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Q32E

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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.

\({\rm{r = 3cos6\theta }}\)

A polar equation is used to draw the curve by sketching a graph.

See the step by step solution

Step by Step Solution

Step 1: Cartesian diagram.

Step 2: Polar curve.

Given\({\rm{r = 3cos(6\theta )}}\).

See from the Cartesian diagram\({\rm{r = cos(6\theta )}}\) that the value of the function decreases from\({\rm{1}}\) to \({\rm{0}}\) in the interval\({\rm{0}} \le {\rm{\theta }} \le \frac{{\rm{\pi }}}{{{\rm{12}}}}\), and then decreases\({\rm{ - 1}}\) when the angle\({\rm{\theta }}\) moves to\(\frac{{\rm{\pi }}}{{\rm{6}}}\). The first leaf of the rose on the right side represents the same curve in polar coordinates. The polar curve is constructed similarly to the\({\rm{12}}\) leafed rose.

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