Sections 3.1 and 3.4 Exponential Functions (Growth and Decay)

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Secions 3.1 and 3.4 Eponenial Funcions (Gowh and Decay) Chape 3. Secions 1 and 4 Page 1 of 5 Wha Would You Rahe Have... $1million, o double you money evey day fo 31 days saing wih 1cen? Day Cens Day Cens 1 16 65536 1 2 17 13172 2 4 18 262144 3 8 19 524288 4 16 2 148576 5 32 21 297152 6 64 22 419434 7 128 23 838868 8 256 24 16777216 9 512 25 33554432 1 124 26 6718864 11 248 27 134217728 12 496 28 2684356 13 8192 29 53687912 14 16384 3 173741824 15 32768 31 2147483648 On day 31, you will have 2,147,483,648 cens, o $21,474,836.48 Q: How can we epess his wih an equaion? day 31 A: # cens( day ) = 2 2 = 2,147, 483, 648. Q: Fom he able, on which day will you supass $1million, o 1 8 cens? A: Fom he cha of values above, his happens beween day 26 and day 27. Eponenial Equaions Aise When: You add he same pecenage o a quaniy each fied ime peiod. You muliply a quaniy by he same amoun each fied ime peiod. Fo he money eample, we wee adding 1% each day, o muliplying by 2. Recall: One fom of he Eponenial Funcion is y = kb b is he base, and k is he iniial quaniy (when = ). Q: Wha ae he condiions on b? A: b has o be posiive (geae han zeo) and no equal o 1. The book uses anohe fom of he eponenial equaion, y = a. Q: Wha is diffeen in hese foms? A: Thee is no value of k in y = a. Ohewise hey ae he same foma. C. Bellomo, evised 14-Oc-9

Chape 3. Secions 1 and 4 Page 2 of 5 k Someimes we alk abou he eponenial fom of he equaion as y () = Pe. Be caeful of he k diffeence beween he foms Pe and kb. Someimes i is sandad when one says pecenage k change ha hey ae eplicily giving he value of k in Pe (as is he case in you ebook) bu his ypically efes o he change in a populaion, which is coninuously gowing. I wouldn be he case fo, say, inees which can gow monhly o weekly. k Q: In he equaions Pe and kb, wha is he elaionship beween P and k, and k and b? A: P = k, and e k = b. Rules: bb = b ( b ) = b 1 b = b y + y y y The Relaionship beween Equaions: When you compae he equaions A 1+ n caeful which is used. n and Pe k and kb, you can see a similaiy bu be n Inees poblems ae of he fom A 1+, whee is he inees ae, A he oiginal amoun of n money, and n is he numbe of imes you compound pe yea. I is assumed in hese poblems ha n is compounded a fied numbe of imes pe yea, no coninuously. When money (o a populaion) gows coninuously he equaion A 1+ changes o Ae. To n n/ 1 see he elaionship hee, noe ha lim 1 n + = e. You gowh ae is hen equal o. ( n / ) n k Povided you ae dealing wih coninual gowh, you can use eihe Pe o is given, Using kb, b= e kb. If he gowh ae k Using Pe, k =. C. Bellomo, evised 14-Oc-9

Chape 3. Secions 1 and 4 Page 3 of 5 Gaphs of Eponenials: y = k b If he base is geae han 1, you have eponenial gowh. If he base is less han 1, you have eponenial decay. The y inecep is k. If k >, he funcion will always be posiive. As ges lage, he funcion will end o (fo decay) and infiniy (fo gowh). If k <, he funcion will always be negaive (i is flipped upside down). If you have an equaion ha is eponenial in fom, bu ae adding o subacing a consan y = k b + c, his is a shif up (c > ) o down (c < ) by c y = k b ( + c), his is a shif lef (c < ) o igh (c > ) by c Eample. Gaph y = 3 1 y = 3 b = 1/3 Eponenial decay k = 1 Cosses a (,1) Eample. Gaph f ( ) = 2 e 1 f( ) = 2 + ( 1) e This has a shif up of 2 k = 1 (flipped upside down) and b = 1.37 e Tesing a Daa Se (OPTIONAL): Deemine if he following se of daa is eponenial by aking divisions of successive y values Fo each fied ime peiod (change in is one) we seem o be muliplying by 3. This means ha b = (y2 / y1) / (change in ) = 3, and k = y() = 2. y () = 23 y y2 / y1 2 1 6 3 2 18 3 3 54 3 4 162 3 C. Bellomo, evised 14-Oc-9

Chape 3. Secions 1 and 4 Page 4 of 5 Anohe Eample, Simple Inees (OPTIONAL): Suppose you savings accoun eans 1.25% inees pe monh, and you sa wih $25. Wie an equaion ha epesens his poblem. Monh Fomula Dollas 25 = 25. 1 25 + 1.25%(25) = 253.13 2 253.13 + 1.25%(253.13) = 256.29 3 256.29 + 1.25%(256.29) = 259.49 4 259.49 + 1.25%(259.49) = 262.73 5 262.74 + 1.25%(262.74) = 266.1 How can we make an equaion ou of his? y () = 25(. 1125 ) Q: Does his epesen gowh o decay? A: Gowh, because he dolla value is geing lage. Q: How much do you ean in he fis monh? A: y(1) 25 = 25(1.125) 25 3.13 Half Life and Doubling Time: Half life is he amoun of ime i akes fo a subsance o decay o half of is oiginal quaniy. I epesens eponenial decay. The half life (T) and decay ae (b) ae elaed 1 T k = k b 2 1/ T ln(.5) 1 T = o b= ln b 2 Doubling ime is he amoun of ime i akes fo a subsance o double is oiginal quaniy. I epesens eponenial gowh. The doubling ime (T) and gowh ae (b) ae also elaed T 2k = k b ln(2) 1/ T T = o b= 2 ln b You can eihe memoize he equaions above, o use you knowledge of he poblem and equaion o deemine he values needed. C. Bellomo, evised 14-Oc-9

Chape 3. Secions 1 and 4 Page 5 of 5 Eample. If a species of baceia has a doubling ime of minues, hen find b in y = kb 2k = kb 2 = b b = 1/ 2 1.15523 kb. k Eample. If a species of baceia has a doubling ime of minues, hen find k in Pe. k y = Pe k () 2P = Pe k () 2 = e ln(2) k =.15 Eample: The wold populaion was 256 million in 195 and 34 million in 196. Model he populaion wih an eponenial equaion. Le = epesen 195 (his makes he mah easie). Mehod 1, using kb 256 = kb k = 256 256b 1 1/1 34 = 256 b b= (34 / 256) 1.173 P = 256(1.173) Mehod 2, using Pe k k 256 = Pe P = 256 k 256e ln(34 / 256) = e k = 1 (.172) 256e k 1 34 256.172 (.172) Noe ha e = 1.173 C. Bellomo, evised 14-Oc-9