Problem 20
Question
\(15-20\) Identify the curve by finding a Cartesian equation for the curve. $$r=\tan \theta \sec \theta$$
Step-by-Step Solution
Verified Answer
The Cartesian equation is \(x^2 = y\), representing a parabola.
1Step 1: Recall Polar to Cartesian Conversion Formulas
To convert the polar equation into a Cartesian form, recall the formulas: \(x = r \cos \theta\) and \(y = r \sin \theta\). Also, remember \(\tan \theta = \frac{y}{x}\) and \(\sec \theta = \frac{1}{\cos \theta}\).
2Step 2: Incorporate Given Polar Equation
Given the equation \(r = \tan \theta \sec \theta\), express \(\tan \theta\) and \(\sec \theta\) in terms of \(x\) and \(y\): \(\tan \theta = \frac{y}{x}\) and \(\sec \theta = \frac{1}{\cos \theta} = \frac{r}{x}\). Substitute these into the equation: \(r = \frac{y}{x} \cdot \frac{r}{x}\).
3Step 3: Simplify the Equation
Simplify the equation from Step 2: \(r = \frac{yr}{x^2}\). Multiply both sides of the equation by \(x^2\) to eliminate the fraction: \(rx^2 = yr\).
4Step 4: Eliminate r from the Equation
Since \(r = \sqrt{x^2 + y^2}\) in polar coordinates, substitute \(r\) with \(\sqrt{x^2 + y^2}\) from the equation: \(\sqrt{x^2 + y^2}x^2 = y\sqrt{x^2 + y^2}\).
5Step 5: Further Simplify
Divide both sides of the equation by \(\sqrt{x^2 + y^2}\) (assuming \(r eq 0\)) to get \(x^2 = y\).
6Step 6: Write the Cartesian Equation
The simplified Cartesian equation is \(x^2 = y\), which represents a parabola opening upwards on the xy-plane.
Key Concepts
Polar CoordinatesCartesian CoordinatesEquation TransformationParabola
Polar Coordinates
Polar coordinates are a flexible way to represent points in a plane through a distance and angle rather than traditional x and y coordinates. Often denoted as \((r, \theta)\), where:
For example, moving around a circle, it's easier to specify an angle and radius than Cartesian coordinates. However, in some cases, it becomes necessary to convert these coordinates into the more commonly used Cartesian form.
- \(r\) is the radial distance from the origin (0,0) to the point.
- \(\theta\) is the angle measured from the positive x-axis counterclockwise to the point.
For example, moving around a circle, it's easier to specify an angle and radius than Cartesian coordinates. However, in some cases, it becomes necessary to convert these coordinates into the more commonly used Cartesian form.
Cartesian Coordinates
Cartesian coordinates use two perpendicular axes to describe a point in the plane by their horizontal and vertical components, called the x-coordinate and y-coordinate, respectively. Imagine superimposing a grid onto a piece of paper; each grid square represents a unit of measurement.
Converting polar coordinates into Cartesian coordinates can facilitate the solving of geometric problems or graphing functions. With transformation formulas such as \(x = r \cos \theta\) and \(y = r \sin \theta\), a polar equation can be expressed in Cartesian form for further mathematical analysis.
- The horizontal axis is typically labeled as the x-axis.
- The vertical axis is known as the y-axis.
- A point is denoted as \((x, y)\).
Converting polar coordinates into Cartesian coordinates can facilitate the solving of geometric problems or graphing functions. With transformation formulas such as \(x = r \cos \theta\) and \(y = r \sin \theta\), a polar equation can be expressed in Cartesian form for further mathematical analysis.
Equation Transformation
Equation transformation often involves re-casting equations from one form to another to simplify operations or gain a new perspective. In polar and Cartesian conversions, this translation is essential for understanding and working with curves and functions across different systems.
To transform the polar equation \(r = \tan \theta \sec \theta\) into a Cartesian equation, we apply several trigonometric identities and transformations:
To transform the polar equation \(r = \tan \theta \sec \theta\) into a Cartesian equation, we apply several trigonometric identities and transformations:
- Recall \(\tan \theta = \frac{y}{x}\) and \(\sec \theta = \frac{1}{\cos \theta}\), resulting in \(\sec \theta = \frac{r}{x}\).
- Substitute these into the equation to express all components in terms of \(x\) and \(y\).
- The equation simplifies and results in \(x^2 = y\), illustrating how radical expressions are eliminated when possible.
Parabola
A parabola is a curve where each point is equidistant from a fixed point, known as the focus, and a fixed line, called the directrix. The equation \(x^2 = y\) represents a parabola in the Cartesian plane, opening upwards.
These attributes make analyzing and sketching the graph manageable and useful for various applications, including projectile motion and optimizing solutions.
- It is symmetric with respect to the y-axis due to the squared term \(x^2\).
- Its vertex is located at the origin (0,0).
These attributes make analyzing and sketching the graph manageable and useful for various applications, including projectile motion and optimizing solutions.
Other exercises in this chapter
Problem 20
Find the vertices, foci, and asymptotes of the hyperbola and sketch its graph. $$\frac{y^{2}}{16}-\frac{x^{2}}{36}=1$$
View solution Problem 20
Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to check your work. $$x=\cos 3 \theta,
View solution Problem 20
Describe the motion of a particle with position \((x, y)\) as \(t\) varies in the given interval. \(x=2 \sin t, \quad y=4+\cos t, \quad 0 \leqslant t \leqslant
View solution Problem 21
Show that a conic with focus at the origin, eccentricity \(e,\) and directrix \(x=-d\) has polar equation $$r=\frac{e d}{1-e \cos \theta}$$
View solution