{"id":329,"date":"2021-02-04T01:12:09","date_gmt":"2021-02-04T01:12:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=329"},"modified":"2022-03-16T05:27:23","modified_gmt":"2022-03-16T05:27:23","slug":"velocities-and-rates-of-change","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/velocities-and-rates-of-change\/","title":{"raw":"Velocities and Rates of Change","rendered":"Velocities and Rates of Change"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Describe the velocity as a rate of change<\/li>\r\n \t<li>Explain the difference between average velocity and instantaneous velocity<\/li>\r\n \t<li>Estimate the derivative from a table of values<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1169739353680\">Now that we can evaluate a derivative, we can use it in velocity applications. Recall that if [latex]s(t)[\/latex] is the position of an object moving along a coordinate axis, the <span class=\"no-emphasis\">average velocity<\/span> of the object over a time interval [latex][a,t][\/latex] if [latex]t&gt;a[\/latex] or [latex][t,a][\/latex] if [latex]t&lt;a[\/latex] is given by the difference quotient<\/p>\r\n\r\n<div id=\"fs-id1169739303202\" class=\"equation\" style=\"text-align: center;\">[latex]v_{\\text{avg}}=\\dfrac{s(t)-s(a)}{t-a}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739189992\">As the values of [latex]t[\/latex] approach [latex]a[\/latex], the values of [latex]v_{\\text{avg}}[\/latex] approach the value we call the <span class=\"no-emphasis\">instantaneous velocity<\/span> at [latex]a[\/latex]. That is, instantaneous velocity at [latex]a[\/latex], denoted [latex]v(a)[\/latex], is given by<\/p>\r\n\r\n<div id=\"fs-id1169739274700\" class=\"equation\" style=\"text-align: center;\">[latex]v(a)=s^{\\prime}(a)=\\underset{t\\to a}{\\lim}\\dfrac{s(t)-s(a)}{t-a}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739064816\">To better understand the relationship between average velocity and instantaneous velocity, see Figure 7. In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time [latex]t=a[\/latex] whose position at time [latex]t[\/latex] is given by the function [latex]s(t)[\/latex]. The slope of the secant line (shown in green) is the average velocity of the object over the time interval [latex][a,t][\/latex].<\/p>\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\/11205202\/CNX_Calc_Figure_03_01_007.jpg\" alt=\"This figure consists of the Cartesian coordinate plane with 0, a, and t1 marked on the t-axis. The function y = s(t) is graphed in the first quadrant along with two lines marked tangent and secant. The tangent line touches y = s(t) at only one point, (a, s(a)). The secant line touches y = s(t) at two points: (a, s(a)) and (t1, s(t1)).\" width=\"487\" height=\"284\" \/> Figure 7. The slope of the secant line is the average velocity over the interval [latex][a,t][\/latex]. The slope of the tangent line is the instantaneous velocity.[\/caption]\r\n<p id=\"fs-id1169736619727\">We can use the definitions to calculate the instantaneous velocity, or we can estimate the velocity of a moving object by using a table of values. We can then confirm the estimate by using the difference quotient.<\/p>\r\n\r\n<div id=\"fs-id1169739179204\" class=\"textbook exercises\">\r\n<h3>Example: Estimating Velocity<\/h3>\r\n<p id=\"fs-id1169739188880\">A lead weight on a spring is oscillating up and down. Its position at time [latex]t[\/latex] with respect to a fixed horizontal line is given by [latex]s(t)= \\sin t[\/latex]. Use a table of values to estimate [latex]v(0)[\/latex]. Check the estimate by using the definition of a derivative.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205205\/CNX_Calc_Figure_03_01_010.jpg\" alt=\"A picture of a spring hanging down with a weight at the end. There is a horizontal dashed line marked 0 a little bit above the weight.\" width=\"325\" height=\"192\" \/> Figure 8. A lead weight suspended from a spring in vertical oscillatory motion.[\/caption]\r\n\r\n[reveal-answer q=\"fs-id1169738960040\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738960040\"]\r\n<p id=\"fs-id1169738960040\">We can estimate the instantaneous velocity at [latex]t=0[\/latex] by computing a table of average velocities using values of [latex]t[\/latex] approaching 0, as shown in the table below.<\/p>\r\n\r\n<table id=\"fs-id1169739242675\" summary=\"This table has seven rows and two columns. The first row is a header row and it labels each column. The first column header is x and the second column is (sin t - sin 0)\/(t \u2212 0) = (sin t)\/t. Under the first column are the values -0.1, -0.01, -0.001, 0.001, 0.01, and 0.1. Under the second column are the values 0.998334166, 0.9999833333, 0.999999833, 0.999999833, 0.9999833333, and 0.998334166.\"><caption>Average velocities using values of [latex]t[\/latex] approaching 0<\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>[latex]t[\/latex]<\/th>\r\n<th>[latex]\\frac{\\sin t - \\sin 0}{t-0}=\\frac{\\sin t}{t}[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>-0.1<\/td>\r\n<td>0.998334166<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>-0.01<\/td>\r\n<td>0.9999833333<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>-0.001<\/td>\r\n<td>0.999999833<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0.001<\/td>\r\n<td>0.999999833<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0.01<\/td>\r\n<td>0.9999833333<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0.1<\/td>\r\n<td>0.998334166<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1169739188902\">From the table we see that the average velocity over the time interval [latex][-0.1,0][\/latex] is 0.998334166, the average velocity over the time interval [latex][-0.01,0][\/latex] is 0.9999833333, and so forth. Using this table of values, it appears that a good estimate is [latex]v(0)=1[\/latex].<\/p>\r\n<p id=\"fs-id1169739298311\">By using the definition of a derivative, we can see that<\/p>\r\n\r\n<div id=\"fs-id1169739302610\" class=\"equation unnumbered\">[latex]v(0)=s^{\\prime}(0)=\\underset{t\\to 0}{\\lim}\\dfrac{\\sin t- \\sin 0}{t-0}=\\underset{t\\to 0}{\\lim}\\dfrac{\\sin t}{t}=1[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739111046\">Thus, in fact, [latex]v(0)=1[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169739220840\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169739220848\">A rock is dropped from a height of 64 feet. Its height above ground at time [latex]t[\/latex] seconds later is given by [latex]s(t)=-16t^2+64, \\, 0\\le t\\le 2[\/latex]. Find its instantaneous velocity 1 second after it is dropped, using the definition of a derivative.<\/p>\r\n[reveal-answer q=\"403817\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"403817\"]\r\n<p id=\"fs-id1169739189236\">[latex]v(t)=s^{\\prime}(t)[\/latex]. Follow the earlier examples of the derivative using the definition of a derivative.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169739298454\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739298454\"]\r\n<p id=\"fs-id1169739298454\">-32 ft\/sec<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/VnDnInldaMM?controls=0&amp;start=1660&amp;end=1763&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\/3.1DefiningTheDerivative1660to1763_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.1 Defining the Derivative\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<p id=\"fs-id1169739305376\">As we have seen throughout this section, the slope of a tangent line to a function and instantaneous velocity are related concepts. Each is calculated by computing a derivative and each measures the instantaneous rate of change of a function, or the rate of change of a function at any point along the function.<\/p>\r\n\r\n<div id=\"fs-id1169739269974\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169739269977\">The<strong> instantaneous rate of change<\/strong> of a function [latex]f(x)[\/latex] at a value [latex]a[\/latex] is its derivative [latex]f^{\\prime}(a)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1169739187343\" class=\"textbook exercises\">\r\n<h3>Example: Rate of Change of Temperature<\/h3>\r\n<p id=\"fs-id1169739187352\">A homeowner sets the thermostat so that the temperature in the house begins to drop from [latex]70^{\\circ}\\text{F}[\/latex] at 9 p.m., reaches a low of [latex]60^{\\circ}[\/latex] during the night, and rises back to [latex]70^{\\circ}[\/latex] by 7 a.m. the next morning. Suppose that the temperature in the house is given by [latex]T(t)=0.4t^2-4t+70[\/latex] for [latex]0\\le t\\le 10[\/latex], where [latex]t[\/latex] is the number of hours past 9 p.m. Find the instantaneous rate of change of the temperature at midnight.