{"id":3212,"date":"2024-06-13T18:03:47","date_gmt":"2024-06-13T18:03:47","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=3212"},"modified":"2024-08-05T02:11:50","modified_gmt":"2024-08-05T02:11:50","slug":"linear-approximations-and-differentials-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/linear-approximations-and-differentials-learn-it-3\/","title":{"raw":"Linear Approximations and Differentials: Learn It 3","rendered":"Linear Approximations and Differentials: Learn It 3"},"content":{"raw":"<h2>Differentials and Amount of Error Cont.<\/h2>\r\n<h3>Calculating the Amount of Error<\/h3>\r\n<p id=\"fs-id1165043392143\">Any type of measurement is prone to a certain amount of error. In many applications, certain quantities are calculated based on measurements. For example, the area of a circle is calculated by measuring the radius of the circle. An error in the measurement of the radius leads to an error in the computed value of the area. Here we examine this type of error and study how differentials can be used to estimate the error.<\/p>\r\n<p id=\"fs-id1165042333071\">Consider a function [latex]f[\/latex] with an input that is a measured quantity. Suppose the exact value of the measured quantity is [latex]a[\/latex], but the measured value is [latex]a+dx[\/latex]. We say the measurement error is [latex]dx[\/latex] (or [latex]\\Delta x[\/latex]). As a result, an error occurs in the calculated quantity [latex]f(x)[\/latex]. This type of error is known as a<strong> propagated error<\/strong> and is given by<\/p>\r\n<div id=\"fs-id1165043257818\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Delta y=f(a+dx)-f(a)[\/latex]<\/div>\r\n<p id=\"fs-id1165043091013\">Since all measurements are prone to some degree of error, we do not know the exact value of a measured quantity, so we cannot calculate the propagated error exactly. However, given an estimate of the accuracy of a measurement, we can use differentials to approximate the propagated error [latex]\\Delta y[\/latex].<\/p>\r\n<p>Specifically, if [latex]f[\/latex] is a differentiable function at [latex]a[\/latex], the propagated error is<\/p>\r\n<div id=\"fs-id1165042328497\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Delta y\\approx dy=f^{\\prime}(a) \\, dx[\/latex]<\/div>\r\n<p id=\"fs-id1165043393184\">Unfortunately, we do not know the exact value [latex]a[\/latex]. However, we can use the measured value [latex]a+dx[\/latex], and estimate<\/p>\r\n<div id=\"fs-id1165042321245\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Delta y\\approx dy\\approx f^{\\prime}(a+dx) \\, dx[\/latex]<\/div>\r\n<p id=\"fs-id1165043013879\">In the next example, we look at how differentials can be used to estimate the error in calculating the volume of a box if we assume the measurement of the side length is made with a certain amount of accuracy.<\/p>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1165043075572\">Suppose the side length of a cube is measured to be [latex]5[\/latex] cm with an accuracy of [latex]0.1[\/latex] cm.<\/p>\r\n<ol id=\"fs-id1165043075575\" style=\"list-style-type: lower-alpha;\">\r\n\t<li>Use differentials to estimate the error in the computed volume of the cube.<\/li>\r\n\t<li>Compute the volume of the cube if the side length is (i) [latex]4.9[\/latex] cm and (ii) [latex]5.1[\/latex] cm to compare the estimated error with the actual potential error.<\/li>\r\n<\/ol>\r\n<p>[reveal-answer q=\"fs-id1165043429548\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165043429548\"]<\/p>\r\n<ol id=\"fs-id1165043429548\" style=\"list-style-type: lower-alpha;\">\r\n\t<li>The measurement of the side length is accurate to within [latex]\\pm 0.1[\/latex] cm. Therefore,<br \/>\r\n<div id=\"fs-id1165043087816\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-0.1\\le dx\\le 0.1[\/latex].<\/div>\r\n<p>The volume of a cube is given by [latex]V=x^3[\/latex], which leads to<\/p>\r\n<div id=\"fs-id1165042948205\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dV=3x^2 \\, dx[\/latex].<\/div>\r\n<p>Using the measured side length of [latex]5[\/latex] cm, we can estimate that<\/p>\r\n<div id=\"fs-id1165043391588\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-3(5)^2(0.1)\\le dV\\le 3(5)^2(0.1)[\/latex].<\/div>\r\n<p>Therefore,<\/p>\r\n<div id=\"fs-id1165042965823\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-7.5\\le dV\\le 7.5[\/latex].