{"id":1637,"date":"2021-07-22T17:19:25","date_gmt":"2021-07-22T17:19:25","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1637"},"modified":"2022-03-21T22:58:54","modified_gmt":"2022-03-21T22:58:54","slug":"applications-of-first-order-linear-differential-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/applications-of-first-order-linear-differential-equations\/","title":{"raw":"Applications of First-order Linear Differential Equations","rendered":"Applications of First-order Linear Differential Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve applied problems involving first-order linear differential equations<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170572558312\">We look at two different applications of first-order linear differential equations. The first involves air resistance as it relates to objects that are rising or falling; the second involves an electrical circuit. Other applications are numerous, but most are solved in a similar fashion.<\/p>\r\n\r\n<section id=\"fs-id1170572558318\" data-depth=\"2\">\r\n<h2 data-type=\"title\">Free fall with air resistance<\/h2>\r\n<p id=\"fs-id1170571829519\">We discussed air resistance at the beginning of this section. The next example shows how to apply this concept for a ball in vertical motion. Other factors can affect the force of air resistance, such as the size and shape of the object, but we ignore them here.<\/p>\r\n\r\n<div id=\"fs-id1170571829525\" data-type=\"example\">\r\n<div id=\"fs-id1170571829527\" data-type=\"exercise\">\r\n<div id=\"fs-id1170572510001\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0A Ball with Air Resistance<\/h3>\r\n<div id=\"fs-id1170572510001\" data-type=\"problem\">\r\n<p id=\"fs-id1170572510006\">A racquetball is hit straight upward with an initial velocity of [latex]2[\/latex] m\/s. The mass of a racquetball is approximately [latex]0.0427[\/latex] kg. Air resistance acts on the ball with a force numerically equal to [latex]0.5v[\/latex], where [latex]v[\/latex] represents the velocity of the ball at time [latex]t[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1170571598283\" type=\"a\">\r\n \t<li>Find the velocity of the ball as a function of time.<\/li>\r\n \t<li>How long does it take for the ball to reach its maximum height?<\/li>\r\n \t<li>If the ball is hit from an initial height of [latex]1[\/latex] meter, how high will it reach?<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n<div id=\"fs-id1170572247894\" data-type=\"solution\">\r\n<ol id=\"fs-id1170572247896\" type=\"a\">\r\n \t<li>The mass [latex]m=0.0427\\text{kg},k=0.5[\/latex], and [latex]g=9.8{\\text{m\/s}}^{2}[\/latex]. The initial velocity is [latex]{v}_{0}=2[\/latex] m\/s. Therefore the initial-value problem is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572379510\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]0.0427\\frac{dv}{dt}=-0.5v - 0.0427\\left(9.8\\right),{v}_{0}=2[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nDividing the differential equation by [latex]0.0427[\/latex] gives<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572547804\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{dv}{dt}=-11.7096v - 9.8,{v}_{0}=2[\/latex].<\/div>\r\n&nbsp;\r\n<p style=\"text-align: left;\"><span data-type=\"newline\">\r\n<\/span>\r\nThe differential equation is linear. Using the problem-solving strategy for linear differential equations:<span data-type=\"newline\">\r\n<\/span>\r\nStep 1. Rewrite the differential equation as [latex]\\frac{dv}{dt}+11.7096v=-9.8[\/latex]. This gives [latex]p\\left(t\\right)=11.7096[\/latex] and [latex]q\\left(t\\right)=-9.8[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\nStep 2. The integrating factor is [latex]\\mu \\left(t\\right)={e}^{\\displaystyle\\int 11.7096dt}={e}^{11.7096t}[\/latex]. <span data-type=\"newline\">\r\n<\/span>\r\nStep 3. Multiply the differential equation by [latex]\\mu \\left(t\\right)\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<div id=\"fs-id1170572644185\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {e}^{11.7096t}\\frac{dv}{dt}+11.7096v{e}^{11.7096t}&amp; =\\hfill &amp; -9.8{e}^{11.7096t}\\hfill \\\\ \\hfill \\frac{d}{dt}\\left[v{e}^{11.7096t}\\right]&amp; =\\hfill &amp; -9.8{e}^{11.7096t}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nStep 4. Integrate both sides:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572415253\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int \\frac{d}{dt}\\left[v{e}^{11.7096t}\\right]dt}&amp; =\\hfill &amp; {\\displaystyle\\int -9.8{e}^{11.7096t}dt}\\hfill \\\\ \\hfill v{e}^{11.7096t}&amp; =\\hfill &amp; \\frac{-9.8}{11.7096}{e}^{11.7096t}+C\\hfill \\\\ \\hfill v\\left(t\\right)&amp; =\\hfill &amp; -0.8369+C{e}^{-11.7096t}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nStep 5. Solve for [latex]C[\/latex] using the initial condition [latex]{v}_{0}=v\\left(0\\right)=2\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572227250\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill v\\left(t\\right)&amp; =\\hfill &amp; -0.8369+C{e}^{-11.7096t}\\hfill \\\\ \\hfill v\\left(0\\right)&amp; =\\hfill &amp; -0.8369+C{e}^{-11.7096\\left(0\\right)}\\hfill \\\\ \\hfill 2&amp; =\\hfill &amp; -0.8369+C\\hfill \\\\ \\hfill C&amp; =\\hfill &amp; 2.8369.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore the solution to the initial-value problem is [latex]v\\left(t\\right)=2.8369{e}^{-11.7096t}-0.8369[\/latex].<\/li>\r\n \t<li>The ball reaches its maximum height when the velocity is equal to zero. The reason is that when the velocity is positive, it is rising, and when it is negative, it is falling. Therefore when it is zero, it is neither rising nor falling, and is at its maximum height:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571553927\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\hfill 2.8369{e}^{-11.7096t}-0.8369&amp; =\\hfill &amp; 0\\hfill \\\\ \\hfill 2.8369{e}^{-11.7096t}&amp; =\\hfill &amp; 0.8369\\hfill \\\\ \\hfill {e}^{-11.7096t}&amp; =\\hfill &amp; \\frac{0.8369}{2.8369}\\approx 0.295\\hfill \\\\ \\hfill \\text{ln}{e}^{-11.7096t}&amp; =\\hfill &amp; \\text{ln}0.295\\approx -1.221\\hfill \\\\ \\hfill -11.7096t&amp; =\\hfill &amp; -1.221\\hfill \\\\ \\hfill t&amp; \\approx \\hfill &amp; 0.104.