{"id":396,"date":"2021-02-04T01:58:20","date_gmt":"2021-02-04T01:58:20","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=396"},"modified":"2022-03-16T05:43:40","modified_gmt":"2022-03-16T05:43:40","slug":"differentials","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/differentials\/","title":{"raw":"Differentials and Amount of Error","rendered":"Differentials and Amount of Error"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Draw a graph that illustrates the use of differentials to approximate the change in a quantity<\/li>\r\n \t<li>Calculate the relative error and percentage error in using a differential approximation<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Computing Differentials<\/h2>\r\n<p id=\"fs-id1165042396173\">We have seen that linear approximations can be used to estimate function values. They can also be used to estimate the amount a function value changes as a result of a small change in the input. To discuss this more formally, we define a related concept: differentials.<strong> Differentials<\/strong> provide us with a way of estimating the amount a function changes as a result of a small change in input values.<\/p>\r\n<p id=\"fs-id1165043094041\">When we first looked at derivatives, we used the Leibniz notation [latex]dy\/dx[\/latex] to represent the derivative of [latex]y[\/latex] with respect to [latex]x[\/latex]. Although we used the expressions [latex]dy[\/latex] and [latex]dx[\/latex] in this notation, they did not have meaning on their own. Here we see a meaning to the expressions [latex]dy[\/latex] and [latex]dx[\/latex]. Suppose [latex]y=f(x)[\/latex] is a differentiable function. Let [latex]dx[\/latex] be an independent variable that can be assigned any nonzero real number, and define the dependent variable [latex]dy[\/latex] by<\/p>\r\n\r\n<div id=\"fs-id1165042369661\" class=\"equation\" style=\"text-align: center;\">[latex]dy=f^{\\prime}(x) \\, dx[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042520672\">It is important to notice that [latex]dy[\/latex] is a function of both [latex]x[\/latex] and [latex]dx[\/latex]. The expressions [latex]dy[\/latex] and [latex]dx[\/latex] are called <strong>differentials<\/strong>. We can divide both sides of the equation by [latex]dx[\/latex], which yields<\/p>\r\n\r\n<div id=\"fs-id1165042610583\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=f^{\\prime}(x)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043191780\">This is the familiar expression we have used to denote a derivative. The first equation is known as the <strong>differential form<\/strong> of the second one.<\/p>\r\n\r\n<div id=\"fs-id1165043166548\" class=\"textbook exercises\">\r\n<h3>Example: Computing differentials<\/h3>\r\n<p id=\"fs-id1165042987985\">For each of the following functions, find [latex]dy[\/latex] and evaluate when [latex]x=3[\/latex] and [latex]dx=0.1[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1165042321504\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]y=x^2+2x[\/latex]<\/li>\r\n \t<li>[latex]y= \\cos x[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1165042926541\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042926541\"]\r\n<p id=\"fs-id1165042926541\">The key step is calculating the derivative. When we have that, we can obtain [latex]dy[\/latex] directly.<\/p>\r\n\r\n<ol id=\"fs-id1165042482219\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Since [latex]f(x)=x^2+2x[\/latex], we know [latex]f^{\\prime}(x)=2x+2[\/latex], and therefore\r\n<div id=\"fs-id1165043352057\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dy=(2x+2) \\, dx[\/latex].<\/div>\r\nWhen [latex]x=3[\/latex] and [latex]dx=0.1[\/latex],\r\n<div id=\"fs-id1165043425324\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dy=(2 \\cdot 3+2)(0.1)=0.8[\/latex].<\/div><\/li>\r\n \t<li>Since [latex]f(x)= \\cos x[\/latex], [latex]f^{\\prime}(x)=\u2212\\sin (x)[\/latex]. This gives us\r\n<div id=\"fs-id1165042330818\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dy=\u2212\\sin x \\, dx[\/latex].<\/div>\r\nWhen [latex]x=3[\/latex] and [latex]dx=0.1[\/latex],\r\n<div id=\"fs-id1165042604734\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dy=\u2212\\sin (3)(0.1)=-0.1 \\sin (3)[\/latex].<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Computing differentials.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/R5C9K3oedGQ?controls=0&amp;start=736&amp;end=851&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]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.