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CHAPTER 6:DIFFERENTIAL EQUATIONS AND
MATHEMATICAL MODELINGSECTION 6.4:
EXPONENTIAL GROWTH AND DECAY
AP CALCULUS AB
What you’ll learn about
Separable Differential EquationsLaw of Exponential ChangeContinuously Compounded InterestModeling Growth with Other BasesNewton’s Law of Cooling
… and whyUnderstanding the differential equationgives us new insight into exponential growth
and decay.
dy kydx
Separable Differential Equation
A differential equation of the form is
called . We by writing it
in the form
1 .
The solution is found by antidifferentiating each s
dyf y g x
dx
dy g x dxf y
separable separate the variables
ide with
respect to its thusly isolated variable.
Example Solving by Separation of Variables
2 2Solve for if and 3 when 0.dy
y x y y xdx
2 2
2 2
2 2
3
1
3
1
3
3Apply the initial conditions to find C.
1 1 3 So, and
3 3 3 1This solution is valid for the continuous section
of the function that goes through the
dyx y
dxy dy x dx
y dy x dx
xy C
xC y y
x
point (0, 3),
that is, on the domain ,1 .
Section 6.4 – Exponential Growth and Decay
Law of Exponential ChangeIf y changes at a rate proportional to the amount present
and y = y0 when t = 0,
then where k>0 represents growth and k<0 represents decay.The number k is the rate constant of the equation.
dyky
dt
0 ,kty y e
Section 6.4 – Exponential Growth and Decay
From Larson: Exponential Growth and Decay ModelIf 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, andk = constant of proportionality
(see proof next slide)
kty Ce
Section 6.4 – Exponential Growth and Decay
Derivation of this formula:
1
lnkt C
kt C
kt
dyky
dtdy
kdty
dykdt
y
y kt C
e y
e e y
C e y
Section 6.4 – Exponential Growth and Decay
This corresponds with the formula for Continuously Compounded Interest
This also corresponds to the formula for radioactive decay
0rtA t A e
0 , 0kty y e k
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:
Differential equation:
Ini
dArA
dt
tial condition: (0)
The amount of money in the account after years
at an annual interest rate :
( ) .
O
rt
O
A A
t
r
A t A e
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 compounded continuously? compounded
monthly?
(a) (b)
0.053 4
12 4
Let 500 and 0.053.
a. (4) 500 618.07
0.053b. (4) 500 1 617.79
12
OA r
A e
A
Example Finding Half-Life
-
Find the half-life of a radioactive substance with decay equation
.ktO
y y e
-
1The half-life is the solution to the equation .
21
Solve algebraically 2
1 - ln
21 1 ln 2
- ln2
Note: The value
kt
O O
kt
y e y
e
kt
tk k
is the half-life of the element. It depends
only on the value of .
t
k
Hint: When will the quantity be half as much?
Section 6.4 – Exponential Growth and Decay
The formula for Derivation:half-life of a radioactive substance is
0 0
0
0 0
0
1
2
1
2
1
21
ln2
ln 2
ln 2
kt
kt
kt
y e y
y e y
y
e
k
t
y
t
kt
k
ln 2half-life
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 is the temperaT
ture of the object at time , and is the
surrounding temperature, then
. (1)
Since ( - ), rewrite (1)
S
S
S
t T
dTk T T
dtdT d T T
d
d
( )
Its solution, by the law of exponential change, is
- ,
Where is the temperature at time 0.
S S
kt
S O S
O
T T k T Tt
T T T T e
T t
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)
1
1
1
to find H as a function of time
lnkt C
Ckt
kt
kt
dHk H T
dt
k H T dtdH
H T H TdH
kdtH TH T kt C
e H T
e e H T
Ce H T
H Ce T
Example Using Newton’s Law of Cooling
A temperature probe is removed from a cup of coffee and placed in water thathas a temperature of T = 4.5 C.Temperature readings T, as recorded in the table below, are takenafter 2 sec, 5 sec, and every 5 sec thereafter.
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
S
o
Example Using Newton’s Law of Cooling
According to Newton's Law of Cooling, - ,
where 4.5 and is the temperature of the coffee at 0.
Use exponential regression to find that - 4.5 61.66 0.9277
is a model for the , - ,
kt
S O S
S O
t
S
T T T T e
T T t
T
t T T t
0
4.5 data. Thus,
4.5 61.66 0.9277 is a model of the , data.
(a) At time 0 the temperature was 4.5 61.66 0.9277 66.16 C.
(b) The figure below shows the graphs of 8 and
4.5 61.66 0.9277
t
t
T
T t T
t T
y
y T
Use time for L1 and T-Ts for L2 to fit an exponential regression equation to the data. This formula is T-Ts.
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
We can express that the resisting force is proportional to velocity by writing
This is a differential equation of exponential change,
Force = mass acceleration = .dv
mdt
or 0 .dv dv k
m kv v kdt dt m
0 .
k tmv v e