# Exponent H.W.

1. Apr 25, 2004

### mustang

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. Apr 26, 2004

### HallsofIvy

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.