1. A woman deposits $10,000 into a bank. In her account, it pays interest at a 5
ID: 2942954 • Letter: 1
Question
1. A woman deposits $10,000 into a bank. In her account, it pays interest at a 5% annual rate, and she also adds money continuously to the account, at a rate of $50 per month.a. Find a formula for the amount of money in the account "t" years after the initial $10,000 deposit.
b.how long will it take until the account balance is $100,000?
c. She arranges for $100 per month to be transfered to her son's checking account, find a formula for the amount of money in the account "t" years after the initial $10,000 deposit.
d. when will the account balance reach $0? (after the initial $100 transfer begins)
Explanation / Answer
a) We will use t in years. So she deposits 50 a month which is 12*50 = 600 a year. So, if we let Y be the balance of the account, the differential equation would look like this: dY/dt = .05Y + 600 This equation is separable, so we solve it like this: dY/(Y+12000) = .05dt Integrating both sides gives: ln(Y+12000) = .05t + c Solving for y gives: Y = Ke^(.05t) - 12000 Knowing that Y(0) = 10000 allows us to solve for K and get: Y = 22000e^(.05t) - 12000 b) To find t we simply plug in 100000 for Y and solve for t: 100000 = 22000e^(.05t) - 12000 112000 = 22000e^(.05t) 56/11 = e^(.05t) ln(56/11) = .05t t = ln(56/11)/.05 which is about 32.549 years. c) Now, with the 100 withdrawal every month, added to the 50 deposit each month, she will be losing 600 dollars every year. So the differential form will look like this: dY/dt = .05Y - 600 We solve this just like the first one to get: Y = -2000e^(.05t) + 12000 d) Plug in 0 for Y and solve for t: 0 = -2000e^(.05t) + 12000 -12000 = -2000e^(.05t) 6 = e^(.05t) ln(6) = .05t t = ln(6)/.05 which is about 35.835 years.
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