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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 $[-1, 1]$
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 $\int \sec x ~ dx = \ln |\sec x + \tan x| + C$.)
Set up the integral to compute the arc length of the function on the given interval: $\displaystyle f(x) = x^2$ on $[0, 1]$.
Let $\displaystyle y = \frac {x^4}{16} + \frac {1}{2x^2}$. Find the arc length for $1 \leq x \leq \sqrt {2}$.

### Sample Quiz Questions

Compute the arc length of the curve between the endpoints $x = \sqrt {3}$ and $x = \sqrt {6}$. (Hints won’t be revealed until after you choose a response.)

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

### 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 $x=1$: (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 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 $1 \leq x \leq 2$.

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