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Homework Help: Show curvature of circle converges to curvature of curve @ 0

  1. Feb 9, 2017 #1
    1. The problem statement, all variables and given/known data
    Let γ : I → ℝ2 be a smooth regular planar curve and assume 0 ∈ I. Take t ≠ 0 in I such that also −t ∈ I and consider the unique circle C(t) (which could also be a line) containing the 3 points γ(0), γ(−t), γ(t). Show that the curvature of C(t) converges to the curvature κ(0) of γ at t = 0 when t → 0. This also shows that the curvature is a geometric quantity, i.e. parametrization invariant

    2. Relevant equations
    κ = det{γ',γ''}/{absval(γ')}3

    3. The attempt at a solution
    Like the previous 2 problems I've posted about tonight I'm unsure whether this should be done in general (taking γ to be as general a representation of a smooth regular planar curve (whatever that may be) as possible) or if I should be specific (i.e., parameterize γ by t, perhaps like an ellipse <acos(t), bsin(t)>.

    I'm not sure where to start exactly if it's the general case but I would imagine I would take C(t) to be a circle? (parameterized like gamma is but with a=b) then find the curvature at each point and then also find the curvature of the curve at t = 0 then take the limit of the curvature of C(t) as t approaches 0 and try to show they are equivalent?

    If anyone can help me either conceptually on whether I can do this problem through a specific example or rather if it needs be proven in a more abstract way ( and how one does that) or in a procedural sense, or in other words if my method seems to be on the right track. Thanks in advance for anyone who can offer me any help. This homework is due on friday so I'll be on much of the day tomorrow (Feb 10th) to answer anyone who responded. Thanks again!
  2. jcsd
  3. Feb 9, 2017 #2


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    I would determine the radius, hence the curvature, of the circle through your three points ##\gamma(t),~\gamma(0),~\gamma(-t)## for the general case. Don't assume the original curve is an ellipse or anything particular. Then check what happens as ##t \to 0##. The case where the three points are collinear would be handled separately.
  4. Feb 9, 2017 #3
    Hmm okay, so in this general case would the radius just be t? since it seems that our interval is symmetric with γ(0) in the middle and γ(±t) on either end of the interval. The curvature can be thought of as 1 over the radius of the circle which inscribes the curve so in this case the curvature of the unique circle would just be 1/t? Hence at t = 0 the curvature blows up to infinity? I must be missing something here or just overly simplifying the process that needs to happen
  5. Feb 9, 2017 #4


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    No. ##t## is just the parameter for the curve.

    I have no idea what you are talking about here. You have a parametric planar curve ##\gamma(t) = \langle x(t),y(t) \rangle## and three points ##\gamma(\pm t),~\gamma(0)##. Three points determine a circle. You need to figure out that circle and get its radius. Then you can worry about what happens as ##t \to 0##. It might get a bit messy with the required algebra. I haven't worked it out and don't intend to, but you asked for guidance and that is my suggestion.
  6. Feb 9, 2017 #5
    Right. I guess where the disconnect is happening with me is I'm unsure how to calculate the radius of a circle in general (when we're saying given the idea that γ(±t) are two points rather than saying something like γ(±2)). Now If what I need to do is put in a point t = "whatever" and then carry on from there I understand but what I dont understand is how I should go about calculating the radius of a circle where the two points outside of 0 are not specified. If we are indeed leaving it in terms of t (like it seems like you're suggesting) then what I was saying was that the radius would then obviously depend on t and the curvature would depend inversely to t.

    I appreciate the information thus far it's definitely been illuminating LCKurtz, I'm just hung up on this one particular notion it seems. Perhaps I'm looking at it from the complete wrong perspective?
  7. Feb 10, 2017 #6


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    If I asked you for the radius of the circle through ##(1,1),~(2,4)## and ##(5,2)##, how would you do it? You need to do the same with your three points. Yes, your answer will involve ##t##, but it's not the radius.
  8. Feb 11, 2017 #7
    Hmm... so I know there are a number of ways to do this but I think I remember one particular way. We take these points and label them P =(1,1), Q = (2,4) and R = (5,2) then we want the slopes of PR and PQ so we find

    Slope PR = (2-1)/(5-1) = 1/4
    Slope PQ = (4-1)/(2-1) = 3

    So now we want two lines to be perpendicular to PQ and PR. So we find two Lines L1 and L2 which have negative reciprocal slopes of PQ and PR and which meet at a point we call S.

    Slope L1 ⊥ PQ = -1/3
    Slope L2 ⊥ PR = -4

    Now we can find equations for L1 nd L2 using point-slope form and the points Q and R respectively

    L1 = y-4 = (-1/3)(x-2) ... (1)
    L2 = y-2 = (-4)(x-5) ... (2)

    Now we can solve for x and y so...

