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Q 44.

Expert-verified
Found in: Page 731

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# In Exercises 32–47 convert the equations given in polar coordinates to rectangular coordinates.$r=\mathrm{sin}4\theta$

The required equation is ${\left({x}^{2}+{y}^{2}\right)}^{\frac{5}{2}}=2xy\left({x}^{2}-{y}^{2}\right)$.

See the step by step solution

## Step 1. Given information.

The given equation in polar coordinates is:

$r=\mathrm{sin}4\theta$

## Step 2. Find the equation in rectangular coordinates.

$r=\mathrm{sin}4\theta \phantom{\rule{0ex}{0ex}}r=2\mathrm{sin}2\theta \mathrm{cos}2\theta \left[\mathrm{Since}\mathrm{sin}4\theta =2\mathrm{sin}2\theta \mathrm{cos}2\theta \right]$

Now substitute $\mathrm{sin}2\theta =2\mathrm{sin}\theta \mathrm{cos}\theta \mathrm{and}\mathrm{cos}2\theta =2{\mathrm{cos}}^{2}\theta -1$,

$r=2\mathrm{sin}\theta \mathrm{cos}\theta ·\left(2{\mathrm{cos}}^{2}\theta -1\right)$

Substitute $\frac{x}{r}=\mathrm{cos}\theta \mathrm{and}\frac{y}{r}=\mathrm{sin}\theta$,

localid="1649319886678" $r=2·\frac{y}{r}·\frac{x}{r}\left(2·\frac{{x}^{2}}{{r}^{2}}-1\right)\phantom{\rule{0ex}{0ex}}r=2·\frac{xy}{{r}^{2}}\left(2·\frac{{x}^{2}}{{r}^{2}}-1\right)\phantom{\rule{0ex}{0ex}}r=\frac{2xy}{{r}^{2}}\left(\frac{2{x}^{2}-{r}^{2}}{{r}^{2}}\right)$

Now cross multiply,

${r}^{5}=2xy\left(2{x}^{2}-{r}^{2}\right)\phantom{\rule{0ex}{0ex}}{\left(\sqrt{{x}^{2}+{y}^{2}}\right)}^{5}=2xy\left(2{x}^{2}-\left({x}^{2}+{y}^{2}\right)\right)\left[{r}^{2}={x}^{2}+{y}^{2}\mathrm{and}r=\sqrt{{x}^{2}+{y}^{2}}\right]\phantom{\rule{0ex}{0ex}}{\left({x}^{2}+{y}^{2}\right)}^{\frac{5}{2}}=2xy\left(2{x}^{2}-\left({x}^{2}+{y}^{2}\right)\right)\phantom{\rule{0ex}{0ex}}{\left({x}^{2}+{y}^{2}\right)}^{\frac{5}{2}}=2xy\left({x}^{2}-{y}^{2}\right)$

Therefore, the equation in rectangular coordinates is ${\left({x}^{2}+{y}^{2}\right)}^{\frac{5}{2}}=2xy\left({x}^{2}-{y}^{2}\right)$.