



Let's
say we have some quantity of stuff, call it "S."



And
as time passes, the amount of stuff changes. 


The
rate that the amount of stuff changes as time passes 


can
be written as a derivative: 








There
are many situations where the rate that stuff grows 


depends
on how much stuff we have at the moment. 





Maybe
sometimes we want to measure time in nanoseconds 


and
sometimes we want to measure time in hours. 


The
way to compensate for that is with a constant. 


Math
types like to use "k" for this constant 


for
some particular reason. 








This
equation says: 


The
rate that the amount of stuff changes as time passes 


is
equal to the amount of stuff we have times some constant. 





The
constant could have been put anywhere in the equation 


but
things work easiest putting it there. 





The
problems usually go something like this: 





The
rate that some particular stuff grows 


is proportional to how much stuff there is at that time. 


We
start with 10 pounds of stuff (S), 


4
hours later we have 15 pounds of stuff. 


When
will we have 20 pounds of stuff? 





What
we have been asked for is the value of time (t). 


Our
equation does not have "t," 


but
it does have the derivative of t (dt). 


We
have to find the integral of our equation to get t. 





This
could be difficult to deal with, 


but
wait! We just saw that: 








So
let's do some algebra to get dS/S by itself 


on
one side of the equation. 











now
find the integral of both sides: 





(remember
that k is a constant, S and t are variables) 











Actually,
a constant would be created on both sides, 


but we can lump it all
together. 


But
to go any further, we need to find out what C is. 





The
natural log of the quantity "stuff we have" (ln S) 


is
equal to the time that's passed (t) 


times
the unit correcting constant (k) plus ... What? 





It's
something to do with the quantity of stuff we originally started
with. 





In
fact it is the quantity of stuff we started with. 


We
call the quantity of stuff we started with S_{o}. 


The
little o subscript means "Original" stuff. 


So
replacing the + C, we have: 





ln S = k t
+ ln S_{o}






When
logs with the same base, in this case e, 


are
on the same side of the equation we can combine them. 


So
let's move this puppy around a bit ... 








Now
a cute little trick, the law of exponents says that ... 





If X = Y,
then e^{X} = e^{Y }






so
we can say ... 








The
left side of this thing is pretty nasty, 


but
a few pages ago we translated the meaning of logs into English. 


Let's
do that here and see what we get: 








It
says, e raised to the power that e must be raised to, 


to
get S/S_{o}. 


That
means: 








Go
through that again, it's really important. 


Go
ahead, we'll wait ... 





So
now we have: 








Generally,
this is written as: 


S = S_{o}e^{kt}



This
is probably the most significant, 


most
often used equation form in all of calculus. 





So
here's what we've got: 


S
= S_{o}e^{kt}






S:
A variable, the amount of stuff we have after some time has passed. 


S_{o}:
A variable, the amount of stuff we originally had 


e:
A constant, 2.71828 ... 


k:
A constant, we don't know the value of yet. 


t:
A variable, time, in particular the amount of time that has passed. 





Variables
can change but constants don't (that's why they are constants) 


If
we know S, So, and t for any particular time interval 


we
can calculate k. 


Then
use that equation for any problem involving that particular
stuff 


that
we ever see! 





OK,
Now we're ready to go. 


The
problem that we had way back when was: 





The
rate that some particular stuff grows 


is
proportional to how much stuff there is at that time. 





STOP! 


That
sentence means that this is the right equation form to use. 


One
where the stuff grows based on how much we have at that time. 





We
start with 10 pounds of stuff (S), 


4
hours later we have 15 pounds of stuff. 





STOP! 


That's
enough information to get us going. 


We
know how much stuff we started with 


and
how much stuff we have later, after a certain time. 


We
have: 



S_{o}
= 10 



S
= 15 



t
= 4 



k
= the thing we want to find 





So
our equation is: 


15
= 10e^{4k}



Now
we solve for k: 


STEP
1: 


Lose
the exponent by taking the log (ln) of both sides of the equation. 





ln
15 = ln 10 + ln e^{4k}



ln
15 = ln 10 + 4k



(Don't
forget that ln e means the power that e must be raised to, to get e.
Which is 1)






STEP
2: 


Get
k by itself and solve for it. 





Now
get out your calculator ... 


k
= 0.10137






The
entire problem read: 


We
start with 10 pounds of stuff, 


4
hours later we have 15 pounds of stuff. 


When
will we have 20 pounds of stuff? 





So
now ... 



S_{o}
= 15 



S
= 20 



k
= 0.10137 



t
= what we need to find 





So
let's do it! 





Don't
forget, ln e = 1 ! 








That
means 2.838 hours after the stuff grew to 15 it grew to 20. 





Notice
that it took 4 hours to grow the 5 from 10 to 15, 


but
only 2.838 hours to grow the 5 from 15 to 20. 


Do
you know why? 


If
you understand what's going on here, you should. 





Next
Example: 


Secret
organism "Q" doubles in size every 6 hours, 


how
long does it take to get 100 times as big? 





Q = Q_{o}e^{kt}






Step
1: Find k ... 



Q_{o}
= 1 



Q
= 2 



t
= 6 



k
= the thing we want to find 





2 =
e^{6k}



ln
2 = 6k












k = 0.11552 





So
the equation for Secret Organism Q growth is: 





Q = Q_{o}e^{0.11552t}






So
to solve the "How long does it take to get 100 times as
big" problem ... 






Q_{o}
= 1 



Q
= 100 



k
= 0.11552 



t
= the thing we want to find 





Q = Q_{o}e^{0.11552t}



100
= e^{0.11552t} 


ln
100 = 0.11552t 











t
= 39.86 hours






Last
Example: 


Quantity
"D" reduces it's amount to half in 3 hours. 


How
long does it take for it to reduce it's amount to 1/10? 





Hey,
this one is getting smaller! 


No
big deal, it works exactly the same way. 


D_{o}
will be bigger than D, but so what? 



D_{o}
= 2 



D
= 1 



t
= 3 



k
= the thing we want to find 





1 =
2e^{3k}



ln
1 = ln 2 + 3k






k
=  0.2310






WOW!
If the stuff is getting smaller, k will be negative. 


THAT
is the only difference. 





Now
let's find the time it takes D to get one tenth as big ... 





D = D_{o}e^{kt}







D_{o}
= 10 



D
= 1 



k
=  0.2310 



t
= the thing we want to find 





1 = 10e^{0.2310t}



ln 1 = ln 10
0.231t






t
= 9.968 hours






So
it takes 9.968 hours for the amount of Quantity D 


to
be reduced to one tenth of the starting amount. 





At
the end of a second 9.968 hours, 


there
will be one tenth as much as there was at the end of the first 9.968
hours. 


And
so on ... 





copyright 2005 Bruce Kirkpatrick 
