Given that the position of a particle as a function of time $t$ in polar coordinates is given by
$s(t) = r(t)\angle\phi(t)$
in which $r(t) = e^{-t}$ is the modulus of position and $\phi(t) = t$ is the angle, how do I find the velocity of the particle $v(t) = s'(t)$? I mean, how do I calculate the derivative of a parameterized vector in polar coordinates?
I tried to transform $s(t)$ to rectangular coordinates, i.e.:
$s(t) = (r(t)\cos[\phi(t)] , r(t)\sin[\phi(t)]) = (e^{-t}\cos(t), e^{-t}\sin(t))$
and then taking the derivative of this and the modulus results in:
$|v(t)|^2 = 2e^{-2t}\,\,\,\,\,\,\,\,\,\,\,\,$(i)
But this is not the same as taking the direct derivative of the modulus of $s(t)$ which is:
$r'(t) = \dfrac{d}{dt}e^{-t} = -e^{-t}$
which squared is:
$(r'(t))^2 = e^{-2t}\,\,\,\,\,\,\,\,\,\,\,\,$ (2)
but (1) $\neq$ (2). Where is my error?
You have $$s(t) = r(t) (\cos\phi(t), \sin\phi(t))$$
Then $$s'(t) = r'(t) (\cos\phi(t), \sin\phi(t)) + r(t) \phi'(t) (-\sin\phi(t), \cos\phi(t))$$ so $$|s'(t)|^2 = s'(t) \cdot s'(t) = r'(t)^2 + r(t)^2 \phi'(t)^2$$ since $$(\cos\phi(t), \sin\phi(t)) \cdot (-\sin\phi(t), \cos\phi(t)) = 0$$
What you forget is thus the term $r(t)^2 \phi'(t)^2$.