x y = 1. The star could have been placed on the negative x-axis, and the answer would still be the same.

Transcription

1 Example The orbit of a planet has the shape of an ellipse, and on one of the foci is the star around which it revolves. The planet is closest to the star when it is at one vertex. It is farthest from the star when it is at the other vertex. Suppose the closest and farthest distances of the planet from this star, are 420 million kilometers and 580 million kilometers, respectively. Find the equation of the ellipse, in standard form, with center at the origin and the star at the x-axis. Assume all units are in millions of kilometers. Solution. In the figure above, the orbit is drawn as a horizontal ellipse with center at the origin. From the planet s distances from the star, at its closest and farthest points, it follows that the major axis is 2a = = 1000 (million kilometers), so a = 500. If we place the star at the positive x-axis, then it is c = = 80 units away from the center. Therefore, we get b 2 = a 2 c 2 = = The equation then is x y = 1. The star could have been placed on the negative x-axis, and the answer would still be the same. Seatwork/Homework The arch of a bridge is in the shape of a semiellipse, with its major axis at the water level. Suppose the arch is 20 ft high in the middle, and 120 ft across its major axis. How high above the water level is the arch, at a point 20 ft from the center (horizontally)? Round off to 2 decimal places. Refer to Example Answer: ft 45

2 Exercises Give the coordinates of the center, vertices, covertices, and foci of the ellipse with the given equation. Sketch the graph, and include these points. (a) x y2 25 = 1 x 2 (b) y2 169 = 1 (c) 4x 2 +13y 2 = 52 (d) (x+7)2 (y 4)2 + = (e) 9x 2 +16y 2 +72x 96y +144 = 0 (f) 36x 2 +20y 2 144x+120y 396 = 0 Answer: Item Center Vertices Covertices Foci (a) (0, 0) (±13, 0) (0, ±5) (±12, 0) (b) (0, 0) (0, ±13) (±12, 0) (0, ±5) (c) (0,0) (± 13,0) (0,±2) (±3,0) (d) ( 7, 4) ( 7, 1) ( 11, 4) ( 7, 1) ( 7, 9) ( 3, 4) ( 7, 7) (e) ( 4,3) ( 8,3) ( 4,0) ( 4± 7,3) (0, 3) ( 4, 6) (f) (2, 3) (2, 9) (2±2 5, 3) (2, 7) (2, 3) (2, 3) (2, 1) (a) (b) (c) 46

3 (d) (e) (f) 2. Find the standard equation of the ellipse which satisfies the given conditions. (a) foci ( 7,6) and ( 1,6), the sum of the distances of any point from the foci is 14 Answer: (x+4)2 (y 6)2 + = (b) center (5,3), horizontal major axis of length 20, minor axis of length 16 Answer: (x 5)2 (y 3)2 + = (c) major axis of length 22, foci 9 units above and below the center (2,4) Answer: (x 2)2 (y 4)2 + = (d) covertices ( 4,8) and (10,8), a focus at (3,12) Answer: (x 3)2 (y 8)2 + = Solution. The midpoint of the covertices is the center, (3, 8). From this point, the given focus is c = 4 units away. Since b = 7 (the distance from the center to a covertex), then a 2 = b 2 + c 2 = 65. The ellipse then has equation (x 3) (y 8)2 65 = 1. (e) focus ( 6, 2), covertex ( 1, 5), horizontal major axis Answer: (x+1) (y +2)2 49 Solution. Make a rough sketch of the points to see that the center is to the right of the given focus, and below the given covertex. The center is thus ( 1, 2). It follows that c = 5, b = 7, so a 2 = b 2 + c 2 = 74. The ellipse then has equation (x+1) (y +2)2 49 = A semielliptical tunnel has height 9 ft and a width of 30 ft. A truck that is about to pass through is 12 ft wide and 8.3 ft high. Will this truck be able to pass through the tunnel? Answer: No = 1 47

4 4. A truck that is about to pass through the tunnel from the previous item is 10 ft wide and 8.3 ft high. Will this truck be able to pass through the tunnel? Answer: Yes 5. An orbit of a satellite around a planet is an ellipse, with the planet at one focus of this ellipse. The distance of the satellite from this star varies from 300,000 km to 500,000 km, attained when the satellite is at each of the two vertices. Find the equation of this ellipse, if its center is at the origin, and the vertices are on the x-axis. Assume all units are in 100,000 km. Answer: x y2 15 = 1 6. The orbit of a planet around a star is described by the equation x 2 y 2 640,000 + = 1, where the star is at one focus, and all units are in millions of 630,000 kilometers. The planet is closest and farthest from the star, when it is at the vertices. How far is the planet when it is closest to the sun? How far is the planet when it is farthest from the sun? Answer: 700 million km, 900 million km Solution. The ellipse has center at the origin, and major axis on the x-axis. Since a 2 = 640,000, then a = 800, so the vertices are V 1 ( 800,0) and V 2 (800,00). Since b 2 = 630,000, then c = a 2 b 2 = 10,000 = 100. Suppose the star is at the focus at the right of the origin (this choice is arbitrary, since we could have chosen instead the focus on the left). Its location is then F(100,0). The closest distance is then V 2 F = 700 (million kilometers) and the farthest distance is V 1 F = 900 (million kilometers). 7. A big room is constructed so that the ceiling is a dome that is semielliptical in shape. If a person stands at one focus and speaks, the sound that is made bounces off the ceiling and gets reflected to the other focus. Thus, if two people stand at the foci (ignoring their heights), they will be able to hear each other. If the room is 34 m long and 8 m high, how far from the center should each of two people stand if they would like to whisper back and forth and hear each other? Answer: 15 m 48

5 Solution. We could put a coordinate system with the floor of the room on the x-axis, and the center of the room at the origin, as shown in the figures. The major axis has length 34, and the height of the room is half of the minor axis. The ellipse that contains the ceiling then has equation x2 + y2 = 1. The distance of a focus from the center is c = a 2 b 2 = = 15. Thus, the two people should stand 15 m away from the center. 8. A whispering gallery has a semielliptical ceiling that is 9 m high and 30 m long. How high is the ceiling above the two foci? Answer: 5.4 m Solution. As in the previous problem, put a coordinate system with the floor of the room on the x-axis, and the center of the room at the origin. The major axis has length 30, and half the minor axis is 9. The ellipse that contains the ceiling then has equation x2 + y2 = 1. The distance of a focus from the center is c = a 2 b 2 = = 12. If we put x = 12 in the equation of the ellipse, we get y2 = 1. Solving for y > 0 yields y = 27 = 5.4. The height of the ceiling above each focus is 5.4 m. 4 Lesson 1.4. Hyperbolas Time Frame: 3 one-hour sessions Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) define a hyperbola; (2) determine the standard form of equation of a hyperbola; (3) graph a hyperbola in a rectangular coordinate system; and (4) solve situational problems involving conic sections (hyperbolas). Lesson Outline (1) Definition of a hyperbola (2) Derivation of the standard equation of a hyperbola (3) Graphing hyperbolas (4) Solving situational problems involving hyperbolas 49

6 Introduction A hyperbola is one of the conic sections that most students have not encountered formally before, unlike circles and parabolas. Its graph consists of two unbounded branches which extend in opposite directions. It is a misconception that each branch is a parabola. This is not true, as parabolas and hyperbolas have very different features. An application of hyperbolas in basic location and navigation schemes are presented in an example and some exercises Definition and Equation of a Hyperbola Consider the points F 1 ( 5,0) and F 2 (5,0) as shown in Figure What is the absolute value of the difference of the distances of A(3.75, 3) from F 1 and from F 2? How about the absolute value of the difference of the distances of B ( ) 5, 16 3 from F 1 and from F 2? AF 1 AF 2 = = 6 BF 1 BF 2 = = 6 There are other points P such that PF 1 PF 2 = 6. The collection of all such points forms a shape called a hyperbola, which consists of two disjoint branches. For points P on the left branch, PF 2 PF 1 = 6; for those on the right branch, PF 1 PF 2 = 6. Figure 1.21 Figure 1.22 Let F 1 and F 2 be two distinct points. The set of all points P, whose distances from F 1 and from F 2 differ by a certain constant, is called a hyperbola. The points F 1 and F 2 are called the foci of the hyperbola. In Figure 1.22, given are two points on the x-axis, F 1 ( c,0) and F 2 (c,0), the foci, both c units away from their midpoint (0,0). This midpoint is the center 50

7 Figure 1.23 Figure 1.24 of the hyperbola. Let P(x,y) be a point on the hyperbola, and let the absolute value of the difference of the distances of P from F 1 and F 2, be 2a (the coefficient 2 will make computations simpler). Thus, PF 1 PF 2 = 2a, and so (x+c) 2 +y 2 (x c) 2 +y 2 = 2a. Algebraic manipulations allow us to rewrite this into the much simpler x 2 a 2 y2 b 2 = 1, where b = c 2 a 2. Whenweletb = c 2 a 2, weassumedc > a. Toseewhythisistrue, suppose that P is closer to F 2, so PF 1 PF 2 = 2a. Refer to Figure Suppose also that P is not on the x-axis, so PF 1 F 2 is formed. From the triangle inequality, F 1 F 2 +PF 2 > PF 1. Thus, 2c > PF 1 PF 2 = 2a, so c > a. Nowwepresentaderivation. Fornow, assumep isclosertof 2 sopf 1 > PF 2, Teaching Notesand PF 1 PF 2 = 2a. If it is assumed that P is closer to F 1, then the same equation will be obtained because of symmetry. PF 1 = 2a+PF 2 (x+c)2 +y 2 = 2a+ (x c) 2 +y 2 ( (x+c)2 +y 2 ) 2 = (2a+ (x c) 2 +y 2 ) 2 cx a 2 = a (x c) 2 +y 2 (cx a 2 ) 2 = (a ) 2 (x c) 2 +y 2 (c 2 a 2 )x 2 a 2 y 2 = a 2 (c 2 a 2 ) b 2 x 2 a 2 y 2 = a 2 b 2 by letting b = c 2 a 2 > 0 x 2 a 2 y2 b 2 = 1 51

8 We collect here the features of the graph of a hyperbola with standard equation x 2 a y2 2 b = 1. 2 Let c = a 2 +b 2. (1) center: origin (0,0) (2) foci: F 1 ( c,0) and F 2 (c,0) Each focus is c units away from the center. For any point on the hyperbola, the absolute value of the difference of its distances from the foci is 2a. (3) vertices: V 1 ( a,0) and V 2 (a,0) The vertices are points on the hyperbola, collinear with the center and foci. If y = 0, then x = ±a. Each vertex is a units away from the center. The segment V 1 V 2 is called the transverse axis. Its length is 2a. (4) asymptotes: y = b a x and y = b a x, the lines l 1 and l 2 in Figure 1.24 The asymptotes of the hyperbola are two lines passing through the center which serve as a guide in graphing the hyperbola: each branch of the hyperbola gets closer and closer to the asymptotes, in the direction towards which the branch extends. (We need the concept of limits from calculus to explain this.) An aid in determining the equations of the asymptotes: in the standard equation, replace 1 by 0, and in the resulting equation x2 y2 = 0, solve a 2 b 2 for y. To help us sketch the asymptotes, we point out that the asymptotes l 1 and l 2 are the extended diagonals of the auxiliary rectangle drawn in Figure This rectangle has sides 2a and 2b with its diagonals intersecting at the center C. Two sides are congruent and parallel to the transverse axis V 1 V 2. The other two sides are congruent and parallel to the conjugate axis, the segment shown which is perpendicular to the transverse axis at the center, and has length 2b. Example Determine the foci, vertices, and asymptotes of the hyperbola with equation x 2 9 y2 7 = 1. Sketch the graph, and include these points and lines, the transverse and conjugate axes, and the auxiliary rectangle. 52

9 Solution. With a 2 = 9 and b 2 = 7, we have a = 3, b = 7, and c = a 2 +b 2 = 4. foci: F 1 ( 4,0) and F 2 (4,0) vertices: V 1 ( 3,0) and V 2 (3,0) asymptotes: y = 7 x and y = 3 7 x 3 The graph is shown at the right. The conjugate axis drawn has its endpoints b = units above and below the center. Example Find the (standard) equation of the hyperbola whose foci are F 1 ( 5,0) and F 2 (5,0), such that for any point on it, the absolute value of the difference of its distances from the foci is 6. See Figure Solution. We have 2a = 6 and c = 5, so a = 3 and b = c 2 a 2 = 4. The hyperbola then has equation x2 9 y2 16 = 1. Seatwork/Homework Determine foci, vertices, and asymptotes of the hyperbola with equation x 2 16 y2 20 = 1. Sketch the graph, and include these points and lines, along with the auxiliary rectangle. Answer: foci F 1 ( 6,0) and F 2 (6,0), vertices V 1 ( 4,0) and V 2 (4,0), asymptotes y = 5 x and y = 2 5 x 2 2. FindtheequationinstandardformofthehyperbolawhosefociareF 1 ( 4 2,0) and F 2 (4 2,0), such that for any point on it, the absolute value of the x 2 difference of its distances from the foci is 8. Answer: 16 y2 16 = 1 53

10 More Properties of Hyperbolas The hyperbolas we considered so far are horizontal and have the origin as their centers. Some hyperbolas have their foci aligned vertically, and some have centers not at the origin. Their standard equations and properties are given in the box. The derivations are more involved, but are similar to the one above, and so are not shown anymore. Center Corresponding Hyperbola (0,0) x 2 a 2 y2 b 2 = 1 y 2 a 2 x2 b 2 = 1 (h,k) (x h) 2 a 2 (y k)2 b 2 = 1 (y k) 2 a 2 (x h)2 b 2 = 1 transverse axis: horizontal conjugate axis: vertical transverse axis: vertical conjugate axis: horizontal 54

