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Material Type: Notes; Class: Calculus II; Subject: Mathematics; University: Pellissippi State Technical Community College; Term: Unknown 1989;
Typology: Study notes
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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
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)
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
.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
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 dm
m dt
dm km dt
kt m(t) =m e 0
X. Example 3
A. Half-life of radium-226 is 1590 yr; initial mass is 100 mg
kt
1590t 50 = 100e ⇒
ln k 1590
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)
B. Compound interest:
nt
0
r A(t) A 1 n
1(20) . A(t) 1000 1 1
2(20) . A(t) 1000 1 2