{"id":445,"date":"2021-02-04T02:08:22","date_gmt":"2021-02-04T02:08:22","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=445"},"modified":"2022-03-16T05:59:09","modified_gmt":"2022-03-16T05:59:09","slug":"initial-value-problems","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/initial-value-problems\/","title":{"raw":"Initial-Value Problems","rendered":"Initial-Value Problems"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use antidifferentiation to solve simple initial-value problems<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165042323634\">We look at techniques for integrating a large variety of functions involving products, quotients, and compositions later in the text. Here we turn to one common use for antiderivatives that arises often in many applications: solving differential equations.<\/p>\r\n<p id=\"fs-id1165042323639\">A <em>differential equation<\/em> is an equation that relates an unknown function and one or more of its derivatives. The equation<\/p>\r\n\r\n<div id=\"fs-id1165042349924\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=f(x)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042349952\">is a simple example of a differential equation. Solving this equation means finding a function [latex]y[\/latex] with a derivative [latex]f[\/latex]. Therefore, the solutions of [latex]\\frac{dy}{dx}[\/latex] are the antiderivatives of [latex]f[\/latex]. If [latex]F[\/latex] is one antiderivative of [latex]f[\/latex], every function of the form [latex]y=F(x)+C[\/latex] is a solution of that differential equation. For example, the solutions of<\/p>\r\n\r\n<div id=\"fs-id1165042323537\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=6x^2[\/latex]<\/div>\r\nare given by\r\n<div id=\"fs-id1165043431000\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\displaystyle\\int 6x^2 dx=2x^3+C[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042407328\">Sometimes we are interested in determining whether a particular solution curve passes through a certain point [latex](x_0,y_0)[\/latex]\u2014that is, [latex]y(x_0)=y_0[\/latex]. The problem of finding a function [latex]y[\/latex] that satisfies a differential equation<\/p>\r\n\r\n<div id=\"fs-id1165042375659\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=f(x)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042368468\">with the additional condition<\/p>\r\n\r\n<div id=\"fs-id1165042368471\" class=\"equation\" style=\"text-align: center;\">[latex]y(x_0)=y_0[\/latex]<\/div>\r\nis an example of an <strong>initial-value problem<\/strong>. The condition [latex]y(x_0)=y_0[\/latex] is known as an <em>initial condition<\/em>. For example, looking for a function [latex]y[\/latex] that satisfies the differential equation\r\n<div id=\"fs-id1165042545822\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=6x^2[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043393696\">and the initial condition<\/p>\r\n\r\n<div id=\"fs-id1165043393699\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y(1)=5[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043393719\">is an example of an initial-value problem. Since the solutions of the differential equation are [latex]y=2x^3+C[\/latex], to find a function [latex]y[\/latex] that also satisfies the initial condition, we need to find [latex]C[\/latex] such that [latex]y(1)=2(1)^3+C=5[\/latex]. From this equation, we see that [latex]C=3[\/latex], and we conclude that [latex]y=2x^3+3[\/latex] is the solution of this initial-value problem as shown in the following graph.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"641\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211402\/CNX_Calc_Figure_04_10_002.jpg\" alt=\"The graphs for y = 2x3 + 6, y = 2x3 + 3, y = 2x3, and y = 2x3 \u2212 3 are shown.\" width=\"641\" height=\"497\" \/> Figure 2. Some of the solution curves of the differential equation [latex]\\frac{dy}{dx}=6x^2[\/latex] are displayed. The function [latex]y=2x^3+3[\/latex] satisfies the differential equation and the initial condition [latex]y(1)=5[\/latex].