Let's solve each part step by step, with explanations at each stage. --- ### (a) Sketch \(\dot{S}_{act}(t)\) vs. \(t\) **Step 1:** The given emission rate is: \[ \dot{S}_{act}(t) = 30t \exp\left(-\frac{t}{10}\right) \] This is the product of a linear term (\(30t\)) and an exponential decay (\(\exp(-t/10)\)). - At \(t=0\), \(\dot{S}_{act}(0) = 0\). - As \(t\) increases, the linear term increases but the exponential term decreases. - The function initially increases, reaches a maximum, and then decreases. **Explanation:** This function is typical for processes where a rate rises initially (due to the \(t\) term) and then falls off (due to the exponential decay). --- ### (b) Maximum Emission Rate and When it Occurs **Step 2:** To find the maximum, set the derivative of \(\dot{S}_{act}(t)\) with respect to \(t\) to zero: \[ \frac{d}{dt}\left[30t \exp\left(-\frac{t}{10}\right)\right] = 0 \] Use the product rule: \[ 30 \exp\left(-\frac{t}{10}\right) + 30t \left(-\frac{1}{10}\right) \exp\left(-\frac{t}{10}\right) = 0 \] \[ 30 \exp\left(-\frac{t}{10}\right) - 3t \exp\left(-\frac{t}{10}\right) = 0 \] \[ 30 - 3t = 0 \] \[ t = 10 \text{ hours} \] Maximum emission rate: \[ \dot{S}_{act}(10) = 30 \times 10 \times \exp\left(-\frac{10}{10}\right) = 300 \times \exp(-1) \approx 300 \times 0.3679 \approx 110.4 \text{ kg}_c/\text{h} \] **Explanation:** The emission rate reaches its maximum at \(t = 10\) hours, with a maximum value of about \(110.4 \text{ kg}_c/\text{h}\). --- ### Final Answer (for parts a and b): - **(a):** The sketch of \(\dot{S}_{act}(t)\) vs. \(t\) starts at 0, rises to a peak at \(t=10\) hours, then falls back toward zero. - **(b):** The maximum emission rate is **110.4 kg\(_c\)/h** at **10 hours**. --- Let me know if you’d like to proceed with parts (c) and (d)!

Let's continue with parts (c) and (d), following the same step-by-step approach. --- ### (c) Total mass of contaminant released and fraction released in first 50 hours **Step 1:** The total mass \(M_{total}\) released is the integral of the emission rate \(\dot{S}_{act}(t)\) over time from 0 to \(\infty\): \[ M_{total} = \int_0^\infty \dot{S}_{act}(t) \, dt = \int_0^\infty 30 t e^{-t/10} \, dt \] This is a standard integral involving \(t e^{-\alpha t}\). **Explanation:** The integral of \(t e^{-\alpha t}\) from 0 to \(\infty\) equals \(\frac{1}{\alpha^2}\). Here, \(\alpha = \frac{1}{10}\). --- **Step 2:** Calculate \(M_{total}\): \[ M_{total} = 30 \times \int_0^\infty t e^{-t/10} dt \] Using the integral result: \[ \int_0^\infty t e^{-\alpha t} dt = \frac{1}{\alpha^2} \] with \(\alpha = \frac{1}{10}\): \[ \int_0^\infty t e^{-t/10} dt = \frac{1}{(1/10)^2} = \frac{1}{1/100} = 100 \] Therefore, \[ M_{total} = 30 \times 100 = 3000 \text{ kg} \] --- **Step 3:** Calculate the mass released in the first 50 hours: \[ M_{50} = \int_0^{50} 30 t e^{-t/10} dt \] This integral can be evaluated using integration by parts or a table. Alternatively, since the indefinite integral of \(t e^{-\beta t}\) is: \[ \int t e^{-\beta t} dt = -\frac{t}{\beta} e^{-\beta t} - \frac{1}{\beta^2} e^{-\beta t} + C \] with \(\beta = 1/10\). Applying limits from 0 to 50: \[ M_{50} = 30 \left[ -\frac{t}{1/10} e^{-t/10} - \frac{1}{(1/10)^2} e^{-t/10} \right]_0^{50} \] Simplify: \[ M_{50} = 30 \left[ -10 t e^{-t/10} - 100 e^{-t/10} \right]_0^{50} \] Calculate at \(t=50\): \[ t=50: \quad e^{-50/10} = e^{-5} \approx 0.0067 \] \[ -10 \times 50 \times 0.0067 = -10 \times 50 \times 0.0067 \approx -3.35 \] \[ -100 \times 0.0067 = -0.67 \] Sum: \(-3.35 - 0.67 = -4.02\) At \(t=0\): \[ e^{0} = 1 \] \[ -10 \times 0 \times 1 = 0 \] \[ -100 \times 1 = -100 \] Difference: \[ -4.02 - (-100) = 95.98 \] Multiply by 30: \[ M_{50} \approx 30 \times 95.98 \approx 2879 \text{ kg} \] --- ### **Final Results:** - **Total mass released over infinite time: \(\boxed{3000 \text{ kg}}\)** - **Mass released in first 50 hours: \(\boxed{2879 \text{ kg}}\)** - **Fraction released in first 50 hours:** \[ \frac{M_{50}}{M_{total}} = \frac{2879}{3000} \approx 0.9597 \quad \text{or} \quad 95.97\% \] --- ### (d) Approximations to the emission rate and their sketches **Step 1:** Write the approximations: (i) **Constant rate approximation:** \[ \dot{S}_1(t) = \dot{S}_{act}(10) = 110.4 \text{ kg/h} \] (since the maximum occurs at \(t=10\), a constant equal to this maximum value is a simple approximation). (ii) **Linear approximation:** From \(t=0\) to \(t=20\), approximate \(\dot{S}_{act}(t)\) as increasing linearly from 0 to the maximum: \[ \dot{S}_2(t) = \frac{\dot{S}_{act}(10)}{10} t = 11.04 t \] for \(0 \leq t \leq 10\), then decreasing linearly to zero at some \(t>10\) (for simplicity, take a symmetric decline to zero at \(t=20\)): \[ \dot{S}_2(t) = 11.04 (20 - t) \quad \text{for} \quad 10 < t \leq 20 \] **Step 2:** Sketch these approximations on the same graph as the original \(\dot{S}_{act}(t)\): - The constant approximation is a horizontal line at 110.4 kg/h. - The linear approximation is a triangle peaking at 110.4 kg/h at \(t=10\), then decreasing to zero at \(t=20\). --- ### **Summary:** - The original emission rate rises to about 110 kg/h at 10 hours and then decreases. - The constant approximation simplifies the emission as a steady rate equal to the maximum. - The linear approximation models a rise and fall over a 20-hour period, matching the peak at 10 hours. --- **Final note:** These approximations help in simplifying the analysis and are useful for quick estimates or initial assessments. --- Let me know if you need further clarification or additional calculations!

**(a) Sketch \(\dot{S}_{act}(t)\) vs. \(t\):** The emission rate \(\dot{S}_{act}(t) = 30 t e^{-t/10}\) starts at zero when \(t=0\), increases to a maximum at \(t=10\) hours, and then decreases. The graph is a unimodal curve peaking at about 110.4 kg/h at 10 hours. **(b) Maximum emission rate and when it occurs:** Maximum occurs at \(t=10\) hours. \[ \dot{S}_{max} = 30 \times 10 \times e^{-10/10} = 300 \times e^{-1} \approx 110.4 \text{ kg/h} \] --- **(c) Total mass released and fraction in first 50 hours:** Total mass: \[ M_{total} = \int_0^\infty 30 t e^{-t/10} dt = 3000 \text{ kg} \] Mass in first 50 hours: \[ M_{50} \approx 2879 \text{ kg} \] Fraction in first 50 hours: \[ \frac{2879}{3000} \approx 0.96 \quad \text{or } 96\% \] --- **(d) Approximations to \(\dot{S}_{act}(t)\):** - **(i) Constant rate approximation:** \[ \dot{S}_1(t) = 110.4 \text{ kg/h} \] - **(ii) Linear approximation:** From 0 to 10 hours: \[ \dot{S}_2(t) = 11.04 t \] from 10 to 20 hours: \[ \dot{S}_2(t) = 110.4 - 11.04 (t - 10) = 221.76 - 11.04 t \] **Sketches:** - The constant approximation is a horizontal line at 110.4 kg/h. - The linear approximation forms a triangle rising to 110.4 kg/h at 10 hours, then decreasing back to zero at 20 hours. --- **Final note:** These steps provide a clear understanding of the emission behavior, maximum rate, total release, and simplified models for analysis.

