{"id":342,"date":"2021-02-04T01:16:18","date_gmt":"2021-02-04T01:16:18","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=342"},"modified":"2022-03-16T05:32:32","modified_gmt":"2022-03-16T05:32:32","slug":"rate-of-change-applications","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/rate-of-change-applications\/","title":{"raw":"Rate of Change Applications","rendered":"Rate of Change Applications"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Apply rates of change to displacement, velocity, and acceleration of an object moving along a straight line.<\/li>\r\n \t<li>Predict the future population from the present value and the population growth rate.<\/li>\r\n \t<li>Use derivatives to calculate marginal cost and revenue in a business situation.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Motion Along a Line<\/h2>\r\n<p id=\"fs-id1169738881072\">Another use for the derivative is to analyze motion along a line. We have described velocity as the rate of change of position. If we take the derivative of the velocity, we can find the acceleration, or the rate of change of velocity. It is also important to introduce the idea of speed, which is the magnitude of velocity. Thus, we can state the following mathematical definitions.<\/p>\r\n\r\n<div id=\"fs-id1169738999142\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169738947283\">Let [latex]s(t)[\/latex] be a function giving the position of an object at time [latex]t[\/latex].<\/p>\r\n<p id=\"fs-id1169739195686\" style=\"padding-left: 60px;\">The <strong>velocity<\/strong> of the object at time [latex]t[\/latex] is given by [latex]v(t)=s^{\\prime}(t)[\/latex].<\/p>\r\n<p id=\"fs-id1169739105154\" style=\"padding-left: 60px;\">The <strong>speed<\/strong> of the object at time [latex]t[\/latex] is given by [latex]|v(t)|[\/latex].<\/p>\r\n<p id=\"fs-id1169739016931\" style=\"padding-left: 60px;\">The <strong>acceleration<\/strong> of the object at [latex]t[\/latex] is given by [latex]a(t)=v^{\\prime}(t)=s''(t)[\/latex].<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\nMany of the problems involving position, velocity and acceleration will require finding zeros of quadratic and higher order polynomial functions. To find these zeros, recall that factoring and setting each factor equal to zero will be the easiest way to solve these functions. If it doesn't factor and it is a quadratic, the quadratic equation will always work.\r\n<div id=\"fs-id1169738906211\" class=\"textbook exercises\">\r\n<h3>Example: Comparing Instantaneous Velocity and Average Velocity<\/h3>\r\n<p id=\"fs-id1169738942237\">A ball is dropped from a height of 64 feet. Its height above ground (in feet) [latex]t[\/latex] seconds later is given by [latex]s(t)=-16t^2+64[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"386\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205359\/CNX_Calc_Figure_03_04_002.jpg\" alt=\"On the Cartesian coordinate plane, the function s(t) = \u221216t2 + 64 is graphed. This function starts at (0, 64) and decreases to (0, 2).\" width=\"386\" height=\"315\" \/> Figure 2. Dropped ball graph, height vs. time.[\/caption]\r\n<ol id=\"fs-id1169738940918\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>What is the instantaneous velocity of the ball when it hits the ground?<\/li>\r\n \t<li>What is the average velocity during its fall?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1169739233455\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739233455\"]\r\n<p id=\"fs-id1169739233455\">The first thing to do is determine how long it takes the ball to reach the ground. To do this, set [latex]s(t)=0[\/latex]. Solving [latex]-16t^2+64=0[\/latex], we get [latex]t=2[\/latex], so it takes 2 seconds for the ball to reach the ground.<\/p>\r\n\r\n<ol id=\"fs-id1169739220828\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>The instantaneous velocity of the ball as it strikes the ground is [latex]v(2)[\/latex]. Since [latex]v(t)=s^{\\prime}(t)=-32t[\/latex] m we obtain [latex]v(t)=-64[\/latex] ft\/s.<\/li>\r\n \t<li>The average velocity of the ball during its fall is\r\n<div id=\"fs-id1169739018761\" class=\"equation unnumbered\">[latex]v_{avg}=\\frac{s(2)-s(0)}{2-0}=\\frac{0-64}{2}=-32[\/latex] ft\/s.<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169738904404\" class=\"textbook exercises\">\r\n<h3>Example: Interpreting the Relationship between [latex]v(t)[\/latex] and [latex]a(t)[\/latex]<\/h3>\r\n<p id=\"fs-id1169738941480\">A particle moves along a coordinate axis in the positive direction to the right. Its position at time [latex]t[\/latex] is given by [latex]s(t)=t^3-4t+2[\/latex]. Find [latex]v(1)[\/latex] and [latex]a(1)[\/latex] and use these values to answer the following questions.<\/p>\r\n\r\n<ol id=\"fs-id1169739274655\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Is the particle moving from left to right or from right to left at time [latex]t=1[\/latex]?<\/li>\r\n \t<li>Is the particle speeding up or slowing down at time [latex]t=1[\/latex]?