
ALGEBRA
 solved problems







Periodic compounding 

118. 
The principal amount of
$2,000 is invested for five years in a compound interest account
paying 6%


compounded quarterly, find the final or accumulated
amount in the account. 
Note
that for any given interest rate the investment grows more if
the compounding period is shorter. 
So,
if P is
an amount of money invested for n years at an interest rate
i, compounded
m times per
year then, the total number of
compounded periods is mn and the interest rate per period is
i/m 
and the accumulated or future value is



Solution: 



Continuous compounding 

119. 
Suppose $5,000 is deposited into an account that pays interest
compounded continuously at an


annual rate of 8%. How
much will the account be worth in 20 years? 
As for any given interest rate the investment grows more if the compounding period is shorter, we let the number of periods in a year
approach infinity to compound the interest continuously, meaning that the
balance grows by a small amount every instant. 
Thus,
to derive a formula for continuous compounding we need to
evaluate the above formula when m
tends to infinity (i.e., the year is divided into infinite
number of periods) 

then
by substituting 

and,
when m ®
oo
then x ®
oo, 



that
is, the limit in the square brackets converges to the number e = 2.71828 . . . ,
thus obtained is 
the
continuous compounding formula A
= P · e ^{i n},
where e
is the base of natural logarithms. 
Solution: 



Exponential
growth and decay, application of the natural exponential
function 

120. 
Suppose that
microorganisms in a culture dish grow exponentially. At the
start of an experiment


there are 8,000 of bacteria, and two
hours later the population has increased to 8,600. 
How long will
it take for the population to reach 20,000? 
Solution: Given the initial
population N_{0}
= 8,000 and for t
=
2 hours
the population increased to N
= 8,600. 
We
first find the growth rate k
and then the time needed the population increases to 20,000. 


121. 
After 800
years a sample of radioisotope radium226 has decayed to 70.71%
of its initial mass, find


the halflife of radium226. 
Radioactive decay is a
typical example to which the exponential decay model can be
applied. 
Solution:
The halflife of a radioactive material is the amount of time
required for half of a given sample to decay. First
we plug the given information into the formula for exponential
decay 
N (t)
= N_{0 }e ^{kt }to
find the decay constant k. 
Since
N_{0}
is the initial quantity, N(t)
is the quantity after time t
and k
is the decay constant then, 
by
substituting
N_{(t = 800)} = 0.7071·N_{0 }into
the formula 
0.7071N_{0
}= N_{0 }e ^{k · }^{800} 
and
solving for k,
0.7071_{ }= _{
}e ^{k · }^{800}
 ln 
ln 0.7071_{ }= 800 · k 
k_{ }= ln (0.7071) / 800 
k_{ }= 
0.0004332217. 
Thus,
the formula for the amount of radium226 present at a time t
is 
N
(t)
= N_{0 }e ^{
0.0004332217 · t} 
As
we want determine the halflife or the time for half of a substance to decay,
we substitute N (t)
= (1/2) N_{0 }into
the formula 











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