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Exponential Growth and Decay - Lecture Notes | MATH 1920, Study notes of Calculus

Material Type: Notes; Class: Calculus II; Subject: Mathematics; University: Pellissippi State Technical Community College; Term: Unknown 1989;

Typology: Study notes

Pre 2010

Uploaded on 08/16/2009

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DIFFERENTIAL EQUATIONS
7.5 Exponential Growth and Decay
Objective: Use differential equations to solve growth and decay problems
I. A population grows at a rate proportional to the size of the population
II. A substance decays at a rate proportional to the size of the mass
III. The value of a savings account at continuously compounded interest increases at a rate
proportional to that value
IV. Law of natural growth: k is a positive constant
dy ky
dt =
V. Law of natural decay: k is a negative constant
dy ky
dt =
VI. Solve the initial-value problem y(0) = y 0
dy ky
dt =
A. dy = (ky)dt dy kdt
y
⇒=
B. C. dy kdt
y=
∫∫
C. ln |y| = kt + C |y| =
ktCCkt
eee
+=
D. where A is an arbitrary constant
kt
yAe=c
(eor0)
E. kt
0
y(t)ye=
VII. The relative growth rate k is constant
A. If then
dP kP,
dt =1dP
kPdt

=


B. If and t is in years, then k = .02
dP .02P,
dt =
1. Population grows at a rate of 2% per year
2. where P0 is the initial population
.02t
0
P(t)Pe=
VIII. Example 2 on p. 530
A. The initial-value problem is P(0) = 2520
dP kP,
dt =
B. The solution is c
kt
P(t)2520e=
pf3

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DIFFERENTIAL EQUATIONS

7.5 Exponential Growth and Decay

Objective: Use differential equations to solve growth and decay problems

I. A population grows at a rate proportional to the size of the population

II. A substance decays at a rate proportional to the size of the mass

III. The value of a savings account at continuously compounded interest increases at a rate

proportional to that value

IV. Law of natural growth: k is a positive constant

dy ky dt

V. Law of natural decay: k is a negative constant

dy ky dt

VI. Solve the initial-value problem y(0) = y (^) 0

dy ky dt

A. dy = (ky)dt

dy kdt y

B. C.

dy kdt y

C. ln |y| = kt + C ⇒ |y| =

kt C C kt e e e

=

D. where A is an arbitrary constant

kt y = Ae

c ( = ±e or 0)

E.

kt y(t) =y e 0

VII. The relative growth rate k is constant

A. If then

dP kP, dt

1 dP k P dt

B. If and t is in years, then k =.

dP .02P, dt

  1. Population grows at a rate of 2% per year
  2. where P 0 is the initial population

.02t P(t) = P e 0

VIII. Example 2 on p. 530

A. The initial-value problem is P(0) = 2520

dP kP, dt

B. The solution is c

kt P(t) = 2520e

C. Estimate k using 1960 data

10t P(10) = 2520e = 3020

k ln. 10 2520

  1. The relative growth rate is about 1. 8% per year

D. Model for world population growth in second half of 20

th century =

.018t P(t) = 2520e

IX. Radioactive decay

A. If m(t) is the mass remaining from an initial mass m (^) 0 of the substance after time t

  1. is constant

1 dm

m dt

  1. k is a negative constant

dm km dt

B.

kt m(t) =m e 0

X. Example 3

A. Half-life of radium-226 is 1590 yr; initial mass is 100 mg

kt

m(t) =100e

1590t 50 = 100e ⇒

ln k 1590

B.

ln2 (^) t 1590 m(t) 100e

XI. Continuously compounded interest: $1000 at 10% for 20 yr

A. Simple interest: I = Prt and A(t) = A 0 (1 + rt)

A = 1000[1 +. 1(20)] = $

B. Compound interest:

nt

0

r A(t) A 1 n

  1. Annually ⇒ n = 1 = $6727.

1(20) . A(t) 1000 1 1

  1. Semiannually ⇒ n = 2 = $7039. 99

2(20) . A(t) 1000 1 2