- Difficulty
- Capstone
- Suggested time
- 9 min
- Marks
- 5
Exponential growth and decay practice question 10
A retiree has $100 000 in an account paying 5 % per year, compounded continuously, and withdraws money continuously at $6000 per year, so the balance satisfies . Solve for , and find how many years it takes for the money to run out.
Give the time in years.
Interest adds to the balance while withdrawals take a fixed amount away. Is there a balance at which the two exactly cancel, and what happens to the gap from it?
Rewrite as . The shifted quantity satisfies the pure growth equation.
Because the account starts below $120 000, the shifted quantity is negative and its magnitude grows; the money runs out when .
An exact expression, or a decimal.
Theory UBC Math 100 §11.4CLP-1 §3.3Key results
Worked solution
Strategy
Shift by the equilibrium balance 6000/0.05 = 120 000, at which interest exactly covers the withdrawals. Below that level the shortfall grows exponentially; solve for the time when the balance reaches zero.
Working
Factor out 0.05.
Initial shortfall of 20 000 grows at 5 %.
The account is empty after about 35.8 years.
Answer
The answer is A of t equals 120,000 minus 20,000 e to the 0.05 t, and the money runs out after 20 natural log 6, about 35.8, years.
Check
A(0) = 120,000 − 20,000 = 100,000, and with no interest the money would last only 100,000/6000 ≈ 16.7 years, so 35.8 is plausible.
Where this goes wrong
Dividing 100 000 by 6000 to get about 16.7 years, ignoring interest, or concluding the money never runs out because the balance is “growing at 5 %”. Interest on 100 000 is only 5000, less than the 6000 withdrawn, so the balance shrinks, and faster every year.
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