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1169738823468\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738823468\"]\r\n<p id=\"fs-id1169738823468\">Begin by finding [latex]v(t)[\/latex] and [latex]a(t)[\/latex].<\/p>\r\n<p id=\"fs-id1169738969278\">[latex]v(t)=s^{\\prime}(t)=3t^2-4[\/latex] and [latex]a(t)=v^{\\prime}(t)=s''(t)=6t[\/latex].<\/p>\r\n<p id=\"fs-id1169739096229\">Evaluating these functions at [latex]t=1[\/latex], we obtain [latex]v(1)=-1[\/latex] and [latex]a(1)=6[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1169738879103\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Because [latex]v(1)&lt;0[\/latex], the particle is moving from right to left.<\/li>\r\n \t<li>Because [latex]v(1)&lt;0[\/latex] and [latex]a(1)&gt;0[\/latex], velocity and acceleration are acting in opposite directions. In other words, the particle is being accelerated in the direction opposite the direction in which it is traveling, causing [latex]|v(t)|[\/latex] to decrease. The particle is slowing down.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169738889577\" class=\"textbook exercises\">\r\n<h3>Example: Position and Velocity<\/h3>\r\n<p id=\"fs-id1169739231121\">The position of a particle moving along a coordinate axis is given by [latex]s(t)=t^3-9t^2+24t+4, \\, t\\ge 0[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1169738912848\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Find [latex]v(t)[\/latex].<\/li>\r\n \t<li>At what time(s) is the particle at rest?<\/li>\r\n \t<li>On what time intervals is the particle moving from left to right? From right to left?<\/li>\r\n \t<li>Use the information obtained to sketch the path of the particle along a coordinate axis.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1169739340338\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739340338\"]\r\n<ol id=\"fs-id1169739340338\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>The velocity is the derivative of the position function:\r\n<div id=\"fs-id1169739198570\" class=\"equation unnumbered\">[latex]v(t)=s^{\\prime}(t)=3t^2-18t+24[\/latex].<\/div><\/li>\r\n \t<li>The particle is at rest when [latex]v(t)=0[\/latex], so set [latex]3t^2-18t+24=0[\/latex]. Factoring the left-hand side of the equation produces [latex]3(t-2)(t-4)=0[\/latex]. Solving, we find that the particle is at rest at [latex]t=2[\/latex] and [latex]t=4[\/latex].<\/li>\r\n \t<li>The particle is moving from left to right when [latex]v(t)&gt;0[\/latex] and from right to left when [latex]v(t)&lt;0[\/latex]. The graph below gives the analysis of the sign of [latex]v(t)[\/latex] for [latex]t\\ge 0[\/latex], but it does not represent the axis along which the particle is moving.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205402\/CNX_Calc_Figure_03_04_003.jpg\" alt=\"A number line marked with 0, 2, and 4. Between 0 and 2, there is a plus sign. Above 2, there is a 0. Between 2 and 4 there is a negative sign. Above 4 there is a 0. After 4 there is a plus sign and v(t).\" width=\"487\" height=\"67\" \/> Figure 3. The sign of v(t) determines the direction of the particle.[\/caption]\r\n\r\n<div class=\"wp-caption-text\" style=\"text-align: center;\"><\/div>\r\n<p style=\"text-align: left;\">Since [latex]3t^2-18t+24&gt;0[\/latex] on [latex][0,2)\\cup (2,+\\infty)[\/latex], the particle is moving from left to right on these intervals.\r\nSince [latex]3t^2-18t+24&lt;0[\/latex] on [latex](2,4)[\/latex], the particle is moving from right to left on this interval.<\/p>\r\n<\/li>\r\n \t<li>Before we can sketch the graph of the particle, we need to know its position at the time it starts moving [latex](t=0)[\/latex] and at the times that it changes direction [latex](t=2,4)[\/latex]. We have [latex]s(0)=4, \\, s(2)=24[\/latex], and [latex]s(4)=20[\/latex]. This means that the particle begins on the coordinate axis at 4 and changes direction at 0 and 20 on the coordinate axis. The path of the particle is shown on a coordinate axis in the graph below.[caption id=\"\" align=\"aligncenter\" width=\"484\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205404\/CNX_Calc_Figure_03_04_005.jpg\" alt=\"A number line is given and above it a line snakes, starting at t = 0 above 4 on the number line. Then the line at t = 2 is above 24 on the number line. Then the line decreases at t = 4 to be above 20 on the number line, at which point the line reverses direction again and increases indefinitely.\" width=\"484\" height=\"109\" \/> Figure 4. The path of the particle can be determined by analyzing [latex]v(t)[\/latex].[\/caption]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Position and Velocity.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/XOgb95xhWF0?controls=0&amp;start=527&amp;end=950&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\/3.4DerivativesAsRatesOfChange527to950_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.4 Derivatives as Rates of Change\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1169736660682\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169739343676\">A particle moves along a coordinate axis. Its position at time [latex]t[\/latex] is given by [latex]s(t)=t^2-5t+1[\/latex]. Is the particle moving from right to left or from left to right at time [latex]t=3[\/latex]?