Findings components of acceleration in $\mathbf{r}(t) = \left(\sin(t),\cos (2t),t^2+2t\right)$.

Question

The position vector of moving point at time $t$ is $\mathbf{r}(t) = \left(\sin(t),\cos (2t),t^2+2t\right)$. Find components of acceleration $\mathbf{a}$ in directions parallel to velocity vector $\mathbf{v}$ and perpendicular to plane of $\mathbf{r}$ and $\mathbf{v}$ at $t=0.$

I know the acceleration vector is the second derivative of the $\mathbf{r}$ vector, but how to proceed?

Answer 1

Given $$ \vec r = (\sin t, \cos 2t, t^2 + 2t)$$ $\Rightarrow \vec v = ( \cos t , -2 \sin 2t ,2t + 2)$ and $ \vec a = ( -\sin t , -4 \cos 2t , 2)$ which at $t = 0$ are $(1,0,0)$ and $( 0,-4,2)$ respectively.

Now,

1. Projection of $\vec a$ on $\vec v$ is $\cfrac{ \vec a \cdot \vec v}{|v|} $ and the component parallel to $\vec v$ is $\cfrac{ \vec a \cdot \vec v}{|v|} \hat v$

2. This part can be proceeded in the exact same way as the first part, you have replace $ \vec v$ with the normal vector of the plane (as that is the vector parallel to which you have to obtain the components) which here is $ \vec v \times \vec r$

Answer 2

$$\vec{r}(t)=(\sin t,\cos2t,t^2+2t)\Rightarrow \vec{r}(0)=(0,1,0)$$ $$\vec{v}(t)=(\cos t,-2\sin2t,2t+2)\Rightarrow \vec{v}(0)=(1,0,2)$$ $$\vec{a}(t)=(-\sin t,-4\cos2t,2)\Rightarrow \vec{a}(0)=(0,-4,2)$$ tangential acceleration is $$a_T(t)=\vec{a}(t)\cdot\vec{T}(t)=\vec{a}(t)\cdot\dfrac{\vec{r'}(t)}{|\vec{r'}(t)|}=\vec{a}(t)\cdot\dfrac{\vec{v}(t)}{|\vec{v}(t)|}$$ then $$a_T(0)=(0,-4,2)\cdot(\dfrac{1}{\sqrt{5}},0,\dfrac{2}{\sqrt{5}})=\dfrac{4}{\sqrt{5}}$$