11 In all four cases above, we let c = a 2 +b 2. The foci F 1 and F 2 are c units away from the center C. The vertices V 1 and V 2 are a units away from the center. The transverse axis V 1 V 2 has length 2a. The conjugate axis has length 2b and is perpendicular to the transverse axis. The transverse and conjugate axes bisect each other at their intersection point, C. Each branch of a hyperbola gets closer and closer to the asymptotes, in the direction towards which the branch extends. The equations of the asymptotes can be determined by replacing 1 in the standard equation by 0. The asymptotes can be drawn as the extended diagonals of the auxiliary rectangle determined by the transverse and conjugate axes. Recall that, foranypointonthehyperbola, theabsolutevalueofthedifferenceofitsdistances from the foci is 2a. In the standard equation, aside from being positive, there are no other restrictions on a and b. In fact, a and b can even be equal. The orientation of the hyperbola is determined by the variable appearing in the first term (the positive term): the corresponding axis is where the two branches will open. For example, if the variable in the first term is x, the hyperbola is horizontal : the transverse axis is horizontal, and the branches open to the left and right in the direction of the x-axis. Example Give the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with the given equation. Sketch the graph, and include these points and lines, the transverse and conjugate axes, and the auxiliary rectangle. (y +2)2 (1) (x 7)2 = (2) 4x 2 5y 2 +32x+30y = 1 Solution. (1) From a 2 = 25 and b 2 = 9, we have a = 5, b = 3, and c = a2 +b 2 = The hyperbola is vertical. To determine the asymptotes, we write (y+2)2 (x 7)2 = 0, which is equivalent to y+2 = ± 5(x 7) We can then solve this for y. center: C(7, 2) foci: F 1 (7, 2 34) (7, 7.8) and F 2 (7, 2+ 34) (7,3.8) vertices: V 1 (7, 7) and V 2 (7,3) asymptotes: y = 5 41 x and y = x+ 3 3 The conjugate axis drawn has its endpoints b = 3 units to the left and right of the center. 55

12 (2) We first change the given equation to standard form. 4(x 2 +8x) 5(y 2 6y) = 1 4(x 2 +8x+16) 5(y 2 6y +9) = 1+4(16) 5(9) 4(x+4) 2 5(y 3) 2 = 20 (x+4) 2 5 (y 3)2 4 We have a = and b = 2. Thus, c = a 2 +b 2 = 3. The hyperbola is horizontal. To determine the asymptotes, we write (x+4)2 (y 3)2 = which is equivalent to y 3 = ± 2 5 (x+4), and solve for y. center: C( 4, 3) foci: F 1 ( 7,3) and F 2 ( 1,3) = 1 vertices: V 1 ( 4 5,3) ( 6.2,3) and V 2 ( 4+ 5,3) ( 1.8,3) asymptotes: y = 2 5 x and y = 2 5 x The conjugate axis drawn has its endpoints b = 2 units above and below the center. 56

13 Example The foci of a hyperbola are ( 5, 3) and (9, 3). For any point on the hyperbola, the absolute value of the difference of its of its distances from the foci is 10. Find the standard equation of the hyperbola. Solution. The midpoint (2, 3) of the foci is the center of the hyperbola. Each focus is c = 7 units away from the center. From the given difference, 2a = 10 so a = 5. Also, b 2 = c 2 a 2 = 24. The hyperbola is horizontal (because the foci are horizontally aligned), so the equation is (x 2) 2 25 (y +3)2 24 = 1. Example A hyperbola has vertices ( 4, 5) and ( 4,9), and one of its foci is ( 4,2 65). Find its standard equation. Solution. The midpoint ( 4, 2) of the vertices is the center of the hyperbola, which is vertical (because the vertices are vertically aligned). Each vertex is a = 7 units away from the center. The given focus is c = 65 units away from the center. Thus, b 2 = c 2 a 2 = 16, and the standard equation is (y 2) 2 49 (x+4)2 16 = 1. 57

14 Seatwork/Homework Give the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with equation 9x 2 4y 2 90x 32y = 305. Sketch the graph, and include these points and lines, along with the auxiliary rectangle. Answer: center C(5, 4), foci F 1 (5, ) and F 2 (5, ), vertices V 1 (5, 10) and V 2 (5,2), asymptotes y = 3x+ 7 and y = 3 23 x A hyperbola has vertices (1,9) and (13,9), and one of its foci is ( 2,9). Find its standard equation. Answer: (x 7)2 (y 9)2 = Situational Problems Involving Hyperbolas We now give an example on an application of hyperbolas. Example An explosion is heard by two stations 1200 m apart, located at F 1 ( 600,0) and F 2 (600,0). If the explosion was heard in F 1 two seconds before it was heard in F 2, identify the possible locations of the explosion. Use 340 m/s as the speed of sound. Solution. Using the given speed of sound, we deduce that the sound traveled 340(2) = 680 m farther in reaching F 2 than in reaching F 1. This is then the difference of the distances of the explosion from the two stations. Thus, the explosion is on a hyperbola with foci are F 1 and F 2, on the branch closer to F 1. 58

15 We have c = 600 and 2a = 680, so a = 340 and b 2 = c 2 a 2 = The explosion could therefore be anywhere on the left branch of the hyperbola x 2 y2 = Seatwork/Homework Two stations, located at M( 1.5,0) and N(1.5,0) (units are in km), simultaneously send sound signals to a ship, with the signal traveling at the speed of 0.33 km/s. If the signal from N was received by the ship four seconds before the signal it received from M, find the equation of the curve containing the x possible location of the ship. Answer: 2 y2 = 1 (right branch) Exercises Give the coordinates of the center, foci, vertices, and the asymptotes of the hyperbola with the given equation. Sketch the graph, and include these points and lines. (a) x2 36 y2 64 = 1 (b) y2 25 x2 16 = 1 (c) (x 1) 2 y 2 = 4 (y +2)2 (d) (x+3)2 = (e) 3x 2 2y 2 42x 16y = 67 (f) 25x 2 39y x+390y =

16 Answer: Item Center Vertices Foci (a) (0, 0) (±6, 0) (±10, 0) (b) (0,0) (0,±5) (0,± 41) (c) (1,0) ( 1,0), (3,0) (1±2 2,0) (d) ( 3, 2) ( 3, 2± 15) ( 3, 7), ( 3,3) (e) (7, 4) (3, 4), (11, 4) (7±2 10, 4) (f) ( 3,5) ( 3,0), ( 3,10) ( 3, 3), ( 3,13) Item (a) (b) Asymptotes y = ± 4 3 x y = ± 5 4 x (c) y = x 1, y = x+1 3 (d) y = ± x± (e) y = ± x (f) y = ± 5 39 x± (a) (b) 60

17 (c) (d) (e) 2. Find the standard equation of the hyperbola which satisfies the given conditions. (a) foci ( 4, 3) and ( 4,13), the absolute value of the difference of the distances of any point from the foci is 14 (y 5)2 Answer: (x+4)2 = (b) vertices ( 2,8) and (8,8), a focus (12,8) Answer: (x 3)2 (y 8)2 = (c) center ( 6,9), a vertex ( 6,15), conjugate axis of length 12 (y 9)2 Answer: (x+6)2 = (d) asymptotes y = 4 3 x+ 1 3 and y = 4 3 x (f), a vertex ( 1,7) Answer: (x 5)2 36 (y 7)2 64 Solution. The asymptotes intersect at (5, 7). This is the center. The distance of the given vertex from the center is a = 6. This vertex and center are aligned horizontally, so the hyperbola has equation of the form (x h) 2 a 2 = 1 (y k)2 b 2 = 1. The asymptotes consequently have the form y k = 61

