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Method of Increments?

  1. Jun 7, 2010 #1
    I am using this book: https://www.amazon.com/Calculus-Int...sr_1_1?ie=UTF8&s=books&qid=1275951079&sr=1-1"
    Calculus: An Intuitive and Physical Approach by: Kline

    I've already finished Calc I and II, but I'm brushing up on my skills because I don't feel entirely confident about them.

    This is what the method says to do to find a derivative:

    [tex]s = 10t^2[/tex] given [tex]t=3[/tex]

    [tex]s_{3} + \Delta s = 10(3 + \Delta t)^2[/tex]

    [tex]s_{3} + \Delta s = 90 + 60\Delta t + 10\Delta t^2[/tex]
    [tex] - (s_{3} = 90) [/tex]


    [tex]\frac{\Delta s}{\Delta t} = \frac{60 \Delta t + 10 \Delta t^2}{\Delta t}[/tex]

    [tex]\displaystyle{\frac{\Delta s}{\Delta t}} = 60 + 10 \Delta t [/tex]

    [tex]lim_{\Delta t \rightarrow 0} \displaystyle{\frac{\Delta s}{\Delta t}} = 60[/tex]

    I understand the method and have finished all of the practice problems in the book, but I'm having trouble linking this with the way I was taught to derive in my calculus class with

    [tex]lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}[/tex]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 7, 2010 #2

    Mark44

    Staff: Mentor

    I really hate his abuse of notation, using s3 for s(3). There seem to be a lot of steps missing, which might make it difficult to follow his logic.

    For example, [itex]\Delta s[/itex] = s(3 + [itex]\Delta t[/itex]) - s(3) = 10(3 + [itex]\Delta t[/itex])2 - 90
    = 90 + 60 [itex]\Delta t[/itex] + 10 ( [itex]\Delta t[/itex])2 - 90
    = 60 [itex]\Delta t[/itex] + 10 ( [itex]\Delta t[/itex])2

    Dividing by delta t gives you this, the same as above:
    [tex]\frac{\Delta s}{\Delta t}} = 60 + 10 \Delta t [/tex]

    Finally, take the limit to get ds/dt.

    Delta t in his exposition is the same as h in what you're used to. Hopefully, my explanation will help you understand the parallels between the two approaches.
     
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