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Essential Calculus: Early Transcendentals
Found in: Page 535
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

To determine,

a) The eccentricity of the equation, \(r = \frac{4}{{5 - 4\sin \theta }}\).

b) To identify the shape of conic.

c) The equation of the directrix.

d) To sketch the conic.

a) The eccentricity of the equation, \(r = \frac{4}{{5 - 4\sin \theta }}\) is \(e = \frac{4}{5}\).

b) The conic shape is Ellipse.

c) The equation of directrix is \(y = - 1\).

d) It is observed that the vertices of the ellipse are \(\left( {4,\frac{\pi }{2}} \right)\) and \(\left( {\frac{4}{9},\frac{{3\pi }}{2}} \right)\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given data

The given polar equation is \(r = \frac{4}{{5 - 4\sin \theta }}\).

Step 2: Theorem of Polar equation

A polar equation of the form \(r = \frac{{ed}}{{{l_{ \pm e}}\sin \theta }}\) or \(r = \frac{{ed}}{{{1_{ \pm e}}\cos \theta }}\) represents a conic section with eccentricity \(e\).

The conic is an ellipse if \(e < 1\), a parabola if \(e = 1\), or a hyperbola if \(e > 1\).

Step 3: Calculation for the eccentricity of the equation

a)

Obtain the eccentricity of the given equation.

The polar equation for the given equation will be either \(r = \frac{{ed}}{{{1_{ \pm e}}\sin \theta }}\) or \(r = \frac{{ed}}{{{1_{ \pm e}}\cos \theta }}\).

Since the given polar equation is \(r = \frac{4}{{5 - 4\sin \theta }}\), divide the given equation by \(5\).

\(\begin{aligned}{l}r = \frac{4}{{5 - 4\sin \theta }}\\r = \frac{{\frac{4}{3}}}{{\frac{3}{5} - \frac{4}{5}\sin \theta }}\\r = \frac{{\frac{4}{3}}}{{1 - \frac{4}{3}\sin \theta }}\end{aligned}\)

From the above-mentioned theorem, it is clear that \(e = \frac{4}{5}\).

Therefore, the eccentricity of the equation, \(r = \frac{4}{{5 - 4\sin \theta }}\) is \(e = \frac{4}{5}\).

Step 4: Identify the shape of conic

b)

From the above-mentioned theorem, there are some conditions to identify the conic. They are, if eccentricity is less than one, then the conic is Ellipse.

If eccentricity is equal to one, then the conic is Parabola.

If eccentricity is greater than one, then the conic is Hyperbola.

From part (a), the calculated eccentricity is \(\frac{4}{5}\), that is less than one.

Hence, the conic is Ellipse.

Step 5: Calculation of the equation of the directrix

c)

Since the eccentricity \(e = \frac{4}{5}\), the polar equation can be rewrite as \(r = \frac{{\frac{4}{5}}}{{1 - \frac{4}{5}\sin \theta }}\).

Compare the equation with the standard equation \(r = \frac{{ed}}{{{l_{ \pm e}}\sin \theta }}\).

Equate the numerator as shown below.

\(\begin{aligned}{c}ed = \frac{4}{5}\\\frac{4}{5}d = \frac{4}{5}\\d = 1\end{aligned}\)

That is, \(d = 1\).

Since the term \( - e\sin \theta \) appears in the denominator, obviously, the directrix must be below the focus at the origin.

Therefore, the equation of directrix is \(y = - 1\).

Step 6: Calculation of the values of \(r\)for different \(\theta \)

d)

The polar co-ordinates are \(x = r\cos \theta \) and \(y = r\sin \theta \).

Substitute 0 for \(\theta \) in \(r = \frac{4}{{5 - 4\sin \theta }}\).

\(\begin{aligned}{l}r = \frac{4}{{5 - 4\sin (0)}}\\r = \frac{4}{5}\end{aligned}\)

Similarly calculate the values of \(r\) for different values of \(\theta \).

Tabulate the calculated values as shown below in the table.

Step 7: Sketch the diagram of an ellipse

Sketch the ellipse \(r = \frac{4}{{5 - 4\sin \theta }}\) as shown below in the figure.

From figure, it is observed that the vertices of the ellipse are \(\left( {4,\frac{\pi }{2}} \right)\) and \(\left( {\frac{4}{9},\frac{{3\pi }}{2}} \right)\).

Most popular questions for Math Textbooks

Sketch the region in the plane consisting of points whose polar coordinates satisfy the given conditions.

\({\rm{r}} \ge {\rm{0,}}\frac{{\rm{\pi }}}{{\rm{4}}} \le {\rm{\theta }} \le \frac{{{\rm{3\pi }}}}{{\rm{4}}}\)

Plot the point whose polar coordinates are given. Then find two other pairs of polar coordinates of this point, one with \(r > 0\)and one with \(r < 0\).

\(\begin{aligned}{l}(a)(1,7\pi /4)\\(b)( - 3,\pi /6)\\(c)(1, - 1)\end{aligned}\)

When recording lives performances, sound engineers often use a microphone with a cardioid pickup pattern because it suppresses noise from the audience. Suppose the microphone is placed\({\rm{4\;m}}\] from the front of the stage (as in the figure) and the boundary of the optimal pickup region is given by the cardioid\({\rm{r = 8 + 8sin\theta }}\], which\({\rm{r}}\] is measured in meters and the microphone is at the pole. The musicians want to know the area they will have on stage within the optimal pickup range of the microphone. Answer their question.

Write a polar equation of a conic with the focus at the origin and the given data.

\({\rm{Hyperbola, eccentricity 3, directrix}}\)\({\rm{r = - 6csc\theta }}\)

Sketch the curve with the given polar equation by first sketching the graph \({\rm{r}}\)as a function of \({\rm{\theta }}\)Cartesian coordinates.

\({{\rm{r}}^{\rm{2}}}{\rm{ = 9sin2\theta }}\)
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