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I know that taking the derivative of position with respect to time defines what we call velocity, but I've never heard of physicist going in the opposite direction with position. Is there any application for the integral of position with respect to time?

Maybe with respect to something else?

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closed as too broad by Kyle Kanos, CuriousOne, Bill N, Gert, ACuriousMind Mar 17 '16 at 12:27

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Example: center of mass. $\endgroup$ – Count Iblis Mar 15 '16 at 20:27
  • $\begingroup$ An integral of position, something like $\int_a^b \textrm{d}x$ is merely the distance $b-a$. Other integrals like $\int_a^b f(x)\textrm{d}x$ is just some physical quantity dependent on $f(x)$. $\endgroup$ – K7PEH Mar 15 '16 at 20:27
  • $\begingroup$ They appear in some autocorrelation functions. $\endgroup$ – lemon Mar 15 '16 at 20:29
  • $\begingroup$ An interesting forum discussion on this topic with one example of an application. scienceforums.net/topic/… $\endgroup$ – Arturo don Juan Mar 15 '16 at 20:40
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    $\begingroup$ This post (v1) seems to be essentially a list question. $\endgroup$ – Qmechanic Mar 15 '16 at 21:29
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One such application is in a PID controller.

In the simplest sense, a PID controller tries to keep the position of some input as close as possible to a known position, by using control input to control the position. As someone who plays Kerbal Space Program, my favourite application is the guidance system which tries to rotate a rocket to point in a specific direction.

A PID controller looks at 3 aspects of the input position in order to control it.

  • Proportional: This is the position itself. For example, if the input is to the left of the desired position, input will be applied to move right.
  • Integral: This is the integral of the position, i.e. where the position has tended to be in the past. For example, if the input has tended to sit to the left of target in the past, input will be applied to move right (to keep it from drifting).
  • Derivative: This is the derivative of the position, i.e. the velocity. For example, if the input is moving to the left, input will be applied to move right (to slow the movement).

These 3 aspects are summed together to provide the final control input.

One way to look at it is:

  • Proportional takes into account where the position is in the present.
  • Integral takes into account where the position was in the past.
  • Derivative takes into account where the position is predicted to be in the future.
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If you consider the mean position of an object over a interval of time to be important, then yes, absolutely. The integral,

$$\int_{t_0}^{t_1} \vec r(t) \ dt=\Delta t \cdot \bar \mu_{\vec r}$$

Where $\Delta t$ is the change in time, $\vec r(t)$ is the posistion as a function of time, and $\mu_{\vec r}$ is the mean position over the time interval. Which is to say,

$$\bar \mu_{\vec r}=\cfrac{1}{\Delta t} \cdot \int_{t_0}^{t_1} \vec r(t) \ dt$$

So the mean position of an object is directly proportional to its integral over a fixed time period.

You said that perhaps your interested in integrals with respect to "something else". Then look no farther than this useful formula,

$$\Delta t=\Delta x \cdot \left(\int_{x_0}^{x_1} \frac{1}{v(x)} \ dx \right)$$

This comes directly from the formula for the Harmonic Mean. For instance, if you travel up a hill a distance of $x$ at a speed $v_1$ and down a hill a distance of $x$ at a speed $v_2$, what was your average speed? You have to use the Harmonic Mean. The above is just a generalization.

Exercise for the reader:

Its commonly told to students that the Conservation of Energy Equation doesn't tell you how much time elapses during a evolution from point $x_0$ to $x_1$. However, this is a slight misnomer. If you know that velocity of the object is at every point of its path, you can figure out how long it took to go from $x_0$ to $x_1$.

$$\Delta T+\Delta U=0$$ $$\Rightarrow \frac{1}{2} \cdot m \cdot v_f^2+\Delta U=0, \quad v_0=0$$ $$\Rightarrow v_f(x)=\sqrt{-\frac{2}{m} \cdot \Delta U}$$

Given the above formula, find the time it takes for a particle dropped from a height of $h$ to fall straight down to a height of zero.

Hint: Notice that this formula gives the velocity as a function of position.

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