Problem 10 - Calculus for Biology and Medicine

Question

In Problems , 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. } $$ r=3, \alpha=150^{\circ} $$

Step-by-Step Solution

Answer

The vector in Cartesian coordinates is \( \left[ \begin{array}{c} -\frac{3\sqrt{3}}{2} \\ \frac{3}{2} \end{array} \right] \).

Step 1: Understanding Polar Coordinates

The vector is given in polar coordinates which consist of a length or magnitude \( r \) and an angle \( \alpha \). Here, we have \( r = 3 \) and \( \alpha = 150^{\circ} \). Polar coordinates describe a vector in terms of how far away it is from the origin and at what angle it is rotated from the positive \( x_1 \)-axis.

Step 2: Convert the Angle to Radians

Vectors often use radians, so first convert the angle. Use the formula: \( \text{radians} = \frac{\pi}{180} \times \text{degrees} \). Substituting in, we get \( \alpha = 150^{\circ} = \frac{\pi}{180} \times 150 = \frac{5\pi}{6} \) radians.

Step 3: Use Conversion Formulas from Polar to Cartesian

To convert from polar to Cartesian coordinates, use the formulas: \( x_1 = r \cos(\alpha) \) and \( x_2 = r \sin(\alpha) \). Substitute the known values: \( r = 3 \) and \( \alpha = \frac{5\pi}{6} \).

Step 4: Calculate the Cartesian Coordinates

First calculate \( x_1 \): \( x_1 = 3 \cos\left(\frac{5\pi}{6}\right) \). Since \( \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} \), we have \( x_1 = 3 \times -\frac{\sqrt{3}}{2} = -\frac{3\sqrt{3}}{2} \). Next calculate \( x_2 \): \( x_2 = 3 \sin\left(\frac{5\pi}{6}\right) \). Since \( \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \), we have \( x_2 = 3 \times \frac{1}{2} = \frac{3}{2} \).

Step 5: Write the Final Cartesian Coordinates

Combine \( x_1 \) and \( x_2 \) to form the vector:\[ \left[ \begin{array}{c} x_1 \ x_2 \end{array} \right] = \left[ \begin{array}{c} -\frac{3\sqrt{3}}{2} \ \frac{3}{2} \end{array} \right] \]. This is the Cartesian representation of the given vector.

Key Concepts

Polar Coordinates

Polar coordinates are a way of representing vectors or points in a plane using a distance and an angle. In polar coordinates, a point is defined by two values: the radial distance, often denoted as \( r \), and the angular coordinate, \( \alpha \). The radial distance \( r \) tells us how far the point is from the origin. The angle \( \alpha \), typically measured in degrees or radians, indicates the direction of the point from the positive x-axis, moving counterclockwise.
For example, given \( r = 3 \) and \( \alpha = 150^{\circ} \), the vector is 3 units away from the origin at a 150-degree angle from the x-axis. This makes it quite useful for circular or rotational problems as they naturally use a center point and a radius.

Cartesian Coordinates

Cartesian coordinates, commonly used in a rectangular coordinate system, represent points through horizontal and vertical distances from a set origin. The system consists of two perpendicular axes, usually labeled \( x \) and \( y \).
The Cartesian coordinates provide a straightforward and scalable way to describe positions on flat surfaces. For example, a point situated 4 units to the left and 2 units up from the origin would be \([-4, 2]\). This forms the basis for most drawings and calculations involving horizontal and vertical components in a grid-like pattern.

Trigonometric Functions

Trigonometric functions such as sine (\( \sin \)) and cosine (\( \cos \)) play a critical role in converting between polar and Cartesian coordinates. They relate the angles in a right triangle to the ratios of its sides.
When converting from polar to Cartesian coordinates, the cosine of the angle (\( \cos(\alpha) \)) gives the horizontal component (\( x_1 \)), while the sine of the angle (\( \sin(\alpha) \)) gives the vertical component (\( x_2 \)). As seen in the example, using \( \alpha = \frac{5\pi}{6} \), \( \cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2} \) and \( \sin(\frac{5\pi}{6}) = \frac{1}{2} \), helping to calculate the Cartesian coordinates efficiently.

Vector Conversion

Vector conversion from polar to Cartesian coordinates involves translating a vector expressed by magnitude and direction into a vector expressed by horizontal and vertical components. This conversion simplifies analysis when dealing with linear pathways or decomposing force vectors.
To convert a vector \( [r, \alpha] \) from polar to Cartesian, use:

• \( x_1 = r \cdot \cos(\alpha) \)
• \( x_2 = r \cdot \sin(\alpha) \)

For instance, using the given \( r = 3 \) and \( \alpha = \frac{5\pi}{6} \):

• \( x_1 = 3 \cdot -\frac{\sqrt{3}}{2} = -\frac{3 \sqrt{3}}{2} \)
• \( x_2 = 3 \cdot \frac{1}{2} = \frac{3}{2} \)

This results in the vector \( \left[ -\frac{3\sqrt{3}}{2}, \frac{3}{2} \right] \), which is its Cartesian form, allowing for easier calculation in many practical scenarios.