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the biggest block of text in this picture

In the image, it says the whole path can be determined by knowing u(t1) and u'(t1) at any point t1. As far as I know, using u(t1) and u'(t1), the best we can do is approximate a nearby point u(t1+del t)= u(t1)+ u'(t1)*del t. Don't we need all the derivatives to calculate the whole path using Taylor polynomial?

Also, what kind of derivative is u'(t) here? The way I understand it, u(t) is the configuration function of the body. The output of u(t) is six numbers (center of mass co-ordinates, and three euler angles). How do we even take the derivative of such a function? Isn't derivative 'change in output' divided by infinitesimal change in input. How is change in output defined here?

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  • $\begingroup$ Please type out the relevant part of the text you want to quote rather than posting a picture. That way the information is easier to find should someone else search for a similar question. Also, it is pretty well known in introductory differential equations that specifying the initial conditions of a function and it's derivative at a point in time is sufficient for determining a unique solution to a second order ordinary differential equation. $\endgroup$ – Aaron Stevens Feb 13 at 5:04
  • $\begingroup$ @AaronStevens It's the big block of text in the pic. Also, I didn't know there was also a diff equation. I thought it was just the value and the derivative. Can you tell me what kind of derivative are we using here? $\endgroup$ – Ryder Rude Feb 13 at 5:13
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    $\begingroup$ @RyderRude Yes, Aaron was aware of that - however, it's general policy here to write out any quoted text rather than posting images or screencaps of it. Not only is it easier to read for users with different screen settings, but it allows the information to be indexed for easier searching. It also forces you to think about the quote word by word, which may help to focus your question or even to answer it for you. $\endgroup$ – J. Murray Feb 13 at 5:28
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I am assuming you are referring to the part that talks about Newton's laws of motion are second order in time. This is talking about the differential equation $$\mathbf F=m\mathbf {\ddot x}$$ where each dot above $\mathbf x$ is shorthand for a time derivative. The equation is second order because the highest order derivative is a second derivative. It is a well known property of second order ordinary differential equations that specifying $\mathbf x(t_1)$ and $\mathbf {\dot x}(t_1)$ is sufficient to determine a unique solution to this differential equation.

Now, in introductory physics classes you just deal with position $\mathbf x$, but when you start dealing with Lagrangian mechanics you can use generalized coordinates $$\mathbf q(t)=(q_1(t), q_2(t),\dots ,q_n(t))$$ But it can be shown that this method is equivalent to Newton's laws, so the idea of a second order differential equation still holds.

These generalized coordinates are the coordinates of our configuration space. So when we talk about "paths" in this context we don't necessarily have to be talking about actual paths in physical space. We are talking about a path in this configuration space defined by the generalized coordinates

Also, all derivatives here are showing respect to time.

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  • $\begingroup$ How does one calculate such derivatives? Each component of u is separately changing with time. Are we talking about something like the gradient or the Jacobian? $\endgroup$ – Ryder Rude Feb 13 at 6:17
  • $\begingroup$ @RyderRude I've been assuming you mean "q" instead of "u", but you keep using "u" so maybe I am missing something here... It's just the usual process of calculating derivatives. Like what you would learn in a normal calculus class $\endgroup$ – Aaron Stevens Feb 13 at 6:20
  • $\begingroup$ I actually meant q. Thanks. Do we separately calculate the derivative of each component of q wrt t? $\endgroup$ – Ryder Rude Feb 13 at 6:24
  • $\begingroup$ @RyderRude Yes. Just like any other vector, $\mathbf{\dot q}=(\dot q_1,\dots,\dot q_n)$ $\endgroup$ – Aaron Stevens Feb 13 at 6:25
  • $\begingroup$ I guess this assumes and initial value theorem to hold? $\endgroup$ – innisfree Feb 13 at 7:49

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