Tangent vector to a curve

Question

I am trying to relate things simply. If a curve is on a flat 2D space represented by the parameter $\lambda$. In polar coordinate system $(r,\theta)$ at any lambda the tangent vector components are $$V^1=\frac{dr(\lambda)}{d\lambda}$$ and $$V^2=\frac{d\theta(\lambda)}{d\lambda}$$ and therefore the tangent vector at any lambda is $$V=V^1\hat r+V^2\hat \theta$$ ( for one end of the curve at the origin where $\lambda=0$ and $\lambda$ increases along the curve).Then if the curve is geodesic then $V^1=1$ and $V^2=0$ in which the tangent vector is simply $V=\hat r$ .Now taking it's covariant derivative along the tangent vector e.g$$\nabla_\hat r V=N$$which is 0 by definition for geodesic.(e.g any $\nu^{th}$ component of N is 0) implying $$(\nabla_\hat r V)^{\nu}=0.....(1).$$Now further how can (1) be proved in general as this expression is known $$(\nabla_\mu A)^{\nu}=\partial_\mu A^{\nu}+\Gamma^\nu_{\mu\lambda}A^\lambda$$ Here as we have assigned 1 to be for $\hat r$ and 2 to be for $\hat \theta$ so (1) it's can be proved. But is there any general expression of proving (1) (e.g where the tangent vector for geodesic is not pointing along any of the coordinate direction unlike the case for here)

Answer 1

There is a tangent space at each point $P$ in the space. It is a vector space consisting of the tangents to all possible smooth curves that pass through $P$. Any particular curve has only a single tangent vector at $P$, so it gives just one of the infinite number of vectors in that tangent space.