The two equivalent parallel velocity vectors

[Dinheiro]

This is an exercise from the textbook Apostol Vol 1 (page 525, second edition), and I do not know how to prove it:
Suppose a curve C is described by two equivalent functions X and Y, where Y(t) = X[u(t)].
Prove that at each point of C the velocity vectors associated with X and Y are parallel, but
that the corresponding acceleration vectors need not be parallel.
I would really appreaciate some enlightenment.

 

[fresh_42]

This is an exercise from the textbook Apostol Vol 1 (page 525, second edition), and I do not know how to prove it:
Suppose a curve C is described by two equivalent functions X and Y, where Y(t) = X[u(t)].
Prove that at each point of C the velocity vectors associated with X and Y are parallel, but
that the corresponding acceleration vectors need not be parallel.
I would really appreaciate some enlightenment.

How are velocity and acceleration vectors defined at a certain point, i.e. how would you compute them?

 

[Dinheiro]

Here's what I tried:
Let r(t) be the position vector and T(t) such that:
T(t) = r'(t)/||r'(t)||
Now, let w(t) be the speed function and v(t) the velocity vector such that:
v(t) = (w(t)*T(t))
Deriving:
v'(t) = w'(t)T(t) + w(t)T'(t) = a(t)
So acceleration is written by the sum of the two components above. if the velocity of Y(t) = X[u(t)], then we can define Y'(t) = X'(u(t))u'(t) = w(t). Substituting this relation in the acceleration equation above, and we get that their acceleration can be defined as:
[X''(u(t))u'(t) + X'(u(t))u''(t)]T(t) + X'(u(t))u'(t)T'(t) = a(t)
Y''(t)T(t) + Y'(t)T'(t) = a(t)
... How should I've written?

 

[fresh_42]

This appears to be a little over complicated to me. The more since you have to deal with the crucial part anyway.
The little given by the text, I assume that $X$ and $Y$ are parametrizations of $C$. That are functions $X\, ,\, Y : \mathbb{R} \rightarrow C$. Wouldn't velocity simply be the change of position related to the change in time at a certain point in time?
But this sounds to me like $\frac{d}{dt} X(t) |_{t=t_0}$ and likewise for $X(u(t))$. The acceleration then is the second derivative.
Note that in the text "at each point" is given which means the differentials have to be evaluated at this point.