18 ± b a (x h), and thus, have slopes ±b a. From the given asymptotes, b a = 4 3. Since a = 6, then b = 8. The standard equation is then (x 5) 2 36 (y 7)2 64 = 1. (e) asymptotes y = 1x+ 5 and y = 3 3 1x+ 7, a focus (1,12) 3 3 (y 2)2 Answer: 10 (x 1)2 90 Solution. The asymptotes intersect at (1, 2). This is the center. The distance of the given focus from the center is c = 10. This focus and center are aligned vertically, so the hyperbola has equation of the form (y k) 2 (x h)2 = 1. The asymptotes consequently have the form y k = 2 b 2 ± a(x h), and thus, have slopes b ±a. From the given asymptotes, a = 1, b b 3 so b = 3a. c 2 = 100 = a 2 +b 2 = a 2 +(3a) 2 = 10a 2 Thus, a 2 = 10, and b 2 = 9a 2 = 90. The standard equation is (y 2) 2 10 (x 1)2 90 = Two control towers are located at points Q( 500,0) and R(500,0), on a straight shore where the x-axis runs through (all distances are in meters). At the same moment, both towers sent a radio signal to a ship out at sea, each travelingat300m/µs. TheshipreceivedthesignalfromQ3µs(microseconds) before the message from R. (a) Find the equation of the curve containing the possible location of the ship. Answer: y2 = 1 (left branch) (b) Find the coordinates (rounded off to two decimal places) of the ship if it is 200 m from the shore (y = 200). Answer: ( ,200) Solution. Since the time delay between the two signals is 3 µs, then the difference between the distances traveled by the two signals is = 900 m. The ship is then on a hyperbola, consisting of points whose distances from Q and R (the foci) differ by 2a = 900. With a = 450 and c = 500 (the distance of each focus from the center, the origin), we have b 2 = c 2 a 2 = = Since a 2 x = , the hyperbola then has equation 2 y2 = 1. Since the signal from Q was received first, the ship is closer to Q than R, so the ship is on the left branch of this hyperbola. Using y = 200, we then solve = 1 for x < 0 (left branch), and we get x x x 2 =

19 Lesson 1.5. More Problems on Conic Sections Time Frame: 2 one-hour sessions Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) recognize the equation and important characteristics of the different types of conic sections; and (2) solve situational problems involving conic sections. Lesson Outline (1) Conic sections with associated equations in general form (2) Problems involving characteristics of various conic sections (3) Solving situational problems involving conic sections Introduction Inspecting the equation can lead us to the right conic section for its graph, and set us on the right step towards analyzing it. We will also look at problems that use the properties of the different conic sections, allowing us to synthesize what has been covered so far Identifying the Conic Section by Inspection The equation of a circle may be written in standard form Ax 2 +Ay 2 +Cx+Dy +E = 0, that is, the coefficients of x 2 and y 2 are the same. However, it does not follow that if the coefficients of x 2 and y 2 are the same, the graph is a circle. General Equation Standard Equation graph ( ) (A) 2x 2 +2y 2 2x+6y +5 = 0 x 1 2 ( 2 + y ) = 0 point (B) x 2 +y 2 6x 8y +50 = 0 (x 3) 2 +(y 4) 2 = 25 empty set For a circle with equation (x h) 2 +(y k) 2 = r 2, we have r 2 > 0. This is not the case for the standard equations of (A) and (B). In (A), because the sum of two squares can only be 0 if and only if each square is ( 0, it follows that x 1 = 0 and y + 3 = 0. The graph is thus the single point ) 2, 3 2. In (B), no real values of x and y can make the nonnegative left side equal to the negative right side. The graph is then the empty set. 63

20 Let us recall the general form of the equations of the other conic sections. We may write the equations of conic sections we discussed in the general form Ax 2 +By 2 +Cx+Dy +E = 0. Some terms may vanish, depending on the kind of conic section. (1) Circle: both x 2 and y 2 appear, and their coefficients are the same Ax 2 +Ay 2 +Cx+Dy +E = 0 Example: 18x 2 +18y 2 24x+48y 5 = 0 Degenerate cases: a point, and the empty set (2) Parabola: exactly one of x 2 or y 2 appears Ax 2 +Cx+Dy +E = 0 (D 0, opens upward or downward) By 2 +Cx+Dy +E = 0 (C 0, opens to the right or left) Examples: 3x 2 12x+2y +26 = 0 (opens downward) 2y 2 +3x+12y 15 = 0 (opens to the right) (3) Ellipse: both x 2 and y 2 appear, and their coefficients A and B have the same sign and are unequal Examples: 2x 2 +5y 2 +8x 10y 7 = 0 (horizontal major axis) 4x 2 +y 2 16x 6y +21 = 0 (vertical major axis) If A = B, we will classify the conic as a circle, instead of an ellipse. Degenerate cases: a point, and the empty set (4) Hyperbola: both x 2 and y 2 appear, and their coefficients A and B have different signs Examples: 5x 2 3y 2 20x 18y 22 = 0 (horizontal transverse axis) 4x 2 +y 2 +24x+4y 36 = 0 (vertical transverse axis) Degenerate case: two intersecting lines The following examples will show the possible degenerate conic (a point, two intersecting lines, or the empty set) as the graph of an equation following a similar pattern as the non-degenerate cases. (1) 4x 2 +9y 2 16x+18y +25 = 0 = (x 2)2 (y +1)2 + = = one point: (2, 1) (2) 4x 2 +9y 2 16x+18y +61 = 0 = (x 2)2 (y +1)2 + = = empty set 64

21 (3) 4x 2 9y 2 16x 18y +7 = 0 = (x 2)2 (y +1)2 = = two lines: y +1 = ± 2 3 (x 2) A Note on Identifying a Conic Section by Its General Equation It is only after transforming a given general equation to standard form that we can identify its graph either as one of the degenerate conic sections (a point, two intersecting lines, or the empty set) or as one of the non-degenerate conic sections (circle, parabola, ellipse, or hyperbola). Seatwork/Homework The graphs of the following equations are(nondegenerate) conic sections. Identify the conic section. (1) 5x 2 3y 2 +10x 12y = 22 Answer: hyperbola (2) 2y 2 5x 12y = 17 Answer: parabola (3) 3x 2 +3y 2 +42x 12y = 154 Answer: circle (4) 3x 2 +6x+4y = 18 Answer: parabola (5) 7x 2 +3y 2 14x+12y = 14 Answer: ellipse (6) 4x 2 +3y 2 +24x 12y = 36 Answer: hyperbola Problems Involving Different Conic Sections The following examples require us to use the properties of different conic sections at the same time. Example A circle has center at the focus of the parabola y 2 +16x+4y = 44, and is tangent to the directrix of this parabola. Find its standard equation. Solution. The standard equation of the parabola is (y + 2) 2 = 16(x 3). Its vertex is V(3, 2). Since 4c = 16 or c = 4, its focus is F( 1, 2) and its directrix is x = 7. The circle has center at ( 1, 2) and radius 8, which is the distance from F to the directrix. Thus, the equation of the circle is (x+1) 2 +(y +2) 2 = 64. Example The vertices and foci of 5x 2 4y x + 16y + 29 = 0 are, respectively, the foci and vertices of an ellipse. Find the standard equation of this ellipse. 65

