- Substitute \(t = \frac{m}{12}\) into the original function \(A = 1000(1.023)^t\) to relate years to months.
- Rewrite the function to identify the monthly growth factor as \(1.023^{\frac{1}{12}}\), resulting in the expression \(A = 1000\left(1.023^{\frac{1}{12}}\right)^{12t}\).
- Calculate the monthly interest rate by finding \(1.023^{\frac{1}{12}} - 1 \approx 0.001897\).
- Convert the decimal to a percentage to get \(\approx 0.19\%\). The function form is \(A = 1000\left(1.023^{\frac{1}{12}}\right)^{12t}\) and the rate is \(\boxed{0.19\%}\).
Explanation
- Understanding the Problem and Variables
We start with the given function for the amount of money in a savings account after \(t\) years:
\[A = 1000(1.023)^t\]
Our objective is to rewrite this function to reflect the compounding period monthly rather than annually, and then determine the monthly interest rate. Since there are 12 months in one year, the number of months \(m\) is related to the number of years \(t\) by the equation \(m = 12t\). We can substitute \(t = \frac{m}{12}\) into the original function.
- Rewriting the Function
Substituting \(t = \frac{m}{12}\) into the formula gives us:
\[A = 1000(1.023)^{\frac{m}{12}} = 1000\left(1.023^{\frac{1}{12}}\right)^m\]
Since \(m = 12t\), we can express the function in terms of \(t\) with a monthly compounding basis:
\[A = 1000\left(1.023^{\frac{1}{12}}\right)^{12t}\]
This form helps us identify the monthly growth factor, which is the base of the expression raised to the power of \(12t\), namely \(1.023^{\frac{1}{12}}\).
- Calculating the Monthly Interest Rate
The monthly growth factor is \(1 + r\), where \(r\) is the monthly interest rate. From our rewritten function, we have:
\[1 + r = 1.023^{\frac{1}{12}}\]
Using the calculation of \(1.023^{\frac{1}{12}}\), we find:
\[1 + r \approx 1.001897\]
Thus, the monthly interest rate is:
\[r = 1.001897 - 1 = 0.001897\]
To express this as a percentage, we multiply by 100:
\[r \approx 0.001897 \times 100\% = 0.1897\% \approx 0.19\%\]
Comparing this to the provided choices, the approximate monthly interest rate is \(0.19\%\).
Examples
This approach to converting annual rates to monthly rates is essential in finance, particularly when dealing with loans, mortgages, or savings accounts. For instance, if you are looking at a bank loan with an annual interest rate, understanding the monthly breakdown allows you to compute the actual cost of borrowing or the growth of your investments on a month-by-month basis, helping you manage your budget more effectively. By using the property of exponents \(a^{t} = (a^{1/12})^{12t}\), you can easily transition between different time units.