Exercises: Arc Length

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We practice computing arc length.

Find the arc length of the function on the given interval: $\displaystyle f(x) = \sqrt {8}x$ on $[-1, 1]$ $L = \answer {6}.$

Find the arc length of the function on the given interval: $\displaystyle f(x) = \ln \left (\cos x\right )$ on $[0, \pi /4]$ . (You may use the fact that $\int \sec x ~ dx = \ln |\sec x + \tan x| + C$ .) $L = \answer {\ln (1 + \sqrt {2})}.$

Set up the integral to compute the arc length of the function on the given interval: $\displaystyle f(x) = x^2$ on $[0, 1]$ . $L = \int _{\answer {0}}^{\answer {1}} \answer {\sqrt {1+4x^2}} \ dx$

Let $\displaystyle y = \frac {x^4}{16} + \frac {1}{2x^2}$ . Find the arc length for $1 \leq x \leq \sqrt {2}$ . $L = \answer {\frac {7}{16}}.$

Sample Quiz Questions

Compute the arc length of the curve $y = \frac {3}{4}x^{-2} + \frac {1}{24}x^{4} - 1$ between the endpoints $x = \sqrt {3}$ and $x = \sqrt {6}$ . (Hints won’t be revealed until after you choose a response.)

$\displaystyle \frac {5}{12}$ $\displaystyle \frac {7}{12}$ $\displaystyle \frac {3}{4}$ $\displaystyle \frac {11}{12}$ $\displaystyle \frac {13}{12}$ $\displaystyle \frac {5}{4}$

Applying the formula for arc length gives that $\begin {aligned} L & = \int _{\sqrt {3}}^{\sqrt {6}} \sqrt {1 + \left ( \frac {dy}{dx} \right )^2}~dx = \int _{\sqrt {3}}^{\sqrt {6}} \sqrt {1 + \left ( -\frac {3}{2}x^{-3} + \frac {1}{6}x^{3} \right )^2}~dx \end {aligned}$

$\begin {aligned} & = \int _{\sqrt {3}}^{\sqrt {6}} \sqrt {1 + \frac {9}{4}x^{-6} - \frac {1}{2} + \frac {1}{36}x^{6}} ~ dx = \int _{\sqrt {3}}^{\sqrt {6}} \sqrt { \left ( \frac {3}{2}x^{-3} + \frac {1}{6}x^{3} \right )^2} ~ dx \\ & = \int _{\sqrt {3}}^{\sqrt {6}} \left ( \frac {3}{2}x^{-3} + \frac {1}{6}x^{3} \right ) ~ dx = \left . \left ( -\frac {3}{4}x^{-2} + \frac {1}{24}x^{4} \right ) \right |_{\sqrt {3}}^{\sqrt {6}} \\ & = \left ( -\frac {1}{8} + \frac {3}{2} \right ) - \left ( -\frac {1}{4} + \frac {3}{8} \right ) = \frac {5}{4}. \end {aligned}$ Note that you must always take the positive square root in going from line two to line three. In particular, if you get a negative answer, you have likely taken the negative square root.

Sample Exam Questions

(2017 Midterm 1) Compute the length of the curve $x = \frac {1}{8} (y^2 + 2y) - \ln (y+1)$ between $y=0$ and $y = 2$ .

$\displaystyle 1 + \ln 3$ $\displaystyle 2 + \ln 6$ $\displaystyle 3 + \ln 9$ $\displaystyle 4 + \ln 12$ $\displaystyle 5 + \ln 15$ none of the above

Find the arc length of the following curve between $x=-1$ and $x=1$ : $y = 3 \cosh \frac {x}{3}.$ (Note: $\cosh x = (e^x + e^{-x})/2$ .)

$\displaystyle \frac {e}{3} - \frac {1}{3e}$ $\displaystyle \frac {e}{2} - \frac {1}{2e}$ $\displaystyle e - \frac {1}{e}$ $\displaystyle 2 e - \frac {2}{e}$ $\displaystyle 3 e - \frac {3}{e}$ none of the above

A certain curve $y = f(x)$ in the plane has the property that its length between the endpoints $x=0$ and $x=a$ is equal to $\int _0^a \sqrt {1 + \sin ^2 t} ~ dt$ for every value of $a > 0$ . Assuming the curve passes through the points $(0,0)$ and $\left (\frac {\pi }{2},1\right )$ , what is $f\left ( \frac {\pi }{4} \right )$ ?

$\displaystyle \frac {1}{2}$ $\displaystyle \frac {1}{\sqrt {2}}$ $\displaystyle 1 - \frac {1}{\sqrt {2}}$ $\displaystyle 0$ $\displaystyle - \frac {1}{\sqrt {2}}$ none of these

Find the length of the part of the curve $\displaystyle y = \frac {3}{16} e^{2x} + \frac {1}{3} e^{-2x}$ for $0 \leq x \leq \ln 2$ .

$\displaystyle \frac {13}{16}$ $\displaystyle \frac {11}{16}$ $\displaystyle \frac {3}{8}$ $\displaystyle \frac {9}{8}$ $\displaystyle \frac {29}{64}$ $\displaystyle \frac {3}{4}$

Find the length of the part of the curve $\displaystyle y = \frac {x^4}{4} + \frac {1}{8x^2}$ for $1 \leq x \leq 2$ .

$\displaystyle \frac {13}{16}$ $\displaystyle \frac {11}{16}$ $\displaystyle \frac {7}{8}$ $\displaystyle \frac {13\sqrt {2}}{16}$ $\displaystyle \frac {11\sqrt {2}}{16}$ $\displaystyle \frac {7\sqrt {2}}{8}$