Find the relative error in Z, if `Z=A^(4)B^(1//3)//CD^(3//2)`.

Options

A. `(DeltaZ)/(Z)=4(DeltaA)/(A)+(1)/(3)(DeltaB)/(B)+(DeltaC)/(C)+(3)/(2)(DeltaD)/(D)`
B. `(DeltaZ)/(Z)=4(DeltaA)/(A)+(1)/(3)(DeltaB)/(B)-(DeltaC)/(C)-(3)/(2)(DeltaD)/(D)`
C. `(DeltaZ)/(Z)=4(DeltaA)/(A)+(1)/(3)(DeltaB)/(B)+(DeltaC)/(C)-(3)/(2)(DeltaD)/(D)`
D. `(DeltaZ)/(Z)=4(DeltaA)/(A)+(1)/(3)(DeltaB)/(B)-(DeltaC)/(C)+(3)/(2)(DeltaD)/(D)`

Explanation

To find the relative error in the quantity \( Z \), given by the equation:

\[ Z = \frac{A^4 B^{1/3}}{C D^{3/2}} \]

we will follow these steps:

### Step 1: Identify the formula for relative error
The relative error in a quantity can be expressed as:

\[ \frac{\Delta Z}{Z} \]

where \( \Delta Z \) is the absolute error in \( Z \).

### Step 2: Apply the error propagation formula
For a function of multiple variables, the relative error can be calculated using the formula:

\[ \frac{\Delta Z}{Z} = \sum \left( \text{power of each variable} \times \frac{\Delta \text{variable}}{\text{variable}} \right) \]

### Step 3: Differentiate the components of \( Z \)
In our case, we have:

- \( A \) raised to the power of 4
- \( B \) raised to the power of \( \frac{1}{3} \)
- \( C \) raised to the power of 1 (in the denominator)
- \( D \) raised to the power of \( \frac{3}{2} \) (in the denominator)

### Step 4: Write the contributions to the relative error
Now we can express the relative error in \( Z \) as follows:

\[ \frac{\Delta Z}{Z} = 4 \frac{\Delta A}{A} + \frac{1}{3} \frac{\Delta B}{B} + 1 \frac{\Delta C}{C} + \frac{3}{2} \frac{\Delta D}{D} \]

### Step 5: Combine the terms
Putting it all together, we have:

\[ \frac{\Delta Z}{Z} = 4 \frac{\Delta A}{A} + \frac{1}{3} \frac{\Delta B}{B} + \frac{\Delta C}{C} + \frac{3}{2} \frac{\Delta D}{D} \]

This expression gives us the relative error in \( Z \).

### Final Answer
Thus, the relative error in \( Z \) is:

\[ \frac{\Delta Z}{Z} = 4 \frac{\Delta A}{A} + \frac{1}{3} \frac{\Delta B}{B} + \frac{\Delta C}{C} + \frac{3}{2} \frac{\Delta D}{D} \]

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