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 \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{r}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace\theta\right\rbrace}}}.\)
Question
\displaystyle{\left(-{1},\frac{{{3}\pi}}{{2}}\right)} lies on the polar graph
Answers (1)
2021-02-27
Relevant Questions
asked 2020-10-21
To solve:
\(\displaystyle{\left(\begin{matrix}{x}-{2}{y}={2}\\{2}{x}+{3}{y}={11}\\{y}-{4}{z}=-{7}\end{matrix}\right)}\)
asked 2021-02-09
To determine:
a) Whether the statement, " The point with Cartesian coordinates \(\displaystyle{\left[\begin{array}{cc} -{2}&\ {2}\end{array}\right]}\) has polar coordinates \(\displaystyle{\left[{b}{f}{\left({2}\sqrt{{{2}}},\ {\frac{{{3}\pi}}{{{4}}}}\right)}\ {\left({2}\sqrt{{{2}}},{\frac{{{11}\pi}}{{{4}}}}\right)}\ {\left({2}\sqrt{{{2}}},\ -{\frac{{{5}\pi}}{{{4}}}}\right)}\ {\quad\text{and}\quad}\ {\left(-{2}\sqrt{{2}},\ -{\frac{{\pi}}{{{4}}}}\right)}\right]}\) " is true or false.
b) Whether the statement, " the graphs of \(\displaystyle{\left[{r}{\cos{\theta}}={4}\ {\quad\text{and}\quad}\ {r}{\sin{\theta}}=\ -{2}\right]}\) intersect exactly once " is true or false.
c) Whether the statement, " the graphs of \(\displaystyle{\left[{r}={4}\ {\quad\text{and}\quad}\ \theta={\frac{{\pi}}{{{4}}}}\right]}\) intersect exactly once ", is true or false.
d) Whether the statement, " the point \(\displaystyle{\left[\begin{array}{cc} {3}&{\frac{{\pi}}{{{2}}}}\end{array}\right]}{l}{i}{e}{s}\ {o}{n}\ {t}{h}{e}\ {g}{r}{a}{p}{h}\ {o}{f}{\left[{r}={3}{\cos{\ }}{2}\ \theta\right]}\) " is true or false.
e) Whether the statement, " the graphs of \(\displaystyle{\left[{r}={2}{\sec{\theta}}\ {\quad\text{and}\quad}\ {r}={3}{\csc{\theta}}\right]}\) are lines " is true or false.
asked 2021-02-27
The system of equation \begin{cases}2x + y = 1\\4x +2y = 3\end{cases} by graphing method and if the system has no solution then the solution is inconsistent.
Given: The linear equations is \begin{cases}2x + y = 1\\4x +2y = 3\end{cases}
asked 2021-01-31
The quadratic function \(\displaystyle{y}={a}{x}^{2}+{b}{x}+{c}\) whose graph passes through the points (1, 4), (2, 1) and (3, 4).
