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# 11.E: Applications of Trigonometry (Exercises)

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## 11.1: Applications of Sinusoids

\begin{enumerate}

\item The sounds we hear are made up of mechanical waves. The note A' above the note middle C' is a sound wave with ordinary frequency $f = 440$ Hertz $= 440 \frac{\text{cycles}}{\text{second}}$. Find a sinusoid which models this note, assuming that the amplitude is $1$ and the phase shift is $0$.

\item The voltage $V$ in an alternating current source has amplitude $220 \sqrt{2}$ and ordinary frequency $f = 60$ Hertz. Find a sinusoid which models this voltage. Assume that the phase is $0$.

\item \label{heightlondoneye} The \href{http://en.wikipedia.org/wiki/London_...derline{London Eye}} is a popular tourist attraction in London, England and is one of the largest Ferris Wheels in the world. It has a diameter of 135 meters and makes one revolution (counter-clockwise) every 30 minutes. It is constructed so that the lowest part of the Eye reaches ground level, enabling passengers to simply walk on to, and off of, the ride. Find a sinsuoid which models the height $h$ of the passenger above the ground in meters $t$ minutes after they board the Eye at ground level.

\item \label{leftrightlondoneye} On page \pageref{equationsforcircularmotion} in Section \ref{cosinesinebeyond}, we found the $x$-coordinate of counter-clockwise motion on a circle of radius $r$ with angular frequency $\omega$ to be $x = r\cos(\omega t)$, where $t=0$ corresponds to the point $(r,0)$. Suppose we are in the situation of Exercise \ref{heightlondoneye} above. Find a sinsusoid which models the horizontal \textit{displacement} $x$ of the passenger from the center of the Eye in meters $t$ minutes after they board the Eye. Here we take $x(t) > 0$ to mean the passenger is to the \textit{right} of the center, while $x(t) < 0$ means the passenger is to the \textit{left} of the center.

\item In Exercise \ref{yoyotrick} in Section \ref{Angles}, we introduced the yo-yo trick Around the World' in which a yo-yo is thrown so it sweeps out a vertical circle. As in that exercise, suppose the yo-yo string is 28 inches and it completes one revolution in 3 seconds. If the closest the yo-yo ever gets to the ground is 2 inches, find a sinsuoid which models the height $h$ of the yo-yo above the ground in inches $t$ seconds after it leaves its lowest point.

\item Suppose an object weighing $10$ pounds is suspended from the ceiling by a spring which stretches $2$ feet to its equilibrium position when the object is attached.

\begin{enumerate}

\item Find the spring constant $k$ in $\frac{\text{lbs.}}{\text{ft.}}$ and the mass of the object in slugs.

\item Find the equation of motion of the object if it is released from $1$ foot \textit{below} the equilibrium position from rest. When is the first time the object passes through the equilibrium position? In which direction is it heading?

\item Find the equation of motion of the object if it is released from $6$ inches \textit{above} the equilibrium position with a \textit{downward} velocity of $2$ feet per second. Find when the object passes through the equilibrium position heading downwards for the third time.

\end{enumerate}

\newpage

\item Consider the pendulum below. Ignoring air resistance, the angular displacement of the pendulum from the vertical position, $\theta$, can be modeled as a sinusoid.\footnote{Provided $\theta$ is kept small.' Carl remembers the Rule of Thumb' as being $20^{\circ}$ or less. Check with your friendly neighborhood physicist to make sure.}

\begin{center}

\begin{mfpic}[15]{-3}{3}{-5}{1}

\polyline{(0,0), (0,-5)}

\dashed \polyline{(0,0), (2.5, -4.33)}

\arrow \parafcn{275, 295, 5}{4*dir(t)}

\tlabel[cc](1.29, -4.83){$\theta$}

\hatchcolor[gray]{.7}

\lhatch \rect{(-3,0), (3,1)}

\fillcolor[gray]{.7}

\gfill \circle{(0,-5),0.25}

\gfill \circle{(2.5, -4.33),0.20}

\penwd{1.025}

\circle{(0,-5),0.25}

\circle{(2.5, -4.33),0.25}

\rect{(-3,0), (3,1)}

\end{mfpic}

\end{center}

The amplitude of the sinusoid is the same as the initial angular displacement, $\theta_{\text{\tiny$0$}}$, of the pendulum and the period of the motion is given by

$T = 2\pi \sqrt{\dfrac{l}{g}}$

where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.

\begin{enumerate}

\item Find a sinusoid which gives the angular displacement $\theta$ as a function of time, $t$. Arrange things so $\theta(0) = \theta_{\text{\tiny$0$}}$.

\item In Exercise \ref{pendulumproblem} section \ref{AlgebraicFunctions}, you found the length of the pendulum needed in Jeff's antique Seth-Thomas clock to ensure the period of the pendulum is $\frac{1}{2}$ of a second. Assuming the initial displacement of the pendulum is $15^{\circ}$, find a sinusoid which models the displacement of the pendulum $\theta$ as a function of time, $t$, in seconds.

\end{enumerate}

\item The table below lists the average temperature of Lake Erie as measured in Cleveland, Ohio on the first of the month for each month during the years 1971 -- 2000.\footnote{See this website: \href{http://www.erh.noaa.gov/cle/climate/...mpcle.html}.}} For example, $t=3$ represents the average of the temperatures recorded for Lake Erie on every March 1 for the years 1971 through 2000.

\medskip

\small

\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|r|r|} \hline

Month & & & & & & & & & & & & \\

Number, $t$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12\\

\hline

Temperature & & & & & & & & & & & & \\

($^{\circ}$ F), $T$ & 36 & 33 & 34 & 38 & 47 & 57 & 67 & 74 & 73 & 67 & 56 & 46 \\ \hline

\end{tabular}

\normalsize

\medskip

\begin{enumerate}

\item \label{LakeErieTempData} Using the techniques discussed in Example \ref{sinusoidsunlight}, fit a sinusoid to these data.

\item Using a graphing utility, graph your model along with the data set to judge the reasonableness of the fit.

\item Use the model you found in part \ref{LakeErieTempData} to predict the average temperature recorded for Lake Erie on April $15^{\text{th}}$ and September $15^{\text{th}}$ during the years 1971--2000.\footnote{The computed average is $41^{\circ}$F for April $15^{\text{th}}$ and $71^{\circ}$F for September $15^{\text{th}}$.}

\item Compare your results to those obtained using a graphing utility.

\end{enumerate}

\item The fraction of the moon illuminated at midnight Eastern Standard Time on the $t^{\text{th}}$ day of June, 2009 is given in the table below.\footnote{See this website: \href{http://www.usno.navy.mil/USNO/astron...c-moon-ill}.}}

\medskip

\small

\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|} \hline

Day of & & & & & & & & & & \\

June, $t$ & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 & 30\\

\hline

Fraction & & & & & & & & & & \\

Illuminated, $F$ & 0.81 & 0.98 & 0.98 & 0.83 & 0.57 & 0.27 & 0.04 & 0.03 & 0.26 & 0.58 \\ \hline

\end{tabular}

\normalsize

\medskip

\begin{enumerate}

\item \label{MoonIllumination} Using the techniques discussed in Example \ref{sinusoidsunlight}, fit a sinusoid to these data.\footnote{You may want to plot the data before you find the phase shift.}

\item Using a graphing utility, graph your model along with the data set to judge the reasonableness of the fit.

\item Use the model you found in part \ref{MoonIllumination} to predict the fraction of the moon illuminated on June 1, 2009. \footnote{The listed fraction is $0.62$.}

\item Compare your results to those obtained using a graphing utility.

\end{enumerate}

\item With the help of your classmates, research the phenomena mentioned in Example \ref{underdampedresonance}, namely \href{http://en.wikipedia.org/wiki/Resonan...ne{resonance}} and \href{http://en.wikipedia.org/wiki/Beat_(a...erline{beats}}.

\item With the help of your classmates, research \href{http://en.wikipedia.org/wiki/Amplitu...line{Amplitude Modulation}} and \href{http://en.wikipedia.org/wiki/Frequen...line{Frequency Modulation}}.

\item What other things in the world might be roughly sinusoidal? Look to see what models you can find for them and share your results with your class.

\end{enumerate}

\newpage

\begin{enumerate}

\begin{multicols}{2}

\item $S(t) = \sin\left(880\pi t\right)$

\item $V(t) = 220 \sqrt{2} \sin\left(120\pi t\right)$

\end{multicols}

\begin{multicols}{2}

\item $h(t) = 67.5 \sin\left(\frac{\pi}{15} t - \frac{\pi}{2} \right) + 67.5$

\item $x(t) = 67.5 \cos\left(\frac{\pi}{15} t - \frac{\pi}{2} \right) = 67.5 \sin\left(\frac{\pi}{15} t \right)$

\end{multicols}

\item $h(t) = 28\sin\left(\frac{2\pi}{3} t - \frac{\pi}{2}\right) + 30$

\item \begin{enumerate} \item $k = 5 \, \frac{\text{lbs.}}{\text{ft.}}$ and $m = \frac{5}{16} \, \text{slugs}$

\item $x(t) = \sin\left(4t + \frac{\pi}{2}\right)$. The object first passes through the equilibrium point when $t = \frac{\pi}{8} \approx 0.39$ seconds after the motion starts. At this time, the object is heading upwards.

\item $x(t) = \frac{\sqrt{2}}{2} \sin\left(4t + \frac{7\pi}{4}\right)$. The object passes through the equilibrium point heading downwards for the third time when $t = \frac{17\pi}{16} \approx 3.34$ seconds.

\end{enumerate}

\item \begin{multicols}{2}

\begin{enumerate}

\item $\theta(t) = \theta_{\text{\tiny$0$}} \sin\left(\sqrt{\frac{g}{l}}\, t + \frac{\pi}{2}\right)$

\item $\theta(t) = \frac{\pi}{12} \sin\left(4\pi t + \frac{\pi}{2}\right)$

\end{enumerate}

\end{multicols}

\item \begin{enumerate} \item $T(t) = 20.5 \sin\left(\frac{\pi}{6} t - \pi\right) + 53.5$

\item Our function and the data set are graphed below. The sinusoid seems to be shifted to the right of our data.

\begin{center}

\includegraphics[width=2in]{./AppExtGraphics/Sinusoid12.jpg}

\end{center}

\item The average temperature on April $15^{\text{th}}$ is approximately $T(4.5) \approx 39.00^{\circ}$F and the average temperature on September $15^{\text{th}}$ is approximately $T(9.5) \approx 73.38^{\circ}$F.

\item Using a graphing calculator, we get the following

\begin{center}

\begin{tabular}{cc}

\includegraphics[width=2in]{./AppExtGraphics/Sinusoid13.jpg} &

\hspace{0.75in} \includegraphics[width=2in]{./AppExtGraphics/Sinusoid14.jpg} \\

\end{tabular}

\end{center}

This model predicts the average temperature for April $15^{\text{th}}$ to be approximately $42.43^{\circ}$F and the average temperature on September $15^{\text{th}}$ to be approximately $70.05^{\circ}$F. This model appears to be more accurate.

\end{enumerate}

\item \begin{enumerate} \item Based on the shape of the data, we either choose $A<0$ or we find the \textit{second} value of $t$ which closely approximates the baseline' value, $F = 0.505$. We choose the latter to obtain $F(t) = 0.475 \sin\left(\frac{\pi}{15} t - 2\pi \right) + 0.505 = 0.475 \sin\left(\frac{\pi}{15} t\right) + 0.505$

\enlargethispage{\baselineskip}

\item Our function and the data set are graphed below. It's a pretty good fit.

\begin{center}

\includegraphics[width=2in]{./AppExtGraphics/Sinusoid15.jpg}

\end{center}

\item The fraction of the moon illuminated on June 1st, 2009 is approximately $F(1) \approx 0.60$

\item Using a graphing calculator, we get the following.

\begin{center}

\begin{tabular}{cc}

\includegraphics[width=2in]{./AppExtGraphics/Sinusoid16.jpg} &

\hspace{0.75in} \includegraphics[width=2in]{./AppExtGraphics/Sinusoid17.jpg} \\

\end{tabular}

\end{center}

This model predicts that the fraction of the moon illuminated on June 1st, 2009 is approximately $0.59$. This appears to be a better fit to the data than our first model.

\end{enumerate}

\end{enumerate}

\closegraphsfile

## 11.2: The Law of Sines

\subsection{Exercises}

In Exercises \ref{firstlawofsines} - \ref{lastlawofsines}, solve for the remaining side(s) and angle(s) if possible. As in the text, $(\alpha, a)$, $(\beta, b)$ and $(\gamma, c)$ are angle-side opposite pairs.

\begin{multicols}{2}

\begin{enumerate}

\item $\alpha = 13^{\circ}, \; \beta = 17^{\circ}, \; a = 5$ \label{firstlawofsines}

\item $\alpha = 73.2^{\circ}, \; \beta = 54.1^{\circ}, \; a = 117$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\alpha = 95^{\circ}, \; \beta = 85^{\circ}, \; a = 33.33$

\item $\alpha = 95^{\circ}, \; \beta = 62^{\circ}, \; a = 33.33$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\alpha = 117^{\circ}, \; a = 35, \; b = 42$

\item $\alpha = 117^{\circ}, \; a = 45, \; b = 42$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\alpha = 68.7^{\circ}, \; a = 88, \; b = 92$

\item $\alpha = 42^{\circ}, \; a = 17, \; b = 23.5$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\alpha = 68.7^{\circ}, \; a = 70, \; b = 90$

\item $\alpha = 30^{\circ}, \; a = 7, \; b = 14$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\alpha = 42^{\circ}, \; a = 39, \; b = 23.5$

\item $\gamma = 53^{\circ}, \; \alpha = 53^{\circ}, \; c = 28.01$ \label{secondarea}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\alpha = 6^{\circ}, \; a = 57, \; b = 100$

\item $\gamma = 74.6^{\circ}, \; c = 3, \; a = 3.05$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\beta = 102^{\circ}, \; b = 16.75, \; c = 13$

\item $\beta = 102^{\circ}, \; b = 16.75, \; c = 18$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\beta = 102^{\circ}, \; \gamma = 35^{\circ}, \; b = 16.75$

\item $\beta = 29.13^{\circ}, \; \gamma = 83.95^{\circ}, \; b = 314.15$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\gamma = 120^{\circ}, \; \beta = 61^{\circ}, \; c = 4$

\item $\alpha = 50^{\circ}, \; a = 25, \; b = 12.5$ \label{lastlawofsines}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\bigskip

\item Find the area of the triangles given in Exercises \ref{firstlawofsines}, \ref{secondarea} and \ref{lastlawofsines} above.

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\phantomsection

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Using a right triangle with a horizontal leg of length 100 and vertical leg with length 7, show that a 7\% grade means that the road (hypotenuse) makes about a $4^{\circ}$ angle with the horizontal. (It will not be exactly $4^{\circ}$, but it's pretty close.) \label{firstroadgrade}

\item What grade is given by a $9.65^{\circ}$ angle made by the road and the horizontal?\footnote{I have friends who live in Pacifica, CA and their road is actually this steep. It's not a nice road to drive.}

\item Along a long, straight stretch of mountain road with a 7\% grade, you see a tall tree standing perfectly plumb alongside the road.\footnote{The word plumb' here means that the tree is perpendicular to the horizontal.} From a point 500 feet downhill from the tree, the angle of inclination from the road to the top of the tree is $6^{\circ}$. Use the Law of Sines to find the height of the tree. (Hint: First show that the tree makes a $94^{\circ}$ angle with the road.) \label{lastroadgrade}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\phantomsection

\label{bearings}

\index{bearings}

(Another Classic Application: Bearings) In the next several exercises we introduce and work with the navigation tool known as bearings. Simply put, a bearing is the direction you are heading according to a compass. The classic nomenclature for bearings, however, is not given as an angle in standard position, so we must first understand the notation. A bearing is given as an acute angle of rotation (to the east or to the west) away from the north-south (up and down) line of a compass rose. For example, N$40^{\circ}$E (read $40^{\circ}$ east of north'') is a bearing which is rotated clockwise $40^{\circ}$ from due north. If we imagine standing at the origin in the Cartesian Plane, this bearing would have us heading into Quadrant I along the terminal side of $\theta = 50^{\circ}$. Similarly, S$50^{\circ}$W would point into Quadrant III along the terminal side of $\theta = 220^{\circ}$ because we started out pointing due south (along $\theta = 270^{\circ}$) and rotated clockwise $50^{\circ}$ back to $220^{\circ}$. Counter-clockwise rotations would be found in the bearings N$60^{\circ}$W (which is on the terminal side of $\theta = 150^{\circ}$) and S$27^{\circ}$E (which lies along the terminal side of $\theta = 297^{\circ}$). These four bearings are drawn in the plane below.

\begin{center}

\begin{mfpic}[14]{-5}{5}{-5}{5}

\axes

\tlabel[cc](0,5.5){N}

\tlabel[cc](5.5,0){E}

\tlabel[cc](0,-5.5){S}

\tlabel[cc](-5.5,0){W}

\arrow[l5pt] \polyline{(0,0), (-5,0)}

\arrow[l5pt] \polyline{(0,0), (0,-5)}

\arrow \reverse \polyline{(3.2139, 3.8302), (0,0)}

\tlabel[cc](3.53, 4.21){N$40^{\circ}$E}

\arrow \arc[c]{(0,0), (0.1,2), -35}

\tlabel[cc](0.86,2.3){\scriptsize $40^{\circ}$}

\arrow \reverse \polyline{(-4.3301,2.5), (0,0)}

\tlabel[cc](-4.76,2.75){N$60^{\circ}$W}

\arrow \arc[c]{(0,0), (-0.1,2), 55}

\tlabel[cc](-1.25,2.17){\scriptsize $60^{\circ}$}

\arrow \reverse \polyline{(-3.83,-3.21), (0,0)}

\tlabel[cc](-4.4,-3.7){S$50^{\circ}$W}

\arrow \arc[c]{(0,0), (-0.1,-2), -45}

\tlabel[cc](-1.04,-2.22){\scriptsize $50^{\circ}$}

\arrow \reverse \polyline{(2.2700,-4.4550), (0,0)}

\tlabel[cc](2.50,-4.9){S$27^{\circ}$E}

\arrow \arc[c]{(0,0), (0.1,-2), 23}

\tlabel[cc](0.57, -2.38){\scriptsize $27^{\circ}$}

\point[3pt]{(0,0)}

\end{mfpic}

\end{center}

The cardinal directions north, south, east and west are usually not given as bearings in the fashion described above, but rather, one just refers to them as due north', due south', due east' and due west', respectively, and it is assumed that you know which quadrantal angle goes with each cardinal direction. (Hint: Look at the diagram above.)

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Find the angle $\theta$ in standard position with $0^{\circ} \leq \theta < 360^{\circ}$ which corresponds to each of the bearings given below.

\enlargethispage{.25in}

\begin{multicols}{4}

\begin{enumerate}

\item due west

\item S$83^{\circ}$E

\item N$5.5^{\circ}$E

\item due south

\setcounter{HWindent}{\value{enumii}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumii}{\value{HWindent}}

\item N$31.25^{\circ}$W

\item S$72^{\circ}41'12''$W\footnote{See Example \ref{degreeex} in Section \ref{Angles} for a review of the DMS system.}

\item N$45^{\circ}$E

\item S$45^{\circ}$W

\end{enumerate}

\end{multicols}

\item The Colonel spots a campfire at a of bearing N$42^{\circ}$E from his current position. Sarge, who is positioned 3000 feet due east of the Colonel, reckons the bearing to the fire to be N$20^{\circ}$W from his current position. Determine the distance from the campfire to each man, rounded to the nearest foot.

\item A hiker starts walking due west from Sasquatch Point and gets to the Chupacabra Trailhead before she realizes that she hasn't reset her pedometer. From the Chupacabra Trailhead she hikes for 5 miles along a bearing of N$53^{\circ}$W which brings her to the Muffin Ridge Observatory. From there, she knows a bearing of S$65^{\circ}$E will take her straight back to Sasquatch Point. How far will she have to walk to get from the Muffin Ridge Observatory to Sasquach Point? What is the distance between Sasquatch Point and the Chupacabra Trailhead?

\item The captain of the SS Bigfoot sees a signal flare at a bearing of N$15^{\circ}$E from her current location. From his position, the captain of the HMS Sasquatch finds the signal flare to be at a bearing of N$75^{\circ}$W. If the SS Bigfoot is 5 miles from the HMS Sasquatch and the bearing from the SS Bigfoot to the HMS Sasquatch is N$50^{\circ}$E, find the distances from the flare to each vessel, rounded to the nearest tenth of a mile.

\item Carl spies a potential Sasquatch nest at a bearing of N$10^{\circ}$E and radios Jeff, who is at a bearing of N$50^{\circ}$E from Carl's position. From Jeff's position, the nest is at a bearing of S$70^{\circ}$W. If Jeff and Carl are 500 feet apart, how far is Jeff from the Sasquatch nest? Round your answer to the nearest foot.

\item A hiker determines the bearing to a lodge from her current position is S$40^{\circ}$W. She proceeds to hike 2 miles at a bearing of S$20^{\circ}$E at which point she determines the bearing to the lodge is S$75^{\circ}$W. How far is she from the lodge at this point? Round your answer to the nearest hundredth of a mile.

\item A watchtower spots a ship off shore at a bearing of N$70^{\circ}$E. A second tower, which is 50 miles from the first at a bearing of S$80^{\circ}$E from the first tower, determines the bearing to the ship to be N$25^{\circ}$W. How far is the boat from the second tower? Round your answer to the nearest tenth of a mile.

\item Skippy and Sally decide to hunt UFOs. One night, they position themselves 2 miles apart on an abandoned stretch of desert runway. An hour into their investigation, Skippy spies a UFO hovering over a spot on the runway directly between him and Sally. He records the angle of inclination from the ground to the craft to be $75^{\circ}$ and radios Sally immediately to find the angle of inclination from her position to the craft is $50^{\circ}$. How high off the ground is the UFO at this point? Round your answer to the nearest foot. (Recall: 1 mile is 5280 feet.)

\item The angle of depression from an observer in an apartment complex to a gargoyle on the building next door is $55^{\circ}$. From a point five stories below the original observer, the angle of inclination to the gargoyle is $20^{\circ}$. Find the distance from each observer to the gargoyle and the distance from the gargoyle to the apartment complex. Round your answers to the nearest foot. (Use the rule of thumb that one story of a building is 9 feet.)

\item Prove that the Law of Sines holds when $\triangle ABC$ is a right triangle.

\item Discuss with your classmates why knowing only the three angles of a triangle is not enough to determine any of the sides.

\item Discuss with your classmates why the Law of Sines cannot be used to find the angles in the triangle when only the three sides are given. Also discuss what happens if only two sides and the angle between them are given. (Said another way, explain why the Law of Sines cannot be used in the SSS and SAS cases.)

\item Given $\alpha = 30^{\circ}$ and $b = 10$, choose four different values for $a$ so that

\begin{enumerate}

\item the information yields no triangle

\item the information yields exactly one right triangle

\item the information yields two distinct triangles

\item the information yields exactly one obtuse triangle

\end{enumerate}

Explain why you cannot choose $a$ in such a way as to have $\alpha = 30^{\circ}$, $b = 10$ and your choice of $a$ yield only one triangle where that unique triangle has three acute angles.

\item Use the cases and diagrams in the proof of the Law of Sines (Theorem \ref{lawofsines}) to prove the area formulas given in Theorem \ref{areaformulasine}. Why do those formulas yield square units when four quantities are being multiplied together?

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\newpage

\begin{multicols}{2}

\begin{enumerate}

\item $\begin{array}{lll} \alpha = 13^{\circ} & \beta = 17^{\circ} & \gamma = 150^{\circ} \\ a = 5 & b \approx 6.50 & c \approx 11.11 \end{array}$

\item $\begin{array}{lll} \alpha = 73.2^{\circ} & \beta = 54.1^{\circ} & \gamma = 52.7^{\circ} \\ a = 117 & b \approx 99.00 & c \approx 97.22 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item \begin{tabular}{l}

Information does not \\

produce a triangle \end{tabular}

\item $\begin{array}{lll} \alpha = 95^{\circ} & \beta = 62^{\circ} & \gamma = 23^{\circ} \\ a = 33.33 & b \approx 29.54 & c \approx 13.07 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item \begin{tabular}{l}

Information does not \\

produce a triangle \end{tabular}

\item $\begin{array}{lll} \alpha = 117^{\circ} & \beta \approx 56.3^{\circ} & \gamma \approx 6.7^{\circ} \\ a = 45 & b = 42 & c \approx 5.89 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\begin{array}{lll} \alpha = 68.7^{\circ} & \beta \approx 76.9^{\circ} & \gamma \approx 34.4^{\circ} \\ a = 88 & b = 92 & c \approx 53.36 \end{array}$

$\begin{array}{lll} \alpha = 68.7^{\circ} & \beta \approx 103.1^{\circ} & \gamma \approx 8.2^{\circ} \\ a = 88 & b = 92 & c \approx 13.47\end{array}$

\item $\begin{array}{lll} \alpha = 42^{\circ} & \beta \approx 67.66^{\circ} & \gamma \approx 70.34^{\circ} \\ a = 17 & b = 23.5 & c \approx 23.93 \end{array}$

$\begin{array}{lll} \alpha = 42^{\circ} & \beta \approx 112.34^{\circ} & \gamma \approx 25.66^{\circ} \\ a = 17 & b = 23.5 & c \approx 11.00 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item \begin{tabular}{l}

Information does not \\

produce a triangle \end{tabular}

\item $\begin{array}{lll} \alpha = 30^{\circ} & \beta = 90^{\circ} & \gamma = 60^{\circ} \\ a = 7 & b = 14 & c = 7\sqrt{3} \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\begin{array}{lll} \alpha = 42^{\circ} & \beta \approx 23.78^{\circ} & \gamma \approx 114.22^{\circ} \\ a = 39 & b = 23.5 & c \approx 53.15 \end{array}$

\item $\begin{array}{lll} \alpha = 53^{\circ} & \beta = 74^{\circ} & \gamma = 53^{\circ} \\ a = 28.01 & b \approx 33.71 & c = 28.01 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\begin{array}{lll} \alpha = 6^{\circ} & \beta \approx 169.43^{\circ} & \gamma \approx 4.57^{\circ} \\ a = 57 & b = 100 & c \approx 43.45 \end{array}$

$\begin{array}{lll} \alpha = 6^{\circ} & \beta \approx 10.57^{\circ} & \gamma \approx 163.43^{\circ} \\ a = 57 & b = 100 & c \approx 155.51 \end{array}$

\item $\begin{array}{lll} \alpha \approx 78.59^{\circ} & \beta \approx 26.81^{\circ} & \gamma = 74.6^{\circ} \\ a = 3.05 & b \approx 1.40 & c = 3 \end{array}$

$\begin{array}{lll} \alpha \approx 101.41^{\circ} & \beta \approx 3.99^{\circ} & \gamma = 74.6^{\circ} \\ a = 3.05 & b \approx 0.217 & c = 3 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\begin{array}{lll} \alpha \approx 28.61^{\circ} & \beta = 102^{\circ} & \gamma \approx 49.39^{\circ} \\ a \approx 8.20 & b = 16.75 & c = 13 \end{array}$

\item \begin{tabular}{l}

Information does not \\

produce a triangle \end{tabular}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\begin{array}{lll} \alpha = 43^{\circ} & \beta = 102^{\circ} & \gamma = 35^{\circ} \\ a \approx 11.68 & b = 16.75 & c \approx 9.82 \end{array}$

\item $\begin{array}{lll} \alpha = 66.92^{\circ} & \beta = 29.13^{\circ} & \gamma = 83.95^{\circ} \\ a \approx 593.69 & b = 314.15 & c \approx 641.75 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item \begin{tabular}{l}

Information does not \\

produce a triangle \end{tabular}

\item $\begin{array}{lll} \alpha = 50^{\circ} & \beta \approx 22.52^{\circ} & \gamma \approx 107.48^{\circ} \\ a = 25 & b = 12.5 & c \approx 31.13 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item The area of the triangle from Exercise \ref{firstlawofsines} is about 8.1 square units.\\

The area of the triangle from Exercise \ref{secondarea} is about 377.1 square units.\\

The area of the triangle from Exercise \ref{lastlawofsines} is about 149 square units.

\item $\arctan\left(\frac{7}{100}\right) \approx 0.699$ radians, which is equivalent to $4.004^{\circ}$

\pagebreak

\item \begin{multicols}{4} \begin{enumerate}

\item $\theta = 180^{\circ}$

\item $\theta = 353^{\circ}$

\item $\theta = 84.5^{\circ}$

\item $\theta = 270^{\circ}$

\setcounter{HWindent}{\value{enumii}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumii}{\value{HWindent}}

\item $\theta = 121.25^{\circ}$

\item $\theta = 197^{\circ} 18' 48''$

\item $\theta = 45^{\circ}$

\item $\theta = 225^{\circ}$

\end{enumerate}

\end{multicols}

\item The Colonel is about 3193 feet from the campfire. \\

Sarge is about 2525 feet to the campfire.

\item The distance from the Muffin Ridge Observatory to Sasquach Point is about 7.12 miles.\\

The distance from Sasquatch Point to the Chupacabra Trailhead is about 2.46 miles.

\item The SS Bigfoot is about 4.1 miles from the flare. \\

The HMS Sasquatch is about 2.9 miles from the flare.

\item Jeff is about 342 feet from the nest.

\item She is about 3.02 miles from the lodge

\item The boat is about 25.1 miles from the second tower.

\item The UFO is hovering about 9539 feet above the ground.

\item The gargoyle is about 44 feet from the observer on the upper floor. \\

The gargoyle is about 27 feet from the observer on the lower floor. \\

The gargoyle is about 25 feet from the other building.

\end{enumerate}

\closegraphsfile

## 11.3: The Law of Cosines

\subsection{Exercises}

In Exercises \ref{firstlawofcosines} - \ref{lastlawofcosines}, use the Law of Cosines to find the remaining side(s) and angle(s) if possible.

\begin{multicols}{2}

\begin{enumerate}

\item $a = 7, \; b = 12, \; \gamma = 59.3^{\circ}$ \label{firstlawofcosines}

\item $\alpha = 104^{\circ}, \; b = 25, \; c = 37$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $a = 153, \; \beta = 8.2^{\circ}, \; c = 153$

\item $a = 3, \; b = 4, \; \gamma = 90^{\circ}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\alpha = 120^{\circ}, \; b = 3, \; c = 4$

\item $a = 7, \; b = 10, \; c = 13$ \label{firstherons}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $a = 1, \; b = 2, \; c = 5$

\item $a = 300, \; b = 302, \; c = 48$ \label{secondherons}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $a = 5, \; b = 5, \; c = 5$

\item $a = 5, \; b = 12,; c = 13$ \label{thirdherons} \label{lastlawofcosines}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{anylawfirst} - \ref{anylawlast}, solve for the remaining side(s) and angle(s), if possible, using any appropriate technique.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $a = 18, \; \alpha = 63^{\circ}, \; b = 20$ \label{ambigfirst} \label{anylawfirst}

\item $a = 37, \; b = 45, \; c = 26$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $a = 16, \; \alpha = 63^{\circ}, \; b = 20$ \label{ambigsecond}

\item $a = 22, \; \alpha = 63^{\circ}, \; b = 20$ \label{ambigthird}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\alpha = 42^{\circ}, \; b = 117, \; c = 88$

\item $\beta = 7^{\circ}, \; \gamma = 170^{\circ}, \; c = 98.6$ \label{anylawlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Find the area of the triangles given in Exercises \ref{firstherons}, \ref{secondherons} and \ref{thirdherons} above.

\item The hour hand on my antique Seth Thomas schoolhouse clock in 4 inches long and the minute hand is 5.5 inches long. Find the distance between the ends of the hands when the clock reads four o'clock. Round your answer to the nearest hundredth of an inch.

\item A geologist wants to measure the diameter of a crater. From her camp, it is 4 miles to the northern-most point of the crater and 2 miles to the southern-most point. If the angle between the two lines of sight is $117^{\circ}$, what is the diameter of the crater? Round your answer to the nearest hundredth of a mile.

\item From the Pedimaxus International Airport a tour helicopter can fly to Cliffs of Insanity Point by following a bearing of N$8.2^{\circ}$E for 192 miles and it can fly to Bigfoot Falls by following a bearing of S$68.5^{\circ}$E for 207 miles.\footnote{Please refer to Page \pageref{bearings} in Section \ref{LawofSines} for an introduction to bearings.} Find the distance between Cliffs of Insanity Point and Bigfoot Falls. Round your answer to the nearest mile. \label{lofcosinesbearingexercise}

\item Cliffs of Insanity Point and Bigfoot Falls from Exericse \ref{lofcosinesbearingexercise} above both lie on a straight stretch of the Great Sasquatch Canyon. What bearing would the tour helicopter need to follow to go directly from Bigfoot Falls to Cliffs of Insanity Point? Round your angle to the nearest tenth of a degree.