<\/div>\r\n<\/li>\r\n\t<li>If the side length is actually [latex]4.9[\/latex] cm, then the volume of the cube is<br \/>\r\n<div id=\"fs-id1165042931789\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V(4.9)=(4.9)^3=117.649 \\, \\text{cm}^3[\/latex].<\/div>\r\n<p>If the side length is actually 5[latex].1[\/latex] cm, then the volume of the cube is<\/p>\r\n<div id=\"fs-id1165042901307\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V(5.1)=(5.1)^3=132.651\\, \\text{cm}^3[\/latex].<\/div>\r\n<p>Therefore, the actual volume of the cube is between [latex]117.649[\/latex] and [latex]132.651[\/latex]. Since the side length is measured to be [latex]5[\/latex] cm, the computed volume is [latex]V(5)=5^3=125[\/latex]. Therefore, the error in the computed volume is<\/p>\r\n<div id=\"fs-id1165042322167\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]117.649-125\\le \\Delta V\\le 132.651-125[\/latex].<\/div>\r\n<p>That is,<\/p>\r\n<div id=\"fs-id1165043425273\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-7.351\\le \\Delta V\\le 7.651[\/latex].<\/div>\r\n<p>We see the estimated error [latex]dV[\/latex] is relatively close to the actual potential error in the computed volume.<\/p>\r\n<\/li>\r\n<\/ol>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<p><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/R5C9K3oedGQ?controls=0&amp;start=1193&amp;end=1461&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\r\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.2LinearApproximationsAndDifferentials1193to1461_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.2 Linear Approximations and Differentials\" here (opens in new window)<\/a>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p>The measurement error [latex]dx \\, (=\\Delta x)[\/latex] and the propagated error [latex]\\Delta y[\/latex] are absolute errors. We are typically interested in the size of an error relative to the size of the quantity being measured or calculated.<\/p>\r\n<p>Given an absolute error [latex]\\Delta q[\/latex] for a particular quantity, we define the <strong>relative error<\/strong> as [latex]\\frac{\\Delta q}{q}[\/latex], where [latex]q[\/latex] is the actual value of the quantity. The <strong>percentage error<\/strong> is the relative error expressed as a percentage.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>relative and percentage error<\/h3>\r\n<p>Relative error ([latex]\\frac{\\Delta q}{q}[\/latex]) and percentage error measure how significant an error is compared to the true value of a quantity. Relative error gives a scale-independent error measure, while percentage error expresses this error as a percentage, providing a clear indication of accuracy in measurements.<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>For example, if we measure the height of a ladder to be [latex]63[\/latex] in. when the actual height is [latex]62[\/latex] in., the absolute error is [latex]1[\/latex] in. but the relative error is [latex]\\frac{1}{62}=0.016[\/latex], or [latex]1.6 \\%[\/latex].<\/p>\r\n<p>By comparison, if we measure the width of a piece of cardboard to be [latex]8.25[\/latex] in. when the actual width is [latex]8[\/latex] in., our absolute error is [latex]\\frac{1}{4}[\/latex] in., whereas the relative error is [latex]\\frac{0.25}{8}=\\frac{1}{32}[\/latex], or [latex]3.1\\%[\/latex].<\/p>\r\n<p>Therefore, the percentage error in the measurement of the cardboard is larger, even though [latex]0.25[\/latex] in. is less than [latex]1[\/latex] in.<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1165043352154\">An astronaut using a camera measures the radius of Earth as [latex]4000[\/latex] mi with an error of [latex]\\pm 80[\/latex] mi. Let\u2019s use differentials to estimate the relative and percentage error of using this radius measurement to calculate the volume of Earth, assuming the planet is a perfect sphere.<\/p>\r\n<p>[reveal-answer q=\"fs-id1165042331913\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165042331913\"]<\/p>\r\n<p id=\"fs-id1165042331913\">If the measurement of the radius is accurate to within [latex]\\pm 80[\/latex], we have<\/p>\r\n<div id=\"fs-id1165042980337\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-80\\le dr\\le 80[\/latex].