<\/p>\r\n[reveal-answer q=\"fs-id1169739297924\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739297924\"]\r\n<p id=\"fs-id1169739297924\">Since midnight is 3 hours past 9 p.m., we want to compute [latex]T^{\\prime }(3)[\/latex]. Refer to the definition of a derivative.<\/p>\r\n\r\n<div id=\"fs-id1169739270561\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll}T^{\\prime}(3) &amp; =\\underset{t\\to 3}{\\lim}\\frac{T(t)-T(3)}{t-3} &amp; &amp; &amp; \\text{Apply the definition.} \\\\ &amp; =\\underset{t\\to 3}{\\lim}\\frac{0.4t^2-4t+70-61.6}{t-3} &amp; &amp; &amp; \\begin{array}{l}\\text{Substitute }T(t)=0.4t^2-4t+70 \\, \\text{and} \\\\ T(3)=61.6. \\end{array} \\\\ &amp; =\\underset{t\\to 3}{\\lim}\\frac{0.4t^2-4t+8.4}{t-3} &amp; &amp; &amp; \\text{Simplify.} \\\\ &amp; =\\underset{t\\to 3}{\\lim}\\frac{0.4(t-3)(t-7)}{t-3} &amp; &amp; &amp; =\\underset{t\\to 3}{\\lim}\\frac{0.4(t-3)(t-7)}{t-3} \\\\ &amp; =\\underset{t\\to 3}{\\lim}0.4(t-7) &amp; &amp; &amp; \\text{Cancel.} \\\\ &amp; =-1.6 &amp; &amp; &amp; \\text{Evaluate the limit.} \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1169739274032\">The instantaneous rate of change of the temperature at midnight is [latex]-1.6^{\\circ}\\text{F}[\/latex] per hour.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169739253544\" class=\"textbook exercises\">\r\n<h3>Example: Rate of Change of Profit<\/h3>\r\n<p id=\"fs-id1169739253553\">A toy company can sell [latex]x[\/latex] electronic gaming systems at a price of [latex]p=-0.01x+400[\/latex] dollars per gaming system. The cost of manufacturing [latex]x[\/latex] systems is given by [latex]C(x)=100x+10,000[\/latex] dollars. Find the rate of change of profit when 10,000 games are produced. Should the toy company increase or decrease production?<\/p>\r\n[reveal-answer q=\"fs-id1169739274608\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739274608\"]\r\n<p id=\"fs-id1169739274608\">The profit [latex]P(x)[\/latex] earned by producing [latex]x[\/latex] gaming systems is [latex]R(x)-C(x)[\/latex], where [latex]R(x)[\/latex] is the revenue obtained from the sale of [latex]x[\/latex] games. Since the company can sell [latex]x[\/latex] games at [latex]p=-0.01x+400[\/latex] per game,<\/p>\r\n\r\n<div id=\"fs-id1169739286461\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]R(x)=xp=x(-0.01x+400)=-0.01x^2+400x[\/latex].<\/div>\r\n<p id=\"fs-id1169739302924\">Consequently,<\/p>\r\n\r\n<div id=\"fs-id1169739302928\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P(x)=-0.01x^2+300x-10,000[\/latex].<\/div>\r\n<p id=\"fs-id1169739190488\">Therefore, evaluating the rate of change of profit gives<\/p>\r\n\r\n<div id=\"fs-id1169739190491\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}P^{\\prime}(10000)&amp; =\\underset{x\\to 10000}{\\lim}\\frac{P(x)-P(10000)}{x-10000} \\\\ &amp; =\\underset{x\\to 10000}{\\lim}\\frac{-0.01x^2+300x-10000-1990000}{x-10000} \\\\ &amp; =\\underset{x\\to 10000}{\\lim}\\frac{-0.01x^2+300x-2000000}{x-10000} \\\\ &amp; =100 \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736587942\">Since the rate of change of profit [latex]P^{\\prime}(10,000)&gt;0[\/latex] and [latex]P(10,000)&gt;0[\/latex], the company should increase production.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169739273077\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169739273085\">A coffee shop determines that the daily profit on scones obtained by charging [latex]s[\/latex] dollars per scone is [latex]P(s)=-20s^2+150s-10[\/latex]. The coffee shop currently charges [latex]\\$3.25[\/latex] per scone. Find [latex]P^{\\prime}(3.25)[\/latex], the rate of change of profit when the price is [latex]\\$3.25[\/latex] and decide whether or not the coffee shop should consider raising or lowering its prices on scones.<\/p>\r\n[reveal-answer q=\"fs-id1169739242330\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739242330\"]\r\n<p id=\"fs-id1169739242330\">[latex]P^{\\prime}(3.25)=20&gt;0[\/latex]; raise prices<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1169736611546\">Use the previous example for a guide.