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.
\(\displaystyle{x}={\left({1}+{\cos{{t}}}\right)}{\cos{{t}}}\)
\(\displaystyle{y}={\left({1}+{\cos{{t}}}\right)}{\sin{{t}}}.{0}\leq{t}\leq{2}\pi\)

Expert Answers (1)

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 \(\displaystyle{r}={1}+{\cos{\theta}}\)

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