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Application of e and Exponential Functions

Application of e and Exponential Functions

In the calculation of interest, the following formula is often used:

where
A is the money accumulated.
P is the principal (beginning) amount
r is the annual interest rate
n is the number of compounding periods per year
t is the number of years
EX. An initial investment of $1,000 is invested at a 12.5% annual interest rate, compounded annually, for 10 years. Using the above formula, we calculate the amount accumulated after 10 years as follows:

P = $1000
r = 12.5% = .125
n = 1
t = 10 years

A = $1000(1 + .125)
A = $1000(1.125)
A = $1000(3.247)
A = $3,247

EX. An initial investment of $2,000 yields $10,000 after 15 years. The interest is compounded quarterly. Using the interest formula, we can calculate the interest rate:

In calculations involving depreciation, the interest formula can also be used, provided that the depreciation rate is taken to be a negative interest rate.

EX. The principal cost of a car is $20,000. It depreciates at a rate of 6% per year. After 5 years, its book value is given by

P = $20,000
r = -6% or -.06
n = 1
t = 5

A = $20,000 (.94)
A = $14,680

In the interest formula

the value of A, given fixed values for P, r (where r> 0) and t, will increase as n increases, i.e., the more often the interest is compounded, the greater the amount of money accumulated. As n approaches infinity, the value of A converges to a finite value. It can be shown that

Therefore

This is known as continuous compounding of interest. The formula used herein is:

A = Pert

whereA is the money accumulated.

P is the principal amount
r is the interest rate compounded continuously
t is the number of years

EX. An investment of $200 is invested at an interest rate of 5% compounded continuously. In 5 years, it will yield an amount computed as

A = Pert

P = $200
r = 5% or .05
t = 5

A = $200e(.05)(5)
A = $200e.25
A = $200(1.285)
A = $257

The number e is also used in applications involving exponential growth and decay. Entities that are subject to exponential growth increase at a rate that is proportional to the amount of substance in the entity; entities subject to exponential decay decrease at a rate proportional to the amount of substance in the entity. The formula often used herein is

Q(t) = Q0ekt

where Q(t) is the amount of the substance at time t, Q0 is the amount of the substance at t = 0, and k is a constant that depends on the characteristics of the entity. If k > 0, then Q(t) increases as t increases; this represents exponential growth. If k < 0, then Q(t) decreases as t increases; this is known as exponential decay.

EX. A certain bacteria culture has a growth rate of k = .03 and an initial count of 10000.

After 3 hours, the population of the bacteria culture is calculated as

Q(t) = Q0ekt
Q0 = 10000
k = .03
t = 3

Q(t) = 10000 e0.03(3)
Q(t) = 10,940

EX. A certain radioactive particle decays at a rate of k = .06. After 5 hours, a 100-gram sample will decay to an amount calculated as .

Q(t) = Q0ekt
Qo = 100 g
k = -0.06
t = 5

Q(t) = 100 g e-0.06(5)
Q(t) = 74.08 g

Subject: 
Algebra [1]
Subject X2: 
Algebra [1]

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