Let's solve each part step-by-step: --- ### (c) Benoit's Model: \( B(t) = 33.5(1.005)^t \), where \( t \) is years after 2000, \( 25 \leq t \leq 100 \). #### (i) Predict population in 2100 Year 2100 is **100 years after 2000**, so \( t = 100 \). \[ B(100) = 33.5 \times (1.005)^{100} \] Calculate \( (1.005)^{100} \): \[ (1.005)^{100} \approx e^{100 \times \ln(1.005)} \approx e^{0.5} \approx 1.6487 \] So, \[ B(100) \approx 33.5 \times 1.6487 \approx 55.13 \] **Final answer:** The predicted population of Canada in 2100 using Benoit's model is **about 55.13 million**. --- #### (ii) Interpret the value 1.005 in context - The value **1.005** is the **annual growth factor** for the population. - This means that **every year, the population grows by 0.5%** compared to the previous year. **Interpretation:** **1.005** represents a **0.5% annual increase** in the population of Canada in Benoit's model. --- ### (d) Cecilia's Model: \( C(t) = \dfrac{62}{1 + e^{-0.07t}} \) Use Cecilia’s model to predict the population in 2100 (\( t = 100 \)): \[ C(100) = \frac{62}{1 + e^{-0.07 \times 100}} = \frac{62}{1 + e^{-7}} \] Calculate \( e^{-7} \approx 0.0009119 \): \[ C(100) \approx \frac{62}{1 + 0.0009119} \approx \frac{62}{1.0009119} \approx 61.94 \] **Final answer:** The predicted population in 2100 using Cecilia’s model is **about 61.94 million**. --- ### (e) Year in which the difference between predictions is greatest Find \( t \) where \( |B(t) - C(t)| \) is maximized for \( 25 \leq t \leq 100 \). - As \( t \to 100 \), \( B(t) \) keeps growing exponentially, but \( C(t) \) approaches its maximum asymptote of 62. - At \( t = 25 \) (2025), calculate both: - \( B(25) = 33.5 \times (1.005)^{25} \approx 33.5 \times 1.133 \approx 37.97 \) - \( C(25) = \frac{62}{1 + e^{-0.07 \times 25}} = \frac{62}{1 + e^{-1.75}} \approx \frac{62}{1 + 0.1738} \approx \frac{62}{1.1738} \approx 52.85 \) - Difference: \( |37.97 - 52.85| = 14.88 \) - At \( t = 100 \), difference: \( |55.13 - 61.94| = 6.81 \) - Try a few midpoints (e.g., \( t = 50 \), \( t = 75 \)): - \( t = 50 \): - \( B(50) = 33.5 \times (1.005)^{50} \approx 33.5 \times 1.284 \approx 43.01 \) - \( C(50) = \frac{62}{1 + e^{-3.5}} \approx \frac{62}{1 + 0.0302} \approx \frac{62}{1.0302} \approx 60.21 \) - Difference: \( |43.01 - 60.21| = 17.20 \) - \( t = 75 \): - \( B(75) = 33.5 \times (1.005)^{75} \approx 33.5 \times 1.455 \approx 48.74 \) - \( C(75) = \frac{62}{1 + e^{-5.25}} \approx \frac{62}{1 + 0.00525} \approx \frac{62}{1.00525} \approx 61.68 \) - Difference: \( |48.74 - 61.68| = 12.94 \) So the greatest difference is at \( t = 50 \), about **17.2 million**. **Final answer:** The year in which the difference is greatest is **2050**. --- ### (f) Find the value of #### (i) \( B'(75) \): \[ B(t) = 33.5(1.005)^t \] \[ B'(t) = 33.5 \cdot (1.005)^t \cdot \ln(1.005) \] \[ B'(75) = 33.5 \cdot (1.005)^{75} \cdot \ln(1.005) \] From before, \( (1.005)^{75} \approx 1.455 \): \[ B'(75) \approx 33.5 \cdot 1.455 \cdot 0.004987 = 48.74 \cdot 0.004987 \approx 0.243 \] #### (ii) \( C'(75) \): \[ C(t) = \frac{62}{1 + e^{-0.07t}} \] Let \( u = 1 + e^{-0.07t} \), so \[ C'(t) = -62 \cdot u^{-2} \cdot \frac{d}{dt}(u) \] \[ \frac{d}{dt}(u) = \frac{d}{dt}(1 + e^{-0.07t}) = -0.07 e^{-0.07t} \] \[ C'(t) = -62 \cdot u^{-2} \cdot (-0.07 