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Sagot :
Alright, let's solve each part step-by-step.
### Part (a)
First, we derive the formula for the investment growth. The problem states that the initial investment is [tex]\(\$ 88,000\)[/tex] with an annual interest rate of [tex]\(4.5\%\)[/tex], compounded quarterly.
1. Quarterly interest rate:
The annual interest rate is [tex]\(4.5\%\)[/tex] which is equivalent to [tex]\(0.045\)[/tex] as a decimal. Since the interest is compounded quarterly, we divide this annual rate by 4 (the number of quarters in a year) to get the quarterly interest rate:
[tex]\[ \text{Quarterly interest rate} = \frac{0.045}{4} = 0.01125 \][/tex]
2. Growth factor per quarter:
The growth factor per quarter is then:
[tex]\[ 1 + 0.01125 = 1.01125 \][/tex]
3. Number of compounding periods in [tex]\(t\)[/tex] years:
If [tex]\(t\)[/tex] is the number of years, then the total number of compounding periods in [tex]\(t\)[/tex] years is [tex]\(4 \cdot t\)[/tex], because interest is compounded quarterly.
4. Formulating the investment function:
Using the compound interest formula [tex]\(A(t) = P \left(1 + \frac{r}{n}\right)^{nt}\)[/tex], we substitute [tex]\(P = 88,000\)[/tex], the quarterly growth factor [tex]\(1.01125\)[/tex], and number of periods [tex]\(4t\)[/tex]:
[tex]\[ A(t) = 88,000 \times (1.01125)^{4t} \][/tex]
Thus, the function for the amount to which the investment grows after [tex]\(t\)[/tex] years is:
[tex]\[ A(t) = 88,000 \times (1.01125)^{4t} \][/tex]
### Part (b)
Next, let's find the amount of money in the account at specific points in time: [tex]\(t = 0\)[/tex], [tex]\(t = 4\)[/tex], [tex]\(t = 5\)[/tex], and [tex]\(t = 10\)[/tex] years. We will use the function derived above and evaluate it at these values of [tex]\(t\)[/tex].
1. At [tex]\(t = 0\)[/tex]:
[tex]\[ A(0) = 88,000 \times (1.01125)^{4 \times 0} = 88,000 \times (1.01125)^0 = 88,000 \times 1 = 88,000 \][/tex]
The amount of money in the account at [tex]\(t = 0\)[/tex] years is [tex]\(\$88,000\)[/tex].
2. At [tex]\(t = 4\)[/tex]:
[tex]\[ A(4) = 88,000 \times (1.01125)^{4 \times 4} = 88,000 \times (1.01125)^{16} \][/tex]
This evaluates to approximately [tex]\(105,249\)[/tex]. The amount of money in the account at [tex]\(t = 4\)[/tex] years is [tex]\(\$105,249\)[/tex].
3. At [tex]\(t = 5\)[/tex]:
[tex]\[ A(5) = 88,000 \times (1.01125)^{4 \times 5} = 88,000 \times (1.01125)^{20} \][/tex]
This evaluates to approximately [tex]\(110,066\)[/tex]. The amount of money in the account at [tex]\(t = 5\)[/tex] years is [tex]\(\$110,066\)[/tex].
4. At [tex]\(t = 10\)[/tex]:
[tex]\[ A(10) = 88,000 \times (1.01125)^{4 \times 10} = 88,000 \times (1.01125)^{40} \][/tex]
This evaluates to approximately [tex]\(137,665\)[/tex]. The amount of money in the account at [tex]\(t = 10\)[/tex] years is [tex]\(\$137,665\)[/tex].
### Summary
Part (a): The function for the amount to which the investment grows after [tex]\(t\)[/tex] years is:
[tex]\[ A(t) = 88,000 \times (1.01125)^{4t} \][/tex]
Part (b): The amounts of money in the account at various times are:
- At [tex]\(t = 0\)[/tex] years: [tex]\(\$88,000\)[/tex]
- At [tex]\(t = 4\)[/tex] years: [tex]\(\$105,249\)[/tex]
- At [tex]\(t = 5\)[/tex] years: [tex]\(\$110,066\)[/tex]
- At [tex]\(t = 10\)[/tex] years: [tex]\(\$137,665\)[/tex]
### Part (a)
First, we derive the formula for the investment growth. The problem states that the initial investment is [tex]\(\$ 88,000\)[/tex] with an annual interest rate of [tex]\(4.5\%\)[/tex], compounded quarterly.
