Question

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

$\text{30.}\mathbf{r}\left(t\right)=⟨t,t^2,t^3⟩$

Short answer

The velocity and acceleration vectors are;

$$\begin{gathered} v\left(t\right)=⟨1,2t,3t^2⟩ \\ a\left(t\right)=⟨0,2,6t⟩ \end{gathered}$$

Step-by-step solution

Step 1. Given data

The given position vector is$\mathbf{r}\left(t\right)=⟨t,t^2,t^3⟩$

We have to find the velocity and acceleration vectors.

Step 2. Velocity vector

The velocity vector v(t) is given by

Therefore,

$\begin{array}{rl}v\left(t\right)& =\frac{d}{dt}r\left(t\right)\\ & =\frac{d}{dt}⟨t,t^2,t^3⟩\\ & =⟨\frac{d}{dt}\left(t\right),\frac{d}{dt}\left(t^2\right),\frac{d}{dt}\left(t^3\right)⟩\\ & =⟨1,2t,3t^2⟩\end{array}$

Hence the velocity vector$v\left(t\right)=⟨1,2t,3t^2⟩\begin{array}{}\end{array}$

Step 3. Acceleration vector

The acceleration vector a(t) is given by,

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

Therefore,

$\begin{array}{rl}a\left(t\right)& =\frac{d}{dt}v\left(t\right)\\ & =\frac{d}{dt}⟨1,2t,3t^2⟩\\ & =⟨\frac{d}{dt}\left(1\right),\frac{d}{dt}\left(2t\right),\frac{d}{dt}\left(3t^2\right)⟩\\ & =⟨0,2,6t⟩\end{array}$

Hence, the acceleration vector is$a\left(t\right)=⟨0,2,6t⟩$