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AP CALCULUS AB
CHAPTER 6:
DIFFERENTIAL EQUATIONS AND
MATHEMATICAL MODELING
SECTION 6.4:
EXPONENTIAL GROWTH AND
DECAY
What you’ll learn about
 Separable Differential Equations
 Law of Exponential Change
 Continuously Compounded Interest
 Modeling Growth with Other Bases
 Newton’s Law of Cooling
… and why
dy  ky
Understanding the differential equation
dx
gives us new insight into exponential growth and
decay.
Separable Differential Equation
dy
 f  y  g  x  is
dx
called separable. We separate the variables by writing it
in the form
1
dy  g  x  dx.
f  y
A differential equation of the form
The solution is found by antidifferentiating each side with
respect to its thusly isolated variable.
Example Solving by Separation of Variables
Solve for y if
dy
 x y and y  3 when x  0.
dx
2
2
dy
x y
dx
y dy  x dx
 y dy   x dx
x
y   C
3
Apply the initial conditions to find C.
1
x 1
3
  C So,  y   and y 
3
3 3
1 x
This solution is valid for the continuous section
of the function that goes through the point (0, 3),
2
2
2
2
2
2
3
1
3
1
3
that is, on the domain  ,1 .
Section 6.4 – Exponential Growth and
Decay
 Law of Exponential Change
If y changes at a rate proportional to the amount
present
 dy
 and y = y0 when t = 0,
  ky 
 dt

then y  y e kt ,
0
where k>0 represents growth and k<0 represents
decay.
The number k is the rate constant of the equation.
Section 6.4 – Exponential Growth and
Decay
 From Larson: Exponential Growth and Decay Model
If y is a differentiable function of t such that y>0 and
y’=kt, for some constant k, then
where C = initial value of y, and
k = constant of proportionality
y  Ce
(see proof next slide)
kt
Section 6.4 – Exponential Growth and
Decay
 Derivation of
this formula:
dy
 ky
dt
dy
 kdt
y
dy
 y   kdt
ln y  kt  C
e kt C  y
e kt  eC  y
C1e kt  y
Section 6.4 – Exponential Growth and
Decay
 This corresponds with the formula for Continuously
Compounded Interest
A  t   A0e
rt
 This also corresponds to the formula for radioactive
decay
 kt
y  y0e ,
k 0
Continuously Compounded Interest
If the interest is added continuously at a rate proportional
to the amount in the account, you can model the growth of
the account with the initial value problem:
dA
Differential equation:
 rA
dt
Initial condition: A(0)  A
O
The amount of money in the account after t years
at an annual interest rate r:
A(t )  A e .
rt
O
Example Compounding Interest
Continuously
Suppose you deposit $500 in an account that pays 5.3%
annual interest. How much will you have 4 years later if
the interest is (a) compounded continuously? (b) compounded
monthly?
Let A  500 and r  0.053.
O
a. A(4)  500e
0.053 4 
 618.07
 0.053 
b. A(4)  500 1 

12 

12  4 
 617.79
Example Finding Half-Life
Find the half-life of a radioactive substance with decay equation
y  y e . Hint: When will the quantity be half as much?
1
The half-life is the solution to the equation y e  y .
2
1
Solve algebraically e 
2
1
- kt  ln
2
1 1 ln 2
t  - ln 
k 2
k
Note: The value t is the half-life of the element. It depends
- kt
O
 kt
O
- kt
only on the value of k .
O
Section 6.4 – Exponential Growth and
Decay
 The formula for Derivation:
half-life of a
radioactive
substance is
1
y0 e  y 0
2
y0 e  kt 1 y0

y0
2 y0
 kt
e
ln 2
half-life 
k
 kt
1

2
1
 kt  ln
2
 kt   ln 2
ln 2
t
k
Newton’s Law of Cooling
The rate at which an object's temerature is changing at any
given time is roughly proportional to the difference between
its temperature and the temperature of the surrounding medium.
If T is the temperature of the object at time t , and T is the
S
surrounding temperature, then
dT
  k T  T  .
dt
Since dT  d (T - T ), rewrite (1)
S
S
d
T  T   k (T  T )
dt
Its solution, by the law of exponential change, is
S
S
T - T  T  T  e ,
 kt
S
O
S
Where T is the temperature at time t  0.
O
(1)
Section 6.4 – Exponential Growth and
Decay
 Another version of
Newton’s Law of
Cooling
(where H=temp of object
& T=temp of outside medium)
dH
 k H T 
dt
to find H as a function of time
k  H  T  dt
dH

H T
H T
dH
 H  T   kdt
ln H  T  kt  C1
e kt C1  H  T
e kt eC1  H  T
Ce kt  H  T
H  Ce kt  T
Example Using Newton’s Law of Cooling
A temperature probe is removed from a cup of coffee and placed in water that
has a temperature of T = 4.5 C.
Temperature readings T, as recorded in the table below, are taken
after 2 sec, 5 sec, and every 5 sec thereafter.
o
S
Estimate
(a) the coffee's temperature at the time
the temperature probe was removed.
(b) the time when the temperature
probe reading will be 8 C.
o
Example Using Newton’s Law of Cooling
Use time for L1 and T-Ts for L2 to fit an exponential regression
equation to the data. This formula is T-Ts.
According to Newton's Law of Cooling, T - T  T  T  e ,
 kt
S
O
S
where T  4.5 and T is the temperature of the coffee at t  0.
S
O
Use exponential regression to find that T - 4.5  61.66  0.9277 
t
is a model for the  t , T - T    t , T  4.5  data. Thus,
S
T  4.5  61.66  0.9277  is a model of the  t , T  data.
t
(a) At time t  0 the temperature was T  4.5  61.66  0.9277   66.16 C.
0
(b) The figure below shows the graphs of y  8 and
y  T  4.5  61.66  0.9277 
t
Section 6.4 – Exponential Growth and
Decay
 Resistance Proportional to Velocity
It is reasonable to assume that, other forces being absent,
the resistance encountered by a moving object, such as a
car coasting to a stop, is proportional to the object’s
velocity.
 The resisting force opposing the motion is
Force = mass  acceleration = m
dv
.
dt
 We can express that the resisting force is proportional to
velocity by writing
m
dv
dv
k
 kv or
 v
dt
dt
m
 k  0.
 This is a differential equation of exponential change,
t

m
v  v0e
.
 k