Math Problem Statement

Second National Bank offers an account that earns interest at an annual rate of 7.1%, com-

pounded continuously. You take advantage of this great interest rate and deposit money into

this account. Your account grows to $4850.45 in three years.

(a) How much did you initially deposit into this account?

(Round your answer to the nearest cent.)

(b) Suppose you want to go on a trip in three years and you need your deposit to grow to $6000,

instead of $4850.45. If you can only afford to deposit $4000 in the account today, what

interest rate do you need the bank to offer instead, so that you to have the $6000 in three

years? (Round your answer to two decimal places, after writing your answer as

a percentage.)

Solution

(a) How much did you initially deposit into this account?

The formula for continuous compounding is:

A=PertA = P e^{rt}

where:

  • AA is the final amount in the account.
  • PP is the principal (initial deposit).
  • rr is the interest rate (in decimal form).
  • tt is the time in years.

Given:

  • A=4850.45A = 4850.45
  • r=0.071r = 0.071 (7.1% annual rate)
  • t=3t = 3 years

We need to find the initial deposit, PP.

Rearranging the formula to solve for PP:

P=AertP = \frac{A}{e^{rt}}

Substitute the given values:

P=4850.45e0.071×3P = \frac{4850.45}{e^{0.071 \times 3}}

Calculating the exponent:

P=4850.45e0.213P = \frac{4850.45}{e^{0.213}}

Now, calculate e0.213e^{0.213} and then divide 4850.454850.45 by that result.

e0.2131.2376e^{0.213} \approx 1.2376

So,

P4850.451.23763920.00P \approx \frac{4850.45}{1.2376} \approx 3920.00

Answer for (a): The initial deposit was approximately $3920.00.


(b) What interest rate do you need for $4000 to grow to $6000 in three years?

Now we need to find the interest rate rr if:

  • The initial deposit P=4000P = 4000
  • The final amount A=6000A = 6000
  • The time t=3t = 3 years

Using the continuous compounding formula:

A=PertA = P e^{rt}

Rearrange to solve for rr:

r=ln(AP)tr = \frac{\ln\left(\frac{A}{P}\right)}{t}

Substitute the values:

r=ln(60004000)3r = \frac{\ln\left(\frac{6000}{4000}\right)}{3}

Calculate the fraction inside the logarithm:

60004000=1.5\frac{6000}{4000} = 1.5

So,

r=ln(1.5)3r = \frac{\ln(1.5)}{3}

Calculate ln(1.5)0.4055\ln(1.5) \approx 0.4055:

r=0.405530.1352r = \frac{0.4055}{3} \approx 0.1352

Convert rr to a percentage by multiplying by 100:

r13.52%r \approx 13.52\%

Answer for (b): You would need an interest rate of approximately 13.52% for your deposit to grow to $6000 in three years.


Would you like more detailed steps or further clarification on any part of this solution?

Here are some additional related questions:

  1. How would the final amount change if the compounding was quarterly instead of continuous?
  2. What would the initial deposit need to be for an account to grow to $7000 in five years at a 7.1% continuous interest rate?
  3. If the interest rate changes after the first year, how would we adjust the formula to calculate the final amount?
  4. What happens to the growth of the account if the compounding frequency is daily instead of continuously?
  5. How long would it take for $4000 to double with a 7.1% continuous interest rate?

Tip: Continuous compounding grows faster than other compounding frequencies due to the mathematical properties of ee, Euler's number.

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Continuous Compounding
Logarithms
Interest Rate Calculations

Formulas

A = P * e^(rt)
r = (ln(A/P)) / t

Theorems

Exponential Growth
Properties of Logarithms

Suitable Grade Level

Grades 11-12