Example: Suppose you buy the latest tablet for N = $400, and you use a
credit card to pay for it, only paying the minimum balance allowed each
month. Your credit card charges an interest of 17% APR (that is, a monthly
interest of 0.013%), and you have to pay at least 2% of your balance each
month. Here are your balance and payments each month (M):
M
0
1
2
3
..
.
Balance
N
((0.98)N)(1.013)
(0.98)2 (1.013)2 N
(0.98)3 (1.013)3 N
..
.
Payment
(0.02)N
(0.02)((0.98)N)(1.013)
(0.02)((0.98)2 (1.013)2 N)
(0.02)((0.98)3 (1.013)3 N)
..
.
Balance after
payment
(0.98)N
(0.98)2 (1.013)N
(0.98)3 (1.013)2 N
(0.98)4 (1.013)3 N
..
.
n
(0.98)n (1.013)n N
(0.02)((0.98)n (1.013)n N)
(0.98)n+1 (1.013)n N
How much have you paid after n months: Put r = (0.98)(1.013) ∼ 0.993.
(0.02)N(1+r +r 2 +r 3 +· · ·+r n ) = N·
(0.02)(1 − r n+1 )
1−r
→ (2.857)N = $1142.80.
AVOID CREDIT CARD DEBT!
MATH 1131Q - Calculus 1.
Álvaro Lozano-Robledo
Department of Mathematics
University of Connecticut
Day 14
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
2 / 40
Exponential Growth
and Decay
Exponential Growth and Decay
Definition
We say that a function f (x) grows (or decays) exponentially if the
growth rate at any x is proportional to the function’s value.
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
4 / 40
Exponential Growth and Decay
Definition
We say that a function f (x) grows (or decays) exponentially if the
growth rate at any x is proportional to the function’s value.
In other words, f 0 (x) is proportional to f (x), or equivalently, there is a
constant k such that
f 0 (x) = k · f (x).
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
4 / 40
Example: Compounded Interest
Example: If $2000 is invested at 4% interest, compounded annually,
how much money will there be in the account after 10 years?
Solution:
$2000 · (1.04) = $2080.00
$2000 · (1.04)2 = $2163.20
..
.
$2000 · (1.04)10 = $2960.48
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
5 / 40
Example: Compounded Interest
Example: If $2000 is invested at 4% interest, compounded
semi-annually, how much money will there be in the account after 10
years?
Solution:
$2000 · 1 +
‚
$2000 ·
1+
0.04
2
2
Œ2
0.04 2
2
= $2080.80
= $2164.86
..
.
‚
$2000 ·
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
1+
0.04
2
2 Œ10
= $2971.89
6 / 40
Example: Compounded Interest
Example: If $2000 is invested at 4% interest, compounded monthly,
how much money will there be in the account after 10 years?
Solution:
$2000 · 1 +
‚
$2000 ·
1+
0.04
12
12
Œ2
0.04 12
12
= $2081.48
= $2166.28
..
.
‚
$2000 ·
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
1+
0.04
12
12 Œ10
= $2981.66
7 / 40
Example: Compounded Interest
Example: If $2000 is invested at 4% interest, compounded daily, how
much money will there be in the account after 10 years?
Solution:
$2000 · 1 +
‚
$2000 ·
1+
0.04
365
365
Œ2
0.04 365
365
= $2081.61
= $2166.56
..
.
‚
$2000 ·
Álvaro Lozano-Robledo (UConn)
1+
MATH 1131Q - Calculus 1
0.04
365
365 Œ10
= $2983.58
8 / 40
Example: Continuously Compounded Interest
Example: If $2000 is invested at 4% interest, compounded
continuously, how much money will there be in the account after 10
years?
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
9 / 40
The number e
e = lim (1 + h)1/ h .
h→0
Or changing h by 1/ n:
‚
e = lim
n→∞
Álvaro Lozano-Robledo (UConn)
1+
MATH 1131Q - Calculus 1
1
Œn
.
n
10 / 40
Example: Continuously Compounded Interest
Example: If $2000 is invested at 4% interest, compounded
continuously, how much money will there be in the account after 10
years?
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
11 / 40
Example: Continuously Compounded Interest
Example: If $2000 is invested at 4% interest, compounded
continuously, how much money will there be in the account after 10
years?
Solution:
$2000 · e(0.04) = $2081.62
$2000 · e(0.04)·2 = $2166.57
..
.
$2000 · e(0.04)·10 = $2983.64
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
11 / 40
Example: Radioactive Decay
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
12 / 40
Example: Radioactive Decay
The Chernobyl disaster was a catastrophic nuclear accident that
occurred on 26 April 1986 at the Chernobyl Nuclear Power Plant in
Ukraine (then officially the Ukrainian SSR), which was under the direct
jurisdiction of the central authorities of the Soviet Union. An explosion
and fire released large quantities of radioactive particles into the
atmosphere, which spread over much of the western USSR and
Europe.
