Tangential and normal vectors in vector calculus

Question

I've understood that if we are given a curve $R(x,y,z)=xi + yj + zk$, $\frac{d}{dt}R(x,y, z) $ is a tangent vector to the curve. If we differentiate the tangent vector again, we obtain a vector that is normal to the curve (As it is normal to the tangent vector). Similarly, if R(x,y,z) defines the position of the body, $\frac{d}{dt}R(x,y, z) $ should give the equation of the velocity vector V(x,y,z) of the particle, which is tangential to the position curve. Using the above concept, shouldn't the acceleration, $\frac{d}{dt}V(x,y, z) $ just be normal to the position curve? Why do we manipulate $\frac{d}{dt}R(x,y, z) $ as $\frac{d}{ds}R(x,y, z) \frac{d}{dt}s $ (where s is the length of curve) and then differentiate it to arrive at two components of acceleration?

Answer 1

Hint: think of a point accelerating in a straight line (e.g. an object falling downwards). What happens if you differentiate the velocity vector?