# Compound Interest (Continuously) - Problem 3

So for this problem what we're going to do is do a combination of interest problems, so we have a bit of a compound continuous problem and we also have a compounded in a set number of time problem built into this so let's take a look at this.

You invest $2000 at 5% interest compounded monthly. After two years you switch your money to an account that gives 6% interest compounded continuously, how much money do you have 10 years after your initial investment?

So what I see when I look at this problem is basically two main steps we have to do. There is the $2000 at 5% that we invested so we have to figure how much that is going to be worth and then that is going to be reinvested at the 6% compounded continuously, so that's going to be another step and then together they're going to give us our answer. So let's take a look at this.

So our first thing is $2000 at 5% monthly, so because it says monthly, we know we have to use our 1 plus r over n equation because there's set number of times that interest is calculated. So we know we have $2000 and our rate for that first part is 5% and the number of times it's monthly which tells us it's calculated 12 times a year into the nt 12 and it says for two years, so that's going to be for two.

So this amount is going to be what we have at the end of two years, so let's go ahead and plug this into our calculator. 2000 times 1 plus .05 divided by 12 to the 20 fourth and that gives us 2209.88. So that's how much we have at the end of two years. So now we are reinvesting this at at another account that compounds continuously at 6%.

So if we are reinvesting this 6% continuously. Continuously tells us we use Pert so we go to a equals Pe to the rt, we are investing this amount into 09.88e and then the rt, so let's go back over this problem and check this out.

So what we have is we have to figure out our rate and our time. Our rate is easy enough it says we change to account that gives us 6%, so we know that our rate is .06. The trick is our time, so we invest in this to compound continuously and it's asking for 10 years after our initial investment. So it's been 10 years from the start, but that already includes the 2 years that we had invested at the initial spot so that leaves us only 8 years compounded continuously, so let's go back in and plug in the 8 and the .06.

So we know that our rate is .06 and our time after those two years is 8. Let's plug this into our calculator then. The 2209 88e to the .06 times 8 leaving us with 3571 and 33 cents.

So what we've really done is taken one problem, combined two different principles, the compounded continuously and compounded a set number of times, neither one was all that hard, but basically doing them together in the same column.

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