1. Quarterly interest rate:
The annual interest rate is [tex]\(4.5\%\)[/tex] which is equivalent to [tex]\(0.045\)[/tex] as a decimal. Since the interest is compounded quarterly, we divide this annual rate by 4 (the number of quarters in a year) to get the quarterly interest rate:
[tex]\[ \text{Quarterly interest rate} = \frac{0.045}{4} = 0.01125 \][/tex]
2. Growth factor per quarter:
The growth factor per quarter is then:
[tex]\[ 1 + 0.01125 = 1.01125 \][/tex]
3. Number of compounding periods in [tex]\(t\)[/tex] years:
If [tex]\(t\)[/tex] is the number of years, then the total number of compounding periods in [tex]\(t\)[/tex] years is [tex]\(4 \cdot t\)[/tex], because interest is compounded quarterly.
4. Formulating the investment function:
Using the compound interest formula [tex]\(A(t) = P \left(1 + \frac{r}{n}\right)^{nt}\)[/tex], we substitute [tex]\(P = 88,000\)[/tex], the quarterly growth factor [tex]\(1.01125\)[/tex], and number of periods [tex]\(4t\)[/tex]:
[tex]\[ A(t) = 88,000 \times (1.01125)^{4t} \][/tex]
Thus, the function for the amount to which the investment grows after [tex]\(t\)[/tex] years is:
[tex]\[ A(t) = 88,000 \times (1.01125)^{4t} \][/tex]
### Part (b)
Next, let's find the amount of money in the account at specific points in time: [tex]\(t = 0\)[/tex], [tex]\(t = 4\)[/tex], [tex]\(t = 5\)[/tex], and [tex]\(t = 10\)[/tex] years. We will use the function derived above and evaluate it at these values of [tex]\(t\)[/tex].
1. At [tex]\(t = 0\)[/tex]:
[tex]\[ A(0) = 88,000 \times (1.01125)^{4 \times 0} = 88,000 \times (1.01125)^0 = 88,000 \times 1 = 88,000 \][/tex]
The amount of money in the account at [tex]\(t = 0\)[/tex] years is [tex]\(\$88,000\)[/tex].
2. At [tex]\(t = 4\)[/tex]:
[tex]\[ A(4) = 88,000 \times (1.01125)^{4 \times 4} = 88,000 \times (1.01125)^{16} \][/tex]
This evaluates to approximately [tex]\(105,249\)[/tex]. The amount of money in the account at [tex]\(t = 4\)[/tex] years is [tex]\(\$105,249\)[/tex].
3. At [tex]\(t = 5\)[/tex]:
[tex]\[ A(5) = 88,000 \times (1.01125)^{4 \times 5} = 88,000 \times (1.01125)^{20} \][/tex]
This evaluates to approximately [tex]\(110,066\)[/tex]. The amount of money in the account at [tex]\(t = 5\)[/tex] years is [tex]\(\$110,066\)[/tex].
4. At [tex]\(t = 10\)[/tex]:
[tex]\[ A(10) = 88,000 \times (1.01125)^{4 \times 10} = 88,000 \times (1.01125)^{40} \][/tex]
This evaluates to approximately [tex]\(137,665\)[/tex]. The amount of money in the account at [tex]\(t = 10\)[/tex] years is [tex]\(\$137,665\)[/tex].
### Summary
Part (a): The function for the amount to which the investment grows after [tex]\(t\)[/tex] years is:
[tex]\[ A(t) = 88,000 \times (1.01125)^{4t} \][/tex]
Part (b): The amounts of money in the account at various times are:
- At [tex]\(t = 0\)[/tex] years: [tex]\(\$88,000\)[/tex]
- At [tex]\(t = 4\)[/tex] years: [tex]\(\$105,249\)[/tex]
- At [tex]\(t = 5\)[/tex] years: [tex]\(\$110,066\)[/tex]
- At [tex]\(t = 10\)[/tex] years: [tex]\(\$137,665\)[/tex]
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