So I am learning the arc length of a function right now and encountered an example from my textbook as follows:
Question: fnd the arc length function for the curve $y=x^2- \frac{ln(x)}{8}$ taking $P_{0}(1,1)$ as a starting point.
The solution:
if $f(x)=x^2- \frac{ln(x)}{8}$ then, $$f'(x)=2x-\frac{1}{8x}$$ $$1+[f'(x)]^2=1+(2x-\frac{1}{8x})^2$$ $$=(2x+\frac{1}{8x})^2$$
Hence, $$\sqrt{1+[f'(x)]^2}= 2x+\frac{1}{8x}$$ Since $x>0$
My question is, why do we take the positive square root if $x>0$? I understand that $x$ must be greater than $0$ because it is according the domain $lnx>0$. But I don't understand why do we take the positive square root because of this.
And is there any occasion where we will take the negative square root? And why?
From $$ 1+f'(x)^2 = (2x+\tfrac{1}{8x})^2 $$ you immediately obtain $$ \sqrt{1+f'(x)^2} = \sqrt{(2x+\tfrac{1}{8x})^2} = |2x+\tfrac{1}{8x}| , $$ where the square root symbol (as always, in real analysis) denotes the nonnegative square root.
The point where the condition $x>0$ enters the argument is when you continue to simplify this: $$ \cdots = |2x+\tfrac{1}{8x}| = 2x+\tfrac{1}{8x}, \quad\text{since $x>0$} . $$ To spell it out explicitly: $x>0$ implies that the expression $2x+\tfrac{1}{8x}$ is positive, so that it equals its own absolute value.
(So this has nothing at all to do with ever using the negative square root.)