Arc length

When $x$ moves by $dx$, $y$ moves by about $y'\,dx$, so a small piece of the curve has length $\sqrt{(dx)^2+(y'\,dx)^2}=\sqrt{1+(y')^2}\,dx$. Adding up these pieces gives the length of the curve.

In textbook problems, the expression under the root often turns into a perfect square. If $y'=\frac{u-\frac{1}{u}}{2}$, then $1+(y')^2=\left(\frac{u+\frac{1}{u}}{2}\right)^2$, and for $u>0$ the square root becomes $\frac{u+\frac{1}{u}}{2}$.

Examples

1. 01The formula for arc length. What goes in □?$\displaystyle L = \int \sqrt{1+(\square)^{2}}\,dx$AnswerHide

   Answer$\displaystyle y'$
2. 02What is $1+(y')^2$?$\displaystyle y'=\frac{e^{x}-\frac{1}{e^{x}}}{2}$AnswerHide

   Answer$\displaystyle \left(\frac{e^{x}+\frac{1}{e^{x}}}{2}\right)^{2}$
3. 03For $x>0$, what is $\sqrt{1+(y')^2}$?$\displaystyle 1+(y')^{2}=\left(\frac{x^{2}+\frac{1}{x^{2}}}{2}\right)^{2}$AnswerHide

   Answer$\displaystyle \frac{x^{2}+\frac{1}{x^{2}}}{2}$
4. 04What is the value of this definite integral?$\displaystyle \int_{1}^{3} \frac{x^{2}+\frac{1}{x^{2}}}{2}\,dx$AnswerHide

   Answer$\displaystyle \frac{14}{3}$
5. 05For $x>0$, what is $\sqrt{1+(y')^2}$?$\displaystyle 1+(y')^{2}=\left(\frac{x+\frac{1}{x}}{2}\right)^{2}$AnswerHide

   Answer$\displaystyle \frac{x+\frac{1}{x}}{2}$
6. 06What is $1+(y')^2$?$\displaystyle y'=\frac{x^{2}-\frac{1}{x^{2}}}{2}$AnswerHide

   Answer$\displaystyle \left(\frac{x^{2}+\frac{1}{x^{2}}}{2}\right)^{2}$

FAQ

Q1Why does $1+(y')^2$ become a perfect square?AnswerHide

A

Because $\left(\frac{u-\frac{1}{u}}{2}\right)^2+1=\frac{u^2-2+\frac{1}{u^2}}{4}+1=\frac{u^2+2+\frac{1}{u^2}}{4}$: the middle $-2$ turns into $+2$. Curves such as $y=\frac{x^3}{6}+\frac{1}{2x}$ and $y=\frac{e^x+e^{-x}}{2}$ have this form.

Q2Why the condition $x>0$ when removing the square root?AnswerHide

A

Because $\sqrt{A^2}=|A|$. For $x>0$, $\frac{x+\frac{1}{x}}{2}>0$, so the root can be removed without an absolute value.