Chapter 11.2, Problem 48E

(a)

To determine

To calculate: The velocity vector for the particle.

Answer to Problem 48E

The required velocity is $\langle \cos t,-2\sin 2t\rangle$ .

Explanation of Solution

Given information:

The position of the particle: $\overrightarrow{x}=\sin t$ , $\overrightarrow{y}=\cos \left(2t\right)$

Calculation:

The given position of the particle is $\overrightarrow{r}\left(t\right)=\langle \sin t,\cos \left(2t\right)\rangle$ .

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

So, $\overrightarrow{v}\left(t\right)=\frac{d\left(\overrightarrow{r}\left(t\right)\right)}{dt}$

Now, $\begin{array}{c}\overrightarrow{v}\left(t\right)=\frac{d}{dt}\langle \sin t,\cos \left(2t\right)\rangle \\ =\langle \frac{d}{dt}\left(\sin t\right),\frac{d}{dt}\left(\cos \left(2t\right)\right)\rangle \\ =\langle \cos t,-2\sin 2t\rangle \end{array}$

Hence, the required velocity is $\langle \cos t,-2\sin 2t\rangle$ .

(b)

To determine

To determine: The values of t for which the particle is at rest.

Answer to Problem 48E

The required values of t are $\frac{\pi }{2}$ and $\frac{3\pi }{2}$ .

Explanation of Solution

Given information:

The position of the particle: $\overrightarrow{x}=\sin t$ , $\overrightarrow{y}=\cos \left(2t\right)$

Calculation:

The given position of the particle is $\overrightarrow{r}\left(t\right)=\langle \sin t,\cos \left(2t\right)\rangle$ .

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

So, $\overrightarrow{v}\left(t\right)=\frac{d\left(\overrightarrow{r}\left(t\right)\right)}{dt}$

Now, $\begin{array}{c}\overrightarrow{v}\left(t\right)=\frac{d}{dt}\langle \sin t,\cos \left(2t\right)\rangle \\ =\langle \frac{d}{dt}\left(\sin t\right),\frac{d}{dt}\left(\cos \left(2t\right)\right)\rangle \\ =\langle \cos t,-2\sin 2t\rangle \end{array}$

Thus, the required velocity is $\langle \cos t,-2\sin 2t\rangle$ .

It is seen that the particle is at rest at $\overrightarrow{v}\left(t\right)=\langle 0,0\rangle$ .

This means, $\cos t=0$ and $-2\sin 2t=0$

For $\cos t=0$ ; $\begin{array}{c}t={\cos }^{-1}\left(0\right)\\ t=\frac{\pi }{2}\end{array}$

Also, for $-2\sin 2t=0$ ; $\begin{array}{c}\sin 2t=0\\ 2t={\sin }^{-1}\left(0\right)\\ t=\frac{3\pi }{2}\end{array}$

Hence, the required values of t are $\frac{\pi }{2}$ and $\frac{3\pi }{2}$ .

(c)

To determine

To write: The equation for the path of the particle in terms of x and y that does not involve trigonometric functions.

Answer to Problem 48E

The required equation is $y=1-2x^2$ .

Explanation of Solution

Given information:

The position of the particle: $\overrightarrow{x}=\sin t$ , $\overrightarrow{y}=\cos \left(2t\right)$

Calculation:

The given position of the particle is $\overrightarrow{r}\left(t\right)=\langle \sin t,\cos \left(2t\right)\rangle$ .

To write an equation for the path of the particle, use the formula $\cos 2t=1-2{\sin }^{2}t$ .

The equation can be written as:

  $\begin{array}{c}\cos 2t=1-2{\sin }^{2}t\\ y=1-2x^2\end{array}$

Hence, the required equation is $y=1-2x^2$ .

(d)

To determine

To graph: The path of the particle.

Explanation of Solution

Given information:

The position of the particle: $\overrightarrow{x}=\sin t$ , $\overrightarrow{y}=\cos \left(2t\right)$

Calculation:

The given position of the particle is $\overrightarrow{r}\left(t\right)=\langle \sin t,\cos \left(2t\right)\rangle$ .

The required for the path of the particle is shown below: