Intuitively predict the direction of the acceleration vector of a single parameter vector valued function The velocity vector of a Vector valued function (or position function), (for example $f(t)= \lt x(t),y(t) \gt$)
 seems to always be tangent to the curve of motion, its pretty intuitive why that is.
However intuition about what the acceleration vector $f ^{\prime \prime}(t)$ looks like or where does it point is hard to find.
Can I somehow predict or get an intuition on what the acceleration vector might be or where does it point to? how?.
 A: The acceleration vector is not uniquely determined by the curve, it depends on the way it is parametrized.  However, when the curve is parametrized by it's arclength, the acceleration vector of the curve is perpendicular to the velocity vector in 2 dimensions. In higher dimensions, this is no longer true.
The arclength of a curve is given by $s=\int_{t_1}^{t_2}dt\sqrt{x'(t)^2+y'(t)^2}$, where $t$ is any given parametrization. If the curve is given by $r(t)=(x(t),y(t))$ and it's arclength parametrization is $r(s)=(X(s), Y(s))$ then 
$$\hat{T}(s)=\Big(\frac{dX}{ds}, \frac{dY}{ds}\Big)=\frac{dr}{ds}=\frac{dr}{dt}\frac{dt}{ds}=\Big(\frac{x'(t)}{\sqrt{x'(t)^2+y'(t)^2}},\frac{y'(t)}{\sqrt{x'(t)^2+y'(t)^2}}\Big)$$
and we conclude that 
$$||\hat{T}(s)||=1$$
Now are allowed to write $\frac{dX}{ds}=\cos\theta(s),\frac{dY}{ds}=\sin\theta(s)$
and the acceleration of the curve can be written as
$$a(s)=\frac{dT(s)}{ds}=\theta'(s)(-\sin\theta, \cos\theta)=\frac{d\theta}{ds}\hat{n}$$
and it's easy to show that $T\cdot \hat{n}=0$ and thus the acceleration is perpendicular to the velocity in this case.
If the curve is parametrized by any other parameter, the direction of the acceleration is not perpendicular to the tangent vector anymore, since we can write $\frac{dx}{dt}=||v(t)||\cos\theta(t), \frac{dy}{dt}=||v(t)||\sin\theta(t)$ and thus 
$$a(t)=\frac{dv}{dt}=\frac{d||v||}{dt}(\cos\theta(t), \sin\theta(t))+||v||\frac{d\theta}{dt}(-\sin\theta(t), \cos\theta(t))=\frac{d||v||}{dt}\hat{T}+||v||\frac{d\theta}{dt}\hat{n}$$
which shows that the acceleration now has a component in the tangent direction as well, which is zero if $t=s$.
EDIT: In Newtonian physics, the trajectory of the particle is determined by the equation of motion $m\frac{d^2r(t)}{dt^2}=F$. In this case it is clear that this equation determines the direction of the acceleration of the particle, and the acceleration is not necessarily the same as the geometrical acceleration of the curve. In other words, the time parameter $t$ has a physical meaning and it's relationship to the arclength $s=s(t)$ is determined by the equations of motion and it's not chosen freely, like in differential geometry. One has to solve the equations of motion to be able to tell the relationship between the curve and the body's acceleration. However, the last equation written above for $a(t)$ is correct, and it represents the decomposition of the particle's acceleration to tangential and normal components.
