Question: The air in a $60$ cubic metre kitchen is initially clean, but when Amanda burns her toast while making breakfast, smoke is mixed with the room's air at a rate of $0.06$ mg per second. An air conditioning system exchanges the mixture of air and smoke with clean air at a rate of $6$ cubic metres per minute. Assume that the pollutant is mixed uniformly throughout the room and that burnt toast is taken outside after $60$ seconds. Let $S(t)$ be the amount of smoke in mg in the room at time $t$ (in seconds) after the toast first began to burn. a. Find a differential equation obeyed by $S(t)$. b. Find $S(t)$ for $0 \leq t \leq 60$ by solving the differential equation in (a) with an appropriate initial condition. c. What is the level of pollution in mg per cubic meter after $60$ seconds? d. How long does it take for the level of pollution to fall to $0.002$ mg per cubic metre after the toast is taken outside?

1. **Problem Statement:** We have a kitchen of volume $60$ cubic metres. Smoke is produced at a rate of $0.06$ mg/s and the air conditioning exchanges air at $6$ cubic metres per minute. We want to model the amount of smoke $S(t)$ in mg at time $t$ seconds. 2. **Formulating the Differential Equation:** - The smoke enters at a constant rate of $0.06$ mg/s. - The air conditioning removes smoke by exchanging $6$ cubic metres per minute, which is $\frac{6}{60} = 0.1$ cubic metres per second. - Since the room volume is $60$ cubic metres, the fraction of smoke removed per second is $\frac{0.1}{60} = \frac{1}{600}$ of the smoke present. Thus, the rate of change of smoke is: $$\frac{dS}{dt} = 0.06 - \frac{S(t)}{600}$$ 3. **Solving the Differential Equation:** - This is a linear first-order ODE. - Initial condition: $S(0) = 0$ since initially the air is clean. The integrating factor is: $$\mu(t) = e^{\int \frac{1}{600} dt} = e^{\frac{t}{600}}$$ Multiply both sides by $\mu(t)$: $$e^{\frac{t}{600}} \frac{dS}{dt} + \frac{1}{600} e^{\frac{t}{600}} S = 0.06 e^{\frac{t}{600}}$$ Left side is derivative of $S e^{\frac{t}{600}}$: $$\frac{d}{dt} \left(S e^{\frac{t}{600}}\right) = 0.06 e^{\frac{t}{600}}$$ Integrate both sides: $$S e^{\frac{t}{600}} = 0.06 \int e^{\frac{t}{600}} dt + C$$ Calculate the integral: $$\int e^{\frac{t}{600}} dt = 600 e^{\frac{t}{600}} + K$$ So: $$S e^{\frac{t}{600}} = 0.06 \times 600 e^{\frac{t}{600}} + C = 36 e^{\frac{t}{600}} + C$$ Divide both sides by $e^{\frac{t}{600}}$: $$S = 36 + C e^{-\frac{t}{600}}$$ Apply initial condition $S(0) = 0$: $$0 = 36 + C \Rightarrow C = -36$$ Final solution: $$S(t) = 36 (1 - e^{-\frac{t}{600}})$$ 4. **Check at $t=13$ seconds:** $$S(13) = 36 (1 - e^{-\frac{13}{600}})$$ 5. **Pollution level after 60 seconds:** Calculate $S(60)$: $$S(60) = 36 (1 - e^{-\frac{60}{600}}) = 36 (1 - e^{-0.1})$$ Numerical value: $$e^{-0.1} \approx 0.9048374180$$ $$S(60) \approx 36 (1 - 0.9048374180) = 36 \times 0.095162582 = 3.426572952$$ Pollution concentration in mg/m$^3$ is: $$\frac{S(60)}{60} = \frac{3.426572952}{60} = 0.0571095492$$ 6. **Time for pollution to fall to $0.002$ mg/m$^3$ after toast is taken outside:** - After $t=60$ seconds, the smoke source stops, so for $t > 60$ the differential equation is: $$\frac{dS}{dt} = - \frac{S}{600}$$ - Initial condition at $t=60$ is $S(60) = 3.426572952$ mg. - Solve: $$\frac{dS}{dt} = - \frac{S}{600} \Rightarrow \frac{dS}{S} = - \frac{dt}{600}$$ Integrate: $$\ln |S| = - \frac{t}{600} + K$$ Exponentiate: $$S = A e^{-\frac{t}{600}}$$ Apply initial condition at $t=60$: $$3.426572952 = A e^{-\frac{60}{600}} = A e^{-0.1}$$ So: $$A = \frac{3.426572952}{e^{-0.1}} = 3.426572952 \times e^{0.1}$$ We want to find $t$ such that pollution concentration is $0.002$ mg/m$^3$: $$\frac{S(t)}{60} = 0.002 \Rightarrow S(t) = 0.002 \times 60 = 0.12$$ Substitute $S(t)$: $$0.12 = A e^{-\frac{t}{600}} = 3.426572952 e^{0.1} e^{-\frac{t}{600}}$$ Divide both sides by $3.426572952 e^{0.1}$: $$\frac{0.12}{3.426572952 e^{0.1}} = e^{-\frac{t}{600}}$$ Calculate denominator: $$3.426572952 \times e^{0.1} \approx 3.426572952 \times 1.105170918 = 3.786$$ So: $$e^{-\frac{t}{600}} = \frac{0.12}{3.786} = 0.0317$$ Take natural log: $$-\frac{t}{600} = \ln(0.0317) = -3.448$$ Solve for $t$: $$t = 600 \times 3.448 = 2068.8 \text{ seconds}$$ - Since $t$ is measured from $t=0$, and the toast was taken outside at $t=60$, the time after toast removal is: $$2068.8 - 60 = 2008.8 \text{ seconds}$$ **Final answers:** - (a) $\frac{dS}{dt} = 0.06 - \frac{S(t)}{600}$ - (b) $S(t) = 36 (1 - e^{-\frac{t}{600}})$ for $0 \leq t \leq 60$ - (c) Pollution level after 60 seconds: $0.0571095492$ mg/m$^3$ - (d) Time for pollution to fall to $0.002$ mg/m$^3$ after toast removal: $2008.8$ seconds