488 Chaper 7 Secion 7.4 Modeling Changing Ampliude and Midline While sinusoidal funcions can model a variey of behaviors, i is ofen necessary o combine sinusoidal funcions wih linear and exponenial curves o model real applicaions and behaviors. We begin his secion by looking a changes o he midline of a sinusoidal funcion. Recall ha he midline describes he middle, or average value, of he sinusoidal funcion. Changing Midlines Example 1 A populaion of elk currenly averages 000 elk, and ha average has been growing by 4% each year. Due o seasonal flucuaion, he populaion oscillaes from 50 below average in he winer up o 50 above average in he summer. Find a funcion ha models he number of elk afer years, saring in he winer. There are wo componens o he behavior of he elk populaion: he changing average, and he oscillaion. The average is an exponenial growh, saring a 000 and growing by 4% each year. Wriing a formula for his: average = iniial(1 + r) = 000(1 + 0.04) For he oscillaion, since he populaion oscillaes 50 above and below average, he ampliude will be 50. Since i akes one year for he populaion o cycle, he period is 1. original period π We find he value of he horizonal srech coefficien B = = = π. new period 1 The funcion sars in winer, so he shape of he funcion will be a negaive cosine, since i sars a he lowes value. Puing i all ogeher, he equaion would be: P( ) = 50cos( π ) + midline Since he midline represens he average populaion, we subsiue in he exponenial funcion ino he populaion equaion o find our final equaion: P( ) = 50cos( π ) + 000(1 + 0.04) This is an example of changing midline in his case an exponenially changing midline.
Secion 7.4 Modeling Changing Ampliude and Midline 489 Changing Midline A funcion of he form f ( ) = Asin( B) + g( ) will oscillae above and below he average given by he funcion g(). Changing midlines can be exponenial, linear, or any oher ype of funcion. Here are some examples: Linear midline Exponenial midline Quadraic midline = ( ) + + ( ) sin ( ) ( f = A B + ab ) f = A ( B ) + a f ( ) Asin B ( m b) ( ) sin ( ) Example Find a funcion wih linear midline of he form hrough he poins given below. 0 1 3 f() 5 10 9 8 f ( ) = Asin + m + b ha will pass Since we are given he value of he horizonal compression coefficien we can calculae original period π he period of his funcion: new period = = = 4. B π Since he sine funcion is a he midline a he beginning of a cycle and halfway hrough a cycle, we would expec his funcion o be a he midline a = 0 and =, since is half he full period of 4. Based on his, we expec he poins (0, 5) and (, 9) o be poins on he midline. We can clearly see ha his is no a consan funcion and so we use he wo poins o calculae a linear funcion: midline = m + b. From hese wo poins we can calculae a slope: 9 5 4 m = = = 0 Combining his wih he iniial value of 5, we have he midline: midline = + 5.
490 Chaper 7 The full funcion will have form f ( ) = Asin + + 5. To find he ampliude, we can plug in a poin we haven already used, such as (1, 10). 10 = A sin (1) + (1) + 5 Evaluae he sine and combine like erms 10 = A + 7 A = 3 A funcion of he form given fiing he daa would be f ( ) = 3sin + + 5 Alernaive Approach Noice we could have aken an alernae approach by plugging poins (0, 5) and (, 9) ino he original equaion. Subsiuing (0, 5), 5 = A sin (0) + m(0) + b Evaluae he sine and simplify 5 = b Subsiuing (, 9) 9 = Asin () + m() + 5 Evaluae he sine and simplify 9 = m + 5 4 = m m =, as we found above. Now we can proceed o find A he same way we did before. Example 3 The number of ouriss visiing a ski and hiking resor averages 4000 people annually and oscillaes seasonally, 1000 above and below he average. Due o a markeing campaign, he average number of ouriss has been increasing by 00 each year. Wrie an equaion for he number of ouriss afer years, beginning a he peak season. Again here are wo componens o his problem: he oscillaion and he average. For he oscillaion, he number of ouriss oscillaes 1000 above and below average, giving an ampliude of 1000. Since he oscillaion is seasonal, i has a period of 1 year. Since we are given a saring poin of peak season, we will model his scenario wih a cosine funcion. So far, his gives an equaion in he form N ( ) = 1000cos( π ) + midline.
