Question

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Plot the point given by the polar coordinates \(\left(-1, \frac{5 \pi}{4}\right)\) and find its rectangular coordinates.

Short answer

The rectangular coordinates are \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\).

Step-by-step solution

- Understanding Polar Coordinates

Polar coordinates are given in the form \(r, \theta\), where \(r\) is the radial distance from the origin, and \(\theta\) is the angle measured counterclockwise from the positive x-axis.

- Plotting the Point

To plot the point given in polar coordinates \(\left(-1, \frac{5 \pi}{4}\right)\), first locate the angle \(\frac{5 \pi}{4}\). This angle is 225 degrees, which is in the third quadrant. Since the radius \(r\) is \(-1\), move 1 unit in the opposite direction (toward the first quadrant).

- Finding Rectangular Coordinates

Rectangular coordinates \((x, y)\) can be found using the formulas: x = r \cos(\theta)\ and y = r \sin(\theta)\. Substitute \(-1\) for \(r\) and \(\frac{5 \pi}{4}\) for \(\theta\).

- Calculate the x-coordinate

\(x = -1 \cos\left( \frac{5 \pi}{4} \right) = -1 \left(-\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{2}\)

- Calculate the y-coordinate

\(y = -1 \sin\left( \frac{5 \pi}{4} \right) = -1 \left(-\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{2}\)

- Combine the Coordinates

Therefore, the rectangular coordinates for the point \(\left(-1, \frac{5 \pi}{4}\right)\) are \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\).

Key concepts

Polar Coordinates

Polar coordinates are a way of representing points in a plane using two values: the radial distance from a reference point (usually the origin) and the angle from a reference direction (usually the positive x-axis). This system is particularly useful for problems involving circular and rotational symmetry.

• The radial distance, denoted by \(r\), indicates how far from the origin the point is.
• The angle, denoted by \(\theta\), specifies the direction of the point relative to the positive x-axis, measured counterclockwise.

For example, in the given exercise, the point \((-1, \frac{5 \pi}{4})\) has:

• \(r = -1\): The point is 1 unit away from the origin, but the negative sign indicates it is in the direction opposite to \(\theta\).
• \( \theta = \frac{5 \pi}{4} \) (or 225 degrees): This angle places the direction in the third quadrant.

By understanding these two components, you can locate any point on a plane using polar coordinates.

Rectangular Coordinates

Rectangular coordinates, also known as Cartesian coordinates, represent points in a plane using two perpendicular axes: the x-axis (horizontal) and the y-axis (vertical). Each point is described by an ordered pair \((x, y)\), where

• \(x\) is the horizontal distance from the y-axis, or how far left or right the point is.
• \(y\) is the vertical distance from the x-axis, or how far up or down the point is.

The transformation between polar and rectangular coordinates involves trigonometric functions:

• \(x = r \cos(\theta)\)
• \(y = r \sin(\theta)\)

For the given exercise:

• \(r = -1\)
• \( \theta = \frac{5 \pi}{4} \)
• Using the trigonometric functions, we calculate: \( x = -1 \cos\left(\frac{5 \pi}{4} \right) = \frac{\sqrt{2}}{2}\) and \(y = -1 \sin\left(\frac{5 \pi}{4} \right) = \frac{\sqrt{2}}{2}\).

So, the rectangular coordinates are \( \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)\).

Trigonometric Functions

Trigonometric functions are essential in converting between polar and rectangular coordinates. These functions relate the angles and sides of right triangles to each other and are fundamental in various areas of mathematics and engineering.

• \( \cos(\theta)\) and \( \sin(\theta)\) are the primary trigonometric functions used in such conversions.
• \( \cos(\theta)\) represents the ratio of the adjacent side to the hypotenuse in a right triangle.
• \( \sin(\theta)\) represents the ratio of the opposite side to the hypotenuse in a right triangle.

For the exercise:

• \( \theta = \frac{5 \pi}{4} \) places the terminal side of the angle in the third quadrant, where both sine and cosine values are negative.
• Given \(r = -1\), we effectively take the opposite direction, resulting in positive coordinates when using the trigonometric functions.
• \( \cos\left( \frac{5 \pi}{4} \right) = -\frac{\sqrt{2}}{2} \) and \( \sin\left( \frac{5 \pi}{4} \right) = -\frac{\sqrt{2}}{2} \). Multiplying by \(-1\) gives us \( \frac{\sqrt{2}}{2} \) for both x and y coordinates.

Understanding trigonometric functions is key to mastering conversions between different coordinate systems.