AP CALCULUS AB

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


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


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


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?


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


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.


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

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