Secion 7.4 Modeling Changing Ampliude and Midline 491 The average is currenly 4000, and is increasing by 00 each year. This is a consan rae of change, so his is linear growh, average = 4000 + 00. This funcion will ac as he midline. Combining hese wo pieces gives a funcion for he number of ouriss: N ( ) = 1000 cos( π ) + 4000 + 00 Try i Now 1. Given he funcion using words. g( x) = ( x 1) + 8cos( x), describe he midline and ampliude Changing Ampliude There are also siuaions in which he ampliude of a sinusoidal funcion does no say consan. Back in Chaper 6, we modeled he moion of a spring using a sinusoidal funcion, bu had o ignore fricion in doing so. If here were fricion in he sysem, we would expec he ampliude of he oscillaion o decrease over ime. In he equaion f ( ) = Asin( B) + k, A gives he ampliude of he oscillaion, we can allow he ampliude o change by replacing his consan A wih a funcion A(). Changing Ampliude A funcion of he form f ( ) = A( )sin( B) + k will oscillae above and below he midline wih an ampliude given by A(). Here are some examples: Linear ampliude Exponenial ampliude Quadraic ampliude f ( ) ( m b)sin( B ) k = + + ( ) ( f = ab )sin ( B ) + k ( ) ( )sin ( ) f = a B + k
49 Chaper 7 When hinking abou a spring wih ampliude decreasing over ime, i is emping o use he simples ool for he job a linear funcion. Bu if we aemp o model he ampliude wih a decreasing linear funcion, such as A( ) = 10, we quickly see he problem when we graph he equaion f ( ) = (10 )sin(4). While he ampliude decreases a firs as inended, he ampliude his zero a = 10, hen coninues pas he inercep, increasing in absolue value, which is no he expeced behavior. This behavior and funcion may model he siuaion on a resriced domain and we migh ry o chalk he res of i up o model breakdown, bu in fac springs jus don behave like his. A beer model, as you will learn laer in physics and calculus, would show he ampliude decreasing by a fixed percenage each second, leading o an exponenial decay model for he ampliude. Damped Harmonic Moion Damped harmonic moion, exhibied by springs subjec o fricion, follows a model of he form r f ( ) = ab sin( B) + k or f ( ) = ae sin( B) + k. Example 4 A spring wih naural lengh of fee inches is pulled back 6 fee and released. I oscillaes once every seconds. Is ampliude decreases by 0% each second. Find a funcion ha models he posiion of he spring seconds afer being released. Since he spring will oscillae on eiher side of he naural lengh, he midline will be a 0 fee. The oscillaion has a period of seconds, and so he horizonal compression coefficien is B = π. Addiionally, i begins a he furhes disance from he wall, indicaing a cosine model. Meanwhile, he ampliude begins a 6 fee, and decreases by 0% each second, giving an ampliude funcion of A ( ) = 6(1 0.0). Combining his wih he sinusoidal informaion gives a funcion for he posiion of he spring: f ( ) = 6(0.80) cos( π ) + 0
Secion 7.4 Modeling Changing Ampliude and Midline 493 Example 5 A spring wih naural lengh of 30 cm is pulled ou 10 cm and released. I oscillaes 4 imes per second. Afer seconds, he ampliude has decreased o 5 cm. Find a funcion ha models he posiion of he spring. The oscillaion has a period of 1 4 second, so π B = = 8π. Since he spring will 1 4 oscillae on eiher side of he naural lengh, he midline will be a 30 cm. I begins a he furhes disance from he wall, suggesing a cosine model. Togeher, his gives f ( ) = A( ) cos(8 π ) + 30. For he ampliude funcion, we noice ha he ampliude sars a 10 cm, and decreases o 5 cm afer seconds. This gives wo poins (0, 10) and (, 5) ha mus be saisfied by an exponenial funcion: A ( 0) = 10 and A ( ) = 5. Since he funcion is exponenial, 0 we can use he form A ( ) = ab. Subsiuing he firs poin, 10 = ab, so a = 10. Subsiuing in he second poin, 5 = 10b Divide by 10 1 = b Take he square roo b = 1 0.707 This gives an ampliude funcion of oscillaion, f ( ) = 10(0.707) cos(8 π) + 30 A ) ( ) = 10(0.707. Combining his wih he Try i Now. A cerain sock sared a a high value of $7 per share, oscillaing monhly above and below he average value, wih he oscillaion decreasing by % per year. However, he average value sared a $4 per share and has grown linearly by 50 cens per year. a. Find a formula for he midline and he ampliude. b. Find a funcion S() ha models he value of he sock afer years. Example 6 In AM (Ampliude Modulaed) radio, a carrier wave wih a high frequency is used o ransmi music or oher signals by applying he o-be-ransmied signal as he ampliude of he carrier signal. A musical noe wih frequency 110 Hz (Herz = cycles per second) is o be carried on a wave wih frequency of KHz (KiloHerz = housands of cycles per second). If he musical wave has an ampliude of 3, wrie a funcion describing he broadcas wave.