\item A naturalist sets off on a hike from a lodge on a bearing of S$80^{\circ}$W. After 1.5 miles, she changes her bearing to S$17^{\circ}$W and continues hiking for 3 miles. Find her distance from the lodge at this point. Round your answer to the nearest hundredth of a mile. What bearing should she follow to return to the lodge? Round your angle to the nearest degree.

\item The HMS Sasquatch leaves port on a bearing of N$23^{\circ}$E and travels for 5 miles. It then changes course and follows a heading of S$41^{\circ}$E for 2 miles. How far is it from port? Round your answer to the nearest hundredth of a mile. What is its bearing to port? Round your angle to the nearest degree.

\item The SS Bigfoot leaves a harbor bound for Nessie Island which is 300 miles away at a bearing of N$32^{\circ}$E. A storm moves in and after 100 miles, the captain of the Bigfoot finds he has drifted off course. If his bearing to the harbor is now S$70^{\circ}$W, how far is the SS Bigfoot from Nessie Island? Round your answer to the nearest hundredth of a mile. What course should the captain set to head to the island? Round your angle to the nearest tenth of a degree.

\item From a point 300 feet above level ground in a firetower, a ranger spots two fires in the Yeti National Forest. The angle of depression\footnote{See Exercise \ref{angleofdepression} in Section \ref{CircularFunctions} for the definition of this angle.} made by the line of sight from the ranger to the first fire is $2.5^{\circ}$ and the angle of depression made by line of sight from the ranger to the second fire is $1.3^{\circ}$. The angle formed by the two lines of sight is $117^{\circ}$. Find the distance between the two fires. Round your answer to the nearest foot. (Hint: In order to use the $117^{\circ}$ angle between the lines of sight, you will first need to use right angle Trigonometry to find the lengths of the lines of sight. This will give you a Side-Angle-Side case in which to apply the Law of Cosines.)

\begin{center}

\begin{mfpic}[15]{-5}{5}{-5}{5}

\plotsymbol[5pt]{Asterisk}{(-4.33,2.5),(2.6, 1.5)}

\dotted \polyline{(0,0), (-4.33,2.5)}

\dotted\polyline{(0,0), (2.6, 1.5)}

\tlabel[cc](0,-0.5){firetower}

\tlabel[cc](4.5,1.5){fire}

\tlabel[cc](-5.5, 2.5){fire}

\arrow \reverse \arrow \parafcn{35, 145, 5}{1.5*dir(t)}

\tlabel[cc](0,2){$117^{\circ}$}

\point[3pt]{(0,0)}

\end{mfpic}

\end{center}

\item If you apply the Law of Cosines to the ambiguous Angle-Side-Side (ASS) case, the result is a quadratic equation whose variable is that of the missing side. If the equation has no positive real zeros then the information given does not yield a triangle. If the equation has only one positive real zero then exactly one triangle is formed and if the equation has two distinct positive real zeros then two distinct triangles are formed. Apply the Law of Cosines to Exercises \ref{ambigfirst}, \ref{ambigsecond} and \ref{ambigthird} above in order to demonstrate this result.

\item Discuss with your classmates why Heron's Formula yields an area in square units even though four lengths are being multiplied together.

\end{enumerate}

\newpage

\begin{multicols}{2}

\begin{enumerate}

\item $\begin{array}{lll} \alpha \approx 35.54^{\circ} & \beta \approx 85.16^{\circ} & \gamma = 59.3^{\circ} \\ a = 7 & b = 12 & c \approx 10.36 \end{array}$

\item $\begin{array}{lll} \alpha = 104^{\circ} & \beta \approx 29.40^{\circ} & \gamma \approx 46.60^{\circ} \\ a \approx 49.41 & b = 25 & c = 37 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\begin{array}{lll} \alpha \approx 85.90^{\circ} & \beta = 8.2^{\circ} & \gamma \approx 85.90^{\circ} \\ a = 153 & b \approx 21.88 & c = 153 \end{array}$

\item $\begin{array}{lll} \alpha \approx 36.87^{\circ} & \beta \approx 53.13^{\circ} & \gamma = 90^{\circ} \\ a = 3 & b = 4 & c = 5 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\begin{array}{lll} \alpha = 120^{\circ} & \beta \approx 25.28^{\circ} & \gamma \approx 34.72^{\circ} \\ a = \sqrt{37} & b = 3 & c = 4 \end{array}$

\item $\begin{array}{lll} \alpha \approx 32.31^{\circ} & \beta \approx 49.58^{\circ} & \gamma \approx 98.21^{\circ} \\ a = 7 & b = 10 & c = 13 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item \begin{tabular}{l}

Information does not \\

produce a triangle \end{tabular}

\item $\begin{array}{lll} \alpha \approx 83.05^{\circ} & \beta \approx 87.81^{\circ} & \gamma \approx 9.14^{\circ} \\ a = 300 & b = 302 & c = 48 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\begin{array}{lll} \alpha = 60^{\circ} & \beta = 60^{\circ} & \gamma = 60^{\circ} \\ a = 5 & b = 5 & c = 5 \end{array}$

\item $\begin{array}{lll} \alpha \approx 22.62^{\circ} & \beta \approx 67.38^{\circ} & \gamma = 90^{\circ} \\ a = 5 & b = 12 & c = 13 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\begin{array}{lll} \alpha = 63^{\circ} & \beta \approx 98.11^{\circ} & \gamma \approx 18.89^{\circ} \\ a = 18 & b = 20 & c \approx 6.54 \end{array}$

$\begin{array}{lll} \alpha = 63^{\circ} & \beta \approx 81.89^{\circ} & \gamma \approx 35.11^{\circ} \\ a = 18 & b = 20 & c \approx 11.62 \end{array}$

\item $\begin{array}{lll} \alpha \approx 55.30^{\circ} & \beta \approx 89.40^{\circ} & \gamma \approx 35.30^{\circ} \\ a = 37 & b = 45 & c = 26 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item \begin{tabular}{l}

Information does not \\

produce a triangle \end{tabular}

\item $\begin{array}{lll} \alpha = 63^{\circ} & \beta \approx 54.1^{\circ} & \gamma \approx 62.9^{\circ} \\ a = 22 & b = 20 & c \approx 21.98 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\begin{array}{lll} \alpha = 42^{\circ} & \beta \approx 89.23^{\circ} & \gamma \approx 48.77^{\circ} \\ a \approx 78.30 & b = 117 & c = 88 \end{array}$

\item $\begin{array}{lll} \alpha \approx 3^{\circ} & \beta = 7^{\circ} & \gamma = 170^{\circ} \\ a \approx 29.72 & b \approx 69.2 & c = 98.6 \end{array}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item The area of the triangle given in Exercise \ref{firstherons} is $\sqrt{1200} = 20\sqrt{3} \approx 34.64$ square units.\\

The area of the triangle given in Exercise \ref{secondherons} is $\sqrt{51764375} \approx 7194.75$ square units.\\

The area of the triangle given in Exercise \ref{thirdherons} is exactly $30$ square units.

\item The distance between the ends of the hands at four o'clock is about $8.26$ inches.

\item The diameter of the crater is about 5.22 miles.

\item N$31.8^{\circ}$W

\item She is about 3.92 miles from the lodge and her bearing to the lodge is N$37^{\circ}$E.

\item It is about 4.50 miles from port and its heading to port is S$47^{\circ}$W.

\item It is about 229.61 miles from the island and the captain should set a course of N$16.4^{\circ}$E to reach the island.

\item The fires are about 17456 feet apart. (Try to avoid rounding errors.)

\end{enumerate}

\closegraphsfile

## 11.4: Polar Coordinates

\subsection{Exercises}

In Exercises \ref{polarpointgraphfirst} - \ref{polarpointgraphlast}, plot the point given in polar coordinates and then give three different expressions for the point such that \hfill (a) $r < 0$ and $0 \leq \theta \leq 2\pi$, \hfill (b) $r > 0$ and $\theta \leq 0$ \hfill (c) $r > 0$ and $\theta \geq 2\pi$

\begin{multicols}{4}

\begin{enumerate}

\item $\left( 2, \dfrac{\pi}{3} \right)$ \vphantom{$\left( \dfrac{3\pi}{2} \right)$} \label{polarpointgraphfirst}

\item $\left( 5, \dfrac{7\pi}{4} \right)$

\item $\left( \dfrac{1}{3}, \dfrac{3\pi}{2} \right)$

\item $\left( \dfrac{5}{2}, \dfrac{5\pi}{6} \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( 12, -\dfrac{7\pi}{6} \right)$

\item $\left( 3, -\dfrac{5\pi}{4} \right)$

\item $\left( 2\sqrt{2}, -\pi \right)$ \vphantom{$\left( \dfrac{3\pi}{2} \right)$}

\item $\left( \dfrac{7}{2}, -\dfrac{13\pi}{6} \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( -20, 3\pi \right)$ \vphantom{$\left( \dfrac{3\pi}{2} \right)$}

\item $\left( -4, \dfrac{5\pi}{4} \right)$

\item $\left( -1, \dfrac{2\pi}{3} \right)$

\item $\left( -3, \dfrac{\pi}{2} \right)$ \vphantom{$\left( \dfrac{3\pi}{2} \right)$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( -3, -\dfrac{11\pi}{6} \right)$

\item $\left( -2.5, -\dfrac{\pi}{4} \right)$ \vphantom{$\left( \dfrac{3\pi}{2} \right)$}

\item $\left( -\sqrt{5}, -\dfrac{4\pi}{3} \right)$

\item $\left( -\pi, -\pi \right)$ \vphantom{$\left( \dfrac{3\pi}{2} \right)$} \label{polarpointgraphlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{polartorectfirst} - \ref{polartorectlast}, convert the point from polar coordinates into rectangular coordinates.

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( 5, \dfrac{7\pi}{4} \right)$ \label{polartorectfirst}

\item $\left( 2, \dfrac{\pi}{3} \right)$ \vphantom{$\left( \dfrac{3\pi}{2} \right)$}

\item $\left( 11, -\dfrac{7\pi}{6} \right)$

\item $\left( -20, 3\pi \right)$ \vphantom{$\left( \dfrac{3\pi}{2} \right)$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( \dfrac{3}{5}, \dfrac{\pi}{2} \right)$

\item $\left( -4, \dfrac{5\pi}{6} \right)$

\item $\left( 9, \dfrac{7\pi}{2} \right)$

\item $\left( -5, -\dfrac{9\pi}{4} \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( 42, \dfrac{13\pi}{6} \right)$

\item $\left( -117, 117\pi \right)$ \vphantom{$\left( \dfrac{3\pi}{2} \right)$}

\item $\left( 6, \arctan(2) \right)$ \vphantom{$\left( \dfrac{3\pi}{2} \right)$}

\item $\left(10, \arctan(3) \right)$ \vphantom{$\left( \dfrac{3\pi}{2} \right)$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( -3, \arctan\left(\dfrac{4}{3}\right) \right)$

\item $\left( 5, \arctan\left(-\dfrac{4}{3}\right) \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( 2, \pi - \arctan\left(\dfrac{1}{2}\right) \right)$

\item $\left( -\dfrac{1}{2}, \pi - \arctan\left(5\right) \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( -1, \pi + \arctan\left(\dfrac{3}{4}\right) \right)$

\item $\left( \dfrac{2}{3}, \pi + \arctan\left(2\sqrt{2}\right) \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( \pi, \arctan(\pi) \right)$ \vphantom{$\left( \dfrac{12}{5} \right)$}

\item $\left( 13, \arctan \left( \dfrac{12}{5} \right) \right)$ \label{polartorectlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{recttopolarfirst} - \ref{recttopolarlast}, convert the point from rectangular coordinates into polar coordinates with $r \geq 0$ and $0 \leq \theta < 2\pi$.

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(0, 5)$ \label{recttopolarfirst}

\item $(3, \sqrt{3})$

\item $(7, -7)$

\item $(-3, -\sqrt{3})$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(-3, 0)$ \vphantom{$\left(\dfrac{\sqrt{3}}{4}\right)$}

\item $\left( -\sqrt{2}, \sqrt{2} \right)$ \vphantom{$\left(\dfrac{\sqrt{3}}{4}\right)$}

\item $\left( -4,-4\sqrt{3} \right)$ \vphantom{$\left(\dfrac{\sqrt{3}}{4}\right)$}

\item $\left( \dfrac{\sqrt{3}}{4}, -\dfrac{1}{4} \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( -\dfrac{3}{10}, -\dfrac{3\sqrt{3}}{10} \right)$

\item $\left( -\sqrt{5}, -\sqrt{5} \right)$ \vphantom{$\left(\dfrac{\sqrt{3}}{4}\right)$}

\item $(6,8)$ \vphantom{$\left(\dfrac{\sqrt{3}}{4}\right)$}

\item $(\sqrt{5},2\sqrt{5})$ \vphantom{$\left(\dfrac{\sqrt{3}}{4}\right)$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(-8,1)$ \vphantom{$\left(\dfrac{\sqrt{3}}{4}\right)$}

\item $(-2\sqrt{10}, 6\sqrt{10})$ \vphantom{$\left(\dfrac{\sqrt{3}}{4}\right)$}

\item $\left(-5, -12 \right)$ \vphantom{$\left(\dfrac{\sqrt{3}}{4}\right)$}

\item $\left(-\dfrac{\sqrt{5}}{15}, -\dfrac{2\sqrt{5}}{15} \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left(24, -7 \right)$ \vphantom{$\left(\dfrac{\sqrt{3}}{4}\right)$}

\item $\left(12, -9\right)$ \vphantom{$\left(\dfrac{\sqrt{3}}{4}\right)$}

\item $\left(\dfrac{\sqrt{2}}{4}, \dfrac{\sqrt{6}}{4}\right)$

\item $\left(-\dfrac{\sqrt{65}}{5}, \dfrac{2\sqrt{65}}{5}\right)$ \label{recttopolarlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{equrecttopolarfirst} - \ref{equrecttopolarlast}, convert the equation from rectangular coordinates into polar coordinates. Solve for $r$ in all but \#\ref{solvethetaone} through \#\ref{solvethetafour}. In Exercises \ref{solvethetaone} - \ref{solvethetafour}, you need to solve for $\theta$

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x = 6$ \label{equrecttopolarfirst}

\item $x = -3$

\item $y = 7$

\item $y = 0$ \label{solvethetaone}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $y = -x$

\item $y = x\sqrt{3}$

\item $y = 2x$ \label{solvethetafour}

\item $x^2 + y^2 = 25$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x^{2} + y^{2} = 117$

\item $y = 4x - 19$

\item $x = 3y + 1$

\item $y = -3x^{2}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $4x = y^2$

\item $x^2 + y^2 - 2y = 0$

\item $x^2 -4x + y^2 = 0$

\item $x^2 + y^2 = x$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $y^2 = 7y - x^2$

\item $(x+2)^2 + y^2 = 4$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x^{2} + (y - 3)^{2} = 9$ \vphantom{$\left( \dfrac{1}{2} \right)$}

\item $4x^2 + 4\left( y - \dfrac{1}{2} \right)^2 = 1$ \label{equrecttopolarlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{equpolartorectfirst} - \ref{equpolartorectlast}, convert the equation from polar coordinates into rectangular coordinates.

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = 7$ \vphantom{$\dfrac{\pi}{3}$} \label{equpolartorectfirst}

\item $r = -3$ \vphantom{$\dfrac{\pi}{3}$}

\item $r = \sqrt{2}$ \vphantom{$\dfrac{\pi}{3}$}

\item $\theta = \dfrac{\pi}{4}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\theta = \dfrac{2\pi}{3}$

\item $\theta = \pi$ \vphantom{$\dfrac{2\pi}{3}$}

\item $\theta = \dfrac{3\pi}{2}$

\item $r = 4\cos(\theta)$ \vphantom{$\dfrac{2\pi}{3}$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $5r = \cos(\theta)$

\item $r = 3\sin(\theta)$

\item $r = -2\sin(\theta)$

\item $r = 7\sec(\theta)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $12r = \csc(\theta)$

\item $r = -2\sec(\theta)$

\item $r = -\sqrt{5} \csc(\theta)$

\item \small $r = 2\sec(\theta)\tan(\theta)$ \normalsize

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item \small $r = -\csc(\theta) \cot(\theta)$ \normalsize

\item $r^{2} = \sin(2\theta)$

\item $r = 1 - 2\cos(\theta)$

\item $r = 1 + \sin(\theta)$ \label{equpolartorectlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Convert the origin $(0,0)$ into polar coordinates in four different ways.

\item With the help of your classmates, use the Law of Cosines to develop a formula for the distance between two points in polar coordinates.

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\newpage

\begin{enumerate}

\item \begin{multicols}{2} \raggedcolumns

$\left( 2, \dfrac{\pi}{3} \right), \, \left( -2, \dfrac{4\pi}{3} \right)$\\

$\left( 2, -\dfrac{5\pi}{3} \right), \, \left( 2, \dfrac{7\pi}{3} \right)$\\

\hspace{.4in} \begin{mfpic}[30]{-1.5}{3}{-0.5}{3}

\drawcolor[gray]{0.7}

\axes

\xmarks{-1,1,2}

\ymarks{1,2}

\tlabel(3,-0.5){\scriptsize $x$}

\tlabel(0.25,3){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\dashed \polyline{(0,0), (1, 1.73205)}

\point[3pt]{(1, 1.73205)}

\arrow \arc[c]{(0,0), (0.5,0.05), 45}

\tlpointsep{5pt}

\scriptsize

\axislabels {x}{{$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2}

\axislabels {y}{{$1$} 1, {$2$} 2}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\left( 5, \dfrac{7\pi}{4} \right), \, \left( -5, \dfrac{3\pi}{4} \right)$\\

$\left( 5, -\dfrac{\pi}{4} \right), \, \left( 5, \dfrac{15\pi}{4} \right)$\\

\hspace{.5in} \begin{mfpic}[25]{-1.5}{4}{-4}{1.5}

\drawcolor[gray]{0.7}

\axes

\xmarks{-1,1,2,3}

\ymarks{-3,-2,-1,1}

\tlabel(4,-0.5){\scriptsize $x$}

\tlabel(0.25,1.5){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\dashed \polyline{(0,0), (3.536, -3.536)}

\point[3pt]{(3.536, -3.536)}

\arrow \arc[c]{(0,0), (0.5,0.07), 295}

\tlpointsep{5pt}

\scriptsize

\axislabels {x}{{$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\axislabels {y}{{$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\left( \dfrac{1}{3}, \dfrac{3\pi}{2} \right), \, \left( -\dfrac{1}{3}, \dfrac{\pi}{2} \right)$\\

$\left( \dfrac{1}{3}, -\dfrac{\pi}{2} \right), \, \left( \dfrac{1}{3}, \dfrac{7\pi}{2} \right)$\\

\hspace{.5in} \begin{mfpic}[75]{-0.5}{1.25}{-0.5}{1.25}

\drawcolor[gray]{0.7}

\axes

\xmarks{1}

\ymarks{1}

\tlabel(1.25,-0.1){\scriptsize $x$}

\tlabel(0.1,1.25){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\point[3pt]{(0, -0.3333)}

\dashed \polyline{(0,0), (0, -0.3333)}

\arrow \arc[c]{(0,0), (0.15,0.03), 250}

\tlpointsep{5pt}

\scriptsize

\axislabels {x}{{$1$} 1}

\axislabels {y}{{$1$} 1}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\left( \dfrac{5}{2}, \dfrac{5\pi}{6} \right), \, \left( -\dfrac{5}{2}, \dfrac{11\pi}{6} \right)$\\

$\left( \dfrac{5}{2}, -\dfrac{7\pi}{6} \right), \, \left( \dfrac{5}{2}, \dfrac{17\pi}{6} \right)$\\

\hspace{.2in} \begin{mfpic}[15]{-4}{4}{-4}{4}

\drawcolor[gray]{0.7}

\axes

\xmarks{-3,-2,-1,1,2,2,3}

\ymarks{-3,-2,-1,1,2,3}

\tlabel(4.25,-0.1){\scriptsize $x$}

\tlabel(0.1,4.25){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\point[3pt]{(-2.16, 1.25)}

\dashed \polyline{(0,0), (-2.16, 1.25)}

\arrow \parafcn{5, 145, 5}{0.5*dir(t)}

\tlpointsep{5pt}

\scriptsize

\axislabels {x}{{$-3 \hspace{7pt}$} -3,{$-2 \hspace{7pt}$} -2,{$-1 \hspace{7pt}$} -1,{$1$} 1, {$2$} 2,{$3$} 3}

\axislabels {y}{{$-1$} -1, {$-2$} -2,{$-3$} -3, {$1$} 1, {$2$} 2,{$3$} 3}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\left( 12, -\dfrac{7\pi}{6} \right), \, \left( -12, \dfrac{11\pi}{6} \right)$\\

$\left( 12, -\dfrac{13\pi}{6} \right), \, \left( 12, \dfrac{17\pi}{6} \right)$\\

\begin{mfpic}[13]{-13}{2}{-2}{7}

\drawcolor[gray]{0.7}

\axes

\xmarks{-12,-9,-6,-3}

\ymarks{3,6}

\tlabel(2,-0.5){\scriptsize $x$}

\tlabel(0.5,7){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\dashed \polyline{(0,0), (-10.392,6)}

\point[3pt]{(-10.392,6)}

\arrow \arc[c]{(0,0), (1,-0.1), -190}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-12 \hspace{7pt}$} -12, {$-9 \hspace{7pt}$} -9, {$-6 \hspace{7pt}$} -6, {$-3 \hspace{7pt}$} -3}

\axislabels {y}{{$3$} 3, {$6$} 6}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\left(3, -\dfrac{5\pi}{4} \right), \, \left( -3, \dfrac{7\pi}{4} \right)$\\

$\left( 3, -\dfrac{13\pi}{4} \right), \, \left( 3, \dfrac{11\pi}{4} \right)$\\

\hspace{.3in} \begin{mfpic}[15]{-4}{4}{-4}{4}

\drawcolor[gray]{0.7}

\axes

\xmarks{-3,-2,-1,1,2,3}

\ymarks{-3,-2,-1,1,2,3}

\tlabel(4,-0.5){\scriptsize $x$}

\tlabel(0.5,4){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\dashed \polyline{(0,0), (-2.12,2.12)}

\point[3pt]{(-2.12,2.12)}

\arrow \arc[c]{(0,0), (0.75,-0.1), -215}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{\scriptsize $-3 \hspace{7pt}$} -3,{\scriptsize $-2 \hspace{7pt}$} -2, {\scriptsize $-1 \hspace{7pt}$} -1, {\scriptsize $1$} 1, {\scriptsize $2$} 2, {\scriptsize $3$} 3}

\axislabels {y}{{\scriptsize $-3$} -3,{\scriptsize $-2$} -2,{\scriptsize $-1$} -1, {\scriptsize $1$} 1, {\scriptsize $2$} 2, {\scriptsize $3$} 3}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\left(2\sqrt{2}, -\pi \right), \, \left( -2\sqrt{2},0 \right)$\\

$\left( 2\sqrt{2}, -3\pi \right), \, \left( 2\sqrt{2}, 3\pi \right)$\\

\hspace{.3in} \begin{mfpic}[15]{-4}{4}{-4}{4}

\drawcolor[gray]{0.7}

\axes

\xmarks{-3,-2,-1,1,2,3}

\ymarks{-3,-2,-1,1,2,3}

\tlabel(4,-0.5){\scriptsize $x$}

\tlabel(0.5,4){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\dashed \polyline{(0,0), (-2.83,0)}

\point[3pt]{(-2.83,0)}

\arrow \parafcn{365,535,5}{t*dir(-t)/750}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{\scriptsize $-3 \hspace{7pt}$} -3,{\scriptsize $-2 \hspace{7pt}$} -2, {\scriptsize $-1 \hspace{7pt}$} -1, {\scriptsize $1$} 1, {\scriptsize $2$} 2, {\scriptsize $3$} 3}

\axislabels {y}{{\scriptsize $-3$} -3,{\scriptsize $-2$} -2,{\scriptsize $-1$} -1, {\scriptsize $1$} 1, {\scriptsize $2$} 2, {\scriptsize $3$} 3}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\left(\dfrac{7}{2}, -\dfrac{13\pi}{6} \right), \, \left( -\dfrac{7}{2}, \dfrac{5\pi}{6} \right)$\\

$\left( \dfrac{7}{2}, -\dfrac{\pi}{6} \right), \, \left( \dfrac{7}{2}, \dfrac{23\pi}{6} \right)$\\

\hspace{.1in} \begin{mfpic}[15]{-5}{5}{-5}{5}

\drawcolor[gray]{0.7}

\axes

\xmarks{-4,-3,-2,-1,1,2,3,4}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\tlabel(5,-0.5){\scriptsize $x$}

\tlabel(0.5,5){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\dashed \polyline{(0,0), (3.03,-1.75)}

\point[3pt]{(3.03,-1.75)}

\arrow \parafcn{365,745,5}{t*dir(-t)/1000}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{\scriptsize $-4 \hspace{7pt}$} -4,{\scriptsize $-3 \hspace{7pt}$} -3,{\scriptsize $-2 \hspace{7pt}$} -2, {\scriptsize $-1 \hspace{7pt}$} -1, {\scriptsize $1$} 1, {\scriptsize $2$} 2, {\scriptsize $3$} 3, {\scriptsize $4$} 4}

\axislabels {y}{{\scriptsize $-4$} -4,{\scriptsize $-3$} -3,{\scriptsize $-2$} -2,{\scriptsize $-1$} -1, {\scriptsize $1$} 1, {\scriptsize $2$} 2, {\scriptsize $3$} 3, {\scriptsize $4$} 4}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$(-20, 3\pi), \, (-20, \pi)$\\

$(20, -2\pi), \, (20, 4\pi)$\\

\begin{mfpic}[15]{-5}{5}{-2}{2}

\drawcolor[gray]{0.7}

\axes

\xmarks{-4,-2,2,4}

\ymarks{-1,1}

\tlabel(4.75,-0.5){\scriptsize $x$}

\tlabel(0.25,2){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\dashed \polyline{(0,0), (4,0)}

\point[3pt]{(4,0)}

\arrow \parafcn{190,715,5}{(t+100)*dir(t)/1000}

\tlpointsep{5pt}

\scriptsize

\axislabels {x}{{$-20 \hspace{7pt}$} -4, {$-10 \hspace{7pt}$} -2, {$10$} 2, {$20$} 4}

\axislabels {y}{{$-1$} -1, {$1$} 1}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\left( -4, \dfrac{5\pi}{4} \right), \, \left( -4, \dfrac{13\pi}{4} \right)$\\

$\left( 4, -\dfrac{7\pi}{4} \right), \, \left( 4, \dfrac{9\pi}{4} \right)$\\

\begin{mfpic}[15]{-5}{5}{-5}{5}

\drawcolor[gray]{0.7}

\axes

\xmarks{-4,-3,-2,-1,1,2,3,4}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\tlabel(5,-0.5){\scriptsize $x$}

\tlabel(0.5,5){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\dashed \polyline{(0,0), (2.828, 2.828)}

\point[3pt]{(2.828, 2.828)}

\arrow \arc[c]{(0,0), (-0.5,-0.07), 205}

\tlpointsep{5pt}

\scriptsize

\axislabels {x}{{$-4 \hspace{7pt}$} -4, {$-3 \hspace{7pt}$} -3, {$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\axislabels {y}{{$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\left( -1, \dfrac{2\pi}{3} \right), \, \left( -1, \dfrac{8\pi}{3} \right)$\\

$\left( 1, -\dfrac{\pi}{3} \right), \, \left( 1, \dfrac{11\pi}{3} \right)$\\

\begin{mfpic}[25]{-3}{3}{-3}{3}

\drawcolor[gray]{0.7}

\axes

\xmarks{-2,-1,1,2}

\ymarks{-2,-1,1,2}

\tlabel(3,-0.5){\scriptsize $x$}

\tlabel(0.5,3){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\dashed \polyline{(0,0), (0.5, -0.87)}

\point[3pt]{(0.5, -0.87)}

\arrow \arc[c]{(0,0), (-0.6,-0.07), 115}

\tlpointsep{5pt}

\scriptsize

\axislabels {x}{{$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2}

\axislabels {y}{{$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\left( -3, \dfrac{\pi}{2} \right), \, \left( -3, \dfrac{5\pi}{2} \right)$\\

$\left( 3, -\dfrac{\pi}{2} \right), \, \left( 3, \dfrac{7\pi}{2} \right)$\\

\hspace{.1in} \begin{mfpic}[15]{-4}{4}{-4}{4}

\drawcolor[gray]{0.7}

\arrow \polyline{(-4,0), (4,0)}

\arrow \polyline{(0,0), (0,4)}

\xmarks{-3,-2,-1,1,2,3}

\ymarks{-3,-2,-1,1,2,3}

\tlabel(4,-0.5){\scriptsize $x$}

\tlabel(0.5,4){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\dashed \polyline{(0,0), (0, -3)}

\point[3pt]{(0, -3)}

\arrow \arc[c]{(0,0), (-1.5,-0.07), 85}

\tlpointsep{5pt}

\scriptsize

\axislabels {x}{{$-3 \hspace{7pt}$} -3,{$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\axislabels {y}{{$-3$} -3,{$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\normalsize

\end{mfpic}

\end{multicols}

\pagebreak

\item \begin{multicols}{2} \raggedcolumns

$\left( -3, -\dfrac{11\pi}{6} \right), \, \left( -3, \dfrac{\pi}{6} \right)$\\

$\left(3, -\dfrac{5\pi}{6} \right), \, \left( 3, \dfrac{19\pi}{6} \right)$\\

\hspace{.3in} \begin{mfpic}[15]{-4}{4}{-4}{4}

\drawcolor[gray]{0.7}

\axes

\xmarks{-3,-2,-1,1,2,3}

\ymarks{-3,-2,-1,1,2,3}

\tlabel(4,-0.5){\scriptsize $x$}

\tlabel(0.5,4){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\dashed \polyline{(0,0), (-2.6, -1.5)}

\point[3pt]{(-2.6, -1.5)}

\arrow \parafcn{175, -145, -5}{1.5*dir(t)}

\tlpointsep{5pt}

\scriptsize

\axislabels {x}{{$-3 \hspace{7pt}$} -3,{$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\axislabels {y}{{$-3$} -3,{$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\left( -2.5, -\dfrac{\pi}{4} \right), \, \left( -2.5, \dfrac{7\pi}{4} \right)$\\

$\left( 2.5, -\dfrac{5\pi}{4} \right), \, \left( 2.5, \dfrac{11\pi}{4} \right)$\\

\hspace{.3in} \begin{mfpic}[15]{-4}{4}{-4}{4}

\drawcolor[gray]{0.7}

\axes

\xmarks{-3,-2,-1,1,2,3}

\ymarks{-3,-2,-1,1,2,3}

\tlabel(4,-0.5){\scriptsize $x$}

\tlabel(0.5,4){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\dashed \polyline{(0,0), (-1.77, 1.77)}

\point[3pt]{(-1.77, 1.77)}

\arrow \parafcn{175, 140, -5}{1.5*dir(t)}

\tlpointsep{5pt}

\scriptsize

\axislabels {x}{{$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2}

\axislabels {y}{{$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\left( -\sqrt{5}, -\dfrac{4\pi}{3} \right), \, \left( -\sqrt{5}, \dfrac{2\pi}{3} \right)$\\

$\left( \sqrt{5}, -\dfrac{\pi}{3} \right), \, \left(\sqrt{5}, \dfrac{11\pi}{3} \right)$\\

\hspace{.3in} \begin{mfpic}[15]{-4}{4}{-4}{4}

\drawcolor[gray]{0.7}

\axes

\xmarks{-3,-2,-1,1,2,3}

\ymarks{-3,-2,-1,1,2,3}

\tlabel(4,-0.5){\scriptsize $x$}

\tlabel(0.5,4){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\dashed \polyline{(0,0), (1.12, -1.94)}

\point[3pt]{(1.12, -1.94)}

\arrow \parafcn{175, -55, -5}{1.5*dir(t)}

\tlpointsep{5pt}

\scriptsize

\axislabels {x}{{$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2}

\axislabels {y}{{$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\left(-\pi, -\pi \right), \, \left( -\pi, \pi \right)$\\

$\left( \pi, -2\pi \right), \, \left( \pi, 2\pi \right)$\\

\hspace{.3in} \begin{mfpic}[15]{-4}{4}{-4}{4}

\drawcolor[gray]{0.7}

\axes

\xmarks{-3,-2,-1,1,2,3}

\ymarks{-3,-2,-1,1,2,3}

\tlabel(4,-0.5){\scriptsize $x$}

\tlabel(0.5,4){\scriptsize $y$}

\drawcolor[rgb]{0.33,0.33,0.33}

\dashed \polyline{(0,0), (3.14,0)}

\point[3pt]{(3.14,0)}

\arrow \parafcn{175, 5, -5}{1.5*dir(t)}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{\scriptsize $-3 \hspace{7pt}$} -3,{\scriptsize $-2 \hspace{7pt}$} -2, {\scriptsize $-1 \hspace{7pt}$} -1, {\scriptsize $1$} 1, {\scriptsize $2$} 2, {\scriptsize $3$} 3}

\axislabels {y}{{\scriptsize $-3$} -3,{\scriptsize $-2$} -2,{\scriptsize $-1$} -1, {\scriptsize $1$} 1, {\scriptsize $2$} 2, {\scriptsize $3$} 3}

\normalsize

\end{mfpic}

\end{multicols}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( \dfrac{5\sqrt{2}}{2}, -\dfrac{5\sqrt{2}}{2} \right)$

\item $\left( 1, \sqrt{3} \right)$ \vphantom{$\left( \dfrac{11\sqrt{3}}{2} \right)$}

\item $\left( -\dfrac{11\sqrt{3}}{2}, \dfrac{11}{2} \right)$

\item $\left( 20, 0 \right)$ \vphantom{$\left( \dfrac{11\sqrt{3}}{2} \right)$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( 0, \dfrac{3}{5} \right)$ \vphantom{$\left( \dfrac{11\sqrt{3}}{2} \right)$}

\item $\left( 2\sqrt{3}, -2 \right)$ \vphantom{$\left( \dfrac{11\sqrt{3}}{2} \right)$}

\item $\left( 0, -9 \right)$ \vphantom{$\left( \dfrac{11\sqrt{3}}{2} \right)$}

\item $\left( -\dfrac{5\sqrt{2}}{2}, \dfrac{5\sqrt{2}}{2} \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( 21\sqrt{3}, 21 \right)$ \vphantom{$\left( \dfrac{11\sqrt{5}}{2} \right)$}

\item $\left(117, 0 \right)$ \vphantom{$\left( \dfrac{11\sqrt{5}}{2} \right)$}

\item $\left( \dfrac{6\sqrt{5}}{5}, \dfrac{12\sqrt{5}}{5} \right)$

\item $\left(\sqrt{10}, 3\sqrt{10} \right)$ \vphantom{$\left( \dfrac{11\sqrt{5}}{2} \right)$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( -\dfrac{9}{5}, -\dfrac{12}{5} \right)$ \vphantom{$\left( \dfrac{4\sqrt{5}}{5} \right)$}

\item $\left( 3,-4 \right)$ \vphantom{$\left( \dfrac{4\sqrt{5}}{5} \right)$}

\item $\left( -\dfrac{4\sqrt{5}}{5}, \dfrac{2\sqrt{5}}{5} \right)$

\item $\left( \dfrac{\sqrt{26}}{52}, -\dfrac{5\sqrt{26}}{52} \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( \dfrac{4}{5}, \dfrac{3}{5} \right)$ \vphantom{$\left( \dfrac{4\sqrt{5}}{5} \right)$}

\item $\left( -\dfrac{2}{9}, -\dfrac{4\sqrt{2}}{9} \right)$

\item $\left( \frac{\pi}{\sqrt{1+\pi^2}}, \frac{\pi^2}{\sqrt{1+\pi^2}} \right)$ \vphantom{$\left( \dfrac{4\sqrt{5}}{5} \right)$}

\item $\left( 5, 12 \right)$ \vphantom{$\left( \dfrac{4\sqrt{5}}{5} \right)$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( 5, \dfrac{\pi}{2} \right)$ \vphantom{$\left( \dfrac{11\pi}{2} \right)$}

\item $\left( 2\sqrt{3}, \dfrac{\pi}{6} \right)$ \vphantom{$\left( \dfrac{11\pi}{2} \right)$}

\item $\left( 7\sqrt{2}, \dfrac{7\pi}{4} \right)$

\item $\left( 2\sqrt{3}, \dfrac{7\pi}{6} \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( 3, \pi \right)$ \vphantom{$\left( \dfrac{11\pi}{2} \right)$}

\item $\left( 2, \dfrac{3\pi}{4} \right)$

\item $\left( 8, \dfrac{4\pi}{3} \right)$

\item $\left( \dfrac{1}{2}, \dfrac{11\pi}{6} \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( \dfrac{3}{5}, \dfrac{2\pi}{3} \right)$

\item $\left( \sqrt{10}, \dfrac{5\pi}{4} \right)$

\item $\left( 10, \arctan\left(\dfrac{4}{3}\right) \right)$

\item $\left( 5, \arctan\left(2\right) \right)$ \vphantom{$\left( \dfrac{11\pi}{2} \right)$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( \sqrt{65}, \pi - \arctan\left(\dfrac{1}{8}\right)\right)$

\item $(20, \pi - \arctan(3))$ \vphantom{$\left( \dfrac{11\pi}{2} \right)$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left(13, \pi + \arctan\left(\dfrac{12}{5}\right) \right)$

\item $\left(\dfrac{1}{3}, \pi + \arctan\left(2\right) \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left(25, 2\pi - \arctan\left(\dfrac{7}{24}\right) \right)$

\item $\left( 15, 2\pi - \arctan\left(\dfrac{3}{4} \right) \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left(\dfrac{\sqrt{2}}{2}, \dfrac{\pi}{3}\right)$

\item $\left(\sqrt{13}, \pi - \arctan(2) \right)$ \vphantom{$\left( \dfrac{11\sqrt{2}}{2} \right)$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = 6\sec(\theta)$

\item $r = -3\sec(\theta)$

\item $r = 7\csc(\theta)$

\item $\theta = 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\theta = \frac{3\pi}{4}$

\item $\theta = \frac{\pi}{3}$

\item $\theta = \arctan(2)$

\item $r = 5$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = \sqrt{117}$

\item $r = \frac{19}{4\cos(\theta) - \sin(\theta)}$

\item $x = \frac{1}{\cos(\theta) - 3\sin(\theta)}$

\item \small $r = \frac{-\sec(\theta)\tan(\theta)}{3}$ \normalsize

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item \small $r = 4\csc(\theta)\cot(\theta)$ \normalsize

\item $r=2\sin(\theta)$

\item $r = 4\cos(\theta)$

\item $r = \cos(\theta)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = 7\sin(\theta)$

\item $r= -4\cos(\theta)$

\item $r = 6\sin(\theta)$

\item $r = \frac{1}{4} \sin(\theta)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x^{2} + y^{2} = 49$

\item $x^{2} + y^{2} = 9$

\item $x^{2} + y^{2} = 2$

\item $y=x$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $y = -\sqrt{3}x$

\item $y=0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x=0$

\item $x^2 + y^2 = 4x$ or $(x-2)^2 + y^2 = 4$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $5x^2 + 5y^2 = x$ or $\left(x - \dfrac{1}{10}\right)^2+y^2 = \dfrac{1}{100}$

\item $x^2 + y^2 = 3y$ or $x^2 + \left(y - \dfrac{3}{2}\right)^2 = \dfrac{9}{4}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x^2 + y^2 = -2y$ or $x^2+(y+1)^2 = 1$

\item $x=7$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $y = \dfrac{1}{12}$

\item $x = -2$ \vphantom{$\dfrac{1}{12}$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $y= -\sqrt{5}$

\item $x^2=2y$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $y^2=-x$

\item $\left( x^{2} + y^{2} \right)^{2} = 2xy$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left( x^{2} + 2x + y^{2} \right)^{2} = x^{2} + y^{2}$

\item $\left( x^{2} + y^{2} + y\right)^{2} = x^{2} + y^{2}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Any point of the form $(0, \theta)$ will work, e.g. $(0, \pi), (0, -117), \left( 0, \dfrac{23\pi}{4} \right)$ and $(0, 0).$

\end{enumerate}

\closegraphsfile

## 11.5: Graphs of Polar Equations

\subsection{Exercises}

In Exercises \ref{polarplotfirst} - \ref{polarplotlast}, plot the graph of the polar equation by hand. Carefully label your graphs.

\begin{multicols}{2}

\begin{enumerate}

\item Circle: $r = 6\sin(\theta)$ \label{polarplotfirst}

\item Circle: $r = 2\cos(\theta)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Rose: $r = 2\sin(2\theta)$

\item Rose: $r = 4\cos(2\theta)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Rose: $r = 5\sin(3\theta)$

\item Rose: $r = \cos(5\theta)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Rose: $r = \sin(4\theta)$

\item Rose: $r = 3\cos(4\theta)$ \label{roseexercise8petal}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Cardioid: $r = 3 - 3\cos(\theta)$

\item Cardioid: $r = 5 + 5\sin(\theta)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Cardioid: $r = 2 + 2\cos(\theta)$

\item Cardioid: $r = 1 - \sin(\theta)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Lima\c{c}on: $r = 1 - 2\cos(\theta)$

\item Lima\c{c}on: $r = 1 - 2\sin(\theta)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Lima\c{c}on: $r = 2\sqrt{3} + 4\cos(\theta)$

\item Lima\c{c}on: $r = 3-5\cos(\theta)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Lima\c{c}on: $r = 3-5\sin(\theta)$

\item Lima\c{c}on: $r = 2 + 7\sin(\theta)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Lemniscate: $r^{2} = \sin(2\theta)$

\item Lemniscate: $r^{2} = 4\cos(2\theta)$ \label{polarplotlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{findpolarintfirst} - \ref{findpolarintlast}, find the exact polar coordinates of the points of intersection of graphs of the polar equations. Remember to check for intersection at the pole (origin).

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = 3\cos(\theta)$ and $r = 1 + \cos(\theta)$ \label{findpolarintfirst}

\item $r = 1 + \sin(\theta)$ and $r = 1 - \cos(\theta)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = 1-2\sin(\theta)$ and $r=2$

\item $r = 1 - 2\cos(\theta)$ and $r = 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = 2\cos(\theta)$ and $r = 2\sqrt{3} \sin(\theta)$

\item $r = 3\cos(\theta)$ and $r = \sin(\theta)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r^2 = 4\cos(2\theta)$ and $r = \sqrt{2}$

\item $r^{2} = 2\sin(2\theta)$ and $r = 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = 4\cos(2\theta)$ and $r=2$

\item $r = 2\sin(2\theta)$ and $r = 1$ \label{findpolarintlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{regionsketchfirst} - \ref{regionsketchlast}, sketch the region in the $xy$-plane described by the given set.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 3, \,0 \leq \theta \leq 2\pi \right\}$ \label{regionsketchfirst}

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 4\sin(\theta), \,0 \leq \theta \leq \pi \right\}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 3\cos(\theta), \, -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \right\}$

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 2\sin(2\theta), \,0 \leq \theta \leq \frac{\pi}{2} \right\}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 4\cos(2\theta), \, -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} \right\}$

\item $\left\{ (r,\theta) \, | \, 1 \leq r \leq 1-2\cos(\theta), \, \frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2} \right\}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ (r,\theta) \, | \, 1 + \cos(\theta) \leq r \leq 3\cos(\theta), \, -\frac{\pi}{3} \leq \theta \leq \frac{\pi}{3} \right\}$

\item $\left\{ (r,\theta) \, | \, 1 \leq r \leq \sqrt{2\sin(2\theta)}, \, \frac{13\pi}{12} \leq \theta \leq \frac{17\pi}{12} \right\}$

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 2\sqrt{3} \sin(\theta), \, 0 \leq \theta \leq \frac{\pi}{6} \right\} \cup \left\{ (r,\theta) \, | \, 0 \leq r \leq 2\cos(\theta), \, \frac{\pi}{6} \leq \theta \leq \frac{\pi}{2} \right\}$

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 2 \sin(2\theta), \, 0 \leq \theta \leq \frac{\pi}{12} \right\} \cup \left\{ (r,\theta) \, | \, 0 \leq r \leq 1, \, \frac{\pi}{12} \leq \theta \leq \frac{\pi}{4} \right\}$ \label{regionsketchlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

In Exercises \ref{setbuildpolarfirst} - \ref{setbuildpolarlast}, use set-builder notation to describe the polar region. Assume that the region contains its bounding curves.

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item The region inside the circle $r = 5$. \label{setbuildpolarfirst}

\item The region inside the circle $r=5$ which lies in Quadrant III.

\item The region inside the left half of the circle $r = 6\sin(\theta)$.

\item The region inside the circle $r = 4\cos(\theta)$ which lies in Quadrant IV.

\item The region inside the top half of the cardioid $r = 3 - 3\cos(\theta)$

\item The region inside the cardioid $r = 2-2\sin(\theta)$ which lies in Quadrants I and IV.

\item The inside of the petal of the rose $r = 3\cos(4\theta)$ which lies on the positive $x$-axis

\item The region inside the circle $r=5$ but outside the circle $r=3$.

\item The region which lies inside of the circle $r = 3\cos(\theta)$ but outside of the circle $r = \sin(\theta)$

\item The region in Quadrant I which lies inside both the circle $r=3$ as well as the rose $r = 6\sin(2\theta)$ \label{setbuildpolarlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\phantomsection

\label{polargraphscalculator}

While the authors truly believe that graphing polar curves by hand is fundamental to your understanding of the polar coordinate system, we would be derelict in our duties if we totally ignored the graphing calculator. Indeed, there are some important polar curves which are simply too difficult to graph by hand and that makes the calculator an important tool for your further studies in Mathematics, Science and Engineering. We now give a brief demonstration of how to use the graphing calculator to plot polar curves. The first thing you must do is switch the MODE of your calculator to POL, which stands for polar''.

\begin{center}

\begin{tabular}{ccc}

\includegraphics[width=1.8in]{./AppExtGraphics/Polar01.jpg} &

\hspace{0.05in} \includegraphics[width=1.8in]{./AppExtGraphics/Polar02.jpg} &

\hspace{0.05in} \includegraphics[width=1.8in]{./AppExtGraphics/Polar03.jpg} \\

\end{tabular}

\end{center}

This changes the Y='' menu as seen above in the middle. Let's plot the polar rose given by $r = 3\cos(4\theta)$ from Exercise \ref{roseexercise8petal} above. We type the function into the r='' menu as seen above on the right. We need to set the viewing window so that the curve displays properly, but when we look at the WINDOW menu, we find three extra lines.

\begin{center}

\begin{tabular}{cc}

\includegraphics[width=1.8in]{./AppExtGraphics/Polar04.jpg} &

\hspace{0.75in} \includegraphics[width=1.8in]{./AppExtGraphics/Polar05.jpg}\\

\end{tabular}

\end{center}

In order for the calculator to be able to plot $r = 3\cos(4\theta)$ in the $xy$-plane, we need to tell it not only the dimensions which $x$ and $y$ will assume, but we also what values of $\theta$ to use. From our previous work, we know that we need $0 \leq \theta \leq 2\pi$, so we enter the data you see above. (I'll say more about the $\theta$-step in just a moment.) Hitting GRAPH yields the curve below on the left which doesn't look quite right. The issue here is that the calculator screen is 96 pixels wide but only 64 pixels tall. To get a true geometric perspective, we need to hit ZOOM SQUARE (seen below in the middle) to produce a more accurate graph which we present below on the right.

\begin{center}

\begin{tabular}{ccc}

\includegraphics[width=1.8in]{./AppExtGraphics/Polar06.jpg} &

\hspace{0.05in} \includegraphics[width=1.8in]{./AppExtGraphics/Polar07.jpg} &

\hspace{0.05in} \includegraphics[width=1.8in]{./AppExtGraphics/Polar08.jpg} \\

\end{tabular}

\end{center}

In function mode, the calculator automatically divided the interval [Xmin, Xmax] into 96 equal subintervals. In polar mode, however, we must specify how to split up the interval [$\theta$min, $\theta$max] using the $\theta$step. For most graphs, a $\theta$step of 0.1 is fine. If you make it too small then the calculator takes a long time to graph. It you make it too big, you get chunky garbage like this.

\begin{center}

\includegraphics[width=1.8in]{./AppExtGraphics/Polar09.jpg}

\end{center}

You will need to experiment with the settings in order to get a nice graph. Exercises \ref{polarcalcfirst} - \ref{polarcalclast} give you some curves to graph using your calculator. Notice that some of them have explicit bounds on $\theta$ and others do not.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = \theta, \, 0 \leq \theta \leq 12\pi$ \label{polarcalcfirst}

\item $r = \ln(\theta), \, 1 \leq \theta \leq 12\pi$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = e^{.1\theta}, \, 0 \leq \theta \leq 12\pi$

\item $r = \theta^{3} - \theta, \, -1.2 \leq \theta \leq 1.2$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = \sin(5\theta) - 3\cos(\theta)$

\item $r = \sin^{3}\left(\frac{\theta}{2}\right) + \cos^{2}\left(\frac{\theta}{3}\right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = \arctan(\theta), \, -\pi \leq \theta \leq \pi$ \vphantom{$\dfrac{1}{1 - \cos(\theta)}$}

\item $r = \dfrac{1}{1 - \cos(\theta)}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = \dfrac{1}{2 - \cos(\theta)}$

\item $r = \dfrac{1}{2 - 3\cos(\theta)}$ \label{polarcalclast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item How many petals does the polar rose $r = \sin(2\theta)$ have? What about $r = \sin(3\theta)$, $r = \sin(4\theta)$ and $r = \sin(5\theta)$? With the help of your classmates, make a conjecture as to how many petals the polar rose $r = \sin(n\theta)$ has for any natural number $n$. Replace sine with cosine and repeat the investigation. How many petals does $r = \cos(n\theta)$ have for each natural number $n$?

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\phantomsection

\label{polarsymmetry}

Looking back through the graphs in the section, it's clear that many polar curves enjoy various forms of symmetry. However, classifying symmetry for polar curves is not as straight-forward as it was for equations back on page \pageref{symmetrytestequations}. In Exercises \ref{sympolarfirst} - \ref{sympolarlast}, we have you and your classmates explore some of the more basic forms of symmetry seen in common polar curves.

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Show that if $f$ is even\footnote{Recall that this means $f(-\theta) = f(\theta)$ for $\theta$ in the domain of $f$.} then the graph of $r = f(\theta)$ is symmetric about the $x$-axis. \label{sympolarfirst}

\begin{enumerate}

\item Show that $f(\theta) = 2 + 4\cos(\theta)$ is even and verify that the graph of $r = 2+4\cos(\theta)$ is indeed symmetric about the $x$-axis. (See Example \ref{polargraphex} number \ref{limacon02}.)

\item Show that $f(\theta) = 3\sin\left(\frac{\theta}{2}\right)$ is \textbf{not} even, yet the graph of $r = 3\sin\left(\frac{\theta}{2}\right)$ \textbf{is} symmetric about the $x$-axis. (See Example \ref{polargraphintex} number \ref{samepolarcurveex}.)

\end{enumerate}

\item Show that if $f$ is odd\footnote{Recall that this means $f(-\theta) = -f(\theta)$ for $\theta$ in the domain of $f$.} then the graph of $r = f(\theta)$ is symmetric about the origin.

\begin{enumerate}

\item Show that $f(\theta) = 5\sin(2\theta)$ is odd and verify that the graph of $r = 5\sin(2\theta)$ is indeed symmetric about the origin. (See Example \ref{polargraphex} number \ref{rose}.)

\item Show that $f(\theta) = 3\cos\left(\frac{\theta}{2}\right)$ is \textbf{not} odd, yet the graph of $r = 3\cos\left(\frac{\theta}{2}\right)$ \textbf{is} symmetric about the origin. (See Example \ref{polargraphintex} number \ref{samepolarcurveex}.)

\end{enumerate}

\item Show that if $f(\pi-\theta)=f(\theta)$ for all $\theta$ in the domain of $f$ then the graph of $r = f(\theta)$ is symmetric about the $y$-axis. \label{sympolarlast}

\begin{enumerate}

\item For $f(\theta) = 4-2\sin(\theta)$, show that $f(\pi - \theta) = f(\theta)$ and the graph of $r = 4-2\sin(\theta)$ is symmetric about the $y$-axis, as required. (See Example \ref{polargraphex} number \ref{limacon01}.)

\item For $f(\theta) = 5\sin(2\theta)$, show that $f\left(\pi - \frac{\pi}{4} \right) \neq f\left( \frac{\pi}{4} \right)$, yet the graph of $r = 5\sin(2\theta)$ \textbf{is} symmetric about the $y$-axis. (See Example \ref{polargraphex} number \ref{rose}.)

\end{enumerate}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

In Section \ref{Transformations}, we discussed transformations of graphs. In Exercise \ref{polargraphtransformations} we have you and your classmates explore transformations of polar graphs.

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item For Exercises \ref{polargraphexercise1} and \ref{polargraphexercise2} below, let $f(\theta) = \cos(\theta)$ and $g(\theta) = 2-\sin(\theta)$. \label{polargraphtransformations}

\begin{enumerate}

\item Using your graphing calculator, compare the graph of $r = f(\theta)$ to each of the graphs of $r = f\left(\theta + \frac{\pi}{4}\right)$, $r = f\left(\theta + \frac{3\pi}{4}\right)$, $r = f\left(\theta - \frac{\pi}{4}\right)$ and $r = f\left(\theta - \frac{3\pi}{4}\right)$. Repeat this process for $g(\theta)$. In general, how do you think the graph of $r = f(\theta + \alpha)$ compares with the graph of $r = f(\theta)$?

\label{polargraphexercise1}

\item Using your graphing calculator, compare the graph of $r = f(\theta)$ to each of the graphs of $r = 2f\left(\theta\right)$, $r = \frac{1}{2} f\left(\theta\right)$, $r = -f\left(\theta\right)$ and $r = -3 f(\theta)$. Repeat this process for $g(\theta)$. In general, how do you think the graph of $r = k \cdot f(\theta)$ compares with the graph of $r = f(\theta)$? (Does it matter if $k>0$ or $k<0$?)

\label{polargraphexercise2}

\end{enumerate}

\item In light of Exercises \ref{sympolarfirst} - \ref{sympolarlast}, how would the graph of $r = f(-\theta)$ compare with the graph of $r = f(\theta)$ for a generic function $f$? What about the graphs of $r = -f(\theta)$ and $r = f(\theta)$? What about $r = f(\theta)$ and $r = f(\pi - \theta)$? Test out your conjectures using a variety of polar functions found in this section with the help of a graphing utility.

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item With the help of your classmates, research cardioid microphones.

\item Back in Section \ref{Relations}, in the paragraph before Exercise \ref{listofcurvesfirst}, we gave you this \href{http://en.wikipedia.org/wiki/List_of...derline{link}} to a fascinating list of curves. Some of these curves have polar representations which we invite you and your classmates to research.

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\newpage

\begin{multicols}{2} %\raggedcolumns

\begin{enumerate}

\item Circle: $r = 6\sin(\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-6 \hspace{6pt}$} -4, {$6$} 4}

\axislabels {y}{{$-6$} -4, {$6$} 4}

\normalsize

\plrfcn{0,180,5}{4*sind t}

\end{mfpic}

\item Circle: $r = 2\cos(\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-2 \hspace{6pt}$} -4, {$2$} 4}

\axislabels {y}{{$-2$} -4, {$2$} 4}

\normalsize

\plrfcn{0,180,5}{4*cosd t}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Rose: $r = 2\sin(2\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-2 \hspace{6pt}$} -4, {$2$} 4}

\axislabels {y}{{$-2$} -4, {$2$} 4}

\normalsize

\plrfcn{0,360,5}{4*sind(2*t)}

\end{mfpic}

\item Rose: $r = 4\cos(2\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-4 \hspace{6pt}$} -4, {$4$} 4}

\axislabels {y}{{$-4$} -4, {$4$} 4}

\normalsize

\plrfcn{0,360,5}{4*cosd(2*t)}

\dashed \polyline{(3,3), (-3,-3)}

\gclear \tlabelrect(3,3){\scriptsize $\theta = \frac{\pi}{4}$}

\dashed \polyline{(3,-3), (-3,3)}

\gclear \tlabelrect(-3,3){\scriptsize $\theta = \frac{3\pi}{4}$}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Rose: $r = 5\sin(3\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-5 \hspace{6pt}$} -4, {$5$} 4}

\axislabels {y}{{$-5$} -4, {$5$} 4}

\normalsize

\plrfcn{0,180,5}{4*sind(3*t)}

\dashed \polyline{(2,3.464), (-2,-3.464)}

\gclear \tlabelrect(2,3.464){\scriptsize $\theta = \frac{\pi}{3}$}

\dashed \polyline{(-2,3.464), (2,-3.464)}

\gclear \tlabelrect(-2,3.464){\scriptsize $\theta = \frac{2\pi}{3}$}

\end{mfpic}

\item Rose: $r = \cos(5\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-1 \hspace{6pt}$} -4, {$1$} 4}

\axislabels {y}{{$-1$} -4, {$1$} 4}

\normalsize

\plrfcn{0,180,5}{4*cosd(5*t)}

\dashed \polyline{(3.804,1.236), (-3.804,-1.236)}

\gclear \tlabelrect(3.804,1.236){\scriptsize $\theta = \frac{\pi}{10}$}

\dashed \polyline{(2.351,3.236), (-2.351,-3.236)}

\gclear \tlabelrect(2.351,3.236){\scriptsize $\theta = \frac{3\pi}{10}$}

\dashed \polyline{(-2.351,3.236), (2.351,-3.236)}

\gclear \tlabelrect(-2.351,3.236){\scriptsize $\theta = \frac{7\pi}{10}$}

\dashed \polyline{(-3.804,1.236), (3.804,-1.236)}

\gclear \tlabelrect(-3.804,1.236){\scriptsize $\theta = \frac{9\pi}{10}$}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Rose: $r = \sin(4\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-1 \hspace{6pt}$} -4, {$1$} 4}

\axislabels {y}{{$-1$} -4, {$1$} 4}

\normalsize

\plrfcn{0,360,5}{4*sind(4*t)}

\dashed \polyline{(3,3), (-3,-3)}

\gclear \tlabelrect(3,3){\scriptsize $\theta = \frac{\pi}{4}$}

\dashed \polyline{(3,-3), (-3,3)}

\gclear \tlabelrect(-3,3){\scriptsize $\theta = \frac{3\pi}{4}$}

\end{mfpic}

\item Rose: $r = 3\cos(4\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-3 \hspace{6pt}$} -4, {$3$} 4}

\axislabels {y}{{$-3$} -4, {$3$} 4}

\normalsize

\plrfcn{0,360,5}{4*cosd(4*t)}

\dashed \polyline{(3.696,1.531), (-3.696,-1.531)}

\gclear \tlabelrect(3.696,1.531){\scriptsize $\theta = \frac{\pi}{8}$}

\dashed \polyline{(1.531,3.696), (-1.531,-3.696)}

\gclear \tlabelrect(1.531,3.696){\scriptsize $\theta = \frac{3\pi}{8}$}

\dashed \polyline{(-1.531,3.696), (1.531,-3.696)}

\gclear \tlabelrect(-1.531,3.696){\scriptsize $\theta = \frac{5\pi}{8}$}

\dashed \polyline{(-3.696,1.531), (3.696,-1.531)}

\gclear \tlabelrect(-3.696,1.531){\scriptsize $\theta = \frac{7\pi}{8}$}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Cardioid: $r = 3 - 3\cos(\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,-2,2,4}

\ymarks{-4,-2,2,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-6 \hspace{6pt}$} -4, {$-3 \hspace{6pt}$} -2, {$3$} 2, {$6$} 4}

\axislabels {y}{{$-6$} -4, {$-3$} -2, {$3$} 2, {$6$} 4}

\normalsize

\plrfcn{0,360,5}{2 - 2*cosd(t)}

\end{mfpic}

\item Cardioid: $r = 5 + 5\sin(\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,-2,2,4}

\ymarks{-4,-2,2,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-10 \hspace{6pt}$} -4, {$-5 \hspace{6pt}$} -2, {$5$} 2, {$10$} 4}

\axislabels {y}{{$-10$} -4, {$-5$} -2, {$5$} 2, {$10$} 4}

\normalsize

\plrfcn{0,360,5}{2 + 2*sind(t)}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Cardioid: $r = 2 + 2\cos(\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,-2,2,4}

\ymarks{-4,-2,2,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-4 \hspace{6pt}$} -4, {$-2 \hspace{6pt}$} -2, {$2$} 2, {$4$} 4}

\axislabels {y}{{$-4$} -4, {$-2$} -2, {$2$} 2, {$4$} 4}

\normalsize

\plrfcn{0,360,5}{2 + 2*cosd(t)}

\end{mfpic}

\item Cardioid: $r = 1 - \sin(\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,-2,2,4}

\ymarks{-4,-2,2,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-2 \hspace{6pt}$} -4, {$-1 \hspace{6pt}$} -2, {$1$} 2, {$2$} 4}

\axislabels {y}{{$-2$} -4, {$-1$} -2, {$1$} 2, {$2$} 4}

\normalsize

\plrfcn{0,360,5}{2 - 2*sind(t)}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Lima\c{c}on: $r = 1 - 2\cos(\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,-1.3333,1.3333,4}

\ymarks{-4,-1.3333,1.3333,4}

\tlpointsep{4pt}

\scriptsize

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\axislabels {y}{{$-3$} -4, {$-1$} -1.3333, {$1$} 1.3333, {$3$} 4}

\normalsize

\plrfcn{0,360,5}{1.3333*(1 - 2*cosd(t))}

\dashed \polyline{(0,0), (2,3.464)}

\gclear \tlabelrect(2,3.464){\scriptsize $\theta = \frac{\pi}{3}$}

\dashed \polyline{(0,0), (2,-3.464)}

\gclear \tlabelrect(2,-3.464){\scriptsize $\theta = \frac{5\pi}{3}$}

\end{mfpic}

\item Lima\c{c}on: $r = 1 - 2\sin(\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,-1.3333,1.3333,4}

\ymarks{-4,-1.3333,1.3333,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-3 \hspace{6pt}$} -4, {$-1 \hspace{6pt}$} -1.3333, {$1$} 1.3333, {$3$} 4}

\axislabels {y}{{$-3$} -4, {$-1$} -1.3333, {$1$} 1.3333, {$3$} 4}

\normalsize

\plrfcn{0,360,5}{1.3333*(1 - 2*sind(t))}

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\gclear \tlabelrect(-3.464,2){\scriptsize $\theta = \frac{5\pi}{6}$}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Lima\c{c}on: $r = 2\sqrt{3} + 4\cos(\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,0.5){$x$}

\tlabel[cc](0.5,5){$y$}

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\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-2\sqrt{3} - 4 \hspace{6pt}$} -4, {$2\sqrt{3} + 4$} 4}

\axislabels {y}{{$-2\sqrt{3} - 4$} -4, {$-2\sqrt{3}$} -1.856, {$2\sqrt{3}$} 1.856, {$2\sqrt{3} + 4$} 4}

\normalsize

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\gclear \tlabelrect(-3.464,2){\scriptsize $\theta = \frac{5\pi}{6}$}

\end{mfpic}

\item Lima\c{c}on: $r = 3-5\cos(\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

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\scriptsize

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\normalsize

\plrfcn{0,360,5}{0.5*(3 - 5*cosd(t))}

\dashed \polyline{(0,0), (2.41,3.19)}

\gclear \tlabelrect(2.9,3.3){\tiny $\theta = \arccos\left(\frac{3}{5}\right)$}

\dashed \polyline{(0,0), (2.41,-3.19)}

\gclear \tlabelrect(2.9,-3.3){\tiny $\theta = 2\pi - \arccos\left(\frac{3}{5}\right)$}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Lima\c{c}on: $r = 3-5\sin(\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,-1.5,1.5,4}

\ymarks{-4,-1,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-8 \hspace{6pt}$} -4, {$-3 \hspace{6pt}$} -1.5, {$3$} 1.5, {$8$} 4}

\axislabels {y}{{$-8$} -4, {$-2$} -1, {$8$} 4}

\normalsize

\plrfcn{0,360,5}{0.5*(3 - 5*sind(t))}

\dashed \polyline{(0,0), (3.2,2.4)}

\gclear \tlabelrect(2.9,3){\tiny $\theta = \arcsin\left(\frac{3}{5}\right)$}

\dashed \polyline{(0,0), (-3.2,2.4)}

\gclear \tlabelrect(-2.9,3){\tiny $\theta = \pi - \arcsin\left(\frac{3}{5}\right)$}

\end{mfpic}

\item Lima\c{c}on: $r = 2 + 7\sin(\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,-0.8888,0.8888,4}

\ymarks{-4,2.2222,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-9 \hspace{6pt}$} -4, {$-2 \hspace{6pt}$} -0.8888, {$2$} 0.8888, {$9$} 4}

\axislabels {y}{{$-9$} -4, {$5$} 2.2222, {$9$} 4}

\normalsize

\plrfcn{0,360,5}{0.4444*(2 + 7*sind(t))}

\dashed \polyline{(0,0), (-3.842,-1.113)}

\gclear \tlabelrect(-2.9,-1.3){\tiny $\theta = \pi + \arcsin\left(\frac{2}{7}\right)$}

\dashed \polyline{(0,0), (3.842,-1.113)}

\gclear \tlabelrect(2.9,-1.3){\tiny $\theta = 2\pi - \arcsin\left(\frac{2}{7}\right)$}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Lemniscate: $r^{2} = \sin(2\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-1 \hspace{6pt}$} -4, {$1$} 4}

