91Ó°ÊÓ

Consider $r\left(t\right)=⟨\sin 3t,\cos 4tâŸ\copyright$

The objective is to find the tangential and normal components of acceleration for $r\left(t\right)$

Acceleration's tangential and normal components

$r\left(t\right)$is a twice - differentiable vector function with $r^{\text{'}}\left(t\right)=v\left(t\right)$and $r^{\text{'}\text{'}}\left(t\right)=a\left(t\right)$.

The tangential component of acceleration, $a_T=\frac{v·a}{‖v‖}$. The normal component of acceleration,

$$\begin{gathered} a_N=\frac{‖v×a‖}{‖v‖} \\ r\left(t\right)=⟨\sin 3t,\cos 4tâŸ\copyright \\ v\left(t\right)=r^{\text{'}}\left(t\right)=⟨3\cos 3t,-4\sin 4tâŸ\copyright \\ a\left(t\right)=r^{\text{'}\text{'}}\left(t\right)=⟨-9\sin 3t,-16\cos 4tâŸ\copyright \end{gathered}$$

$v×a=⟨3\cos 3t,-4\sin 4tâŸ\copyright ×⟨-9\sin 3t,-16\cos 4tâŸ\copyright$

$=\left|\begin{array}{c}ijk\\ 3cso3t-4\sin 4t0\\ -9\sin 3t-16\cos 4t0\end{array}\right|$

$=i\left(0\right)-j\left(0\right)+k(-48\cos 3t\cos 4t-36\sin 3t\sin 4t)$

$=⟨0,0,-48\cos 3t\cos 4t-36\sin 3t\sin 4t)$

$‖v×a‖=\sqrt{(-(48\cos 3t\cos 4t+36\sin 3t\sin 4t){)}^2}$

$=|48\cos 3t\cos 4t+36\sin 3t\sin 4t|$

The tangential component of acceleration

thus $a_r=\frac{-27\sin 3t\cos 3t+64\sin 4t\cos 4t}{\sqrt{9{\cos }^{2}3t+16{\sin }^{2}4t}}$

The normal component of acceleration,

thus $a_N=\frac{|48\cos 3t\cos 4t+36\sin 3t\sin 4t|}{\sqrt{9{\cos }^{2}3t+16{\sin }^{2}4t}}$