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## MATH-M343/S343 Section 2.3 Notes

by: Kathryn Brinser

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# MATH-M343/S343 Section 2.3 Notes MATH-S343

Marketplace > Indiana University > Mathematics > MATH-S343 > MATH M343 S343 Section 2 3 Notes
Kathryn Brinser
IU
GPA 4.0

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Covers modeling with linear differential equations.
COURSE
Honors Differential Equations
PROF.
Michael Jolly
TYPE
Class Notes
PAGES
6
WORDS
CONCEPTS
math-s343, math-m343, Differential Equations, modeling
KARMA
25 ?

## Popular in Mathematics

This 6 page Class Notes was uploaded by Kathryn Brinser on Saturday September 10, 2016. The Class Notes belongs to MATH-S343 at Indiana University taught by Michael Jolly in Fall 2016. Since its upload, it has received 5 views. For similar materials see Honors Differential Equations in Mathematics at Indiana University.

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Date Created: 09/10/16
S343 Section 2.3 Notes- Modeling with First Order Equations 8-30-16  Ex. A pipe is pouring liquid into a tank at ???? gallons per minute. The tank is stirring and has an exit rate, through another pipe, of ???? gallons per minute. Suppose the tank has a capacity of 100 gallons. The 1 liquid coming into the tank is water with lb ???????????????? per gallon, and the amount of ???????????????? in the tank at 4 time ???? is ???? ???? in lbs. o Model this situation with a differential equation. ???????? = ???????????????? − ???????????????? ???????? ???????? ???????????? = ( 1 ????????)(???? ????????????) − ( ???? ????????)(???? ????????????) 4 ???????????? ???????????? 100???????????? ???????????? ???? ???????? ???????? ???????? = 4 ????????????− 100 ???????????? 1 ???? = ????( − ) 4 100 o What is the limiting concentration/amount of ???????????????? (ie. How much is there as ???? → ∞)? ????????  ???????? = 0 when amount stops changing 1 ????  0 = ????( 4 100) ???? ???????? = 1 ????00 = 4 100 ???? = 25 ???????????? o Assuming ???? is given and constant, equation can be solved as linear or separable o Linear: ???????? ???? ????  + ???? = ???????? 100 4 ???? ???? Let ???? ???? = ???? ∫100= ???? 100 ???? ???? ???? ???????? ????100????+ ???? ???????? 100????= ???????? 100???? ???????? 100 4 ???? ???????? ???? ???? ???? ???????? (???????? 100) = ????4 100 ???? ???? ???? ???? ∫ (???????? 100) = ∫ ???? 100 ???????? ???????????? 4???? ???????? 100????= (???? 100)???? 100????+ ???? 4 ???? ???? ???? ???????? ???????? 100 = 25???? 100 + ???? − ???????? ????(????) = 25 + ???????? 100  Given ???? 0 = ???? : 0 ????  ???? 0 = ???? = 25 + ???????? − 1000) 0 0 = 25 + ???????? = 25 + ???? ???? = ???? 0 25 ???? −100????  ???? ???? = 25 + ???? ???? 20 ???? ) ???? −100 −100???? = ???? 0 + 25(1 − ???? )  As ???? → ∞:  If ????0> 25, negative progress (?)  If ????0< 25, positive progress (?)  Compound Interest as a Differential Equation o ????0= initial amount (invested or a loan borrowed) o ????(????) = amount at time ???? years o ???? = rate of interest o Frequency of Compounding  Biannually 1 ????  ????( 2 = ???? (0 + ) 2 ???? ????  ???? 1 = ???? (0 + )(1 + ) 2 ???? 2  ???? ???? = ???? (1 + ) ???? 0 2  Monthly 1 ????  ????( ) = ???? (10+ ) 12 12 ???? 12  ???? 1 = ???? (0 + 12) ???? 12????  ???? ???? = ???? (0 + ) 12  Daily 365????  ???? ???? = ???? (0 + ???? ) assuming no leap years 365 ???? ???????? o General formula: ???? ???? = ???? 01 + ) ???? when compounded ???? times per year  If compounded continuously (ie. ???? → ∞): ???? ???????? ????→∞m ((1 + )????) = ???? ???????? lim (ln((1 + ) )) = ln???? clever logarithmic manipulation ????→∞ ???? ???? lim (????????ln(1 + ))????= ln???? ????→∞ ???? ln(1???? ) 0 lim ( 1 ) = ln???? now have limit approaching 0 use L’Hôpital ????→∞ ???????? ????(ln(1+ )) lim (???????? ???? 1 ???? ) = ln???? ????→∞ ???????? ???????? 1 −???? 1+????(????2) lim ( −1 ) = ln???? ????→∞ ???? ???? −1 ???? ????21+????) lim ( ???? ) = ln???? ????→∞ −2( ) ???? ???? ???? 1+???? lim ( 1 ) = ln???? ????→∞ ???? lim ( ???????? ) = ln???? ????→∞ 1+ ???? ???????? ???? = ln???? 1+0 ???????? = ln???? ???? = ???? ???????? ???????? o Final formula for continuously compounded interest: ???? ???? = ????0????  Must include initial condition ???? 0 = ???? ???????? ???? 0 = (????0???? ????) ???????? ???????? = ???? 0 ????????(????) = ???????? o Suppose we deposit (or withdraw) ???? dollars per year: ???????? ???????? = ???????? + ???? ???? 0 = ???? 0 ???????? ???????? − ???????? = ???? Let ???? ???? = ???? ∫−???? = ???? −???????? ???????? ????−????????− ???????????? −???????? = ???????? −???????? ???????? ???????? −????????) = ???????? −???????? ???????????? −???????? −???????? ∫ ????????(???????? )= ∫???????? ???????? −???????? −???????? −1 ???????? = ???????? ( ???? + ???? −???????? −???? −???????? ???????? = ???? ???? + ???? −???? ???????? ???? ???? = ???? + ???????? −???? 0 ???? 0 = ???? =0 ???? + ???????? −???? ????0= ???? + ???? ???? ???? = ???? 0 ???? −???? ???? ???????? So ???? ???? = ???? + (????0+ )???? ???????? ???? ???????? = ???? 0 + ???? (???? − 1 ) o Ex. You borrow \$8000 for a car at 10% interest (compounded continuously). How much should you pay each year to pay it off in 3 years?  ???? 0 = ???? =08000, ???? = 0.10 ???????? ???? ????????  ???? ???? = ???? ????0 + ???? (???? − 1 ) Find ???? < 0 (taking money out) such that ???? 3 = 0 0.1 3 ???? 0.1 3)  0 = 8000???? + 0.1(???? − 1) = 8000???? 0.3+ 10???? ???? 0.3− 1) 0.3 0.3 −8000???? 0.30???? ???? − 1) ???? = −8000???? ≈ −\$3086.64 per year 10 ????0.−1)  Find the total interest paid on the car: ( ) 3 3086.64 − 8000 = \$1295.92  Ex. At 7:00 pm on Wednesday, a professor is found dead in his office. At 7:00, his body temperature is measured at 90℉. At 8:00, it is measured at 88℉. The temperature of the surroundings remains ???????? constant at 70℉. Using Newton’s Law of Cooling ( = −???? ???? − ???? ????, establish the time of death. ???????? o ???? = temperature of object ???? 2 o ???? = combination of heat transfer coefficien???? ????) and heat transfer surface area ???? )  Units not important  If ???? > 0, depending on initial temperature, ???? will approa????h ???? o ????????= temperature of surroundings o ???????? = −???? ???? − ???? ) ???? 0 = ???? ???????? ???? 0 ???????? = −???????? + ???????? ???????? ???? ???????? ???????? + ???????? = ???????? ???? ( ) ∫ ???? ???????? Let ???? ???? = ???? = ???? ???????? ????????????+ ???????????? ????????= ???????? ???? ???????? ???????? ???? ???? (???????? ????????) = ???????? ???? ???????? ???????? ???? ???? ???????? ???????? = ???????? ???? ???????????????? ∫ ????????( ) ∫ ???? ???????? ???????? 1 ???????? = ???????? ???? (????) + ???? ???????? ????????