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 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_oe^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_oe^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_oe^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 =	ln 2

	6

k = 0.11552

So the equation for Secret Organism Q growth is:

Q = Q_oe^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_oe^0.11552t

100 = e^0.11552t

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.

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 =	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 = D_oe^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 =	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 ...

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