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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 of as a function of\({\rm{\theta }}\) in Cartesian coordinates.

\({\rm{r = \theta ,\theta > 0}}\)

Make a Cartesian graph with the horizontal axis \({\rm{\theta }}\)and the vertical axis \({\rm{r}}{\rm{.}}\)This is the same as \({\rm{y = x,}}\)which is a line.

See the step by step solution

Step by Step Solution

Step 1:  The following equation is.

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

To begin sketching the curve, identify several points by calculating \({\rm{r}}\)values for various\({\rm{\theta }}\)values. The curve will then be created by connecting the points \(\left( {{\rm{r, \theta }}} \right){\rm{.}}\)

Step 2: Points will be the first plot. 

\({\rm{(0,0),}}\frac{{\rm{\pi }}}{{\rm{3}}}{\rm{,}}\frac{{\rm{\pi }}}{{\rm{3}}}{\rm{,}}\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{,}}\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{,}}\frac{{{\rm{2\pi }}}}{{\rm{3}}}{\rm{,}}\frac{{{\rm{2\pi }}}}{{\rm{3}}}{\rm{,(\pi ,\pi ),}}\frac{{{\rm{4\pi }}}}{{\rm{3}}}{\rm{,}}\frac{{{\rm{4\pi }}}}{{\rm{3}}}{\rm{,}}\frac{{{\rm{3\pi }}}}{{\rm{2}}}{\rm{,}}\frac{{{\rm{3\pi }}}}{{\rm{2}}}\frac{{{\rm{5\pi }}}}{{\rm{3}}}{\rm{,}}\frac{{{\rm{5\pi }}}}{{\rm{3}}}{\rm{,(2\pi ,2\pi )}}\)

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