<\/p>\r\n[reveal-answer q=\"789345\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"789345\"]\r\n<p id=\"fs-id1169739236686\">Find [latex]v(3)[\/latex] and look at the sign.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169739299934\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739299934\"]\r\n<p id=\"fs-id1169739299934\">left to right<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]205568[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Population Change<\/h2>\r\n<p id=\"fs-id1169736660717\">In addition to analyzing velocity, speed, acceleration, and position, we can use derivatives to analyze various types of populations, including those as diverse as bacteria colonies and cities. We can use a current population, together with a growth rate, to estimate the size of a population in the future. The <strong>population growth rate<\/strong> is the rate of change of a population and consequently can be represented by the derivative of the size of the population.<\/p>\r\n\r\n<div id=\"fs-id1169738909068\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169739274980\">If [latex]P(t)[\/latex] is the number of entities present in a population, then the population growth rate of [latex]P(t)[\/latex] is defined to be [latex]P^{\\prime}(t)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1169739270775\" class=\"textbook exercises\">\r\n<h3>Example: Estimating a Population<\/h3>\r\n<p id=\"fs-id1169739300220\">The population of a city is tripling every 5 years. If its current population is 10,000, what will be its approximate population 2 years from now?<\/p>\r\n[reveal-answer q=\"fs-id1169739299610\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739299610\"]\r\n<p id=\"fs-id1169739299610\">Let [latex]P(t)[\/latex] be the population (in thousands) [latex]t[\/latex] years from now. Thus, we know that [latex]P(0)=10[\/latex] and based on the information, we anticipate [latex]P(5)=30[\/latex]. Now estimate [latex]P^{\\prime}(0)[\/latex], the current growth rate, using<\/p>\r\n\r\n<div id=\"fs-id1169736589175\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P^{\\prime}(0)\\approx \\frac{P(5)-P(0)}{5-0}=\\frac{30-10}{5}=4[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739201103\">By applying the average rate of change formula to [latex]P(t)[\/latex], we can estimate the population 2 years from now by writing<\/p>\r\n\r\n<div id=\"fs-id1169736662465\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P(2)\\approx P(0)+(2)P^{\\prime}(0)\\approx 10+2(4)=18[\/latex];<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736662775\">thus, in 2 years the population will be approximately 18,000.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Estimating a Population.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/XOgb95xhWF0?controls=0&amp;start=981&amp;end=1123&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\/3.4DerivativesAsRatesOfChange981to1123_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.4 Derivatives as Rates of Change\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1169739202491\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169736656696\">The current population of a mosquito colony is known to be 3,000; that is, [latex]P(0)=3,000[\/latex]. If [latex]P^{\\prime}(0)=100[\/latex], estimate the size of the population in 3 days, where [latex]t[\/latex] is measured in days.<\/p>\r\n[reveal-answer q=\"fs-id1169739019878\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739019878\"]\r\n<p id=\"fs-id1169739188550\">Use [latex]P(3)\\approx P(0)+3P^{\\prime}(0)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169739019878\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739019878\"]\r\n<p id=\"fs-id1169739019878\">3,300<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Changes in Cost and Revenue<\/h2>\r\n<p id=\"fs-id1169736616122\">In addition to analyzing motion along a line and population growth, derivatives are useful in analyzing changes in cost, revenue, and profit. The concept of a marginal function is common in the fields of business and economics and implies the use of derivatives. The<strong> marginal cost<\/strong> is the derivative of the cost function. The <strong>marginal revenue<\/strong> is the derivative of the revenue function. The<strong> marginal profit <\/strong>is the derivative of the profit function, which is based on the cost function and the revenue function.<\/p>\r\n\r\n<div id=\"fs-id1169736618493\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169736618496\">If [latex]C(x)[\/latex] is the cost of producing [latex]x[\/latex] items, then the <strong>marginal cost<\/strong> [latex]MC(x)[\/latex] is [latex]MC(x)=C^{\\prime}(x)[\/latex].