{"id":1372,"date":"2024-04-16T15:27:34","date_gmt":"2024-04-16T15:27:34","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&p=1372"},"modified":"2024-08-30T19:38:36","modified_gmt":"2024-08-30T19:38:36","slug":"a-preview-of-calculus-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/a-preview-of-calculus-learn-it-3\/","title":{"raw":"A Preview of Calculus: Learn It 3","rendered":"A Preview of Calculus: Learn It 3"},"content":{"raw":"

The Tangent Problem and Differential Calculus Cont.<\/h2>\r\n

Calculating Instantaneous Velocity<\/h3>\r\n

Let's continue our investigation by exploring a related application, instantaneous velocity. Velocity may be thought of as the rate of change of position.<\/p>\r\n

Suppose that we have a function, [latex]s(t)[\/latex], that gives the position of an object along a coordinate axis at any given time [latex]t[\/latex]. Can we use these same ideas to create a reasonable definition of the instantaneous velocity at a given time [latex]t=a[\/latex]?<\/p>\r\n

We start by approximating the instantaneous velocity with an average velocity.\u00a0<\/p>\r\n

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Recall that the speed of an object traveling at a constant rate is the ratio of the distance traveled to the length of time it has traveled.<\/p>\r\n<\/section>\r\n

We define the average velocity<\/strong> of an object over a time period to be the change in its position divided by the length of the time period.<\/p>\r\n

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average velocity<\/h3>\r\n

Let [latex]s(t)[\/latex] be the position of an object moving along a coordinate axis at time [latex]t[\/latex]. The average velocity<\/strong> of the object over a time interval [latex][a,t][\/latex] where [latex]a < t[\/latex] (or [latex][t,a][\/latex] if [latex]t < a[\/latex]) is<\/p>\r\n

[latex]v_{\\text{avg}}=\\dfrac{s(t)-s(a)}{t-a}[\/latex]\u00a0<\/div>\r\n<\/section>\r\n

As [latex]t[\/latex] is chosen closer to [latex]a[\/latex], the average velocity becomes closer to the instantaneous velocity.<\/p>\r\n

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Note that finding the average velocity of a position function over a time interval is essentially the same as finding the slope of a secant line to a function. Furthermore, to find the slope of a tangent line at a point [latex]a[\/latex], we let the [latex]x[\/latex]-values approach [latex]a[\/latex] in the slope of the secant line.<\/p>\r\n<\/section>\r\n

To find the instantaneous velocity at time [latex]a[\/latex], we let the [latex]t[\/latex]-values approach [latex]a[\/latex] in the average velocity. This process of letting [latex]x[\/latex] or [latex]t[\/latex] approach [latex]a[\/latex] in an expression is called taking a limit<\/strong>.\u00a0<\/p>\r\n

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instantaneous velocity<\/h3>\r\n

For a position function [latex]s(t)[\/latex], the instantaneous velocity<\/strong> at a time [latex]t=a[\/latex] is the value that the average velocities approach on intervals of the form [latex][a,t][\/latex] and [latex][t,a][\/latex] as the values of [latex]t[\/latex] become closer to [latex]a[\/latex], provided such a value exists.<\/p>\r\n<\/section>\r\n

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A rock is dropped from a height of [latex]64[\/latex] ft. It is determined that its height (in feet) above ground [latex]t[\/latex] seconds later (for [latex]0\\le t\\le 2[\/latex]) is given by [latex]s(t)=-16t^2+64[\/latex].<\/p>\r\n

Find the average velocity of the rock over each of the given time intervals. Use this information to guess the instantaneous velocity of the rock at time [latex]t=0.5[\/latex].<\/p>\r\n

    \r\n\t
  1. [latex][0.49,0.5][\/latex]<\/li>\r\n\t
  2. [latex][0.5,0.51][\/latex]<\/li>\r\n<\/ol>\r\n
    [reveal-answer q=\"fs-id1170573428397\"]Show Solution[\/reveal-answer]
    \r\n[hidden-answer a=\"fs-id1170573428397\"]\r\n\r\n

    Substitute the data into the formula for the definition of average velocity.<\/p>\r\n

    [latex]\\begin{array}{rcl} s(0.5) &=& -16(0.5)^2 + 64 = 60 \\\\ s(0.49) &=& -16(0.49)^2 + 64 = 60.1584 \\\\ s(0.51) &=& -16(0.51)^2 + 64 = 59.8384 \\\\ v_{\\text{avg}} \\text{ for interval [0.49, 0.5]} &=& \\frac{s(0.5) - s(0.49)}{0.5 - 0.49} = \\frac{60 - 60.1584}{0.01} = -15.84 \\\\ v_{\\text{avg}} \\text{ for interval [0.5, 0.51]} &=& \\frac{s(0.51) - s(0.5)}{0.51 - 0.5} = \\frac{59.8384 - 60}{0.01} = -16.16 \\end{array}[\/latex]<\/center>
    <\/center>\r\n

    The instantaneous velocity is somewhere between [latex]\u221215.84[\/latex] and [latex]\u221216.16[\/latex] ft\/sec. A good guess might be [latex]\u221216[\/latex] ft\/sec.<\/p>\r\n\r\nWatch the following video to see the worked solution to this example.