Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Graph the following equations. Then use arrows and labeled points to indicate how the curve is generated as \(\theta\) increases from 0 to \(2 \pi\). $$r=\frac{1}{1+2 \cos \theta}$$

Short Answer

Expert verified
Based on the solution provided, the short answer can be written as follows: The polar equation \(r = \frac{1}{1+2 \cos \theta}\) creates a curve formed by two loops as \(\theta\) increases from 0 to \(2 \pi\). The first smaller loop is formed when \(\theta\) ranges from 0 to \(\pi\), and the second larger loop, symmetric to the first one, is formed when \(\theta\) ranges from \(\pi\) to \(2 \pi\). The curve is generated in a counterclockwise direction, as indicated by arrows on the graph.

Step by step solution

01

Analyze the Polar Equation

We begin by analyzing the equation \(r = \frac{1}{1+2 \cos \theta}\). The main element here is the cosine function, which varies between -1 and 1 as \(\theta\) increases from 0 to \(2 \pi\). Let's analyze the equation as follows: - For \(\theta = 0\): \(\cos \theta = 1\), so \(r = \frac{1}{1+2 (1)} = \frac{1}{3}\) - For \(\theta = \frac{\pi}{2}\): \(\cos \theta = 0\), so \(r = \frac{1}{1+2 (0)} = 1\) - For \(\theta = \pi\): \(\cos \theta = -1\), so \(r = \frac{1}{1+2(-1)} = -1\) - For \(\theta = \frac{3 \pi}{2}\): \(\cos \theta = 0\), so \(r = \frac{1}{1+2 (0)} = 1\) Since the cosine function repeats itself in the range from 0 to \(2 \pi\), the curve will have a loop shape.
02

Plot the Polar Equation

To plot the polar equation \(r = \frac{1}{1+2 \cos \theta}\), we first create a polar grid. Next, we can choose values for \(\theta\) from 0 to \(2 \pi\) and calculate the corresponding \(r\) values using the polar equation. As a result, we obtain a series of points in polar coordinates. Connect the points one by one, following the order in which they appear as \(\theta\) increases. Observe that for \(\theta\) in the range from 0 to \(\pi\), the curve forms a smaller loop, while for \(\theta\) in the range from \(\pi\) to \(2 \pi\), the curve forms a larger, symmetric loop in relation to the first one.
03

Add Arrows and Labeled Points

Now that we have graphed the curve and identified how it evolves as \(\theta\) increases from 0 to \(2 \pi\), we can add arrows to indicate the direction in which the curve is generated. Following the sequence of points, place an arrowhead at the endpoint of each segment, pointing towards the next point. Label specific points on the curve corresponding to key angles, such as \(\theta = 0, \frac{\pi}{2}, \pi, \frac{3 \pi}{2}\), and \(2 \pi\). For each labeled point, indicate its polar coordinates, \((r, \theta)\), alongside the label. This completes the graphical representation of the polar equation, with arrows indicating how the curve is generated as \(\theta\) increases from 0 to \(2 \pi\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the curve \(r=f(\theta)=\cos a^{\theta}-1.5\) where \(a=(1+12 \pi)^{1 /(2 \pi)} \approx 1.78933\) (see figure). a. Show that \(f(0)=f(2 \pi)\) and find the point on the curve that corresponds to \(\theta=0\) and \(\theta=2 \pi\) b. Is the same curve produced over the intervals \([-\pi, \pi]\) and \([0,2 \pi] ?\) c. Let \(f(\theta)=\cos a^{\theta}-b,\) where \(a=(1+2 k \pi)^{1 /(2 \pi)}, k\) is an integer, and \(b\) is a real number. Show that \(f(0)=f(2 \pi)\) and that the curve closes on itself. d. Plot the curve with various values of \(k\). How many fingers can you produce?

A simplified model assumes that the orbits of Earth and Mars are circular with radii of 2 and \(3,\) respectively, and that Earth completes one orbit in one year while Mars takes two years. When \(t=0,\) Earth is at (2,0) and Mars is at (3,0) both orbit the Sun (at (0,0) ) in the counterclockwise direction. The position of Mars relative to Earth is given by the parametric equations \(x=(3-4 \cos \pi t) \cos \pi t+2, \quad y=(3-4 \cos \pi t) \sin \pi t\) a. Graph the parametric equations, for \(0 \leq t \leq 2\) b. Letting \(r=(3-4 \cos \pi t),\) explain why the path of Mars relative to Earth is a limaçon (Exercise 89).

Consider the parametric equations $$ x=a \cos t+b \sin t, \quad y=c \cos t+d \sin t $$ where \(a, b, c,\) and \(d\) are real numbers. a. Show that (apart from a set of special cases) the equations describe an ellipse of the form \(A x^{2}+B x y+C y^{2}=K,\) where \(A, B, C,\) and \(K\) are constants. b. Show that (apart from a set of special cases), the equations describe an ellipse with its axes aligned with the \(x\) - and \(y\) -axes provided \(a b+c d=0\) c. Show that the equations describe a circle provided \(a b+c d=0\) and \(c^{2}+d^{2}=a^{2}+b^{2} \neq 0\)

The region bounded by the parabola \(y=a x^{2}\) and the horizontal line \(y=h\) is revolved about the \(y\) -axis to generate a solid bounded by a surface called a paraboloid (where \(a>0\) and \(h>0\) ). Show that the volume of the solid is \(\frac{3}{2}\) the volume of the cone with the same base and vertex.

Graph the following equations. Then use arrows and labeled points to indicate how the curve is generated as \(\theta\) increases from 0 to \(2 \pi\). $$r=\frac{1}{1-2 \cos \theta}$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free