<\/p>\r\n<p id=\"fs-id1169739343667\">If [latex]R(x)[\/latex] is the revenue obtained from selling [latex]x[\/latex] items, then the <strong>marginal revenue<\/strong> [latex]MR(x)[\/latex] is [latex]MR(x)=R^{\\prime}(x)[\/latex].<\/p>\r\n<p id=\"fs-id1169739194208\">If [latex]P(x)=R(x)-C(x)[\/latex] is the profit obtained from selling [latex]x[\/latex] items, then the <strong>marginal profit<\/strong> [latex]MP(x)[\/latex] is defined to be [latex]MP(x)=P^{\\prime}(x)=MR(x)-MC(x)=R^{\\prime}(x)-C^{\\prime}(x)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1169739299256\">We can roughly approximate<\/p>\r\n\r\n<div id=\"fs-id1165040641522\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]MC(x)=C^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\dfrac{C(x+h)-C(x)}{h}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739007166\">by choosing an appropriate value for [latex]h[\/latex]. Since [latex]x[\/latex] represents objects, a reasonable and small value for [latex]h[\/latex] is 1. Thus, by substituting [latex]h=1[\/latex], we get the approximation [latex]MC(x)=C^{\\prime}(x)\\approx C(x+1)-C(x)[\/latex]. Consequently, [latex]C^{\\prime}(x)[\/latex] for a given value of [latex]x[\/latex] can be thought of as the change in cost associated with producing one additional item. In a similar way, [latex]MR(x)=R^{\\prime}(x)[\/latex] approximates the revenue obtained by selling one additional item, and [latex]MP(x)=P^{\\prime}(x)[\/latex] approximates the profit obtained by producing and selling one additional item.<\/p>\r\n\r\n<div id=\"fs-id1169739302508\" class=\"textbook exercises\">\r\n<h3>Example: Applying Marginal Revenue<\/h3>\r\n<p id=\"fs-id1169739274279\">Assume that the number of barbeque dinners that can be sold, [latex]x[\/latex], can be related to the price charged, [latex]p[\/latex], by the equation [latex]p(x)=9-0.03x, \\, 0\\le x\\le 300[\/latex].<\/p>\r\n<p id=\"fs-id1169739305211\">In this case, the revenue in dollars obtained by selling [latex]x[\/latex] barbeque dinners is given by<\/p>\r\n\r\n<div id=\"fs-id1169739212801\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]R(x)=xp(x)=x(9-0.03x)=-0.03x^2+9x[\/latex] for [latex]0\\le x\\le 300[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739194211\">Use the marginal revenue function to estimate the revenue obtained from selling the 101st barbeque dinner. Compare this to the actual revenue obtained from the sale of this dinner.<\/p>\r\n[reveal-answer q=\"fs-id1169739351550\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739351550\"]\r\n<p id=\"fs-id1169739351550\">First, find the marginal revenue function: [latex]MR(x)=R^{\\prime}(x)=-0.06x+9[\/latex].<\/p>\r\n<p id=\"fs-id1169739305060\">Next, use [latex]R^{\\prime}(100)[\/latex] to approximate [latex]R(101)-R(100)[\/latex], the revenue obtained from the sale of the 101st dinner. Since [latex]R^{\\prime}(100)=3[\/latex], the revenue obtained from the sale of the 101st dinner is approximately $3.<\/p>\r\n<p id=\"fs-id1169736611332\">The actual revenue obtained from the sale of the 101st dinner is<\/p>\r\n\r\n<div id=\"fs-id1169737285423\" class=\"equation unnumbered\">[latex]R(101)-R(100)=602.97-600=2.97[\/latex], or [latex]\\$2.97[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739298517\">The marginal revenue is a fairly good estimate in this case and has the advantage of being easy to compute.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Applying Marginal Revenue.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/XOgb95xhWF0?controls=0&amp;start=1168&amp;end=1320&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266835\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266835\"]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\/3.4DerivativesAsRatesOfChange1168to1320_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.4 Derivatives as Rates of Change\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1169739054962\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169736589106\">Suppose that the profit obtained from the sale of [latex]x[\/latex] fish-fry dinners is given by [latex]P(x)=-0.03x^2+8x-50[\/latex]. Use the marginal profit function to estimate the profit from the sale of the 101st fish-fry dinner.<\/p>\r\n[reveal-answer q=\"554428\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"554428\"]\r\n<p id=\"fs-id1169739273788\">Use [latex]P^{\\prime}(100)[\/latex] to approximate [latex]P(101)-P(100)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169739208505\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739208505\"]\r\n<p id=\"fs-id1169739208505\">$2<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Apply rates of change to displacement, velocity, and acceleration of an object moving along a straight line.<\/li>\n<li>Predict the future population from the present value and the population growth rate.<\/li>\n<li>Use derivatives to calculate marginal cost and revenue in a business situation.