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\begin{document}


%%%(change to appropriate class and semester)
MATH 2222HJ Spring 2016

%%%(change to appropriate quiz type and date)
Quiz 2 \hspace{1.9in} SOLUTION
\vspace{2pc}

%%%(modify rules, time, points as appropriate)
Show all work clearly and in order, and circle your final answers.  

Justify your answers algebraically whenever possible. Unjustified work may not receive full credit.
\vspace{2pc}

\begin{problem}{2}
Find a parametrization of the tangent line of the helix $\vec{H}(t)=\left<\cos(t),\sin(t),t \right>$ at the point $\left(0,1,\dfrac{\pi}{2} \right)$.
\end{problem} \\
\textit{Solution:} First note that the value of $t_0$ such that $\vec{H}(t_0) = \left< 0, 1, \dfrac{\pi}{2} \right>$ is $t_0=\dfrac{\pi}{2}$. Now find the tangent vector function
$$\vec{H}^{\hspace{2pt} \prime}(t)=\left< - \sin(t), \cos(t), 1 \right>.$$
Thus the tangent vector at the point in question is $\vec{H}^{\hspace{2pt}\prime}\left(\dfrac{\pi}{2} \right)=<-1,0,1>.$ The tangent line we are seeking is the line that goes through the point $\left( 0,1, \dfrac{\pi}{2} \right)$ parallel to the tangent vector $<-1,0,1>$, hence the tangent line is given by
$$\vec{r}(t)=\left< 0,1,\dfrac{\pi}{2} \right> + t <-1,0,1>.$$
\begin{problem}{3}
Find the arc length of the curve $\left\{ \begin{array}{ll} \vec{r}(t)=\left< t, -\log \left( \cos(t) \right) \right> \\
-\dfrac{\pi}{4} \leq t \leq \dfrac{\pi}{4}.
\end{array} \right.$ \\
Note: you may find the trigonometric identity $\tan^2(t)+1=\sec^2(t)$ and the following integral \href{http://www.wolframalpha.com/input/?i=antiderivative+of+sec%28t%29}{formula} useful:
$$\displaystyle\int \sec(t) \mathrm{d}t = \log(\tan(t)+\sec(t))+C$$
\end{problem} \\
\textit{Solution:} Compute the tangent vector function $\vec{r}'(t)=<1,\tan(t)>.$ Now \href{http://www.wolframalpha.com/input/?i=arclength+of+y%3D-log%28cos%28t%29%29+on+-pi%2F4+to+pi%2F4}{compute the arc length} as follows:
$$\begin{array}{ll}
L &= \displaystyle\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \lVert <1, \tan(t)> \rVert dt \\
&= \displaystyle\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sqrt{1+\tan^2(t)} dt \\
&= \displaystyle\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sqrt{\sec^2(t)} dt \\
&= \displaystyle\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec(t) dt \\
&= \log(\tan(t)+\sec(t)) \Bigg|_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \\
&= \log\left( \tan\left( \dfrac{\pi}{4} \right) + \sec\left( \dfrac{\pi}{4} \right) \right) - \log \left( \tan\left( -\dfrac{\pi}{4} \right) + \sec\left( -\dfrac{\pi}{4} \right) \right) \\
&=\log(1+\sqrt{2})-\log(-1+\sqrt{2}) \\
&= \log \left( \dfrac{\sqrt{2}+1}{\sqrt{2}-1} \right) \\
&= \log(3+2\sqrt{2}),
\end{array}$$
where on the last line we rationalized the denominator.

\showpoints
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