    (2)-(1) ⇒ x = 52/11 ≅ 4.72727272

    now sub x = 52/11 into, say, (2) and find...

    y = 34/11

    Thus we know the coordinates of the point S where the two perpendicular bisectors meet is S = (52/11, 34/11)

    And based on the way we constructed this whole argument we can say that PS is the diameter of the circle and then we can write the eq...

    P = (1,1), S = (52/11, 34/11)

    (x-1)(x-{52/11}) + (y-1)(y-{34/11}) = 0 and then you basically multiply through and group by like power and you have your equation, correct?

    Also if this is indeed correct this is then basically something that I carry out at the points γ(0) = <x(0), y(0)> and γ(±t) = <x(±t), y(±t)> and then I'll end up with an equation (which most likely involves lots of t's) which I can then extract the radius from and then the curvature?
  9. Feb 11, 2017 #8


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    There are some mistakes in your analysis above. You need the point where the perpendicular bisectors of ##PQ## and ##QR## meet to find the center. When you get it correct you should find the center of circle in that example to be ##(\frac {63}{22},\frac{45}{22})##. And I have to tell you, given the algebra involved in this example, I would expect it to be quite messy when you do your general problem. But, in principle, when you do it with the ##t##'s, it should work.
  10. Feb 11, 2017 #9
    Are you sure my analysis is wrong? I only ask because upon reinspection of the procedure and having drawn out the picture for myself as well I'm pretty sure its right. I've included a design of the process here for your inspection.. I don't see anything wrong. You take each line from one point (here being P) to each of the other two points (Q and R) so P is the tail of both then you take the perpendicular bisector (or maybe not bisector but you take a ray that's perpendicular to PQ(PR) at the point Q(R)) and then the point where those two intersect is the other point which satisfied and equation for the diameter of the circle when paired with P. Is this not correct?

    But yes you're right about the huge mess part. I've basically set it up so that my 3 points are P = γ(0) = <x(0), y(0)> and Q(+)/R(-) = γ(±t) = <x(±t), y(±t)>
    then I end up getting this for my slopes:

    - ∇PQ = {y(t)-y(0)}/{x(t)-x(0)} and
    - ∇PR = {y(-t)-y(0)}/{x(-t)-x(0)}

    and then I found the slopes for L1 and L2 by taking the neg reciprocal slope of each and found the equations...

    - L1 = y - y(t) = -[{x(t)-x(0)}/y(t)-y(0)](x-x(t)) [1]
    - L2 = Same as L1 just with (-t) instead of (t) [2]

    So when I subtracted [1]-[2] I found:

    - L1-L2 = y(-t) - y(t) = [{x(-t)2 - x(-t)x(0) - x(-t)x + x(0)x}/{y(-t)-y(0)}] - [{x(t)2 - x(t)x(0) - x(t)x + x(0)x}/{y(t)-y(0)}]

    Where I want to solve for x;

    I put it in wolframalpha to solve cuz (F THAT) and it did give me an answer but it was so long and disgusting that I almost dont believe that its the right answer let alone want to substitute it back into an equation to find y. This just seems too convoluted to be able to analyze when t→0 and I'm getting a little discouraged. Can anyone let me know if I am indeed on the right track and that this is just par for the course with some of these problems or should I be seeking a more elegant solution?

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  11. Feb 12, 2017 #10


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    We are talking about this example, right?
    I don't follow that, and I don't see what the picture you enclosed has to do with the above example. Are you saying that PS is a diameter in your picture? If so, I don't see why it would be.

    [Edit, added later] OK, I do see why PS is a diameter. Still...

    And yes, I am sure your answer is wrong for my example. Here is a Maple plot showing the points, center, and the circle with center at ##(\frac {63}{22},\frac{45}{22})##. If you don't get the correct answer for this example, no sense in going for the general case yet. See the picture below.


    So you first job is to get this example correct. Then go for the general case. When you get to that stage, I'm guessing you are going to need the mean value theorem to get the derivatives in your answer. For example, when you have an expression like ##x(t)-x(0)##, by the MVT you could write that as ##(t - 0)x'(c)## where ##c## is a point between ##0## and ##t##. And in the limit, terms like that will go to ##0##. Still, I give you the caveat that I haven't worked it out myself, and there may in fact be some more clever way to work it out.
    Last edited: Feb 12, 2017
  12. Feb 13, 2017 #11
    I ended up doing it my way and did get the correct answer though it seemed a bit wonky and less straight-forward than I would have hoped. Perhaps your way was more elegant however I wanted to just work with what I was positive worked, I'm sure you understand. I really appreciate your continued assistance in helping to think differently about this problem and consider all perspectives, LCKurtz. Have a good one!
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