Page 192 - Calculus Demystified

CHAPTER 6
Transcendental Functions
dollars at the end of the year. Similarly, at the end of t years, the money accumulated 179
will be
p
nt
P · 1 + .
100n
Let us set
n · 100
k =
p
and rewrite (∗)as
! " p/100
kp/100 k
1 1
P · 1 + = P · 1 + .
k k
It is useful to know the behavior of the account if the number of times the inter-
est is compounded per year becomes arbitrarily large (this is called continuous
compounding of interest). Continuous compounding corresponds to calculating the
limit of the last formula as k (or, equivalently, n), tends to infinity.
We know from the discussion in Subsection 6.2.3 that the expression (1+1/k) k
tends to e. Therefore the size of the account after one year of continuous
compounding of interest is
P · e p/100 .

After t years of continuous compounding of interest the total money is
P · e pt/100 . (..)
EXAMPLE 6.33
If $6000 isplaced in a savingsaccount with 5% annual interest compounded
continuously, then how large is the account after four and one half years?
SOLUTION
If M(t) is the amount of money in the account at time t, then the preceding
discussion guarantees that
M(t) = 6000 · e 5t/100 .
After four and one half years the size of the account is therefore

M(9/2) = 6000 · e 5·(9/2)/100 ≈ $7513.94.

EXAMPLE 6.34
A wealthy woman wishes to set up an endowment for her nephew. She
wantsthe endowment to pay the young man $100,000 in cash on the day of
histwenty-first birthday.The endowment isset up on the day of the nephew’s