Question

# Find an equation of the following curve in polar coordinates and describe the curve.x=(1+\cost)\costy=(1+\cost)\sint. 0\leqt\leq2\pi

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

2021-08-31

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