Model Comparison

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Model Comparison · Grade 11 · Worksheet 2

1. Isabella is a materials scientist comparing the performance of three different cooling systems for an industrial reactor. System L (linear) reduces temperature at a constant rate: T(t) = 327 - 17t, where T is temperature in degrees Celsius and t is time in minutes. System Q (quadratic) follows the model: T(t) = 327 - 2t². System E (exponential) follows the model: T(t) = 327(0.92)^t. After 12 minutes, which system will have reduced the temperature the most? Answer: System Q (quadratic) Solution: Evaluate System L (linear) at t = 12. T(12) = 327 - 17(12) = 327 - 204 = 123 degrees Celsius. Evaluate System Q (quadratic) at t = 12.
   Full step-by-step solution

   Step 1: Evaluate System L (linear) at t = 12. T(12) = 327 - 17(12) = 327 - 204 = 123 degrees Celsius. Step 2: Evaluate System Q (quadratic) at t = 12. T(12) = 327 - 2(12)² = 327 - 2(144) = 327 - 288 = 39 degrees Celsius. Step 3: Evaluate System E (exponential) at t = 12. T(12) = 327(0.92)^12. First compute (0.92)^12. Since 0.92² = 0.8464, 0.92⁴ = (0.8464)² ≈ 0.7164, 0.92⁸ = (0.7164)² ≈ 0.5132, and 0.92¹² = 0.92⁸ × 0.92⁴ ≈ 0.5132 × 0.7164 ≈ 0.3677. Then T(12) = 327 × 0.3677 ≈ 120.2 degrees Celsius. Step 4: Compare the final temperatures. System L: 123°C System Q: 39°C System E: 120.2°C System Q gives the lowest temperature, meaning it has reduced the temperature the most. The answer is System Q (quadratic).

2. Mere is comparing three functions: f(x) = 6x + 4 (linear), g(x) = 2x² + 8 (quadratic), and h(x) = 4^x (exponential). For large x, which function grows fastest? Answer: h(x) = 4^x Solution: Analyze f(x) = 6x + 4 (linear). As x increases, f(x) grows by a constant 6 for each unit increase in x. For large x, f(x) is approximately 6x.
   Full step-by-step solution

   Step 1: Analyze f(x) = 6x + 4 (linear). As x increases, f(x) grows by a constant 6 for each unit increase in x. For large x, f(x) is approximately 6x. Step 2: Analyze g(x) = 2x² + 8 (quadratic). As x increases, g(x) grows proportionally to x². For large x, g(x) is approximately 2x², which grows faster than 6x. Step 3: Analyze h(x) = 4^x (exponential). As x increases, h(x) multiplies by 4 for each unit increase in x. For large x, h(x) grows much faster than any polynomial function. Step 4: Compare growth rates. For large x, exponential functions with base > 1 always outpace polynomial functions. Since 4^x grows faster than 2x² and 6x for sufficiently large x, h(x) = 4^x grows fastest. The answer is h(x) = 4^x.

3. Aroha is comparing three functions: f(x) = 9x + 5 (linear), g(x) = 3x² + 7 (quadratic), and h(x) = 3^x (exponential). For large values of x, which function grows the fastest? Answer: h(x) = 3^x Solution: Consider f(x) = 9x + 5 (linear). For large x, f(x) grows at a constant rate of 9 per unit increase in x. Consider g(x) = 3x² + 7 (quadratic).
   Full step-by-step solution

   Step 1: Consider f(x) = 9x + 5 (linear). For large x, f(x) grows at a constant rate of 9 per unit increase in x. For example, at x = 10, f(10) = 95; at x = 100, f(100) = 905. Step 2: Consider g(x) = 3x² + 7 (quadratic). For large x, g(x) grows proportionally to x². At x = 10, g(10) = 307; at x = 100, g(100) = 30,007. This is much faster than linear growth. Step 3: Consider h(x) = 3^x (exponential). For large x, h(x) grows by multiplying by 3 each time x increases by 1. At x = 10, h(10) = 59,049; at x = 20, h(20) = 3,486,784,401. This far exceeds the quadratic growth. Step 4: Compare growth rates. For any polynomial (like linear or quadratic) and any exponential with base > 1, the exponential eventually grows faster. Here, 3^x grows faster than 3x² + 7 and 9x + 5 for sufficiently large x. Therefore, h(x) = 3^x grows the fastest for large x.

