/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Find the area inside the larger ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the area inside the larger loop and outside the smaller loop of the limacon \( r = \dfrac{1}{2} + \cos \theta \).

Short Answer

Expert verified
The area between the larger loop and the smaller loop of the limacon is found by subtracting the area of the smaller loop from the larger loop.

Step by step solution

01

Understand the Limacon Equation

The given polar equation is a limacon of the form \( r = \frac{1}{2} + \cos \theta \). This type of limacon can have inner loops, and our task is to find the area between the larger and the smaller loop.
02

Find Inner Loop Condition

First, we find where the inner loop exists. We know that the inner loop occurs when \( r \) becomes zero. Solve \( 0 = \frac{1}{2} + \cos \theta \). This gives \( \cos \theta = -\frac{1}{2} \). This occurs at \( \theta = \frac{2\pi}{3} \) and \( \theta = \frac{4\pi}{3} \).
03

Define Boundaries for the Areas

The outer loop range is from \( \theta = 0 \) to \( \theta = 2\pi \). The inner loop exists between \( \theta = \frac{2\pi}{3} \) to \( \theta = \frac{4\pi}{3} \). We will integrate to find the areas below.
04

Find the Area of the Outer Loop

To find the area of the outer loop, use the formula for polar areas: \[ A = \frac{1}{2} \int_0^{2\pi} (\frac{1}{2} + \cos \theta)^2 d\theta \]. Evaluate this integral to find the area of the entire curve.
05

Find the Area of the Inner Loop

To find the area of the inner loop, use the same formula but over the range where the inner loop exists: \[ A = \frac{1}{2} \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} (\frac{1}{2} + \cos \theta)^2 d\theta \]. Calculate this integral to find the area of the inner loop.
06

Calculate the Net Area Outside Inner Loop Inside Outer Loop

Subtract the area of the inner loop from the area of the outer loop to find the area between the loops: \[ A_{net} = A_{outer} - A_{inner} \]. This will yield the area that lies outside the inner loop, but inside the outer loop.

