Question

two thousand dollars is deposited into a savings account at 8 5% interest compounded continuously
(a) what is the formula for a(t), the balance after t years?
(b) what differential equation is satisfied by a(t), the balance after t years?
(c) how much money will be in the account after 2 years?
(d) when will the balance reach $6000?
(e) how fast is the balance growing when it reaches $6000?
(a) a(t) = 2000e^{(085t)}
(b) a(t) = 0.85a
(c) $ (round to the nearest cent as needed )

Explanation:

Step1: Substitute values into the formula

We know the formula for continuous - compounding is \(A(t)=P e^{rt}\), where \(P = 2000\) (principal amount), \(r=0.085\) (annual interest rate). For \(t = 2\) years, we substitute into \(A(t)=2000e^{0.085t}\).
So \(A(2)=2000e^{0.085\times2}\).

Step2: Calculate the exponent

First, calculate \(0.085\times2=0.17\). Then \(A(2)=2000e^{0.17}\).
Since \(e^{0.17}\approx1.18527\) (using a calculator, \(e^x=\sum_{n = 0}^{\infty}\frac{x^n}{n!}\), or simply using the \(e^x\) function on a scientific calculator).

Step3: Calculate the final amount

Multiply \(2000\times1.18527 = 2370.54\).

Answer:

\(2370.54\)