{"id":1743,"date":"2024-04-24T17:56:14","date_gmt":"2024-04-24T17:56:14","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=1743"},"modified":"2025-08-18T01:56:00","modified_gmt":"2025-08-18T01:56:00","slug":"physical-applications-of-integration-background-youll-need-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/physical-applications-of-integration-background-youll-need-1\/","title":{"raw":"Physical Applications of Integration: Background You'll Need 1","rendered":"Physical Applications of Integration: Background You&#8217;ll Need 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Understand and use direct variation to solve problems<\/li>\r\n<\/ul>\r\n<\/section>\r\n<p>Direct variation describes a simple relationship between two variables where one variable is a constant multiple of the other. This concept is crucial in various physical applications, such as calculating mass from density functions and determining work done by variable forces.<\/p>\r\n<h2>Write Direct Variation Equations<\/h2>\r\n<p>A used-car company has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance if she sells a vehicle for $4,600, she will earn $736. She wants to evaluate the offer, but she is not sure how.\u00a0<\/p>\r\n<p>In the example above, Nicole\u2019s earnings can be found by multiplying her sales by her commission. The formula [latex]e = 0.16s[\/latex] tells us her earnings, [latex]e[\/latex], come from the product of 0.16, her commission, and the sale price of the vehicle, [latex]s[\/latex]. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive.<\/p>\r\n<table>\r\n<thead>\r\n<tr>\r\n<th style=\"width: 20%;\">[latex]s[\/latex] (Sales Prices)<\/th>\r\n<th style=\"width: 20%;\">[latex]e = 0.16s[\/latex]<\/th>\r\n<th style=\"width: 60%;\">Interpretation<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>$4,600<\/td>\r\n<td>[latex]\\begin{array}{rcl} e &amp; = &amp; 0.16(4,600) \\\\ &amp; = &amp; 736 \\end{array} [\/latex]<\/td>\r\n<td>A sale of a $4,600 vehicle results in $736 earnings.<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>$9,200<\/td>\r\n<td>[latex]\\begin{array}{rcl} e &amp; = &amp; 0.16(9,200) \\\\ &amp; = &amp; 1,472 \\end{array} [\/latex]<\/td>\r\n<td>A sale of a $9,200 vehicle results in $1472 earnings.<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>$18,400<\/td>\r\n<td>[latex]\\begin{array}{rcl} e &amp; = &amp; 0.16(18,400) \\\\ &amp; = &amp; 2,944 \\end{array} [\/latex]<\/td>\r\n<td>A sale of a $18,400 vehicle results in $2944 earnings.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called <strong>direct variation<\/strong>. Each variable in this type of relationship <strong>varies directly <\/strong>with the other.<\/p>\r\n<p>The graph below\u00a0represents the data for Nicole\u2019s potential earnings. We say that earnings vary directly with the sales price of the car. The formula [latex]y=k{x}^{n}[\/latex] is used for direct variation. The value [latex]k[\/latex] is a nonzero constant greater than zero and is called the <strong>constant of variation<\/strong>. In this case, [latex]k=0.16[\/latex]\u00a0and [latex]n=1[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02222950\/CNX_Precalc_Figure_03_09_0012.jpg\" alt=\"Graph of y=(0.16)x where the horizontal axis is labeled,\" width=\"487\" height=\"459\" \/> Graph of Nicole's potential earnings[\/caption]\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>direct variation<\/h3>\r\n<p>If [latex]x[\/latex]<em>\u00a0<\/em>and [latex]y[\/latex]\u00a0are related by an equation of the form<\/p>\r\n<p style=\"text-align: center;\">[latex]y=k{x}^{n}[\/latex]<\/p>\r\n<p>then we say that the relationship is <strong>direct variation<\/strong> and [latex]y[\/latex]\u00a0<strong>varies directly<\/strong> with the [latex]n[\/latex]th power of [latex]x[\/latex]. <br \/>\r\n<br \/>\r\nIn direct variation relationships, there is a nonzero constant ratio [latex]k=\\dfrac{y}{{x}^{n}}[\/latex], where [latex]k[\/latex]\u00a0is called the <strong>constant of variation<\/strong>, which help defines the relationship between the variables.<\/p>\r\n<\/section>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>How To: Given a description of a direct variation problem, solve for an unknown<br \/>\r\n<\/strong><\/p>\r\n<ol id=\"fs-id1165137724401\">\r\n\t<li>Identify the input, [latex]x[\/latex], and the output, [latex]y[\/latex].<\/li>\r\n\t<li>Determine the constant of variation. You may need to divide [latex]y[\/latex]\u00a0by the specified power of [latex]x[\/latex]\u00a0to determine the constant of variation.<\/li>\r\n\t<li>Use the constant of variation to write an equation for the relationship.<\/li>\r\n\t<li>Substitute known values into the equation to find the unknown.