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Chapter 10: Problem 53

A point in rectangular coordinates is given. Convert the point to polar coordinates. $$(-\sqrt{3},-\sqrt{3})$$

Short Answer

Expert verified
The polar coordinate of the point \(-\sqrt{3},-\sqrt{3}\) in rectangular coordinates is \(\sqrt{6}, \frac{5\pi}{4}\).

Step by step solution

01

Identify given coordinates

The point given is \(-\sqrt{3},-\sqrt{3}\) in rectangular coordinates. Let's call the x-coordinate as 'x' and the y-coordinate as 'y'. So, x = -\sqrt{3} and y = -\sqrt{3}.
02

Calculate the radial coordinate r

In polar coordinates, a point is described using a radial distance 'r' and an angle '\(\theta\)'. The radial coordinate 'r' is calculated as the square root of the sum of the squares of x and y. According to Pythagorean theorem, \(r= \sqrt{x^{2} + y^{2}}\). So, plugging the given x and y into the formula: \(r= \sqrt{(-\sqrt{3})^{2} + (-\sqrt{3})^{2}}\). This simplifies to \(r= \sqrt{6}\).
03

Find the angle \(\theta\)

Now, the angle '\(\theta\)' in polar coordinates can be found using the equation: \(\theta = \arctan\frac{y}{x}\) . But since the coordinates lie in the third quadrant (as both x and y are negative), we have to add PI to the result to get the correct angle. Plugging in the given y and x into the formula gives: \(\theta = \arctan\frac{-\sqrt{3}}{-\sqrt{3}}\). Solving this yields: \(\theta = \arctan(1) + \pi\). This simplifies to \(\theta = \frac{\pi}{4} + \pi = \frac{5\pi}{4}\).
04

Write down the polar coordinates

Finally, the polar coordinates are given as the radial distance 'r' and the angle '\(\theta\)'. Hence, the polar coordinate of the point \(-\sqrt{3},-\sqrt{3}\) is \(\sqrt{6}, \frac{5\pi}{4}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rectangular Coordinates
Rectangular coordinates, also known as Cartesian coordinates, describe a point in a two-dimensional plane using two numbers. These numbers represent distances along two perpendicular axes: the x-axis and the y-axis. For the point \(-\sqrt{3}, -\sqrt{3}\), the value of \(-\sqrt{3}\) for both coordinates shows its position relative to the origin (0,0). The x-coordinate \(-\sqrt{3}\) tells us how far the point is from the origin in the horizontal direction, while the y-coordinate \(-\sqrt{3}\) indicates its vertical distance from the origin.
Radial Distance
Radial distance, often denoted as \(r\) in polar coordinates, represents how far a point is from the origin, regardless of direction. This distance is always non-negative and can be calculated using the formula derived from the Pythagorean theorem: \[r = \sqrt{x^2 + y^2}\].
  • In our example, for the coordinates \(-\sqrt{3}, -\sqrt{3}\), this formula becomes \[r = \sqrt{(-\sqrt{3})^2 + (-\sqrt{3})^2}\].
  • This simplifies to \[r = \sqrt{6}\], indicating the point's radial distance from the origin.
Angle Conversion
In polar coordinates, the angle \(\theta\) describes the direction of the point relative to the positive x-axis. This angle is measured in radians.
  • To find \(\theta\), use the arctangent function: \(\theta = \arctan\frac{y}{x}\).
  • For the point \(-\sqrt{3}, -\sqrt{3}\), both \(x\) and \(y\) are negative, placing the point in the third quadrant where angles range from \(\pi\) to \(\frac{3\pi}{2}\).
It's crucial to add \(\pi\) to the basic angle found for points in this quadrant to correctly convert to polar coordinates. The correct angle in our situation is \[\theta = \arctan(1) + \pi = \frac{\pi}{4} + \pi = \frac{5\pi}{4}\].This adjustment ensures that the angle accounts for its location in the coordinate plane.
Pythagorean Theorem
The Pythagorean theorem plays a vital role in converting rectangular coordinates to polar coordinates. This theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
  • When applying this to rectangular coordinates, the theorem helps calculate the radial distance \(r\) as \(r = \sqrt{x^2 + y^2}\).
  • This formula comes directly from the idea that every point on the plane can form a right triangle with the origin, where \(x\) and \(y\) are the triangle's legs, and \(r\) is the hypotenuse.
Through this understanding, the transition from the rectangular to polar system becomes quite intuitive, with the theorem providing the foundation for this geometric interpretation.

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Most popular questions from this chapter

Determine whether the statement is true or false. Justify your answer. For values of \(e>1\) and \(0 \leq \theta \leq 2 \pi,\) the graphs of the following equations are the same. $$r=\frac{e x}{1-e \cos \theta} \quad \text { and } \quad r=\frac{e(-x)}{1+e \cos \theta}$$

Use a graphing utility to graph the polar equation. Identify the graph. $$r=\frac{4}{3-\cos \theta}$$

Use a graphing utility to graph the polar equation. Identify the graph. $$r=\frac{3}{-4+2 \cos \theta}$$

Write the polar equation of the conic for \(e=1, e=0.5,\) and \(e=1.5\) Identify the conic for each equation. Verify your answers with a graphing utility. $$r=\frac{2 e}{1+e \cos \theta}$$

Convert the polar equation to rectangular form. $$r=\frac{5}{\sin \theta-4 \cos \theta}$$
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