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Q35E

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Essential Calculus: Early Transcendentals
Found in: Page 523
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

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

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

A form of the equation is \({{\rm{r}}^{\rm{2}}}{\rm{ = }}{{\rm{a}}^{\rm{2}}}{\rm{sin2\theta ,}}\)and it lies on \({\rm{x = y}}\).

See the step by step solution

Step by Step Solution

Step 1: Simplest way.

A form of the equation is given by\({{\rm{r}}^{\rm{2}}}{\rm{ = }}{{\rm{a}}^{\rm{2}}}{\rm{sin2\theta ,}}\)and it lies on \({\rm{x = y}}\).

Polar Coordinate

\(\begin{array}{l}r = \sqrt a ,\\\theta = \frac{\pi }{4}\end{array}\)

Therefore,

\(\begin{aligned}{l}{\rm{r = 3,}}\\{\rm{\theta = }}\frac{{\rm{\pi }}}{{\rm{4}}} \cdot \left( {{\rm{ \pm 3, \pm }}\frac{{\rm{\pi }}}{{\rm{4}}}} \right)\end{aligned}\)

Step 2:Alternaye Way.

Periodicity

\(\begin{aligned}{l}{\rm{9sin2\theta }}\\\frac{{{\rm{2\pi }}}}{{\rm{2}}}{\rm{ = \pi }}{\rm{.}}\end{aligned}\)

Now, \({{\rm{r}}^{\rm{2}}}{\rm{ = 9sin2\theta }}\)can be written as \({\rm{r = \pm 3}}\sqrt {{\rm{sin2\theta }}} {\rm{.}}\)

\({\rm{sin2\theta }}\)cannot be negative. So, any values \(\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ < \theta < \pi }}\)would be discarded.

An increment of \(\frac{{\rm{\pi }}}{{\rm{4}}}\) into \({\rm{\theta }}\)to solve for\({\rm{r}}\).

\({\rm{\theta = 0}}\;\;\;{\rm{r = 0}}\)

\({\rm{\theta = }}\frac{{\rm{\pi }}}{{\rm{4}}}\;\;\;{\rm{r = \pm 3}}\)(Here, \({\rm{\theta }}\)is \( - 3\), then \(\left( {{\rm{ - 3,}}\frac{{\rm{\pi }}}{{\rm{4}}}} \right)\)would be located in III quadrants.)

\({\rm{\theta = }}\frac{{\rm{\pi }}}{{\rm{2}}}\;\;\;{\rm{r = 0}}\)

Discard \(\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ < \theta < \pi \ldots }}\)

\({\rm{\theta = \pi }}\;\;\;{\rm{r = 0}}\)

Step 3:Cartesian Graph.

Step 4:Polar Graph.

Thus, a form of the equation is \({{\rm{r}}^{\rm{2}}}{\rm{ = }}{{\rm{a}}^{\rm{2}}}{\rm{sin2\theta ,}}\)and it lies on \({\rm{x = y}}\).

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