Problem 45 - College Trigonometry

Question

Convert the point from rectangular coordinates into polar coordinates with \(r \geq 0\) and \(0 \leq \theta<2 \pi\). $$ \left(-\frac{3}{10},-\frac{3 \sqrt{3}}{10}\right) $$

Step-by-Step Solution

Answer

Polar coordinates: \(\left(\frac{3}{5}, \frac{4\pi}{3}\right)\).

Step 1: Identify the Rectangular Coordinates

The given rectangular coordinates are \(x = -\frac{3}{10}\) and \(y = -\frac{3\sqrt{3}}{10}\).

Step 2: Calculate the Radius

Use the formula for the radius \( r \) in polar coordinates: \[ r = \sqrt{x^2 + y^2} \]Substitute the values:\[ r = \sqrt{\left(-\frac{3}{10}\right)^2 + \left(-\frac{3\sqrt{3}}{10}\right)^2} \] \[ r = \sqrt{\frac{9}{100} + \frac{27}{100}} \] \[ r = \sqrt{\frac{36}{100}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

Step 3: Calculate the Angle

Use the formula for the angle \( \theta \) in polar coordinates: \[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \] Substitute the values:\[ \theta = \tan^{-1}\left(\frac{-\frac{3\sqrt{3}}{10}}{-\frac{3}{10}}\right) \] \[ \theta = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \] Since the point is in the third quadrant (both coordinates are negative), adjust theta by adding \( \pi \): \[ \theta = \frac{\pi}{3} + \pi = \frac{4\pi}{3} \]

Step 4: Express in Polar Coordinates

The polar coordinates \((r, \theta)\) for the point are \(\left(\frac{3}{5}, \frac{4\pi}{3}\right)\).

Key Concepts

Rectangular Coordinates

Rectangular coordinates are a way of representing the position of a point in a two-dimensional plane using two values, typically denoted as \(x\) and \(y\). These coordinates are part of the Cartesian coordinate system, named after the mathematician René Descartes, who helped popularize this concept.
The coordinate \(x\) determines the horizontal position, while \(y\) determines the vertical position. This system is precise as every point in the plane has a unique set of \((x, y)\) coordinates. For example, the given point in the problem \(\left(-\frac{3}{10}, -\frac{3\sqrt{3}}{10}\right)\) is represented in rectangular coordinates. Here, the values are both negative, indicating that the point is located in the third quadrant of the Cartesian plane, where both 'x' and 'y' are negative.
Understanding the nature of the rectangular coordinates is the first step in converting them into polar coordinates.

Radius Calculation

The radius in polar coordinates, denoted as \(r\), represents the distance from the origin (0,0) to the point in question. Calculating this value is essential, as it translates the length across the Cartesian plane into polar terms.
To find \(r\), use the Pythagorean theorem, expressed in the equation:

• \[ r = \sqrt{x^2 + y^2} \]

Applying this formula to our given point \(\left(-\frac{3}{10}, -\frac{3\sqrt{3}}{10}\right)\), we find:

• \[ r = \sqrt{\left(-\frac{3}{10}\right)^2 + \left(-\frac{3\sqrt{3}}{10}\right)^2} \]
• \[ r = \sqrt{\frac{9}{100} + \frac{27}{100}} \]
• \[ r = \frac{3}{5} \]

Calculating \(r\) accurately ensures that we've properly accounted for the magnitude of our vector in the transition from rectangular to polar form.

Angle Calculation

Determining the angle \(\theta\) is crucial when converting to polar coordinates because it tells us the exact direction of the radius vector relative to the positive x-axis. This calculation must consider the specific quadrant of the point's location.
Typically, the angle \(\theta\) can be found using the arctangent function:

• \[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]

In our example, for the point \(\left(-\frac{3}{10}, -\frac{3\sqrt{3}}{10}\right)\), substitute the \(x\) and \(y\) into the equation:

• \[ \theta = \tan^{-1}\left(\frac{-\frac{3\sqrt{3}}{10}}{-\frac{3}{10}}\right) \]
• \[ \theta = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]

Since the point is in the third quadrant, add \(\pi\) to \(\theta\) to adjust for this:

• \[ \theta = \frac{\pi}{3} + \pi = \frac{4\pi}{3} \]

Trigonometric Functions

Trigonometric functions play a pivotal role in relating rectangular and polar coordinates. Not only are they used in the computation of the angle \(\theta\), but they also offer insight into the fundamental links between various coordinate systems.
Two critical trigonometric identities to keep in mind are:

• \( \sin \theta = \frac{y}{r} \)
• \( \cos \theta = \frac{x}{r} \)

These relationships are derived from the basic definitions of sine and cosine, relating the opposite and adjacent sides of a right triangle (respectively) to the hypotenuse (which is \(r\) here). They allow direct conversion back and forth between the coordinate systems by deriving either \(x\) and \(y\) from a known \(r\) and \(\theta\), or vice versa.
Understanding these functions and their applications helps solidify one’s comprehension of coordinate transformations and geometric interpretations in general.