Secton 8. Polar Form of Complex Numbers 1 Chapter 8: Further Applcatons of Trgonometry In ths chapter, we wll explore addtonal applcatons of trgonometry. We wll begn wth an extenson of the rght trangle trgonometry we explored n Chapter 5 to stuatons nvolvng non-rght trangles. We wll explore the polar coordnate system and parametrc equatons as new ways of descrbng curves n the plane. In the process, we wll ntroduce vectors and an alternatve way of wrtng complex numbers, two mportant mathematcal tools we use when analyzng and modelng the world around us. Secton 8. Polar Form of Complex Numbers... Secton 8.4 Vectors... 1 Secton 8.5 Parametrc Equatons... 6
Secton 8. Polar Form of Complex Numbers From prevous classes, you may have encountered magnary numbers the square roots of negatve numbers and, more generally, complex numbers whch are the sum of a real number and an magnary number. Whle these are useful for expressng the solutons to quadratc equatons, they have much rcher applcatons n electrcal engneerng, sgnal analyss, and other felds. Most of these more advanced applcatons rely on propertes that arse from lookng at complex numbers from the perspectve of polar coordnates. We wll begn wth a revew of the defnton of complex numbers. Imagnary Number The most basc complex number s, defned to be 1, commonly called an magnary number. Any real multple of s also an magnary number. Example 1 Smplfy 9. We can separate 9 as 9 1. We can take the square root of 9, and wrte the square root of -1 as. 9 = 9 1 A complex number s the sum of a real number and an magnary number. Complex Number A complex number s a number z a b, where a and b are real numbers a s the real part of the complex number b s the magnary part of the complex number 1 Plottng a complex number We can plot real numbers on a number lne. For example, f we wanted to show the number, we plot a pont:
Secton 8. Polar Form of Complex Numbers To plot a complex number lke 4, we need more than just a number lne snce there are two components to the number. To plot ths number, we need two number lnes, crossed to form a complex plane. Complex Plane In the complex plane, the horzontal axs s the real axs and the vertcal axs s the magnary axs. magnary real Example Plot the number 4 on the complex plane. The real part of ths number s, and the magnary part s -4. To plot ths, we draw a pont unts to the rght of the orgn n the horzontal drecton and 4 unts down n the vertcal drecton. magnary real Because ths s analogous to the Cartesan coordnate system for plottng ponts, we can thnk about plottng our complex number z a b as f we were plottng the pont (a, b) n Cartesan coordnates. Sometmes people wrte complex numbers as z x y to hghlght ths relaton. Arthmetc on Complex Numbers Before we dve nto the more complcated uses of complex numbers, let s make sure we remember the basc arthmetc nvolved. To add or subtract complex numbers, we smply add the lke terms, combnng the real parts and combnng the magnary parts. Example Add 4 and 5. Addng ( 4) ( 5), we add the real parts and the magnary parts 4 5 5 Try t Now 1. Subtract 5 from 4. We can also multply and dvde complex numbers.
Example 4 Multply: 4( 5). To multply the complex number by a real number, we smply dstrbute as we would when multplyng polynomals. 4( 5) = 4 45 8 0 Example 5 Dvde ( 5 ). (4 ) To dvde two complex numbers, we have to devse a way to wrte ths as a complex number wth a real part and an magnary part. We start ths process by elmnatng the complex number n the denomnator. To do ths, we multply the numerator and denomnator by a specal complex number so that the result n the denomnator s a real number. The number we need to multply by s called the complex conjugate, n whch the sgn of the magnary part s changed. Here, 4+ s the complex conjugate of 4. Of course, obeyng our algebrac rules, we must multply by 4+ on both the top and bottom. ( 5 ) (4 ) (4 ) (4 ) To multply two complex numbers, we expand the product as we would wth polynomals (the process commonly called FOIL frst outer nner last ). In the numerator: ( 5 )(4 ) Expand 8 0 5 Snce 1, 1 8 0 5( 1) Smplfy Followng the same process to multply the denomnator (4 )(4 ) Expand (16 4 4 ) Snce 1, 1 (16 ( 1)) =17 Combnng ths we get 17 17 17
Secton 8. Polar Form of Complex Numbers 5 Try t Now. Multply 4 and. Wth the nterpretaton of complex numbers as ponts n a plane, whch can be related to the Cartesan coordnate system, you mght be startng to guess our next step to refer to ths pont not by ts horzontal and vertcal components, but usng ts polar locaton, gven by the dstance from the orgn and an angle. Polar Form of Complex Numbers Remember, because the complex plane s analogous to the Cartesan plane that we can thnk of a complex number z x y as analogous to the Cartesan pont (x, y) and recall how we converted from (x, y) to polar (r, θ) coordnates n the last secton. Brngng n all of our old rules we remember the followng: x cos( ) r x r cos( ) y sn( ) r y r sn( ) y tan( ) x y r x magnary r θ x x + y y real Wth ths n mnd, we can wrte z x y r cos( ) r sn( ). Example 6 Express the complex number 4 usng polar coordnates. On the complex plane, the number 4 s a dstance of 4 from the orgn at an angle of, so 4 4cos 4sn Note that the real part of ths complex number s 0. 4 4 In the 18 th century, Leonhard Euler demonstrated a relatonshp between exponental and trgonometrc functons that allows the use of complex numbers to greatly smplfy some trgonometrc calculatons. Whle the proof s beyond the scope of ths class, you wll lkely see t n a later calculus class.
Polar Form of a Complex Number and Euler s Formula The polar form of a complex number s z re, where Euler s Formula holds: re r cos( ) r sn( ) Smlar to plottng a pont n the polar coordnate system we need r and to fnd the polar form of a complex number. Example 7 Fnd the polar form of the complex number -7. Treatng ths s a complex number, we can consder the unsmplfed verson -7+0. Plotted n the complex plane, the number -7 s on the negatve horzontal axs, a dstance of 7 from the orgn at an angle of π from the postve horzontal axs. The polar form of the number -7 s 7 e. Pluggng r = 7 and θ = π back nto Euler s formula, we have: 7e 7cos( ) 7 sn( ) 7 0 7 as desred. Example 8 Fnd the polar form of 4 4. On the complex plane, ths complex number would correspond to the pont (-4, 4) on a Cartesan plane. We can fnd the dstance r and angle θ as we dd n the last secton. r r x y ( 4) 4 r 4 To fnd θ, we can use cos( ) 4 cos( ) 4 Ths s one of known cosne values, and snce the pont s n the second quadrant, we can conclude that. 4 The polar form of ths complex number s x r 4 e 4. -4+4 4 4
Secton 8. Polar Form of Complex Numbers 7 Note we could have used check the quadrant. y tan( ) nstead to fnd the angle, so long as we remember to x Try t Now. Wrte n polar form. Example 9 Wrte e 6 n complex a b form. e 6 cos sn 6 6 1 Smplfy Evaluate the trg functons The polar form of a complex number provdes a powerful way to compute powers and roots of complex numbers by usng exponent rules you learned n algebra. To compute a power of a complex number, we: 1) Convert to polar form ) Rase to the power, usng exponent rules to smplfy ) Convert back to a + b form, f needed Example 10 Evaluate 4 4 6. Whle we could multply ths number by tself fve tmes, that would be very tedous. To compute ths more effcently, we can utlze the polar form of the complex number. In an earler example, we found that 4 4 6 4 4 4 4 e. Usng ths, Wrte the complex number n polar form 4 4 e 6 4 e 6 4 6 Utlze the exponent rule ( m m ab) a m n On the second factor, use the rule ( a ) a b m mn
4 6 768 e 6 4 9 e Smplfy At ths pont, we have found the power as a complex number n polar form. If we want the answer n standard a + b form, we can utlze Euler s formula. 9 768e 9 768cos 9 768sn 9 Snce s cotermnal wth, we can use our specal angle knowledge to evaluate the sne and cosne. 9 768cos 9 768sn 6 We have found that 4 4 768. 768(0) 768(1) 768 The result of the process can be summarzed by DeMovre s Theorem. DeMovre s Theorem z r cos sn n n If, then for any nteger n, z r cos n sn n We omt the proof, but note we can easly verfy t holds n one case usng Example 10: 6 6 9 9 ( 4 4) 4 cos 6 sn6 768 cos sn 768 4 4 Example 11 Evaluate 9. To evaluate the square root of a complex number, we can frst note that the square root 1 s the same as havng an exponent of : 1/ 9 (9). To evaluate the power, we frst wrte the complex number n polar form. Snce 9 has no real part, we know that ths value would be plotted along the vertcal axs, a dstance of 9 from the orgn at an angle of. Ths gves the polar form: 9e 9.
Secton 8. Polar Form of Complex Numbers 9 9 = 9e 1/ (9 ) Use the polar form 1/ 9 9 1/ e e 1/ 1 1/ Use exponent rules to smplfy Smplfy e 4 Rewrte usng Euler s formula f desred cos sn 4 4 Evaluate the sne and cosne Usng the polar form, we were able to fnd a square root of a complex number. 9 Alternatvely, usng DeMovre s Theorem we can wrte 9e 1/ cos sn 4 4 and smplfy Try t Now 4. Wrte 6 n polar form. You may remember that equatons lke x 4 have two solutons, and - n ths case, though the square root 4 only gves one of those solutons. Lkewse, the square root we found n Example 11 s only one of two complex numbers whose square s 9. Smlarly, the equaton z 8 would have three solutons where only one s gven by the cube root. In ths case, however, only one of those solutons, z =, s a real value. To fnd the others, we can use the fact that complex numbers have multple representatons n polar form. Example 1 Fnd all complex solutons to z 8.
1/ Snce we are tryng to solve z 8, we can solve for x as z 8. Certanly one of these solutons s the basc cube root, gvng z =. To fnd others, we can turn to the polar representaton of 8. Snce 8 s a real number, s would st n the complex plane on the horzontal axs at an angle of 0, gvng the polar form 8 e 0. Takng the 1/ power of ths gves the real soluton: 0 8 1/ 1/ 0 0 e 8 e 1/ e cos(0) sn(0) However, snce the angle π s cotermnal wth the angle of 0, we could also represent the number 8 as 8 e. Takng the 1/ power of ths gves a frst complex soluton: 1/ 1/ 1 8e 8 1/ e e cos sn 1 To fnd the thrd root, we use the angle of 4π, whch s also cotermnal wth an angle of 0. 4 1/ 1/ 4 4 1 4 8e 8 1/ 4 e e cos sn 1 Altogether, we found all three complex solutons to z 8, 1 z, 1, 1 Graphed, these three numbers would be equally spaced on a crcle about the orgn at a radus of. 1 Important Topcs of Ths Secton Complex numbers Imagnary numbers Plottng ponts n the complex coordnate system Basc operatons wth complex numbers Euler s Formula DeMovre s Theorem Fndng complex solutons to equatons Try t Now Answers 1. ( 4 ) ( 5 ) 1 9. ( 4 )( ) 18 6. n polar form s 4. 64 e
Secton 8. Polar Form of Complex Numbers 11 Secton 8. Exercses Smplfy each expresson to a sngle complex number. 1. 9. 16. 6 4 4. 75 5. 1 6. 4 0 Smplfy each expresson to a sngle complex number. (5 ) 4 1 6 7. 8. 9. 5 (6 ) 10. 11. (4 ) 1. ( ) 5 ( ) 1. 6 (5) 14. 48 15. (4 ) 16. 1 ( ) 17. 4 (4 ) 18. 4 4 19. 4 1. 5. 4 0. 6. 6 4 4. 4 5. 6 6. 11 7. 17 8. 4 Rewrte each complex number from polar form nto a b form. 9. e 0. 4 4 6 6 e 1. e 8e.. e 5 4 4. 5e 7 4 Rewrte each complex number nto polar 5. 6 6. 8 re form. 7. 4 8. 6 9. 40. 4 4 41. 4. 4 4 4. 5 44. 4 7 45. 46. 47. 1 4 48. 6 49. 5 50. 1
Compute each of the followng, leavng the result n polar 6 4 51. e e 5 5. e 4e re 5. form. 6e e 4 6 54. 4 4e 55. 6e 4 e 10 56. 6 e 4 57. 16 e 58. 9e Compute each of the followng, smplfyng the result nto a b form. 59. 8 60. 4 4 6 61. 6. 4 4 6. 5 64. 4 4 7 Solve each of the followng equatons for all complex solutons. 5 7 6 8 65. z 66. z 67. z 1 68. z 1
Secton 8.4 Vectors 1 Secton 8.4 Vectors A woman leaves home, walks mles north, then mles southeast. How far s she from home, and n whch drecton would she need to walk to return home? How far has she walked by the tme she gets home? Ths queston may seem famlar ndeed we dd a smlar problem wth a boat n the frst secton of ths chapter. In that secton, we solved the problem usng trangles. In ths secton, we wll nvestgate another way to approach the problem usng vectors, a geometrc entty that ndcates both a dstance and a drecton. We wll begn our nvestgaton usng a purely geometrc vew of vectors. A Geometrc Vew of Vectors Vector A vector s an object that has both a length and a drecton. Geometrcally, a vector can be represented by an arrow that has a fxed length and ndcates a drecton. If, startng at the pont A, a vector, whch means carrer n Latn, moves toward pont B, we wrte AB to represent the vector. A vector may also be ndcated usng a sngle letter n boldface type, lke u, or by cappng the letter representng the vector wth an arrow, lke u. Example 1 Fnd a vector that represents the movement from the pont P(-1, ) to the pont Q(,) By drawng an arrow from the frst pont to the second, we can construct a vector PQ. P Q Usng ths geometrc representaton of vectors, we can vsualze the addton and scalng of vectors. To add vectors, we envson a sum of two movements. To fnd u v, we frst draw the vector u, then from the end of u we drawn the vector v. Ths corresponds to the noton that frst we move along the frst vector, and then from that end poston we move along the second vector. The sum u v s the new vector that travels drectly from the begnnng of u to the end of v n a straght path.
Addng Vectors Geometrcally To add vectors geometrcally, draw v startng from the end of u. The sum u v s the vector from the begnnng of u to the end of v. u v u v Example Gven the two vectors shown below, draw u v u v v We draw v startng from the end of u, then draw n the sum u v from the begnnng of u to the end of v. u u v Notce that path of the walkng woman from the begnnng of the secton could be vsualzed as the sum of two vectors. The resultng sum vector would ndcate her end poston relatve to home. Try t Now 1. Draw a vector, v, that travels from the orgn to the pont (, 5). Note that although vectors can exst anywhere n the plane, f we put the startng pont at the orgn t s easy to understand ts sze and drecton relatve to other vectors. To scale vectors by a constant, such as u, we can magne addng u u u. The result wll be a vector three tmes as long n the same drecton as the orgnal vector. If we were to scale a vector by a negatve number, such as u, we can envson ths as the opposte of u ; the vector so that u ( u) returns us to the startng pont. Ths vector u would pont n the opposte drecton as u but have the same length. Another way to thnk about scalng a vector s to mantan ts drecton and multply ts length by a constant, so that u would pont n the same drecton but wll be tmes as long.
