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Chapter 11: Problem 38

Convert the point from rectangular coordinates into polar coordinates with \(r \geq 0\) and \(0 \leq \theta<2 \pi\) \((3, \sqrt{3})\)

Short Answer

Expert verified
The polar coordinates are \((2\sqrt{3}, \frac{\pi}{6})\).

Step by step solution

01

Determine the Radius

To find the radius \( r \), we use the formula for converting rectangular coordinates to polar coordinates: \( r = \sqrt{x^2 + y^2} \). This gives us:\[ r = \sqrt{3^2 + (\5\u202664;3)^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3} \].Thus, the radius is \( r = 2\sqrt{3} \).
02

Determine the Angle

Next, we calculate the angle \( \theta \) using the formula \( \theta = \arctan\left(\frac{y}{x}\right) \). Substituting the given values, we have:\[ \theta = \arctan\left(\frac{\sqrt{3}}{3}\right) \].Since \( \tan(\frac{\pi}{6}) = \frac{\sqrt{3}}{3} \), it follows that \( \theta = \frac{\pi}{6} \). As both \( x \) and \( y \) are positive, the point is in the first quadrant, and thus the angle is still \( \frac{\pi}{6} \).
03

Verify the Quadrant

Verify that the calculated angle \( \theta \) falls in the correct quadrant. Since the point \((3, \sqrt{3})\) lies in the first quadrant where both \( x \) and \( y \) are positive, the angle \( \theta = \frac{\pi}{6} \) is valid for this quadrant.
04

Write the Polar Coordinates

Finally, we can express the point \((3, \sqrt{3})\) in polar coordinates as \((r, \theta)\). From our calculations, we know \( r = 2\sqrt{3} \) and \( \theta = \frac{\pi}{6} \). Thus, the polar coordinates are:\[ (2\sqrt{3}, \frac{\pi}{6}) \].

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Key Concepts

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

Polar Coordinates
Polar coordinates offer a different way of looking at locations on a plane. Instead of using horizontal and vertical axes as in rectangular (or Cartesian) coordinates, polar coordinates use a distance and an angle. The distance is from a fixed reference point, typically called the origin, and the angle is measured from the positive x-axis.
  • Distance is denoted by \( r \), which represents how far the point is from the origin. It is always non-negative (\( r \geq 0 \)) to ensure a valid location in space.
  • Angle is represented as \( \theta \), which describes the rotation needed from the positive x-axis to the line connecting the origin to the point. Angles are measured in radians, with a complete circle being \( 2\pi \).
This system is especially useful in situations involving circular motion or systems where direction and distance from a central point are more relevant than horizontal and vertical positioning.
Rectangular Coordinates
Rectangular coordinates are probably the most familiar way to locate a point on a plane. This system uses two perpendicular axes - typically called the x-axis and the y-axis - to describe the position of a point.
  • X-coordinate tells you how far left or right the point is from the origin.
  • Y-coordinate describes how far up or down the point is from the origin.
Given by a pair \((x, y)\), these coordinates are straightforward but might not be the most efficient in certain problems, especially those involving rotation and angles. That's where converting to polar coordinates can come in handy, as seen in the exercise with the point \((3, \sqrt{3})\).
Trigonometric Functions
Trigonometric functions play a crucial role in converting between rectangular and polar coordinates. They relate the angles and sides of triangles, which is essential in this conversion.For polar conversion, the primary functions used are:
  • Sine (\( \sin \)) and Cosine (\( \cos \)): Used in polar coordinates to express the vertical and horizontal component of a radius.
  • Tangent (\( \tan \)): Specifically useful for determining the angle \( \theta \) when converting from rectangular to polar coordinates using \( \theta = \arctan\left(\frac{y}{x}\right) \).
In our example, we found the angle \( \theta = \frac{\pi}{6} \) since \( \tan(\frac{\pi}{6}) = \frac{\sqrt{3}}{3} \). Understanding these functions helps us seamlessly switch between the coordinate systems, allowing us to solve various geometric problems more effectively.
Quadrants in Coordinate Plane
The coordinate plane is divided into four quadrants, each one representing a unique combination of positive and negative x and y values. This division is crucial when determining the correct angle \( \theta \) in polar coordinates.
  • First Quadrant: Both \( x \) and \( y \) are positive, \((+,+)\).
  • Second Quadrant: \( x \) is negative and \( y \) is positive, \((- ,+)\).
  • Third Quadrant: Both \( x \) and \( y \) are negative, \((-,-)\).
  • Fourth Quadrant: \( x \) is positive and \( y \) is negative, \((+, -)\).
Recognizing the quadrant where a point falls helps verify if the calculated angle \( \theta \) is appropriate. In our exercise, with the point \((3, \sqrt{3})\) residing in the first quadrant, the angle \( \theta = \frac{\pi}{6} \) was confirmed as correct, because points here are located with positive x and y values.

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Most popular questions from this chapter

Plot the set of parametric equations by hand. Be sure to indicate the orientation imparted on the curve by the parametrization. $$ \left\\{\begin{array}{l} x=t^{2}+2 t+1 \\ y=t+1 \end{array}\right. \text { for } t \leq 1 $$

Find a parametric description for the given oriented curve. the ellipse \((x-1)^{2}+9 y^{2}=9,\) oriented counter-clockwise

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Plot the set of parametric equations with the help of a graphing utility. \(\mathrm{Be}\) sure to indicate the orientation imparted on the curve by the parametrization. $$ \left\\{\begin{array}{l} x=e^{t}+e^{-t} \\ y=e^{t}-e^{-t} \end{array}\right. \text { for }-2 \leq t \leq 2 $$

60 through #63. In Exercises \(60-63,\) you need to solve for \(\theta\) 鈥� # Convert the equation from rectangular coordinates into polar coordinates. Solve for \(r\) in all but #60 through #63. In Exercises \(60-63,\) you need to solve for \(\theta\) \(x^{2}+y^{2}-2 y=0\)
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