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
12 / 40
Example: Radioactive Decay
The Chernobyl disaster was a catastrophic nuclear accident that
occurred on 26 April 1986 at the Chernobyl Nuclear Power Plant in
Ukraine (then officially the Ukrainian SSR), which was under the direct
jurisdiction of the central authorities of the Soviet Union. An explosion
and fire released large quantities of radioactive particles into the
atmosphere, which spread over much of the western USSR and
Europe.
The Chernobyl disaster was the worst nuclear power plant accident in
history in terms of cost and casualties, and is one of only two classified
as a level 7 event (the maximum classification) on the International
Nuclear Event Scale (the other being the Fukushima Daiichi nuclear
disaster in 2011). The battle to contain the contamination and avert a
greater catastrophe ultimately involved over 500,000 workers and cost
an estimated 18 billion rubles (18 billion $USD). During the accident
itself, 31 people died, and long-term effects such as cancers and
deformities are still being accounted for.
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
12 / 40
View of Chernobyl taken from the abandoned city of Pripyat.
Radiation levels (millisieverts) and their effects
Event
Single dose, fatal within weeks
Typical dosage recorded in those
Chernobyl workers who died within a month
Single does which would kill half
of those exposed to it within a month
Single dosage which would cause radiation sickness
including nausea, lower white blood cell count.
Not fatal.
Accumulated dosage estimated to cause a fatal
cancer many years later in 5% of people
Max radiation levels recorded at Fukushima
plant 15 March 2011, per hour
Exposure of Chernobyl residents who were
relocated after the blast in 1986
Recommended limit for radiation workers
every five years
Radiation
10,000 mSv
6,000 mSv
5,000 mSv
1,000 mSv
1,000 mSv
400 mSv
350 mSv
100 mSv
Radiation levels (millisieverts) and their effects
Event
Lowest annual dose at which any increase
in cancer is clearly evident
CT scan: heart
CT scan: abdomen and pelvis
Dose in full-body CT scan
Airline crew flying New York to Tokyo
polar route, annual exposure
Natural radiation we are all exposed to, per year
CT scan: head
Spine x-ray
Radiation per hour detected at Fukushima site,
12 March 2011, one day after disaster
Mammogram breast x-ray
Chest x-ray
Radiation
100 mSv
16 mSv
15 mSv
10 mSv
9 mSv
2 mSv
2 mSv
1.50 mSv
1.02 mSv
0.40 mSv
0.10 mSv
Radiation levels (millisieverts) and their effects
Event
Lowest annual dose at which any increase
in cancer is clearly evident
CT scan: heart
CT scan: abdomen and pelvis
Dose in full-body CT scan
Airline crew flying New York to Tokyo
polar route, annual exposure
Natural radiation we are all exposed to, per year
CT scan: head
Spine x-ray
Radiation per hour detected at Fukushima site,
12 March 2011, one day after disaster
Mammogram breast x-ray
Chest x-ray
Radiation from eating a banana, for scale
Radiation
100 mSv
16 mSv
15 mSv
10 mSv
9 mSv
2 mSv
2 mSv
1.50 mSv
1.02 mSv
0.40 mSv
0.10 mSv
0.000098 mSv
Current radiation levels at Chernobyl and Pripyat
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
20 / 40
Current radiation levels at Chernobyl and Pripyat
Radiation levels of up to 0.336mSv per hour at the power plant.
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
21 / 40
Current radiation levels at Chernobyl and Pripyat
Radiation levels from 0.0002 up to 0.022mSv per hour at Pripyat.
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
22 / 40
Current radiation levels at Chernobyl and Pripyat
0.022mSv per hour means
0.022 · 24 · 365 = 192.72mSv
per year.
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
23 / 40
Current radiation levels at Chernobyl and Pripyat
0.022mSv per hour means
0.022 · 24 · 365 = 192.72mSv
per year. Recall that 5% of a population receiving a dose of 1000mSv
will suffer a fatal form of cancer.
Fukushima evacuees will receive radiation doses from 0.7 to 3
millisieverts per year after they return to their homes, following the lift
of an evacuation order in surrounding areas in April 2014.
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
23 / 40
Example: Radioactive Decay
Caesium-137 is the principal source of radiation in the exclusion zone
around the Chernobyl nuclear power plant. The half-life of Cs-137 is
approximately 30 years.
Example
One gram of Caesium-137 outputs 43345 Sv of radiation per year, or
4.94 Sv per hour. If we want the radiation level to go down to a safe
level of 20 mSv per year, how long should we wait?
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
24 / 40
This slide left intentionally blank
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
25 / 40
Example: Radioactive Decay
Example
One gram of Caesium-137 outputs 43345 Sv of radiation per year, or
4.94 Sv per hour. If we want the radiation level to go down to a safe
level of 20 mSv per year, how long should we wait?
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
26 / 40
Example: Radioactive Decay
Example
One gram of Caesium-137 outputs 43345 Sv of radiation per year, or
4.94 Sv per hour. If we want the radiation level to go down to a safe
level of 20 mSv per year, how long should we wait?
If 1 g of Cs-137 outputs 43345000 mSv per year, then
x=
1 · 20
43345000
∼
= 0.000000461
grams of Cs-137 output 20 mSv per year.