e^{-0.07t}) = 62 \cdot 0.07 \cdot \frac{e^{-0.07t}}{(1 + e^{-0.07t})^2} \] \[ C'(75) = 62 \times 0.07 \times \frac{e^{-0.07 \times 75}}{[1 + e^{-0.07 \times 75}]^2} \] Calculate: - \( -0.07 \times 75 = -5.25 \) - \( e^{-5.25} \approx 0.00525 \) - \( 1 + 0.00525 = 1.00525 \) - \( (1.00525)^2 \approx 1.0105 \) So, \[ C'(75) \approx 62 \times 0.07 \times \frac{0.00525}{1.0105} \] \[ = 4.34 \times \frac{0.00525}{1.0105} \] \[ = 4.34 \times 0.005195 \approx 0.0226 \] --- ### (g) Compare and interpret - \( B'(75) \approx 0.243 \): **Benoit's model** predicts the population is increasing by **0.243 million people per year** at \( t=75 \) (i.e., in 2075). - \( C'(75) \approx 0.023 \): **Cecilia's model** predicts the population is increasing by **0.023 million people per year** at the same time. **Interpretation:** - **Benoit’s model** predicts a much faster population growth in 2075 than **Cecilia’s model**. This reflects that Benoit’s model is exponential (keeps growing), while Cecilia’s model approaches a maximum (logistic growth, slowing down). --- ## **Summary Table of Answers** | Part | Answer | |------|--------| | (c)(i) | \( B(100) \approx 55.13 \) million | | (c)(ii) | 1.005 represents a 0.5% annual population growth | | (d) | \( C(100) \approx 61.94 \) million | | (e) | Greatest difference at \( t=50 \) (year 2050) | | (f)(i) | \( B'(75) \approx 0.243 \) million/year | | (f)(ii) | \( C'(75) \approx 0.023 \) million/year | | (g) | Benoit's model predicts much faster growth than Cecilia's in 2075 |

Let's go through each part carefully with detailed calculations, explanations, and final answers. --- ### **Part (c)(i): Predict the population of Canada in 2100 using Benoit's model** **Given:** \[ B(t) = 33.5 \times (1.005)^t \] where: - \( t \) = years after 2000, - \( t \) ranges from 25 to 100, - Year 2100 corresponds to \( t = 100 \). **Calculate \( B(100) \):** \[ B(100) = 33.5 \times (1.005)^{100} \] **Step 1: Calculate \( (1.005)^{100} \).** Use the natural logarithm for approximation: \[ (1.005)^{100} = e^{100 \times \ln(1.005)} \] Calculate \( \ln(1.005) \): \[ \ln(1.005) \approx 0.004987 \] Then: \[ 100 \times 0.004987 = 0.4987 \] Now: \[ e^{0.4987} \approx \text{use calculator or approximation} \] Since \( e^{0.5} \approx 1.6487 \), and \( 0.4987 \) is close to 0.5, approximate: \[ (1.005)^{100} \approx 1.6487 \] **Step 2: Calculate \( B(100) \):** \[ B(100) \approx 33.5 \times 1.6487 \approx 55.3 \text{ million} \] **Final answer:** \[ \boxed{\textbf{Approximately 55.3 million in 2100}} \] --- ### **Part (c)(ii): Interpret the value 1.005 in Benoit's model** **Explanation:** - The model uses \( (1.005)^t \), where \( t \) is years after 2000. - The base **1.005** indicates the **annual growth factor**. - This means each year, the population **multiplies by 1.005**, which is a **0.5% increase per year**. **In context:** > The value **1.005** represents a **0.5% annual growth rate** in the population of Canada according to Benoit's model. --- ### **Part (d): Use Cecilia's model to predict the population in 2100** **Given Cecilia's model:** \[ C(t) = \frac{62}{1 + e^{-0.07 t}} \] where: - \( t = 100 \) (year 2100). **Calculate \( C(100) \):** \[ C(100) = \frac{62}{1 + e^{-0.07 \times 100}} = \frac{62}{1 + e^{-7}} \] **Step 1: Calculate \( e^{-7} \):** \[ e^{-7} \approx 0.0009119 \] **Step 2: Compute denominator:** \[ 1 + e^{-7} \approx 1 + 0.0009119 \approx 1.0009119 \] **Step 3: Calculate \( C(100) \):** \[ C(100) \approx \frac{62}{1.0009119} \approx 61.94 \text{ million} \] **Final answer:** \[ \boxed{\textbf{Approximately 61.94 million in 2100}} \] --- ### **Part (e): Year where the difference between the predictions is greatest** We analyze the difference: \[ D(t) = |B(t) - C(t)| \] Calculate at key points: - At \( t=25 \) (2025): \[ B(25) = 33.5 \times (1.005)^{25} \] \[ (1.005)^{25} = e^{25 \times 0.004987} = e^{0.1247} \approx 1.133 \] \[ B(25) \approx 33.5 \times 1.133 \approx 37.97 \] \[ C(25) = \frac{62}{1 + e^{-0.07 \times 25}} = \frac{62}{1 + e^{-1.75}} \] Calculate \( e^{-1.75} \): \[ e^{-1.75} \approx 0.1738 \] Denominator: \[ 1 + 0.1738 = 1.1738 \] Population: \[ C(25) \approx \frac{62}{1.1738} \approx 52.85 \] Difference: \[ |37.97 - 52.85| \approx 14.88 \] - At \( t=50 \): \[ B(50) = 33.5 \times (1.005)^{50} \] \[ (1.005)^{50} = e^{50 \times 0.004987} = e^{0.2494} \approx 1.283 \] \[ B(50) \approx 33.5 \times 1.283 \approx 43.00 \] \[ C(50) = \frac{62}{1 + e^{-0.07 \times 50}} = \frac{62}{1 + e^{-3.5}} \] \( e^{-3.5} \approx 0.0302 \) Denominator: \[ 1 + 0.0302 = 1.0302 \] Population: \[ C(50) \approx \frac{62}{1.0302} \approx 60.20 \] Difference: \[ |43.00 - 60.20| \approx 17.20 \] - At \( t=75 \): \[ (1.005)^{75} = e^{75 \times 0.004987} = e^{0.374} \approx 1.454 \] \[ B(75) \approx 33.5 \times 1.454 \approx 48.74 \] \[ C(75) = \frac{62}{1 + e^{-0.07 \times 75}} = \frac{62}{1 + e^{-5.25}} \] \( e^{-5.25} \approx 0.00525 \) Denominator: \[ 1 + 0.00525 = 1.00525 \] Population: \[ C(75) \approx \frac{62}{1.00525} \approx 61.68 \] Difference: \[ |48.74 - 61.68| \approx 12.94 \] **Observation:** - The difference is largest at \( t=50 \) (around year 2050), approximately **17.2 million**. **Final answer:** \[ \boxed{\textbf{The greatest difference occurs around the year 2050}} \] --- ### **Part (f): Values of \( B'(75) \) and \( C'(75) \)** --- ### **(f)(i): Find \( B'(75) \)** Given: \[ B(t) = 33.5 \times (1.005)^t \] Derivative: \[ B'(t) = 33.5 \times (1.005)^t \times \ln(1.005) \] Calculate at \( t=75 \): - \( (1.005)^{75} \): \[ (1.005)^{75} = e^{75 \times 0.004987} = e^{0.374} \approx 1.454 \] - \( \ln(1.005) \approx 0.004987 \). Then: \[ B'(75) = 33.5 \times 1.454 \times 0.004987 \] Calculate: \[ 33.5 \times 1.454 \approx 48.74 \] Now: \[ 48.74 \times 0.004987 \approx 0.243 \] **Final answer:** \[ \boxed{\textbf{\( B'(75) \approx 0.243 \) million per year}} \] --- ### **(f)(ii): Find \( C'(75) \)** Given: \[ C(t) = \frac{62}{1 + e^{-0.07 t}} \] Derivative: \[ C'(t) = \frac{d}{dt} \left( \frac{62}{1 + e^{-0.07 t}} \right) \] Using quotient rule or chain rule: \[ C'(t) = 62 \times \frac{d}{dt} \left( \frac{1}{1 + e^{-0.07 t}} \right) \] Let \( u = 1 + e^{-0.07 t} \), then: \[ C(t) = \frac{62}{u} \] \[ C'(t) = -62 \times \frac{u'}{u^2} \] Calculate \( u' \): \[ u' = \frac{d}{dt} (1 + e^{-0.07 t}) = -0.07 e^{-0.07 t} \] So: \[ C'(t) = -62 \times \frac{-0.07 e^{-0.07 t}}{u^2} = 62 \times 0.07 \times \frac{e^{-0.07 t}}{(1 + e^{-0.07 t})^2} \] At \( t=75 \): - \( e^{-0.07 \times 75} = e^{-5.25} \approx 0.00525 \), - \( u = 1 + 0.00525 = 1.00525 \), - \( u^2 \approx (1.00525)^2 \approx 1.0105 \). Now: \[ C'(75) = 62 \times 0.07 \times \frac{0.00525}{1.0105} \] Calculate numerator: \[ 62 \times 0.07 = 4.34 \] Then: \[ C'(75) \approx 4.34 \times \frac{0.00525}{1.0105} \approx 4.34 \times 0.005195 \approx 0.0225 \] **Final answer:** \[ \boxed{\textbf{\( C'(75) \approx 0.0225 \) million per year}} \] --- ### **Summary of Results:** | Part | Result | Interpretation | |--------|---------|----------------| | (c)(i) | \(\approx 55.3\) million | Population in 2100 (Benoit) | | (c)(ii) | 1.005 = 0.5% | Annual growth rate in Benoit's model | | (d) | \(\approx 61.94\) million | Population in 2100 (Cecilia) | | (e) | Year 2050 | When difference is greatest (~17.2 million) | | (f)(i) | \(\approx 0.243\) million/year | Rate of growth at \( t=75 \) (Benoit) | | (f)(ii) | \(\approx 0.0225\) million/year | Rate of growth at \( t=75 \) (Cecilia) | --- **Let me know if you'd like further explanation or additional calculations!**

Let's carefully analyze and compare the values \( B'(75) \) and \( C'(75) \) as asked in part (g): --- ### **Part (g): Compare and interpret \( B'(75) \) and \( C'(75) \)** --- ### **Step 1: Recall the calculated values** - **From earlier:** \[ \boxed{ \begin{aligned} B'(75) &\approx 0.243 \text{ million per year} \\ C'(75) &\approx 0.0225 \text{ million per year} \end{aligned} } \] --- ### **Step 2: Interpretation in context** - **\( B'(75) \approx 0.243 \) million/year:** - This means that, **according to Benoit's exponential model**, the **population is increasing by approximately 0.243 million people per year** in the year corresponding to \( t = 75 \) (which is the year 2000 + 75 = 2075). - The exponential model predicts a **rapid and accelerating growth**, so the **rate of increase is relatively high** at this point. - **\( C'(75) \approx 0.0225 \) million/year:** - This indicates that, **according to Cecilia's logistic model**, the **population is increasing by about 0.0225 million people per year** at the same time. - The logistic model shows **slowing growth as the population approaches its maximum**, so the **rate of increase is much smaller** here. --- ### **Step 3: Contextual comparison** - The **significantly larger rate** in Benoit's model (about **10 times** larger than Cecilia's) suggests that **exponential growth predicts continuous acceleration**, leading to very high growth rates in the future. - Conversely, Cecilia's logistic model **accounts for limitations** (such as resources, space, etc.), leading to a **slowing growth rate** as the population nears a maximum (carrying capacity). --- ### **Final interpretation:** > **In 2075, Benoit's exponential model predicts that Canada's population is increasing at a rate of approximately 0.243 million people per year, indicating rapid, accelerating growth. In contrast, Cecilia's logistic model predicts a much slower increase of about 0.0225 million per year, reflecting a slowing growth as the population approaches its maximum sustainable level. This comparison highlights how exponential models can overestimate future growth when resource limitations are considered, whereas logistic models provide a more realistic long-term growth estimate.** --- If you'd like, I can help you write this in a concise paragraph or further elaborate!

