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Chapter 6: Problem 17

Test for symmetry and then graph each polar equation. $$r=2+2 \cos \theta$$

Short Answer

Expert verified
The polar equation \(r=2+2 \cos \theta\) is symmetric about the x-axis. Its Cartesian form is an ellipse centered at the origin, with a semi-minor axis along the y-axis of length 1, and a semi-major axis along the x-axis of length 2.

Step by step solution

01

Testing for Symmetry

For a polar equation \(r=f(\theta)\), we can test for symmetry about the x-axis by verifying if \(f(\theta)=f(-\theta)\), symmetry about the y-axis by checking if \(f(\theta) = f(\pi-\theta)\), and symmetry about the origin by checking if \(f(\theta) = f(\pi + \theta)\). When we plug \(-\theta\), \(\pi-\theta\), and \(\pi+\theta\) in place of \(\theta\) in the given equation \(r = 2 + 2 \cos \theta\), we get: \(r = 2 + 2 \cos(-\theta) = 2 + 2 \cos \theta\), \(r = 2 + 2 \cos(\pi-\theta)\), and \(r = 2 + 2 \cos(\pi+\theta)\)}. Notice that only the first equation matches our given equation. Thus, the polar equation is symmetric about the x-axis.
02

Converting the Polar Equation to Cartesian Form

In order to graph the polar equation, convert it to Cartesian form. The conversion formulas are \(x = r \cos \theta\) and \(y = r \sin \theta\). From the given polar equation, we know that \(r = 2 + 2 \cos \theta\). Substituting \(r\) into the Cartesian conversion formulas, we get \(x = (2 + 2 \cos \theta) \cos \theta\) and \(y = (2 + 2 \cos \theta) \sin \theta\). Simplifying, we get the Cartesian form of the polar equation: \(x^2 + (y^2)/4 = 1\).
03

Graphing the Polar Equation

Use the Cartesian form of the equation to draw the graph. Note that this equation is the equation for an ellipse, centered at the origin, with a semi-minor axis along the y-axis of length 1, and a semi-major axis along the x-axis of length 2. The ellipse is elongated along the x-axis, indicative of the positive cosine function in the initial polar equation.

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Most popular questions from this chapter

Determine whether each statement makes sense or does not make sense, and explain your reasoning. I'm working with a polar equation that failed the symmetry test with respect to \(\theta=\frac{\pi}{2},\) so my graph will not have this kind of symmetry.

Find the angle, in degrees, between \(\mathbf{v}\) and \(\mathbf{w}.\) $$\mathbf{v}=2 \cos \frac{4 \pi}{3} \mathbf{i}+2 \sin \frac{4 \pi}{3} \mathbf{j}, \quad \mathbf{w}=3 \cos \frac{3 \pi}{2} \mathbf{i}+3 \sin \frac{3 \pi}{2} \mathbf{j}$$

Find the angle between \(\mathbf{v}\) and \(\mathbf{w} .\) Round to the nearest tenth of a degree. $$\mathbf{v}=3 \mathbf{j}, \quad \mathbf{w}=4 \mathbf{i}+5 \mathbf{j}$$

Use a graphing utility to graph \(r=1+2 \sin n \theta\) for \(n=1,2,3,4,5,\) and \(6 .\) Use a separate viewing screen for each of the six graphs. What is the pattern for the number of large and small petals that occur corresponding to each value of \(n\) ? How are the large and small petals related when \(n\) is odd and when \(n\) is even?

Determine whether v and w are parallel, orthogonal, or neither. $$\mathbf{v}=3 \mathbf{i}-5 \mathbf{j}, \quad \mathbf{w}=6 \mathbf{i}+10 \mathbf{j}$$
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