[\/caption]\r\n<div id=\"fs-id1165043327648\" class=\"textbook exercises\">\r\n<h3>Example: Solving an Initial-Value Problem<\/h3>\r\n<p id=\"fs-id1165043327658\">Solve the initial-value problem<\/p>\r\n\r\n<div id=\"fs-id1165043327661\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}= \\sin x, \\,\\,\\, y(0)=5[\/latex]<\/div>\r\n<div id=\"fs-id1165043327650\" class=\"exercise\">[reveal-answer q=\"fs-id1165043430929\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043430929\"]\r\n<p id=\"fs-id1165043430929\">First we need to solve the differential equation. If [latex]\\frac{dy}{dx}= \\sin x,[\/latex] then<\/p>\r\n\r\n<div id=\"fs-id1165043286683\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\displaystyle\\int \\sin (x) dx=\u2212 \\cos x+C[\/latex]<\/div>\r\n<p id=\"fs-id1165042318817\">Next we need to look for a solution [latex]y[\/latex] that satisfies the initial condition. The initial condition [latex]y(0)=5[\/latex] means we need a constant [latex]C[\/latex] such that [latex]\u2212 \\cos x+C=5[\/latex]. Therefore,<\/p>\r\n\r\n<div id=\"fs-id1165043424804\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]C=5+ \\cos (0)=6[\/latex]<\/div>\r\n<p id=\"fs-id1165043424835\">The solution of the initial-value problem is [latex]y=\u2212 \\cos x+6[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042327334\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the initial value problem [latex]\\frac{dy}{dx}=3x^{-2}, \\,\\,\\, y(1)=2[\/latex].\r\n\r\n[reveal-answer q=\"7073552\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"7073552\"]\r\n\r\nFind all antiderivatives of [latex]f(x)=3x^{-2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165043327482\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1165043327482\"]\r\n\r\n[latex]y=-\\frac{3}{x}+5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Solving an Initial-Value Problem and the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/j81IZAEfwhI?controls=0&amp;start=696&amp;end=840&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\/4.10Antiderivatives696to840_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.10 Antiderivatives\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5304[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165043174043\">Initial-value problems arise in many applications. Next we consider a problem in which a driver applies the brakes in a car. We are interested in how long it takes for the car to stop. Recall that the velocity function [latex]v(t)[\/latex] is the derivative of a position function [latex]s(t)[\/latex], and the acceleration [latex]a(t)[\/latex] is the derivative of the velocity function. In earlier examples in the text, we could calculate the velocity from the position and then compute the acceleration from the velocity. In the next example, we work the other way around. Given an acceleration function, we calculate the velocity function. We then use the velocity function to determine the position function.<\/p>\r\n\r\n<div id=\"fs-id1165042640762\" class=\"textbook exercises\">\r\n<h3>Example: Decelerating Car<\/h3>\r\n<p id=\"fs-id1165042640772\">A car is traveling at the rate of [latex]88[\/latex] ft\/sec ([latex]60[\/latex] mph) when the brakes are applied. The car begins decelerating at a constant rate of [latex]15[\/latex] ft\/sec<sup>2<\/sup>.<\/p>\r\n\r\n<ol id=\"fs-id1165043430821\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>How many seconds elapse before the car stops?<\/li>\r\n \t<li>How far does the car travel during that time?<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1165042640764\" class=\"exercise\">[reveal-answer q=\"fs-id1165043430836\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165043430836\"]\r\n<ol id=\"fs-id1165043430836\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>First we introduce variables for this problem. Let [latex]t[\/latex] be the time (in seconds) after the brakes are first applied. Let [latex]a(t)[\/latex] be the acceleration of the car (in feet per seconds squared) at time [latex]t[\/latex]. Let [latex]v(t)[\/latex] be the velocity of the car (in feet per second) at time [latex]t[\/latex]. Let [latex]s(t)[\/latex] be the car\u2019s position (in feet) beyond the point where the brakes are applied at time [latex]t[\/latex].\r\nThe car is traveling at a rate of 88 ft\/sec. Therefore, the initial velocity is [latex]v(0)=88[\/latex] ft\/sec. Since the car is decelerating, the acceleration is\r\n<div id=\"fs-id1165043323894\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]a(t)=-15[\/latex] ft\/sec<sup>2<\/sup><\/div>\r\nThe acceleration is the derivative of the velocity,\r\n<div id=\"fs-id1165043219161\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v^{\\prime}(t)=-15[\/latex].<\/div>\r\nTherefore, we have an initial-value problem to solve:\r\n<div id=\"fs-id1165043219190\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v^{\\prime}(t)=-15, \\, v(0)=88[\/latex].<\/div>\r\nIntegrating, we find that\r\n<div id=\"fs-id1165043395175\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(t)=-15t+C[\/latex].<\/div>\r\nSince [latex]v(0)=88, \\, C=88[\/latex]. Thus, the velocity function is\r\n<div id=\"fs-id1165042373710\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(t)=-15t+88[\/latex].<\/div>\r\nTo find how long it takes for the car to stop, we need to find the time [latex]t[\/latex] such that the velocity is zero. Solving [latex]-15t+88=0[\/latex], we obtain [latex]t=\\frac{88}{15}[\/latex] sec.