Q12Q - Fundamentals Of Physics

Question

The x and y components of four vectors, $\overline{a}$,$\overrightarrow{b}$,$\overrightarrow{c}$  and $\overrightarrow{d}$are given below. For which vectors will your calculator give you the correct angle$\theta$ when you use it to find $\theta$ with Eq. 3-6? Answer first by examining Fig. 3-12, and then check your answers with your calculator.

$$\begin{gathered} a_x=3 a_y=3 c_x=-3 c_y=-3 \\ {b}_x=-3 b_y=3 d_x=3 d_y=-3 \end{gathered}$$

Step-by-Step Solution

Answer

for all vectors, the calculator gives you the correct angle $\theta$ when you use it to find $\theta$

Step 1: Given information:

$$\begin{gathered} a_x=\text{3} a_y=\text{3} c_x=-\text{3} c_y=-\text{3} \\ {b}_x=-\text{3} b_y=\text{3} d_x=\text{3} d_y=-\text{3} \end{gathered}$$

Step 2: Understanding the given information

The problem involves a simple calculation of the direction of the vector. A vector's orientation, or the angle it forms with the x-axis, determines its direction. A line with an arrow on top and a fixed point at the other end is used to represent a vector. The vector's direction can be determined by looking at the direction of its arrowhead.

Step 3: Explanation for correct angle

The angle theta is defined as the tan inverse of the ratio of the y component to the x component of the vector. 

For the vector $\overrightarrow{a}$,

$\theta = {\tan }^{-1}\frac{a_y}{a_x}$

$$\begin{gathered} \theta = {\tan }^{-1}\frac{3}{3} \\ = {\tan }^{-1}\left(1\right) \\ ={45}^{\text{o}} \end{gathered}$$

As indicated by the darker portion of the graph from figure 3.12 the value of ${\tan }^{-1}\left(1\right)$ which is angle 45^o fits the calculator range. 

For the vector $\overrightarrow{b}$,

$\theta = {\tan }^{-1}\frac{b_y}{b_x}$

$$\begin{gathered} \theta = {\tan }^{-1}\frac{3}{-3} \\ = {\tan }^{-1}\left(-1\right) \\ =-{45}^o \end{gathered}$$

As indicated by the darker portion of the graph from figure 3.12 the value of ${\tan }^{-1}\left(1\right)$ which is angle - 45^o fits the calculator range. 

 

For the vector $\overrightarrow{c}$, 

$\theta = {\tan }^{-1}\frac{c_y}{c_x}$

$$\begin{gathered} \theta = {\tan }^{-1}\frac{-3}{-3} \\ = {\tan }^{-1}\left(1\right) \\ ={45}^o \end{gathered}$$

As indicated by the darker portion of the graph from figure 3.12 the value of ${\tan }^{-1}\left(1\right)$ which is angle 45^o fits the calculator range. 

For vector 4, 

$\theta = {\tan }^{-1}\frac{d_y}{d_x}$

$$\begin{gathered} \theta = {\tan }^{-1}\frac{-3}{3} \\ = {\tan }^{-1}\left(-1\right) \\ =-{45}^o \end{gathered}$$

As indicated by the darker portion of the graph from figure 3.12 the value of  ${\tan }^{-1}\left(1\right)$which is angle - 45^o fits the calculator range. 

Thus, from the figure, it can be said that for all vectors the calculator give you the correct angle when you use it to find $\theta$