The Natural Base, e Applications

The Formula for continuously compounded interest is:

rtA Pe

A

P

r

t

Total amount

Principal (amount invested)

Interest rate

time

Example 1What is the total amount for an investment of $500 invested at 5.25% for 40 years and compounded continuously?

rtA Pe .0525 40500A e

A = $4083.08

Example 2What is the total amount for an investment of $100 invested at 3.5% for 8 years and compounded continuously ?

rtA Pe .035 8100A e

A = $132.31

The half-life of a substance is the time it takes for half of the substance to breakdown or convert to another substance during the process of decay. Natural decay is modeled by the function below.

0( ) ktN t N e

N(t)

k

t

The amount remaining

Initial amount (at t=(0)

Decay constant

time

0N

Example 3Pluonium-239 has a half-life of 24,110 years. How long does it take for a 1 g sample of Pu-239 to decay to .1 g ?

0( ) ktN t N eStep 1:Find decay constant ‘k’

2411011

2ke

1 24110ln 2 ln ke ln 2 24110k ln 2

24110k

.0000287k

Example 3--continuedPluonium-239 has a half-life of 24,110 years. How long does it take for a 1 g sample of Pu-239 to decay to .1 g ?

0( ) ktN t N eStep 2: Solve for ‘t’

.0000287.10 1 te.0000287ln .1 ln te

ln .1 .0000287tln .1

.0000287t

80,229t

Pluonium takes approximately 80,229 years to decay.

Example 4The half-life of carbon-14 is 5730 years. What is the age of a fossil that only has 8% of its original carbon-14?

Step 1:Find decay constant ‘k’ln 2

5730k

.00012k

Step 2: Solve for ‘t’

0( ) ktN t N e.000128 100 te

.00012.08 te.00012ln .08 ln te

ln .08 .00012tln .08

.00012t

21,048t The age of the fossil is approximately 21,048 years.