Find the arc length of the curve y = a ln (a²)/(a² - x²)

Find the arc length of the curve $y = a\ln\frac{a^2}{a^2 - x^2}$ between the points with abscissas $x = 0$ and $x = \frac{a}{2}$.

The key step is recognizing that $1+(y')^2$ simplifies to a perfect square, reducing the square root to a simple rational function.

Step 1: Differentiate

\[y = 2a\ln a - a\ln(a^2-x^2) \implies \frac{dy}{dx} = \frac{2ax}{a^2-x^2}\]

Step 2: Simplify $1+(y')^2$

\[1+\left(\frac{2ax}{a^2-x^2}\right)^2 = \frac{(a^2-x^2)^2+4a^2x^2}{(a^2-x^2)^2} = \frac{a^4+2a^2x^2+x^4}{(a^2-x^2)^2} = \frac{(a^2+x^2)^2}{(a^2-x^2)^2}\]

Since $0\le x\le\frac a2$ keeps $a^2-x^2>0$:

\[\sqrt{1+(y')^2} = \frac{a^2+x^2}{a^2-x^2}\]

Step 3: Decompose via partial fractions

\[\frac{a^2+x^2}{a^2-x^2} = -1+\frac{2a^2}{a^2-x^2} = -1+a\left(\frac{1}{a-x}+\frac{1}{a+x}\right)\]

Step 4: Integrate

\[L = \int_0^{a/2}\left[-1+a\left(\frac{1}{a-x}+\frac{1}{a+x}\right)\right]dx = \left[-x+a\ln\frac{a+x}{a-x}\right]_0^{a/2}\]

Step 5: Evaluate at the limits

At $x=\frac a2$: $-\dfrac a2+a\ln\dfrac{3a/2}{a/2} = -\dfrac a2+a\ln3$. At $x=0$: $0$.

\[L = a\ln3-\frac a2 = a\left(\ln3-\frac12\right)\]

\[\boxed{L = a\left(\ln3-\frac12\right) \text{ units}}\]

Step 6: Numerical value

\[L \approx 0.599\,a \text{ units}\]

Summary

\[\boxed{L = a\left(\ln3-\frac12\right) \approx 0.599\,a \text{ units}}\]

Key insight: The logarithmic curve is engineered so that $1+(y')^2$ is an exact perfect square — a hallmark of curves whose arc length was designed to come out elementary; without this, the square root would not simplify so cleanly.