Find length of the arc of $y^2=x$ I am embarrassingly stuck on this example. My textbook provides the answer (picture below) and the steps, but I am unable to follow the math. I have been stuck for more then an hour.
Find the length of the arc of the parabola $y^2=x$ from (0,0) to (1,1). 
Since $x=y^2$, we have $\frac{dx}{dy} = 2y$.
$$\begin{align}
L &= \int_{0}^{1} \sqrt{1+(\frac{dx}{dy})^2}dy \\
&= \int_{0}^{1} \sqrt{1+(2y)^2}dy
\end{align}$$
Here is where I am getting stuck. I am supposed to make a trigonometric where $y = \frac{1}{2}\tan\theta$, but based off the trig substitution equation I have, $\sqrt{a^2+y^2} \rightarrow y = \tan\theta$ I am getting confused how I am supposed to get $y = \frac{1}{2}\tan\theta$. I keep thinking it should be $ y = \frac{1}{4}\tan\theta$.
Here is where I get my $y$ value from: 
$$\sqrt{a^2 + y^2} = \sqrt{1+ 2^2 y^2} \\
= \sqrt{\frac{1}{4} + y^2}$$ Which gives me $y = \frac{1}{4} tan\theta$
Here is a picture of the example in the textbook. I am not able to follow the math :(


When you say:

"Here is where I get my $y$ value from:  $$\sqrt{a^2 + y^2} = \sqrt{1+
> 2^2 y^2} \\
> = \sqrt{\frac{1}{4} + y^2}$$ Which gives me $y = \frac{1}{4} tan\theta$"

From here you want not $y = \frac{1}{4} \tan\theta$ but $y^2 = \frac{1}{4} \tan^2\theta$. Taking square roots gives $y = \frac{1}{2} \tan\theta$.
