Curvature kappa with known acceleration, unit normal and unit tangent vectors The acceleration of a particle is $a(t) = (4\sin t \cos t)T(t) + (4e^t \sin^2 (t/6))N(t)$,
where $T(t)$ is the unit tangent vector and $N(t)$ is the unit normal vector.
At $t = \pi/2$, the speed of the particle is $4$. What is $\kappa(2\pi)$?
acceleration using the formula $a = (a_T)T + (a_N)N$
 A: Cute!  A very cute problem indeed!
For the sake of definiteness, just to give it a name, let us suppose $\vec \gamma:I \to \Bbb R^3$ is the curve the particle follows; here $I \subset \Bbb R$ is an open interval such that $\pi/2, 2\pi \in I$.  Then $\vec v(t) = \dot{\vec \gamma}(t)$ is the velocity of the particle at time $t$; the speed of the particle, which is in fact the rate of change of the arc-length $s$ with respect to $t$, $\dot s(t) = ds(t)/dt = \vert \vec v(t) \vert$, is related to the velocity $\vec v(t)$ and the unit tangent vector $T(t)$ through the equation
$\vec v(t) = \dot s(t) T(t), \tag{1}$
and thus we have for the acceleration $\vec a(t) = \dot{\vec v}(t)$,
$\vec a(t) = \dot {\vec v}(t) = \dfrac{d}{dt}(\dot s(t) T(t)) = \ddot s(t) T(t) + \dot s(t) \dot T(t).  \tag{2}$
We consider the last term $\dot s(t) \dot T(t)$ of (2);  we have, by the chain rule,
$\dot T(t) = \dfrac{ds(t)}{dt} \dfrac{dT(s)}{ds}, \tag{3}$
and by the Frenet-Serret equation for $dT(s)/ds$,
$\dfrac{dT(s)}{ds} = \kappa(s) N(s); \tag{4}$
inserting (4) into (3) yields
$\dot T(t) = \dfrac{ds(t)}{dt} \kappa(t) N(t), \tag{5}$
and thus (2) may be written
$\vec a(t) = \ddot s(t) T(t) + (\dot s(t))^2 \kappa(t) N(t). \tag{6}$
Comparing (6) to the given acceleration,
$\vec a(t) = (4 \cos t \sin t) T + (4e^t \sin^2 (\dfrac{t}{6})) N(t), \tag{7}$
we first conclude that
$\ddot s(t) = 4 \cos t \sin t = (2 \sin^2 t)^\prime, \tag{8}$
where ${}^\prime$, like $\dot {}$, is used to denote the $t$-derivative.  Integrating (8), we find
$\dot s(t) -  \dot s(t_0) = \int_{t_0}^t \ddot s(u) du = \int_{t_0}^t (2\sin^2 u)^\prime du = 2(\sin^2 t - \sin^2 t_0), \tag{9}$
which leads to
$\dot s(t) = 2(\sin^2 t - \sin^2 t_0) + \dot s(t_0). \tag{10}$
If we take $t_0 = \pi / 2$, then $\dot s(t_0) = 4$, so (10) becomes
$\dot s(t) = 2(\sin^2 t - \sin^2 \dfrac{\pi}{2}) + 4 = 2(\sin^2 t + 1). \tag{11}$
We next observe that, comparing the coefficients of $N(t)$ from (6) and (7),
$(\dot s(t))^2 \kappa(t) = 4e^t \sin^2 (\dfrac{t}{6}); \tag{12}$
setting $t = 2\pi$ in (11), (12) yields
$\dot s(2\pi) = 2, \tag{13}$
whence
$4\kappa(2\pi) = 4e^{2\pi} \sin^2 \dfrac{\pi}{3}; \tag{14}$
finally, since $\sin (\pi/3) = \sqrt{3}/2$, 
$\kappa(2\pi) = \dfrac{3e^{2\pi}}{4}. \tag{15}$
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