= ???? ???? ???????? + ???? ???? ???? ???? = ???? +???????????? −???????? o ???? 0 = ???? = ???? + ???????? 0 0 ???? ????0= ???? ???? ???? ???? = ???? − ???? 0 ???? −???????? So ???? ???? = ???? +???????? − ????0???? ????) o Let time 0 be time of death o Let 1 be 7:00 pm − ???? 0 o Let 8:00 pm be ????2= ???? 1 1 o Assume ???? at time of death is 98.6℉ and find ????:  90 = 70 + 98.6 − 70 ???? ) −????????1 at 7:00 pm −???????? 1 20 = 28.6???? 20 = ???? −????????1 substitute into second equation 28.6  88 = 70 + 98.6 − 70 ???? ) −???? ????1+1) at 8:00 pm −???? ???? +1 ) 18 = 28.6???? 1 18 −????????1 −???? 28.6= ???? ???? 18 20 −???? 28.6= 28.6???? substitution from first equation 18 28.6 −???? 20 28.6= ???? 9 = ????−???? 10 9 ln = −???? 10 10 ???? = ln 9  Go back to first equation to find1???? and0???? : 20 (−ln )???? = ???? 9 1 28.6 9 20 = ???? (l10)????1 28.6 20 (ln )????1 ln28.6= ln(???? 10 ) 20 9 ln = ln( )???? 1 28.6 20 10 ln28.6 ????1= 9 ≈ 3.52 ln10 7:00 − ???? 0 3:31.20 ℎ???????? ????0= 7:00 − 3:31.20 = 4:00 − 0:31.20 = 3:28.80 ????????  Ex. Escape velocity of a rocket (using Σ???? = ????????) ???????????? 2 o Force due to gravity (weight) = 2 where ???? = mass, ???? = acceleration of gravity, ???? = radius ????+???? ) of Earth, ???? = distance from surface of Earth  Pulling rocket back toward Earth, rocket trying to overcome and go outward o Initial condition: ???? 0 =0???? o ???? = ???????? → ???????? = ???? 2  ????( ) = −???????????? make force of weight negative from pulling backward ???????? (????+????)2 ???????? −???????? 2  = ( ) let ????,???? be constant ???????? ????+???? o Note that ???? = ????(???? ????)) ???? ???????? ????????  (????(???? ???? ) = ???????? ???????? ???????? ???????? ????????  = ???? is exactly the velocity (change in displacement) ???????? 2 ???????? o ???????????? = −???????? ???????? (????+???? ) 2 1 ∫ ???????????? = −???????? ∫(????+???? ) ???????? 1 2 2 1 ???? = ???????? ( ) + ???? 2 2 ????+???? ???? = 2???????? + ???? ????+???? o Solve for ???? before taking square root 2????????2 ???? = ???? 0 ????+0 2 = ???? − 2???????? 0 ???? = ???? − 2???????? 2 0 o ???? = 2???????? + ????0− 2???????? ????+???? 2????????2 o ???? = ± ???? +0 − 2???????? keep ± because rocket could come back down ????+???? o If ???? = 0, then ???? 0 = ±√???? = ???? as it should 0 0 o If escape velocity not reached, rocket goes up, pauses for an instant, then comes back; find ???? = ???? such that ???? ???? = 0: 2 0 = ???? + 2???????? − 2???????? 0 ????+???? 2 2????????2 2???????? = ???? +0 ????+???? 2???????? 2 = 2???????? − ???? 0 ????+???? 2 2 2???????? = 2???????? − ???? 0)(???? + ???? ) 2???????? 2 2????????−???? 2= ???? + ???? 0 2 ???? = 2???????? − ???? 2????????−????0 2???????? −????(2????????−????0) = 2 22????????−20 2 2???????? −2???????? +????0???? = 2????????−???? 0 2 ????0???? = 2????????−????0 o Solve for ???? in terms of ???? as ???? → ∞: 0 (2???????? − ????0???? = ???? 0 2???????????? − ???? ???? = ???? ???? 02 0 2???????????? = ????0???? + ???? ) 2 2???????????? ????0= ????+???? 2???????????? ????0= √ can disregard ± because we know0???? > 0 ????+???? o Escape velocity: 2???????????? ????→∞ (√ ????+????) 1 √ 2???????????????? ) = ????→∞ ( 1(????+????)) ???? = lim ( 2????) ????→∞ √ +1 ???? 2???????? = √ 0+1 = √ 2????????

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