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore it takes approximately [latex]0.104[\/latex] second to reach maximum height.<\/li>\r\n \t<li>To find the height of the ball as a function of time, use the fact that the derivative of position is velocity, i.e., if [latex]h\\left(t\\right)[\/latex] represents the height at time [latex]t[\/latex], then [latex]{h}^{\\prime }\\left(t\\right)=v\\left(t\\right)[\/latex]. Because we know [latex]v\\left(t\\right)[\/latex] and the initial height, we can form an initial-value problem:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571825011\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]{h}^{\\prime }\\left(t\\right)=2.8369{e}^{-11.7096t}-0.8369,h\\left(0\\right)=1[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nIntegrating both sides of the differential equation with respect to [latex]t[\/latex] gives<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571595078\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int {h}^{\\prime }\\left(t\\right)dt}&amp; =\\hfill &amp; {\\displaystyle\\int 2.8369{e}^{-11.7096t}-0.8369dt}\\hfill \\\\ \\hfill h\\left(t\\right)&amp; =\\hfill &amp; -\\frac{2.8369}{11.7096}{e}^{-11.7096t}-0.8369t+C\\hfill \\\\ \\hfill h\\left(t\\right)&amp; =\\hfill &amp; -0.2423{e}^{-11.7096t}-0.8369t+C.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSolve for [latex]C[\/latex] by using the initial condition:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572140765\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill h\\left(t\\right)&amp; =\\hfill &amp; -0.2423{e}^{-11.7096t}-0.8369t+C\\hfill \\\\ \\hfill h\\left(0\\right)&amp; =\\hfill &amp; -0.2423{e}^{-11.7096\\left(0\\right)}-0.8369\\left(0\\right)+C\\hfill \\\\ \\hfill 1&amp; =\\hfill &amp; -0.2423+C\\hfill \\\\ \\hfill C&amp; =\\hfill &amp; 1.2423.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572208058\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]h\\left(t\\right)=-0.2423{e}^{-11.7096t}-0.8369t+1.2423[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nAfter [latex]0.104[\/latex] second, the height is given by<span data-type=\"newline\">\r\n<\/span>\r\n[latex]h\\left(0.2\\right)=-0.2423{e}^{-11.7096t}-0.8369t+1.2423\\approx 1.0836[\/latex] meter.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example:\u00a0A Ball with Air Resistance.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/N6wo00P3ab0?controls=0&amp;start=28&amp;end=834&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">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\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.5.3_28to834_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.5.3\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1170572481763\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1170572481767\" data-type=\"exercise\">\r\n<div id=\"fs-id1170572481770\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1170572481770\" data-type=\"problem\">\r\n<p id=\"fs-id1170572481772\">The weight of a penny is [latex]2.5[\/latex] grams (United States Mint, \"Coin Specifications,\" accessed April 9, 2015, http:\/\/www.usmint.gov\/about_the_mint\/?action=coin_specifications), and the upper observation deck of the Empire State Building is [latex]369[\/latex] meters above the street. Since the penny is a small and relatively smooth object, air resistance acting on the penny is actually quite small. We assume the air resistance is numerically equal to [latex]0.0025v[\/latex]. Furthermore, the penny is dropped with no initial velocity imparted to it.<\/p>\r\n\r\n<ol id=\"fs-id1170571532540\" type=\"a\">\r\n \t<li>Set up an initial-value problem that represents the falling penny.<\/li>\r\n \t<li>Solve the problem for [latex]v\\left(t\\right)[\/latex].<\/li>\r\n \t<li>What is the terminal velocity of the penny (i.e., calculate the limit of the velocity as [latex]t[\/latex] approaches infinity)?<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558879\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558879\"]\r\n<div id=\"fs-id1170572523011\" data-type=\"commentary\" data-element-type=\"hint\">\r\n\r\nSet up the differential equation the same way as the example: Writing First-Order Linear Equations in Standard Form Remember to convert from grams to kilograms.\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n<div id=\"fs-id1170571555522\" data-type=\"solution\">\r\n<ol id=\"fs-id1170571711999\" type=\"a\">\r\n \t<li>[latex]\\begin{array}{ccc}\\hfill \\frac{dv}{dt}&amp; =\\hfill &amp; \\text{-}v - 9.8\\hfill \\\\ \\hfill v\\left(0\\right)&amp; =\\hfill &amp; 0\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>[latex]v\\left(t\\right)=9.8\\left({e}^{\\text{-}t}-1\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\underset{t\\to \\infty }{\\text{lim}}v\\left(t\\right)=\\underset{t\\to \\infty }{\\text{lim}}\\left(9.8\\left({e}^{\\text{-}t}-1\\right)\\right)=-9.8\\text{m\/s}\\approx -21.922\\text{mph}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1170572404173\" data-depth=\"2\">\r\n<h2 data-type=\"title\">Electrical Circuits<\/h2>\r\n<p id=\"fs-id1170572404179\">A source of electromotive force (e.g., a battery or generator) produces a flow of current in a closed circuit, and this current produces a voltage drop across each resistor, inductor, and capacitor in the circuit. Kirchhoff\u2019s Loop Rule states that the sum of the voltage drops across resistors, inductors, and capacitors is equal to the total electromotive force in a closed circuit. We have the following three results:<\/p>\r\n\r\n<ol id=\"fs-id1170571674159\" type=\"1\">\r\n \t<li>The voltage drop across a resistor is given by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571674169\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{E}_{R}=Ri[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwhere [latex]R[\/latex] is a constant of proportionality called the <em data-effect=\"italics\">resistance,<\/em> and [latex]i[\/latex] is the current.<\/li>\r\n \t<li>The voltage drop across an inductor is given by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572375868\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{E}_{L}=L{i}^{\\prime }[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwhere [latex]L[\/latex] is a constant of proportionality called the <em data-effect=\"italics\">inductance<\/em>, and [latex]i[\/latex] again denotes the current.