<\/li>\r\n<\/ol>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>The quantity [latex]y[\/latex]\u00a0varies directly with the cube of [latex]x[\/latex]. If [latex]y=25[\/latex]\u00a0when [latex]x=2[\/latex], write the equation that represents this relationship. Then, find [latex]y[\/latex]\u00a0when [latex]x[\/latex]\u00a0is 6.<\/p>\r\n<p>[reveal-answer q=\"647220\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"647220\"]<\/p>\r\n<p>The general formula for direct variation with a cube is [latex]y=k{x}^{3}[\/latex]. The constant can be found by dividing [latex]y[\/latex]\u00a0by the cube of [latex]x[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align} k&amp;=\\dfrac{y}{{x}^{3}} \\\\[1mm] &amp;=\\dfrac{25}{{2}^{3}}\\\\[1mm] &amp;=\\dfrac{25}{8}\\end{align}[\/latex]<\/p>\r\n<p>Now use the constant to write an equation that represents this relationship.<\/p>\r\n<p style=\"text-align: center;\">[latex]y=\\dfrac{25}{8}{x}^{3}[\/latex]<\/p>\r\n<p>Substitute [latex]x=6[\/latex]\u00a0and solve for [latex]y[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}y&amp;=\\dfrac{25}{8}{\\left(6\\right)}^{3} \\\\[1mm] &amp;=675\\hfill \\end{align}[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>The quantity [latex]y[\/latex]\u00a0varies directly with the square of [latex]y[\/latex]. If [latex]y=24[\/latex]\u00a0when [latex]x=3[\/latex], find [latex]y[\/latex]\u00a0when [latex]x[\/latex]\u00a0is 4.<\/p>\r\n<p>[reveal-answer q=\"536994\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"536994\"]<\/p>\r\n<p>[latex]\\dfrac{128}{3}[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]288442[\/ohm_question]<\/p>\r\n<\/section>\r\n<section class=\"textbox watchIt\">\r\n<p>Watch this video to see a quick lesson about direct variation. You will see more worked examples.<\/p>\r\n<p><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6454977&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=plFOq4JaEyI&amp;video_target=tpm-plugin-ncnoy4d6-plFOq4JaEyI\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/DirectVariationApplications_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Direct Variation Applications\" here (opens in new window)<\/a>.<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Understand and use direct variation to solve problems<\/li>\n<\/ul>\n<\/section>\n<p>Direct variation describes a simple relationship between two variables where one variable is a constant multiple of the other. This concept is crucial in various physical applications, such as calculating mass from density functions and determining work done by variable forces.<\/p>\n<h2>Write Direct Variation Equations<\/h2>\n<p>A used-car company has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance if she sells a vehicle for $4,600, she will earn $736. She wants to evaluate the offer, but she is not sure how.\u00a0<\/p>\n<p>In the example above, Nicole\u2019s earnings can be found by multiplying her sales by her commission. The formula [latex]e = 0.16s[\/latex] tells us her earnings, [latex]e[\/latex], come from the product of 0.16, her commission, and the sale price of the vehicle, [latex]s[\/latex]. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive.<\/p>\n<table>\n<thead>\n<tr>\n<th style=\"width: 20%;\">[latex]s[\/latex] (Sales Prices)<\/th>\n<th style=\"width: 20%;\">[latex]e = 0.16s[\/latex]<\/th>\n<th style=\"width: 60%;\">Interpretation<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>$4,600<\/td>\n<td>[latex]\\begin{array}{rcl} e & = & 0.16(4,600) \\\\ & = & 736 \\end{array}[\/latex]<\/td>\n<td>A sale of a $4,600 vehicle results in $736 earnings.<\/td>\n<\/tr>\n<tr>\n<td>$9,200<\/td>\n<td>[latex]\\begin{array}{rcl} e & = & 0.16(9,200) \\\\ & = & 1,472 \\end{array}[\/latex]<\/td>\n<td>A sale of a $9,200 vehicle results in $1472 earnings.<\/td>\n<\/tr>\n<tr>\n<td>$18,400<\/td>\n<td>[latex]\\begin{array}{rcl} e & = & 0.16(18,400) \\\\ & = & 2,944 \\end{array}[\/latex]<\/td>\n<td>A sale of a $18,400 vehicle results in $2944 earnings.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called <strong>direct variation<\/strong>. Each variable in this type of relationship <strong>varies directly <\/strong>with the other.<\/p>\n<p>The graph below\u00a0represents the data for Nicole\u2019s potential earnings. We say that earnings vary directly with the sales price of the car. The formula [latex]y=k{x}^{n}[\/latex] is used for direct variation. The value [latex]k[\/latex] is a nonzero constant greater than zero and is called the <strong>constant of variation<\/strong>. In this case, [latex]k=0.16[\/latex]\u00a0and [latex]n=1[\/latex].