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\problemgrouptitle{Arc Length}

\addtoanswerkey{\begin{multicols}{2}}

\begin{problemblock}{Find the length of the given curve over the given interval.}

\item $\displaystyle y=x^{3/2}+1$ \ over $1 \le x \le 4$
\answer{
$\displaystyle \frac 1{27}[40^{3/2}-13^{3/2}]$
}

\item $\displaystyle y=\frac 23 (x-2)^{3/2}$ \ over $2\le x\le 5$
\answer{
$\displaystyle \frac{14}{3}$
}

\item $\displaystyle y=\ln(\sin x)$ \ over $\frac{\pi}{6}\le x\le \frac{\pi}{4}$
\answer{
$\displaystyle \ln | \sqrt 2-1|-\ln |2-\sqrt 3 |$
}

\item $\displaystyle x=\frac 1 {32} y^4+y^{-2}$ \ over $1\le x\le 3$
\answer{
$\displaystyle \frac {61}{18}$
}

\item $\displaystyle x=\frac 1 {5} y^{5/2}+y^{-1/2}$ \ over $1\le x\le 4$
\answer{
$\displaystyle \frac{67}{10}$
}

\item $\displaystyle y=e^{x/2}+e^{-x/2}$ \ over $0\le x\le 1$
\answer{
$\displaystyle \frac{e-1}{\sqrt e}$
}

\end{problemblock}

\begin{problemblock}{Find the arc length function $s(x)$ of the given curve.}

\item $\displaystyle y=1+x^2$ \  with starting point $P(0, 1)$ 
\answer{
$\displaystyle s(x)=\frac{1}{4}\left(2x\sqrt{1+4x^2}+\ln|\sqrt{1+4x^2}+2x|\right)$
}

\item $\displaystyle y=2x^{1/2}$ \ with starting point $P(1, 2)$
\answer{
$\displaystyle s(x)=\sqrt x \,\sqrt{1+x}+\ln |\sqrt{1+x}+\sqrt x|-\sqrt2-\ln |\sqrt 2+1|$
} 

\item $\displaystyle y=\frac{1}{3} (x-2)^{3/2}-(x-2)^{1/2}$ \ with starting point $P(3, -\frac{2}{3})$ 
\answer{
$\displaystyle \frac{1}{3}(x-2)^{3/2}+(x-2)^{1/2}-\frac{4}{3}$
} 

\item $\displaystyle y=\frac{4}{3} (x+4)^{3/2}+5$ \ with starting point $P\left(0, \frac {47}{3}\right)$ 
\answer{
$\displaystyle \frac{1}{6} (4x+17)^{3/2}-\frac {1}{6} (17)^{3/2}$
} 

\end{problemblock}

\addtoanswerkey{\end{multicols}}

\item \label{itm:arclengthodd} Show that the arc length function $s(x)=\ln|\sec x+\tan x|$ from Example~\ref{exam:arclength1} is an odd function of $x$.
\answer{
  Let $s(x)=\ln|\sec x+\tan x|$.  Then
  \begin{align*}
    s(-x)&=\ln|\sec(-x)+\tan(-x)|=\ln|\sec x-\tan x|=\ln\left|\frac{1}{\cos x}-\frac{\sin x}{\cos x}\right|=\ln\left|\frac{1-\sin x}{\cos x}\right|\\
         &=\ln\left|\left(\frac{\cos x}{1-\sin x}\right)^{-1}\right|=\ln\left(\left|\frac{\cos x}{1-\sin x}\right|^{-1}\right)=-\ln\left|\frac{\cos x}{1-\sin x}\right|\\
         &=-\ln\left|\left(\frac{\cos x}{1-\sin x}\right)\left(\frac{1+\sin x}{1+\sin x}\right)\right|
           =-\ln\left|\left(\frac{\cos x}{1-\sin^2\! x}\right)\left(1+\sin x\right)\right|\\
         &=-\ln\left|\left(\frac{\cos x}{\cos^2\! x}\right)\left(1+\sin x\right)\right|=-\ln\left|\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right|
           =-\ln|\sec x+\tan x|=-s(x)
  \end{align*}
  Therefore $s(x)$ is an odd function of $x$\,.
}

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