\axislabels {y}{{$-1$} -4, {$1$} 4}

\normalsize

\plrfcn{0,90,5}{4*sqrt(sind(2*t))}

\plrfcn{180,270,5}{4*sqrt(sind(2*t))}

\end{mfpic}

\item Lemniscate: $r^{2} = 4\cos(2\theta)$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-2 \hspace{6pt}$} -4, {$2$} 4}

\axislabels {y}{{$-2$} -4, {$2$} 4}

\normalsize

\plrfcn{-45,45,5}{4*sqrt(cosd(2*t))}

\plrfcn{135,225,5}{4*sqrt(cosd(2*t))}

\dashed \polyline{(3,3), (-3,-3)}

\gclear \tlabelrect(3,3){\scriptsize $\theta = \frac{\pi}{4}$}

\dashed \polyline{(3,-3), (-3,3)}

\gclear \tlabelrect(-3,3){\scriptsize $\theta = \frac{3\pi}{4}$}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item \begin{multicols}{2} \raggedcolumns

$r = 3\cos(\theta)$ and $r = 1 + \cos(\theta)$

\begin{mfpic}[17]{-4}{4}{-4}{4}

\axes

\tlabel[cc](4,-0.5){$x$}

\tlabel[cc](0.5,4){$y$}

\xmarks{-3,-2,-1,1,2,3}

\ymarks{-3,-2,-1,1,2,3}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\axislabels {y}{{$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\normalsize

\plrfcn{0,360,5}{1 + cosd(t)}

\plrfcn{0,360,5}{3*cosd(t)}

\end{mfpic}

$\left( \dfrac{3}{2}, \dfrac{\pi}{3} \right)$, $\left( \dfrac{3}{2}, \dfrac{5\pi}{3} \right)$, pole

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$r = 1 + \sin(\theta)$ and $r = 1 - \cos(\theta)$

\begin{mfpic}[23]{-2.9}{3}{-3}{3}

\axes

\tlabel[cc](3,-0.5){$x$}

\tlabel[cc](0.5,3){$y$}

\xmarks{-2,-1,1,2}

\ymarks{-2,-1,1,2}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2}

\axislabels {y}{{$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2}

\normalsize

\plrfcn{0,360,5}{1 - cosd(t)}

\plrfcn{0,360,5}{1 + sind(t)}

\end{mfpic}

$\left( \dfrac{2 + \sqrt{2}}{2}, \dfrac{3\pi}{4} \right)$, $\left( \dfrac{2 - \sqrt{2}}{2}, \dfrac{7\pi}{4} \right)$, pole

\end{multicols}

\pagebreak

\item \begin{multicols}{2} \raggedcolumns

$r = 1 - 2\sin(\theta)$ and $r = 2$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,-1.3333,1.3333,4}

\ymarks{-4,-1.3333,1.3333,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-3 \hspace{6pt}$} -4, {$-1 \hspace{6pt}$} -1.3333, {$1$} 1.3333, {$3$} 4}

\axislabels {y}{{$-3$} -4, {$-1$} -1.3333, {$1$} 1.3333, {$3$} 4}

\normalsize

\plrfcn{0,360,5}{1.3333*(1 - 2*sind(t))}

\circle{(0,0),2}

\end{mfpic}

$\left( 2, \dfrac{7\pi}{6} \right)$, $\left( 2, \dfrac{11\pi}{6} \right)$

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$r = 1 - 2\cos(\theta)$ and $r = 1$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,-1.3333,1.3333,4}

\ymarks{-4,-1.3333,1.3333,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-3 \hspace{6pt}$} -4, {$-1 \hspace{6pt}$} -1.3333, {$1$} 1.3333, {$3$} 4}

\axislabels {y}{{$-3$} -4, {$-1$} -1.3333, {$1$} 1.3333, {$3$} 4}

\normalsize

\plrfcn{0,360,5}{1.3333*(1 - 2*cosd(t))}

\plrfcn{0,360,5}{1.3333}

\end{mfpic}

$\left( 1, \dfrac{\pi}{2} \right)$, $\left( 1, \dfrac{3\pi}{2} \right)$, $(-1, 0)$

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$r = 2\cos(\theta)$ and $r = 2\sqrt{3} \sin(\theta)$

\begin{mfpic}[19]{-4}{4}{-5}{5}

\axes

\tlabel[cc](4,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-3,-2,-1,1,2,3}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\axislabels {y}{{$-4$} -4,{$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\normalsize

\plrfcn{0,360,5}{3.46*sind(t)}

\plrfcn{0,360,5}{2*cosd(t)}

\end{mfpic}

$\left(\sqrt{3}, \dfrac{\pi}{6} \right)$, pole

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$r = 3\cos(\theta)$ and $r = \sin(\theta)$

\begin{mfpic}[19]{-4}{4}{-4}{4}

\axes

\tlabel[cc](4,-0.5){$x$}

\tlabel[cc](0.5,4){$y$}

\xmarks{-3,-2,-1,1,2,3}

\ymarks{-3,-2,-1,1,2,3}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\axislabels {y}{{$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\normalsize

\plrfcn{0,360,5}{sind(t)}

\plrfcn{0,360,5}{3*cosd(t)}

\end{mfpic}

$\left(\dfrac{3\sqrt{10}}{10}, \arctan(3)\right)$, pole

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$r^2 = 4\cos(2\theta)$ and $r = \sqrt{2}$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-2 \hspace{6pt}$} -4, {$2$} 4}

\axislabels {y}{{$-2$} -4, {$2$} 4}

\normalsize

\plrfcn{-45,45,5}{4*sqrt(cosd(2*t))}

\plrfcn{135,225,5}{4*sqrt(cosd(2*t))}

\circle{(0,0), 1.414}

\end{mfpic}

$\left(\sqrt{2}, \dfrac{\pi}{6}\right)$, $\left(\sqrt{2}, \dfrac{5\pi}{6}\right)$, $\left(\sqrt{2}, \dfrac{7\pi}{6}\right)$, $\left(\sqrt{2}, \dfrac{11\pi}{6}\right)$

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$r^{2} = 2\sin(2\theta)$ and $r = 1$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{\tiny $-\sqrt{2} \hspace{6pt}$} -4, {$-1 \hspace{6pt}$} -2.828, {$1$} 2.828, {\tiny $\sqrt{2}$} 4}

\axislabels {y}{{\tiny $-\sqrt{2}$} -4, {$-1$} -2.828, {$1$} 2.828, {\tiny $\sqrt{2}$} 4}

\normalsize

\plrfcn{0,90,5}{4*sqrt(sind(2*t))}

\plrfcn{180,270,5}{4*sqrt(sind(2*t))}

\plrfcn{0,360,5}{2.828}

\end{mfpic}

$\left(1, \dfrac{\pi}{12}\right)$, $\left(1, \dfrac{5\pi}{12}\right)$, $\left(1, \dfrac{13\pi}{12}\right)$, $\left(1, \dfrac{17\pi}{12}\right)$

\end{multicols}

\pagebreak

\item \begin{multicols}{2} \raggedcolumns

$r = 4\cos(2\theta)$ and $r = 2$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-4 \hspace{6pt}$} -4, {$4$} 4}

\axislabels {y}{{$-4$} -4, {$4$} 4}

\normalsize

\plrfcn{0,360,5}{4*cosd(2*t)}

\circle{(0,0),2}

\end{mfpic}

$\left( 2, \dfrac{\pi}{6} \right)$, $\left( 2, \dfrac{5\pi}{6} \right)$, $\left( 2, \dfrac{7\pi}{6} \right)$,

$\left( 2, \dfrac{11\pi}{6} \right)$, $\left( -2, \dfrac{\pi}{3} \right)$, $\left( -2, \dfrac{2\pi}{3} \right)$,

$\left( -2, \dfrac{4\pi}{3} \right)$, $\left( -2, \dfrac{5\pi}{3} \right)$

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$r = 2\sin(2\theta)$ and $r = 1$

\begin{mfpic}[15]{-4.5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-2 \hspace{6pt}$} -4, {$2$} 4}

\axislabels {y}{{$-2$} -4, {$2$} 4}

\normalsize

\plrfcn{0,360,5}{4*sind(2*t)}

\plrfcn{0,360,5}{2}

\end{mfpic}

$\left( 1, \dfrac{\pi}{12} \right)$, $\left( 1, \dfrac{5\pi}{12} \right)$, $\left( 1, \dfrac{13\pi}{12} \right)$,

$\left( 1, \dfrac{17\pi}{12} \right)$, $\left( -1, \dfrac{7\pi}{12} \right)$, $\left( -1, \dfrac{11\pi}{12} \right)$,

$\left( -1, \dfrac{19\pi}{12} \right)$, $\left( -1, \dfrac{23\pi}{12} \right)$

\end{multicols}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\pagebreak

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 3, \,0 \leq \theta \leq 2\pi \right\}$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\fillcolor[gray]{0.7}

\gfill \circle{(0,0),3}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-3,-2,-1,1,2,3}

\ymarks{-3,-2,-1,1,2,3}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\axislabels {y}{{$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\normalsize

\circle{(0,0),3}

\end{mfpic}

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 4\sin(\theta), \,0 \leq \theta \leq \pi \right\}$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\fillcolor[gray]{0.7}

\gfill \plrregion{0,180,1}{4*sind(t)}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,-3,-2,-1,1,2,3,4}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-4 \hspace{6pt}$} -4,{$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\axislabels {y}{{$-4$} -4,{$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\normalsize

\plrfcn{0,180,5}{4*sind(t)}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 3\cos(\theta), \, -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \right\}$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\fillcolor[gray]{0.7}

\gfill \plrregion{-90,90,1}{3*cosd(t)}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-3,-2,-1,1,2,3}

\ymarks{-3,-2,-1,1,2,3}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\axislabels {y}{{$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\normalsize

\plrfcn{-90,90,5}{3*cosd(t)}

\end{mfpic}

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 2\sin(2\theta), \,0 \leq \theta \leq \frac{\pi}{2} \right\}$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\fillcolor[gray]{0.7}

\gfill \plrregion{0,90,5}{4*sind(2*t)}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-2 \hspace{6pt}$} -4, {$2$} 4}

\axislabels {y}{{$-2$} -4, {$2$} 4}

\normalsize

\plrfcn{0,360,5}{4*sind(2*t)}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 4\cos(2\theta), \, -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} \right\}$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\fillcolor[gray]{0.7}

\gfill \plrregion{-45,45,5}{4*cosd(2*t)}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-4 \hspace{6pt}$} -4, {$4$} 4}

\axislabels {y}{{$-4$} -4, {$4$} 4}

\normalsize

\plrfcn{0,360,5}{4*cosd(2*t)}

\end{mfpic}

\item $\left\{ (r,\theta) \, | \, 1 \leq r \leq 1-2\cos(\theta), \, \frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2} \right\}$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\fillcolor[gray]{0.7}

\gfill \plrregion{90,270,5}{1.3333*(1 - 2*cosd(t))}

\gclear \plrregion{90,270,5}{1.3333}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,-1.3333,1.3333,4}

\ymarks{-4,-1.3333,1.3333,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-3 \hspace{6pt}$} -4, {$-1 \hspace{6pt}$} -1.3333, {$1$} 1.3333, {$3$} 4}

\axislabels {y}{{$-3$} -4, {$-1$} -1.3333, {$1$} 1.3333, {$3$} 4}

\normalsize

\plrfcn{0,360,5}{1.3333*(1 - 2*cosd(t))}

\plrfcn{0,360,5}{1.3333}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\pagebreak

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ (r,\theta) \, | \, 1 + \cos(\theta) \leq r \leq 3\cos(\theta), \, -\frac{\pi}{3} \leq \theta \leq \frac{\pi}{3} \right\}$

\begin{mfpic}[17]{-4}{4}{-4}{4}

\fillcolor[gray]{0.7}

\gfill \plrregion{-60,60,1}{3*cosd(t)}

\gclear \plrregion{-60,60,1}{1 + cosd(t)}

\axes

\tlabel[cc](4,-0.5){$x$}

\tlabel[cc](0.5,4){$y$}

\xmarks{-3,-2,-1,1,2,3}

\ymarks{-3,-2,-1,1,2,3}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\axislabels {y}{{$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\normalsize

\plrfcn{0,360,5}{1 + cosd(t)}

\plrfcn{0,360,5}{3*cosd(t)}

\end{mfpic}

\item $\left\{ (r,\theta) \, | \, 1 \leq r \leq \sqrt{2\sin(2\theta)}, \, \frac{13\pi}{12} \leq \theta \leq \frac{17\pi}{12} \right\}$

\begin{mfpic}[15]{-5}{5}{-5}{5}

\fillcolor[gray]{0.7}

\gfill \plrregion{195,255,1}{4*sqrt(sind(2*t))}

\gclear \plrregion{195,255,1}{2.828}

\axes

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{\tiny $-\sqrt{2} \hspace{6pt}$} -4, {$-1 \hspace{6pt}$} -2.828, {$1$} 2.828, {\tiny $\sqrt{2}$} 4}

\axislabels {y}{{\tiny $-\sqrt{2}$} -4, {$-1$} -2.828, {$1$} 2.828, {\tiny $\sqrt{2}$} 4}

\normalsize

\plrfcn{0,90,5}{4*sqrt(sind(2*t))}

\plrfcn{180,270,5}{4*sqrt(sind(2*t))}

\plrfcn{0,360,5}{2.828}

\end{mfpic}

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 2\sqrt{3} \sin(\theta), \, 0 \leq \theta \leq \frac{\pi}{6} \right\} \cup \left\{ (r,\theta) \, | \, 0 \leq r \leq 2\cos(\theta), \, \frac{\pi}{6} \leq \theta \leq \frac{\pi}{2} \right\}$

\begin{mfpic}[19]{-4}{4}{-5}{5}

\fillcolor[gray]{0.7}

\gfill \plrregion{0,30,5}{3.46*sind(t)}

\gfill \plrregion{30,90,5}{2*cosd(t)}

\axes

\tlabel[cc](4,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-3,-2,-1,1,2,3}

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\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

\axislabels {y}{{$-4$} -4,{$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\normalsize

\plrfcn{0,360,5}{3.46*sind(t)}

\plrfcn{0,360,5}{2*cosd(t)}

\end{mfpic}

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 2 \sin(2\theta), \, 0 \leq \theta \leq \frac{\pi}{12} \right\} \cup \left\{ (r,\theta) \, | \, 0 \leq r \leq 1, \, \frac{\pi}{12} \leq \theta \leq \frac{\pi}{4} \right\}$

\begin{mfpic}[15]{-4.5}{5}{-5}{5}

\axes

\fillcolor[gray]{0.7}

\gfill \plrregion{0,15,5}{4*sind(2*t)}

\gfill \plrregion{15,45,5}{2}

\tlabel[cc](5,-0.5){$x$}

\tlabel[cc](0.5,5){$y$}

\xmarks{-4,4}

\ymarks{-4,4}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-2 \hspace{6pt}$} -4, {$2$} 4}

\axislabels {y}{{$-2$} -4, {$2$} 4}

\normalsize

\plrfcn{0,360,5}{4*sind(2*t)}

\plrfcn{0,360,5}{2}

\polyline{(0,0), (1.414, 1.414)}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 5, \, 0\leq \theta \leq 2\pi \right\}$

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 5, \, \pi \leq \theta \leq \frac{3\pi}{2} \right\}$

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 6\sin(\theta), \, \frac{\pi}{2} \leq \theta \leq \pi \right\}$

\item $\left\{ (r,\theta) \, | \, 4\cos(\theta) \leq r \leq 0, \, \frac{\pi}{2} \leq \theta \leq \pi \right\}$

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 3 - 3\cos(\theta), \, 0 \leq \theta \leq \pi \right\}$

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 2-2\sin(\theta), \, 0 \leq \theta \leq \frac{\pi}{2} \right\} \cup$

$\left\{ (r,\theta) \, | \, 0 \leq r \leq 2-2\sin(\theta), \, \frac{3\pi}{2} \leq \theta \leq 2\pi \right\}$

or $\left\{ (r,\theta) \, | \, 0 \leq r \leq 2-2\sin(\theta), \, \frac{3\pi}{2} \leq \theta \leq \frac{5\pi}{2} \right\}$

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 3\cos(4\theta), \,0 \leq \theta \leq \frac{\pi}{8} \right\} \cup$ $\left\{ (r,\theta) \, | \, 0 \leq r \leq 3\cos(4\theta), \,\frac{15\pi}{8} \leq \theta \leq 2\pi \right\}$

or $\left\{ (r,\theta) \, | \, 0 \leq r \leq 3\cos(4\theta), \, -\frac{\pi}{8} \leq \theta \leq \frac{\pi}{8} \right\}$

\item $\left\{ (r,\theta) \, | \, 3 \leq r \leq 5, \, 0 \leq \theta \leq 2\pi \right\}$

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 3\cos(\theta), \, -\frac{\pi}{2} \leq \theta \leq 0 \right\} \cup$

$\left\{ (r,\theta) \, | \, \sin(\theta) \leq r \leq 3\cos(\theta), \, 0 \leq \theta \leq \arctan(3) \right\}$

\item $\left\{ (r,\theta) \, | \, 0 \leq r \leq 6\sin(2\theta), \,0 \leq \theta \leq \frac{\pi}{12} \right\} \cup$ $\left\{ (r,\theta) \, | \, 0 \leq r \leq 3, \,\frac{\pi}{12} \leq \theta \leq \frac{5\pi}{12} \right\} \cup$ \\ $\left\{ (r,\theta) \, | \, 0 \leq r \leq 6\sin(2\theta), \, \frac{5\pi}{12} \leq \theta \leq \frac{\pi}{2} \right\}$

\end{enumerate}

\closegraphsfile

## 11.6: Hooked on Conics Again

\subsection{Exercises}

Graph the following equations.

\begin{multicols}{2}

\begin{enumerate}

\item $x^2+2xy+y^2 -x\sqrt{2}+y\sqrt{2} -6= 0$

\item $7x^2-4xy\sqrt{3}+3y^2-2x-2y\sqrt{3}-5= 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $5x^2+6xy+5y^2 - 4\sqrt{2}x+4\sqrt{2}y = 0$

\item $x^2+ 2\sqrt{3}xy+3y^2+ 2\sqrt{3}x-2y-16 = 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $13x^2-34xy\sqrt{3}+47y^2 - 64=0$

\item $x^2-2\sqrt{3} xy-y^2+8=0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x^2-4xy+4y^2-2x\sqrt{5}-y\sqrt{5}=0$

\item $8x^2+12xy+17y^2 - 20 = 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

Graph the following equations.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = \dfrac{2}{1-\cos(\theta)}$

\item $r = \dfrac{3}{2 + \sin(\theta)}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = \dfrac{3}{2-\cos(\theta)}$

\item $r = \dfrac{2}{1 + \sin(\theta)}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = \dfrac{4}{1+3\cos(\theta)}$

\item $r = \dfrac{2}{1-2\sin(\theta)}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = \dfrac{2}{1 + \sin(\theta - \frac{\pi}{3})}$

\item $r = \dfrac{6}{3 - \cos\left(\theta + \frac{\pi}{4}\right)}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

The matrix $A(\theta) = \left[ \begin{array}{rr} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \\ \end{array} \right]$ is called a \textbf{rotation matrix}\index{matrix ! rotation}\index{rotation matrix}. We've seen this matrix most recently in the proof of used in the proof of Theorem \ref{rotatecoordinatesthm}.

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Show the matrix from Example \ref{rotationmatrixex} in Section \ref{MatArithmetic} is none other than $A\left(\frac{\pi}{4}\right)$.

\item Discuss with your classmates how to use $A(\theta)$ to rotate points in the plane.

\item Using the even / odd identities for cosine and sine, show $A(\theta)^{-1} = A(-\theta)$. Interpret this geometrically.

\end{enumerate}

\newpage

\begin{multicols}{2}

\begin{enumerate}

\item $x^2+2xy+y^2 -x\sqrt{2}+y\sqrt{2} -6= 0$ \\ becomes $(x')^2 = -(y'-3)$ after rotating \\ counter-clockwise through $\theta = \frac{\pi}{4}$.

\begin{mfpic}[18]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\tlabel[cc](4,3.5){\scriptsize $x'$}

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\xmarks{-4,-3,-2,-1,1,2,3,4}

\ymarks{-4,-3,-2,-1,1,2,3,4}

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\rotatepath{(0,0),45} \polyline{(4,-0.15),(4,0.15)}

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\rotatepath{(0,0),135} \polyline{(2,-0.15),(2,0.15)}

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\rotatepath{(0,0),225} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),225} \polyline{(4,-0.15),(4,0.15)}

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\rotatepath{(0,0),315} \polyline{(2,-0.15),(2,0.15)}

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\rotatepath{(0,0),315} \polyline{(4,-0.15),(4,0.15)}

\point[3pt]{(0,0)}

\arrow \parafcn{5,40,5}{4*dir(t)}

\gclear \tlabelrect[cc](4,1.5){\scriptsize $\theta = \frac{\pi}{4}$}

\tcaption{ $x^2+2xy+y^2 -x\sqrt{2}+y\sqrt{2} -6= 0$}

\end{mfpic}

\item $7x^2-4xy\sqrt{3}+3y^2-2x-2y\sqrt{3}-5= 0$ \\ becomes $\frac{(x'-2)^2}{9}+(y')^2 = 1$ after rotating \\ counter-clockwise through $\theta = \frac{\pi}{3}$

\begin{mfpic}[18]{-4}{6}{-5}{5}

\axes

\tlabel[cc](6,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\tlabel[cc](3.5,5){\scriptsize $x'$}

\tlabel[cc](-4,3){\scriptsize $y'$}

\xmarks{-3,-2,-1,1,2,3,4,5}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\point[3pt]{\plr{(-1,60)}, \plr{(5,60)}, (1.87,1.23), (0.13,2.23)}

\dashed \arrow \rotatepath{(0,0),60} \polyline{(-4,0), (6,0)}

\dashed \arrow \rotatepath{(0,0),60} \polyline{(0,-5), (0,5)}

\rotatepath{(0,0),60} \ellipse{(2,0),3,1}

\rotatepath{(0,0),60} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),60} \polyline{(2,-0.15),(2,0.15)}

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\rotatepath{(0,0),60} \polyline{(4,-0.15),(4,0.15)}

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\rotatepath{(0,0),150} \polyline{(2,-0.15),(2,0.15)}

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\rotatepath{(0,0),150} \polyline{(4,-0.15),(4,0.15)}

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\rotatepath{(0,0),240} \polyline{(2,-0.15),(2,0.15)}

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\rotatepath{(0,0),330} \polyline{(2,-0.15),(2,0.15)}

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\rotatepath{(0,0),330} \polyline{(4,-0.15),(4,0.15)}

\point[3pt]{(0,0)}

\arrow \parafcn{5,55,5}{5.5*dir(t)}

\gclear \tlabelrect[cc](4.5,3){\scriptsize $\theta = \frac{\pi}{3}$}

\tcaption{$7x^2-4xy\sqrt{3}+3y^2-2x-2y\sqrt{3}-5= 0$}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $5x^2+6xy+5y^2 - 4\sqrt{2}x+4\sqrt{2}y = 0$ \\ becomes $(x')^2+\frac{(y'+2)^2}{4} = 1$ after rotating \\ counter-clockwise through $\theta = \frac{\pi}{4}$.

\begin{mfpic}[18]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\tlabel[cc](4,3.5){\scriptsize $x'$}

\tlabel[cc](-3,4){\scriptsize $y'$}

\xmarks{-4,-3,-2,-1,1,2,3,4}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\point[3pt]{\plr{(0,0)}, \plr{(-2,135)}, \plr{(-4,135)}, \plr{(2.24,288)}, \plr{(2.24,341)}}

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\rotatepath{(0,0),135} \polyline{(4,-0.15),(4,0.15)}

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\rotatepath{(0,0),225} \polyline{(2,-0.15),(2,0.15)}

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\rotatepath{(0,0),315} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),315} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),315} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),315} \polyline{(4,-0.15),(4,0.15)}

\point[3pt]{(0,0)}

\arrow \parafcn{5,40,5}{4*dir(t)}

\gclear \tlabelrect[cc](4,1.5){\scriptsize $\theta = \frac{\pi}{4}$}

\tcaption{ $5x^2+6xy+5y^2 - 4\sqrt{2}x+4\sqrt{2}y = 0$}

\end{mfpic}

\item $x^2+ 2\sqrt{3}xy+3y^2+ 2\sqrt{3}x-2y-16 = 0$ \\ becomes$(x')^2 = y'+4$ after rotating \\ counter-clockwise through $\theta = \frac{\pi}{3}$

\begin{mfpic}[18]{-4}{6}{-5}{5}

\axes

\tlabel[cc](6,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\tlabel[cc](3.5,5){\scriptsize $x'$}

\tlabel[cc](-4,3){\scriptsize $y'$}

\xmarks{-3,-2,-1,1,2,3,4,5}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\point[3pt]{\plr{(-4,150)}, \plr{(-2,60)}, \plr{(2,60)}}

\dashed \arrow \rotatepath{(0,0),60} \polyline{(-4,0), (6,0)}

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\rotatepath{(0,0),330} \polyline{(4,-0.15),(4,0.15)}

\point[3pt]{(0,0)}

\arrow \parafcn{5,55,5}{5.5*dir(t)}

\gclear \tlabelrect[cc](4.5,3){\scriptsize $\theta = \frac{\pi}{3}$}

\tcaption{$x^2+ 2\sqrt{3}xy+3y^2+ 2\sqrt{3}x-2y-16 = 0$}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $13x^2-34xy\sqrt{3}+47y^2 - 64=0$ \\ becomes $(y')^2 - \frac{(x')^2}{16} =1$ after rotating \\ counter-clockwise through $\theta = \frac{\pi}{6}$.

\begin{mfpic}[18]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\tlabel[cc](4.75,2.25){\scriptsize $x'$}

\tlabel[cc](-2.25,4.5){\scriptsize $y'$}

\xmarks{-4,-3,-2,-1,1,2,3,4}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\point[3pt]{\plr{(1,120)}, \plr{(1,300)}}

\dashed \arrow \rotatepath{(0,0),30} \polyline{(-5,0), (5,0)}

\dashed \arrow \rotatepath{(0,0),30} \polyline{(0,-5), (0,5)}

\arrow \reverse \arrow \rotatepath{(0,0),30} \parafcn{-45,45,5}{(4*tand(t), 1/cosd(t))}

\arrow \reverse \arrow \rotatepath{(0,0),30} \parafcn{-45,45,5}{(4*tand(t), 0-1/cosd(t))}

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\rotatepath{(0,0),210} \polyline{(2,-0.15),(2,0.15)}

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\rotatepath{(0,0),210} \polyline{(4,-0.15),(4,0.15)}

\rotatepath{(0,0),300} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),300} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),300} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),300} \polyline{(4,-0.15),(4,0.15)}

\point[3pt]{(0,0)}

\arrow \parafcn{95,115,5}{4*dir(t)}

\gclear \tlabelrect[cc](-1,4.5){\scriptsize $\theta = \frac{\pi}{6}$}

\tcaption{$13x^2-34xy\sqrt{3}+47y^2 - 64=0$}

\end{mfpic}

\item $x^2-2\sqrt{3} xy-y^2+8=0$ \\ becomes $\frac{(x')^2}{4} - \frac{(y')^2}{4} = 1$ after rotating \\ counter-clockwise through $\theta = \frac{\pi}{3}$

\begin{mfpic}[18]{-4}{6}{-5}{5}

\axes

\tlabel[cc](6,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\tlabel[cc](3.5,5){\scriptsize $x'$}

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\rotatepath{(0,0),150} \polyline{(4,-0.15),(4,0.15)}

\rotatepath{(0,0),240} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),240} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),240} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),330} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),330} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),330} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),330} \polyline{(4,-0.15),(4,0.15)}

\point[3pt]{(0,0)}

\arrow \parafcn{5,55,5}{5.5*dir(t)}

\gclear \tlabelrect[cc](4.5,3){\scriptsize $\theta = \frac{\pi}{3}$}

\tcaption{$x^2-2\sqrt{3} xy-y^2+8=0$}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x^2-4xy+4y^2-2x\sqrt{5}-y\sqrt{5}=0$ \\ becomes $(y')^2=x$ after rotating \\ counter-clockwise through $\theta = \arctan\left(\frac{1}{2}\right)$.

\begin{mfpic}[18]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\tlabel[cc](4.75,2.25){\scriptsize $x'$}

\tlabel[cc](-2.25,4.5){\scriptsize $y'$}

\xmarks{-4,-3,-2,-1,1,2,3,4}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\point[3pt]{\plr{(4.47,53.12)}, \plr{(4.47,0)}}

\dashed \arrow \rotatepath{(0,0),26.56} \polyline{(-5,0), (5,0)}

\dashed \arrow \rotatepath{(0,0),26.56} \polyline{(0,-5), (0,5)}

\arrow \reverse \arrow \rotatepath{(0,0),26.56} \parafcn{-2.25,2.25,0.1}{(t**2,t)}

\rotatepath{(0,0),26.56} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),26.56} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),26.56} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),26.56} \polyline{(4,-0.15),(4,0.15)}

\rotatepath{(0,0),116.56} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),116.56} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),116.56} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),116.56} \polyline{(4,-0.15),(4,0.15)}

\rotatepath{(0,0),206.56} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),206.56} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),206.56} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),206.56} \polyline{(4,-0.15),(4,0.15)}

\rotatepath{(0,0),296.56} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),296.56} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),296.56} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),296.56} \polyline{(4,-0.15),(4,0.15)}

\point[3pt]{(0,0)}

\arrow \parafcn{5,20,5}{4*dir(t)}

\tlabel(4,.75){\scriptsize $\theta = \arctan\left(\frac{1}{2}\right)$}

\tcaption{$x^2-4xy+4y^2-2x\sqrt{5}-y\sqrt{5}=0$}

\end{mfpic}

\item $8x^2+12xy+17y^2 - 20 = 0$ \\ becomes $(x')^2 + \frac{(y')^2}{4} = 1$ after rotating \\ counter-clockwise through $\theta = \arctan(2)$

\begin{mfpic}[18]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\tlabel[cc](2.75,4.25){\scriptsize $x'$}

\tlabel[cc](-4.25,2.75){\scriptsize $y'$}

\xmarks{-4,-3,-2,-1,1,2,3,4}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\point[3pt]{\plr{(2,153)}, \plr{(-2,153)},\plr{(1,63)}, \plr{(-1,63)} }

\dashed \arrow \rotatepath{(0,0),63} \polyline{(-5,0), (5,0)}

\dashed \arrow \rotatepath{(0,0),63} \polyline{(0,-5), (0,5)}

\rotatepath{(0,0),63} \ellipse{(0,0),1,2}

\rotatepath{(0,0),63} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),63} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),63} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),63} \polyline{(4,-0.15),(4,0.15)}

\rotatepath{(0,0),153} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),153} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),153} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),153} \polyline{(4,-0.15),(4,0.15)}

\rotatepath{(0,0),243} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),243} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),243} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),243} \polyline{(4,-0.15),(4,0.15)}

\rotatepath{(0,0),333} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),333} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),333} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),333} \polyline{(4,-0.15),(4,0.15)}

\point[3pt]{(0,0)}

\arrow \parafcn{5,58,5}{2.5*dir(t)}

\gclear \tlabelrect[cc](2.5,1){\scriptsize $\theta = \arctan(2)$}

\tcaption{$8x^2+12xy+17y^2 - 20 = 0$}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\newpage

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = \frac{2}{1-\cos(\theta)}$ is a parabola \\ directrix $x = -2$ , vertex $(-1,0)$ \\ focus $(0,0)$, focal diameter $4$ \\

\begin{mfpic}[18]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\xmarks{-4,-3,-2,-1,1,2,3,4}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\plotsymbol[3pt]{Asterisk}{(0, 0)}

\point[3pt]{(0,2), (0,-2), (-1,0)}

\arrow \reverse \arrow \polyline{(-2,-5), (-2,5)}

\arrow \reverse \arrow \plrfcn{60,300,5}{2/(1-cosd(t))}

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\scriptsize

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\axislabels {y}{{$-4$} -4,{$-3$} -3,{$-2$} -2,{$-1$} -1,{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\normalsize

\end{mfpic}

\vspace{.15in}

\item $r = \frac{3}{2 + \sin(\theta)} = \frac{\frac{3}{2}}{1 + \frac{1}{2} \sin(\theta)}$ is an ellipse \\ directrix $y = 3$ , vertices $(0,1)$, $(0,-3)$ \\ center $(0,-2)$ , foci $(0,0)$, $(0,-2)$ \\ minor axis length $2\sqrt{3}$ \\

\begin{mfpic}[18]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\xmarks{-4,-3,-2,-1,1,2,3,4}

