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

Question
Alternate coordinate systems
The reason ehy 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 polar graph $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{r}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace\theta\right\rbrace}}}$$ 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 $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{r}\right\rbrace}={\left\lbrace{1}\right\rbrace}+{\cos{{\left\lbrace\theta\right\rbrace}}}.$$

### Relevant Questions

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.
Interraption: To show that the system $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{r}\right\rbrace}}}={\left\lbrace{r}\right\rbrace}{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}-{\left\lbrace{r}\right\rbrace}^{{{2}}}\right)}\right\rbrace},\dot{{\theta}}={\left\lbrace{1}\right\rbrace}$$ is equivalent to $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{x}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}-{\left\lbrace{y}\right\rbrace}-{\left\lbrace{x}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}\right)}\right\rbrace},\dot{{{\left\lbrace{y}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}+{\left\lbrace{y}\right\rbrace}-{\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}\right)}\right\rbrace}$$ for polar to Cartesian coordinates.
A limit cycle is a closed trajectory. Isolated means that neighboring trajectories are not closed.
A limit cycle is said to be unstable or half stable, if all neighboring trajectories approach the lemin cycle.
These systems oscillate even in the absence of external periodic force.
To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
$$\displaystyle\ {x}=\ {r}{\cos{\theta}}$$
$$\displaystyle\ {y}=\ {r}{\sin{\theta}}$$
b. From rectangular to polar:
$$\displaystyle{r}=\pm\sqrt{{{x}^{{{2}}}\ +\ {y}^{{{2}}}}}$$
$$\displaystyle{\cos{\theta}}={\frac{{{x}}}{{{r}}}},{\sin{\theta}}={\frac{{{y}}}{{{r}}}},{\tan{\theta}}={\frac{{{x}}}{{{y}}}}$$
Calculation:
Given: equation in rectangular-coordinate is $$\displaystyle{y}={x}$$.
Converting into equivalent polar equation -
$$\displaystyle{y}={x}$$
Put $$\displaystyle{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}},$$
$$\displaystyle\Rightarrow\ {r}{\sin{\theta}}={r}{\cos{\theta}}$$
$$\displaystyle\Rightarrow\ {\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}={1}$$
$$\displaystyle\Rightarrow\ {\tan{\theta}}={1}$$
Thus, desired equivalent polar equation would be $$\displaystyle\theta={1}$$
To plot: Thepoints which has polar coordinate $$\displaystyle{\left({2},\frac{{{7}\pi}}{{4}}\right)}$$ also two alternaitve sets for the same.
To determine:
a) The origin $$\displaystyle{\left[\begin{array}{cc} {0}&\ {0}\end{array}\right]}$$ is a critical point of the systems.
$$\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}{\quad\text{and}\quad}{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}$$. Futhermore, it is a center of the corresponding linear system.
b) The systems $$\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}{\quad\text{and}\quad}{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}$$ are almost linear.
c) To prove: $$\displaystyle{\left[{\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}\ {<}\ {0}\right]}{\quad\text{and}\quad}{\left[{r}\rightarrow\ {0}\ {a}{s}\ {t}\rightarrow\ \infty\right]},$$ hence the critical point for the system $$\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}$$ is asymptotically stable and the solution of the initial value problem for $$\displaystyle{\left[{r}\ {w}{i}{t}{h}\ {r}={r}_{{{0}}}\ {a}{t}\ {t}={0}\right]}$$ becomes unbounded as $$\displaystyle{\left[{t}\rightarrow{\frac{{{1}}}{{{2}}}}\ {r}{\frac{{{2}}}{{{0}}}}\right]}$$, hence the critical point for the system $$\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}$$ is unstable.
