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Q 44.

Expert-verifiedFound in: Page 731

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

The given equation in polar coordinates is:

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

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

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

$r=2\mathrm{sin}\theta \mathrm{cos}\theta \xb7\left(2{\mathrm{cos}}^{2}\theta -1\right)$

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

localid="1649319886678" $r=2\xb7\frac{y}{r}\xb7\frac{x}{r}\left(2\xb7\frac{{x}^{2}}{{r}^{2}}-1\right)\phantom{\rule{0ex}{0ex}}r=2\xb7\frac{xy}{{r}^{2}}\left(2\xb7\frac{{x}^{2}}{{r}^{2}}-1\right)\phantom{\rule{0ex}{0ex}}r=\frac{2xy}{{r}^{2}}\left(\frac{2{x}^{2}-{r}^{2}}{{r}^{2}}\right)$

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

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