The question asks for the exponential growth function given continuous compounding of interest at a rate of 3% per year. The formula for exponential growth is:

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

Where:

• $P(t)$ is the balance at time $t$.
• $P_0$ (or $P$ in this case) is the initial principal amount.
• $k$ is the continuous growth rate (in decimal form).
• $t$ is the time in years.
• $e$ is the base of the natural logarithm.

The problem specifies that the interest rate is 3% per year, which means $k = 0.03$.

Therefore, the exponential growth function in terms of $P$ and $0.03$ is:

$P(t) = P e^{0.03t}$

Let me know if you'd like further clarification or have additional questions!

Here are some related questions:

1. How do you compute the balance after 5 years if the initial amount is $1000?
2. What would the formula look like for a different interest rate, say 5%?
3. How do you solve for the time when the balance doubles in continuous compounding?
4. What is the difference between continuous compounding and compound interest?
5. How can you derive the continuous compounding formula from the compound interest formula?

Tip: When solving for time in growth problems, you can use logarithms to isolate $t$.