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Chapter 10: Problem 38

Replace the polar equations in Exercises \(23-48\) by equivalent Cartesian equations. Then describe or identify the graph. $$ r \sin \theta=\ln r+\ln \cos \theta $$

Short Answer

Expert verified
The polar equation converts to the Cartesian equation \( y = \ln x \), representing a logarithmic function graph for \( x > 0 \).

Step by step solution

01

Convert Polar Equations to Cartesian Coordinates

To convert a polar equation to Cartesian form, recall that the relationships between polar and Cartesian coordinates are: - For polar coordinates \( (r, \theta) \), the corresponding Cartesian coordinates are \( (x, y) \) where \( x = r \cos \theta \) and \( y = r \sin \theta \).- The polar radius \( r \) can be found as \( r = \sqrt{x^2 + y^2} \), and \( \theta \) as \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \).With this information, consider the polar equation: \[ r \sin \theta = \ln r + \ln \cos \theta \]
02

Substitute Polar Terms into Cartesian Terms

Express each term in the equation using Cartesian coordinates:- Replace \( r \sin \theta \) with \( y \).- Recognize \( \ln r + \ln \cos \theta = \ln(r \cos \theta) \) using the property of logarithms \( \ln a + \ln b = \ln(ab) \).- Substitute \( r \cos \theta \) with \( x \) (since \( x = r \cos \theta \)).This gives us:\[ y = \ln x \]
03

Identify the Graph

The Cartesian equation \( y = \ln x \) corresponds to a natural logarithm function.- The graph of \( y = \ln x \) is defined for \( x > 0 \).- It forms a curve passing through the point \( (1, 0) \), increases slowly, and approaches negative infinity as \( x \) approaches zero from the positive side.This is the curve of a logarithmic function.

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Key Concepts

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

Cartesian Coordinates
Cartesian coordinates help us describe the position of a point in a plane using a pair of numerical values, representing distances from fixed reference lines. Here are the key things you need to know about them:
  • The Cartesian coordinate system consists of two axes: the x-axis (horizontal) and the y-axis (vertical).
  • Each point in the plane is represented with a pair of coordinates \((x, y)\), where \(x\) is the horizontal distance from the origin, and \(y\) is the vertical distance.
  • Unlike polar coordinates, which use a radius and an angle, Cartesian coordinates make it easy to plot points and functions directly on graphs.
To convert from polar to Cartesian coordinates, use:
  • \(x = r \cos \theta\)
  • \(y = r \sin \theta\)
Here, \((r, \theta)\) are the polar coordinates, with \(r\) being the radius and \(\theta\) the angle in radians.
Polar Equations
Polar equations represent relationships using polar coordinates, where each point is identified by a radius and an angle. Here's how they work:
  • Polar coordinates \((r, \theta)\) consist of the distance \(r\) from the origin and the angle \(\theta\).The angle is measured from the positive x-axis in a counter-clockwise direction.
  • Polar equations can describe curves that are not easily represented in Cartesian coordinates, like spirals and circles.
  • To solve a polar equation in Cartesian form, transform it using the relationships \(x = r \cos \theta\) and \(y = r \sin \theta\), then simplify.
For the given equation \( r \sin \theta = \ln r + \ln \cos \theta \), it's converted to the simpler Cartesian form \( y = \ln x \).
Natural Logarithm Function
The natural logarithm function, denoted as \(\ln x\), is a special logarithmic function with the base of Euler's number \(e\) (approximately 2.718). Here are its important characteristics:
  • \(\ln x\) is defined only for positive values of \(x\).
  • It signifies the power to which the base \(e\) must be raised to obtain the number \(x\).
  • The natural logarithm grows logarithmically, increasing at a slower rate as \(x\) becomes larger.
  • It's useful in solving equations involving exponential growth or decay and frequently appears in continuous compounding problems.
For the equation \(y = \ln x\), the function forms a curve that passes through the point \((1, 0)\), meaning when \(x = 1\), \(y\) is zero.
Graph Identification
Identifying a graph involves recognizing the visual representation of equations or functions in a coordinate system. Here's how you can understand it better:
  • The graph of a function gives a visual way to see all solutions of the equation.
  • For \(y = \ln x\), the graph is a natural logarithmic curve which is steep near \(x = 0\) and flattens as \(x\) increases.
  • It鈥檚 defined only for \(x > 0\), and approaches negative infinity as \(x\) goes to zero but stays positive.
  • This curve emerges prominently in the first quadrant of a Cartesian plane.
When identifying graphs, focus on key features such as asymptotes (lines that the graph approaches but never touches), intercepts, and the general shape of the curve.

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Most popular questions from this chapter

Exercises \(53-56\) give equations for hyperbolas and tell how many units up or down and to the right or left each hyperbola is to be shifted. Find an equation for the new hyperbola, and find the new center, foci, vertices, and asymptotes. $$ \frac{x^{2}}{16}-\frac{y^{2}}{9}=1, \quad \text { left } 2, \text { down } 1 $$

The hyperbola \(\left(x^{2} / 16\right)-\left(y^{2} / 9\right)=1\) is shifted 2 units to the right to generate the hyperbola $$\frac{(x-2)^{2}}{16}-\frac{y^{2}}{9}=1$$ a. Find the center, foci, vertices, and asymptotes of the new hyperbola. b. Plot the new center, foci, vertices, and asymptotes, and sketch in the hyperbola.

Find the areas of the regions Inside the lemniscate \(r^{2}=6 \cos 2 \theta\) and outside the circle \(r=\sqrt{3}\)

Exercises \(45-48\) give equations for parabolas and tell how many units up or down and to the right or left each parabola is to be shifted. Find an equation for the new parabola, and find the new vertex, focus, and directrix. $$ y^{2}=4 x, \quad \text { left } 2, \text { down } 3 $$

Sketch the regions in the \(x y\) -plane whose coordinates satisfy the inequalities or pairs of inequalities in Exercises \(69-74 .\) $$ \left(x^{2}+y^{2}-4\right)\left(x^{2}+9 y^{2}-9\right) \leq 0 $$
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