Arc Length

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We practice setting up and executing arc length calculations.

(Video) Calculus: Single Variable

Online Texts

Examples

Compute the arc length of the curve $x = -\frac {1}{8}y^{-2} - \frac {1}{4}y^{4} - 2$ between the endpoints $y = 1/\sqrt {2}$ and $y = 1$ .

First we compute the derivative: $\frac {dx}{dy} = \answer {\frac {1}{4} y^{-3} - y^3}.$
Next, we write the arc length element: $ds = \sqrt { 1 + \left (\frac {dx}{dy}\right )^2} dy = \sqrt {1 + \left ( \answer { \frac {1}{4} y^{-3} - y^3} \right )^2} dy.$
The key to integrating is to first fully simplify the quantity inside the square root: $\begin {aligned} 1 + \left ( \answer {\frac {1}{4} y^{-3} - y^3} \right )^2 = & 1 + \frac {1}{16} y^{-6} + 2 \left ( \frac {1}{4} y^{-3} \right ) \left ( - y^3 \right ) + y^6 \\ & = 1 + \frac {1}{16} y^{-6} - \frac {1}{2} + y^6 \end {aligned}$
When the middle term in a binomial (i.e., FOIL) expansion is $-1/2$ adding one always gives a perfect square. In this case, the algebraic fact is $1 + \frac {1}{16} y^{-6} - \frac {1}{2} + y^6 = \frac {1}{16} y^{-6} + \frac {1}{2} + y^6 = \left ( \answer {\frac {1}{4} y^{-3} + y^3} \right )^2.$ The general phenomenon is that what was formerly the square of a difference—in this case, $\left ( \frac {1}{4} y^{-3} - y^3 \right )^2$ —becomes the square of the corresponding sum.
Therefore $\begin {aligned} L & = \int _{1/\sqrt {2}}^{1} \sqrt {1 + \left ( \frac {dx}{dy} \right )^2}~dy = \int _{1/\sqrt {2}}^{1} \sqrt {1 + \left ( \answer {\frac {1}{4}y^{-3} - y^{3}} \right )^2}~dy \\ & = \int _{1/\sqrt {2}}^{1} \sqrt {1 + \frac {1}{16}y^{-6} - \frac {1}{2} + y^{6}} ~ dy = \int _{1/\sqrt {2}}^{1} \sqrt { \left ( \answer {\frac {1}{4}y^{-3} + y^{3}} \right )^2} ~ dy \\ & = \int _{1/\sqrt {2}}^{1} \answer {\frac {1}{4}y^{-3} + y^{3}} ~ dy = \left . \left ( \answer {-\frac {1}{8}y^{-2} + \frac {1}{4}y^{4}} \right ) \right |_{1/\sqrt {2}}^{1} \\ & = \left ( -\frac {1}{8} + \frac {1}{4} \right ) - \left ( -\frac {1}{4} + \frac {1}{16} \right ) = \frac {5}{16}. \end {aligned}$