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.2LinearApproximationsAndDifferentials736to851_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.2 Linear Approximations and Differentials\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1165043395020\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165042367595\">For [latex]y=e^{x^2}[\/latex], find [latex]dy[\/latex].<\/p>\r\n[reveal-answer q=\"9076558\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"9076558\"]\r\n<p id=\"fs-id1165043256948\">[latex]dy=f^{\\prime}(x)dx[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043326695\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043326695\"]\r\n<p id=\"fs-id1165043326695\">[latex]dy=2xe^{x^2} \\, dx[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165043392916\">We now connect differentials to linear approximations. Differentials can be used to estimate the change in the value of a function resulting from a small change in input values. Consider a function [latex]f[\/latex] that is differentiable at point [latex]a[\/latex]. Suppose the input [latex]x[\/latex] changes by a small amount. We are interested in how much the output [latex]y[\/latex] changes. If [latex]x[\/latex] changes from [latex]a[\/latex] to [latex]a+dx[\/latex], then the change in [latex]x[\/latex] is [latex]dx[\/latex] (also denoted [latex]\\Delta x[\/latex]), and the change in [latex]y[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1165043089453\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Delta y=f(a+dx)-f(a)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043187497\">Instead of calculating the exact change in [latex]y[\/latex], however, it is often easier to approximate the change in [latex]y[\/latex] by using a linear approximation. For [latex]x[\/latex] near [latex]a[\/latex], [latex]f(x)[\/latex] can be approximated by the linear approximation<\/p>\r\n\r\n<div id=\"fs-id1165043318352\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]L(x)=f(a)+f^{\\prime}(a)(x-a)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043428364\">Therefore, if [latex]dx[\/latex] is small,<\/p>\r\n\r\n<div id=\"fs-id1165042925818\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(a+dx)\\approx L(a+dx)=f(a)+f^{\\prime}(a)(a+dx-a)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043178215\">That is,<\/p>\r\n\r\n<div id=\"fs-id1165042528283\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(a+dx)-f(a)\\approx L(a+dx)-f(a)=f^{\\prime}(a) \\, dx.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043395432\">In other words, the actual change in the function [latex]f[\/latex] if [latex]x[\/latex] increases from [latex]a[\/latex] to [latex]a+dx[\/latex] is approximately the difference between [latex]L(a+dx)[\/latex] and [latex]f(a)[\/latex], where [latex]L(x)[\/latex] is the linear approximation of [latex]f[\/latex] at [latex]a[\/latex]. By definition of [latex]L(x)[\/latex], this difference is equal to [latex]f^{\\prime}(a)dx[\/latex]. In summary,<\/p>\r\n\r\n<div id=\"fs-id1165042367953\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Delta y=f(a+dx)-f(a)\\approx L(a+dx)-f(a)=f^{\\prime}(a) \\, dx=dy[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043379826\">Therefore, we can use the differential [latex]dy=f^{\\prime}(a) \\, dx[\/latex] to approximate the change in [latex]y[\/latex] if [latex]x[\/latex] increases from [latex]x=a[\/latex] to [latex]x=a+dx[\/latex]. We can see this in the following graph.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"642\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210757\/CNX_Calc_Figure_04_02_005.jpg\" alt=\"A function y = f(x) is shown along with its tangent line at (a, f(a)). The tangent line is denoted L(x). The x axis is marked with a and a + dx, with a dashed line showing the distance between a and a + dx as dx. The points (a + dx, f(a + dx)) and (a + dx, L(a + dx)) are marked on the curves for y = f(x) and y = L(x), respectively. The distance between f(a) and L(a + dx) is marked as dy = f\u2019(a) dx, and the distance between f(a) and f(a + dx) is marked as \u0394y = f(a + dx) \u2013 f(a).\" width=\"642\" height=\"308\" \/> Figure 5. The differential [latex]dy=f^{\\prime}(a) \\, dx[\/latex] is used to approximate the actual change in [latex]y[\/latex] if [latex]x[\/latex] increases from [latex]a[\/latex] to [latex]a+dx[\/latex].[\/caption]\r\n<p id=\"fs-id1165043395105\">We now take a look at how to use differentials to approximate the change in the value of the function that results from a small change in the value of the input. Note the calculation with differentials is much simpler than calculating actual values of functions and the result is very close to what we would obtain with the more exact calculation.