<\/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]162461[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Describe the velocity as a rate of change<\/li>\n<li>Explain the difference between average velocity and instantaneous velocity<\/li>\n<li>Estimate the derivative from a table of values<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1169739353680\">Now that we can evaluate a derivative, we can use it in velocity applications. Recall that if [latex]s(t)[\/latex] is the position of an object moving along a coordinate axis, the <span class=\"no-emphasis\">average velocity<\/span> of the object over a time interval [latex][a,t][\/latex] if [latex]t>a[\/latex] or [latex][t,a][\/latex] if [latex]t<a[\/latex] is given by the difference quotient<\/p>\n<div id=\"fs-id1169739303202\" class=\"equation\" style=\"text-align: center;\">[latex]v_{\\text{avg}}=\\dfrac{s(t)-s(a)}{t-a}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739189992\">As the values of [latex]t[\/latex] approach [latex]a[\/latex], the values of [latex]v_{\\text{avg}}[\/latex] approach the value we call the <span class=\"no-emphasis\">instantaneous velocity<\/span> at [latex]a[\/latex]. That is, instantaneous velocity at [latex]a[\/latex], denoted [latex]v(a)[\/latex], is given by<\/p>\n<div id=\"fs-id1169739274700\" class=\"equation\" style=\"text-align: center;\">[latex]v(a)=s^{\\prime}(a)=\\underset{t\\to a}{\\lim}\\dfrac{s(t)-s(a)}{t-a}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739064816\">To better understand the relationship between average velocity and instantaneous velocity, see Figure 7. In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time [latex]t=a[\/latex] whose position at time [latex]t[\/latex] is given by the function [latex]s(t)[\/latex]. The slope of the secant line (shown in green) is the average velocity of the object over the time interval [latex][a,t][\/latex].<\/p>\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\/11205202\/CNX_Calc_Figure_03_01_007.jpg\" alt=\"This figure consists of the Cartesian coordinate plane with 0, a, and t1 marked on the t-axis. The function y = s(t) is graphed in the first quadrant along with two lines marked tangent and secant. The tangent line touches y = s(t) at only one point, (a, s(a)). The secant line touches y = s(t) at two points: (a, s(a)) and (t1, s(t1)).\" width=\"487\" height=\"284\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. The slope of the secant line is the average velocity over the interval [latex][a,t][\/latex]. The slope of the tangent line is the instantaneous velocity.<\/p>\n<\/div>\n<p id=\"fs-id1169736619727\">We can use the definitions to calculate the instantaneous velocity, or we can estimate the velocity of a moving object by using a table of values. We can then confirm the estimate by using the difference quotient.<\/p>\n<div id=\"fs-id1169739179204\" class=\"textbook exercises\">\n<h3>Example: Estimating Velocity<\/h3>\n<p id=\"fs-id1169739188880\">A lead weight on a spring is oscillating up and down. Its position at time [latex]t[\/latex] with respect to a fixed horizontal line is given by [latex]s(t)= \\sin t[\/latex]. Use a table of values to estimate [latex]v(0)[\/latex]. Check the estimate by using the definition of a derivative.<\/p>\n<div style=\"width: 335px\" 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\/11205205\/CNX_Calc_Figure_03_01_010.jpg\" alt=\"A picture of a spring hanging down with a weight at the end. There is a horizontal dashed line marked 0 a little bit above the weight.\" width=\"325\" height=\"192\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8. A lead weight suspended from a spring in vertical oscillatory motion.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738960040\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738960040\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738960040\">We can estimate the instantaneous velocity at [latex]t=0[\/latex] by computing a table of average velocities using values of [latex]t[\/latex] approaching 0, as shown in the table below.<\/p>\n<table id=\"fs-id1169739242675\" summary=\"This table has seven rows and two columns. The first row is a header row and it labels each column. The first column header is x and the second column is (sin t - sin 0)\/(t \u2212 0) = (sin t)\/t. Under the first column are the values -0.1, -0.01, -0.001, 0.001, 0.01, and 0.1. Under the second column are the values 0.998334166, 0.9999833333, 0.999999833, 0.999999833, 0.9999833333, and 0.998334166.