<\/li>\r\n \t<li>The voltage drop across a capacitor is given by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571683191\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{E}_{C}=\\frac{1}{C}q[\/latex],<\/div>\r\n&nbsp;<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1170572472586\">where [latex]C[\/latex] is a constant of proportionality called the <em data-effect=\"italics\">capacitance<\/em>, and [latex]q[\/latex] is the instantaneous charge on the capacitor. The relationship between [latex]i[\/latex] and [latex]q[\/latex] is [latex]i={q}^{\\prime }[\/latex].<\/p>\r\n<p id=\"fs-id1170571714388\">We use units of volts [latex]\\left(\\text{V}\\right)[\/latex] to measure voltage [latex]E[\/latex], amperes [latex]\\left(\\text{A}\\right)[\/latex] to measure current [latex]i[\/latex], coulombs [latex]\\left(\\text{C}\\right)[\/latex] to measure charge [latex]q[\/latex], ohms [latex]\\left(\\Omega \\right)[\/latex] to measure resistance [latex]R[\/latex], henrys [latex]\\left(\\text{H}\\right)[\/latex] to measure inductance [latex]L[\/latex], and farads [latex]\\left(\\text{F}\\right)[\/latex] to measure capacitance [latex]C[\/latex]. Consider the circuit in Figure 2.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_08_05_002\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"392\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234225\/CNX_Calc_Figure_08_05_003.jpg\" alt=\"A diagram of an electric circuit in a rectangle. The top has a capacitor C, the left has a voltage generator Vs, the bottom was a resistor R, and the right has an inductor L.\" width=\"392\" height=\"242\" data-media-type=\"image\/jpeg\" \/> Figure 2. A typical electric circuit, containing a voltage generator [latex]\\left({V}_{S}\\right)[\/latex], capacitor [latex]\\left(C\\right)[\/latex], inductor [latex]\\left(L\\right)[\/latex], and resistor [latex]\\left(R\\right)[\/latex].[\/caption]<\/figure>\r\n<p id=\"fs-id1170571783881\">Applying Kirchhoff\u2019s Loop Rule to this circuit, we let [latex]E[\/latex] denote the electromotive force supplied by the voltage generator. Then<\/p>\r\n\r\n<div id=\"fs-id1170571733859\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{E}_{L}+{E}_{R}+{E}_{C}=E[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572378413\">Substituting the expressions for [latex]{E}_{L},{E}_{R}[\/latex], and [latex]{E}_{C}[\/latex] into this equation, we obtain<\/p>\r\n\r\n<div id=\"fs-id1170572177052\" style=\"text-align: center;\" data-type=\"equation\">[latex]L{i}^{\\prime }+Ri+\\frac{1}{C}q=E[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572241888\">If there is no capacitor in the circuit, then the equation becomes<\/p>\r\n\r\n<div id=\"fs-id1170572241891\" style=\"text-align: center;\" data-type=\"equation\">[latex]L{i}^{\\prime }+Ri=E[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572585910\">This is a first-order differential equation in [latex]i[\/latex]. The circuit is referred to as an [latex]LR[\/latex] circuit.<\/p>\r\n<p id=\"fs-id1170572585927\">Next, suppose there is no inductor in the circuit, but there is a capacitor and a resistor, so [latex]L=0,R\\ne 0[\/latex], and [latex]C\\ne 0[\/latex]. Then [latex]y=\\dfrac{1}{\\mu \\left(x\\right)}\\left[\\displaystyle\\int \\mu \\left(x\\right)q\\left(x\\right)dx+C\\right][\/latex] can be rewritten as<\/p>\r\n\r\n<div id=\"fs-id1170571653410\" style=\"text-align: center;\" data-type=\"equation\">[latex]R{q}^{\\prime }+\\frac{1}{C}q=E[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571553379\">which is a first-order linear differential equation. This is referred to as an <span class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">RC<\/em> circuit<\/span>. In either case, we can set up and solve an initial-value problem.<\/p>\r\n\r\n<div id=\"fs-id1170571553391\" data-type=\"example\">\r\n<div id=\"fs-id1170571553393\" data-type=\"exercise\">\r\n<div id=\"fs-id1170571553396\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Finding Current in an <em data-effect=\"italics\">RL<\/em> Electric Circuit<\/h3>\r\n<div id=\"fs-id1170571553396\" data-type=\"problem\">\r\n<p id=\"fs-id1170571610452\">A circuit has in series an electromotive force given by [latex]E=50\\sin{20t}\\text{ V}[\/latex], a resistor of [latex]5 \\Omega[\/latex], and an inductor of [latex]0.4\\text{H}\\text{.}[\/latex] If the initial current is [latex]0[\/latex], find the current at time [latex]t&gt;0[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558799\"]\r\n<div id=\"fs-id1170571667084\" data-type=\"solution\">\r\n<p id=\"fs-id1170571667086\">We have a resistor and an inductor in the circuit, so we use [latex]L{i}^{\\prime }+Ri+\\frac{1}{C}q=E[\/latex].\u00a0The voltage drop across the resistor is given by [latex]{E}_{R}=Ri=5i[\/latex]. The voltage drop across the inductor is given by [latex]{E}_{L}=L{i}^{\\prime }=0.4{i}^{\\prime }[\/latex]. The electromotive force becomes the right-hand side of the equation. Therefore the equation becomes<\/p>\r\n\r\n<div id=\"fs-id1170572146954\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]0.4{i}^{\\prime }+5i=50\\sin{20t}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571546018\">Dividing both sides by [latex]0.4[\/latex] gives the equation<\/p>\r\n\r\n<div id=\"fs-id1170571546026\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{i}^{\\prime }+12.5i=125\\sin{20t}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571780476\">Since the initial current is 0, this result gives an initial condition of [latex]i\\left(0\\right)=0[\/latex]. We can solve this initial-value problem using the five-step strategy for solving first-order differential equations.<\/p>\r\n<p id=\"fs-id1170571728237\" style=\"text-align: left;\">Step 1. Rewrite the differential equation as [latex]{i}^{\\prime }+12.5i=125\\sin{20t}[\/latex]. This gives [latex]p\\left(t\\right)=12.5[\/latex] and [latex]q\\left(t\\right)=125\\sin20t[\/latex].<\/p>\r\n<p id=\"fs-id1170572451340\">Step 2. The integrating factor is [latex]\\mu \\left(t\\right)={e}^{\\displaystyle\\int 12.5dt}={e}^{12.5t}[\/latex].<\/p>\r\n<p id=\"fs-id1170572420925\">Step 3. Multiply the differential equation by [latex]\\mu \\left(t\\right)\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1170572420942\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {e}^{12.5t}{i}^{\\prime }+12.5{e}^{12.5t}i&amp; =\\hfill &amp; 125{e}^{12.5t}\\sin20t\\hfill \\\\ \\hfill \\frac{d}{dt}\\left[i{e}^{12.5t}\\right]&amp; =\\hfill &amp; 125{e}^{12.5t}\\sin20t.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572375908\">Step 4. Integrate both sides:<\/p>\r\n\r\n<div id=\"fs-id1170572375911\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int \\frac{d}{dt}\\left[i{e}^{12.5t}\\right]dt}&amp; =\\hfill &amp; {\\displaystyle\\int 125{e}^{12.5t}\\sin{20t}dt}\\hfill \\\\ \\hfill i{e}^{12.5t}&amp; =\\hfill &amp; \\left(\\frac{250\\sin{20t} - 400\\cos{20t}}{89}\\right){e}^{12.5t}+C\\hfill \\\\ \\hfill i\\left(t\\right)&amp; =\\hfill &amp; \\frac{250\\sin20t - 400\\cos20t}{89}+C{e}^{-12.5t}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571728146\">Step 5. Solve for [latex]C[\/latex] using the initial condition [latex]v\\left(0\\right)=2\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1170572278803\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill i\\left(t\\right)&amp; =\\hfill &amp; \\frac{250\\sin{20t} - 400\\cos{20t}}{89}+C{e}^{-12.5t}\\hfill \\\\ \\hfill i\\left(0\\right)&amp; =\\hfill &amp; \\frac{250\\sin{20}\\left(0\\right)-400\\cos{20}\\left(0\\right)}{89}+C{e}^{-12.5\\left(0\\right)}\\hfill \\\\ \\hfill 0&amp; =\\hfill &amp; -\\frac{400}{89}+C\\hfill \\\\ \\hfill C&amp; =\\hfill &amp; \\frac{400}{89}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572398033\">Therefore the solution to the initial-value problem is [latex]i\\left(t\\right)=\\frac{250\\sin{20t} - 400\\cos{20t}+400{e}^{-12.5t}}{89}=\\frac{250\\sin{20t} - 400\\cos{20t}}{89}+\\frac{400{e}^{-12.5t}}{89}[\/latex].<\/p>\r\n<p id=\"fs-id1170572305727\">The first term can be rewritten as a single cosine function. First, multiply and divide by [latex]\\sqrt{{250}^{2}+{400}^{2}}=50\\sqrt{89}\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1170571591219\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{250\\sin{20t} - 400\\cos{20t}}{89}&amp; =\\frac{50\\sqrt{89}}{89}\\left(\\frac{250\\sin{20t} - 400\\cos{20t}}{50\\sqrt{89}}\\right)\\hfill \\\\ &amp; =-\\frac{50\\sqrt{89}}{89}\\left(\\frac{8\\cos{20t}}{\\sqrt{89}}-\\frac{5\\sin{20t}}{\\sqrt{89}}\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571671371\">Next, define [latex]\\phi [\/latex] to be an acute angle such that [latex]\\cos\\phi =\\frac{8}{\\sqrt{89}}[\/latex]. Then [latex]\\sin\\phi =\\frac{5}{\\sqrt{89}}[\/latex] and<\/p>\r\n\r\n<div id=\"fs-id1170572638752\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill -\\frac{50\\sqrt{89}}{89}\\left(\\frac{8\\cos20t}{\\sqrt{89}}-\\frac{5\\sin20t}{\\sqrt{89}}\\right)&amp; =-\\frac{50\\sqrt{89}}{89}\\left(\\cos\\phi \\cos20t-\\sin\\phi \\sin20t\\right)\\hfill \\\\ &amp; =-\\frac{50\\sqrt{89}}{89}\\cos\\left(20t+\\phi \\right).\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572347331\">Therefore the solution can be written as<\/p>\r\n\r\n<div id=\"fs-id1170571814765\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]i\\left(t\\right)=-\\frac{50\\sqrt{89}}{89}\\cos\\left(20t+\\phi \\right)+\\frac{400{e}^{-12.5t}}{89}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572173047\">The second term is called the <em data-effect=\"italics\">attenuation<\/em> term, because it disappears rapidly as <em data-effect=\"italics\">t<\/em> grows larger. The phase shift is given by [latex]\\phi [\/latex], and the amplitude of the steady-state current is given by [latex]\\frac{50\\sqrt{89}}{89}[\/latex]. The graph of this solution appears in Figure 3:<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_08_05_003\"><span data-type=\"media\" data-alt=\"A graph of the given solution over [0, 6] on the x axis. It is an oscillating function, rapidly going from just below -5 to just above 5.\"><span data-type=\"media\" data-alt=\"A graph of the given solution over [0, 6] on the x axis. It is an oscillating function, rapidly going from just below -5 to just above 5.\">\r\n<\/span><\/span>[caption id=\"\" align=\"aligncenter\" width=\"464\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234229\/CNX_Calc_Figure_08_05_004.jpg\" alt=\"A graph of the given solution over [0, 6] on the x axis. It is an oscillating function, rapidly going from just below -5 to just above 5.\" width=\"464\" height=\"497\" data-media-type=\"image\/jpeg\" \/> Figure 3.[\/caption]<\/figure>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572375807\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1170572375811\" data-type=\"exercise\">\r\n<div id=\"fs-id1170572375813\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1170572375813\" data-type=\"problem\">\r\n<p id=\"fs-id1170572375815\">A circuit has in series an electromotive force given by [latex]E=20\\sin5t[\/latex] V, a capacitor with capacitance [latex]0.02\\text{F}[\/latex], and a resistor of [latex]8 \\Omega[\/latex]. If the initial charge is [latex]4\\text{C}[\/latex], find the charge at time [latex]t&gt;0[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"42558899\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"42558899\"]\r\n<div id=\"fs-id1170572517552\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1170572517560\">Use [latex]L{i}^{\\prime }+Ri=E[\/latex] for an [latex]RC[\/latex] circuit to set up an initial-value problem.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"43558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"43558899\"]\r\n<div id=\"fs-id1170572227498\" data-type=\"solution\">\r\n<p id=\"fs-id1170572227500\">Initial-value problem:<\/p>\r\n<p id=\"fs-id1170572227503\">[latex]8{q}^{\\prime }+\\frac{1}{0.02}q=20\\sin5t,q\\left(0\\right)=4[\/latex]<\/p>\r\n<p id=\"fs-id1170571581832\">[latex]q\\left(t\\right)=\\frac{10\\sin5t - 8\\cos5t+172{e}^{-6.25t}}{41}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1170572517580\" class=\"key-concepts\" data-depth=\"1\"><\/section>\r\n<div data-type=\"glossary\"><\/div>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve applied problems involving first-order linear differential equations<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170572558312\">We look at two different applications of first-order linear differential equations. The first involves air resistance as it relates to objects that are rising or falling; the second involves an electrical circuit. Other applications are numerous, but most are solved in a similar fashion.<\/p>\n<section id=\"fs-id1170572558318\" data-depth=\"2\">\n<h2 data-type=\"title\">Free fall with air resistance<\/h2>\n<p id=\"fs-id1170571829519\">We discussed air resistance at the beginning of this section. The next example shows how to apply this concept for a ball in vertical motion. Other factors can affect the force of air resistance, such as the size and shape of the object, but we ignore them here.<\/p>\n<div id=\"fs-id1170571829525\" data-type=\"example\">\n<div id=\"fs-id1170571829527\" data-type=\"exercise\">\n<div id=\"fs-id1170572510001\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0A Ball with Air Resistance<\/h3>\n<div id=\"fs-id1170572510001\" data-type=\"problem\">\n<p id=\"fs-id1170572510006\">A racquetball is hit straight upward with an initial velocity of [latex]2[\/latex] m\/s. The mass of a racquetball is approximately [latex]0.0427[\/latex] kg. Air resistance acts on the ball with a force numerically equal to [latex]0.5v[\/latex], where [latex]v[\/latex] represents the velocity of the ball at time [latex]t[\/latex].<\/p>\n<ol id=\"fs-id1170571598283\" type=\"a\">\n<li>Find the velocity of the ball as a function of time.<\/li>\n<li>How long does it take for the ball to reach its maximum height?<\/li>\n<li>If the ball is hit from an initial height of [latex]1[\/latex] meter, how high will it reach?<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558890\">Show Solution<\/span><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572247894\" data-type=\"solution\">\n<ol id=\"fs-id1170572247896\" type=\"a\">\n<li>The mass [latex]m=0.0427\\text{kg},k=0.5[\/latex], and [latex]g=9.8{\\text{m\/s}}^{2}[\/latex]. The initial velocity is [latex]{v}_{0}=2[\/latex] m\/s. Therefore the initial-value problem is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572379510\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]0.0427\\frac{dv}{dt}=-0.5v - 0.0427\\left(9.8\\right),{v}_{0}=2[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nDividing the differential equation by [latex]0.0427[\/latex] gives<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572547804\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{dv}{dt}=-11.7096v - 9.8,{v}_{0}=2[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p style=\"text-align: left;\"><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe differential equation is linear. Using the problem-solving strategy for linear differential equations:<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 1. Rewrite the differential equation as [latex]\\frac{dv}{dt}+11.7096v=-9.8[\/latex]. This gives [latex]p\\left(t\\right)=11.7096[\/latex] and [latex]q\\left(t\\right)=-9.8[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 2. The integrating factor is [latex]\\mu \\left(t\\right)={e}^{\\displaystyle\\int 11.7096dt}={e}^{11.7096t}[\/latex]. <span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 3. Multiply the differential equation by [latex]\\mu \\left(t\\right)\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572644185\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {e}^{11.7096t}\\frac{dv}{dt}+11.7096v{e}^{11.7096t}& =\\hfill & -9.8{e}^{11.7096t}\\hfill \\\\ \\hfill \\frac{d}{dt}\\left[v{e}^{11.7096t}\\right]& =\\hfill & -9.8{e}^{11.7096t}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 4. Integrate both sides:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572415253\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int \\frac{d}{dt}\\left[v{e}^{11.7096t}\\right]dt}& =\\hfill & {\\displaystyle\\int -9.8{e}^{11.7096t}dt}\\hfill \\\\ \\hfill v{e}^{11.7096t}& =\\hfill & \\frac{-9.8}{11.7096}{e}^{11.7096t}+C\\hfill \\\\ \\hfill v\\left(t\\right)& =\\hfill & -0.8369+C{e}^{-11.7096t}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 5. Solve for [latex]C[\/latex] using the initial condition [latex]{v}_{0}=v\\left(0\\right)=2\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572227250\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill v\\left(t\\right)& =\\hfill & -0.8369+C{e}^{-11.7096t}\\hfill \\\\ \\hfill v\\left(0\\right)& =\\hfill & -0.8369+C{e}^{-11.7096\\left(0\\right)}\\hfill \\\\ \\hfill 2& =\\hfill & -0.8369+C\\hfill \\\\ \\hfill C& =\\hfill & 2.8369.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore the solution to the initial-value problem is [latex]v\\left(t\\right)=2.8369{e}^{-11.7096t}-0.8369[\/latex].<\/li>\n<li>The ball reaches its maximum height when the velocity is equal to zero. The reason is that when the velocity is positive, it is rising, and when it is negative, it is falling. Therefore when it is zero, it is neither rising nor falling, and is at its maximum height:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571553927\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\hfill 2.8369{e}^{-11.7096t}-0.8369& =\\hfill & 0\\hfill \\\\ \\hfill 2.8369{e}^{-11.7096t}& =\\hfill & 0.8369\\hfill \\\\ \\hfill {e}^{-11.7096t}& =\\hfill & \\frac{0.8369}{2.8369}\\approx 0.295\\hfill \\\\ \\hfill \\text{ln}{e}^{-11.7096t}& =\\hfill & \\text{ln}0.295\\approx -1.221\\hfill \\\\ \\hfill -11.7096t& =\\hfill & -1.221\\hfill \\\\ \\hfill t& \\approx \\hfill & 0.104.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore it takes approximately [latex]0.104[\/latex] second to reach maximum height.<\/li>\n<li>To find the height of the ball as a function of time, use the fact that the derivative of position is velocity, i.e., if [latex]h\\left(t\\right)[\/latex] represents the height at time [latex]t[\/latex], then [latex]{h}^{\\prime }\\left(t\\right)=v\\left(t\\right)[\/latex]. Because we know [latex]v\\left(t\\right)[\/latex] and the initial height, we can form an initial-value problem:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571825011\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]{h}^{\\prime }\\left(t\\right)=2.8369{e}^{-11.7096t}-0.8369,h\\left(0\\right)=1[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nIntegrating both sides of the differential equation with respect to [latex]t[\/latex] gives<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571595078\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int {h}^{\\prime }\\left(t\\right)dt}& =\\hfill & {\\displaystyle\\int 2.8369{e}^{-11.7096t}-0.8369dt}\\hfill \\\\ \\hfill h\\left(t\\right)& =\\hfill & -\\frac{2.8369}{11.7096}{e}^{-11.7096t}-0.8369t+C\\hfill \\\\ \\hfill h\\left(t\\right)& =\\hfill & -0.2423{e}^{-11.7096t}-0.8369t+C.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSolve for [latex]C[\/latex] by using the initial condition:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572140765\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill h\\left(t\\right)& =\\hfill & -0.2423{e}^{-11.7096t}-0.8369t+C\\hfill \\\\ \\hfill h\\left(0\\right)& =\\hfill & -0.2423{e}^{-11.7096\\left(0\\right)}-0.8369\\left(0\\right)+C\\hfill \\\\ \\hfill 1& =\\hfill & -0.2423+C\\hfill \\\\ \\hfill C& =\\hfill & 1.2423.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572208058\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]h\\left(t\\right)=-0.2423{e}^{-11.7096t}-0.8369t+1.2423[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nAfter [latex]0.104[\/latex] second, the height is given by<span data-type=\"newline\"><br \/>\n<\/span><br \/>\n[latex]h\\left(0.2\\right)=-0.2423{e}^{-11.7096t}-0.8369t+1.2423\\approx 1.0836[\/latex] meter.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example:\u00a0A Ball with Air Resistance.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/N6wo00P3ab0?controls=0&amp;start=28&amp;end=834&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">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+II\/Transcripts\/4.5.3_28to834_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.5.3&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1170572481763\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1170572481767\" data-type=\"exercise\">\n<div id=\"fs-id1170572481770\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1170572481770\" data-type=\"problem\">\n<p id=\"fs-id1170572481772\">The weight of a penny is [latex]2.5[\/latex] grams (United States Mint, &#8220;Coin Specifications,&#8221; accessed April 9, 2015, http:\/\/www.usmint.gov\/about_the_mint\/?action=coin_specifications), and the upper observation deck of the Empire State Building is [latex]369[\/latex] meters above the street. Since the penny is a small and relatively smooth object, air resistance acting on the penny is actually quite small. We assume the air resistance is numerically equal to [latex]0.0025v[\/latex]. Furthermore, the penny is dropped with no initial velocity imparted to it.<\/p>\n<ol id=\"fs-id1170571532540\" type=\"a\">\n<li>Set up an initial-value problem that represents the falling penny.<\/li>\n<li>Solve the problem for [latex]v\\left(t\\right)[\/latex].<\/li>\n<li>What is the terminal velocity of the penny (i.e., calculate the limit of the velocity as [latex]t[\/latex] approaches infinity)?<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558879\">Hint<\/span><\/p>\n<div id=\"q44558879\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572523011\" data-type=\"commentary\" data-element-type=\"hint\">\n<p>Set up the differential equation the same way as the example: Writing First-Order Linear Equations in Standard Form Remember to convert from grams to kilograms.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558889\">Show Solution<\/span><\/p>\n<div id=\"q44558889\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571555522\" data-type=\"solution\">\n<ol id=\"fs-id1170571711999\" type=\"a\">\n<li>[latex]\\begin{array}{ccc}\\hfill \\frac{dv}{dt}& =\\hfill & \\text{-}v - 9.8\\hfill \\\\ \\hfill v\\left(0\\right)& =\\hfill & 0\\hfill \\end{array}[\/latex]<\/li>\n<li>[latex]v\\left(t\\right)=9.8\\left({e}^{\\text{-}t}-1\\right)[\/latex]<\/li>\n<li>[latex]\\underset{t\\to \\infty }{\\text{lim}}v\\left(t\\right)=\\underset{t\\to \\infty }{\\text{lim}}\\left(9.8\\left({e}^{\\text{-}t}-1\\right)\\right)=-9.8\\text{m\/s}\\approx -21.922\\text{mph}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1170572404173\" data-depth=\"2\">\n<h2 data-type=\"title\">Electrical Circuits<\/h2>\n<p id=\"fs-id1170572404179\">A source of electromotive force (e.g., a battery or generator) produces a flow of current in a closed circuit, and this current produces a voltage drop across each resistor, inductor, and capacitor in the circuit. Kirchhoff\u2019s Loop Rule states that the sum of the voltage drops across resistors, inductors, and capacitors is equal to the total electromotive force in a closed circuit. We have the following three results:<\/p>\n<ol id=\"fs-id1170571674159\" type=\"1\">\n<li>The voltage drop across a resistor is given by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571674169\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{E}_{R}=Ri[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwhere [latex]R[\/latex] is a constant of proportionality called the <em data-effect=\"italics\">resistance,<\/em> and [latex]i[\/latex] is the current.<\/li>\n<li>The voltage drop across an inductor is given by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572375868\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{E}_{L}=L{i}^{\\prime }[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwhere [latex]L[\/latex] is a constant of proportionality called the <em data-effect=\"italics\">inductance<\/em>, and [latex]i[\/latex] again denotes the current.<\/li>\n<li>The voltage drop across a capacitor is given by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571683191\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{E}_{C}=\\frac{1}{C}q[\/latex],<\/div>\n<p>&nbsp;<\/li>\n<\/ol>\n<p id=\"fs-id1170572472586\">where [latex]C[\/latex] is a constant of proportionality called the <em data-effect=\"italics\">capacitance<\/em>, and [latex]q[\/latex] is the instantaneous charge on the capacitor. The relationship between [latex]i[\/latex] and [latex]q[\/latex] is [latex]i={q}^{\\prime }[\/latex].