Geometric proof of orthogonality of tangent vector and position vector.

Let $\gamma: I \to \mathbb{R}^2$ be a smooth curve from an open interval $I\subset \mathbb{R}$, which does not meet the origin. Let $\gamma(t_0)$ be the point on the curve closest to the origin. Then $\gamma(t_0)$ and $\gamma'(t_0)$ are orthogonal.

I know the proof which follows from observing that the function $t \mapsto \lvert\lvert(\gamma(t))\rvert\rvert$ is minimized at $t=t_0$ and the using the formula for the derivative. However, if one makes a sketch of the situation the theorem seems "obvious" and the above proof does not really supply the intuition for why the theorem is true. So I started thinking about a more geometric proof.

We can without loss of generality assume that $\gamma'(t_0)\neq 0$, since $0$ is orthogonal to any vector. Then we can construct a circle with radius $\lvert\lvert\gamma(t_0)\rvert\rvert$. This (if you sketch it) should seem like the tangent vector $\gamma'(t_0)$ should also be a tangent vector for the circle, and we know the result is true for the circle.

The part of this proof I haven't been able to iron out is verifying that the tangent vectors really are the same (or atleast parallel). Since we can assume that $\gamma'(t_0)\neq 0$ there should be a neighborhood around this point where the curve is regular and therefore reasonably well-behaved and then show that the difference between the tangent vectors get arbitrarily small; I have however been unable to do this.

So my question, can this proof be finished to show the (normalized) tangent vectors really are the same, or if there is some other geometric way to see the theorem?

• Sorry for responding so late. So writing $\gamma$ in polar gives $\gamma(t)=(x(t),y(t))=(r(t)\cos(\theta(t)),r(t)\sin(\theta(t))$ and should I then compute the tangent vector at $t_0$ using polar coordinates? – ThePuix May 10 '17 at 19:44
• I would write things as $\gamma(t)=r(t)\cdot (cos(\theta(t)),sin(\theta(t))$. Given this, just think about what your assumptions mean for the function $r(t)$ and then a little computation will finish things off. – Andreas Cap May 12 '17 at 6:27