<\/div>\r\n<p id=\"fs-id1165043392096\">Since the volume of a sphere is given by [latex]V=(\\frac{4}{3})\\pi r^3[\/latex], we have<\/p>\r\n<div id=\"fs-id1165042960149\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dV=4\\pi r^2 \\, dr[\/latex].<\/div>\r\n<p id=\"fs-id1165043352072\">Using the measured radius of [latex]4000[\/latex] mi, we can estimate<\/p>\r\n<div id=\"fs-id1165043166554\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-4\\pi (4000)^2(80)\\le dV\\le 4\\pi (4000)^2(80)[\/latex].<\/div>\r\n<p id=\"fs-id1165043422551\">To estimate the relative error, consider [latex]\\frac{dV}{V}[\/latex]. Since we do not know the exact value of the volume [latex]V[\/latex], use the measured radius [latex]r=4000[\/latex] mi to estimate [latex]V[\/latex]. We obtain [latex]V\\approx (\\frac{4}{3})\\pi (4000)^3[\/latex]. Therefore the relative error satisfies<\/p>\r\n<div id=\"fs-id1165043380581\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{-4\\pi (4000)^2(80)}{4\\pi (4000)^3 \/ 3}\\le \\dfrac{dV}{V}\\le \\dfrac{4\\pi (4000)^2(80)}{4\\pi (4000)^3 \/3}[\/latex],<\/div>\r\n<p id=\"fs-id1165043135038\">which simplifies to<\/p>\r\n<div id=\"fs-id1165042513675\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-0.06\\le \\frac{dV}{V}\\le 0.06[\/latex].<\/div>\r\n<p id=\"fs-id1165043315322\">The relative error is [latex]0.06[\/latex] and the percentage error is [latex]6 \\%[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1165043315347\">Determine the percentage error if the radius of Earth is measured to be [latex] 3950[\/latex] mi with an error of [latex]\\pm 100[\/latex] mi.<\/p>\r\n<p>[reveal-answer q=\"11870355\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"11870355\"]<\/p>\r\n<p id=\"fs-id1165042647114\">Use the fact that [latex]dV=4\\pi r^2 \\, dr[\/latex] to find [latex]dV\/V[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<p>[reveal-answer q=\"fs-id1165043309919\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165043309919\"]<\/p>\r\n<p id=\"fs-id1165043309919\">[latex]7.6\\%[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>","rendered":"<h2>Differentials and Amount of Error Cont.<\/h2>\n<h3>Calculating the Amount of Error<\/h3>\n<p id=\"fs-id1165043392143\">Any type of measurement is prone to a certain amount of error. In many applications, certain quantities are calculated based on measurements. For example, the area of a circle is calculated by measuring the radius of the circle. An error in the measurement of the radius leads to an error in the computed value of the area. Here we examine this type of error and study how differentials can be used to estimate the error.<\/p>\n<p id=\"fs-id1165042333071\">Consider a function [latex]f[\/latex] with an input that is a measured quantity. Suppose the exact value of the measured quantity is [latex]a[\/latex], but the measured value is [latex]a+dx[\/latex]. We say the measurement error is [latex]dx[\/latex] (or [latex]\\Delta x[\/latex]). As a result, an error occurs in the calculated quantity [latex]f(x)[\/latex]. This type of error is known as a<strong> propagated error<\/strong> and is given by<\/p>\n<div id=\"fs-id1165043257818\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Delta y=f(a+dx)-f(a)[\/latex]<\/div>\n<p id=\"fs-id1165043091013\">Since all measurements are prone to some degree of error, we do not know the exact value of a measured quantity, so we cannot calculate the propagated error exactly. However, given an estimate of the accuracy of a measurement, we can use differentials to approximate the propagated error [latex]\\Delta y[\/latex].<\/p>\n<p>Specifically, if [latex]f[\/latex] is a differentiable function at [latex]a[\/latex], the propagated error is<\/p>\n<div id=\"fs-id1165042328497\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Delta y\\approx dy=f^{\\prime}(a) \\, dx[\/latex]<\/div>\n<p id=\"fs-id1165043393184\">Unfortunately, we do not know the exact value [latex]a[\/latex]. However, we can use the measured value [latex]a+dx[\/latex], and estimate<\/p>\n<div id=\"fs-id1165042321245\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Delta y\\approx dy\\approx f^{\\prime}(a+dx) \\, dx[\/latex]<\/div>\n<p id=\"fs-id1165043013879\">In the next example, we look at how differentials can be used to estimate the error in calculating the volume of a box if we assume the measurement of the side length is made with a certain amount of accuracy.