A college professor contributes $5,000 per year into her retirement fund by making many small deposits throughout the year. The fund grows at a rate of 7% per year compounded continuously. After 30 years, she retires and begins withdrawing from her fund at a rate of $3,000 per month. If she does not make any deposits after retirement, how long will the money last? {Hint: Solve this in two steps, before retirement and after retirement.}

Before retirement:

dy/dt = .07y+5000
separate and solve:
1/(.07y+5000)dy = dt
(1/.07)*ln(.07y+5000) = t+c
ln(.07y+5000) = .07t+c
.07y+5000 = ce^(.07t)
y = [c*e^(.07t)-5000]/.07
plug in y(0)=0 to find c. c=5000 so
y = [5000e^(.07t)-5000]/.07
plug in t=30 to get y=511869.27 (this is the amount in the fund after 30 yrs)

After retirement:
dy/dt = .07y-3000(12)
dy/dt = .07y-36000
separate and solve (almost the same as above)
you get y = [ce^(.07t)+36000]/.07
to solve for c, plug in y=511869.28 when t=0 since that is our starting amount. You get c = -169.15 so
y = [-169.15e^(.07t)+36000]/.07
When y=0, t = 76.58

So the money will last 76.58 years after she retires

Step 1: Calculate the balance of the retirement fund before retirement.

To calculate the balance before retirement, we can use the formula for compounding continuously:

A = P * e^(rt)

Where:
A = Final amount
P = Principal amount (initial deposit)
e = Euler's number (approximately 2.71828)
r = Annual interest rate (as a decimal)
t = Time in years

In this case, the annual interest rate is 7% and the time is 30 years. The principal amount will be the sum of all the small deposits made throughout the year, which is $5,000 per year.

So, the formula becomes:

A = 5,000 * e^(0.07 * 30)

Calculating this, we get:

A ≈ 5,000 * 2.71828^(0.07 * 30)
A ≈ 5,000 * 2.71828^2.1
A ≈ 5,000 * 8.166
A ≈ 40,830

Therefore, the balance of the retirement fund before retirement is approximately $40,830.

Step 2: Calculate how long the money will last after retirement.

To calculate how long the money will last after retirement, we divide the remaining balance by the monthly withdrawal amount.

Remaining balance = $40,830
Monthly withdrawal amount = $3,000

Time = Remaining balance / Monthly withdrawal amount

Time = $40,830 / $3,000
Time ≈ 13.61

Therefore, the money will last approximately 13.61 years after retirement.

To solve this problem, we need to calculate two periods separately: before retirement and after retirement.

Step 1: Calculate the amount accumulated before retirement

We can use the formula for compound interest with continuous compounding to calculate the total amount accumulated before retirement:

A = P * e^(rt)

Where:
A = final amount
P = principal (initial deposit)
e = Euler's number (approximately 2.71828)
r = interest rate
t = time in years

Given:
Principal (initial deposit) = $5,000
Interest rate (r) = 7% = 0.07 (in decimal form)
Time (t) = 30 years

Using the formula, we can calculate the final amount accumulated before retirement:

A = $5,000 * e^(0.07 * 30)

Step 2: Calculate the duration the money will last

After retirement, the professor will withdraw $3,000 per month. To find out how long the money will last, we need to divide the total amount accumulated before retirement by the monthly withdrawal:

Duration = A / Withdrawal per month

Given:
Total amount accumulated before retirement (A) = Calculated in Step 1
Withdrawal per month = $3,000

Using the formula, we can find the duration:

Duration = A / $3,000

By evaluating these steps, we can determine how long the money will last after retirement.

".. by making many small deposits throughout the year. "

I don't know what you mean by that.
- how often are they made ?
- are the payments the same ?