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Chapter 6: Problem 16

Identify the conic and sketch its graph. \(r=\frac{7}{1+\sin \theta}\)

Short Answer

Expert verified
The conic represented by the given polar equation is a parabola

Step by step solution

01

Identify The Conic

Look at the equation and compare it with the general forms for conics in polar coordinates, \(r=\frac{p}{1+e\sin \theta}\) and \(r=\frac{p}{1+e\cos \theta}\). Consider the coefficient of sine in the denominator. If \(e<1\), the conic is an ellipse. If \(e=1\), the conic is a parabola. If \(e>1\), the conic is a hyperbola. In this case, \(r=\frac{7}{1+\sin \theta}\) matches the form \(r=\frac{p}{1+e\sin \theta}\) with \(p=7\) and \(e=1\), so it is a parabola.
02

Get Key Points For Sketch

In polar coordinates, get key points by plugging in simple angles for \(\theta\). For instance, \(\theta = 0\), \(\pi/2\), \(\pi\), \(3\pi/2\), \(2\pi\). For \(\theta = 0\) and \(2\pi\), \(r = 3.5\). For \(\theta = \pi\), \(r = 7\). For \(\theta = \pi/2\) and \(3\pi/2\), \(r\) is undefined. This suggests a kind of symmetry, particularly about the line \(\theta = \pi/2\).
03

Sketch The Graph

Plot these points, apply the symmetry, and connect them to form the conic section. Start with the theta values that define the extremes of the graph, note discontinuities (where \(r\) is undefined), and sketch the parabola from this. Without the actual plot, it's described as: from the main point at \(r=3.5\), \(\theta=0\) (which is also \(\theta=2\pi\)), the graph extends to a maximum point at \(r=7\), \(\theta=\pi\), then comes back to the same radius but at \(\theta=2\pi\). The graph has a discontinuity at \(\theta=\pi/2\) and \(3\pi/2\).

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