Some arithmetic sequences are defined in terms of the previous term using a recursive formula. The formula provides an algebraic rule for determining the terms of the sequence. A recursive formula allows us to find any term of an arithmetic sequence using a function of the preceding term. Each term is the sum of the previous term and the 954 CHAPTER 11 seQuences, proBaBility and counting theory common difference. For example, if the common difference is 5, then each term is the previous term plus 5. As with any recursive formula, the first term must be given. an = an − 1 + d n ≥ 2 recursive formula for an arithmetic sequence The recursive formula for an arithmetic sequence with common difference d is: an = an − 1 + d n ≥ 2 How To… Given an arithmetic sequence, write its recursive formula. 1. Subtract any term from the subsequent term to find the common difference. 2. State the initial term and substitute the common difference into the recursive formula for arithmetic sequences. Example 4 Writing a Recursive Formula for an Arithmetic Sequence Write a recursive formula for the arithmetic sequence. { −18, −7, 4, 15, 26, …} Solution The first term is given as −18. The common difference can be found by subtracting the first term from the second term. Substitute the initial term and the common difference into the recursive formula for arithmetic sequences. d = −7 − (−18) = 11 a1 = −18 an = an − 1 + 11, for n ≥ 2 Analysis We see that the common difference is the slope of the line formed when we graph the terms of the sequence, as shown in Figure 3. The growth pattern of the sequence shows the constant difference of 11 units. an 30 20 10 0 −10 −20 1 2 3 4 5 6 n Figure 3 Q & A… Do we have to subtract the first term from the second term to find the common difference? No. We can subtract any term in the sequence from the subsequent term. It is, however, most common to subtract the first term from the second term because it is often the easiest method of finding the common difference. Try It #5 Write a recursive formula for the arithmetic sequence. {25, 37, 49, 61, …} SECTION 11.2 arithmetic seQuences 955 Using explicit Formulas for Arithmetic Sequences We can think of an arithmetic sequence as a function on the domain of the natural numbers; it is a linear function because it has a constant rate of change. The common difference is the constant rate of change, or the slope of the function. We can construct the linear function if we know the slope and the vertical intercept. an = a1 + d(n − 1) To find the y-intercept of the function, we can subtract the common difference from the first term of the sequence. Consider the following sequence. −50 −50 −50 −50 {200, 150, 100, 50, 0, ...} The common difference is −50, so the sequence represents a linear function with a slope of −50. To find the y-intercept, we subtract −50 from 200: 200 − ( −50) = 200 + 50 = 250. You can also find the y -intercept by graphing the function and determining where a line that connects the points would intersect the vertical axis. The graph is shown in Figure 4. an 250 200 150 100 50 0 1 2 4 3 Figure 4 56 n Recall the slope-intercept form of a line is y = mx + b. When dealing with sequences, we use an in place of y and n in place of x. If we know the slope and vertical intercept of the function, we can substitute them for m and b in the slopeintercept form of a line. Substituting − 50 for the slope and 250 for the vertical intercept, we get the following equation: an = −50n + 250 We do not need to find the vertical intercept to write an explicit formula for an arithmetic sequence. Another explicit formula for this sequence is an = 200 − 50(n − 1), which simplifies to an = −50n + 250. explicit formula for an arithmetic sequence An explicit formula for the nth term of an arithmetic sequence is given by an = a1 + d (n − 1) How To… Given the first several terms for an arithmetic sequence, write an explicit formula. 1. Find the common difference, a2 − a1. 2. Substitute the common difference and the first term into an = a1 + d(n − 1). Example 5 Writing the nth Term Explicit Formula for an Arithmetic Sequence Write an explicit formula for the arithmetic sequence. {2, 12, 22, 32, 42, …} 956 CHAPTER 11 seQuences, proBaBility and counting theory Solution The common difference can be found by subtracting the first term from the second term. d = a2 − a1 = 12 − 2 = 10 The common difference is 10. Substitute the common difference and the first term of the sequence into the formula and simplify. an = 2 + 10(n − 1) an = 10n − 8 Analysis The graph of this sequence, represented in Figure 5, shows a slope of 10 and a vertical intercept of −8. an 50 40 30 20 10 0 –10 1 2 3 4 5 6 7 8 9 10 n Figure 5 Try It #6 Write an explicit formula for the following arithmetic sequence. {50, 47, 44, 41, … } Finding the Number of Terms in a Finite Arithmetic Sequence Explicit formulas can be used to determine the number of terms in a finite arithmetic sequence. We need to find the common difference, and then determine how many times the common difference must be added to the first term to obtain the final term of the sequence. How To… Given the first three terms and the last term of a finite arithmetic sequence, find the total number of terms. 1. Find the common difference d. 2. Substitute the common difference and the first term into an = a1 + d(n − 1). 3. Substitute the last term for an and solve for n. Example 6 Finding the Number of Terms in a Finite Arithmetic Sequence Find the number of terms in the finite arithmetic sequence. {8, 1, −6, ... , −41} Solution The common difference can be found by subtracting the first term from the second term. 1 − 8 = −7 SECTION 11.2 arithmetic seQuences 957 The common difference is −7. Substitute the common difference and the initial term of the sequence into the nth term formula and simplify. Substitute −41 for an and solve for n There are eight terms in the sequence. an = a1 + d(n − 1) an = 8 + (−7)(n − 1) an = 15 − 7n −41 = 15 − 7n 8 = n Try It #7 Find the number of terms in the f
inite arithmetic sequence. {6, 11, 16, ... , 56} Solving Application Problems with Arithmetic Sequences In many application problems, it often makes sense to use an initial term of a0 instead of a1. In these problems, we alter the explicit formula slightly to account for the difference in initial terms. We use the following formula: an = a0 + dn Example 7 Solving Application Problems with Arithmetic Sequences A five-year old child receives an allowance of $1 each week. His parents promise him an annual increase of $2 per week. a. Write a formula for the child’s weekly allowance in a given year. b. What will the child’s allowance be when he is 16 years old? Solution a. The situation can be modeled by an arithmetic sequence with an initial term of 1 and a common difference of 2. Let A be the amount of the allowance and n be the number of years after age 5. Using the altered explicit formula for an arithmetic sequence we get: b. We can find the number of years since age 5 by subtracting. 16 − 5 = 11 An = 1 + 2n We are looking for the child’s allowance after 11 years. Substitute 11 into the formula to find the child’s allowance at age 16. The child’s allowance at age 16 will be $23 per week. A11 = 1 + 2(11) = 23 Try It #8 A woman decides to go for a 10-minute run every day this week and plans to increase the time of her daily run by 4 minutes each week. Write a formula for the time of her run after n weeks. How long will her daily run be 8 weeks from today? Access this online resource for additional instruction and practice with arithmetic sequences. • Arithmetic Sequences (http://openstaxcollege.org/l/arithmeticseq) 958 CHAPTER 11 seQuences, proBaBility and counting theory 11.2 SeCTIOn exeRCISeS VeRBAl 1. What is an arithmetic sequence? 2. How is the common difference of an arithmetic sequence found? 3. How do we determine whether a sequence is 4. What are the main differences between using a arithmetic? recursive formula and using an explicit formula to describe an arithmetic sequence? 5. Describe how linear functions and arithmetic sequences are similar. How are they different? AlGeBRAIC For the following exercises, find the common difference for the arithmetic sequence provided. 6. {5, 11, 17, 23, 29, ... } 3 1 , 2, ...  7.  0, __ __ , 1, 2 2 For the following exercises, determine whether the sequence is arithmetic. If so find the common difference. 8. {11.4, 9.3, 7.2, 5.1, 3, ... } 9. {4, 16, 64, 256, 1024, ... } For the following exercises, write the first five terms of the arithmetic sequence given the first term and common difference. 10. a1 = −25, d = −9 2 __ 11. a1 = 0, d = 3 For the following exercises, write the first five terms of the arithmetic series given two terms. 12. a1 = 17, a7 = −31 13. a13 = −60, a33 = −160 For the following exercises, find the specified term for the arithmetic sequence given the first term and common difference. 14. First term is 3, common difference is 4, find the 15. First term is 4, common difference is 5, find the 5th term. 4th term. 16. First term is 5, common difference is 6, find the 17. First term is 6, common difference is 7, find the 8th term. 18. First term is 7, common difference is 8, find the 6th term. 7th term. For the following exercises, find the first term given two terms from an arithmetic sequence. 19. Find the first term or a1 of an arithmetic sequence if 20. Find the first term or a1 of an arithmetic sequence if a6 = 12 and a14 = 28. a7 = 21 and a15 = 42. 21. Find the first term or a1 of an arithmetic sequence if a8 = 40 and a23 = 115. 22. Find the first term or a1 of an arithmetic sequence if a9 = 54 and a17 = 102. 23. Find the first term or a1 of an arithmetic sequence if a11 = 11 and a21 = 16. For the following exercises, find the specified term given two terms from an arithmetic sequence. 24. a1 = 33 and a7 = −15. Find a4. 25. a3 = −17.1 and a10 = −15.7. Find a21. For the following exercises, use the recursive formula to write the first five terms of the arithmetic sequence. 26. a1 = 39; an = an − 1 −3 27. a1 = −19; an = an − 1 −1.4 SECTION 11.2 section exercises 959 For the following exercises, write a recursive formula for each arithmetic sequence. 28. a = {40, 60, 80, ... } 29. a = {17, 26, 35, ... } 30. a = {−1, 2, 5, ... } 31. a = {12, 17, 22, ... } 34. a = {−0.52, −1.02, −1.52, ... } 32. a = {−15, −7, 1, ... } 1 , ...  35. a =  __ , 5 7 9 ___ ___ , 20 10 33. a = {8.9, 10.3, 11.7, ... } 5 1 , −2, ...  36. a =  − __ __ , − 4 2 11 ___ 12 , −2, ...  1 37. a =  __ , − 6 For the following exercises, use the recursive formula to write the first five terms of the arithmetic sequence. 38. a = {7, 4, 1, ... }; Find the 17th term. 40. a = {2, 6, 10, ... }; Find the 12th term. 39. a = {4, 11, 18, ... }; Find the 14th term. For the following exercises, use the recursive formula to write the first five terms of the arithmetic sequence. 41. a = 24 − 4n 1 1 __ __ n − 42. a = 2 2 For the following exercises, write an explicit formula for each arithmetic sequence. 43. a = {3, 5, 7, ... } 46. a = {−17, −217, −417, ... } 44. a = {32, 24, 16, ... } 47. a = {1.8, 3.6, 5.4, ... } 4 1 , −3, ...  50. a =  __ __ , − 3 3 45. a = {−5, 95, 195, ... } 48. a = {−18.1, −16.2, −14.3, ... } 2 1 , ...  51. a =  0, __ __ , 3 3 49. a = {15.8, 18.5, 21.2, ... } 10 52. a =  −5, − ___ 3 5 , …  __ , − 3 For the following exercises, find the number of terms in the given finite arithmetic sequence. 53. a = {3, −4, −11, ... , −60} 54. a = {1.2, 1.4, 1.6, ... , 3.8} 7 1 , ... , 8  55. a =  __ __ , 2, 2 2 GRAPHICAl For the following exercises, determine whether the graph shown represents an arithmetic sequence. 56. an 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 −0.5 −0.5 −1 −1.5 −2 −2.5 −3 −3.5 −4 −4.5 −5 −5.5 (5, 4) (4, 2) (3, 0) 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 n (2, −2) (1, −4) 57. an 8.5 8 7.5 7 6.5 6 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 (5, 7.5938) (4, 5.0625) (3, 3.375) (2, 2.25) (1, 1.5) 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 n −0.5 0 −0.5 960 CHAPTER 11 seQuences, proBaBility and counting theory For the following exercises, use the information provided to graph the first 5 terms of the arithmetic sequence. 58. a1 = 0, d = 4 59. a1 = 9; an = an − 1 − 10 60. an = −12 + 5n TeCHnOlOGY For the following exercises, follow the steps to work with the arithmetic sequence an = 3n − 2 using a graphing calculator: • Press [MODE] › Select [SEQ] in the fourth line › Select [DOT] in the fifth line › Press [ENTER] • Press [Y=] › nMin is the first counting number for the sequence. Set nMin = 1 › u(n) is the pattern for the sequence. Set u(n) = 3n − 2 › u(nMin) is the first number in the sequence. Set u(nMin) = 1 • Press [2ND] then [WINDOW] to go to TBLSET › Set TblStart = 1 › Set ΔTbl = 1 › Set Indpnt: Auto and Depend: Auto • Press [2ND] then [GRAPH] to go to the [TABLE] 61. What are the first seven terms shown in the column 62. Use the scroll-down arrow to scroll to n = 50. What with the heading u(n)? value is given for u(n)? 63. Press [WINDOW]. Set nMin = 1, nMax = 5, xMin = 0, xMax = 6, yMin = −1, and yMax = 14. Then press [GRAPH]. Graph the sequence as it appears on the graphing calculator. 1 __ For the following exercises, follow the steps given above to work with the arithmetic sequence an = n + 5 using a 2 graphing calculator. 64. What are the first seven terms shown in the column with the heading u(n) in the [TABLE] feature? 65. Graph the sequence as it appears on the graphing calculator. Be sure to adjust the [WINDOW] settings as needed. exTenSIOnS 66. Give two examples of arithmetic sequences whose 67. Give two examples of arithmetic sequences whose 4th terms are 9. 10th terms are 206. 68. Find the 5th term of the arithmetic sequence 69. Find the 11th term of the arithmetic sequence {9b, 5b, b, … }. 70. At which term does the sequence {5.4, 14.5, 23.6, ...} exceed 151? {3a − 2b, a + 2b, −a + 6b, … }. 17 71. At which term does the sequence  _ , 3 begin to have negative values? 31 _ , 6 14 _ ,...  3 72. For which terms does the finite arithmetic sequence 73. Write an arithmetic sequence using a recursive 5  _ , 2 9 19 1 _ _  have integer values? ___ , ... , , 8 4 8 formula. Show the first 4 terms, and then find the 31st term. 74. Write an arithmetic sequence using an explicit formula. Show the first 4 terms, and then find the 28th term. SECTION 11.3 geometric seQuences 961 leARnInG OBjeCTIVeS In this section, you will: • Find the common ratio for a geometric sequence. • List the terms of a geometric sequence. • Use a recursive formula for a geometric sequence. • Use an explicit formula for a geometric sequence. 11. 3 GeOMeTRIC SeQUenCeS Many jobs offer an annual cost-of-living increase to keep salaries consistent with inflation. Suppose, for example, a recent college graduate finds a position as a sales manager earning an annual salary of $26,000. He is promised a 2% cost of living increase each year. His annual salary in any given year can be found by multiplying his salary from the previous year by 102%. His salary will be $26,520 after one year; $27,050.40 after two years; $27,591.41 after three years; and so on. When a salary increases by a constant rate each year, the salary grows by a constant factor. In this section, we will review sequences that grow in this way. Finding Common Ratios The yearly salary values described form a geometric sequence because they change by a constant factor each year. Each term of a geometric sequence increases or decreases by a constant factor called the common ratio. The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term. ×6 ×6 ×6 ×6 {1, 6, 36, 216, 1,296, ...} definition of a geometric sequence A geometric sequence is one in which any term divided by the previous term is a constant. This constant is called the common ratio of the sequence. The common ratio can be found by dividing any term in the seque
