WARM-UP:
1. Use Euler’s Method to find 𝑓(3) given that
𝑓 1 = 3. Use ∆𝑥 = 0.5.
(𝒙, 𝒚)
∆𝒙
(𝒐𝒓 𝒅𝒙)
(1,3)
0.5
𝑑𝑦
𝑑𝑥
3
(1.5, 4.5)
0.5
(2,7.875)
(2.5, 15.75)
𝒅𝒚 = 𝒅𝒙 ∙
𝒅𝒚
𝒅𝒙
𝑑𝑦
𝑑𝑥
= 𝑥𝑦 and
(𝒙 + 𝒅𝒙, 𝒚 + 𝒅𝒚)
1.5
(1.5,4.5)
6.75
3.375
(2,7.875)
0.5
15.75
7.875
(2.5, 15.75)
0.5
39.375
19.6875
(3,35.4375)
Euler’s Method approximates 𝑓 3 = 35.4375.
2. Use separation of variables to find 𝑓(3) given that
= 𝑥𝑦 and 𝑓 1 = 3. Compare your answer to #1.
𝑑𝑦
𝑑𝑥
Exponential
Growth and
Decay
Sec. 6.4
The number of bighorn sheep in a population increases at
a rate that is proportional to the number of sheep present
(at least for awhile.)
So does any population of living creatures. Other things
that increase or decrease at a rate proportional to the
amount present include radioactive material and money in
an interest-bearing account.
If the rate of change is proportional to the amount present,
the change can be modeled by:
dy
 ky
dt

dy
 ky
dt
1
dy  k dt
y
1
 y dy   k dt
Rate of change is proportional
to the amount present.
Divide both sides by y.
Integrate both sides.
ln y  kt  C

1
 y dy   k dt
Integrate both sides.
ln y  kt  C
ln y
e
e
kt C
y  e e
C
kt
Exponentiate both sides.
When multiplying like bases, add
exponents. So added exponents
can be written as multiplication.

ln y
e
e
kt C
y  e e
C
kt
Exponentiate both sides.
When multiplying like bases, add
exponents. So added exponents
can be written as multiplication.
y  e e
C kt
y  Ae
kt
Since
eC
is a constant, let e
C
 A.

y  e e
C kt
y  Ae
kt
y0  Ae
Since
eC
is a constant, let e
C
 A.
1
k 0
At
t  0 , y  y0
.
y0  A
y  y0e
kt
This is the solution to our original initial
value problem.

Exponential Change:
y  y0e
kt
If the constant k is positive then the equation
represents growth. If k is negative then the equation
represents decay.

Continuously Compounded Interest
If money is invested in a fixed-interest account where the
interest is added to the account k times per year, the
amount present after t years is:
 r
A  t   A0 1  
 k
kt
If the money is added back more frequently, you will make
a little more money.
The best you can do is if the
interest is added continuously.

Of course, the bank does not employ some clerk to
continuously calculate your interest with an adding machine.
 r
lim A0 1  
k 
 k
We could calculate:
kt
but we won’t learn how to find this limit until chapter 8.
Since the interest is proportional to the amount present,
the equation becomes:
Continuously Compounded
Interest:
A  A0e
rt
You may also use:
A  Pe
rt
which is the same thing.

EXAMPLE: A man invests $10,000 in an account that
pays 8.5% interest per year. What is the amount of money
that he will have after 3 years if the interest is
compounded…
$12,772.89
a) annually?
$12,870.19
b) quarterly?
$12,901.92
c) weekly?
$12,904.23
d) daily?
e) continuously?
$12,904.61
Radioactive Decay
The equation for the amount of
a radioactive element left after
time t is:
 kt
y  y0e
This allows the decay constant, k,
to be positive.
The half-life is the time required for half the material to decay.

