We are tasked with finding the amount of chlorine in a tank as a function of time, $y(t)$, where $t$ is the time in seconds, given the following conditions:

• The initial capacity of the tank is $1600 \, \text{L}$.
• The initial concentration of chlorine is $0.0125 \, \text{g/L}$.
• Fresh water is pumped in at a rate of $16 \, \text{L/s}$.
• The mixture is pumped out at a rate of $40 \, \text{L/s}$.

Step 1: Determine the initial amount of chlorine

The initial amount of chlorine in the tank is: $y(0) = \text{initial concentration} \times \text{volume of the tank} = 0.0125 \, \frac{\text{g}}{\text{L}} \times 1600 \, \text{L} = 20 \, \text{g}.$

Step 2: Set up the differential equation

Let $y(t)$ be the amount of chlorine in grams at time $t$. The rate at which chlorine leaves the tank depends on both the rate of the outflow and the concentration of chlorine in the tank at time $t$.

The volume in the tank is decreasing since more liquid is pumped out than in. The net outflow rate is: $\text{Net outflow rate} = 40 \, \text{L/s} - 16 \, \text{L/s} = 24 \, \text{L/s}.$ So, the volume of liquid in the tank at time $t$ is: $V(t) = 1600 \, \text{L} - 24t \, \text{L}.$

Now, the chlorine is being removed from the tank at the outflow rate $40 \, \text{L/s}$ with a concentration of $\frac{y(t)}{V(t)}$. Hence, the rate of change of chlorine in the tank is: $\frac{dy}{dt} = -40 \times \frac{y(t)}{V(t)}.$

Substituting $V(t) = 1600 - 24t$, we get the differential equation: $\frac{dy}{dt} = -\frac{40y(t)}{1600 - 24t}.$

Step 3: Solve the differential equation

This is a separable differential equation. We can rearrange it as: $\frac{dy}{y(t)} = -\frac{40 \, dt}{1600 - 24t}.$ Integrating both sides: $\int \frac{1}{y(t)} \, dy = -40 \int \frac{1}{1600 - 24t} \, dt.$ The left side becomes $\ln |y(t)|$, and the right side can be solved using a substitution. Let $u = 1600 - 24t$, so $du = -24 \, dt$. This gives: $\int \frac{1}{1600 - 24t} \, dt = -\frac{1}{24} \ln |1600 - 24t|.$ Thus, the solution becomes: $\ln |y(t)| = \frac{40}{24} \ln |1600 - 24t| + C,$ where $C$ is the constant of integration. Simplifying: $\ln |y(t)| = \frac{5}{3} \ln |1600 - 24t| + C.$

Exponentiating both sides to solve for $y(t)$: $y(t) = e^C (1600 - 24t)^{5/3}.$ Let $e^C = k$ (a new constant), so: $y(t) = k (1600 - 24t)^{5/3}.$

Step 4: Apply the initial condition

We know that at $t = 0$, $y(0) = 20$. Substituting this into the equation: $20 = k (1600)^{5/3}.$ Solving for $k$: $k = \frac{20}{1600^{5/3}}.$ Now we have the complete solution for $y(t)$: $y(t) = \frac{20}{1600^{5/3}} (1600 - 24t)^{5/3}.$

This is the amount of chlorine in the tank as a function of time.

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Would you like more details or clarifications on any of the steps?

Here are 5 related questions for further exploration:

1. How would the solution change if the outflow rate were reduced?
2. How long does it take for the tank to be completely empty of liquid?
3. What happens to the concentration of chlorine as time approaches the point where the tank empties?
4. How would the equation change if the inflow and outflow rates were equal?
5. Can you determine the time at which half of the initial chlorine is removed?

Tip: When dealing with mixing problems, the differential equation often arises from the relationship between the inflow/outflow rates and the concentration of the substance being studied.