Continuous Compounding over 12 months (finding the rate)

[tomadom]

This problem is actually for a program I'm writing and I've forgotten my basic maths.I have an initial value starting at period 0 . That value is 20.
At the end of 12 periods (period 12) I have a value of 33.

So I know the last value is 33 and the first value is 20 and I want to find the
continuous compounding rate so that I can apply it at any of the periods 1 to 11 and chart the output. I need to find r.

I know the formula is A = Pe^rt but I get stuck.

33 = 20*e^r at the 12 period, but if it's continuously compounded then I'm not sure what to do with t because:

log(33/20) = r gives me the total amount of interest accumed for that 12 months (21.75%) I think. So I divide that rate by 12 = 1.81% for each month on top of the previous month. But by the time that 12th month is reached I'm at 66.15 not 33.

So conceptually I'm getting mixed up somewhere? Could someone show me?..
Thanks

0 - 20
1 - ?
2 - ?
3 - ?
4 - ?
5 - ?
6 - ?
7 - ?
8 - ?
9 - ?
10 - ?
11 - ?
12 - 33

 

[MarkFL]

Your formula is correct:

\(\displaystyle A=Pe^{rt}\)

Plug in known values:

\(\displaystyle 33=20e^{12r}\)

Solve for $r$:

\(\displaystyle r=\frac{1}{12}\ln\left(\frac{33}{20}\right)\approx0.04173127399270744\)

Thus, we find:

\(\displaystyle A(t)=20\left(\frac{33}{20}\right)^{\frac{t}{12}}\)

Now we can fill in the table (rounded to the nearest penny):

Month​	Amount​
0	\$20.00
1	\$20.85
2	\$21.74
3	\$22.67
4	\$23.63
5	\$24.64
6	\$25.69
7	\$26.79
8	\$27.93
9	\$29.12
10	\$30.36
11	\$31.65
12	\$33.00

 

[tomadom]

Thanks very much for that.. Very neat explanation.