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

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

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Ezra Herbert

Given that the cylindrical coordinates,
To convert cylindrical coordinates to Cartesian coordinates,
$$\displaystyle{r}={t}^{{{2}}},\theta={\frac{{\pi}}{{{2}}}}$$
Apply the rule of conversion
$$\displaystyle{x}={r}{\cos{\theta}},{z}={t}$$
Substituting $$\displaystyle{r}={t}^{{{2}}},\theta={\frac{{\pi}}{{{2}}}}$$ in Cartesian coordinates, $$\displaystyle{x}={\cos{\theta}}$$
$$\displaystyle{x}={t}^{{{2}}}{\cos{{\frac{{\pi}}{{{2}}}}}}$$
$$\displaystyle={t}^{{{2}}}$$
x=0
and
$$\displaystyle{y}={r}{\sin{\theta}}$$
$$\displaystyle={t}^{{{2}}}{\sin{{\frac{{\pi}}{{{2}}}}}}$$
$$=t^{2}*(1) \because sin\frac{\pi}{2}=1$$
$$y=t^{2}$$
Substitute z=t in equation (1)
$$\displaystyle{y}={t}^{{{2}}}$$
$$\displaystyle{y}={\left({z}\right)}^{{{2}}}$$
$$\displaystyle{y}={z}^{{{2}}}$$
Hence the parametric equation of Cartesian coordinates is,
$$\displaystyle{y}={z}^{{{2}}}$$ And the curve is a parabola in plane