K

Exercise Set 1.1 Notes (Patterns and Exponential Growth)

Pattern Problems (EXERCISE SET 1.1)

  • 1. A, C, E, G, I, -> Next: K

    • Reason: These are every other letter of the alphabet, increasing by 2. A (1), C (3), E (5), G (7), I (9), so the next is K (11).

  • 2. 15, 10, 14, 10, 13, 10, -> Next: 12

    • Reason: The sequence alternates between a decreasing-by-1 series (15, 14, 13,
      ) and a constant 10. After 13, 10, the next odd term should be 12, followed by 10 again.
    • Extended pattern: 15, 10, 14, 10, 13, 10, 12, 10,

  • 3. 3, 6, 12, 24, 48, 96, -> Next: 192

    • Reason: Each term is multiplied by 2 (3 \times 2 = 6, 6 \times 2 = 12, etc.). Hence next is 96 \times 2 = 192.

  • 4. 27, 30, 33, 36, 39, -> Next: 42

    • Reason: Arithmetic progression with common difference +3.

  • 5. 41, 39, 37, 35, 33, -> Next: 31

    • Reason: Arithmetic progression with common difference \text{-}2.
Exponential Growth Problems (A = P e^{rt})
  • Concept note:
    • For continuous compounding, the amount after time t is given by A = P e^{rt} where P is the principal, r is the annual interest rate (in decimal), and t is time in years.
    • To solve for a missing quantity, rearrange as needed and use natural logarithms: e.g., for P, r, t given and A unknown, A = P e^{rt}; for P = A e^{-rt}; for r given A, P, t: r = \frac{1}{t} \ln\left(\frac{A}{P}\right); for t given A, P, r: t = \frac{1}{r} \ln\left(\frac{A}{P}\right).
  • 6. P = 680,000; r = 12% per year; t = 8 years
    • Formula: A = P e^{rt}
    • Convert r to decimal: r = 0.12; compute rt: rt = 0.12 \times 8 = 0.96
    • Amount: A = 680{,}000 \times e^{0.96}
    • Numerical evaluation: e^{0.96} \approx 2.6117, so A \approx 680{,}000 \times 2.6117 \approx 1{,}775{,}953
    • Result: A \approx 1{,}775{,}953\$ (about 1.78 million)
  • 7. A = 1,240,000; r = 8% per year; t = 30 years
    • Solve for P: P = \frac{A}{e^{rt}} = A e^{-rt}
    • Compute rt: rt = 0.08 \times 30 = 2.4
    • Exponential factor: e^{-rt} = e^{-2.4} \approx 0.090718
    • Principal: P \approx 1{,}240{,}000 \times 0.090718 \approx 112{,}490
    • Result: P \approx 112{,}490\$ (about 112.5 thousand)
  • 8. A = 786,000; P = 247,000; t = 17 years
    • Solve for r: r = \frac{1}{t} \ln\left(\frac{A}{P}\right)
    • Ratio: \frac{A}{P} = \frac{786{,}000}{247{,}000} \approx 3.183
    • Natural log: \ln(3.183) \approx 1.157
    • Rate: r \approx \frac{1.157}{17} \approx 0.0680
    • Result: r \approx 6.80\%\text{ per year}
  • 9. A = 731,093; P = 525,600; r = 3% per year
    • Solve for t: t = \frac{1}{r} \ln\left(\frac{A}{P}\right)
    • Ratio: \frac{A}{P} = \frac{731{,}093}{525{,}600} \approx 1.391
    • Natural log: \ln(1.391) \approx 0.330
    • Time: t \approx \frac{0.330}{0.03} \approx 11.0\text{ years}
  • Result: t \approx 11\text{ years}
Population Growth (Bacteria)
  • Scenario: A population doubles every 6 hours. Model with continuous growth: P(t) = P_0 e^{rt}
  • Doubling condition: P(6) = 2 P_0 \Rightarrow e^{r\cdot 6} = 2 \Rightarrow r = \frac{\ln 2}{6} \approx 0.1155\ \text{hour}^{-1}
  • Growth rate: r \approx 0.1155 \text{ per hour} \quad (\approx 11.55\% \text{ per hour})
  • How many after two days (48 hours)?
    • Using the model: P(48) = P_0 e^{r \cdot 48} = P_0 e^{\frac{\ln 2}{6} \cdot 48} = P_0 e^{8 \ln 2} = P_0 \cdot 2^{8} = 100 \cdot 256 = 25{,}600
    • Result: after 2 days, there are 25,600 bacteria (assuming continuous growth model aligns with To get the answers for these problems, different mathematical reasoning and formulas are applied based on the type of problem:
Pattern Problems

For pattern problems (like questions 1-5), the key is to identify the rule governing the sequence. This often involves looking at the differences between consecutive terms, the ratios between them, or specific operations (addition, subtraction, multiplication, division, alternation).