Secton 8.4 Vectors 15 Scalng a Vector Geometrcally To geometrcally scale a vector by a constant, scale the length of the vector by the constant. Scalng a vector by a negatve constant wll reverse the drecton of the vector. Example Gven the vector shown, draw u, u, and u. u The vector u wll be three tmes as long. The vector u wll have the same length but pont n the opposte drecton. The vector u wll pont n the opposte drecton and be twce as long. u u u By combnng scalng and addton, we can fnd the dfference between vectors geometrcally as well, snce u v u ( v). Example 4 Gven the vectors shown, draw u v u From the end of u we draw v v, then draw n the result. u v v u Notce that the sum and dfference of two vectors are the two dagonals of a parallelogram wth the vectors u and v as edges. u u v v u v v u Try t Now. Usng vector v from Try t Now #1, draw v.
u θ Component Form of Vectors Whle the geometrc nterpretaton of vectors gves us an ntutve understandng of vectors, t does not provde us a convenent way to do calculatons. For that, we need a handy way to represent vectors. Snce a vector nvolves a length and drecton, t would be logcal to want to represent a vector usng a length and an angle θ, usually measured from standard poston. Magntude and Drecton of a Vector A vector u can be descrbed by ts magntude, or length, u, and an angle θ. A vector wth length 1 s called unt vector. Whle ths s very reasonable, and a common way to descrbe vectors, t s often more convenent for calculatons to represent a vector by horzontal and vertcal components. Component Form of a Vector The component form of a vector represents the vector usng two components. u x, y ndcates the vector represents a dsplacement of x unts horzontally and y unts vertcally. u θ x y Notce how we can see the magntude of the vector as the length of the hypotenuse of a rght trangle, or n polar form as the radus, r. Alternate Notaton for Vector Components Sometmes you may see vectors wrtten as the combnaton of unt vectors and j, where ponts the rght and j ponts up. In other words, 1, 0 and j 0, 1. In ths notaton, the vector u, 4 would be wrtten as u 4 j snce both ndcate a dsplacement of unts to the rght, and 4 unts down. Whle t can be convenent to thnk of the vector u x, y as an arrow from the orgn to the pont (x, y), be sure to remember that most vectors can be stuated anywhere n the plane, and smply ndcate a dsplacement (change n poston) rather than a poston.
Secton 8.4 Vectors 17 It s common to need to convert from a magntude and angle to the component form of the vector and vce versa. Happly, ths process s dentcal to convertng from polar coordnates to Cartesan coordnates, or from the polar form of complex numbers to the a+b form. Example 5 Fnd the component form of a vector wth length 7 at an angle of 15 degrees. Usng the converson formulas x r cos( ) and y r sn( ), we can fnd the components 7 x 7cos(15) 7 y 7sn(15) Ths vector can be wrtten n component form as 7 7,. Example 6 Fnd the magntude and angle representng the vector u,. Frst we can fnd the magntude by rememberng the relatonshp between x, y and r: r ( ) 1 r 1 Second we can fnd the angle. Usng the tangent, tan( ) tan 1.69, or wrtten as a cotermnal postve angle, 6.1, because we know our pont les n the 4 th quadrant. Try t Now. Usng vector v from Try t Now #1, the vector that travels from the orgn to the pont (, 5), fnd the components, magntude and angle that represent ths vector.
In addton to representng dstance movements, vectors are commonly used n physcs and engneerng to represent any quantty that has both drecton and magntude, ncludng veloctes and forces. Example 7 An object s launched wth ntal velocty 00 meters per second at an angle of 0 degrees. Fnd the ntal horzontal and vertcal veloctes. By vewng the ntal velocty as a vector, we can resolve the vector nto horzontal and vertcal components. x 00cos(0) 00 17.05 m/sec 00 m/s 1 y 00sn(0) 00 100 m/sec 0 17 m/s 100 m/s Ths tells us that, absent wnd resstance, the object wll travel horzontally at about 17 meters each second. Gravty wll cause the vertcal velocty to change over tme we ll leave a dscusson of that to physcs or calculus classes. Addng and Scalng Vectors n Component Form To add vectors n component form, we can smply add the correspondng components. To scale a vector by a constant, we scale each component by that constant. Combnng Vectors n Component Form To add, subtract, or scale vectors n component form If u u,u 1, v v,v 1, and c s any constant, then u v u1 v1, u v u v u1 v1, u v cu cu,cu 1 Example 8 Gven u, and v 1, 4, fnd a new vector w u v Usng the vectors gven, w u v, 1,4 Scale each vector 9, 6,8 Subtract correspondng components 11, 14
Secton 8.4 Vectors 19 By representng vectors n component form, we can fnd the resultng dsplacement vector after a multtude of movements wthout needng to draw a lot of complcated nonrght trangles. For a smple example, we revst the problem from the openng of the secton. The general procedure we wll follow s: 1) Convert vectors to component form ) Add the components of the vectors ) Convert back to length and drecton f needed to sut the context of the queston Example 9 A woman leaves home, walks mles north, then mles southeast. How far s she from home, and what drecton would she need to walk to return home? How far has she walked by the tme she gets home? Let s begn by understandng the queston n a lttle more depth. When we use vectors to descrbe a travelng drecton, we often poston thngs so north ponts n the upward drecton, east ponts to the rght, and so on, as pctured here. NW W N NE E Consequently, travellng NW, SW, NE or SE, means we are travellng through the quadrant bordered by the gven drectons at a 45 degree angle. SW S SE Wth ths n mnd, we begn by convertng each vector to components. A walk mles north would, n components, be 0,. A walk of mles southeast would be at an angle of 45 South of East, or measurng from standard poston the angle would be 15. Convertng to components, we choose to use the standard poston angle so that we do not have to worry about whether the sgns are negatve or postve; they wll work out automatcally. cos(15 ), sn(15), 1.414, 1.414 Addng these vectors gves the sum of the movements n component form 0, 1.414, 1.414 1.414,1.586 To fnd how far she s from home and the drecton she would need to walk to return home, we could fnd the magntude and angle of ths vector. Length = 1.414 1.586. 15
To fnd the angle, we can use the tangent 1.586 tan( ) 1.414 1.586 tan 1 48.7 north of east 1.414 Of course, ths s the angle from her startng pont to her endng pont. To return home, she would need to head the opposte drecton, whch we could ether descrbe as 180 +48.7 = 8.7 measured n standard poston, or as 48.7 south of west (or 41.77 west of south). She has walked a total dstance of + +.15 = 7.15 mles. Keep n mnd that total dstance travelled s not the same as the length of the resultng dsplacement vector or the return vector. Try t Now 4. In a scavenger hunt, drectons are gven to fnd a bured treasure. From a startng pont at a flag pole you must walk 0 feet east, turn 0 degrees to the north and travel 50 feet, and then turn due south and travel 75 feet. Sketch a pcture of these vectors, fnd ther components, and calculate how far and n what drecton you must travel to go drectly to the treasure from the flag pole wthout followng the map. Whle usng vectors s not much faster than usng law of cosnes wth only two movements, when combnng three or more movements, forces, or other vector quanttes, usng vectors quckly becomes much more effcent than tryng to use trangles. Example 10 Three forces are actng on an object as shown below, each measured n Newtons (N). What force must be exerted to keep the object n equlbrum (where the sum of the forces s zero)? 7 N 6 N 0 00 4 N We start by resolvng each vector nto components.