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1165043075572\">Suppose the side length of a cube is measured to be [latex]5[\/latex] cm with an accuracy of [latex]0.1[\/latex] cm.<\/p>\n<ol id=\"fs-id1165043075575\" style=\"list-style-type: lower-alpha;\">\n<li>Use differentials to estimate the error in the computed volume of the cube.<\/li>\n<li>Compute the volume of the cube if the side length is (i) [latex]4.9[\/latex] cm and (ii) [latex]5.1[\/latex] cm to compare the estimated error with the actual potential error.<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165043429548\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165043429548\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165043429548\" style=\"list-style-type: lower-alpha;\">\n<li>The measurement of the side length is accurate to within [latex]\\pm 0.1[\/latex] cm. Therefore,\n<div id=\"fs-id1165043087816\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-0.1\\le dx\\le 0.1[\/latex].<\/div>\n<p>The volume of a cube is given by [latex]V=x^3[\/latex], which leads to<\/p>\n<div id=\"fs-id1165042948205\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dV=3x^2 \\, dx[\/latex].<\/div>\n<p>Using the measured side length of [latex]5[\/latex] cm, we can estimate that<\/p>\n<div id=\"fs-id1165043391588\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-3(5)^2(0.1)\\le dV\\le 3(5)^2(0.1)[\/latex].<\/div>\n<p>Therefore,<\/p>\n<div id=\"fs-id1165042965823\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-7.5\\le dV\\le 7.5[\/latex].<\/div>\n<\/li>\n<li>If the side length is actually [latex]4.9[\/latex] cm, then the volume of the cube is\n<div id=\"fs-id1165042931789\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V(4.9)=(4.9)^3=117.649 \\, \\text{cm}^3[\/latex].<\/div>\n<p>If the side length is actually 5[latex].1[\/latex] cm, then the volume of the cube is<\/p>\n<div id=\"fs-id1165042901307\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V(5.1)=(5.1)^3=132.651\\, \\text{cm}^3[\/latex].<\/div>\n<p>Therefore, the actual volume of the cube is between [latex]117.649[\/latex] and [latex]132.651[\/latex]. Since the side length is measured to be [latex]5[\/latex] cm, the computed volume is [latex]V(5)=5^3=125[\/latex]. Therefore, the error in the computed volume is<\/p>\n<div id=\"fs-id1165042322167\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]117.649-125\\le \\Delta V\\le 132.651-125[\/latex].<\/div>\n<p>That is,<\/p>\n<div id=\"fs-id1165043425273\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-7.351\\le \\Delta V\\le 7.651[\/latex].<\/div>\n<p>We see the estimated error [latex]dV[\/latex] is relatively close to the actual potential error in the computed volume.<\/p>\n<\/li>\n<\/ol>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<p><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/R5C9K3oedGQ?controls=0&amp;start=1193&amp;end=1461&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.2LinearApproximationsAndDifferentials1193to1461_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.2 Linear Approximations and Differentials&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p>The measurement error [latex]dx \\, (=\\Delta x)[\/latex] and the propagated error [latex]\\Delta y[\/latex] are absolute errors. We are typically interested in the size of an error relative to the size of the quantity being measured or calculated.<\/p>\n<p>Given an absolute error [latex]\\Delta q[\/latex] for a particular quantity, we define the <strong>relative error<\/strong> as [latex]\\frac{\\Delta q}{q}[\/latex], where [latex]q[\/latex] is the actual value of the quantity. The <strong>percentage error<\/strong> is the relative error expressed as a percentage.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>relative and percentage error<\/h3>\n<p>Relative error ([latex]\\frac{\\Delta q}{q}[\/latex]) and percentage error measure how significant an error is compared to the true value of a quantity. Relative error gives a scale-independent error measure, while percentage error expresses this error as a percentage, providing a clear indication of accuracy in measurements.