<\/p>\n<p id=\"fs-id1170571714388\">We use units of volts [latex]\\left(\\text{V}\\right)[\/latex] to measure voltage [latex]E[\/latex], amperes [latex]\\left(\\text{A}\\right)[\/latex] to measure current [latex]i[\/latex], coulombs [latex]\\left(\\text{C}\\right)[\/latex] to measure charge [latex]q[\/latex], ohms [latex]\\left(\\Omega \\right)[\/latex] to measure resistance [latex]R[\/latex], henrys [latex]\\left(\\text{H}\\right)[\/latex] to measure inductance [latex]L[\/latex], and farads [latex]\\left(\\text{F}\\right)[\/latex] to measure capacitance [latex]C[\/latex]. Consider the circuit in Figure 2.<\/p>\n<figure id=\"CNX_Calc_Figure_08_05_002\"><figcaption><\/figcaption><div style=\"width: 402px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234225\/CNX_Calc_Figure_08_05_003.jpg\" alt=\"A diagram of an electric circuit in a rectangle. The top has a capacitor C, the left has a voltage generator Vs, the bottom was a resistor R, and the right has an inductor L.\" width=\"392\" height=\"242\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. A typical electric circuit, containing a voltage generator [latex]\\left({V}_{S}\\right)[\/latex], capacitor [latex]\\left(C\\right)[\/latex], inductor [latex]\\left(L\\right)[\/latex], and resistor [latex]\\left(R\\right)[\/latex].<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1170571783881\">Applying Kirchhoff\u2019s Loop Rule to this circuit, we let [latex]E[\/latex] denote the electromotive force supplied by the voltage generator. Then<\/p>\n<div id=\"fs-id1170571733859\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{E}_{L}+{E}_{R}+{E}_{C}=E[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572378413\">Substituting the expressions for [latex]{E}_{L},{E}_{R}[\/latex], and [latex]{E}_{C}[\/latex] into this equation, we obtain<\/p>\n<div id=\"fs-id1170572177052\" style=\"text-align: center;\" data-type=\"equation\">[latex]L{i}^{\\prime }+Ri+\\frac{1}{C}q=E[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572241888\">If there is no capacitor in the circuit, then the equation becomes<\/p>\n<div id=\"fs-id1170572241891\" style=\"text-align: center;\" data-type=\"equation\">[latex]L{i}^{\\prime }+Ri=E[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572585910\">This is a first-order differential equation in [latex]i[\/latex]. The circuit is referred to as an [latex]LR[\/latex] circuit.<\/p>\n<p id=\"fs-id1170572585927\">Next, suppose there is no inductor in the circuit, but there is a capacitor and a resistor, so [latex]L=0,R\\ne 0[\/latex], and [latex]C\\ne 0[\/latex]. Then [latex]y=\\dfrac{1}{\\mu \\left(x\\right)}\\left[\\displaystyle\\int \\mu \\left(x\\right)q\\left(x\\right)dx+C\\right][\/latex] can be rewritten as<\/p>\n<div id=\"fs-id1170571653410\" style=\"text-align: center;\" data-type=\"equation\">[latex]R{q}^{\\prime }+\\frac{1}{C}q=E[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571553379\">which is a first-order linear differential equation. This is referred to as an <span class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">RC<\/em> circuit<\/span>. In either case, we can set up and solve an initial-value problem.<\/p>\n<div id=\"fs-id1170571553391\" data-type=\"example\">\n<div id=\"fs-id1170571553393\" data-type=\"exercise\">\n<div id=\"fs-id1170571553396\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Finding Current in an <em data-effect=\"italics\">RL<\/em> Electric Circuit<\/h3>\n<div id=\"fs-id1170571553396\" data-type=\"problem\">\n<p id=\"fs-id1170571610452\">A circuit has in series an electromotive force given by [latex]E=50\\sin{20t}\\text{ V}[\/latex], a resistor of [latex]5 \\Omega[\/latex], and an inductor of [latex]0.4\\text{H}\\text{.}[\/latex] If the initial current is [latex]0[\/latex], find the current at time [latex]t>0[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558799\">Show Solution<\/span><\/p>\n<div id=\"q44558799\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571667084\" data-type=\"solution\">\n<p id=\"fs-id1170571667086\">We have a resistor and an inductor in the circuit, so we use [latex]L{i}^{\\prime }+Ri+\\frac{1}{C}q=E[\/latex].\u00a0The voltage drop across the resistor is given by [latex]{E}_{R}=Ri=5i[\/latex]. The voltage drop across the inductor is given by [latex]{E}_{L}=L{i}^{\\prime }=0.4{i}^{\\prime }[\/latex]. The electromotive force becomes the right-hand side of the equation. Therefore the equation becomes<\/p>\n<div id=\"fs-id1170572146954\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]0.4{i}^{\\prime }+5i=50\\sin{20t}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571546018\">Dividing both sides by [latex]0.4[\/latex] gives the equation<\/p>\n<div id=\"fs-id1170571546026\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{i}^{\\prime }+12.5i=125\\sin{20t}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571780476\">Since the initial current is 0, this result gives an initial condition of [latex]i\\left(0\\right)=0[\/latex]. We can solve this initial-value problem using the five-step strategy for solving first-order differential equations.<\/p>\n<p id=\"fs-id1170571728237\" style=\"text-align: left;\">Step 1. Rewrite the differential equation as [latex]{i}^{\\prime }+12.5i=125\\sin{20t}[\/latex]. This gives [latex]p\\left(t\\right)=12.5[\/latex] and [latex]q\\left(t\\right)=125\\sin20t[\/latex].<\/p>\n<p id=\"fs-id1170572451340\">Step 2. The integrating factor is [latex]\\mu \\left(t\\right)={e}^{\\displaystyle\\int 12.5dt}={e}^{12.5t}[\/latex].<\/p>\n<p id=\"fs-id1170572420925\">Step 3. Multiply the differential equation by [latex]\\mu \\left(t\\right)\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1170572420942\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {e}^{12.5t}{i}^{\\prime }+12.5{e}^{12.5t}i& =\\hfill & 125{e}^{12.5t}\\sin20t\\hfill \\\\ \\hfill \\frac{d}{dt}\\left[i{e}^{12.5t}\\right]& =\\hfill & 125{e}^{12.5t}\\sin20t.