Page 170 - How To Solve Word Problems In Calculus

ln 2 ln 2
To find k, we recall that k = = .
T 5730

ln 2 t
r(t) = r 0 e − 5730
r(t) − ln 2
t
= e 5730
r 0
r(t)
Since = 10% = 0.1, we have
r 0

ln 2 t
0.1 = e − 5730

ln 2
ln 0.1 =− t
5730
−5730(ln 0.1)
t = ≈ 19,034.6
ln 2
The fossil is approximately 19,000 years old.

Continuous Compounding of Interest
If P dollars are invested in a bank account at an annual rate of
interest r compounded n times per year, the amount of money
A in the account after t years is

nt
r
A = P 1 +
n
Quarterly compounding, monthly compounding, and daily
compounding use values of n = 4, 12, and 365, respectively.
(Some banks use a 360-day business year, but there is virtually
no difference in the interest between 360 and 365 days.)
If we let n →∞, we get what is called continuous com-
pounding of interest, and the amount of money in the account
after t years becomes

nt
r
A = lim P 1 +
n→∞ n
t
n
r
= P lim 1 +
n→∞ n
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