4. Compare f(x)=15x+20, g(x)=5x²+10, h(x)=4^x for large x. Which function grows fastest? Answer: h(x)=4^x Solution: Analyze f(x)=15x+20 (linear function). As x increases, f(x) grows by a constant 15 for each unit increase in x. For large x, f(x) is approximately 15x.
   Full step-by-step solution

   Step 1: Analyze f(x)=15x+20 (linear function). As x increases, f(x) grows by a constant 15 for each unit increase in x. For large x, f(x) is approximately 15x. Step 2: Analyze g(x)=5x²+10 (quadratic function). As x increases, g(x) grows proportionally to x². For large x, g(x) is approximately 5x². Since x² grows faster than x for large x, g(x) will eventually exceed f(x). Step 3: Analyze h(x)=4^x (exponential function with base 4 > 1). As x increases, h(x) multiplies by 4 for each unit increase in x. For large x, h(x) grows much faster than any polynomial function. Step 4: Compare growth rates. For sufficiently large x, exponential functions always outpace polynomial functions (linear or quadratic). Therefore, h(x)=4^x grows fastest. The answer is h(x)=4^x.

5. Dr. Rodriguez is comparing three different models for predicting city population growth over time. Model A is linear: P(t) = 5,000t + 100,000, Model B is quadratic: P(t) = 200t² + 100,000, and Model C is exponential: P(t) = 100,000(1.04)^t, where t represents years and P(t) represents population. After how many years will the exponential model first predict a higher population than both the linear and quadratic models? Answer: 12 Solution: Step 1: Set up the comparison equations We need to find when C(t) > A(t) and C(t) > B(t) Step 2: Compare exponential and linear models 100,000(1.04)^t > 5,000t + 100,000 Divide both sides by 100,000: (1.04)^t > 0.05t + 1 Test values: t = 10: (1.04)^10 = 1.480, 0.05(10) + 1 = 1.5 → 1.480 < 1.5 t…
   Full step-by-step solution

   Step 1: Set up the comparison equations We need to find when C(t) > A(t) and C(t) > B(t) Step 2: Compare exponential and linear models 100,000(1.04)^t > 5,000t + 100,000 Divide both sides by 100,000: (1.04)^t > 0.05t + 1 Test values: t = 10: (1.04)^10 = 1.480, 0.05(10) + 1 = 1.5 → 1.480 < 1.5 t = 11: (1.04)^11 = 1.539, 0.05(11) + 1 = 1.55 → 1.539 < 1.55 t = 12: (1.04)^12 = 1.601, 0.05(12) + 1 = 1.6 → 1.601 > 1.6 ✓ Step 3: Compare exponential and quadratic models 100,000(1.04)^t > 200t² + 100,000 Divide both sides by 100,000: (1.04)^t > 0.002t² + 1 Test values: t = 10: (1.04)^10 = 1.480, 0.002(100) + 1 = 1.2 → 1.480 > 1.2 ✓ t = 11: (1.04)^11 = 1.539, 0.002(121) + 1 = 1.242 → 1.539 > 1.242 ✓ t = 12: (1.04)^12 = 1.601, 0.002(144) + 1 = 1.288 → 1.601 > 1.288 ✓ Step 4: Verify both conditions are met at t = 12 At t = 12: C(12) > A(12) and C(12) > B(12) At t = 11: C(11) > B(11) but C(11) < A(11) Therefore, the exponential model first exceeds both other models at t = 12 years.

