Chapter 11.2, Problem 47E

(a)

To determine

To calculate: The velocity vector.

Answer to Problem 47E

The required velocity is $\langle -\frac{4t}{{\left(1+t^2\right)}^2},\frac{2-2t}{{\left(1+t^2\right)}^2}\rangle$ .

Explanation of Solution

Given information:

The position of the particle: $\langle \frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\rangle$

Calculation:

The given position of the particle is $\overrightarrow{r}\left(t\right)=\langle \frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\rangle$ .

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, $\begin{array}{c}\overrightarrow{v}\left(t\right)=\frac{d}{dt}\langle \frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\rangle \\ =\langle \frac{d}{dt}\left(\frac{1-t^2}{1+t^2}\right),\frac{d}{dt}\left(\frac{2t}{1+t^2}\right)\rangle \\ =\langle -\frac{4t}{{\left(1+t^2\right)}^2},\frac{2-2t}{{\left(1+t^2\right)}^2}\rangle \end{array}$

Hence, the required velocity is $\langle -\frac{4t}{{\left(1+t^2\right)}^2},\frac{2-2t}{{\left(1+t^2\right)}^2}\rangle$ .

(b)

To determine

To determine: Whether the particle is ever at rest.

Answer to Problem 47E

The particle will never be at rest.

Explanation of Solution

Given information:

The position of the particle: $\langle \frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\rangle$

Calculation:

The given position of the particle is $\overrightarrow{r}\left(t\right)=\langle \frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\rangle$ .

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, $\begin{array}{c}\overrightarrow{v}\left(t\right)=\frac{d}{dt}\langle \frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\rangle \\ =\langle \frac{d}{dt}\left(\frac{1-t^2}{1+t^2}\right),\frac{d}{dt}\left(\frac{2t}{1+t^2}\right)\rangle \\ =\langle -\frac{4t}{{\left(1+t^2\right)}^2},\frac{2-2t}{{\left(1+t^2\right)}^2}\rangle \end{array}$

Thus, the velocity is $\langle -\frac{4t}{{\left(1+t^2\right)}^2},\frac{2-2t}{{\left(1+t^2\right)}^2}\rangle$ .

It is seen that at $t=0$ ; the x -component of the velocity vector is 0 and at $t=1$ ; the y -component of the velocity vector is 0.

This means, the velocity vector will not be zero at any time.

Hence, the particle will never be at rest.

(c)

To determine

To determine: The coordinates of the point that the particle approaches as t increases without bound.

Answer to Problem 47E

The required coordinates of the point is $\langle -1,0\rangle$ .

Explanation of Solution

Given information:

The position of the particle: $\langle \frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\rangle$

Calculation:

The given position of the particle is $\overrightarrow{r}\left(t\right)=\langle \frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\rangle$ .

To find the required coordinates of the point, take limit of the position vector as t approaches without bound.

  $\underset{x\to \infty }{\lim }\langle \frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\rangle =\langle -1.0\rangle$

Hence, the required coordinates of the point is $\langle -1,0\rangle$ .