#set document(title: "10.2 Polar Graphs", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 10.2#h(0.6em)Polar Graphs === Graphing in Polar Coordinates When we plot points in Cartesian coordinates, we start at the origin and move a distance right or left given by the #math.equation(block: false, alt: "x")[$x$]-coordinate of the point, then move up or down according to the #math.equation(block: false, alt: "y")[$y$]-coordinate. When we sketch the graph of an equation or function, we think of drawing the graph from left to right, with the "height" of the graph at each #math.equation(block: false, alt: "x")[$x$]-value given by the function, as shown in figure (a). #figure(figph[Cartesian and polar graphs], alt: "Cartesian and polar graphs", caption: none) In polar coordinates, however, the dependent variable, #math.equation(block: false, alt: "r")[$r$], gives not a height but a distance from the pole in direction #math.equation(block: false, alt: "θ")[$θ$], as shown in figure (b). When graphing an equation in polar coordinates, we think of sweeping around the pole in the counterclockwise direction, and at each angle #math.equation(block: false, alt: "θ")[$θ$] the #math.equation(block: false, alt: "r")[$r$]-value tells us how far the graph is from the pole. #examplebox("Example 1")[][ Graph the polar equation #math.equation(block: false, alt: "r equals 2 sin ⁡ open parenthesis θ close parenthesis")[$r = 2 sin ( θ )$]. #solutionbox[ We make a table of values, choosing the special values for #math.equation(block: false, alt: "θ")[$θ$]. For each value of #math.equation(block: false, alt: "θ")[$θ$], we evaluate #math.equation(block: false, alt: "r equals 2 sin ⁡ open parenthesis θ close parenthesis")[$r = 2 sin ( θ )$]. #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$]], [#math.equation(block: false, alt: "π")[$π$]]), [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$]], [#math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$]], [#math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], )) First we'll plot the points in the first quadrant. Observe that as #math.equation(block: false, alt: "θ")[$θ$] increases from #math.equation(block: false, alt: "0")[$0$] to #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$], #math.equation(block: false, alt: "r")[$r$] increases from #math.equation(block: false, alt: "0")[$0$] to #math.equation(block: false, alt: "2")[$2$]. Starting at the pole, we connect the points in order of increasing #math.equation(block: false, alt: "θ")[$θ$]. Imagine a radial line sweeping around the graph through the first quadrant: as the angle increases, the length of the segment increases, so that its tip traces out the graph shown in figure (a). #figure(figph[polar graphs], alt: "polar graphs", caption: none) Now continue plotting the points in the table as #math.equation(block: false, alt: "θ")[$θ$] increases from #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$] to #math.equation(block: false, alt: "π")[$π$]. In the second quadrant, #math.equation(block: false, alt: "r")[$r$] decreases as #math.equation(block: false, alt: "θ")[$θ$] increases, as shown in figure (b). The graph we obtain is, in fact, a circle, which we will prove algebraically shortly. However, we have not yet plotted points for #math.equation(block: false, alt: "θ")[$θ$] between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$]. Because #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis")[$sin ( θ )$] is negative in the third and fourth quadrants, all the #math.equation(block: false, alt: "r")[$r$]-values for these angles are negative. #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 7 π over 6")[$display(frac(7 π, 6))$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 4 π over 3")[$display(frac(4 π, 3))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 5 π over 3")[$display(frac(5 π, 3))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "the fraction 11 π over 6")[$display(frac(11 π, 6))$]], [#math.equation(block: false, alt: "π")[$π$]]), [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus the square root of 2")[$− sqrt(2)$]], [#math.equation(block: false, alt: "minus the square root of 3")[$− sqrt(3)$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus the square root of 3")[$− sqrt(3)$]], [#math.equation(block: false, alt: "minus the square root of 2")[$− sqrt(2)$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], )) When we plot the points in this table, we see that the original graph is traced out again. For example, the point #math.equation(block: false, alt: "open parenthesis the fraction 7 π over 6 , minus 1 close parenthesis")[$( display(frac(7 π, 6)) , − 1 )$] is the same as the point #math.equation(block: false, alt: "open parenthesis the fraction π over 6 , 1 close parenthesis")[$( display(frac(π, 6)) , 1 )$], the point #math.equation(block: false, alt: "open parenthesis the fraction 5 π over 4 , minus the square root of 2 close parenthesis")[$( display(frac(5 π, 4)) , − sqrt(2) )$] is the same as the point #math.equation(block: false, alt: "open parenthesis the fraction π over 4 , the square root of 2 close parenthesis")[$( display(frac(π, 4)) , sqrt(2) )$], and so on, around the circle. Thus, the graph of #math.equation(block: false, alt: "r equals 2 sin ⁡ θ")[$r = 2 sin θ$] is a circle, traced twice for #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 2 π")[$0 ≤ θ ≤ 2 π$]. ] ] To prove that the graph in the previous Example is really a circle, we convert the equation #math.equation(block: false, alt: "r equals 2 sin ⁡ open parenthesis θ close parenthesis")[$" " r = 2 sin ( θ ) " "$] to Cartesian form. First, multiply both sides by #math.equation(block: false, alt: "r")[$r$] to obtain #math.equation(block: true, alt: "r squared equals 2 r sin ⁡ open parenthesis θ close parenthesis")[$r^(2) = 2 r sin ( θ )$] Next, replace #math.equation(block: false, alt: "r squared")[$r^(2)$] by #math.equation(block: false, alt: "x squared plus y squared")[$x^(2) + y^(2)$] and #math.equation(block: false, alt: "r sin ⁡ open parenthesis θ close parenthesis")[$r sin ( θ )$] by #math.equation(block: false, alt: "y")[$y$], to get #math.equation(block: true, alt: "x squared plus y squared equals 2 y")[$x^(2) + y^(2) = 2 y$] This equation is quadratic in two variables, so its graph is a conic section. We put the equation in standard form by completing the square in #math.equation(block: false, alt: "y")[$y$]. #math.equation(block: true, alt: "x squared plus y squared minus 2 y, equals 0, Add 1 to both sides.; x squared plus open parenthesis y squared minus 2 y plus 1 close parenthesis, equals 1, Write the equation in standard form.; open parenthesis x minus 0 close parenthesis squared plus open parenthesis y minus 1 close parenthesis squared, equals 1")[$x^(2) + y^(2) − 2 y & = 0 & & "Add 1 to both sides." \ x^(2) + ( y^(2) − 2 y " " + 1 ) & = 1 & & "Write the equation in standard form." \ ( x − 0 )^(2) + ( y − 1 )^(2) & = 1$] We have the equation of a circle with center #math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$] and radius #math.equation(block: false, alt: "1")[$1$]. Graph the polar equation #math.equation(block: false, alt: "r equals 4 cos ⁡ open parenthesis θ close parenthesis")[$" " r = 4 cos ( θ )$]. #figure(figph[polar graph], alt: "polar graph", caption: none) === Using a Graphing Calculator You are familiar with the graphs of many equations in Cartesian coordinates, including lines, parabolas and other conic sections, and the graphs of basic functions. You should now become familiar with some standard graphs in polar coordinates. These include circles and roses, cardioids and limaçons, lemniscates, and spirals. At the end of this section you will find a Catalog of the basic polar graphs and their properties. You can use your calculator, set in Polar mode, to experiment with these graphs. #examplebox("Example 2")[][ Graph the Archimedean spiral #math.equation(block: false, alt: "r equals 2 θ")[$" " r = 2 θ$], for #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 6 π")[$" " 0 ≤ θ ≤ 6 π$]. #solutionbox[ After setting the calculator in Polar mode, we enter the equation #math.equation(block: false, alt: "r equals 2 θ")[$r = 2 θ$] and enter the window settings #math.equation(block: true, alt: "θ min, equals 0 θ max equals 6 π θ step equals 0.1; Xmin, equals minus 25 Xmax equals 25 Xscl equals 5; Ymin, equals minus 25 Ymax equals 25 Yscl equals 5")[$θ "min" & = 0 " " " " " " " " " " " " " " " " " " " " θ "max" = 6 π " " " " " " " " θ "step" = 0.1 \ "Xmin" & = − 25 " " " " " " " " "Xmax" = 25 " " " " " " " " "Xscl" = 5 \ "Ymin" & = − 25 " " " " " " " " "Ymax" = 25 " " " " " " " " "Yscl" = 5$] We then press Zoom 5 to set a square window. The graph is shown at right. #figure(figph[polar graph], alt: "polar graph", caption: none) Studying a table of values can help us understand the shape of the graph. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0.25")[$0.25$]], [#math.equation(block: false, alt: "0.50")[$0.50$]], [#math.equation(block: false, alt: "0.75")[$0.75$]], [#math.equation(block: false, alt: "1.00")[$1.00$]], [#math.equation(block: false, alt: "1.25")[$1.25$]]), [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "0.50")[$0.50$]], [#math.equation(block: false, alt: "1.00")[$1.00$]], [#math.equation(block: false, alt: "1.50")[$1.50$]], [#math.equation(block: false, alt: "2.00")[$2.00$]], [#math.equation(block: false, alt: "2.50")[$2.50$]], )) As #math.equation(block: false, alt: "θ")[$θ$] increases, #math.equation(block: false, alt: "r")[$r$] increases also, at a constant rate. We wind our way around the pole, steadily increasing our distance from the pole as we go. We spiral outwards, tracing the graph shown above. ] ] Use a calculator to graph the polar equation #math.equation(block: false, alt: "r equals e to the power 0.1 θ")[$" " r = e^(0.1 θ)$]. Use the window settings #math.equation(block: true, alt: "θ min, equals 0 θ max equals 12 π θ step equals 0.5; Xmin, equals minus 45 Xmax equals 45 Xscl equals 10; Ymin, equals minus 35 Ymax equals 35 Yscl equals 10")[$θ "min" & = 0 " " " " " " " " " " " " " " " " " " " " θ "max" = 12 π " " " " " " " " " " θ "step" = 0.5 \ "Xmin" & = − 45 " " " " " " " " "Xmax" = 45 " " " " " " " " " " " " " " "Xscl" = 10 \ "Ymin" & = − 35 " " " " " " " " "Ymax" = 35 " " " " " " " " " " " " " " " " "Yscl" = 10$] #figure(figph[polar graph spiral], alt: "polar graph spiral", caption: none) It is important to connect the points on the graph in order of increasing #math.equation(block: false, alt: "θ")[$θ$]. #examplebox("Example 3")[][ Graph the polar equation #math.equation(block: false, alt: "r equals 2 sin ⁡ open parenthesis 3 θ close parenthesis")[$" " r = 2 sin ( 3 θ )$]. #solutionbox[ We'll graph the equation in stages in order to see how the graph is traced out. Begin with the window settings #math.equation(block: true, alt: "θ min, equals 0 θ max equals the fraction π over 3 θ step equals 0.5; Xmin, equals minus 4.5 Xmax equals 4.5 Xscl equals 1; Ymin, equals minus 3 Ymax equals 3 Yscl equals 1")[$θ "min" & = 0 " " " " " " " " " " " " " " " " " " " " " " θ "max" = display(frac(π, 3)) " " " " " " " " " " " " θ "step" = 0.5 \ "Xmin" & = − 4.5 " " " " " " " " "Xmax" = 4.5 " " " " " " " " " " " " "Xscl" = 1 \ "Ymin" & = − 3 " " " " " " " " " " " " " " "Ymax" = 3 " " " " " " " " " " " " " " " " " " " " " " "Yscl" = 1$] Watch as your calculator produces the graph shown in figure (a). #figure(figph[one loop], alt: "one loop", caption: none) #figure(figph[two loops], alt: "two loops", caption: none) #figure(figph[three loops], alt: "three loops", caption: none) Observe that, as #math.equation(block: false, alt: "θ")[$θ$] increases from #math.equation(block: false, alt: "0")[$0$] to #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$], #math.equation(block: false, alt: "r")[$r$] first increases, reaching its maximum value of 2 at #math.equation(block: false, alt: "θ equals the fraction π over 6")[$θ = display(frac(π, 6))$] and then decreases back to 0. You can verify the values in the table below, which shows the points at multiples of #math.equation(block: false, alt: "the fraction π over 18")[$display(frac(π, 18))$]. These points create the first loop of the graph. #figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 18")[$display(frac(π, 18))$]], [#math.equation(block: false, alt: "the fraction π over 9")[$display(frac(π, 9))$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction 2 π over 9")[$display(frac(2 π, 9))$]], [#math.equation(block: false, alt: "the fraction 5 π over 18")[$display(frac(5 π, 18))$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]]), [#math.equation(block: false, alt: "3 θ")[$3 θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "the square root of 3")[$sqrt(3)$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], )) Next, change #math.equation(block: false, alt: "θ max")[$θ "max"$] to #math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$], and graph again. This time the calculator traces out two loops, as shown in figure (b). For #math.equation(block: false, alt: "θ")[$θ$] between #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] and #math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$], #math.equation(block: false, alt: "r")[$r$] is negative, so the second loop lies in the third and fourth quadrants. #figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction 7 π over 18")[$display(frac(7 π, 18))$]], [#math.equation(block: false, alt: "the fraction 4 π over 9")[$display(frac(4 π, 9))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 5 π over 9")[$display(frac(5 π, 9))$]], [#math.equation(block: false, alt: "the fraction 11 π over 18")[$display(frac(11 π, 18))$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]]), [#math.equation(block: false, alt: "3 θ")[$3 θ$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 7 π over 6")[$display(frac(7 π, 6))$]], [#math.equation(block: false, alt: "the fraction 4 π over 3")[$display(frac(4 π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 5 π over 3")[$display(frac(5 π, 3))$]], [#math.equation(block: false, alt: "the fraction 11 π over 6")[$display(frac(11 π, 6))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus the square root of 3")[$− sqrt(3)$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus the square root of 3")[$− sqrt(3)$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], )) Finally, change #math.equation(block: false, alt: "θ max")[$θ "max"$] to #math.equation(block: false, alt: "π")[$π$], and graph again. For #math.equation(block: false, alt: "θ")[$θ$] between #math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$] and #math.equation(block: false, alt: "π")[$π$], the graph traces out a third loop, as shown in figure (c). For #math.equation(block: false, alt: "θ")[$θ$] between #math.equation(block: false, alt: "π")[$π$] and #math.equation(block: false, alt: "2 π")[$2 π$], the entire graph is traced a second time. The finished graph is a rose with 3 petals of length 2. ] ] + Use your calculator to graph the polar equation #math.equation(block: false, alt: "r equals 2 cos ⁡ open parenthesis 3 θ close parenthesis")[$" " r = 2 cos ( 3 θ )$]. + Complete the table of values for the function.#figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$]], [#math.equation(block: false, alt: "π")[$π$]]), [#math.equation(block: false, alt: "r")[$r$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + Sketch the graph by hand on the grid at right. #figure(figph[polar grid], alt: "polar grid", caption: none) #figure(figph[polar graph: three petals], alt: "polar graph: three petals", caption: none) === Sketching Familiar Equations You should also be able to sketch the standard polar graphs by hand. Once you recognize an equation as a particular type of graph, say a rose or a limaçon, you can sketch the graph quickly by finding just a few well-chosen guidepoints. The next example demonstrates a technique for sketching a rose. The rose family r = a cos(nθ), live. The slider a sets the petal length — each petal reaches out to distance a from the pole. The slider n sets the petal count, and it follows the section's rule exactly: an odd n gives n petals, an even n gives 2n. Step n from 3 to 4 and watch the count jump from three petals to eight — with n odd the curve traces each petal twice as θ runs from 0 to 2π, so half the petals coincide, while with n even all 2n petals are distinct. At n = 1 the “rose” has a single petal: a circle of diameter a through the pole, the standard circle graph from earlier in this section.{"functions":\[{"color":"\#1f4e79","expression":"a\*cos(n\*theta)","expression2":"","id":"rose","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"polar","tMax":6.2832,"tMin":0,"variable":"x","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-a","label":"Petal length a","max":3,"min":1,"name":"a","step":0.1,"value":2},{"id":"p-n","label":"Petal number n","max":6,"min":1,"name":"n","step":1,"value":4}\],"title":"Polar rose r = a cos(n\\u03b8)","version":1,"viewport":{"centerX":0,"centerY":0,"scale":76}}#examplebox("Example 4")[][ Graph the polar equation #math.equation(block: false, alt: "r equals 3 sin ⁡ open parenthesis 2 θ close parenthesis")[$" " r = 3 sin ( 2 θ )$]. #solutionbox[ Referring to the Catalog of Polar Graphs, we see that the graph of this equation is a #strong[rose], with petal length #math.equation(block: false, alt: "a equals 3")[$a = 3$] and four petals, because #math.equation(block: false, alt: "2 n equals 4")[$2 n = 4$]. If we can locate the tips of the petals, we can use them as guidepoints to sketch the graph. Now, the points at the tips of the petals have #math.equation(block: false, alt: "r equals 3")[$r = 3$], so we substitute #math.equation(block: false, alt: "r equals 3")[$r = 3$] into the equation of the rose to find the values of #math.equation(block: false, alt: "θ")[$θ$] at those points. #math.equation(block: true, alt: "3, equals 3 sin ⁡ open parenthesis 2 θ close parenthesis, Divide both sides by 3.; 1, equals sin ⁡ open parenthesis 2 θ close parenthesis, Apply the arcsine function to both sides.; the fraction π over 2, equals 2 θ, Divide both sides by 2.; the fraction π over 4, equals θ")[$3 & = 3 sin ( 2 θ ) & & "Divide both sides by " 3. \ 1 & = sin ( 2 θ ) & & "Apply the arcsine function to both sides." \ display(frac(π, 2)) & = 2 θ & & "Divide both sides by " 2. \ display(frac(π, 4)) & = θ$] Thus, one of the petal tips is located at #math.equation(block: false, alt: "θ equals the fraction π over 4")[$θ = display(frac(π, 4))$]. Because the 4 petals are evenly spaced around the pole, the angle between the petals is #math.equation(block: false, alt: "the fraction 2 π over 4 equals the fraction π over 2")[$display(frac(2 π, 4)) = display(frac(π, 2))$], and the other petal tips occur at #math.equation(block: true, alt: "θ, equals the fraction π over 4 plus the fraction π over 2 equals the fraction 3 π over 4; θ, equals the fraction π over 4 plus 2 open parenthesis the fraction π over 2 close parenthesis equals the fraction 5 π over 4; θ, equals the fraction π over 4 plus 3 open parenthesis the fraction π over 2 close parenthesis equals the fraction 7 π over 4")[$θ & = display(frac(π, 4)) + display(frac(π, 2)) = display(frac(3 π, 4)) \ θ & = display(frac(π, 4)) + 2 ( display(frac(π, 2)) ) = display(frac(5 π, 4)) \ θ & = display(frac(π, 4)) + 3 ( display(frac(π, 2)) ) = display(frac(7 π, 4))$] We plot the tips of the petals as guidepoints, at #math.equation(block: false, alt: "open parenthesis 3 , the fraction π over 4 close parenthesis")[$( 3 , display(frac(π, 4)) )$],#math.equation(block: false, alt: "open parenthesis 3 , the fraction 3 