nce by the previous term. If a1 is the initial term of a geometric sequence and r is the common ratio, the sequence will be {a1, a1r, a1r 2, a1r 3, ... }. How To… Given a set of numbers, determine if they represent a geometric sequence. 1. Divide each term by the previous term. 2. Compare the quotients. If they are the same, a common ratio exists and the sequence is geometric. Example 1 Finding Common Ratios Is the sequence geometric? If so, find the common ratio. a. 1, 2, 4, 8, 16, ... b. 48, 12, 4, 2,... Solution Divide each term by the previous term to determine whether a common ratio exists. 2 _ = 2 a. 1 The sequence is geometric because there is a common ratio. The common ratio is 2 16 _ 8 = 2 b. 1 __ = 4 12 _ 48 The sequence is not geometric because there is not a common ratio 12 962 CHAPTER 11 seQuences, proBaBility and counting theory Analysis The graph of each sequence is shown in Figure 1. It seems from the graphs that both (a) and (b) appear have the form of the graph of an exponential function in this viewing window. However, we know that (a) is geometric and so this interpretation holds, but (b) is not. an 20 16 12 8 4 0 an 60 48 36 24 12 1 2 3 4 5 6 (a) n 0 Figure 1 1 2 3 4 5 6 n (b) Q & A… If you are told that a sequence is geometric, do you have to divide every term by the previous term to find the common ratio? No. If you know that the sequence is geometric, you can choose any one term in the sequence and divide it by the previous term to find the common ratio. Try It #1 Is the sequence geometric? If so, find the common ratio. 5, 10, 15, 20, ... Try It #2 Is the sequence geometric? If so, find the common ratio. 4 __ , ... 100, 20, 4, 5 Writing Terms of Geometric Sequences Now that we can identify a geometric sequence, we will learn how to find the terms of a geometric sequence if we are given the first term and the common ratio. The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. For instance, if the first term of a geometric sequence is a1 = −2 and the common ratio is r = 4, we can find subsequent terms by multiplying −2 ⋅ 4 to get −8 then multiplying the result −8 ⋅ 4 to get −32 and so on. a1 = −2 a2 = (−2 ⋅ 4) = −8 a3 = (−8 ⋅ 4) = −32 a4 = (−32 ⋅ 4) = −128 The first four terms are {−2, −8, −32, −128}. How To… Given the first term and the common factor, find the first four terms of a geometric sequence. 1. Multiply the initial term, a1, by the common ratio to find the next term, a2. 2. Repeat the process, using an = a2 to find a3 and then a3 to find a4, until all four terms have been identified. 3. Write the terms separated by commons within brackets. SECTION 11.3 geometric seQuences 963 Example 2 Writing the Terms of a Geometric Sequence List the first four terms of the geometric sequence with a1 = 5 and r = −2. Solution Multiply a1 by −2 to find a2. Repeat the process, using a2 to find a3, and so on. a1 = 5 a2 = −2a1 = −10 a3 = −2a2 = 20 a4 = −2a3 = −40 The first four terms are {5, −10, 20, −40}. Try It #3 1 __ List the first five terms of the geometric sequence with a1 = 18 and r = . 3 Using Recursive Formulas for Geometric Sequences A recursive formula allows us to find any term of a geometric sequence by using the previous term. Each term is the product of the common ratio and the previous term. For example, suppose the common ratio is 9. Then each term is nine times the previous term. As with any recursive formula, the initial term must be given. recursive formula for a geometric sequence The recursive formula for a geometric sequence with common ratio r and first term a1 is an = r an − 1, n ≥ 2 How To… Given the first several terms of a geometric sequence, write its recursive formula. 1. State the initial term. 2. Find the common ratio by dividing any term by the preceding term. 3. Substitute the common ratio into the recursive formula for a geometric sequence. Example 3 Using Recursive Formulas for Geometric Sequences Write a recursive formula for the following geometric sequence. {6, 9, 13.5, 20.25, ... } Solution The first term is given as 6. The common ratio can be found by dividing the second term by the first term. 9 _ = 1.5 r = 6 Substitute the common ratio into the recursive formula for geometric sequences and define a1. an = ran − 1 an = 1.5an − 1 for n ≥ 2 a1 = 6 Analysis The sequence of data points follows an exponential pattern. The common ratio is also the base of an exponential function as shown in Figure 2. an 24 18 12 6 0 1 2 3 4 5 n Figure 2 964 CHAPTER 11 seQuences, proBaBility and counting theory Q & A… Do we have to divide the second term by the first term to find the common ratio? No. We can divide any term in the sequence by the previous term. It is, however, most common to divide the second term by the first term because it is often the easiest method of finding the common ratio. Try It #4 Write a recursive formula for the following geometric sequence. 8  2, 4 _ _ , , 9 3 16 _ 27 ,...  Using explicit Formulas for Geometric Sequences Because a geometric sequence is an exponential function whose domain is the set of positive integers, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms. an = a1 r n − 1 Let’s take a look at the sequence {18, 36, 72, 144, 288,...}. This is a geometric sequence with a common ratio of 2 and an exponential function with a base of 2. An explicit formula for this sequence is The graph of the sequence is shown in Figure 3. an = 18 · 2n − 1 an 360 324 288 252 216 180 144 108 72 36 0 1 4 5 6 2 3 Figure 3 n explicit formula for a geometric sequence The nth term of a geometric sequence is given by the explicit formula: an = a1 r n − 1 Example 4 Writing Terms of Geometric Sequences Using the Explicit Formula Given a geometric sequence with a1 = 3 and a4 = 24, find a2. Solution The sequence can be written in terms of the initial term and the common ratio r. 3, 3r, 3r 2, 3r 3,... SECTION 11.3 geometric seQuences 965 Find the common ratio using the given fourth term. an = a1r n − 1 a4 = 3r 3 24 = 3r 3 8 = r 3 r = 2 Write the fourth term of sequence in terms of a1 and r Substitute 24 for a4 Divide Solve for the common ratio Find the second term by multiplying the first term by the common ratio. a2 = 2a1 = 2(3) = 6 Analysis The common ratio is multiplied by the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by multiplying the first term by the common ratio nine times or by multiplying by the common ratio raised to the ninth power. Try It #5 Given a geometric sequence with a2 = 4 and a3 = 32, find a6. Example 5 Writing an Explicit Formula for the nth Term of a Geometric Sequence Write an explicit formula for the nth term of the following geometric sequence. {2, 10, 50, 250,...} Solution The first term is 2. The common ratio can be found by dividing the second term by the first term. 10 ___ = 5 2 The common ratio is 5. Substitute the common ratio and the first term of the sequence into the formula. The graph of this sequence in Figure 4 shows an exponential pattern. an = a1r (n − 1) an = 2 ⋅ 5n − 1 an 300 250 200 150 100 50 0 1 2 3 4 5 n Figure 4 Try It #6 Write an explicit formula for the following geometric sequence. {–1, 3, –9, 27, ...} 966 CHAPTER 11 seQuences, proBaBility and counting theory Solving Application Problems with Geometric Sequences In real-world scenarios involving arithmetic sequences, we may need to use an initial term of a0 instead of a1. In these problems, we can alter the explicit formula slightly by using the following formula: an = a0 r n Example 6 Solving Application Problems with Geometric Sequences In 2013, the number of students in a small school is 284. It is estimated that the student population will increase by 4% each year. a. Write a formula for the student population. b. Estimate the student population in 2020. Solution a. The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04. Let P be the student population and n be the number of years after 2013. Using the explicit formula for a geometric sequence we get b. We can find the number of years since 2013 by subtracting. 2020 − 2013 = 7 Pn = 284 ⋅ 1.04n We are looking for the population after 7 years. We can substitute 7 for n to estimate the population in 2020. The student population will be about 374 in 2020. P7 = 284 ⋅ 1.047 ≈ 374 Try It #7 A business starts a new website. Initially the number of hits is 293 due to the curiosity factor. The business estimates the number of hits will increase by 2.6% per week. a. Write a formula for the number of hits. b. Estimate the number of hits in 5 weeks. Access these online resources for additional instruction and practice with geometric sequences. • Geometric Sequences (http://openstaxcollege.org/l/geometricseq) • Determine the Type of Sequence (http://openstaxcollege.org/l/sequencetype) • Find the Formula for a Sequence (http://openstaxcollege.org/l/sequenceformula) SECTION 11.3 section exercises 967 11.3 SeCTIOn exeRCISeS VeRBAl 1. What is a geometric sequence? 2. How is the common ratio of a geometric sequence found? 3. What is the procedure for determining whether a 4. What is the difference between an arithmetic sequence is geometric? sequence and a geometric sequence? 5. Describe how exponential functions and geometric sequences are similar. How are they different? AlGeBRAIC For the following exercises, find the common ratio for the geometric sequence. 6. 1, 3, 9, 27, 81, ... 7. −0.125, 0.25, −0.5, 1, −2, ... 1 1 ___ ___ , − , − 8. −2, − 8 2 1 ___ 32 , − 1 ___ 128 , ... For the following exercises, determine whether the sequence is geometric. If so, find the common ratio. 1 ___ 16 1 1 1 __ ___ __ , − , − 11. −1, , 8 4 2 9. −6, −12, −24, −48, −96, ... 10. 5, 5.2, 5.4,
5.6, 5.8, ... , ... 12. 6, 8, 11, 15, 20, ... 13. 0.8, 4, 20, 100, 500, ... For the following exercises, write the first five terms of the geometric sequence, given the first term and common ratio. 14. a1 = 8, r = 0.3 1 __ 15. a1 = 5, r = 5 For the following exercises, write the first five terms of the geometric sequence, given any two terms. 16. a7 = 64, a10 = 512 17. a6 = 25, a8 = 6.25 For the following exercises, find the specified term for the geometric sequence, given the first term and common ratio. 18. The first term is 2, and the common ratio is 3. Find the 5th term. 1 ___ 19. The first term is 16 and the common ratio is − . 3 Find the 4th term. For the following exercises, find the specified term for the geometric sequence, given the first four terms. 20. an = {−1, 2, −4, 8, ...}. Find a12. 2 2 21. an =  −2, __ __ , − , 9 3 2 ___ 27 , ....  Find a 7 . For the following exercises, write the first five terms of the geometric sequence. 1 ___ 22. a1 = −486, an = − an − 1 3 23. a1 = 7, an = 0.2an − 1 For the following exercises, write a recursive formula for each geometric sequence. 24. an = {−1, 5, −25, 125, ...} 26. an = {14, 56, 224, 896, ...} 25. an = {−32, −16, −8, −4, ...} 27. an = {10, −3, 0.9, −0.27, ...} 28. an = {0.61, 1.83, 5.49, 16.47, ...} 3 29. an =  __ , 5 1 ___ , 10 1 ___ , 60 1 ___ 360 , ...  8 4 30. an =  −2, __ __ , − , 9 3 16 ___ 27 , ...  31. an =  1 ___ 512 , − 1 ___ , 128 1 ___ 32 1 , ...  ___ , − 8 For the following exercises, write the first five terms of the geometric sequence. 1 32. an = −4 ⋅ 5n − 1  33. an = 12 ⋅  − ___ 2 n − 1 968 CHAPTER 11 seQuences, proBaBility and counting theory For the following exercises, write an explicit formula for each geometric sequence. 35. an = {1, 3, 9, 27, ...} 34. an = {−2, −4, −8, −16, ...} 37. an = {0.8, −4, 20, −100, ...} 36. an = {−4, −12, −36, −108, ...} 4 16 64 38. an = {−1.25, −5, −20, −80, ...} 39. an =  −1, − ___ __ ___ , − 5 125 25 1 1 , ...  41. an =  3, −1, __ __ , − 9 3 1 40. an =  2, __ , 3 1 ___ 108 1 ___ , 18 , ...  , − , ...  For the following exercises, find the specified term for the geometric sequence given. n − 1 42. Let a1 = 4, an = −3an − 1. Find a8. 1  43. Let an = −  − __ 3 . Find a12. For the following exercises, find the number of terms in the given finite geometric sequence. 44. an = {−1, 3, −9, ... , 2187} 1 45. an =  2, 1, __ , ... , 2 1  ____ 1024 GRAPHICAl For the following exercises, determine whether the graph shown represents a geometric sequence. 46. an 47. an 6 5 4 3 2 1 0 −0.5 −1 −2 −3 −4 (5, 5) (4, 3) (3, 1) 3.5 4 4.5 5 5.5 n 0.5 1 1.5 2 2.5 3 (2, −1) (1, −3) 6 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 −0.5 −0.5 −1 1.5 0.5 1 (1, −0.5) (5, 5.5938) (4, 3.0625) (3, 1.375) (2, 0.25) 2 2.5 3 3.5 4 4.5 5 5.5 n For the following exercises, use the information provided to graph the first five terms of the geometric sequence. 1 __ 48. a1 = 1, r = 2 exTenSIOnS 49. a1 = 3, an = 2an − 1 50. an = 27 ⋅ 0.3n − 1 51. Use recursive formulas to give two examples of geometric sequences whose 3rd terms are 200. 52. Use explicit formulas to give two examples of geometric sequences whose 7th terms are 1024. 53. Find the 5th term of the geometric sequence 54. Find the 7th term of the geometric sequence {b, 4b, 16b, ...}. {64a(−b), 32a(−3b), 16a(−9b), ...}. 55. At which term does the sequence {10, 12, 14.4, 17.28, ...} exceed 100? 56. At which term does the sequence 1 ___ , 243 1  ____ , 2187 1 ___ , 729 1 ___ 81 ...  begin to have integer values? 57. For which term does the geometric sequence 58. Use the recursive formula to write a geometric 2  an = −36  __ 3 n − 1 first have a non-integer value? sequence whose common ratio is an integer. Show the first four terms, and then find the 10th term. 59. Use the explicit formula to write a geometric 60. Is it possible for a sequence to be both arithmetic and sequence whose common ratio is a decimal number between 0 and 1. Show the first 4 terms, and then find the 8th term. geometric? If so, give an example. SECTION 11.4 series and their notations 969 leARnInG OBjeCTIVeS In this section, you will: • Use summation notation. • Use the formula for the sum of the first n terms of an arithmetic series. • Use the formula for the sum of the first n terms of a geometric series. • Use the formula for the sum of an infinite geometric series. • Solve annuity problems. 11. 4 SeRIeS AnD THeIR nOTATIOnS A couple decides to start a college fund for their daughter. They plan to invest $50 in the fund each month. The fund pays 6% annual interest, compounded monthly. How much money will they have saved when their daughter is ready to start college in 6 years? In this section, we will learn how to answer this question. To do so, we need to consider the amount of money invested and the amount of interest earned. Using Summation notation To find the total amount of money in the college fund and the sum of the amounts deposited, we need to add the amounts deposited each month and the amounts earned monthly. The sum of the terms of a sequence is called a series. Consider, for example, the following series. 3 + 7 + 11 + 15 + 19 + ... The nth partial sum of a series is the sum of a finite number of consecutive terms beginning with the first term. The notation Sn represents the partial sum. S1 = 3 S2 = 3 + 7 = 10 S3 = 3 + 7 + 11 = 21 S4 = 3 + 7 + 11 + 15 = 36 Summation notation is used to represent series. Summation notation is often known as sigma notation because it uses the Greek capital letter sigma, Σ, to represent the sum. Summation notation includes an explicit formula and specifies the first and last terms in the series. An explicit formula for each term of the series is given to the right of the sigma. A variable called the index of summation is written below the sigma. The index of summation is set equal to the lower limit of summation, which is the number used to generate the first term in the series. The number above the sigma, called the upper limit of summation, is the number used to generate the last term in a series. Upper limit of summation → 5 ∑ 2k Index of summation → ← Lower limit of summation ← Explicit formula for kth term of series k = 1 If we interpret the given notation, we see that it asks us to find the sum of the terms in the series ak = 2k for k = 1 through k = 5. We can begin by substituting the terms for k and listing out the terms of this series. a1 = 2(1) = 2 a2 = 2(2) = 4 a3 = 2(3) = 6 a4 = 2(4) = 8 a5 = 2(5) = 10 We can find the sum of the series by adding the terms: 5 ∑ 2k = 2 + 4 + 6 + 8 + 10 = 30 k = 1 970 CHAPTER 11 seQuences, proBaBility and counting theory summation notation The sum of the first n terms of a series can be expressed in summation notation as follows: n ∑ ak k = 1 This notation tells us to find the sum of ak from k = 1 to k = n. k is called the index of summation, 1 is the lower limit of summation, and n is the upper limit of summation. Q & A… Does the lower limit of summation have to be 1? No. The lower limit of summation can be any number, but 1 is frequently used. We will look at examples with lower limits of summation other than 1. How To… Given summation notation for a series, evaluate the value. 1. Identify the lower limit of summation. 2. Identify the upper limit of summation. 3. Substitute each value of k from the lower limit to the upper limit into the formula. 4. Add to find the sum. Example 1 Using Summation Notation 7 Evaluate ∑ k2 . k = 3 Solution According to the notation, the lower limit of summation is 3 and the upper limit is 7. So we need to find the sum of k 2 from k = 3 to k = 7. We find the terms of the series by substituting k = 3, 4, 5, 6, and 7 into the function k 2. We add the terms to find the sum. 7 ∑ k 2 = 32 + 42 + 52 + 62 + 72 k = 3 = 9 + 16 + 25 + 36 + 49 = 135 Try It #1 Evaluate ∑ k = 2 5 (3k − 1) . Using the Formula for Arithmetic Series Just as we studied special types of sequences, we will look at special types of series. Recall that an arithmetic sequence is a sequence in which the difference between any two consecutive terms is the common difference, d. The sum of the terms of an arithmetic sequence is called an arithmetic series. We can write the sum of the first n terms of an arithmetic series as: We can also reverse the order of the terms and write the sum as Sn = a1 + (a1 + d) + (a1 + 2d) + ... + (an − d) + an. Sn = an + (an − d) + (an − 2d) + ... + (a1 + d) + a1. If we add these two expressions for the sum of the first n terms of an arithmetic series, we can derive a formula for the sum of the first n terms of any arithmetic series. SECTION 11.4 series and their notations 971 Sn = a1 + (a1 + d) + (a1 + 2d) + ... + (an − d) + an + Sn = an + (an − d) + (an − 2d) + ... + (a1 + d) + a1 2Sn = (a1 + an) + (a1 + an) + ... + (a1 + an) Because there are n terms in the series, we can simplify this sum to 2Sn = n(a1 + an). We divide by 2 to find the formula for the sum of the first n terms of an arithmetic series. Sn = n(a1 + an) __________ 2 formula for the sum of the first n terms of an arithmetic series An arithmetic series is the sum of the terms of an arithmetic sequence. The formula for the sum of the first n terms of an arithmetic sequence is Sn = n(a1 + an) __________ 2 How To… Given terms of an arithmetic series, find the sum of the first n terms. 1. Identify a1 and an. 2. Determine n. 3. Substitute values for a1, an, and n into the formula Sn = 4. Simplify to find Sn. n(a1 + an) _ . 2 Example 2 Finding the First n Terms of an Arithmetic Series Find the sum of each arithmetic series. a. 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32 b. 20 + 15 + 10 + … + −50 c. ∑ 3k − 8 12 k = 1 Solution a. We are given a1 = 5 and an = 32. Count the number of terms in the sequence to find n = 10. Substitute values for a1, an, and n into the formula and simplify. n(a1 + an) __________ 2 10(5 + 32) _________ 2 S10 = Sn = = 185 b. We are given a1 = 20 and an = −50. Use the formula for the general term of an arithmetic s
equence to find n. an = a1 + (n − 1)d −50 = 20 + (n − 1)( −5) −70 = (n − 1)( −5) 14 = n − 1 15 = n Substitute values for a1, an, n into the formula and simplify. Sn = S15 = n(a1 + an) __________ 2 15(20 − 50) __________ 2 = −225 972 CHAPTER 11 seQuences, proBaBility and counting theory c. To find a1, substitute k = 1 into the given explicit formula. ak = 3k − 8 a1 = 3(1) − 8 = −5 We are given that n = 12. To find a12, substitute k = 12 into the given explicit formula. ak = 3k − 8 a12 = 3(12) − 8 = 28 Substitute values for a1, an, and n into the formula and simplify. Sn = n(a1 + an) _ 2 S12 = 12( −5 + 28) __ 2 = 138 Try It #2 Use the formula to find the sum of the arithmetic series. 1.4 + 1.6 + 1.8 + 2.0 + 2.2 + 2.4 + 2.6 + 2.8 + 3.0 + 3.2 + 3.4 Try It #3 Use the formula to find the sum of the arithmetic series. 13 + 21 + 29 + … + 69 Try It #4 Use the formula to find the sum of the arithmetic series. 10 ∑ k = 1 5 − 6k Example 3 Solving Application Problems with Arithmetic Series On the Sunday after a minor surgery, a woman is able to walk a half-mile. Each Sunday, she walks an additional quarter-mile. After 8 weeks, what will be the total number of miles she has walked? 1 1 __ __ Solution This problem can be modeled by an arithmetic series with a1 = and d = . We are looking for the total 4 2 number of miles walked after 8 weeks, so we know that n = 8, and we are looking for S8. To find a8, we can use the explicit formula for an arithmetic sequence. We can now use the formula for arithmetic series. an = a1 + d(n − 1) 9 1 1 __ __ __ (8 − 1) = + a8 = 4 4 2 Sn = S8 = n(a1 + an) _ 2 9 1  8  __ __ + 4 2 __ = 11 2 She will have walked a total of 11 miles. Try It #5 A man earns $100 in the first week of June. Each week, he earns $12.50 more than the previous week. After 12 weeks, how much has he earned? SECTION 11.4 series and their notations 973 Using the Formula for Geometric Series Just as the sum of the terms of an arithmetic sequence is called an arithmetic series, the sum of the terms in a geometric sequence is called a geometric series. Recall that a geometric sequence is a sequence in which the ratio of any two consecutive terms is the common ratio, r. We can write the sum of the first n terms of a geometric series as Just as with arithmetic series, we can do some algebraic manipulation to derive a formula for the sum of the first n terms of a geometric series. We will begin by multiplying both sides of the equation by r. Sn = a1 + ra1 + r2a1 + ... + rn – 1a1. Next, we subtract this equation from the original equation. rSn = ra1 + r 2a1 + r 3a1 + ... + r na1 Sn = a1 + ra1 + r 2a1 + ... + r n – 1 a1. −rSn = −(ra1 + r 2a1 + r 3a1 + ... + r na1) (1 − r)Sn = a1 − r n a1 Notice that when we subtract, all but the first term of the top equation and the last term of the bottom equation cancel out. To obtain a formula for Sn, divide both sides by (1 − r). Sn = a1(1 − rn) _________ 1 − r r ≠ 1 formula for the sum of the first n terms of a geometric series A geometric series is the sum of the terms in a geometric sequence. The formula for the sum of the first n terms of a geometric sequence is represented as Sn = a1(1 − r n) _________ 1 − r r ≠ 1 How To… Given a geometric series, find the sum of the first n terms. 1. Identify a1, r, and n. 2. Substitute values for a1, r, and n into the formula Sn = 3. Simplify to find Sn. a1(1 − r n) _________ 1 − r . Example 4 Finding the First n Terms of a Geometric Series Use the formula to find the indicated partial sum of each geometric series. a. S11 for the series 8 + (−4) + 2 + … b. ∑ 6 3 ⋅ 2k k = 1 Solution a. a1 = 8, and we are given that n = 11. We can find r by dividing the second term of the series by the first. Substitute values for a1, r, and n into the formula and simplify. r = −4 1 ___ ___ = − 2 8 Sn = S11 = a1(1 − rn −  − __ 2 __ 1  1 −  − __ 2 11  ≈ 5.336 974 CHAPTER 11 seQuences, proBaBility and counting theory b. Find a1 by substituting k = 1 into the given explicit formula. a1 = 3 ⋅ 21 = 6 We can see from the given explicit formula that r = 2. The upper limit of summation is 6, so n = 6. Substitute values for a1, r, and n into the formula, and simplify. Sn = a1(1 − r n) _________ 1 − r S6 = 6(1 − 26) ________ 1 − 2 = 378 Try It #6 Use the formula to find the indicated partial sum of each geometric series. S20 for the series 1,000 + 500 + 250 + … Try It #7 Use the formula to find the indicated partial sum of each geometric series. 8 ∑ 3k k = 1 Example 5 Solving an Application Problem with a Geometric Series At a new job, an employee’s starting salary is $26,750. He receives a 1.6% annual raise. Find his total earnings at the end of 5 years. Solution The problem can be represented by a geometric series with a1 = 26, 750; n = 5; and r = 1.016. Substitute values for a1, r, and n into the formula and simplify to find the total amount earned at the end of 5 years. Sn = a1(1 − rn) _________ 1 − r S5 = 26,750(1 − 1.0165) ________________ 1 − 1.016 ≈ 138,099.03 He will have earned a total of $138,099.03 by the end of 5 years. Try It #8 At a new job, an employee’s starting salary is $32,100. She receives a 2% annual raise. How much will she have earned by the end of 8 years? Using the Formula for the Sum of an Infinite Geometric Series Thus far, we have looked only at finite series. Sometimes, however, we are interested in the sum of the terms of an infinite sequence rather than the sum of only the first n terms. An infinite series is the sum of the terms of an infinite sequence. An example of an infinite series is 2 + 4 + 6 + 8 + ... This series can also be written in summation notation as ∑ the terms are not tending to zero, the sum of the series increases without bound as we add more terms. Therefore, the sum of this infinite series is not defined. When the sum is not a real number, we say the series diverges. 2k , where the upper limit of summation is infinity. Because k = 1 ∞ Determining Whether the Sum of an Infinite Geometric Series is Defined If the terms of an infinite geometric series approach 0, the sum of an infinite geometric series can be defined. The terms in this series approach 0: 1 + 0.2 + 0.04 + 0.008 + 0.0016 + ... SECTION 11.4 series and their notations 975 The common ratio r = 0.2. As n gets very large, the values of rn get very small and approach 0. Each successive term affects the sum less than the preceding term. As each succeeding term gets closer to 0, the sum of the terms approaches a finite value. The terms of any infinite geometric series with −1 < r < 1 approach 0; the sum of a geometric series is defined when −1 < r < 1. determining whether the sum of an infinite geometric series is defined The sum of an infinite series is defined if the series is geometric and −1 < r < 1. How To… Given the first several terms of an infinite series, determine if the sum of the series exists. 1. Find the ratio of the second term to the first term. 2. Find the ratio of the third term to the second term. 3. Continue this process to ensure the ratio of a term to the preceding term is constant throughout. If so, the series is geometric. 4. If a common ratio, r, was found in step 3, check to see if −1 < r < 1 . If so, the sum is defined. If not, the sum is not defined. Example 6 Determining Whether the Sum of an Infinite Series is Defined Determine whether the sum of each infinite series is defined. 1 1 3 __ __ __ + ... + + b. 3 2 4 ∞ c. ∑ k 1  27 ⋅  __ 3 k = 1 ∞ d. ∑ 5k k = 1 a. 12 + 8 + 4 + … Solution 1 2 __ __ . , which is not the same as the ratio of the third term to the second, a. The ratio of the second term to the first is 2 3 The series is not geometric. b. The ratio of the second term to the first is the same as the ratio of the third term to the second. The series is 2 __ . The sum of the infinite series is defined. geometric with a common ratio of 3 1 1 __ __ . The sum ; the series is geometric with a common ratio of c. The given formula is exponential with a base of 3 3 of the infinite series is defined. d. The given formula is not exponential; the series is not geometric because the terms are increasing, and so cannot yield a finite sum. Try It #9 Determine whether the sum of the infinite series is defined. 9 3 1 1 __ __ __ __ + ... + + + 8 4 2 3 Try It #10 Determine whether the sum of the infinite series is defined. 24 + (−12) + 6 + (−3) + ... Try It #11 Determine whether the sum of the infinite series is defined. ∞ ∑ k = 1 15 ⋅ (−0.3)k Finding Sums of Infinite Series When the sum of an infinite geometric series exists, we can calculate the sum. The formula for the sum of an infinite series is related to the formula for the sum of the first n terms of a geometric series. Sn = a1(1 − r n) _________ 1 − r 976 CHAPTER 11 seQuences, proBaBility and counting theory 1 __ We will examine an infinite series with r = . What happens to r n as n increases __ = 4 1 3 __ = 8 4 = 1 ___ 16 The value of rn decreases rapidly. What happens for greater values of n 10 = 1 _____ 1,024 20 = 1 ________ 1,048,576 30 = 1 ____________ 1,073,741,824 As n gets very large, r n gets very small. We say that, as n increases without bound, r n approaches 0. As r n approaches 0, 1 − r n approaches 1. When this happens, the numerator approaches a1. This give us a formula for the sum of an infinite geometric series. formula for the sum of an infinite geometric series The formula for the sum of an infinite geometric series with −1 < r < 1 is S = a1 _ 1 − r How To… Given an infinite geometric series, find its sum. 1. Identify a1 and r. 2. Confirm that −1 < r < 1. 3. Substitute values for a1 and r into the formula, S = 4. Simplify to find S. a1 _ . 1 − r Example 7 Finding the Sum of an Infinite Geometric Series Find the sum, if it exists, for the following: a. 10 + 9 + 8 + 7 + … b. 248.6 + 99.44 + 39.776 + … Solution ∞ c. ∑ 1  4,374 ⋅  − __ 3 k = 1 k − 1 ∞ d. ∑ k = 1 k 4 1  ⋅  __ __ 3 9 a. There is not a constant ratio; the series is not geometric. b. There is a constant ratio;
the series is geometric. a1 = 248.6 and r = 99.44 _____ 248.6 = 0.4, so the sum exists. Substitute a1 = 248.6 and r = 0.4 into the formula and simplify to find the sum: S = a1 _ 1 − r S = _ = 414. 3 248.6 _ 1 − 0.4 1 __ . Find a1 by substituting k = 1 into the given c. The formula is exponential, so the series is geometric with r = − 3 1 − 1 1  a1 = 4,374 ⋅  − __ 3 explicit formula: = 4,374 SECTION 11.4 series and their notations 977 1 __ Substitute a1 = 4,374 and r = − into the formula, and simplify to find the sum: 3 S = a1 _ 1 − r d. The formula is exponential, so the series is geometric, but r > 1. The sum does not exist. S = 4,374 __ = 3,280.5 1  1 −  − __ 3 Example 8 Finding an Equivalent Fraction for a Repeating Decimal _ 3 Find an equivalent fraction for the repeating decimal 0. _ 3 = 0.333... so we can rewrite the repeating decimal as a sum of terms. Solution We notice the repeating decimal 0. _ 3 = 0.3 + 0.03 + 0.003 + ... 0. Looking for a pattern, we rewrite the sum, noticing that we see the first term multiplied to 0.1 in the second term, and the second term multiplied to 0.1 in the third term. _ 3 = 0.3 + (0.1) 0. (0.3) + (0.1) (0.1)(0.3) } Notice the pattern; we multiply each consecutive term by a common ratio of 0.1 starting with the first term of 0.3. So, substituting into our formula for an infinite geometric sum, we have First Term Second Term S = a1 _ = 1 − r 0.3 ______ 1 − 0.1 = 0.3 ___ 0.9 1 ___ = . 3 Try It #12 Find the sum, if it exists. Try It #13 Find the sum, if it exists. Try It #14 Find the sum, if it exists. Solving Annuity Problems 2 2 __ __ + ... + .76k + 1 ∞ k ∑ 3   − __ 8 k = 1 At the beginning of the section, we looked at a problem in which a couple invested a set amount of money each month into a college fund for six years. An annuity is an investment in which the purchaser makes a sequence of periodic, equal payments. To find the amount of an annuity, we need to find the sum of all the payments and the interest earned. In the example, the couple invests $50 each month. This is the value of the initial deposit. The account paid 6% annual interest, compounded monthly. To find the interest rate per payment period, we need to divide the 6% annual percentage interest (APR) rate by 12. So the monthly interest rate is 0.5%. We can multiply the amount in the account each month by 100.5% to find the value of the account after interest has been added. We can find the value of the annuity right after the last deposit by using a geometric series with a1 = 50 and r = 100.5% = 1.005. After the first deposit, the value of the annuity will be $50. Let us see if we can determine the amount in the college fund and the interest earned. We can find the value of the annuity after n deposits using the formula for the sum of the first n terms of a geometric series. In 6 years, there are 72 months, so n = 72. We can substitute a1 = 50, r = 1.005, and n = 72 into the formula, and simplify to find the value of the annuity after 6 years. S72 = 50(1 − 1.00572) _____________ 1 − 1.005 ≈ 4,320.44 After the last deposit, the couple will have a total of $4,320.44 in the account. Notice, the couple made 72 payments of $50 each for a total of 72(50) = $3,600. This means that because of the annuity, the couple earned $720.44 interest in their college fund. 978 CHAPTER 11 seQuences, proBaBility and counting theory How To… Given an initial deposit and an interest rate, find the value of an annuity. 1. Determine a1, the value of the initial deposit. 2. Determine n, the number of deposits. 3. Determine r. a. Divide the annual interest rate by the number of times per year that interest is compounded. b. Add 1 to this amount to find r. 4. Substitute values for a1, r, and n into the formula for the sum of the first n terms of a geometric series, Sn = a1(1 – rn) _ . 1 – r 5. Simplify to find Sn, the value of the annuity after n deposits. Example 9 Solving an Annuity Problem A deposit of $100 is placed into a college fund at the beginning of every month for 10 years. The fund earns 9% annual interest, compounded monthly, and paid at the end of the month. How much is in the account right after the last deposit? Solution The value of the initial deposit is $100, so a1 = 100. A total of 120 monthly deposits are made in the 10 years, so n = 120. To find r, divide the annual interest rate by 12 to find the monthly interest rate and add 1 to represent the new monthly deposit. Substitute a1 = 100, r = 1.0075, and n = 120 into the formula for the sum of the first n terms of a geometric series, and simplify to find the value of the annuity. r = 1 + = 1.0075 0.09 ____ 12 S120 = 100(1 − 1.0075120) _______________ 1 − 1.0075 ≈ 19,351.43 So the account has $19,351.43 after the last deposit is made. Try It #15 At the beginning of each month, $200 is deposited into a retirement fund. The fund earns 6% annual interest, compounded monthly, and paid into the account at the end of the month. How much is in the account if deposits are made for 10 years? Access these online resources for additional instruction and practice with series. • Arithmetic Series (http://openstaxcollege.org/l/arithmeticser) • Geometric Series (http://openstaxcollege.org/l/geometricser) • Summation notation (http://openstaxcollege.org/l/sumnotation) SECTION 11.4 section exercises 979 2. What is the difference between an arithmetic sequence and an arithmetic series? 4. How is finding the sum of an infinite geometric series different from finding the nth partial sum? 11.4 SeCTIOn exeRCISeS VeRBAl 1. What is an nth partial sum? 3. What is a geometric series? 5. What is an annuity? AlGeBRAIC For the following exercises, express each description of a sum using summation notation. 6. The sum of terms m2 + 3m from m = 1 to m = 5 7. The sum from of n = 0 to n = 4 of 5n 8. The sum of 6k − 5 from k = −2 to k = 1 9. The sum that results from adding the number 4 five times For the following exercises, express each arithmetic sum using summation notation. 10. 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50 11. 10 + 18 + 26 + … + 162 3 1 ___ __ + 2 + … + 4 + 1 + 12. 2 2 For the following exercises, use the formula for the sum of the first n terms of each arithmetic sequence. 7 5 3 __ __ __ + 3 + + 2 + 13. 2 2 2 14. 19 + 25 + 31 + … + 73 15. 3.2 + 3.4 + 3.6 + … + 5.6 For the following exercises, express each geometric sum using summation notation. 16. 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 17.125 1 __ + 18. − 6 1 ___ 12 1 ___ 24 − + … + 1 ___ 768 For the following exercises, use the formula for the sum of the first n terms of each geometric sequence, and then state the indicated sum. 1 1 _ _ + 19. 9 + 3 + 1 + 9 3 11 64 ⋅ 0.2a − 1 9 5 ⋅ 2n − 1 21. ∑ 20. ∑ n = 1 a = 1 For the following exercises, determine whether the infinite series has a sum. If so, write the formula for the sum. If not, state the reason. ∞ 22. 12 + 18 + 24 + 30 + ... 23. 2 + 1.6 + 1.28 + 1.024 + ... 24. ∑ m = 1 4m − 1 ∞ 25. ∑ 1  −  − __ 2 k = 1 k − 1 980 CHAPTER 11 seQuences, proBaBility and counting theory GRAPHICAl For the following exercises, use the following scenario. Javier makes monthly deposits into a savings account. He opened the account with an initial deposit of $50. Each month thereafter he increased the previous deposit amount by $20. 26. Graph the arithmetic sequence showing one year of 27. Graph the arithmetic series showing the monthly Javier’s deposits. sums of one year of Javier’s deposits. For the following exercises, use the geometric series ∑ 1   __ 2 k . k = 1 ∞ 28. Graph the first 7 partial sums of the series. nUMeRIC For the following exercises, find the indicated sum. 29. What number does Sn seem to be approaching in the graph? Find the sum to explain why this makes sense. 14 30. ∑ a a = 1 6 n(n − 2) 31. ∑ n = 1 17 32. ∑ k2 k = 1 7 33. ∑ 2k k = 1 For the following exercises, use the formula for the sum of the first n terms of an arithmetic series to find the sum. 34. −1.7 + −0.4 + 0.9 + 2.2 + 3.5 + 4.8 36. −1 + 3 + 7 + ... + 31 35. 6 + + 9 + + 12 + 15 ___ 2 21 ___ 2 27 ___ + 15 2 11 37. ∑ 1 k   __ __ − 2 2 k = 1 For the following exercises, use the formula for the sum of the first n terms of a geometric series to find the partial sum. 38. S6 for the series −2 − 10 − 50 − 250 ... 39. S7 for the series 0.4 − 2 + 10 − 50 ... 40. ∑ 9 2k − 1 k = 1 10 41. ∑ n = 1 1  −2 ⋅  __ 2 n − 1 For the following exercises, find the sum of the infinite geometric series. 1 __ 42. 4 + 2 + 1 + ... 2 ∞ 45. ∑ n = 1 4.6 ⋅ 0.5n − 1 1 __ − 43. −1 − 4 1 ___ 16 − 1 ___ ... 64 k = 1 44. ∑ ∞ 1  3 ⋅  __ 4 k − 1 For the following exercises, determine the value of the annuity for the indicated monthly deposit amount, the number of deposits, and the interest rate. 46. Deposit amount: $50; total deposits: 60; interest rate: 47. Deposit amount: $150; total deposits: 24; interest 5%, compounded monthly rate: 3%, compounded monthly 48. Deposit amount: $450; total deposits: 60; interest 49. Deposit amount: $100; total deposits: 120; interest rate: 4.5%, compounded quarterly rate: 10%, compounded semi-annually exTenSIOnS 50. The sum of terms 50 − k 2 from k = x through 7 is 115. What is x? 52. Find the smallest value of n such that n (3k − 5) > 100 . ∑ k = 1 51. Write an explicit formula for ak such that ∑ ak = 189 . Assume this is an arithmetic series. 6 k = 0 53. How many terms must be added before the series −1 − 3 − 5 − 7.... has a sum less than −75? SECTION 11.4 section exercises 981 54. Write 0. _ 65 as an infinite geometric series using summation notation. Then use the formula for finding the sum of an infinite geometric series to convert 0.65 to a fraction. 55. The sum of an infinite geometric series is five times the value of the first term. What is the common ratio of the series? 56. To get the best loan rates available, the Riches 57. Karl has two years to save $10,000 to buy a used car when he graduates. To the nearest dollar, what would his monthly deposits need to be if he invests in an account o
