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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 $${\rm{r}}$$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 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 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 $${\rm{x = y}}$$.