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What does the plot of $f'(x)$ against $f(x)$ represent?

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  • $\begingroup$ In one-dimensional motion, take x to represent time, and f position, so you are discussing a plot of velocity versus position. $\endgroup$ Dec 7, 2020 at 21:50
  • $\begingroup$ I understand that, but what exactly does the velocity against position plot describe? $\endgroup$
    – GitGoodCodes
    Dec 7, 2020 at 21:56
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    $\begingroup$ Describe? A portal to undefined profundity? How did you define "describe"? $\endgroup$ Dec 7, 2020 at 21:58
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    $\begingroup$ This is not a physics question. $\endgroup$
    – G. Smith
    Dec 7, 2020 at 22:04
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    $\begingroup$ FWIW, it can be quite illuminating to plot speed vs displacement. Try it with a body in an elliptical orbit. Feynman has a nice discussion about this (maybe in Six Easy Pieces). $\endgroup$
    – PM 2Ring
    Dec 7, 2020 at 22:11

3 Answers 3

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In conservative and non-conservative dynamical system analysis such as ( Limit cycles, Van der Pol equation of oscillation, Lissajou's figures involving vibration in two directions/frequencies etc.) DEs of two, three variables are together considered coupled or decoupled. They occur in mechanical or electrical dynamic systems.

They can be represented as phase portraits which are displacement/velocity plots. We see how much deviation is there from simplest harmonic model situation (ellipse) due to oscillations in-phase, out of phase or partial offset at start points of prescribed boundary values.The plot can visually display effects of stiffness or damping.

In Lyapunev Stability $( \dot x, \ddot x) $ of state variables we see elliptic points, saddle points etc., characterizing the dynamics coming out of the governing ODE or transfer function.

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Phase portraits are a characteristic of the entire dynamical process.

Van der Pol Oscillator

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It represents the derivative in terms of the original function. Without context, that's as much as I can answer.

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  • $\begingroup$ Lets suppose $f(x)$ is displacement and $x$ is time. $\endgroup$
    – GitGoodCodes
    Dec 7, 2020 at 21:34
  • $\begingroup$ Well assuming that the displacement vs time function is either monotonically increasing of decreasing, then $f$ vs $f'$ would represent the linear position vs velocity graph. I say monotonically because if your function $f(x)$ increases then decreases, then for the same displacement, it will have 2 different possible values of velocity. So you'd have to figure out which speed you need. $\endgroup$
    – user256872
    Dec 7, 2020 at 21:39
  • $\begingroup$ I see! That makes sense. Thank you. $\endgroup$
    – GitGoodCodes
    Dec 7, 2020 at 21:57
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In general, the only thing you can tell from a velocity-displacement graph is what the velocity is given the displacement, and vice-versa (although if the graph isn't one-to-one, then we may have various different values of displacement for a given value of velocity).

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