Problem 52
Question
Find a polar equation in the form \(r \cos \left(\theta-\theta_{0}\right)=r_{0}\) for each of the lines in Exercises \(49-52 .\) $$x=-4$$
Step-by-Step Solution
Verified Answer
The polar equation is \( r \cos(\theta - 0) = -4 \).
1Step 1: Identify Line Equation
The given line equation is in Cartesian coordinates: \(x = -4\). This represents a vertical line passing through the point \((-4, y)\) for any value of \(y\).
2Step 2: Convert to Polar Form
The polar coordinates \((r, \theta)\) relate to Cartesian coordinates \((x, y)\) by the equations \(x = r \cos \theta\) and \(y = r \sin \theta\). Since \(x = -4\), we have the equation \(r \cos \theta = -4\).
3Step 3: Express in Desired Form
We need to express the equation in the form \(r \cos(\theta - \theta_0) = r_0\). Notice that the equation we obtained, \(r \cos \theta = -4\), is already in this format with \(\theta_0 = 0\) and \(r_0 = -4\).
Key Concepts
Cartesian coordinatespolar equationsconversion of coordinates
Cartesian coordinates
Cartesian coordinates are a coordinate system that specifies each point uniquely in a plane by a pair of numerical coordinates. These are usually expressed as
This system is particularly helpful for describing locations in a 2-D space and is commonly used in everyday applications, such as maps. In this exercise, the line equation \(x = -4\) is a typical Cartesian representation, denoting a vertical line crossing the x-axis at -4 and extending infinitely upwards and downwards. When looking at this line in the Cartesian plane, any point on it, such as \((-4, y)\), shares the same x-coordinate of -4. This consistent x-value defines all points of our line in the Cartesian system.
- \(x\): which is the horizontal distance from the origin
- \(y\): which is the vertical distance from the origin
This system is particularly helpful for describing locations in a 2-D space and is commonly used in everyday applications, such as maps. In this exercise, the line equation \(x = -4\) is a typical Cartesian representation, denoting a vertical line crossing the x-axis at -4 and extending infinitely upwards and downwards. When looking at this line in the Cartesian plane, any point on it, such as \((-4, y)\), shares the same x-coordinate of -4. This consistent x-value defines all points of our line in the Cartesian system.
polar equations
Polar equations use a different system to define points on a plane. Instead of using two perpendicular axes, polar coordinates use
Polar equations are particularly useful in scenarios where symmetry about a point is more relevant than symmetry about two perpendicular lines, like in circles or spirals. In this exercise, we find a polar equation equivalent to the Cartesian line \(x = -4\), resulting in \(r \cos \theta = -4\). This expresses each point on the line as a function of the radial distance and angle, encompassing the essence of polar coordinates. Polar equations can often reveal unique properties of curves that Cartesian equations may not, due to the inherently different way they map position.
- \(r\): the distance from the origin (or pole)
- \(\theta\): the angle measured from the positive x-axis (or polar axis)
Polar equations are particularly useful in scenarios where symmetry about a point is more relevant than symmetry about two perpendicular lines, like in circles or spirals. In this exercise, we find a polar equation equivalent to the Cartesian line \(x = -4\), resulting in \(r \cos \theta = -4\). This expresses each point on the line as a function of the radial distance and angle, encompassing the essence of polar coordinates. Polar equations can often reveal unique properties of curves that Cartesian equations may not, due to the inherently different way they map position.
conversion of coordinates
Conversion between Cartesian and polar coordinates is a vital skill in mathematics, allowing different approaches to problem-solving. The transformation between these systems is given by the equations \(x = r \cos \theta\) and \(y = r \sin \theta\).
To switch from Cartesian to polar, you can compute \(r = \sqrt{x^2 + y^2}\) and \(\theta = \arctan\left(\frac{y}{x}\right)\), while adjusting for the correct quadrant as necessary. Conversely, converting polar equations back to Cartesian involves using trigonometric identities to express \(x\) and \(y\) in terms of \(r\) and \(\theta\).
In the exercise, starting from the Cartesian equation \(x = -4\), we converted it into a polar equation \(r \cos \theta = -4\). This conversion process highlights how lines in the Cartesian plane can be expressed in terms of radius and angle, offering a new perspective and potential strategies for mathematical problems. Such conversions are particularly handy when dealing with complex shapes in physics and engineering. They facilitate analyses in which direction and magnitude relate more naturally than perpendicular distances.
To switch from Cartesian to polar, you can compute \(r = \sqrt{x^2 + y^2}\) and \(\theta = \arctan\left(\frac{y}{x}\right)\), while adjusting for the correct quadrant as necessary. Conversely, converting polar equations back to Cartesian involves using trigonometric identities to express \(x\) and \(y\) in terms of \(r\) and \(\theta\).
In the exercise, starting from the Cartesian equation \(x = -4\), we converted it into a polar equation \(r \cos \theta = -4\). This conversion process highlights how lines in the Cartesian plane can be expressed in terms of radius and angle, offering a new perspective and potential strategies for mathematical problems. Such conversions are particularly handy when dealing with complex shapes in physics and engineering. They facilitate analyses in which direction and magnitude relate more naturally than perpendicular distances.
Other exercises in this chapter
Problem 51
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