Let's approach each part systematically, with detailed explanations: --- ### **(a) Sketch \(\dot{S}_{act}(t)\) vs. \(t\)** **Step 1:** Given emission rate: \[ \dot{S}_{act}(t) = 30 t \exp\left(-\frac{t}{10}\right) \] - At \(t=0\), \(\dot{S}_{act}(0)=0\). - For small \(t\), the function increases because \(t\) increases faster than the exponential decay. - The function reaches a maximum at some \(t>0\), then decreases towards zero as \(t \to \infty\). **Explanation:** This is a *product of a linearly increasing term* and an *exponentially decreasing* function, which produces a bell-shaped curve with a peak at some finite \(t\). --- ### **(b) Find the maximum emission rate and when it occurs** **Step 2:** To find the maximum, differentiate \(\dot{S}_{act}(t)\): \[ \frac{d}{dt} \left[ 30 t e^{-t/10} \right] = 0 \] Apply the product rule: \[ 30 e^{-t/10} + 30 t \left(-\frac{1}{10}\right) e^{-t/10} = 0 \] Simplify: \[ 30 e^{-t/10} - 3 t e^{-t/10} = 0 \] Divide both sides by \(e^{-t/10}\): \[ 30 - 3 t = 0 \Rightarrow t=10 \text{ hours} \] **Maximum emission rate:** \[ \dot{S}_{max} = 30 \times 10 \times e^{-10/10} = 300 \times e^{-1} \approx 300 \times 0.3679 \approx 110.4\, \text{kg/h} \] --- ### **Final answer for (a) and (b):** - The plot starts at zero, rises to approximately 110.4 kg/h at 10 hours, then declines. - The maximum emission rate is **110.4 kg/h** at **10 hours**. --- ### **(c) Total mass released and fraction in first 50 hours** **Step 3:** Total mass: \[ M_{total} = \int_0^{\infty} 30 t e^{-t/10} dt \] Recall the integral: \[ \int_0^{\infty} t e^{-\alpha t} dt = \frac{1}{\alpha^2} \] Here, \(\alpha = 1/10\), so: \[ M_{total} = 30 \times \frac{1}{(1/10)^2} = 30 \times 100 = \boxed{3000\, \text{kg}} \] **Step 4:** Mass released in the first 50 hours: \[ M_{50} = \int_0^{50} 30 t e^{-t/10} dt \] Using the indefinite integral: \[ \int t e^{-\beta t} dt = -\frac{t}{\beta} e^{-\beta t} - \frac{1}{\beta^2} e^{-\beta t} + C \] with \(\beta=1/10\): \[ M_{50} = 30 \left[ -10 t e^{-t/10} - 100 e^{-t/10} \right]_0^{50} \] Evaluate at \(t=50\): \[ e^{-50/10} = e^{-5} \approx 0.0067 \] \[ -10 \times 50 \times 0.0067 \approx -3.35 \] \[ -100 \times 0.0067 \approx -0.67 \] Sum: \(-3.35 - 0.67 = -4.02\) At \(t=0\): \[ e^{0} = 1 \] \[ -10 \times 0 \times 1 = 0 \] \[ -100 \times 1 = -100 \] Difference: \[ -4.02 - (-100) = 95.98 \] Multiply by 30: \[ M_{50} \approx 30 \times 95.98 \approx \boxed{2879\, \text{kg}} \] --- ### **Fraction released in first 50 hours:** \[ \frac{M_{50}}{M_{total}} \approx \frac{2879}{3000} \approx 0.96 \text{ or } 96\% \] --- ### **(d) Approximations to the emission rate** **Step 5:** Construct two simplified representations: - **(i) Constant rate approximation:** Use the peak rate at \(t=10\) hours: \[ \boxed{\dot{S}_1(t) = 110.4\, \text{kg/h}} \] - **(ii) Linear approximation:** Model as rising from 0 at \(t=0\) to maximum at \(t=10\): \[ \dot{S}_2(t) = \frac{110.4}{10} t = 11.04 t \quad \text{for } 0 \le t \le 10 \] and then decreasing linearly from 110.4 at \(t=10\) to zero at \(t=20\): \[ \dot{S}_2(t) = 110.4 - 11.04 (t - 10) = 221.76 - 11.04 t \quad \text{for } 10 \le t \le 20 \] **Sketches:** - The constant approximation is a flat line at 110.4 kg/h. - The linear approximation rises linearly to the maximum at 10 hours, then declines linearly to zero at 20 hours. --- ### **Final summary:** - The original emission rate peaks at about 110.4 kg/h at 10 hours. - Total contaminant released over infinite time is approximately 3000 kg. - About 97% of the total is released in the first 50 hours. - The constant approximation simplifies the rate to 110.4 kg/h. - The linear approximation models a rise and fall, peaking at 10 hours. --- Let me know if you'd like further elaboration!