Chapter 1.7, Problem 41E

(a)

To determine

To find: The graph for the paths of both the particle and the number of intersection in the path.

Answer to Problem 41E

The number of intersection is two and the required graph is shown in Figure 1

Explanation of Solution

Given:

The position of the particle at time $t$ is,

  $\begin{array}{l}{x}_1=3\sin t\\ y_1=2\cos t\end{array}$

The range of the time $t$ is,

  $0\le t\le 2\pi$

The position of the second particle is,

  $\begin{array}{l}{x}_2=-3+\cos t\\ y_2=1+\sin t\end{array}$

Calculation:

Consider the position of the first particle is,

  $\begin{array}{l}{x}_1=3\sin t\\ y_1=2\cos t\end{array}$

Consider the position of the second particle is,

  $\begin{array}{l}{x}_2=-3+\cos t\\ y_2=1+\sin t\end{array}$

The sketch for the path of both the particle is shown in Figure 1

  

Figure 1

From the above figure it is clear that the number of intersection is $2$

(b)

To determine

To find: Whether any of the points are the point of collision, if yes than find the collision points.

Answer to Problem 41E

There is only one collision point s $\left(-3,0\right)$ at time $\frac{3\pi }{2}$ .

Explanation of Solution

Given:

The position of the particle at time $t$ is,

  $\begin{array}{l}{x}_1=3\sin t\\ y_1=2\cos t\end{array}$

The range of the time $t$ is,

  $0\le t\le 2\pi$

The position of the second particle is,

  $\begin{array}{l}{x}_2=-3+\cos t\\ y_2=1+\sin t\end{array}$

Calculation:

From figure 1, the points of intersection are at,

  $\left(-3,0\right)$

Consider the position $x$ is,

  $x=3\sin t$

Then, the time for the particle for collision is,

  $\begin{array}{l}x=3\sin t\\ -3=3\sin t\\ t=\frac{3\pi }{2}\end{array}$

Consider the position of the particle is,

  $x_2=-3+\cos t$

Then,

  $\begin{array}{l}{x}_2=-3+\cos t\\ -3=-3+\cos t\\ \cos t=0\\ t=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2}\end{array}$

From figure 1, the other points of intersection are at,

  $\left(-2.1,1.4\right)$

Consider the position of the first particle is,

  $x_1=3\sin t$

Then,

  $\begin{array}{l}-2.1=3\sin t\\ \sin t=\left(\frac{-2.1}{2}\right)\\ t=-0.78\end{array}$

Consider the position of the second particle is,

  $x_2=-3+\cos t$

Then,

  $\begin{array}{l}-2.7=-3+\cos t\\ \cos t=0.9\\ t=\frac{3\pi }{2}\end{array}$

The time for both the particle is not same then there is only one collision point that is,

  $\left(-3,0\right)$ at time $\frac{3\pi }{2}$

(b)

To determine

To find: The scenario for the give path of the second particle.

Answer to Problem 41E

The position of the particle changes to $\left(3,1\right)$ and the point of intersection are $\left(3,0\right),\left(2.1,1.4\right)$ .

Explanation of Solution

Given:

The range of the time $t$ is,

  $0\le t\le 2\pi$

The position of the second particle is,

  $\begin{array}{l}{x}_2=3+\cos t\\ y_2=1+\sin t\end{array}$

Calculation:

Consider position of the second particle is,

  $\begin{array}{l}{x}_2=3+\cos t\\ y_2=1+\sin t\end{array}$

For the above position the centre of the particle is,

  $\left(3,1\right)$

The point of intersection of the particle is,

  $\left(3,0\right),\left(2.1,1.4\right)$