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Math Help - Is the midpoint of an elipse's "cord", through the center, the midpoint of the curve?

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    Member mfetch22's Avatar
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    Is the midpoint of an elipse's "cord", through the center, the midpoint of the curve?

    Consider an elipse centered at (0, 0). It intersects the x-axis at (a, 0) and (-a, 0), and it intersects the y-axis at (0, b) and (0, -b). Construct a line segment from point (a, 0) to point (0, b), and call it segment A. Now, mark off the point that is the midpoint of this segment, call it M. Finnally, draw a line segment starting at the origin (0, 0), the center of the elipse, which intersects A at M, and draw this segement until it intersects the elipse and call this point on the elipse P. Is it neccesarily true, or not neccesarily true, that the point P is the "midpoint" of the distance along the elipse from (0, b) to (a, 0)?
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    Quote Originally Posted by mfetch22 View Post
    Consider an elipse centered at (0, 0). It intersects the x-axis at (a, 0) and (-a, 0), and it intersects the y-axis at (0, b) and (0, -b).

    Is it necessarily true, or not necessarily true, that the point P is the "midpoint" of the distance along the ellipse from (0, b) to (a, 0)?
    Consider what you are really asking: “the "midpoint" of the distance along the ellipse from (0, b) to (a, 0)”
    Written in parametric form the ellipse is <a\cos(t),b\sin(t)>.
    So when t=0 we get (a,0) while t=\frac{\pi}{2} we get (0,b).
    Therefore, the midpoint P is when t=\frac{\pi}{4}

    Now the midpoint between (a,0)~\&~(0,b) is \left( {\frac{a}{2},\frac{b}{2}} \right).

    Now you work out the answer to this question.
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    Member mfetch22's Avatar
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    Quote Originally Posted by Plato View Post
    Consider what you are really asking: “the "midpoint" of the distance along the ellipse from (0, b) to (a, 0)”
    Written in parametric form the ellipse is <a\cos(t),b\sin(t)>.
    So when t=0 we get (a,0) while t=\frac{\pi}{2} we get (0,b).
    Therefore, the midpoint P is when t=\frac{\pi}{4}

    Now the midpoint between (a,0)~\&~(0,b) is \left( {\frac{a}{2},\frac{b}{2}} \right).

    Now you work out the answer to this question.
    Wait, I may have confused you with my wording. When I said the "midpoint" along the elipse from (0, b) to (a, 0), I meant the point G along the elipse such that if the arc-length along the elipse from (0, b) to point G is Q, and the arclength along the elipse from G to (a, 0) is W; then such that Q = W. Now, this may be exactly the midpoint you described with t = \frac{\pi}{4}, I was simply trying to ensure that we were talking about the same "midpoint".

    In other words, I didn't know weather or not (in a parametrically described equation of a curve) the following was true:


    If a parametric representation of an elipse obtains the value of:

    (x_1, y_1)

    at time t_1, and also obtains the value:

    (x_2, y_2)

    at time t_2

    Then the arc-length midpoint (midpoint of the distance along the elipse) from point (x_1, y_1) to (x_2, y_2) is the point returned from the parametric representation by letting:

    t = \frac{t_1 + t_2}{2}
    Now, I'm not sure if the above is true, but I'm showing you what I thought your were saying. I could've interpreted it wrong. I'm starting to see this problem more clearly now though. Don't worry, I'm not just gona throw my hands up and let you do all the work. I think I know how to check weather the above is correct. So, I'll do some work, and you let me know if its correct, then once you post back, I'll then apply it to the original problem (I also interested now in the above statement in quotation).



    Okay, soooo basically if the arclength midpoint can be found by diving up the time intervals, then this would imply that a point tracing out an elipse over time t would have to travel along the elipse with a constant speed, correct? So, to show weather the arclength midpoint between to points along the elipse can be found this way (you probably know weather it can or cannot, but I'm doing the work myself because I don't want to be one of those posters who simply expects others to do the work for them), then I need only show that the point tracing out the elipse moves at a constant speed. I know how to find the derivitve of functions....

    Given the parametric representation of an elipse, I know how to find the rate at which it moves in the x direction with respect to t, or the y direction ...... but how can I find the rate at which it moves "in the direction of x and y"? What I mean is, how can I find the rate, with respect to time, at which the point travles along the arc length of the elipse? Please don't simply give me the answer, I like figuring these things out, but I do need some sort of direction at this point.

    Wait,.... I 've got a thought... the distance something travles along a line, can be found with perpediculars of x, and y, correct? Obviously:

    D^2 = x^2 + y^2

    Now, I don't know how this works with the calculus... maybe:

    (\frac{dD}{dt})^2 = (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2

    Is that correct, or no?
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    Quote Originally Posted by mfetch22 View Post
    I 've got a thought... the distance something travels along a line, can be found with perpendiculars of x, and y, correct? Obviously:

    D^2 = x^2 + y^2

    Now, I don't know how this works with the calculus... maybe:

    (\frac{dD}{dt})^2 = (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2

    Is that correct, or no?
    That is the correct formula for using calculus to compute arc length. Unfortunately, when you try to do this with an ellipse, you get an elliptic integral, which cannot be computed in terms of manageable functions. So you will not be able to get a formula for the length of an elliptic arc.

    The answer to the original question is that the two arcs on the ellipse will not have the same length. Here is a heuristic argument to explain why.

    The ellipse with semi-axes a and b is obtained from a circle of radius a by compressing the y-axis by a factor of b/a, so that the point (a\cos\theta,a\sin\theta) on the circle becomes the point (a\cos\theta,b\sin\theta) on the ellipse. On the circle, the arc from \theta=0 to \theta=\pi/4 (the green arc in the attachment) has the same length as the red arc from \theta=\pi/4 to \theta=\pi/2. But the green arc is oriented in a more vertical direction than the red arc. So when the y-axis is scrunched down to transform the circle into the ellipse, the green arc gets shortened more than the red one. Therefore the arc AP on the ellipse is shorter than the arc BP.

    That is not a rigorous mathematical argument, of course, but I think it's pretty obvious from the picture that AP is indeed shorter than BP.
    Attached Thumbnails Attached Thumbnails Is the midpoint of an elipse's &quot;cord&quot;, through the center, the midpoint of the curve?-ellipse.gif  
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