#set document(title: "3.1 Variation", 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")) == 3.1#h(0.6em)Variation Two types of functions are widely used in modeling and are known by special names: #strong[direct variation] and #strong[inverse variation]. === Direct Variation Two variables are #strong[directly proportional] (or just #strong[proportional]) if the ratios of their corresponding values are always equal. Consider the functions described in the tables below. The first table shows the price of gasoline as a function of the number of gallons purchased. #figure(table( columns: 3, align: left, inset: 6pt, table.header([Gallons of #linebreak() gasoline #linebreak()], [Total #linebreak() price #linebreak()], [Price/ #linebreak() Gallons #linebreak()]), [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "$ 9.60")[$upright(\$) 9.60$]], [#math.equation(block: false, alt: "the fraction 9.60 over 4 equals 2.40")[$display(frac(9.60, 4)) = 2.40$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "$ 14.40")[$upright(\$) 14.40$]], [#math.equation(block: false, alt: "the fraction 14.40 over 6 equals 2.40")[$display(frac(14.40, 6)) = 2.40$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "$ 19.20")[$upright(\$) 19.20$]], [#math.equation(block: false, alt: "the fraction 19.20 over 8 equals 2.40")[$display(frac(19.20, 8)) = 2.40$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "$ 28.80")[$upright(\$) 28.80$]], [#math.equation(block: false, alt: "the fraction 28.80 over 12 equals 2.40")[$display(frac(28.80, 12)) = 2.40$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "$ 36.00")[$upright(\$) 36.00$]], [#math.equation(block: false, alt: "the fraction 36.00 over 15 equals 2.40")[$display(frac(36.00, 15)) = 2.40$]], )) #figure(table( columns: 3, align: left, inset: 6pt, table.header([Years], [Population], [People/Years]), [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "432")[$432$]], [#math.equation(block: false, alt: "the fraction 432 over 10 approximately equals 43")[$display(frac(432, 10)) ≈ 43$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "932")[$932$]], [#math.equation(block: false, alt: "the fraction 932 over 20 approximately equals 47")[$display(frac(932, 20)) ≈ 47$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "2013")[$2013$]], [#math.equation(block: false, alt: "the fraction 2013 over 30 approximately equals 67")[$display(frac(2013, 30)) ≈ 67$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "4345")[$4345$]], [#math.equation(block: false, alt: "the fraction 4345 over 40 approximately equals 109")[$display(frac(4345, 40)) ≈ 109$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "9380")[$9380$]], [#math.equation(block: false, alt: "the fraction 9380 over 50 approximately equals 188")[$display(frac(9380, 50)) ≈ 188$]], [#math.equation(block: false, alt: "60")[$60$]], [#math.equation(block: false, alt: "20 , 251")[$20 , 251$]], [#math.equation(block: false, alt: "the fraction 20 , 251 over 60 approximately equals 338")[$display(frac(20 "," 251, 60)) ≈ 338$]], )) The ratio #math.equation(block: false, alt: "the fraction total price over number of gallons")[$display(frac("total price", "number of gallons"))$], or price per gallon, is the same for each pair of values in the first table. This agrees with everyday experience: The price per gallon of gasoline is the same no matter how many gallons you buy. Thus, the total price of a gasoline purchase is directly proportional to the number of gallons purchased. The second table shows the population of a small town as a function of the town’s age. The ratio #math.equation(block: false, alt: "the fraction number of people over number of years")[$display(frac("number of people", "number of years"))$] gives the average rate of growth of the population in people per year. You can see that this ratio is #emph[not] constant; in fact, it increases as time goes on. Thus, the population of the town is #emph[not] proportional to its age. When does a table represent direct variation? \_\_\_\_\_ #solutionbox[ A table can represent direct variation if the ratio output/input is constant ] When does a table represent direct variation? + If it has a constant slope. + If it includes the point #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]. + If the ratio output/input is constant. + If each output is double the previous one. The graphs of these two functions are shown below. #figure(figph[linear graph and population growth graph], alt: "linear graph and population growth graph", caption: none) We see that the price, #math.equation(block: false, alt: "P")[$P$], of a fill-up is a linear function of the number of gallons, #math.equation(block: false, alt: "g")[$g$], purchased. This should not be surprising if we write an equation relating the variables #math.equation(block: false, alt: "g")[$g$] and #math.equation(block: false, alt: "P")[$P$]. Because the ratio of their values is constant, we can write #math.equation(block: true, alt: "the fraction P over g equals k")[$frac(P, g) = k$] where #math.equation(block: false, alt: "k")[$k$] is a constant. In this example, the constant #math.equation(block: false, alt: "k")[$k$] is #math.equation(block: false, alt: "2.40")[$2.40$], the price of gasoline per gallon. Solving for #math.equation(block: false, alt: "P")[$P$] in terms of #math.equation(block: false, alt: "g")[$g$], we have #math.equation(block: true, alt: "P equals k g equals 2.40 g")[$P = k g = 2.40 g$] which we recognize as the equation of a line through the origin. In general, we make the following definition. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Direct Variation] #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x")[$x$] if #math.equation(block: true, alt: "y equals k x")[$y = k x$] where #math.equation(block: false, alt: "k")[$k$] is a positive constant called the #strong[constant of variation]. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ From the preceding discussion, we see that #emph[vary directly] means exactly the same thing as #emph[are directly proportional]. The two phrases are interchangeable. ] #examplebox("Example 1")[][ + The circumference, #math.equation(block: false, alt: "C")[$C$], of a circle varies directly with its radius, #math.equation(block: false, alt: "r")[$r$], because #math.equation(block: true, alt: "C equals 2 π r")[$C = 2 π r$] The constant of variation is #math.equation(block: false, alt: "2 π")[$2 π$], or about 6.28. + The amount of interest, #math.equation(block: false, alt: "I")[$I$], earned in one year on an account paying 7% simple interest, varies directly with the principal, #math.equation(block: false, alt: "P")[$P$], invested, because #math.equation(block: true, alt: "I equals 0.07 P")[$I = 0.07 P$] ] Direct variation defines a linear function of the form #math.equation(block: true, alt: "y equals f open parenthesis x close parenthesis equals k x")[$y = f ( x ) = k x$] The positive constant #math.equation(block: false, alt: "k")[$k$] in the equation #math.equation(block: false, alt: "y equals k x")[$y = k x$] is just the slope of the graph, so it tells us how rapidly the graph increases. Compared to the standard form for a linear function, #math.equation(block: false, alt: "y equals b plus m x")[$y = b + m x$], the constant term, #math.equation(block: false, alt: "b")[$b$], is zero, so the graph of a direct variation passes through the origin. #figure(figph[graphs of three lines], alt: "graphs of three lines", caption: none) Which of the graphs above could represent direct variation? Explain why. \_\_\_\_\_ #solutionbox[ (b): The graph is a straight line through the origin. ] #figure(figph[graphs of three lines], alt: "graphs of three lines", caption: none) Which of the graphs above could represent direct variation? Explain why. #solutionbox[ (b): The graph is a straight line through the origin. ] If \\(y\\) varies directly with \\(x\\), what does the constant of variation tell us? \_\_\_\_\_ #solutionbox[ If \\(y\\) varies directly with \\(x\\), the constant of variation is the slope of the graph ] If #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x")[$x$], what does the constant of variation tell us? + The #math.equation(block: false, alt: "y")[$y$]-intercept of the graph. + The slope of the graph. + What happens to #math.equation(block: false, alt: "y")[$y$] when you double #math.equation(block: false, alt: "x")[$x$]. + The size of the input values. === The Scaling Property of Direct Variation The fact that the constant term is zero in a direct variation is significant: If we double the value of #math.equation(block: false, alt: "x")[$x$], then the value of #math.equation(block: false, alt: "y")[$y$] will double also. In fact, increasing #math.equation(block: false, alt: "x")[$x$] by any factor causes #math.equation(block: false, alt: "y")[$y$] to increase by the same factor. For example, in the table of gasoline prices, doubling the number of gallons of gas purchased, say, from #math.equation(block: false, alt: "4")[$4$] gallons to #math.equation(block: false, alt: "8")[$8$] gallons or from #math.equation(block: false, alt: "6")[$6$] gallons to #math.equation(block: false, alt: "12")[$12$] gallons, causes the total price to double also. Or, consider investing \$#math.equation(block: false, alt: "800")[$800$] for one year at 7% simple interest, as in Exampleb. The interest earned is #math.equation(block: true, alt: "I equals 0.07 open parenthesis 800 close parenthesis equals $ 56")[$I = 0.07 ( 800 ) = upright(\$) 56$] If we increase the investment by a factor of #math.equation(block: false, alt: "1.6")[$1.6$] to #math.equation(block: false, alt: "1.6 open parenthesis 800 close parenthesis")[$1.6 ( 800 )$], or \$#math.equation(block: false, alt: "1280")[$1280$], the interest will be #math.equation(block: true, alt: "I equals 0.07 open parenthesis 1280 close parenthesis equals $ 89.60")[$I = 0.07 ( 1280 ) = upright(\$) 89.60$] You can check that multiplying the original interest of \$#math.equation(block: false, alt: "56")[$56$] by a factor of #math.equation(block: false, alt: "1.6")[$1.6$] does give the same figure for the new interest, \$#math.equation(block: false, alt: "89.60")[$89.60$]. #examplebox("Example 2")[][ + Tuition at Woodrow University is \$400 plus \$30 per unit. Is the tuition proportional to the number of units you take? + Imogen makes a 15% commission on her sales of environmentally friendly products marketed by her co-op. Do her earnings vary directly with her sales? #solutionbox[ + Let #math.equation(block: false, alt: "u")[$u$] represent the number of units you take, and let #math.equation(block: false, alt: "T open parenthesis u close parenthesis")[$T ( u )$] represent your tuition. Then #math.equation(block: true, alt: "T open parenthesis u close parenthesis equals 400 plus 30 u")[$T ( u ) = 400 + 30 u$] Thus, #math.equation(block: false, alt: "T open parenthesis u close parenthesis")[$T ( u )$] is a linear function of #math.equation(block: false, alt: "u")[$u$], but the #math.equation(block: false, alt: "T")[$T$]-intercept is #math.equation(block: false, alt: "400")[$400$], not #math.equation(block: false, alt: "0")[$0$]. Your tuition is #emph[not] proportional to the number of units you take, so this is not an example of direct variation. You can check that doubling the number of units does not double the tuition. For example, #math.equation(block: true, alt: "T open parenthesis 6 close parenthesis equals 400 plus 30 open parenthesis 6 close parenthesis equals 580")[$T ( 6 ) = 400 + 30 ( 6 ) = 580$] and #math.equation(block: true, alt: "T open parenthesis 12 close parenthesis equals 400 plus 30 open parenthesis 12 close parenthesis equals 760")[$T ( 12 ) = 400 + 30 ( 12 ) = 760$] Tuition for #math.equation(block: false, alt: "12")[$12$] units is not double the tuition for #math.equation(block: false, alt: "6")[$6$] units. The graph of #math.equation(block: false, alt: "T open parenthesis u close parenthesis")[$T ( u )$] in figure (a) does not pass through the origin.