\">\n<caption>Average velocities using values of [latex]t[\/latex] approaching 0<\/caption>\n<thead>\n<tr valign=\"top\">\n<th>[latex]t[\/latex]<\/th>\n<th>[latex]\\frac{\\sin t - \\sin 0}{t-0}=\\frac{\\sin t}{t}[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>-0.1<\/td>\n<td>0.998334166<\/td>\n<\/tr>\n<tr>\n<td>-0.01<\/td>\n<td>0.9999833333<\/td>\n<\/tr>\n<tr>\n<td>-0.001<\/td>\n<td>0.999999833<\/td>\n<\/tr>\n<tr>\n<td>0.001<\/td>\n<td>0.999999833<\/td>\n<\/tr>\n<tr>\n<td>0.01<\/td>\n<td>0.9999833333<\/td>\n<\/tr>\n<tr>\n<td>0.1<\/td>\n<td>0.998334166<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1169739188902\">From the table we see that the average velocity over the time interval [latex][-0.1,0][\/latex] is 0.998334166, the average velocity over the time interval [latex][-0.01,0][\/latex] is 0.9999833333, and so forth. Using this table of values, it appears that a good estimate is [latex]v(0)=1[\/latex].<\/p>\n<p id=\"fs-id1169739298311\">By using the definition of a derivative, we can see that<\/p>\n<div id=\"fs-id1169739302610\" class=\"equation unnumbered\">[latex]v(0)=s^{\\prime}(0)=\\underset{t\\to 0}{\\lim}\\dfrac{\\sin t- \\sin 0}{t-0}=\\underset{t\\to 0}{\\lim}\\dfrac{\\sin t}{t}=1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739111046\">Thus, in fact, [latex]v(0)=1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169739220840\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169739220848\">A rock is dropped from a height of 64 feet. Its height above ground at time [latex]t[\/latex] seconds later is given by [latex]s(t)=-16t^2+64, \\, 0\\le t\\le 2[\/latex]. Find its instantaneous velocity 1 second after it is dropped, using the definition of a derivative.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q403817\">Hint<\/span><\/p>\n<div id=\"q403817\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739189236\">[latex]v(t)=s^{\\prime}(t)[\/latex]. Follow the earlier examples of the derivative using the definition of a derivative.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739298454\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739298454\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739298454\">-32 ft\/sec<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/VnDnInldaMM?controls=0&amp;start=1660&amp;end=1763&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\/3.1DefiningTheDerivative1660to1763_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.1 Defining the Derivative&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<p id=\"fs-id1169739305376\">As we have seen throughout this section, the slope of a tangent line to a function and instantaneous velocity are related concepts. Each is calculated by computing a derivative and each measures the instantaneous rate of change of a function, or the rate of change of a function at any point along the function.<\/p>\n<div id=\"fs-id1169739269974\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1169739269977\">The<strong> instantaneous rate of change<\/strong> of a function [latex]f(x)[\/latex] at a value [latex]a[\/latex] is its derivative [latex]f^{\\prime}(a)[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1169739187343\" class=\"textbook exercises\">\n<h3>Example: Rate of Change of Temperature<\/h3>\n<p id=\"fs-id1169739187352\">A homeowner sets the thermostat so that the temperature in the house begins to drop from [latex]70^{\\circ}\\text{F}[\/latex] at 9 p.m., reaches a low of [latex]60^{\\circ}[\/latex] during the night, and rises back to [latex]70^{\\circ}[\/latex] by 7 a.m. the next morning. Suppose that the temperature in the house is given by [latex]T(t)=0.4t^2-4t+70[\/latex] for [latex]0\\le t\\le 10[\/latex], where [latex]t[\/latex] is the number of hours past 9 p.m. Find the instantaneous rate of change of the temperature at midnight.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739297924\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739297924\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739297924\">Since midnight is 3 hours past 9 p.m., we want to compute [latex]T^{\\prime }(3)[\/latex]. Refer to the definition of a derivative.<\/p>\n<div id=\"fs-id1169739270561\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll}T^{\\prime}(3) & =\\underset{t\\to 3}{\\lim}\\frac{T(t)-T(3)}{t-3} & & & \\text{Apply the definition.} \\\\ & =\\underset{t\\to 3}{\\lim}\\frac{0.4t^2-4t+70-61.6}{t-3} & & & \\begin{array}{l}\\text{Substitute }T(t)=0.4t^2-4t+70 \\, \\text{and} \\\\ T(3)=61.6. \\end{array} \\\\ & =\\underset{t\\to 3}{\\lim}\\frac{0.4t^2-4t+8.4}{t-3} & & & \\text{Simplify.} \\\\ & =\\underset{t\\to 3}{\\lim}\\frac{0.4(t-3)(t-7)}{t-3} & & & =\\underset{t\\to 3}{\\lim}\\frac{0.4(t-3)(t-7)}{t-3} \\\\ & =\\underset{t\\to 3}{\\lim}0.4(t-7) & & & \\text{Cancel.