Problem 51
Question
In Exercises 37-54, a point in rectangular coordinates is given. Convert the point to polar coordinates. \(\left(6, 9\right)\)
Step-by-Step Solution
Verified Answer
The polar coordinates for the given rectangular coordinates \((6, 9)\) are \(\(\sqrt{117}\), 56.31 degrees\).
1Step 1: Calculating r
The radial coordinate r is the distance from the origin to the point. It can be calculated using the Pythagorean Theorem on the x and y coordinates. Thus, \(r = \sqrt{x^2 + y^2}\). Substituting the given values, \(r = \sqrt{6^2 + 9^2} = \sqrt{36 + 81} = \sqrt{117}\).
2Step 2: Calculating Theta
The angle \(\theta\) is created with respect to the x-axis. Calculating \(\theta\) involves the use of trigonometry, more specifically the tangent function. Given the formula \(tan(\theta) = \frac{y}{x}\), rearranging for \(\theta\) gives us \(\theta = tan^{-1}\left(\frac{y}{x}\right)\). Substituting the given values, \(\theta = tan^{-1}\left(\frac{9}{6}\right) = tan^{-1}(1.5)\). Using a calculator to find the inverse tangent of 1.5 gives approximately 56.31 degrees.
3Step 3: Final Conversion
The polar coordinates are typically written as (r, \(\theta\)). So, the polar coordinates for the given rectangular coordinates (6,9) are \(\sqrt{117}\), 56.31 degrees.
Key Concepts
Polar CoordinatesPythagorean TheoremInverse Trigonometric FunctionsTrigonometry
Polar Coordinates
Polar coordinates provide an alternative to rectangular coordinates for describing the position of points in a plane. While rectangular coordinates use a base pair of perpendicular lines, polar coordinates represent each point based on its distance from a reference point (called the pole, comparable to the origin in rectangular coordinates) and the angle it makes with a reference direction (usually the positive x-axis).
The representation of a point in polar coordinates is given as a pair \(r, \theta\), where \(r\) is the radial distance from the pole to the point, and \(\theta\) is the angle measured in degrees or radians from the reference direction. Transitioning between the two systems enables problem-solving in more than one framework, which can be advantageous depending on the scenario.
The representation of a point in polar coordinates is given as a pair \(r, \theta\), where \(r\) is the radial distance from the pole to the point, and \(\theta\) is the angle measured in degrees or radians from the reference direction. Transitioning between the two systems enables problem-solving in more than one framework, which can be advantageous depending on the scenario.
Pythagorean Theorem
The Pythagorean Theorem is a fundamental principle in geometry. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This theorem is mathematically expressed as \(c^2 = a^2 + b^2\), where \(c\) is the hypotenuse and \(a\) and \(b\) are the other two sides.
In the context of polar coordinates, the Pythagorean Theorem can be utilized to calculate the radial distance \(r\) from the rectangular coordinates. For example, if a point in rectangular coordinates is given as \(x, y\), the distance \(r\) from the origin can be found by imagining a right triangle with \(x\) and \(y\) as legs, yielding \(r = \sqrt{x^2 + y^2}\).
In the context of polar coordinates, the Pythagorean Theorem can be utilized to calculate the radial distance \(r\) from the rectangular coordinates. For example, if a point in rectangular coordinates is given as \(x, y\), the distance \(r\) from the origin can be found by imagining a right triangle with \(x\) and \(y\) as legs, yielding \(r = \sqrt{x^2 + y^2}\).
Inverse Trigonometric Functions
Finding Angles in Polar Coordinates
The inverse trigonometric functions are the reverse of the regular trigonometric functions, such as sine, cosine, and tangent. They are used to determine the angle when the value of the trigonometric function is known. The most commonly used inverse trigonometric functions are arcsine (\(sin^{-1}\)), arccosine (\(cos^{-1}\)), and arctangent (\(tan^{-1}\)).In polar coordinates, the arctangent function, also denoted as tan inverse or \(tan^{-1}\), is particularly useful. Given a point with rectangular coordinates \(x, y\), the angle \(\theta\) of the polar coordinates can be found using \(\theta = tan^{-1}\left(\frac{y}{x}\right)\). It is important to note that calculators may give the primary solution for \(\theta\), so understanding the full range of angles and their corresponding points on the plane is critical to obtaining the correct polar coordinates.
Trigonometry
Converting Coordinates Using Trigonometry
Trigonometry is the branch of mathematics that deals with the relationships between the sides and angles of triangles. When converting from rectangular to polar coordinates, trigonometry is used to find the angle \(\theta\) that a point makes with the x-axis, hence defining its direction. In this process, we often use the trigonometric ratios of sine, cosine, and tangent.For rectangular coordinates \(x, y\), the tangent of the angle \(\theta\) can be expressed as \(\tan(\theta) = \frac{y}{x}\). To find \(\theta\), we take the inverse tangent of the ratio \(\frac{y}{x}\). It's important to consider that the arctangent function has a range of \( -90\degree \), \( 90\degree \) for its output, which may require additional adjustments to identify the correct quadrant for the angle in the polar coordinate system.
Other exercises in this chapter
Problem 51
In Exercises 39-54, find a polar equation of the conic with its focus at the pole. \(\textit{Conic}\) Ellipse \(\textit{Vertex or Vertices}\) \((20, 0), (4, \pi
View solution Problem 51
In Exercises 49-58, use a graphing utility to graph the polar equation. Describe your viewing window. \(r=\dfrac{5\pi}{8}\)
View solution Problem 51
In Exercises 49-56, use a graphing utility to graph the curve represented by the parametric equations. Prolate cycloid: \(\quad x= \theta - \frac{3}{2} \sin\ \t
View solution Problem 51
In Exercises 51-56, sketch (if possible) the graph of the degenerate conic. \(y^2-16x^2=0\)
View solution