A physical quantity ( is calculated using the formula (² = ẞ2 x y³ where ẞ is a known standard constant. If the percentage errors in measuring x and y are 2% and 4% respectively, what is the maximum percentage error in the calculated value of ( (in %)?

We are given the formula for a physical quantity \( \phi \) (represented by the symbol \( \phi \)):

\[
\phi^2 = \beta^2 \cdot x \cdot y^3
\]
where \( \beta \) is a known constant, and the percentage errors in measuring \( x \) and \( y \) are 2% and 4%, respectively. We are asked to find the maximum percentage error in the calculated value of \( \phi \).

### Step 1: Rewrite the Formula
First, let's solve for \( \phi \):

\[
\phi = \sqrt{\beta^2 \cdot x \cdot y^3}
\]

Since \( \beta \) is a constant and has no error, it does not contribute to the percentage error. So, we simplify the formula to:

\[
\phi = \beta \cdot \sqrt{x \cdot y^3}
\]

### Step 2: Use Logarithmic Differentiation
To find the percentage error in \( \phi \), we need to take the logarithmic differential of the equation. This will give us the relationship between the percentage errors in \( x \), \( y \), and \( \phi \). Taking the natural logarithm of both sides:

\[
\ln(\phi) = \ln(\beta) + \frac{1}{2} \ln(x) + \frac{3}{2} \ln(y)
\]

Now, differentiate both sides:

\[
\frac{d\phi}{\phi} = \frac{1}{2} \cdot \frac{dx}{x} + \frac{3}{2} \cdot \frac{dy}{y}
\]

### Step 3: Apply Percentage Errors
The percentage error in \( x \) is 2%, so:

\[
\frac{dx}{x} = 0.02
\]

The percentage error in \( y \) is 4%, so:

\[
\frac{dy}{y} = 0.04
\]

Now, substitute these into the differentiated equation:

\[
\frac{d\phi}{\phi} = \frac{1}{2} \cdot 0.02 + \frac{3}{2} \cdot 0.04
\]

### Step 4: Calculate the Total Percentage Error
Simplify the expression:

\[
\frac{d\phi}{\phi} = 0.01 + 0.06 = 0.07
\]

Thus, the maximum percentage error in \( \phi \) is:

\[
\text{Percentage error in } \phi = 7\%
\]

### Final Answer:
The maximum percentage error in the calculated value of \( \phi \) is **7%**.

### Tip
When dealing with percentage errors, constants like \( \beta \) do not contribute to the error. Focus on the variables that directly affect the result.

Would you like more details or clarification?

### Further Questions:
1. How would the result change if the percentage error in \( y \) was 6% instead of 4%?
2. How does the form of the formula affect how percentage errors propagate?
3. If the percentage error in \( x \) were 0%, what would be the new maximum percentage error in \( \phi \)?
4. How would you calculate the percentage error if \( \beta \) also had an error?
5. Can you apply the same method if the relationship between \( x \) and \( y \) was nonlinear (e.g., \( \phi = \sqrt{x^2 + y^2} \))?

Learn how to calculate the maximum percentage error in a physical quantity given the formula φ² = β² * x * y³. This problem applies logarithmic differentiation to handle percentage errors in measurements, with a detailed solution for students in Grades 11-12.

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