While I was reading about motion with an emphasis on vectors I noticed that if we took the displacement vector from any point to another, (that is $\vec \Delta r = \vec r2 - \vec r1$ where $\vec r2$ is the position vector of the end point and $\vec r1$ is the position vector of the start point), we get something like a secant line. Now I thought perhaps to get the distance instead of the vector, we could differentiate the curve then integrate it from wherever to wherever. Another way of putting it is we make the time intervals over which we take $\vec \Delta r$ so small that $\vec \Delta r$ eventually traces the path of the particle instead of giving a secant line. Then we could add all these $\Delta r$'s together. Basically that would be like differentiating it then integrating it. But I'm not sure about this because I don't know what value the y-axis takes on because I'm not sure how to define "the path of a particle" in numerical values like meters etc. So I'd really like it if someone could tell me what the y-axis is in this case (x is time obviously) and what we would get upon differentiating it, and if the method I outlined to calculate the distance of the particle (differentiating then integrating) makes sense. If it is wrong, I'd like it if you could tell me exactly how to calculate the total distance travelled by a particle from A to B, given only the path of the particle.

  • $\begingroup$ I think you are talking about arclength, which, if given a rectifiable curve $\vec \gamma (t)$, is defined as $\int_{t_0}^{t_f} |\vec \gamma' (t)| dt$. $\endgroup$ – Rellek Dec 7 '15 at 15:48

The setup I think you're discussing is the idea that if we consider a particle with position $(x_1(t), x_2(t), x_3(t))$ at time $t$, then integrating $(\dot x_1, \dot x_2, \dot x_3)$ should give you $x$. That's true: $\int_{t_0}^{t_1} \dot x_i = x_i(t_1) - x_i(t_0)$, which is precisely the displacement along the $x_i$ axis between time $t_0$ and $t_1$. (If the multidimensional setup confuses you, just consider a point travelling on the real line.) That isn't the same as the distance travelled by the particle, though, unless it's moving along a straight line. For that, you want the length $L$ of the path $(x_i(t))$ from time $t_0$ to $t_1$, which is given by \begin{align*} L = \int_{t_0}^{t_1} dt\, \sqrt{\dot x_1^2 + \dot x_2^2 + \dot x_3^2} \end{align*} To derive that equation of $L$, divide $x$ into small line segments and note that the displacement of $x$ after a short time $\Delta t$ is approximately $(\dot x_1, \dot x_2, \dot x_3)\Delta t$. (There are some technical considerations I'm sweeping under the rug, but since this question was asked in the context of mechanics rather than analysis, I'm going to happily ignore them.)

  • $\begingroup$ A little confused but I think you've confirmed what I was thinking. Basically you're taking the displacement vector over every small $\Delta t$ and summing them up right? Do you know somewhere I can read more about this type of analysis (I just thought of the problem while reading some physics and I don't know if there's a field dealing with it or not) $\endgroup$ – Airdish Dec 8 '15 at 10:31
  • 1
    $\begingroup$ It's pretty standard multivariable calculus; any textbook should have the formula for arc-length, and most will give at least a heuristic justification for it, if not a full proof. On the physics side, this sort of thing is usually covered in a generic Newtonian mechanics class (actually, it sounds a bit similar to some ideas from Lagrangian mechanics, but that's a bit removed from Newtonain mechanics). $\endgroup$ – anomaly Dec 8 '15 at 15:28
  • $\begingroup$ Ah, multivariable calculus, haha well I'm in 10th grade so I guess I'll have to wait. Thanks for telling me where to look though. $\endgroup$ – Airdish Dec 8 '15 at 17:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.