Calculus Exponential Growth and Decay Word Problems
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Applied Problems (It's About Time!)
Exponential Growth and Decay Word Problems

 

 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 So.
 The little o subscript means "Original" stuff.
 So replacing the + C, we have:
 

 ln S = k t + ln So

 
 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 eX = e

 
 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/So.
 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 = Soekt

 This is probably the most significant, 
 most often used equation form in all of calculus. 
 
 So here's what we've got:

 S = Soekt

 
 S: A variable, the amount of stuff we have after some time has passed.
 So: 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:
  So = 10
  S = 15
  t = 4
  k = the thing we want to find 
 
 So our equation is:

 15 = 10e4k

 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 e4k

 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 ...
  So = 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 = Qoekt

 
 Step 1: Find k ...
  Qo = 1
  Q = 2
  t = 6
  k = the thing we want to find 
 

 2 = e6k

 ln 2 = 6k

k =  

ln 2

6

 

k = 0.11552

 
 So the equation for Secret Organism Q growth is:

 

 Q = Qoe0.11552t

 
 So to solve the "How long does it take to get 100 times as big" problem ...
 
  Qo = 1
  Q = 100
  k = 0.11552
  t = the thing we want to find 
 

 Q = Qoe0.11552t

100 = e0.11552t
ln 100 = 0.11552t
 

t =  

ln 100

0.11552

  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.
 Do will be bigger than D, but so what?
  Do = 2
  D = 1
  t = 3
  k = the thing we want to find 
 

1 = 2e3k

 ln 1 = ln 2 + 3k

k =   ln 1 - ln 2

3

 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 = Doekt

 
  Do = 10
  D = 1
  k = - 0.2310
  t = the thing we want to find 
 

 1 = 10e-0.2310t

ln 1 = ln 10 -0.231t

t =   ln 1 - ln 10

-0.231

 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

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