#set document(title: "6.1 Arclength and Radians", 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")) == 6.1#h(0.6em)Arclength and Radians Imagine that you are riding on a Ferris wheel of radius 100 feet, and each rotation takes eight minutes. We can use angles in standard position to describe your location as you travel around the wheel. The figure at right shows the locations indicated by #math.equation(block: false, alt: "θ equals 0 degrees , 90 degrees , 180 degrees ,")[$θ = 0^(∘) , " " 90^(∘) , " " 180^(∘) ,$] and #math.equation(block: false, alt: "270 degrees")[$270^(∘)$]. But degrees are not the only way to specify location on a circle. #figure(figph[circle with quadrantal angles labeled], alt: "circle with quadrantal angles labeled", caption: none) We could use percent of one complete rotation and label the same locations by #math.equation(block: false, alt: "p equals 0 , p equals 25 , p equals 50 , and p equals 75")[$" " p = 0 , " " p = 25 , " " p = 50 , " " "and" " " p = 75$]. Or we could use the time elapsed, so that for this example we would have #math.equation(block: false, alt: "t equals 0 , t equals 2 , t equals 4 , and t equals 6")[$" " t = 0 , " " t = 2 , " " t = 4 , " " "and" " " t = 6$] minutes. Another useful method to describe your location uses the distance traveled, or #strong[arclength], along the circle. How far have you traveled around the Ferris wheel at each of the locations shown? Before we consider that question, let's agree on some vocabulary. An #strong[arc] is a portion of a circle, and its length, quite naturally, is called #strong[arclength]. An angle with vertex at the center of the circle is called a #strong[central angle], and a central angle whose sides meet the endpoints of an arc is said to #strong[subtend] the arc. Or we may say that the angle #strong[spans] the arc. If the arc represents a distance traveled, we sometimes refer to such an angle as the #strong[angle of displacement]. === Arclength Recall that the circumference of a circle is proportional to its radius, #math.equation(block: true, alt: "C equals 2 π r")[$bold(italic(C)) = 2 bold(italic(π)) bold(italic(r))$] If we walk around the entire circumference of a circle, the distance we travel is #math.equation(block: false, alt: "2 π")[$2 π$] times the length of the radius, or about 6.28 times the radius. If we walk only part of the way around the circle, then the distance we travel depends also on the angle of displacement. For example, an angle of #math.equation(block: false, alt: "45 degrees")[$45^(∘)$] is #math.equation(block: false, alt: "the fraction 1 over 8")[$display(frac(1, 8))$] of a complete revolution, so the the length of the arc from point #math.equation(block: false, alt: "A")[$A$] to point #math.equation(block: false, alt: "B")[$B$], called #math.equation(block: false, alt: "s")[$s$] in the figure at right, is #math.equation(block: false, alt: "the fraction 1 over 8")[$display(frac(1, 8))$] of the circumference. Thus #figure(figph[arc on circle], alt: "arc on circle", caption: none) #math.equation(block: true, alt: "s equals the fraction 1 over 8 open parenthesis 2 π r close parenthesis equals the fraction π over 4 r")[$s = display(frac(1, 8)) ( 2 π r ) = display(frac(π, 4)) r$] Similarly, the angle of displacement from point #math.equation(block: false, alt: "A")[$A$] to point #math.equation(block: false, alt: "C")[$C$] is #math.equation(block: false, alt: "the fraction 3 over 4")[$display(frac(3, 4))$] of a complete revolution, so the arclength #math.equation(block: false, alt: "s")[$s$] along the circle from #math.equation(block: false, alt: "A")[$A$] to #math.equation(block: false, alt: "C")[$C$], shown at right, is #figure(figph[arc on circle], alt: "arc on circle", caption: none) #math.equation(block: true, alt: "s equals the fraction 3 over 4 open parenthesis 2 π r close parenthesis equals the fraction 3 π over 2 r")[$s = display(frac(3, 4)) ( 2 π r ) = display(frac(3 π, 2)) r$] In general, for a given circle the length of the arc spanned by an angle is proportional to the size of the angle. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Arclength on a Circle] #math.equation(block: true, alt: "Arclength equals open parenthesis fraction of one revolution close parenthesis times open parenthesis 2 π r close parenthesis")[$"Arclength" " " = " " ( "fraction of one revolution" ) ⋅ ( 2 bold(italic(π)) bold(italic(r)) )$] ] The Ferris wheel in the introduction has circumference #math.equation(block: true, alt: "C equals 2 π open parenthesis 100 close parenthesis equals 628 feet")[$C = 2 π ( 100 ) = 628 " " "feet"$] so in half a revolution you travel 314 feet around the edge, and in one-quarter revolution you travel 157 feet. To indicate the same four locations on the wheel by distance traveled, we would use #math.equation(block: true, alt: "s equals 0 , s equals 157 , s equals 314 , and s equals 471")[$s = 0 , " " s = 157 , " " s = 314 , " " "and" " " s = 471$] , as shown at right. #figure(figph[circle], alt: "circle", caption: none) #examplebox("Example 1")[][ What length of arc is spanned by an angle of #math.equation(block: false, alt: "120 degrees")[$120^(∘)$] on a circle of radius 12 centimeters? #solutionbox[ Because #math.equation(block: false, alt: "the fraction 120 over 360 equals the fraction 1 over 3")[$display(frac(120, 360)) = display(frac(1, 3))$], an angle of #math.equation(block: false, alt: "120 degrees")[$120^(∘)$] is #math.equation(block: false, alt: "the fraction 1 over 3")[$display(frac(1, 3))$] of a complete revolution, as shown at right. Using the formula above with #math.equation(block: false, alt: "r equals 12")[$r = 12$], we find that #math.equation(block: true, alt: "s equals the fraction 1 over 3 open parenthesis 2 π times 12 close parenthesis equals the fraction 2 π over 3 times 12 equals 8 π cm")[$s = display(frac(1, 3)) ( 2 π ⋅ 12 ) = display(frac(2 π, 3)) ⋅ 12 = 8 π " " "cm"$] or about 25.1 cm. #figure(figph[arc on circle], alt: "arc on circle", caption: none) ] ] How far have you traveled around the edge of a Ferris wheel of radius 100 feet when you have turned through an angle of #math.equation(block: false, alt: "150 degrees")[$150^(∘)$]? #math.equation(block: false, alt: "261.8")[$261.8$] ft === Measuring Angles in Radians If you think about measuring arclength, you will see that the degree measure of the spanning angle is not as important as the fraction of one revolution it covers. This observation suggests a new unit of measurement for angles, one that is better suited to calculations involving arclength. We'll make one change in our formula for arclength, from #math.equation(block: true, alt: "Arclength equals open parenthesis fraction of one revolution close parenthesis times open parenthesis 2 π r close parenthesis")[$"Arclength" " " = " " ( "fraction of one revolution" ) ⋅ ( 2 π r )$] to #math.equation(block: true, alt: "Arclength equals open parenthesis fraction of one revolution times 2 π close parenthesis times r")[$"Arclength" " " = " " ( "fraction of one revolution" × 2 bold(italic(π)) ) ⋅ bold(italic(r))$] We'll call the quantity in parentheses, (fraction of one revolution #math.equation(block: false, alt: "times 2 π")[$× 2 π$]), the #strong[radian measure] of the angle that spans the arc. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Radians] The #strong[radian measure] of an angle is given by #math.equation(block: true, alt: "open parenthesis fraction of one revolution times 2 π close parenthesis")[$( "fraction of one revolution" × 2 bold(italic(π)) )$] ] For example, one complete revolution, or #math.equation(block: false, alt: "360 degrees")[$360^(∘)$], is equal to #math.equation(block: false, alt: "2 π")[$2 π$] radians, and one-quarter revolution, or #math.equation(block: false, alt: "90 degrees")[$90^(∘)$], is equal to #math.equation(block: false, alt: "the fraction 1 over 4 open parenthesis 2 π close parenthesis")[$display(frac(1, 4)) ( 2 π )$] or #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$] radians. The figure below shows the radian measure of the quadrantal angles. #figure(figph[angle on unit circle from positive x-axis to positive y-axis], alt: "angle on unit circle from positive x-axis to positive y-axis", caption: none) #figure(figph[angle on unit circle from positive x-axis to negative x-axis], alt: "angle on unit circle from positive x-axis to negative x-axis", caption: none) #figure(figph[angle on unit circle from positive x-axis to negative y-axis], alt: "angle on unit circle from positive x-axis to negative y-axis", caption: none) #figure(figph[angle on unit circle from positive x-axis counterclockwise back to positive x-axis], alt: "angle on unit circle from positive x-axis counterclockwise back to positive x-axis", caption: none) #examplebox("Example 2")[][ What is the radian measure of an angle of #math.equation(block: false, alt: "120 degrees")[$120^(∘)$]? #solutionbox[ An angle of #math.equation(block: false, alt: "120 degrees")[$120^(∘)$] is #math.equation(block: false, alt: "the fraction 1 over 3")[$display(frac(1, 3))$] of a complete revolution, as we saw in the previous example. Thus, an angle of #math.equation(block: false, alt: "120 degrees")[$120^(∘)$] has a radian measure of #math.equation(block: false, alt: "the fraction 1 over 3 open parenthesis 2 π close parenthesis")[$display(frac(1, 3)) ( 2 π )$], or #math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]. ] ] What fraction of a revolution is #math.equation(block: false, alt: "π")[$π$] radians? How many degrees is that? Half a revolution, or #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] Radian measure does not have to be expressed in multiples of #math.equation(block: false, alt: "π")[$π$]. Remember that #math.equation(block: false, alt: "π approximately equals 3.14")[$π ≈ 3.14$], so one complete revolution is about 6.28 radians, and one-quarter revolution is #math.equation(block: false, alt: "the fraction 1 over 4 open parenthesis 2 π close parenthesis equals the fraction π over 2")[$display(frac(1, 4)) ( 2 π ) = display(frac(π, 2))$], or about 1.57 radians. The figure below shows decimal approximations for the quadrantal angles. #figure(table( columns: 3, align: left, inset: 6pt, table.header([Degrees], [Radians: #linebreak() Exact Values #linebreak()], [Radians: Decimal #linebreak() Approximations #linebreak()]), [#math.equation(block: false, alt: "0 degrees")[$0^(∘)$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "90 degrees")[$90^(∘)$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "1.57")[$1.57$]], [#math.equation(block: false, alt: "180 degrees")[$180^(∘)$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "3.14")[$3.14$]], [#math.equation(block: false, alt: "270 degrees")[$270^(∘)$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "4.71")[$4.71$]], [#math.equation(block: false, alt: "360 degrees")[$360^(∘)$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "6.28")[$6.28$]], )) #figure(figph[Quadrantal angles on unit circle radian approximation], alt: "Quadrantal angles on unit circle radian approximation", caption: none) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Because they are "benchmarks" for comparing angles, you should be very familiar with both the exact values of these angles in radians and their approximations! ] #examplebox("Example 3")[][ In which quadrant would you find an angle of 2 radians? An angle of 5 radians? #solutionbox[ Look at the figure above. The second quadrant includes angles between #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$] and #math.equation(block: false, alt: "π")[$π$], or 1.57 and 3.14 radians, so 2 radians lies in the second quadrant. An angle of 5 radians is between 4.71 and 6.28, or between #math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$] and #math.equation(block: false, alt: "2 π")[$2 π$] radians, so it lies in the fourth quadrant. ] ] Draw circles centered at the origin and sketch (in standard position) angles of approximately 3 radians, 4 radians, and 6 radians. #figure(figph[circle with 3 radians], alt: "circle with 3 radians", caption: none) #figure(figph[circle with 4 radians], alt: "circle with 4 radians", caption: none) #figure(figph[circle with 6 radians], alt: "circle with 6 radians", caption: none) It turns out that measuring angles in radians is useful for many applications besides calculating arclengths, so we need to start thinking in radians. To help that process, we'll first learn to convert between degrees and radians. === Converting Between Degrees and Radians It is not difficult to convert the measure of an angle in degrees to its measure in radians, or vice versa. One complete revolution is equal to #math.equation(block: false, alt: "2 π")[$2 π$] radians or to #math.equation(block: false, alt: "360 degrees")[$360^(∘)$], so #math.equation(block: true, alt: "360 degrees equals 2 π radians")[$360^(∘) = 2 π " " "radians"$] If we divide both sides of this equation by #math.equation(block: false, alt: "360 degrees")[$360^(∘)$], we get a fraction that is equal to 1: #math.equation(block: true, alt: "1 equals the fraction 2 π radians over 360 degrees equals the fraction π radians over 180 degrees")[$1 = display(frac(2 π " " "radians", 360^(∘))) = display(frac(π " " "radians", 180^(∘)))$] And of course it is also true that #math.equation(block: true, alt: "1 equals the fraction 180 degrees over π radians")[$1 = display(frac(180^(∘), π " " "radians"))$] Because multiplying by 1 does not change the value of a number, we can use these fractions to convert between degrees and radians. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[ Converting between degrees and radians] - To convert from radians to degrees we multiply the radian measure by #math.equation(block: false, alt: "the fraction 180 degrees over π radians")[$" " display(frac(180^(∘), π " " "radians"))$]. - To convert from degrees to radians we multiply the degree measure by #math.equation(block: false, alt: "the fraction π radians over 180 degrees")[$" " display(frac(π " " "radians", 180^(∘)))$]. ] #examplebox("Example 4")[][ + Convert 3 radians to degrees. + Convert 3 degrees to radians. #solutionbox[ + #math.equation(block: false, alt: "open parenthesis 3 radians close parenthesis times open parenthesis the fraction 180 degrees over π close parenthesis equals the fraction 540 degrees over π approximately equals 171.9 degrees")[$( 3 " " "radians" ) × ( display(frac(180^(∘), π)) ) = display(frac(540^(∘), π)) ≈ 171.9^(∘)$] + #math.equation(block: false, alt: "open parenthesis 3 degrees close parenthesis times open parenthesis the fraction π over 180 degrees close parenthesis equals the fraction π over 60 approximately equals 0.05 radians.")[$( 3^(∘) ) × ( display(frac(π, 180^(∘))) ) = display(frac(π, 60)) ≈ 0.05 " " "radians."$] ] ] + Convert #math.equation(block: false, alt: "60 degrees")[$60^(∘)$] to radians. Give both an exact answer and an approximation to three decimal places. + Convert #math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$] radians to degrees. + #math.equation(block: false, alt: "the fraction π over 3 approximately equals 1.047")[$display(frac(π, 3)) ≈ 1.047$] radians + #math.equation(block: false, alt: "135 degrees")[$135^(∘)$] Thus, the fraction #math.equation(block: false, alt: "the fraction 180 degrees over π")[$display(frac(180^(∘), π))$] (or its reciprocal #math.equation(block: false, alt: "the fraction π over 180 degrees")[$display(frac(π, 180^(∘)))$]) gives us a #emph[conversion factor] between degrees and radians: #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Unit Conversion for Angles] #math.equation(block: true, alt: "the fraction 180 degrees over π radians equals 1")[$display(frac(180^(∘), bold(italic(π)) " " "radians")) = 1$] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ You can review the use of conversion factors in the Algebra Refresher at the end of this Section. ] From our conversion factor we also learn that #math.equation(block: true, alt: "1 radian equals the fraction 180 degrees over π approximately equals 57.3 degrees")[$1 " " "radian" = display(frac(180^(∘), bold(italic(π)))) ≈ 57.3^(∘)$] So while #math.equation(block: false, alt: "1 degrees")[$1^(∘)$] is a relatively small angle, 1 radian is much larger — nearly #math.equation(block: false, alt: "60 degrees")[$60^(∘)$], in fact. #figure(figph[circle with 1 radian], alt: "circle with 1 radian", caption: none) But this is reasonable, because there are only a little more than 6 radians in an entire revolution. An angle of 1 radian is shown above. We'll soon see that, for many applications, it is easier to work entirely in radians. For reference, the figure below shows a radian protractor. #figure(figph[radian protractor], alt: "radian protractor", caption: none) === Arclength Formula Now let us return to our calculation of arclength, and we'll see the first instance in which measuring angles in radians is useful: To calculate an arclength we need only multiply the radius of the circle by the radian measure of the spanning angle, #math.equation(block: false, alt: "θ")[$θ$]. Look again at our formula for arclength: #math.equation(block: true, alt: "Arclength equals open parenthesis fraction of one revolution times 2 π close parenthesis times r")[$"Arclength" " " = " " ( "fraction of one revolution" × 2 bold(italic(π)) ) ⋅ bold(italic(r))$] The quantity in parentheses, fraction of one revolution #math.equation(block: false, alt: "times 2 π")[$× 2 π$], is just the measure of the spanning angle in radians. Thus, if #math.equation(block: false, alt: "θ")[$θ$] is measured in radians, we have the following simple formula for arclength, #math.equation(block: false, alt: "s")[$s$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Arclength Formula] On a circle of radius #math.equation(block: false, alt: "r")[$r$], the #strong[length #math.equation(block: false, alt: "s")[$s$] of an arc] spanned by an angle #math.equation(block: false, alt: "θ")[$θ$] in radians is #math.equation(block: true, alt: "s equals r θ")[$bold(italic(s)) = bold(italic(r)) bold(italic(θ))$] ] Thus, there is a special relationship between arclength and radians. An angle of 1 radian spans an arc on a circle equal to the radius of the circle, as shown at right. And the length of any arc is just the measure of its spanning angle in radians times the radius of the circle. The formula s = rθ with both letters on sliders. The red arc runs counterclockwise from the horizontal spoke through the angle θ, measured in radians, on the dashed circle of radius r; its length is exactly r·θ. Set θ = 1 and you are looking at the sentence above: an angle of one radian spans an arc exactly as long as the spoke — whatever r you choose, since both scale together. Doubling r doubles the arc without touching the angle; doubling θ doubles the arc without touching the circle. Drag θ out to 6.25 and the arc nearly closes the circle — one full turn is 2π ≈ 6.28 radians, just beyond the slider's reach, which is the circumference formula C = 2πr read as arclength.{"functions":\[{"color":"\#6a6a6a","expression":"r\*cos(t)","expression2":"r\*sin(t)","id":"circle","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true},{"color":"\#6a6a6a","expression":"t\*r","expression2":"0","id":"spoke-start","inequality":"lt","lineDash":\[\],"lineWidth":2,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#1f4e79","expression":"t\*r\*cos(th)","expression2":"t\*r\*sin(th)","id":"spoke-end","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#b8442a","expression":"r\*cos(th\*t)","expression2":"r\*sin(th\*t)","id":"arc","inequality":"lt","lineDash":\[\],"lineWidth":3.