Chapter 2 - First Order Differential Equations - 2.3 Modeling with First Order Equations - Problems - Page 59: 1

$460.52 \min$

Step 1: Define the Variables and the Differential Equation Let: • V=200 L (volume of the tank, constant since inflow = outflow) • $C(t)$ : concentration of dye (g/L) at time $t$ (min) • Initial concentration: $C(0) = 1$ g/L • Inflow: fresh water at $2$ L/min (so rate of dye in = $0$ g/min) • Outflow: well-stirred mixture at $2$ L/min, so rate of dye out = $C(t) \times 2$ g/min The rate of change of dye in the tank is: \[ \frac{dQ}{dt} = \text{rate in} - \text{rate out} = 0 - 2C(t) \] Since $Q(t) = V \cdot C(t) = 200 C(t)$, we have: \[ \frac{dQ}{dt} = 200 \frac{dC}{dt} \] Thus: \[ 200 \frac{dC}{dt} = -2C(t) \] Divide both sides by 200: \[ \frac{dC}{dt} = -\frac{2}{200} C(t) = -\frac{1}{100} C(t) \] Step 2: Solve the Differential Equation The equation is: \[ \frac{dC}{dt} = -\frac{1}{100} C \] This is separable: \[ \frac{dC}{C} = -\frac{1}{100} dt \] Integrate both sides: \[ \int \frac{1}{C} dC = -\frac{1}{100} \int dt \] \[ \ln|C| = -\frac{t}{100} + k \] Exponentiate: \[ C(t) = e^{k} e^{-t/100} \] Let $C(0) = e^k = 1$, so: \[ C(t) = e^{-t/100} \] Step 3: Find Time for Concentration to Reach 1% of Original We want: \[ C(t) = 0.01 \cdot C(0) = 0.01 \] So: \[ e^{-t/100} = 0.01 \] Take natural logarithm: \[ -\frac{t}{100} = \ln(0.01) \] \[ \frac{t}{100} = -\ln(0.01) = \ln\left(\frac{1}{0.01}\right) = \ln(100) \] \[ t = 100 \ln(100) \] Since $\ln(100) = \ln(10^2) = 2 \ln(10) \approx 2 \times 2.302585 = 4.60517$, we get: \[ t \approx 100 \times 4.60517 \approx 460.52 \text{ minutes} \] Step 4: Final Answer \[ \boxed{t = 100 \ln(100)} \] or approximately $\boxed{460.52}$ minutes.