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 xe
2x
dx
Drill
u  x, dv  e dx
2x
2x
2x
e
du  dx, v 
2
2x
e
e
xe
dx

x


dx


2
2
2x
2x
xe
e
2x
xe
dx



C

2
4
2x
Exponential Growth and Decay
Lesson 6.4
Objectives
• Students will be able to:
– solve problems involving exponential growth and
decay in a variety of applications.
Definition:
Separable Differential
Equation
A differential equation of the form dy  f  y g x 
dx
is called separable. We separate the variable by
writing it in the form
dy
 g x dx.
f y
The solution is found by integrating each side with
respect to its variable.
Example
Solve by Separation
of Variables
dy
1
2
Solve for y if
when x = 1.
 6 y x and y 
dx
25
dy
 6 x dx
2
y
2
y dy  6 x dx
y
2
dy   6 x dx
1
 y  3x  C
2
1
 1
2
    31  C
 25 
 25  3  C
 28  C
1
 y  3x  28
2
Example
Solve by Separation
of Variables
dy
1
2
Solve for y if
when x = 1.
 6 y x and y 
dx
25
1
2
  3x  28
y
1
2
 28  3 x
y
1
y
2
28  3x
1
 y  3x  28
2
Example
Solve by Separation
of Variables
dy
2
Solve for y if
  xy  and y  1 when x = 1.
dx
dy
2
 x dx
2
y
2
2
y dy  x dx
y
dy   x dx
1 3
1
 y  x C
3
2
2
1 2
 1  1  C
3
1
1   C
3
4
C
3
1 3 4
1
y  x 
3
3
1
Example
Solve by Separation
of Variables
dy
Solve for y if
 (xy ) 2 and y  1 when x = 1.
dx
1 x 4
 
y
3
3
1 4 x

y
3
3
y
3
4 x
3
y
1
3
x
4


3 3
The Law of Exponential Change
• If y changes at a rate proportional to the amount
present (that is, if dy/dt = kt), and if y = y0 (initial
amount) when t = 0, then
y = y0ekt
The constant k is the growth constant if k > 0 or
the decay constant if k< 0.
Interest Formulas
• Continuously Compounded Interest
A(t) = A0ert
A : initial amount
t: time
r: continuous interest rate
• Compounded interest for n compounding
periods
A(t) = A0(1+r/n)nt
Example
Compounding Interest
Continuously
• Suppose you deposit $800 is an account that
pays 6.3% annual interest. How much money
will you have 8 years later if:
– Compounded continuously?
• A(t) = 800e.063(8)
• $1324.26
– Compounded quarterly?
• A(t) = 800(1+.063/4)4(8)
• $1319.07
Half-Life
• dy/dt: decay of a radioactive element over time.
• dy/dt=-ky, k>0
• Half-life = ln 2/k, where k is a rate constant
• Also, A(t) = A0(.5)t/h
• t = time, h = half-life time period, A0 = original
amount
Example
Half-Life
• An isotope of neptunium (Np-240) has a half-life of 65
minutes. If the decay of Np-240 is modeled by the differential
equation dy/dt = -ky, where t is measured in minutes, what is
the decay constant k?
• Half-life = ln 2/k
• 65=ln2/k
• 65k = ln2
• k=ln(2)/65=.01066
Example
Choosing a Base
• At the beginning of the summer, the
population of a hive of a bald-faced hornets
(which are actually wasps) is growing at a rate
proportional to the population. From a
population of 10 on May 1, the number of
hornets grows to 50 in 30 days. If the growth
continues to follow the same model, how
many days after May 1 will the population
reach 100?
Example
Solution 1
• Since dy/dt = -ky, the grown
is exponential.
• The population grows by a
factor of 5 in 30 days: 10 X 5
= 50; therefore we model
the growth in base 5
• y = 10(5)(t/30)
• 100 = 10(5)t/30
• 10 = (5)t/30
Choosing a Base
Solution 2
• Using the two points of (0,
10) and (30, 50), create an
exponential equation to find
b:
• 50 = 10b30
• 5 = b30
• 1.055= b
• 100=10(1.055)t
• 10= (1.055)t
Example Using Carbon-14 Dating
Scientists who use carbon-14 dating use 5700 years for
its half-life. Find the age of a sample in which 10% of
the radioactive nuclei originally present have decayed.
.90A0  A0 (.5)
ln .90  ln(.5)
ln .90
t

ln .5 5700
t / 5700
.90  (.5)
t / 5700
ln .90 
ln(.5)
ln .90
5700 
t
ln .5
t / 5700
t
5700
Using Newton’s Law
of Cooling
Example
dT
 k (T  Ts )
dt
It’s solution, by the law of exponential
T is temperature of the object at time t
Ts is the surrounding temperature.
change: (T  Ts )  (T0  Ts )e
 kt
where T0 is the temperature at t = 0.
A hard-boiled egg at 980 C is put in a pan under running 180 C water to
cool. After 5 minutes, the egg’s temperature is found to be 380 C. How
much longer will it take to reach 200 C?
20  80e 5k
(38  18)  (98  18)e k 5
ln .25  5k ln e
(20  18)  (98  18)e .277t
.25  e 5k
ln .25  5k (1)
2  80e .277 t
ln .025
 13 .3 years  t
 .277
ln .25  ln e 5k
ln .25
 .277  k
5
.025  e
.277 t
ln .025  ln e .277 t
Homework
• Page 357: 5, 8, 9, 10, 21, 23, 25, 31

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