what continuous yearly interest rate r% is needed in their savings account if their deposits of $4200 per year are to grow to $80,437 in 13 years. They make deposits continuously throughout the year

To find the continuous yearly interest rate $ r \% $ that will allow continuous deposits of $4200 per year to grow to $80,437 in 13 years, we use the formula for the future value of continuously compounded interest with continuous payments.

The formula is:

\[
A = P \left( e^{rt} - 1 \right) / r
\]

Where:
- $ A $ is the future value ($80,437),
- $ P $ is the annual deposit amount ($4200),
- $ r $ is the continuous interest rate,
- $ t $ is the time in years (13 years),
- $ e $ is the base of the natural logarithm (approximately 2.71828).

Rearrange the formula to solve for $ r $:

\[
80,437 = 4200 \left( \frac{e^{13r} - 1}{r} \right)
\]

Let's solve for $ r $.The continuous yearly interest rate needed is approximately **5.62%**.

Would you like further details on the calculation or have any other questions?

Here are some related questions you might find interesting:
1. How would the result change if the annual deposit amount increased?
2. What is the effect of extending the investment period to 15 years?
3. How does continuous compounding differ from annual compounding?
4. What is the formula for calculating the future value with annual compounding?
5. How does the future value change if the interest rate increases by 1%?
6. What happens to the final amount if deposits are made at the beginning of each year instead of continuously?
7. How can you calculate the present value of these continuous deposits?
8. What factors should be considered when choosing an interest rate for investments?

**Tip:** When making continuous deposits, the compounding effect can significantly increase the future value, especially with a higher interest rate and longer investment duration.

Learn how to calculate the continuous yearly interest rate needed for savings to grow to $80,437 in 13 years with continuous annual deposits of $4,200. Includes a step-by-step solution using the continuous compounding formula.

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