\ymarks{-4,-3,-2,1,2,3,4}

\plotsymbol[3pt]{Asterisk}{(0, 0), (0,-2)}

\plotsymbol[3pt]{Cross}{(0,-1)}

\point[3pt]{(0,1), (0,-3), (1.73,-1), (-1.73,-1)}

\arrow \reverse \arrow \polyline{(-5,3), (5,3)}

\plrfcn{0,360,5}{3/(2+sind(t))}

\tlabelsep{5pt}

\scriptsize

\axislabels {x}{{$-4 \hspace{7pt}$} -4,{$-3 \hspace{7pt}$} -3, {$-2\hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\axislabels {y}{{$-4$} -4,{$-2$} -2,{$-1$} -1,{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\normalsize

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = \frac{3}{2 - \cos(\theta)} = \frac{\frac{3}{2}}{1 - \frac{1}{2} \cos(\theta)}$ is an ellipse \\ directrix $x = -3$ , vertices $(-1,0)$, $(3,0)$ \\ center $(1,0)$ , foci $(0,0)$, $(2,0)$ \\ minor axis length $2\sqrt{3}$ \\

\begin{mfpic}[18]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\xmarks{-4,-3,-2,-1,2,3,4}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\plotsymbol[3pt]{Asterisk}{(0, 0), (2,0)}

\plotsymbol[3pt]{Cross}{(1,0)}

\point[3pt]{(-1,0), (3,0), (1,1.73), (1,-1.73)}

\arrow \reverse \arrow \polyline{(-3,-5), (-3,5)}

\plrfcn{0,360,5}{3/(2-cosd(t))}

\tlabelsep{5pt}

\scriptsize

\axislabels {x}{{$-4 \hspace{7pt}$} -4,{$-3 \hspace{7pt}$} -3, {$-2\hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\axislabels {y}{{$-4$} -4,{$-2$} -2,{$-1$} -1,{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\normalsize

\end{mfpic}

\item $r = \frac{2}{1+\sin(\theta)}$ is a parabola \\ directrix $y=2$ , vertex $(0,1)$ \\ focus $(0,0)$, focal diameter $4$ \\

\begin{mfpic}[18]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\xmarks{-4,-3,-2,-1,1,2,3,4}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\plotsymbol[3pt]{Asterisk}{(0, 0)}

\point[3pt]{(-2,0), (2,0), (0,1)}

\arrow \reverse \arrow \polyline{(-5,2), (5,2)}

\arrow \reverse \arrow \plrfcn{-30,210,5}{2/(1+sind(t))}

\tlabelsep{5pt}

\scriptsize

\axislabels {x}{{$-4 \hspace{7pt}$} -4,{$-3 \hspace{7pt}$} -3, {$-2\hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\axislabels {y}{{$-4$} -4,{$-3$} -3,{$-2$} -2,{$-1$} -1,{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\normalsize

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = \frac{4}{1+3\cos(\theta)}$ is a hyperbola \\ directrix $x = \frac{4}{3}$, vertices $(1,0)$, $(2,0)$ \\ center $\left(\frac{3}{2}, 0\right)$, foci $(0,0)$, $(3,0)$ \\ conjugate axis length $2\sqrt{2}$ \\

\begin{mfpic}[18]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\xmarks{-4,-3,-2,-1,1,2,3,4}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\plotsymbol[3pt]{Asterisk}{(0, 0), (3,0)}

\plotsymbol[3pt]{Cross}{(1.5,0)}

\point[3pt]{(1,0), (2,0)}

\arrow \reverse \arrow \polyline{(1.33,-5), (1.33,5)}

\arrow \reverse \arrow \plrfcn{-88,88,5}{4/(1+3*cosd(t))}

\arrow \reverse \arrow \plrfcn{129,231.5,5}{4/(1+3*cosd(t))}

\dotted \function{0,3,0.1}{2.828*(x-1.5)}

\dotted \function{0,3,0.1}{0-2.828*(x-1.5)}

\tlabelsep{5pt}

\scriptsize

\axislabels {x}{{$-4 \hspace{7pt}$} -4,{$-3 \hspace{7pt}$} -3, {$-2\hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\axislabels {y}{{$-4$} -4,{$-3$} -3,{$-2$} -2,{$-1$} -1,{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\normalsize

\end{mfpic}

\item $r = \frac{2}{1-2\sin(\theta)}$ is a hyperbola \\ directrix $y = -1$, vertices $\left(0,-\frac{2}{3}\right)$, $(0,-2)$ \\ center $\left(0, -\frac{4}{3} \right)$, foci $(0,0)$, $\left(0, -\frac{8}{3}\right)$ \\ conjugate axis length $\frac{2\sqrt{3}}{3}$ \\

\begin{mfpic}[18]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\xmarks{-4,-3,-2,-1,1,2,3,4}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\plotsymbol[3pt]{Asterisk}{(0, 0), (0,-2.67)}

\plotsymbol[3pt]{Cross}{(0, -1.33)}

\point[3pt]{(0, -0.67), (0,-2)}

\arrow \reverse \arrow \polyline{(-5,-1), (5,-1)}

\arrow \reverse \arrow \plrfcn{45,135,5}{2/(1-2*sind(t))}

\arrow \reverse \arrow \plrfcn{168,372,5}{2/(1-2*sind(t))}

\dotted \function{-3,3,0.1}{0.577*x-1.33}

\dotted \function{-3,3,0.1}{-1.33-0.577x}

\tlabelsep{5pt}

\scriptsize

\axislabels {x}{{$-4 \hspace{7pt}$} -4,{$-3 \hspace{7pt}$} -3, {$-2\hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\axislabels {y}{{$-4$} -4,{$-3$} -3,{$-2$} -2,{$-1$} -1,{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\normalsize

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $r = \frac{2}{1 + \sin(\theta - \frac{\pi}{3})}$ is \\

the parabola $r = \frac{2}{1 + \sin(\theta)}$ \\

rotated through $\phi = \frac{\pi}{3}$ \\

\begin{mfpic}[18]{-4}{6}{-5}{5}

\axes

\tlabel[cc](6,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\tlabel[cc](3.5,5){\scriptsize $x'$}

\tlabel[cc](-4,3){\scriptsize $y'$}

\xmarks{-3,-2,-1,1,2,3,4,5}

\ymarks{-4,-3,-2,-1,1,2,3,4}

\point[3pt]{\plr{(-2,60)}, \plr{(2,60)}}

\plotsymbol[3pt]{Asterisk}{(0, 0)}

\dashed \arrow \rotatepath{(0,0),60} \polyline{(-4,0), (6,0)}

\dashed \arrow \rotatepath{(0,0),60} \polyline{(0,-5), (0,5)}

\arrow \reverse \arrow \rotatepath{(0,0),60} \polyline{(-4,2), (4,2)}

\arrow \reverse \arrow \plrfcn{25,275,5}{2/(1+sind(t-60))}

\rotatepath{(0,0),60} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),60} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),60} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),60} \polyline{(4,-0.15),(4,0.15)}

\rotatepath{(0,0),60} \polyline{(5,-0.15),(5,0.15)}

\rotatepath{(0,0),150} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),150} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),150} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),150} \polyline{(4,-0.15),(4,0.15)}

\rotatepath{(0,0),240} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),240} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),240} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),330} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),330} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),330} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),330} \polyline{(4,-0.15),(4,0.15)}

\point[3pt]{(0,0)}

\arrow \parafcn{5,55,5}{5.5*dir(t)}

\gclear \tlabelrect[cc](4.5,3){\scriptsize $\phi = \frac{\pi}{3}$}

\end{mfpic}

\item $r = \frac{6}{3 - \cos\left(\theta + \frac{\pi}{4}\right)}$ is the ellipse \\

$r = \frac{6}{3 - \cos\left(\theta \right)} = \frac{2}{1 - \frac{1}{3} \cos\left(\theta \right)}$ \\

rotated through $\phi = -\frac{\pi}{4}$ \\

\begin{mfpic}[18]{-5}{5}{-5}{5}

\axes

\tlabel[cc](5,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,5){\scriptsize $y$}

\tlabel[cc](4,-3.5){\scriptsize $x'$}

\tlabel[cc](4,3){\scriptsize $y'$}

\xmarks{-4,-3,-2,-1,1,2,3,4}

\ymarks{-5,-4,-3,-2,-1,1,2,3,4}

\point[3pt]{\plr{(-1.5,-45)}, \plr{(3,-45)}, (-0.97, -2.03), (2.03, 0.97) }

\plotsymbol[3pt]{Asterisk}{(0, 0), \plr{(1.5,-45)}}

\plrfcn{0,360,5}{6/(3-cosd(t+45))}

\tlabelsep{5pt}

\scriptsize

\axislabels {x}{{$-4 \hspace{7pt}$} -4, {$-3 \hspace{7pt}$} -3, {$-2\hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\axislabels {y}{{$-4$} -4,{$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\normalsize

\dashed \arrow \rotatepath{(0,0),-45} \polyline{(-5,0), (5,0)}

\dashed \arrow \rotatepath{(0,0),-45} \polyline{(0,-5), (0,5)}

\rotatepath{(0,0),-45} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),-45} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),-45} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),-45} \polyline{(4,-0.15),(4,0.15)}

\rotatepath{(0,0),-135} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),-135} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),-135} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),-135} \polyline{(4,-0.15),(4,0.15)}

\rotatepath{(0,0),-225} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),-225} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),-225} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),-225} \polyline{(4,-0.15),(4,0.15)}

\rotatepath{(0,0),-315} \polyline{(1,-0.15),(1,0.15)}

\rotatepath{(0,0),-315} \polyline{(2,-0.15),(2,0.15)}

\rotatepath{(0,0),-315} \polyline{(3,-0.15),(3,0.15)}

\rotatepath{(0,0),-315} \polyline{(4,-0.15),(4,0.15)}

\arrow \parafcn{-5,-40,-5}{4*dir(t)}

\gclear \tlabelrect[cc](4.5,-1.75){\scriptsize $\phi = -\frac{\pi}{4}$}

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\closegraphsfile

## 11.7: Polar Form of Complex Numbers

\subsection{Exercises}

In Exercises \ref{polarcompbasicfirst} - \ref{polarcompbasiclast}, find a polar representation for the complex number $z$ and then identify $\text{Re}(z)$, $\text{Im}(z)$, $|z|$, $\text{arg}(z)$ and $\text{Arg}(z)$.

\begin{multicols}{4}

\begin{enumerate}

\item $z = 9 + 9i$ \label{polarcompbasicfirst}

\item $z = 5 + 5i\sqrt{3}$

\item $z = 6i$

\item $z = -3\sqrt{2} + 3i\sqrt{2}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = -6\sqrt{3} + 6i$ \vphantom{$\dfrac{\sqrt{3}}{2}$}

\item $z = -2$ \vphantom{$\dfrac{\sqrt{3}}{2}$}

\item $z = -\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}i$

\item $z = -3-3i$ \vphantom{$\dfrac{\sqrt{3}}{2}$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = -5i$

\item $z = 2\sqrt{2} - 2i\sqrt{2}$

\item $z = 6$

\item $z = i\sqrt[3]{7}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 3 + 4i$

\item $z = \sqrt{2} + i$

\item $z = -7 + 24i$

\item $z = -2+6i$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = -12-5i$

\item $z = -5-2i$

\item $z = 4-2i$

\item $z = 1-3i$ \label{polarcompbasiclast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{rectcompfirst} - \ref{rectcomplast}, find the rectangular form of the given complex number. Use whatever identities are necessary to find the exact values.

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 6\text{cis}(0)$ \vphantom{$\left(\dfrac{\pi}{6}\right)$} \label{rectcompfirst}

\item $z = 2\text{cis}\left(\dfrac{\pi}{6}\right)$

\item $z = 7\sqrt{2}\text{cis}\left(\dfrac{\pi}{4}\right)$

\item $z = 3\text{cis}\left(\dfrac{\pi}{2}\right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 4\text{cis}\left(\dfrac{2\pi}{3}\right)$

\item $z = \sqrt{6}\text{cis}\left(\dfrac{3\pi}{4}\right)$

\item $z = 9\text{cis}\left(\pi\right)$ \vphantom{$\left(\dfrac{7\pi}{6}\right)$}

\item $z = 3\text{cis}\left(\dfrac{4\pi}{3}\right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 7\text{cis}\left(-\dfrac{3\pi}{4}\right)$

\item \small $z = \sqrt{13}\text{cis}\left(\dfrac{3\pi}{2}\right)$ \normalsize

\item $z = \dfrac{1}{2}\text{cis}\left(\dfrac{7\pi}{4}\right)$

\item $z = 12\text{cis}\left(-\dfrac{\pi}{3}\right)$ \vphantom{$\left(\dfrac{7\pi}{6}\right)$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 8\text{cis}\left(\dfrac{\pi}{12}\right)$ \vphantom{$\left(\dfrac{7\pi}{6}\right)$}

\item $z = 2\text{cis}\left(\dfrac{7\pi}{8}\right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 5\text{cis}\left(\arctan\left(\dfrac{4}{3}\right)\right)$

\item $z = \sqrt{10}\text{cis}\left(\arctan\left(\dfrac{1}{3}\right)\right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 15\text{cis}\left(\arctan\left(-2\right)\right)$

\item $z= \sqrt{3}\left(\arctan\left(-\sqrt{2}\right)\right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 50\text{cis}\left(\pi-\arctan\left(\dfrac{7}{24}\right)\right)$

\item $z = \dfrac{1}{2}\text{cis}\left(\pi+\arctan\left(\dfrac{5}{12}\right)\right)$ \label{rectcomplast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

For Exercises \ref{polarcomparithfirst} - \ref{polarcomparithlast}, use $z = -\dfrac{3\sqrt{3}}{2} + \dfrac{3}{2}i$ and $w = 3\sqrt{2} - 3i\sqrt{2}$ to compute the quantity. Express your answers in polar form using the principal argument.

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $zw$ \vphantom{$\dfrac{z}{w}$} \label{polarcomparithfirst}

\item $\dfrac{z}{w}$

\item $\dfrac{w}{z}$

\item $z^{4}$ \vphantom{$\dfrac{z}{w}$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

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\item $w^{3}$ \vphantom{$\dfrac{z^{2}}{w}$}

\item $z^{5}w^{2}$ \vphantom{$\dfrac{z^{2}}{w}$}

\item $z^{3}w^{2}$ \vphantom{$\dfrac{z^{2}}{w}$}

\item $\dfrac{z^{2}}{w}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{w}{z^2}$ \vphantom{$\dfrac{z^{2}}{w^{3}}$}

\item $\dfrac{z^3}{w^2}$

\item $\dfrac{w^2}{z^3}$

\item $\left(\dfrac{w}{z}\right)^6$ \vphantom{$\dfrac{z^{2}}{w^{3}}$} \label{polarcomparithlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{demoivrefirst} - \ref{demoivrelast}, use DeMoivre's Theorem to find the indicated power of the given complex number. Express your final answers in rectangular form.

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left(-2 + 2i\sqrt{3}\right)^3$ \label{demoivrefirst}

\item $(-\sqrt{3} - i)^3$

\item $(-3+3i)^{4}$

\item $(\sqrt{3} + i)^4$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left(\dfrac{5}{2} + \dfrac{5}{2} i\right)^3$ \vphantom{$\left(\dfrac{\sqrt{2}}{2}\right)$}

\item $\left(-\dfrac{1}{2} - \dfrac{\sqrt{3}}{2} i\right)^{6}$

\item $\left(\dfrac{3}{2} - \dfrac{3}{2} i\right)^3$ \vphantom{$\left(\dfrac{\sqrt{2}}{2}\right)$}

\item $\left(\dfrac{\sqrt{3}}{3} - \dfrac{1}{3} i\right)^4$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left(\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} i\right)^4$

\item $(2+2i)^5$ \vphantom{$\left(\dfrac{\sqrt{2}}{2}\right)$}

\item $(\sqrt{3} - i)^{5}$ \vphantom{$\left(\dfrac{\sqrt{2}}{2}\right)$}

\item $(1-i)^8$ \vphantom{$\left(\dfrac{\sqrt{2}}{2}\right)$} \label{demoivrelast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{polarrootsfirst} - \ref{polarrootslast}, find the indicated complex roots. Express your answers in polar form and then convert them into rectangular form.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item the two square roots of $z = 4i$ \label{polarrootsfirst}

\item the two square roots of $z = -25i$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item the two square roots of $z = 1 + i\sqrt{3}$

\item the two square roots of $\frac{5}{2} - \frac{5\sqrt{3}}{2}i$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item the three cube roots of $z=64$

\item the three cube roots of $z = -125$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item the three cube roots of $z = i$

\item the three cube roots of $z = -8i$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item the four fourth roots of $z=16$

\item the four fourth roots of $z=-81$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item the six sixth roots of $z = 64$

\item the six sixth roots of $z = -729$ \label{polarrootslast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Use the Sum and Difference Identities in Theorem \ref{circularsumdifference} or the Half Angle Identities in Theorem \ref{halfangle} to express the three cube roots of $z=\sqrt{2} + i\sqrt{2}$ in rectangular form. (See Example \ref{nthrootscomplexex}, number \ref{halfanglecuberoot}.)

\item Use a calculator to approximate the five fifth roots of $1$. (See Example \ref{nthrootscomplexex}, number \ref{calculatorfifthroot}.)

\item According to Theorem \ref{realfactorization} in Section \ref{ComplexZeros}, the polynomial $p(x) = x^{4} + 4$ can be factored into the product linear and irreducible quadratic factors. In Exercise \ref{factorpolywithnonlinear} in Section \ref{NonLinear}, we showed you how to factor this polynomial into the product of two irreducible quadratic factors using a system of non-linear equations. Now that we can compute the complex fourth roots of $-4$ directly, we can simply apply the Complex Factorization Theorem, Theorem \ref{complexfactorization}, to obtain the linear factorization $p(x) = (x - (1 + i))(x - (1 - i))(x - (-1 + i))(x - (-1 - i))$. By multiplying the first two factors together and then the second two factors together, thus pairing up the complex conjugate pairs of zeros Theorem \ref{conjugatepairsthm} told us we'd get, we have that $p(x) = (x^{2} - 2x + 2)(x^{2} + 2x + 2)$. Use the 12 complex $12^{\text{th}}$ roots of 4096 to factor $p(x) = x^{12} - 4096$ into a product of linear and irreducible quadratic factors.

\item Complete the proof of Theorem \ref{modprops} by showing that if $w \neq 0$ than $\left| \frac{1}{w}\right| = \frac{1}{|w|}$.

\item Recall from Section \ref{ComplexZeros} that given a complex number $z = a+bi$ its complex conjugate, denoted $\overline{z}$, is given by $\overline{z} = a - bi$.

\begin{enumerate}

\item Prove that $\left| \overline{z} \right| = |z|$.

\item Prove that $|z| = \sqrt{z \overline{z}}$

\item Show that $\text{Re}(z) = \dfrac{z + \overline{z}}{2}$ and $\text{Im}(z) = \dfrac{z - \overline{z}}{2i}$

\item Show that if $\theta \in \text{arg}(z)$ then $-\theta \in \text{arg}\left(\overline{z}\right)$. Interpret this result geometrically.

\item Is it always true that $\text{Arg}\left(\overline{z}\right) = -\text{Arg}(z)$?

\end{enumerate}

\item Given any natural number $n \geq 2$, the $n$ complex $n^{\text{th}}$ roots of the number $z = 1$ are called the \textbf{\boldmath $n^{\mbox{\textbf{\scriptsize th}}}$ Roots of Unity}. \index{$n^{\textrm{th}}$ Roots of Unity} \index{complex number ! $n^{\textrm{th}}$ Roots of Unity} \index{Roots of Unity} In the following exercises, assume that $n$ is a fixed, but arbitrary, natural number such that $n \geq 2$.

\begin{enumerate}

\item Show that $w = 1$ is an $n^{\text{th}}$ root of unity.

\item Show that if both $w_{\text{\tiny$j$}}$ and $w_{\text{\tiny$k$}}$ are $n^{\text{th}}$ roots of unity then so is their product $w_{\text{\tiny$j$}}w_{\text{\tiny$k$}}$.

\item Show that if $w_{\text{\tiny$j$}}$ is an $n^{\text{th}}$ root of unity then there exists another $n^{\text{th}}$ root of unity $w_{\text{\tiny$j'$}}$ such that $w_{\text{\tiny$j$}}w_{\text{\tiny$j'$}} = 1$. Hint: If $w_{\text{\tiny$j$}} = \text{cis}(\theta)$ let $w_{\text{\tiny$j'$}} = \text{cis}(2\pi - \theta)$. You'll need to verify that $w_{\text{\tiny$j'$}} = \text{cis}(2\pi - \theta)$ is indeed an $n^{\text{th}}$ root of unity.

\end{enumerate}

\item \label{eulerformulaexercise} Another way to express the polar form of a complex number is to use the exponential function. For real numbers $t$, \href{http://en.wikipedia.org/wiki/Leonhar...line{Euler}}'s Formula defines $e^{it} = \cos(t) + i \sin(t)$.

\begin{enumerate}

\item Use Theorem \ref{prodquotpolarcomplex} to show that $e^{ix} e^{iy} = e^{i(x+y)}$ for all real numbers $x$ and $y$.

\item Use Theorem \ref{prodquotpolarcomplex} to show that $\left(e^{ix}\right)^{n} = e^{i(nx)}$ for any real number $x$ and any natural number $n$.

\item Use Theorem \ref{prodquotpolarcomplex} to show that $\dfrac{e^{ix}}{e^{iy}} = e^{i(x-y)}$ for all real numbers $x$ and $y$.

\item If $z = r\text{cis}(\theta)$ is the polar form of $z$, show that $z = re^{it}$ where $\theta = t$ radians.

\item Show that $e^{i\pi} + 1 = 0$. (This famous equation relates the five most important constants in all of Mathematics with the three most fundamental operations in Mathematics.)

\item \label{expformcosandsin} Show that $\cos(t) = \dfrac{e^{it} + e^{-it}}{2}$ and that $\sin(t) = \dfrac{e^{it} - e^{-it}}{2i}$ for all real numbers $t$.

\end{enumerate}

\end{enumerate}

\newpage

\begin{enumerate}

\item $z = 9 + 9i = 9\sqrt{2}\text{cis}\left(\frac{\pi}{4}\right)$, \, $\text{Re}(z) = 9$, \, $\text{Im}(z) = 9$, \, $|z| = 9\sqrt{2}$

$\text{arg}(z) = \left\{\frac{\pi}{4} + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) = \frac{\pi}{4}$.

\item $z = 5+5i\sqrt{3} = 10\text{cis}\left(\frac{\pi}{3}\right)$, \, $\text{Re}(z) = 5$, \, $\text{Im}(z) = 5\sqrt{3}$, \, $|z| = 10$

$\text{arg}(z) = \left\{\frac{\pi}{3} + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) = \frac{\pi}{3}$.

\item $z = 6i = 6\text{cis}\left(\frac{\pi}{2}\right)$, \, $\text{Re}(z) = 0$, \, $\text{Im}(z) = 6$, \, $|z| = 6$

$\text{arg}(z) = \left\{\frac{\pi}{2} + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) = \frac{\pi}{2}$.

\item $z = -3\sqrt{2} + 3i\sqrt{2} = 6\text{cis}\left(\frac{3\pi}{4}\right)$, \, $\text{Re}(z) = -3\sqrt{2}$, \, $\text{Im}(z) =3\sqrt{2}$, \, $|z| = 6$

$\text{arg}(z) = \left\{\frac{3\pi}{4} + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) = \frac{3\pi}{4}$.

\item $z = -6\sqrt{3} + 6i = 12\text{cis}\left(\frac{5\pi}{6}\right)$, \, $\text{Re}(z) = -6\sqrt{3}$, \, $\text{Im}(z) =6$, \, $|z| = 12$

$\text{arg}(z) = \left\{\frac{5\pi}{6} + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) = \frac{5\pi}{6}$.

\item $z = -2 = 2\text{cis}\left(\pi\right)$, \, $\text{Re}(z) = -2$, \, $\text{Im}(z) =0$, \, $|z| = 2$

$\text{arg}(z) = \left\{(2k+1)\pi \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) = \pi$.

\item $z = -\frac{\sqrt{3}}{2} - \frac{1}{2}i = \text{cis}\left(\frac{7\pi}{6}\right)$, \, $\text{Re}(z) = -\frac{\sqrt{3}}{2}$, \, $\text{Im}(z) = -\frac{1}{2}$, \, $|z| = 1$

$\text{arg}(z) = \left\{\frac{7\pi}{6} + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) = -\frac{5\pi}{6}$.

\item $z = -3-3i = 3\sqrt{2}\text{cis}\left(\frac{5\pi}{4}\right)$, \, $\text{Re}(z) = -3$, \, $\text{Im}(z) =-3$, \, $|z| = 3\sqrt{2}$

$\text{arg}(z) = \left\{\frac{5\pi}{4} + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) = -\frac{3\pi}{4}$.

\item $z = -5i = 5\text{cis}\left(\frac{3\pi}{2}\right)$, \, $\text{Re}(z) = 0$, \, $\text{Im}(z) = -5$, \, $|z| = 5$

$\text{arg}(z) = \left\{\frac{3\pi}{2} + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) = -\frac{\pi}{2}$.

\item $z = 2\sqrt{2} - 2i\sqrt{2} = 4\text{cis}\left(\frac{7\pi}{4}\right)$, \, $\text{Re}(z) = 2\sqrt{2}$, \, $\text{Im}(z) = -2\sqrt{2}$, \, $|z| = 4$

$\text{arg}(z) = \left\{\frac{7\pi}{4} + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) = -\frac{\pi}{4}$.

\item $z =6 = 6\text{cis}\left(0\right)$, \, $\text{Re}(z) = 6$, \, $\text{Im}(z) = 0$, \, $|z| = 6$

$\text{arg}(z) = \left\{2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) =0$.

\item $z = i \sqrt[3]{7} = \sqrt[3]{7}\text{cis}\left(\frac{\pi}{2}\right)$, \, $\text{Re}(z) =0$, \, $\text{Im}(z) = \sqrt[3]{7}$, \, $|z| = \sqrt[3]{7}$

$\text{arg}(z) = \left\{\frac{\pi}{2} + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) = \frac{\pi}{2}$.

\item $z = 3+4i = 5\text{cis}\left(\arctan\left(\frac{4}{3}\right)\right)$, \, $\text{Re}(z) = 3$, \, $\text{Im}(z) = 4$, \, $|z| = 5$

$\text{arg}(z) = \left\{\arctan\left(\frac{4}{3}\right) + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) =\arctan\left(\frac{4}{3}\right)$.

\item $z = \sqrt{2}+i = \sqrt{3}\text{cis}\left(\arctan\left(\frac{\sqrt{2}}{2}\right)\right)$, \, $\text{Re}(z) = \sqrt{2}$, \, $\text{Im}(z) = 1$, \, $|z| = \sqrt{3}$

$\text{arg}(z) = \left\{\arctan\left(\frac{\sqrt{2}}{2}\right) + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) =\arctan\left(\frac{\sqrt{2}}{2}\right)$.

\item $z = -7 + 24i = 25\text{cis}\left(\pi - \arctan\left(\frac{24}{7}\right)\right)$, \, $\text{Re}(z) = -7$, \, $\text{Im}(z) = 24$, \, $|z| = 25$

$\text{arg}(z) = \left\{\pi - \arctan\left(\frac{24}{7}\right) + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) =\pi - \arctan\left(\frac{24}{7}\right)$.

\item $z = -2 + 6i = 2\sqrt{10}\text{cis}\left(\pi - \arctan\left(3\right)\right)$, \, $\text{Re}(z) = -2$, \, $\text{Im}(z) = 6$, \, $|z| =2\sqrt{10}$

$\text{arg}(z) = \left\{\pi - \arctan\left(3\right) + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) =\pi - \arctan\left(3\right)$.

\item $z = -12 -5i = 13\text{cis}\left(\pi + \arctan\left(\frac{5}{12}\right)\right)$, \, $\text{Re}(z) = -12$, \, $\text{Im}(z) = -5$, \, $|z| = 13$

$\text{arg}(z) = \left\{\pi +\arctan\left(\frac{5}{12}\right) + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) = \arctan\left(\frac{5}{12}\right) -\pi$.

\item $z = -5-2i = \sqrt{29}\text{cis}\left(\pi + \arctan\left(\frac{2}{5}\right)\right)$, \, $\text{Re}(z) = -5$, \, $\text{Im}(z) = -2$, \, $|z| = \sqrt{29}$

$\text{arg}(z) = \left\{\pi +\arctan\left(\frac{2}{5}\right) + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) = \arctan\left(\frac{2}{5}\right) -\pi$.

\item $z =4-2i = 2\sqrt{5}\text{cis}\left(\arctan\left(-\frac{1}{2}\right)\right)$, \, $\text{Re}(z) =4$, \, $\text{Im}(z) = -2$, \, $|z| = 2\sqrt{5}$

$\text{arg}(z) = \left\{\arctan\left(-\frac{1}{2}\right) + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) = \arctan\left(-\frac{1}{2}\right) = -\arctan\left(\frac{1}{2}\right)$.

\item $z =1-3i = \sqrt{10}\text{cis}\left(\arctan\left(-3\right)\right)$, \, $\text{Re}(z) =1$, \, $\text{Im}(z) = -3$, \, $|z| =\sqrt{10}$

$\text{arg}(z) = \left\{\arctan\left(-3\right) + 2\pi k \, | \, \text{$k$is an integer} \right\}$ and $\text{Arg}(z) = \arctan\left(-3\right) = -\arctan(3)$.

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 6\text{cis}(0) = 6$

\item $z = 2\text{cis}\left(\frac{\pi}{6}\right) = \sqrt{3} + i$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 7\sqrt{2}\text{cis}\left(\frac{\pi}{4}\right) = 7+7i$

\item $z = 3\text{cis}\left(\frac{\pi}{2}\right) = 3i$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 4\text{cis}\left(\frac{2\pi}{3}\right) = -2+2i\sqrt{3}$

\item $z = \sqrt{6}\text{cis}\left(\frac{3\pi}{4}\right) = -\sqrt{3}+i\sqrt{3}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 9\text{cis}\left(\pi\right) = -9$

\item $z = 3\text{cis}\left(\frac{4\pi}{3}\right) = -\frac{3}{2} - \frac{3i\sqrt{3}}{2}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 7\text{cis}\left(-\frac{3\pi}{4}\right) = -\frac{7\sqrt{2}}{2} - \frac{7\sqrt{2}}{2}i$

\item $z = \sqrt{13}\text{cis}\left(\frac{3\pi}{2}\right) = -i\sqrt{13}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = \frac{1}{2}\text{cis}\left(\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{4} - i\frac{\sqrt{2}}{4}$

\item $z = 12\text{cis}\left(-\frac{\pi}{3}\right) = 6 - 6i\sqrt{3}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 8\text{cis}\left(\frac{\pi}{12}\right) = 4\sqrt{2+\sqrt{3}}+4i\sqrt{2-\sqrt{3}}$

\item $z = 2\text{cis}\left(\frac{7\pi}{8}\right) = -\sqrt{2 + \sqrt{2}} + i\sqrt{2 - \sqrt{2}}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 5\text{cis}\left(\arctan\left(\frac{4}{3}\right)\right) = 3 + 4i$

\item $z = \sqrt{10}\text{cis}\left(\arctan\left(\frac{1}{3}\right)\right) = 3+i$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 15\text{cis}\left(\arctan\left(-2\right)\right) = 3\sqrt{5} -6i\sqrt{5}$

\item $z= \sqrt{3}\text{cis}\left(\arctan\left(-\sqrt{2}\right)\right) = 1-i\sqrt{2}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z = 50\text{cis}\left(\pi-\arctan\left(\frac{7}{24}\right)\right) = -48 + 14i$

\item $z = \frac{1}{2}\text{cis}\left(\pi+\arctan\left(\frac{5}{12}\right)\right) = -\frac{6}{13} - \frac{5i}{26}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\pagebreak

In Exercises \ref{polarcomparithfirst} - \ref{polarcomparithlast}, we have that $z = -\frac{3\sqrt{3}}{2} + \frac{3}{2}i = 3\text{cis}\left(\frac{5\pi}{6}\right)$ and $w = 3\sqrt{2} - 3i\sqrt{2} = 6\text{cis}\left(-\frac{\pi}{4}\right)$ so we get the following.