The amount of Cs-137 left is given by
C(t) = e(−Ln(2)/ 30)t = e(−0.0231)t .
C(t) = 0.000000461 at t =
Álvaro Lozano-Robledo (UConn)
ln(0.000000461)
−0.0231
MATH 1131Q - Calculus 1
∼
= 631.59 years.
26 / 40
Example: Newton’s Law of Cooling
Newton’s Law of Cooling (or Warming) states that the rate of cooling
of an object is proportional to the difference between the object and
the room temperature (provided that difference is not too large).
Let T (t) be the temperature of the object, and let R be the room
temperature. Newton’s law says:
dT
dt
= k · (T (t) − R),
for some constant k .
If we put y = T (t) − R, then dy / dt = dT / dt − 0 = dT / dt. Hence:
dy
dt
= k · y,
and y follows the differential equation of exponential growth/decay.
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
27 / 40
Example: Newton’s Law of Cooling
Thus, if T (t) satisfies
dT
dt
= k · (T (t) − R),
then y (t) = T (t) − R satisfies
dy
dt
Álvaro Lozano-Robledo (UConn)
= k · y,
MATH 1131Q - Calculus 1
28 / 40
Example: Newton’s Law of Cooling
Thus, if T (t) satisfies
dT
= k · (T (t) − R),
dt
then y (t) = T (t) − R satisfies
dy
dt
= k · y,
and therefore
y (t) = C · ekt
for some constants C and k .
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
28 / 40
Example: Newton’s Law of Cooling
Thus, if T (t) satisfies
dT
dt
= k · (T (t) − R),
then y (t) = T (t) − R satisfies
dy
dt
= k · y,
and therefore
y (t) = C · ekt
for some constants C and k . In particular,
T (t) = R + y (t) = R + C · ekt .
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
28 / 40
Example: Newton’s Law of Cooling
Thus, if T (t) satisfies
dT
dt
= k · (T (t) − R),
then
T (t) = R + C · ekt .
What is C?
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
29 / 40
Example: Newton’s Law of Cooling
Thus, if T (t) satisfies
dT
dt
= k · (T (t) − R),
then
T (t) = R + C · ekt .
What is C?
T (0) = R + C · e0 = R + C
so C = T (0) − R, i.e., the initial difference in temperature between the
object and the room temperature.
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
29 / 40
Example: Newton’s Law of Cooling
Example
According to the internet, coffee is best served at a temperature of
160F. If the coffee shop where you are having your coffee is at a room
temperature of 65F, how long will it take for your coffee to be lukewarm
at 80F?
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
30 / 40
Example: Newton’s Law of Cooling
Example
According to the internet, coffee is best served at a temperature of 160F. You
are having a cup at a coffee shop that the keep at a frigid room temperature
of 65F. After 10 minutes, the coffee is already at 120F. How long will it take for
your coffee to be lukewarm at 80F?
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
31 / 40
Example: Newton’s Law of Cooling
Example
According to the internet, coffee is best served at a temperature of 160F. You
are having a cup at a coffee shop that the keep at a frigid room temperature
of 65F. After 10 minutes, the coffee is already at 120F. How long will it take for
your coffee to be lukewarm at 80F?
Solution: The temperature of the coffee T (t) follows Newton’s law of cooling,
so
T (t) = R + C · ekt
where R = 65 and C = T (0) − R = 160 − 65 = 95. Moreover, T (10) = 120.
Thus,
120 = 65 + 95e10k
and so k = Ln(55/ 95)/ 10 ∼
= −0.05465. We want T (t) = 80, so
80 = 65 + 95e(−0.05465)t
and so t = Ln(15/ 95)/ (−0.05465) ∼
= 33.775 minutes.
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
31 / 40
Example: Population Growth
The population of Lionfish (Pterois Volitans) in the Caribbean is out of control.
If the population in the
Caribbean in 2014 is
estimated at 600 million, and
in 2000 the population was
estimated at 70000
individuals, when was the first
pair of lionfish introduced,
approximately?
If the population in the
Caribbean in 2014 is
estimated at 600 million, and
in 2000 the population was
estimated at 70000
individuals, when was the first
pair of lionfish introduced,
approximately?
If the population in the
Caribbean in 2014 is
estimated at 600 million, and
in 2000 the population was
estimated at 70000
individuals, when was the first
pair of lionfish introduced,
approximately?
Let us set L(t) to be the population of lionfish in the Caribbean, with
t = 0 at the year 2000. Then L(t) = 70000 · ekt and
L(14) = 600000000, so
k = Ln(600000000/ 70000)/ 14 ∼
= 0.6469
Thus, L(t) = 70000 · e(0.6469)t and we are looking for L(t) = 2. Thus,
t = Ln(2/ 70000)/ (0.6469) ∼
= −16.17
Hence, they were introduced in ∼
= 2000 − 16 = 1984.
Example: Population Growth
The population of Lionfish (Pterois Volitans) in the Caribbean is out of control.