Lecture 12: Reference frames, Coriolis force, and Hurricanes

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Sasha Tchekhovskoy

Sasha Tchekhovskoy

Күн бұрын

Пікірлер: 6
@aname1362
@aname1362 Жыл бұрын
Can someone advise me what is meaning of full-time derivative and why it doesn't affect Equations of Motions? (at 18.28) (BTW these are great lectures)
@diegobautistaaviles4182
@diegobautistaaviles4182 4 ай бұрын
I hope you get to read this comment even if it’s one year later hehehe. A full-time derivative is a derivative that contemplates both explicit and implicit change in time. Imagine a function that describes the temperature of a room at each point, and at each moment. Thus, the function for the temperature of the room is something of the form T(x,t). As an example, take T = x^2 + t. So, in a pure calculus context, if I asked you how that function changes in time, the only logical answer would be to give me the partial derivative of the function with respect to time: ∂T/∂t. In this specific example: ∂T/∂t = 1. Now, in mechanics, we have an extra ingredient: motion. For example, consider we describe the motion of a bug through the room by a function x(t). The temperature experienced by this bug is then T = T(x(t),t). Note that the position parameter of the function also depends on t. For example, if x(t) = sin t, then T = sin^2 t + t. If I asked you for the time derivative of this temperature, you can’t just simply give me the partial derivative as before (∂T/∂t) = 1), because as time passes the temperature experienced by the bug may change both through the sole time depence of t and by the motion of the bug. It is necessary to introduce a new derivative that contemplates the changes through motion. We call this new derivative the total time derivative. We denote it by dT/dt. One way to calculate it is to take the time derivative of the function you are studying after substituting the explicit time dependence of position. I’ve done that before, so in this case, it follows that dT/dt = 2 sin t * cos t + 1. Now, the reason why you take out the time derivative of the Lagrangian is the theorem proved before in the video.
@Vagran
@Vagran 2 жыл бұрын
I have to know, were you writing backwards or is this a mirrored reflection?
@atchekho
@atchekho 2 жыл бұрын
Hey, I was mirroring it in a video editor
@Vagran
@Vagran 2 жыл бұрын
@@atchekho Ah excellent, it's a great presentation by the way.
@HelloWorlds__JTS
@HelloWorlds__JTS 2 жыл бұрын
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