Calculating Interest with Compounding: Problem 9c

[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?

 

[HallsofIvy]

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?

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.

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?

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.

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?

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.

 

[M98Ranger]

For problem 9c, your calculation is correct so far. The continuous compounding formula is A = Pe^(rt), where P is the principal amount, r is the annual interest rate, and t is the time in years. In this case, P = $1, r = 0.06, and t = 1. So the calculation is A = 1e^(0.06)(1) = $1.06.

For problem 11b, your calculation is also correct. A(t) represents the population at time t, so to find the population in a week (7 days), we plug in t = 7 into the formula A(t) = 3000e^(0.01)(t) to get A(7) = 3000e^(0.01)(7) = 3000.56, which means there will be approximately 3000.56 ladybugs present in a week.

For problem 12, your calculation is also correct. The formula for continuously compounded interest is P(t) = Pe^(rt), where P is the initial investment, r is the annual interest rate, and t is the time in years. In this case, P = $10,000, r = 0.09, and t = 1 (since the time is given in days, we need to convert it to years by dividing by 365). So the calculation is P(1) = 10,000e^(0.09)(1) = $26,141.16. This means that at the end of one year, the investment will be worth $26,141.16. Comparing this to the calculation using the formula A = Pe^(rt), we can see that the answer is the same, which is expected since both formulas represent continuous compounding.