Problem 17

Question

Exer. 13-26: Find a polar equation that has the same graph as the equation in \(x\) and \(y\). $$ y^{2}=6 x $$

Step-by-Step Solution

Verified
Answer
The polar equation is \(r = \frac{6\cos(\theta)}{\sin^2(\theta)}\).
1Step 1: Recall conversion formulas
Recall the conversion formulas between Cartesian and polar coordinates: - In polar coordinates, a point is represented as \((r, \theta)\), where \(r\) is the radius and \(\theta\) is the angle.- The conversion formulas are: \[ x = r \cos(\theta) \] \[ y = r \sin(\theta) \]
2Step 2: Substitute Cartesian formulas into the given equation
The given equation is \[ y^2 = 6x \] Substitute \(y = r\sin(\theta)\) and \(x = r\cos(\theta)\) into the equation: \[ (r\sin(\theta))^2 = 6(r\cos(\theta)) \]
3Step 3: Simplify the trigonometric equation
Simplify the equation: \[ r^2\sin^2(\theta) = 6r\cos(\theta) \] Since \(r\) is part of both terms, assume \(r eq 0\) and divide by \(r\): \[ r\sin^2(\theta) = 6\cos(\theta) \]
4Step 4: Solve for r
Rearrange the equation to solve for \(r\): \[ r = \frac{6\cos(\theta)}{\sin^2(\theta)} \] This equation represents the same graph in polar coordinates.

Key Concepts

Polar EquationConversion FormulasCartesian CoordinatesTrigonometric Simplification
Polar Equation
Polar equations are an essential representation of mathematical graphs in polar coordinates. Unlike Cartesian equations that involve the usual x and y-coordinates, polar equations use the radius \(r\) and angle \(\theta\) to describe a point in 2D space.
This representation can be particularly handy when dealing with curves or shapes that have a natural symmetry centered around a point, like circles or spirals.
In this exercise, we convert a typical Cartesian equation, \(y^2 = 6x\), into a polar equation, demonstrating how the same graph can be expressed in these alternative coordinates. By understanding polar equations, you're opening doors to analyzing complex shapes with rotational characteristics.
Conversion Formulas
To transition from Cartesian coordinates, where equations are typically expressed as \(x\) and \(y\), to polar coordinates, we use conversion formulas:
  • \( x = r \cos(\theta) \)
  • \( y = r \sin(\theta) \)
These formulas connect each point's representation in the Cartesian system to its equivalent in the polar system.
When solving a problem like finding the polar equation that matches \(y^2 = 6x\), we substitute \(x\) and \(y\) with these conversion formulas, leading to modifications in the original equation that make it suitable for polar coordinates.
Understanding these conversions forms the backbone of transforming graphical representations.
Cartesian Coordinates
The Cartesian coordinate system is a familiar way to represent points in a two-dimensional plane using two values, \(x\) and \(y\). In our original equation, \(y^2 = 6x\), the Cartesian plane is a straightforward setting where one can graph this relationship as a parabola.
In mathematical problem-solving, starting with Cartesian coordinates is often more intuitive, given that most basic equations are learned in this format. However, knowing how to switch to polar coordinates allows us to view the graph from different perspectives, giving us more tools for problem analysis and solving.
Ultimately, being proficient in both coordinate systems enriches our understanding of mathematical graphs and relationships.
Trigonometric Simplification
Trigonometric simplification plays a critical role when converting and solving equations as we bridge Cartesian and polar coordinates. After substituting the conversion formulas \(y = r\sin(\theta)\) and \(x = r\cos(\theta)\) into the Cartesian equation \(y^2 = 6x\), we achieve an intermediate form: \(r^2\sin^2(\theta) = 6r\cos(\theta)\).
This is where simplification becomes key. By assuming \(r eq 0\), we divide both sides by \(r\), simplifying the expression to \(r\sin^2(\theta) = 6\cos(\theta)\).
This resulting equation can then be further rearranged to solve for \(r\), yielding \(r = \frac{6\cos(\theta)}{\sin^2(\theta)}\). Simplifying trigonomic functions in this manner enables the expression of the graph in a clear polar form.