I have a homework assignment in which I am asked to find the arc length of $y=\frac{1} {4}x^2 - \frac 1 2 ln(x)$ over the interval [1, 3e]. I know that to find arc length I can take the integral of $\sqrt[]{1 + {\frac {dy} {dx}}^2}$, however the answer key to the homework takes a different approach:
If $(y')^2 = (\frac x 2 - \frac 1 {2x})^2 = (\frac {x^2-1} {2x})^2 = \frac {x^4-2x^2+1} {4x^2}$
Then $(y')^2 + 1 = \frac {x^4-2x^2+1} {4x^2} + 1 = \frac {x^4+2x^2+1} {4x^2}$, right?
They find $y'$, and then equate $(y')^2 + 1$ to $(y')^2$. Why can they do that? It certainly makes the problem a lot easier.
They are using $(y^{\prime})^2+1=(\frac{x}{2}-\frac{1}{2x})^2+1=(\frac{x^2}{4}-\frac{1}{2}+\frac{1}{4x^2})+1=\frac{x^2}{4}+\frac{1}{2}+\frac{1}{4x^2}=(\frac{x}{2}+\frac{1}{2x})^2$