Problem 53
Question
In Exercises 37-54, a point in rectangular coordinates is given. Convert the point to polar coordinates. \(\left(5, 12\right)\)
Step-by-Step Solution
Verified Answer
The polar coordinates of the point (5, 12) are (13, 4.318).
1Step 1: Calculate magnitude 'r'
Using rectangular coordinates (5, 12), we can calculate the magnitude r using the formula \(r = \sqrt{x^2 + y^2}\). Substituting the given values, we get \(r = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\)
2Step 2: Calculate angle 'θ'
We use the formula \(\theta = \arctan(\frac{y}{x})\) to find the angle. However, since the point (5,12) is in the second quadrant, we need to add π to the calculated angle. Substituting the given coordinate values we get \(\theta = \arctan(\frac{12}{5})\) which yields an angle approximately equal to 1.176. Since we are in the second quadrant we add π to our angle: \(\theta = 1.176 + \pi = 4.318\) radians.
3Step 3: Write the result in Polar coordinates
Now we have the magnitude 'r' and angle 'θ', we can write the point in polar coordinates as (r, θ). Thus, the polar coordinates of the given point (5,12) are (13, 4.318).
Key Concepts
Polar CoordinatesMagnitude CalculationAngle Calculation
Polar Coordinates
Polar coordinates are an alternative to the more common Cartesian or rectangular coordinate system most students learn first. In polar coordinates, each point on a plane is determined by a distance from a reference point, known as the pole (similar to the origin in the rectangular system), and an angle from a reference direction, typically the positive x-axis.
In polar coordinates, the point is represented as \( (r, \theta) \), where \(r\) stands for the radial distance from the pole and \(\theta\) represents the angular component measured in radians. One key aspect of polar coordinates is that they are particularly useful in situations involving circular or spiral shapes where measurements are more naturally expressed as a function of angles and radii rather than horizontal and vertical distances.
In polar coordinates, the point is represented as \( (r, \theta) \), where \(r\) stands for the radial distance from the pole and \(\theta\) represents the angular component measured in radians. One key aspect of polar coordinates is that they are particularly useful in situations involving circular or spiral shapes where measurements are more naturally expressed as a function of angles and radii rather than horizontal and vertical distances.
Magnitude Calculation
The magnitude calculation in the context of converting rectangular coordinates to polar coordinates is a measure of how far a point is from the origin of the coordinate system. To find the magnitude, designated as \(r\), we use the Pythagorean theorem since a point \( (x, y) \) in rectangular coordinates can be thought of as forming a right triangle with its position vectors.
The formula to calculate magnitude is \( r = \sqrt{x^2 + y^2} \). For our specific exercise example with the point \( (5, 12) \), this gives us \( r = \sqrt{5^2 + 12^2} = \sqrt{169} = 13 \). The magnitude is a crucial part of the polar coordinate, as it directly tells us the radius of the circle on which our point lies in polar space.
The formula to calculate magnitude is \( r = \sqrt{x^2 + y^2} \). For our specific exercise example with the point \( (5, 12) \), this gives us \( r = \sqrt{5^2 + 12^2} = \sqrt{169} = 13 \). The magnitude is a crucial part of the polar coordinate, as it directly tells us the radius of the circle on which our point lies in polar space.
Angle Calculation
After determining the magnitude, the next step is to calculate the polar angle, \(\theta\). The angle tells us the direction of the radius vector starting from the positive x-axis and sweeping towards the point. To find \(\theta\), we use the arctangent function, which calculates the angle whose tangent is a given number.
For our point \( (5, 12) \), the procedure is to use the formula \(\theta = \arctan(\frac{y}{x})\), leading us to \(\theta = \arctan(\frac{12}{5})\). Since \(\arctan\) returns values between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), an adjustment is necessary if the point lies in a quadrant where \(\theta\) would be greater than \( \frac{\pi}{2} \) or less than \( -\frac{\pi}{2} \). For points in the second or third quadrant, like our example, we add \(\pi\) to the result of \(\arctan\) to rotate the angle accordingly to its correct position. Hence, the corrected angle for \( (5, 12) \) would be \( \theta = 1.176 + \pi = 4.318 \) radians.
Understanding the properties of arctangent and the adjustments needed based on which quadrant the point resides in is essential for correct angle calculation in polar coordinates.
For our point \( (5, 12) \), the procedure is to use the formula \(\theta = \arctan(\frac{y}{x})\), leading us to \(\theta = \arctan(\frac{12}{5})\). Since \(\arctan\) returns values between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), an adjustment is necessary if the point lies in a quadrant where \(\theta\) would be greater than \( \frac{\pi}{2} \) or less than \( -\frac{\pi}{2} \). For points in the second or third quadrant, like our example, we add \(\pi\) to the result of \(\arctan\) to rotate the angle accordingly to its correct position. Hence, the corrected angle for \( (5, 12) \) would be \( \theta = 1.176 + \pi = 4.318 \) radians.
Understanding the properties of arctangent and the adjustments needed based on which quadrant the point resides in is essential for correct angle calculation in polar coordinates.
Other exercises in this chapter
Problem 53
In Exercises 39-54, find a polar equation of the conic with its focus at the pole. \(\textit{Conic}\) Hyperbola \(\textit{Vertex or Vertices}\) \((1, 3\pi/2), (
View solution Problem 53
In Exercises 49-58, use a graphing utility to graph the polar equation. Describe your viewing window. \(r=8\ \cos\ \theta\)
View solution Problem 53
In Exercises 49-56, use a graphing utility to graph the curve represented by the parametric equations. Hypocycloid: \(\quad x= 3 \cos^{3}\ \theta, \quad y= 3 \s
View solution Problem 53
In Exercises 51-56, sketch (if possible) the graph of the degenerate conic. \(x^2-2xy+y^2=0\)
View solution