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $zw = 18\text{cis}\left(\frac{7\pi}{12}\right)$

\item $\frac{z}{w} = \frac{1}{2}\text{cis}\left(-\frac{11\pi}{12}\right)$

\item $\frac{w}{z} = 2\text{cis}\left(\frac{11\pi}{12}\right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z^{4} = 81\text{cis}\left(-\frac{2\pi}{3}\right)$

\item $w^{3} = 216\text{cis}\left(-\frac{3\pi}{4}\right)$

\item $z^{5}w^{2} = 8748\text{cis}\left(-\frac{\pi}{3}\right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $z^3w^2 = 972 \text{cis}(0)$

\item $\frac{z^2}{w} =\frac{3}{2}\text{cis}\left(-\frac{\pi}{12}\right)$

\item $\frac{w}{z^2} =\frac{2}{3}\text{cis}\left(\frac{\pi}{12}\right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\frac{z^3}{w^2} =\frac{3}{4}\text{cis}(\pi)$

\item $\frac{w^2}{z^3} =\frac{4}{3}\text{cis}(\pi)$

\item $\left(\frac{w}{z}\right)^6 =64\text{cis}\left(-\frac{\pi}{2} \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left(-2 + 2i\sqrt{3}\right)^3 = 64$

\item $(-\sqrt{3} - i)^3 =-8i$

\item $(-3+3i)^{4}=-324$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(\sqrt{3} + i)^4 =-8 + 8i\sqrt{3}$ \vphantom{$\left(\frac{\sqrt{2}}{2}\right)^{2}$}

\item $\left(\frac{5}{2} + \frac{5}{2} i\right)^3=-\frac{125}{4}+\frac{125}{4} i$ \vphantom{$\left(\frac{\sqrt{2}}{2}\right)^{2}$}

\item $\left(-\frac{1}{2} - \frac{i \sqrt{3}}{2}\right)^{6}=1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left(\frac{3}{2} - \frac{3}{2} i\right)^3=-\frac{27}{4}-\frac{27}{4} i$ \vphantom{$\left(\frac{\sqrt{2}}{2}\right)^{2}$}

\item $\left(\frac{\sqrt{3}}{3} - \frac{1}{3} i\right)^4 =-\frac{8}{81} - \frac{8i\sqrt{3}}{81}$

\item $\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i\right)^4=-1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(2+2i)^5 -128-128i$

\item $(\sqrt{3} - i)^{5} = -16\sqrt{3} - 16i$

\item $(1-i)^8=16$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Since $z=4i = 4\text{cis}\left(\frac{\pi}{2}\right)$ we have

\begin{multicols}{2}

$w_{\text{\tiny$0$}} = 2\text{cis}\left(\frac{\pi}{4}\right) = \sqrt{2} +i\sqrt{2}$

$w_{\text{\tiny$1$}} = 2\text{cis}\left(\frac{5\pi}{4}\right) = -\sqrt{2} - i\sqrt{2}$

\end{multicols}

\item Since $z=-25i = 25\text{cis}\left(\frac{3\pi}{2}\right)$ we have

\begin{multicols}{2}

$w_{\text{\tiny$0$}} = 5\text{cis}\left(\frac{3\pi}{4}\right) = -\frac{5\sqrt{2}}{2} +\frac{5\sqrt{2}}{2} i$

$w_{\text{\tiny$1$}} = 5\text{cis}\left(\frac{7\pi}{4}\right) = \frac{5\sqrt{2}}{2} - \frac{5\sqrt{2}}{2} i$

\end{multicols}

\item Since $z=1 + i\sqrt{3} = 2\text{cis}\left(\frac{\pi}{3}\right)$ we have

\begin{multicols}{2}

$w_{\text{\tiny$0$}} = \sqrt{2}\text{cis}\left(\frac{\pi}{6}\right) = \frac{\sqrt{6}}{2} +\frac{\sqrt{2}}{2} i$

$w_{\text{\tiny$1$}} = \sqrt{2}\text{cis}\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2} i$

\end{multicols}

\item Since $z=\frac{5}{2} - \frac{5\sqrt{3}}{2}i = 5\text{cis}\left(\frac{5\pi}{3}\right)$ we have

\begin{multicols}{2}

$w_{\text{\tiny$0$}} =\sqrt{5}\text{cis}\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{15}}{2} + \frac{\sqrt{5}}{2}i$

$w_{\text{\tiny$1$}} = \sqrt{5}\text{cis}\left(\frac{11\pi}{6}\right) = \frac{\sqrt{15}}{2} - \frac{\sqrt{5}}{2}i$

\end{multicols}

\item Since $z = 64 = 64\text{cis}\left(0\right)$ we have

\begin{multicols}{3}

$w_{\text{\tiny$0$}} = 4\text{cis}\left(0\right) = 4$

$w_{\text{\tiny$1$}} =4\text{cis}\left(\frac{2\pi}{3}\right) = -2 + 2i\sqrt{3}$

$w_{\text{\tiny$2$}} = 4\text{cis}\left(\frac{4\pi}{3}\right) = -2 - 2i\sqrt{3}$

\end{multicols}

\pagebreak

\item Since $z = -125 = 125\text{cis}\left(\pi\right)$ we have

\begin{multicols}{3}

$w_{\text{\tiny$0$}} = 5\text{cis}\left(\frac{\pi}{3}\right) = \frac{5}{2} + \frac{5\sqrt{3}}{2} i$

$w_{\text{\tiny$1$}} =5\text{cis}\left(\pi\right) = -5$

$w_{\text{\tiny$2$}} = 5\text{cis}\left(\frac{5\pi}{3}\right) = \frac{5}{2} - \frac{5\sqrt{3}}{2} i$

\end{multicols}

\item Since $z = i = \text{cis}\left(\frac{\pi}{2}\right)$ we have

\begin{multicols}{3}

$w_{\text{\tiny$0$}} = \text{cis}\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} + \frac{1}{2}i$

$w_{\text{\tiny$1$}} = \text{cis}\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} + \frac{1}{2}i$

$w_{\text{\tiny$2$}} = \text{cis}\left(\frac{3\pi}{2}\right) = -i$

\end{multicols}

\item Since $z = -8i = 8\text{cis}\left(\frac{3\pi}{2}\right)$ we have

\begin{multicols}{3}

$w_{\text{\tiny$0$}} = 2\text{cis}\left(\frac{\pi}{2}\right) = 2i$

$w_{\text{\tiny$1$}} = 2\text{cis}\left(\frac{7\pi}{6}\right) = -\sqrt{3} -i$

$w_{\text{\tiny$2$}} = \text{cis}\left(\frac{11\pi}{6}\right) = \sqrt{3}-i$

\end{multicols}

\item Since $z=16 = 16\text{cis}\left(0 \right)$ we have

\begin{multicols}{2}

$w_{\text{\tiny$0$}} =2\text{cis}\left(0\right) =2$

$w_{\text{\tiny$1$}} = 2\text{cis}\left(\frac{\pi}{2}\right) = 2i$

\end{multicols}

\begin{multicols}{2}

$w_{\text{\tiny$2$}} = 2\text{cis}\left(\pi\right) = -2$

$w_{\text{\tiny$3$}} = 2\text{cis}\left(\frac{3\pi}{2}\right) = -2i$

\end{multicols}

\item Since $z=-81 = 81\text{cis}\left(\pi \right)$ we have

\begin{multicols}{2}

$w_{\text{\tiny$0$}} =3\text{cis}\left(\frac{\pi}{4}\right) = \frac{3\sqrt{2}}{2} + \frac{3\sqrt{2}}{2}i$

$w_{\text{\tiny$1$}} = 3\text{cis}\left(\frac{3\pi}{4}\right) =-\frac{3\sqrt{2}}{2} + \frac{3\sqrt{2}}{2}i$

\end{multicols}

\begin{multicols}{2}

$w_{\text{\tiny$2$}} = 3\text{cis}\left(\frac{5\pi}{4}\right) =-\frac{3\sqrt{2}}{2} - \frac{3\sqrt{2}}{2}i$

$w_{\text{\tiny$3$}} = 3\text{cis}\left(\frac{7\pi}{4}\right) =\frac{3\sqrt{2}}{2} - \frac{3\sqrt{2}}{2}i$

\end{multicols}

\item Since $z = 64 = 64\text{cis}(0)$ we have

\begin{multicols}{3}

$w_{\text{\tiny$0$}} = 2\text{cis}(0) = 2$

$w_{\text{\tiny$1$}} = 2\text{cis}\left(\frac{\pi}{3}\right) = 1 + \sqrt{3}i$

$w_{\text{\tiny$2$}} = 2\text{cis}\left(\frac{2\pi}{3}\right) = -1 + \sqrt{3}i$

\end{multicols}

\begin{multicols}{3}

$w_{\text{\tiny$3$}} = 2\text{cis}\left(\pi\right) = -2$

$w_{\text{\tiny$4$}} = 2\text{cis}\left(-\frac{2\pi}{3}\right) = -1 - \sqrt{3}i$

$w_{\text{\tiny$5$}} = 2\text{cis}\left(-\frac{\pi}{3}\right) = 1 - \sqrt{3}i$

\end{multicols}

\item Since $z = -729 = 729 \text{cis}(\pi)$ we have

\begin{multicols}{3}

$w_{\text{\tiny$0$}} = 3\text{cis}\left(\frac{\pi}{6}\right) = \frac{3\sqrt{3}}{2} + \frac{3}{2}i$

$w_{\text{\tiny$1$}} = 3\text{cis}\left(\frac{\pi}{2}\right) = 3i$

$w_{\text{\tiny$2$}} = 3\text{cis}\left(\frac{5\pi}{6}\right) = -\frac{3\sqrt{3}}{2} + \frac{3}{2}i$

\end{multicols}

\begin{multicols}{3}

$w_{\text{\tiny$3$}} = 3\text{cis}\left(\frac{7\pi}{6}\right) = -\frac{3\sqrt{3}}{2}-\frac{3}{2}i$

$w_{\text{\tiny$4$}} = 3\text{cis}\left(-\frac{3\pi}{2}\right) = -3i$

$w_{\text{\tiny$5$}} = 3\text{cis}\left(-\frac{11\pi}{6}\right) = \frac{3\sqrt{3}}{2} - \frac{3}{2}i$

\end{multicols}

\item Note: In the answers for $w_{\text{\tiny$0$}}$ and $w_{\text{\tiny$2$}}$ the first rectangular form comes from applying the appropriate Sum or Difference Identity ($\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$ and $\frac{17\pi}{12} = \frac{2\pi}{3} + \frac{3\pi}{4}$, respectively) and the second comes from using the Half-Angle Identities.

$w_{\text{\tiny$0$}} = \sqrt[3]{2} \text{cis}\left(\frac{\pi}{12}\right) = \sqrt[3]{2}\left( \frac{\sqrt{6} + \sqrt{2}}{4} + i\left( \frac{\sqrt{6} - \sqrt{2}}{4} \right) \right) = \sqrt[3]{2}\left( \frac{\sqrt{2 + \sqrt{3}}}{2} + i\frac{\sqrt{2 - \sqrt{3}}}{2} \right)$

$w_{\text{\tiny$1$}} = \sqrt[3]{2} \text{cis}\left(\frac{3\pi}{4}\right) = \sqrt[3]{2} \left( -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \right)$

$w_{\text{\tiny$2$}} = \sqrt[3]{2} \text{cis}\left(\frac{17\pi}{12}\right) = \sqrt[3]{2}\left( \frac{\sqrt{2} - \sqrt{6}}{4} + i\left( \frac{-\sqrt{2} - \sqrt{6}}{4} \right) \right) = \sqrt[3]{2}\left( \frac{\sqrt{2 - \sqrt{3}}}{2} + i\frac{\sqrt{2 + \sqrt{3}}}{2} \right)$

\item $w_{\text{\tiny$0$}} = \text{cis}(0) = 1$

$w_{\text{\tiny$1$}} = \text{cis}\left(\frac{2\pi}{5}\right) \approx 0.309 + 0.951i$

$w_{\text{\tiny$2$}} = \text{cis}\left(\frac{4\pi}{5}\right) \approx -0.809 + 0.588i$

$w_{\text{\tiny$3$}} = \text{cis}\left(\frac{6\pi}{5}\right) \approx -0.809 - 0.588i$

$w_{\text{\tiny$4$}} = \text{cis}\left(\frac{8\pi}{5}\right) \approx 0.309 - 0.951i$

\item $p(x) = x^{12} - 4096 = (x - 2)(x + 2)(x^{2} + 4)(x^{2} - 2x + 4)(x^{2} + 2x + 4)(x^{2} - 2\sqrt{3}x + 4)(x^{2} + 2\sqrt{3} + 4)$

\end{enumerate}

\closegraphsfile

## 11.8: Vectors

\subsection{Exercises}

In Exercises \ref{vectorbasicfirst} - \ref{vectorbasiclast}, use the given pair of vectors $\vec{v}$ and $\vec{w}$ to find the following quantities. State whether the result is a vector or a scalar.

\medskip

\hspace{.15in} $\bullet \, \vec{v} + \vec{w} \;\;\;$ \hfill $\bullet \, \vec{w} - 2\vec{v} \;\;\;$ \hfill $\bullet \, \| \vec{v} + \vec{w} \| \;\;\;$ \hfill $\bullet \, \| \vec{v} \| + \| \vec{w} \|$ \hfill $\bullet \, \| \vec{v} \| \vec{w} - \| \vec{w} \| \vec{v}$ \hfill $\bullet \, \|w\| \hat{v}$

\medskip

Finally, verify that the vectors satisfy the \href{http://en.wikipedia.org/wiki/Paralle...{Parallelogram Law}}}

$\|\vec{v}\|^2 + \|\vec{w}\|^2 = \dfrac{1}{2}\left[ \| \vec{v} + \vec{w}\|^2 + \|\vec{v} - \vec{w}\|^2\right]$

\begin{multicols}{2}

\begin{enumerate}

\item $\vec{v} = \left<12, -5\right>$, $\vec{w} = \left<3, 4\right>$ \label{vectorbasicfirst}

\item $\vec{v} = \left<-7, 24 \right>$, $\vec{w} = \left<-5, -12\right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left<2, -1 \right>$, $\vec{w} = \left<-2, 4 \right>$

\item $\vec{v} = \left<10, 4 \right>$, $\vec{w} = \left<-2, 5 \right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left<-\sqrt{3}, 1\right>$, $\vec{w} = \left<2\sqrt{3}, 2\right>$

\item $\vec{v} = \left<\frac{3}{5}, \frac{4}{5}\right>$, $\vec{w} = \left<-\frac{4}{5}, \frac{3}{5}\right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left<\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right>$, $\vec{w} = \left<-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right>$

\item $\vec{v} = \left<\frac{1}{2}, \frac{\sqrt{3}}{2} \right>$, $\vec{w} = \left< -1, -\sqrt{3} \right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = 3\hat{\imath} + 4\hat{\jmath}$, $\vec{w} = -2\hat{\jmath}$

\item $\vec{v} =\frac{1}{2} \left(\hat{\imath} + \hat{\jmath}\right)$, $\vec{w} = \frac{1}{2} \left(\hat{\imath} - \hat{\jmath}\right)$ \label{vectorbasiclast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{vectorcompfirst} - \ref{vectorcomplast}, find the component form of the vector $\vec{v}$ using the information given about its magnitude and direction. Give exact values.

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\|\vec{v}\| = 6$; when drawn in standard position $\vec{v}$ lies in Quadrant I and makes a $60^{\circ}$ angle with the positive $x$-axis \label{vectorcompfirst}

\item $\|\vec{v}\| = 3$; when drawn in standard position $\vec{v}$ lies in Quadrant I and makes a $45^{\circ}$ angle with the positive $x$-axis

\item $\|\vec{v}\| = \frac{2}{3}$; when drawn in standard position $\vec{v}$ lies in Quadrant I and makes a $60^{\circ}$ angle with the positive $y$-axis

\item $\|\vec{v}\| = 12$; when drawn in standard position $\vec{v}$ lies along the positive $y$-axis

\item $\|\vec{v}\| = 4$; when drawn in standard position $\vec{v}$ lies in Quadrant II and makes a $30^{\circ}$ angle with the negative $x$-axis

\item $\|\vec{v}\| = 2\sqrt{3}$; when drawn in standard position $\vec{v}$ lies in Quadrant II and makes a $30^{\circ}$ angle with the positive $y$-axis

\item $\|\vec{v}\| = \frac{7}{2}$; when drawn in standard position $\vec{v}$ lies along the negative $x$-axis

\item $\|\vec{v}\| = 5\sqrt{6}$; when drawn in standard position $\vec{v}$ lies in Quadrant III and makes a $45^{\circ}$ angle with the negative $x$-axis

\item $\|\vec{v}\| = 6.25$; when drawn in standard position $\vec{v}$ lies along the negative $y$-axis

\item $\|\vec{v}\| = 4\sqrt{3}$; when drawn in standard position $\vec{v}$ lies in Quadrant IV and makes a $30^{\circ}$ angle with the positive $x$-axis

\item $\|\vec{v}\| = 5\sqrt{2}$; when drawn in standard position $\vec{v}$ lies in Quadrant IV and makes a $45^{\circ}$ angle with the negative $y$-axis

\item $\| \vec{v}\| = 2\sqrt{5}$; when drawn in standard position $\vec{v}$ lies in Quadrant I and makes an angle measuring $\arctan(2)$ with the positive $x$-axis

\item $\| \vec{v}\| = \sqrt{10}$; when drawn in standard position $\vec{v}$ lies in Quadrant II and makes an angle measuring $\arctan(3)$ with the negative $x$-axis

\item $\| \vec{v}\| = 5$; when drawn in standard position $\vec{v}$ lies in Quadrant III and makes an angle measuring $\arctan\left(\frac{4}{3}\right)$ with the negative $x$-axis

\item $\| \vec{v}\| = 26$; when drawn in standard position $\vec{v}$ lies in Quadrant IV and makes an angle measuring $\arctan\left(\frac{5}{12}\right)$ with the positive $x$-axis \label{vectorcomplast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

In Exercises \ref{vectorcompcalcfirst} - \ref{vectorcompcalclast}, approximate the component form of the vector $\vec{v}$ using the information given about its magnitude and direction. Round your approximations to two decimal places.

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\|\vec{v}\| = 392$; when drawn in standard position $\vec{v}$ makes a $117^{\circ}$ angle with the positive $x$-axis \label{vectorcompcalcfirst}

\item $\|\vec{v}\| = 63.92$; when drawn in standard position $\vec{v}$ makes a $78.3^{\circ}$ angle with the positive $x$-axis

\item $\|\vec{v}\| = 5280$; when drawn in standard position $\vec{v}$ makes a $12^{\circ}$ angle with the positive $x$-axis

\item $\|\vec{v}\| = 450$; when drawn in standard position $\vec{v}$ makes a $210.75^{\circ}$ angle with the positive $x$-axis

\item $\|\vec{v}\| = 168.7$; when drawn in standard position $\vec{v}$ makes a $252^{\circ}$ angle with the positive $x$-axis

\item $\| \vec{v}\| = 26$; when drawn in standard position $\vec{v}$ makes a $304.5^{\circ}$ angle with the positive $x$-axis \label{vectorcompcalclast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

In Exercises \ref{findmaganglefirst} - \ref{findmaganglelast}, for the given vector $\vec{v}$, find the magnitude $\|\vec{v}\|$ and an angle $\theta$ with $0 \leq \theta < 360^{\circ}$ so that $\vec{v} = \|\vec{v}\| \left<\cos(\theta), \sin(\theta) \right>$ (See Definition \ref{polarformvector}.) Round approximations to two decimal places.

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left<1,\sqrt{3}\right>$ \label{findmaganglefirst}

\item $\vec{v} = \left<5,5\right>$

\item $\vec{v} = \left<-2\sqrt{3}, 2 \right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left<-\sqrt{2}, \sqrt{2} \right>$

\item $\vec{v} = \left<-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right>$

\item $\vec{v} = \left<-\frac{1}{2}, -\frac{\sqrt{3}}{2} \right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left<6, 0\right>$

\item $\vec{v} = \left<-2.5, 0\right>$

\item $\vec{v} = \left<0, \sqrt{7} \right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = -10 \hat{\jmath}$

\item $\vec{v} = \left< 3,4\right>$

\item $\vec{v} = \left<12, 5\right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left<-4, 3 \right>$

\item $\vec{v} = \left<-7, 24\right>$

\item $\vec{v} = \left<-2, -1 \right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left<-2, -6\right>$

\item $\vec{v} = \hat{\imath} + \hat{\jmath}$

\item $\vec{v} = \hat{\imath} - 4\hat{\jmath}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left<123.4, -77.05\right>$

\item $\vec{v} = \left<965.15, 831.6\right>$

\item $\vec{v} = \left<-114.1, 42.3\right>$ \label{findmaganglelast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item A small boat leaves the dock at Camp DuNuthin and heads across the Nessie River at 17 miles per hour (that is, with respect to the water) at a bearing of S$68^{\circ}$W. The river is flowing due east at 8 miles per hour. What is the boat's true speed and heading? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree.

\item \label{HMSSasquatchVectorBearing} The HMS Sasquatch leaves port with bearing S$20^{\circ}$E maintaining a speed of 42 miles per hour (that is, with respect to the water). If the ocean current is 5 miles per hour with a bearing of N$60^{\circ}$E, find the HMS Sasquatch's true speed and bearing. Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree.

\item If the captain of the HMS Sasquatch in Exercise \ref{HMSSasquatchVectorBearing} wishes to reach Chupacabra Cove, an island 100 miles away at a bearing of S$20^{\circ}$E from port, in three hours, what speed and heading should she set to take into account the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree.

\textbf{HINT:} If $\vec{v}$ denotes the velocity of the HMS Sasquatch and $\vec{w}$ denotes the velocity of the current, what does $\vec{v} + \vec{w}$ need to be to reach Chupacabra Cove in three hours?

\item In calm air, a plane flying from the Pedimaxus International Airport can reach Cliffs of Insanity Point in two hours by following a bearing of N$8.2^{\circ}$E at 96 miles an hour. (The distance between the airport and the cliffs is 192 miles.) If the wind is blowing from the southeast at 25 miles per hour, what speed and bearing should the pilot take so that she makes the trip in two hours along the original heading? Round the speed to the nearest hundredth of a mile per hour and your angle to the nearest tenth of a degree.

\item The SS Bigfoot leaves Yeti Bay on a course of N$37^{\circ}$W at a speed of 50 miles per hour. After traveling half an hour, the captain determines he is 30 miles from the bay and his bearing back to the bay is S$40^{\circ}$E. What is the speed and bearing of the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree.

\item A $600$ pound Sasquatch statue is suspended by two cables from a gymnasium ceiling. If each cable makes a $60^{\circ}$ angle with the ceiling, find the tension on each cable. Round your answer to the nearest pound.

\item Two cables are to support an object hanging from a ceiling. If the cables are each to make a $42^{\circ}$ angle with the ceiling, and each cable is rated to withstand a maximum tension of $100$ pounds, what is the heaviest object that can be supported? Round your answer down to the nearest pound.

\item A $300$ pound metal star is hanging on two cables which are attached to the ceiling. The left hand cable makes a $72^{\circ}$ angle with the ceiling while the right hand cable makes a $18^{\circ}$ angle with the ceiling. What is the tension on each of the cables? Round your answers to three decimal places.

\item Two drunken college students have filled an empty beer keg with rocks and tied ropes to it in order to drag it down the street in the middle of the night. The stronger of the two students pulls with a force of 100 pounds at a heading of N$77^{\circ}$E and the other pulls at a heading of S$68^{\circ}$E. What force should the weaker student apply to his rope so that the keg of rocks heads due east? What resultant force is applied to the keg? Round your answer to the nearest pound.

\label{kegpull}

\item Emboldened by the success of their late night keg pull in Exercise \ref{kegpull} above, our intrepid young scholars have decided to pay homage to the chariot race scene from the movie Ben-Hur' by tying three ropes to a couch, loading the couch with all but one of their friends and pulling it due west down the street. The first rope points N$80^{\circ}$W, the second points due west and the third points S$80^{\circ}$W. The force applied to the first rope is 100 pounds, the force applied to the second rope is 40 pounds and the force applied (by the non-riding friend) to the third rope is 160 pounds. They need the resultant force to be at least 300 pounds otherwise the couch won't move. Does it move? If so, is it heading due west?

\item Let $\vec{v} = \langle v_{\text{\tiny$1$}}, v_{\text{\tiny$2$}} \rangle$ be any non-zero vector. Show that $\dfrac{1}{\|\vec{v}\|} \vec{v}$ has length 1.

\item We say that two non-zero vectors $\vec{v}$ and $\vec{w}$ are {\bf parallel}\index{vector ! parallel}\index{parallel vectors} if they have same or opposite directions. That is, $\vec{v} \neq \vec{0}$ and $\vec{w} \neq \vec{0}$ are parallel if either $\hat{v} = \hat{w}$ or $\hat{v} = -\hat{w}$. Show that this means $\vec{v} = k\vec{w}$ for some non-zero scalar $k$ and that $k > 0$ if the vectors have the same direction and $k < 0$ if they point in opposite directions.

\label{parallelvectorexercise}

\item The goal of this exercise is to use vectors to describe non-vertical lines in the plane. To that end, consider the line $y = 2x - 4$. Let $\vec{v}_{\text{\tiny$0$}} = \langle 0, -4 \rangle$ and let $\vec{s} = \langle 1, 2 \rangle$. Let $t$ be any real number. Show that the vector defined by $\vec{v} = \vec{v}_{\text{\tiny$0$}} + t\vec{s}$, when drawn in standard position, has its terminal point on the line $y = 2x - 4$. (Hint: Show that $\vec{v}_{\text{\tiny$0$}} + t\vec{s} = \langle t, 2t - 4 \rangle$ for any real number $t$.) Now consider the non-vertical line $y = mx + b$. Repeat the previous analysis with $\vec{v}_{\text{\tiny$0$}} = \langle 0, b \rangle$ and let $\vec{s} = \langle 1, m \rangle$. Thus any non-vertical line can be thought of as a collection of terminal points of the vector sum of $\langle 0, b \rangle$ (the position vector of the $y$-intercept) and a scalar multiple of the slope vector $\vec{s} = \langle 1, m \rangle$.

\label{2dvectorsgiveuslines}

\item Prove the associative and identity properties of vector addition in Theorem \ref{vectoradditionprops}.

\item Prove the properties of scalar multiplication in Theorem \ref{vectorscalarmultprops}.

\end{enumerate}

\newpage

\begin{enumerate}

\item

\begin{multicols}{2}

\begin{itemize}

\item $\vec{v} + \vec{w} = \left<15,-1 \right>$, vector

\item $\vec{w} - 2\vec{v} = \left<-21,14 \right>$, vector

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} + \vec{w} \| = \sqrt{226}$, scalar

\item $\| \vec{v} \| + \| \vec{w}\| = 18$, scalar

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} \| \vec{w} - \| \vec{w} \| \vec{v} = \left<-21,77\right>$, vector

\item $\|w\| \hat{v}= \left<\frac{60}{13}, -\frac{25}{13} \right>$, vector

\end{itemize}

\end{multicols}

\item

\begin{multicols}{2}

\begin{itemize}

\item $\vec{v} + \vec{w} = \left<-12,12 \right>$, vector

\item $\vec{w} - 2\vec{v} = \left<9,-60 \right>$, vector

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} + \vec{w} \| = 12\sqrt{2}$, scalar

\item $\| \vec{v} \| + \| \vec{w}\| = 38$, scalar

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} \| \vec{w} - \| \vec{w} \| \vec{v} = \left<-34,-612\right>$, vector

\item $\|w\| \hat{v}= \left<-\frac{91}{25}, \frac{312}{25} \right>$, vector

\end{itemize}

\end{multicols}

\item

\begin{multicols}{2}

\begin{itemize}

\item $\vec{v} + \vec{w} = \left<0,3\right>$, vector

\item $\vec{w} - 2\vec{v} = \left<-6,6 \right>$, vector

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} + \vec{w} \| = 3$, scalar

\item $\| \vec{v} \| + \| \vec{w}\| = 3\sqrt{5}$, scalar

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} \| \vec{w} - \| \vec{w} \| \vec{v} = \left<-6\sqrt{5},6\sqrt{5}\right>$, vector

\item $\|w\| \hat{v}= \left<4, -2 \right>$, vector

\end{itemize}

\end{multicols}

\item

\begin{multicols}{2}

\begin{itemize}

\item $\vec{v} + \vec{w} = \left<8,9\right>$, vector

\item $\vec{w} - 2\vec{v} = \left<-22, -3 \right>$, vector

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} + \vec{w} \| = \sqrt{145}$, scalar

\item $\| \vec{v} \| + \| \vec{w}\| = 3\sqrt{29}$, scalar

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} \| \vec{w} - \| \vec{w} \| \vec{v} = \left<-14\sqrt{29},6\sqrt{29}\right>$, vector

\item $\|w\| \hat{v}= \left<5, 2 \right>$, vector

\end{itemize}

\end{multicols}

\item

\begin{multicols}{2}

\begin{itemize}

\item $\vec{v} + \vec{w} = \left<\sqrt{3},3\right>$, vector

\item $\vec{w} - 2\vec{v} = \left<4\sqrt{3}, 0 \right>$, vector

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} + \vec{w} \| = 2\sqrt{3}$, scalar

\item $\| \vec{v} \| + \| \vec{w}\| = 6$, scalar

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} \| \vec{w} - \| \vec{w} \| \vec{v} = \left<8\sqrt{3},0\right>$, vector

\item $\|w\| \hat{v}= \left<-2\sqrt{3}, 2 \right>$, vector

\end{itemize}

\end{multicols}

\item

\begin{multicols}{2}

\begin{itemize}

\item $\vec{v} + \vec{w} = \left<-\frac{1}{5},\frac{7}{5}\right>$, vector

\item $\vec{w} - 2\vec{v} = \left<-2, -1 \right>$, vector

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} + \vec{w} \| = \sqrt{2}$, scalar

\item $\| \vec{v} \| + \| \vec{w}\| = 2$, scalar

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} \| \vec{w} - \| \vec{w} \| \vec{v} = \left<-\frac{7}{5},-\frac{1}{5}\right>$, vector

\item $\|w\| \hat{v}= \left<\frac{3}{5}, \frac{4}{5} \right>$, vector

\end{itemize}

\end{multicols}

\item

\begin{multicols}{2}

\begin{itemize}

\item $\vec{v} + \vec{w} = \left<0,0\right>$, vector

\item $\vec{w} - 2\vec{v} = \left<-\frac{3\sqrt{2}}{2}, \frac{3\sqrt{2}}{2} \right>$, vector

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} + \vec{w} \| = 0$, scalar

\item $\| \vec{v} \| + \| \vec{w}\| = 2$, scalar

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} \| \vec{w} - \| \vec{w} \| \vec{v} = \left<-\sqrt{2},\sqrt{2}\right>$, vector

\item $\|w\| \hat{v}= \left<\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right>$, vector

\end{itemize}

\end{multicols}

\pagebreak

\item

\begin{multicols}{2}

\begin{itemize}

\item $\vec{v} + \vec{w} = \left<-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right>$, vector

\item $\vec{w} - 2\vec{v} = \left<-2, -2\sqrt{3} \right>$, vector

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} + \vec{w} \| = 1$, scalar

\item $\| \vec{v} \| + \| \vec{w}\| = 3$, scalar

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} \| \vec{w} - \| \vec{w} \| \vec{v} = \left<-2,-2\sqrt{3}\right>$, vector

\item $\|w\| \hat{v}= \left<1, \sqrt{3} \right>$, vector

\end{itemize}

\end{multicols}

\item

\begin{multicols}{2}

\begin{itemize}

\item $\vec{v} + \vec{w} = \left<3,2\right>$, vector

\item $\vec{w} - 2\vec{v} = \left<-6, -10 \right>$, vector

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} + \vec{w} \| = \sqrt{13}$, scalar

\item $\| \vec{v} \| + \| \vec{w}\| = 7$, scalar

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} \| \vec{w} - \| \vec{w} \| \vec{v} = \left<-6,-18\right>$, vector

\item $\|w\| \hat{v}= \left<\frac{6}{5}, \frac{8}{5}\right>$, vector

\end{itemize}

\end{multicols}

\item

\begin{multicols}{2}

\begin{itemize}

\item $\vec{v} + \vec{w} = \left<1,0\right>$, vector

\item $\vec{w} - 2\vec{v} = \left<-\frac{1}{2}, -\frac{3}{2} \right>$, vector

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} + \vec{w} \| = 1$, scalar

\item $\| \vec{v} \| + \| \vec{w}\| = \sqrt{2}$, scalar

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{itemize}

\item $\| \vec{v} \| \vec{w} - \| \vec{w} \| \vec{v} = \left<0,-\frac{\sqrt{2}}{2}\right>$, vector

\item $\|w\| \hat{v}= \left<\frac{1}{2}, \frac{1}{2}\right>$, vector

\end{itemize}

\end{multicols}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left<3,3\sqrt{3}\right>$

\item $\vec{v} = \left<\frac{3\sqrt{2}}{2},\frac{3\sqrt{2}}{2}\right>$

\item $\vec{v} = \left< \frac{\sqrt{3}}{3}, \frac{1}{3}\right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left<0,12\right>$

\item $\vec{v} = \left<-2\sqrt{3}, 2\right>$

\item $\vec{v} = \left<-\sqrt{3}, 3\right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left<-\frac{7}{2}, 0\right>$

\item $\vec{v} = \left<-5\sqrt{3}, -5\sqrt{3}\right>$

\item $\vec{v} = \left<0, -6.25\right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left<6, -2\sqrt{3}\right>$

\item $\vec{v} = \left<5, -5\right>$

\item $\vec{v} = \left<2,4\right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left<-1, 3\right>$

\item $\vec{v} = \left<-3, -4\right>$

\item $\vec{v} = \left<24, -10\right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} \approx \left<-177.96, 349.27\right>$

\item $\vec{v} \approx \left<12.96, 62.59\right>$

\item $\vec{v} \approx \left<5164.62, 1097.77\right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} \approx \left<-386.73, -230.08\right>$

\item $\vec{v} \approx \left<-52.13, -160.44\right>$

\item $\vec{v} \approx \left<14.73, -21.43\right>$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\|\vec{v}\| = 2$, $\theta = 60^{\circ}$

\item $\|\vec{v}\| = 5\sqrt{2}$, $\theta = 45^{\circ}$

\item $\|\vec{v}\| = 4$, $\theta = 150^{\circ}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\|\vec{v}\| = 2$, $\theta = 135^{\circ}$

\item $\|\vec{v}\| = 1$, $\theta = 225^{\circ}$

\item $\|\vec{v}\| = 1$, $\theta = 240^{\circ}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\|\vec{v}\| = 6$, $\theta = 0^{\circ}$

\item $\|\vec{v}\| = 2.5$, $\theta = 180^{\circ}$

\item $\|\vec{v}\| = \sqrt{7}$, $\theta = 90^{\circ}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\|\vec{v}\| = 10$, $\theta = 270^{\circ}$

\item $\|\vec{v}\| = 5$, $\theta \approx 53.13^{\circ}$

\item $\|\vec{v}\| = 13$, $\theta \approx 22.62^{\circ}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\|\vec{v}\| = 5$, $\theta \approx 143.13^{\circ}$

\item $\|\vec{v}\| = 25$, $\theta \approx 106.26^{\circ}$

\item $\|\vec{v}\| = \sqrt{5}$, $\theta \approx 206.57^{\circ}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\|\vec{v}\| = 2\sqrt{10}$, $\theta \approx 251.57^{\circ}$

\item $\|\vec{v}\| = \sqrt{2}$, $\theta \approx 45^{\circ}$

\item $\|\vec{v}\| = \sqrt{17}$, $\theta \approx 284.04^{\circ}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item \small $\|\vec{v}\| \approx 145.48$, $\theta \approx 328.02^{\circ}$ \normalsize

\item \small $\|\vec{v}\| \approx 1274.00$, $\theta \approx 40.75^{\circ}$ \normalsize

\item \small $\|\vec{v}\| \approx 121.69$, $\theta \approx 159.66^{\circ}$ \normalsize

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item The boat's true speed is about 10 miles per hour at a heading of S$50.6^{\circ}$W.

