Finding the velocity of a position vector Let $\{\tilde{i}, \tilde{j}\}$ be the standard basis vectors for IR2. Define two paths in IR2 by
$\tilde{v1}$(θ) = cosθ$\tilde{i}$ + sinθ$\tilde{j}$
$\tilde{v2}$(θ) = −sinθ$\tilde{i}$ + cosθ$\tilde{j}$.
Show that $\tilde{v1}$ and $\tilde{v2}$ are orthonormal, and that $\tilde{v'1}$ = $\tilde{v2}$, $\tilde{v'2}$ = -$\tilde{v'1}$.
The position vector $\tilde{x}$(t) at time t of a particle moving in a plane is
$\tilde{x}$(t) = r(t)cosθ(t)$\tilde{i}$ + r(t)sinθ(t)$\tilde{j}$ = r(t)$\tilde{v1}$(θ(t));
so that (r(t),θ(t)) are the polar coordinates of the particle at time t. Show that the velocity of the particle
is 
$\tilde{x'}$ = r'$\tilde{v1}$ + rθ'$\tilde{v2}$
It's the second part of this question I'm having difficulty with, I can't seem to derive the velocity.
Apologies if it's hard to read, I'm new to this forum.
Thanks for any help in advance!!

Start with your formula $\vec{x} = r \, \cos(\theta) \, \vec{i} + r \, \sin(\theta) \, \vec{j}$. I'm leaving off the $t$'s for the sake of brevity. Now differentiate, using the product rule (and chain rule):
$$
\vec{x} \, ' = \left[ r' \cos(\theta) - r \sin(\theta) \, \theta' \right] \, \vec{i}
+ \left[ r' \, \sin(\theta) + r \cos(\theta) \, \theta' \right] \, \vec{j}
$$
Now just regroup the terms: factor the $r'$ out of the first and third terms, and factor the $r \, \theta'$ out of the second and fourth.