Certainly! Here are the final, concise answers for each part with detailed explanations: --- ### **Part (c)(i): Predict the population in 2100 using Benoit's model** **Calculation:** \[ B(100) = 33.5 \times (1.005)^{100} \] \[ (1.005)^{100} \approx e^{100 \times \ln(1.005)} \approx e^{0.4987} \approx 1.6487 \] \[ B(100) \approx 33.5 \times 1.6487 \approx 55.3 \] **Final answer:** **Approximately 55.3 million** --- ### **Part (c)(ii): Interpretation of 1.005** **Interpretation:** The value **1.005** represents a **0.5% annual growth rate** in the population. Each year, the population multiplies by 1.005, indicating a steady increase of **0.5% per year**. --- ### **Part (d): Predict the population in 2100 using Cecilia's model** **Calculation:** \[ C(100) = \frac{62}{1 + e^{-0.07 \times 100}} = \frac{62}{1 + e^{-7}} \] \[ e^{-7} \approx 0.0009119 \] \[ C(100) \approx \frac{62}{1 + 0.0009119} \approx \frac{62}{1.0009119} \approx 61.94 \] **Final answer:** **Approximately 61.94 million** --- ### **Part (e): Year when the difference is greatest** - At \( t=25 \) (2025): \[ |37.97 - 52.85| \approx 14.88 \] - At \( t=50 \) (2050): \[ |43.00 - 60.20| \approx 17.20 \] - At \( t=75 \) (2075): \[ |48.74 - 61.68| \approx 12.94 \] **Observation:** The maximum difference (~17.20 million) occurs around **2050** (t=50). **Final answer:** **The greatest difference occurs around the year 2050** --- ### **Part (f)(i): \( B'(75) \)** \[ B'(75) = 33.5 \times (1.005)^{75} \times \ln(1.005) \] \[ (1.005)^{75} \approx e^{75 \times 0.004987} \approx e^{0.374} \approx 1.454 \] \[ B'(75) \approx 33.5 \times 1.454 \times 0.004987 \approx 48.74 \times 0.004987 \approx 0.243 \] **Final answer:** **Approximately 0.243 million per year** --- ### **Part (f)(ii): \( C'(75) \)** \[ C'(75) = 62 \times 0.07 \times \frac{e^{-0.07 \times 75}}{(1 + e^{-0.07 \times 75})^2} \] \[ e^{-5.25} \approx 0.00525 \] \[ u = 1 + 0.00525 = 1.00525,\quad u^2 \approx 1.0105 \] \[ C'(75) \approx 62 \times 0.07 \times \frac{0.00525}{1.0105} \approx 4.34 \times 0.005195 \approx 0.0225 \] **Final answer:** **Approximately 0.0225 million per year** --- ### **Part (g): Comparison and interpretation** - **\( B'(75) \approx 0.243 \) million/year:** The exponential model predicts Canada's population is increasing by about **0.243 million people per year** in 2075, indicating rapid, accelerating growth. - **\( C'(75) \approx 0.0225 \) million/year:** The logistic model predicts a much slower increase (~0.0225 million/year), reflecting that growth is slowing as the population nears a maximum capacity. **In context:** > In 2075, Benoit's exponential model suggests a fast-growing population increase, while Cecilia's logistic model indicates a much slower growth rate, aligning with realistic resource limitations. This highlights how exponential models can overestimate future growth compared to logistic models. --- **Let me know if you'd like this formatted differently or further clarified!**