22 Solution. We first write the equation of the hyperbola in standard form: (x+5) 2 16 (y 2)2 20 = 1. For this hyperbola, using the notations a h, b h, and c h to refer to a, b, and c of the standard equation of the hyperbola, respectively, we have a h = 4, b h = 2 5, c h = a 2 h +b2 h = 6, so we have the following points: center: ( 5, 2) vertices: ( 9,2) and ( 1,2) foci: ( 11,2) and (1,2). It means that, for the ellipse, we have these points: center: ( 5, 2) vertices: ( 11,2) and (1,2) foci: ( 9,2) and ( 1,2). In this case, c e = 4 and a e = 6, so that b e = a 2 e c 2 e = 20. The standard equation of the ellipse is (x+5) (y 2)2 20 = 1. Seatwork/Homework Find the standard equation of all circles having center at a focus of 21x 2 4y 2 +84x 24y = 36 and passing through the farther vertex. Answer: (x+7) 2 +(y +3) 2 = 49, (x 3) 2 +(y +3) 2 = Findthestandardequationofthehyperbolaonebranchofwhichhasfocusand vertex that are the same as those of x 2 6x+8y = 23, and whose conjugate axis is on the directrix of the same parabola. Exercises 1.5 Answer: (y 6)2 4 (x 3) The graphs of the following equations are non-degenerate conic sections. Identify the conic section. = 1 (a) 5x 2 +7y 2 40x 28y = 73 (b) 5y 2 +2x 30y = 49 (c) 3x 2 3y 2 +12x 12y = 5 (d) 3x 2 +3y 2 +12x+12y = 4 Answer: ellipse Answer: parabola Answer: hyperbola Answer: circle 66

23 (e) 2x 2 +24x 5y = 57 Answer: parabola 2. The graphs of the following equations are degenerate conic sections. What are the specific graphs? (a) x 2 +3y 2 4x+24y = 52 (b) 9x 2 4y 2 +18x 16y = 7 (c) 3x 2 +5y 2 6x 20y = 25 Answer: point: (2, 4) Answer: lines: y +2 = ± 3 2 (x+1) Answer: empty set 3. An ellipse has equation 25x 2 +16y x 32y = 159. Find the standard equations of all parabolas whose vertex is a focus of this ellipse and whose focus is a vertex of this ellipse. Answer: (x+3) 2 = 8(y +2), (x+3) 2 = 32(y +2), (x+3) 2 = 32(y 4), and (x+3) 2 = 8(y 4) Solution. The standard equation of the ellipse is (x+3) (y 1)2 25 = 1. Its center is ( 3,1). Since a = 5 and b = 4, we get c = 3, so the vertices are P( 3, 4) and S( 3,6), while its foci are Q( 3, 2) and R( 3,4). We then get four parabolas satisfying the conditions of the problem. The focal distance indicated below is the distance from the vertex to the focus. vertex focus focal distance standard equation Q( 3, 2) P( 3, 4) 2 (x+3) 2 = 8(y +2) Q( 3, 2) S( 3,6) 8 (x+3) 2 = 32(y +2) R( 3,4) P( 3, 4) 8 (x+3) 2 = 32(y 4) R( 3,4) S( 3,6) 2 (x+3) 2 = 8(y 4) 4. Find the standard equation of the hyperbola whose conjugate axis is on the directrix of the parabola y 2 +12x+6y = 39, having the focus of the parabola as one of its foci, and the vertex of the parabola as one of its vertices. Answer: (x 7)2 9 (y +3)2 27 = 1 Solution. The standard equation of the parabola is (y +3) 2 = 12(x 4), so its vertex is V(4, 3), and it opens to the left. With 4c = 12, or c = 3, its focus is F(1, 3), and its directrix is x = 7. The hyperbola has its center on 67

24 x = 7, its conjugate axis, and a vertex at (4, 3). Its center is then C(7, 3). The conjugate axis is vertical so the hyperbola is horizontal, with constants a h = CV = 3 and c h = CF = 6, so b 2 h = c2 h a2 h = 27. The standard equation of the required hyperbola is (x 7) 2 9 (y +3)2 27 = Find the standard equation of the parabola opening to the left whose axis contains the major axis of the ellipse x 2 + 4y 2 10x 24y + 45 = 0, whose focus is the center of the ellipse, and which passes through the covertices of this ellipse. Answer: (y 3) 2 = 4(x 6) Solution. The standard form of the ellipse is (x 5) (y 3)2 4 = 1. Its center (5,3) is the focus of the parabola. Since b = 2, its covertices are W 1 (5,1) and W 2 (5,5). The vertex of the parabola, c units to the right of (5,3), is (5+c,3). Its equation can be written as (y 3) 2 = 4c(x (5+c)). Since (5,5) is a point on this parabola, we have (5 3) 2 = 4c(5 (5+c)). Solving this equation for c > 0 yields c = 1. Therefore, the standard equation of the required parabola is (y 3) 2 = 4(x 6). 6. Find the standard equation of the ellipse whose major and minor axes are the transverse and conjugate axes (not necessarily in that order) of the hyperbola 4x 2 9y 2 16x 54y = 29. Answer: (x 2)2 (y +3)2 + = Solution. The standard equation of the hyperbola is (y +3) 2 4 (x 2)2 9 = 1, with center (2, 3), and constants a h = 2 and b h = 3. Since its conjugate axis (which is horizontal and has length 2b h = 6) is longer than its transverse axis (length 2a h = 4), the ellipse is horizontal. Its major axis has length 2a e = 6 and its minor axis has length 2b e = 4, so a e = 3 and b e = 2. The ellipse shares the same center as the hyperbola. Thus, the standard equation of the required ellipse is (x 2) 2 (y +3)2 + = If m 3,2, find the value(s) of m so that the graph of is (2m 4)x 2 +(m+3)y 2 = (m+3)(2m 4) 68

25 (a) a circle, (b) a horizontal ellipse, (c) a vertical ellipse, (d) a hyperbola (is it horizontal or vertical?), or (e) the empty set. Answer: (a) m = 7, (b) 2 < m < 7, (c) m > 7, (d) 3 < m < 2 (horizontal), (e) m < 3 Solution. It might be helpful to observe that the equation is equivalent to x 2 m+3 + y2 2m 4 = 1. (a) The graph is a circle if m+3 = 2m 4 > 0 (positive, so the graph is not a point or the empty set). This happens if m = 7. (b) We require 0 < 2m 4 < m+3. Thus, 2 < m < 7. (c) We require 0 < m+3 < 2m 4. Thus, m > 7. (d) We need m+3 and 2m 4 to have different signs. We consider two cases. i. If m + 3 < 0 < 2m 4, then m < 3 AND m > 2, which cannot happen. ii. If 2m 4 < 0 < m+3, then 3 < m < 2. In this case, the equation can be written, with positive denominators, as The hyperbola is horizontal. x 2 m+3 y2 4 2m = 1. (e) The remaining case is when m < 3. In this case, m + 3 < 0 and 2m 4 < 0. This makes the expression x 2 m+3 + y2 2m 4 negative, and never equal to 1. The graph is then the empty set. 4 69