π over 4 close parenthesis")[$" " ( 3 , display(frac(3 π, 4)) )$], #math.equation(block: false, alt: "open parenthesis 3 , the fraction 5 π over 4 close parenthesis")[$" " ( 3 , display(frac(5 π, 4)) )$]and #math.equation(block: false, alt: "open parenthesis 3 , the fraction 7 π over 4 close parenthesis")[$( 3 , display(frac(7 π, 4)) )$]. Now we can sketch a rose with 4 petals of length 3, as shown at right. #figure(figph[rose], alt: "rose", caption: none) ] ] Graph the polar equation #math.equation(block: false, alt: "r equals 4 cos ⁡ open parenthesis 5 θ close parenthesis")[$" " r = 4 cos ( 5 θ )$]. #figure(figph[rose], alt: "rose", caption: none) The #strong[limaçons], #math.equation(block: false, alt: "r equals a plus or minus sin ⁡ open parenthesis θ close parenthesis")[$" " r = a ± sin ( θ ) " "$] and #math.equation(block: false, alt: "r equals a plus or minus cos ⁡ open parenthesis θ close parenthesis")[$" " r = a ± cos ( θ ) " "$], are another family of polar graphs. In particular, the #strong[cardioid] is a special case of a limaçon with #math.equation(block: false, alt: "a equals b")[$a = b$]. #examplebox("Example 5")[][ Graph the polar equation #math.equation(block: false, alt: "r equals 3 plus 2 cos ⁡ open parenthesis θ close parenthesis")[$" " r = 3 + 2 cos ( θ )$]. #solutionbox[ This is the equation for a limaçon, with #math.equation(block: false, alt: "a equals 3")[$a = 3$] and #math.equation(block: false, alt: "b equals 2")[$b = 2$]. Because #math.equation(block: false, alt: "a greater than b")[$a > b$], the limaçon will have a dent, like a lima bean, rather than a loop. As guidepoints, we locate the points at the four quadrantal angles. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "r equals 3 plus 2 cos ⁡ open parenthesis θ close parenthesis")[$r = 3 + 2 cos ( θ )$]]), [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "3")[$3$]], )) #figure(figph[limacon], alt: "limacon", caption: none) We plot the guidepoints and connect them in order with a smooth curve, as shown above. Note that this limaçon involving cosine is symmetric about the #math.equation(block: false, alt: "x")[$x$]-axis; limaçons involving sine are symmetric about the #math.equation(block: false, alt: "y")[$y$]-axis. ] ] Graph the polar equation #math.equation(block: false, alt: "r equals 1 minus 3 sin ⁡ open parenthesis θ close parenthesis")[$" " r = 1 − 3 sin ( θ )$]. #figure(figph[limacon], alt: "limacon", caption: none) You should also be able to identify a polar graph and write its equation. #examplebox("Example 6")[][ Give a polar equation for each of the graphs below. #figure(figph[circle], alt: "circle", caption: none) #figure(figph[cardioid], alt: "cardioid", caption: none) #solutionbox[ + The graph is a circle centered at #math.equation(block: false, alt: "open parenthesis minus 2 , 0 close parenthesis")[$( − 2 , 0 )$] and with radius 2, so we choose the equation #math.equation(block: false, alt: "r equals 2 a cos ⁡ open parenthesis θ close parenthesis")[$" " r = 2 a cos ( θ )$], with #math.equation(block: false, alt: "a equals minus 2")[$a = − 2$]. Thus, #math.equation(block: false, alt: "r equals minus 4 cos ⁡ open parenthesis θ close parenthesis")[$" " r = − 4 cos ( θ )$]. + The graph is a cardioid with its axis of symmetry on the #math.equation(block: false, alt: "x")[$x$]-axis, and the "bottom" of the heart points in the positive#math.equation(block: false, alt: "x")[$x$]direction, so its equation has the form #math.equation(block: false, alt: "r equals a plus a cos ⁡ open parenthesis θ close parenthesis")[$" " r = a + a cos ( θ )$]. At #math.equation(block: false, alt: "θ equals 0 , r equals 2")[$θ = 0 , " " r = 2$], so we can solve for #math.equation(block: false, alt: "a")[$a$]: #math.equation(block: true, alt: "2, equals a plus a cos ⁡ open parenthesis 0 close parenthesis; 2, equals a plus a open parenthesis 1 close parenthesis equals 2 a")[$2 & = a + a cos ( 0 ) \ 2 & = a + a ( 1 ) = 2 a$] Thus, #math.equation(block: false, alt: "a equals 1")[$a = 1$] and we choose the equation #math.equation(block: false, alt: "r equals 1 plus cos ⁡ open parenthesis θ close parenthesis")[$" " r = 1 + cos ( θ )$]. ] ] Give a polar equation for each of the graphs below. #figure(figph[lemniscate], alt: "lemniscate", caption: none) #figure(figph[cardioid], alt: "cardioid", caption: none) + #math.equation(block: false, alt: "r squared equals 4 sin ⁡ open parenthesis 2 θ close parenthesis")[$r^(2) = 4 sin ( 2 θ )$] + #math.equation(block: false, alt: "r equals 1 minus sin ⁡ open parenthesis θ close parenthesis")[$r = 1 − sin ( θ )$] === Finding Intersection Points To find the intersection points of two graphs, we solve the system made up of their equations. If the equations are #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] and #math.equation(block: false, alt: "y equals g open parenthesis x close parenthesis")[$y = g ( x )$], we simply solve the equation #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals g open parenthesis x close parenthesis")[$f ( x ) = g ( x )$]. For example, we find the intersection points of #math.equation(block: false, alt: "y equals x squared")[$" " y = x^(2) " "$] and #math.equation(block: false, alt: "y equals x plus 2")[$" " y = x + 2 " "$] by solving the equation #math.equation(block: false, alt: "x squared minus x minus 2 equals 0")[$x^(2) − x − 2 = 0$] to get #math.equation(block: true, alt: "x squared minus x minus 2, equals 0; open parenthesis x minus 2 close parenthesis open parenthesis x plus 1 close parenthesis, equals 0; x minus 2 equals 0 , x plus 1, equals 0; x equals 2 , x, equals minus 1")[$x^(2) − x − 2 & = 0 \ ( x − 2 ) ( x + 1 ) & = 0 \ x − 2 = 0 , " " " " " " x + 1 & = 0 \ x = 2 , " " " " " " " " " " " " x & = − 1$] These are the #math.equation(block: false, alt: "x")[$x$]-coordinates of the intersection points, and we can find the #math.equation(block: false, alt: "y")[$y$]-coordinates by substituting these values into either equation. - For #math.equation(block: false, alt: "x equals 2 ,")[$x = 2 ,$] we find #math.equation(block: false, alt: "y equals 2 squared equals 4.")[$y = 2^(2) = 4.$] - For #math.equation(block: false, alt: "x equals minus 1 ,")[$x = − 1 ,$] we find #math.equation(block: false, alt: "y equals open parenthesis minus 1 close parenthesis squared equals 1.")[$y = ( − 1 )^(2) = 1.$] Thus, the intersection points are #math.equation(block: false, alt: "open parenthesis 2 , 4 close parenthesis")[$( 2 , 4 )$] and #math.equation(block: false, alt: "open parenthesis minus 1 , 1 close parenthesis")[$( − 1 , 1 )$], as shown at right. #figure(figph[line intersecting parabola], alt: "line intersecting parabola", caption: none) To find the intersection points of the polar graphs #math.equation(block: false, alt: "r equals f open parenthesis θ close parenthesis")[$r = f ( θ )$] and #math.equation(block: false, alt: "r equals g open parenthesis θ close parenthesis")[$r = g ( θ )$] we solve the equation #math.equation(block: false, alt: "f open parenthesis θ close parenthesis equals g open parenthesis θ close parenthesis")[$f ( θ ) = g ( θ )$]. #examplebox("Example 7")[][ Find all intersection points of the graphs of #math.equation(block: false, alt: "r equals 2 plus 2 sin ⁡ open parenthesis θ close parenthesis")[$" " " " r = 2 + 2 sin ( θ ) " " " "$] and #math.equation(block: false, alt: "r equals 2 plus 2 cos ⁡ open parenthesis θ close parenthesis")[$" " " " r = 2 + 2 cos ( θ )$], for #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 2 π")[$" " 0 ≤ θ ≤ 2 π$]. #solutionbox[ We equate the two expressions for #math.equation(block: false, alt: "r")[$r$] and solve for #math.equation(block: false, alt: "θ")[$θ$]. #math.equation(block: true, alt: "r equals 2 plus 2 sin ⁡ open parenthesis θ close parenthesis, equals r equals 2 plus 2 cos ⁡ open parenthesis θ close parenthesis; sin ⁡ open parenthesis θ close parenthesis, equals cos ⁡ open parenthesis θ close parenthesis, Divide both sides by cos ⁡ open parenthesis θ close parenthesis .; tan ⁡ open parenthesis θ close parenthesis, equals 1, Solve for θ .; θ, equals the fraction π over 4 , the fraction 5 π over 4")[$r = 2 + 2 sin ( θ ) & = r = 2 + 2 cos ( θ ) \ sin ( θ ) & = cos ( θ ) & & "Divide both sides by " bold(cos) ( bold(italic(θ)) ) . \ tan ( θ ) & = 1 & & "Solve for " bold(italic(θ)) . \ θ & = display(frac(π, 4)) , " " display(frac(5 π, 4))$] We evaluate either expression for #math.equation(block: false, alt: "r")[$r$] to find the other coordinate of the intersection point. #math.equation(block: true, alt: "when θ equals the fraction π over 4 , r equals 2 plus 2 sin ⁡ open parenthesis the fraction π over 4 close parenthesis equals 2 plus open parenthesis the fraction 1 over the square root of 2 close parenthesis equals 2 plus the square root of 2; when θ equals the fraction 5 π over 4 , r equals 2 plus 2 sin ⁡ open parenthesis the fraction 5 π over 4 close parenthesis equals 2 plus open parenthesis the fraction minus 1 over the square root of 2 close parenthesis equals 2 minus the square root of 2")[$"when" " " " " " " " " θ = display(frac(π, 4)) , " " " " " " " " r = 2 + 2 sin ( display(frac(π, 4)) ) = 2 + ( display(frac(1, sqrt(2))) ) = 2 + sqrt(2) \ "when" " " " " " " " " θ = display(frac(5 π, 4)) , " " " " " " " " r = 2 + 2 sin ( display(frac(5 π, 4)) ) = 2 + ( display(frac(− 1, sqrt(2))) ) = 2 − sqrt(2)$] Thus, two of the intersection points are #math.equation(block: true, alt: "open parenthesis 2 plus the square root of 2 , the fraction π over 4 close parenthesis and open parenthesis 2 minus the square root of 2 , the fraction 5 π over 4 close parenthesis")[$( 2 + sqrt(2) , display(frac(π, 4)) ) " " " " "and" " " " " ( 2 − sqrt(2) , display(frac(5 π, 4)) )$] as shown at right. #figure(figph[cardioids], alt: "cardioids", caption: none) However, you can see in the figure that the graphs also appear to