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?

  • 1
    $\begingroup$ My take is that in the proof of the arc length, one really considers the Pythagorean theorem to determine a distance, so the notion of taking the negative square root becomes obsolete, as length is +ve $\endgroup$
    – asymptotic
    Mar 10, 2023 at 12:12

1 Answer 1


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.)

  • 1
    $\begingroup$ I see, so is it because x>0, the function $|2x+ \frac{1}{8x}|$ would equal to $2x+ \frac{1}{8x}$? And would that also mean that if $x<0$, the resulting function would instead be $-2x- \frac{1}{8x}$ as we want the nonnegative answer? $\endgroup$
    – Willow
    Mar 11, 2023 at 5:43
  • 1
    $\begingroup$ @Willow: Yes, supposing you had $\ln|x|$ instead of just $\ln x$ in your function $f(x)$, so that it's defined for $x<0$, then if you were computing the arc length of some part of the curve $y=f(x)$ that lies in the left half-plane $x<0$, you would indeed use $-2x-\frac{1}{8x}$ (since that is the nonnegative square root in that region). $\endgroup$ Mar 11, 2023 at 7:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .