Hi fredewq,
I've followed pretty much the same strategy as Maniac-ga in the answer
to your last question
(https://answers.google.com/answers/main?cmd=threadview&id=119411).
I entered the data in Excel, generated an X/Y Scatter Plot, added a
trend line and displayed the R^2 value and formula. This time,
however, the trend line an exponential function. You'll see in part d)
why I did this.
Here's what Excel spit out:
a) A(t) = 105,071 * e^(0.0598 * t) where t is in years and A(t) is in
dollars.
The R^2 I got was 1, which means that the model is a "perfect fit."
However, this does not mean the same here as in your last question
(https://answers.google.com/answers/main?cmd=threadview&id=119411),
where you could plug in your x-values and got out exactly the y-value
you wanted. Here it means that A(t) is pretty much the best possible
approximation you can get with such an exponential model.
b) The derivative, A'(t), is calculated the same way you've shown in
the clarification to this question. It's just going to be a bit
messier...
A'(t) = 0.0598 * 105,071 * e^(0.0598 * t) = 6283.25 * e^(0.0598 * t)
where t is again in years and A'(t) is in dollars.
c) to complete the table, you plug in your values for t into the A(t)
equation and you'll get
A(25) = 105,071 * e^(0.0598 * 25) = 468,547
and
A(55) = 105,071 * e^(0.0598 * 55) = 2,817,588 (rounded off to the
nearest whole dollar)
So you'd put 468,547 and the row with the 25 and 2,817,588 in the row
with the 55 years in your table.
d) To calculate the interest with such an exponential model now is a
breeze. All you do is to take the exponent and multiply it by 100%.
That'd be 0.0598 * 100% = 5.98%
e) The value the account started with in mathematical terms is equal
to A(t) where t = 0; i.e. A(0).
That's
A(0) = 105,071 * e^(0.0598 * 0) = 105,071 * e^0 = 105,071 * 1 =
105,071
Note that e^0 = 1!
That'd be it. Good luck!
gwagner-ga |