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Math Help - Particle/polar question

  1. #1
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    Particle/polar question

    How do you find the rate of a particle that travels along a polar graph? I know that the slope of a polar graph is found through parametrics (and in some case in a rectagular form) adjusted into dy/dx. But the "rate" wouldn't make sense. For example, at a location where the slope is exactly vertical, you find the slope (aka rate) of the particle. The slope, of course, is infinite. But does that also mean that the particle's rate is infinite? If so, then the particle should basically jump off the polar graph but its distance from the origin is barely changing. What's up? Am I missing something?
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  2. #2
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    Quote Originally Posted by Kaitosan View Post
    How do you find the rate of a particle that travels along a polar graph? I know that the slope of a polar graph is found through parametrics (and in some case in a rectagular form) adjusted into dy/dx. But the "rate" wouldn't make sense. For example, at a location where the slope is exactly vertical, you find the slope (aka rate) of the particle. The slope, of course, is infinite. But does that also mean that the particle's rate is infinite? If so, then the particle should basically jump off the polar graph but its distance from the origin is barely changing. What's up? Am I missing something?
    This is a good question. I will try to explain it but let me know what parts of my explanation still don't make sense.

    When thinking of polar coordinates, like you said we find dy/dx through manipulation of parametric forms of x and y. But what does this mean? Just like with a normal derivative in Cartesian coordinates, dy/dx at x=a represents the slope at that point. If the slope becomes infinite for a point or even for multiple points, that means that the graph is like you said going straight up towards infinity at those points. But that doesn't mean that it will reach infinity immediately. Once the slope changes to something finite again, the graph will move a different direction.

    I would remember these two distinctions.

    1)f(a) is the position of "a" on the actual graph.
    2)f'(a) represents how fast the graph is increasing/decreasing at that point, but only at that point - not necessarily at points close to "a"
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    Quote Originally Posted by Jameson View Post
    If the slope becomes infinite for a point or even for multiple points, that means that the graph is like you said going straight up towards infinity at those points. But that doesn't mean that it will reach infinity immediately.
    Hm ok. So, an object with an infinite rate doesn't necessarily travel to infinity, but from a certain distance (even though it may be infinitely small at, say, a point) without any time used up. Maybe that's it.
    Last edited by Kaitosan; April 27th 2009 at 05:24 PM.
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    Quote Originally Posted by Kaitosan View Post
    Hm ok. So, an object with an infinite rate doesn't necessarily travel to infinity, but from a certain distance (even though it may be infinitely small at, say, a point) without any time used up. Maybe that's it.
    Right.

    The derivative is at a fixed point in time that is infinitely small. If you think of a normal x-y Cartesian plane where x is representing time and y is some function of time, then an infinite slope means that some event is occurring at different degrees at the same time (multiple y-values for one x-value). I think intuitively it makes sense that if something is traveling at an infinite speed it would immediately cover all possible points, but this way of thinking about infinity and slopes isn't how we apply this concept in this case. Usually we deal with infinity as a limit that isn't reached all at once, as a vertical asymptote.
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