Problem 46 - College Trigonometry

Question

Convert the point from rectangular coordinates into polar coordinates with \(r \geq 0\) and \(0 \leq \theta<2 \pi\). $$ (-\sqrt{5},-\sqrt{5}) $$

Step-by-Step Solution

Answer

The polar coordinates are \((\sqrt{10}, \frac{5\pi}{4})\).

Step 1: Understand the Rectangular Coordinates

The given point is \((-\sqrt{5}, -\sqrt{5})\) in rectangular coordinates, where \(x = -\sqrt{5}\) and \(y = -\sqrt{5}\). These represent the point's position on the Cartesian plane.

Step 2: Calculate the Distance from the Origin

We use the formula for the polar radius \(r = \sqrt{x^2 + y^2}\). Substituting in the given values, we have:\[r = \sqrt{(-\sqrt{5})^2 + (-\sqrt{5})^2} = \sqrt{5 + 5} = \sqrt{10}\]

Step 3: Find the Angle Using Inverse Tangent

The formula to find the angle \(\theta\) is \(\tan^{-1}\left(\frac{y}{x}\right)\). For our point, \[\theta = \tan^{-1}\left(\frac{-\sqrt{5}}{-\sqrt{5}}\right) = \tan^{-1}(1)\]since both \(x\) and \(y\) are negative, the point is in the third quadrant where the actual angle is \(\theta = \tan^{-1}(1) + \pi = \frac{\pi}{4} + \pi = \frac{5\pi}{4}\).

Step 4: Assemble the Polar Coordinates

Combine the calculated \(r\) and \(\theta\) to express the polar coordinates as \((r, \theta) = (\sqrt{10}, \frac{5\pi}{4})\).

Key Concepts

Rectangular Coordinates

Rectangular coordinates are a way to represent a point on the Cartesian plane using two numbers, often denoted as \((x, y)\). These numbers correspond to distances along the x-axis and y-axis. For our example, the rectangular coordinates \((-\sqrt{5}, -\sqrt{5})\) mean the point is positioned with \(x\) and \(y\) distances of \(-\sqrt{5}\) from the origin.Understanding rectangular coordinates is crucial because they provide a simple way to graphically represent any point in two-dimensional space. On a grid, each coordinate tells you how far and in what direction to move from the origin (0,0).

• The first value \(x\) moves horizontally.
• The second value \(y\) moves vertically.

This coordinate system is foundational in mathematics and is widely used for graphing equations and finding distances between points.

Inverse Tangent

The inverse tangent function (often written as \(\tan^{-1}\) or \(\arctan\)) helps find an angle when you know the opposite and adjacent sides of a right triangle. In the context of converting rectangular coordinates to polar coordinates, it is used to determine the angle \(\theta\) formed with the positive x-axis.For example, given a point with coordinates \((-\sqrt{5}, -\sqrt{5})\), we calculate the angle using:\[\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-\sqrt{5}}{-\sqrt{5}}\right) = \tan^{-1}(1)\]Using inverse tangent is particularly useful because it's a direct method to find the principal angle related to a set of rectangular coordinates. However, remember, the inverse tangent alone doesn't account for which quadrant the point is in. Here, it gives the angle as \(\frac{\pi}{4}\), but since both coordinates are negative, it must be adjusted to the third quadrant by adding \(\pi\). So, adjusted \(\theta = \frac{5\pi}{4}\).

Cartesian Plane

The Cartesian plane, devised by René Descartes, allows us to pinpoint any location in two-dimensional space using a set of coordinates. It's composed of two perpendicular lines called axes: the horizontal line (x-axis) and the vertical line (y-axis).When a point like \((-\sqrt{5}, -\sqrt{5})\) is plotted on this plane:

• The x-axis measures horizontal distance.
• The y-axis measures vertical distance.

Its layout helps visualize algebraic equations, geometric shapes, and other mathematical concepts like slope and intersection points. With the Cartesian plane, you see how different points relate to each other, making it a powerful tool in both math and science for graphing and analysis.

Angle Calculation

Calculating angles, especially in a coordinate system, is an essential skill. When converting to polar coordinates, identifying the correct angle \(\theta\) is crucial.To calculate the angle from the origin to the point with given coordinates, use:\[\theta = \tan^{-1}\left(\frac{y}{x}\right)\]For our point \((-\sqrt{5}, -\sqrt{5})\), calculating gives \(\theta = \tan^{-1}(1)\). Since both x and y are negative, indicating the point is in the third quadrant, the proper angle is found by adjusting:\[\theta = \tan^{-1}(1) + \pi = \frac{\pi}{4} + \pi = \frac{5\pi}{4}\]Remember, the initial output from \(\tan^{-1}\) must sometimes be adjusted to reflect the correct direction in the Cartesian plane depending on the location of your point (quadrant). This ensures you have the right angle when working with math problems involving rotation or conversion.