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Homework Help: Plotting Derivatives

  1. Jan 19, 2010 #1
    The problem statement.

    Suppose x''(t) = 1 for [tex]1\leq(t)\leq2[/tex], and x''(t) = 0 for all other (t)

    (a) Plot x''(t) for [tex]0\leq(t)\leq3[/tex]
    (b) Plot x'(t) for [tex]0\leq(t)\leq3[/tex]. Assume x'(0) = 0
    (c) Plot x(t) for [tex]0\leq(t)\leq3[/tex]. Assume x(0) = 0

    The attempt at a solution

    I assumed 'x' being the vertical axis and 't' being the horizontal axis.

    For (a) I know that there are going to be two points at 1 and two points at 0.

    My main question is when I graph these plots should I treat points 0 and 3 on the t-axis as discontinuities, and just put a point of where they're at and not include them when connecting the non-zero points, or should I connect all the points together, despite the discontinuity?
  2. jcsd
  3. Jan 19, 2010 #2


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    Hi jayeffarr! :smile:

    (have a ≤ :wink:)
    Personally, I'd leave them disconnected …

    x''(t) has only one value for each t, so why pretend it has more? :wink:
  4. Jan 19, 2010 #3
    Would you say to do the same thing when it comes to plotting x' and x...

    Since x'(t) = t

    and the plot will be (0,0), (1,1), (2,2), (3,0).

    And x(t) = (1/2)t²

    and the plot will be (0,0), (1,0.5), (2,2), (3,0).

    ...and just draw a line between 1 & 2, while having solid circles and 1 & 3?

    Or are you saying just to plot the points without connecting ANY of them?
    Last edited: Jan 19, 2010
  5. Jan 20, 2010 #4


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    Not following you :redface:

    x'(t) is the area under the graph of x''(t).​
  6. Jan 20, 2010 #5


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    I don't know what you mean by "two points". The graph is the horizontal straight line at y= 1 between 1 and 2, the horizontal straight line at y= 0, the x-axis, for all other t.

    "at 3 on the t-xis"? x"(3)= 0 and is 0 for every point near t= 3. Did you mean t= 2? In any case, it really doesn't matter. Technically, the vertical line is not part of the graph but if you can notice that this forms a rectangle under the graph, that will help.
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