ffering a 4.2% annual interest rate that compounds monthly? 59. A boulder rolled down a mountain, traveling 6 feet in the first second. Each successive second, its distance increased by 8 feet. How far did the boulder travel after 10 seconds? 61. A pendulum travels a distance of 3 feet on its first 3 __ the swing. On each successive swing, it travels 4 distance of the previous swing. What is the total distance traveled by the pendulum when it stops swinging? want to save enough money to place 20% down on a $160,000 home. They plan to make monthly deposits of $125 in an investment account that offers 8.5% annual interest compounded semiannually. Will the Riches have enough for a 20% down payment after five years of saving? How much money will they have saved? ReAl-WORlD APPlICATIOnS 58. Keisha devised a week-long study plan to prepare for finals. On the first day, she plans to study for 1 hour, and each successive day she will increase her study time by 30 minutes. How many hours will Keisha have studied after one week? 60. A scientist places 50 cells in a petri dish. Every hour, the population increases by 1.5%. What will the cell count be after 1 day? 62. Rachael deposits $1,500 into a retirement fund each year. The fund earns 8.2% annual interest, compounded monthly. If she opened her account when she was 19 years old, how much will she have by the time she is 55? How much of that amount will be interest earned? 982 CHAPTER 11 seQuences, proBaBility and counting theory leARnInG OBjeCTIVeS In this section, you will: • Solve counting problems using the Addition Principle. • Solve counting problems using the Multiplication Principle. • Solve counting problems using permutations involving n distinct objects. • Solve counting problems using combinations. • Find the number of subsets of a given set. • Solve counting problems using permutations involving n non-distinct objects. 11. 5 COUnTInG PRInCIPleS A new company sells customizable cases for tablets and smartphones. Each case comes in a variety of colors and can be personalized for an additional fee with images or a monogram. A customer can choose not to personalize or could choose to have one, two, or three images or a monogram. The customer can choose the order of the images and the letters in the monogram. The company is working with an agency to develop a marketing campaign with a focus on the huge number of options they offer. Counting the possibilities is challenging! We encounter a wide variety of counting problems every day. There is a branch of mathematics devoted to the study of counting problems such as this one. Other applications of counting include secure passwords, horse racing outcomes, and college scheduling choices. We will examine this type of mathematics in this section. Using the Addition Principle The company that sells customizable cases offers cases for tablets and smartphones. There are 3 supported tablet models and 5 supported smartphone models. The Addition Principle tells us that we can add the number of tablet options to the number of smartphone options to find the total number of options. By the Addition Principle, there are 8 total options, as we can see in Figure 1. Figure 1 the Addition Principle According to the Addition Principle, if one event can occur in m ways and a second event with no common outcomes can occur in n ways, then the first or second event can occur in m + n ways. Example 1 Using the Addition Principle There are 2 vegetarian entrée options and 5 meat entrée options on a dinner menu. What is the total number of entrée options? SECTION 11.5 counting principles 983 Solution We can add the number of vegetarian options to the number of meat options to find the total number of entrée options. Vegetarian + Vegetarian + Meat + Meat + Meat + Meat + Meat ↓ ↓ Option 1 + Option 2 + Option 3 + Option 4 + Option 5 + Option 6 + Option 7 ↓ ↓ ↓ ↓ ↓ There are 7 total options. Try It #1 A student is shopping for a new computer. He is deciding among 3 desktop computers and 4 laptop computers. What is the total number of computer options? Using the Multiplication Principle The Multiplication Principle applies when we are making more than one selection. Suppose we are choosing an appetizer, an entrée, and a dessert. If there are 2 appetizer options, 3 entrée options, and 2 dessert options on a fixedprice dinner menu, there are a total of 12 possible choices of one each as shown in the tree diagram in Figure 2. (Appetizers) Soup Salad (Entress) Chicken Fish Steak Chicken Fish Steak (Dessert) Cake Pudding Cake Pudding Cake Pudding Cake Pudding Cake Pudding Cake Pudding Figure 2 The possible choices are: 1. soup, chicken, cake 2. soup, chicken, pudding 3. soup, fish, cake 4. soup, fish, pudding 5. soup, steak, cake 6. soup, steak, pudding 7. salad, chicken, cake 8. salad, chicken, pudding 9. salad, fish, cake 10. salad, fish, pudding 11. salad, steak, cake 12. salad, steak, pudding We can also find the total number of possible dinners by multiplying. We could also conclude that there are 12 possible dinner choices simply by applying the Multiplication Principle. # of appetizer options × # of entree options × # of dessert options 2 × 3 × 2 = 12 984 CHAPTER 11 seQuences, proBaBility and counting theory the Multiplication Principle According to the Multiplication Principle, if one event can occur in m ways and a second event can occur in n ways after the first event has occurred, then the two events can occur in m × n ways. This is also known as the Fundamental Counting Principle. Example 2 Using the Multiplication Principle Diane packed 2 skirts, 4 blouses, and a sweater for her business trip. She will need to choose a skirt and a blouse for each outfit and decide whether to wear the sweater. Use the Multiplication Principle to find the total number of possible outfits. Solution To find the total number of outfits, find the product of the number of skirt options, the number of blouse options, and the number of sweater options. # of skirt options × # of blouse options × # of sweater options 2 × 4 × 2 = 16 There are 16 possible outfits. Try It #2 A restaurant offers a breakfast special that includes a breakfast sandwich, a side dish, and a beverage. There are 3 types of breakfast sandwiches, 4 side dish options, and 5 beverage choices. Find the total number of possible breakfast specials. Finding the number of Permutations of n Distinct Objects The Multiplication Principle can be used to solve a variety of problem types. One type of problem involves placing objects in order. We arrange letters into words and digits into numbers, line up for photographs, decorate rooms, and more. An ordering of objects is called a permutation. Finding the Number of Permutations of n Distinct Objects Using the Multiplication Principle To solve permutation problems, it is often helpful to draw line segments for each option. That enables us to determine the number of each option so we can multiply. For instance, suppose we have four paintings, and we want to find the number of ways we can hang three of the paintings in order on the wall. We can draw three lines to represent the three places on the wall. There are four options for the first place, so we write a 4 on the first line. × × 4 × × After the first place has been filled, there are three options for the second place so we write a 3 on the second line. 4 × 3 × After the second place has been filled, there are two options for the third place so we write a 2 on the third line. Finally, we find the product. 4 × 3 × 2 = 24 There are 24 possible permutations of the paintings. How To… Given n distinct options, determine how many permutations there are. 1. Determine how many options there are for the first situation. 2. Determine how many options are left for the second situation. 3. Continue until all of the spots are filled. 4. Multiply the numbers together. SECTION 11.5 counting principles 985 Example 3 Finding the Number of Permutations Using the Multiplication Principle At a swimming competition, nine swimmers compete in a race. a. How many ways can they place first, second, and third? b. How many ways can they place first, second, and third if a swimmer named Ariel wins first place? (Assume there is only one contestant named Ariel.) c. How many ways can all nine swimmers line up for a photo? Solution a. Draw lines for each place. options for 1st place × options for 2nd place × options for 3rd place There are 9 options for first place. Once someone has won first place, there are 8 remaining options for second place. Once first and second place have been won, there are 7 remaining options for third place. 9 × 8 × 7 = 504 Multiply to find that there are 504 ways for the swimmers to place. b. Draw lines for describing each place. options for 1st place × options for 2nd place × options for 3rd place We know Ariel must win first place, so there is only 1 option for first place. There are 8 remaining options for second place, and then 7 remaining options for third place. 1 × 8 × 7 = 56 Multiply to find that there are 56 ways for the swimmers to place if Ariel wins first. c. Draw lines for describing each place in the photo. × × × × × × × × There are 9 choices for the first spot, then 8 for the second, 7 for the third, 6 for the fourth, and so on until only 1 person remains for the last spot × × × × × = 362,880 × × × There are 362,880 possible permutations for the swimmers to line up. Analysis Note that in part c, we found there were 9! ways for 9 people to line up. The number of permutations of n distinct objects can always be found by n!. Try It #3 A family of five is having portraits taken. Use the Multiplication Principle to find how many ways the family can line up for the portrait. Try It #4 A family of five is having portraits taken. Use the Multiplication Principle to find how many ways the photographer can line up 3 of the family members. Try It #5 A family of five is having portraits taken. Use the Mu
ltiplication Principle to find how many ways the family can line up for the portrait if the parents are required to stand on each end. 986 CHAPTER 11 seQuences, proBaBility and counting theory Finding the Number of Permutations of n Distinct Objects Using a Formula For some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. Fortunately, we can solve these problems using a formula. Before we learn the formula, let’s look at two common notations for permutations. If we have a set of n objects and we want to choose r objects from the set in order, we write P (n, r). Another way to write this is nPr, a notation commonly seen on computers and calculators. To calculate P(n, r), we begin by finding n!, the number of ways to line up all n objects. We then divide by (n − r)! to cancel out the (n − r) items that we do not wish to line up. Let’s see how this works with a simple example. Imagine a club of six people. They need to elect a president, a vice president, and a treasurer. Six people can be elected president, any one of the five remaining people can be elected vice president, and any of the remaining four people could be elected treasurer. The number of ways this may be done is 6 × 5 × 4 = 120. Using factorials, we get the same result. 6 · 5 · 4 · 3! _________ 3! = 6 · 5 · 4 = 120 6! __ 3! = There are 120 ways to select 3 officers in order from a club with 6 members. We refer to this as a permutation of 6 taken 3 at a time. The general formula is as follows. P(n, r) = n! _______ (n − r)! Note that the formula stills works if we are choosing all n objects and placing them in order. In that case we would be dividing by (n − n)! or 0!, which we said earlier is equal to 1. So the number of permutations of n objects taken n at a time is n! __ or just n!. 1 formula for permutations of n distinct objects Given n distinct objects, the number of ways to select r objects from the set in order is P(n, r) = n! _______ (n − r)! How To… Given a word problem, evaluate the possible permutations. 1. Identify n from the given information. 2. Identify r from the given information. 3. Replace n and r in the formula with the given values. 4. Evaluate. Example 4 Finding the Number of Permutations Using the Formula A professor is creating an exam of 9 questions from a test bank of 12 questions. How many ways can she select and arrange the questions? Solution Substitute n = 12 and r = 9 into the permutation formula and simplify. P(n, r) = n! _______ (n − r)! 12! ________ = (12 − 9)! 12! ___ 3! P(12, 9) = = 79,833,600 There are 79,833,600 possible permutations of exam questions! Analysis We can also use a calculator to find permutations. For this problem, we would enter 15, press the [nPr function], enter [12], and then press the equal sign. The [nPr function] may be located under the [MATH] menu with probability commands. Q & A… Could we have solved Example 4 using the Multiplication Principle? Yes. We could have multiplied 15 ⋅ 14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ to find the same answer. SECTION 11.5 counting principles 987 Try It #6 A play has a cast of 7 actors preparing to make their curtain call. Use the permutation formula to find how many ways the 7 actors can line up. Try It #7 A play has a cast of 7 actors preparing to make their curtain call. Use the permutation formula to find how many ways 5 of the 7 actors can be chosen to line up. Find the number of Combinations Using the Formula So far, we have looked at problems asking us to put objects in order. There are many problems in which we want to select a few objects from a group of objects, but we do not care about the order. When we are selecting objects and the order does not matter, we are dealing with combinations. A selection of r objects from a set of n objects where the order does not matter can be written as C (n, r). Just as with permutations, C(n, r) can also be written as nCr. In this case, the general formula is as follows. C (n, r) = n! _________ r!(n − r)! An earlier problem considered choosing 3 of 4 possible paintings to hang on a wall. We found that there were 24 ways to select 3 of the 4 paintings in order. But what if we did not care about the order? We would expect a smaller number because selecting paintings 1, 2, 3 would be the same as selecting paintings 2, 3, 1. To find the number of ways to select 3 of the 4 paintings, disregarding the order of the paintings, divide the number of permutations by the number of ways to order 3 paintings. There are 3! = 3 · 2 · 1 = 6 ways to order 3 paintings. There are 3 of the 4 paintings. , or 4 ways to select 24 ___ 6 This number makes sense because every time we are selecting 3 paintings, we are not selecting 1 painting. There are 4 paintings we could choose not to select, so there are 4 ways to select 3 of the 4 paintings. formula for combinations of n distinct objects Given n distinct objects, the number of ways to select r objects from the set is C (n, r) = n! _________ r!(n − r)! How To… Given a number of options, determine the possible number of combinations. 1. Identify n from the given information. 2. Identify r from the given information. 3. Replace n and r in the formula with the given values. 4. Evaluate. Example 5 Finding the Number of Combinations Using the Formula A fast food restaurant offers five side dish options. Your meal comes with two side dishes. a. How many ways can you select your side dishes? b. How many ways can you select 3 side dishes? Solution a. We want to choose 2 side dishes from 5 options. C(5, 2) = 5! ________ 2!(5 − 2)! = 10 b. We want to choose 3 side dishes from 5 options. C(5, 3) = 5! ________ 3!