Question

Find the velocity and acceleration vectors for the position vectors given in Exercises 30–34

$\text{33.}\mathbf{r}\left(t\right)=⟨\sin t,\cos t,2\sin 2t⟩$

Short answer

The velocity and acceleration vectors are;

$$\begin{gathered} v\left(t\right)=⟨\cos t,-\sin t,4\cos 2t⟩ \\ a\left(t\right)=⟨-\sin t,-\cos t,-8\sin 2t⟩ \end{gathered}$$

Step-by-step solution

Step 1. Given data

The given position vector is$\mathbf{r}\left(t\right)=⟨\sin t,\cos t,2\sin 2t⟩$

We have to find the velocity and acceleration vectors.

Step 2. Velocity vector

The velocity vector v(t) is given by

$v\left(t\right)=\frac{d}{dt}r\left(t\right)=⟨\frac{d}{dt}x\left(t\right),\frac{d}{dt}y\left(t\right),\frac{d}{dt}z\left(t\right)⟩$

Therefore,

$\begin{array}{rl}v\left(t\right)& =\frac{d}{dt}r\left(t\right)\\ & =\frac{d}{dt}⟨\sin t,\cos t,2\sin 2t⟩\\ & =⟨\frac{d}{dt}\sin t,\frac{d}{dt}\cos t,\frac{d}{dt}2\sin 2t⟩\\ & =⟨\cos t,-\sin t,4\cos 2t⟩\end{array}$

Hence the velocity vector$v\left(t\right)=⟨\cos t,-\sin t,4\cos 2t⟩$

Step 3. Acceleration vector

The acceleration vector a(t) is given by

$a\left(t\right)=\frac{d}{dt}v\left(t\right)$

Therefore,

$\begin{array}{rl}a\left(t\right)& =\frac{d}{dt}v\left(t\right)\\ & =\frac{d}{dt}⟨\cos t,-\sin t,4\cos 2t⟩\\ & =⟨\frac{d}{dt}\left(\cos t\right),\frac{d}{dt}-\sin t,\frac{d}{dt}4\cos 2t⟩\\ & =⟨-\sin t,-\cos t,-8\sin 2t⟩\end{array}$

Hence the acceleration vector$a\left(t\right)=⟨-\sin t,-\cos t,-8\sin 2t⟩$