Braxton Pugh

2021-08-30

Find an equation of the following curve in polar coordinates and describe the curve.
$x=\left(1+\mathrm{cos}t\right)\mathrm{cos}t$
$y=\left(1+\mathrm{cos}t\right)\mathrm{sin}t.0\le t\le 2\pi$

### Answer & Explanation

Ayesha Gomez

Consider the given curves, $x=\left(1+\mathrm{cos}t\right)\mathrm{cos}t$
$y=\left(1+\mathrm{cos}t\right)\mathrm{sin}t$ Square and add the above equations, ${x}^{2}+{y}^{2}={\left(1+\mathrm{cos}t\right)}^{2}$ Divide the above equations, $\frac{y}{x}=\frac{\left(1+\mathrm{cos}t\right)\mathrm{sin}t}{\left(1+\mathrm{cos}t\right)\mathrm{cos}t}=\mathrm{tan}t$
${x}^{2}+{y}^{2}={\left(1+\mathrm{cos}t\right)}^{2}$
${\left(1+\mathrm{cos}t\right)}^{2}={\left(1+\frac{1}{\mathrm{sec}t}\right)}^{2}$
${\left(1+\frac{1}{\mathrm{sec}t}\right)}^{2}={\left(1+\frac{1}{\sqrt{1+{\mathrm{tan}}^{2}t}}\right)}^{2}$
Therefore, ${x}^{2}+{y}^{2}={\left(1+\frac{1}{\sqrt{1+{\mathrm{tan}}^{2}t}}\right)}^{2}$
$=\left(1+{\frac{1}{\sqrt{1+{\left(\frac{y}{x}\right)}^{2}}}}^{2}\right).$
$\left(1+{\frac{1}{\sqrt{\frac{{x}^{2}+{y}^{2}}{{x}^{2}}}}}^{2}\right).$
$\left(1+\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}\right).$
For polar form, substitute $x=r\mathrm{cos}\theta ,y=r\mathrm{sin}\theta ,{r}^{2}={x}^{2}+{y}^{2},\theta ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)$
${x}^{2}+{y}^{2}={\left(1+\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}\right)}^{2}$
${r}^{2}={\left(1+\frac{r\mathrm{cos}\theta }{\sqrt{{r}^{2}}}\right)}^{2}$
${r}^{2}={\left(1+\mathrm{cos}\theta \right)}^{2}$ Thus, required solution is $r=1+\mathrm{cos}\theta$

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