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qr code generator visual basic 2010 ln( 1/5) = ln in .NET framework
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Drawing GS1 DataBar14 In Java Using Barcode generator for Java Control to generate, create GS1 DataBar image in Java applications. UCC128 Printer In Java Using Barcode creator for Java Control to generate, create USS128 image in Java applications. amount of matter at time t Then our physical law is expressed as dM = C M 2 dt Here C is a (negative) constant of proportionality We apply the method of ``separation of variables'' described earlier in the section Thus dM/dt =C M2 so that dM/dt dt = M2 Evaluating the integrals, we find that 1 = C t + D M C dt Make Barcode In C#.NET Using Barcode creator for Visual Studio .NET Control to generate, create bar code image in Visual Studio .NET applications. European Article Number 13 Generator In None Using Barcode encoder for Software Control to generate, create GTIN  13 image in Software applications. We have combined the constants from the two integrations In summary, M( t) = 1 Ct + D
Scanning Code39 In .NET Using Barcode scanner for .NET framework Control to read, scan read, scan image in Visual Studio .NET applications. Make Barcode In None Using Barcode drawer for Word Control to generate, create bar code image in Office Word applications. For the problem to be realistic, we will require that C < 0 (so that M > 0 for large values of t) and we see that the population decays like the reciprocal of a linear function when t becomes large GS1 128 Maker In Java Using Barcode generator for BIRT reports Control to generate, create USS128 image in Eclipse BIRT applications. Encode EAN13 In Java Using Barcode creation for Java Control to generate, create GS1  13 image in Java applications. YOU TRY IT Recalculate Example 632 using this new law of exponential
decay
654 Compound Interest
Yet a third illustration of exponential growth is in the compounding of interest If principal P dollars is put in the bank at p percent simple interest per year then after one year the account has P 1+ p 100 dollars [Here we assume, of course, that all interest is reinvested in the account] But if the interest is compounded n times during the year then the year 6 T R A N S C E N D E N T A L F U N C T I O N S
is divided into n equal pieces and at each time interval of length 1/n an interest payment of percent p/n is added to the account Each time this fraction of the interest is added to the account, the money in the account is multiplied by 1+ p/n 100 Since this is done n times during the year, the result at the end of the year is that the account holds P 1+ p 100n dollars at the end of the year Similarly, at the end of t years, the money accumulated will be p P 1+ 100n Let us set k= and rewrite ( ) as 1 P 1+ k kp/100 nt
n 100 p
1 1+ k
p/100 It is useful to know the behavior of the account if the number of times the interest 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 623 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 ( ) CALCULUS DeMYSTiFieD
EXAMPLE
If $6000 is placed in a savings account 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 e5t/100 After four and one half years the size of the account is therefore M( 9/2) = 6000 e5 ( 9/2) /100 $751394 EXAMPLE
A wealthy woman wishes to set up an endowment for her nephew She wants the endowment to pay the young man $100,000 in cash on the day of his twentyfirst birthday The endowment is set up on the day of the nephew's birth and is locked in at 11% interest compounded continuously How much principal should be put into the account to yield the desired payoff SOLUTION
Let P be the initial principal deposited in the account on the day of the nephew's birth Using our compound interest equation ( ) , we have 100000 = P e11 21/100 , expressing the fact that after 21 years at 11% interest compounded continuously we want the value of the account to be $100,000 Solving for P gives P = 100000 e 011 21 = 100000 e 231 = 992613 The aunt needs to endow the fund with an initial $992613

