Sure! Let’s proceed step by step with explanations and clear answers for each part. --- ## Step 1: Answer parts (a), (b), and (c) ### (a) **At what time(s) was the CO₂ level the lowest? Highest?** **Explanation:** - The CO₂ level is lowest when its rate of change \( f(t) \) switches from negative to positive (i.e., the minimum point on the graph, where the curve crosses from below to above the t-axis). - The CO₂ level is highest when \( f(t) \) switches from positive to negative (i.e., the maximum point on the graph, where the curve crosses from above to below the t-axis). **From the graph:** - Lowest CO₂: Where \( f(t) \) goes from negative to positive, which is at around **t ≈ 10 hours**. - Highest CO₂: Where \( f(t) \) goes from positive to negative, which is at around **t ≈ 18 hours**. --- ### (b) **Write a definite integral for the CO₂ added during the night (t = 12 to t = 24).** **Explanation:** - The total CO₂ added is the integral of the rate over the interval \([12, 24]\) hours. **Integral:** \[ \int_{12}^{24} f(t) \, dt \] --- ### (c) **Estimate the value of your definite integral in (b) using a Riemann sum. Include units.** **Explanation:** - Use the table values for \( f(t) \) from \( t = 12 \) to \( t = 24 \). - Let’s use the left-endpoint Riemann sum with \(\Delta t = 2\) hours for intervals: \([12, 14], [14, 16], [16, 18], [18, 20], [20, 22]\), and \([22, 24]\). - Sum: \(\sum f(t_i) \cdot \Delta t\), where \( t_i = 12, 14, 16, 18, 20, 22 \). Values: - \( f(12) = .045 \) - \( f(14) = .035 \) - \( f(16) = .025 \) - \( f(18) = .013 \) - \( f(20) = .010 \) - \( f(22) = .010 \) Calculation: \[ \int_{12}^{24} f(t) dt \approx 2 \times (.045 + .035 + .025 + .013 + .010 + .010) = 2 \times .138 = .276 \text{ mmol/L} \] **Units:** mmol/L --- ## Step 2: Answer parts (d) and (e) ### (d) **Estimate the CO₂ level at dusk (t = 12). Include units.** **Explanation:** - The CO₂ level at dusk is the initial value plus the net change up to \( t = 12 \). - The net change is \(\int_{}^{12} f(t) dt\), which we estimate with a left-endpoint Riemann sum using the table for \( t = , 2, ..., 10 \). Values: - \( f() = .000 \) - \( f(2) = -.044 \) - \( f(4) = -.039 \) - \( f(6) = -.035 \) - \( f(8) = -.026 \) - \( f(10) = -.020 \) Sum: \[ \int_{}^{12} f(t) dt \approx 2 \times (.000 - .044 - .039 - .035 - .026 - .020) \] \[ = 2 \times (-.164) = -.328 \text{ mmol/L} \] So, CO₂ at dusk: \[ \text{CO}_2 \text{ at dusk} = 2.600 + (-.328) = 2.272 \text{ mmol/L} \] --- ### (e) **Does the CO₂ level return to approximately the same amount each day? Explain.** **Explanation:** - The CO₂ level at the next dawn would be: \[ \text{CO}_2 \text{ at next dawn} = \text{CO}_2 \text{ at dusk} + \text{CO}_2 \text{ added during night} \] \[ = 2.272 + .276 = 2.548 \text{ mmol/L} \] - This is slightly less than the starting value of 2.600 mmol/L. **Conclusion:** No, the CO₂ level does not return exactly to the starting amount each day; there is a small net decrease, suggesting it is **not exactly in equilibrium**, but it is close. --- ## **Final Structured Answers** --- ### (a) - **Lowest CO₂:** Around \( t = 10 \) hours. - **Highest CO₂:** Around \( t = 18 \) hours. - **Explanation:** These are the points where the rate \( f(t) \) crosses the t-axis (rate changes sign). --- ### (b) - **Definite Integral:** \[ \int_{12}^{24} f(t)\,dt \] This gives the total CO₂ added during the night. --- ### (c) - **Estimated Value:** \[ \int_{12}^{24} f(t)\,dt \approx .276 \text{ mmol/L} \] --- ### (d) - **Estimated CO₂ at dusk:** \[ 2.272 \text{ mmol/L} \] --- ### (e) - **Equilibrium?** No, the CO₂ level does not return exactly to its starting value at dawn. There is a small net decrease each cycle, so it is not in perfect equilibrium, but is close. - **Explanation:** The next dawn value is about 2.548 mmol/L, less than the starting 2.600 mmol/L.