Math  /  Algebra

QuestionSuppose that $12,251\$ 12,251 is invested at an interest rate of 6.6%6.6 \% per year, compounded continuously. a) Find the exponential function that describes the amount in the account after time t , in years. b) What is the balance after 1 year? 2 years? 5 years? 10 years? c) What is the doubling time? a) The exponential growth function is P(t)=12251e0.066t\mathrm{P}(\mathrm{t})=12251 e^{0.066 t} (Type exponential notation with positive exponents. Do not simplify. Use integers or decimals for any numbers in the equation.) b) The balance after 1 year is $\$ \square (Simplify your answers. Round to two decimal places as needed.)

Studdy Solution

STEP 1

1. The principal amount invested is \$12,251.
2. The annual interest rate is 6.6%, compounded continuously.
3. We need to find the exponential growth function for the investment.
4. We need to calculate the balance after 1, 2, 5, and 10 years.
5. We need to determine the doubling time for the investment.

STEP 2

1. Derive the exponential growth function for continuous compounding.
2. Calculate the balance after 1 year, 2 years, 5 years, and 10 years.
3. Determine the doubling time using the exponential growth function.

STEP 3

The formula for continuous compounding is given by:
P(t)=P0ert P(t) = P_0 e^{rt}
where P0 P_0 is the initial principal, r r is the annual interest rate, and t t is the time in years.

STEP 4

Substitute the given values into the formula:
P(t)=12251e0.066t P(t) = 12251 e^{0.066t}
This is the exponential function that describes the amount in the account after time t t .

STEP 5

Calculate the balance after 1 year by substituting t=1 t = 1 into the exponential function:
P(1)=12251e0.066×1 P(1) = 12251 e^{0.066 \times 1}

STEP 6

Compute the value:
P(1)=12251e0.06612251×1.0682713086.56 P(1) = 12251 e^{0.066} \approx 12251 \times 1.06827 \approx 13086.56
The balance after 1 year is approximately \$13,086.56.

STEP 7

Calculate the balance after 2 years by substituting t=2 t = 2 into the exponential function:
P(2)=12251e0.066×2 P(2) = 12251 e^{0.066 \times 2}

STEP 8

Compute the value:
P(2)=12251e0.13212251×1.1416313990.79 P(2) = 12251 e^{0.132} \approx 12251 \times 1.14163 \approx 13990.79
The balance after 2 years is approximately \$13,990.79.

STEP 9

Calculate the balance after 5 years by substituting t=5 t = 5 into the exponential function:
P(5)=12251e0.066×5 P(5) = 12251 e^{0.066 \times 5}

STEP 10

Compute the value:
P(5)=12251e0.3312251×1.3919217043.06 P(5) = 12251 e^{0.33} \approx 12251 \times 1.39192 \approx 17043.06
The balance after 5 years is approximately \$17,043.06.

STEP 11

Calculate the balance after 10 years by substituting t=10 t = 10 into the exponential function:
P(10)=12251e0.066×10 P(10) = 12251 e^{0.066 \times 10}

STEP 12

Compute the value:
P(10)=12251e0.6612251×1.9348423717.88 P(10) = 12251 e^{0.66} \approx 12251 \times 1.93484 \approx 23717.88
The balance after 10 years is approximately \$23,717.88.

STEP 13

To find the doubling time, use the formula for doubling time in continuous compounding:
t=ln(2)r t = \frac{\ln(2)}{r}
where r=0.066 r = 0.066 .

STEP 14

Compute the doubling time:
t=ln(2)0.0660.69310.06610.50 t = \frac{\ln(2)}{0.066} \approx \frac{0.6931}{0.066} \approx 10.50
The doubling time is approximately 10.50 years.
The exponential function is P(t)=12251e0.066t P(t) = 12251 e^{0.066t} . The balance after 1 year is approximately \$13,086.56. The balance after 2 years is approximately \$13,990.79. The balance after 5 years is approximately \$17,043.06. The balance after 10 years is approximately \$23,717.88. The doubling time is approximately 10.50 years.

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