#figure(figph[calculator graphs of two linear functions], alt: "calculator graphs of two linear functions", caption: none) + Let #math.equation(block: false, alt: "S")[$S$] represent Imogen’s sales, and let #math.equation(block: false, alt: "C open parenthesis S close parenthesis")[$C ( S )$] represent her commission. Then #math.equation(block: true, alt: "C open parenthesis S close parenthesis equals 0.15 S")[$C ( S ) = 0.15 S$] Thus, #math.equation(block: false, alt: "C open parenthesis S close parenthesis")[$C ( S )$] is a linear function of #math.equation(block: false, alt: "S")[$S$] with a #math.equation(block: false, alt: "C")[$C$]-intercept of zero, so Imogen's earnings do vary directly with her sales. This is an example of direct variation. (See figure (b).) ] ] What does the scaling property of direct variation mean? \_\_\_\_\_ #solutionbox[ If we multiply the input by #math.equation(block: false, alt: "p")[$p$], the output will also be multiplied by #math.equation(block: false, alt: "p")[$p$]. ] What does the scaling property of direct variation mean? + The output is found by scaling the input. + The output can be regarded as a scale. + The constant of variation can be multiplied by any factor. + If we multiply the input by #math.equation(block: false, alt: "p")[$p$], the output will also be multiplied by #math.equation(block: false, alt: "p")[$p$]. Which tables could represent direct variation? Explain why. (Hint: What happens to #math.equation(block: false, alt: "y")[$y$] if you multiply #math.equation(block: false, alt: "x")[$x$] by a constant?) + #figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#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: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "9")[$9$]]), [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "2.5")[$2.5$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "7.5")[$7.5$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "22.5")[$22.5$]], )) + #figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#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: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "9")[$9$]]), [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3.5")[$3.5$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "8.5")[$8.5$]], [#math.equation(block: false, alt: "10")[$10$]], )) \_\_\_\_\_ #solutionbox[ (a): If we multiply #math.equation(block: false, alt: "x")[$x$] by #math.equation(block: false, alt: "c")[$c$], #math.equation(block: false, alt: "y")[$y$] is also multiplied by #math.equation(block: false, alt: "c")[$c$]. ] Which table could represent direct variation? Explain why. (Hint: What happens to #math.equation(block: false, alt: "y")[$y$] if you multiply #math.equation(block: false, alt: "x")[$x$] by a constant?) + #figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$" " x " "$]], [#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: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "9")[$9$]]), [#math.equation(block: false, alt: "y")[$" " y " "$]], [#math.equation(block: false, alt: "2.5")[$2.5$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "7.5")[$7.5$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "22.5")[$22.5$]], )) + #figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$" " x " "$]], [#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: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "9")[$9$]]), [#math.equation(block: false, alt: "y")[$" " y " "$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3.5")[$3.5$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "8.5")[$8.5$]], [#math.equation(block: false, alt: "10")[$10$]], )) #solutionbox[ Table (a): If we multiply #math.equation(block: false, alt: "x")[$x$] by #math.equation(block: false, alt: "c")[$c$], #math.equation(block: false, alt: "y")[$y$] is also multiplied by #math.equation(block: false, alt: "c")[$c$]. ] === Finding a Formula for Direct Variation If we know any one pair of values for the variables in a direct variation, we can find the constant of variation. We can then use the constant to write a formula for one of the variables as a function of the other. #examplebox("Example 3")[][ If an object is dropped from a great height, its speed, #math.equation(block: false, alt: "v")[$v$], varies directly with the time, #math.equation(block: false, alt: "t")[$t$], the object has been falling. A rock dropped off the rim of the Grand Canyon is falling at a speed of #math.equation(block: false, alt: "39.2")[$39.2$] meters per second when it passes a lizard on a ledge #math.equation(block: false, alt: "4")[$4$] seconds later. + Express #math.equation(block: false, alt: "v")[$v$] as a function of #math.equation(block: false, alt: "t")[$t$]. + What is the speed of the rock after it has fallen for 6 seconds? + Sketch a graph of #math.equation(block: false, alt: "v open parenthesis t close parenthesis")[$v ( t )$] versus #math.equation(block: false, alt: "t")[$t$]. #solutionbox[ + Because #math.equation(block: false, alt: "v")[$v$] varies directly with #math.equation(block: false, alt: "t")[$t$], there is a positive constant #math.equation(block: false, alt: "k")[$k$] for which #math.equation(block: false, alt: "v equals k t")[$v = k t$]. We substitute #math.equation(block: false, alt: "v equals 39.2")[$v = 39.2$] when #math.equation(block: false, alt: "t equals 4")[$t = 4$] and solve for #math.equation(block: false, alt: "k")[$k$] to find #math.equation(block: true, alt: "39.2, equals k open parenthesis 4 close parenthesis, Divide both sides by 4.; k, equals 9.8")[$39.2 & = k ( 4 ) & & "Divide both sides by 4." \ k & = 9.8$] Thus, #math.equation(block: false, alt: "v open parenthesis t close parenthesis equals 9.8 t")[$v ( t ) = 9.8 t$]. + We evaluate the function found in part (a) for #math.equation(block: false, alt: "t equals 6")[$t = 6$]. #math.equation(block: true, alt: "v open parenthesis 6 close parenthesis equals 9.8 open parenthesis 6 close parenthesis equals 58.8")[$v ( 6 ) = 9.8 ( 6 ) = 58.8$] At #math.equation(block: false, alt: "t equals 6")[$t = 6$] seconds, the rock is falling at a speed of #math.equation(block: false, alt: "58.8")[$58.8$] meters per second. + You can use your calculator to graph the function #math.equation(block: false, alt: "v open parenthesis t close parenthesis equals 9.8 t")[$v ( t ) = 9.8 t$]. The graph is shown below.#figure(figph[calculator graph of line with point labeled], alt: "calculator graph of line with point labeled", caption: none) ] ] Is the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 0.3 x plus 5")[$f ( x ) = 0.3 x + 5$] an example of direct variation? \_\_\_\_\_ #solutionbox[ No, the graph does not pass through the origin. ] Is the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 0.3 x plus 5")[$" " f ( x ) = 0.3 x + 5 " "$] an example of direct variation? + Yes, it is linear. + Yes, the scale factor is 0.3. + No, the graph does not pass through the origin. + No, its slope is less than 1. The volume of a bag of rice, in cups, is directly proportional to the weight of the bag. A 2-pound bag contains 3.5 cups of rice. + Express the volume, #math.equation(block: false, alt: "V")[$V$], of a bag of rice as a function of its weight, #math.equation(block: false, alt: "w")[$w$]. #linebreak() \_\_\_\_\_#math.equation(block: false, alt: "equals")[$=$]\_\_\_\_\_ + How many cups of rice are in a #math.equation(block: false, alt: "15")[$15$]-pound bag? #linebreak() \_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "f1 equals f2")[$"f1" = "f2"$] + #math.equation(block: false, alt: "26.25")[$26.25$] ] The volume of a bag of rice, in cups, is directly proportional to the weight of the bag. A 2-pound bag contains 3.5 cups of rice. + Express the volume, #math.equation(block: false, alt: "V")[$V$], of a bag of rice as a function of its weight, #math.equation(block: false, alt: "w")[$w$]. + How many cups of rice are in a #math.equation(block: false, alt: "15")[$15$]-pound bag? #solutionbox[ + #math.equation(block: false, alt: "V equals 1.75 w")[$V = 1.75 w$] + #math.equation(block: false, alt: "26.25")[$26.25 " "$]cups ] === Direct Variation with a Power of #math.equation(block: false, alt: "x")[$x$] We can generalize the notion of direct variation to include situations in which #math.equation(block: false, alt: "y")[$y$] is proportional to a power of #math.equation(block: false, alt: "x")[$x$], instead of #math.equation(block: false, alt: "x")[$x$] itself. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Direct Variation with a Power] #math.equation(block: false, alt: "y")[$y$] #strong[varies directly] with a power of #math.equation(block: false, alt: "x")[$x$] if #math.equation(block: true, alt: "y equals k x to the power n")[$y = k x^(n)$] where #math.equation(block: false, alt: "k")[$k$] and #math.equation(block: false, alt: "n")[$n$] are positive constants. ] #examplebox("Example 4")[][ The surface area of a sphere varies directly with the #emph[square] of its radius. A balloon of radius #math.equation(block: false, alt: "5")[$5$] centimeters has surface area #math.equation(block: false, alt: "100 π")[$100 π$] square centimeters, or about #math.equation(block: false, alt: "314")[$314$] square centimeters. Find a formula for the surface area of a sphere as a function of its radius. #solutionbox[ If #math.equation(block: false, alt: "S")[$S$] stands for the surface area of a sphere of radius #math.equation(block: false, alt: "r")[$r$], then #math.equation(block: true, alt: "S equals f open parenthesis r close parenthesis equals k r squared")[$S = f ( r ) = k r^(2)$] To find the constant of variation, #math.equation(block: false, alt: "k")[$k$], we substitute the values of #math.equation(block: false, alt: "S")[$S$] and #math.equation(block: false, alt: "r")[$r$]. #math.equation(block: true, alt: "100 π, equals k open parenthesis 5 close parenthesis squared; 4 π, equals k")[$100 π & = k ( 5 )^(2) \ 4 π & = k$] Thus, #math.equation(block: false, alt: "S equals f open parenthesis r close parenthesis equals 4 π r squared")[$S = f ( r ) = 4 π r^(2)$]. ] ] The volume of a sphere varies directly with the #emph[cube] of its radius. A balloon of radius 5 centimeters has volume #math.equation(block: false, alt: "the fraction 500 π over 3")[$display(frac(500 π, 3))$] cubic centimeters, or about 524 cubic centimeters. Find a formula for the volume, #math.equation(block: false, alt: "V")[$V$], of a sphere as a function of its radius, #math.equation(block: false, alt: "r")[$r$]. \_\_\_\_\_#math.equation(block: false, alt: "equals")[$=$]\_\_\_\_\_ Use "pi" for #math.equation(block: false, alt: "π")[$π$]. #solutionbox[ #math.equation(block: false, alt: "V equals k r cubed")[$V = k r^(3)$], and solving for #math.equation(block: false, alt: "k")[$k$] we find that #math.equation(block: false, alt: "k equals the fraction 4 over 3 π")[$k = display(frac(4, 3)) π$], so #math.equation(block: false, alt: "V equals the fraction 4 over 3 π r cubed")[$V = display(frac(4, 3)) π r^(3)$] ] The volume of a sphere varies directly with the #emph[cube] of its radius. A balloon of radius 5 centimeters has volume #math.equation(block: false, alt: "the fraction 500 π over 3")[$display(frac(500 π, 3))$] cubic centimeters, or about 524 cubic centimeters. Find a formula for the volume, #math.equation(block: false, alt: "V")[$V$], of a sphere as a function of its radius, #math.equation(block: false, alt: "r")[$r$]. #solutionbox[ #math.equation(block: false, alt: "V equals k r cubed")[$V = k r^(3)$], and solving for #math.equation(block: false, alt: "k")[$k$] we find that #math.equation(block: false, alt: "k equals the fraction 4 over 3 π")[$" " k = display(frac(4, 3)) π$], so #math.equation(block: false, alt: "V equals the fraction 4 over 3 π r cubed")[$" " V = display(frac(4, 3)) π r^(3)$]. ] In any example of direct variation, as the input variable increases through positive values, the output variable increases also. Thus, a direct variation is an increasing function, as we can see when we consider the graphs of some typical direct variations shown below. The curve is y = k·xⁿ and the red dot sits at (1, k). Every setting passes through the origin — put x = 0 and the output is 0 for any k and any n — and every setting rises as x increases through positive values, which is exactly the claim above that a direct variation is an increasing function. Drag n on its own: at n = 1 you get the straight line from the gasoline table, and n = 2 and n = 3 are the powers in the sphere laws later in the section. Now watch the dot rather than the curve while you drag n — it never moves, because 1ⁿ = 1 for every power, so k is precisely the output when the input is 1. Double the input and the output multiplies by 2ⁿ: twice over at n = 1, four times at n = 2, eight times at n = 3.