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#b8442a","expression":"r\*cos(th) + 0.15\*cos(t)","expression2":"r\*sin(th) + 0.15\*sin(t)","id":"arc-end","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-r","label":"Radius r","max":4,"min":1,"name":"r","step":0.5,"value":2},{"id":"p-th","label":"Angle \\u03b8 in radians","max":6.25,"min":0.25,"name":"th","step":0.25,"value":1}\],"title":"The arc spanned by \\u03b8 radians on a circle of radius r","version":1,"viewport":{"centerX":0,"centerY":0,"scale":40}}#figure(figph[arcs on circles], alt: "arcs on circles", caption: none) So for instance, we can use the formula to find the arclength spanned by an angle of 2.5 radians on a circle of radius six inches: #math.equation(block: true, alt: "s equals r θ equals 6 open parenthesis 2.5 close parenthesis equals 15 inches")[$s = r θ = 6 ( 2.5 ) = 15 " " "inches"$] #figure(figph[arclength on circle], alt: "arclength on circle", caption: none) We can also use the formula in the form #math.equation(block: false, alt: "θ equals the fraction s over r")[$θ = display(frac(s, r))$] to find an angle that spans a given arc. For example, an arclength equal to one radius determines a central angle of one radian, or about #math.equation(block: false, alt: "57.3 degrees")[$57.3^(∘)$]. In the next example, we compute a change in latitude on the Earth's surface. #examplebox("Example 5")[][ The radius of the Earth is about 3960 miles. If you travel 500 miles due north, how many degrees of latitude will you traverse? (Latitude is measured in degrees north or south of the equator.) #solutionbox[ We think of the distance 500 miles as an arclength on the surface of the Earth, as shown at right. Substituting #math.equation(block: false, alt: "s equals 500")[$s = 500$] and #math.equation(block: false, alt: "r equals 3960")[$r = 3960$] into the arclength formula gives #math.equation(block: true, alt: "500, equals 3960 θ; θ, equals the fraction 500 over 3960 equals 0.1263 radians")[$500 & = 3960 " " θ \ θ & = display(frac(500, 3960)) = 0.1263 " " "radians"$] #figure(figph[circle], alt: "circle", caption: none) To convert the angle measure to degrees, we multiply by #math.equation(block: false, alt: "the fraction 180 degrees over π")[$display(frac(180^(∘), π))$] to get #math.equation(block: true, alt: "0.1263 open parenthesis the fraction 180 degrees over π close parenthesis equals 7.23 degrees")[$0.1263 ( display(frac(180^(∘), π)) ) = 7.23^(∘)$] Your latitude has changed by about #math.equation(block: false, alt: "7.23 degrees")[$7.23^(∘)$]. ] ] The distance around the face of a large clock from 2 to 3 is five feet. What is the radius of the clock? #math.equation(block: false, alt: "9.55")[$9.55$] ft === Unit Circle In the rest of this chapter we will see how to use the trigonometric functions sine, cosine, and tangent when the input variable is measured in radians instead of degrees, and why making that change greatly increases the utility of those functions. In Section 4.1 we connected the sine and cosine to the coordinates of points on a #strong[unit circle], a circle of radius 1. Here is an important observation that will inform our study: On a unit circle, #math.equation(block: false, alt: "r equals 1")[$r = 1$], so the arclength formula becomes #math.equation(block: false, alt: "s equals θ")[$s = θ$]. Thus, on a unit circle, an arc of length 1 determines a central angle of 1 radian, or about #math.equation(block: false, alt: "57.3 degrees")[$57.3^(∘)$]. And the measure of a (positive) angle in radians is equal to the length of the arc it spans. In other words, if we walk around the circle, each time we travel over an arclength of one radius, we turn through an angle of one radian. #examplebox("Example 6")[][ You are walking on a trail around a circular pond of radius one mile. You have have walked 4 miles from the trailhead. Sketch your location now. #solutionbox[ The pond is a unit circle, so you have traversed an angle in radians equal to the arc length traveled, 4 miles. An angle of 4 radians is in the middle of the third quadrant relative to your starting point, more than halfway but less than three-quarters around the pond. #figure(figph[walk along circle pond], alt: "walk along circle pond", caption: none) ] ] An ant walks around the rim of a circular birdbath of diameter 2 feet. How far has the ant walked when it has turned through an angle of #math.equation(block: false, alt: "210 degrees")[$210^(∘)$]? #figure(figph[walk along circular birdbath], alt: "walk along circular birdbath", caption: none) #math.equation(block: false, alt: "3.67")[$3.67$] ft Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use the appropriate conversion factor to convert units. + #math.equation(block: false, alt: "the fraction 1 mile over 1.609 kilometers equals 1")[$display(frac(1 " " "mile", 1.609 " " "kilometers")) = 1$] + 10 miles = \_\_\_\_\_\_ km + 50 km = \_\_\_\_\_\_ miles + #math.equation(block: false, alt: "the fraction 1 acre over 0.405 hectare equals 1")[$display(frac(1 " " "acre", 0.405 " " "hectare")) = 1$] + 40 acres = \_\_\_\_\_\_ hectares + 5 hectares = \_\_\_\_\_\_acres + #math.equation(block: false, alt: "the fraction 1 horsepower over 746 watts equals 1")[$display(frac(1 " " "horsepower", 746 " " "watts")) = 1$] + 250 horsepower = \_\_\_\_\_\_ watts + 1000 watts = \_\_\_\_\_\_ horsepower + #math.equation(block: false, alt: "the fraction 1 troy ounce over 480 grains equals 1")[$display(frac(1 " " "troy ounce", 480 " " "grains")) = 1$] + 0.5 troy oz = \_\_\_\_\_\_ grains + 100 grains = \_\_\_\_\_\_ troy oz #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + a. #math.equation(block: false, alt: "16.09")[$16.09$] km b. #math.equation(block: false, alt: "31.08")[$31.08$] mi + a. #math.equation(block: false, alt: "16.2")[$16.2$] hectares b.#math.equation(block: false, alt: "12.35")[$12.35$] acres + a. #math.equation(block: false, alt: "186 , 500")[$186 , 500$] watts b. #math.equation(block: false, alt: "1.34")[$1.34$] horsepower + a. #math.equation(block: false, alt: "240")[$240$] grains b. #math.equation(block: false, alt: "0.21")[$0.21$] troy oz ] === Section 6.1 Summary ==== Vocabulary - Arclength - Radian - Conversion factor - Latitude - Unit circle ==== Concepts + The distance we travel around a circle of radius is proportional to the angle of displacement. #math.equation(block: true, alt: "Arclength equals open parenthesis fraction of one revolution close parenthesis times open parenthesis 2 π r close parenthesis")[$"Arclength" " " = " " ( "fraction of one revolution" ) ⋅ ( 2 π r )$] + We measure angles in radians when we work with arclength.