<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>For example, if we measure the height of a ladder to be [latex]63[\/latex] in. when the actual height is [latex]62[\/latex] in., the absolute error is [latex]1[\/latex] in. but the relative error is [latex]\\frac{1}{62}=0.016[\/latex], or [latex]1.6 \\%[\/latex].<\/p>\n<p>By comparison, if we measure the width of a piece of cardboard to be [latex]8.25[\/latex] in. when the actual width is [latex]8[\/latex] in., our absolute error is [latex]\\frac{1}{4}[\/latex] in., whereas the relative error is [latex]\\frac{0.25}{8}=\\frac{1}{32}[\/latex], or [latex]3.1\\%[\/latex].<\/p>\n<p>Therefore, the percentage error in the measurement of the cardboard is larger, even though [latex]0.25[\/latex] in. is less than [latex]1[\/latex] in.<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1165043352154\">An astronaut using a camera measures the radius of Earth as [latex]4000[\/latex] mi with an error of [latex]\\pm 80[\/latex] mi. Let\u2019s use differentials to estimate the relative and percentage error of using this radius measurement to calculate the volume of Earth, assuming the planet is a perfect sphere.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165042331913\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165042331913\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042331913\">If the measurement of the radius is accurate to within [latex]\\pm 80[\/latex], we have<\/p>\n<div id=\"fs-id1165042980337\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-80\\le dr\\le 80[\/latex].<\/div>\n<p id=\"fs-id1165043392096\">Since the volume of a sphere is given by [latex]V=(\\frac{4}{3})\\pi r^3[\/latex], we have<\/p>\n<div id=\"fs-id1165042960149\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dV=4\\pi r^2 \\, dr[\/latex].<\/div>\n<p id=\"fs-id1165043352072\">Using the measured radius of [latex]4000[\/latex] mi, we can estimate<\/p>\n<div id=\"fs-id1165043166554\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-4\\pi (4000)^2(80)\\le dV\\le 4\\pi (4000)^2(80)[\/latex].<\/div>\n<p id=\"fs-id1165043422551\">To estimate the relative error, consider [latex]\\frac{dV}{V}[\/latex]. Since we do not know the exact value of the volume [latex]V[\/latex], use the measured radius [latex]r=4000[\/latex] mi to estimate [latex]V[\/latex]. We obtain [latex]V\\approx (\\frac{4}{3})\\pi (4000)^3[\/latex]. Therefore the relative error satisfies<\/p>\n<div id=\"fs-id1165043380581\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{-4\\pi (4000)^2(80)}{4\\pi (4000)^3 \/ 3}\\le \\dfrac{dV}{V}\\le \\dfrac{4\\pi (4000)^2(80)}{4\\pi (4000)^3 \/3}[\/latex],<\/div>\n<p id=\"fs-id1165043135038\">which simplifies to<\/p>\n<div id=\"fs-id1165042513675\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-0.06\\le \\frac{dV}{V}\\le 0.06[\/latex].<\/div>\n<p id=\"fs-id1165043315322\">The relative error is [latex]0.06[\/latex] and the percentage error is [latex]6 \\%[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1165043315347\">Determine the percentage error if the radius of Earth is measured to be [latex]3950[\/latex] mi with an error of [latex]\\pm 100[\/latex] mi.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q11870355\">Hint<\/button><\/p>\n<div id=\"q11870355\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042647114\">Use the fact that [latex]dV=4\\pi r^2 \\, dr[\/latex] to find [latex]dV\/V[\/latex].<\/p>\n<\/div>\n<\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165043309919\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165043309919\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043309919\">[latex]7.6\\%[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":12,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":652,"module-header":"learn_it","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\/3212"}],"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":4,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3212\/revisions"}],"predecessor-version":[{"id":3831,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3212\/revisions\/3831"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/652"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3212\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=3212"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=3212"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=3212"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=3212"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}