26 Lesson 1.6. Systems of Nonlinear Equations Time Frame: 4 one-hour sessions Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) illustrate systems of nonlinear equations; (2) determine the solutions of systems of nonlinear equations using techniques such as substitution, elimination, and graphing; and (3) solve situational problems involving systems of nonlinear equations. Lesson Outline (1) Review systems of linear equations (2) Solving a system involving one linear and one quadratic equation (3) Solving a system involving two quadratic equations (4) Applications of systems of nonlinear equations Introduction After recalling the techniques used in solving systems of linear equations in Grade 8, we extend these methods to solving a system of equations to systems in which the equations are not necessarily linear. In this lesson, the equations are restricted to linear and quadratic types, although it is possible to adapt the methodology so systems with other types of equations. We focus on quadratic equations for two reasons: to include a graphical representation of the solution and to ensure that either a solution is obtained or it is determined that there is no solution. The latter is possible because of the quadratic formula Review of Techniques in Solving Systems of Linear Equations Recall the methods we used to solve systems of linear equations. three methods used: substitution, elimination, and graphical. There were Example Use the substitution method to solve the system, and sketch the graphs in one Cartesian plane showing the point of intersection. 4x+y = 6 5x+3y = 4 Solution. Isolate the variable y in the first equation, and then substitute into the second equation. 70 Teaching Notes Recall that the task of solving a system of equations is equivalent to finding points of intersection. Teaching Notes Systems of linear equations and solving them were introduced and studied in Grade 8 at the last part of Quarter I.

27 4x+y = 6 = y = 6 4x 5x+3y = 4 5x+3(6 4x) = 4 7x+18 = 4 x = 2 y = 6 4(2) = 2 Example Use the elimination method to solve the system, and sketch the graphs in one Cartesian plane showing the point of intersection. 2x+7 = 3y 4x+7y = 12 Solution. We eliminate first the variable x. Rewrite the first equation wherein only the constant term is on the right-hand side of the equation, then multiply it by 2, and then add the resulting equation to the second equation. 2x 3y = 7 ( 2)(2x 3y) = ( 2)( 7) 4x+6y = 14 4x+6y = 14 4x+7y = 12 13y = 26 y = 2 x = 1 2 Seatwork/Homework Use either substitution or elimination method to solve the system, and sketch the graphs in one Cartesian plane showing the point of intersection. x 3y = x+5y = 1 71

28 Answer: (2, 1) 2. 5x+3y = 4 3x+5y = 9 Answer: ( 1, ) Solving Systems of Equations Using Substitution We begin our extension with a system involving one linear equation and one quadratic equation. In this case, it is always possible to use substitution by solving the linear equation for one of the variables. Example Solve the following system, and sketch the graphs in one Cartesian plane. x y +2 = 0 y 1 = x 2 72

29 Solution. We solve for y in terms of x in the first equation, and substitute this expression to the second equation. x y +2 = 0 = y = x+2 y 1 = x 2 (x+2) 1 = x 2 x 2 x 1 = 0 x = 1± 5 2 Solutions: ( x = x = = y = = y = = = ) ( 5, ) 5 and, The first equation represents a line with x-intercept 2 and y-intercept 2, while the second equation represents a parabola with vertex at (0, 1) and which opens upward. Seatwork/Homework Solve each system, and sketch the graphs in one Cartesian plane showing the point(s) of intersection. 1. x 2 +y 2 = 16 x y = 4 Answer: (4,0) and (0, 4) Solution. Solving for x in the second equation, we get x = y+4. Substitute 73

30 this expression into the first equation. x 2 +y 2 = 16 = (y +4) 2 +y 2 = 16 y 2 +8y +16+y 2 = 16 2y 2 +8y = 0 y = 0 or y = 4 y = 0 = x = 4 and y = 4 = x = 0 Solutions: (4,0) and (0, 4) Teaching Notes We substitute each value of y (0 and 4) to the second equation x y = 4 (or x = y +4). 2. y = x 2 x = y 2 Answer: (0,0) and (1,1) Solution. Since the equations represent parabolas, we can use either of them to isolate one variable. This is in fact the form in which both equations are given. Substituting y = x 2 into x = y 2, we get x = y 2 = x = (x 2 ) 2 74 x 4 x = 0 x(x 3 1) = 0 x = 0 or x = 1

31 x = 0 = y = 0 and x = 1 = y = 1 Solutions: (0,0) and (1,1) Solving Systems of Equations Using Elimination Elimination method is also useful in systems of nonlinear equations. Sometimes, some systems need both techniques (substitution and elimination) to solve them. Example Solve the following system: y 2 4x 6y = 11 4(3 x) = (y 3) 2. Solution 1. We expand the second equation, and eliminate the variable x by Teaching Notes adding the equations. The variable y could also be eliminated first by subtracting the second equation from the first. 4(3 x) = (y 3) 2 = 12 4x = y 2 6y +9 = y 2 +4x 6y = 3 y 2 4x 6y = 11 Adding these equations, we get y 2 +4x 6y = 3 2y 2 12y = 14 = y 2 6y 7 = 0 = (y 7)(y+1) = 0 = y = 7 or y = 1. Teaching Notes We may actually substitute y = 7 and y = 1 (one at a time) into any of the two given equations, and then solve for x. Solving for x in the second equation, we have x = 3 (y 3)2. 4 y = 7 = x = 1 and y = 1 = x = 1 Solutions: ( 1, 7) and ( 1, 1) 75

32 The graphs of the equations in the preceding example with the points of intersection are shown below. Sometimes the solution can be simplified by writing the equations in standard form, although it is usually the general form which is more convenient to use in solving systems of equations. Moreover, the standard form is best for graphing. We solve again the previous example in a different way. Solution 2. By completing the square, we can change the first equation into standard form: y 2 4x 6y = 11 = 4(x+5) = (y 3) 2. 4(x+5) = (y 3) 2 4(3 x) = (y 3) 2 Using substitution or the transitive property of equality, we get 4(x+5) = 4(3 x) = x = 1. Substituting this value of x into the second equation, we have 4[3 ( 1)] = (y 3) 2 = 16 = (y 3) 2 = y = 7 or y = 1. The solutions are ( 1,7) and ( 1, 1), same as Solution 1. Example Solve the system and graph the curves: (x 3) 2 +(y 5) 2 = 10 x 2 +(y +1) 2 =

