Problem 9 - Calculus for Biology and Medicine

Question

In Problems 9-12, vectors are given in their polar coordinate representation (length \(r\), and angle \(\alpha\) measured counterclockwise from the positive \(x_{1}-\) axis \() .\) Find the representation of the vector \(\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right]\)\text { in Cartesian coordinates. } $$ r=2, \alpha=30^{\circ} $$

Step-by-Step Solution

Answer

The Cartesian coordinates are \( \begin{bmatrix} \sqrt{3} \\ 1 \end{bmatrix} \).

Step 1: Understand Polar Coordinates

Polar coordinates represent a point in terms of its distance from the origin (radius \( r \)) and the angle \( \alpha \) from the positive \( x_1 \)-axis. The given problem provides \( r = 2 \) and \( \alpha = 30^{\circ} \). Our task is to convert these polar coordinates into Cartesian coordinates \( (x_1, x_2) \).

Step 2: Convert Angle to Radians

In mathematical operations involving trigonometric functions, angles are usually required in radians. Use the conversion formula, where \(1^{\circ} = \frac{\pi}{180^{\circ}}\). Therefore, \(30^{\circ} = 30 \times \frac{\pi}{180} = \frac{\pi}{6}\) radians.

Step 3: Calculate Cartesian Coordinates

Use the formulas for converting polar coordinates to Cartesian coordinates: \( x_1 = r \cdot \cos(\alpha) \) and \( x_2 = r \cdot \sin(\alpha) \). Substitute the values: - \( x_1 = 2 \cdot \cos \left( \frac{\pi}{6} \right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \) - \( x_2 = 2 \cdot \sin \left( \frac{\pi}{6} \right) = 2 \cdot \frac{1}{2} = 1 \)

Step 4: Write Final Cartesian Coordinates

The final Cartesian representation of the vector is \( \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} \sqrt{3} \ 1 \end{bmatrix} \).

Key Concepts

Polar Coordinates

Polar coordinates are a two-dimensional method of representing positions using a radius and an angle. Imagine you are standing at the center of a circle, which is called the origin. A point in this system is determined by how far away it is from you (this is the radius, or \( r \)) and what direction you are facing when looking at that point (this is the angle, or \( \alpha \)).

In our exercise, the given polar coordinates are \( r = 2 \) and \( \alpha = 30^{\circ} \). This means the point is 2 units away from the origin and located at a 30-degree angle. When visualizing the coordinates, remember that the angle is measured counterclockwise from the positive x-axis.

Cartesian Coordinates

Cartesian coordinates, sometimes called rectangular coordinates, are one of the most common ways to plot points on a plane. It uses two numbers, usually referred to as \( x \) and \( y \), to specify the location. Think of it as a way to tell you how far to go horizontally and vertically from a fixed center point (the origin) to reach a particular point.

In our task, we are converting from polar to Cartesian, seeking \( x_1 \) and \( x_2 \) coordinates. These represent distances along the horizontal and vertical axes respectively. This conversion helps to express the vector in terms of the familiar x and y layout most of us are accustomed to.

Angle Conversion

Converting angles from degrees to radians is a critical step in problems involving trigonometric functions. Degrees and radians are both units for measuring angles, with radians being the standard unit used in mathematical calculations.

To convert degrees to radians, use the formula:

• \( 1^{\circ} = \frac{\pi}{180^{\circ}} \)
• For \( 30^{\circ} \), the conversion would be: \( 30 \times \frac{\pi}{180} = \frac{\pi}{6} \) radians.

This conversion is important because trigonometric functions like cosine and sine typically work with radian measures. In our exercise, converting the angle ensures accuracy in calculating the Cartesian coordinates.

Trigonometric Functions

Trigonometric functions, namely sine and cosine, are fundamental to converting between polar and Cartesian coordinates. These functions work as a bridge, translating the radius and angle into horizontal and vertical distances.

To convert:

• The cosine function calculates the horizontal distance: \( x_1 = r \cdot \cos(\alpha) \).
• The sine function calculates the vertical distance: \( x_2 = r \cdot \sin(\alpha) \).

In our example, using \( r = 2 \) and \( \alpha = \frac{\pi}{6} \):

• \( x_1 = 2 \cdot \cos\left(\frac{\pi}{6}\right) = \sqrt{3} \)
• \( x_2 = 2 \cdot \sin\left(\frac{\pi}{6}\right) = 1 \)

By employing these functions, we successfully convert the polar coordinates into the Cartesian coordinates \( \begin{bmatrix} \sqrt{3} \, 1 \end{bmatrix} \). This highlights the importance of trigonometric functions in translating spatial information between different coordinate systems.