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brinethery Group Title

Quick question about derivatives involving Newton's dot notation and the chain rule: I was just wondering if, say you have a velocity function in terms of t, would that be considered x-dot? So for example, if they give x=2t and that's in terms of velocity, it's x-dot? And then the derivative of that for acceleration would be x-double dot?

  • one year ago
  • one year ago

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  1. brinethery Group Title
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    Furthermore, say you have a velocity position function y=x^2. Would the chain rule for that be y = 2x*(x-dot)? If you're confused by when I say x-dot, I'm talking about this: http://web.mst.edu/~reflori/be150/Dyn%20Lecture%20Videos/Particle%20Kinem%20

    • one year ago
  2. inkyvoyd Group Title
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    uhh, it's currently being updated. I'll read up on x-dot and see what I think is the answer.

    • one year ago
  3. brinethery Group Title
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    Oh, it is? Interesting. I'm studying dynamics this quarter and the notation is very new and confusing to me.

    • one year ago
  4. inkyvoyd Group Title
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    What do you mean by x=2t is given in terms of velocity?

    • one year ago
  5. brinethery Group Title
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    It's a parametric equation

    • one year ago
  6. inkyvoyd Group Title
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    okay, is that the full details?

    • one year ago
  7. inkyvoyd Group Title
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    if you have a y parameter and you are looking for the velocity it's defined differently.

    • one year ago
  8. brinethery Group Title
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    So you'll have an x-function in terms of t, and then a y-function in terms of x. The xy-function is position.

    • one year ago
  9. brinethery Group Title
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    Just a sec, I can post the few pages out of Hibbeler's book. Hold on.

    • one year ago
  10. inkyvoyd Group Title
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    Yes. Then, the velocity is actually defined to be sqrt ((x dot)^2+(y dot)^2). Although I probalby have less qualification than you to say this.

    • one year ago
  11. brinethery Group Title
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    • one year ago
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  12. brinethery Group Title
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    Example 12.10 gives a lot of insight, but I'm still not completely clear on everything.

    • one year ago
  13. brinethery Group Title
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    12.9 I mean

    • one year ago
  14. inkyvoyd Group Title
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    ohh - what are you having trouble with?

    • one year ago
  15. brinethery Group Title
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    I was wondering if I was given a parametric equation in terms of x and t that is velocity, if that would be considered x-dot.

    • one year ago
  16. inkyvoyd Group Title
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    if there was just x and t. If there were more variables, such as in this example, it is not just x-dot.

    • one year ago
  17. inkyvoyd Group Title
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    What is important to know is that in this example y is given in terms of x, but in reality we see y as a function of t. In a way, it's a composite function. If you are talking about just moving on say a number line, you would only have one direction to move in, and x-dot would be the velocity.

    • one year ago
  18. inkyvoyd Group Title
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    but the example has y, which is really afunction of t, although it is given in terms of x. So you need to use the two-dimensional definition of velocity.

    • one year ago
  19. brinethery Group Title
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    what I got stuck on is after you do the chain rule for the y-function in terms of x... and then you're plugging in your x, x-dot, and x-double-dot if you were to accidentally plug in x-dot for x. Sorry if it sounds like I'm speaking in riddles!

    • one year ago
  20. brinethery Group Title
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    I think I'll go in and bug my instructor tomorrow morning :-)

    • one year ago
  21. inkyvoyd Group Title
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    instead of using x-dot try to use \(V_x\) if you are allowed to,

    • one year ago
  22. inkyvoyd Group Title
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    x-dot is a simple way of saying the change of x with relation to t.

    • one year ago
  23. inkyvoyd Group Title
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    just have to be clear on what the variables are and what they mean. Y is the vertical *position* of the point, and X is the horizontal *position* of the point. t is time. We are interested in measuring the change of both x and y with respect to t in a way such that it makes sense.

    • one year ago
  24. brinethery Group Title
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    Exactly

    • one year ago
  25. brinethery Group Title
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    I thank you for your help and I'll have something to think about on this now. But I've gotta go make dinner and work on the physics that's been piling up. I'll be back later to check this thread. Thank you again.

    • one year ago
  26. inkyvoyd Group Title
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    Alright. Good luck!

    • one year ago
  27. brinethery Group Title
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    Let me know if you have any new insight :-)

    • one year ago
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