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ALGEBRA
  Interest calculation
 ::  Simple interest

 An amount of money deposited into a bank for a given period of time brings to the depositor a profit called interest.

Amount of interest, principal (initial amount), interest rate and amount after n years

The amount of interest ( I ) the bank pays you, depends on the interest rate ( i % ), the amount of money deposited, denoted as principal P also called original balance (or initial investment), and the period of time n the money is deposited,

since   P : I = 1 : (i n)  then,
Therefore,

Simple interest is calculated on a yearly basis (annually, n = 1) in which balance grows linearly with time (as opposed to compound interest).

If P is the principal (or initial value of investment) and I is the interest amount, the accumulated value A (or final value of investment) at the end of n investment periods is given by
A = P + I = P + i P n = P (1 + i n)
and

Example:  Somebody deposits $20,000 into a savings account where the rate of interest is 4.8% annually.  How much money in interest will earn after nine months?

Solution:

Example:  A bank lends a company money for the six months period at a rate of 8% annually. How much was lent if the company should pay $12,000 of interest?

Solution:

Example:  At what an interest rate was borrowed $75,000 for one year if $3,000 to interest is charged?

Solution:

Example:  For what period of time should be deposited $200,000 at a 6% interest rate to earn $6,000 of interest?

Solution:
 ::  Compound interest
In compound interest calculations, the interest earned in each period is added at the end of a period to the principal of the previous period, to become the principal for the next period.
The compounding periods can be yearly, semiannually, quarterly, or the interest can be compounded more frequently even continuously.
Thus, if P is the principal or initial value of investment and the compound interest rate is i %, then
                              I1 = i P  is the amount of interest earned in the first period.
So, at the end of the first period, the accumulated amount is
                                           A1 = P + I1 = P  + i P = P(1 + i).
Thus, at the end of n periods (or after n years) the accumulated amount or final value of investment is
                                           An  = A = P(1 + i)(1 + i) (1 + i) = P(1 + i)n
or, An = A = P r n,   where  r = 1 + i
therefore,
Example:  After how many years will deposit double at an interest rate of 6%.
Solution:
Example:  At what annual interest has to be deposited $5,000 for four years to grow to $8,000.
Solution:
Periodic compounding
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

Example:  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.

Solution:
Continuous compounding
The formula for continuous compounding, A = Pein .

Example:  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?

Solution:
 :: Exponential growth and decay, application of the natural exponential function
We use the same formula as for continuous compound interest in many natural processes where the rate of change of a quantity through time is proportional to the current amount of the quantity.
A quantity is said to be subject to exponential growth if it increases at a rate proportional to its current value.
The constant k is called the growth rate and in exponential growth k > 0. The rate constant k depends only on the process and the conditions under which it is carried out.
A quantity is said to be subject to exponential decay if it decreases at a rate proportional to its current value.
The constant k is called the decay rate and in exponential decay k < 0.

Example:  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 N0 = 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.

Radioactive decay is a typical example to which the exponential decay model can be applied.
Example:  After 800 years a sample of radioisotope radium-226 has decayed to 70.71% of its initial mass, find the half-life of radium-226.
Solution:  The half-life 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) = N0 ekt    to find the decay constant k.
Since N0 is the initial quantity, N(t) is the quantity after time t and k is the decay constant then, by substituting   N(t = 800) = 0.7071Ninto the formula
0.7071N0 = N0 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 radium-226 present at a time t is
N(t) = N0 e - 0.0004332217 t .
As we want determine the half-life or the time for half of a substance to decay, we substitute                   
 N(t) = (1/2) N0   into the formula
 
 
 
 
 
 
 
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