Let $\alpha(t)$ be a parameterized curve which does not pass through the origin. If $\alpha(t_0)$ is a point of the trace of $\alpha$ closest to the origin and $\alpha'(t_0)\ne 0$, show that the position vector $\alpha(t_0)$ is orthogonal to $\alpha'(t_0)$
Now here was my thought pattern:
- What does it mean to move away from the origin?
Well it means that euclidean distance is increasing. When does this happen? When the curve moves into the circle with radius given by the euclidean distance of $\alpha(t_0)$ from the origin.
- How does this geometrically yield the answer?
Well it seems that the position vector points to our point, and $\alpha'(t_0)$ is the tangent to our circle, which definitely looks orthogonal to our position vector.
- How do I make this rigorous?
Compute a dot product for it? Use the circle some how, not really sure to be honest, this is what I want help with, hints are fine, as long as they are helpful :).
Best Answer
Take care that you speak of a circle while $\alpha(t)$ is a more general curve.
What I would answer:
Than the distance of a point $\alpha(t)$ of the curve to the origin is increasing. Precisely that $\Vert \alpha(t) \Vert$ is increasing.
If $\alpha(t_0)$ is a point of the trace of $\alpha$ closest to the origin, then $\Vert \alpha(t) \Vert^2$ is minimum.
Hence the derivative vanishes. And the derivative of $\Vert \alpha(t) \Vert^2$ is $2\alpha(t) \cdot \alpha^\prime(t)$.