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Q 46.

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{cos}4\theta $

The required equation is ${\left({x}^{2}+{y}^{2}\right)}^{\frac{3}{2}}={x}^{2}-3{y}^{2}$.

The given equation in polar coordinates is:

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

$r=\mathrm{cos}4\theta \phantom{\rule{0ex}{0ex}}r=2{\mathrm{cos}}^{2}2\theta -1\left[\mathrm{cos}4\theta =\left(2{\mathrm{cos}}^{2}2\theta -1\right)\right]\phantom{\rule{0ex}{0ex}}r=2\left(2{\mathrm{cos}}^{2}2\theta -1\right)-1\left[\mathrm{cos}2\theta =2{\mathrm{cos}}^{2}\theta -1\right]\phantom{\rule{0ex}{0ex}}r=4{\mathrm{cos}}^{2}\theta -2-1\phantom{\rule{0ex}{0ex}}r=4{\mathrm{cos}}^{2}\theta -3$

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

role="math" localid="1649323840721" $r=4\xb7\frac{{x}^{2}}{{r}^{2}}-3\phantom{\rule{0ex}{0ex}}r=\frac{4{x}^{2}}{{r}^{2}}-3\phantom{\rule{0ex}{0ex}}r=\frac{4{x}^{2}-3{r}^{2}}{{r}^{2}}$

Cross multiply,

${r}^{3}=4{x}^{2}-3{r}^{2}\phantom{\rule{0ex}{0ex}}{\left(\sqrt{{x}^{2}+{y}^{2}}\right)}^{3}=4{x}^{2}-3\left({x}^{2}+{y}^{2}\right)\left[{r}^{2}={x}^{2}+{y}^{2}\mathrm{and}r=\sqrt{{x}^{2}+{y}^{2}}\right]\phantom{\rule{0ex}{0ex}}{\left({x}^{2}+{y}^{2}\right)}^{\frac{3}{2}}=4{x}^{2}-3{x}^{2}-3{y}^{2}\phantom{\rule{0ex}{0ex}}{\left({x}^{2}+{y}^{2}\right)}^{\frac{3}{2}}={x}^{2}-3{y}^{2}$

Therefore, the equation in rectangular coordinates is ${\left({x}^{2}+{y}^{2}\right)}^{\frac{3}{2}}={x}^{2}-3{y}^{2}$.

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