Secton 8.4 Vectors 1 The frst vector wth magntude 6 Newtons at an angle of 0 degrees wll have components 1 6cos(0 ),6sn(0) 6,6, The second vector s only n the horzontal drecton, so can be wrtten as 7, 0. The thrd vector wth magntude 4 Newtons at an angle of 00 degrees wll have components 1 4cos(00 ),4sn(00) 4,4, To keep the object n equlbrum, we need to fnd a force vector the four vectors s the zero vector, 0, 0., 7, 0, x, y 0, 0 Add component-wse 7, 0 x, y 0, 0 Smplfy 5, x, y 0, 0 Solve x, y 0, 0 5, x, y 5, 0.196, 0.464 x, y so the sum of Ths vector gves n components the force that would need to be appled to keep the object n equlbrum. If desred, we could fnd the magntude of ths force and drecton t would need to be appled n. Magntude = ( 0.196) 0.464 0. 504 N Angle: 0.464 tan( ) 0.196 0.464 tan 1 67.089. 0.196 Ths s n the wrong quadrant, so we adjust by fndng the next angle wth the same tangent value by addng a full perod of tangent: 67.089 180 11. 911 To keep the object n equlbrum, a force of 0.504 Newtons would need to be appled at an angle of 11.911.
Important Topcs of Ths Secton Vectors, magntude (length) & drecton Addton of vectors Scalng of vectors Components of vectors Vectors as velocty Vectors as forces Addng & Scalng vectors n component form Total dstance travelled vs. total dsplacement Try t Now Answers v v 1. 1 5 1. v,5 magntude 4 tan 59.04. 0 ft 50 ft 75 ft v 1 v f 0,0 v 50cos(0),50sn(0) v 0, 75 0 50cos(0),50sn(0) 75 7.01, 50 Magntude = 88.7 feet at an angle of 4. south of east.
Secton 8.4 Vectors Secton 8.4 Exercses Wrte the vector shown n component form. 1.. Gven the vectors shown, sketch u v, u v, and u.. 4. Wrte each vector below as a combnaton of the vectors u and v from queston #. 5. 6. From the gven magntude and drecton n standard poston, wrte the vector n component form. 7. Magntude: 6, Drecton: 45 8. Magntude: 10, Drecton: 10 9. Magntude: 8, Drecton: 0 10. Magntude: 7, Drecton: 05 Fnd the magntude and drecton of the vector. 11. 0, 4 1., 0 1. 6, 5 14., 7 15., 1 16. 10, 1 17., 5 18. 8, 4 19. 4, 6 0. 1, 9 Usng the vectors gven, compute u v, u v, and u v. 1. u,, v 1, 5. u,4, v, 1
. A woman leaves home and walks mles west, then mles southwest. How far from home s she, and n what drecton must she walk to head drectly home? 4. A boat leaves the marna and sals 6 mles north, then mles northeast. How far from the marna s the boat, and n what drecton must t sal to head drectly back to the marna? 5. A person starts walkng from home and walks 4 mles east, mles southeast, 5 mles south, 4 mles southwest, and mles east. How far have they walked? If they walked straght home, how far would they have to walk? 6. A person starts walkng from home and walks 4 mles east, 7 mles southeast, 6 mles south, 5 mles southwest, and mles east. How far have they walked? If they walked straght home, how far would they have to walk? 7. Three forces act on an object: F, 5, F 0,1, F 4, 7 force actng on the object. 1 8. Fnd the net 8. Three forces act on an object: F,5, F 8,, F 0, 7 actng on the object. 1. Fnd the net force 9. A person starts walkng from home and walks mles at 0 north of west, then 5 mles at 10 west of south, then 4 mles at 15 north of east. If they walked straght home, how far would they have to walk, and n what drecton? 0. A person starts walkng from home and walks 6 mles at 40 north of east, then mles at 15 east of south, then 5 mles at 0 south of west. If they walked straght home, how far would they have to walk, and n what drecton? 1. An arplane s headng north at an arspeed of 600 km/hr, but there s a wnd blowng from the southwest at 80 km/hr. How many degrees off course wll the plane end up flyng, and what s the plane s speed relatve to the ground?. An arplane s headng north at an arspeed of 500 km/hr, but there s a wnd blowng from the northwest at 50 km/hr. How many degrees off course wll the plane end up flyng, and what s the plane s speed relatve to the ground?. An arplane needs to head due north, but there s a wnd blowng from the southwest at 60 km/hr. The plane fles wth an arspeed of 550 km/hr. To end up flyng due north, the plot wll need to fly the plane how many degrees west of north?
