# Finding arc length, why does $(y')^2 + 1= (y')^2$?

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.

• Because $\frac{x^4-2x^2+1}{4x^2} \neq \frac{x^4+2x^2+1}{4x^2}$. One has $-2x^2$ in the numerator, the other has $+2x^2$. (Also, nothing is ever equal to itself plus $1$ because $1\neq 0$.) – kccu Feb 27 '16 at 23:48
• "They find $y'$, and then equate $(y')^2 + 1$ to $(y')^2$. Why can they do that?" They can't do that, and they don't, where do you see them do this? – Adam Francey Feb 28 '16 at 0:27

## 1 Answer

They are using $(y^{\prime})^2+1=(\frac{x}{2}-\frac{1}{2x})^2+1=(\frac{x^2}{4}-\frac{1}{2}+\frac{1}{4x^2})+1=\frac{x^2}{4}+\frac{1}{2}+\frac{1}{4x^2}=(\frac{x}{2}+\frac{1}{2x})^2$