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Key Concepts

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

³¢¾±³¾²¹Ã§´Ç²Ô
A ³¢¾±³¾²¹Ã§´Ç²Ô is a type of polar graph that is plotted with a specific set of equations. Typically, it comes in the form of an equation like \( r = a + b \cos \theta \) or \( r = a + b \sin \theta \). These graphs can have various shapes, such as loops, dimples, or cardiods, depending on the values of \( a \) and \( b \). In the given exercise, the equation \( r = \frac{1}{2} + \cos \theta \) represents a ³¢¾±³¾²¹Ã§´Ç²Ô with an inner loop because the value of \( a \) is less than \( b \).Some key properties include:
  • If \( |a| < |b| \), the ³¢¾±³¾²¹Ã§´Ç²Ô has an inner loop, like the one we are working on.
  • If \( |a| = |b| \), the ³¢¾±³¾²¹Ã§´Ç²Ô is a "cardioid," resembling a heart shape without an inner loop.
  • If \( |a| > |b| \), the ³¢¾±³¾²¹Ã§´Ç²Ô has no inner loop and might have a dimple instead.
Understanding these properties helps in identifying the features of the graph, including loops which are integral in calculating areas enclosed by such curves.
Polar Coordinates
Polar coordinates are a different way of expressing points in the plane compared to the usual Cartesian coordinates. Instead of using \(x\) and \(y\) to indicate position, they use a radius \(r\) and an angle \(\theta\). The distance from the origin (radius) and the angle relative to the positive x-axis describe a point's location.Polar coordinates are especially useful when dealing with curves that display circular or angular symmetry, such as circles, spirals, and of course, ³¢¾±³¾²¹Ã§´Ç²Ôs. This coordinate system simplifies integration and differentiation processes because it aligns more naturally with the types of shapes and areas these curves describe. For instance, in our exercise, the ³¢¾±³¾²¹Ã§´Ç²Ô equation \( r = \frac{1}{2} + \cos \theta \) directly relates the radius at any angle to its length.Switching between polar and Cartesian coordinates is possible with these formulas:
  • From Polar to Cartesian: \( x = r \cos \theta \), \( y = r \sin \theta \).
  • From Cartesian to Polar: \( r = \sqrt{x^2 + y^2} \), \( \theta = \tan^{-1}(\frac{y}{x}) \).
Integration
Integration in the context of polar coordinates can help calculate the area within curves like ³¢¾±³¾²¹Ã§´Ç²Ôs. To find the area within these curves, we use integral calculus to sum infinitely many tiny slices of the area.The formula for finding the area enclosed by a polar curve \( r(\theta) \) from \( \theta = a \) to \( \theta = b \) is:\[A = \frac{1}{2} \int_a^b [r(\theta)]^2 d\theta\]In this exercise, calculating the areas for both the inner and outer loop involves using this formula twice, but with different integration limits for each part.
  • Outer Loop Area: Integrate from \( 0 \) to \( 2\pi \), as specified.
  • Inner Loop Area: Integrate from \( \frac{2\pi}{3} \) to \( \frac{4\pi}{3} \), where the inner loop occurs.
The understanding of integration in polar coordinates thus becomes vital, as it allows us to find the specific area between the loops by subtracting their respective integrated areas.
Trigonometry
Trigonometry plays a key role in solving problems involving polar coordinates. It helps to analyze and solve the exercise involving ³¢¾±³¾²¹Ã§´Ç²Ôs by providing tools to evaluate angles and calculate distances along the curves.For example, trigonometric identities help convert between angles and their cosines or sines, such as finding where \( \cos \theta = -\frac{1}{2} \) results in specific critical points for our ³¢¾±³¾²¹Ã§´Ç²Ô: \( \theta = \frac{2\pi}{3} \) and \( \theta = \frac{4\pi}{3} \).Important points about trigonometry in this context include:
  • Understanding the unit circle and specific values of trigonometric functions based on angle locations.
  • Utilizing trigonometric identities for simplifying integrals.
  • Identifying specific angles that lead to values like the ones seen in the inner loop condition solution.
In our exercise, recognizing trigonometric solutions allows for correctly setting up integrals and discerning where loops form in the ³¢¾±³¾²¹Ã§´Ç²Ô graph.

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Most popular questions from this chapter

(a) Eliminate the parameter to find a Cartesian equation of the curve. (b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as parameter increases. \( x = \dfrac{1}{2}\cos\theta \), \( \quad y = 2\sin\theta \), \( \quad 0 \leqslant \theta \leqslant \pi \)

Find the area between a large loop and the enclosed small loop of the curve \( r = 1 + 2 \cos 3 \theta \).

The orbit of Mars around the sun is an ellipse with eccentricity 0.093 and semi major axis \(2.28 \times 10^{8} \mathrm{km} .\) Find a polar equation for the orbit.

Find an equation for the conic that satisfies the given conditions. Ellipse, center \( (-1, 4) \), vertex \( (-1, 0) \), focus \( (-1, 6) \)

The LORAN (LOng RAnge Navigation) radio navigation system was widely used until the 1990s when it was superseded by the GPS system. In the LORAN system, two radio stations located at \( A \) and \( B \) transmit simultaneous signals to a ship or an aircraft located at \( P \). The onboard computer converts the time difference in receiving these signals into a distance difference \( | PA | - | PB | \), and this, according to the definition of a hyperbola, locates the ship or aircraft on one branch of a hyperbola (see the figure). Suppose that station \( B \) is located \( 400\;mi \) due east of station \( A \) on a coastline. A ship received the signal from \( B \) \( 1200 \) microseconds \( (\mu s) \) before it received the signal from \( A \). (a) Assuming that radio signals travel at a speed of \( 980\; ft/\mu s \), find an equation of the hyperbola on which the ship lies. (b) If the ship is due north of \( B \). how far off the coastline is the ship?

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