\item The HMS Sasquatch's true speed is about 41 miles per hour at a heading of S$26.8^{\circ}$E.

\item She should maintain a speed of about 35 miles per hour at a heading of S$11.8^{\circ}$E.

\item She should fly at 83.46 miles per hour with a heading of N$22.1^{\circ}$E

\item The current is moving at about 10 miles per hour bearing N$54.6^{\circ}$W.

\item The tension on each of the cables is about $346$ pounds.

\item The maximum weight that can be held by the cables in that configuration is about $133$ pounds.

\item The tension on the left hand cable is $285.317$ lbs. and on the right hand cable is $92.705$ lbs.

\item The weaker student should pull about 60 pounds. The net force on the keg is about 153 pounds.

\item The resultant force is only about 296 pounds so the couch doesn't budge. Even if it did move, the stronger force on the third rope would have made the couch drift slightly to the south as it traveled down the street.

\end{enumerate}

\closegraphsfile

## 11.9: The Dot Product and Projection

\subsection{Exercises}

In Exercises \ref{dotprodbasicfirst} - \ref{dotprodbasiclast}, use the pair of vectors $\vec{v}$ and $\vec{w}$ to find the following quantities.

\begin{multicols}{2} \raggedcolumns

\begin{itemize}

\item $\vec{v} \cdot \vec{w}$

\item The angle $\theta$ (in degrees) between $\vec{v}$ and $\vec{w}$

\item $\text{proj}_{\vec{w}}(\vec{v})$

\item $\vec{q} = \vec{v} - \text{proj}_{\vec{w}}(\vec{v})$ (Show that $\vec{q} \cdot \vec{w} = 0$.)

\end{itemize}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\item $\vec{v} = \left\langle -2, -7 \right\rangle$ and $\vec{w} = \left\langle 5, -9 \right\rangle$ \label{dotprodbasicfirst}

\item $\vec{v} = \left\langle -6, -5 \right\rangle$ and $\vec{w} = \left\langle 10, -12 \right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left\langle 1, \sqrt{3} \right\rangle$ and $\vec{w} = \left\langle 1, -\sqrt{3} \right\rangle$

\item $\vec{v} = \left\langle 3, 4 \right\rangle$ and $\vec{w} = \left\langle -6, -8 \right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left\langle -2,1 \right\rangle$ and $\vec{w} = \left\langle 3,6 \right\rangle$

\item $\vec{v} = \left\langle -3\sqrt{3}, 3\right\rangle$ and $\vec{w} = \left\langle -\sqrt{3}, -1 \right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left\langle 1, 17 \right\rangle$ and $\vec{w} = \left\langle -1, 0 \right\rangle$

\item $\vec{v} = \left\langle 3, 4 \right\rangle$ and $\vec{w} = \left\langle 5, 12 \right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left\langle -4, -2 \right\rangle$ and $\vec{w} = \left\langle 1, -5 \right\rangle$

\item $\vec{v} = \left\langle -5, 6 \right\rangle$ and $\vec{w} = \left\langle 4, -7 \right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left\langle -8, 3 \right\rangle$ and $\vec{w} = \left\langle 2, 6 \right\rangle$

\item $\vec{v} = \left\langle 34, -91 \right\rangle$ and $\vec{w} = \left\langle 0, 1 \right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} =3\hat{\imath} - \hat{\jmath}$ and $\vec{w} = 4\hat{\jmath}$

\item $\vec{v} = -24\hat{\imath} + 7\hat{\jmath}$ and $\vec{w} = 2\hat{\imath}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} =\frac{3}{2} \hat{\imath} + \frac{3}{2} \hat{\jmath}$ and $\vec{w} = \hat{\imath} - \hat{\jmath}$

\item $\vec{v} = 5\hat{\imath} +12\hat{\jmath}$ and $\vec{w} = -3\hat{\imath} + 4\hat{\jmath}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left\langle \frac{1}{2}, \frac{\sqrt{3}}{2} \right\rangle$ and $\vec{w} = \left\langle -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle$

\item $\vec{v} = \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle$ and $\vec{w} = \left\langle \frac{1}{2}, -\frac{\sqrt{3}}{2} \right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left\langle \frac{\sqrt{3}}{2}, \frac{1}{2} \right\rangle$ and $\vec{w} = \left\langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle$

\item $\vec{v} = \left\langle \frac{1}{2}, -\frac{\sqrt{3}}{2} \right\rangle$ and $\vec{w} = \left\langle \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle$ \label{dotprodbasiclast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item A force of $1500$ pounds is required to tow a trailer. Find the work done towing the trailer along a flat stretch of road $300$ feet. Assume the force is applied in the direction of the motion.

\item Find the work done lifting a $10$ pound book $3$ feet straight up into the air. Assume the force of gravity is acting straight downwards.

\item Suppose Taylor fills her wagon with rocks and must exert a force of 13 pounds to pull her wagon across the yard. If she maintains a $15^{\circ}$ angle between the handle of the wagon and the horizontal, compute how much work Taylor does pulling her wagon 25 feet. Round your answer to two decimal places.

\item In Exercise \ref{kegpull} in Section \ref{Vectors}, two drunken college students have filled an empty beer keg with rocks which they drag down the street by pulling on two attached ropes. The stronger of the two students pulls with a force of 100 pounds on a rope which makes a $13^{\circ}$ angle with the direction of motion. (In this case, the keg was being pulled due east and the student's heading was N$77^{\circ}$E.) Find the work done by this student if the keg is dragged 42 feet.

\item Find the work done pushing a 200 pound barrel 10 feet up a $12.5^{\circ}$ incline. Ignore all forces acting on the barrel except gravity, which acts downwards. Round your answer to two decimal places.

\textbf{HINT:} Since you are working to overcome gravity only, the force being applied acts directly upwards. This means that the angle between the applied force in this case and the motion of the object is \textit{not} the $12.5^{\circ}$ of the incline!

\item Prove the distributive property of the dot product in Theorem \ref{dotprodprops}.

\item Finish the proof of the scalar property of the dot product in Theorem \ref{dotprodprops}.

\item Use the identity in Example \ref{dotprodpropex} to prove the \href{http://en.wikipedia.org/wiki/Paralle...{Parallelogram Law}}}

$\|\vec{v}\|^2 + \|\vec{w}\|^2 = \dfrac{1}{2}\left[ \| \vec{v} + \vec{w}\|^2 + \|\vec{v} - \vec{w}\|^2\right]$

\item We know that $|x + y| \leq |x| + |y|$ for all real numbers $x$ and $y$ by the Triangle Inequality established in Exercise \ref{triangleinequalityreals} in Section \ref{AbsoluteValueFunctions}. We can now establish a Triangle Inequality for vectors. In this exercise, we prove that $\| \vec{u} + \vec{v} \| \leq \| \vec{u} \| + \| \vec{v} \|$ for all pairs of vectors $\vec{u}$ and $\vec{v}$. \index{vector ! triangle inequality}

\begin{enumerate}

\item (Step 1) Show that $\| \vec{u} + \vec{v} \|^{2} = \| \vec{u} \|^{2} + 2\vec{u} \cdot \vec{v} + \| \vec{v} \|^{2}$.

\item (Step 2) Show that $|\vec{u} \cdot \vec{v}| \leq \| \vec{u} \| \| \vec{v} \|$. This is the celebrated Cauchy-Schwarz Inequality.\footnote{It is also known by other names. Check out this \href{http://en.wikipedia.org/wiki/Cauchy-...derline{site}} for details.} (Hint: To show this inequality, start with the fact that $|\vec{u} \cdot \vec{v}| = |\; \| \vec{u} \| \| \vec{v} \|\cos(\theta) \;|$ and use the fact that $|\cos(\theta)| \leq 1$ for all $\theta$.)

\item (Step 3) Show that $\| \vec{u} + \vec{v} \|^{2} = \| \vec{u} \|^{2} + 2\vec{u} \cdot \vec{v} + \| \vec{v} \|^{2} \leq \| \vec{u} \|^{2} + 2|\vec{u} \cdot \vec{v}| + \| \vec{v} \|^{2} \leq \| \vec{u} \|^{2} + 2\| \vec{u} \| \| \vec{v} \| + \| \vec{v} \|^{2} = (\| \vec{u} \| + \| \vec{v} \|)^{2}$.

\item (Step 4) Use Step 3 to show that $\| \vec{u} + \vec{v} \| \leq \| \vec{u} \| + \| \vec{v} \|$ for all pairs of vectors $\vec{u}$ and $\vec{v}$.

\item As an added bonus, we can now show that the Triangle Inequality $|z + w| \leq |z| + |w|$ holds for all complex numbers $z$ and $w$ as well. Identify the complex number $z = a + bi$ with the vector $u = \langle a, b \rangle$ and identify the complex number $w = c + di$ with the vector $v = \langle c, d \rangle$ and just follow your nose!

\end{enumerate}

\end{enumerate}

\newpage

\begin{multicols}{2} \raggedcolumns

\begin{enumerate}

\item $\vec{v} = \left\langle -2, -7 \right\rangle$ and $\vec{w} = \left\langle 5, -9 \right\rangle$

$\vec{v} \cdot \vec{w} = 53$

$\theta = 45^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle \frac{5}{2}, -\frac{9}{2} \right\rangle$

$\vec{q} = \left\langle -\frac{9}{2}, -\frac{5}{2} \right\rangle$

\vfill

\item $\vec{v} = \left\langle -6, -5 \right\rangle$ and $\vec{w} = \left\langle 10, -12 \right\rangle$

$\vec{v} \cdot \vec{w} = 0$

$\theta = 90^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle 0, 0 \right\rangle$

$\vec{q} = \left\langle -6, -5 \right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left\langle 1, \sqrt{3} \right\rangle$ and $\vec{w} = \left\langle 1, -\sqrt{3} \right\rangle$

$\vec{v} \cdot \vec{w} = -2$

$\theta = 120^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle -\frac{1}{2}, \frac{\sqrt{3}}{2} \right\rangle$

$\vec{q} = \left\langle \frac{3}{2}, \frac{\sqrt{3}}{2} \right\rangle$

\vfill

\item $\vec{v} = \left\langle 3,4 \right\rangle$ and $\vec{w} = \left\langle -6, -8 \right\rangle$

$\vec{v} \cdot \vec{w} = -50$

$\theta = 180^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle 3, 4 \right\rangle$

$\vec{q} = \left\langle0, 0\right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left\langle -2,1 \right\rangle$ and $\vec{w} = \left\langle 3,6 \right\rangle$

$\vec{v} \cdot \vec{w} = 0$

$\theta = 90^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle 0, 0 \right\rangle$

$\vec{q} = \left\langle -2, 1 \right\rangle$

\vfill

\item $\vec{v} = \left\langle -3\sqrt{3}, 3\right\rangle$ and $\vec{w} = \left\langle -\sqrt{3}, -1 \right\rangle$

$\vec{v} \cdot \vec{w} = 6$

$\theta = 60^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle -\frac{3\sqrt{3}}{2}, -\frac{3}{2} \right\rangle$

$\vec{q} = \left\langle -\frac{3\sqrt{3}}{2}, \frac{9}{2} \right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left\langle 1, 17 \right\rangle$ and $\vec{w} = \left\langle -1, 0 \right\rangle$

$\vec{v} \cdot \vec{w} = -1$

$\theta \approx 93.37^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle 1, 0 \right\rangle$

$\vec{q} = \left\langle 0, 17 \right\rangle$

\vfill

\item $\vec{v} = \left\langle 3, 4 \right\rangle$ and $\vec{w} = \left\langle 5, 12 \right\rangle$

$\vec{v} \cdot \vec{w} = 63$

$\theta \approx 14.25^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle \frac{315}{169}, \frac{756}{169} \right\rangle$

$\vec{q} = \left\langle \frac{192}{169}, -\frac{80}{169} \right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left\langle -4, -2 \right\rangle$ and $\vec{w} = \left\langle 1, -5 \right\rangle$

$\vec{v} \cdot \vec{w} = 6$

$\theta \approx 74.74^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle \frac{3}{13}, -\frac{15}{13} \right\rangle$

$\vec{q} = \left\langle -\frac{55}{13}, -\frac{11}{13} \right\rangle$

\vfill

\item $\vec{v} = \left\langle -5, 6 \right\rangle$ and $\vec{w} = \left\langle 4, -7 \right\rangle$

$\vec{v} \cdot \vec{w} = -62$

$\theta \approx 169.94^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle -\frac{248}{65}, \frac{434}{65} \right\rangle$

$\vec{q} = \left\langle -\frac{77}{65}, -\frac{44}{65} \right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\pagebreak

\begin{multicols}{2} \raggedcolumns

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left\langle -8, 3 \right\rangle$ and $\vec{w} = \left\langle 2, 6 \right\rangle$

$\vec{v} \cdot \vec{w} = 2$

$\theta \approx 87.88^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle \frac{1}{10}, \frac{3}{10} \right\rangle$

$\vec{q} = \left\langle -\frac{81}{10}, \frac{27}{10} \right\rangle$

\vfill

\item $\vec{v} = \left\langle 34, -91 \right\rangle$ and $\vec{w} = \left\langle 0, 1 \right\rangle$

$\vec{v} \cdot \vec{w} = -91$

$\theta \approx 159.51^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle 0, -91 \right\rangle$

$\vec{q} = \left\langle 34, 0 \right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} =3\hat{\imath} - \hat{\jmath}$ and $\vec{w} = 4\hat{\jmath}$

$\vec{v} \cdot \vec{w} = -4$

$\theta \approx 108.43^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle 0,-1 \right\rangle$

$\vec{q} = \left\langle 3,0 \right\rangle$

\vfill

\item $\vec{v} = -24\hat{\imath} + 7\hat{\jmath}$ and $\vec{w} = 2\hat{\imath}$

$\vec{v} \cdot \vec{w} = -48$

$\theta \approx 163.74^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle -24,0 \right\rangle$

$\vec{q} = \left\langle 0,7\right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} =\frac{3}{2} \hat{\imath} + \frac{3}{2} \hat{\jmath}$ and $\vec{w} = \hat{\imath} - \hat{\jmath}$

$\vec{v} \cdot \vec{w} = 0$

$\theta = 90^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle 0,0 \right\rangle$

$\vec{q} = \left\langle \frac{3}{2},\frac{3}{2} \right\rangle$

\vfill

\item $\vec{v} = 5\hat{\imath} + 12\hat{\jmath}$ and $\vec{w} = -3\hat{\imath} + 4\hat{\jmath}$

$\vec{v} \cdot \vec{w} = 33$

$\theta \approx 59.49^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle -\frac{99}{25}, \frac{132}{25} \right\rangle$

$\vec{q} = \left\langle \frac{224}{25},\frac{168}{25} \right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left\langle \frac{1}{2}, \frac{\sqrt{3}}{2} \right\rangle$ and $\vec{w} = \left\langle -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle$

$\vec{v} \cdot \vec{w} = \frac{\sqrt{6} - \sqrt{2}}{4}$

$\theta = 75^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle \frac{1-\sqrt{3}}{4}, \frac{\sqrt{3} - 1}{4} \right\rangle$

$\vec{q} = \left\langle \frac{1+\sqrt{3}}{4}, \frac{1 +\sqrt{3}}{4} \right\rangle$

\vfill

\item $\vec{v} = \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle$ and $\vec{w} = \left\langle \frac{1}{2}, -\frac{\sqrt{3}}{2} \right\rangle$

$\vec{v} \cdot \vec{w} = \frac{\sqrt{2} - \sqrt{6}}{4}$

$\theta = 105^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle \frac{\sqrt{2}-\sqrt{6}}{8}, \frac{3\sqrt{2} - \sqrt{6}}{8} \right\rangle$

$\vec{q} = \left\langle \frac{3\sqrt{2}+\sqrt{6}}{8}, \frac{\sqrt{2} + \sqrt{6}}{8} \right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\vec{v} = \left\langle \frac{\sqrt{3}}{2}, \frac{1}{2} \right\rangle$ and $\vec{w} = \left\langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle$

$\vec{v} \cdot \vec{w} = -\frac{\sqrt{6} + \sqrt{2}}{4}$

$\theta = 165^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle \frac{\sqrt{3} + 1}{4}, \frac{\sqrt{3} + 1}{4} \right\rangle$

$\vec{q} = \left\langle \frac{\sqrt{3} - 1}{4}, \frac{1 - \sqrt{3}}{4} \right\rangle$

\vfill

\item $\vec{v} = \left\langle \frac{1}{2}, -\frac{\sqrt{3}}{2} \right\rangle$ and $\vec{w} = \left\langle \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right\rangle$

$\vec{v} \cdot \vec{w} = \frac{\sqrt{6} + \sqrt{2}}{4}$

$\theta = 15^{\circ}$

$\text{proj}_{\vec{w}}(\vec{v}) = \left\langle \frac{\sqrt{3} + 1}{4}, -\frac{\sqrt{3} + 1}{4} \right\rangle$

$\vec{q} = \left\langle \frac{1 - \sqrt{3}}{4}, \frac{1 - \sqrt{3}}{4} \right\rangle$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(1500 \, \text{pounds})(300 \, \text{feet})\cos\left(0^{\circ}\right) = 450,000$ foot-pounds

\item $(10 \, \text{pounds})(3 \, \text{feet})\cos\left(0^{\circ}\right) = 30$ foot-pounds

\item $(13 \, \text{pounds})(25 \, \text{feet}) \cos\left(15^{\circ}\right) \approx 313.92$ foot-pounds

\item $(100 \, \text{pounds})(42 \, \text{feet}) \cos\left(13^{\circ}\right) \approx 4092.35$ foot-pounds

\item $(200 \, \text{pounds})(10 \, \text{feet}) \cos\left(77.5^{\circ}\right) \approx 432.88$ foot-pounds

\end{enumerate}

\closegraphsfile

## 11.10: Parametric Equations

\subsection{Exercises}

In Exercises \ref{paraplotfirst} - \ref{paraplotlast}, plot the set of parametric equations by hand. Be sure to indicate the orientation imparted on the curve by the parametrization.

\begin{multicols}{2} \raggedcolumns

\begin{enumerate}

\item ${\displaystyle \left\{ \begin{array}{l} x = 4t-3 \\ y = 6t-2 \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 1}$ \label{paraplotfirst}

\item ${\displaystyle \left\{ \begin{array}{l} x = 4t-1 \\ y = 3-4t \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 1}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item ${\displaystyle \left\{ \begin{array}{l} x = 2t \\ y = t^2 \end{array} \right. \vspace{.25in} \mbox{for } -1 \leq t \leq 2}$

\item ${\displaystyle \left\{ \begin{array}{l} x = t-1 \\ y = 3+2t-t^2 \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 3}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2} \raggedcolumns