<\/p>\r\n\r\n<div id=\"fs-id1165043349128\" class=\"textbook exercises\">\r\n<h3>Example: Approximating Change with Differentials<\/h3>\r\n<p id=\"fs-id1165043109832\">Let [latex]y=x^2+2x[\/latex].<\/p>\r\nCompute [latex]\\Delta y[\/latex] and [latex]dy[\/latex] at [latex]x=3[\/latex] if [latex]dx=0.1[\/latex].\r\n\r\n[reveal-answer q=\"fs-id1165043286816\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043286816\"]\r\n<p id=\"fs-id1165043286816\">The actual change in [latex]y[\/latex] if [latex]x[\/latex] changes from [latex]x=3[\/latex] to [latex]x=3.1[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1165042713824\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Delta y=f(3.1)-f(3)=[(3.1)^2+2(3.1)]-[3^2+2(3)]=0.81[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043319960\">The approximate change in [latex]y[\/latex] is given by [latex]dy=f^{\\prime}(3) \\, dx[\/latex]. Since [latex]f^{\\prime}(x)=2x+2[\/latex], we have<\/p>\r\n\r\n<div id=\"fs-id1165042326840\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dy=f^{\\prime}(3) \\, dx=(2(3)+2)(0.1)=0.8[\/latex].[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042987991\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165042637207\">For [latex]y=x^2+2x[\/latex], find [latex]\\Delta y[\/latex] and [latex]dy[\/latex] at [latex]x=3[\/latex] if [latex]dx=0.2[\/latex].<\/p>\r\n[reveal-answer q=\"33228801\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"33228801\"]\r\n<p id=\"fs-id1165042925900\">[latex]dy=f^{\\prime}(3) \\, dx[\/latex], [latex]\\Delta y=f(3.2)-f(3)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043033503\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043033503\"]\r\n<p id=\"fs-id1165043033503\">[latex]dy=1.6[\/latex], [latex]\\Delta y=1.64[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Calculating the Amount of Error<\/h2>\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\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&nbsp;\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]. Specifically, if [latex]f[\/latex] is a differentiable function at [latex]a[\/latex], the propagated error is<\/p>\r\n\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&nbsp;\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\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&nbsp;\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\r\n<div id=\"fs-id1165043397423\" class=\"textbook exercises\">\r\n<h3>Example: Volume of a Cube<\/h3>\r\n<p id=\"fs-id1165043075572\">Suppose the side length of a cube is measured to be 5 cm with an accuracy of 0.1 cm.<\/p>\r\n\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) 4.9 cm and (ii) 5.1 cm to compare the estimated error with the actual potential error.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1165043429548\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043429548\"]\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,\r\n<div id=\"fs-id1165043087816\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-0.1\\le dx\\le 0.1[\/latex].<\/div>\r\nThe volume of a cube is given by [latex]V=x^3[\/latex], which leads to\r\n<div id=\"fs-id1165042948205\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dV=3x^2 \\, dx[\/latex].<\/div>\r\nUsing the measured side length of 5 cm, we can estimate that\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\nTherefore,\r\n<div id=\"fs-id1165042965823\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-7.5\\le dV\\le 7.5[\/latex].<\/div><\/li>\r\n \t<li>If the side length is actually 4.9 cm, then the volume of the cube is\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\nIf the side length is actually 5.1 cm, then the volume of the cube is\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\nTherefore, the actual volume of the cube is between 117.649 and 132.651. Since the side length is measured to be 5 cm, the computed volume is [latex]V(5)=5^3=125[\/latex]. Therefore, the error in the computed volume is\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\nThat is,\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\nWe see the estimated error [latex]dV[\/latex] is relatively close to the actual potential error in the computed volume.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Volume of a Cube.\r\n\r\n<center><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><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]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.