# Exponential Growth and Decay - PowerPoint - PowerPoint

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```					Exponential Growth and
Decay

Dr. Dillon
Calculus II
SPSU
Fall 1999
Today’s Goals

Identify growth and decay problems

Learn to solve growth and decay
problems
Which is which?

“Growth’’ refers to exponential growth

f ( x)  k  a x , where a  1

“Decay” refers to exponential decay
f ( x)  k  a x , where 0  a  1
Recognize the Problem

Key words
– population
– Newton’s Law of Cooling
 One differential equation says it all
dy
 k  y, k a constant
dx
A differential equation?

 Information about the derivative of a function.
 To solve a differential equation, find the
function.
 We want y, a function of x, which satisfies

dy
 k  y, k a constant
dx
Technicalities

The TI-89 solves differential equations.
This one is easy to solve by hand.
One diff. eq. underlies every growth &
decay problem.
We only have to solve it once.
The Solution

dy       dy
ky     k dx 
dx        y
dy
 y  k  dx  ln ( y)  kx  C 
ye kxC  y  kx  eC  kx  C ' , where C'  C
e           e                 e

Thus y  C ' e kx
Are we there yet?

No.

So far, we just have an outline of the type
of problem we want to solve.
Example

A population of bacteria doubles in twenty
minutes. How long will it take to triple
in size?
Where is the diff. eq.?

If P is the size of any population at time
t then P grows at a rate proportional to
itself, i.e.

dP
 k  P, for some constant k
dt
Thus we know...

P(t )  Ce kt

where P(t) is the size of the population at
time t
It’s always true...

that C is the value of the function (in this
case P) when the variable (in this case
t) is zero
that k is a feature of the situation at
hand (in this case, the bacteria in your
petri dish)
Showing that C=P(0)

P(t )  C  e k t  P(0)  C k 0  C 0  C
e        e

which gives us…

t k e)0(P  ) t(P

dy
If     k  y, then y  y0e kx , where y0 is the value of y when x  0.
dx
dP
If     k  P, then P  P0e kt , where P0 is the value of P when t  0.
dt

If f ' ( x)  k  f ( x), then f ( x)  f (0)e kx.

dP(t )
If         k  t , then P(t )  P(0)e kt .
dt

What do we know?

What are we looking for?
Find k

k 20
P  2P(0)  P(0)e                   P(0)e   20k

2e   20k
 ln( 2)  20k  k  ln( 2) / 20
thus, for this population the model is

 ln( 2) 
ln(2 )
t

P  P(0)e      20
 P(0) exp        t 
 20       
Finally...

Find t when P=3P(0) as follows
 ln( 2)                ln( 2) 
3P(0)  P(0) exp         t   3  exp        t 
 20                    20       
ln( 2)      20 ln( 3)
 ln( 3)         t            t
20         ln( 2)
The Moral

A diff eq underlies every problem

The solution is always of the form

P(t )  P(0)e kt

k is different in every problem

Work with what you know to find what you seek.

```
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