intersect at the pole. To verify that the pole indeed lies on both graphs, we can solve for #math.equation(block: false, alt: "θ")[$θ$] in each equation when #math.equation(block: false, alt: "r equals 0")[$r = 0$]. #math.equation(block: true, alt: "0, equals 2 plus 2 sin ⁡ open parenthesis θ close parenthesis, 0, equals 2 plus 2 cos ⁡ open parenthesis θ close parenthesis; minus 1, equals sin ⁡ open parenthesis θ close parenthesis, minus 1, equals cos ⁡ θ; θ, equals the fraction 3 π over 2, θ, equals π")[$0 & = 2 + 2 sin ( θ ) #h(2em) #h(2em) & 0 & = 2 + 2 cos ( θ ) #h(2em) #h(2em) \ − 1 & = sin ( θ ) & − 1 & = cos θ \ θ & = display(frac(3 π, 2)) & θ & = π$] Both points, #math.equation(block: false, alt: "open parenthesis 0 , the fraction 3 π over 2 close parenthesis")[$( 0 , display(frac(3 π, 2)) )$] and #math.equation(block: false, alt: "open parenthesis 0 , π close parenthesis")[$( 0 , π )$], represent the pole. Thus, the graphs intersect at three points: #math.equation(block: false, alt: "open parenthesis 2 plus the square root of 2 , the fraction π over 4 close parenthesis")[$( 2 + sqrt(2) , display(frac(π, 4)) )$], #math.equation(block: false, alt: "open parenthesis 2 minus the square root of 2 , the fraction 5 π over 4 close parenthesis")[$( 2 − sqrt(2) , display(frac(5 π, 4)) )$], and the pole. ] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ As we saw in the previous Example, solving a system of equations #math.equation(block: false, alt: "r equals f open parenthesis θ close parenthesis")[$" " r = f ( θ ) " "$] and #math.equation(block: false, alt: "r equals g open parenthesis θ close parenthesis")[$" " r = g ( θ ) " "$] will not always reveal an intersection at the pole, because #math.equation(block: false, alt: "r")[$r$] may be equal to zero for different values of #math.equation(block: false, alt: "θ")[$θ$] in the two equations. In fact, every point has multiple representation in polar coordinates, for example #math.equation(block: false, alt: "open parenthesis r , θ close parenthesis")[$( r , θ )$] and #math.equation(block: false, alt: "open parenthesis minus r , π plus θ close parenthesis")[$( − r , π + θ )$]. Solving the system of equations #math.equation(block: false, alt: "r equals f open parenthesis θ close parenthesis")[$" " r = f ( θ ) " "$] and #math.equation(block: false, alt: "r equals g open parenthesis θ close parenthesis")[$" " r = g ( θ ) " "$] could miss intersection points besides the pole. We should check the graphs to identify how many points of intersection there are. ] Find all intersection points of the graphs of #math.equation(block: false, alt: "r equals 1")[$" " r = 1 " "$] and #math.equation(block: false, alt: "r equals 2 cos ⁡ open parenthesis θ close parenthesis")[$" " r = 2 cos ( θ ) " "$], for #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 2 π")[$" " 0 ≤ θ ≤ 2 π$]. #math.equation(block: true, alt: "open parenthesis 1 , the fraction π over 3 close parenthesis")[$( 1 , display(frac(π, 3)) )$],#math.equation(block: true, alt: "open parenthesis 1 , the fraction 5 π over 3 close parenthesis")[$" " ( 1 , display(frac(5 π, 3)) )$] Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve each equation. Give two solutions, #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 2 π")[$0 ≤ θ ≤ 2 π$]. + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction the square root of 3 over 2")[$cos ( θ ) = display(frac(sqrt(3), 2))$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction 1 over the square root of 2")[$sin ( θ ) = display(frac(1, sqrt(2)))$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the square root of 3")[$tan ( θ ) = sqrt(3)$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction 1 over the square root of 3")[$tan ( θ ) = display(frac(1, sqrt(3)))$] + #math.equation(block: false, alt: "sin ⁡ open parenthesis θ close parenthesis equals the fraction minus 1 over the square root of 2")[$sin ( θ ) = display(frac(− 1, sqrt(2)))$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals the fraction minus 1 over 2")[$cos ( θ ) = display(frac(− 1, 2))$] + #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis equals minus 1")[$tan ( θ ) = − 1$] + #math.equation(block: false, alt: "cos ⁡ open parenthesis θ close parenthesis equals 0")[$cos ( θ ) = 0$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Skills Refresher Answers + #math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$], #math.equation(block: false, alt: "the fraction 11 π over 6")[$display(frac(11 π, 6))$] + #math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$], #math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$] + #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$], #math.equation(block: false, alt: "the fraction 4 π over 3")[$display(frac(4 π, 3))$] + #math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$], #math.equation(block: false, alt: "the fraction 7 π over 6")[$display(frac(7 π, 6))$] + #math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$], #math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$] + #math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$], #math.equation(block: false, alt: "the fraction 4 π over 3")[$display(frac(4 π, 3))$] + #math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$], #math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$] + #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$], #math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$] ] === A Catalog of Polar Curves ==== The Coordinate Curves - #strong[#math.equation(block: false, alt: "θ equals k")[$θ = k " " " "$]] (#math.equation(block: false, alt: "k")[$k$] a constant) A #strong[line] through the pole. #math.equation(block: false, alt: "k")[$k$] gives the angle of inclination of the line (in radians). #figure(figph[line], alt: "line", caption: none) - #strong[#math.equation(block: false, alt: "r equals k")[$r = k " " " "$]] (#math.equation(block: false, alt: "k")[$k$] a constant) A #strong[circle] centered at the pole. #math.equation(block: false, alt: "k")[$k$] is the radius of the circle. #figure(figph[circle], alt: "circle", caption: none) ==== Circles - #strong[#math.equation(block: false, alt: "r equals 2 a sin ⁡ open parenthesis θ close parenthesis")[$r = 2 a sin ( θ )$]] A #strong[circle] with center #math.equation(block: false, alt: "open parenthesis 0 , a close parenthesis")[$( 0 , a )$] on the #math.equation(block: false, alt: "y")[$y$]-axis, and radius #math.equation(block: false, alt: "vertical bar a vertical bar")[$| a |$]. #figure(figph[circle], alt: "circle", caption: none) - #strong[#math.equation(block: false, alt: "r equals 2 a cos ⁡ open parenthesis θ close parenthesis")[$r = 2 a cos ( θ )$]] A #strong[circle] with center #math.equation(block: false, alt: "open parenthesis a , 0 close parenthesis")[$( a , 0 )$] on the #math.equation(block: false, alt: "x")[$x$]-axis, and radius #math.equation(block: false, alt: "vertical bar a vertical bar")[$| a |$]. #figure(figph[circle], alt: "circle", caption: none) ==== Roses - #strong[#math.equation(block: false, alt: "r equals a sin ⁡ open parenthesis n θ close parenthesis")[$r = a sin ( n θ )$]] A #strong[rose] with petal length #math.equation(block: false, alt: "a")[$a$]. #math.equation(block: false, alt: "n")[$n$] petals if #math.equation(block: false, alt: "n")[$n$] is odd; #math.equation(block: false, alt: "2 n")[$" " " " 2 n$] petals if #math.equation(block: false, alt: "n")[$n$] is even. #figure(figph[rose], alt: "rose", caption: none) - #strong[#math.equation(block: false, alt: "r equals a cos ⁡ open parenthesis n θ close parenthesis")[$r = a cos ( n θ )$]] A #strong[rose] with petal length #math.equation(block: false, alt: "a")[$a$]. #math.equation(block: false, alt: "n")[$n$] petals if #math.equation(block: false, alt: "n")[$n$] is odd; #math.equation(block: false, alt: "2 n")[$" " " " 2 n$] petals if #math.equation(block: false, alt: "n")[$n$] is even. #figure(figph[rose], alt: "rose", caption: none) ==== Limaçons #strong[#math.equation(block: false, alt: "r equals a plus or minus b sin ⁡ open parenthesis θ close parenthesis")[$r = a ± b sin ( θ ) " " " " " " " " " "$]] or #strong[#math.equation(block: false, alt: "r equals a plus or minus b cos ⁡ open parenthesis θ close parenthesis")[$" " " " " " " " " " r = a ± b cos ( θ )$]] - #math.equation(block: false, alt: "a greater than b")[$a > b$], with a dent. #figure(figph[limacon with dent], alt: "limacon with dent", caption: none) - #math.equation(block: false, alt: "a less than b")[$a < b$], with a loop. #figure(figph[limacon with loop], alt: "limacon with loop", caption: none) - #math.equation(block: false, alt: "a greater than b")[$a > b$], a #strong[cardioid]. #figure(figph[cardioid], alt: "cardioid", caption: none) ==== Lemniscates - #strong[#math.equation(block: false, alt: "r squared equals a squared cos ⁡ open parenthesis 2 θ close parenthesis")[$r^(2) = a^(2) cos ( 2 θ )$]] #figure(figph[lemniscate], alt: "lemniscate", caption: none) - #strong[#math.equation(block: false, alt: "r squared equals a squared sin ⁡ open parenthesis 2 θ close parenthesis")[$r^(2) = a^(2) sin ( 2 θ )$]] #figure(figph[lemniscate], alt: "lemniscate", caption: none) ==== Spirals - #strong[Archimedean spiral] #strong[#math.equation(block: false, alt: "r equals a θ")[$r = a θ$]] #figure(figph[Archimedean spiral], alt: "Archimedean spiral", caption: none) - #strong[Logarithmic spiral] #strong[#math.equation(block: false, alt: "r equals e to the power a θ")[$r = e^(a θ)$]] #figure(figph[Logarithmic spiral], alt: "Logarithmic spiral", caption: none) === Section 10.2 Summary ==== Vocabulary - Rose - Limaçon - Cardioid - Lemniscate ==== Concepts + When graphing an equation in polar coordinates, we think of sweeping around the pole in the counterclockwise direction, and at each angle #math.equation(block: false, alt: "θ")[$θ$] the #math.equation(block: false, alt: "r")[$r$]-value tells us how far the graph is from the pole. + Standard graphs in polar coordinates include circles and roses, cardioids and limaçons, lemniscates, and spirals. + To find the intersection points of the polar graphs #math.equation(block: false, alt: "r equals f open parenthesis θ close parenthesis")[$r = f ( θ )$] and #math.equation(block: false, alt: "r equals g open parenthesis θ close parenthesis")[$r = g ( θ )$] we solve the equation #math.equation(block: false, alt: "f open parenthesis θ close parenthesis equals g open parenthesis θ close parenthesis")[$f ( θ ) = g ( θ )$]. In addition, we should always check whether the pole is a point on both graphs. ==== Study Questions + Delbert says that the graph of #math.equation(block: false, alt: "r equals 2 sin ⁡ open parenthesis θ close parenthesis")[$r = 2 sin ( θ )$] in the first example cannot be correct, because there are no points on the graph for angles between #math.equation(block: false, alt: "π")[$π$]and #math.equation(block: false, alt: "2 π")[$2 π$]. How do you respond? + Is it possible to have a rose with only two petals? What would its equation be? + Francine says that a circle of the form #math.equation(block: false, alt: "r equals 2 a cos ⁡ open parenthesis θ close parenthesis")[$r = 2 a cos ( θ )$] is just a special case of a limaçon. Support or refute her statement. + There are no points on the graph of #math.equation(block: false, alt: "r squared equals 4 cos ⁡ open parenthesis 2 θ close parenthesis")[$r^(2) = 4 cos ( 2 θ )$] for angles between #math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$] and #math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$] Why is that? ==== Skills + Describe the effect of parameters in polar curves \#1–16, 83–84 + Compare polar and Cartesian graphs \#21–24 + Sketch standard polar graphs \#17–20, 25–42, 75–82 + Identify standard polar graphs \#43–58 + Write equations for standard polar graphs \#59–66 + Find intersection points of polar graphs \#67–74 === Homework 10-2 In Problems 1-4, use graphing technology to graph the equations. + Graph #math.equation(block: false, alt: "r equals k")[$r = k$], for #math.equation(block: false, alt: "k equals 1 , 2 , 3")[$k = 1 , 2 , 3$]. How does the graph change for different values of #math.equation(block: false, alt: "k")[$k$]? + Write a Cartesian equation for each graph in part (a). + #figure(figph[circles], alt: "circles", caption: none) #linebreak() #math.equation(block: false, alt: "k")[$k$] is the radius + #math.equation(block: false, alt: "x squared plus y squared equals 1 , x squared plus y squared equals 4 , x squared plus y squared equals 9")[$x^(2) + y^(2) = 1 , " " x^(2) + y^(2) = 4 , " " x^(2) + y^(2) = 9$] + Graph #math.equation(block: false, alt: "r equals k")[$r = k$], for #math.equation(block: false, alt: "k equals minus 1 , minus 2 , minus 3")[$k = − 1 , − 2 , − 3$]. How does these graphs compare to the graphs in Problem 1? + Write a Cartesian equation for each graph in part (a). + Graph #math.equation(block: false, alt: "θ equals k")[$θ = k$], for #math.equation(block: false, alt: "k equals the fraction π over 6 , the fraction π over 3 , the fraction 2 π over 3 , the fraction 5 π over 6")[$k = display(frac(π, 6)) , " " display(frac(π, 3)) , " " display(frac(2 π, 3)) , " " display(frac(5 π, 6))$]. How does the graph change for different values of #math.equation(block: false, alt: "k")[$k$]? + Write a Cartesian equation for each graph in part (a). + #figure(figph[lines on polar grid], alt: "lines on polar grid", caption: none) #linebreak() #math.equation(block: false, alt: "tan ⁡ k")[$tan k$] is the slope + #math.equation(block: false, alt: "y equals the fraction x over the square root of 3 , y equals the square root of 3 x , y equals minus the square root of 3 x , y equals the fraction x over the square root of 3")[$y = display(frac(x, sqrt(3))) , " " y = sqrt(3) x , " " y = − sqrt(3) x , " " y = display(frac(x, sqrt(3)))$] + Graph #math.equation(block: false, alt: "θ equals k")[$θ = k$], for #math.equation(block: false, alt: "k equals the fraction 7 π over 6 , the fraction 4 π over 3 , the fraction 5 π over 3 , the fraction 11 π over 6")[$k = display(frac(7 π, 6)) , " " display(frac(4 π, 3)) , " " display(frac(5 π, 3)) , " " display(frac(11 π, 6))$]. How does the graph change for different values of #math.equation(block: false, alt: "k")[$k$]? + Write a Cartesian equation for each graph in part (a). Complete the table of values for each equation. Plot the points in order of increasing #math.equation(block: false, alt: "θ")[$θ$]. What is different about the two graphs? Equation 1: #math.equation(block: false, alt: "r equals 2")[$" " " " r = 2$] #figure(table( columns: 9, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]]), [#math.equation(block: false, alt: "r equals 2")[$r = 2$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) Equation 2: #math.equation(block: false, alt: "r equals minus 2")[$" " " " r = − 2$] #figure(table( columns: 9, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]]), [#math.equation(block: false, alt: "r equals minus 2")[$r = − 2$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) #figure(table( columns: 9, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]]), [#math.equation(block: false, alt: "r equals 2")[$r = 2$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2")[$2$]], )) #figure(table( columns: 9, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]]), [#math.equation(block: false, alt: "r equals minus 2")[$r = − 2$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], )) The graph of #math.equation(block: false, alt: "r equals 2")[$r = 2$] begins at the right-most point (and proceeds counter-clockwise); the graph of #math.equation(block: false, alt: "r equals minus 2")[$r = − 2$] begins at the left-most point. #figure(figph[polar points on circle], alt: "polar points on circle", caption: none) Graph each line, and label the points with their coordinates. How are the points on the two lines related? Equation 1: #math.equation(block: false, alt: "θ equals the fraction π over 4")[$" " " " θ = display(frac(π, 4))$] #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ equals the fraction π over 4")[$θ = display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]]), [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], )) Equation 1: #math.equation(block: false, alt: "θ equals the fraction 5 π over 4")[$" " " " θ = display(frac(5 π, 4))$] #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ equals the fraction 5 π over 4")[$θ = display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]]), [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], )) + Graph the circle #math.equation(block: false, alt: "r equals 4 cos ⁡ open parenthesis θ close parenthesis")[$r = 4 cos ( θ )$]. Label the points corresponding to #math.equation(block: false, alt: "θ equals 0 , the fraction π over 4 , the fraction π over 2 , the fraction 3 π over 4 ,")[$θ = 0 , " " display(frac(π, 4)) , " " display(frac(π, 2)) , " " display(frac(3 π, 4)) ,$] and #math.equation(block: false, alt: "π")[$π$]. + Complete the table of values. What happens to the graph as #math.equation(block: false, alt: "θ")[$θ$] increases from #math.equation(block: false, alt: "π")[$π$] to #math.equation(block: false, alt: "2 π")[$2 π$]?#figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + Find the center and radius of the circle. + Give the Cartesian equation of the circle. + #figure(figph[circle on polar grid], alt: "circle on polar grid", caption: none) + #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "minus 4")[$− 4$]], [#math.equation(block: false, alt: "minus 2 the square root of 2")[$− 2 sqrt(2)$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "2 the square root of 2")[$2 sqrt(2)$]], [#math.equation(block: false, alt: "4")[$4$]], )) #linebreak() The graph is traced again. + center: #math.equation(block: false, alt: "open parenthesis 2 , 0 close parenthesis")[$( 2 , 0 )$], radius: #math.equation(block: false, alt: "2")[$2$] + #math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis squared plus y squared equals 4")[$( x − 2 )^(2) + y^(2) = 4$] + Graph the circle #math.equation(block: false, alt: "r equals 4 sin ⁡ open parenthesis θ close parenthesis")[$r = 4 sin ( θ )$]. Label the points corresponding to #math.equation(block: false, alt: "θ equals 0 , the fraction π over 4 , the fraction π over 2 , the fraction 3 π over 4 ,")[$θ = 0 , " " display(frac(π, 4)) , " " display(frac(π, 2)) , " " display(frac(3 π, 4)) ,$] and #math.equation(block: false, alt: "π")[$π$]. + Complete the table of values. What happens to the graph as #math.equation(block: false, alt: "θ")[$θ$] increases from #math.equation(block: false, alt: "π")[$π$] to #math.equation(block: false, alt: "2 π")[$2 π$]?#figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + Find the center and radius of the circle. + Give the Cartesian equation of the circle. + Graph #math.equation(block: false, alt: "r equals 2 a sin ⁡ open parenthesis θ close parenthesis")[$r = 2 a sin ( θ )$] for #math.equation(block: false, alt: "a equals minus 2 , minus 1 , 1 , 2")[$a = − 2 , − 1 , 1 , 2$]. + How do the graphs change for different values of #math.equation(block: false, alt: "a")[$a$]? + #figure(figph[circcles on polar grid], alt: "circcles on polar grid", caption: none) + For #math.equation(block: false, alt: "a greater than 0")[$a > 0$], #math.equation(block: false, alt: "a")[$a$] is the radius of a circle centered on the positive #math.equation(block: false, alt: "y")[$y$]-axis; for #math.equation(block: false, alt: "a less than 0")[$a < 0$], #math.equation(block: false, alt: "vertical bar a vertical bar")[$| a |$] is the radius of a circle centerd on the negative #math.equation(block: false, alt: "y")[$y$]-axis. + Graph #math.equation(block: false, alt: "r equals 2 a cos ⁡ open parenthesis θ close parenthesis")[$r = 2 a cos ( θ )$] for #math.equation(block: false, alt: "a equals minus 2 , minus 1 , 1 , 2")[$a = − 2 , − 1 , 1 , 2$]. + How do the graphs change for different values of #math.equation(block: false, alt: "a")[$a$]? Complete the table of values for each cardioid, and graph the equation. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + #math.equation(block: false, alt: "r equals 1 plus sin ⁡ open parenthesis θ close parenthesis")[$r = 1 + sin ( θ )$] + #math.equation(block: false, alt: "r equals minus 1 plus sin ⁡ open parenthesis θ close parenthesis")[$r = − 1 + sin ( θ )$] + #math.equation(block: false, alt: "r equals 1 minus sin ⁡ open parenthesis θ close parenthesis")[$r = 1 − sin ( θ )$] + #math.equation(block: false, alt: "r equals minus 1 minus sin ⁡ open parenthesis θ close parenthesis")[$r = − 1 − sin ( θ )$] + #figure(figph[cardioid], alt: "cardioid", caption: none) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], )) + #figure(figph[cardioid], alt: "cardioid", caption: none) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], )) + #figure(figph[cardioid], alt: "cardioid", caption: none) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "1")[$1$]], )) + #figure(figph[cardioid], alt: "cardioid", caption: none) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], )) Complete the table of values for each cardioid, and graph the equation. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + #math.equation(block: false, alt: "r equals 1 plus cos ⁡ open parenthesis θ close parenthesis")[$r = 1 + cos ( θ )$] + #math.equation(block: false, alt: "r equals minus 1 plus cos ⁡ open parenthesis θ close parenthesis")[$r = − 1 + cos ( θ )$] + #math.equation(block: false, alt: "r equals 1 minus cos ⁡ open parenthesis θ close parenthesis")[$r = 1 − cos ( θ )$] + #math.equation(block: false, alt: "r equals minus 1 minus cos ⁡ open parenthesis θ close parenthesis")[$r = − 1 − cos ( θ )$] Complete the table of values for each limaçon, and graph the equation. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + #math.equation(block: false, alt: "r equals 2 plus cos ⁡ open parenthesis θ close parenthesis")[$r = 2 + cos ( θ )$] + #math.equation(block: false, alt: "r equals 2 minus cos ⁡ open parenthesis θ close parenthesis")[$r = 2 − cos ( θ )$] + #math.equation(block: false, alt: "r equals 1 plus 2 cos ⁡ open parenthesis θ close parenthesis")[$r = 1 + 2 cos ( θ )$] + #math.equation(block: false, alt: "r equals 1 minus 2 cos ⁡ open parenthesis θ close parenthesis")[$r = 1 − 2 cos ( θ )$] + #figure(figph[limacon], alt: "limacon", caption: none) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], )) + #figure(figph[limacon], alt: "limacon", caption: none) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "1")[$1$]], )) + #figure(figph[limacon], alt: "limacon", caption: none) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "3")[$3$]], )) + #figure(figph[limacon], alt: "limacon", caption: none) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], )) Complete the table of values for each limaçon, and graph the equation. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + #math.equation(block: false, alt: "r equals 2 plus sin ⁡ open parenthesis θ close parenthesis")[$r = 2 + sin ( θ )$] + #math.equation(block: false, alt: "r equals 2 minus sin ⁡ open parenthesis θ close parenthesis")[$r = 2 − sin ( θ )$] + #math.equation(block: false, alt: "r equals 1 plus 2 sin ⁡ open parenthesis θ close parenthesis")[$r = 1 + 2 sin ( θ )$] + #math.equation(block: false, alt: "r equals 1 minus 2 sin ⁡ open parenthesis θ close parenthesis")[$r = 1 − 2 sin ( θ )$] + Graph the following roses and compare. How is the number of petals related to the value of #math.equation(block: false, alt: "n")[$n$] in the equation #math.equation(block: false, alt: "r equals a sin ⁡ open parenthesis n θ close parenthesis")[$r = a sin ( n θ )$]? #math.equation(block: true, alt: "r equals sin ⁡ open parenthesis 2 θ close parenthesis , r equals sin ⁡ open parenthesis 3 θ close parenthesis , r equals sin ⁡ open parenthesis 4 θ close parenthesis , r equals sin ⁡ open parenthesis 5 θ close parenthesis")[$r = sin ( 2 θ ) , " " " " r = sin ( 3 θ ) , " " " " r = sin ( 4 θ ) , " " " " r = sin ( 5 θ )$] + For each graph above, list the values of #math.equation(block: false, alt: "θ")[$θ$] where the tips of the petals occur. + Graph #math.equation(block: false, alt: "r equals a sin ⁡ open parenthesis 3 θ close parenthesis")[$r = a sin ( 3 θ ) " "$] for #math.equation(block: false, alt: "a equals 1 , 2 ,")[$a = 1 , 2 ,$] and #math.equation(block: false, alt: "3")[$3$]. How does the value of #math.equation(block: false, alt: "a")[$a$] affect the graph? + #figure(figph[4-petal rose], alt: "4-petal rose", caption: none) #figure(figph[3-petal rose], alt: "3-petal rose", caption: none) #figure(figph[8-petal rose], alt: "8-petal rose", caption: none) #figure(figph[5 petal rose], alt: "5 petal rose", caption: none) #linebreak() There are #math.equation(block: false, alt: "n")[$n$] petals if #math.equation(block: false, alt: "n")[$n$] is odd, and #math.equation(block: false, alt: "2 n")[$2 n$] petals if #math.equation(block: false, alt: "n")[$n$] is even. + #math.equation(block: false, alt: "n equals 2 : the fraction π over 4 ,")[$n = 2 : " " display(frac(π, 4)) ,$] #math.equation(block: false, alt: "the fraction 3 π over 4 ,")[$" " display(frac(3 π, 4)) , " "$]#math.equation(block: false, alt: "the fraction 5 π over 4 ,")[$display(frac(5 π, 4)) , " "$]#math.equation(block: false, alt: "the fraction 7 π over 4 ;")[$display(frac(7 π, 4)) ; " "$] #math.equation(block: false, alt: "n equals 3 : the fraction π over 6 ,")[$n = 3 : " " display(frac(π, 6)) , " "$]#math.equation(block: false, alt: "the fraction 5 π over 6 ,")[$display(frac(5 π, 6)) , " "$]#math.equation(block: false, alt: "the fraction 3 π over 2 ;")[$display(frac(3 π, 2)) ;$] #math.equation(block: false, alt: "n equals 4 : the fraction π over 8 ,")[$n = 4 : " " display(frac(π, 8)) , " "$]#math.equation(block: false, alt: "the fraction 3 π over 8 ,")[$display(frac(3 π, 8)) , " "$]#math.equation(block: false, alt: "the fraction 5 π over 8 ,")[$display(frac(5 π, 8)) , " "$]#math.equation(block: false, alt: "the fraction 7 π over 8 ,")[$display(frac(7 π, 8)) , " "$]#math.equation(block: false, alt: "the fraction 9 π over 8 ,")[$display(frac(9 π, 8)) , " "$]#math.equation(block: false, alt: "the fraction 11 π over 8 ,")[$display(frac(11 π, 8)) , " "$]#math.equation(block: false, alt: "the fraction 13 π over 8 ,")[$display(frac(13 π, 8)) , " "$]#math.equation(block: false, alt: "the fraction 15 π over 8 ;")[$display(frac(15 π, 8)) ;$] #math.equation(block: false, alt: "n equals 5 : the fraction π over 10 ,")[$n = 5 : " " " " display(frac(π, 10)) , " "$]#math.equation(block: false, alt: "the fraction π over 2 ,")[$display(frac(π, 2)) , " "$]#math.equation(block: false, alt: "the fraction 9 π over 10 ,")[$display(frac(9 π, 10)) , " "$]#math.equation(block: false, alt: "the fraction 13 π over 10 ,")[$display(frac(13 π, 10)) , " "$]#math.equation(block: false, alt: "the fraction 17 π over 10")[$display(frac(17 π, 10))$] + #figure(figph[3-petal rose], alt: "3-petal rose", caption: none) #figure(figph[3-petal rose], alt: "3-petal rose", caption: none) #figure(figph[3-petal rose], alt: "3-petal rose", caption: none) #linebreak() #math.equation(block: false, alt: "a")[$a$] is the length of the petal. + Graph the following roses and compare. How is the number of petals related to the value of #math.equation(block: false, alt: "n")[$n$] in the equation #math.equation(block: false, alt: "r equals a cos ⁡ open parenthesis n θ close parenthesis")[$r = a cos ( n θ )$]? #math.equation(block: true, alt: "r equals cos ⁡ open parenthesis 2 θ close parenthesis , r equals cos ⁡ open parenthesis 3 θ close parenthesis , r equals cos ⁡ open parenthesis 4 θ close parenthesis , r equals cos ⁡ open parenthesis 5 θ close parenthesis")[$r = cos ( 2 θ ) , " " " " r = cos ( 3 θ ) , " " " " r = cos ( 4 θ ) , " " " " r = cos ( 5 θ )$] + For each graph above, list the values of #math.equation(block: false, alt: "θ")[$θ$] where the tips of the petals occur. + Graph #math.equation(block: false, alt: "r equals a cos ⁡ open parenthesis 3 θ close parenthesis")[$r = a cos ( 3 θ ) " "$] for #math.equation(block: false, alt: "a equals 1 , 2 ,")[$a = 1 , 2 ,$] and #math.equation(block: false, alt: "3")[$3$]. How does the value of #math.equation(block: false, alt: "a")[$a$] affect the graph? + Solve #math.equation(block: false, alt: "r squared equals 9 cos ⁡ open parenthesis 2 θ close parenthesis")[$r^(2) = 9 cos ( 2 θ ) " "$] for #math.equation(block: false, alt: "r")[$r$]. (You should get two equations for #math.equation(block: false, alt: "r")[$r$].) + Graph both equations together. Change #math.equation(block: false, alt: "θ")[$θ$]step to 0.02 to see the whole graph. + How does the value of #math.equation(block: false, alt: "a")[$a$] affect the graph of #math.equation(block: false, alt: "r squared equals a squared cos ⁡ open parenthesis 2 θ close parenthesis")[$r^(2) = a^(2) cos ( 2 θ )$]? + #math.equation(block: false, alt: "r equals plus or minus 3 the square root of cos ⁡ 2 θ")[$r = ± 3 sqrt(cos 2 θ)$] + #figure(figph[lemniscate], alt: "lemniscate", caption: none) + #math.equation(block: false, alt: "a")[$a$] is the length of the loop. + Solve #math.equation(block: false, alt: "r squared equals 9 sin ⁡ open parenthesis 2 θ close parenthesis")[$r^(2) = 9 sin ( 2 θ ) " "$] for #math.equation(block: false, alt: "r")[$r$]. (You should get two equations for #math.equation(block: false, alt: "r")[$r$].) + Graph both equations together. Change #math.equation(block: false, alt: "θ")[$θ$]step to 0.02 to see the whole graph. + How does this graph differ from the graph in Problem 17? Graph the Archimedean spiral #math.equation(block: false, alt: "r equals θ")[$r = θ$]. Set your window to #math.equation(block: true, alt: "θ min, equals 0 θ max equals 8 π; Xmin, equals minus 20 Xmax equals 20; Ymin, equals minus 20 Ymax equals 20")[$θ "min" & = 0 " " " " " " " " " " " " " " " " " " " " θ "max" = 8 π \ "Xmin" & = − 20 " " " " " " " " "Xmax" = 20 \ "Ymin" & = − 20 " " " " " " " " "Ymax" = 20$] Then graph by pressing Zoom 5. #figure(figph[Archimedean spiral], alt: "Archimedean spiral", caption: none) Graph the logarithmic spiral #math.equation(block: false, alt: "r equals e to the power 0.2 θ")[$r = e^(0.2 θ)$]. Set your window to #math.equation(block: true, alt: "θ min, equals 0 θ max equals 8 π; Xmin, equals minus 100 Xmax equals 100; Ymin, equals minus 100 Ymax equals 100")[$θ "min" & = 0 " " " " " " " " " " " " " " " " " " " " θ "max" = 8 π \ "Xmin" & = − 100 " " " " " " " " "Xmax" = 100 \ "Ymin" & = − 100 " " " " " " " " "Ymax" = 100$] Then graph by pressing Zoom 5. + Complete the table and graph the equation #math.equation(block: false, alt: "y equals sin ⁡ open parenthesis 3 θ close parenthesis")[$y = sin ( 3 θ )$] in Cartesian coordinates, for #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 2 π")[$0 ≤ θ ≤ 2 π$].#figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$]), [#math.equation(block: false, alt: "3 θ")[$3 θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "y")[$y$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + Complete the table and graph the equation #math.equation(block: false, alt: "r equals sin ⁡ open parenthesis 3 θ close parenthesis")[$r = sin ( 3 θ )$] in polar coordinates, for #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 2 π")[$0 ≤ θ ≤ 2 π$].#figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$]), [#math.equation(block: false, alt: "3 θ")[$3 θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "r")[$r$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 12")[$display(frac(π, 12))$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction 5 π over 12")[$display(frac(5 π, 12))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 12")[$display(frac(7 π, 12))$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]]), [#math.equation(block: false, alt: "3 θ")[$3 θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction the square root of 2 over 2")[$display(frac(sqrt(2), 2))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction the square root of 2 over 2")[$display(frac(sqrt(2), 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction minus the square root of 2 over 2")[$display(frac(− sqrt(2), 2))$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction minus the square root of 2 over 2")[$display(frac(− sqrt(2), 2))$]], [#math.equation(block: false, alt: "0")[$0$]], )) #figure(figph[sinusoidal curve], alt: "sinusoidal curve", caption: none) + #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 12")[$display(frac(π, 12))$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction 5 π over 12")[$display(frac(5 π, 12))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 12")[$display(frac(7 π, 12))$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]]), [#math.equation(block: false, alt: "3 θ")[$3 θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction the square root of 2 over 2")[$display(frac(sqrt(2), 2))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction the square root of 2 over 2")[$display(frac(sqrt(2), 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction minus the square root of 2 over 2")[$display(frac(− sqrt(2), 2))$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction minus the square root of 2 over 2")[$display(frac(− sqrt(2), 2))$]], [#math.equation(block: false, alt: "0")[$0$]], )) #figure(figph[three-petal rose], alt: "three-petal rose", caption: none) + Complete the table and graph the equation #math.equation(block: false, alt: "y equals cos ⁡ open parenthesis 2 θ close parenthesis")[$y = cos ( 2 θ )$] in Cartesian coordinates, for #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 2 π")[$0 ≤ θ ≤ 2 π$].#figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$]), [#math.equation(block: false, alt: "2 θ")[$2 θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "y")[$y$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + Complete the table and graph the equation #math.equation(block: false, alt: "r equals cos ⁡ open parenthesis 2 θ close parenthesis")[$r = cos ( 2 θ )$] in polar coordinates, for #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 2 π")[$0 ≤ θ ≤ 2 π$].#figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$]), [#math.equation(block: false, alt: "2 θ")[$2 θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "r")[$r$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + Complete the table and graph the equation #math.equation(block: false, alt: "y equals 2 plus 2 cos ⁡ open parenthesis θ close parenthesis")[$y = 2 + 2 cos ( θ )$] in Cartesian coordinates, for #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 2 π")[$0 ≤ θ ≤ 2 π$].#figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "y")[$y$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + Complete the table and graph the equation #math.equation(block: false, alt: "r equals 2 plus 2 cos ⁡ open parenthesis θ close parenthesis")[$r = 2 + 2 cos ( θ )$] in polar coordinates, for #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 2 π")[$0 ≤ θ ≤ 2 π$].#figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "2 plus the square root of 2")[$2 + sqrt(2)$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2 minus the square root of 2")[$2 − sqrt(2)$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "2 minus the square root of 2")[$2 − sqrt(2)$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2 plus the square root of 2")[$2 + sqrt(2)$]], [#math.equation(block: false, alt: "4")[$4$]], )) #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "2 plus the square root of 2")[$2 + sqrt(2)$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2 minus the square root of 2")[$2 − sqrt(2)$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "2 minus the square root of 2")[$2 − sqrt(2)$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2 plus the square root of 2")[$2 + sqrt(2)$]], [#math.equation(block: false, alt: "4")[$4$]], )) #figure(figph[cardioid], alt: "cardioid", caption: none) + Complete the table and graph the equation #math.equation(block: false, alt: "y equals 1 minus sin ⁡ open parenthesis θ close parenthesis")[$y = 1 − sin ( θ )$] in Cartesian coordinates, for #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 2 π")[$0 ≤ θ ≤ 2 π$].#figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "y")[$y$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + Complete the table and graph the equation #math.equation(block: false, alt: "r equals 1 minus sin ⁡ open parenthesis θ close parenthesis")[$r = 1 − sin ( θ )$] in polar coordinates, for #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 2 π")[$0 ≤ θ ≤ 2 π$].#figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "r")[$r$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) For Problems 25–42, use the catalog of polar graphs to help you identify and sketch the following curves. Check your work by graphing with a graphing utility. #math.equation(block: true, alt: "r equals 3 cos ⁡ open parenthesis θ close parenthesis")[$r = 3 cos ( θ )$] circle #figure(figph[circle on polar grid], alt: "circle on polar grid", caption: none) #math.equation(block: true, alt: "r equals 2 sin ⁡ open parenthesis θ close parenthesis")[$r = 2 sin ( θ )$] #math.equation(block: true, alt: "θ equals the fraction π over 4")[$θ = display(frac(π, 4))$] line #figure(figph[line on polar grid], alt: "line on polar grid", caption: none) #math.equation(block: true, alt: "θ equals the fraction 4 π over 3")[$θ = display(frac(4 π, 3))$] #math.equation(block: true, alt: "r equals 4")[$r = 4$] circle #figure(figph[circle], alt: "circle", caption: none) #math.equation(block: true, alt: "r equals 2")[$r = 2$] #math.equation(block: true, alt: "r equals 2 plus 2 sin ⁡ open parenthesis θ close parenthesis")[$r = 2 + 2 sin ( θ )$] cardioid #figure(figph[cardioid], alt: "cardioid", caption: none) #math.equation(block: true, alt: "r equals 3 plus 3 cos ⁡ open parenthesis θ close parenthesis")[$r = 3 + 3 cos ( θ )$] #math.equation(block: true, alt: "r equals 2 minus cos ⁡ open parenthesis θ close parenthesis")[$r = 2 − cos ( θ )$] limaçon #figure(figph[limacon], alt: "limacon", caption: none) #math.equation(block: true, alt: "r equals 1 minus 3 sin ⁡ open parenthesis θ close parenthesis")[$r = 1 − 3 sin ( θ )$] #math.equation(block: true, alt: "r equals 3 sin ⁡ open parenthesis 2 θ close parenthesis")[$r = 3 sin ( 2 θ )$] rose #figure(figph[four-petal rose], alt: "four-petal rose", caption: none) #math.equation(block: true, alt: "r equals 2 cos ⁡ open parenthesis 3 θ close parenthesis")[$r = 2 cos ( 3 θ )$] #math.equation(block: true, alt: "r equals 2 cos ⁡ open parenthesis 5 θ close parenthesis")[$r = 2 cos ( 5 θ )$] rose #figure(figph[rose], alt: "rose", caption: none) #math.equation(block: true, alt: "r equals 4 sin ⁡ open parenthesis 4 θ close parenthesis")[$r = 4 sin ( 4 θ )$] #math.equation(block: true, alt: "r equals 2 plus 3 sin ⁡ open parenthesis θ close parenthesis")[$r = 2 + 3 sin ( θ )$] limaçon #figure(figph[limacon], alt: "limacon", caption: none) #math.equation(block: true, alt: "r equals 3 plus 2 sin ⁡ open parenthesis θ close parenthesis")[$r = 3 + 2 sin ( θ )$] #math.equation(block: true, alt: "r squared equals cos ⁡ open parenthesis 2 θ close parenthesis")[$r^(2) = cos ( 2 θ )$] lemniscate #figure(figph[lemniscate], alt: "lemniscate", caption: none) #math.equation(block: true, alt: "r squared equals 4 sin ⁡ open parenthesis 2 θ close parenthesis")[$r^(2) = 4 sin ( 2 θ )$] For