\r\n\r\nYou 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>.[\/hidden-answer]\r\n<div id=\"fs-id1165042708694\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165043253780\">Estimate the error in the computed volume of a cube if the side length is measured to be 6 cm with an accuracy of 0.2 cm.<\/p>\r\n[reveal-answer q=\"8011542\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"8011542\"]\r\n<p id=\"fs-id1165042561334\">[latex]dV=3x^2 \\, dx[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042946479\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042946479\"]\r\n<p id=\"fs-id1165042946479\">The volume measurement is accurate to within [latex]21.6 \\, \\text{cm}^3[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165043061852\">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. 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. For example, if we measure the height of a ladder to be 63 in. when the actual height is 62 in., the absolute error is 1 in. but the relative error is [latex]\\frac{1}{62}=0.016[\/latex], or [latex]1.6 \\%[\/latex]. By comparison, if we measure the width of a piece of cardboard to be 8.25 in. when the actual width is 8 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]. Therefore, the percentage error in the measurement of the cardboard is larger, even though 0.25 in. is less than 1 in.<\/p>\r\n\r\n<div id=\"fs-id1165043087612\" class=\"textbook exercises\">\r\n<h3>Example: Relative and Percentage Error<\/h3>\r\n<p id=\"fs-id1165043352154\">An astronaut using a camera measures the radius of Earth as 4000 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[reveal-answer q=\"fs-id1165042331913\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042331913\"]\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\r\n<div id=\"fs-id1165042980337\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-80\\le dr\\le 80[\/latex].<\/div>\r\n&nbsp;\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\r\n<div id=\"fs-id1165042960149\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dV=4\\pi r^2 \\, dr[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043352072\">Using the measured radius of 4000 mi, we can estimate<\/p>\r\n\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&nbsp;\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\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&nbsp;\r\n<p id=\"fs-id1165043135038\">which simplifies to<\/p>\r\n\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&nbsp;\r\n<p id=\"fs-id1165043315322\">The relative error is 0.06 and the percentage error is [latex]6 \\%[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165043315340\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165043315347\">Determine the percentage error if the radius of Earth is measured to be 3950 mi with an error of [latex]\\pm 100[\/latex] mi.<\/p>\r\n[reveal-answer q=\"11870355\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"11870355\"]\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[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043309919\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043309919\"]\r\n<p id=\"fs-id1165043309919\">7.6%<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Draw a graph that illustrates the use of differentials to approximate the change in a quantity<\/li>\n<li>Calculate the relative error and percentage error in using a differential approximation<\/li>\n<\/ul>\n<\/div>\n<h2>Computing Differentials<\/h2>\n<p id=\"fs-id1165042396173\">We have seen that linear approximations can be used to estimate function values. They can also be used to estimate the amount a function value changes as a result of a small change in the input. To discuss this more formally, we define a related concept: differentials.<strong> Differentials<\/strong> provide us with a way of estimating the amount a function changes as a result of a small change in input values.