{"id":3011,"date":"2025-08-15T20:05:13","date_gmt":"2025-08-15T20:05:13","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=3011"},"modified":"2025-08-15T20:29:39","modified_gmt":"2025-08-15T20:29:39","slug":"3011","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/3011\/","title":{"raw":"Derivatives: Learn It 3","rendered":"Derivatives: Learn It 3"},"content":{"raw":"

Finding Instantaneous Rates of Change<\/h2>\r\nMany applications of the derivative involve determining the rate of change at a given instant of a function with the independent variable time\u2014which is why the term instantaneous<\/em> is used. Consider the height of a ball tossed upward with an initial velocity of 64 feet per second, given by [latex]s\\left(t\\right)=-16{t}^{2}+64t+6[\/latex], where [latex]t[\/latex] is measured in seconds and [latex]s\\left(t\\right)[\/latex] is measured in feet. We know the path is that of a parabola. The derivative will tell us how the height is changing at any given point in time. The height of the ball is shown as a function of time. In physics, we call this the \"s<\/em>-t<\/em> graph.\"\r\n\r\n\"Graph\r\n\r\n
Using the function, [latex]s\\left(t\\right)=-16{t}^{2}+64t+6[\/latex], what is the instantaneous velocity of the ball at 1 second and 3 seconds into its flight?[reveal-answer q=\"14141\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"14141\"]\r\n\r\nThe velocity at [latex]t=1[\/latex] and [latex]t=3[\/latex] is the instantaneous rate of change of distance per time, or velocity. Notice that the initial height is 6 feet. To find the instantaneous velocity, we find the derivative<\/strong> and evaluate it at [latex]t=1[\/latex] and [latex]t=3:[\/latex]\r\n

[latex]\\begin{align}{f}^{\\prime }\\left(a\\right)&=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h} \\\\ &=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{-16{\\left(t+h\\right)}^{2}+64\\left(t+h\\right)+6-\\left(-16{t}^{2}+64t+6\\right)}{h}&& \\text{Substitute }s\\left(t+h\\right)\\text{ and }s\\left(t\\right). \\\\ &=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{-16{t}^{2}-32ht-{h}^{2}+64t+64h+6+16{t}^{2}-64t - 6}{h}&& \\text{Distribute}. \\\\ &=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{-32ht-{h}^{2}+64h}{h}&& \\text{Simplify}. \\\\ &=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{\\cancel{h}\\left(-32t-h+64\\right)}{\\cancel{h}}&& \\text{Factor the numerator}. \\\\ &=\\underset{h\\to 0}{\\mathrm{lim}}-32t-h+64&& \\text{Cancel out the common factor }h. \\\\ {s}^{\\prime }\\left(t\\right)&=-32t+64&& \\text{Evaluate the limit by letting }h=0. \\end{align}[\/latex]<\/p>\r\nFor any value of [latex]\\begin{align}t,{s}^{\\prime }\\left(t\\right)\\end{align}[\/latex] tells us the velocity at that value of [latex]t[\/latex].\r\n\r\nEvaluate [latex]t=1[\/latex] and [latex]t=3[\/latex].\r\n

[latex]\\begin{align}&{s}^{\\prime }\\left(1\\right)=-32\\left(1\\right)+64=32 \\\\ &{s}^{\\prime }\\left(3\\right)=-32\\left(3\\right)+64=-32 \\end{align}[\/latex]<\/p>\r\nThe velocity of the ball after 1 second is 32 feet per second, as it is on the way up.\r\n\r\nThe velocity of the ball after 3 seconds is [latex]-32[\/latex] feet per second, as it is on the way down.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

The position of the ball is given by [latex]s\\left(t\\right)=-16{t}^{2}+64t+6[\/latex]. What is its velocity 2 seconds into flight?[reveal-answer q=\"327969\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"327969\"]0[\/hidden-answer]<\/section>\r\n

Using Graphs to Find Instantaneous Rates of Change<\/h2>\r\nWe can estimate an instantaneous rate of change at [latex]x=a[\/latex] by observing the slope of the curve of the function [latex]f\\left(x\\right)[\/latex] at [latex]x=a[\/latex]. We do this by drawing a line tangent to the function at [latex]x=a[\/latex] and finding its slope.\r\n\r\n
How To: Given a graph of a function [latex]f\\left(x\\right)[\/latex], find the instantaneous rate of change of the function at [latex]x=a[\/latex].<\/strong>\r\n
    \r\n \t
  1. Locate [latex]x=a[\/latex] on the graph of the function [latex]f\\left(x\\right)[\/latex].<\/li>\r\n \t
  2. Draw a tangent line, a line that goes through [latex]x=a[\/latex] at [latex]a[\/latex] and at no other point in that section of the curve. Extend the line far enough to calculate its slope as\r\n
    [latex]\\frac{\\text{change in }y}{\\text{change in }x}[\/latex].<\/div><\/li>\r\n<\/ol>\r\n<\/section>
    From the graph of the function [latex]y=f\\left(x\\right)[\/latex], estimate each of the following:\r\n
      \r\n \t
    1. [latex]f\\left(0\\right)[\/latex]<\/li>\r\n \t
    2. [latex]f\\left(2\\right)[\/latex]<\/li>\r\n \t
    3. [latex]\\begin{align}f^{\\prime}\\left(0\\right)\\end{align}[\/latex]<\/li>\r\n \t
    4. [latex]\\begin{align}f^{\\prime}\\left(2\\right)\\end{align}[\/latex]<\/li>\r\n<\/ol>\r\n\"Graph\r\n\r\n[reveal-answer q=\"606117\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"606117\"]\r\n\r\nTo find the functional value, [latex]f\\left(a\\right)[\/latex], find the y<\/em>-coordinate at [latex]x=a[\/latex].\r\n\r\nTo find the derivative<\/strong> at [latex]x=a[\/latex], [latex]\\begin{align}{f}^{\\prime }\\left(a\\right)\\end{align}[\/latex], draw a tangent line at [latex]x=a[\/latex], and estimate the slope of that tangent line.\r\n\r\n\"Graph\r\n
        \r\n \t
      1. [latex]f\\left(0\\right)[\/latex] is the y<\/em>-coordinate at [latex]x=0[\/latex]. The point has coordinates [latex]\\left(0,1\\right)[\/latex], thus [latex]f\\left(0\\right)=1[\/latex].<\/li>\r\n \t
      2. [latex]f\\left(2\\right)[\/latex] is the y<\/em>-coordinate at [latex]x=2[\/latex]. The point has coordinates [latex]\\left(2,1\\right)[\/latex], thus [latex]f\\left(2\\right)=1[\/latex].<\/li>\r\n \t
      3. [latex]\\begin{align}{f}^{\\prime }\\left(0\\right)\\end{align}[\/latex] is found by estimating the slope of the tangent line to the curve at [latex]x=0[\/latex]. The tangent line to the curve at [latex]x=0[\/latex] appears horizontal. Horizontal lines have a slope of 0, thus [latex]{f}^{\\prime }\\left(0\\right)=0[\/latex].<\/li>\r\n \t
      4. [latex]\\begin{align}{f}^{\\prime }\\left(2\\right)\\end{align}[\/latex] is found by estimating the slope of the tangent line to the curve at [latex]x=2[\/latex]. Observe the path of the tangent line to the curve at [latex]x=2[\/latex]. As the [latex]x[\/latex] value moves one unit to the right, the [latex]y[\/latex] value moves up four units to another point on the line. Thus, the slope is 4, so [latex]\\begin{align}{f}^{\\prime }\\left(2\\right)\\end{align}=4[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>
        Using the graph of the function [latex]f\\left(x\\right)={x}^{3}-3x[\/latex], estimate: [latex]f\\left(1\\right)[\/latex], [latex]\\begin{align}{f}^{\\prime }\\left(1\\right)\\end{align}[\/latex], [latex]f\\left(0\\right)[\/latex], and [latex]\\begin{align}{f}^{\\prime }\\left(0\\right)\\end{align}[\/latex].\"Graph[reveal-answer q=\"107267\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"107267\"][latex]-2,0,0,-3[\/latex][\/hidden-answer]\r\n\r\n<\/section>","rendered":"

        Finding Instantaneous Rates of Change<\/h2>\n

        Many applications of the derivative involve determining the rate of change at a given instant of a function with the independent variable time\u2014which is why the term instantaneous<\/em> is used. Consider the height of a ball tossed upward with an initial velocity of 64 feet per second, given by [latex]s\\left(t\\right)=-16{t}^{2}+64t+6[\/latex], where [latex]t[\/latex] is measured in seconds and [latex]s\\left(t\\right)[\/latex] is measured in feet. We know the path is that of a parabola. The derivative will tell us how the height is changing at any given point in time. The height of the ball is shown as a function of time. In physics, we call this the “s<\/em>–t<\/em> graph.”<\/p>\n

        \"Graph<\/p>\n

        Using the function, [latex]s\\left(t\\right)=-16{t}^{2}+64t+6[\/latex], what is the instantaneous velocity of the ball at 1 second and 3 seconds into its flight?<\/p>\n