Certainly! Let's break down the problem into two parts.
### Part (a): Finding the Function for the Amount After [tex]\( t \)[/tex] Years
Given:
- Initial investment ([tex]\( P \)[/tex]) = \[tex]$64,000
- Annual interest rate (\( r \)) = 5.5%
- Compounded quarterly, so the number of times interest is compounded per year (\( n \)) = 4
We need to find the function \( A(t) \) which represents the amount of money in the account after \( t \) years.
The formula for the amount \( A \) after \( t \) years when interest is compounded \( n \) times per year is given by:
\[
A(t) = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Substitute the given values:
- \( P = 64,000 \)
- \( r = \frac{5.5}{100} = 0.055 \)
- \( n = 4 \)
Thus, the function becomes:
\[
A(t) = 64,000 \left(1 + \frac{0.055}{4}\right)^{4t}
\]
Simplify the expression inside the parentheses:
\[
1 + \frac{0.055}{4} = 1 + 0.01375 = 1.01375
\]
So, the function for the amount after \( t \) years is:
\[
A(t) = 64,000 \left(1.01375\right)^{4t}
\]
### Part (b): Finding the Amount at \( t = 0, 2, 6, \) and \( 10 \) Years
Now we need to calculate the amount at different times using the function we derived.
1. At \( t = 0 \) years:
\[
A(0) = 64,000 \left(1.01375\right)^{4 \cdot 0} = 64,000 \left(1.01375\right)^{0} = 64,000 \cdot 1 = 64,000
\]
2. At \( t = 2 \) years:
\[
A(2) = 64,000 \left(1.01375\right)^{4 \cdot 2} = 64,000 \left(1.01375\right)^{8} \approx 71,388.2789095907
\]
3. At \( t = 6 \) years:
\[
A(6) = 64,000 \left(1.01375\right)^{4 \cdot 6} = 64,000 \left(1.01375\right)^{24} \approx 88,822.04894429815
\]
4. At \( t = 10 \) years:
\[
A(10) = 64,000 \left(1.01375\right)^{4 \cdot 10} = 64,000 \left(1.01375\right)^{40} \approx 110,513.32934717093
\]
So, the amounts at different times are:
- At \( t = 0 \) years: \$[/tex]64,000.00
- At [tex]\( t = 2 \)[/tex] years: \[tex]$71,388.28
- At \( t = 6 \) years: \$[/tex]88,822.05
- At [tex]\( t = 10 \)[/tex] years: \[tex]$110,513.33
To summarize:
### a) The function for the amount to which the investment grows after \( t \) years is:
\[
A(t) = 64,000 \left(1.01375\right)^{4t}
\]
### b) The amounts of money in the account at different times are:
- At \( t = 0 \): \$[/tex]64,000
- At [tex]\( t = 2 \)[/tex]: \[tex]$71,388.28
- At \( t = 6 \): \$[/tex]88,822.05
- At [tex]\( t = 10 \)[/tex]: \$110,513.33