<\/p>\n<p id=\"fs-id1165043094041\">When we first looked at derivatives, we used the Leibniz notation [latex]dy\/dx[\/latex] to represent the derivative of [latex]y[\/latex] with respect to [latex]x[\/latex]. Although we used the expressions [latex]dy[\/latex] and [latex]dx[\/latex] in this notation, they did not have meaning on their own. Here we see a meaning to the expressions [latex]dy[\/latex] and [latex]dx[\/latex]. Suppose [latex]y=f(x)[\/latex] is a differentiable function. Let [latex]dx[\/latex] be an independent variable that can be assigned any nonzero real number, and define the dependent variable [latex]dy[\/latex] by<\/p>\n<div id=\"fs-id1165042369661\" class=\"equation\" style=\"text-align: center;\">[latex]dy=f^{\\prime}(x) \\, dx[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042520672\">It is important to notice that [latex]dy[\/latex] is a function of both [latex]x[\/latex] and [latex]dx[\/latex]. The expressions [latex]dy[\/latex] and [latex]dx[\/latex] are called <strong>differentials<\/strong>. We can divide both sides of the equation by [latex]dx[\/latex], which yields<\/p>\n<div id=\"fs-id1165042610583\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=f^{\\prime}(x)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043191780\">This is the familiar expression we have used to denote a derivative. The first equation is known as the <strong>differential form<\/strong> of the second one.<\/p>\n<div id=\"fs-id1165043166548\" class=\"textbook exercises\">\n<h3>Example: Computing differentials<\/h3>\n<p id=\"fs-id1165042987985\">For each of the following functions, find [latex]dy[\/latex] and evaluate when [latex]x=3[\/latex] and [latex]dx=0.1[\/latex].<\/p>\n<ol id=\"fs-id1165042321504\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]y=x^2+2x[\/latex]<\/li>\n<li>[latex]y= \\cos x[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042926541\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042926541\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042926541\">The key step is calculating the derivative. When we have that, we can obtain [latex]dy[\/latex] directly.<\/p>\n<ol id=\"fs-id1165042482219\" style=\"list-style-type: lower-alpha;\">\n<li>Since [latex]f(x)=x^2+2x[\/latex], we know [latex]f^{\\prime}(x)=2x+2[\/latex], and therefore\n<div id=\"fs-id1165043352057\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dy=(2x+2) \\, dx[\/latex].<\/div>\n<p>When [latex]x=3[\/latex] and [latex]dx=0.1[\/latex],<\/p>\n<div id=\"fs-id1165043425324\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dy=(2 \\cdot 3+2)(0.1)=0.8[\/latex].<\/div>\n<\/li>\n<li>Since [latex]f(x)= \\cos x[\/latex], [latex]f^{\\prime}(x)=\u2212\\sin (x)[\/latex]. This gives us\n<div id=\"fs-id1165042330818\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dy=\u2212\\sin x \\, dx[\/latex].<\/div>\n<p>When [latex]x=3[\/latex] and [latex]dx=0.1[\/latex],<\/p>\n<div id=\"fs-id1165042604734\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dy=\u2212\\sin (3)(0.1)=-0.1 \\sin (3)[\/latex].<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Computing differentials.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/R5C9K3oedGQ?controls=0&amp;start=736&amp;end=851&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">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\/4.2LinearApproximationsAndDifferentials736to851_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>.<\/div>\n<\/div>\n<div id=\"fs-id1165043395020\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165042367595\">For [latex]y=e^{x^2}[\/latex], find [latex]dy[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q9076558\">Hint<\/span><\/p>\n<div id=\"q9076558\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043256948\">[latex]dy=f^{\\prime}(x)dx[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043326695\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043326695\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043326695\">[latex]dy=2xe^{x^2} \\, dx[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165043392916\">We now connect differentials to linear approximations. Differentials can be used to estimate the change in the value of a function resulting from a small change in input values. Consider a function [latex]f[\/latex] that is differentiable at point [latex]a[\/latex]. Suppose the input [latex]x[\/latex] changes by a small amount. We are interested in how much the output [latex]y[\/latex] changes. If [latex]x[\/latex] changes from [latex]a[\/latex] to [latex]a+dx[\/latex], then the change in [latex]x[\/latex] is [latex]dx[\/latex] (also denoted [latex]\\Delta x[\/latex]), and the change in [latex]y[\/latex] is given by<\/p>\n<div id=\"fs-id1165043089453\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Delta y=f(a+dx)-f(a)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043187497\">Instead of calculating the exact change in [latex]y[\/latex], however, it is often easier to approximate the change in [latex]y[\/latex] by using a linear approximation. For [latex]x[\/latex] near [latex]a[\/latex], [latex]f(x)[\/latex] can be approximated by the linear approximation<\/p>\n<div id=\"fs-id1165043318352\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]L(x)=f(a)+f^{\\prime}(a)(x-a)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043428364\">Therefore, if [latex]dx[\/latex] is small,<\/p>\n<div id=\"fs-id1165042925818\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(a+dx)\\approx L(a+dx)=f(a)+f^{\\prime}(a)(a+dx-a)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043178215\">That is,<\/p>\n<div id=\"fs-id1165042528283\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(a+dx)-f(a)\\approx L(a+dx)-f(a)=f^{\\prime}(a) \\, dx.