Exponential Models

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Answer Key & Explanations

Exponential Models · Grade 11 · Worksheet 3

1. Sophia analyzed data from a chemical reaction experiment. The concentration of substance A was measured every 6 minutes: (0, 81), (6, 27), (12, 9), (18, 3). Is an exponential model appropriate for this data? Justify your answer by calculating ratios. Answer: A. yes Solution: Check if x-values increase by a constant amount. Time intervals: 6-0 = 6, 12-6 = 6, 18-12 = 6 ✓ (constant interval of 6 minutes) 27 ÷ 81 = 1/3 9 ÷ 27 = 1/3 3 ÷ 9 = 1/3 Since all ratios equal 1/3 (a constant), the data shows exponential decay.
   Full step-by-step solution

   Step 1: Check if x-values increase by a constant amount. Time intervals: 6-0 = 6, 12-6 = 6, 18-12 = 6 ✓ (constant interval of 6 minutes) Step 2: Calculate ratios between consecutive y-values: 27 ÷ 81 = 1/3 9 ÷ 27 = 1/3 3 ÷ 9 = 1/3 Step 3: Since all ratios equal 1/3 (a constant), the data shows exponential decay. Step 4: The exponential model is appropriate because there is a constant ratio between y-values when x increases by a constant amount. Conclusion: Yes, an exponential model is appropriate.

2. Mere analyzed data from a chemical reaction: time (min) = 0, 2, 4, 6; concentration (mg/L) = 64, 16, 4, 1. Is an exponential model appropriate? Justify your answer. Answer: A. yes Solution: Check the time intervals: 0 to 2 is +2 min, 2 to 4 is +2 min, 4 to 6 is +2 min. The time increases by a constant difference of 2 minutes.
   Full step-by-step solution

   Step 1: Check the time intervals: 0 to 2 is +2 min, 2 to 4 is +2 min, 4 to 6 is +2 min. The time increases by a constant difference of 2 minutes. Step 2: Check the concentration ratios between consecutive measurements: From 0 to 2 min: 16 / 64 = 0.25 From 2 to 4 min: 4 / 16 = 0.25 From 4 to 6 min: 1 / 4 = 0.25 Step 3: Since the concentration is multiplied by the same factor (0.25) each time the time increases by 2 minutes, this indicates a constant ratio. Step 4: A constant ratio with equally spaced x-values means an exponential model is appropriate. The answer is Yes.

3. A research team is studying the spread of information through social media. They model the number of people who have seen a viral post using the function P(t) = 5000 / (1 + 49e^(-0.3t)), where t is time in hours after the post was first shared. How many people had seen the post when it was first shared (at t = 0)? Answer: 100 Solution: P(t) = 5000 / (1 + 49 * e^(-0.3 * t)) Substitute t = 0 into the function. P(0) = 5000 / (1 + 49 * e^(-0.3 * 0)) Simplify the exponent. e^(-0.3 * 0) = e^0 We know e^0 = 1.
   Full step-by-step solution

   Let's find how many people had seen the post at t = 0. We are given: P(t) = 5000 / (1 + 49 * e^(-0.3 * t)) Step 1: Substitute t = 0 into the function. P(0) = 5000 / (1 + 49 * e^(-0.3 * 0)) Step 2: Simplify the exponent. e^(-0.3 * 0) = e^0 We know e^0 = 1. Step 3: Replace e^0 with 1. P(0) = 5000 / (1 + 49 * 1) P(0) = 5000 / (1 + 49) Step 4: Add the numbers in the denominator. 1 + 49 = 50 So P(0) = 5000 / 50 Step 5: Perform the division. 5000 / 50 = 100 Therefore, when the post was first shared (t = 0), 100 people had seen it. Final answer: 100

4. A research scientist is studying the growth of a bacteria culture in a lab. The initial population is 500 bacteria, and the culture grows at a continuous rate of 8% per hour. The scientist models the population using the function P(t) = P₀e^(kt), where t is time in hours. Determine how many hours it will take for the bacteria population to reach 2,000 individuals. Answer: 17.33 Solution: P(t) = P₀ * e^(k t) Initial population P₀ = 500 Growth rate k = 8% per hour = 0.08 Final population P(t) = 2000 2000 = 500 * e^(0.08 t) Divide both sides by 500 2000 / 500 = e^(0.08 t) 4 = e^(0.08 t) ln(4) = ln(e^(0.08 t)) ln(4) = 0.08 t t = ln(4) / 0.08 Calculate ln(4) ln(4) = ln(2^2) = 2 *…
   Full step-by-step solution

   We are given the continuous growth model: P(t) = P₀ * e^(k t) Initial population P₀ = 500 Growth rate k = 8% per hour = 0.08 Final population P(t) = 2000 --- **Step 1: Write the equation with given numbers** 2000 = 500 * e^(0.08 t) --- **Step 2: Divide both sides by 500** 2000 / 500 = e^(0.08 t) 4 = e^(0.08 t) --- **Step 3: Take the natural logarithm of both sides** ln(4) = ln(e^(0.08 t)) ln(4) = 0.08 t --- **Step 4: Solve for t** t = ln(4) / 0.08 --- **Step 5: Calculate ln(4)** ln(4) = ln(2^2) = 2 * ln(2) ln(2) ≈ 0.693147 So ln(4) ≈ 2 * 0.693147 = 1.386294 --- **Step 6: Divide by 0.08** t ≈ 1.386294 / 0.08 t ≈ 17.328675 --- **Step 7: Round to two decimal places** t ≈ 17.33 hours --- **Final answer:** 17.33

5. Sophia is analyzing the growth of a certain species of plant in a controlled environment. She records the height of the plant (in centimeters) at the end of each week for five weeks. The data is shown in the table below: Week (x): 0, 1, 2, 3, 4 Height (y): 6, 21, 36, 51, 66 Determine whether a linear model or an exponential model is more appropriate for this data. Justify your answer by checking for a constant difference or a constant ratio between consecutive y-values. Then, if linear, write the linear equation; if exponential, write the exponential equation in the form y = ab^x. Answer: Linear model; y = 15x + 6 Solution: Check for constant difference (linear model). Differences: 21 - 6 = 15, 36 - 21 = 15, 51 - 36 = 15, 66 - 51 = 15. The differences are all 15, so the data has a constant difference.
   Full step-by-step solution

   Step 1: Check for constant difference (linear model). Differences: 21 - 6 = 15, 36 - 21 = 15, 51 - 36 = 15, 66 - 51 = 15. The differences are all 15, so the data has a constant difference. Step 2: Check for constant ratio (exponential model). Ratios: 21/6 = 3.5, 36/21 = 1.714..., 51/36 = 1.416..., 66/51 = 1.294... The ratios are not constant. Step 3: Since the differences are constant, a linear model is appropriate. Step 4: Find the linear equation y = mx + b. The slope m = 15 (the constant difference). When x = 0, y = 6, so the y-intercept b = 6. Step 5: The linear equation is y = 15x + 6. The answer is linear model; y = 15x + 6.

6. A biologist is studying bacterial growth in a lab culture. The initial population is 500 bacteria, and observations show the population doubles every 3 hours. The biologist models this growth with the function P(t) = P₀ * e^(kt), where t is time in hours. Determine the value of the continuous growth rate k for this bacterial population. Answer: ln(2)/3 Solution: We are given that the initial population is 500 bacteria and that the population doubles every 3 hours. P(t) = P₀ * e^(k t) where P₀ = 500, t is in hours, and k is the continuous growth rate we need to find.
   Full step-by-step solution

   We are given that the initial population is 500 bacteria and that the population doubles every 3 hours. The model is: P(t) = P₀ * e^(k t) where P₀ = 500, t is in hours, and k is the continuous growth rate we need to find. --- **Step 1: Use the doubling time information.** If the population doubles every 3 hours, then: P(3) = 2 * P₀ Substitute into the model: 500 * e^(k * 3) = 2 * 500 --- **Step 2: Simplify the equation.** Divide both sides by 500: e^(3k) = 2 --- **Step 3: Solve for k.** Take the natural logarithm of both sides: ln(e^(3k)) = ln(2) 3k = ln(2) Divide both sides by 3: k = ln(2) / 3 --- **Final Answer:** k = ln(2)/3