#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Radians] The #strong[radian measure] of an angle is given by #math.equation(block: true, alt: "open parenthesis fraction of one revolution times 2 π close parenthesis")[$( "fraction of one revolution" × 2 π )$] ] + An arclength equal to one radius determines a central angle of one radian. + Radian measure can be expressed as multiples of #math.equation(block: false, alt: "π")[$π$] or as decimals.#figure(table( columns: 3, align: left, inset: 6pt, table.header([Degrees], [#math.equation(block: false, alt: "the fraction Radians: over Exact Values")[$display(frac("Radians:", "Exact Values"))$]], [#math.equation(block: false, alt: "the fraction Radians: Decimal over Approximations")[$display(frac("Radians: Decimal", "Approximations"))$]]), [#math.equation(block: false, alt: "0 degrees")[$0^(∘)$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "90 degrees")[$90^(∘)$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "1.57")[$1.57$]], [#math.equation(block: false, alt: "180 degrees")[$180^(∘)$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "3.14")[$3.14$]], [#math.equation(block: false, alt: "270 degrees")[$270^(∘)$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "4.71")[$4.71$]], [#math.equation(block: false, alt: "360 degrees")[$360^(∘)$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "6.28")[$6.28$]], )) #figure(figph[Quadrantal angles on unit circle radian approximation], alt: "Quadrantal angles on unit circle radian approximation", caption: none) + We multiply by the appropriate conversion factor to convert between degrees and radians.#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Unit Conversion for Angles] #math.equation(block: true, alt: "the fraction 180 degrees over π radians equals 1")[$display(frac(180^(∘), π " " "radians")) = 1$] ] #linebreak() To convert from radians to degrees we multiply the radian measure by #math.equation(block: false, alt: "the fraction 180 degrees over π")[$display(frac(180^(∘), π))$]. #linebreak() To convert from degrees to radians we multiply the degree measure by #math.equation(block: false, alt: "the fraction π over 180")[$display(frac(π, 180))$]. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Arclength Formula] On a circle of radius #math.equation(block: false, alt: "r")[$r$], the #strong[length #math.equation(block: false, alt: "s")[$s$] of an arc] spanned by an angle #math.equation(block: false, alt: "θ")[$θ$] in radians is #math.equation(block: true, alt: "s equals r θ")[$s = r θ$] ] + On a #strong[unit circle], the measure of a (positive) angle in radians is equal to the length of the arc it spans. ==== Study Questions + The length of a circular arc depends on what two variables? + Define the radian measure of an angle. + What is the conversion factor from radians to degrees? + On a unit circle, the length of an arc is equal to what other quantity? ==== Skills + Express angles in degrees and radians \#1–8, 25–32 + Sketch angles given in radians \#1 and 2, 11 and 12 + Estimate angles in radians \#9–10, 13–24 + Use the arclength formula \#33–46 + Find coordinates of a point on a unit circle \#47–52 + Calculate angular velocity and area of a sector \#55–60 === Homework 6.1 #figure(table( columns: 10, align: left, inset: 6pt, table.header([Radians], [#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")[$2 π$]]), [Degrees], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Convert each angle to degrees. + Sketch each angle on a circle like this one, and label in radians. #figure(figph[circle], alt: "circle", caption: none) #figure(table( columns: 10, align: left, inset: 6pt, table.header([Radians], [#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")[$2 π$]]), [Degrees], [#math.equation(block: false, alt: "0 degrees")[$0^(∘)$]], [#math.equation(block: false, alt: "45 degrees")[$45^(∘)$]], [#math.equation(block: false, alt: "90 degrees")[$90^(∘)$]], [#math.equation(block: false, alt: "135 degrees")[$135^(∘)$]], [#math.equation(block: false, alt: "180 degrees")[$180^(∘)$]], [#math.equation(block: false, alt: "225 degrees")[$225^(∘)$]], [#math.equation(block: false, alt: "270 degrees")[$270^(∘)$]], [#math.equation(block: false, alt: "315 degrees")[$315^(∘)$]], [#math.equation(block: false, alt: "360 degrees")[$360^(∘)$]], )) #figure(figph[circle], alt: "circle", caption: none) #figure(table( columns: 14, align: left, inset: 6pt, table.header([Radians], [#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: "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 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 11 π over 6")[$display(frac(11 π, 6))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [Degrees], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Convert each angle to degrees. + Sketch each angle on a circle like this one, and label in radians. #figure(figph[circle], alt: "circle", caption: none) For Problems 3–6, express each fraction of one complete rotation in degrees and in radians. + #math.equation(block: false, alt: "the fraction 1 over 3")[$display(frac(1, 3))$] + #math.equation(block: false, alt: "the fraction 2 over 3")[$display(frac(2, 3))$] + #math.equation(block: false, alt: "the fraction 4 over 3")[$display(frac(4, 3))$] + #math.equation(block: false, alt: "the fraction 5 over 3")[$display(frac(5, 3))$] #figure(figph[circle], alt: "circle", caption: none) + #math.equation(block: false, alt: "120 degrees equals the fraction 2 π over 3 radians")[$120^(∘) = display(frac(2 π, 3)) "radians"$] + #math.equation(block: false, alt: "240 degrees equals the fraction 4 π over 3 radians")[$240^(∘) = display(frac(4 π, 3)) "radians"$] + #math.equation(block: false, alt: "480 degrees equals the fraction 8 π over 3 radians")[$480^(∘) = display(frac(8 π, 3)) "radians"$] + #math.equation(block: false, alt: "600 degrees equals the fraction 10 π over 3 radians")[$600^(∘) = display(frac(10 π, 3)) "radians"$] + #math.equation(block: false, alt: "the fraction 1 over 5")[$display(frac(1, 5))$] + #math.equation(block: false, alt: "the fraction 2 over 5")[$display(frac(2, 5))$] + #math.equation(block: false, alt: "the fraction 3 over 5")[$display(frac(3, 5))$] + #math.equation(block: false, alt: "the fraction 4 over 5")[$display(frac(4, 5))$] #figure(figph[circle], alt: "circle", caption: none) + #math.equation(block: false, alt: "the fraction 1 over 8")[$display(frac(1, 8))$] + #math.equation(block: false, alt: "the fraction 3 over 8")[$display(frac(3, 8))$] + #math.equation(block: false, alt: "the fraction 5 over 8")[$display(frac(5, 8))$] + #math.equation(block: false, alt: "the fraction 7 over 8")[$display(frac(7, 8))$] #figure(figph[circle], alt: "circle", caption: none) + #math.equation(block: false, alt: "45 degrees equals the fraction π over 4 radians")[$45^(∘) = display(frac(π, 4)) "radians"$] + #math.equation(block: false, alt: "135 degrees equals the fraction 3 π over 4 radians")[$135^(∘) = display(frac(3 π, 4)) "radians"$] + #math.equation(block: false, alt: "225 degrees equals the fraction 5 π over 4 radians")[$225^(∘) = display(frac(5 π, 4)) "radians"$] + #math.equation(block: false, alt: "315 degrees equals the fraction 7 π over 4 radians")[$315^(∘) = display(frac(7 π, 4)) "radians"$] + #math.equation(block: false, alt: "the fraction 1 over 12")[$display(frac(1, 12))$] + #math.equation(block: false, alt: "the fraction 1 over 6")[$display(frac(1, 6))$] + #math.equation(block: false, alt: "the fraction 5 over 12")[$display(frac(5, 12))$] + #math.equation(block: false, alt: "the fraction 5 over 6")[$display(frac(5, 6))$] #figure(figph[circle], alt: "circle", caption: none) For Problems 7–8, label each angle in standard position with radian measure. Rotate counter-clockwise from 0. #figure(figph[circle], alt: "circle", caption: none) #figure(figph[circle], alt: "circle", caption: none) Rotate clockwise from 0. #figure(figph[circle], alt: "circle", caption: none) For Problems 9–10, give a decimal approximation to hundredths for each angle in radians. + #math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$] + #math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$] + #math.equation(block: false, alt: "the fraction 7 π over 6")[$display(frac(7 π, 6))$] + #math.equation(block: false, alt: "the fraction 11 π over 6")[$display(frac(11 π, 6))$] + #math.equation(block: false, alt: "0.52")[$0.52$] + #math.equation(block: false, alt: "2.62")[$2.62$] + #math.equation(block: false, alt: "3.67")[$3.67$] + #math.equation(block: false, alt: "5.76")[$5.76$] + #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 5 π over 4")[$display(frac(5 π, 4))$] + #math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$] Locate and label each angle from Problem 9 on the unit circle below. (The circle is marked off in tenths of a radian.) #figure(figph[circle], alt: "circle", caption: none) #figure(figph[circle], alt: "circle", caption: none) Locate and label each angle from Problem 10 on the unit circle below. (The circle is marked off in tenths of a radian.) #figure(figph[circle], alt: "circle", caption: none) From the list below, choose the best decimal approximation for each angle in radians in Problems 13–20. Do not use a calculator; use the fact that #math.equation(block: false, alt: "π")[$π$] is a little greater than 3. #math.equation(block: true, alt: "0.52 , 0.79 , 2.09 , 2.36 , 2.62 , 3.67 , 5.24 , 5.50")[$0.52 , " " " " 0.79 , " " " " 2.09 , " " " " 2.36 , " " " " 2.62 , " " " " 3.67 , " " " " 5.24 , " " " " 5.50$] #math.equation(block: true, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$] #math.equation(block: true, alt: "2.09")[$2.09$] #math.equation(block: true, alt: "the fraction π over 4")[$display(frac(π, 4))$] #math.equation(block: true, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$] #math.equation(block: true, alt: "2.62")[$2.62$] #math.equation(block: true, alt: "the fraction 5 π over 3")[$display(frac(5 π, 3))$] #math.equation(block: true, alt: "the fraction π over 6")[$display(frac(π, 6))$] #math.equation(block: true, alt: "0.52")[$0.52$] #math.equation(block: true, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$] #math.equation(block: true, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$] #math.equation(block: true, alt: "2.36")[$2.36$] #math.equation(block: true, alt: "the fraction 7 π over 6")[$display(frac(7 π, 6))$] For Problems 21–24, say in which quadrant each angle lies. + #math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$] + #math.equation(block: false, alt: "the fraction 11 π over 6")[$display(frac(11 π, 6))$] + #math.equation(block: false, alt: "the fraction minus π over 6")[$display(frac(− π, 6))$] + #math.equation(block: false, alt: "the fraction minus 7 π over 4")[$display(frac(− 7 π, 4))$] + II + IV + IV + I + #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 8")[$display(frac(3 π, 8))$] + #math.equation(block: false, alt: "the fraction minus 5 π over 6")[$display(frac(− 5 π, 6))$] + #math.equation(block: false, alt: "the fraction minus 2 π over 3")[$display(frac(− 2 π, 3))$] + #math.equation(block: false, alt: "3.5")[$3.5$] + #math.equation(block: false, alt: "1.9")[$1.9$] + #math.equation(block: false, alt: "0.8")[$0.8$] + #math.equation(block: false, alt: "5.5")[$5.5$] + III + II + I + IV + #math.equation(block: false, alt: "4.0")[$4.0$] + #math.equation(block: false, alt: "2.6")[$2.6$] + #math.equation(block: false, alt: "6.1")[$6.1$] + #math.equation(block: false, alt: "1.5")[$1.5$] For Problems 25–28, complete the table. #figure(table( columns: 4, align: left, inset: 6pt, table.header([Radians], [#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")[$display(frac(π, 3))$]]), [Degrees], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(table( columns: 4, align: left, inset: 6pt, table.header([Radians], [#math.equation(block: false, alt: "the fraction π over 6")[$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))$]]), [Degrees], [#math.equation(block: false, alt: "30 degrees")[$30^(∘)$]], [#math.equation(block: false, alt: "45 degrees")[$45^(∘)$]], [#math.equation(block: false, alt: "60 degrees")[$60^(∘)$]], )) #figure(table( columns: 4, align: left, inset: 6pt, table.header([Radians], [#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")[$display(frac(5 π, 6))$]]), [Degrees], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(table( columns: 4, align: left, inset: 6pt, table.header([Radians], [#math.equation(block: false, alt: "the fraction 7 π over 6")[$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))$]]), [Degrees], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(table( columns: 4, align: left, inset: 6pt, table.header([Radians], [#math.equation(block: false, alt: "the fraction 7 π over 6")[$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))$]]), [Degrees], [#math.equation(block: false, alt: "210 degrees")[$210^(∘)$]], [#math.equation(block: false, alt: "225 degrees")[$225^(∘)$]], [#math.equation(block: false, alt: "240 