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Chapter 8: Problem 29

Convert the polar equation to rectangular form and sketch its graph. $$ r=\sin \theta $$

Short Answer

Expert verified
The rectangular form of the equation is \(y = r\). The graph of this equation in polar form is a semi-circle in the right half of the coordinate system.

Step by step solution

01

Convert the Polar Equation to Rectangular Form

To convert from polar coordinates \((r,\theta)\) to rectangular coordinates \((x,y)\), we need to use the relationships \(r^2 = x^2 + y^2\) and \(y = r \sin \theta\). As the equation is \(r=\sin \theta\), the conversion will give \(r = r \sin \theta\). This simplifies, through division by \(r\) (assuming \(r \neq 0\)), to \(1 = \sin \theta\). Also, by substituting \(y/r\) for \(\sin \theta\), we get \(1 = y/r\) which further gives the rectangular equation as \(y = r\).
02

Sketch the graph

According to the rectangular equation obtained, \(y = r\), the curve is not a usual function in rectangular coordinates because it doesn't represent \(y\) as a function of \(x\). It represents a whole circle centered at the origin. But in the polar form of \(r=\sin\theta\), we observe that for each \(\theta\), the radius \(r\) is non-negative. So, it only represents the right semi-circle from \((0,0)\) to \((0,2)\).
03

Completion and Review

Upon reviewing the process, we see the conversion from polar to rectangular form has been correctly achieved and the graph accurately represents our final equation. This graph shows that the rectangular equation does not fully capture the nature of the polar equation, as it can only represent \(y\) as a function of \(x\) while the polar equation \(r = \sin \theta\) represents a semi-circle.

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Most popular questions from this chapter

Which integral yields the arc length of \(r=3(1-\cos 2 \theta)\) ? State why the other integrals are incorrect. (a) \(3 \int_{0}^{2 \pi} \sqrt{(1-\cos 2 \theta)^{2}+4 \sin ^{2} 2 \theta} d \theta\) (b) \(12 \int_{0}^{\pi / 4} \sqrt{(1-\cos 2 \theta)^{2}+4 \sin ^{2} 2 \theta} d \theta\) (c) \(3 \int_{0}^{\pi} \sqrt{(1-\cos 2 \theta)^{2}+4 \sin ^{2} 2 \theta} d \theta\) (d) \(6 \int_{0}^{\pi / 2} \sqrt{(1-\cos 2 \theta)^{2}+4 \sin ^{2} 2 \theta} d \theta\)

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

Use a graphing utility to graph the curve represented by the parametric equations (indicate the orientation of the curve). Eliminate the parameter and write the corresponding rectangular equation. $$ x=4 \sec \theta, \quad y=3 \tan \theta $$

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. \(r=\frac{-1}{1-\sin \theta}\)

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.) \(\frac{\text { Conic }}{\text { Ellipse }} \quad \frac{\text { Eccentricity }}{e=\frac{3}{4}} \quad \frac{\text { Directrix }}{x=-2}\)
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