Chapter 10.2, Problem 37E

$\text{(a)}$

To determine

To find: The velocity and the speed of the particle.

Answer to Problem 37E

The velocity:

  

And the speed:

  

Explanation of Solution

Given information:

  $\begin{array}{c}y=\sin 4t\cos t\\ y=\sin 2t\end{array}$

Calculation:

Know that,

  

Then,

  

In this exercise,

  

Then,

  $\begin{array}{c}\frac{dx}{dt}=\cos (4t)\cdot 4\cdot \cos t+\sin (4t)\cdot (-\sin t)\\ =4\cos (4t)\cos t-\sin (4t)\sin t\\ \frac{dy}{dt}=\cos (2t)\cdot 2\\ =2\cos (2t)\end{array}$

To use the previous result,

For $t=5\pi /4$

Then,

  $\begin{array}{c}\frac{dx}{dt}(5\pi /4)=4\cos (4\cdot (5\pi /4))\cos (5\pi /4)-\sin (4\cdot (5\pi /4))\sin (5\pi /4)\\ =4\cos (5\pi )\cos (5\pi /4)-\sin (5\pi )\sin (5\pi /4)\\ =4\cdot (-1)\cdot \left(-\frac{\sqrt{2}}{2}\right)-0\cdot \left(-\frac{\sqrt{2}}{2}\right)\\ =2\sqrt{2}\\ \frac{dy}{dt}(5\pi /4)=2\cos (2\cdot (5\pi /4))\\ =2\cos (5\pi /2)\\ =2\cdot 0\\ =0\end{array}$

The previous results show that

  

Know that the magnitude of a vector.

  

To use the previous result,

  

Therefore, the required velocity of the particle is

And the speed is .

$\text{(b)}$

To determine

To draw: The path of the particle and show that the velocity of vector at $t=5\pi /4$ .

Answer to Problem 37E

The answer: Draw the vector from the point $(x(5\pi /4),y(5\pi /4))$ .

Explanation of Solution

Given information:

  $\begin{array}{c}y=\sin 4t\cos t\\ y=\sin 2t\end{array}$

Calculation:

From part $\text{(a)}$ ,

  

To use the previous result,

  $\begin{array}{c}x(5\pi /4)=\sin (4\cdot (5\pi /4))\cos (5\pi /4)\\ =\sin (5\pi )\cos (5\pi /4)\\ =0\cdot \left(-\frac{\sqrt{2}}{2}\right)\\ =0\\ y(5\pi /4)=\sin (2t)\\ =\sin (2\cdot (5\pi /4))\\ =\sin (5\pi /2)\\ =1\end{array}$

Then,

  

Know that,

  $\begin{array}{c}\sin \theta =\frac{b}{|(a,b)|}\\ \cos \theta =\frac{a}{|(a,b)|}\\ \Rightarrow \tan \theta =\frac{b}{a}\end{array}$

To use the previous result,

  $\begin{array}{c}\tan \theta =\frac{0}{2\sqrt{2}}\\ =0\\ \Rightarrow \theta =0^{\circ }\end{array}$

The following graph is the graph of the path of the particle with the velocity vector of the particle at time $t=5\pi /4$:

  

Therefore, the required draw the vector from the point $(x(5\pi /4),y(5\pi /4))$ .

$(c)$

To determine

To find: The particle moving to left or right when $t=5\pi /4$ .

Answer to Problem 37E

The answer: The particle moving to the right when $t=5\pi /4$ .

Explanation of Solution

Given information:

  $\begin{array}{c}y=\sin 4t\cos t\\ y=\sin 2t\end{array}$

Calculation:

Aware that a particle's velocity vector indicates the speed at which a particle's position is changing. Next, a particle's direction is represented by the direction of its velocity vector.

The direction angle of the particle's velocity vector with the positive $x$ -axis at time $t=5\pi /4$ is equal to $0^{\circ }$ , which indicates that the direction of the particle's velocity vector at time $t=5\pi /4$ is to the right.

Therefore, the required The particle moving to the right when $t=5\pi /4$ .