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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]

Let's break down the problem step by step.

### Given:
The concentration of the pollutant, \( c(t) \), at time \( t \) (in minutes) is given by the equation:

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

where \( c \) is measured in kilograms per cubic meter.

### Part (a):
We need to find the time \( t \) when the concentration \( c(t) \) is 6 kg/m\(^3\).

Set \( c(t) = 6 \):

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

Now, solve for \( t \).

### Part (b):
We need to determine what happens to the concentration \( c(t) \) as time \( t \) increases indefinitely.

I'll start by solving part (a) and then discuss part (b).

### Part (a) Solution:

First, isolate the exponential term:

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

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

Next, divide both sides by 90,000:

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

Simplify the right side:

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

Now take the natural logarithm of both sides to solve for \( t \):

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

Now, solve for \( t \):

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

Finally, I'll calculate the value of \( t \).

### Part (b) Discussion:
To understand what happens to the concentration \( c(t) \) as time \( t \) increases indefinitely, we can analyze the equation's behavior as \( t \) tends to infinity. Notice that as \( t \) increases, the term \( \left(\frac{1}{10,000}\right)^{\frac{t}{3}} \) tends to zero because it is a very small positive number raised to a positive power. This implies that the concentration \( c(t) \) will approach 9 kg/m\(^3\) over time.

I’ll now calculate the value for \( t \).### Part (a):
The concentration of the pollutant will reach 6 kg/m³ after approximately \( t = 3.36 \) minutes.

### Part (b):
As time \( t \) increases indefinitely, the concentration \( c(t) \) will approach 9 kg/m³. This is because the exponential term \( \left(\frac{1}{10,000}\right)^{\frac{t}{3}} \) will get closer to zero, leaving the concentration to approach its maximum value of 9 kg/m³.

Would you like any more details or have any questions?

Here are some related questions to consider:
1. How would the time change if the concentration needed was 8 kg/m³ instead of 6 kg/m³?
2. What if the initial concentration was different? How would that affect the time calculation?
3. How does the factor \( 10,000 \) influence the speed at which the concentration changes?
4. What would happen if the exponent \( \frac{t}{3} \) was modified to a different fraction?
5. Can this model be applied to other types of pollutants, and what adjustments would be needed?
6. How does the concentration function behave for very small and very large values of \( t \)?
7. What is the significance of the initial concentration being 9 kg/m³?
8. How can we verify this solution experimentally or using a different mathematical approach?

**Tip:** Always ensure to check the limits of the function or model when dealing with real-world scenarios to predict long-term behavior accurately.

Explore how to analyze pollutant concentration over time using an exponential decay model. Learn how to find specific concentrations and predict long-term behavior. Suitable for advanced high school students.

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