Arc Length (College Board AP® Calculus BC): Study Guide

Questions on arc lengths may ask you to leave your answer as a definite integral.

Show that the length of the curve $y=1+\ln x$ from $x=1$ to $x=10$ can be written as

${\int }_1^{10}\sqrt{\frac{x^2+1}{x^2}}dx$

Answer:

In this question, you do not need to evaluate the definite integral

Start by finding $\frac{dy}{dx}$

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{1}{x}\end{array}$

Substitute this derivative and the limits $a=1$ and $b=10$ into the formula $L={\int }_a^b\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$

$L={\int }_1^{10}\sqrt{1+{\left(\frac{1}{x}\right)}^2}dx$

This answer is not yet in the form required by the question

Square the $\frac{1}{x}$ and add it to the $1$ (using a lowest common denominator of $x^2$)

$\begin{array}{rcl}L& =& {\int }_1^{10}\sqrt{1+\frac{1}{x^2}}dx\\ & =& {\int }_1^{10}\sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}}dx\\ & =& {\int }_1^{10}\sqrt{\frac{x^2+1}{x^2}}dx\end{array}$

This is now in the form required

${\int }_1^{10}\sqrt{\frac{x^2+1}{x^2}}dx$

Find the length of the curve $y=\frac{2}{3}{x}^{\frac{3}{2}}$ from $x=0$ to $x=8$.

Answer:

Use the formula $L={\int }_a^b\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$

First find $\frac{dy}{dx}$

$\begin{array}{rcl}\frac{dy}{dx}& =& x^{\frac{1}{2}}\end{array}$

Substitute the derivative and the limits $a=0$ and $b=8$ into the formula $L={\int }_a^b\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$

$L={\int }_0^8\sqrt{1+{\left(x^{\frac{1}{2}}\right)}^2}dx$

Method 1

If calculators are allowed, evaluate this definite integral on your calculator

$L={\int }_0^8\sqrt{1+{\left(x^{\frac{1}{2}}\right)}^2}dx=\frac{52}{3}$

Method 2

If calculators are not allowed, continue by simplifying under the square root

$\begin{array}{rcl}L& =& {\int }_0^8\sqrt{1+x}dx\end{array}$

There are many ways to evaluate this definite integral, for example integration by substitution using $u=1+x$

Note that $\frac{du}{dx}=1$ so $du=dx$, and also that $x=0\Rightarrow u=1$, $x=8\Rightarrow u=9$

$\begin{array}{rcl}L& =& {\int }_1^9\sqrt{u}du\\ & =& {\int }_1^9{u}^{\frac{1}{2}}du\\ & =& {\left[\frac{2}{3}{u}^{\frac{3}{2}}\right]}_1^9\\ & =& \frac{2}{3}{\left[u^{\frac{3}{2}}\right]}_1^9\\ & =& \frac{2}{3}\left({\left(\sqrt{9}\right)}^3-{\left(\sqrt{1}\right)}^3\right)\\ & =& \frac{2}{3}\times \left(27-1\right)\end{array}$

The length of the curve is $\frac{52}{3}$ units