<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02222950\/CNX_Precalc_Figure_03_09_0012.jpg\" alt=\"Graph of y=(0.16)x where the horizontal axis is labeled,\" width=\"487\" height=\"459\" \/><figcaption class=\"wp-caption-text\">Graph of Nicole&#8217;s potential earnings<\/figcaption><\/figure>\n<section class=\"textbox keyTakeaway\">\n<h3>direct variation<\/h3>\n<p>If [latex]x[\/latex]<em>\u00a0<\/em>and [latex]y[\/latex]\u00a0are related by an equation of the form<\/p>\n<p style=\"text-align: center;\">[latex]y=k{x}^{n}[\/latex]<\/p>\n<p>then we say that the relationship is <strong>direct variation<\/strong> and [latex]y[\/latex]\u00a0<strong>varies directly<\/strong> with the [latex]n[\/latex]th power of [latex]x[\/latex]. <\/p>\n<p>In direct variation relationships, there is a nonzero constant ratio [latex]k=\\dfrac{y}{{x}^{n}}[\/latex], where [latex]k[\/latex]\u00a0is called the <strong>constant of variation<\/strong>, which help defines the relationship between the variables.<\/p>\n<\/section>\n<section class=\"textbox questionHelp\">\n<p><strong>How To: Given a description of a direct variation problem, solve for an unknown<br \/>\n<\/strong><\/p>\n<ol id=\"fs-id1165137724401\">\n<li>Identify the input, [latex]x[\/latex], and the output, [latex]y[\/latex].<\/li>\n<li>Determine the constant of variation. You may need to divide [latex]y[\/latex]\u00a0by the specified power of [latex]x[\/latex]\u00a0to determine the constant of variation.<\/li>\n<li>Use the constant of variation to write an equation for the relationship.<\/li>\n<li>Substitute known values into the equation to find the unknown.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>The quantity [latex]y[\/latex]\u00a0varies directly with the cube of [latex]x[\/latex]. If [latex]y=25[\/latex]\u00a0when [latex]x=2[\/latex], write the equation that represents this relationship. Then, find [latex]y[\/latex]\u00a0when [latex]x[\/latex]\u00a0is 6.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q647220\">Show Solution<\/button><\/p>\n<div id=\"q647220\" class=\"hidden-answer\" style=\"display: none\">\n<p>The general formula for direct variation with a cube is [latex]y=k{x}^{3}[\/latex]. The constant can be found by dividing [latex]y[\/latex]\u00a0by the cube of [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} k&=\\dfrac{y}{{x}^{3}} \\\\[1mm] &=\\dfrac{25}{{2}^{3}}\\\\[1mm] &=\\dfrac{25}{8}\\end{align}[\/latex]<\/p>\n<p>Now use the constant to write an equation that represents this relationship.<\/p>\n<p style=\"text-align: center;\">[latex]y=\\dfrac{25}{8}{x}^{3}[\/latex]<\/p>\n<p>Substitute [latex]x=6[\/latex]\u00a0and solve for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}y&=\\dfrac{25}{8}{\\left(6\\right)}^{3} \\\\[1mm] &=675\\hfill \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>The quantity [latex]y[\/latex]\u00a0varies directly with the square of [latex]y[\/latex]. If [latex]y=24[\/latex]\u00a0when [latex]x=3[\/latex], find [latex]y[\/latex]\u00a0when [latex]x[\/latex]\u00a0is 4.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q536994\">Show Solution<\/button><\/p>\n<div id=\"q536994\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\dfrac{128}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288442\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288442&theme=lumen&iframe_resize_id=ohm288442&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<section class=\"textbox watchIt\">\n<p>Watch this video to see a quick lesson about direct variation. You will see more worked examples.<\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6454977&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=plFOq4JaEyI&amp;video_target=tpm-plugin-ncnoy4d6-plFOq4JaEyI\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/DirectVariationApplications_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Direct Variation Applications&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n","protected":false},"author":15,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":770,"module-header":"background_you_need","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1743"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":12,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1743\/revisions"}],"predecessor-version":[{"id":4804,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1743\/revisions\/4804"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/770"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1743\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=1743"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=1743"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=1743"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=1743"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}