6. A biologist is studying bacterial growth in a lab culture. The initial population is 500 bacteria. After 3 hours, the population reaches 4,000 bacteria. Assuming exponential growth, determine the time it will take for the bacterial population to reach 50,000 bacteria. Round your answer to the nearest tenth of an hour. Answer: 6.6 Solution: Write the exponential growth formula. P(t) = P_0 \cdot e^{kt} - \( P_0 \) = initial population - \( k \) = growth constant - \( t \) = time in hours P_0 = 500 P(3) = 4000 Find \( k \) using the data after 3 hours.
   Full step-by-step solution

   Let's solve this step-by-step. --- **Step 1: Write the exponential growth formula.** For exponential growth: \[ P(t) = P_0 \cdot e^{kt} \] where: - \( P_0 \) = initial population - \( k \) = growth constant - \( t \) = time in hours Given: \[ P_0 = 500 \] \[ P(3) = 4000 \] --- **Step 2: Find \( k \) using the data after 3 hours.** \[ 4000 = 500 \cdot e^{k \cdot 3} \] Divide both sides by 500: \[ 8 = e^{3k} \] Take natural log of both sides: \[ \ln(8) = 3k \] \[ k = \frac{\ln(8)}{3} \] Since \( \ln(8) = \ln(2^3) = 3 \ln(2) \): \[ k = \frac{3 \ln(2)}{3} = \ln(2) \] So \( k = \ln(2) \). --- **Step 3: Write the population function.** \[ P(t) = 500 \cdot e^{t \ln(2)} \] But \( e^{t \ln(2)} = 2^t \), so: \[ P(t) = 500 \cdot 2^t \] --- **Step 4: Find \( t \) when \( P(t) = 50000 \).** \[ 50000 = 500 \cdot 2^t \] Divide both sides by 500: \[ 100 = 2^t \] Take log (base 2 or natural log) of both sides: \[ \ln(100) = t \cdot \ln(2) \] \[ t = \frac{\ln(100)}{\ln(2)} \] --- **Step 5: Calculate \( t \).** \[ \ln(100) = \ln(10^2) = 2 \ln(10) \] Numerically: \[ \ln(100) \approx 4.605170 \] \[ \ln(2) \approx 0.693147 \] \[ t \approx \frac{4.605170}{0.693147} \approx 6.644 \] --- **Step 6: Round to nearest tenth of an hour.** \[ t \approx 6.6 \ \text{hours} \] --- **Final Answer:** 6.6

7. Compare f(x) = 14x + 22, g(x) = 2x² + 18, h(x) = 6^x for large x. Which function grows fastest? Answer: h(x) = 6^x Solution: Identify the function types. f(x) = 14x + 22 is linear (degree 1). g(x) = 2x² + 18 is quadratic (degree 2).
   Full step-by-step solution

   Step 1: Identify the function types. f(x) = 14x + 22 is linear (degree 1). g(x) = 2x² + 18 is quadratic (degree 2). h(x) = 6^x is exponential (base 6 > 1). Step 2: Compare linear and quadratic. For large x, the quadratic term 2x² dominates the linear term 14x, so g(x) > f(x) for sufficiently large x. Step 3: Compare quadratic and exponential. For large x, an exponential function with base > 1 grows faster than any polynomial function. Since 6^x grows much faster than 2x² for large x, h(x) > g(x) for sufficiently large x. Step 4: Order the functions by growth rate for large x: f(x) < g(x) < h(x). Therefore, h(x) = 6^x grows fastest for large x.

8. Compare f(x)=17x+22, g(x)=7x²+12, h(x)=2^x for large x. Which function grows fastest? Answer: h(x)=2^x Solution: Analyze f(x)=17x+22 (linear function). For large x, f(x) increases by a constant 17 for each unit increase in x. Its growth is steady but slow compared to other types.
   Full step-by-step solution

   Step 1: Analyze f(x)=17x+22 (linear function). For large x, f(x) increases by a constant 17 for each unit increase in x. Its growth is steady but slow compared to other types. Step 2: Analyze g(x)=7x²+12 (quadratic function). For large x, g(x) grows like 7x². The increase per unit x is roughly 14x, which itself grows with x, making it faster than linear. Step 3: Analyze h(x)=2^x (exponential function). For large x, h(x) doubles with each unit increase in x. This multiplicative growth is far more powerful than any polynomial growth. Step 4: Compare for large x. For x=10: f(10)=17(10)+22=192, g(10)=7(100)+12=712, h(10)=2^10=1024. For x=20: f(20)=17(20)+22=362, g(20)=7(400)+12=2812, h(20)=2^20=1,048,576. The exponential function h(x) far exceeds the others for large x. The answer is h(x)=2^x.