Where k is either given or determined from the data and c is an arbitrary constant.

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1 Exponntial growth and dcay applications W wish to solv an quation that has a drivativ. dy ky k > dx This quation says that th rat of chang of th function is proportional to th function. Th solution is kx y c. W can show this by taking th drivativ of y. dy dx d kt kx kx ( c ) c( k ) k( c ) dx ky Whr k is ithr givn or dtrmind from th data and c is an arbitrary constant. MATH 138 Lctur 23 1 of 23

2 Suppos y is rplacd by P which rprsnts th population of som spcis. Th assumption P kp maks prfct sns for smallr populations. It says th rat of chang of th population is proportional to th population. A spcific problm is: With initial condition which is calld an initial valu problm. P kp P ( ) P Th initial condition stats what th starting population is at tim. Exampl: Suppos Find th solution givn W know th solution is P P. 1P so k.1. P ( ) 1..1t c. To gt th valu of c, plug in. So, MATH 138 Lctur 23 2 of 23

3 P ().1( ) c c c(1) c 1 So th spcific solution is P(t) P( t) 1.1t t MATH 138 Lctur 23 3 of 23

4 If k is not spcifid som othr pic of information is ndd. Suppos P kp P ( ) 1 Th last quation again givs an initial population of 1 popl. If, in addition, w know W can figur out k. P ( 1) 1 Th solution from bfor is P( t) 1 kt MATH 138 Lctur 23 4 of 23

5 To gt k plug in th scond point P ( 1) 1 k(1) k P (1) or k 1 Solving for k by taking th natural logarithm of both sid. ln( k ) ln1 k ln1 And th solution is P( t) 1 (ln1) t MATH 138 Lctur 23 5 of 23

6 kt Lt P( t) P, and suppos th xtra pic of information is th tim T whn th initial population doubls. That is: P ( T ) 2P i.. Th tim T is also calld th gnration tim. So w nd to solv for T: P kt kt 2P or 2 Notic that th initial population is no long in th quation. kt Logging both sids givs ln( ) kt This is a crucial quation that rlats th growth constant k to th doubling tim T. MATH 138 Lctur 23 6 of 23

7 Th growth rat k and th gnration (doubling tim) ar linkd by th formula kt Dividing by T givs k T Dividing by k givs T k Exampl: What is th growth rat k if th doubling tim is T ? k Exampl: What is th doubling tim if th growth rat is.24. T MATH 138 Lctur 23 7 of 23

8 Exampl: Suppos a population has a doubling tim T yars, and an initial population of 25. What is th population aftr 1 yars. First comput k: k Plugging into th solution quation givs P( t) 25 (.399) t P (1) 25 (.399) Exampl: A crtain town has an initial population of 1,231 and it doubls in 13 yars. Find th solution and dtrmin whn th population is 17,. So th solution is kt ( t) 1,. To gt k w nd to solv P 231 k k so T MATH 138 Lctur 23 8 of 23

9 And so P( t) 1,231 (.5332) t Lt t t b th tim whn th population is 17,, thn P (.5332) t t ) 1,231 17, ( Which givs Logging both sids: Which givs t (.5332) t (.5332) t 17, ln 1, , 1,231 17, ln 1, yars MATH 138 Lctur 23 9 of 23

10 Logistic Population Modl Th problm with th xponntial solution w just obtaind is that th population gos to as tim gos to. W all know that limitd spac and or food limits th population. A bttr modl is calld th Logistic modl. This population modl is L P( t) 1+ kt b L L Notic that whn t, P ( ) P, and whn t gos to, 1+ b 1+ b lim P( t) L lim t 1+ b L 1+ limb L 1+ t kt kt t L This mans th limiting population is L. Dpnding on initial condition th solution looks lik: MATH 138 Lctur 23 1 of 23

11 P(t) L t Notic that in th uppr curv, th initial population is gratr than L, and in th lowr curv, th initial population is lss than th limiting population MATH 138 Lctur of 23

