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Chapter 8: Problem 59

Use a graphing utility to graph the polar equation and find all points of horizontal tangency. $$ r=2 \csc \theta+5 $$

Short Answer

Expert verified
The points of horizontal tangency for the given polar equation occur at \( \theta = \frac{\pi}{2} + k\pi \) where k is an integer.

Step by step solution

01

Graph the Polar Equation

First the given polar equation should be graphed using a graphing utility. \(r = 2 \csc(\theta) + 5\). This will give a general understanding of the shape and where tangent lines might be.
02

Derive the equation

Next, differential calculus is performed on the polar equation to find the derivative. This requires knowledge on how to derive csc function. The derivative of \( \csc(\theta) \) is \( -\csc(\theta) \cot(\theta) \). Therefore, after applying the derivative, the equation becomes \( r' = -2\csc(\theta) \cot(\theta) \)
03

Look for Horizontal Tangency

In a polar coordinate graph, horizontal tangency occurs when r' = 0. So, setting \( -2\csc(\theta) \cot(\theta) \) equal to 0 will give the points of horizontal tangency.
04

Solve for Tangency Points

In order for \( -2\csc(\theta) \cot(\theta) \) to be 0, either \( \csc(\theta) \) or \( \cot(\theta) \) needs to be 0. However, csc function is never 0, hence we set \( \cot(\theta) = 0 \) which gives \( \theta = \frac{\pi}{2} + k\pi \) where k is an integer. These are the all the angles for which horizontal tangency occurs. To find the corresponding 'r' value we substitute \(\theta = \frac{\pi}{2} + k\pi \) into original equation \( r = 2 \csc(\theta) + 5 \)

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Most popular questions from this chapter

Use the integration capabilities of a graphing utility to approximate to two decimal places the area of the surface formed by revolving the curve about the polar axis. $$ r=\theta, \quad 0 \leq \theta \leq \pi $$

Sketch the curve represented by the parametric equations (indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the parameter. $$ x=e^{t}, \quad y=e^{3 t}+1 $$

In Exercises \(27-38,\) find a polar equation for the conic with its focus at the pole. (For convenience, the equation for the directrix is given in rectangular form.) \(\frac{\text { Conic }}{\text { Hyperbola }} \quad \frac{\text { Eccentricity }}{e=\frac{3}{2}} \quad \frac{\text { Directrix }}{x=-1}\)

Identify each conic. (a) \(r=\frac{5}{1-2 \cos \theta}\) (b) \(r=\frac{5}{10-\sin \theta}\) (c) \(r=\frac{5}{3-3 \cos \theta}\) (d) \(r=\frac{5}{1-3 \sin (\theta-\pi / 4)}\)

Sketch the curve represented by the parametric equations (indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the parameter. $$ x=e^{-t}, \quad y=e^{2 t}-1 $$
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