Question: Question 6, 9.6.7 Part 1 of 6 10 points Points: 0 of 1 Save Suppose that $17,585 is invested at an…

Question 6, 9.6.7 Part 1 of 6 10 points Points: 0 of 1 Save

Suppose that $17,585 is invested at an interest rate of 5.5% 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) = [ ] (Type exponential notation with positive exponents. Do not simplify. Use integers or decimals for any numbers in the equation.)

Solution

Suppose that $17,585 is invested at an interest rate of 5.5% per year, compounded continuously. a) Find the exponential function that describes the amount in the account after time \( t \), in years. The formula for continuous compounding is: \[ P(t) = P_0 e^{rt} \] where \( P_0 \) is the initial amount, \( r \) is the interest rate, and \( t \) is time in years. Substitute the given values: \[ P(t) = 17585 \cdot e^{0.055t} \] b) What is the balance after 1 year? 2 years? 5 years? 10 years? Substitute \( t = 1 \), \( t = 2 \), \( t = 5 \), and \( t = 10 \) into the exponential function. For \( t = 1 \): \[ P(1) = 17585 \cdot e^{0.055 \cdot 1} \] For \( t = 2 \): \[ P(2) = 17585 \cdot e^{0.055 \cdot 2} \] For \( t = 5 \): \[ P(5) = 17585 \cdot e^{0.055 \cdot 5} \] For \( t = 10 \): \[ P(10) = 17585 \cdot e^{0.055 \cdot 10} \] c) What is the doubling time? The doubling time is the time it takes for the investment to double in value. Use the formula for doubling time in continuous compounding: \[ t = \frac{\ln(2)}{0.055} \]