Precalculus An Investigation of Functions

Transcription

1 Precalculus An Invesigaion of Funcions David Lippman Melonie Rasmussen Ediion.3 This book is also available o read free online a hp:// If you wan a prined copy, buying from he booksore is cheaper han prining yourself.

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3 70 Chaper 4 Secion 4.6 Exponenial and Logarihmic Models While we have explored some basic applicaions of exponenial and logarihmic funcions, in his secion we explore some imporan applicaions in more deph. Radioacive Decay In an earlier secion, we discussed radioacive decay he idea ha radioacive isoopes change over ime. One of he common erms associaed wih radioacive decay is halflife. Half Life The half-life of a radioacive isoope is he ime i akes for half he subsance o decay. Given he basic exponenial growh/decay equaion h = solving for when half he original amoun remains; by solving simply = b ( ) ab, half-life can be found by a a( b) =, or more. Noice how he iniial amoun is irrelevan when solving for half-life. Example Bismuh-0 is an isoope ha decays by abou 3% each day. Wha is he half-life of Bismuh-0? We were no given a saring quaniy, so we could eiher make up a value or use an unknown consan o represen he saring amoun. To show ha saring quaniy does no affec he resul, le us denoe he iniial quaniy by he consan a. Then he decay d of Bismuh-0 can be described by he equaion Q ( d) = a(0.87). To find he half-life, we wan o deermine when he remaining quaniy is half he original: a. Solving, d a = a(0.87) Dividing by a, = d 0.87 Take he log of boh sides d log = log( 0.87 ) Use he exponen propery of logs log = d log( 0.87) Divide o solve for d

4 Secion 4.6 Exponenial and Logarihmic Models 7 log d = days log 0.87 ( ) This ells us ha he half-life of Bismuh-0 is approximaely 5 days. Example Cesium-37 has a half-life of abou 30 years. If you begin wih 00mg of cesium-37, how much will remain afer 30 years? 60 years? 90 years? Since he half-life is 30 years, afer 30 years, half he original amoun, 00mg, will remain. Afer 60 years, anoher 30 years have passed, so during ha second 30 years, anoher half of he subsance will decay, leaving 50mg. Afer 90 years, anoher 30 years have passed, so anoher half of he subsance will decay, leaving 5mg. Example 3 Cesium-37 has a half-life of abou 30 years. Find he annual decay rae. Since we are looking for an annual decay rae, we will use an equaion of he form Q ( ) = a( + r). We know ha afer 30 years, half he original amoun will remain. Using his informaion 30 a = a( + r) Dividing by a 30 = ( + r) Taking he 30 h roo of boh sides 30 = + r Subracing one from boh sides, r = This ells us cesium-37 is decaying a an annual rae of.84% per year. Try i Now Chlorine-36 is eliminaed from he body wih a biological half-life of 0 days 3. Find he daily decay rae. 3 hp://

5 7 Chaper 4 Example 4 Carbon-4 is a radioacive isoope ha is presen in organic maerials, and is commonly used for daing hisorical arifacs. Carbon-4 has a half-life of 5730 years. If a bone fragmen is found ha conains 0% of is original carbon-4, how old is he bone? To find how old he bone is, we firs will need o find an equaion for he decay of he carbon-4. We could eiher use a coninuous or annual decay formula, bu op o use he coninuous decay formula since i is more common in scienific exs. The half life ells us ha afer 5730 years, half he original subsance remains. Solving for he rae, r5730 a = ae Dividing by a r5730 = e Taking he naural log of boh sides r5730 ln = ln( e ) Use he inverse propery of logs on he righ side ln = 5730r Divide by 5730 ln r = Now we know he decay will follow he equaion Q( ) = ae. To find how old he bone fragmen is ha conains 0% of he original amoun, we solve for so ha Q() = 0.0a a = ae 0.0 = e ln(0.0) = ln( e ) ln( 0.0) = ln(0.0) = 330 years The bone fragmen is abou 3,300 years old. Try i Now. In Example, we learned ha Cesium-37 has a half-life of abou 30 years. If you begin wih 00mg of cesium-37, will i ake more or less han 30 years unil only milligram remains?

6 Secion 4.6 Exponenial and Logarihmic Models 73 Doubling Time For decaying quaniies, we asked how long i akes for half he subsance o decay. For growing quaniies we migh ask how long i akes for he quaniy o double. Doubling Time The doubling ime of a growing quaniy is he ime i akes for he quaniy o double. Given he basic exponenial growh equaion h = solving for when he original quaniy has doubled; by solving ( ) ab, doubling ime can be found by x a = a( b), or more x simply = b. Again noice how he iniial amoun is irrelevan when solving for doubling ime. Example 5 Cancer cells someimes increase exponenially. If a cancerous growh conained 300 cells las monh and 360 cells his monh, how long will i ake for he number of cancer cells o double? Defining o be ime in monhs, wih = 0 corresponding o his monh, we are given wo pieces of daa: his monh, (0, 360), and las monh, (-, 300). From his daa, we can find an equaion for he growh. Using he form C ( ) = ab, we know immediaely a = 360, giving C( ) = 360b. Subsiuing in (-, 300), 300 = 360b = b 360 b = =. 300 This gives us he equaion C ( ) = 360(.) To find he doubling ime, we look for he ime unil we have wice he original amoun, so when C() = a. a = a(.) = (.) log ( ) = log(. ) log( ) = log(.) log( ) = 3.80 monhs. log(.) I will ake abou 3.8 monhs for he number of cancer cells o double.

7 74 Chaper 4 Example 6 Use of a new social neworking websie has been growing exponenially, wih he number of new members doubling every 5 monhs. If he sie currenly has 0,000 users and his rend coninues, how many users will he sie have in year? We can use he doubling ime o find a funcion ha models he number of sie users, and hen use ha equaion o answer he quesion. While we could use an arbirary a as we have before for he iniial amoun, in his case, we know he iniial amoun was 0,000. r If we use a coninuous growh equaion, i would look like N( ) = 0e, measured in housands of users afer monhs. Based on he doubling ime, here would be 40 housand users afer 5 monhs. This allows us o solve for he coninuous growh rae: r5 40 = 0e r5 = e ln = 5r ln r = Now ha we have an equaion, N( ) = 0e, we can predic he number of users afer monhs: 0.386() N () = 0e = housand users. So afer year, we would expec he sie o have around 633,40 users. Try i Now 3. If uiion a a college is increasing by 6.6% each year, how many years will i ake for uiion o double? Newon s Law of Cooling When a ho objec is lef in surrounding air ha is a a lower emperaure, he objec s emperaure will decrease exponenially, leveling off owards he surrounding air emperaure. This "leveling off" will correspond o a horizonal asympoe in he graph of he emperaure funcion. Unless he room emperaure is zero, his will correspond o a verical shif of he generic exponenial decay funcion.

8 8 Chaper 4 Imporan Topics of his Secion Radioacive decay Half life Doubling ime Newon s law of cooling Logarihmic Scales Orders of Magniude Momen Magniude scale Try i Now Answers. r = or 6.7% is he daily rae of decay.. Less han 30 years, o be exac 3. I will ake years, or approximaely years, for uiion o double hours 5. Broadcas Conversaion Sofes room Sof Sound Whisper Train Symphony Space Shule x = 0 The sound pressure in µpa creaed by launching he space shule is 8 x0 orders of magniude greaer han he sound pressure in µpa creaed by he sofes sound a human ear can hear.