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item ${\displaystyle \left\{ \begin{array}{l} x = t^2+2t+1 \$3pt] y = t+1 \end{array} \right. \vspace{.25in} \mbox{for } t \leq 1} \item {\displaystyle \left\{ \begin{array}{l} x = \frac{1}{9}\left(18-t^2\right) \\[3pt] y = \frac{1}{3} t \end{array} \right. \vspace{.25in} \mbox{for } t \geq -3} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \raggedcolumns \begin{enumerate} \setcounter{enumi}{\value{HW}} \item {\displaystyle \left\{ \begin{array}{l} x = t \\ y = t^3 \end{array} \right. \vspace{.25in} \mbox{for } -\infty < t < \infty} \item {\displaystyle \left\{ \begin{array}{l} x = t^3 \\ y = t \end{array} \right. \vspace{.25in} \mbox{for } -\infty < t < \infty} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \raggedcolumns \begin{enumerate} \setcounter{enumi}{\value{HW}} \item {\displaystyle \left\{ \begin{array}{l} x = \cos(t) \\ y = \sin(t) \end{array} \right. \vspace{.25in} \mbox{for } -\dfrac{\pi}{2} \leq t \leq \dfrac{\pi}{2}} \item {\displaystyle \left\{ \begin{array}{l} x = 3\cos(t) \\ y = 3\sin(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq \pi} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \raggedcolumns \begin{enumerate} \setcounter{enumi}{\value{HW}} \item {\displaystyle \left\{ \begin{array}{l} x = -1+ 3\cos(t) \\ y = 4\sin(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 2\pi} \item {\displaystyle \left\{ \begin{array}{l} x = 3\cos(t) \\ y = 2\sin(t)+1 \end{array} \right. \vspace{.25in} \mbox{for } \dfrac{\pi}{2} \leq t \leq 2\pi} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \raggedcolumns \begin{enumerate} \setcounter{enumi}{\value{HW}} \item {\displaystyle \left\{ \begin{array}{l} x = 2\cos(t) \\ y = \sec(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 < t < \dfrac{\pi}{2}} \item {\displaystyle \left\{ \begin{array}{l} x = 2\tan(t) \\ y = \cot(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 < t < \dfrac{\pi}{2}} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \raggedcolumns \begin{enumerate} \setcounter{enumi}{\value{HW}} \item {\displaystyle \left\{ \begin{array}{l} x = \sec(t) \\ y = \tan(t) \end{array} \right. \vspace{.25in} \mbox{for } -\dfrac{\pi}{2} < t < \dfrac{\pi}{2}} \item {\displaystyle \left\{ \begin{array}{l} x = \sec(t) \\ y = \tan(t) \end{array} \right. \vspace{.25in} \mbox{for } \dfrac{\pi}{2} < t < \dfrac{3\pi}{2}} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \raggedcolumns \begin{enumerate} \setcounter{enumi}{\value{HW}} \item {\displaystyle \left\{ \begin{array}{l} x = \tan(t) \\ y = 2\sec(t) \end{array} \right. \vspace{.25in} \mbox{for } -\dfrac{\pi}{2} < t < \dfrac{\pi}{2}} \item {\displaystyle \left\{ \begin{array}{l} x = \tan(t) \\ y = 2\sec(t) \end{array} \right. \vspace{.25in} \mbox{for } \dfrac{\pi}{2} < t < \dfrac{3\pi}{2}} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \raggedcolumns \begin{enumerate} \setcounter{enumi}{\value{HW}} \item {\displaystyle \left\{ \begin{array}{l} x = \cos(t) \\ y = t \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq \pi} \item {\displaystyle \left\{ \begin{array}{l} x = \sin(t) \\ y = t \end{array} \right. \vspace{.25in} \mbox{for } -\dfrac{\pi}{2} \leq t \leq \dfrac{\pi}{2}} \label{paraplotlast} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} In Exercises \ref{paracalcfirst} - \ref{paracalclast}, plot the set of parametric equations with the help of a graphing utility. Be sure to indicate the orientation imparted on the curve by the parametrization. \begin{multicols}{2} \raggedcolumns \begin{enumerate} \setcounter{enumi}{\value{HW}} \item {\displaystyle \left\{ \begin{array}{l} x = t^{3} - 3t \\ y = t^{2} - 4 \end{array} \right. \vspace{.25in} \mbox{for } -2 \leq t \leq 2} \label{paracalcfirst} \item {\displaystyle \left\{ \begin{array}{l} x = 4\cos^{3}(t) \\ y = 4\sin^{3}(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 2\pi} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \raggedcolumns \begin{enumerate} \setcounter{enumi}{\value{HW}} \item {\displaystyle \left\{ \begin{array}{l} x = e^{t} + e^{-t} \\ y = e^{t} - e^{-t} \end{array} \right. \vspace{.25in} \mbox{for } -2 \leq t \leq 2} \item {\displaystyle \left\{ \begin{array}{l} x = \cos(3t) \\ y = \sin(4t) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 2\pi} \label{paracalclast} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \pagebreak In Exercises \ref{findparamfirst} - \ref{findparamlast}, find a parametric description for the given oriented curve. \begin{enumerate} \setcounter{enumi}{\value{HW}} \item the directed line segment from (3,-5) to (-2,2) \label{findparamfirst} \item the directed line segment from (-2,-1) to (3, -4) \item the curve y = 4-x^2 from (-2,0) to (2,0). \item the curve y = 4-x^2 from (-2,0) to (2,0) \\ (Shift the parameter so t=0 corresponds to (-2,0).) \item the curve x = y^2 - 9 from (-5,-2) to (0,3). \item the curve x = y^2 - 9 from (0,3) to (-5,-2).\\ (Shift the parameter so t=0 corresponds to (0,3).) \item the circle x^2 + y^2 = 25, oriented counter-clockwise \item the circle (x-1)^2 + y^2 = 4, oriented counter-clockwise \item the circle x^2 + y^2 - 6y = 0, oriented counter-clockwise \item the circle x^2 + y^2 - 6y = 0, oriented \emph{clockwise}\\ (Shift the parameter so t begins at 0.) \item the circle (x-3)^2 + (y+1)^2 = 117, oriented counter-clockwise \item the ellipse (x-1)^2 + 9y^2 = 9, oriented counter-clockwise \item the ellipse 9x^2 + 4y^2 + 24y =0, oriented counter-clockwise \item the ellipse 9x^2 + 4y^2 + 24y =0, oriented clockwise \\ (Shift the parameter so t=0 corresponds to (0,0).) \item the triangle with vertices (0,0), (3,0), (0,4), oriented counter-clockwise \\ (Shift the parameter so t=0 corresponds to (0,0).) \label{findparamlast} \setcounter{HW}{\value{enumi}} \end{enumerate} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item Use parametric equations and a graphing utility to graph the inverse of f(x) = x^{3} + 3x - 4. \item Every polar curve r = f(\theta) can be translated to a system of parametric equations with parameter \theta by \left\{ x = r\cos(\theta) = f(\theta) \cos(\theta), \, y = r \sin(\theta) = f(\theta) \sin(\theta) \right.. Convert r = 6\cos(2\theta) to a system of parametric equations. Check your answer by graphing r = 6\cos(2\theta) by hand using the techniques presented in Section \ref{PolarGraphs} and then graphing the parametric equations you found using a graphing utility. \item Use your results from Exercises \ref{heightlondoneye} and \ref{leftrightlondoneye} in Section \ref{Sinusoid} to find the parametric equations which model a passenger's position as they ride the \href{http://en.wikipedia.org/wiki/London_...derline{London Eye}}. \setcounter{HW}{\value{enumi}} \end{enumerate} \phantomsection \label{projectoilemotion} Suppose an object, called a projectile, is launched into the air. Ignoring everything except the force gravity, the path of the projectile is given by\footnote{A nice mix of vectors and Calculus are needed to derive this.} \[ \left\{ \begin{array}{l} x = v_{\text{\tiny 0}} \cos(\theta) \, t \\ [3pt] y = -\dfrac{1}{2} g t^2 + v_{\text{\tiny 0}} \sin(\theta) \, t + s_{\text{\tiny 0}} \\ \end{array} \right. \; \text{for} \; 0 \leq t \leq T$ where$v_{\text{\tiny $0$}}$is the initial speed of the object,$\theta$is the angle from the horizontal at which the projectile is launched,\footnote{We've seen this before. It's the angle of elevation which was defined on page \pageref{angleofelevation}.}$g$is the acceleration due to gravity,$s_{\text{\tiny $0$}}$is the initial height of the projectile above the ground and$T$is the time when the object returns to the ground. (See the figure below.) \begin{center} \begin{mfpic}[20]{-1}{9}{-1}{7} \axes \tlabel[cc](9,-0.5){\scriptsize$x$} \tlabel[cc](0.5,7){\scriptsize$y$} \dashed \polyline{(0,4), (1.5,4)} \tlabelsep{5pt} \axislabels{y}{{\scriptsize$s_{\text{\tiny $0$}}$} 4} \point[3pt]{(0,4), (5.98,0)} \arrow \parafcn{0,0.75,0.1}{(3.5*t, 6.062*t+4-(4.9*(t**2)))} \parafcn{0.75,1.71,0.1}{(3.5*t, 6.062*t+4-(4.9*(t**2)))} \arrow \shiftpath{(0,4)} \parafcn{5, 55, 5}{0.75*dir(t)} \tlabel[cc](1,4.5){\scriptsize$\theta$} \tlabel[cc](5.98,-0.5){\scriptsize$(x(T), 0)$} \end{mfpic} \end{center} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item Carl's friend Jason competes in Highland Games Competitions across the country. In one event, the hammer throw', he throws a 56 pound weight for distance. If the weight is released$6$feet above the ground at an angle of$42^{\circ}$with respect to the horizontal with an initial speed of$33$feet per second, find the parametric equations for the flight of the hammer. (Here, use$g = 32 \frac{\text{ft.}}{s^2}$.) When will the hammer hit the ground? How far away will it hit the ground? Check your answer using a graphing utility. \item \label{projectileeliminate} Eliminate the parameter in the equations for projectile motion to show that the path of the projectile follows the curve $y = -\dfrac{g \sec^{2}(\theta)}{2 v_{\text{\tiny0}}^2} x^2 + \tan(\theta) x + s_{\text{\tiny 0}}$ Use the vertex formula (Equation \ref{vertexofquadraticfunctions}) to show the maximum height of the projectile is $y = \dfrac{v_{\text{\tiny0}}^2 \sin^{2}(\theta)}{2g} + s_{\text{\tiny 0}} \quad \text{when} \quad x = \dfrac{v_{\text{\tiny0}}^2 \sin(2\theta) }{2g}$ \item In another event, the sheaf toss', Jason throws a 20 pound weight for height. If the weight is released 5 feet above the ground at an angle of$85^{\circ}$with respect to the horizontal and the sheaf reaches a maximum height of 31.5 feet, use your results from part \ref{projectileeliminate} to determine how fast the sheaf was launched into the air. (Once again, use$g = 32 \frac{\text{ft.}}{s^2}$.) \item Suppose$\theta = \frac{\pi}{2}$. (The projectile was launched vertically.) Simplify the general parametric formula given for$y(t)$above using$g = 9.8 \, \frac{m}{s^2}$and compare that to the formula for$s(t)$given in Exercise \ref{whatgoesup} in Section \ref{QuadraticFunctions}. What is$x(t)$in this case? \setcounter{HW}{\value{enumi}} \end{enumerate} \phantomsection \label{hyperboliccosinesine} In Exercises \ref{hyperbolicfirst} - \ref{hyperboliclast}, we explore the \textbf{hyperbolic cosine}\index{hyperbolic cosine} function, denoted$\cosh(t)$, and the \textbf{hyperbolic sine}\index{hyperbolic sine} function, denoted$\sinh(t)$, defined below: $\begin{array}{ccc} \cosh(t) = \dfrac{e^{t} + e^{-t}}{2} & \text{and} & \sinh(t) = \dfrac{e^{t} - e^{-t}}{2} \\ \end{array}$ \begin{enumerate} \setcounter{enumi}{\value{HW}} \item Using a graphing utility as needed, verify that the domain of$\cosh(t)$is$(-\infty, \infty)$and the range of$\cosh(t)$is$[1,\infty)$. \label{hyperbolicfirst} \item Using a graphing utility as needed, verify that the domain and range of$\sinh(t)$are both$(-\infty, \infty)$. \item Show that$\left\{ x(t) = \cosh(t), \, y(t) = \sinh(t) \right.$parametrize the right half of the unit' hyperbola$x^2 - y^2 = 1$. (Hence the use of the adjective hyperbolic.') \item Compare the definitions of$\cosh(t)$and$\sinh(t)$to the formulas for$\cos(t)$and$\sin(t)$given in Exercise \ref{expformcosandsin} in Section \ref{PolarComplex}. \item \label{andtheresthyperbolic} Four other hyperbolic functions are waiting to be defined: the hyperbolic secant$\text{sech}(t)$, the hyperbolic cosecant$\text{csch}(t)$, the hyperbolic tangent$\tanh(t)$and the hyperbolic cotangent$\coth(t)$. Define these functions in terms of$\cosh(t)$and$\sinh(t)$, then convert them to formulas involving$e^{t}$and$e^{-t}$. Consult a suitable reference (a Calculus book, or this entry on the \href{http://en.wikipedia.org/wiki/Hyperbo...ine{hyperbolic functions}}) and spend some time reliving the thrills of trigonometry with these `hyperbolic' functions. \item If these functions look familiar, they should. Enjoy some nostalgia and revisit Exercise \ref{catenary} in Section \ref{ExpLogApplications}, Exercise \ref{hyperbolicsine} in Section \ref{ExpEquations} and the answer to Exercise \ref{inversehyptangent} in Section \ref{LogEquations}. \label{hyperboliclast} \end{enumerate} \newpage \subsection{Answers} \begin{multicols}{2} \raggedcolumns \begin{enumerate} \item${\displaystyle \left\{ \begin{array}{l} x = 4t-3 \\ y = 6t-2 \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 1}$\begin{mfpic}[15]{-4}{2}{-3}{5} \axes \tlabel[cc](2,-0.5){\scriptsize$x$} \tlabel[cc](0.5,5){\scriptsize$y$} \xmarks{-3,-2,-1,1} \ymarks{-2,-1,1,2,3,4} \point[3pt]{(-3,-2), (1,4)} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-3 \hspace{6pt}$} -3,{$-2 \hspace{6pt}$} -2,{$-1 \hspace{6pt}$} -1, {$1$} 1} \axislabels {y}{{$-1$} -1,{$-2$} -2, {$1$} 1,{$2$} 2,{$3$} 3,{$4$} 4} \normalsize \arrow \parafcn{0,0.5,0.1}{(4*t-3,6*t-2)} \parafcn{0.5,1,0.1}{(4*t-3,6*t-2)} \end{mfpic} \item${\displaystyle \left\{ \begin{array}{l} x = 4t-1 \\ y = 3-4t \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 1}$\begin{mfpic}[15]{-2}{4}{-2}{4} \axes \tlabel[cc](4,-0.5){\scriptsize$x$} \tlabel[cc](0.5,4){\scriptsize$y$} \xmarks{-1,1,2,3} \ymarks{-1,1,2,3} \point[3pt]{(-1,3), (3,-1)} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2,{$3$} 3} \axislabels {y}{{$-1$} -1, {$1$} 1,{$2$} 2,{$3$} 3} \normalsize \arrow \parafcn{0,0.5,0.1}{(4*t-1,3-4*t)} \parafcn{0.5,1,0.1}{(4*t-1,3-4*t)} \end{mfpic} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \raggedcolumns \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = 2t \\ y = t^2 \end{array} \right. \vspace{.25in} \mbox{for } -1 \leq t \leq 2}$\begin{mfpic}[15]{-3}{5}{-1}{5} \axes \tlabel[cc](5,-0.5){\scriptsize$x$} \tlabel[cc](0.5,5){\scriptsize$y$} \xmarks{-2,-1,1,2,3,4} \ymarks{1,2,3,4} \point[3pt]{(-2,1), (0,0), (4,4)} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-3 \hspace{6pt}$} -3,{$-2 \hspace{6pt}$} -2,{$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2,{$3$} 3,{$4$} 4} \axislabels {y}{{$1$} 1,{$2$} 2,{$3$} 3,{$4$} 4} \normalsize \arrow \parafcn{-1,-0.5,0.1}{(2*t,(t**2))} \arrow \parafcn{-0.5,1,0.1}{(2*t,(t**2))} \parafcn{1,2,0.1}{(2t,t**2)} \end{mfpic} \item${\displaystyle \left\{ \begin{array}{l} x = t-1 \\ y = 3+2t-t^2 \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 3}$\begin{mfpic}[15]{-2}{3}{-1}{5} \axes \tlabel[cc](4,-0.5){\scriptsize$x$} \tlabel[cc](0.5,5){\scriptsize$y$} \xmarks{-1,1,2} \ymarks{1,2,3,4} \point[3pt]{(-1,3), (0,4), (2,0)} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2} \axislabels {y}{{$1$} 1,{$2$} 2,{$3$} 3,{$4$} 4} \normalsize \arrow \parafcn{0,0.5,0.1}{(t-1,3+2*t-(t**2))} \arrow \parafcn{0.5,2,0.1}{(t-1,3+2*t-(t**2))} \parafcn{2,3,0.1}{(t-1,3+2*t-(t**2))} \end{mfpic} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \raggedcolumns \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = t^2+2t+1 \\ y = t+1 \end{array} \right. \vspace{.25in} \mbox{for } t \leq 1}$\begin{mfpic}[15]{-1}{6}{-3}{3} \axes \tlabel[cc](6,-0.5){\scriptsize$x$} \tlabel[cc](0.5,3){\scriptsize$y$} \xmarks{1,2,3,4,5} \ymarks{-2,-1,1,2} \point[3pt]{(4,2), (0,0), (4,-2)} \tlpointsep{4pt} \scriptsize \axislabels {x}{ {$1$} 1, {$2$} 2,{$3$} 3,{$4$} 4,{$5$} 5} \axislabels {y} ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Precalculus/Book:_Precalculus_(Stitz-Zeager)/11:_Applications_of_Trigonometry/11.E:_Applications_of_Trigonometry_(Exercises)), /content/body/div[10]/p[272]/span, line 1, column 4 \normalsize \parafcn{0,1,0.1}{((t**2)+(2*t)+1,t+1)} \arrow \parafcn{-2,0,0.1}{((t**2)+(2*t)+1,t+1)} \arrow \parafcn{-3.25,-2,0.1}{((t**2)+(2*t)+1,t+1)} \arrow \parafcn{-3.44,-3.25,0.1}{((t**2)+(2*t)+1,t+1)} \end{mfpic} \item${\displaystyle \left\{ \begin{array}{l} x = \frac{1}{9}\left(18-t^2\right) \\ y = \frac{1}{3} t \end{array} \right. \vspace{.25in} \mbox{for } t \geq -3}$\begin{mfpic}[15]{-4}{3}{-2}{3} \axes \tlabel[cc](3,-0.5){\scriptsize$x$} \tlabel[cc](0.5,3){\scriptsize$y$} \xmarks{-3,-2,-1,1,2} \ymarks{-1,1,2} \point[3pt]{(1,-1), (2,0), (-2,2)} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-3 \hspace{6pt}$} -3,{$-2 \hspace{6pt}$} -2,{$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2} \axislabels {y}{{$-1$} -1,{$1$} 1,{$2$} 2} \normalsize \arrow \parafcn{-3,-1.5,0.1}{((18-(t**2))/9,t/3)} \arrow \parafcn{-1.5,3,0.1}{((18-(t**2))/9,t/3)} \arrow \parafcn{3,6.5,0.1}{((18-(t**2))/9,t/3)} \end{mfpic} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \pagebreak \begin{multicols}{2} \raggedcolumns \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = t \\ y = t^3 \end{array} \right. \vspace{.25in} \mbox{for } -\infty < t < \infty}$\begin{mfpic}[15]{-2}{2}{-5}{5} \axes \tlabel[cc](2,-0.5){\scriptsize$x$} \tlabel[cc](0.5,5){\scriptsize$y$} \xmarks{-1,1} \ymarks{-4,-3,-2,-1,1,2,3,4} \point[3pt]{(-1,-1), (0,0), (1,1)} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-1 \hspace{6pt}$} -1, {$1$} 1} \axislabels {y}{{$-4$} -4,{$-3$} -3,{$-2$} -2,{$-1$} -1,{$1$} 1,{$2$} 2,{$3$} 3,{$4$} 4} \normalsize \arrow \parafcn{-1.7,-1.25,0.1}{(t, t**3)} \arrow \parafcn{-1.25,1.25,0.1}{(t, t**3)} \arrow \parafcn{1.25,1.7,0.1}{(t, t**3)} \end{mfpic} \item${\displaystyle \left\{ \begin{array}{l} x = t^3 \\ y = t \end{array} \right. \vspace{.25in} \mbox{for } -\infty < t < \infty}$\begin{mfpic}[15]{-5}{5}{-2}{2} \axes \tlabel[cc](5,-0.5){\scriptsize$x$} \tlabel[cc](0.5,2){\scriptsize$y$} \xmarks{-4,-3,-2,-1,1,2,3,4} \ymarks{-1,1} \point[3pt]{(-1,-1), (0,0), (1,1)} \tlpointsep{4pt} \scriptsize \axislabels {y}{{$-1$} -1, {$1$} 1} \axislabels {x}{{$-4 \hspace{6pt}$} -4,{$-3 \hspace{6pt}$} -3,{$-2 \hspace{6pt}$} -2,{$-1 \hspace{6pt}$} -1,{$1$} 1,{$2$} 2,{$3$} 3,{$4$} 4} \normalsize \arrow \parafcn{-1.7,-1.25,0.1}{(t**3, t)} \arrow \parafcn{-1.25,1.25,0.1}{(t**3, t)} \arrow \parafcn{1.25,1.7,0.1}{(t**3, t)} \end{mfpic} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = \cos(t) \\ y = \sin(t) \end{array} \right. \vspace{.25in} \mbox{for } -\dfrac{\pi}{2} \leq t \leq \dfrac{\pi}{2}}$\begin{mfpic}[10]{-5}{5}{-5}{5} \axes \tlabel[cc](5,-0.5){\scriptsize$x$} \tlabel[cc](0.5,5){\scriptsize$y$} \point[3pt]{(0,-4), (4,0), (0,4)} \xmarks{-4,4} \ymarks{-4,4} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-1 \hspace{6pt}$} -4, {$1$} 4} \axislabels {y}{{$-1$} -4, {$1$} 4} \normalsize \arrow \parafcn{-1.57,-0.78,0.1}{(4*cos(t),4*sin(t))} \arrow \parafcn{-0.78,0.78,0.1}{(4*cos(t),4*sin(t))} \parafcn{0.78,1.57,0.1}{(4*cos(t),4*sin(t))} \end{mfpic} \item${\displaystyle \left\{ \begin{array}{l} x = 3\cos(t) \\ y = 3\sin(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq \pi}$\begin{mfpic}[15]{-4}{4}{-1}{4} \axes \tlabel[cc](4,-0.5){\scriptsize$x$} \tlabel[cc](0.5,4){\scriptsize$y$} \point[3pt]{(-3,0), (3,0), (0,3)} \xmarks{-3,-2,-1,1,2,3} \ymarks{1,2,3} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2,{$-1 \hspace{6pt}$} -1,{$1$} 1,{$2$} 2,{$3$} 3} \axislabels {y}{{$1$} 1,{$2$} 2,{$3$} 3} \normalsize \arrow \parafcn{0,0.78,0.1}{(3*cos(t),3*sin(t))} \arrow \parafcn{0.78,2.36,0.1}{(3*cos(t),3*sin(t))} \parafcn{2.36,3.14,0.1}{(3*cos(t),3*sin(t))} \end{mfpic} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = -1+3\cos(t) \\ y = 4\sin(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 2\pi}$\begin{mfpic}[15]{-5}{3}{-5}{5} \axes \tlabel[cc](3,-0.5){\scriptsize$x$} \tlabel[cc](0.5,5){\scriptsize$y$} \point[3pt]{(2,0), (-1,4), (-4,0), (-1,-4)} \xmarks{-4,-3,-2,-1,1,2} \ymarks{-4,-3,-2,-1,1,2,3,4} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-4 \hspace{6pt}$} -4,{$-3 \hspace{6pt}$} -3,{$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1,{$1$} 1,{$2$} 2} \axislabels {y}{{$-4$} -4, {$-3$} -3,{$-2$} -2,{$-1$} -1,{$1$} 1,{$2$} 2,{$3$} 3,{$4$} 4} \normalsize \arrow \parafcn{0,0.78,0.1}{(3*cos(t)-1,4*sin(t))} \arrow \parafcn{0.78,2.36, 0.1}{(3*cos(t)-1,4*sin(t))} \arrow \parafcn{2.36,3.93, 0.1}{(3*cos(t)-1,4*sin(t))} \arrow \parafcn{3.93,5.5, 0.1}{(3*cos(t)-1,4*sin(t))} \parafcn{5.5,6.28, 0.1}{(3*cos(t)-1,4*sin(t))} \end{mfpic} \item${\displaystyle \left\{ \begin{array}{l} x = 3\cos(t) \\ y = 2\sin(t)+1 \end{array} \right. \vspace{.25in} \mbox{for } \dfrac{\pi}{2} \leq t \leq 2\pi}$\begin{mfpic}[15]{-4}{4}{-2}{4} \axes \tlabel[cc](4,-0.5){\scriptsize$x$} \tlabel[cc](0.5,4){\scriptsize$y$} \point[3pt]{(0,3), (-3,1), (0,-1), (3,1)} \xmarks{-3,-2,-1,1,2,3} \ymarks{-1,1,2,3} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-3 \hspace{6pt}$} -3, {$-1 \hspace{6pt}$} -1,{$1$} 1,{$3$} 3} \axislabels {y}{{$-1$} -1,{$1$} 1,{$2$} 2,{$3$} 3} \normalsize \arrow \parafcn{1.57,2.36,0.1}{(3*cos(t),1+2*sin(t))} \arrow \parafcn{2.36,3.93,0.1}{(3*cos(t),1+2*sin(t))} \arrow \parafcn{3.93,5.50,0.1}{(3*cos(t),1+2*sin(t))} \parafcn{5.50,6.28,0.1}{(3*cos(t),1+2*sin(t))} \end{mfpic} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \raggedcolumns \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = 2\cos(t) \\ y = \sec(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 < t < \dfrac{\pi}{2}}$\begin{mfpic}[25]{-1}{5}{-1}{5} \axes \tlabel[cc](5,-0.5){\scriptsize$x$} \tlabel[cc](0.5,5){\scriptsize$y$} \point[3pt]{(2,1)} \xmarks{1,2,3,4} \ymarks{1,2,3,4} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$1$} 1,{$2$} 2,{$3$} 3,{$4$} 4} \axislabels {y}{{$1$} 1,{$2$} 2,{$3$} 3,{$4$} 4} \normalsize \arrow \parafcn{0,1,0.1}{(2*cos(t),sec(t))} \arrow \parafcn{1,1.31,0.1}{(2*cos(t),sec(t))} \end{mfpic} \item${\displaystyle \left\{ \begin{array}{l} x = 2\tan(t) \\ y = \cot(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 < t < \dfrac{\pi}{2}}$\begin{mfpic}[25]{-1}{5}{-1}{5} \axes \tlabel[cc](5,-0.5){\scriptsize$x$} \tlabel[cc](0.5,5){\scriptsize$y$} \xmarks{1,2,3,4} \ymarks{1,2,3,4} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$1$} 1,{$2$} 2,{$3$} 3,{$4$} 4} \axislabels {y}{{$1$} 1,{$2$} 2,{$3$} 3,{$4$} 4} \normalsize \arrow \parafcn{0.25,0.4,0.1}{(2*tan(t),cot(t))} \arrow \parafcn{0.4,0.7,0.1}{(2*tan(t),cot(t))} \arrow \parafcn{0.7,1.1,0.1}{(2*tan(t),cot(t))} \end{mfpic} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = \sec(t) \\ y = \tan(t) \end{array} \right. \vspace{.25in} \mbox{for } -\dfrac{\pi}{2} < t < \dfrac{\pi}{2}}$\begin{mfpic}[15]{-1}{5}{-5}{5} \axes \tlabel[cc](5,-0.5){\scriptsize$x$} \tlabel[cc](0.5,5){\scriptsize$y$} \point[3pt]{(1,0)} \xmarks{1,2,3,4} \ymarks{-4,-3,-2,-1,1,2,3,4} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$1$} 1,{$2$} 2,{$3$} 3,{$4$} 4} \axislabels {y}{{$-4$} -4, {$-3$} -3,{$-2$} -2,{$-1$} -1,{$1$} 1,{$2$} 2,{$3$} 3,{$4$} 4} \normalsize \arrow \parafcn{1.85,2,0.1}{(0-sec(t),tan(t))} \arrow \parafcn{2,4.25,0.1}{(0-sec(t),tan(t))} \arrow \parafcn{4.25,4.4,0.1}{(0-sec(t),tan(t))} \dashed \polyline{(4,4),(-1,-1)} \dashed \polyline{(4,-4),(-1,1)} \end{mfpic} \item${\displaystyle \left\{ \begin{array}{l} x = \sec(t) \\ y = \tan(t) \end{array} \right. \vspace{.25in} \mbox{for } \dfrac{\pi}{2} < t < \dfrac{3\pi}{2}}$\begin{mfpic}[15]{-5}{1}{-5}{5} \axes \tlabel[cc](1,-0.5){\scriptsize$x$} \tlabel[cc](0.5,5){\scriptsize$y$} \point[3pt]{(-1,0)} \xmarks{-4,-3,-2,-1} \ymarks{-4,-3,-2,-1,1,2,3,4} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-4 \hspace{6pt}$} -4,{$-3 \hspace{6pt}$} -3,{$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1} \axislabels {y}{{$-4$} -4, {$-3$} -3,{$-2$} -2,{$-1$} -1,{$1$} 1,{$2$} 2,{$3$} 3,{$4$} 4} \normalsize \arrow \parafcn{1.85,2,0.1}{(sec(t),tan(t))} \arrow \parafcn{2,4.25,0.1}{(sec(t),tan(t))} \arrow \parafcn{4.25,4.4,0.1}{(sec(t),tan(t))} \dashed \polyline{(-4,4),(1,-1)} \dashed \polyline{(-4,-4),(1,1)} \end{mfpic} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \pagebreak \begin{multicols}{2} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = \tan(t) \\ y = 2\sec(t) \end{array} \right. \vspace{.25in} \mbox{for } -\dfrac{\pi}{2} < t < \dfrac{\pi}{2}}$\begin{mfpic}[25]{-3}{3}{-1}{5} \axes \tlabel[cc](3,-0.5){\scriptsize$x$} \tlabel[cc](0.5,5){\scriptsize$y$} \point[3pt]{(0,2)} \xmarks{-2,-1,1,2} \ymarks{1,2,3,4} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-2 \hspace{6pt}$} -2,{$-1 \hspace{6pt}$} -1, {$1$} 1,{$2$} 2} \axislabels {y}{{$1$} 1,{$2$} 2,{$3$} 3,{$4$} 4} \normalsize \arrow \parafcn{-1,-0.75,0.1}{(tan(t),2*sec(t))} \arrow \parafcn{-0.75,1,0.1}{(tan(t),2*sec(t))} \dashed \polyline{(-0.5,-1),(2,4)} \dashed \polyline{(0.5,-1),(-2,4)} \end{mfpic} \item${\displaystyle \left\{ \begin{array}{l} x = \tan(t) \\ y = 2\sec(t) \end{array} \right. \vspace{.25in} \mbox{for } \dfrac{\pi}{2} < t < \dfrac{3\pi}{2}}$\begin{mfpic}[25]{-3}{3}{-5}{1} \axes \tlabel[cc](3,-0.5){\scriptsize$x$} \tlabel[cc](0.5,1){\scriptsize$y$} \point[3pt]{(0,-2)} \xmarks{-2,-1,1,2} \ymarks{-1,-2,-3,-4} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-2 \hspace{6pt}$} -2,{$-1 \hspace{6pt}$} -1, {$1$} 1,{$2$} 2} \axislabels {y}{{$-1$} -1,{$-2$} -2,{$-3$} -3,{$-4$} -4} \normalsize \arrow \parafcn{-1,-0.75,0.1}{(tan(t),0-2*sec(t))} \arrow \parafcn{-0.75,1,0.1}{(tan(t),0-2*sec(t))} \dashed \polyline{(-0.5,1),(2,-4)} \dashed \polyline{(0.5,1),(-2,-4)} \end{mfpic} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \raggedcolumns \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = \cos(t) \\ y = t \end{array} \right. \vspace{.25in} \mbox{for } 0 < t < \pi}$\begin{mfpic}[25]{-2.25}{2.25}{-0.5}{4} \point[3pt]{(1,0), (0,1.5708), (-1,3.1416)} \axes \tlabel[cc](2.25,-0.25){\scriptsize$x$} \tlabel[cc](0.25,4){\scriptsize$y$} \xmarks{-1,1} \ymarks{1.5708, 3.1416} \tlpointsep{4pt} \axislabels {y}{{\scriptsize$\frac{\pi}{2}$} 1.5708, {\scriptsize$\pi$} 3.1416} \axislabels {x}{{\scriptsize$-1 \hspace{7pt}$} -1, {\scriptsize$1$} 1} \arrow \parafcn{0, 0.78, 0.1}{(cos(t), t)} \arrow \parafcn{0.78, 2.36, 0.1}{(cos(t), t)} \parafcn{2.36, 3.14, 0.1}{(cos(t), t)} \end{mfpic} \item${\displaystyle \left\{ \begin{array}{l} x = \sin(t) \\ y = t \end{array} \right. \vspace{.25in} \mbox{for } -\dfrac{\pi}{2} < t < \dfrac{\pi}{2}}$\begin{mfpic}[25]{-2}{2}{-2}{2} \point[3pt]{(-1,-1.5708), (0,0), (1,1.5708)} \axes \tlabel[cc](2,-0.25){\scriptsize$x$} \tlabel[cc](0.25,2){\scriptsize$y$} \ymarks{-1.5708, 1.5708} \xmarks{-1,1} \tlpointsep{4pt} \axislabels {y}{{\scriptsize$-\frac{\pi}{2}$} -1.5708, {\scriptsize$\frac{\pi}{2}$} 1.5708} \axislabels {x}{{\scriptsize$-1 \hspace{7pt}$} -1, {\scriptsize$1$} 1} \arrow \parafcn{-1.57, -0.78, 0.1}{(sin(t),t)} \arrow \parafcn{-0.78,0.78, 0.1}{(sin(t),t)} \arrow \parafcn{0.78, 1.57, 0.1}{(sin(t),t)} \end{mfpic} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = t^{3} - 3t \\ y = t^{2} - 4 \end{array} \right. \vspace{.25in} \mbox{for } -2 \leq t \leq 2}$\begin{mfpic}[15]{-3}{3}{-5}{1} \axes \tlabel[cc](3,-0.5){\scriptsize$x$} \tlabel[cc](0.5,1){\scriptsize$y$} \point[3pt]{(-2,0), (2,0), (0,-1), (0,-4)} \xmarks{-2,-1,1,2} \ymarks{-4,-3,-2,-1} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1,{$1$} 1, {$2$} 2} \axislabels {y}{{$-4$} -4, {$-3$} -3,{$-2$} -2,{$-1$} -1} \normalsize \arrow \parafcn{-2,-1,0.1}{(t**3 - 3*t,t**2 - 4)} \arrow \parafcn{-1,1,0.1}{(t**3 - 3*t,t**2 - 4)} \parafcn{1,2,0.1}{(t**3 - 3*t,t**2 - 4)} \end{mfpic} \vspace{1in} \item${\displaystyle \left\{ \begin{array}{l} x = 4\cos^{3}(t) \\ y = 4\sin^{3}(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 2\pi}$\begin{mfpic}[12]{-5}{5}{-5}{5} \axes \tlabel[cc](5,-0.5){\scriptsize$x$} \tlabel[cc](0.5,5){\scriptsize$y$} \point[3pt]{(-4,0), (0,4), (4,0), (0,-4)} \xmarks{-4,-3,-2,-1,1,2,3,4} \ymarks{-4,-3,-2,-1,1,2,3,4} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-4 \hspace{6pt}$} -4, {$-3 \hspace{6pt}$} -3,{$-2 \hspace{6pt}$} -2,{$-1 \hspace{6pt}$} -1,{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4} \axislabels {y}{{$-4$} -4,{$-3$} -3,{$-2$} -2,{$-1$} -1,{$1$} 1,{$2$} 2,{$3$} 3, {$4$} 4} \normalsize \arrow \parafcn{0,0.78,0.1}{(4*((cos(t))**3),4*((sin(t))**3))} \arrow \parafcn{0.78, 2.36,0.1}{(4*((cos(t))**3),4*((sin(t))**3))} \arrow \parafcn{2.36, 3.93 ,0.1}{(4*((cos(t))**3),4*((sin(t))**3))} \arrow \parafcn{3.93, 5.5 ,0.1}{(4*((cos(t))**3),4*((sin(t))**3))} \parafcn{5.5, 6.28 ,0.1}{(4*((cos(t))**3),4*((sin(t))**3))} \end{mfpic} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \pagebreak \begin{multicols}{2} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = e^{t} + e^{-t} \\ y = e^{t} - e^{-t} \end{array} \right. \vspace{.25in} \mbox{for } -2 \leq t \leq 2}$\begin{mfpic}[15][8.5]{-1}{8}{-8}{8} \axes \tlabel[cc](8,-0.5){\scriptsize$x$} \tlabel[cc](0.5,8){\scriptsize$y$} \point{(2,0), (7.52, 7.25), (7.52, -7.25)} \xmarks{1,2,3,4,5,6,7} \ymarks{-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6,7} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$1$} 1, {$2$} 2, {$3$} 3,{$4$} 4,{$5$} 5,{$6$} 6, {$7$} 7} \axislabels {y}{{$-7$} -7,{$-5$} -5,{$-3$} -3,{$-1$} -1, {$1$} 1, {$3$} 3, {$5$} 5, {$7$} 7} \normalsize \arrow \parafcn{-2,-1.5,0.1}{(exp(t) + exp(-t),exp(t) - exp(-t))} \arrow \parafcn{-1.5,1.5,0.1}{(exp(t) + exp(-t),exp(t) - exp(-t))} \parafcn{1.5,2,0.1}{(exp(t) + exp(-t),exp(t) - exp(-t))} \end{mfpic} \item${\displaystyle \left\{ \begin{array}{l} x = \cos(3t) \\ y = \sin(4t) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 2\pi}$\begin{mfpic}[15]{-5}{5}{-5}{5} \axes \tlabel[cc](5,-0.5){\scriptsize$x$} \tlabel[cc](0.5,5){\scriptsize$y$} \xmarks{-4,4} \ymarks{-4,4} \tlpointsep{4pt} \scriptsize \axislabels {x}{{$-1 \hspace{6pt}$} -4, {$1$} 4} \axislabels {y}{{$-1$} -4, {$1$} 4} \normalsize \arrow \parafcn{0,22.5,5}{(4*cosd(3*t),4*sind(4*t))} \arrow \parafcn{22.5,67.5,5}{(4*cosd(3*t),4*sind(4*t))} \arrow \parafcn{67.5,112.5,5}{(4*cosd(3*t),4*sind(4*t))} \arrow \parafcn{112.5,157.5,5}{(4*cosd(3*t),4*sind(4*t))} \arrow \parafcn{157.5,202.5,5}{(4*cosd(3*t),4*sind(4*t))} \arrow \parafcn{202.5,247.5,5}{(4*cosd(3*t),4*sind(4*t))} \arrow \parafcn{247.5,292.5,5}{(4*cosd(3*t),4*sind(4*t))} \arrow \parafcn{292.5,337.5,5}{(4*cosd(3*t),4*sind(4*t))} \parafcn{337.5, 360,5}{(4*cosd(3*t),4*sind(4*t))} \end{mfpic} \setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = 3-5t \\ y =-5+7t \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 1}$\item${\displaystyle \left\{ \begin{array}{l} x = 5t-2 \\ y =-1-3t \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 1}$\setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = t \\ y = 4-t^2 \end{array} \right. \vspace{.25in} \mbox{for } -2 \leq t \leq 2}$\item${\displaystyle \left\{ \begin{array}{l} x = t-2 \\ y = 4t-t^2 \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 4}$\setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = t^2-9 \\ y = t \end{array} \right. \vspace{.25in} \mbox{for } -2 \leq t \leq 3}$\item${\displaystyle \left\{ \begin{array}{l} x = t^2-6t \\ y = 3-t \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t \leq 5}$\setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = 5\cos(t) \\ y = 5\sin(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t < 2\pi}$\item${\displaystyle \left\{ \begin{array}{l} x = 1+2\cos(t) \\ y = 2\sin(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t < 2\pi}$\setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = 3\cos(t) \\ y = 3 + 3\sin(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t < 2\pi}$\item${\displaystyle \left\{ \begin{array}{l} x = 3\cos(t) \\ y = 3 - 3\sin(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t < 2\pi }$\setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{2} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = 3+\sqrt{117} \, \cos(t) \\ y = -1 + \sqrt{117} \, \sin(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t < 2\pi}$\item${\displaystyle \left\{ \begin{array}{l} x = 1+3\cos(t) \\ y = \sin(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t < 2\pi}$\setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item${\displaystyle \left\{ \begin{array}{l} x = 2\cos(t) \\ y = 3\sin(t)-3 \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t < 2\pi }$\item${\displaystyle \left\{ \begin{array}{l} x = 2\cos\left(t-\dfrac{\pi}{2}\right) = 2\sin(t) \\ y = -3 - 3\sin\left(t-\dfrac{\pi}{2}\right) = -3+3\cos(t) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq t < 2\pi}$\item$\left\{ x(t), \, y(t) \right.$where: $\begin{array}{cc} x(t) = \left\{ \begin{array}{rr} 3t,& 0 \leq t \leq 1 \\ 6-3t, & 1 \leq t \leq 2 \\ 0, & 2 \leq t \leq 3 \\ \end{array} \right. & y(t) = \left\{ \begin{array}{rr} 0,& 0 \leq t \leq 1 \\ 4t-4, & 1 \leq t \leq 2 \\ 12-4t, & 2 \leq t \leq 3 \\ \end{array} \right. \end{array}$ \setcounter{HW}{\value{enumi}} \end{enumerate} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item The parametric equations for the inverse are${\displaystyle \left\{ \begin{array}{l} x = t^3+3t-4 \\ y = t \end{array} \right. \vspace{.25in} \mbox{for } -\infty < t < \infty}$\item$r = 6\cos(2\theta)$translates to${\displaystyle \left\{ \begin{array}{l} x = 6\cos(2\theta)\cos(\theta) \\ y = 6\cos(2\theta)\sin(\theta) \end{array} \right. \vspace{.25in} \mbox{for } 0 \leq \theta < 2\pi}$. \item The parametric equations which describe the locations of passengers on the London Eye are${\displaystyle \left\{ \begin{array}{l} x = 67.5 \cos\left(\frac{\pi}{15} t - \frac{\pi}{2} \right) = 67.5 \sin\left(\frac{\pi}{15} t \right) \\ y = 67.5 \sin\left(\frac{\pi}{15} t - \frac{\pi}{2} \right) + 67.5 = 67.5 - 67.5 \cos\left(\frac{\pi}{15} t \right) \end{array} \right. \vspace{.25in} \mbox{for } -\infty < t < \infty}$\setcounter{HW}{\value{enumi}} \end{enumerate} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item The parametric equations for the hammer throw are${\displaystyle \left\{ \begin{array}{l} x = 33 \cos(42^{\circ}) t \\ y =-16t^2 + 33 \sin(42^{\circ}) t + 6 \end{array} \right.}$for$t \geq 0$. To find when the hammer hits the ground, we solve$y(t) = 0$and get$t \approx -0.23$or$1.61$. Since$t \geq 0$, the hammer hits the ground after approximately$t = 1.61$seconds after it was launched into the air. To find how far away the hammer hits the ground, we find$x(1.61) \approx 39.48$feet from where it was thrown into the air. \addtocounter{enumi}{1} \item We solve$y = \dfrac{v_{\text{\tiny$0$}}^2 \sin^{2}(\theta)}{2g} + s_{\text{\tiny $0$}} = \dfrac{v_{\text{\tiny$0$}}^2 \sin^{2}(85^{\circ})}{2(32)} + 5 = 31.5$to get$v_{\text{\tiny$0$}} = \pm 41.34$. The initial speed of the sheaf was approximately$41.34\$ feet per second.

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