Problems 43–52, identify each curve, and graph it. #math.equation(block: true, alt: "r csc ⁡ open parenthesis θ close parenthesis equals 2")[$r csc ( θ ) = 2$] circle #figure(figph[circle], alt: "circle", caption: none) #math.equation(block: true, alt: "r equals 2 sec ⁡ open parenthesis θ close parenthesis")[$r = 2 sec ( θ )$] #math.equation(block: true, alt: "r squared equals 4 , 0 less than or equal to θ less than or equal to the fraction 3 π over 4")[$r^(2) = 4 , " " 0 ≤ θ ≤ display(frac(3 π, 4))$] arcs of a circle #figure(figph[arcs of circle], alt: "arcs of circle", caption: none) #math.equation(block: true, alt: "θ equals the fraction π over 4 , vertical bar r vertical bar less than 2")[$θ = display(frac(π, 4)) , " " | r | < 2$] #math.equation(block: true, alt: "r equals sin ⁡ open parenthesis θ close parenthesis , the fraction 3 π over 4 less than or equal to θ less than or equal to the fraction 5 π over 4")[$r = sin ( θ ) , " " display(frac(3 π, 4)) ≤ θ ≤ display(frac(5 π, 4))$] semicircle #figure(figph[semicircle], alt: "semicircle", caption: none) #math.equation(block: true, alt: "r equals cos ⁡ open parenthesis θ close parenthesis , 0 less than or equal to θ less than or equal to the fraction π over 2")[$r = cos ( θ ) , " " 0 ≤ θ ≤ display(frac(π, 2))$] #math.equation(block: true, alt: "r equals 2 sin ⁡ open parenthesis 2 θ close parenthesis cos ⁡ open parenthesis 2 θ close parenthesis")[$r = 2 sin ( 2 θ ) cos ( 2 θ )$] rose #figure(figph[eight-petal rose], alt: "eight-petal rose", caption: none) #math.equation(block: true, alt: "r equals cos squared ⁡ open parenthesis θ close parenthesis minus sin squared ⁡ open parenthesis θ close parenthesis")[$r = cos^(2) ( θ ) − sin^(2) ( θ )$] #math.equation(block: true, alt: "r open parenthesis 1 minus cos ⁡ open parenthesis θ close parenthesis close parenthesis equals sin squared ⁡ open parenthesis θ close parenthesis")[$r ( 1 − cos ( θ ) ) = sin^(2) ( θ )$] cardioid #figure(figph[cardioid], alt: "cardioid", caption: none) #math.equation(block: true, alt: "r sec ⁡ open parenthesis θ close parenthesis equals sec ⁡ open parenthesis θ close parenthesis minus tan ⁡ open parenthesis θ close parenthesis")[$r sec ( θ ) = sec ( θ ) − tan ( θ )$] For Problems 53–58, graph the following polar curves. Do you recognize them? #math.equation(block: true, alt: "r equals the fraction 2 over 1 minus cos ⁡ open parenthesis θ close parenthesis")[$r = display(frac(2, 1 − cos ( θ )))$] parabola #figure(figph[parabola], alt: "parabola", caption: none) #math.equation(block: true, alt: "r equals the fraction 6 over 2 plus sin ⁡ open parenthesis θ close parenthesis")[$r = display(frac(6, 2 + sin ( θ )))$] #math.equation(block: true, alt: "r equals the fraction 2 over 2 minus cos ⁡ open parenthesis θ close parenthesis")[$r = display(frac(2, 2 − cos ( θ )))$] ellipse #figure(figph[ellipse], alt: "ellipse", caption: none) #math.equation(block: true, alt: "r equals the fraction 1 over 1 plus sin ⁡ open parenthesis θ close parenthesis")[$r = display(frac(1, 1 + sin ( θ )))$] #math.equation(block: true, alt: "r equals the fraction 1 over 1 plus 2 sin ⁡ open parenthesis θ close parenthesis")[$r = display(frac(1, 1 + 2 sin ( θ )))$] hyperbola #figure(figph[hyperbola], alt: "hyperbola", caption: none) #math.equation(block: true, alt: "r equals the fraction 3 over 2 minus 3 cos ⁡ open parenthesis θ close parenthesis")[$r = display(frac(3, 2 − 3 cos ( θ )))$] For Problems 59–66, write a polar equation for the graph. #figure(figph[cardioid], alt: "cardioid", caption: none) #math.equation(block: true, alt: "r equals 2 plus 2 cos ⁡ open parenthesis θ close parenthesis")[$r = 2 + 2 cos ( θ )$] #figure(figph[cardioid], alt: "cardioid", caption: none) #figure(figph[five-petal rose], alt: "five-petal rose", caption: none) #math.equation(block: true, alt: "r equals 3 sin ⁡ open parenthesis 5 θ close parenthesis")[$r = 3 sin ( 5 θ )$] #figure(figph[four-petal rose], alt: "four-petal rose", caption: none) #figure(figph[circle], alt: "circle", caption: none) #math.equation(block: true, alt: "r equals 5 sin ⁡ open parenthesis θ close parenthesis")[$r = 5 sin ( θ )$] #figure(figph[circle], alt: "circle", caption: none) #figure(figph[limacon], alt: "limacon", caption: none) #math.equation(block: true, alt: "r equals 1 plus 2 cos ⁡ open parenthesis θ close parenthesis")[$r = 1 + 2 cos ( θ )$] #figure(figph[limacon], alt: "limacon", caption: none) For Problems 67–74, find the coordinates of the intersection points of the two curves analytically. Then graph the curves to verify your answers. #math.equation(block: true, alt: "r equals cos ⁡ open parenthesis θ close parenthesis , r equals 1 minus cos ⁡ open parenthesis θ close parenthesis")[$r = cos ( θ ) , " " r = 1 − cos ( θ )$] #math.equation(block: true, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$],#math.equation(block: true, alt: "open parenthesis the fraction 1 over 2 , the fraction π over 3 close parenthesis")[$" " ( display(frac(1, 2)) , display(frac(π, 3)) )$],#math.equation(block: true, alt: "open parenthesis the fraction 1 over 2 , the fraction 5 π over 3 close parenthesis")[$" " ( display(frac(1, 2)) , display(frac(5 π, 3)) )$] #math.equation(block: true, alt: "r equals sin ⁡ open parenthesis θ close parenthesis , r equals cos ⁡ open parenthesis θ close parenthesis")[$r = sin ( θ ) , " " r = cos ( θ )$] #math.equation(block: true, alt: "r equals 3 sin ⁡ open parenthesis θ close parenthesis , r equals 3 cos ⁡ open parenthesis θ close parenthesis")[$r = 3 sin ( θ ) , " " r = 3 cos ( θ )$] #math.equation(block: true, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$], #math.equation(block: true, alt: "open parenthesis the fraction 3 over the square root of 2 , the fraction π over 4 close parenthesis")[$" " ( display(frac(3, sqrt(2))) , display(frac(π, 4)) )$], #math.equation(block: true, alt: "open parenthesis the fraction minus 3 over the square root of 2 , the fraction 5 π over 4 close parenthesis")[$" " ( display(frac(− 3, sqrt(2))) , display(frac(5 π, 4)) )$] #math.equation(block: true, alt: "r equals sin ⁡ open parenthesis 2 θ close parenthesis , r equals cos ⁡ open parenthesis 2 θ close parenthesis")[$r = sin ( 2 θ ) , " " r = cos ( 2 θ )$] #math.equation(block: true, alt: "r equals 1 , r equals 1 minus cos ⁡ open parenthesis θ close parenthesis")[$r = 1 , " " r = 1 − cos ( θ )$] #math.equation(block: true, alt: "open parenthesis 1 , the fraction π over 2 close parenthesis")[$( 1 , display(frac(π, 2)) )$], #math.equation(block: true, alt: "open parenthesis 1 , the fraction 3 π over 2 close parenthesis")[$" " ( 1 , display(frac(3 π, 2)) )$] #math.equation(block: true, alt: "r equals 3 cos ⁡ open parenthesis θ close parenthesis , r equals 1 plus cos ⁡ open parenthesis θ close parenthesis")[$r = 3 cos ( θ ) , " " r = 1 + cos ( θ )$] #math.equation(block: true, alt: "r equals 2 plus sin ⁡ open parenthesis θ close parenthesis , r equals 2 minus cos ⁡ open parenthesis θ close parenthesis")[$r = 2 + sin ( θ ) , " " r = 2 − cos ( θ )$] #math.equation(block: true, alt: "open parenthesis the fraction 4 plus the square root of 2 over 2 , the fraction 3 π over 4 close parenthesis")[$( display(frac(4 + sqrt(2), 2)) , display(frac(3 π, 4)) )$], #math.equation(block: true, alt: "open parenthesis the fraction 4 minus the square root of 2 over 2 , the fraction 7 π over 4 close parenthesis")[$" " ( display(frac(4 − sqrt(2), 2)) , display(frac(7 π, 4)) )$] #math.equation(block: true, alt: "r equals sin ⁡ open parenthesis θ close parenthesis , r equals sin ⁡ open parenthesis 2 θ close parenthesis")[$r = sin ( θ ) , " " r = sin ( 2 θ )$] For Problems 75–82, graph the polar curve. #math.equation(block: true, alt: "r squared equals tan ⁡ open parenthesis θ close parenthesis")[$r^(2) = tan ( θ )$] #figure(figph[polar plot], alt: "polar plot", caption: none) #math.equation(block: true, alt: "r squared equals cot ⁡ open parenthesis θ close parenthesis")[$r^(2) = cot ( θ )$] #math.equation(block: false, alt: "r equals csc ⁡ open parenthesis θ close parenthesis minus 2")[$r = csc ( θ ) − 2$] (conchoid) #figure(figph[conchoid], alt: "conchoid", caption: none) #math.equation(block: false, alt: "r equals tan ⁡ open parenthesis θ close parenthesis")[$r = tan ( θ )$] (kappa curve) #math.equation(block: false, alt: "r equals cos ⁡ open parenthesis 2 θ close parenthesis sec ⁡ open parenthesis θ close parenthesis")[$r = cos ( 2 θ ) sec ( θ )$] (strophoid) #figure(figph[strophoid], alt: "strophoid", caption: none) #math.equation(block: false, alt: "r equals sin ⁡ open parenthesis θ close parenthesis tan ⁡ open parenthesis θ close parenthesis")[$r = sin ( θ ) tan ( θ )$] (cissoid) #math.equation(block: true, alt: "r equals the fraction 1 over the square root of θ")[$r = display(frac(1, sqrt(θ)))$] #figure(figph[polar plot], alt: "polar plot", caption: none) #math.equation(block: true, alt: "r equals cos ⁡ open parenthesis the fraction θ over 2 close parenthesis , 0 less than or equal to θ less than or equal to 4 π")[$r = cos " " ( display(frac(θ, 2)) ) , " " 0 ≤ θ ≤ 4 π$] Graph the polar curves #math.equation(block: false, alt: "r equals 1 minus 2 sin ⁡ open parenthesis n θ close parenthesis")[$r = 1 − 2 sin ( n θ )$] for #math.equation(block: false, alt: "n equals 2 , 3 , 4 , 5 , 6")[$n = 2 , 3 , 4 , 5 , 6$]. Explain how the value of the parameter #math.equation(block: false, alt: "n")[$n$] affects the curve. The curve has #math.equation(block: false, alt: "n")[$n$] large loops and #math.equation(block: false, alt: "n")[$n$] small loops. Graph the polar curves #math.equation(block: false, alt: "r equals 1 minus 3 cos ⁡ open parenthesis n θ close parenthesis")[$r = 1 − 3 cos ( n θ )$] for #math.equation(block: false, alt: "n equals 2 , 3 , 4 , 5 , 6")[$n = 2 , 3 , 4 , 5 , 6$]. Explain how the value of the parameter #math.equation(block: false, alt: "n")[$n$] affects the curve.