} \\\\ & =-1.6 & & & \\text{Evaluate the limit.} \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1169739274032\">The instantaneous rate of change of the temperature at midnight is [latex]-1.6^{\\circ}\\text{F}[\/latex] per hour.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169739253544\" class=\"textbook exercises\">\n<h3>Example: Rate of Change of Profit<\/h3>\n<p id=\"fs-id1169739253553\">A toy company can sell [latex]x[\/latex] electronic gaming systems at a price of [latex]p=-0.01x+400[\/latex] dollars per gaming system. The cost of manufacturing [latex]x[\/latex] systems is given by [latex]C(x)=100x+10,000[\/latex] dollars. Find the rate of change of profit when 10,000 games are produced. Should the toy company increase or decrease production?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739274608\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739274608\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739274608\">The profit [latex]P(x)[\/latex] earned by producing [latex]x[\/latex] gaming systems is [latex]R(x)-C(x)[\/latex], where [latex]R(x)[\/latex] is the revenue obtained from the sale of [latex]x[\/latex] games. Since the company can sell [latex]x[\/latex] games at [latex]p=-0.01x+400[\/latex] per game,<\/p>\n<div id=\"fs-id1169739286461\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]R(x)=xp=x(-0.01x+400)=-0.01x^2+400x[\/latex].<\/div>\n<p id=\"fs-id1169739302924\">Consequently,<\/p>\n<div id=\"fs-id1169739302928\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P(x)=-0.01x^2+300x-10,000[\/latex].<\/div>\n<p id=\"fs-id1169739190488\">Therefore, evaluating the rate of change of profit gives<\/p>\n<div id=\"fs-id1169739190491\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}P^{\\prime}(10000)& =\\underset{x\\to 10000}{\\lim}\\frac{P(x)-P(10000)}{x-10000} \\\\ & =\\underset{x\\to 10000}{\\lim}\\frac{-0.01x^2+300x-10000-1990000}{x-10000} \\\\ & =\\underset{x\\to 10000}{\\lim}\\frac{-0.01x^2+300x-2000000}{x-10000} \\\\ & =100 \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736587942\">Since the rate of change of profit [latex]P^{\\prime}(10,000)>0[\/latex] and [latex]P(10,000)>0[\/latex], the company should increase production.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169739273077\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169739273085\">A coffee shop determines that the daily profit on scones obtained by charging [latex]s[\/latex] dollars per scone is [latex]P(s)=-20s^2+150s-10[\/latex]. The coffee shop currently charges [latex]\\$3.25[\/latex] per scone. Find [latex]P^{\\prime}(3.25)[\/latex], the rate of change of profit when the price is [latex]\\$3.25[\/latex] and decide whether or not the coffee shop should consider raising or lowering its prices on scones.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739242330\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739242330\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739242330\">[latex]P^{\\prime}(3.25)=20>0[\/latex]; raise prices<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1169736611546\">Use the previous example for a guide.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm162461\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=162461&theme=oea&iframe_resize_id=ohm162461&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\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-329\">\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>3.1 Defining the Derivative. <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":4,"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\":\"3.1 Defining the Derivative\",\"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-329","chapter","type-chapter","status-publish","hentry"],"part":35,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/329","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\/329\/revisions"}],"predecessor-version":[{"id":4802,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/329\/revisions\/4802"}],"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\/329\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=329"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=329"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=329"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=329"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}