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Showing that a function does not have a strong tangent

  1. Sep 25, 2011 #1
    my book defines a weak tangent as one where the line through [itex]\alpha(t_0 + h) [/itex] and [itex] \alpha(t_0) [/itex] has a limit position when [itex] h \rightarrow 0 [/itex]. they define a strong tangent as one where the line through [itex]\alpha(t_0 + h) [/itex] and [itex] \alpha(t_0 + k) [/itex] has a limit position when [itex] h, k \rightarrow 0 [/itex].

    i am trying to show that the curve [itex] \alpha(t) = (t^3, t^2) [/itex] has a weak tangent at t = 0 but no strong tangent there.

    i have that [itex] \alpha(0) = (0, 0) [/itex] and [itex] \alpha(h) = (h^3, h^2) [/itex]. i then have that [itex] \alpha(h) - \alpha(0) = (h^3, h^2) [/itex] and that [itex] \lim_{h \to 0} \frac{1}{h} (h^3, h^2) = \lim_{h \to 0} (h^2, h) = (0, 0) [/itex]. so a weak tangent exists.

    analogously i try for the strong tangent with: [itex] \alpha(h) = (h^3, h^2) [/itex] and [itex] \alpha(k) = (k^3, k^2) [/itex] and [itex] \alpha(h) - \alpha(k) = (h^3 - k^3, h^2 - k^2) [/itex]. Then [itex] \lim_{h,k \to 0} \frac{1}{h-k} (h^3 - k^3, h^2 - k^2) = \lim_{h,k \to 0} (h^2 + hk + k^2, h + k) = (0, 0) [/itex].

    but it seems like there is a strong tangent there since the limit exists. have i made a mistake somewhere?
    Last edited: Sep 26, 2011
  2. jcsd
  3. Sep 25, 2011 #2
    differential geometry of curves and surfaces by Do Carmo?

    i think that the error is that you're dividing [itex]\alpha (h)- \alpha (k)[/itex] by "the norm of h" instead of the norm of the vector [itex](h,k)[/itex]
  4. Sep 26, 2011 #3
    haha yes i am using do carmo's book. when i use the norm of (h, k) it seems like i still get 0 after using the squeeze theorem for both components though.

    i am not sure what to divide [itex] \alpha(h) - \alpha(k) [/itex] by. in the case of [itex] \alpha(h) - \alpha(0) [/itex] i sort of intuitively chose to divide by h. but for the case of the strong tangent i am not sure what to do. i know that [itex] (h-k, h^3 - k^3, h^2 - k^2) [/itex] is the difference vector between the 2 points and is kind of like the "slope" which characterizes a line. in the first case i divided by h because that was the change in x and slope is change in y over change in x. however in the 2nd case i'm not sure why i should divide by the norm of (h, k).
  5. Sep 26, 2011 #4
    I'm sorry, i made a mistake. I don't remember so good the definition of strong tangent. I googled for it, and i found that a function has strong tangent is the following limit exists:

    [tex]\lim_{h,k\to 0} \dfrac{\alpha(t_0+h)-\alpha(t_0+k)}{||\alpha(t_0+h)-\alpha(t_0+k)||}[/tex]

    if we think [itex]\alpha(t)[/itex] as a vector, the expression above is the difference vector divided by his norm. The intuitive aproach of that expression, is to think that the unitary tangent vector has a limit position (the fraction is not the unitary tangent, but it aproach's to it) It's difficult to me see geometrically why that curve doesn't have strong tangent.

    I hope that the last expression is correct hahaha
  6. Sep 26, 2011 #5
    this may be related. for the next part of this same question he asks to prove that for a C1 curve that is regular at the point t = t0 then it must have a strong tangent there. In the hint he provides he uses [itex] \frac{\alpha(t_0 + h) - \alpha(t0 + k)}{h-k} [/itex] and says that using the mean value theorem it approaches [itex] \alpha'(t) [/itex]. Because [itex] \alpha'(t_0) \not= 0 [/itex] the line determined by [itex] \alpha(t_0 + h) [/itex] and [itex] \alpha(t_0 + k) [/itex] converge to the line determined by [itex] \alpha'(t) [/itex].

    for the curve [itex] \alpha(t) = (t^3, t^2) [/itex] it is clear that [itex] \alpha'(t) = (0, 0) [/itex] but it seems like [itex] \frac{\alpha(h) - \alpha(k)}{h-k} = \frac{1}{h-k} (h^3 - k^3, h^2 - k^2) = (h^2 + hk + k^2, h + k) [/itex] and when h, k approach 0 it seems to me that [itex] (h^2 + hk + k^2, h + k) [/itex] approaches (0, 0). but according to his hint, since [itex] \alpha'(0) [/itex] then [itex] \frac{\alpha(t_0 + h) - \alpha(t0 + k)}{h-k} [/itex] should not converge to (0, 0) but in this case it seems like it does. i am very confused on this.
  7. Dec 31, 2013 #6
    strong and weak tangents

    it may be a bit late to offer insights on this but if anyone is interested mine are below. (this problem had bugged me for some time as I was unable to prove it using the definition given in Q7 of the Do Carmo book - Diff Geom of Curves and Surfaces.) The following is a proof that α(t)=(t3,t2) does not have a strong tangent at t=0 using the definition given in the question. I hope the proof makes sense.

    Since this is only a 2D parameterization I have used Cauchy's Mean-Value Theorem in 2D (see http://mathworld.wolfram.com/CauchysMean-ValueTheorem.html). This is:

    f'(t0)/g'(t0) = (f(t0+h) - f(t0+k)) / (g(t0+h) - g(t0+k))
    where in our case:f(t) = t^3 and g(t) is t^2.
    h and k are small values away from the point t=0. It is helpful to think of h being in the positive direction and k being in the negative direction from t0. Hence the interval between t0+h and t0+k includes the point t0 which in our case is 0 (this means k is negative)

    expanding out [ f(t0+h) - f(t0+k) ] and setting t0=0 we get [f(t0+h) - f(t0+k)] = (h^3 - k^3)

    now, since both f'(t0) and g'(t0) are zero at t0 = 0, [f(t0+h) - f(t0+k)] is indeterminate, which means (h^3 - k^3) is also indeterminate and we can choose h and k as whatever we like. for example h can approach zero as t^-10 and k can approach as t^-2 or any other possibility. Hence there is no unique tangent line at t=0. (NOTE here that indeterminate here does not mean infinity, since if it did, then it means that
    [g(t0+h) - g(t0+k)] = h^2 - k^2 =0 and (h^3 - k^3)!=0 thus we must choose h=-k which gives a unique tangent line as h,k approach 0)

    If however, the original question was α′(t) = (t^3,t) for example, now g'(t0=0) = 1. Then [f(t0+h) - f(t0+k)] = h^3 - k^3 = 0. Now h and k can only converge in a single specific way (since h=k) and the tangent line is unique at t=0. (By the way all this could have been made much simpler by recognizing that that α′(t) is singular at t=0 and the curve is y = x^2/3 which has a cusp at t=0. Then the tangent line can be either horizontal or vertical)

    in question 7b this idea is extended to 3 dimensions. If we get back to 2D for a bit it is easy to see that the for any regular curve that is differentiable at least once (i.e. C1 curve) at t0, we can always have some interval around t0 where the Cauchy Mean-Value Theorem is valid since f'(t0)/g'(t0) is always determinate for regular curves.

    Hope this was useful. I would like to know what you think about this line of reasoning (i.e. correct/incorrect, maybe, etc....)
    Last edited: Dec 31, 2013
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