Chapter 11.2, Problem 36E

To determine

To calculate: the velocity and acceleration vectors of the particle. Also, find the path of the particle.

Answer to Problem 36E

The required velocity vector and acceleration vector are $\langle 4\cos 4t,-3\sin 3t\rangle$ and $\langle -16\sin 4t,-9\cos 3t\rangle$ respectively.

Explanation of Solution

Given information:

The position vector of the particle $=\langle \sin 4t,\cos 3t\rangle$

Calculation:

Consider the position vector be $\overrightarrow{r}\left(t\right)$ .

So, $\overrightarrow{r}\left(t\right)=\langle \sin 4t,\cos 3t\rangle$

It is known that for a position vector $\overrightarrow{r}\left(t\right)$ of a particle, the first derivative of $\overrightarrow{r}\left(t\right)$ and the second derivative of $\overrightarrow{r}\left(t\right)$ are the velocity vector and acceleration vector of the particle respectively.

So, $\overrightarrow{v}\left(t\right)=\frac{d\left(\overrightarrow{r}\left(t\right)\right)}{dt}$ and $\begin{array}{c}\overrightarrow{a}\left(t\right)=\frac{d^2\left(\overrightarrow{r}\left(t\right)\right)}{d{t}^2}\\ \overrightarrow{a}\left(t\right)=\frac{d\left(\overrightarrow{v}\left(t\right)\right)}{dt}\end{array}$

Now, $\begin{array}{c}\overrightarrow{v}\left(t\right)=\frac{d}{dt}\langle \sin 4t,\cos 3t\rangle \\ =\langle 4\cos 4t,-3\sin 3t\rangle \end{array}$

Also, $\begin{array}{c}\overrightarrow{a}\left(t\right)=\frac{d\left(\overrightarrow{v}\left(t\right)\right)}{dt}\\ =\frac{d}{dt}\langle 4\cos 4t,-3\sin 3t\rangle \\ =\langle \frac{d}{dt}\left(4\cos 4t\right),\frac{d}{dt}\left(-3\sin 3t\right)\rangle \\ =\langle -16\sin 4t,-9\cos 3t\rangle \end{array}$

The graph of the position vector is shown below:

  

Hence, the required velocity vector and acceleration vector are $\langle 4\cos 4t,-3\sin 3t\rangle$ and $\langle -16\sin 4t,-9\cos 3t\rangle$ respectively.