coexpennan

2021-08-11

Find an equation of the following curve in polar coordinates and describe the curve.

davonliefI

Solution:
1.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,
$\frac{y}{x}=\frac{\left(1+\mathrm{cos}t\right)\mathrm{sin}t}{\left(1+\mathrm{cos}t\right)\mathrm{cos}t}=\mathrm{tan}t$
2.
${x}^{2}+{y}^{2}={\left(1+\mathrm{cos}t\right)}^{2}$
${\left(1+\mathrm{cos}t\right)}^{2}={\left(1+\frac{1}{\mathrm{sec}t}\right)}^{2}$
${\left(1+\frac{1}{\mathrm{sec}t}\right)}^{2}=\left(1+\frac{1}{\sqrt{1+{\mathrm{tan}}^{2}t}}\right)$
Therefore,
${x}^{2}+{y}^{2}={\left(1+\frac{1}{\sqrt{1+{\mathrm{tan}}^{2}t}}\right)}^{2}$
=${\left(1+\frac{1}{\sqrt{1+{\left(\frac{y}{x}\right)}^{2}}}\right)}^{2}$
=${\left(1+\frac{1}{\sqrt{1+\frac{1}{\sqrt{\frac{{x}^{2}+{y}^{2}}{{x}^{2}}}}}}\right)}^{2}$
=${\left(1+\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}\right)}^{2}$
3. For polar form, substitute$x=r\mathrm{cos}\theta ,y=r\mathrm{cos}\theta ,{r}^{2}={x}^{2}+{y}^{2},\theta ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right){\left(1+\frac{1}{\sqrt{1+\frac{1}{\sqrt{\frac{{x}^{2}+{y}^{2}}{{x}^{2}}}}}}\right)}^{2}$
${r}^{2}={\left(1+\frac{r\mathrm{cos}\theta }{{\sqrt{r}}^{2}}\right)}^{2}$
${r}^{2}=\left(1+\mathrm{cos}\theta \right)$
Thus, the required solution is $r=1+\mathrm{cos}\theta$

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