Word Problems: Functions & Modeling
Learn
Word problems require you to translate real-world scenarios into mathematical models. This lesson focuses on identifying the correct function type and setting up equations from context.
Problem Types You Will Encounter
- Population Growth/Decay: Bacteria, populations, radioactive decay
- Financial Models: Compound interest, investments, depreciation
- Scientific Applications: Half-life, pH levels, sound intensity
- Temperature Models: Newton's Law of Cooling
- Logistic Growth: Population with carrying capacity
Key Phrases to Recognize
| Phrase | Model Type | Formula Hint |
|---|---|---|
| "doubles every..." | Exponential Growth | P(t) = P_0 * 2^(t/d) |
| "half-life of..." | Exponential Decay | A(t) = A_0 * (1/2)^(t/h) |
| "compounded continuously" | Continuous Growth | A = Pe^(rt) |
| "decreases by x% per year" | Exponential Decay | A(t) = A_0 * (1-r)^t |
| "increases by x% per year" | Exponential Growth | A(t) = A_0 * (1+r)^t |
Worked Examples
Example 1: Investment Problem
Problem: Sarah invests $5,000 in an account that pays 4.5% annual interest compounded monthly. How much will she have after 10 years? When will her investment double?
Show Solution
Part A: Find the amount after 10 years
Formula: A = P(1 + r/n)^(nt) where n = 12 (monthly)
A = 5000(1 + 0.045/12)^(12*10)
A = 5000(1.00375)^120
A = 5000 * 1.5669 = $7,834.50
Part B: Find doubling time
10000 = 5000(1.00375)^(12t)
2 = (1.00375)^(12t)
ln(2) = 12t * ln(1.00375)
t = ln(2) / (12 * ln(1.00375)) = 15.44 years
Example 2: Carbon Dating
Problem: Carbon-14 has a half-life of 5,730 years. A fossil contains 30% of its original Carbon-14. How old is the fossil?
Show Solution
Use the half-life formula: A = A_0 * (1/2)^(t/5730)
0.30A_0 = A_0 * (1/2)^(t/5730)
0.30 = (0.5)^(t/5730)
ln(0.30) = (t/5730) * ln(0.5)
t = 5730 * ln(0.30) / ln(0.5)
t = 5730 * (-1.204) / (-0.693) = 9,953 years old
Example 3: Earthquake Intensity
Problem: The Richter scale is defined as M = log(I/I_0). An earthquake measures 6.2 on the Richter scale. How many times more intense is this than a 4.5 magnitude earthquake?
Show Solution
For M = 6.2: 6.2 = log(I_1/I_0), so I_1 = I_0 * 10^6.2
For M = 4.5: 4.5 = log(I_2/I_0), so I_2 = I_0 * 10^4.5
Ratio: I_1/I_2 = 10^6.2 / 10^4.5 = 10^(6.2-4.5) = 10^1.7
= 50.12 times more intense
Practice Problems
Solve these word problems. Show your work and set up equations before solving.
Problem 1: A town had a population of 12,000 in 2010. The population grows at 2.5% per year. What will the population be in 2025?
Show Answer
P(15) = 12000 * (1.025)^15 = 12000 * 1.4483 = 17,380 people
Problem 2: A patient is given 200 mg of medication. Every 4 hours, 40% of the medication in the bloodstream is eliminated. How much remains after 12 hours?
Show Answer
After 12 hours = 3 periods of 4 hours. A = 200 * (0.6)^3 = 200 * 0.216 = 43.2 mg
Problem 3: Sound intensity is measured in decibels: dB = 10 log(I/I_0). A concert is 110 dB and normal conversation is 60 dB. How many times more intense is the concert?
Show Answer
Difference = 110 - 60 = 50 dB. Ratio = 10^(50/10) = 10^5 = 100,000 times more intense
Problem 4: A car worth $35,000 depreciates at 18% per year. In how many years will it be worth less than $10,000?
Show Answer
10000 = 35000 * (0.82)^t, ln(10000/35000) = t * ln(0.82), t = ln(0.2857)/ln(0.82) = 6.32 years
Problem 5: A cup of coffee cools according to T(t) = 20 + 70e^(-0.05t), where T is in Celsius and t is in minutes. How long until the coffee reaches 50 degrees?
Show Answer
50 = 20 + 70e^(-0.05t), 30 = 70e^(-0.05t), ln(30/70) = -0.05t, t = -ln(0.4286)/0.05 = 16.95 minutes
Problem 6: An investment of $8,000 grows to $12,500 in 8 years with continuous compounding. What is the annual interest rate?
Show Answer
12500 = 8000 * e^(8r), ln(12500/8000) = 8r, r = ln(1.5625)/8 = 0.0558 = 5.58%
Problem 7: The pH of a solution is given by pH = -log[H+]. If one solution has pH 3 and another has pH 7, how many times more acidic is the first solution?
Show Answer
Difference in [H+] = 10^(7-3) = 10^4 = 10,000 times more acidic
Problem 8: Strontium-90 has a half-life of 29 years. How long will it take for 90% of a sample to decay?
Show Answer
0.1 = (0.5)^(t/29), ln(0.1) = (t/29)*ln(0.5), t = 29 * ln(0.1)/ln(0.5) = 96.3 years
Problem 9: A colony of 250 rabbits grows according to P(t) = 250e^(0.15t). After how many months will the population reach 1,000?
Show Answer
1000 = 250e^(0.15t), 4 = e^(0.15t), ln(4) = 0.15t, t = ln(4)/0.15 = 9.24 months
Problem 10: A savings account offers 3.2% interest compounded quarterly. How much must you deposit now to have $20,000 in 15 years?
Show Answer
20000 = P(1 + 0.032/4)^(60), P = 20000/(1.008)^60 = 20000/1.6140 = $12,392.40
Check Your Understanding
Reflect on these questions to assess your mastery.
- How do you recognize whether a problem involves growth or decay?
- What is the difference between compounding monthly and compounding continuously?
- When solving for time in an exponential equation, why must you use logarithms?
- How does the half-life relate to the decay constant k in the formula A = A_0 * e^(-kt)?
Next Steps
- Practice identifying function types from word problem descriptions
- Review the Common Mistakes lesson to avoid typical errors
- Create a reference sheet of formulas for different problem types
- Work through SAT/ACT practice problems involving exponential functions