(5 − 3)! = 10 988 CHAPTER 11 seQuences, proBaBility and counting theory Analysis We can also use a graphing calculator to find combinations. Enter 5, then press nCr, enter 3, and then press the equal sign. The nCr function may be located under the MATH menu with probability commands. Q & A… Is it a coincidence that parts ( a) and ( b) in Example 5 have the same answers? No. When we choose r objects from n objects, we are not choosing (n − r) objects. Therefore, C(n, r) = C(n, n − r). Try It #8 An ice cream shop offers 10 flavors of ice cream. How many ways are there to choose 3 flavors for a banana split? Finding the number of Subsets of a Set We have looked only at combination problems in which we chose exactly r objects. In some problems, we want to consider choosing every possible number of objects. Consider, for example, a pizza restaurant that offers 5 toppings. Any number of toppings can be ordered. How many different pizzas are possible? To answer this question, we need to consider pizzas with any number of toppings. There is C(5, 0) = 1 way to order a pizza with no toppings. There are C(5, 1) = 5 ways to order a pizza with exactly one topping. If we continue this process, we get C(5, 0) + C(5, 1) + C(5, 2) + C(5, 3) + C(5, 4) + C(5, 5) = 32 There are 32 possible pizzas. This result is equal to 25. We are presented with a sequence of choices. For each of the n objects we have two choices: include it in the subset or not. So for the whole subset we have made n choices, each with two options. So there are a total of 2 · 2 · 2 · … · 2 possible resulting subsets, all the way from the empty subset, which we obtain when we say “no” each time, to the original set itself, which we obtain when we say “yes” each time. formula for the number of subsets of a set A set containing n distinct objects has 2n subsets. Example 6 Finding the Number of Subsets of a Set A restaurant offers butter, cheese, chives, and sour cream as toppings for a baked potato. How many different ways are there to order a potato? Solution We are looking for the number of subsets of a set with 4 objects. Substitute n = 4 into the formula. There are 16 possible ways to order a potato. 2n = 24 = 16 Try It #9 A sundae bar at a wedding has 6 toppings to choose from. Any number of toppings can be chosen. How many different sundaes are possible? Finding the number of Permutations of n non-Distinct Objects We have studied permutations where all of the objects involved were distinct. What happens if some of the objects are indistinguishable? For example, suppose there is a sheet of 12 stickers. If all of the stickers were distinct, there would be 12! ways to order the stickers. However, 4 of the stickers are identical stars, and 3 are identical moons. Because all of the objects are not distinct, many of the 12! permutations we counted are duplicates. The general formula for this situation is as follows. SECTION 11.5 counting principles 989 n! __________ r1! r2 ! … rk ! In this example, we need to divide by the number of ways to order the 4 stars and the ways to order the 3 moons to find the number of unique permutations of the stickers. There are 4! ways to order the stars and 3! ways to order the moon. There are 3,326,400 ways to order the sheet of stickers. 12! ____ 4!3! = 3,326,400 formula for finding the number of permutations of n non-distinct objects If there are n elements in a set and r1 are alike, r2 are alike, r3 are alike, and so on through rk , the number of permutations can be found by n! __________ r1! r2! … rk ! Example 7 Finding the Number of Permutations of n Non-Distinct Objects Find the number of rearrangements of the letters in the word DISTINCT. Solution There are 8 letters. Both I and T are repeated 2 times. Substitute n = 8, r1 = 2, and r2 = 2 into the formula. There are 10,080 arrangements. 8! ____ 2!2! = 10,080 Try It #10 Find the number of rearrangements of the letters in the word CARRIER. Access these online resources for additional instruction and practice with combinations and permutations. • Combinations (http://openstaxcollege.org/l/combinations) • Permutations (http://openstaxcollege.org/l/permutations) 990 CHAPTER 11 seQuences, proBaBility and counting theory 11.5 SeCTIOn exeRCISeS VeRBAl For the following exercises, assume that there
are n ways an event A can happen, m ways an event B can happen, and that A and B are non-overlapping. 1. Use the Addition Principle of counting to explain 2. Use the Multiplication Principle of counting to how many ways event A or B can occur. explain how many ways event A and B can occur. Answer the following questions. 3. When given two separate events, how do we know whether to apply the Addition Principle or the Multiplication Principle when calculating possible outcomes? What conjunctions may help to determine which operations to use? 5. What is the term for the arrangement that selects r objects from a set of n objects when the order of the r objects is not important? What is the formula for calculating the number of possible outcomes for this type of arrangement? 4. Describe how the permutation of n objects differs from the permutation of choosing r objects from a set of n objects. Include how each is calculated. nUMeRIC For the following exercises, determine whether to use the Addition Principle or the Multiplication Principle. Then perform the calculations. 6. Let the set A = { −5, −3, −1, 2, 3, 4, 5, 6}. How many ways are there to choose a negative or an even number from A? 7. Let the set B = { −23, −16, −7, −2, 20, 36, 48, 72}. How many ways are there to choose a positive or an odd number from A? 8. How many ways are there to pick a red ace or a club from a standard card playing deck? 9. How many ways are there to pick a paint color from 5 shades of green, 4 shades of blue, or 7 shades of yellow? 10. How many outcomes are possible from tossing a pair 11. How many outcomes are possible from tossing a coin of coins? and rolling a 6-sided die? 12. How many two-letter strings—the first letter from A and the second letter from B—can be formed from the sets A = {b, c, d } and B = {a, e, i, o, u}? 14. How many ways are there to construct a string of 3 digits if numbers cannot be repeated? 13. How many ways are there to construct a string of 3 digits if numbers can be repeated? For the following exercises, compute the value of the expression. 15. P(5, 2) 20. C(8, 5) 16. P(8, 4) 17. P(3, 3) 21. C(12, 4) 22. C(26, 3) 18. P(9, 6) 23. C(7, 6) 19. P(11, 5) 24. C(10, 3) For the following exercises, find the number of subsets in each given set. 25. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 27. A set containing 5 distinct numbers, 4 distinct 26. {a, b, c, … , z} 28. The set of even numbers from 2 to 28 letters, and 3 distinct symbols 29. The set of two-digit numbers between 1 and 100 containing the digit 0 For the following exercises, find the distinct number of arrangements. 30. The letters in the word “juggernaut” 32. The letters in the word “academia” that begin and 31. The letters in the word “academia” 33. The symbols in the string #,#,#,@,@,$,$,$,%,%,%,% end in “a” 34. The symbols in the string #,#,#,@,@,$,$,$,%,%,%,% that begin and end with “%” SECTION 11.5 section exercises 991 exTenSIOnS 35. The set, S consists of 900,000,000 whole numbers, each being the same number of digits long. How many digits long is a number from S? (Hint: use the fact that a whole number cannot start with the digit 0.) 36. The number of 5-element subsets from a set containing n elements is equal to the number of 6-element subsets from the same set. What is the value of n? (Hint: the order in which the elements for the subsets are chosen is not important.) 37. Can C(n, r) ever equal P(n, r)? Explain. 38. Suppose a set A has 2,048 subsets. How many distinct objects are contained in A? 39. How many arrangements can be made from the letters of the word “mountains” if all the vowels must form a string? ReAl-WORlD APPlICATIOnS 40. A family consisting of 2 parents and 3 children is 41. A cell phone company offers 6 different voice to pose for a picture with 2 family members in the front and 3 in the back. a. How many arrangements are possible with no restrictions? b. How many arrangements are possible if the parents must sit in the front? c. How many arrangements are possible if the parents must be next to each other? packages and 8 different data packages. Of those, 3 packages include both voice and data. How many ways are there to choose either voice or data, but not both? 42. In horse racing, a “trifecta” occurs when a bettor 43. A wholesale T-shirt company offers sizes small, wins by selecting the first three finishers in the exact order (1st place, 2nd place, and 3rd place). How many different trifectas are possible if there are 14 horses in a race? medium, large, and extra-large in organic or nonorganic cotton and colors white, black, gray, blue, and red. How many different T-shirts are there to choose from? 44. Hector wants to place billboard advertisements throughout the county for his new business. How many ways can Hector choose 15 neighborhoods to advertise in if there are 30 neighborhoods in the county? 46. How many ways can a committee of 3 freshmen and 4 juniors be formed from a group of 8 freshmen and 11 juniors? 48. A conductor needs 5 cellists and 5 violinists to play at a diplomatic event. To do this, he ranks the orchestra’s 10 cellists and 16 violinists in order of musical proficiency. What is the ratio of the total cellist rankings possible to the total violinist rankings possible? 50. A skateboard shop stocks 10 types of board decks, 3 types of trucks, and 4 types of wheels. How many different skateboards can be constructed? 52. A car wash offers the following optional services to the basic wash: clear coat wax, triple foam polish, undercarriage wash, rust inhibitor, wheel brightener, air freshener, and interior shampoo. How many washes are possible if any number of options can be added to the basic wash? 54. How many unique ways can a string of Christmas lights be arranged from 9 red, 10 green, 6 white, and 12 gold color bulbs? 45. An art store has 4 brands of paint pens in 12 different colors and 3 types of ink. How many paint pens are there to choose from? 47. How many ways can a baseball coach arrange the order of 9 batters if there are 15 players on the team? 49. A motorcycle shop has 10 choppers, 6 bobbers, and 5 café racers—different types of vintage motorcycles. How many ways can the shop choose 3 choppers, 5 bobbers, and 2 café racers for a weekend showcase? 51. Just-For-Kicks Sneaker Company offers an online customizing service. How many ways are there to design a custom pair of Just-For-Kicks sneakers if a customer can choose from a basic shoe up to 11 customizable options? 53. Susan bought 20 plants to arrange along the border of her garden. How many distinct arrangements can she make if the plants are comprised of 6 tulips, 6 roses, and 8 daisies? 992 CHAPTER 11 seQuences, proBaBility and counting theory leARnInG OBjeCTIVeS In this section, you will: • Apply the Binomial Theorem. 11. 6 BInOMIAl THeOReM A polynomial with two terms is called a binomial. We have already learned to multiply binomials and to raise binomials to powers, but raising a binomial to a high power can be tedious and time-consuming. In this section, we will discuss a shortcut that will allow us to find (x + y)n without multiplying the binomial by itself n times. Identifying Binomial Coefficients In Counting Principles, we studied combinations. In the shortcut to finding (x + y)n, we will need to use combinations n to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation  _ r  instead of C(n, r), but it can be calculated in the same way. So n 5 r  is called a binomial coefficient. An example of a binomial coefficient is  The combination  _  = C(5, 2) = 10. __ 2 n  _ r  = C(n, r) = n! ________ r!(n − r)! binomial coefficients If n and r are integers greater than or equal to 0 with n ≥ r, then the binomial coefficient is n  _ r  = C(n, r) = n! ________ r!(n − r)! Q & A… Is a binomial coefficient always a whole number? Yes. Just as the number of combinations must always be a whole number, a binomial coefficient will always be a whole number. Example 1 Finding Binomial Coefficients Find each binomial coefficient. 9  b.  __ 2 5  a.  __ 3 Solution 9  c.  __ 7 Use the formula to calculate each binomial coefficient. You can also use the nCr function on your calculator. n  _ r  = C(n, r) = n! ________ r!(n − r)! 5  = a.  __ 3 5! ________ = 3!(5 − 3)! 9  = b.  __ 2 9  = c.  __ 7 9! ________ = 2!(9 − 2)! 9! ________ = 7!(9 − 7)! = 10 5 ⋅ 4 ⋅ 3! _______ 3!2! 9 ⋅ 8 ⋅ 7! _______ 2!7! 9 ⋅ 8 ⋅ 7! _______ 7!2! = 36 = 36 Analysis Notice that we obtained the same result for parts (b) and (c). If you look closely at the solution for these two parts, you will see that you end up with the same two factorials in the denominator, but the order is reversed, just as with combinations  SECTION 11.6 Binomial theorem 993 Try It #1 Find each binomial coefficient. 7  a.  __ 3 b.  11  __ 4 Using the Binomial Theorem When we expand (x + y)n by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. If we wanted to expand (x + y)52, we might multiply (x + y) by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions. (x + y)2 = x 2 + 2xy + y 2 (x + y)3 = x 3 + 3x 2 y + 3xy 2 + y 3 (x + y)4 = x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + y 4 First, let’s examine the exponents. With each successive term, the exponent for x decreases and the exponent for y increases. The sum of the two exponents is n for each term. Next, let’s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern  .  , ...,   ,   ,   2 1 0 These patterns lead us to the Binomial Theorem, which can be used to expand any binomial. n (x + y)n = ∑ n  xn − kyk  _ k k = 0 n n _ _  xn − 2 y 2 + ... +   xn − 1 y +  = xn +  2 1 n _ n − 1  xy n − 1 + yn Another way to see the co