Secton 8.4 Vectors 5 4. An arplane needs to head due north, but there s a wnd blowng from the northwest at 80 km/hr. The plane fles wth an arspeed of 500 km/hr. To end up flyng due north, the plot wll need to fly the plane how many degrees west of north? 5. As part of a vdeo game, the pont (5, 7) s rotated counterclockwse about the orgn through an angle of 5 degrees. Fnd the new coordnates of ths pont. 6. As part of a vdeo game, the pont (7, ) s rotated counterclockwse about the orgn through an angle of 40 degrees. Fnd the new coordnates of ths pont. 7. Two chldren are throwng a ball back and forth straght across the back seat of a car. The ball s beng thrown 10 mph relatve to the car, and the car s travellng 5 mph down the road. If one chld doesn't catch the ball and t fles out the wndow, n what drecton does the ball fly (gnorng wnd resstance)? 8. Two chldren are throwng a ball back and forth straght across the back seat of a car. The ball s beng thrown 8 mph relatve to the car, and the car s travellng 45 mph down the road. If one chld doesn't catch the ball and t fles out the wndow, n what drecton does the ball fly (gnorng wnd resstance)?
Secton 8.5 Parametrc Equatons Many shapes, even ones as smple as crcles, cannot be represented as an equaton where y s a functon of x. Consder, for example, the path a moon follows as t orbts around a planet, whch smultaneously rotates around a sun. In some cases, polar equatons provde a way to represent such a path. In others, we need a more versatle approach that allows us to represent both the x and y coordnates n terms of a thrd varable, or parameter. Parametrc Equatons A system of parametrc equatons s a par of functons x( and y( n whch the x and y coordnates are the output, represented n terms of a thrd nput parameter, t. Example 1 Movng at a constant speed, an object moves at a steady rate along a straght path from coordnates (-5, ) to the coordnates (, -1) n 4 seconds, where the coordnates are measured n meters. Fnd parametrc equatons for the poston of the object. The x coordnate of the object starts at -5 meters, and goes to + meters, ths means the x drecton has changed by 8 meters n 4 seconds, gvng us a rate of meters per second. We can now wrte the x coordnate as a lnear functon wth respect to tme, t, x( 5 t. Smlarly, the y value starts at and goes to -1, gvng a change n y value of 4 meters, meanng the y values have decreased by 4 meters n 4 seconds, for a rate of -1 meter per second, t x y gvng equaton y( t. Together, these are the 0-5 parametrc equatons for the poston of the object: 1 - x( 5 t -1 1 y( t 1 0 4-1 Usng these equatons, we can buld a table of t, x, and y values. Because of the context, we lmted ourselves to non-negatve t values for ths example, but n general you can use any values. From ths table, we could create three possble graphs: a graph of x vs. t, whch would show the horzontal poston over tme, a graph of y vs. t, whch would show the vertcal poston over tme, or a graph of y vs. x, showng the poston of the object n the plane.
Secton 8.5 Parametrc Equatons 7 Poston of x as a functon of tme x Poston of y as a functon of tme y t t Poston of y relatve to x y x Notce that the parameter t does not explctly show up n ths thrd graph. Sometmes, when the parameter t does represent a quantty lke tme, we mght ndcate the drecton of movement on the graph usng an arrow, as shown above. There s often no sngle parametrc representaton for a curve. In Example 1 we assumed the object was movng at a steady rate along a straght lne. If we kept the assumpton about the path (straght lne) but dd not assume the speed was constant, we mght get a system lke: x( 5 t y( x( t y( t Ths starts at (-5, ) when t = 0 and ends up at (, -1) when t =. If we graph the x( and y( functon separately, we can see that those are no longer lnear, but f we graph x vs. y we wll see that we stll get a straght-lne path. t=0 t=1 y x t=
Example Sketch a graph of x( t 1 y( t We can begn by creatng a table of values. From ths table, we can plot the (x, y) ponts n the plane, sketch n a rough graph of the curve, and ndcate the drecton of moton wth respect to tme by usng arrows. y t x y - 10-1 - 5 0-1 1 0 1 1 5 4 x Notce that here the parametrc equatons descrbe a shape for whch y s not a functon of x. Ths s an example of why usng parametrc equatons can be useful snce they can represent such a graph as a set of functons. Ths partcular graph also appears to be a parabola where x s a functon of y, whch we wll soon verfy. Whle creatng a t-x-y table, plottng ponts and connectng the dots wth a smooth curve usually works to gve us a rough dea of what the graph of a system of parametrc equatons looks lke, t's generally easer to use technology to create these tables and (smultaneously) much ncer-lookng graphs. Example Sketch a graph of x( cos(. y( sn( y Usng technology we can generate a graph of ths equaton, producng an ellpse. Smlar to graphng polar equatons, you must change the MODE on your calculator (or select parametrc equatons on your graphng technology) before graphng a system of parametrc equatons. You wll know you have successfully entered parametrc mode when the equaton nput has changed to ask for a x(= and y(= par of equatons. x
Secton 8.5 Parametrc Equatons 9 Try t Now 1. Sketch a graph of x( 4cos(t ). Ths s an example of a Lssajous fgure. y( sn( Example 4 The populatons of rabbts and wolves on an sland over tme are gven by the graphs below. Use these graphs to sketch a graph n the r-w plane showng the relatonshp between the number of rabbts and number of wolves. Wolves 0 15 10 5 0 0 1 4 5 6 7 Years Rabbts 100 80 60 40 0 0 0 1 4 5 6 7 Years For each nput t, we can determne the number of rabbts, r, and the number of wolves, w, from the respectve graphs, and then plot the correspondng pont n the r-w plane. 100 80 Rabbts 60 40 0 0 0 5 10 15 0 Wolves Ths graph helps reveal the cyclcal nteracton between the two populatons. Convertng from Parametrc to Cartesan In some cases, t s possble to elmnate the parameter t, allowng you to wrte a par of parametrc equatons as a Cartesan equaton. It s easest to do ths f one of the x( or y( functons can easly be solved for t, allowng you to then substtute the remanng expresson nto the second part.