<\/li>\n<\/ul>\n<\/div>\n<h2>Motion Along a Line<\/h2>\n<p id=\"fs-id1169738881072\">Another use for the derivative is to analyze motion along a line. We have described velocity as the rate of change of position. If we take the derivative of the velocity, we can find the acceleration, or the rate of change of velocity. It is also important to introduce the idea of speed, which is the magnitude of velocity. Thus, we can state the following mathematical definitions.<\/p>\n<div id=\"fs-id1169738999142\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1169738947283\">Let [latex]s(t)[\/latex] be a function giving the position of an object at time [latex]t[\/latex].<\/p>\n<p id=\"fs-id1169739195686\" style=\"padding-left: 60px;\">The <strong>velocity<\/strong> of the object at time [latex]t[\/latex] is given by [latex]v(t)=s^{\\prime}(t)[\/latex].<\/p>\n<p id=\"fs-id1169739105154\" style=\"padding-left: 60px;\">The <strong>speed<\/strong> of the object at time [latex]t[\/latex] is given by [latex]|v(t)|[\/latex].<\/p>\n<p id=\"fs-id1169739016931\" style=\"padding-left: 60px;\">The <strong>acceleration<\/strong> of the object at [latex]t[\/latex] is given by [latex]a(t)=v^{\\prime}(t)=s''(t)[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<p>Many of the problems involving position, velocity and acceleration will require finding zeros of quadratic and higher order polynomial functions. To find these zeros, recall that factoring and setting each factor equal to zero will be the easiest way to solve these functions. If it doesn&#8217;t factor and it is a quadratic, the quadratic equation will always work.<\/p>\n<div id=\"fs-id1169738906211\" class=\"textbook exercises\">\n<h3>Example: Comparing Instantaneous Velocity and Average Velocity<\/h3>\n<p id=\"fs-id1169738942237\">A ball is dropped from a height of 64 feet. Its height above ground (in feet) [latex]t[\/latex] seconds later is given by [latex]s(t)=-16t^2+64[\/latex].<\/p>\n<div style=\"width: 396px\" 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\/11205359\/CNX_Calc_Figure_03_04_002.jpg\" alt=\"On the Cartesian coordinate plane, the function s(t) = \u221216t2 + 64 is graphed. This function starts at (0, 64) and decreases to (0, 2).\" width=\"386\" height=\"315\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. Dropped ball graph, height vs. time.<\/p>\n<\/div>\n<ol id=\"fs-id1169738940918\" style=\"list-style-type: lower-alpha;\">\n<li>What is the instantaneous velocity of the ball when it hits the ground?<\/li>\n<li>What is the average velocity during its fall?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739233455\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739233455\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739233455\">The first thing to do is determine how long it takes the ball to reach the ground. To do this, set [latex]s(t)=0[\/latex]. Solving [latex]-16t^2+64=0[\/latex], we get [latex]t=2[\/latex], so it takes 2 seconds for the ball to reach the ground.<\/p>\n<ol id=\"fs-id1169739220828\" style=\"list-style-type: lower-alpha;\">\n<li>The instantaneous velocity of the ball as it strikes the ground is [latex]v(2)[\/latex]. Since [latex]v(t)=s^{\\prime}(t)=-32t[\/latex] m we obtain [latex]v(t)=-64[\/latex] ft\/s.<\/li>\n<li>The average velocity of the ball during its fall is\n<div id=\"fs-id1169739018761\" class=\"equation unnumbered\">[latex]v_{avg}=\\frac{s(2)-s(0)}{2-0}=\\frac{0-64}{2}=-32[\/latex] ft\/s.<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738904404\" class=\"textbook exercises\">\n<h3>Example: Interpreting the Relationship between [latex]v(t)[\/latex] and [latex]a(t)[\/latex]<\/h3>\n<p id=\"fs-id1169738941480\">A particle moves along a coordinate axis in the positive direction to the right. Its position at time [latex]t[\/latex] is given by [latex]s(t)=t^3-4t+2[\/latex]. Find [latex]v(1)[\/latex] and [latex]a(1)[\/latex] and use these values to answer the following questions.<\/p>\n<ol id=\"fs-id1169739274655\" style=\"list-style-type: lower-alpha;\">\n<li>Is the particle moving from left to right or from right to left at time [latex]t=1[\/latex]?<\/li>\n<li>Is the particle speeding up or slowing down at time [latex]t=1[\/latex]?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738823468\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738823468\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738823468\">Begin by finding [latex]v(t)[\/latex] and [latex]a(t)[\/latex].<\/p>\n<p id=\"fs-id1169738969278\">[latex]v(t)=s^{\\prime}(t)=3t^2-4[\/latex] and [latex]a(t)=v^{\\prime}(t)=s''(t)=6t[\/latex].<\/p>\n<p id=\"fs-id1169739096229\">Evaluating these functions at [latex]t=1[\/latex], we obtain [latex]v(1)=-1[\/latex] and [latex]a(1)=6[\/latex].<\/p>\n<ol id=\"fs-id1169738879103\" style=\"list-style-type: lower-alpha;\">\n<li>Because [latex]v(1)<0[\/latex], the particle is moving from right to left.<\/li>\n<li>Because [latex]v(1)<0[\/latex] and [latex]a(1)>0[\/latex], velocity and acceleration are acting in opposite directions. In other words, the particle is being accelerated in the direction opposite the direction in which it is traveling, causing [latex]|v(t)|[\/latex] to decrease. The particle is slowing down.