Sure, let's go through the solution step by step.
### Step 1: Determine the Relationship Formula
Given that [tex]\( P \propto t^{-3} \)[/tex], we can express this relationship as:
[tex]\[ P = k \cdot t^{-3} \][/tex]
where [tex]\( k \)[/tex] is a proportionality constant.
### Step 2: Calculate the Proportionality Constant [tex]\( k \)[/tex]
We are given that [tex]\( P = 4 \)[/tex] when [tex]\( t = 2 \)[/tex]. Using the relationship formula, we can solve for [tex]\( k \)[/tex]:
[tex]\[ 4 = k \cdot 2^{-3} \][/tex]
First, calculate [tex]\( 2^{-3} \)[/tex]:
[tex]\[ 2^{-3} = \frac{1}{2^3} = \frac{1}{8} \][/tex]
So, the equation becomes:
[tex]\[ 4 = k \cdot \frac{1}{8} \][/tex]
Now solve for [tex]\( k \)[/tex]:
[tex]\[ k = 4 \cdot 8 = 32 \][/tex]
Thus, the constant [tex]\( k \)[/tex] is 32.
### Step 3: Derive the Formula for [tex]\( P \)[/tex] in Terms of [tex]\( t \)[/tex]
Substitute [tex]\( k = 32 \)[/tex] back into the original relationship formula:
[tex]\[ P = 32 \cdot t^{-3} \][/tex]
or equivalently,
[tex]\[ P = \frac{32}{t^3} \][/tex]
### Step 4: Find the Value of [tex]\( t \)[/tex] when [tex]\( P = \frac{1}{2} \)[/tex]
We are asked to find the value of [tex]\( t \)[/tex] when [tex]\( P = \frac{1}{2} \)[/tex]. Using the derived formula, set [tex]\( P = \frac{1}{2} \)[/tex]:
[tex]\[ \frac{1}{2} = \frac{32}{t^3} \][/tex]
Now solve for [tex]\( t^3 \)[/tex]:
[tex]\[ t^3 = \frac{32}{\frac{1}{2}} \][/tex]
[tex]\[ t^3 = 32 \cdot 2 \][/tex]
[tex]\[ t^3 = 64 \][/tex]
Finally, take the cube root of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \sqrt[3]{64} \][/tex]
[tex]\[ t = 4 \][/tex]
Therefore, the value of [tex]\( t \)[/tex] when [tex]\( P = \frac{1}{2} \)[/tex] is approximately [tex]\( 4 \)[/tex] (or more precisely, 3.9999999999999996 due to the limitations of floating-point arithmetic in calculations).
