Chapter 3, Problem 70RE

(a).

To determine

Find parametric equations that can be used to simulate the motion of the particle.

Answer to Problem 70RE

The ordered pairs of the parametric equation are:

  $\left(t,s\left(t\right)\right)=\left(t,10\cos \left(t+\frac{\pi }{4}\right)\right)$

Explanation of Solution

Given:

  $s\left(t\right)=10\cos \left(t+\frac{\pi }{4}\right)$

Concept Used:

The coordinate of the x can be taken as coordinates of t and coordinates of the y can be taken as coordinate of $s\left(t\right)$ .

Calculation:

By position vector of the particle the values of:

  $x\left(t\right)=t,y\left(t\right)=10\cos \left(t+\frac{\pi }{4}\right)$

The ordered pairs of the parametric equation are:

  $\left(t,s\left(t\right)\right)=\left(t,10\cos \left(t+\frac{\pi }{4}\right)\right)$

Conclusion:

The ordered pairs of the parametric equation are:

  $\left(t,s\left(t\right)\right)=\left(t,10\cos \left(t+\frac{\pi }{4}\right)\right)$

(b).

To determine

The initial motion of the particle.

Answer to Problem 70RE

The initial position of the particle at $t=0$ is $5\sqrt{2}$

Explanation of Solution

Given:

  $s\left(t\right)=10\cos \left(t+\frac{\pi }{4}\right)$

Concept Used:

The particle’s initial motion is the value at $t=0$ .

Calculation:

Put $t=0$ in $s\left(t\right)$ is:

  $\begin{array}{c}s\left(0\right)=10\cos \left(0+\frac{\pi }{4}\right)\\ =10\cos \left(\frac{\pi }{4}\right)\\ =10\left(\frac{\sqrt{2}}{2}\right)\\ =5\sqrt{2}\end{array}$

Conclusion:

The initial position of the particle at $t=0$ is $5\sqrt{2}$

(c).

To determine

To find, what points reached by the particle are fastest to the left and right of the origin.

Answer to Problem 70RE

The particle is fastest to the left to the origin is $-10$

And the particle right of the origin is $10$ .

Explanation of Solution

Given:

  $s\left(t\right)=10\cos \left(t+\frac{\pi }{4}\right)$

Concept Used:

Calculation:

The derivative of the function is:

  $\begin{array}{c}s\text{'}\left(t\right)=\frac{d}{dt}\left[s\left(t\right)\right]\\ =\frac{d}{dt}\left[10\cos \left(t+\frac{\pi }{4}\right)\right]\\ =10\cdot \frac{d}{dt}\left[\cos \left(t+\frac{\pi }{4}\right)\right]\\ =-10\sin \left(t+\frac{\pi }{4}\right)\cdot \frac{d}{dt}\left[t+\frac{\pi }{4}\right]\\ =-10\sin \left(t+\frac{\pi }{4}\right)\cdot \left(1+0\right)\\ =-10\sin \left(t+\frac{\pi }{4}\right)\end{array}$

Put $s\text{'}\left(t\right)=0$ :

  $\begin{array}{c}-10\sin \left(t+\frac{\pi }{4}\right)=0\\ \sin \left(t+\frac{\pi }{4}\right)=0\\ t+\frac{\pi }{4}={\sin }^1\left(0\right)\\ t+\frac{\pi }{4}=n\pi \\ t=n\pi -\frac{\pi }{4},n\ge 0\end{array}$

Put $n=0$ :

  $t=-\frac{\pi }{4}$ which is negative, so eliminate this.

Put $n=1$ :

  $\begin{array}{c}t=\pi -\frac{\pi }{4}\\ =\frac{3\pi }{4}\end{array}$

Substitute the value of $t$ in $s\left(t\right)$ :

  $\begin{array}{c}s\left(\frac{3\pi }{4}\right)=10\cos \left(\frac{3\pi }{4}+\frac{\pi }{4}\right)\\ =10\cos \left(\pi \right)\\ =10\left(-1\right)\\ =-10\end{array}$

Put $n=2$ in $t$ :

  $\begin{array}{c}t=2\pi -\frac{\pi }{4}\\ =\frac{7\pi }{4}\end{array}$

Substitute the value of $t$ in $s\left(t\right)$ :

  $\begin{array}{c}s\left(\frac{7\pi }{4}\right)=10\cos \left(\frac{7\pi }{4}+\frac{\pi }{4}\right)\\ =10\cos \left(\frac{8\pi }{4}\right)\\ =10\left(2\pi \right)\\ =10\left(1\right)\\ =10\end{array}$

Conclusion:

The particle is fastest to the left to the origin is $-10$ .

And the particle right of the origin is $10$ .

(d).

To determine

To find when does the particle first reach the origin, also the velocity, speed and acceleration at that point.

Answer to Problem 70RE

  $\begin{array}{l}\text{velocity=-10}\\ \text{acceleration}=0\end{array}$

Explanation of Solution

Given:

  $s\left(t\right)=10\cos \left(t+\frac{\pi }{4}\right)$

Concept Used:

The value of particle first reach at origin is $s\left(t\right)=0$ .

Calculation:

Put $s\left(t\right)=0$ :

  $\begin{array}{c}0=10\cos \left(t+\frac{\pi }{4}\right)\\ 0=\cos \left(t+\frac{\pi }{4}\right)\\ 0=\cos \left(t+\frac{\pi }{4}\right)\\ {\cos }^{-1}\left(0\right)=t+\frac{\pi }{4}\\ \frac{\pi }{2}+k\pi =t+\frac{\pi }{4},k\in \mathbb{Z}\\ t=\frac{\pi }{4}\end{array}$

The velocity and acceleration is:

  $\begin{array}{c}v\left(t\right)=\frac{d}{dt}\left[10\cos \left(t+\frac{\pi }{4}\right)\right]\\ =-10\sin \left(t+\frac{\pi }{4}\right)\cdot \frac{d}{dt}\left[\cos \left(t+\frac{\pi }{4}\right)\right]\\ =-10\sin \left(t+\frac{\pi }{4}\right)\left(1+0\right)\\ =-10\sin \left(t+\frac{\pi }{4}\right)\\ a\left(t\right)=\frac{d}{dt}\left[-10\sin \left(t+\frac{\pi }{4}\right)\right]\\ =-10\frac{d}{dt}\left[\sin \left(t+\frac{\pi }{4}\right)\right]\\ =-10\cos \left(t+\frac{\pi }{4}\right)\frac{d}{dt}\left[t+\frac{\pi }{4}\right]\\ =-10\cos \left(t+\frac{\pi }{4}\right)\end{array}$

At $t=\frac{\pi }{4}$ :

  $\begin{array}{c}v\left(\frac{\pi }{4}\right)=-10\sin \left(\frac{\pi }{4}+\frac{\pi }{4}\right)\\ =-10\sin \left(\frac{\pi }{2}\right)\\ =-10\left(1\right)\\ =-10\\ a\left(\frac{\pi }{4}\right)=-10\cos \left(\frac{\pi }{4}+\frac{\pi }{4}\right)\\ =-10\cos \left(\frac{\pi }{2}\right)\\ =-10\left(0\right)\\ =0\end{array}$

Conclusion:

  $\begin{array}{l}\text{velocity}=-10\\ \text{acceleration}=0\end{array}$