12 Exponntial Dcay: W again wish to solv an quation that has a drivativ. dy ky k > dx This quation says that th rat of chang of th function is proportional to th function, but now thr is a ngativ on th right hand sid. Th solution is y kx. W can show this by taking th drivativ c dy dx ck kx k( c kx ) ky Whr k is ithr givn or dtrmind from th data and c is an arbitrary constant dtrmind by th initial condition. MATH 138 Lctur of 23

13 Suppos y is rplacd by N which rprsnts th amount of radioactiv matrial in som objct. Th assumption N kn maks sns. It says th rat of chang of th amount is proportional to th amount prsnt. A spcific problm is: N kn k > With initial condition N ( ) N Whos solution is N( t) N kt MATH 138 Lctur of 23

14 Exampl: Suppos Find th solution givn W know th solution is N. 52N thn k.52. N N ( ) N..52t. To s this: c N d ( c dx.52t.52t.52t ) c(.52 ) c( k ) kn To gt th valu of c, plug in. So, N.52( ) ( ) c c N So th spcific solution is N( t) N.52t MATH 138 Lctur of 23

15 N(t) N MATH 138 Lctur of 23 t

16 If k is not spcifid som othr pic of information is ndd. Suppos N kn N ( ) N 1 Which is th initial population. If in addition, w know W can thn figur out k. N ( 2) 75 First, th solution is N kt ( t) N 1 kt MATH 138 Lctur of 23

17 To gt k plug in th scond point k( 2) 2k N (2) or 1 2k 2k Solving for k by taking th natural logarithm of both sid. ln( 2 k ) ln.75 2k ln.75 1 Not: ln(.75) ln(3/ 4) ln[(4 / 3) ] ln(4 / 3) k ln.75 2 > And th solution is N( t) 1 ln 4 / 3 t 2 MATH 138 Lctur of 23

18 If N kt ( t) N and th xtra pic of information is N ( T ) (1/ 2) N Thn w want th tim T whn th initial amount halvs. or N kt 1/ 2) kt ( N 1/ 2 Notic that th quation dosn t hav N in it. Logging both sids givs ln( kt ) ln(1/ 2) kt kt This is just lik th quation bfor. th dcay constant k is rlatd to th halving tim T, also calld th half lif. MATH 138 Lctur of 23

19 Th dcay rat k and th half lif T, ar rlatd by In which cas kt ~ k and T T k Exampl: What is th dcay rat k if th halving tim if T 36.45? k Exampl: What is th half-lif if th growth rat is.24? T MATH 138 Lctur of 23

20 Exampl: Carbon-14 has a half-lif of 575 yars. Suppos an objct has lost 2% of its carbon-14. How old is it? Th dcay rat is k T So th amount prsnt is N( t) N.125t So w want to know how old (th tim t ) whn thr is 8% lft..125t N( t ) N. N Solving for t so MATH 138 Lctur 23 2 of t.8 ln.8 t 1785yrs.125

21 Nwton s Law of Cooling. Suppos th tmpratur of an objct changs at a rat proportional to th diffrnc of th objct s tmpratur and th surrounding mdium. T(t) C Th quation to solv is T ( t) k( T C) with T ( ) T Notic. If T > C thn T ( t) < and th objct cools. If T < C thn T ( t) > thn th objct warms up. MATH 138 Lctur of 23

22 Suppos w lt Thn Or Which w just solvd so P( t) T ( t) C T ( t) P ( t) k( T ( t) C) kp( t) P ( t) kp( t) Thn And thn P( t) P( ) T ( t) ( T P P T C) MATH 138 Lctur of 23 kt C kt + Suppos th objct is initially 75 dgrs, and th surrounding mdium is 25 dgrs. Thn C

23 Which has graph: T(t) kt T ( t) T 75 o C 25 o t MATH 138 Lctur of 23

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