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572375908\">Step 4. Integrate both sides:<\/p>\n<div id=\"fs-id1170572375911\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int \\frac{d}{dt}\\left[i{e}^{12.5t}\\right]dt}& =\\hfill & {\\displaystyle\\int 125{e}^{12.5t}\\sin{20t}dt}\\hfill \\\\ \\hfill i{e}^{12.5t}& =\\hfill & \\left(\\frac{250\\sin{20t} - 400\\cos{20t}}{89}\\right){e}^{12.5t}+C\\hfill \\\\ \\hfill i\\left(t\\right)& =\\hfill & \\frac{250\\sin20t - 400\\cos20t}{89}+C{e}^{-12.5t}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571728146\">Step 5. Solve for [latex]C[\/latex] using the initial condition [latex]v\\left(0\\right)=2\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1170572278803\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill i\\left(t\\right)& =\\hfill & \\frac{250\\sin{20t} - 400\\cos{20t}}{89}+C{e}^{-12.5t}\\hfill \\\\ \\hfill i\\left(0\\right)& =\\hfill & \\frac{250\\sin{20}\\left(0\\right)-400\\cos{20}\\left(0\\right)}{89}+C{e}^{-12.5\\left(0\\right)}\\hfill \\\\ \\hfill 0& =\\hfill & -\\frac{400}{89}+C\\hfill \\\\ \\hfill C& =\\hfill & \\frac{400}{89}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572398033\">Therefore the solution to the initial-value problem is [latex]i\\left(t\\right)=\\frac{250\\sin{20t} - 400\\cos{20t}+400{e}^{-12.5t}}{89}=\\frac{250\\sin{20t} - 400\\cos{20t}}{89}+\\frac{400{e}^{-12.5t}}{89}[\/latex].<\/p>\n<p id=\"fs-id1170572305727\">The first term can be rewritten as a single cosine function. First, multiply and divide by [latex]\\sqrt{{250}^{2}+{400}^{2}}=50\\sqrt{89}\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1170571591219\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{250\\sin{20t} - 400\\cos{20t}}{89}& =\\frac{50\\sqrt{89}}{89}\\left(\\frac{250\\sin{20t} - 400\\cos{20t}}{50\\sqrt{89}}\\right)\\hfill \\\\ & =-\\frac{50\\sqrt{89}}{89}\\left(\\frac{8\\cos{20t}}{\\sqrt{89}}-\\frac{5\\sin{20t}}{\\sqrt{89}}\\right).\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571671371\">Next, define [latex]\\phi[\/latex] to be an acute angle such that [latex]\\cos\\phi =\\frac{8}{\\sqrt{89}}[\/latex]. Then [latex]\\sin\\phi =\\frac{5}{\\sqrt{89}}[\/latex] and<\/p>\n<div id=\"fs-id1170572638752\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill -\\frac{50\\sqrt{89}}{89}\\left(\\frac{8\\cos20t}{\\sqrt{89}}-\\frac{5\\sin20t}{\\sqrt{89}}\\right)& =-\\frac{50\\sqrt{89}}{89}\\left(\\cos\\phi \\cos20t-\\sin\\phi \\sin20t\\right)\\hfill \\\\ & =-\\frac{50\\sqrt{89}}{89}\\cos\\left(20t+\\phi \\right).\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572347331\">Therefore the solution can be written as<\/p>\n<div id=\"fs-id1170571814765\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]i\\left(t\\right)=-\\frac{50\\sqrt{89}}{89}\\cos\\left(20t+\\phi \\right)+\\frac{400{e}^{-12.5t}}{89}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572173047\">The second term is called the <em data-effect=\"italics\">attenuation<\/em> term, because it disappears rapidly as <em data-effect=\"italics\">t<\/em> grows larger. The phase shift is given by [latex]\\phi[\/latex], and the amplitude of the steady-state current is given by [latex]\\frac{50\\sqrt{89}}{89}[\/latex]. The graph of this solution appears in Figure 3:<\/p>\n<figure id=\"CNX_Calc_Figure_08_05_003\"><span data-type=\"media\" data-alt=\"A graph of the given solution over &#091;0, 6&#093; on the x axis. It is an oscillating function, rapidly going from just below -5 to just above 5.\"><span data-type=\"media\" data-alt=\"A graph of the given solution over &#091;0, 6&#093; on the x axis. It is an oscillating function, rapidly going from just below -5 to just above 5.\"><br \/>\n<\/span><\/span><\/p>\n<div style=\"width: 474px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234229\/CNX_Calc_Figure_08_05_004.jpg\" alt=\"A graph of the given solution over &#091;0, 6&#093; on the x axis. It is an oscillating function, rapidly going from just below -5 to just above 5.\" width=\"464\" height=\"497\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3.<\/p>\n<\/div>\n<\/figure>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572375807\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1170572375811\" data-type=\"exercise\">\n<div id=\"fs-id1170572375813\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1170572375813\" data-type=\"problem\">\n<p id=\"fs-id1170572375815\">A circuit has in series an electromotive force given by [latex]E=20\\sin5t[\/latex] V, a capacitor with capacitance [latex]0.02\\text{F}[\/latex], and a resistor of [latex]8 \\Omega[\/latex]. If the initial charge is [latex]4\\text{C}[\/latex], find the charge at time [latex]t>0[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q42558899\">Hint<\/span><\/p>\n<div id=\"q42558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572517552\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1170572517560\">Use [latex]L{i}^{\\prime }+Ri=E[\/latex] for an [latex]RC[\/latex] circuit to set up an initial-value problem.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q43558899\">Show Solution<\/span><\/p>\n<div id=\"q43558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572227498\" data-type=\"solution\">\n<p id=\"fs-id1170572227500\">Initial-value problem:<\/p>\n<p id=\"fs-id1170572227503\">[latex]8{q}^{\\prime }+\\frac{1}{0.02}q=20\\sin5t,q\\left(0\\right)=4[\/latex]<\/p>\n<p id=\"fs-id1170571581832\">[latex]q\\left(t\\right)=\\frac{10\\sin5t - 8\\cos5t+172{e}^{-6.25t}}{41}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1170572517580\" class=\"key-concepts\" data-depth=\"1\"><\/section>\n<div data-type=\"glossary\"><\/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-1637\">\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>4.5.3. <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 2. <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\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/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-2\/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":19,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.5.3\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1637","chapter","type-chapter","status-publish","hentry"],"part":159,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1637","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1637\/revisions"}],"predecessor-version":[{"id":2094,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1637\/revisions\/2094"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/159"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1637\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1637"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1637"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1637"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1637"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}