# Parametric to polar equations Find an equation of the following curve in polar coordinates and describe the curve. x = (1 + cos t) cos t, y = (1 + cos t) sin t, 0 leq t leq 2pi

Parametric to polar equations Find an equation of the following curve in polar coordinates and describe the curve. $x=\left(1+cost\right)cost,y=\left(1+cost\right)sint,0\le t\le 2\pi$
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Step 1

Consider the given curves, $x=\left(1+cost\right)cost$
$y=\left(1+cost\right)sint$ Square and add the above equations, ${x}^{2}+{y}^{2}=\left(1+cost{\right)}^{2}$ Divide the above equations, $\frac{x}{y}=\frac{\left(1+cost\right)sint}{\left(1+cost\right)cost}=tant$

Step 2

${x}^{2}+{y}^{2}=\left(1+cost{\right)}^{2}$
$\left(1+cost{\right)}^{2}={\frac{1+\left(1}{sect\right)}}^{2}$
$\left(1+\frac{1}{sect}{\right)}^{2}=\left(1+\frac{1}{\sqrt{\left(}1+ta{n}_{t}^{2}\right)}{\right)}^{2}$ Therefore, ${x}^{2}+{y}^{2}=\left(1+\frac{1}{\sqrt{\left(}1+ta{n}_{t}^{2}\right)}{\right)}^{2}$
$=\left(1+\frac{1}{\left(\sqrt{1}+\frac{1}{\sqrt{1}+\left(\frac{x}{y}{\right)}^{2}\right)}{\right)}^{2}}\right)$
$=\left(1+\frac{1}{\sqrt{\left(}\left(\frac{{x}^{2}+{y}^{2}}{{x}^{2}}\right)}\right){\right)}^{2}$
$=\left(1+\frac{x}{\sqrt{\left({x}^{2}+{y}^{2}\right)}}{\right)}^{2}$

Step 3

For polar form, substitute $x=rcos\theta ,y=rsin\theta ,{r}^{2}={x}^{2}+{y}^{2},\theta =ta{n}^{\left(}-1\right)\left(\frac{y}{x}\right)$
${x}^{2}+{y}^{2}\left(1+\frac{x}{\sqrt{\left(}\left({x}^{2}+{y}^{2}\right)\right)}{\right)}^{2}$
${r}^{2}=\frac{\left(}{+\left(rcos\theta }sqrt\left({r}^{2}\right){\right)}^{2}$
${r}^{2}=\left(1+cos\theta {\right)}^{2}$ Thus, the required solution is $r=1+cos\theta$

Jeffrey Jordon