showing that a function does not have a strong tangent

[demonelite123]

my book defines a weak tangent as one where the line through \alpha(t_0 + h) and \alpha(t_0) has a limit position when h \rightarrow 0 . they define a strong tangent as one where the line through \alpha(t_0 + h) and \alpha(t_0 + k) has a limit position when h, k \rightarrow 0 .

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

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

analogously i try for the strong tangent with: \alpha(h) = (h^3, h^2) and \alpha(k) = (k^3, k^2) and \alpha(h) - \alpha(k) = (h^3 - k^3, h^2 - k^2) . Then \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) .

but it seems like there is a strong tangent there since the limit exists. have i made a mistake somewhere?

[felper]

differential geometry of curves and surfaces by Do Carmo?

i think that the error is that you're dividing \alpha (h)- \alpha (k) by "the norm of h" instead of the norm of the vector (h,k)

[demonelite123]

differential geometry of curves and surfaces by Do Carmo?

i think that the error is that you're dividing \alpha (h)- \alpha (k) by "the norm of h" instead of the norm of the vector (h,k)

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 \alpha(h) - \alpha(k) by. in the case of \alpha(h) - \alpha(0) 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 (h-k, h^3 - k^3, h^2 - k^2) 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).

[felper]

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 \alpha(h) - \alpha(k) by. in the case of \alpha(h) - \alpha(0) 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 (h-k, h^3 - k^3, h^2 - k^2) 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).

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:

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

if we think \alpha(t) 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

[demonelite123]

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 \frac{\alpha(t_0 + h) - \alpha(t0 + k)}{h-k} and says that using the mean value theorem it approaches \alpha'(t) . Because \alpha'(t_0) \not= 0 the line determined by \alpha(t_0 + h) and \alpha(t_0 + k) converge to the line determined by \alpha'(t) .

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