Let's tackle the problem part by part.
### i) Finding the Formula Connecting [tex]\(P\)[/tex] and [tex]\(t\)[/tex]:
We are given the relationship [tex]\(P \propto t^{-3}\)[/tex]. This means that [tex]\(P\)[/tex] is inversely proportional to the cube of [tex]\(t\)[/tex].
In mathematical terms, we can write this relationship as:
[tex]\[ P = C \cdot t^{-3} \][/tex]
where [tex]\(C\)[/tex] is a constant of proportionality.
Since we are given that [tex]\(P = 4\)[/tex] at some specific time [tex]\(t\)[/tex], we can use this information to determine [tex]\(C\)[/tex].
Let's assume at [tex]\(t = t_0\)[/tex], [tex]\(P = 4\)[/tex]. Thus:
[tex]\[ 4 = C \cdot t_0^{-3} \][/tex]
Solving for [tex]\(C\)[/tex], we have:
[tex]\[ C = 4 \cdot t_0^{3} \][/tex]
So the formula that connects [tex]\(P\)[/tex] and [tex]\(t\)[/tex] is:
[tex]\[ P = 4 \cdot t_0^{3} \cdot t^{-3} \][/tex]
However, since [tex]\(t_0\)[/tex] is a specific constant associated with the initial condition, we can generalize our formula to:
[tex]\[ P = 4 \cdot \left(t / t_0\right)^{-3} \][/tex]
If we set [tex]\(t_0 = 1\)[/tex] for simplicity, then:
[tex]\[ P = 4 \cdot t^{-3} \][/tex]
### ii) Finding the Value of [tex]\(t\)[/tex] When [tex]\(P = \frac{1}{2}\)[/tex]:
We need to find [tex]\(t\)[/tex] when [tex]\(P = \frac{1}{2}\)[/tex] using the formula we derived.
Starting from:
[tex]\[ P = 4 \cdot t^{-3} \][/tex]
Substitute [tex]\(P\)[/tex] with [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ \frac{1}{2} = 4 \cdot t^{-3} \][/tex]
Now, solve for [tex]\(t\)[/tex]:
[tex]\[ \frac{1}{2} = \frac{4}{t^{3}} \][/tex]
To isolate [tex]\(t^3\)[/tex], we multiply both sides by [tex]\(t^3\)[/tex]:
[tex]\[ \frac{1}{2} t^3 = 4 \][/tex]
Next, multiply both sides by 2:
[tex]\[ t^3 = 8 \][/tex]
Finally, take the cube root of both sides to solve for [tex]\(t\)[/tex]:
[tex]\[ t = \sqrt[3]{8} \][/tex]
[tex]\[ t = 2 \][/tex]
So, when [tex]\(P = \frac{1}{2}\)[/tex], the value of [tex]\(t\)[/tex] is 2.