{"functions":\[{"color":"\#1f4e79","expression":"k\*x^n","expression2":"","id":"variation","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"cartesian","tMax":6,"tMin":-6,"variable":"x","visible":true},{"color":"\#b8442a","expression":"1 + 0.12\*cos(t)","expression2":"k + 0.12\*sin(t)","id":"at-one","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-k","label":"Constant of variation k (the output at x = 1)","max":3,"min":0.5,"name":"k","step":0.5,"value":2},{"id":"p-n","label":"Power n (1 = linear, 2 = square, 3 = cube)","max":4,"min":1,"name":"n","step":1,"value":2}\],"title":"Direct variation y = k x^n","version":1,"viewport":{"centerX":1,"centerY":4,"scale":40}}#figure(figph[graphs of three variations with power], alt: "graphs of three variations with power", caption: none) #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The graph of a direct variation always passes through the origin, so when the input is zero, the output is also zero. Thus, the functions #math.equation(block: false, alt: "y equals 3 x plus 2")[$" " y = 3 x + 2 " "$] and #math.equation(block: false, alt: "y equals 0.4 x squared minus 2.3")[$" " y = 0.4 x^(2) − 2.3$], for example, are not direct variation, even though they are increasing functions for positive #math.equation(block: false, alt: "x")[$x$]. ] Even without an equation, we can check whether a table of data describes direct variation or merely an increasing function. If #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x to the power n")[$x^(n)$] , then #math.equation(block: false, alt: "y equals k x to the power n")[$" " y = k x^(n)$] , or, equivalently,#math.equation(block: false, alt: "the fraction y over x to the power n equals k")[$" " display(frac(y, x^(n))) = k$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Test for Direct Variation] If the ratio #math.equation(block: false, alt: "the fraction y over x to the power n")[$display(frac(y, x^(n)))$] is constant, then #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x to the power n")[$x^(n)$]. ] #examplebox("Example 5")[][ Delbert collects the following data and would like to know if #math.equation(block: false, alt: "y")[$y$] varies directly with the square of #math.equation(block: false, alt: "x")[$x$]. What should he calculate? #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$" " x " "$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "12")[$12$]]), [#math.equation(block: false, alt: "y")[$" " y " "$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "16.5")[$16.5$]], [#math.equation(block: false, alt: "36")[$36$]], [#math.equation(block: false, alt: "54")[$54$]], [#math.equation(block: false, alt: "76")[$76$]], )) #solutionbox[ If #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x squared")[$x^(2)$], then #math.equation(block: false, alt: "y equals k x squared")[$y = k x^(2)$], or #math.equation(block: false, alt: "the fraction y over x squared equals k")[$display(frac(y, x^(2))) = k$]. Delbert should calculate the ratio #math.equation(block: false, alt: "the fraction y over x squared")[$display(frac(y, x^(2)))$] for each data point. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$" " x " "$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "12")[$12$]]), [#math.equation(block: false, alt: "y")[$" " y " "$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "16.5")[$16.5$]], [#math.equation(block: false, alt: "36")[$36$]], [#math.equation(block: false, alt: "54")[$54$]], [#math.equation(block: false, alt: "76")[$76$]], [#math.equation(block: false, alt: "the fraction y over x squared")[$display(frac(y, x^(2)))$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "0.66")[$0.66$]], [#math.equation(block: false, alt: "0.56")[$0.56$]], [#math.equation(block: false, alt: "0.54")[$0.54$]], [#math.equation(block: false, alt: "0.53")[$0.53$]], )) Because the ratio #math.equation(block: false, alt: "the fraction y over x squared")[$display(frac(y, x^(2)))$] is not constant, #math.equation(block: false, alt: "y")[$y$] does not vary directly with #math.equation(block: false, alt: "x squared")[$x^(2)$]. ] ] How can you test whether a table for #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] represents direct variation with a power? \_\_\_\_\_ #solutionbox[ Check whether #math.equation(block: false, alt: "y / x to the power n")[$y / x^(n)$] is a constant. ] How can you test whether a table for #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] represents direct variation with a power? + Plot the points. + Check whether #math.equation(block: false, alt: "y to the power n")[$y^(n)$] is a constant. + Check whether #math.equation(block: false, alt: "the fraction y over x to the power n")[$display(frac(y, x^(n)))$] is a constant. + Check whether #math.equation(block: false, alt: "the fraction y to the power n over x to the power n")[$display(frac(y^(n), x^(n)))$] is a constant. Does #math.equation(block: false, alt: "B")[$B$] vary directly with the cube of #math.equation(block: false, alt: "r")[$r$]? Explain your decision. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "0.3")[$0.3$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "0.8")[$0.8$]], [#math.equation(block: false, alt: "1.2")[$1.2$]]), [#math.equation(block: false, alt: "B")[$B$]], [#math.equation(block: false, alt: "0.072")[$0.072$]], [#math.equation(block: false, alt: "1.944")[$1.944$]], [#math.equation(block: false, alt: "9.0")[$9.0$]], [#math.equation(block: false, alt: "36.864")[$36.864$]], [#math.equation(block: false, alt: "124.416")[$124.416$]], )) \_\_\_\_\_ #solutionbox[ Yes, #math.equation(block: false, alt: "the fraction B over r cubed")[$display(frac(B, r^(3)))$] is constant. ] Does #math.equation(block: false, alt: "B")[$B$] vary directly with the cube of #math.equation(block: false, alt: "r")[$r$]? Explain your decision. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "r")[$" " r " "$]], [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "0.3")[$0.3$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "0.8")[$0.8$]], [#math.equation(block: false, alt: "1.2")[$1.2$]]), [#math.equation(block: false, alt: "B")[$" " B " "$]], [#math.equation(block: false, alt: "0.072")[$0.072$]], [#math.equation(block: false, alt: "1.944")[$1.944$]], [#math.equation(block: false, alt: "9.0")[$9.0$]], [#math.equation(block: false, alt: "36.864")[$36.864$]], [#math.equation(block: false, alt: "124.416")[$124.416$]], )) #solutionbox[ Yes, #math.equation(block: false, alt: "the fraction B over r cubed")[$display(frac(B, r^(3)))$] is constant. ] === Scaling Recall that if #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x")[$x$], then doubling #math.equation(block: false, alt: "x")[$x$] causes #math.equation(block: false, alt: "y")[$y$] to double also. But: - Is the area of a #math.equation(block: false, alt: "16")[$16$]-inch circular pizza double the area of an #math.equation(block: false, alt: "8")[$8$]-inch pizza? - If you double the dimensions of a model of a skyscraper, will its weight double also? You probably know that the answer to both of these questions is #emph[No]. The area of a circle is proportional to the #emph[square] of its radius, and the volume (and hence the weight) of an object is proportional to the #emph[cube] of its linear dimension. Variation with a power of #math.equation(block: false, alt: "x")[$x$] produces a different scaling effect. #examplebox("Example 6")[][ The Taipei 101 building is 1671 feet tall, and in 2006 it was the tallest skyscraper in the world. Show that doubling the dimensions of a model of the Taipei 101 building produces a model that weighs #math.equation(block: false, alt: "8")[$8$] times as much. #solutionbox[ The Taipei 101 skyscraper is approximately box shaped, so its volume is given by the product of its linear dimensions, #math.equation(block: false, alt: "V equals l w h")[$" " V = l w h$]. The weight of an object is proportional to its volume, so the weight, #math.equation(block: false, alt: "W")[$W$], of the model is #math.equation(block: true, alt: "W equals k l w h")[$W = k l w h$] where the constant #math.equation(block: false, alt: "k")[$k$] depends on the material of the model. If we double the length, width, and height of the model, then #math.equation(block: true, alt: "W sub new, equals k open parenthesis 2 l close parenthesis open parenthesis 2 w close parenthesis open parenthesis 2 h close parenthesis; equals 2 cubed open parenthesis k l w h close parenthesis equals 8 W sub old")[$W_("new") & = k ( 2 l ) ( 2 w ) ( 2 h ) \ & = 2^(3) ( k l w h ) = 8 W_("old")$] The weight of the new model is #math.equation(block: false, alt: "2 cubed equals 8")[$2^(3) = 8$] times the weight of the original model. ] ] Use the formula for the area of a circle to show that doubling the diameter of a pizza quadruples its area. The formula for the area of a circle of radius #math.equation(block: false, alt: "r")[$r$] is #math.equation(block: false, alt: "A equals")[$A =$]\_\_\_\_\_ Enter "pi" to get #math.equation(block: false, alt: "π")[$π$]. If we double the diameter, the new radius is \_\_\_\_\_ Substitute this expression into the area formula to get the area of the new circle #math.equation(block: false, alt: "A sub new equals")[$A_("new") =$]\_\_\_\_\_ So #math.equation(block: false, alt: "A sub new")[$A_("new")$] is \_\_\_\_\_ times the original area. #solutionbox[ #math.equation(block: true, alt: "A equals π r squared")[$A = π r^(2)$] Doubling the diameter means doubling the radius. #math.equation(block: true, alt: "A sub new, equals π open parenthesis 2 r close parenthesis squared; equals 4 π r squared; equals 4 A sub old")[$A_("new") & = π ( 2 r )^(2) \ & = 4 π r^(2) \ & = 4 A_("old")$] ] Use the formula for the area of a circle to show that doubling the diameter of a pizza quadruples its area. #solutionbox[ #math.equation(block: true, alt: "A equals π r squared")[$A = π r^(2)$] Doubling the diameter means doubling the radius. #math.equation(block: true, alt: "A equals π open parenthesis 2 r close parenthesis squared equals 4 π r squared equals 4 A sub old")[$A = π ( 2 r )^(2) = 4 π r^(2) = 4 A_("old")$] ] In general, if #math.equation(block: false, alt: "y")[$y$] varies directly with a power of #math.equation(block: false, alt: "x")[$x$], that is, if #math.equation(block: false, alt: "y equals k x to the power n")[$" " y = k x^(n)$], then doubling the value of #math.equation(block: false, alt: "x")[$x$] causes #math.equation(block: false, alt: "y")[$y$] to increase by a factor of #math.equation(block: false, alt: "2 to the power n")[$2^(n)$]. In fact, if we multiply #math.equation(block: false, alt: "x")[$x$] by any positive number #math.equation(block: false, alt: "c")[$c$], then #math.equation(block: true, alt: "y sub new, equals k open parenthesis c x close parenthesis to the power n; equals c to the power n open parenthesis k x to the power n close parenthesis equals c to the power n open parenthesis y sub old close parenthesis")[$y_("new") & = k ( c x )^(n) \ & = c^(n) ( k x^(n) ) = c^(n) ( y_("old") )$] so the value of #math.equation(block: false, alt: "y")[$y$] is multiplied by #math.equation(block: false, alt: "c to the power n")[$c^(n)$]. We will call #math.equation(block: false, alt: "n")[$n$] the #strong[scaling exponent], and you will often see variation described in terms of scaling. For example, we might say that "the area of a circle scales as the square of its radius." (In many applications, the power #math.equation(block: false, alt: "n")[$n$] is called the #emph[scale factor], even though it is not a factor but an exponent.) === Inverse Variation How long does it take to travel a distance of #math.equation(block: false, alt: "600")[$600$] miles? The answer depends on your average speed. If you are on a bicycle trip, your average speed might be #math.equation(block: false, alt: "15")[$15$] miles per hour. In that case, your traveling time will be #math.equation(block: true, alt: "T equals the fraction D over R equals the fraction 600 over 15 equals 40 hours")[$T = frac(D, R) = frac(600, 15) = 40 " hours"$] (Of course, you will have to add time for rest stops; the #math.equation(block: false, alt: "40")[$40$] hours are just your travel time.) If you are driving your car, you might average #math.equation(block: false, alt: "50")[$50$] miles per hour. Your travel time is then #math.equation(block: true, alt: "T equals the fraction D over R equals the fraction 600 over 50 equals 12 hours")[$T = frac(D, R) = frac(600, 50) = 12 " hours"$] If you take a commercial air flight, the plane’s speed might be #math.equation(block: false, alt: "400")[$400$] miles per hour, and the flight time would be #math.equation(block: true, alt: "T equals the fraction D over R equals the fraction 600 over 400 equals 1.5 hours")[$T = frac(D, R) = frac(600, 400) = 1.5 " hours"$] You can see that for higher average speeds, the travel time is shorter. In other words, the time needed for a #math.equation(block: false, alt: "600")[$600$]-mile journey is a decreasing function of average speed. In fact, a formula for the function is #math.equation(block: true, alt: "T equals f open parenthesis R close parenthesis equals the fraction 600 over R")[$T = f ( R ) = frac(600, R)$] This function is an example of #strong[inverse variation]. A table of values and a graph of the function are shown below. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "R")[$R$]], [#math.equation(block: false, alt: "T")[$T$]]), [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "60")[$60$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "200")[$200$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "400")[$400$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], )) #figure(figph[graph of time vs average speed], alt: "graph of time vs average speed", caption: none) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Inverse Variation] #math.equation(block: false, alt: "y")[$y$] #strong[varies inversely] with #math.equation(block: false, alt: "x")[$x$] if #math.equation(block: true, alt: "y equals the fraction k over x , x not equal to 0")[$y = display(frac(k, x)) ", " x ≠ 0$] where #math.equation(block: false, alt: "k")[$k$] is a positive constant. ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Inverse variation describes a decreasing function, but not every decreasing function represents inverse variation. People sometimes mistakenly use the phrase #emph[varies inversely] to describe any decreasing function, but if #math.equation(block: false, alt: "y")[$y$] varies inversely with #math.equation(block: false, alt: "x")[$x$], the variables must satisfy an equation of the form #math.equation(block: false, alt: "y equals the fraction k over x")[$y = display(frac(k, x))$], or #math.equation(block: false, alt: "x y equals k")[$x y = k$]. ] To decide whether two variables truly vary inversely, we can check whether their product is constant. For instance, in the preceding travel-time example, we see from the table that #math.equation(block: false, alt: "R T equals 600")[$R T = 600$]. #figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "R")[$R$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "200")[$200$]], [#math.equation(block: false, alt: "400")[$400$]]), [#math.equation(block: false, alt: "T")[$T$]], [#math.equation(block: false, alt: "60")[$60$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "R T")[$R T$]], [#math.equation(block: false, alt: "600")[$600$]], [#math.equation(block: false, alt: "600")[$600$]], [#math.equation(block: false, alt: "600")[$600$]], [#math.equation(block: false, alt: "600")[$600$]], [#math.equation(block: false, alt: "600")[$600$]], [#math.equation(block: false, alt: "600")[$600$]], )) How can you test whether a table for #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] represents inverse variation? \_\_\_\_\_ #solutionbox[ Check whether #math.equation(block: false, alt: "x y")[$x y$] is a constant. ] How can you test whether a table for #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$" " y = f ( x ) " "$] represents inverse variation? + Check whether #math.equation(block: false, alt: "x y")[$x y$] is a constant. + Check whether the function is decreasing. + Check whether #math.equation(block: false, alt: "y")[$y$] is the reciprocal of #math.equation(block: false, alt: "x")[$x$]. + Check whether #math.equation(block: false, alt: "y / x")[$y / x$] is a constant. We can also define inverse variation with a power of the variable. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Inverse Variation with a power] #math.equation(block: false, alt: "y")[$y$] #strong[varies inversely] with #math.equation(block: false, alt: "x to the power n")[$x^(n)$] if #math.equation(block: true, alt: "y equals the fraction k over x to the power n , x not equal to 0")[$y = frac(k, x^(n)) ", " x ≠ 0$] where #math.equation(block: false, alt: "k")[$k$] and #math.equation(block: false, alt: "n")[$n$] are positive constants. ] We may also say that #math.equation(block: false, alt: "y")[$y$] is #strong[inversely proportional] to #math.equation(block: false, alt: "x to the power n")[$x^(n)$]. #examplebox("Example 7")[][ The weight, #math.equation(block: false, alt: "w")[$w$], of an object varies inversely with the square of its distance, #math.equation(block: false, alt: "d")[$d$], from the center of the Earth. Thus, #math.equation(block: true, alt: "w equals the fraction k over d squared")[$w = display(frac(k, d^(2)))$] If you double your distance from the center of the Earth, what happens to your weight? What if you triple the distance? #solutionbox[ Suppose you weigh #math.equation(block: false, alt: "W")[$W$] pounds at distance #math.equation(block: false, alt: "D")[$D$] from the center of the Earth. Then #math.equation(block: false, alt: "W equals the fraction k over D squared")[$W = display(frac(k, D^(2)))$]. At distance #math.equation(block: false, alt: "2 D")[$2 D$], your weight will be #math.equation(block: true, alt: "w equals the fraction k over open parenthesis 2 D close parenthesis squared equals the fraction k over 4 D squared equals the fraction 1 over 4 times the fraction k over D squared equals the fraction 1 over 4 W")[$w = frac(k, ( 2 D )^(2)) = frac(k, 4 D^(2)) = frac(1, 4) ⋅ frac(k, D^(2)) = frac(1, 4) W$] Your new weight will be #math.equation(block: false, alt: "the fraction 1 over 4")[$display(frac(1, 4))$] of your old weight. By a similar calculation, you can check that by tripling the distance, your weight will be reduced to #math.equation(block: false, alt: "the fraction 1 over 9")[$display(frac(1, 9))$] of its original value. ] ] The amount of force, #math.equation(block: false, alt: "F")[$F$], (in pounds) needed to loosen a rusty bolt with a wrench is inversely proportional to the length, #math.equation(block: false, alt: "l")[$l$], of the wrench. Thus, #math.equation(block: true, alt: "F equals the fraction k over l")[$F = frac(k, l)$] If you increase the length of the wrench by 50% so that the new length is #math.equation(block: false, alt: "1.5 l")[$1.5 l$], what happens to the amount of force required to loosen the bolt? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "F sub new equals the fraction 2 over 3 F sub old")[$F_("new") = display(frac(2, 3)) F_("old")$] ] The amount of force, #math.equation(block: false, alt: "F")[$F$], (in pounds) needed to loosen a rusty bolt with a wrench is inversely proportional to the length, #math.equation(block: false, alt: "l")[$l$], of the wrench. Thus, #math.equation(block: true, alt: "F equals the fraction k over l")[$F = frac(k, l)$] If you increase the length of the wrench by 50% so that the new length is #math.equation(block: false, alt: "1.5 l")[$1.5 l$], what happens to the amount of force required to loosen the bolt? #solutionbox[ #math.equation(block: true, alt: "F sub new equals the fraction 2 over 3 F sub old")[$F_("new") = display(frac(2, 3)) F_("old")$] ] In Example and Practice 7, as the independent variable increases through positive values, the dependent variable decreases. An inverse variation is an example of a decreasing function. The graphs of some typical inverse variations are shown below. Notice that both graphs have a vertical asymptote at #math.equation(block: false, alt: "x equals 0")[$x = 0$]. #figure(figph[inverse variation graphs], alt: "inverse variation graphs", caption: none) In Example and Practice 7, as the independent variable increases through positive values, the dependent variable decreases. An inverse variation is an example of a decreasing function. The graphs of some typical inverse variations are shown below. Notice that both graphs have a vertical asymptote at #math.equation(block: false, alt: "x equals 0")[$x = 0$]. #figure(figph[inverse variation graphs], alt: "inverse variation graphs", caption: none) === Finding a Formula for Inverse Variation If we know that two variables vary inversely and we can find one pair of corresponding values for the variables, we can determine #math.equation(block: false, alt: "k")[$k$], the constant of variation. #examplebox("Example 8")[][ The intensity of electromagnetic radiation, such as light or radio waves, varies inversely with the square of the distance from its source. Radio station KPCC broadcasts a signal that is measured at #math.equation(block: false, alt: "0.016")[$0.016$] watt per square meter by a receiver #math.equation(block: false, alt: "1")[$1$] kilometer away. + Write a formula that gives signal strength as a function of distance. + If you live #math.equation(block: false, alt: "5")[$5$] kilometers from the station, what is the strength of the signal you will receive? #solutionbox[ + Let #math.equation(block: false, alt: "I")[$I$] stand for the intensity of the signal in watts per square meter, and #math.equation(block: false, alt: "d")[$d$] for the distance from the station in kilometers. Then #math.equation(block: false, alt: "I equals the fraction k over d squared")[$I = display(frac(k, d^(2)))$]. To find the constant #math.equation(block: false, alt: "k")[$k$], we substitute #math.equation(block: false, alt: "0.016")[$0.016$] for #math.equation(block: false, alt: "I")[$I$] and #math.equation(block: false, alt: "1")[$1$] for #math.equation(block: false, alt: "d")[$d$]. Solving for #math.equation(block: false, alt: "k")[$k$] gives us #math.equation(block: true, alt: "0.016, equals the fraction k over 1 squared; k, equals 0.016 open parenthesis 1 squared close parenthesis equals 0.016")[$0.016 & = display(frac(k, 1^(2))) \ k & = 0.016 ( 1^(2) ) = 0.016$] Thus, #math.equation(block: false, alt: "I equals the fraction 0.016 over d squared")[$I = display(frac(0.016, d^(2)))$]. + Now we can substitute #math.equation(block: false, alt: "5")[$5$] for #math.equation(block: false, alt: "d")[$d$] and solve for #math.equation(block: false, alt: "I")[$I$]. #math.equation(block: true, alt: "I equals the fraction 0.016 over 5 squared equals 0.00064")[$I = frac(0.016, 5^(2)) = 0.00064$] At a distance of #math.equation(block: false, alt: "5")[$5$] kilometers from the station, the signal strength is #math.equation(block: false, alt: "0.00064")[$0.00064$] watt per square meter. ] ] Delbert's officemates want to buy a \$120 gold watch for a colleague who is retiring. The cost per person is inversely proportional to the number of people who contribute. + Express the cost per person, #math.equation(block: false, alt: "C")[$C$], as a function of the number of people, #math.equation(block: false, alt: "p")[$p$], who contribute. #linebreak() \_\_\_\_\_#math.equation(block: false, alt: "equals")[$=$]\_\_\_\_\_ + Sketch the function on the domain #math.equation(block: false, alt: "0 less than or equal to p less than or equal to 20")[$0 ≤ p ≤ 20$]. #linebreak() The graph of cost per person is #linebreak() \_\_\_\_\_ #solutionbox[ + A graph is also shown below. + #math.equation(block: false, alt: "x equals minus 9")[$x = − 9$] or #math.equation(block: false, alt: "x equals 2")[$x = 2$] ] #figure(figph[reciprocal], alt: "reciprocal", caption: none) Delbert's officemates want to buy a \$120 gold watch for a colleague who is retiring. The cost per person is inversely proportional to the number of people who contribute. + Express the cost per person, #math.equation(block: false, alt: "C")[$C$], as a function of the number of people, #math.equation(block: false, alt: "p")[$p$], who contribute. + Sketch the function on the domain #math.equation(block: false, alt: "0 less than or equal to p less than or equal to 20")[$0 ≤ p ≤ 20$]. #solutionbox[ + #math.equation(block: false, alt: "C equals the fraction 120 over p")[$C = display(frac(120, p))$] + #figure(figph[reciprocal], alt: "reciprocal", caption: none) ] What distinguishes inverse variation from other decreasing functions? Mention properties of the graph and the formula. \_\_\_\_\_ What distinguishes inverse variation from other decreasing functions? Mention properties of the graph and the formula. === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Direct variation - Directly proportional - Constant of variation - Scaling exponent - Inverse variation - Inversely proportional ==== CONCEPTS + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Direct and Inverse Variation] - #strong[#math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x")[$x$]] if the ratio #math.equation(block: false, alt: "the fraction y over x")[$display(frac(y, x))$] is constant, that is, if #math.equation(block: false, alt: "y equals k x")[$y = k x$] . - #strong[#math.equation(block: false, alt: "y")[$y$] varies directly with a power of #math.equation(block: false, alt: "x")[$x$]] if the ratio #math.equation(block: false, alt: "the fraction y over x to the power n")[$display(frac(y, x^(n)))$] is constant, that is, if #math.equation(block: false, alt: "y equals k x to the power n")[$y = k x^(n)$] . - #strong[#math.equation(block: false, alt: "y")[$y$] varies inversely with #math.equation(block: false, alt: "x")[$x$]] if the product #math.equation(block: false, alt: "x y")[$x y$] is constant, that is, if #math.equation(block: false, alt: "y equals the fraction k over x")[$y = display(frac(k, x))$] . - #strong[#math.equation(block: false, alt: "y")[$y$] varies inversely with a power of #math.equation(block: false, alt: "x")[$x$]] if the product #math.equation(block: false, alt: "x to the power n y")[$x^(n) y$] is constant, that is, if #math.equation(block: false, alt: "y equals the fraction k over x to the power n")[$y = display(frac(k, x^(n)))$] . ] + The graph of a direct variation passes through the origin. The graph of an inverse variation has a vertical asymptote at the origin. + If #math.equation(block: false, alt: "y equals k x to the power n")[$y = k x^(n)$] , we say that #math.equation(block: false, alt: "y")[$y$] #strong[scales] as #math.equation(block: false, alt: "x to the power n")[$x^(n)$]. ==== STUDY QUESTIONS + Describe the graph of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] if #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x")[$x$]. + What is true about the ratio of two variables if they are directly proportional? + If #math.equation(block: false, alt: "y")[$y$] is inversely proportional to #math.equation(block: false, alt: "x")[$x$], then the graph of #math.equation(block: false, alt: "y")[$y$] versus #math.equation(block: false, alt: "x")[$x$] is a transformation of which basic graph? + If #math.equation(block: false, alt: "y")[$y$] varies directly with a power of #math.equation(block: false, alt: "x")[$x$], write a formula for #math.equation(block: false, alt: "y")[$y$] as a function of #math.equation(block: false, alt: "x")[$x$]. + If #math.equation(block: false, alt: "y")[$y$] varies inversely with a power of #math.equation(block: false, alt: "x")[$x$], write a formula for #math.equation(block: false, alt: "y")[$y$] as a function of #math.equation(block: false, alt: "x")[$x$]. + If #math.equation(block: false, alt: "y equals k x to the power 4")[$y = k x^(4)$], what happens to #math.equation(block: false, alt: "y")[$y$] if you double #math.equation(block: false, alt: "x")[$x$]? + State a test to determine whether #math.equation(block: false, alt: "y")[$y$] varies inversely with #math.equation(block: false, alt: "x to the power n")[$x^(n)$]. + If #math.equation(block: false, alt: "y equals the fraction k over x squared")[$y = display(frac(k, x^(2)))$], and we double the value of #math.equation(block: false, alt: "x")[$x$], what happens to the value of #math.equation(block: false, alt: "y")[$y$]? ==== SKILLS Practice each skill in the Homework problems listed. + Find the constant of variation: \#1–4, 13–26 + Write a formula for direct or inverse variation: \#1–4, 13–26, 35–46 + Recognize direct and inverse variation from a table of values: \#27–34, 39–42 + Recognize direct or inverse variation from a graph: \#9–12, 35–38 + Use scaling in direct and inverse variation: \#13–20, 43–46 === Homework 3.1 Delbert's credit card statement lists three purchases he made while on a business trip in the Midwest. His company's accountant would like to know the sales tax rate on the purchases. #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "Price of item")[$"Price of item"$]], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "28")[$28$]], [#math.equation(block: false, alt: "12")[$12$]]), [#math.equation(block: false, alt: "Tax")[$"Tax"$]], [#math.equation(block: false, alt: "1.17")[$1.17$]], [#math.equation(block: false, alt: "1.82")[$1.82$]], [#math.equation(block: false, alt: "0.78")[$0.78$]], [#math.equation(block: false, alt: "Tax / Price")[$"Tax" / "Price"$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$], )) + Compute the ratio of the tax to the price of each item. Is the tax proportional to the price? What is the tax rate? + Express the tax, #math.equation(block: false, alt: "T")[$T$], as a function of the price, #math.equation(block: false, alt: "p")[$p$], of the item. + Sketch a graph of the function by hand, and label the scales on the axes. + #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "Price of item")[$"Price of item"$]], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "28")[$28$]], [#math.equation(block: false, alt: "12")[$12$]]), [#math.equation(block: false, alt: "Tax")[$"Tax"$]], [#math.equation(block: false, alt: "1.17")[$1.17$]], [#math.equation(block: false, alt: "1.82")[$1.82$]], [#math.equation(block: false, alt: "0.78")[$0.78$]], [#math.equation(block: false, alt: "Tax / Price")[$"Tax" / "Price"$]], [#math.equation(block: false, alt: "0.065")[$0.065$]], [#math.equation(block: false, alt: "0.065")[$0.065$]], [#math.equation(block: false, alt: "0.065")[$0.065$]], )) #linebreak() Yes; #math.equation(block: false, alt: "6.5 %")[$6.5 upright(%)$] + #math.equation(block: false, alt: "T equals 0.065 p")[$T = 0.065 p$] + #figure(figph[direct variation], alt: "direct variation", caption: none) At constant acceleration from rest, the distance traveled by a race car is proportional to the square of the time elapsed. The highest recorded road-tested acceleration in 2023 was #math.equation(block: false, alt: "0")[$0$] to #math.equation(block: false, alt: "60")[$60$] miles per hour in #math.equation(block: false, alt: "1.7")[$1.7$] seconds, which produces the following data. #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "Time (seconds)")[$"Time (seconds)"$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2.5")[$2.5$]]), [#math.equation(block: false, alt: "Distance (feet)")[$"Distance (feet)"$]], [#math.equation(block: false, alt: "58.235")[$58.235$]], [#math.equation(block: false, alt: "103.529")[$103.529$]], [#math.equation(block: false, alt: "161.765")[$161.765$]], [#math.equation(block: false, alt: "Distance/Time squared")[$"Distance/Time"^(2)$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$], )) + Compute the ratios of the distance traveled to the square of the time elapsed. What was the acceleration, in feet per second squared? + Express the distance traveled, #math.equation(block: false, alt: "d")[$d$], as a function of time in seconds, #math.equation(block: false, alt: "t")[$t$]. + Sketch a graph of the function by hand, and label the scales on the axes. The marketing department for a paper company is testing wrapping paper rolls in various dimensions to see which shape consumers prefer. All the rolls contain the same amount of wrapping paper. #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "Width (feet)")[$"Width (feet)"$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2.5")[$2.5$]], [#math.equation(block: false, alt: "3")[$3$]]), [#math.equation(block: false, alt: "Length (feet)")[$"Length (feet)"$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "9.6")[$9.6$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "Length times Width")[$"Length" × "Width"$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$], )) + Compute the product of the length and width for each roll of wrapping paper. What is the constant of inverse proportionality? + Express the length, #math.equation(block: false, alt: "L")[$L$], of the paper as a function of the width, #math.equation(block: false, alt: "w")[$w$], of the roll. + Sketch a graph of the function by hand, and label the scales on the axes. + #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "Width (feet)")[$"Width (feet)"$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2.5")[$2.5$]], [#math.equation(block: false, alt: "3")[$3$]]), [#math.equation(block: false, alt: "Length (feet)")[$"Length (feet)"$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "9.6")[$9.6$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "Length times width")[$"Length" × "width"$]], [#math.equation(block: false, alt: "24")[$24$]], [#math.equation(block: false, alt: "24")[$24$]], [#math.equation(block: false, alt: "24")[$24$]], )) #linebreak() #math.equation(block: false, alt: "24")[$24$] square feet + #math.equation(block: false, alt: "L equals the fraction 24 over w")[$L = display(frac(24, w))$] + #figure(figph[inverse variation], alt: "inverse variation", caption: none) The force of gravity on a #math.equation(block: false, alt: "1")[$1$]-kilogram mass is inversely proportional to the square of the object's distance from the center of the Earth. The table shows the force on the object, in newtons, at distances that are multiples of the Earth's radius. #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "Distance (Earth radii)")[$"Distance (Earth radii)"$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "Force (newtons)")[$"Force (newtons)"$]], [#math.equation(block: false, alt: "9.8")[$9.8$]], [#math.equation(block: false, alt: "2.45")[$2.45$]], [#math.equation(block: false, alt: "0.6125")[$0.6125$]], [#math.equation(block: false, alt: "Force times distance squared")[$"Force" × "distance"^(2)$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$], )) + Compute the products of the force and the square of the distance. What is the constant of inverse proportionality? + Express the gravitational force, #math.equation(block: false, alt: "F ,")[$F ,$] on a #math.equation(block: false, alt: "1")[$1$]-kilogram mass as a function of its distance, #math.equation(block: false, alt: "r ,")[$r ,$] from the Earth's center, measured in Earth radii. + Sketch a graph of the function by hand, and label the scales on the axes. + How can you tell from a table of values whether #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x")[$x$]? + How can you tell from a table of values whether #math.equation(block: false, alt: "y")[$y$] varies inversely with #math.equation(block: false, alt: "x")[$x$]? + The ratio #math.equation(block: false, alt: "the fraction y over x")[$display(frac(y, x))$] is a constant. + The product #math.equation(block: false, alt: "x y")[$x y$] is a constant. + How can you tell from a table of values whether #math.equation(block: false, alt: "y")[$y$] varies directly with a power of #math.equation(block: false, alt: "x")[$x$]? + How can you tell from a table of values whether #math.equation(block: false, alt: "y")[$y$] varies inversely with a power of #math.equation(block: false, alt: "x")[$x$]? The length of a rectangle is #math.equation(block: false, alt: "10")[$10$] inches, and its width is #math.equation(block: false, alt: "8")[$8$] inches. Suppose we increase the length of the rectangle while holding the width constant. + Fill in the table.