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Chapter 11: Problem 59

Replace the Cartesian equations in Exercises \(53-66\) with equivalent polar equations. $$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$$

Short Answer

Expert verified
The polar equation is \( r = \sqrt{\frac{1}{\frac{\cos^{2} \theta}{9} + \frac{\sin^{2} \theta}{4}}} \).

Step by step solution

01

Understand the Cartesian Equation

We start with the equation \( \frac{x^{2}}{9}+\frac{y^{2}}{4}=1 \). This is the equation of an ellipse in Cartesian coordinates, where the semi-major axis is 3 (along the x-axis) and the semi-minor axis is 2 (along the y-axis).
02

Convert Cartesian Coordinates to Polar Coordinates

In polar coordinates, \( x = r \cos \theta \) and \( y = r \sin \theta \). Substitute these into the equation \( \frac{x^{2}}{9}+\frac{y^{2}}{4}=1 \).
03

Substitute Values into the Equation

Replace \( x \) and \( y \) in the equation with their polar equivalents: \[\frac{(r \cos \theta)^{2}}{9} + \frac{(r \sin \theta)^{2}}{4} = 1.\] Simplify this to get: \[\frac{r^{2} \cos^{2} \theta}{9} + \frac{r^{2} \sin^{2} \theta}{4} = 1.\]
04

Factor Out \( r^{2} \)

Factor out \( r^{2} \) from the left side of the equation: \[r^{2} \left( \frac{\cos^{2} \theta}{9} + \frac{\sin^{2} \theta}{4} \right) = 1.\]
05

Isolate \( r^{2} \)

Divide both sides by \( \left( \frac{\cos^{2} \theta}{9} + \frac{\sin^{2} \theta}{4} \right) \) to isolate \( r^{2} \): \[r^{2} = \frac{1}{\frac{\cos^{2} \theta}{9} + \frac{\sin^{2} \theta}{4}}.\]
06

Simplify the Polar Equation

Simplify the expression: \[r = \sqrt{\frac{1}{\frac{\cos^{2} \theta}{9} + \frac{\sin^{2} \theta}{4}}}.\] This is the equivalent polar equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cartesian Coordinates
The concept of Cartesian Coordinates is fundamental in understanding how we represent points in a two-dimensional plane. They are named after the French mathematician Ren茅 Descartes. Here, every point on the plane is specified by a pair of numerical coordinates.
  • The first number, commonly called \(x\), is the position along the horizontal axis.
  • The second number, \(y\), represents the position along the vertical axis.
Together, these define the coordinate \((x, y)\). For example, if you have a point \( (3, 2) \), it tells us that the point is located 3 units along the x-axis (horizontal) and 2 units along the y-axis (vertical).
Cartesian coordinates are very intuitive and provide a straightforward way to plot points, analyze shapes, and solve geometry problems.
Ellipse Equation
An ellipse is a smooth, closed curve, which is essentially an elongated circle. The equation of an ellipse in Cartesian coordinates is given by \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where:
  • \(a\) is the length of the semi-major axis (along the x-axis if \(a > b\)).
  • \(b\) is the length of the semi-minor axis (along the y-axis if \(a > b\)).
In our original exercise, the ellipse is defined by \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \). Here, \(a = 3\) and \(b = 2\), meaning the ellipse is centered at the origin (0,0) and extends 3 units along the x-axis and 2 units along the y-axis. Ellipses are important in many areas of mathematics, physics, and engineering due to their unique properties and the way they model the presence of two focal points.
Coordinate Conversion
Converting between Cartesian and polar coordinates allows us to leverage different systems for solving problems more easily. In polar coordinates, a point is defined by \((r, \theta)\), where:
  • \(r\) is the radius or distance from the origin.
  • \(\theta\) is the angle from the positive x-axis.
To convert from Cartesian \((x, y)\) to polar \((r, \theta)\):
  • Use \(x = r \cos\theta\).
  • Use \(y = r \sin\theta\).
  • Find \(r\) as \(\sqrt{x^2 + y^2}\).
  • Determine \(\theta\) using \(\tan^{-1}(y/x)\).
In our exercise, the coordinate conversion helps transform the ellipse equation from Cartesian to polar form. This is done by substituting polar equivalents, simplifying, and rearranging terms to get the equation in terms of \(r\) and \(\theta\). Learning these conversions is key to solving complex equations in a variety of mathematical contexts.

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Most popular questions from this chapter

Cycloid a. Find the length of one arch of the cycloid \(x=a(t-\sin t), \quad y=a(1-\cos t)\) b. Find the area of the surface generated by revolving one arch of the cycloid in part (a) about the \(x\) -axis for \(a=1 .\)

Replace the Cartesian equations in Exercises \(53-66\) with equivalent polar equations. $$x=7$$

The witch of Maria Agnesi The bell-shaped witch of Maria Agnesi can be constructed in the following way. Start with a circle of radius \(1,\) centered at the point \((0,1),\) as shown in the accompanying figure. Choose a point \(A\) on the line \(y=2\) and connect it to the origin with a line segment. Call the point where the segment crosses the circle \(B\) . Let \(P\) be the point where the vertical line through \(A\) crosses the horizontal line through \(B\) . The witch is the curve traced by \(P\) as \(A\) moves along the line \(y=2\) . Find parametric equations and a parameter interval for the witch by expressing the coordinates of \(P\) in terms of \(t\) the radian measure of the angle that segment \(O A\) makes with the positive \(x\) -axis. The following equalities (which you may assume) will help. $$\begin{array}{ll}{\text { a. } x=A Q} & {\text { b. } y=2-A B \sin t} \\\ {\text { c. } A B \cdot O A=(A Q)^{2}}\end{array}$$

A cone The line segment joining the origin to the point \((h, r)\) is revolved about the \(x\) -axis to generate a cone of height \(h\) and base radius \(r .\) Find the cone's surface area with the parametric equations \(x=h t, y=r t, 0 \leq t \leq 1 .\) Check your result with the geometry formula: Area \(=\pi r(\) slant height\().\)

Exercises \(35-38\) give information about the foci, vertices, and asymptotes of hyperbolas centered at the origin of the \(x y\) -plane. In each case, find the hyperbola's standard-form equation from the information given. $$ \begin{array}{l}{\text { Vertices: }( \pm 3,0)} \\ {\text { Asymptotes: } y=\pm \frac{4}{3} x}\end{array} $$
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