I'm not sure why this is tripping me up but I'm not sure what tools to use.
Let $\alpha(t)$ be a parametrized curve which does not pass through the origin. If $\alpha(t_0)$ is the point of the trace of $\alpha$ closest to the origin and $\alpha'(t_0) \neq 0$, show that the position vector $\alpha(t_0)$ is orthogonal to $\alpha'(t_0)$.
So I need to show that $\alpha(t_0) \cdot \alpha'(t_0) = 0$, but I don't see where to go from here.
Thanks!
Best Answer
Let $f(t)=\alpha(t).\alpha(t)$. Notice this is the distance to the origin squared. We know the derivative of a function is zero at its minimum (if the derivative exists). So $f'(t_0)=2(\alpha'(t_0).\alpha(t_0))=0$