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043395432\">In other words, the actual change in the function [latex]f[\/latex] if [latex]x[\/latex] increases from [latex]a[\/latex] to [latex]a+dx[\/latex] is approximately the difference between [latex]L(a+dx)[\/latex] and [latex]f(a)[\/latex], where [latex]L(x)[\/latex] is the linear approximation of [latex]f[\/latex] at [latex]a[\/latex]. By definition of [latex]L(x)[\/latex], this difference is equal to [latex]f^{\\prime}(a)dx[\/latex]. In summary,<\/p>\n<div id=\"fs-id1165042367953\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Delta y=f(a+dx)-f(a)\\approx L(a+dx)-f(a)=f^{\\prime}(a) \\, dx=dy[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043379826\">Therefore, we can use the differential [latex]dy=f^{\\prime}(a) \\, dx[\/latex] to approximate the change in [latex]y[\/latex] if [latex]x[\/latex] increases from [latex]x=a[\/latex] to [latex]x=a+dx[\/latex]. We can see this in the following graph.<\/p>\n<div style=\"width: 652px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210757\/CNX_Calc_Figure_04_02_005.jpg\" alt=\"A function y = f(x) is shown along with its tangent line at (a, f(a)). The tangent line is denoted L(x). The x axis is marked with a and a + dx, with a dashed line showing the distance between a and a + dx as dx. The points (a + dx, f(a + dx)) and (a + dx, L(a + dx)) are marked on the curves for y = f(x) and y = L(x), respectively. The distance between f(a) and L(a + dx) is marked as dy = f\u2019(a) dx, and the distance between f(a) and f(a + dx) is marked as \u0394y = f(a + dx) \u2013 f(a).\" width=\"642\" height=\"308\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. The differential [latex]dy=f^{\\prime}(a) \\, dx[\/latex] is used to approximate the actual change in [latex]y[\/latex] if [latex]x[\/latex] increases from [latex]a[\/latex] to [latex]a+dx[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1165043395105\">We now take a look at how to use differentials to approximate the change in the value of the function that results from a small change in the value of the input. Note the calculation with differentials is much simpler than calculating actual values of functions and the result is very close to what we would obtain with the more exact calculation.<\/p>\n<div id=\"fs-id1165043349128\" class=\"textbook exercises\">\n<h3>Example: Approximating Change with Differentials<\/h3>\n<p id=\"fs-id1165043109832\">Let [latex]y=x^2+2x[\/latex].<\/p>\n<p>Compute [latex]\\Delta y[\/latex] and [latex]dy[\/latex] at [latex]x=3[\/latex] if [latex]dx=0.1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043286816\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043286816\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043286816\">The actual change in [latex]y[\/latex] if [latex]x[\/latex] changes from [latex]x=3[\/latex] to [latex]x=3.1[\/latex] is given by<\/p>\n<div id=\"fs-id1165042713824\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Delta y=f(3.1)-f(3)=[(3.1)^2+2(3.1)]-[3^2+2(3)]=0.81[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043319960\">The approximate change in [latex]y[\/latex] is given by [latex]dy=f^{\\prime}(3) \\, dx[\/latex]. Since [latex]f^{\\prime}(x)=2x+2[\/latex], we have<\/p>\n<div id=\"fs-id1165042326840\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]dy=f^{\\prime}(3) \\, dx=(2(3)+2)(0.1)=0.8[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042987991\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165042637207\">For [latex]y=x^2+2x[\/latex], find [latex]\\Delta y[\/latex] and [latex]dy[\/latex] at [latex]x=3[\/latex] if [latex]dx=0.2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q33228801\">Hint<\/span><\/p>\n<div id=\"q33228801\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042925900\">[latex]dy=f^{\\prime}(3) \\, dx[\/latex], [latex]\\Delta y=f(3.2)-f(3)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043033503\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043033503\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043033503\">[latex]dy=1.6[\/latex], [latex]\\Delta y=1.64[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Calculating the Amount of Error<\/h2>\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>&nbsp;<\/p>\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]. 