Given the elow bases for R^2 and the point at the specified coordinate in the standard basis as below, (40 points) $$\displaystyle{B}{1}=\le{f}{t}{\left\lbrace{\left({1},{0}\right)},{\left({0},{1}\right)}{r}{i}{g}{h}{t}\right\rbrace}&{M}{S}{K}{B}{2}={\left({1},{2}\right)},{\left({2},-{1}\right)}{r}{i}{g}{h}{t}\rbrace{\left({1},{7}\right)}={3}^{\cdot}{\left({1},{2}\right)}-{\left({2},{1}\right)}{N}{S}{K}{B}{2}={\left({1},{1}\right)},{\left(-{1},{1}\right)}{\left({3},{7}={5}^{\cdot}{\left({1},{1}\right)}+{2}^{\cdot}{\left(-{1},{1}\right)}{N}{S}{K}{B}{2}={\left({1},{2}\right)},{\left({2},{1}\right)}{\left({0},{3}\right)}={2}^{\cdot}{\left({1},{2}\right)}-{2}^{\cdot}{\left({2},{1}\right)}{N}{S}{K}{\left({8},{10}\right)}={4}^{\cdot}{\left({1},{2}\right)}+{2}^{\cdot}{\left({2},{1}\right)}{N}{S}{K}{B}{2}={\left({1},{2}\right)},{\left(-{2},{1}\right)}{\left({0},{5}\right)}={N}{S}{K}{\left({1},{7}\right)}=\right.}$$ a. Use graph technique to find the coordinate in the second basis. (10 points) b. Show that each basis is orthogonal. (5 points) c. Determine if each basis is normal. (5 points) d. Find the transition matrix from the standard basis to the alternate basis. (15 points)
$$\displaystyle{r}{\cos{\theta}}=\ -{1}$$
The system of equation $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{b}\right\rbrace}{\left\lbrace{e}\right\rbrace}{\left\lbrace{g}\right\rbrace}\in{\left\lbrace\le{f}{t}{\left\lbrace{\left\lbrace{c}\right\rbrace}{\left\lbrace{a}\right\rbrace}{\left\lbrace{s}\right\rbrace}{\left\lbrace{e}\right\rbrace}{\left\lbrace{s}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace}\right\rbrace}{\frac{{{n}}}{{{\left\lbrace{2}\right\rbrace}}}}\backslash-{\frac{{{\left\lbrace{\left\lbrace{n}\right\rbrace}+{\left\lbrace{1}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{2}\right\rbrace}}}}{\left\lbrace{e}\right\rbrace}{\left\lbrace{n}\right\rbrace}{\left\lbrace{d}\right\rbrace}{\left\lbrace\le{f}{t}{\left\lbrace{\left\lbrace{c}\right\rbrace}{\left\lbrace{a}\right\rbrace}{\left\lbrace{s}\right\rbrace}{\left\lbrace{e}\right\rbrace}{\left\lbrace{s}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace}\right\rbrace}$$ by graphing method and if the system has no solution then the solution is inconsistent.
Given: The linear equations is $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{y}\right\rbrace}={\left\lbrace{1}\right\rbrace}{\left\lbrace\quad\text{and}\quad\right\rbrace}{\left\lbrace{4}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{2}\right\rbrace}{\left\lbrace{y}\right\rbrace}={\left\lbrace{3}\right\rbrace}.$$
$$\displaystyle{d}={{\cos}^{ -{{1}}}{\left( \sin{{\left({L}{T}_{{1}}\right)}} \sin{{\left({L}{T}_{{2}}\right)}}+ \cos{{\left({L}{T}_{{1}}\right)}} \cos{{\left({L}{T}_{{2}}\right)}} \cos{{\left({{\ln}_{{1}}-}{\ln}_{{2}}\right)}}\right)}}$$
$$\displaystyle{d}={2}{{\sin}^{ -{{1}}}{\left(\sqrt{{{{\sin}^{2}{\left(\frac{{{L}{T}_{{1}}-{L}{T}_{{2}}}}{{2}}\right)}}+ \cos{{\left({L}{T}_{{1}}\right)}} \cos{{\left({L}{T}_{{2}}\right)}}{{\sin}^{2}{\left(\frac{{{{\ln}_{{1}}-}{\ln}_{{2}}}}{{2}}\right)}}}}\right)}}$$
To calculate: The sections represented by the polar equation $$\displaystyle{r}=\frac{18}{{{6}-{6} \cos{\theta}}}$$ and graph it by the use of graphing utility.