Sure, let's tackle the problem step-by-step.
### Part (a)
To find the function for the amount to which the investment grows after [tex]\( t \)[/tex] years, we will use the compound interest formula:
[tex]\[ A(t) = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( A(t) \)[/tex] is the amount after [tex]\( t \)[/tex] years.
- [tex]\( P \)[/tex] is the principal amount (initial investment).
- [tex]\( r \)[/tex] is the annual interest rate (expressed as a decimal).
- [tex]\( n \)[/tex] is the number of times interest is compounded per year.
- [tex]\( t \)[/tex] is the time in years.
Given:
- [tex]\( P = \$78{,}000 \)[/tex]
- [tex]\( r = 4.5\% = 4.5/100 = 0.045 \)[/tex]
- [tex]\( n = 4 \)[/tex] (since interest is compounded quarterly)
Substitute these values into the formula to get the function [tex]\( A(t) \)[/tex]:
[tex]\[ A(t) = 78{,}000 \left(1 + \frac{0.045}{4}\right)^{4t} \][/tex]
Simplify inside the parentheses:
[tex]\[ A(t) = 78{,}000 \left(1 + 0.01125\right)^{4t} \][/tex]
[tex]\[ A(t) = 78{,}000 \left(1.01125\right)^{4t} \][/tex]
So, the function for the amount to which the investment grows after [tex]\( t \)[/tex] years is:
[tex]\[ A(t) = 78{,}000 (1.01125)^{4t} \][/tex]
### Part (b)
Now, we need to find the amount of money in the account at [tex]\( t = 0, 4, 8, \)[/tex] and [tex]\( 10 \)[/tex] years.
1. At [tex]\( t = 0 \)[/tex]:
[tex]\[ A(0) = 78{,}000 \left(1.01125\right)^{4 \times 0} \][/tex]
[tex]\[ A(0) = 78{,}000 \left(1.01125\right)^0 \][/tex]
[tex]\[ A(0) = 78{,}000 \times 1 \][/tex]
[tex]\[ A(0) = 78{,}000 \][/tex]
2. At [tex]\( t = 4 \)[/tex]:
[tex]\[ A(4) = 78{,}000 \left(1.01125\right)^{4 \times 4} \][/tex]
[tex]\[ A(4) = 78{,}000 \left(1.01125\right)^{16} \][/tex]
By evaluating this expression:
[tex]\[ A(4) \approx 93{,}289.1544200772 \][/tex]
3. At [tex]\( t = 8 \)[/tex]:
[tex]\[ A(8) = 78{,}000 \left(1.01125\right)^{4 \times 8} \][/tex]
[tex]\[ A(8) = 78{,}000 \left(1.01125\right)^{32} \][/tex]
By evaluating this expression:
[tex]\[ A(8) \approx 111{,}575.20938991036 \][/tex]
4. At [tex]\( t = 10 \)[/tex]:
[tex]\[ A(10) = 78{,}000 \left(1.01125\right)^{4 \times 10} \][/tex]
[tex]\[ A(10) = 78{,}000 \left(1.01125\right)^{40} \][/tex]
By evaluating this expression:
[tex]\[ A(10) \approx 122{,}021.3955005038 \][/tex]
### Summary
(a) The function for the amount to which the investment grows after [tex]\( t \)[/tex] years is:
[tex]\[ A(t) = 78{,}000 (1.01125)^{4t} \][/tex]
(b) The amounts of money in the account are:
- At [tex]\( t = 0 \)[/tex]: \[tex]$78,000.00
- At \( t = 4 \): \$[/tex]93,289.15
- At [tex]\( t = 8 \)[/tex]: \[tex]$111,575.21
- At \( t = 10 \): \$[/tex]122,021.40
