- Difficulty
- Challenging
- Suggested time
- 7 min
- Marks
- 4
Exponential growth and decay practice question 9
A savings account pays 4 % per year, compounded continuously, and its owner deposits money continuously at $3000 per year. The balance therefore satisfies . If the account opens with $20 000, what is the balance after 10 years?
Give the answer in dollars, to the nearest dollar.
The constant 3000 is what stops from being plain exponential growth. Which constant shift of would absorb it?
The equation is not because of the constant. Let ; then .
So grows exponentially: . Convert back to at the end.
A number, to the nearest dollar.
Theory UBC Math 100 §11.4CLP-1 §3.3Key results
Worked solution
Strategy
A constant term can be absorbed by a shift: adding 3000/0.04 to A produces a quantity that satisfies the pure growth equation. Solve that, then shift back.
Working
Factor out 0.04; 3000/0.04 = 75000.
B(0) = 20000 + 75000.
Shift back and evaluate.
Answer
The answer is about 66,723 dollars.
Check
The balance lies between 59,836, if the deposits earned no interest, and 74,591, if all 30,000 went in at the start, as it must.
Where this goes wrong
Adding the deposits separately, 20000e0.4 + 30000 = 59,836. The deposits also earn interest from the moment they arrive, which is what the shifted exponential accounts for.
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