33 Solution. Expanding both equations, we obtain x 2 +y 2 6x 10y +24 = 0 x 2 +y 2 +2y 24 = 0. Teaching Notes Subtracting these two equations, we get Because the equation x+2y 8 = 0 is obtained by combining the two equations (through substraction), this equation also contains the solutions of the original system. In fact, this is the line passing through the common points of the two circles. 6x 12y +48 = 0 = x+2y 8 = 0 x = 8 2y. We can substitute x = 8 2y to either the first equation or the second equation. For convenience, we choose the second equation. x 2 +y 2 +2y 24 = 0 (8 2y) 2 +y 2 +2y 24 = 0 y 2 6y +8 = 0 y = 2 or y = 4 y = 2 = x = 8 2(2) = 4 and y = 4 = x = 8 2(4) = 0 The solutions are (4,2) and (0,4). The graphs of both equations are circles. One has center (3,5) and radius 10, while the other has center (0, 1) and radius 5. The graphs with the points of intersection are show below. 77

34 Seatwork/Homework Solve the system, and graph the curves in one Cartesian plane showing the point(s) of intersection. 1. x 2 +y 2 = 25 x y2 32 = 1 Answer: (3,4), ( 3,4), (3, 4), and ( 3, 4) 2. x 2 +2y 12 = 0 x 2 +y 2 = 36 Answer: (0,6), ( 2 5, 4 ), and ( 2 5, 4 ) 78

35 (x 1) 2 +(y 3) 2 = x 2 +(y 1) 2 = 5 Answer: ( 2,2) and (2,0) Applications of Systems of Nonlinear Equations As we expect, systems of equations are important in applications. In this session, we consider some of them. Example The screen size of television sets is given in inches. This indicatesthelengthofthediagonal. Screensofthesamesizecancomeindifferent shapes. Wide-screen TV s usually have screens with aspect ratio 16 : 9, indicating the ratio of the width to the height. Older TV models often have aspect ratio 4 : 3. A 40-inch LED TV has screen aspect ratio 16 : 9. Find the length and the width of the screen. Solution. Let w represent the width and h the height of the screen. Then, by Pythagorean Theorem, we have the system w 2 +h 2 = 40 2 = w 2 +h 2 = 1600 w h = 16 9 = h = 9w 16 79

36 w 2 +h 2 = 1600 = w 2 + ( ) 2 9w = w = w = h = 19x 16 19(34.86) 16 = Therefore, a 40-inch TV with aspect ratio 16 : 9 is about inches wide and inches high. Seatwork/Homework From a circular piece of metal sheet with diameter 20 cm, a rectangular piece with perimeter 28 cm is to be cut as shown. Find the dimensions of the rectangular piece. Answer: 6 cm 8 cm Exercises Solve the system, and graph the curves. y = 2x+4 (a) y = 2x 2 Answer: ( 1,2) and (2,8) 80

37 x 2 +y 2 = 25 (b) 2x 3y = 6 Answer: (3,4) and ( 63 13, ) (c) x 2 +y 2 = 12 x 2 y 2 = 4 Answer: ( 2 2,2 ), ( 2 2,2 ), ( 2 2, 2 ), and ( 2 2, 2 ) 81

38 x 2 4y 2 = 200 (d) x+2y = 100 Answer: ( ) 51, 49 2 (e) 1 4 (x+1)2 (y +2) 2 = 1 (y +2) 2 = 1(x 1) 4 ( Answer: (1, 2), 4, ) (, and 4, ) 82

39 2. A laptop has screen size 13 inches with aspect ratio 5 : 4. Find the length and the width of the screen. Answer: in 8.12 in 3. What are the dimensions of a rectangle whose perimeter is 50 cm and diagonal 18 cm? Answer: 14.9 cm 10.1 cm 4. Thegraphof2xy y 2 +5x+20 = 0isarotatedhyperbola. Findtheintersection of this hyperbola with the graph of 3x+2y = 3. (The graph is not required.) Answer: ( 1,3), ( ) 71 21, For what values of a will the system x 2 +y 2 +2x 1 = 0 x y +a = 0 have only one solution? Answer: a = 1 or a =

40 Unit 2 Mathematical Induction rice terraces in Ifugao.jpg By Ericmontalban (Own work) [CC BY-SA 3.0 ( via Wikimedia Commons Listed as one of the UNESCO World Heritage sites since 1995, the two-millenniumold Rice Terraces of the Philippine Cordilleras by the Ifugaos is a living testimony of mankind s creative engineering to adapt with physically-challenging environment in nature. One of the five clusters of terraces inscribed in the UNESCO list is the majestic Batad terrace cluster (shown above), which is characterized by its amphitheater-like semi-circular terraces with a village at its base.

41 Lesson 2.1. Review of Sequences and Series Time Frame: 1 one-hour session Learning Outcomes of the Lesson At the end of the lesson, the student is able to: (1) illustrate a series; and (2) differentiate a series from a sequence. Lesson Outline (1) Sequences and series (2) Different types of sequences and series (Fibonacci sequence, arithmetic and geometric sequence and series, and harmonic series) (3) Difference between sequence and series Introduction Pose the following problem to the class: Jason s classroom is on the second floor of the school. He can take one or two steps of the stairs in one leap. In how many ways can Jason climb the stairs if it has 16 steps? Get students to suggest strategies they can use to solve this problem. Lead or encourage them to try out smaller number of steps and find a pattern. Work with the class to complete the following table (on the board): 85

42 Number of Steps in the Stairs Number of Ways to Climb the Stairs The students should be able to recognize the Fibonacci sequence. Ask the students to recall what Fibonacci sequences are and where they had encountered this sequence before. In this lesson, we will review the definitions and different types of sequences and series.. Teaching Notes This is equivalent to the number of ways to express a number (number of steps in the stairs) as a sum of 1 s and 2 s. For example, we can write 3 as a sum of 1 s and 2 s in three ways: 2+1, 1+2, and In 2+1, it means Jason leaps 2 steps first, then 1 step to finish the three-step stairs. Lesson Proper Recall the following definitions: A sequence is a function whose domain is the set of positive integers or the set {1,2,3,...,n}. A series represents the sum of the terms of a sequence. If a sequence is finite, we will refer to the sum of the terms of the sequence as the series associated with the sequence. If the sequence has infinitely many terms, the sum is defined more precisely in calculus. A sequence is a list of numbers (separated by commas), while a series is a sum of numbers (separated by + or sign). As an illustration, 1, 1, 1 2 3, 1 4 is a sequence, and = 7 is its associated series The sequence with nth term a n is usually denoted by {a n }, and the associated series is given by S = a 1 +a 2 +a 3 + +a n. 86

43 Example Determine the first five terms of each defined sequence, and give their associated series. (1) {2 n} (3) {( 1) n } (2) {1+2n+3n 2 } (4) { n} Solution. We denote the nth term of a sequence by a n, and S = a 1 +a 2 +a 3 + a 4 +a 5. (1) a n = 2 n First five terms: a 1 = 2 1 = 1, a 2 = 2 2 = 0, a 3 = 1, a 4 = 2, a 5 = 3 Associated series: S = a 1 +a 2 +a 3 +a 4 +a 5 = = 5 (2) a n = 1+2n+3n 2 First five terms: a 1 = = 6, a 2 = 17, a 3 = 34, a 4 = 57, a 5 = 86 Associated series: S = = 200 (3) a n = ( 1) n First five terms: a 1 = ( 1) 1 = 1, a 2 = ( 1) 2 = 1, a 3 = 1, a 4 = 1, a 5 = 1 Associated series: S = = 1 (4) a n = n First five terms: a 1 = 1, a 2 = 1+2 = 3, a 3 = = 6, a 4 = = 10, a 5 = = 15 Associated series: S = = 35 An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant (called the common difference) to the preceding term. If the nth term of an arithmetic sequence is a n and the common difference is d, then a n = a 1 +(n 1)d. The associated arithmetic series with n terms is given by S n = n(a 1 +a n ) 2 = n[2a 1 +(n 1)d]. 2 87

