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.

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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., $\phi = \sqrt{x^2 + y^2}$)?