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738889577\" class=\"textbook exercises\">\n<h3>Example: Position and Velocity<\/h3>\n<p id=\"fs-id1169739231121\">The position of a particle moving along a coordinate axis is given by [latex]s(t)=t^3-9t^2+24t+4, \\, t\\ge 0[\/latex].<\/p>\n<ol id=\"fs-id1169738912848\" style=\"list-style-type: lower-alpha;\">\n<li>Find [latex]v(t)[\/latex].<\/li>\n<li>At what time(s) is the particle at rest?<\/li>\n<li>On what time intervals is the particle moving from left to right? From right to left?<\/li>\n<li>Use the information obtained to sketch the path of the particle along a coordinate axis.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739340338\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739340338\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1169739340338\" style=\"list-style-type: lower-alpha;\">\n<li>The velocity is the derivative of the position function:\n<div id=\"fs-id1169739198570\" class=\"equation unnumbered\">[latex]v(t)=s^{\\prime}(t)=3t^2-18t+24[\/latex].<\/div>\n<\/li>\n<li>The particle is at rest when [latex]v(t)=0[\/latex], so set [latex]3t^2-18t+24=0[\/latex]. Factoring the left-hand side of the equation produces [latex]3(t-2)(t-4)=0[\/latex]. Solving, we find that the particle is at rest at [latex]t=2[\/latex] and [latex]t=4[\/latex].<\/li>\n<li>The particle is moving from left to right when [latex]v(t)>0[\/latex] and from right to left when [latex]v(t)<0[\/latex]. The graph below gives the analysis of the sign of [latex]v(t)[\/latex] for [latex]t\\ge 0[\/latex], but it does not represent the axis along which the particle is moving.\n\n\n\n<div style=\"width: 497px\" 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\/11205402\/CNX_Calc_Figure_03_04_003.jpg\" alt=\"A number line marked with 0, 2, and 4. Between 0 and 2, there is a plus sign. Above 2, there is a 0. Between 2 and 4 there is a negative sign. Above 4 there is a 0. After 4 there is a plus sign and v(t).\" width=\"487\" height=\"67\" \/>\n<p class=\"wp-caption-text\">Figure 3. The sign of v(t) determines the direction of the particle.<\/p>\n<\/div>\n<div class=\"wp-caption-text\" style=\"text-align: center;\"><\/div>\n<p style=\"text-align: left;\">Since [latex]3t^2-18t+24>0[\/latex] on [latex][0,2)\\cup (2,+\\infty)[\/latex], the particle is moving from left to right on these intervals.<br \/>\nSince [latex]3t^2-18t+24<0[\/latex] on [latex](2,4)[\/latex], the particle is moving from right to left on this interval.<\/p>\n<\/li>\n<li>Before we can sketch the graph of the particle, we need to know its position at the time it starts moving [latex](t=0)[\/latex] and at the times that it changes direction [latex](t=2,4)[\/latex]. We have [latex]s(0)=4, \\, s(2)=24[\/latex], and [latex]s(4)=20[\/latex]. This means that the particle begins on the coordinate axis at 4 and changes direction at 0 and 20 on the coordinate axis. The path of the particle is shown on a coordinate axis in the graph below.\n<div style=\"width: 494px\" 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\/11205404\/CNX_Calc_Figure_03_04_005.jpg\" alt=\"A number line is given and above it a line snakes, starting at t = 0 above 4 on the number line. Then the line at t = 2 is above 24 on the number line. Then the line decreases at t = 4 to be above 20 on the number line, at which point the line reverses direction again and increases indefinitely.\" width=\"484\" height=\"109\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. The path of the particle can be determined by analyzing [latex]v(t)[\/latex].<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Position and Velocity.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/XOgb95xhWF0?controls=0&amp;start=527&amp;end=950&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\/3.4DerivativesAsRatesOfChange527to950_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.4 Derivatives as Rates of Change&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1169736660682\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169739343676\">A particle moves along a coordinate axis. Its position at time [latex]t[\/latex] is given by [latex]s(t)=t^2-5t+1[\/latex]. Is the particle moving from right to left or from left to right at time [latex]t=3[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q789345\">Hint<\/span><\/p>\n<div id=\"q789345\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739236686\">Find [latex]v(3)[\/latex] and look at the sign.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739299934\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739299934\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739299934\">left to right<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm205568\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=205568&theme=oea&iframe_resize_id=ohm205568&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Population Change<\/h2>\n<p id=\"fs-id1169736660717\">In addition to analyzing velocity, speed, acceleration, and position, we can use derivatives to analyze various types of populations, including those as diverse as bacteria colonies and cities. We can use a current population, together with a growth rate, to estimate the size of a population in the future. The <strong>population growth rate<\/strong> is the rate of change of a population and consequently can be represented by the derivative of the size of the population.