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Exponent H.W.

  1. Apr 25, 2004 #1
    Problem 9c. Suppose you invest $1.00 at 6% annual interest. Calcualte the amount that is compounded continuosly.
    This is what I have done:
    A=Pe^rt
    =1.00e^(0.06)(1) Is this right so far?

    Problem 11b.
    A population of ladybugs rapidly multiplies so that population t days form now is given by A(t)=3000e^(0.01)(t). How many will be present in a week?
    A(t)=3000e^(0.01)(7)
    = 3000.56 Is this right?

    Problem 12. Suppose that $10,00 is invested at an annual rate of 9% and that interest is compounded every second for 365 days. Find the value of this investment at the end of one year. Compare this answer with the value of 10,000e^0.09.
    This is what I have done:
    P(t)=Pe^rt
    =10,000e^(0.09)(31536000)
    =10,000e^2838240
    =26,141,156.46
    Is this right?
     
  2. jcsd
  3. Apr 26, 2004 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I started to say "yes, that is correct" but then I looked at 11.b.
    IF you mean 1.00 e^{(0.06)(1)}, yes it is correct. If you meant (as I would guess from 11.b) 1.00 {e^(0.06)}(1) then you would get the same answer but the concept is wrong.

    No, it isn't. I'm not at all sure how you got "0.56". At first I thought you had calculated e^(0.01) then subtracted 1 then multiplied by 7 and finally added 3000 but that doesn't quite give the same thing.
    A(t)= 3000 e^{(0.01)(7)}= 3000 e^(0.07)= 3000(1.0725)= 3217.52, according to my calculator.

    Oh, my God! I want you working at my local bank (and I'll withdraw my saving fast before the bank goes bust!).

    Yes, you have correctly calculated that, since there are 60 seconds in a minute, 60 minutes in an hour and 24 hours in a day, there are (60)(60)(24)(365)= 31536000 seconds in 365 days and so t= 31536000. However, "9%" is the annual rate of interest. Since this is "compounded each second", your r should be 0.09/(31536000). Notice that if you put both r= 0.09/31536000 and t= 31536000 into your formula, the whole "31536000" calculation cancels! That's because "Pe^(rt)" only applies to continuous compounding. For non-continuous compounding, you need to use
    A(t)= P(t)(1+r)t where r is the interest rate per compounding interval and t is the number of compounding intervals. Here,
    A(t)= 3000(1+ 0.09/31536000)31536000.
     
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