Half-life
1
 kt
y0  y0 e
2
1
ln    ln e kt
2
0
ln1  ln 2  kt
 
ln 2  kt
ln 2
t
k
Half-life:
ln 2
half-life 
k

EXAMPLE: Carbon-14 is a radioactive isotope of carbon
that has a half life of 5600 years. It is used extensively in
dating organic material that is tens of thousands of years
old. What fraction of the original amount of Carbon-14 in a
sample would be present after 10,000 years?
 kt
y  y0e
𝑦
= 𝑒 −𝑘𝑡
𝑦0
𝑦
−4 )(10,000)
−(1.2×10
=𝑒
≈ 0.301
𝑦0
So approximately 30% of the original Carbon-14 would be
present in 10,000 years.
EXAMPLE:
How long would it take for Cs137 to decay to 0.1% of its initial level, just after the explosion at
Chernobyl?
0.001𝑦0 = 𝑦0 𝑒 −0.023𝑡
0.001 = 𝑒 −0.023𝑡
ln(0.001) = −0.023t
𝑡=
ln(0.001)
≈ 300.337
−0.023
Therefore, it would take about 330 YEARS for the level of cesium-137 to decay to 0.1% of its
initial level!!!!!!!!!!!!!!!
Newton’s Law of Cooling
• An object’s temperature over time will approach the
temperature of its surroundings (the medium)
• The greater the difference between the object’s temperature
and the medium’s temperature, the greater the rate of change
of the object’s temperature
• This change is a form of exponential decay
T0
Tm
Newton’s Law of Cooling
Espresso left in a cup will cool to the temperature of the
surrounding air. The rate of cooling is proportional to the
difference in temperature between the liquid and the air.
(It is assumed that the air temperature is constant.)
dT
 k T  Ts 
If we solve the differential equation:
dt
we get:
Newton’s Law of Cooling
T  Ts  T0  Ts  ekt
where Ts is the temperature
of the surrounding medium,
which is a constant.
p
Example:
The great detective Sherlock Holmes and his assistant, Dr.
Watson, are discussing the murder of actor Cornelius McHam. McHam was
shot in the head, and his understudy, Barry Moore, was found standing
over the body with the murder weapon in hand. Let’s listen in:
Watson:Open-and-shut case, Holmes. Moore is the murderer.
Holmes:Not so fast, Watson – you are forgetting Newton’s Law of Cooling!
Watson:Huh?
Holmes:Elementary, my dear Watson. Moore was found standing over
McHam at 10:06 p.m., at which time the coroner recorded a body
temperature of 77.9°F and noted that the room thermostat was set to 72°F.
At 11:06 p.m. the coroner took another reading and recorded a body
temperature of 75.6°F. Since McHam’s normal temperature was 98.6°F, and
since Moore was on stage between 6:00 p.m. and 8:00 p.m., Moore is
obviously innocent. Ask any calculus student to figure it out for you.
How did Holmes know that Moore was innocent?
Solution:
NOTE: m is used in place
of s in this example.
EXAMPLE: A detective is called to the scene of a crime where a dead body
has just been found. She arrives on the scene at 10:23 pm and begins her
investigation. Immediately, the temperature of the body is taken and is found
to be 80o F. The detective checks the programmable thermostat and finds that
the room has been kept at a constant 68o F for the past 3 days. After
evidence from the crime scene is collected, the temperature of the body is
taken once more and found to be 78.5o F. This last temperature reading was
taken exactly one hour after the first one. The next day the detective is asked
by another investigator, “What time did our victim die?” Assuming that the
victim’s body temperature was normal (98.6o F) prior to death, what is her
answer to this question?
To organize our thinking about this problem, let’s be explicit about what we are trying
to solve for. We would like to know the time at which a person died.
In particular, we know the investigator arrived on the scene at 10:23 pm, which we
will call τ hours after death.
At 10:23 (i.e. τ hours after death), the temperature of the body was found to be 80o F.
One hour later, τ + 1 hours after death, the body was found to be 78.5o F.
Our known constants for this problem are, Ts = 68o F and T0 = 98.6o F.
Click HERE to submit your answer!
Homework:
p. 357-359 #1-10, 15-18, 21, 3032, 34, 45 +