<\/p>\n<div id=\"fs-id1169738909068\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1169739274980\">If [latex]P(t)[\/latex] is the number of entities present in a population, then the population growth rate of [latex]P(t)[\/latex] is defined to be [latex]P^{\\prime}(t)[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1169739270775\" class=\"textbook exercises\">\n<h3>Example: Estimating a Population<\/h3>\n<p id=\"fs-id1169739300220\">The population of a city is tripling every 5 years. If its current population is 10,000, what will be its approximate population 2 years from now?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739299610\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739299610\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739299610\">Let [latex]P(t)[\/latex] be the population (in thousands) [latex]t[\/latex] years from now. Thus, we know that [latex]P(0)=10[\/latex] and based on the information, we anticipate [latex]P(5)=30[\/latex]. Now estimate [latex]P^{\\prime}(0)[\/latex], the current growth rate, using<\/p>\n<div id=\"fs-id1169736589175\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P^{\\prime}(0)\\approx \\frac{P(5)-P(0)}{5-0}=\\frac{30-10}{5}=4[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739201103\">By applying the average rate of change formula to [latex]P(t)[\/latex], we can estimate the population 2 years from now by writing<\/p>\n<div id=\"fs-id1169736662465\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P(2)\\approx P(0)+(2)P^{\\prime}(0)\\approx 10+2(4)=18[\/latex];<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736662775\">thus, in 2 years the population will be approximately 18,000.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Estimating a Population.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/XOgb95xhWF0?controls=0&amp;start=981&amp;end=1123&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\/3.4DerivativesAsRatesOfChange981to1123_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.4 Derivatives as Rates of Change&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1169739202491\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169736656696\">The current population of a mosquito colony is known to be 3,000; that is, [latex]P(0)=3,000[\/latex]. If [latex]P^{\\prime}(0)=100[\/latex], estimate the size of the population in 3 days, where [latex]t[\/latex] is measured in days.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739019878\">Hint<\/span><\/p>\n<div id=\"qfs-id1169739019878\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739188550\">Use [latex]P(3)\\approx P(0)+3P^{\\prime}(0)[\/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-id1169739019878\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739019878\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739019878\">3,300<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Changes in Cost and Revenue<\/h2>\n<p id=\"fs-id1169736616122\">In addition to analyzing motion along a line and population growth, derivatives are useful in analyzing changes in cost, revenue, and profit. The concept of a marginal function is common in the fields of business and economics and implies the use of derivatives. The<strong> marginal cost<\/strong> is the derivative of the cost function. The <strong>marginal revenue<\/strong> is the derivative of the revenue function. The<strong> marginal profit <\/strong>is the derivative of the profit function, which is based on the cost function and the revenue function.<\/p>\n<div id=\"fs-id1169736618493\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1169736618496\">If [latex]C(x)[\/latex] is the cost of producing [latex]x[\/latex] items, then the <strong>marginal cost<\/strong> [latex]MC(x)[\/latex] is [latex]MC(x)=C^{\\prime}(x)[\/latex].<\/p>\n<p id=\"fs-id1169739343667\">If [latex]R(x)[\/latex] is the revenue obtained from selling [latex]x[\/latex] items, then the <strong>marginal revenue<\/strong> [latex]MR(x)[\/latex] is [latex]MR(x)=R^{\\prime}(x)[\/latex].