math266% % % review%problemsfor%quiz%%part2% answer… · 8....

Math 266 Review Problems for Quiz Part 2 Answer Key Do the following problems as indicated. 6. The bacterium E. coli undergoes cell division approximately every 20 minutes. Starting with 100,000 cells, determine the number of cells after 2 hours. Given: ! = ! ! ! !" ; ! ! = 100,000 !"##$; ! 20 !"#$%&’ = 200,000 (Note: Cell division means that a cell divides itself to become 2 cells every 20 minutes.) Find: ! 2 ℎ!"#$ = !(120 !"#$%&’) Solution: ! ! = 100,000! !" ! 20 = 200,000 ⇒ 100,000! !∙!" = 200,000 ⇒ ! !∙!" = 2 ⟹ ! !∙!" ! !" = 2 ! !" ⟹ ! ! = 2 ! !" ⟹ ! ! = 100,000 ∙ 2 ! !" ! ! 120 = 100,000 ∙ 2 ! !" ∙!"# = 100,000 ∙ 2 ! = 6,400,000 There will be 6,400,000 cells after 2 hours. 7. A physicist finds that an unknown radioactive substance registers 2000 counts per minute on a Geiger counter. Ten days later the substance registers 1500 counts per minute. Approximate its halflife. Given: ! = ! ! ! !" ; ! ! = 2000; ! 10 !"#$ = 1500 Find: Halflife T such that ! ! = ! ! ! ! Solution: ! ! = 2000! !" ! 10 = 1500 ⇒ 2000! !∙!" = 1500 ⇒ ! !∙!" = 1500 2000 = 3 4 ⟹ ! !∙!" ! !" = 3 4 ! !" ⟹ ! ! = 3 4 ! !" ⟹ ! ! = 2000 ∙ 3 4 ! !" ! ! ! = 2000 ∙ 3 4 ! !" ! = 1000 ⟹ 3 4 ! !" ! = 1 2 ⟹ ln 3 4 ! !" ! = ln 1 2 ⟹ ! !" T ∙ ln ! ! = ln ! ! ⟹ ! = !" !".! !".!" = 24 ⟹ Half life is 24 days; that is, it takes 24 days to half itself.

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Page 1: Math266% % % Review%Problemsfor%Quiz%%Part2% Answer… · 8. Ametal!plate!that!has!been!heated!cools!from180℉!to150℉!in!20!minutes!when! surrounded!by!air!at!a!temperature!of!60℉.!!Use!Newton’s!law!of!cooling!to

Math 266 Review Problems for Quiz Part 2 Answer Key Do the following problems as indicated.

6. The bacterium E. coli undergoes cell division approximately every 20 minutes. Starting with 100,000 cells, determine the number of cells after 2 hours.

Given: ! = !!!!"; !! = 100,000 !"##$; ! 20 !"#$%&' = 200,000 (Note: Cell division means that a cell divides itself to become 2 cells every 20 minutes.) Find: ! 2 ℎ!"#$ = !(120 !"#$%&') Solution: ! ! = 100,000!!" ! 20 = 200,000 ⇒ 100,000!!∙!" = 200,000 ⇒ !!∙!" = 2⟹ !!∙!"

!!" = 2

!!" ⟹

!! = 2!!" ⟹ ! ! = 100,000 ∙ 2

!!"!

! 120 = 100,000 ∙ 2!!"∙!"# = 100,000 ∙ 2! = 6,400,000

There will be 6,400,000 cells after 2 hours. 7. A physicist finds that an unknown radioactive substance registers 2000 counts per minute

on a Geiger counter. Ten days later the substance registers 1500 counts per minute. Approximate its half-‐life.

Given: ! = !!!!"; !! = 2000; ! 10 !"#$ = 1500 Find: Half-‐life T such that ! ! = !

!!!

Solution: ! ! = 2000!!"

! 10 = 1500 ⇒ 2000!!∙!" = 1500 ⇒ !!∙!" =15002000 =

34⟹ !!∙!"

!!" =

!!" ⟹

!! = 34

!!"⟹ ! ! = 2000 ∙

!!"!

! ! = 2000 ∙34

!!"!

= 1000 ⟹34

!!"!

=12 ⟹ ln

!!"!

= ln12 ⟹

!!"T ∙ ln !

!= ln !

! ⟹ ! = !" !".!

!".!"= 24 ⟹ Half life is 24 days; that is, it takes 24 days to half itself.

Page 2: Math266% % % Review%Problemsfor%Quiz%%Part2% Answer… · 8. Ametal!plate!that!has!been!heated!cools!from180℉!to150℉!in!20!minutes!when! surrounded!by!air!at!a!temperature!of!60℉.!!Use!Newton’s!law!of!cooling!to

8. A metal plate that has been heated cools from 180℉ to 150℉ in 20 minutes when surrounded by air at a temperature of 60℉. Use Newton’s law of cooling to approximate its temperature at the end of one hour of cooling. When will the temperature be 100℉?

Given: ! ! = !! + !! − !! !!"; !! = 180°!; !! = 60°!;! 20 !"#$%&' = 150°! Find: ! 1 ℎ!"# = ! 60 !"#$%&' ; ! !ℎ!" ! ! = 100°! Solution: ! ! = 60+ 180− 60 !!" ⟹ ! ! = 60+ 120!!" ! 20 = 150 ⟹ 60+ 120!!∙!" = 150 ⟹ 120!!∙!" = 90 ⟹ !!∙!" = !"

!"#!.!" ⟹

!!∙!"!!" = . 75

!!" ⟹ ! ! = 60+ 120 . 75

!!"! ⟹

! 60 = 60+ 120 . 75!!"∙!" = 60+ 120 . 75 ! = 110.625°! Temperature after 1 hour

! ! = 100 ⟹ 60+ 120 . 75!!"! = 100 ⟹ 120 . 75

!!"! = 40 ⟹ . 75

!!"! = !"

!"#= !

! ⟹

ln . 75!!"! = ln !

! ⟹ !