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

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?

• By "... perpendicular to plane $\mathbf{r}$ and $\mathbf{v}$ ..." is the problem asking for the component of $\mathbf{a}$ that is in the plane containing $\mathbf{r}$ and $\mathbf{v}?$ – Adrian Keister Jun 26 '19 at 18:09
• Aren't you able to compute this second derivative ? Please show us. – Yves Daoust Jun 26 '19 at 18:36
• i have difficulty in understanding point of question – J. Deff Jun 27 '19 at 5:20

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 $$\vec v \times \vec r$$

• I think you mean $\vec{v}\times\vec{r},$ not $\vec{v}\times\vec{a}.$ – Adrian Keister Jun 26 '19 at 18:18
• can anyone explain what is going on – J. Deff Jun 26 '19 at 18:19
• Which part @J.Deff ? – XRFXLP Jun 26 '19 at 18:19
• @J.Deff Ajay is basically determining a unit vector in the required direction. Then finding the projection of the acceleration vector onto that unit vector. The second part involves finding a unit normal vector to the plane, which can be done by taking the cross product between two vectors lying on the plane, then dividing by its own magnitude. – Deepak Jun 26 '19 at 18:21
• I have edited the part in which I thought you had the problem. – XRFXLP Jun 26 '19 at 18:22

$$\vec{r}(t)=(\sin t,\cos2t,t^2+2t)\Rightarrow \vec{r}(0)=(0,1,0)$$ $$\vec{v}(t)=(\cos t,-2\sin2t,2t+2)\Rightarrow \vec{v}(0)=(1,0,2)$$ $$\vec{a}(t)=(-\sin t,-4\cos2t,2)\Rightarrow \vec{a}(0)=(0,-4,2)$$ 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 $$a_T(0)=(0,-4,2)\cdot(\dfrac{1}{\sqrt{5}},0,\dfrac{2}{\sqrt{5}})=\dfrac{4}{\sqrt{5}}$$