Example 6 x( t 1 Wrte y( t as a Cartesan equaton, f possble. Here, the equaton for y s lnear, so s relatvely easy to solve for t. Snce the resultng Cartesan equaton wll lkely not be a functon, and for convenence, we drop the functon notaton. y t Solve for t y t Substtute ths for t n the x equaton x ( y ) 1 Notce that ths s the equaton of a parabola wth x as a functon of y, wth vertex at (1,), openng to the rght. Comparng ths wth the graph from Example, we see (unsurprsngly) that t yelds the same graph n the x-y plane as dd the orgnal parametrc equatons. Try t Now. Wrte x( t y( t 6 as a Cartesan equaton, f possble. Example 7 x( t Wrte y( log( as a Cartesan equaton, f possble. We could solve ether the frst or second equaton for t. Solvng the frst, x t x t Square both sdes x t y x Substtute nto the y equaton log Snce the parametrc equaton s only defned for t 0, ths Cartesan equaton s equvalent to the parametrc equaton on the correspondng doman. The parametrc equatons show that when t > 0, x > and y > 0, so the doman of the Cartesan equaton should be lmted to x >. To ensure that the Cartesan equaton s as equvalent as possble to the orgnal parametrc equaton, we try to avod usng doman-restrcted nverse functons, such as the nverse trg functons, when possble. For equatons nvolvng trg functons, we often try to fnd an dentty to utlze to avod the nverse functons.
Secton 8.5 Parametrc Equatons 1 Example 8 x( cos( Wrte y( sn( as a Cartesan equaton, f possble. To rewrte ths, we can utlze the Pythagorean dentty cos ( t ) sn ( 1. x x cos( so cos( y y sn( so sn( Startng wth the Pythagorean Identty cos ( t ) sn ( 1 Substtute n the expressons from the parametrc form x y x y 1 4 9 1 Smplfy Ths s a Cartesan equaton for the ellpse we graphed earler. Parameterzng Curves Whle convertng from parametrc form to Cartesan can be useful, t s often more useful to parameterze a Cartesan equaton convertng t nto parametrc form. If the Cartesan equaton gves one varable as a functon of the other, then parameterzaton s trval the ndependent varable n the functon can smply be defned as t. Example 9 Parameterze the equaton x y y. In ths equaton, x s expressed as a functon of y. By defnng y t we can then substtute that nto the Cartesan equaton, yeldng x t t. Together, ths produces the parametrc form: x( t t y( t Try t Now. Wrte x y n parametrc form, f possble.