efficients is to examine the expansion of a binomial in general form, x + y, to successive powers 1, 2, 3, and 4. (x + y)1 = x + y (x + y)2 = x2 + 2xy + y2 (x + y)3 = x3 + 3x2 y + 3xy2 + y3 (x + y)4 = x4 + 4x3 y + 6x2 y2 + 4xy3 + y4 Can you guess the next expansion for the binomial (x + y)5? Pascal’s Triangle Exponent (x + y) 1 = x + y (x + y) 2 = x 2 + 2xy + y2 (x + y) 3 = x 3 + 3x 2y + 3xy2 + y3 (x + y) 4 = x 4 + 4x 3y + 6x 2y2 + 4xy3 + y4 1 2 3 4 n Pattern # of Terms 2 3 4 5 n + 1 Exponent sum: Exponents on x : Exponents on y: 4+0 xy 4 0 3+1 xy 3 1 2+2 xy 2 2 1+3 xy 1 3 0+4 xy 0 4 Figure 1 994 CHAPTER 11 seQuences, proBaBility and counting theory See Figure 1, which illustrates the following: • There are n + 1 terms in the expansion of (x + y)n. • The degree (or sum of the exponents) for each term is n. • The powers on x begin with n and decrease to 0. • The powers on y begin with 0 and increase to n. • The coefficients are symmetric. To determine the expansion on (x + y)5, we see n = 5, thus, there will be 5 + 1 = 6 terms. Each term has a combined degree of 5. In descending order for powers of x, the pattern is as follows: • Introduce x 5, and then for each successive term reduce the exponent on x by 1 until x 0 = 1 is reached. • Introduce y 0 = 1, and then increase the exponent on y by 1 until y 5 is reached. The next expansion would be x5, x 4y, x3y 2, x 2y 3, xy 4, y 5 (x + y)5 = x5 + 5x 4y + 10x 3y2 + 10x 2y 3 + 5xy 4 + y 5. But where do those coefficients come from? The binomial coefficients are symmetric. We can see these coefficients in an array known as Pascal's Triangle, shown in Figure 2 Pascal’s Triangle 10 10 Figure 2 1 1 5 1 6 + 4 = 10 To generate Pascal’s Triangle, we start by writing a 1. In the row below, row 2, we write two 1’s. In the 3rd row, flank the ends of the rows with 1’s, and add 1 + 1 to find the middle number, 2. In the nth row, flank the ends of the row with 1’s. Each element in the triangle is the sum of the two elements immediately above it. To see the connection between Pascal’s Triangle and binomial coefficients, let us revisit the expansion of the binomials in general form. 1 → (x + y)0 = 1 1 → (x + y)x + y)2 = x 2 + 2xy + y 2 1 → (x + y)3 = x 3 + 3x 2y + 3xy2 + y 3 1 → (x + y)4 = x4 + 4x 3y + 6x 2y 2 + 4xy 3 + y 4 10 10 5 1 → (x + y)5 = x 5 + 5x 4y + 10x 3y 2 + 10x 2 y 3 + 5xy 4 + y 5 the Binomial Theorem The Binomial Theorem is a formula that can be used to expand any binomial. n (x + y)n = ∑ n  x n − k yk  − 2y 2 + ... +    xy n − 1 + y n SECTION 11.6 Binomial theorem 995 How To… Given a binomial, write it in expanded form. 1. Determine the value of n according to the exponent. 2. Evaluate the k = 0 through k = n using the Binomial Theorem formula. 3. Simplify. Example 2 Expanding a Binomial Write in expanded form. a. (x + y)5 b. (3x − y)4 Solution a. Substitute n = 5 into the formula. Evaluate the k = 0 through k = 5 terms. Simplify  x1y 4 +   x 2y 3 +   x 3y 2 +   x 4y1 +   x 5y 0 +   x 0y 5 (x + y)x + y)5 = x 5 + 5x 4y + 10x 3y 2 + 10x 2y 3 + 5xy 4 + y 5 b. Substitute n = 4 into the formula. Evaluate the k = 0 through k = 4 terms. Notice that 3x is in the place that was occupied by x and that −y is in the place that was occupied by y. So we substitute them. Simplify3x)0(−y)4  (3x)1(−y)3 +   (3x)2(−y)2 +   (3x)3(−y)1 +   (3x)4(−y)0 +  (3x − y)4 =  4 3 2 1 0 (3x − y)4 = 81x 4 − 108x 3y + 54x 2y 2 − 12xy 3 + y 4 Analysis Notice the alternating signs in part b. This happens because (−y) raised to odd powers is negative, but (−y) raised to even powers is positive. This will occur whenever the binomial contains a subtraction sign. Try It #2 Write in expanded form. a. (x − y)5 b. (2x + 5y)3 Using the Binomial Theorem to Find a Single Term Expanding a binomial with a high exponent such as (x + 2y)16 can be a lengthy process. Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term. Note the pattern of coefficients in the expansion of (x + y)5. 5 5 5 5 _ _ _ _  xy 4 + y 5  x 2y 3 +   x3y2 +   x 4y +  (x + y) 3y 2. We can generalize this result.  x 4y. The third term is  The second term is  2 1 n _ r  xn − ry r  the (r + 1)th term of a binomial expansion The (r + 1)th term of the binomial expansion of (x + y)n is: n _ r  x n − ry r  996 CHAPTER 11 seQuences, proBaBility and counting theory How To… Given a binomial, write a specific term without fully expanding. 1. Determine the value of n according to the exponent. 2. Determine (r + 1). 3. Determine r. 4. Replace r in the formula for the (r + 1)th term of the binomial expansion. Example 3 Writing a Given Term of a Binomial Expansion Find the tenth term of (x + 2y)16 without fully expanding the binomial. Solution Because we are looking for the tenth term, r + 1 = 10, we will use r = 9 in our calculations. n _ r  xn − r yr  16 _  x16 − 9(2y)9 = 5,857,280x7y 9  9 Try It #3 Find the sixth term of (3x − y)9 without fully expanding the binomial. Access these online resources for additional instruction and practice with binomial expansion. • The Binomial Theorem (http://openstaxcollege.org/l/binomialtheorem) • Binomial Theorem example (http://openstaxcollege.org/l/btexample) SECTION 11.6 section exercises 997 11.6 SeCTIOn exeRCISeS VeRBAl 1. What is a binomial coefficient, and how it is 2. What role do binomial coefficients play in a calculated? binomial expansion? Are they restricted to any type of number? 3. What is the Binomial Theorem and what is its use? 4. When is it an advantage to use the Binomial Theorem? Explain. AlGeBRAIC For the following exercises, evaluate the binomial coefficient. 6 _  5.  2 25 _  10.  11 5 _  6.  3 17 _  11.  6 7 _  7.  4 200 _  12.  199 9 _  8.  7 9.  10 _  9 For the following exercises, use the Binomial Theorem to expand each binomial. 13. (4a − b)3 14. (5a + 2)3 18. (3x − 2y)4 19. (4x − 3y)5 15. (3a + 2b)3 1 x + 3y  20.  _ 5 16. (2x + 3y)4 17. (4x + 2y)5 21. (x−1 + 2y −1)4 225 For the following exercises, use the Binomial Theorem to write the first three terms of each binomial. 23. (a + b)17 28. (2a + 4b)7 24. (x − 1)18 25. (a − 2b)15 26. (x − 2y)8 27. (3a + b)20 29. (x3 − √ — y )8 For the following exercises, find the indicated term of each binomial without fully expanding the binomial. 30. The fourth term of (2x − 3y)4 32. The third term of (6x − 3y)7 34. The seventh term of (a + b)11 36. The tenth term of (x − 1)12 1  38. The fourth term of  x 3 − __ 2 10 GRAPHICAl 31. The fourth term of (3x − 2y)5 33. The eighth term of (7 + 5y)14 35. The fifth term of (x − y)7 37. The ninth term of (a − 3b 2)11 y 9 x  39. The eighth term of  2 _ _ + 2 For the following exercises, use the Binomial Theorem to expand the binomial f (x) = (x + 3)4. Then find and graph each indicated sum on one set of axes. 40. Find and graph f1(x), such that f1(x) is the first term 41. Find and graph f2(x), such that f2(x) is the sum of the of the expansion. first two terms of the expansion. 42. Find and graph f3(x), such that f3(x) is the sum of the 43. Find and graph f4(x), such that f4(x) is the sum of the first three terms of the expansion. first four terms of the expansion. 44. Find and graph f5(x), such that f5(x) is the sum of the first five terms of the expansion. 998 CHAPTER 11 seQuences, proBaBility and counting theory exTenSIOnS 45. In the expansion of (5x + 3y)n, each term has the n _  an − kbk, where k successively takes on form  k 7 n _ _  , what is the  =  the value 0, 1, 2, ..., n. If  2 k corresponding term? 47. Consider the expansion of (x + b)40. What is the exponent of b in the kth term? 49. Which expression cannot be expanded using the 46. In the expansion of (a + b)n, the coefficient of a n − kbk is the same as the coefficient of which other term? 48. Find  n _ k − 1 n _  +   and write the answer as a k n _ binomial coefficient in the form   . Prove it. Hint: k Use the fact that, for any integer p, such that p ≥ 1, p! = p(p − 1)!. — Binomial Theorem? Explain. a. (x 2 − 2x + 1) b. ( √ c. (x 3 + 2y 2 − z)5 2y 3 )12 d. (3x 2 − √ a − 5)8 a + 4 √ — — SECTION 11.7 proBaBility 999 leARnInG OBjeCTIVeS In this section, you will: • Construct probability models. • Compute probabilities of equally likely outcomes. • Compute probabilities of the union of two events. • Use the complement rule to find probabilities. • Compute probability using counting theory. 11.7 PROBABIlITY Figure 1 An example of a “spaghetti model,” which can be used to predict possible paths of a tropical storm.[34] Residents of the Southeastern United States are all too familiar with charts, known as spaghetti models, such as the one in Figure 1. They combine a collection of weather data to predict the most likely path of a hurricane. Each colored line represents one possible path. The group of squiggly lines can begin to resemble strands of spaghetti, hence the name. In this section, we will investigate methods for making these types of predictions. Constructing Probability Models Suppose we roll a six-sided number cube. Rolling a number cube is an example of an experiment, or an activity with an observable result. The numbers on the cube are possible results, or outcomes, of this experiment. The set of all possible outcomes of an experiment is called the sample space of the experiment. The sample space for this experiment is {1, 2, 3, 4, 5, 6}. An event is any subset of a sample space. The likelihood of an event is known as probability. The probability of an event p is a number that always satisfies 0 ≤ p ≤ 1, where 0 indicates an impossible event and 1 indicates a certain event. A probability model is a mathematical description of an experiment listing all possible outcomes and their associated probabilities. For instance, if there is a 1% chance of winning a raffle and a 99% chance of losing the raffle, a probability model would look much like Table 1. Outcome Winning the raffle L
osing the raffle Probability 1% 99% Table 1 The sum of the probabilities listed in a probability model must equal 1, or 100%. 34 The figure is for illustrative purposes only and does not model any particular storm. 1000 CHAPTER 11 seQuences, proBaBility and counting theory How To… Given a probability event where each event is equally likely, construct a probability model. 1. Identify every outcome. 2. Determine the total number of possible outcomes. 3. Compare each outcome to the total number of possible outcomes. Example 1 Constructing a Probability Model Construct a probability model for rolling a single, fair die, with the event being the number shown on the die. Solution Begin by making a list of all possible outcomes for the experiment. The possible outcomes are the numbers that can be rolled: 1, 2, 3, 4, 5, and 6. There are six possible outcomes that make up the sample space. Assign probabilities to each outcome in the sample space by determining a ratio of the outcome to the number of possible outcomes. There is one of each of the six numbers on the cube, and there is no reason to think that any 1 __ . particular face is more likely to show up than any other one, so the probability of rolling any number is 6 Outcome Probability Roll of 1 1 __ 6 Roll of 2 1 __ 6 Roll of 3 1 __ 6 Roll of 4 1 __ 6 Roll of 5 1 __ 6 Roll of 6 1 __ 6 Table 2 Q & A… Do probabilities always have to be expressed as fractions? No. Probabilities can be expressed as fractions, decimals, or percents. Probability must always be a number between 0 and 1, inclusive of 0 and 1. Try It #1 Construct a probability model for tossing a fair coin. Computing Probabilities of equally likely Outcomes Let S be a sample space for an experiment. When investigating probability, an event is any subset of S. When the outcomes of an experiment are all equally likely, we can find the probability of an event by dividing the number of outcomes in the event by the total number of outcomes in S. Suppose a number cube is rolled, and we are interested in finding the probability of the event “rolling a number less than or equal to 4.” There are 4 possible outcomes in the 2 4 __ __ = . event and 6 possible outcomes in S, so the probability of the event is 3 6 computing the probability of an event with equally likely outcomes The probability of an event E in an experiment with sample space S with equally likely outcomes is given by P(E) = number of elements in E ____________________ = number of elements in S n(E) ____ n(S) E is a subset of S, so it is always true that 0 ≤ P(E) ≤ 1. Example 2 Computing the Probability of an Event with Equally Likely Outcomes A six-sided number cube is rolled. Find the probability of rolling an odd number. Solution The event “rolling an odd number” contains three outcomes. There are 6 equally likely outcomes in the sample space. Divide to find the probability of the event. 3 1 __ __ P(E) = = 2 6 SECTION 11.7 proBaBility 1001 Try It #2 A six-sided number cube is rolled. Find the probability of rolling a number greater than 2. Computing the Probability of the Union of Two events We are often interested in finding the probability that one of multiple events occurs. Suppose we are playing a card game, and we will win if the next card drawn is either a heart or a king. We would be interested in finding the probability of the next card being a heart or a king. The union of two events E and F, written E ∪ F, is the event that occurs if either or both events occur. P(E ∪ F) = P(E) + P(F) − P(E ∩ F) Suppose the spinner in Figure 2 is spun. We want to find the probability of spinning orange or spinning a b. a b d a c b Figure 2 1 3 __ __ = . There There are a total of 6 sections, and 3 of them are orange. So the probability of spinning orange is 2 6 1 2 __ __ = . If we added these two are a total of 6 sections, and 2 of them have a b. So the probability of spinning a b is 3 6 probabilities, we would be counting the sector that is both orange and a b twice. To find the probability of spinning an orange or a b, we need to subtract the probability that the sector is both orange and has a b. 2 __ The probability of spinning orange or a b is . 3 2 1 1 1 __ __ __ __ = − + 3 6 3 2 probability of the union of two events The probability of the union of two events E and F (written E ∪ F) equals the sum of the probability of E and the probability of F minus the probability of E and F occurring together (which is called the intersection of E and F and is written as E ∩ F). P(E ∪ F) = P(E) + P(F) − P(E ∩ F) Example 3 Computing the Probability of the Union of Two Events A card is drawn from a standard deck. Find the probability of drawing a heart or a 7. Solution A standard deck contains an equal number of hearts, diamonds, clubs, and spades. So the probability 1 __ of drawing a heart is . There are four 7s in a standard deck, and there are a total of 52 cards. So the probability of 4 drawing a 7 is 1 ___ . 13 The only card in the deck that is both a heart and a 7 is the 7 of hearts, so the probability of drawing both a heart and a 7 is 1 ___ 52 1 __ , P(7) = . Substitute P(H) = 4 1 ___ 13 , and P(H ∩ 7) = 1 ___ 52 P(E ∪ F) = P(E) + P(F) − P(E ∩ F) into the formula. The probability of drawing a heart or a 7 is 4 ___ . 