{"id":245,"date":"2023-09-20T22:48:32","date_gmt":"2023-09-20T22:48:32","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/rate-of-change-applications\/"},"modified":"2024-08-05T12:43:24","modified_gmt":"2024-08-05T12:43:24","slug":"derivatives-as-rates-of-change-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/derivatives-as-rates-of-change-learn-it-2\/","title":{"raw":"Derivatives as Rates of Change: Learn It 2","rendered":"Derivatives as Rates of Change: Learn It 2"},"content":{"raw":"

Rate of Change Applications<\/h2>\r\n

Motion Along a Line<\/h3>\r\n

Another use for the derivative is to analyze motion along a line.<\/p>\r\n

Velocity is the rate of change of position. By taking the derivative of velocity, we can find acceleration, which is the rate of change of velocity. Additionally, it's important to introduce the concept of speed, which is the magnitude of velocity. Thus, we can state the following mathematical definitions:<\/p>\r\n

\r\n

velocity, speed and acceleration<\/h3>\r\n

Let [latex]s(t)[\/latex] be a function giving the position of an object at time [latex]t[\/latex].<\/p>\r\n

    \r\n\t
  • The velocity<\/strong> of the object at time [latex]t[\/latex] is given by [latex]v(t)=s^{\\prime}(t)[\/latex].<\/li>\r\n\t
  • The speed<\/strong> of the object at time [latex]t[\/latex] is given by [latex]|v(t)|[\/latex].<\/li>\r\n\t
  • The acceleration<\/strong> of the object at [latex]t[\/latex] is given by [latex]a(t)=v^{\\prime}(t)=s''(t)[\/latex].<\/li>\r\n<\/ul>\r\n<\/section>\r\n
    \r\n

    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't factor and it is a quadratic, the quadratic equation will always work.<\/p>\r\n<\/section>\r\n

    \r\n
    \r\n

    A ball is dropped from a height of [latex]64[\/latex] 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[caption id=\"\" align=\"aligncenter\" width=\"386\"]\"On Figure 2. Dropped ball graph, height vs. time.[\/caption]\r\n\r\n

      \r\n\t
    1. What is the instantaneous velocity of the ball when it hits the ground?<\/li>\r\n\t
    2. 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\"]<\/p>\r\n

      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 [latex]2[\/latex] seconds for the ball to reach the ground.<\/p>\r\n

        \r\n\t
      1. 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
      2. The average velocity of the ball during its fall is
        \r\n
        [latex]v_{avg}=\\frac{s(2)-s(0)}{2-0}=\\frac{0-64}{2}=-32[\/latex] ft\/s.<\/div>\r\n<\/li>\r\n<\/ol>\r\n

        [\/hidden-answer]<\/p>\r\n<\/section>\r\n

        \r\n

        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\t
        1. Find [latex]v(t)[\/latex].<\/li>\r\n\t
        2. At what time(s) is the particle at rest?<\/li>\r\n\t
        3. On what time intervals is the particle moving from left to right? From right to left?<\/li>\r\n\t
        4. 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\"]<\/p>\r\n

            \r\n\t
          1. The velocity is the derivative of the position function:
            \r\n
            [latex]v(t)=s^{\\prime}(t)=3t^2-18t+24[\/latex].<\/div>\r\n<\/li>\r\n\t
          2. 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
          3. 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.\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Figure 3. The sign of v(t) determines the direction of the particle.[\/caption]\r\n\r\n
            \u00a0<\/div>\r\n

            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.
            \r\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>\r\n<\/li>\r\n\t

          4. 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.\r\n[caption id=\"\" align=\"aligncenter\" width=\"484\"]\"A Figure 4. The path of the particle can be determined by analyzing [latex]v(t)[\/latex].[\/caption]\r\n<\/li>\r\n<\/ol>\r\n

            Watch the following video to see the worked solution to this example.<\/p>\r\n