Question

# \displaystyle{\left(-{1},\frac{{{3}\pi}}{{2}}\right)} lies on the polar graph

Alternate coordinate systems

The reason ehy the point $$(-1, \frac{3\pi}{2})$$ lies on the polar graph $$r=1+\cos \theta$$ even though it does not satisfy the equation.

2021-02-27

The given equation of the polar curve is $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{r}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace\theta\right\rbrace}}}.$$
By substituing the point $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left(-{\left\lbrace{1}\right\rbrace},{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\pi\right\rbrace}}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}\in{\left\lbrace{r}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace\theta\right\rbrace}}},$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{r}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace\theta\right\rbrace}}}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le-{\left\lbrace{1}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace{\left\lbrace{\left({\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\pi\right\rbrace}}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}\right\rbrace}}}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le-{\left\lbrace{1}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\left\lbrace{0}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le-{\left\lbrace{1}\right\rbrace}\ne{\left\lbrace{1}\right\rbrace}$$
Therefore, the point $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left(-{\left\lbrace{1}\right\rbrace},{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\pi\right\rbrace}}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}$$ does not satisfy the equation of the polar curve $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{r}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace\theta\right\rbrace}}}.$$
Use online graphing calculator and draw the graph of $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{r}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace\theta\right\rbrace}}}$$ as shown below in Figure 1 and identify the point $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left(-{\left\lbrace{1}\right\rbrace},{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\pi\right\rbrace}}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}.$$

From Figure 1 it can be noted that the point $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left(-{\left\lbrace{1}\right\rbrace},{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\pi\right\rbrace}}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}$$ lies on the graph.
Consider the point $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left(-{\left\lbrace{1}\right\rbrace},{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\pi\right\rbrace}}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}$$ and change its radial coordinate to 1 and subtract $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\pi$$ from the directional coordinate to obtain the alternate representation of the point $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left(-{\left\lbrace{1}\right\rbrace},{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\pi\right\rbrace}}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}.$$
Therefore, the point $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left(-{\left\lbrace{1}\right\rbrace},{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\pi\right\rbrace}}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}$$ can also be represented by $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left({\left\lbrace{1}\right\rbrace},{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\pi\right\rbrace}}}{{{\left\lbrace{2}\right\rbrace}}}}-\pi\right)}\right\rbrace}={\left\lbrace{\left({\left\lbrace{1}\right\rbrace},{\frac{{\pi}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}.$$
By substituing the point $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left({\left\lbrace{1}\right\rbrace},{\frac{{\pi}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}\in{\left\lbrace{r}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace\theta\right\rbrace}}},$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{r}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace\theta\right\rbrace}}}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{1}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace{\left\lbrace{\left({\frac{{\pi}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}\right\rbrace}}}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{1}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\left\lbrace{0}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{1}\right\rbrace}={\left\lbrace{1}\right\rbrace}$$
Thus, the point $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left({\left\lbrace{1}\right\rbrace},{\frac{{\pi}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}$$ satisfies the polar equation $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{r}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace\theta\right\rbrace}}}.$$
And from Figure 1 it can be seen that the points $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left({\left\lbrace{1}\right\rbrace},{\frac{{\pi}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}{\left\lbrace\quad\text{and}\quad\right\rbrace}{\left\lbrace{\left(-{\left\lbrace{1}\right\rbrace},{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\pi\right\rbrace}}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}$$ are the same points.
Therefore, due to this multiple identity, the point $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left(-{\left\lbrace{1}\right\rbrace},{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\pi\right\rbrace}}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}$$ lies on the graph even if does not satisfy the polar equation $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{r}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace\theta\right\rbrace}}}.$$