  1. **A, C, E, G, I,
    **: The pattern is every other letter of the alphabet, increasing by 2 positions. So, after I (9th letter), the next is K (11th letter).
  2. **15, 10, 14, 10, 13, 10,
    **: This is an alternating pattern. One sub-sequence decreases by 1 (15, 14, 13,
    ), and the other is constant (10, 10, 10,
    ). Following 13, 10, the next term in the decreasing sequence is 12.
  3. **3, 6, 12, 24, 48, 96,
    **: The pattern is multiplication by 2. Each term is twice the previous term (3 \times 2 = 6, 6 \times 2 = 12, etc.). So, the next term is 96 \times 2 = 192.
  4. **27, 30, 33, 36, 39,
    **: This is an arithmetic progression where each term increases by 3. So, 39 + 3 = 42.
  5. **41, 39, 37, 35, 33,
    **: This is also an arithmetic progression, but each term decreases by 2. So, 33 - 2 = 31.
Exponential Growth Problems (A = P e^{rt})

For problems involving continuous exponential growth or compounding (like questions 6-9 and population growth), the formula A = P e^{rt} is used.

  • A is the final amount/value.
  • P is the principal amount/initial value.
  • e is Euler's number (approximately 2.71828).
  • r is the annual interest rate or growth rate (expressed as a decimal).
  • t is the time in years (or hours, depending on the rate unit).
    To find a missing variable, the formula is rearranged and natural logarithms (ln) are often used.
  • To find A (future value): Use A = P e^{rt} directly. Convert the percentage rate to a decimal.
    • Example: For P = 680,000, r = 12% (0.12), t = 8 years, calculate rt = 0.12 \times 8 = 0.96, then A = 680{,}000 \times e^{0.96}.
  • To find P (initial principal): Rearrange to P = A e^{-rt}.
    • Example: For A = 1,240,000, r = 8% (0.08), t = 30 years, calculate rt = 0.08 \times 30 = 2.4, then P = 1{,}240{,}000 \times e^{-2.4}.
  • To find r (growth rate): Rearrange to r = \frac{1}{t} \times \text{ln}(\frac{A}{P}).
    • Example: For A = 786,000, P = 247,000, t = 17 years, calculate \frac{A}{P} = \frac{786{,}000}{247{,}000} . Then find the natural logarithm of this ratio, and divide by t.
  • To find t (time): Rearrange to t = \frac{1}{r} \times \text{ln}(\frac{A}{P}).
    • Example: For A = 731,093, P = 525,600, r = 3% (0.03), calculate \frac{A}{P} = \frac{731{,}093}{525{,}600} . Then find the natural logarithm of this ratio, and divide by r.
Population Growth (Bacteria)

This is a specific case of exponential growth. When a doubling time is given, you first need to find the continuous growth rate r. If a population doubles in time T, then P(T) = 2 P_0. Using P(T) = P_0 e^{rT}, we get 2 P_0 = P_0 e^{rT} , which simplifies to 2 = e^{rT} . Taking the natural logarithm of both sides gives \text{ln}(2) = rT, so r = \frac{\text{ln}(2)}{T} .

  • For bacteria doubling every 6 hours, T = 6, so r = \frac{\text{ln}(2)}{6} . This value of r is then used in the general formula P(t) = P_0 e^{rt} to predict the population at any other time t.
  • Example: To find the population after 2 days (48 hours), use t = 48 and the calculated r. So, P(48) = P_0 e^{(\frac{\text{ln}(2)}{6}) \times 48 = P_0 e^{8 \ln 2} = P_0 \cdot 2^8 = 256 P_0}.