Let's break down the problem step-by-step:

Part (a)

The general formula for continuous compounding is:

$P(t) = P_0 e^{rt}$

where:

• $P(t)$ is the amount after time $t$ years,
• $P_0$ is the initial principal,
• $r$ is the annual interest rate (as a decimal), and
• $t$ is the time in years.

Given:

• $P_0 = 14,152$
• $r = 5.95\% = 0.059$

The exponential function describing the amount in the account after time $t$ is:

$P(t) = 14,152 e^{0.059t}$

Part (b)

We need to calculate the balance after 1, 2, 5, and 10 years using the formula $P(t) = 14,152 e^{0.059t}$.

1. After 1 year: $P(1) = 14,152 e^{0.059 \times 1} = 14,152 \times e^{0.059}$

2. After 2 years: $P(2) = 14,152 e^{0.059 \times 2} = 14,152 \times e^{0.118}$

3. After 5 years: $P(5) = 14,152 e^{0.059 \times 5} = 14,152 \times e^{0.295}$

4. After 10 years: $P(10) = 14,152 e^{0.059 \times 10} = 14,152 \times e^{0.59}$

Let's calculate these values.

Part (c)

The doubling time $T$ for continuous compounding can be found using the formula:

$T = \frac{\ln(2)}{r}$

where:

• $\ln(2) \approx 0.693$
• $r = 0.059$

Now, I will compute all these values.### Results

Part (b): The balance after:

1. 1 year: $15,012.09
2. 2 years: $15,924.45
3. 5 years: $19,007.92
4. 10 years: $25,530.04

Part (c): The doubling time is approximately 11.7 years.

If you have any questions or need further details, feel free to ask!

Related Questions:

1. How would the formula change if the interest rate were compounded quarterly instead of continuously?
2. What is the formula to calculate the amount after $n$ years if the interest is compounded monthly?
3. How would you find the time required to triple the investment with the same interest rate?
4. What is the effective annual rate corresponding to a continuous compounding rate of 5.9%?
5. How does the doubling time change if the interest rate is increased to 6.5%?

Tip: For continuous compounding, the formula $P(t) = P_0 e^{rt}$ can be used to easily find the amount at any time $t$ without needing complex periodic calculations.