494 Chaper 7 1 The carrier wave, wih a frequency of 000 cycles per second, would have period 000 of a second, giving an equaion of he form sin(4000 π ). Our choice of a sine funcion here was arbirary i would have worked jus was well o use a cosine. The musical one, wih a frequency of 110 cycles per second, would have a period of 1 of a second. Wih an ampliude of 3, his would correspond o a funcion of he 110 form 3sin(0 π ). Again our choice of using a sine funcion is arbirary. The musical wave is acing as he ampliude of he carrier wave, so we will muliply he musical one s funcion by he carrier wave funcion, resuling in he funcion f ( ) = 3sin(0 π ) sin(4000 π ) Imporan Topics of This Secion Changing midline Changing ampliude Linear Changes Exponenial Changes Damped Harmonic Moion Try i Now Answers 1. The midline follows he pah of he quadraic x 1and he ampliude is a consan value of 8.. m( ) = 4 + 0.5 A( ) = 7(0.98) S()= 7 (0.98) cos( 4 ) + 4 + 0.5 π
Secion 7.4 Modeling Changing Ampliude and Midline 495 Secion 7.4 Exercises Find a possible formula for he rigonomeric funcion whose values are given in he following ables. 1. x 0 3 6 9 1 15 18 y -4-1 -1-4 -1. x 0 4 6 8 10 1 y 5 1-3 1 5 1-3 3. The displacemen h( ), in cenimeers, of a mass suspended by a spring is modeled h by he funcion ( ) 8sin(6 ) = π, where is measured in seconds. Find he ampliude, period, and frequency of his displacemen. 4. The displacemen h( ), in cenimeers, of a mass suspended by a spring is modeled h by he funcion ( ) 11sin(1 ) = π, where is measured in seconds. Find he ampliude, period, and frequency of his displacemen. 5. A populaion of rabbis oscillaes 19 above and below average during he year, reaching he lowes value in January. The average populaion sars a 650 rabbis and increases by 160 each year. Find a funcion ha models he populaion, P, in erms of he monhs since January,. 6. A populaion of deer oscillaes 15 above and below average during he year, reaching he lowes value in January. The average populaion sars a 800 deer and increases by 110 each year. Find a funcion ha models he populaion, P, in erms of he monhs since January,. 7. A populaion of muskras oscillaes 33 above and below average during he year, reaching he lowes value in January. The average populaion sars a 900 muskras and increases by 7% each monh. Find a funcion ha models he populaion, P, in erms of he monhs since January,. 8. A populaion of fish oscillaes 40 above and below average during he year, reaching he lowes value in January. The average populaion sars a 800 fish and increases by 4% each monh. Find a funcion ha models he populaion, P, in erms of he monhs since January,. 9. A spring is aached o he ceiling and pulled 10 cm down from equilibrium and released. The ampliude decreases by 15% each second. The spring oscillaes 18 imes each second. Find a funcion ha models he disance, D, he end of he spring is below equilibrium in erms of seconds,, since he spring was released.
496 Chaper 7 10. A spring is aached o he ceiling and pulled 7 cm down from equilibrium and released. The ampliude decreases by 11% each second. The spring oscillaes 0 imes each second. Find a funcion ha models he disance, D, he end of he spring is below equilibrium in erms of seconds,, since he spring was released. 11. A spring is aached o he ceiling and pulled 17 cm down from equilibrium and released. Afer 3 seconds he ampliude has decreased o 13 cm. The spring oscillaes 14 imes each second. Find a funcion ha models he disance, D he end of he spring is below equilibrium in erms of seconds,, since he spring was released. 1. A spring is aached o he ceiling and pulled 19 cm down from equilibrium and released. Afer 4 seconds he ampliude has decreased o 14 cm. The spring oscillaes 13 imes each second. Find a funcion ha models he disance, D he end of he spring is below equilibrium in erms of seconds,, since he spring was released. Mach each equaion form wih one of he graphs. x ab sin 5x sin 5x + mx + b 13. a. + ( ) b. ( ) x 14. a. sin ( 5 ) + ab x b. ( mx b)sin(5 x) I II III IV x Find a funcion of he form y = ab + csin x 15. x 0 1 3 y 6 9 96 379 ha fis he daa given. 16. x 0 1 3 y 6 34 150 746 Find a funcion of he form y a sin π = x + m + bx ha fis he daa given. 17. x 0 1 3 18. x 0 1 3 y 7 6 11 16 y - 6 4 Find a funcion of he form 19. x 0 1 3 y 11 3 1 3 y = ab x cos π x + c ha fis he daa given. 0. x 0 1 3 y 4 1-11 1