Problem:
Solution:
The squared distance between $ \alpha(t) $ and the origin is $ \alpha(t) \cdot \alpha(t) $. It is a differentiable function of $ t $ and it reaches minimum at $ t_0 $, so $ (\alpha(t_0) \cdot \alpha(t_0))' = 0 $
Using the product rule, we get $ \alpha'(t_0) \cdot \alpha(t_0) = 0 $, now both vectors are non zero so they must be orthogonal.
Solution:
The squared distance between $ \alpha(t) $ and the origin is $ \alpha(t) \cdot \alpha(t) $. It is a differentiable function of $ t $ and it reaches minimum at $ t_0 $, so $ (\alpha(t_0) \cdot \alpha(t_0))' = 0 $
Using the product rule, we get $ \alpha'(t_0) \cdot \alpha(t_0) = 0 $, now both vectors are non zero so they must be orthogonal.
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