In addton to parameterzng Cartesan equatons, we also can parameterze behavors and movements. Example 10 A robot follows the path shown. Create a table of values for the x( and y( functons, assumng the robot takes one second to make each movement. Snce we know the drecton of moton, we can ntroduce consecutve values for t along the path of the robot. Usng these values wth the x and y coordnates of the robot, we can create the tables. For example, we desgnate the startng pont, at (1, 1), as the poston at t = 0, the next pont at (, 1) as the poston at t = 1, and so on. t 0 1 4 5 6 x 1 4 1 1 t 0 1 4 5 6 y 1 1 4 5 4 Notce how ths also tes back to vectors. The journey of the robot as t moves through the Cartesan plane could also be dsplayed as vectors and total dstance traveled and dsplacement could be calculated. Example 11 A lght s placed on the edge of a bcycle tre as shown and the bcycle starts rollng down the street. Fnd a parametrc equaton for the poston of the lght after the wheel has rotated through an angle of θ. r θ Startng Rotated by θ Relatve to the center of the wheel, the poston of the lght can be found as the coordnates of a pont on a crcle, but snce the x coordnate begns at 0 and moves n the negatve drecton, whle the y coordnate starts at the lowest value, the coordnates of the pont wll be gven by: x r sn( ) y r cos( )
Secton 8.5 Parametrc Equatons The center of the wheel, meanwhle, s movng horzontally. It remans at a constant heght of r, but the horzontal poston wll move a dstance equvalent to the arclength of the crcle drawn out by the angle, s r. The poston of the center of the crcle s then x r y r Combnng the poston of the center of the wheel wth the poston of the lght on the wheel relatve to the center, we get the followng parametrc equatonw, wth θ as the parameter: x r r sn( ) r sn( ) y r r cos( ) r 1 cos( ) The result graph s called a cyclod. Example 1 A moon travels around a planet as shown, orbtng once every 10 days. The planet travels around a sun as shown, orbtng once every 100 days. Fnd a parametrc equaton for the poston of the moon, relatve to the center of the sun, after t days. For ths example, we ll assume the orbts are crcular, though n real lfe they re actually ellptcal. The coordnates of a pont on a crcle can always be wrtten n the form x r cos( ) y r sn( ) 0 6
Snce the orbt of the moon around the planet has a perod of 10 days, the equaton for the poston of the moon relatve to the planet wll be x( 6cos t 6cos t 10 5 y( 6sn t 6sn t 10 5 Wth a perod of 100 days, the equaton for the poston of the planet relatve to the sun wll be x( 0cos t 0cos t 100 50 y( 0sn t 0sn t 100 50 Combnng these together, we can fnd the poston of the moon relatve to the sun as the sum of the components. x( 6cos t 0cos t 5 50 y( 6sn t 0sn t 5 50 The resultng graph s shown here. Try t Now 4. A wheel of radus 4 s rolled around the outsde of a crcle of radus 7. Fnd a parametrc equaton for the poston of a pont on the boundary of the smaller wheel. Ths shape s called an epcyclod. Important Topcs of Ths Secton Parametrc equatons Graphng x(, y( and the correspondng x-y graph Sketchng graphs and buldng a table of values Convertng parametrc to Cartesan Convertng Cartesan to parametrc (parameterzng curves)
Secton 8.5 Parametrc Equatons 5 Try t Now Answers 1... 4. y x x( cos( y( sn( 11 x( 11cos t 4cos t 4 11 y( 11sn t 4sn t 4
Secton 8.5 Exercses Match each set of equatons wth one of the graphs below. xt t xt t 1 1.. y t t 1 y t t 4. x t y t sn( 4cos( 5. x t t y t t. 6. x t y t x t y t 4sn cos t t t t A B C D E F From each par of graphs n the t-x and t-y planes shown, sketch a graph n the x-y plane. 7. 8.
Secton 8.5 Parametrc Equatons 7 From each graph n the x-y plane shown, sketch a graph of the parameter functons n the t-x and t-y planes. 9. 10. Sketch the parametrc equatons for t. xt 1t 11. 1. y t t t t x t y t Elmnate the parameter t to rewrte the parametrc equaton as a Cartesan equaton xt 5 t xt 6t 1. 14. y t 8t y t 10 t 15. 17. 19. 1.. x t t 1 y t t x t y t e t 15t y t t x t t t x t y t x t y t e e t 6t 4cos 5sn t t x t t 1 16. y t t xt 4log t 18. y t t 4 xt t t 0. y t t 5 xt t. 10 y t t xt sn t 4. y t 6cos t
Parameterze (wrte a parametrc equaton for) each Cartesan equaton 5. yx x y x sn x 1 6. 7. x y log y y 8. x y y y 9. x y 1 0. 4 9 x y 1 16 6 Parameterze the graphs shown. 1... 4. 5. Parameterze the lne from ( 1,5) to (,) so that the lne s at ( 1,5) at t = 0, and at (, ) at t = 1. 6. Parameterze the lne from (4,1) to (6, ) so that the lne s at (4,1) at t = 0, and at (6, ) at t = 1.
Secton 8.5 Parametrc Equatons 9 The graphs below are created by parameterc equatons of the form Fnd the values of a, b, c, and d to acheve each graph. cos sn x t a bt y t c dt. 7. 8. 9. 40. 41. An object s thrown n the ar wth vertcal velocty 0 ft/s and horzontal velocty 15 ft/s. The object s heght can be descrbed by the equaton y t 16t 0t, whle the object moves horzontally wth constant velocty 15 ft/s. Wrte parametrc equatons for the object s poston, then elmnate tme to wrte heght as a functon of horzontal poston. 4. A skateboarder rdng on a level surface at a constant speed of 9 ft/s throws a ball n the ar, the heght of whch can be descrbed by the equaton y t 16t 10t 5. Wrte parametrc equatons for the ball s poston, then elmnate tme to wrte heght as a functon of horzontal poston.
4. A carnval rde has a large rotatng arm wth dameter 40 feet centered 5 feet off the ground. At each end of the large arm are two smaller rotatng arms wth dameter 16 feet each. The larger arm rotates once every 5 seconds, whle the smaller arms rotate once every seconds. If you board the rde when the pont P s closest to the ground, fnd parametrc equatons for your poston over tme. P 44. A hypocyclod s a shape generated by trackng a fxed pont on a small crcle as t rolls around the nsde of a larger crcle. If the smaller crcle has radus 1 and the large crcle has radus 6, fnd parametrc equatons for the poston of the pont P as the smaller wheel rolls n the drecton ndcated. P