Compound Interest

Deﬁnition: Interest is money paid for the use of money. The total amount borrowed is called principle. The rate of interest, expressed as a percent, is the amount charged for the use of the principle for a given amount of time.

Theorem: If a principle P dollars is borrowed for a period of t years on a yearly interest rate r, the interest I charged is: I = P rt The interest charged according to this formula is called simple interest.

Deﬁnition: When the interest due at the end of a payment period is added to the principle so that the interest computed at the end of the next payment period is based on this new amount (old principle + interest), the interest is said to have been compounded.

Example 1: A great bank pays 10% interest per year compounded semi-annually on a certain savings account. If $5000 is deposited in such an account and the interest is left to accumulate, how much is in the account after 1 year?

1st Period: After the ﬁrst period, the interest =

1 1 I = P rt = (5000)(.1)( 2 ) = 500( 2 ) = $250. The amount in the account is the principle + interest = $5000 + 250 = $5250 2nd Period: After the second period (1 year from start), the interest =

1 1 525 I = P rt = (5250)(.1)( 2 ) = 525( 2 ) = $ 2 525 So after 1 year there is $5250 + 2 in the account.

From the simple interest formula we can develop the following theorem.

Theorem: The amount A after t years due to principle P invested at an annual interest rate r compounded n times per year is given by the formula:

r nt A = P · (1 + n )

Example 2: We can rework example 1. Given P = $5000, r = .1, n = 2 (semi-annual compounding), and t = 1:

r nt .1 2 2.1 2 A = P · (1 + n ) = 5000(1 + 2 ) = $5000( 2 ) Since you are not allowed calculator on an exam, this is an acceptable answer.

Deﬁnition: When interest is compounded so that the amount after 1 year is P er, we say the interest is compounded continuously.

Theorem: The amount A after t years due to a principle P invested at an annual interest rate r compounded continuously is

A = P ert

1 Example 3: The amount A that results from investing P = 2000 at an annual rate of r = 5% compounded continuously after 3 years is: A = 2000e(.05·3)

We can also determine the rate of interest or time require to double a given amount of money: Example 5: What annual rate of interest, compounded annually, would double your investment in 3 years?

If P is the principle and we want P to double ⇒ A = 2P . Since we’re only compounded the interest 1 time per year (n = 1), and t = 3 r nt ⇒ A = P (1 + n ) r 3 ⇒ 2P = P (1 + 1 ) ⇒ 2 = (1 + r)3 1 ⇒ 1 + r = (2) 3 1 ⇒ r = (2) 3 − 1

Example 6: How long will it take for an investment to triple in value if it earns 10% compounded continuously each year? Suppose P is the initial investment. Then A = 3P and we apply the formula for continuously compounded interest with r = .1: ⇒ A = P ert ⇒ 3P = P e.1t ⇒ 3 = e.1t ⇒ ln(3) = ln(e.1t) ⇒ ln(3) = .1t ⇒ t = 10 · ln(3) years

2 Exponential Growth and Decay

Exponential functions measure the size of a growing population, the amount of money in a compound interest account, the number of atoms left after radioactive decay, ect... Notice that for every real number a > 0, we can write a as a power of e: a = eln(a).

Lemma: Every exponential function can be written in the form:

kt N(t) = N0e

N0 = N(0) is the initial amount. k is the growth constant If k > 0, N(t) measures exponential growth. If k < 0, N(t) measures exponential decay.

Example 1: A colony of bacteria grows according to the equation N(t) = 100e.1t, where N is measured in grams and t is measured in days. (a) Determine the initial amount of bacteria. (b) What is the growth rate of the bacteria? (c) What is the population after 10 days? (d) How long does it take for the population to triple? 0 (a) N0 = N(0) = 100e = 100 grams.

(b) k = .1 = 10% growth rate.

(d) 3 · (100) = 100e.1t ⇒ 3 = e.1t ⇒ ln(3) = .1t ⇒ t = 10 · ln(3) days.

3 Example 2: A bacteria colony starts with 103 bugs. Four hours later it has 5 × 103 bugs. (a) Find the growth constant k. (b) What is the population two hours after the start? (c) How long will it take for the population to triple? First we want to ﬁnd the equation for the number of bugs after t hours, N(t):

3 3 kt Given: N0 = 10 , N(4) = 5 , and N(t) = N0e , ⇒ N(4) = 5 · 103 = 103ek4 ⇒ 5 = e4k ⇒ ln(5) = ln(e4k) ⇒ ln(5) = 4k ln(5) ⇒ k = 4

3 ln(5) ·t So the number of bugs after t hours is N(t) = 10 e 4 . ln(5) (a) The growth constant is k = 4 (b) After two hours the population is:

3 ln(5) ·2 3 ln(5) N(2) = 10 e 4 = 10 e 2 bugs.

3 N(t) = 3 × N0 = 3 × 10 3 3 ln(5) t ⇒ N(t) = 3 × 10 = 10 e 4 ln(5) t ⇒ 3 = e 4 ln(5) ⇒ ln(3) = 4 t 4 ln(3) ⇒ t = ln(5) hours.

Example 3: Initially a sample has 8 pounds of a radioactive substance with a half-life of 5 days (i.e. half of the substance decays after 5 days). How many pounds remain after 3 days? First we want to ﬁnd the equation for the number of lbs left after t days, N(t):

kt Given: N0 = 8, N(5) = 4, and N(t) = N0e , ⇒ N(5) = 4 = 8ek5 1 5k ⇒ 2 = e 1 5k ⇒ ln( 2 ) = ln(e ) 1 ⇒ ln( 2 ) = 5k 1 ln( 2 ) ⇒ k = 5

(ln( 1 ) 2 t So N(t) = 8e 5 .

ln( 1 ) 2 ·3 3 ·ln( 1 ) ⇒ N(3) = 8e 5 = 8e 5 2

3 ·( 1 ) Thus 8e 5 2 pounds remain after 3 days.