<\/li>\r\n \t<li>To find how far the car travels during this time, we need to find the position of the car after [latex]\\frac{88}{15}[\/latex] sec. We know the velocity [latex]v(t)[\/latex] is the derivative of the position [latex]s(t)[\/latex]. Consider the initial position to be [latex]s(0)=0[\/latex]. Therefore, we need to solve the initial-value problem\r\n<div id=\"fs-id1165043380494\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]s^{\\prime}(t)=-15t+88, \\, s(0)=0[\/latex].<\/div>\r\nIntegrating, we have\r\n<div id=\"fs-id1165043317270\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]s(t)=-\\frac{15}{2}t^2+88t+C[\/latex].<\/div>\r\nSince [latex]s(0)=0[\/latex], the constant is [latex]C=0[\/latex]. Therefore, the position function is\r\n<div id=\"fs-id1165043250997\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]s(t)=-\\frac{15}{2}t^2+88t[\/latex].<\/div>\r\nAfter [latex]t=\\frac{88}{15}[\/latex] sec, the position is [latex]s(\\frac{88}{15})\\approx 258.133[\/latex] ft.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Decelerating Car.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/j81IZAEfwhI?controls=0&amp;start=849&amp;end=1094&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\/4.10Antiderivatives849to1094_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.10 Antiderivatives\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1165042708223\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSuppose the car is traveling at the rate of 44 ft\/sec. How long does it take for the car to stop? How far will the car travel?\r\n\r\n[reveal-answer q=\"7660112\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"7660112\"]\r\n\r\n[latex]v(t)=-15t+44[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"923849\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"923849\"]\r\n\r\n2.93 sec, 64.5 ft\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use antidifferentiation to solve simple initial-value problems<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165042323634\">We look at techniques for integrating a large variety of functions involving products, quotients, and compositions later in the text. Here we turn to one common use for antiderivatives that arises often in many applications: solving differential equations.<\/p>\n<p id=\"fs-id1165042323639\">A <em>differential equation<\/em> is an equation that relates an unknown function and one or more of its derivatives. The equation<\/p>\n<div id=\"fs-id1165042349924\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=f(x)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042349952\">is a simple example of a differential equation. Solving this equation means finding a function [latex]y[\/latex] with a derivative [latex]f[\/latex]. Therefore, the solutions of [latex]\\frac{dy}{dx}[\/latex] are the antiderivatives of [latex]f[\/latex]. If [latex]F[\/latex] is one antiderivative of [latex]f[\/latex], every function of the form [latex]y=F(x)+C[\/latex] is a solution of that differential equation. For example, the solutions of<\/p>\n<div id=\"fs-id1165042323537\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=6x^2[\/latex]<\/div>\n<p>are given by<\/p>\n<div id=\"fs-id1165043431000\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\displaystyle\\int 6x^2 dx=2x^3+C[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042407328\">Sometimes we are interested in determining whether a particular solution curve passes through a certain point [latex](x_0,y_0)[\/latex]\u2014that is, [latex]y(x_0)=y_0[\/latex]. The problem of finding a function [latex]y[\/latex] that satisfies a differential equation<\/p>\n<div id=\"fs-id1165042375659\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=f(x)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042368468\">with the additional condition<\/p>\n<div id=\"fs-id1165042368471\" class=\"equation\" style=\"text-align: center;\">[latex]y(x_0)=y_0[\/latex]<\/div>\n<p>is an example of an <strong>initial-value problem<\/strong>. The condition [latex]y(x_0)=y_0[\/latex] is known as an <em>initial condition<\/em>. For example, looking for a function [latex]y[\/latex] that satisfies the differential equation<\/p>\n<div id=\"fs-id1165042545822\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=6x^2[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043393696\">and the initial condition<\/p>\n<div id=\"fs-id1165043393699\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y(1)=5[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043393719\">is an example of an initial-value problem. Since the solutions of the differential equation are [latex]y=2x^3+C[\/latex], to find a function [latex]y[\/latex] that also satisfies the initial condition, we need to find [latex]C[\/latex] such that [latex]y(1)=2(1)^3+C=5[\/latex]. From this equation, we see that [latex]C=3[\/latex], and we conclude that [latex]y=2x^3+3[\/latex] is the solution of this initial-value problem as shown in the following graph.