Math-3. Lesson 5-6 Euler s Number e Logarithmic and Exponential Modeling (Newton s Law of Cooling)

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1 Math-3 Lsson 5-6 Eulr s Numbr Logarithmic and Eponntial Modling (Nwton s Law of Cooling)

2 f ( ) What is th numbr? is th horizontal asymptot of th function: 1 1 ~

3 On my 3rd submarin (USS Springfild, SSN 761) w usd to jok that th answr to any math problm is 2 (raisd to som ponnt) How would you find? ln 2 ln 2 ln 3 ln 3 is btwn 1 and 2. ln ln

4 Writ th following numbrs as powrs of 2: log 2 log5 log log 2 2 Not possibl: Why? log(2) dosn' t ist No input will giv a ngativ output valu for f(). g( ) log f ( ) 2

5 W can writ any numbr as (raisd to som ponnt) is btwn 1 and How would you find? ln ln 3 ln ln 3 ln

6 Rwrit th following numbrs as a powr of 2 2 ln Not possibl: Why? ln(3) dosn' t ist y ln

7 W can rwrit th bas of any ponntial as a powr of. y 2 y k k y k 2 k ln 2 y k

8 y 4 y 1. 1 y y y How can you tll if a bas B ponntial is growth or dcay? y B Growth: B > 1 Dcay: 0 < B < 1

9 Rwrit th following ponntial quations as bas ponntials. y 4 y 1. 1 y y y

10 y 4 y 1. 1 y y 1 y y y B (0) look at th pattrn of th ponnts of k y Growth: k > 0 Dcay: k < 0 Growth: B > 1 Dcay: 0 < B < 1

11 f ( ) g( ) 2 Sinc all bas ponntial functions hav a bas of, w say that thir growth rat is k h( ) k

12 Bas 2 Eponntial Function f ( ) 2 Bas Eponntial Function g( ) 2 W say this function has a growth factor of 2. W say this function has a growth rat of 2.

13 Quantity: a catgory of masurmnts in th ral world. Vocabulary Unit of Masur: th unit that is usd to masur a quantity. Eampls of quantitis: Hight Wight Tmpratur Eampls of units: Hight: inchs, ft, mils Wight: pounds, kilograms Tmpratur: dgrs Fahrnhit or Cntigrad (Clsius)

14 Rat: th chang of on quantity compard to th chang in anothr quantity using a fraction. In pur mathmatics w would call this a slop. y slop Rat: (a ratio of quantitis) bcoms a nw quantity. tmp tim hatup/cooldown rat

15 Graphs compar th chang in quantitis. Each ais is labld with th quantity (with th units of masur for th quantity in parnthss).

16 Suppos boiling watr (100⁰ C) is takn off th stov to cool. Your turn: draw a graph of what you think th tmpratur will look lik as tim passs by (tmpratur as a function of tim). Labl th -ais and y-ais of your graph with th quantity that th ais rprsnts. Labl th -ais and y-ais of your graph with th units of masur (in parnthss) for th quantity th ais rprsnts. Dos th tmpratur go down forvr? At what tmpratur dos it start and nd up at? Will it tak hours, or minuts, or sconds to cool down?

17 y AB Growth (dcay) factor Initial Valu Initial Valu dos not man y-intrcpt if th graph has bn shiftd up/down. y A k m Growth (dcay) rat Horizontal asymptot

18 Tmp (⁰ C) (0, 100) Tim (min.) T = 15 Boiling watr (100⁰ C) is takn off th stov to cool in a room at 15⁰ C. y A k m Cool down rat kt T ( t) k =? Positiv or ngativ? y 100 M =? A =? k A k A M = (0, 100) (0) 100 A A 85

19 Nwton s Law of Cooling A high tmpratur itm will cool off in a lowr tmpratur nvironmnt in which it is placd. This cooling off procss can b modld by th following quation. Initial Tmp of th objct T ( t) Rat of chang of Tmp of th objct ( T T ) o s kt T s Tim Tmpratur (as a function of tim) Surrounding (final) Tmpratur Boiling watr (100⁰ C) is takn off th stov to cool in a room at 15⁰ C. T ( t) (100 15) kt 15 W nd a (tim, tmpratur) pair that th graph passs through in ordr to find k!

20 A hard-boild gg at tmpratur 100 º C is placd in 15 º C watr to cool. Fiv minuts latr th tmpratur of th gg is 55 º C. Whn will th gg b 25 º C? T ( t) ( T T ) kt o s Th first sntnc giv th initial and final tmpraturs. A hard-boild gg at tmpratur 100 º C is placd in 15 º C watr to cool. T T 100 T 15 o Th scond sntnc will hlp us find th cool down rat k. Fiv minuts latr th tmpratur of th gg is 55 º C. ( t, T ) (5 min, s 55 C s )

21 A hard-boild gg at tmpratur 100 º C is placd in 15 º C watr to cool. T ( t) ( T T ) o s kt T s T s T o kt T ( t) (100 15) k (0) chcks! kt T ( t) min,100 C

22 Tim (min.) Fiv minuts latr th tmpratur of th gg is 55 º C. T ( t) 85 kt 15 0 min,100 5 min, 55 C C t, T ( t) Each graph has a diffrnt cool down rat. mp (⁰ C) 5 min, 55 C W will us th ordrd pair (5, 55) to find th cool down rat k.

23 Fiv minuts latr th tmpratur of th gg is 55 º C. T(5) = 55 5 min, 55 C k 15 Isolat th ponntial k 5k Convrt to a log ln( ) 5k k T ( t) t 15 This is th quation that modls our problm.

24 A hard-boild gg at tmpratur 100 º C is placd in 15 º C watr to cool. Fiv minuts latr th tmpratur of th gg is 55 º C. Whn will th gg b 25 º C? T ( t) t t ln( ) (0.1508) t t t This is th quation that modls our problm. Now w solv our problm t min, Convrt to a log 25 t C 14.2 min

25 A hard-boild gg at tmpratur 100 º C is placd in 15 º C watr to cool. Fiv minuts latr th tmpratur of th gg is 55 º C. Whn will th gg b 25 º C?

26 A cak takn out of th ovn at tmpratur of 350 º F. It is placd on in a room with an ambint tmpratur of 70 º F to cool. Tn minuts latr th tmpratur of th cak is 150 º F. Whn will th cak b cool nough to put th frosting on (90 º F)? T( t) T ( T T ) m o m kt

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