The equation is this  𝑐(𝑡) = 9 − 90 000 ( 1 ), where c is

 10 000 􏰂3􏰁 a) When will the concentration of pollutant in the pond reach 6 𝑘𝑔/𝑚3 ?

measured in kilograms per cubic metre.

b) What will happen to the concentration of pollutant over time?

[2 marks]

Thank you for the clarification. Based on the correct equation:

\[
c(t) = 9 - 90,000 \left(\frac{1}{10,000}\right)^{\frac{t}{3}}
\]

I'll walk through the problem again to ensure that we correctly solve for \( t \) when the concentration \( c(t) = 6 \) kg/m³, and then discuss the long-term behavior of the concentration.

### Part (a):
Set \( c(t) = 6 \) kg/m³:

\[
6 = 9 - 90,000 \left(\frac{1}{10,000}\right)^{\frac{t}{3}}
\]

Isolate the exponential term:

\[
90,000 \left(\frac{1}{10,000}\right)^{\frac{t}{3}} = 3
\]

Simplify and solve for \( t \):

\[
\left(\frac{1}{10,000}\right)^{\frac{t}{3}} = \frac{1}{30,000}
\]

Taking the natural logarithm of both sides:

\[
\frac{t}{3} \ln\left(\frac{1}{10,000}\right) = \ln\left(\frac{1}{30,000}\right)
\]

Now solve for \( t \):

\[
t = 3 \times \frac{\ln\left(\frac{1}{30,000}\right)}{\ln\left(\frac{1}{10,000}\right)}
\]

### Part (b):
As \( t \) increases indefinitely, the term \( \left(\frac{1}{10,000}\right)^{\frac{t}{3}} \) will tend toward zero, causing the concentration \( c(t) \) to approach 9 kg/m³.

I'll now recalculate \( t \) using this equation.### Part (a):
The time \( t \) when the concentration of the pollutant in the pond will reach 6 kg/m³ is approximately **3.36 minutes**.

### Part (b):
As time \( t \) increases, the concentration \( c(t) \) will approach **9 kg/m³** because the exponential term \( \left(\frac{1}{10,000}\right)^{\frac{t}{3}} \) tends toward zero over time.

Would you like any further details or have any other questions?

Here are some related questions to consider:
1. How would the result change if the initial concentration was less than 9 kg/m³?
2. What is the impact on the time \( t \) if the concentration required is 7 kg/m³ instead of 6 kg/m³?
3. How does the value \( 90,000 \) in the equation affect the concentration behavior over time?
4. What are the practical implications of the concentration asymptotically approaching 9 kg/m³?
5. How would the result be different if the time factor \( \frac{t}{3} \) was instead \( \frac{t}{4} \)?
6. Could this model apply to a scenario where the pollutant is being removed rather than added?
7. What happens to the concentration \( c(t) \) if \( t = 0 \)?
8. How can we graphically represent the concentration \( c(t) \) over time?

**Tip:** Understanding the asymptotic behavior of functions is crucial in predicting the long-term effects of processes modeled by exponential functions.

Explore how to solve the exponential decay problem involving the concentration of a pollutant in a pond over time. Learn how to calculate the time when the concentration reaches 6 kg/m³ and analyze its long-term behavior approaching 9 kg/m³. Suitable for advanced high school students.

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