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>&nbsp;<\/p>\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>&nbsp;<\/p>\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<div id=\"fs-id1165043397423\" class=\"textbook exercises\">\n<h3>Example: Volume of a Cube<\/h3>\n<p id=\"fs-id1165043075572\">Suppose the side length of a cube is measured to be 5 cm with an accuracy of 0.1 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) 4.9 cm and (ii) 5.1 cm to compare the estimated error with the actual potential error.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043429548\">Show Solution<\/span><\/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 5 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 4.9 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 5.1 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 117.649 and 132.651. Since the side length is measured to be 5 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.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Volume of a Cube.<\/p>\n<div style=\"text-align: center;\"><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><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">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>.<\/div>\n<\/div>\n<div id=\"fs-id1165042708694\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165043253780\">Estimate the error in the computed volume of a cube if the side length is measured to be 6 cm with an accuracy of 0.2 cm.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q8011542\">Hint<\/span><\/p>\n<div id=\"q8011542\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042561334\">[latex]dV=3x^2 \\, dx[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042946479\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042946479\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042946479\">The volume measurement is accurate to within [latex]21.6 \\, \\text{cm}^3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165043061852\">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. 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. For example, if we measure the height of a ladder to be 63 in. when the actual height is 62 in., the absolute error is 1 in. but the relative error is [latex]\\frac{1}{62}=0.016[\/latex], or [latex]1.6 \\%[\/latex]. By comparison, if we measure the width of a piece of cardboard to be 8.25 in. when the actual width is 8 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]. Therefore, the percentage error in the measurement of the cardboard is larger, even though 0.25 in. is less than 1 in.<\/p>\n<div id=\"fs-id1165043087612\" class=\"textbook exercises\">\n<h3>Example: Relative and Percentage Error<\/h3>\n<p id=\"fs-id1165043352154\">An astronaut using a camera measures the radius of Earth as 4000 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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042331913\">Show Solution<\/span><\/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>&nbsp;<\/p>\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>&nbsp;<\/p>\n<p id=\"fs-id1165043352072\">Using the measured radius of 4000 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>&nbsp;<\/p>\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>&nbsp;<\/p>\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>&nbsp;<\/p>\n<p id=\"fs-id1165043315322\">The relative error is 0.06 and the percentage error is [latex]6 \\%[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165043315340\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165043315347\">Determine the percentage error if the radius of Earth is measured to be 3950 mi with an error of [latex]\\pm 100[\/latex] mi.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q11870355\">Hint<\/span><\/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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043309919\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043309919\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043309919\">7.6%<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-396\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>4.2 Linear Approximations and Differentials. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.2 Linear Approximations and Differentials\",\"author\":\"Ryan 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