Here is(are) the question(s).
Let $f(t): (-t_0-\epsilon, t_0 +\epsilon)\to \Bbb R^3$ be a unit speed parametrized curve in $\Bbb R^3$. Suppose that $k(f(t_0)) > 0$, where $k$ is the curvature function. EDIT: I have a feeling the domain should be $(t_0-\epsilon,t_0+\epsilon)$.
Prove there exists a unique unit speed circle $g(t)$ such that $f(t_0) = g(t_0)$ and $ \lim_{t\to t_0} \frac{|f(t)-g(t)|}{(t-t_0)^2}=0$. EDIT: The homework problem says it should be $f(t_0) = g(t_1)$, but it was pointed out that that is probably wrong.
Conclude that for this circle $k(f(t_0)) = k(g(t_0))$.
Find $g(t)$ for $f(t)=(cos(t),sin(t),t)$ and $t_0=\frac \pi 4$.
Really three questions in one. I'm at a total loss. Best I have is that, since curvature at $t_0$ isn't $0$ then $f'(t)$ and $f''(t)$ are not co-incident. EDIT: In fact, they are orthogonal, because $f(t)$ is unit speed. This is much stronger. I think it should be possible to show that the circle must lie in the plane spanned by $f'(t)$ and $f''(t)$.
My gut tells me that the circle must lie in the same plane as the neighbourhood of $t_0$.
I'm really way out of my depth here. Any sort of insight will help me out!
Thanks.