44 A geometric sequence is a sequence in which each term after the first is obtained by multiplying the preceding term by a constant (called the common ratio). If the nth term of a geometric sequence is a n and the common ratio is r, then a n = a 1 r n 1. The associated geometric series with n terms is given by na 1 if r = 1 S n = a 1 (1 r n ) if r 1. (1 r) The proof of this sum formula is an example in Lesson 2.3. When 1 < r < 1, the infinite geometric series has a sum, and is given by a 1 +a 1 r+a 1 r 2 + +a 1 r n 1 + S = a 1 1 r. If {a n } is an arithmetic sequence, then the sequence with nth term b n = 1 a n is a harmonic sequence. Teaching Notes The proof of the fact that the infinite geometric series a 1 +a 1 r + has a sum when r < 1 is beyond the scope of Precalculus, and can be found in university Calculus. Seatwork/Homework 1. Write SEQ if the given item is a sequence, and write SER if it is a series. (a) 1,2,4,8,... (b) 2,8,10,18,... (c) (d) 1, 2, 3, 4, (e) (f) Answer: SEQ Answer: SEQ Answer: SER Answer: SEQ Answer: SER Answer: SER 2. Write A if the sequence is arithmetic, G if it is geometric, F if Fibonacci, and O if it is not one of the mentioned types. (a) 3,5,7,9,11,... Answer: A 88

45 (b) 2,4,9,16,25,... Answer: O (c) 1, 1 1, Answer: G (d) 1, 2, 3, Answer: O (e) 1, 1, 1, Answer: A (f) 4,6,10,16,26,... Answer: F (g) 3, 4, 5, 6,... (h) 0.1,0.01,0.001,0.0001,... Answer: O Answer: G 3. Determine the first five terms of each defined sequence, and give their associated series. (a) {1+n n 2 } Answer: a 1 = 1, a 2 = 1, a 3 = 5, a 4 = 11, a 5 = 19 Associated series: = 35 (b) {1 ( 1) n+1 } Answer: a 1 = 0, a 2 = 2, a 3 = 0, a 4 = 2, a 5 = 0 Associated series: = 4 (c) a 1 = 3 and a n = 2a n 1 +3 for n 2 Answer: a 1 = 3, a 2 = 9, a 3 = 21, a 4 = 45, a 5 = 93 Associated series: = 35 (d) {1 2 3 n} Answer: a 1 = 1, a 2 = 1 2 = 2, a 3 = = 6, a 4 = 24, a 5 = 120 Associated series: = Identify the series (and write NAGIG if it is not arithmetic, geometric, and infinite geometric series), and determine the sum (and write NO SUM if it cannot be summed up). (a) Answer: Arithmetic, 442 (b) Answer: Geometric, (c) Answer: NAGIG, 220 (d) Answer: Arithmetic, 144 (e) Answer: Infinite geometric, 12.5 (f) Answer: NAGIG, NO SUM (g) Answer: Infinite geometric,

46 Lesson 2.2. Sigma Notation Time Frame: 2 one-hour sessions Learning Outcomes of the Lesson At the end of the lesson, the student is able to use the sigma notation to represent a series. Lesson Outline (1) Definition of and writing in sigma notation (2) Evaluate sums written in sigma notation (3) Properties of sigma notation (4) Calculating sums using the properties of sigma notation Introduction The sigma notation is a shorthand for writing sums. In this lesson, we will see the power of this notation in computing sums of numbers as well as algebraic expressions Writing and Evaluating Sums in Sigma Notation Mathematicians use the sigma notation to denote a sum. The uppercase Greek letter Σ (sigma) is used to indicate a sum. The notation consists of several components or parts. Let f(i) be an expression involving an integer i. The expression f(m)+f(m+1)+f(m+2)+ +f(n) can be compactly written in sigma notation, and we write it as n f(i), i=m Teaching Notes Emphasize that the value of i starts at m, increases by 1, and ends at n. which is read the summation of f(i) from i = m to n. Here, m and n are integers with m n, f(i) is a term (or summand) of the summation, and the letter i is the index, m the lower bound, and n the upper bound. 90

47 Example Expand each summation, and simplify if possible. 4 n (1) (2i+3) (3) a i (2) i=2 5 i=0 2 i (4) i=1 6 n=1 n n+1 Solution. We apply the definition of sigma notation. (1) (2) (3) (4) 4 (2i+3) = [2(2)+3]+[2(3)+3]+[2(4)+3] = 27 i=2 5 2 i = = 63 i=0 n a i = a 1 +a 2 +a 3 + +a n i=1 6 n=1 n n+1 = Example Write each expression in sigma notation. (1) (2) (3) a 2 +a 4 +a 6 +a 8 + +a 20 (4) Solution. (1) = 1 n (2) n=1 = ( 1) 1 1+( 1) 2 2+( 1) 3 3+( 1) 4 4 +( 1) ( 1) = ( 1) j j j=1 91

48 (3) a 2 +a 4 +a 6 +a 8 + +a 20 = a 2(1) +a 2(2) +a 2(3) +a 2(4) + +a 2(10) 10 = i=1 a 2i (4) = 1 2 k The sigma notation of a sum expression is not necessarily unique. For example, the last item in the preceding example can also be expressed in sigma notation as follows: k= = 1 2 k 1. However, this last sigma notation is equivalent to the one given in the example. Seatwork/Homework Expand each summation, and simplify if possible. (a) (b) (c) (d) (e) 5 (2 3k) Answer: 28 k= 1 n j=1 x j k=1 Answer: x+x 2 +x 3 + +x n 6 (j 2 j) Answer: 68 j=3 4 ( 1) k+1 k k=1 Answer: 2 3 (a n+1 a n ) Answer: a 4 a 1 n=1 2. Write each expression in sigma notation. (a) x+2x 2 +3x 3 +4x 4 +5x 5 (b) Answer: 5 Answer: kx k 10 k=1 k=1 ( 1) k+1 k 92

49 Teaching Notes Equivalent answer: = 51 (2k 1) k=1 (c) (d) a 4 +a 8 +a 12 +a 16 (e) Answer: 50 k=0 Answer: Answer: 4 k=0 (2k +1) 4 k=1 a 4k ( 1) k 2k Properties of Sigma Notation We start with finding a formula for the sum of n i = n i=1 in terms of n. The sum can be evaluated in different ways. A simple, though informal, approach is pictorial. Teaching Notes This illustration can be done with manipulatives, and allow the students to guess. n i=1 i = n = n(n+1) 2 Another way is to use the formula for an arithmetic series with a 1 = 1 and a n = n: S = n(a 1 +a n ) = n(n+1). 2 2 We now derive some useful summation facts. They are based on the axioms of arithmetic addition and multiplication. 93