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Found in: Page 523

### Essential Calculus: Early Transcendentals

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

# 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 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.

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