<\/p>\n<p id=\"fs-id1169739194208\">If [latex]P(x)=R(x)-C(x)[\/latex] is the profit obtained from selling [latex]x[\/latex] items, then the <strong>marginal profit<\/strong> [latex]MP(x)[\/latex] is defined to be [latex]MP(x)=P^{\\prime}(x)=MR(x)-MC(x)=R^{\\prime}(x)-C^{\\prime}(x)[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1169739299256\">We can roughly approximate<\/p>\n<div id=\"fs-id1165040641522\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]MC(x)=C^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\dfrac{C(x+h)-C(x)}{h}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739007166\">by choosing an appropriate value for [latex]h[\/latex]. Since [latex]x[\/latex] represents objects, a reasonable and small value for [latex]h[\/latex] is 1. Thus, by substituting [latex]h=1[\/latex], we get the approximation [latex]MC(x)=C^{\\prime}(x)\\approx C(x+1)-C(x)[\/latex]. Consequently, [latex]C^{\\prime}(x)[\/latex] for a given value of [latex]x[\/latex] can be thought of as the change in cost associated with producing one additional item. In a similar way, [latex]MR(x)=R^{\\prime}(x)[\/latex] approximates the revenue obtained by selling one additional item, and [latex]MP(x)=P^{\\prime}(x)[\/latex] approximates the profit obtained by producing and selling one additional item.<\/p>\n<div id=\"fs-id1169739302508\" class=\"textbook exercises\">\n<h3>Example: Applying Marginal Revenue<\/h3>\n<p id=\"fs-id1169739274279\">Assume that the number of barbeque dinners that can be sold, [latex]x[\/latex], can be related to the price charged, [latex]p[\/latex], by the equation [latex]p(x)=9-0.03x, \\, 0\\le x\\le 300[\/latex].<\/p>\n<p id=\"fs-id1169739305211\">In this case, the revenue in dollars obtained by selling [latex]x[\/latex] barbeque dinners is given by<\/p>\n<div id=\"fs-id1169739212801\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]R(x)=xp(x)=x(9-0.03x)=-0.03x^2+9x[\/latex] for [latex]0\\le x\\le 300[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739194211\">Use the marginal revenue function to estimate the revenue obtained from selling the 101st barbeque dinner. Compare this to the actual revenue obtained from the sale of this dinner.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739351550\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739351550\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739351550\">First, find the marginal revenue function: [latex]MR(x)=R^{\\prime}(x)=-0.06x+9[\/latex].<\/p>\n<p id=\"fs-id1169739305060\">Next, use [latex]R^{\\prime}(100)[\/latex] to approximate [latex]R(101)-R(100)[\/latex], the revenue obtained from the sale of the 101st dinner. Since [latex]R^{\\prime}(100)=3[\/latex], the revenue obtained from the sale of the 101st dinner is approximately $3.<\/p>\n<p id=\"fs-id1169736611332\">The actual revenue obtained from the sale of the 101st dinner is<\/p>\n<div id=\"fs-id1169737285423\" class=\"equation unnumbered\">[latex]R(101)-R(100)=602.97-600=2.97[\/latex], or [latex]\\$2.97[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739298517\">The marginal revenue is a fairly good estimate in this case and has the advantage of being easy to compute.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Applying Marginal Revenue.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/XOgb95xhWF0?controls=0&amp;start=1168&amp;end=1320&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=\"q266835\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266835\" 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\/3.4DerivativesAsRatesOfChange1168to1320_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.4 Derivatives as Rates of Change&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1169739054962\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169736589106\">Suppose that the profit obtained from the sale of [latex]x[\/latex] fish-fry dinners is given by [latex]P(x)=-0.03x^2+8x-50[\/latex]. Use the marginal profit function to estimate the profit from the sale of the 101st fish-fry dinner.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q554428\">Hint<\/span><\/p>\n<div id=\"q554428\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739273788\">Use [latex]P^{\\prime}(100)[\/latex] to approximate [latex]P(101)-P(100)[\/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-id1169739208505\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739208505\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739208505\">$2<\/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-342\">\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>3.4 Derivatives as Rates of Change. <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":18,"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\":\"3.4 Derivatives as Rates of Change\",\"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-342","chapter","type-chapter","status-publish","hentry"],"part":35,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/342","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":22,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/342\/revisions"}],"predecessor-version":[{"id":4810,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/342\/revisions\/4810"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/35"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/342\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=342"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=342"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=342"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=342"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}