/*! 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 53 Find the points of horizontal an... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 53

Find the points of horizontal and vertical tangency (if any) to the polar curve. $$ r=1-\sin \theta $$

Short Answer

Expert verified
The horizontal tangents occur at (r, \(\theta\)) coordinates where \(r = 1 - \sin(\theta)\), \(\theta = \frac{\pi}{2} + k\pi\) and k is an integer. The vertical tangents occur at (r, \(\theta\)) points where r = 0, \(\theta = \frac{\pi}{2} + 2k\pi\) and k is an integer.

Step by step solution

01

- Differentiate the Polar curve with respect to \(\theta\)

Differentiation of \(r = 1 - \sin(\theta)\) with respect to \(\theta\) gives: \[ \frac{dr}{d\theta} = - \cos(\theta) \] This derivative represents the rate of change of the radius \(r\) with respect to the angle \(\theta\).
02

- Find the points where derivative equals zero

We solve the equation \(- \cos(\theta) = 0\).From trigonometry, this occurs when \(\theta = \frac{\pi}{2} + k\pi\), where \(k\) is an integer. These are the angles at which horizontal tangents occur. Note, to get the corresponding points (r, \(\theta\)) on the curve, substitute \(\theta\) into \(r = 1 - \sin(\theta)\).
03

- Set r = 0

As vertical tangents occur at points where \(r = 0\), set \(r = 1 - \sin(\theta) = 0\). Solving the equation gives \(\sin(\theta) = 1\). From trigonometry, we know this happens at \(\theta = \frac{\pi}{2} + 2k\pi\), where k is an integer. These are the angles at which vertical tangents occur. We get the points (r ,\(\theta\)) by substituting \(\theta\) into \(r = 1 - \sin(\theta)\). Please note that set r=0 only gives us \(\theta\), as the corresponding r is zero.

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Most popular questions from this chapter

In Exercises \(7-16,\) find the eccentricity and the distance from the pole to the directrix of the conic. Then sketch and identify the graph. Use a graphing utility to confirm your results. \(r=\frac{4}{1+2 \cos \theta}\)

Use the formula for the arc length of a curve in parametric form to derive the formula for the arc length of a polar curve.

In Exercises 43-46, find the area of the surface formed by revolving the curve about the given line. $$ \begin{array}{lll} \underline{\text { Polar Equation }} & \underline{\text { Interval }} & \underline{\text { Axis of Revolution }} \\ r=a \cos \theta & 0 \leq \theta \leq \frac{\pi}{2} & \theta=\frac{\pi}{2} \end{array} $$

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. $$ r=\theta, \quad 0 \leq \theta \leq \pi $$

In Exercises \(7-16,\) find the eccentricity and the distance from the pole to the directrix of the conic. Then sketch and identify the graph. Use a graphing utility to confirm your results. \(r=\frac{6}{2+\cos \theta}\)
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