Question

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

Alternate coordinate systems
ANSWERED
asked 2021-02-26

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.

Answers (1)

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 \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left(-{\left\lbrace{1}\right\rbrace},{\frac{{{\left\lbrace{\left\lbrace{3}\right\rbrace}\pi\right\rbrace}}}{{{\left\lbrace{2}\right\rbrace}}}}\right)}\right\rbrace}.\)
image
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 \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{r}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace\theta\right\rbrace}}}.\)

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