13 1 ___ 13 − 1 ___ 52 1 __ = + 4 4 ___ 13 = 1002 CHAPTER 11 seQuences, proBaBility and counting theory Try It #3 A card is drawn from a standard deck. Find the probability of drawing a red card or an ace. Computing the Probability of Mutually exclusive events Suppose the spinner in Figure 2 is spun again, but this time we are interested in the probability of spinning an orange or a d. There are no sectors that are both orange and contain a d, so these two events have no outcomes in common. Events are said to be mutually exclusive events when they have no outcomes in common. Because there is no overlap, there is nothing to subtract, so the general formula is P(E ∪ F) = P(E) + P(F) Notice that with mutually exclusive events, the intersection of E and F is the empty set. The probability of spinning 1 1 3 __ __ __ = . We can find the probability of spinning an orange or a and the probability of spinning a d is an orange is 6 2 6 d simply by adding the two probabilities. P(E ∪ F) = P(E) + P(F) 2 __ The probability of spinning an orange or a d is . 3 1 1 __ __ = + 2 6 2 __ = 3 probability of the union of mutually exclusive events The probability of the union of two mutually exclusive events E and F is given by P(E ∪ F) = P(E) + P(F) How To… Given a set of events, compute the probability of the union of mutually exclusive events. 1. Determine the total number of outcomes for the first event. 2. Find the probability of the first event. 3. Determine the total number of outcomes for the second event. 4. Find the probability of the second event. 5. Add the probabilities. Example 4 Computing the Probability of the Union of Mutually Exclusive Events A card is drawn from a standard deck. Find the probability of drawing a heart or a spade. Solution The events “drawing a heart” and “drawing a spade” are mutually exclusive because they cannot occur 1 1 __ __ , so the , and the probability of drawing a spade is also at the same time. The probability of drawing a heart is 4 4 probability of drawing a heart or a spade is 1 1 1 __ __ __ = + 2 4 4 Try It #4 A card is drawn from a standard deck. Find the probability of drawing an ace or a king. SECTION 11.7 proBaBility 1003 Using the Complement Rule to Compute Probabilities We have discussed how to calculate the probability that an event will happen. Sometimes, we are interested in finding the probability that an event will not happen. The complement of an event E, denoted E′, is the set of outcomes in the sample space that are not in E. For example, suppose we are interested in the probability that a horse will lose a race. If event W is the horse winning the race, then the complement of event W is the horse losing the race. To find the probability that the horse loses the race, we need to use the fact that the sum of all probabilities in a probability model must be 1. P(E′) = 1 − P(E) The probability of the horse winning added to the probability of the horse losing must be equal to 1. Therefore, if the 1 __ probability of the horse winning the race is , the probability of the horse losing the race is simply 9 8 1 __ __ = 1 − 9 9 the complement rule The probability that the complement of an event will occur is given by P(E′) = 1 − P(E) Example 5 Using the Complement Rule to Calculate Probabilities Two six-sided number cubes are rolled. a. Find the probability that the sum of the numbers rolled is less than or equal to 3. b. Find the probability that the sum of the numbers rolled is greater than 3. Solution The first step is to identify the sample space, which consists of all the possible outcomes. There are two number cubes, and each number cube has six possible outcomes. Using the Multiplication Principle, we find that there are 6 × 6, or 36 total possible outcomes. So, for example, 1-1 represents a 1 rolled on each number cube. 1-1 2-1 3-1 4-1 5-1 6-1 1-2 2-2 3-2 4-2 5-2 6-2 1-3 2-3 3-3 4-3 5-3 6-3 1-4 2-4 3-4 4-4 5-4 6-4 1-5 2-5 3-5 4-5 5-5 6-5 1-6 2-6 3-6 4-6 5-6 6-6 Table 3 a. We need to count the number of ways to roll a sum of 3 or less. These would include the following outcomes: 1-1, 1-2, and 2-1. So there are only three ways to roll a sum of 3 or less. The probability is 3 ___ 36 = 1 ___ 12 b. Rather than listing all the possibilities, we can use the Complement Rule. Because we have already found the probability of the complement of this event, we can simply subtract that probability from 1 to find the probability that the sum of the numbers rolled is greater than 3. P(E′) = 1 − P(E) 1 ___ 12 = 1 − = 11 ___ 12 Try It #5 Two number cubes are rolled. Use the Complement Rule to find the probability that the sum is less than 10. 1004 CHAPTER 11 seQuences, proBaBility and counting theory Computing Probabi
lity Using Counting Theory Many interesting probability problems involve counting principles, permutations, and combinations. In these problems, we will use permutations and combinations to find the number of elements in events and sample spaces. These problems can be complicated, but they can be made easier by breaking them down into smaller counting problems. Assume, for example, that a store has 8 cellular phones and that 3 of those are defective. We might want to find the probability that a couple purchasing 2 phones receives 2 phones that are not defective. To solve this problem, we need to calculate all of the ways to select 2 phones that are not defective as well as all of the ways to select 2 phones. There are 5 phones that are not defective, so there are C(5, 2) ways to select 2 phones that are not defective. There are 8 phones, so there are C(8, 2) ways to select 2 phones. The probability of selecting 2 phones that are not defective is: ways to select 2 phones that are not defective ____ = ways to select 2 phones C(5, 2) _ C(8, 2) = = 10 ___ 28 5 ___ 14 Example 6 Computing Probability Using Counting Theory A child randomly selects 5 toys from a bin containing 3 bunnies, 5 dogs, and 6 bears. a. Find the probability that only bears are chosen. b. Find the probability that 2 bears and 3 dogs are chosen. c. Find the probability that at least 2 dogs are chosen. Solution a. We need to count the number of ways to choose only bears and the total number of possible ways to select 5 toys. There are 6 bears, so there are C(6, 5) ways to choose 5 bears. There are 14 toys, so there are C(14, 5) ways to choose any 5 toys. C(6, 5) _______ = C(14, 5) 6 _____ 2,002 = 3 _____ 1,001 b. We need to count the number of ways to choose 2 bears and 3 dogs and the total number of possible ways to select 5 toys. There are 6 bears, so there are C(6, 2) ways to choose 2 bears. There are 5 dogs, so there are C(5, 3) ways to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are C(6, 2) ⋅ C(5, 3) ways to choose 2 bears and 3 dogs. We can use this result to find the probability. C(6, 2)C(5, 3) ____________ = C(14, 5) 15 ⋅ 10 ______ 2,002 = 75 _____ 1,001 c. It is often easiest to solve “at least” problems using the Complement Rule. We will begin by finding the probability that fewer than 2 dogs are chosen. If less than 2 dogs are chosen, then either no dogs could be chosen, or 1 dog could be chosen. When no dogs are chosen, all 5 toys come from the 9 toys that are not dogs. There are C(9, 5) ways to choose toys from the 9 toys that are not dogs. Since there are 14 toys, there are C(14, 5) ways to choose the 5 toys from all of the toys. C(9, 5) _______ C(14,5) = 63 _____ 1,001 If there is 1 dog chosen, then 4 toys must come from the 9 toys that are not dogs, and 1 must come from the 5 dogs. Since we are choosing both dogs and other toys at the same time, we will use the Multiplication Principle. There are C(5, 1) ⋅ C(9, 4) ways to choose 1 dog and 1 other toy. C(5, 1)C(9, 4) ____________ = C(14, 5) 5 ⋅ 126 ______ 2,002 = 315 _____ 1,001 SECTION 11.7 proBaBility 1005 Because these events would not occur together and are therefore mutually exclusive, we add the probabilities to find the probability that fewer than 2 dogs are chosen. 63 _____ 1,001 + 315 _____ 1,001 = 378 _____ 1,001 We then subtract that probability from 1 to find the probability that at least 2 dogs are chosen. 1 − 378 _____ 1,001 = 623 _____ 1,001 Try It #6 A child randomly selects 3 gumballs from a container holding 4 purple gumballs, 8 yellow gumballs, and 2 green gumballs. a. Find the probability that all 3 gumballs selected are purple. b. Find the probability that no yellow gumballs are selected. c. Find the probability that at least 1 yellow gumball is selected. Access these online resources for additional instruction and practice with probability. • Introduction to Probability (http://openstaxcollege.org/l/introprob) • Determining Probability (http://openstaxcollege.org/l/determineprob) 1006 CHAPTER 11 seQuences, proBaBility and counting theory 11.7 SeCTIOn exeRCISeS VeRBAl 1. What term is used to express the likelihood of an 2. What is a sample space? event occurring? Are there restrictions on its values? If so, what are they? If not, explain. 3. What is an experiment? 5. The union of two sets is defined as a set of elements that are present in at least one of the sets. How is this similar to the definition used for the union of two events from a probability model? How is it different? 4. What is the difference between events and outcomes? Give an example of both using the sample space of tossing a coin 50 times. nUMeRIC For the following exercises, use the spinner shown in Figure 3 to find the probabilities indicated. F A O C B I E D Figure 3 6. Landing on red 7. Landing on a vowel 8. Not landing on blue 9. Landing on purple or a vowel 10. Landing on blue or a vowel 11. Landing on green or blue 12. Landing on yellow or a consonant 13. Not landing on yellow or a consonant For the following exercises, two coins are tossed. 14. What is the sample space? 15. Find the probability of tossing two heads. 16. Find the probability of tossing exactly one tail. 17. Find the probability of tossing at least one tail. For the following exercises, four coins are tossed. 18. What is the sample space? 19. Find the probability of tossing exactly two heads. 20. Find the probability of tossing exactly three heads. 21. Find the probability of tossing four heads or four tails. 22. Find the probability of tossing all tails. 23. Find the probability of tossing not all tails. 24. Find the probability of tossing exactly two heads or 25. Find the probability of tossing either two heads or at least two tails. three heads. For the following exercises, one card is drawn from a standard deck of 52 cards. Find the probability of drawing the following: 26. A club 30. An ace or a diamond 27. A two 31. A non-ace 28. Six or seven 32. A heart or a non-jack 29. Red six For the following exercises, two dice are rolled, and the results are summed. 33. Construct a table showing the sample space of 34. Find the probability of rolling a sum of 3. outcomes and sums. SECTION 11.7 section exercises 1007 35. Find the probability of rolling at least one four or a 36. Find the probability of rolling an odd sum less sum of 8. than 9. 37. Find the probability of rolling a sum greater than or 38. Find the probability of rolling a sum less than 15. equal to 15. 39. Find the probability of rolling a sum less than 6 or 40. Find the probability of rolling a sum between 6 and greater than 9. 9, inclusive. 41. Find the probability of rolling a sum of 5 or 6. 42. Find the probability of rolling any sum other than 5 or 6. For the following exercises, a coin is tossed, and a card is pulled from a standard deck. Find the probability of the following: 43. A head on the coin or a club 44. A tail on the coin or red ace 45. A head on the coin or a face card 46. No aces For the following exercises, use this scenario: a bag of M&Ms contains 12 blue, 6 brown, 10 orange, 8 yellow, 8 red, and 4 green M&Ms. Reaching into the bag, a person grabs 5 M&Ms. 47. What is the probability of getting all blue M&Ms? 48. What is the probability of getting 4 blue M&Ms? 49. What is the probability of getting 3 blue M&Ms? 50. What is the probability of getting no brown M&Ms? exTenSIOnS Use the following scenario for the exercises that follow: In the game of Keno, a player starts by selecting 20 numbers from the numbers 1 to 80. After the player makes his selections, 20 winning numbers are randomly selected from numbers 1 to 80. A win occurs if the player has correctly selected 3, 4, or 5 of the 20 winning numbers. (Round all answers to the nearest hundredth of a percent.) 51. What is the percent chance that a player selects 52. What is the percent chance that a player selects exactly 3 winning numbers? exactly 4 winning numbers? 53. What is the percent chance that a player selects all 54. What is the percent chance of winning? 5 winning numbers? 55. How much less is a player’s chance of selecting 3 winning numbers than the chance of selecting either 4 or 5 winning numbers? ReAl-WORlD APPlICATIOnS Use this data for the exercises that follow: In 2013, there were roughly 317 million citizens in the United States, and about 40 million were elderly (aged 65 and over).[35] 56. If you meet a U.S. citizen, what is the percent chance that the person is elderly? (Round to the nearest tenth of a percent.) 57. If you meet five U.S. citizens, what is the percent chance that exactly one is elderly? (Round to the nearest tenth of a percent.) 58. If you meet five U.S. citizens, what is the percent 59. If you meet five U.S. citizens, what is the percent chance that three are elderly? (Round to the nearest tenth of a percent.) chance that four are elderly? (Round to the nearest thousandth of a percent.) 60. It is predicted that by 2030, one in five U.S. citizens will be elderly. How much greater will the chances of meeting an elderly person be at that time? What policy changes do you foresee if these statistics hold true? 35 United States Census Bureau. http://www.census.gov 1008 CHAPTER 11 seQuences, proBaBility and counting theory CHAPTeR 11 ReVIeW Key Terms Addition Principle if one event can occur in m ways and a second event with no common outcomes can occur in n ways, then the first or second event can occur in m + n ways annuity an investment in which the purchaser makes a sequence of periodic, equal payments arithmetic sequence a sequence in which the difference between any two consecutive terms is a constant arithmetic series the sum of the terms in an arithmetic sequence binomial coefficient the number of ways to choose r objects from n objects where order does not matter; equivalent to n _ r  C(n, r), denoted  binomial expansion the result of expanding (x + y)n by multiplying B

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