degrees")[$240^(∘)$]], )) #figure(table( columns: 4, align: left, inset: 6pt, table.header([Radians], [#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")[$display(frac(11 π, 6))$]]), [Degrees], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) For Problems 29–30, convert to radians. Round to hundredths. + #math.equation(block: false, alt: "75 degrees")[$75^(∘)$] + #math.equation(block: false, alt: "236 degrees")[$236^(∘)$] + #math.equation(block: false, alt: "327 degrees")[$327^(∘)$] + #math.equation(block: false, alt: "1.31")[$1.31$] + #math.equation(block: false, alt: "4.12")[$4.12$] + #math.equation(block: false, alt: "5.71")[$5.71$] + #math.equation(block: false, alt: "138 degrees")[$138^(∘)$] + #math.equation(block: false, alt: "194 degrees")[$194^(∘)$] + #math.equation(block: false, alt: "342 degrees")[$342^(∘)$] For Problems 31–32, convert to degrees. Round to tenths. + #math.equation(block: false, alt: "0.8")[$0.8$] + #math.equation(block: false, alt: "3.5")[$3.5$] + #math.equation(block: false, alt: "5.1")[$5.1$] + #math.equation(block: false, alt: "45.8 degrees")[$45.8^(∘)$] + #math.equation(block: false, alt: "200.5 degrees")[$200.5^(∘)$] + #math.equation(block: false, alt: "292.2 degrees")[$292.2^(∘)$] + #math.equation(block: false, alt: "1.1")[$1.1$] + #math.equation(block: false, alt: "2.6")[$2.6$] + #math.equation(block: false, alt: "4.6")[$4.6$] For Problems 33–37, use the arclength formula to answer the questions. Round answers to hundredths. Find the arclength spanned by an angle of #math.equation(block: false, alt: "80 degrees")[$80^(∘)$] on a circle of radius 4 inches. #math.equation(block: true, alt: "5.86 in")[$5.86 " " "in"$] Find the arclength spanned by an angle of #math.equation(block: false, alt: "200 degrees")[$200^(∘)$] on a circle of radius 18 feet. Find the radius of a cricle if an angle of #math.equation(block: false, alt: "250 degrees")[$250^(∘)$] spans an arclength of 18 meters. #math.equation(block: true, alt: "4.13 m")[$4.13 " " "m"$] Find the radius of a cricle if an angle of #math.equation(block: false, alt: "20 degrees")[$20^(∘)$] spans an arclength of 0.5 kilometers. Find the angle subtended by an arclength of 28 centimeters on a circle of diameter 20 centimeters. #math.equation(block: true, alt: "160.42 degrees")[$160.42^(∘)$] Find the angle subtended by an arclength of 1.6 yards on a circle of diameter 2 yards. For Problems 39–46, use the arclength formula to answer the questions. + Through how many radians does the minute hand of a clock sweep between 9:05 pm and 9:30 pm? + The dial of Big Ben's clock in London is 23 feet in diameter. How long is the arc traced by the minute hand between 9:05 pm and 9:30 pm? + #math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$] + #math.equation(block: false, alt: "32.72 ft")[$32.72 " " "ft"$] The largest clock ever constructed was the Floral Clock in the garden of the 1904 World's Fair in St. Louis. The hour hand was 50 feet long, the minute hand was 75 feet long, and the radius of the clockface was 112 feet. + If you started at the 12 and walked 500 feet clockwise around the clockface, through how many radians would you walk? + If you started your walk at noon, how long would it take the minute hand to reach your position? How far did the tip of the minute hand move in its arc? In 1851 Jean-Bernard Foucault demonstrated the rotation of the earth with a pendulum installed in the Pantheon in Paris. Foucault's pendulum consisted of a cannonball suspended on a 67 meter wire, and it swept out an arc of 8 meters on each swing. Through what angle did the pendulum swing? Give your answer in radians and then in degrees, rounded to the nearest hundredth. #math.equation(block: true, alt: "the fraction 8 over 67 radians approximately equals 6.84 degrees")[$display(frac(8, 67)) " " "radians" " " ≈ 6.84^(∘)$] A wheel with radius 40 centimeters is rolled a distance of 1000 centimeters on a flat surface. Through what angle has the wheel rotated? Give your answer in radians and then in degrees, rounded to one decimal place. Clothes dryers draw 3.5 times as much power as washing machines, so newer machines have been engineered for greater efficiency. A vigorous spin cycle reduces the time needed for drying, and some front-loading models spin at a rate of 1500 rotations per minute. + If the radius of the drum is 11 inches, how far do your socks travel in one minute? + How fast are your socks traveling during the spin cycle? + #math.equation(block: false, alt: "33 , 000 π approximately equals 103 , 672.6 in")[$33 , 000 π ≈ 103 , 672.6 " " "in"$] + #math.equation(block: false, alt: "33 , 000 π approximately equals 103.672.6 in per min")[$33 , 000 π ≈ 103.672 .6 " " "in per min"$] The Hubble telescope is in orbit around the earth at an altitude of 600 kilometers, and completes one orbit in 97 minutes. + How far does the telescope travel in one hour? (The radius of the earth is 6400 kilometers.) + What is the speed of the Hubble telescope? The first large windmill used to generate electricity was built in Cleveland, Ohio in 1888. Its sails were 17 meters in diameter, and moved at 10 rotations per minute. How fast did the ends of the sails travel? #math.equation(block: true, alt: "170 π approximately equals 534.1 m per min")[$170 π ≈ 534.1 " " "m per min"$] The largest windmill operating today has wings 54 meters in length. To be most efficient, the tips of the wings must travel at 50 meters per second. How fast must the wings rotate? For Problems 47–52, find two points on the unit circle with the given coordinate.Sketch the approximate location of the points on the circle. (Hint: what is the equation for the unit circle?) #math.equation(block: true, alt: "x equals 0.2")[$x = 0.2$] #figure(figph[unit circle], alt: "unit circle", caption: none) #math.equation(block: true, alt: "open parenthesis 0.2 , 0.98 close parenthesis")[$( 0.2 , 0.98 )$], #math.equation(block: true, alt: "open parenthesis 0.2 , minus 0.98 close parenthesis")[$" " ( 0.2 , − 0.98 )$] #math.equation(block: true, alt: "x equals minus 0.6")[$x = − 0.6$] #math.equation(block: true, alt: "y equals minus 0.35")[$y = − 0.35$] #figure(figph[unit circle], alt: "unit circle", caption: none) #math.equation(block: true, alt: "open parenthesis 0.94 , minus 0.35 close parenthesis")[$( 0.94 , − 0.35 )$], #math.equation(block: true, alt: "open parenthesis minus 0.94 , minus 0.35 close parenthesis")[$" " ( − 0.94 , − 0.35 )$] #math.equation(block: true, alt: "y equals 0.7")[$y = 0.7$] #math.equation(block: true, alt: "x equals the fraction minus the square root of 3 over 2")[$x = display(frac(− sqrt(3), 2))$] #figure(figph[unit circle], alt: "unit circle", caption: none) #math.equation(block: true, alt: "open parenthesis the fraction minus the square root of 3 over 2 , the fraction 1 over 2 close parenthesis")[$( display(frac(− sqrt(3), 2)) , display(frac(1, 2)) )$], #math.equation(block: true, alt: "open parenthesis the fraction minus the square root of 3 over 2 , the fraction minus 1 over 2 close parenthesis")[$" " ( display(frac(− sqrt(3), 2)) , display(frac(− 1, 2)) )$] #math.equation(block: true, alt: "y equals the fraction 1 over the square root of 2")[$y = display(frac(1, sqrt(2)))$] + Sketch a circle of radius 4 units, and mark the positions of 1, 2, 3, 4, 5, and 6 radians on the circle. + On a circle of radius 4 feet, find the arclength determined by each angle in radians.#figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#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: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]]), [#math.equation(block: false, alt: "s")[$s$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Graph #math.equation(block: false, alt: "θ")[$θ$] against #math.equation(block: false, alt: "s")[$s$]. What is the slope of the graph? + If you double the angle , what happens to the arclength? What happens if you triple #math.equation(block: false, alt: "θ")[$θ$] ? #figure(figph[grid], alt: "grid", caption: none) + #figure(figph[circle], alt: "circle", caption: none) + #figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "θ")[$θ$]], [#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: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]]), [#math.equation(block: false, alt: "s")[$s$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "24")[$24$]], )) + #figure(figph[linear graph of arclength vs angle], alt: "linear graph of arclength vs angle", caption: none) #linebreak() #math.equation(block: false, alt: "m equals 4")[$m = 4$] + Arclength doubles; arclength triples + Sketch several concentric circles with increasing radius, and draw an angle of 2 radians through all of them. + Find the arclength determined by an angle of 2 radians on circles of given radius.#figure(table( columns: 7, align: left, inset: 6pt, table.header([#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: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]]), [#math.equation(block: false, alt: "s")[$s$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Graph #math.equation(block: false, alt: "s")[$s$] against #math.equation(block: false, alt: "r")[$r$]. What is the slope of the graph? + If you double the radus #math.equation(block: false, alt: "r")[$r$], what happens to the arclength? What happens if you triple #math.equation(block: false, alt: "r")[$r$]? #figure(figph[grid], alt: "grid", caption: none) The angular velocity, #math.equation(block: false, alt: "ω")[$ω$], of a rotating object is given in radians per unit time. Thus, an object that rotates through #math.equation(block: false, alt: "θ")[$θ$] radians in time #math.equation(block: false, alt: "t")[$t$] has angular velocity given by #math.equation(block: false, alt: "ω equals the fraction θ over t")[$ω = display(frac(θ, t))$]. Find the angular velocity of the following objects. + The London Eye ferris wheel, which makes one revolution every 20 minutes. + An old-fashioned long-playing record, which revolved #math.equation(block: false, alt: "33 the fraction 1 over 3")[$33 frac(1, 3)$] times in 60 seconds. + #math.equation(block: false, alt: "the fraction π over 10 radians per min")[$display(frac(π, 10)) " " "radians per min"$] + #math.equation(block: false, alt: "the fraction 10 π over 9 radians per sec")[$display(frac(10 π, 9)) " " "radians per sec"$] Use the arclength formula to derive a formula relating linear velocity, #math.equation(block: false, alt: "v")[$v$], and angular velocity, #math.equation(block: false, alt: "ω")[$ω$]. (See Problem 55 for the definition of angular velocity.) Start with the formula for linear velocity: #math.equation(block: true, alt: "velocity equals the fraction distance over time , or v equals the fraction s over t")[$"velocity" = display(frac("distance", "time")) , " " " " " " "or" " " " " " " v = display(frac(s, t))$] and substitute the arclength formula for #math.equation(block: false, alt: "s")[$s$]. Recall that to calculate a fraction of a revolution in degrees we divide the angle by #math.equation(block: false, alt: "360 degrees")[$360^(∘)$]. For example, #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] is #math.equation(block: false, alt: "the fraction 1 over 4")[$display(frac(1, 4))$] of a revolution because #math.equation(block: false, alt: "the fraction 90 degrees over 360 degrees equals the fraction 1 over 4")[$display(frac(90^(∘), 360^(∘))) = display(frac(1, 4))$]. + Write an expression that gives the fraction of a revolution for an angle #math.equation(block: false, alt: "θ")[$θ$] in radians. + Use your expression to calculate what fraction of a revolution is represented by each of the following angles: #math.equation(block: false, alt: "θ equals the fraction 3 π over 4 , θ equals the fraction 5 π over 3 , θ equals the fraction 7 π over 6")[$" " θ = display(frac(3 π, 4)) , " " " " θ = display(frac(5 π, 3)) , " " " " θ = display(frac(7 π, 6))$]. + #math.equation(block: false, alt: "the fraction θ over 2 π")[$display(frac(θ, 2 π))$] + #math.equation(block: false, alt: "the fraction 3 over 8 , the fraction 5 over 6 , the fraction 7 over 12")[$display(frac(3, 8)) , " " display(frac(5, 6)) , " " display(frac(7, 12))$] Use your result from Problem 57a to write each statement as a mathematical formula. #figure(figph[circle], alt: "circle", caption: none) #figure(figph[circle], alt: "circle", caption: none) + The length of the arc, #math.equation(block: false, alt: "s")[$s$], is equal to: #math.equation(block: true, alt: "(the fraction of a revolution) times (circumference of the circle)")[$"(the fraction of a revolution)" × "(circumference of the circle)"$] + The area of the sector, #math.equation(block: false, alt: "A")[$A$], is equal to: #math.equation(block: true, alt: "(the fraction of a revolution) times (area of the circle)")[$"(the fraction of a revolution)" × "(area of the circle)"$] For Problems 59–60, Use the formula for the area of a sector from Problem 58. #figure(figph[circle], alt: "circle", caption: none) #math.equation(block: true, alt: "32.5 cm squared")[$32.5 " " "cm"^(2)$] #figure(figph[circle], alt: "circle", caption: none)