How can I prove the tangential acceleration equation? This is my assignment for this weekend.
$$a=a_{TT}+a_{NN} = \frac{d^2 s}{dt^2}\vec{T} + \kappa \frac{ds}{dt} 2\vec{N}$$
Actually, I want to why $a_N=\kappa (ds/dt)$ hold
Please help me.
 A: The equation as presented in the text of the question still has errors.  It should read
$\vec a = (d^2s/dt^2)\vec T(s) + κ (ds/dt)^2 \vec N(s), \tag{0}$
where I have used $\vec a$ for the OP's $a$ to achieve a consistent notation for vectors.  The demonstration of ($0$) follows.
Let $\gamma(t) = (x_1(t), x_2(t))$ be a regular curve in $\Bbb R^2$; we consider the parameter $t$ as a time-variable.  Then the velocity vector $\vec v(t)$ is given by
$\vec v(t) = \dfrac{d}{dt}\gamma(t) = \dot \gamma(t) = (\dot x_1(t), \dot x_2(t)); \tag{1}$
we express $\vec v(t)$ in terms of the unit tangent vector $\vec T(s)$, where $s$ is the arc-length along $\gamma(t)$, thus:
since
$\vec T(s) = \dfrac{d}{ds}\gamma(s), \tag{2}$
we have
$\vec v(t) = \dfrac{d}{dt}\gamma(t) = \dfrac{ds}{dt} \dfrac{d}{ds}\gamma(s) = \dfrac{ds}{dt} \vec T(s), \tag{3}$
where the chain rule is used in (3).  The acceleration vector $\vec a(t)$ is given by
$\vec a(t) = \dot {\vec v}(t) = \dfrac{d}{dt}(\dfrac{ds}{dt} \vec T(s)) = \dfrac{d^2s}{dt^2} \vec T(s) + \dfrac{ds}{dt} \dfrac{d}{dt}\vec T(s), \tag{4}$
where, again using the chain rule,
$\dfrac{d}{dt}\vec T(s) = \dfrac{ds}{dt} \dfrac{d}{ds} \vec T(s) = \dfrac{ds}{dt} \kappa \vec N(s); \tag{5}$
the Frenet-Serret equation
$\dfrac{d}{ds} \vec T(s) = \kappa \vec N(s) \tag{6}$
has been deployed in (5).  Incorporating (5) into (4) yields
$\vec a = (d^2s/dt^2)\vec T(s) + κ (ds/dt)^2 \vec N(s) \tag{7}$
the requisite result; if we set
$a_{\vec T} = \dfrac{d^2s}{dt^2} \tag{8}$
and
$a_{\vec N} = \kappa (\dfrac{ds}{dt})^2, \tag{9}$
then we see that (7) may be written
$\vec a = a_{\vec T} \vec T(s) + a_{\vec N} \vec N(s). \tag{7}$
It should be noted that, by (1), 
$\dot s = \dfrac{ds}{dt} = \vert \vec {v}(t) \vert = \sqrt{(\dot x_1(t))^2 +  (\dot x_2(t))^2}, \tag{8}$
i.e., $\dot s$ is the rate of change of distance with time along $\gamma(t)$, also known as the speed of the point $\gamma(t)$; likewise $\ddot s = \dot {\vert \vec v \vert}$ is the rate of change of speed with time, again along $\gamma(t)$.  From (7), $\ddot s$ is the tangential component of the acceleration vector $\vec a$.  The normal component, $\kappa \dot s^2$, is analogous to the centripetal acceleration familiar to us from elementary kinematics; indeed, since $\kappa = \rho^{-1}$, where $\rho$ is the radius of curvature, it corresponds to what we often see as $\vert \vec v \vert^2/r$ in the mechanics of circular motion of radius $r$.  Finally, $s$ and $t$ are used somewhat interchangably as parameters for $\gamma$ in the above; one should remember that, though the parameters are different, $\gamma(s)$ and $\gamma(t)$ always refer to the same point.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
