The problem you're working on involves modeling the decay of a chemical, Trioxin, using exponential decay.

We know that:

• The amount of Trioxin decays exponentially.
• At $t = 0$ days (when it was first dumped), the initial amount is unknown.
• After 4 days, the amount was 42.5 barrels.
• After 7 days, the amount was 20.5 barrels.

The general formula for exponential decay is:

$A(t) = A_0 e^{-kt}$ Where:

• $A(t)$ is the amount at time $t$,
• $A_0$ is the initial amount,
• $k$ is the decay constant,
• $t$ is the time in days.

Step 1: Find the Initial Amount $A_0$

From the information provided:

• $A(4) = 42.5$ barrels (at $t = 4$ days),
• $A(7) = 20.5$ barrels (at $t = 7$ days).

Using the exponential decay formula, we can set up two equations:

$42.5 = A_0 e^{-4k}$ $20.5 = A_0 e^{-7k}$

By dividing these two equations, we can eliminate $A_0$:

$\frac{20.5}{42.5} = \frac{e^{-7k}}{e^{-4k}} = e^{-3k}$

Now solve for $k$:

$\frac{20.5}{42.5} = e^{-3k} \quad \Rightarrow \quad e^{-3k} = 0.4824$

Take the natural logarithm of both sides:

$-3k = \ln(0.4824) \quad \Rightarrow \quad k = -\frac{\ln(0.4824)}{3}$

Calculating $k$:

$k \approx 0.2435$

Step 2: Solve for $A_0$

Now that we have $k$, substitute it into one of the original equations to solve for $A_0$. Using $A(4) = 42.5$:

$42.5 = A_0 e^{-4(0.2435)} \quad \Rightarrow \quad 42.5 = A_0 e^{-0.974}$

Calculate $e^{-0.974}$:

$42.5 = A_0 (0.3779) \quad \Rightarrow \quad A_0 = \frac{42.5}{0.3779}$

$A_0 \approx 112.5 \text{ barrels}$

Step 3: Find the Rate of Decay After 4 Days

Now, we need to calculate the rate of decay at $t = 4$ days. The rate of decay is given by the derivative of the decay function:

$\frac{dA}{dt} = -kA_0 e^{-kt}$

Substitute $t = 4$, $k = 0.2435$, and $A_0 = 112.5$:

$\frac{dA}{dt} \bigg|_{t=4} = -0.2435 \times 112.5 \times e^{-4(0.2435)}$

$\frac{dA}{dt} \bigg|_{t=4} = -0.2435 \times 112.5 \times 0.3779 \approx -10.4 \text{ barrels per day}$

Step 4: Find When the Amount Drops to 2 Barrels

Set $A(t) = 2$ in the exponential decay formula:

$2 = 112.5 e^{-0.2435t}$

Solve for $t$:

$\frac{2}{112.5} = e^{-0.2435t} \quad \Rightarrow \quad e^{-0.2435t} = 0.0178$

Take the natural logarithm of both sides:

$-0.2435t = \ln(0.0178) \quad \Rightarrow \quad t = \frac{\ln(0.0178)}{-0.2435}$

$t \approx 16.3 \text{ days}$

Final Answers:

1. About 112.5 barrels were initially dumped into the swamp.
2. The Trioxin was decreasing at a rate of 10.4 barrels per day after 4 days.
3. It will take about 16.3 days for the Trioxin to drop to 2 barrels.

Would you like more details on any step or have further questions?

Here are some related questions for practice:

1. How does changing the decay constant $k$ affect the rate of decay?
2. What would happen if the initial amount of Trioxin was halved?
3. How many days would it take for the amount of Trioxin to drop to half its original value?
4. What is the half-life of Trioxin based on this model?
5. How would the model change if Trioxin was added gradually instead of all at once?

Tip: Exponential decay models are used in many fields, such as radioactive decay, population decline, and even cooling rates. Understanding how the parameters affect the outcome is key to analyzing real-world scenarios.