<\/p>\n<div style=\"width: 651px\" 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\/11211402\/CNX_Calc_Figure_04_10_002.jpg\" alt=\"The graphs for y = 2x3 + 6, y = 2x3 + 3, y = 2x3, and y = 2x3 \u2212 3 are shown.\" width=\"641\" height=\"497\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. Some of the solution curves of the differential equation [latex]\\frac{dy}{dx}=6x^2[\/latex] are displayed. The function [latex]y=2x^3+3[\/latex] satisfies the differential equation and the initial condition [latex]y(1)=5[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165043327648\" class=\"textbook exercises\">\n<h3>Example: Solving an Initial-Value Problem<\/h3>\n<p id=\"fs-id1165043327658\">Solve the initial-value problem<\/p>\n<div id=\"fs-id1165043327661\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}= \\sin x, \\,\\,\\, y(0)=5[\/latex]<\/div>\n<div id=\"fs-id1165043327650\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043430929\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043430929\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043430929\">First we need to solve the differential equation. If [latex]\\frac{dy}{dx}= \\sin x,[\/latex] then<\/p>\n<div id=\"fs-id1165043286683\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\displaystyle\\int \\sin (x) dx=\u2212 \\cos x+C[\/latex]<\/div>\n<p id=\"fs-id1165042318817\">Next we need to look for a solution [latex]y[\/latex] that satisfies the initial condition. The initial condition [latex]y(0)=5[\/latex] means we need a constant [latex]C[\/latex] such that [latex]\u2212 \\cos x+C=5[\/latex]. Therefore,<\/p>\n<div id=\"fs-id1165043424804\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]C=5+ \\cos (0)=6[\/latex]<\/div>\n<p id=\"fs-id1165043424835\">The solution of the initial-value problem is [latex]y=\u2212 \\cos x+6[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042327334\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the initial value problem [latex]\\frac{dy}{dx}=3x^{-2}, \\,\\,\\, y(1)=2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q7073552\">Hint<\/span><\/p>\n<div id=\"q7073552\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find all antiderivatives of [latex]f(x)=3x^{-2}[\/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-id1165043327482\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043327482\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=-\\frac{3}{x}+5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Solving an Initial-Value Problem and the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/j81IZAEfwhI?controls=0&amp;start=696&amp;end=840&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\/4.10Antiderivatives696to840_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.10 Antiderivatives&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5304\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5304&theme=oea&iframe_resize_id=ohm5304&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1165043174043\">Initial-value problems arise in many applications. Next we consider a problem in which a driver applies the brakes in a car. We are interested in how long it takes for the car to stop. Recall that the velocity function [latex]v(t)[\/latex] is the derivative of a position function [latex]s(t)[\/latex], and the acceleration [latex]a(t)[\/latex] is the derivative of the velocity function. In earlier examples in the text, we could calculate the velocity from the position and then compute the acceleration from the velocity. In the next example, we work the other way around. Given an acceleration function, we calculate the velocity function. We then use the velocity function to determine the position function.<\/p>\n<div id=\"fs-id1165042640762\" class=\"textbook exercises\">\n<h3>Example: Decelerating Car<\/h3>\n<p id=\"fs-id1165042640772\">A car is traveling at the rate of [latex]88[\/latex] ft\/sec ([latex]60[\/latex] mph) when the brakes are applied. The car begins decelerating at a constant rate of [latex]15[\/latex] ft\/sec<sup>2<\/sup>.<\/p>\n<ol id=\"fs-id1165043430821\" style=\"list-style-type: lower-alpha;\">\n<li>How many seconds elapse before the car stops?<\/li>\n<li>How far does the car travel during that time?