#figure(table( columns: 4, align: left, inset: 6pt, table.header([Length], [Width], [Perimeter], [Area]), [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "8")[$8$]], [], [], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "8")[$8$]], [], [], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "8")[$8$]], [], [], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "8")[$8$]], [], [], )) + Does the perimeter vary directly with the length? + Write a formula for the perimeter of the rectangle in terms of its length. + Does the area vary directly with the length? + Write a formula for the area of the rectangle in terms of its length. + #figure(table( columns: 4, align: left, inset: 6pt, table.header([Length], [Width], [Perimeter], [Area]), [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "36")[$36$]], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "96")[$96$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "46")[$46$]], [#math.equation(block: false, alt: "120")[$120$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "56")[$56$]], [#math.equation(block: false, alt: "160")[$160$]], )) + No + #math.equation(block: false, alt: "P equals 16 plus 2 l")[$P = 16 + 2 l$] + Yes + #math.equation(block: false, alt: "A equals 8 l")[$A = 8 l$] The base of an isosceles triangle is #math.equation(block: false, alt: "12")[$12$] centimeters, and the equal sides have length #math.equation(block: false, alt: "15")[$15$] centimeters. Suppose we increase the base of the triangle while holding the sides constant. + Fill in the table.#figure(table( columns: 5, align: left, inset: 6pt, table.header([Base], [Sides], [Height], [Perimeter], [Area]), [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "15")[$15$]], [], [], [], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "15")[$15$]], [], [], [], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "15")[$15$]], [], [], [], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "15")[$15$]], [], [], [], )) + Does the perimeter vary directly with the base? + Write a formula for the perimeter of the triangle in terms of its base. + Write a formula for the area of the triangle in terms of its base. + Does the area vary directly with the base? Which of the graphs could describe direct variation? Explain your answer. #figure(figph[four curves], alt: "four curves", caption: none) (b) could be direct variation with a power of #math.equation(block: false, alt: "x")[$x$]. Which of the graphs could describe direct variation? Explain your answer. #figure(figph[four curves], alt: "four curves", caption: none) Which of the graphs could describe inverse variation? Explain your answer. #figure(figph[four curves], alt: "four curves", caption: none) (c) Which of the graphs could describe inverse variation? Explain your answer. #figure(figph[four curves], alt: "four curves", caption: none) The weight of an object on the Moon varies directly with its weight on Earth. A person who weighs #math.equation(block: false, alt: "150")[$150$] pounds on Earth would weigh only #math.equation(block: false, alt: "24.75")[$24.75$] pounds on the Moon. + Find a function that gives the weight #math.equation(block: false, alt: "m")[$m$] of an object on the Moon in terms of its weight #math.equation(block: false, alt: "w")[$w$] on Earth. Complete the table and graph your function in a suitable window.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "w")[$w$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "200")[$200$]], [#math.equation(block: false, alt: "400")[$400$]]), [#math.equation(block: false, alt: "m")[$m$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + How much would a person weigh on the Moon if she weighs #math.equation(block: false, alt: "120")[$120$] pounds on Earth? + A piece of rock weighs #math.equation(block: false, alt: "50")[$50$] pounds on the Moon. How much will it weigh back on Earth? + If you double the weight of an object on Earth, what will happen to its weight on the Moon? + #math.equation(block: false, alt: "m equals 0.165 w")[$m = 0.165 w$]#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "w")[$w$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "200")[$200$]], [#math.equation(block: false, alt: "400")[$400$]]), [#math.equation(block: false, alt: "m")[$m$]], [#math.equation(block: false, alt: "8.25")[$8.25$]], [#math.equation(block: false, alt: "16.5")[$16.5$]], [#math.equation(block: false, alt: "33")[$33$]], [#math.equation(block: false, alt: "66")[$66$]], )) + #math.equation(block: false, alt: "19.8")[$19.8$] lb + #math.equation(block: false, alt: "303.03")[$303.03$] lb + It will double. Hubble's law says that distant galaxies are receding from us at a rate that varies directly with their distance. (The speeds of the galaxies are measured using a phenomenon called redshifting.) A galaxy in the constellation Ursa Major is #math.equation(block: false, alt: "980")[$980$] million light-years away and is receding at a speed of #math.equation(block: false, alt: "15 , 000")[$15 , 000$] kilometers per second. + Find a function that gives the speed, #math.equation(block: false, alt: "v")[$v$], of a galaxy in terms of its distance, #math.equation(block: false, alt: "d")[$d$], from Earth. Complete the table and graph your function in a suitable window. (Distances are given in millions of light-years.)#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "d")[$d$]], [#math.equation(block: false, alt: "500")[$500$]], [#math.equation(block: false, alt: "1000")[$1000$]], [#math.equation(block: false, alt: "2000")[$2000$]], [#math.equation(block: false, alt: "4000")[$4000$]]), [#math.equation(block: false, alt: "v")[$v$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + How far away is a galaxy in the constellation Hydra that is receding at #math.equation(block: false, alt: "61 , 000")[$61 , 000$] kilometers per second? + A galaxy in Leo is #math.equation(block: false, alt: "1240")[$1240$] million light-years away. How fast is it receding from us? + If one constellation is twice as distant as another, how do their speeds compare? The length, #math.equation(block: false, alt: "L")[$L$], of a pendulum varies directly with the square of its period, #math.equation(block: false, alt: "T")[$T$], the time required for the pendulum to make one complete swing back and forth. The pendulum on a grandfather clock is #math.equation(block: false, alt: "3.25")[$3.25$] feet long and has a period of #math.equation(block: false, alt: "2")[$2$] seconds. + Express #math.equation(block: false, alt: "L")[$L$] as a function of #math.equation(block: false, alt: "T")[$T$]. Complete the table and graph your function in a suitable window.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "T")[$T$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]]), [#math.equation(block: false, alt: "L")[$L$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + How long is the Foucault pendulum in the Pantheon in Paris, which has a period of #math.equation(block: false, alt: "17")[$17$] seconds? + A hypnotist uses a gold pendant as a pendulum to mesmerize his clients. If the chain on the pendant is #math.equation(block: false, alt: "9")[$9$] inches long, what is the period of its swing? + In order to double the period of a pendulum, how must you vary its length? + #math.equation(block: false, alt: "L equals 0.8125 T squared")[$L = 0.8125 T^(2)$]#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "T")[$T$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]]), [#math.equation(block: false, alt: "L")[$L$]], [#math.equation(block: false, alt: "0.8125")[$0.8125$]], [#math.equation(block: false, alt: "20.3")[$20.3$]], [#math.equation(block: false, alt: "81.25")[$81.25$]], [#math.equation(block: false, alt: "325")[$325$]], )) + #math.equation(block: false, alt: "234.8125")[$234.8125$] ft + #math.equation(block: false, alt: "0.96")[$0.96$] sec + It must be four times as long. The load, #math.equation(block: false, alt: "L")[$L$], that a beam can support varies directly with the square of its vertical thickness, #math.equation(block: false, alt: "h")[$h$]. A beam that is #math.equation(block: false, alt: "4")[$4$] inches thick can support a load of #math.equation(block: false, alt: "2000")[$2000$] pounds. + Express #math.equation(block: false, alt: "L")[$L$] as a function of #math.equation(block: false, alt: "h")[$h$]. Complete the table and graph your function in a suitable window.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "h")[$h$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "8")[$8$]]), [#math.equation(block: false, alt: "L")[$L$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + What size load can be supported by a beam that is #math.equation(block: false, alt: "6")[$6$] inches thick? + How thick a beam is needed to support a load of #math.equation(block: false, alt: "100")[$100$] pounds? + If you double the thickness of a beam, how will the load it can support change? Computer monitors produce a magnetic field. The effect of the field, #math.equation(block: false, alt: "B")[$B$], on the user varies inversely with his or her distance, #math.equation(block: false, alt: "d")[$d$], from the screen. The field from a certain color monitor was measured at #math.equation(block: false, alt: "22")[$22$] milligauss #math.equation(block: false, alt: "4")[$4$] inches from the screen. + Express the field strength as a function of distance from the screen. Complete the table and graph your function in a suitable window.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "d")[$d$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "24")[$24$]]), [#math.equation(block: false, alt: "B")[$B$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + What is the field strength #math.equation(block: false, alt: "10")[$10$] inches from the screen? + An elevated risk of cancer can result from exposure to field strengths of #math.equation(block: false, alt: "4.3")[$4.3$] milligauss. How far from the screen should the computer user sit to keep the field level below #math.equation(block: false, alt: "4.3")[$4.3$] milligauss? + If you double your distance from the screen, how does the field strength change? + #math.equation(block: false, alt: "B equals the fraction 88 over d")[$B = display(frac(88, d))$]#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "d")[$d$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "24")[$24$]]), [#math.equation(block: false, alt: "B")[$B$]], [#math.equation(block: false, alt: "88")[$88$]], [#math.equation(block: false, alt: "44")[$44$]], [#math.equation(block: false, alt: "7.3")[$7.3$]], [#math.equation(block: false, alt: "3.7")[$3.7$]], )) + #math.equation(block: false, alt: "8.8")[$8.8$] milligauss + More than #math.equation(block: false, alt: "20.47")[$20.47$] in + It is one half as strong. The amount of current, #math.equation(block: false, alt: "I")[$I$], that flows through a circuit varies inversely with the resistance, #math.equation(block: false, alt: "R")[$R$], on the circuit. An iron with a resistance of #math.equation(block: false, alt: "12")[$12$] ohms draws #math.equation(block: false, alt: "10")[$10$] amps of current. + Express the current as a function of the resistance. Complete the table and graph your function in a suitable window.#figure(table( columns: 5, 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: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]]), [#math.equation(block: false, alt: "I")[$I$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + How much current is drawn by a light bulb with a resistance of #math.equation(block: false, alt: "533.3")[$533.3$] ohms? + What is the resistance of a toaster that draws #math.equation(block: false, alt: "12.5")[$12.5$] amps of current? + If the resistance of one appliance is double the resistance of a second appliance, how does the current they draw compare? The amount of power, #math.equation(block: false, alt: "P")[$P$], generated by a windmill varies directly with the cube of the wind speed, #math.equation(block: false, alt: "w")[$w$]. A windmill on Oahu, Hawaii, produces #math.equation(block: false, alt: "7300")[$7300$] kilowatts of power when the wind speed is #math.equation(block: false, alt: "32")[$32$] miles per hour. + Express the power as a function of wind speed. Complete the table and graph your function in a suitable window.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "w")[$w$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "80")[$80$]]), [#math.equation(block: false, alt: "P")[$P$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + How much power would the windmill produce in a light breeze of #math.equation(block: false, alt: "15")[$15$] miles per hour? + What wind speed is needed to produce #math.equation(block: false, alt: "10 , 000")[$10 , 000$] kilowatts of power? + If the wind speed doubles, what happens to the amount of power generated? + #math.equation(block: false, alt: "P equals the fraction 1825 over 8192 w cubed approximately equals 0.228 w cubed")[$P = display(frac(1825, 8192)) w^(3) ≈ 0.228 w^(3)$]#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "w")[$w$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "80")[$80$]]), [#math.equation(block: false, alt: "P")[$P$]], [#math.equation(block: false, alt: "223")[$223$]], [#math.equation(block: false, alt: "1782")[$1782$]], [#math.equation(block: false, alt: "14 , 259")[$14 , 259$]], [#math.equation(block: false, alt: "114 , 074")[$114 , 074$]], )) + #math.equation(block: false, alt: "752")[$752$] kilowatts + #math.equation(block: false, alt: "33.54")[$33.54$] mph + It is multiplied by #math.equation(block: false, alt: "8")[$8$]. A crystal form of pyrite (a compound of iron and sulfur) has the shape of a regular solid with #math.equation(block: false, alt: "12")[$12$] faces. Each face is a regular pentagon. This compound is called pyritohedron, and its mass, #math.equation(block: false, alt: "M")[$M$], varies directly with the cube of the length, #math.equation(block: false, alt: "L")[$L$], of one edge. If each edge is #math.equation(block: false, alt: "1.1")[$1.1$] centimeters, then the mass is #math.equation(block: false, alt: "51")[$51$] grams. + Express the mass of pyritohedron as a function of the length of one edge. Complete the table and graph your function in a suitable window.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "L")[$L$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "M")[$M$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + What is the mass of a chunk of pyritohedron if each edge is #math.equation(block: false, alt: "2.2")[$2.2$] centimeters? + How long would each edge be for a #math.equation(block: false, alt: "1643")[$1643$]-gram piece of pyritohedron? + If one chunk has double the length of a second chunk, how do their masses compare? For Problems 21–26, + Use the values in the table to find the constant of variation, #math.equation(block: false, alt: "k")[$k$], and write #math.equation(block: false, alt: "y")[$y$] as a function of #math.equation(block: false, alt: "x")[$x$]. + Fill in the rest of the table with the correct values. + What happens to #math.equation(block: false, alt: "y")[$y$] when you double the value of #math.equation(block: false, alt: "x")[$x$]? #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x")[$x$]. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "2")[$2$]], [$#hide($000$)$], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [$#hide($000$)$], [#math.equation(block: false, alt: "2.4")[$2.4$]], [#math.equation(block: false, alt: "12")[$12$]], [$#hide($000$)$], [$#hide($000$)$], [#math.equation(block: false, alt: "4.5")[$4.5$]], )) + #math.equation(block: false, alt: "y equals 0.3 x")[$y = 0.3 x$] + #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "0.6")[$0.6$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "2.4")[$2.4$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "3.6")[$3.6$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "4.5")[$4.5$]], )) + #math.equation(block: false, alt: "y")[$y$] doubles. #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x")[$x$]. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "0.8")[$0.8$]], [$#hide($000$)$], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "54")[$54$]], [$#hide($000$)$], [#math.equation(block: false, alt: "108")[$108$]], [$#hide($000$)$], [#math.equation(block: false, alt: "126")[$126$]], [#math.equation(block: false, alt: "6")[$6$]], [$#hide($000$)$], )) #math.equation(block: false, alt: "y")[$y$] varies directly with the square of #math.equation(block: false, alt: "x")[$x$]. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "3")[$3$]], [$#hide($000$)$], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "24")[$24$]], [$#hide($000$)$], [#math.equation(block: false, alt: "54")[$54$]], [#math.equation(block: false, alt: "12")[$12$]], [$#hide($000$)$], [$#hide($000$)$], [#math.equation(block: false, alt: "150")[$150$]], )) + #math.equation(block: false, alt: "y equals the fraction 2 over 3 x squared")[$y = display(frac(2, 3)) x^(2)$] + #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "24")[$24$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "54")[$54$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "96")[$96$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "150")[$150$]], )) + #math.equation(block: false, alt: "y")[$y$] is quadrupled. #math.equation(block: false, alt: "y")[$y$] varies directly with the cube of #math.equation(block: false, alt: "x")[$x$]. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "120")[$120$]], [#math.equation(block: false, alt: "3")[$3$]], [$#hide($000$)$], [$#hide($000$)$], [#math.equation(block: false, alt: "1875")[$1875$]], [#math.equation(block: false, alt: "6")[$6$]], [$#hide($000$)$], [$#hide($000$)$], [#math.equation(block: false, alt: "15 , 000")[$15 , 000$]], )) #math.equation(block: false, alt: "y")[$y$] varies inversely with #math.equation(block: false, alt: "x")[$x$]. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "4")[$4$]], [$#hide($000$)$], [$#hide($000$)$], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "30")[$30$]], [$#hide($000$)$], [$#hide($000$)$], [#math.equation(block: false, alt: "3")[$3$]], )) + #math.equation(block: false, alt: "y equals the fraction 120 over x")[$y = display(frac(120, x))$] + #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "30")[$30$]]), [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "3")[$3$]], )) + #math.equation(block: false, alt: "y")[$y$] is halved. #math.equation(block: false, alt: "y")[$y$] varies inversely with the square of #math.equation(block: false, alt: "x")[$x$]. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "0.2")[$0.2$]], [$#hide($000$)$], [$#hide($000$)$], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "2")[$2$]], [$#hide($000$)$], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "1.25")[$1.25$]], [$#hide($000$)$], [#math.equation(block: false, alt: "0.8")[$0.8$]], )) For Problems 27–30, decide whether + #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x")[$x$], + #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x squared")[$x^(2)$], or + #math.equation(block: false, alt: "y")[$y$] does not vary directly with a power of #math.equation(block: false, alt: "x")[$x$]. Explain why your choice is correct. If your choice is (a) or (b), find the constant of variation. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2.0")[$2.0$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4.5")[$4.5$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "12.5")[$12.5$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "32.0")[$32.0$]], )) (b) #math.equation(block: false, alt: "y equals 0.5 x squared")[$y = 0.5 x^(2)$] #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "28")[$28$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "44")[$44$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "68")[$68$]], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "3.0")[$3.0$]], [#math.equation(block: false, alt: "2.4")[$2.4$]], [#math.equation(block: false, alt: "5.3")[$5.3$]], [#math.equation(block: false, alt: "5.5")[$5.5$]], [#math.equation(block: false, alt: "33")[$33$]], [#math.equation(block: false, alt: "8.2")[$8.2$]], [#math.equation(block: false, alt: "73.8")[$73.8$]], )) (c) #math.equation(block: false, alt: "the fraction y over x to the power p")[$display(frac(y, x^(p)))$] is not constant for any exponent #math.equation(block: false, alt: "p")[$p$]. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "1.2")[$1.2$]], [#math.equation(block: false, alt: "7.20")[$7.20$]], [#math.equation(block: false, alt: "2.5")[$2.5$]], [#math.equation(block: false, alt: "31.25")[$31.25$]], [#math.equation(block: false, alt: "6.4")[$6.4$]], [#math.equation(block: false, alt: "204.80")[$204.80$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "720.00")[$720.00$]], )) For Problems 31–34, decide whether + #math.equation(block: false, alt: "y")[$y$] varies inversely with #math.equation(block: false, alt: "x")[$x$], + #math.equation(block: false, alt: "y")[$y$] varies inversely with #math.equation(block: false, alt: "x squared")[$x^(2)$], or + #math.equation(block: false, alt: "y")[$y$] does not vary inversely with a power of #math.equation(block: false, alt: "x")[$x$]. Explain why your choice is correct. If your choice is (a) or (b), find the constant of variation. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "288")[$288$]], [#math.equation(block: false, alt: "2.0")[$2.0$]], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "3.0")[$3.0$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "6.0")[$6.0$]], [#math.equation(block: false, alt: "2")[$2$]], )) (b) #math.equation(block: false, alt: "y equals the fraction 72 over x squared")[$y = display(frac(72, x^(2)))$] #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "100.0")[$100.0$]], [#math.equation(block: false, alt: "2.0")[$2.0$]], [#math.equation(block: false, alt: "25.0")[$25.0$]], [#math.equation(block: false, alt: "4.0")[$4.0$]], [#math.equation(block: false, alt: "12.5")[$12.5$]], [#math.equation(block: false, alt: "5.0")[$5.0$]], [#math.equation(block: false, alt: "10.0")[$10.0$]], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "1.0")[$1.0$]], [#math.equation(block: false, alt: "4.0")[$4.0$]], [#math.equation(block: false, alt: "1.3")[$1.3$]], [#math.equation(block: false, alt: "3.7")[$3.7$]], [#math.equation(block: false, alt: "3.0")[$3.0$]], [#math.equation(block: false, alt: "2.0")[$2.0$]], [#math.equation(block: false, alt: "4.0")[$4.0$]], [#math.equation(block: false, alt: "1.0")[$1.0$]], )) (c) #math.equation(block: false, alt: "x to the power p y")[$x^(p) y$] is not constant for any exponent #math.equation(block: false, alt: "p")[$p$]. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "180.00")[$180.00$]], [#math.equation(block: false, alt: "2.0")[$2.0$]], [#math.equation(block: false, alt: "11.25")[$11.25$]], [#math.equation(block: false, alt: "3.0")[$3.0$]], [#math.equation(block: false, alt: "5.00")[$5.00$]], [#math.equation(block: false, alt: "5.0")[$5.0$]], [#math.equation(block: false, alt: "1.80")[$1.80$]], )) The functions described by a table of data or by a graph in Problems 35–42 are examples of direct or inverse variation. + Find an algebraic formula for the function, including the constant of variation, #math.equation(block: false, alt: "k")[$k$]. + Answer the question in the problem. The faster a car moves, the more difficult it is to stop. The graph shows the distance, #math.equation(block: false, alt: "d")[$d$], required to stop a car as a function of its velocity, #math.equation(block: false, alt: "v")[$v$], before the brakes were applied. What distance is needed to stop a car moving at #math.equation(block: false, alt: "100")[$100$] kilometers per hour? #figure(figph[increasing concave up], alt: "increasing concave up", caption: none) + #math.equation(block: false, alt: "d equals 0.005 v squared")[$d = 0.005 v^(2)$] + #math.equation(block: false, alt: "50")[$50$] m A wide pipe can handle a greater water flow than a narrow pipe. The graph shows the water flow through a pipe, #math.equation(block: false, alt: "w")[$w$], as a function of its radius, #math.equation(block: false, alt: "r")[$r$]. How great is the water flow through a pipe of radius of #math.equation(block: false, alt: "10")[$10$] inches? #figure(figph[increasing concave up], alt: "increasing concave up", caption: none) If the price of mushrooms goes up, the amount consumers are willing to buy goes down. The graph shows the number of tons of shiitake mushrooms, #math.equation(block: false, alt: "m")[$m$], sold in California each week as a function of their price, #math.equation(block: false, alt: "p")[$p$]. If the price of shiitake mushrooms rises to #math.equation(block: false, alt: "$ 10")[$upright(\$) 10$] per pound, how many tons will be sold? #figure(figph[decreasing concave up], alt: "decreasing concave up", caption: none) + #math.equation(block: false, alt: "m equals the fraction 8 over p")[$m = display(frac(8, p))$] + #math.equation(block: false, alt: "0.8")[$0.8$] ton When an adult plays with a small child on a seesaw, the adult must sit closer to the pivot point to balance the seesaw. The graph shows this distance, #math.equation(block: false, alt: "d")[$d$], as a function of the adult's weight, #math.equation(block: false, alt: "w")[$w$]. How far from the pivot must Kareem sit if he weighs #math.equation(block: false, alt: "280")[$280$] pounds? #figure(figph[decreasing concave up], alt: "decreasing concave up", caption: none) Ocean temperatures are generally colder at the greater depths. The table shows the temperature of the water as a function of depth. What is the ocean temperature at a depth of #math.equation(block: false, alt: "6")[$6$] kilometers? #figure(table( columns: 2, align: left, inset: 6pt, table.header([Depth (km)], [Temperature (#math.equation(block: false, alt: "∘")[$∘$]C)]), [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "2")[$2$]], )) + #math.equation(block: false, alt: "T equals the fraction 6 over d")[$T = display(frac(6, d))$] + #math.equation(block: false, alt: "1 degrees")[$1^(∘)$]C The shorter the length of a vibrating guitar string, the higher the frequency of the vibrations. The fifth string is #math.equation(block: false, alt: "65")[$65$] centimeters long and is tuned to A (with a frequency of #math.equation(block: false, alt: "220")[$220$] vibrations per second). The placement of the fret relative to the bridge changes the effective length of the guitar string. The table shows frequency as a function of effective length. How far from the bridge should the fret be placed for the note C (#math.equation(block: false, alt: "256")[$256$] vibrations per second)? #figure(table( columns: 2, align: left, inset: 6pt, table.header([Length (cm)], [Frequency]), [#math.equation(block: false, alt: "55")[$55$]], [#math.equation(block: false, alt: "260")[$260$]], [#math.equation(block: false, alt: "57.2")[$57.2$]], [#math.equation(block: false, alt: "250")[$250$]], [#math.equation(block: false, alt: "65")[$65$]], [#math.equation(block: false, alt: "220")[$220$]], [#math.equation(block: false, alt: "71.5")[$71.5$]], [#math.equation(block: false, alt: "200")[$200$]], )) The strength of a cylindrical rod depends on its diameter. The greater the diameter of the rod, the more weight it can support before collapsing. The table shows the maximum weight supported by a rod as a function of its diameter. How much weight can a #math.equation(block: false, alt: "1.2")[$1.2$]-centimeter rod support before collapsing? #figure(table( columns: 2, align: left, inset: 6pt, table.header([Diameter (cm)], [Weight (newtons)]), [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "150")[$150$]], [#math.equation(block: false, alt: "1.0")[$1.0$]], [#math.equation(block: false, alt: "600")[$600$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "1350")[$1350$]], [#math.equation(block: false, alt: "2.0")[$2.0$]], [#math.equation(block: false, alt: "2400")[$2400$]], )) + #math.equation(block: false, alt: "W equals 600 d squared")[$W = 600 d^(2)$] + #math.equation(block: false, alt: "864")[$864$] newtons The maximum height attained by a cannonball depends on the speed at which it was shot. The table shows maximum height as a function of initial speed. What height is attained by a cannonball whose initial upward speed was #math.equation(block: false, alt: "240")[$240$] feet per second? #figure(table( columns: 2, align: left, inset: 6pt, table.header([Speed (ft/sec)], [Height (ft)]), [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "120")[$120$]], [#math.equation(block: false, alt: "225")[$225$]], [#math.equation(block: false, alt: "160")[$160$]], [#math.equation(block: false, alt: "400")[$400$]], [#math.equation(block: false, alt: "200")[$200$]], [#math.equation(block: false, alt: "625")[$625$]], )) The wind resistance, #math.equation(block: false, alt: "W")[$W$], experienced by a vehicle on the freeway varies directly with the square of its speed, #math.equation(block: false, alt: "v")[$v$]. + If you double your speed, what happens to the wind resistance? + If you drive one-third as fast, what happens to the wind resistance? + If you decrease your speed by #math.equation(block: false, alt: "10 %")[$10 upright(%)$], what happens to the wind resistance? + Wind resistance quadruples. + It is one-ninth as great. + It is decreased by #math.equation(block: false, alt: "19 %")[$19 upright(%)$] because it is #math.equation(block: false, alt: "81 %")[$81 upright(%)$] of the original. The weight, #math.equation(block: false, alt: "w")[$w$], of a bronze statue varies directly with the cube of its height, #math.equation(block: false, alt: "h")[$h$]. + If you double the height of the statue, what happens to its weight? + If you make the statue one-fourth as tall, what happens to its weight? + If you increase the height of the statue by #math.equation(block: false, alt: "50 %")[$50 upright(%)$], what happens to its weight? The intensity of illumination, #math.equation(block: false, alt: "I")[$I$], from a lamp varies inversely with the square of your distance, #math.equation(block: false, alt: "d")[$d$], from the lamp. + If you double your distance from a reading lamp, what happens to the illumination? + If you triple the distance, what happens to the illumination? + If you increase the distance by #math.equation(block: false, alt: "25 %")[$25 upright(%)$], what happens to the illumination? + It is one-fourth the original illumination. + It is one-ninth the illumination. + It is #math.equation(block: false, alt: "64 %")[$64 upright(%)$] of the illumination. The resistance, #math.equation(block: false, alt: "R")[$R$], of an electrical wire varies inversely with the square of its diameter, #math.equation(block: false, alt: "d")[$d$]. + If you replace an old wire with a new one whose diameter is half that of the old one, what happens to the resistance? + If you replace an old wire with a new one whose diameter is two-thirds of the old one, what happens to the resistance? + If you decrease the diameter of the wire by #math.equation(block: false, alt: "30 %")[$30 upright(%)$], what happens to the resistance? The quoted material in Problems 47–50 is taken from the article "Quantum Black Holes," by Bernard J. Carr and Steven B. Giddings, in the May 2005 issue of #emph[Scientific American]. (See Algebra Skills Refresher Scientific Notation to review scientific notation.) “The density to which matter must be squeezed \[to create a black hole\] scales as the inverse square of the mass. For a hole with the mass of the Sun, the density is about #math.equation(block: false, alt: "10 to the power 9")[$10^(9)$] kilograms per cubic meter, higher than that of an atomic nucleus.” + Recall that the density of an object is its mass per unit volume. Given that the mass of the sun is about #math.equation(block: false, alt: "2 times 10 to the power 30")[$2 × 10^(30)$] kilograms, write a formula for the density, #math.equation(block: false, alt: "D")[$D$], of a black hole as a function of its mass, #math.equation(block: false, alt: "m")[$m$]. + “The known laws of physics allow for a matter density up to the so-called Planck value of #math.equation(block: false, alt: "10 to the power 97")[$10^(97)$] kilograms per cubic meter.” If a black hole with this density could be created, it would be the smallest possible black hole. What would its mass be? + Assuming that a black hole is spherical, what would be the radius of the smallest possible black hole? + #math.equation(block: false, alt: "D equals open parenthesis 2.5 times 10 to the power minus 52 close parenthesis m squared")[$D = ( 2.5 × 10^(− 52) ) m^(2)$] + #math.equation(block: false, alt: "2 times 10 to the power 74")[$2 × 10^(74)$] kg + #math.equation(block: false, alt: "1.7 times 10 to the power minus 8 m")[$1.7 × 10^(− 8) " " m$] “A black hole radiates thermally, like a hot coal, with a temperature inversely proportional to its mass. For a solar-mass black hole, the temperature is around a millionth of a kelvin.” + The solar mass is given in Problem 47. Write a formula for the temperature, #math.equation(block: false, alt: "T")[$T$], of a black hole as a function of its mass, #math.equation(block: false, alt: "m")[$m$]. + What is the temperature of a black hole of mass #math.equation(block: false, alt: "10 to the power 12")[$10^(12)$] kilograms, about the mass of a mountain? “The total time for a black hole to evaporate away is proportional to the cube of its initial mass. For a solar-mass hole, the lifetime is an unobservably long #math.equation(block: false, alt: "10 to the power 64")[$10^(64)$] years.” + The solar mass is given in Problem 47. Write a formula for the lifetime, #math.equation(block: false, alt: "L")[$L$], of a black hole as a function of its mass, #math.equation(block: false, alt: "m")[$m$]. + The present age of the universe is about #math.equation(block: false, alt: "10 to the power 10")[$10^(10)$] years. What would be the mass of a black hole as old as the universe? + #math.equation(block: false, alt: "L equals open parenthesis 1.25 times 10 minus 27 close parenthesis m cubed")[$L = ( 1.25 × 10 − 27 ) m^(3)$] + #math.equation(block: false, alt: "2 times 10 to the power 12")[$2 × 10^(12)$] kg “String theory #math.equation(block: false, alt: "and so on")[$…$] predicts that space has dimensions beyond the usual three. In three dimensions, the force of gravity quadruples as you halve the distance between two objects. But in nine dimensions, gravity would get #math.equation(block: false, alt: "256")[$256$] times as strong.” In three dimensions, the force of gravity varies inversely with the square of distance. Write a formula for the force of gravity in nine dimensions. Use algebra to support your answers to Problems 51–56. Begin with a formula for direct or inverse variation. Suppose #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x")[$x$]. Show that if you multiply #math.equation(block: false, alt: "x")[$x$] by any constant #math.equation(block: false, alt: "c")[$c$], then #math.equation(block: false, alt: "y")[$y$] will be multiplied by the same constant. #math.equation(block: false, alt: "y equals k x")[$y = k x$] implies that #math.equation(block: false, alt: "k open parenthesis c x close parenthesis equals c open parenthesis k x close parenthesis equals c y")[$k ( c x ) = c ( k x ) = c y$]. Suppose #math.equation(block: false, alt: "y")[$y$] varies inversely with #math.equation(block: false, alt: "x")[$x$]. Show that if you multiply #math.equation(block: false, alt: "x")[$x$] by any constant #math.equation(block: false, alt: "c")[$c$], then #math.equation(block: false, alt: "y")[$y$] will be divided by the same constant. Explain why the ratio #math.equation(block: false, alt: "the fraction y over x squared")[$display(frac(y, x^(2)))$] is a constant when #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x squared")[$x^(2)$]. If #math.equation(block: false, alt: "y equals k x squared")[$y = k x^(2)$], then dividing both sides of the equation by #math.equation(block: false, alt: "x squared")[$x^(2)$] gives #math.equation(block: false, alt: "the fraction y over x squared equals k")[$display(frac(y, x^(2))) = k$]. Explain why the product #math.equation(block: false, alt: "y x squared")[$y x^(2)$] is a constant when #math.equation(block: false, alt: "y")[$y$] varies inversely with #math.equation(block: false, alt: "x squared")[$x^(2)$]. If #math.equation(block: false, alt: "x")[$x$] varies directly with #math.equation(block: false, alt: "y")[$y$] and #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "z")[$z$], does #math.equation(block: false, alt: "x")[$x$] vary directly with #math.equation(block: false, alt: "z")[$z$]? Yes If #math.equation(block: false, alt: "x")[$x$] varies inversely with #math.equation(block: false, alt: "y")[$y$] and #math.equation(block: false, alt: "y")[$y$] varies inversely with #math.equation(block: false, alt: "z")[$z$], does #math.equation(block: false, alt: "x")[$x$] vary invesely with #math.equation(block: false, alt: "z")[$z$]?