showing that a function has a weak tangent but no strong tangent at t = 0

**oblixps** · September 27th 2011, 07:54 AM

in do carmo's differential geometry of curves and surfaces he has the following problem:

let $\alpha$ be a continuous and simple (one to one) curve. we say $\alpha$ has a weak tangent at $t_0$ if the line determined by $\alpha (t_0 + h)$ and $\alpha (t_0)$ has a limit position as h approaches 0. we say $\alpha$ has a strong tangent at $t_0$ if the line determined by $\alpha (t_0 + h)$ and $\alpha (t_0 + k)$ has a limit position as h, k approach 0. show that $\alpha(t) = (t^3, t^2)$ has a weak tangent but no strong tangent at t = 0.

so for the case of a weak tangent, i have that $\alpha(0) = (0, 0), \alpha(h) = (h^3, h^2)$ and $\alpha(h) - \alpha(0) = (h^3, h^2)$ . so $\lim_{h \to 0} \frac{1}{h} (h^3, h^2) = \lim_{h \to 0} (h^2, h) = (0, 0)$

in the case for a strong tangent i have $\alpha(h) = (h^3, h^2), \alpha(k) = (k^3, k^2)$ 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 to me that both have limit positions so both the weak and strong tangent exist at t = 0? i am not too clear on the definition he gives in the problem of weak and strong tangent. what does he exactly mean by "limiting position". if he means that the line will obtain a fixed "direction" then wouldn't neither a weak nor a strong tangent exist in this case since both lines in this case just go to (0, 0) so actually there isn't even a line at all, but just a point or the zero vector?

**Opalg** · September 27th 2011, 12:17 PM

Originally Posted by oblixps

in do carmo's differential geometry of curves and surfaces he has the following problem:

let $\alpha$ be a continuous and simple (one to one) curve. we say $\alpha$ has a weak tangent at $t_0$ if the line determined by $\alpha (t_0 + h)$ and $\alpha (t_0)$ has a limit position as h approaches 0. we say $\alpha$ has a strong tangent at $t_0$ if the line determined by $\alpha (t_0 + h)$ and $\alpha (t_0 + k)$ has a limit position as h, k approach 0. show that $\alpha(t) = (t^3, t^2)$ has a weak tangent but no strong tangent at t = 0.

so for the case of a weak tangent, i have that $\alpha(0) = (0, 0), \alpha(h) = (h^3, h^2)$ and $\alpha(h) - \alpha(0) = (h^3, h^2)$ . so $\lim_{h \to 0} \frac{1}{h} (h^3, h^2) = \lim_{h \to 0} (h^2, h) = (0, 0)$

in the case for a strong tangent i have $\alpha(h) = (h^3, h^2), \alpha(k) = (k^3, k^2)$ 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 to me that both have limit positions so both the weak and strong tangent exist at t = 0? i am not too clear on the definition he gives in the problem of weak and strong tangent. what does he exactly mean by "limiting position". if he means that the line will obtain a fixed "direction" then wouldn't neither a weak nor a strong tangent exist in this case since both lines in this case just go to (0, 0) so actually there isn't even a line at all, but just a point or the zero vector?

The limits that you have calculated are not the relevant ones. The weak and strong tangents are the limiting positions of lines, not points. What you should be looking at is whether the gradient of the line joining the given points tends to a limit.

The gradient of the line joining $\alpha(0)$ to $\alpha(h)$ is $\frac{h^2}{h^3} = \frac1h$ , which tends to infinity as h goes to 0. So the weak tangent is the vertical line through the origin.

The gradient of the line joining $\alpha(h)$ to $\alpha(k)$ is $\frac{h^2-k^2}{h^3-k^3} = \frac{h+k}{h^2+hk+k^2}.$ If for example k=–h then that fraction is 0. Use that fact to convince yourself that the limit as $(h,k)\to(0,0)$ does not exist.

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