Solve the problem.
-An office contains $1000 \mathrm { ft } ^ { 3 }$ of air initially free of carbon monoxide. Starting at time $= 0$, cigarette smoke containing $4 \%$ carbon monoxide is blown into the room at the rate of $0.5 \mathrm { ft } 3 / \mathrm { min }$. A ceiling fan keeps the air in the room well circulated and the air leaves the room at the same rate of $0.5 \mathrm { ft } ^ { 3 } / \mathrm { min } ^ { \text {. Find } }$ the time when the concentration of carbon monoxide reaches $0.01 \%$.
A) 5.01 min
B) 6.01 min
C) 8.01 min
D) 7.01 min

Question

Quizplus · Accepted Answer

The concentration of carbon monoxide in the room at time $t$ is given by the differential equation$$\frac{dQ}{dt} = 0.04(0.5) - \frac{Q}{1000}(0.5)$$where $Q$ is the amount of carbon monoxide in the room at time $t$. Solving this equation gives$$Q(t) = 2 - 2e^{-t/1000}$$The concentration of carbon monoxide in the room at time $t$ is given by$$C(t) = \frac{Q(t)}{1000} = 0.002 - 0.002e^{-t/1000}$$Setting $C(t) = 0.0001$ and solving for $t$ gives$$0.0001 = 0.002 - 0.002e^{-t/1000}$$$$e^{-t/1000} = 0.995$$$$t = -1000 \ln(0.995) \approx 5.01 \text{ min}$$

Solve the Problem 1000ft31000 \mathrm { ft } ^ { 3 }1000ft3

Solve the Problem $1000 \mathrm { ft } ^ { 3 }$