Suppose that P0 is invested in a savings account in which interest is pounded co

Question

Suppose that P0 is invested in a savings account in which interest is pounded continuously at 6.4% per year. That is, the balance P grows at the rate given by the following equation. dt (a)Find the function P(t) that satisfies the equation. Write it in terms of Po and 0.064 (b)Suppose that $1500 is invested. What is the balance after 3 years? (c)When wil an investment of $1500 double rseif? (a) Choose the corect answer below. 0.004 (b) The balance after 3 year is s (Type an integer or decimal rounded to two decimal places as needed.) (c) The doubling time is year (Type an integer or decimal rounded to two decimal places as needed)

Explanation / Answer

dP/dt = 0.064P

dP / P = 0.064dt

Integrating both sides

ln P = 0.064t + P0

P(t) = P0e0.064t

b. P(3) = 1500e0.064(3)=$1817.51

c. 3000 = 1500e0.064t

    ln 2 = 0.064t

    t = 10.83 year

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