# Math Help - Continuous deposits

1. ## Continuous deposits

a) Tiffany deposits $12 into an account at the end of each day in 1989 and 1990 and$15 at the end of each day in 1991. The account earns an annual effective interest rate of 9% in 1989 and 1990 and an annual effective interest rate of 12% in 1991. Find the amount in Tiffany’s account on December 31, 1991. (Answer: $16,502.58) b) Rework part (a) using the approximation that the deposits are made continuously. (Answer:$16,504.75)

I think I'm pretty close, just missing something somewhere. I have
12[(1.000236)^730 - 1 / .000236] + 15[(1.000311)^365 - 1 / .000311]
= 15355.5
(because (1+j)^4=1.045 and (1+j)^4=(1.025)^5)

And also I'm not sure for part (b) how you would work it out differently if the deposits are continuous.

Thank you!

2. For part (a) you forgot to multiply the first part (The 2 years of \$12 deposits) by 1.12 in order to bring it to the end of 1991, then you will get 16502.58
ie.
[12[(1.000236)^730 - 1 / .000236]](1.12) + 15[(1.000311)^365 - 1 / .000311] = 16502.58

3. Ok awesome thank you!

And for part (b), I think it's setup like:
(12)(730)(1.12)(S bar angle 1) + (15)(365)(S bar angle 1) , where (S bar angle 1) = [(1+i)^1 - 1] / ln(1+i)

but again I'm just slightly off...

4. Originally Posted by tbl9301
I think I'm pretty close, just missing something somewhere. I have
12[(1.000236)^730 - 1 / .000236] + 15[(1.000311)^365 - 1 / .000311]
= 15355.5
You forgot the interest on the balance at Dec.31/90 for year 1991. OK?