Problem 41
Question
Convert the point from rectangular coordinates into polar coordinates with \(r \geq 0\) and \(0 \leq \theta<2 \pi\). $$ (-3,0) $$
Step-by-Step Solution
Verified Answer
The polar coordinates are \((3, \pi)\).
1Step 1: Determine the Radius
To find the radius \(r\), use the formula \(r = \sqrt{x^2 + y^2}\). For the point \((-3, 0)\), substitute \(x = -3\) and \(y = 0\): \[r = \sqrt{(-3)^2 + 0^2} = \sqrt{9} = 3\] Thus, the radius \(r = 3\).
2Step 2: Calculate the Angle
The angle \(\theta\) in polar coordinates is found using \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\). Substitute \(x = -3\) and \(y = 0\): \[\theta = \tan^{-1}\left(\frac{0}{-3}\right) = \tan^{-1}(0)\] The tangent inverse of 0 is 0 or \(\pi\), but the point is on the negative x-axis, so \(\theta = \pi\).
3Step 3: Verify the Angle
Since \(0 \leq \theta < 2\pi\), the angle \(\theta = \pi\) lies within the required range. Therefore, no adjustment is needed.
Key Concepts
Rectangular CoordinatesRadius CalculationAngle CalculationCoordinate Conversion
Rectangular Coordinates
Rectangular coordinates, also known as Cartesian coordinates, use two perpendicular lines (axes) to uniquely identify the position of a point in the plane. These axes are usually labeled as the "x-axis," which is horizontal, and the "y-axis," which is vertical. To locate a point, you simply move horizontally to the appropriate x-value and then vertically to the corresponding y-value. For example, the point (-3, 0) means you move 3 units left along the x-axis and do not move up or down along the y-axis.
This system is intuitive because it directly relates to our usual perception of distance and direction. However, for some applications, especially those involving circles and angles, polar coordinates might be more appropriate.
This system is intuitive because it directly relates to our usual perception of distance and direction. However, for some applications, especially those involving circles and angles, polar coordinates might be more appropriate.
Radius Calculation
The radius in polar coordinates represents the distance of the point from the origin (0, 0). In our exercise, this gives a clear picture of how far the point is from the center of the axes. To find this radius, you apply the formula:
- \( r = \sqrt{x^2 + y^2} \)
- \( r = \sqrt{(-3)^2 + 0^2} = \sqrt{9} = 3 \)
Angle Calculation
Calculating the angle \( \theta \) in polar coordinates involves using the arctan function to find the angle between the positive x-axis and the line joining the origin to the point. The formula used is:
Since the point is situated on the negative x-axis, the correct angle reflecting this position is \( \theta = \pi \). Ensuring the correct angle is crucial for accurately representing the direction in which the point lies relative to the origin.
- \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \)
- \( \theta = \tan^{-1}\left(\frac{0}{-3}\right) = \tan^{-1}(0) \)
Since the point is situated on the negative x-axis, the correct angle reflecting this position is \( \theta = \pi \). Ensuring the correct angle is crucial for accurately representing the direction in which the point lies relative to the origin.
Coordinate Conversion
Converting from rectangular to polar coordinates involves two main steps: calculating the radius and finding the correct angle. Once you have these, the conversion is straightforward: the point \((-3, 0)\) in rectangular coordinates becomes \((3, \pi)\) in polar coordinates.
The polar coordinate system is especially useful in situations where direction and distance from a central point (like a radar system) are more relevant than traditional x and y positioning. Providing both types of coordinates gives you flexibility in applications and broader understanding of geometric concepts.
The polar coordinate system is especially useful in situations where direction and distance from a central point (like a radar system) are more relevant than traditional x and y positioning. Providing both types of coordinates gives you flexibility in applications and broader understanding of geometric concepts.
Other exercises in this chapter
Problem 41
For the given vector \(\vec{v}\), find the magnitude \(\|\vec{v}\|\) and an angle \(\theta\) with \(0 \leq \theta
View solution Problem 41
Use \(z=-\frac{3 \sqrt{3}}{2}+\frac{3}{2} i\) and \(w=3 \sqrt{2}-3 i \sqrt{2}\) to compute the quantity. Express your answers in polar form using the principal
View solution Problem 42
In Exercises \(41-50\), use set-builder notation to describe the polar region. Assume that the region contains its bounding curves. The region inside the circle
View solution Problem 42
For the given vector \(\vec{v}\), find the magnitude \(\|\vec{v}\|\) and an angle \(\theta\) with \(0 \leq \theta
View solution