<\/li>\n<\/ol>\n<div id=\"fs-id1165042640764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165043430836\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165043430836\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165043430836\" style=\"list-style-type: lower-alpha;\">\n<li>First we introduce variables for this problem. Let [latex]t[\/latex] be the time (in seconds) after the brakes are first applied. Let [latex]a(t)[\/latex] be the acceleration of the car (in feet per seconds squared) at time [latex]t[\/latex]. Let [latex]v(t)[\/latex] be the velocity of the car (in feet per second) at time [latex]t[\/latex]. Let [latex]s(t)[\/latex] be the car\u2019s position (in feet) beyond the point where the brakes are applied at time [latex]t[\/latex].<br \/>\nThe car is traveling at a rate of 88 ft\/sec. Therefore, the initial velocity is [latex]v(0)=88[\/latex] ft\/sec. Since the car is decelerating, the acceleration is<\/p>\n<div id=\"fs-id1165043323894\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]a(t)=-15[\/latex] ft\/sec<sup>2<\/sup><\/div>\n<p>The acceleration is the derivative of the velocity,<\/p>\n<div id=\"fs-id1165043219161\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v^{\\prime}(t)=-15[\/latex].<\/div>\n<p>Therefore, we have an initial-value problem to solve:<\/p>\n<div id=\"fs-id1165043219190\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v^{\\prime}(t)=-15, \\, v(0)=88[\/latex].<\/div>\n<p>Integrating, we find that<\/p>\n<div id=\"fs-id1165043395175\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(t)=-15t+C[\/latex].<\/div>\n<p>Since [latex]v(0)=88, \\, C=88[\/latex]. Thus, the velocity function is<\/p>\n<div id=\"fs-id1165042373710\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(t)=-15t+88[\/latex].<\/div>\n<p>To find how long it takes for the car to stop, we need to find the time [latex]t[\/latex] such that the velocity is zero. Solving [latex]-15t+88=0[\/latex], we obtain [latex]t=\\frac{88}{15}[\/latex] sec.<\/li>\n<li>To find how far the car travels during this time, we need to find the position of the car after [latex]\\frac{88}{15}[\/latex] sec. We know the velocity [latex]v(t)[\/latex] is the derivative of the position [latex]s(t)[\/latex]. Consider the initial position to be [latex]s(0)=0[\/latex]. Therefore, we need to solve the initial-value problem\n<div id=\"fs-id1165043380494\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]s^{\\prime}(t)=-15t+88, \\, s(0)=0[\/latex].<\/div>\n<p>Integrating, we have<\/p>\n<div id=\"fs-id1165043317270\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]s(t)=-\\frac{15}{2}t^2+88t+C[\/latex].<\/div>\n<p>Since [latex]s(0)=0[\/latex], the constant is [latex]C=0[\/latex]. Therefore, the position function is<\/p>\n<div id=\"fs-id1165043250997\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]s(t)=-\\frac{15}{2}t^2+88t[\/latex].<\/div>\n<p>After [latex]t=\\frac{88}{15}[\/latex] sec, the position is [latex]s(\\frac{88}{15})\\approx 258.133[\/latex] ft.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Decelerating Car.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/j81IZAEfwhI?controls=0&amp;start=849&amp;end=1094&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\/4.10Antiderivatives849to1094_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.10 Antiderivatives&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1165042708223\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Suppose the car is traveling at the rate of 44 ft\/sec. How long does it take for the car to stop? How far will the car travel?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q7660112\">Hint<\/span><\/p>\n<div id=\"q7660112\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]v(t)=-15t+44[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q923849\">Show Solution<\/span><\/p>\n<div id=\"q923849\" class=\"hidden-answer\" style=\"display: none\">\n<p>2.93 sec, 64.5 ft<\/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-445\">\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>4.10 Antiderivatives. <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":38,"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\":\"4.10 Antiderivatives\",\"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-445","chapter","type-chapter","status-publish","hentry"],"part":48,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/445","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":18,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/445\/revisions"}],"predecessor-version":[{"id":4859,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/445\/revisions\/4859"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/48"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/445\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=445"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=445"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=445"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=445"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}