Steps to calculate the tangential acceleration and normal acceleration of a vector I am given $r(t) = \langle\cos(3) , 1 − \sin(3) , 3 \sin(3) + \cos(3)\rangle$ and $t= \pi /4$
I have calculated $r'(t)$ and  $r''(t)$.
I don't know what to do next to get either tangential or normal acceleration component. Any pointers?
 A: For any regular curve  $r(t)$, the velocity is $\dot r(t)$; this vector is clearly tangent to $r(t)$ for every $t$; the acceleration vector is evidently $\ddot r(t)$, but it is not in general tangent to $r(t)$.  We may find both the tangential and normal components of $\ddot r(t)$; first, we form the unit tangent vector field $T(t)$ to $r(t)$:
$T(t) = \dfrac{\dot r(t)}{\vert \dot r(t) \vert}, \tag 1$
where we have used the assumed regularity if $r(t)$ that is,
$\dot r(t) \ne 0, \tag 2$
in order to form $T(t)$, since (2) is equivalent to
$\vert \dot r(t) \vert \ne 0. \tag 3$
The tangential component of any vector field $v(t)$ along $r(t)$ is then
$v_T(t) = (v(t) \cdot T(t)) T(t); \tag 4$
in particular, taking
$v(t) = \ddot r(t), \tag 5$
we see that the tangential component of the acceleration is
$\ddot r_T(t) = (\ddot r(t) \cdot T(t)) T(t); \tag 6$
the component of acceleration normal to $r(t)$ is thus
$\ddot r_N(t) = \ddot r(t) - \ddot r_T(t) = \ddot r(t) - (\ddot r(t) \cdot T(t)) T(t); \tag 7$
indeed we see that
$\ddot r(t) = \ddot r_T(t) + \ddot r_N(t), \tag 8$
and
$\ddot r_N(t) \cdot T(t) = \ddot r(t) \cdot T(t) - (\ddot r(t) \cdot T(t)) T(t) \cdot T(t)$
$= \ddot r(t) \cdot T(t) - (\ddot r(t) \cdot T(t))  = 0.  \tag 9$
We note that $\ddot r_T(t)$ and $\ddot r_N(t)$ may both be expressed solely in terms of $r(t)$, that is, without explicit reference to $T(t)$.  We substitute (1) into (6):
$\ddot r_T(t) = \left (\ddot r(t) \cdot \dfrac{\dot r(t)}{\vert \dot r(t) \vert} \right ) \dfrac{\dot r(t)}{\vert \dot r(t) \vert} = \dfrac{\ddot r(t) \cdot \dot r(t)}{\vert \dot r(t) \vert^2}\dot r(t), \tag{10}$
and also into (7):
$\ddot r_N(t) = \ddot r(t) - \ddot r_T(t) = \ddot r(t) - \left (\ddot r(t) \cdot \dfrac{\dot r(t)}{\vert \dot r(t) \vert} \right ) \dfrac{\dot r(t)}{\vert \dot r(t) \vert}$
$= \ddot r(t) - \dfrac{\ddot r(t) \cdot \dot r(t)}{\vert \dot r(t) \vert^2}\dot r(t). \tag{11}$
I leave calculating $\ddot r_T(t)$ and $\ddot r_N(t)$ for the specific curve given in the text of the question itself to the sufficiently engaged amongst my readers; it is not difficult.
