Chapter 4, Problem 4.1P

A position vector is defined as having a length equal to your height in inches (or centimeters). The tangent of its angle is defined as your weight in pounds (or kilograms) divided by your age in years. Calculate the data for this vector and:

1. Draw the position vector to scale on cartesian axes.
2. Write an expression for the position vector using unit vector notation.
3. Write an expression for the position vector using complex number notation, in both polar and cartesian forms.

To determine

To draw: The position vector to scale on Cartesian axes.

Answer to Problem 4.1P

The position vector to scale on Cartesian axesisshown in Figure-1.

Explanation of Solution

Given information:The tangent is defined as the weight in pound divided by age and the length is equal to the height.

Calculation:

The magnitude of the vector is,

  $\theta =\arctan \left(\frac{w}{a}\right)$

Here, $w$ is the weight and $a$ is age.

Substitute $140$ for $w$ and $18$ for $a$ in above equation.

  $\begin{array}{c}\theta =\arctan \left(\frac{140}{18}\right)\\ =\arctan \left(7.778\right)\\ =82.67°\end{array}$

The graph is shown below.

  

Figure-1

To determine

To write: The expression for the position vector.

Answer to Problem 4.1P

The expression for the position vector is

  $6.35i+49.59j$ .

Explanation of Solution

Given information:The tangent is defined as the weight in pound divided by age and the length is equal to the height.

Calculation:

The expression for the position vector is,

  $R=r\left(\begin{array}{c}\cos \left(\theta \right)\\ \sin \left(\theta \right)\end{array}\right)$

Substitute $50$ for $r$ and $82.67°$ for $\theta$ in above equation.

  $\begin{array}{c}R=50\left(\begin{array}{c}\cos \left(82.67\right)\\ \sin \left(82.67\right)\end{array}\right)\\ =50\left(\begin{array}{c}0.127\\ 0.991\end{array}\right)\\ =6.35i+49.59j\end{array}$

To determine

To write: The expression for the position vector in polar and Cartesian coordinates.

Answer to Problem 4.1P

The expression for the position vector in polar and Cartesian coordinates is

  $R=50e^{j-1.442}$ and $6.35+49.59j$ .

Explanation of Solution

Given information:The tangent is defined as the weight in pound divided by age and the length is equal to the height.

Calculation:

The expression for the position vector is,

  $R=r{e}^{j-\theta }$

Substitute $50$ for $r$ and $1.442\text{rad}$ for $\theta$ in above equation.

  $R=50e^{j-1.442}$

The expression for the position vector is,

  $R=r\cos \left(\theta \right)+jr\sin \left(\theta \right)$

Substitute $50$ for $r$ and $82.67$ for $\theta$ in above equation.

  $\begin{array}{c}R=50\cos \left(82.67\right)+j50\sin \left(82.67\right)\\ =6.35+49.59j\end{array}$