# To find: The equivalent polar equation for the given rectangular-coordinate equation. Given: x = r cos theta y = r sin theta b. From rectangular to polar: r= pm sqrt{x^{2} + y^{2}} cos theta= frac{x}{r}, sin theta= frac{y}{r}, tan theta= frac{x}{y} Calculation: Given: equation in rectangular-coordinate is y = x. Converting into equivalent polar equation - y=x Put x=r cos theta, y=r sin theta, Rightarrow r sin theta=r cos theta Rightarrow frac{sin theta}{cos theta}=1 Rightarrow tan theta=1 Thus, desired equivalent polar equation would be theta=1

Question
Alternate coordinate systems
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 $$\displaystyle\theta={1}$$

2020-11-23
Concept used:
Conversion formulafor coordinate systems are given as -
a. From polar to rectangular:
$$\displaystyle\ {x}=\ {r}{\cos{\theta}}$$
$$\displaystyle\ {y}=\ {r}{\sin{\theta}}$$
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 $$\displaystyle\theta={1}$$

### Relevant Questions

Find the Equivalent Polar Equation for a given Equation with Rectangular Coordinates:
$$\displaystyle{r}{\cos{\theta}}=\ -{1}$$
To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}+{\left\lbrace{8}\right\rbrace}{\left\lbrace{x}\right\rbrace}={\left\lbrace{0}\right\rbrace}$$
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 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.
The equivalent polar equation for the given rectangular-coordinate equation:
$$\displaystyle{x}^{{{2}}}\ +\ {y}^{{{2}}}\ +\ {8}{x}={0}$$
The equivalent polar equation for the given rectangular - coordinate equation.
$$\displaystyle{y}=\ -{3}$$
Consider the elliptical-cylindrical coordinate system (eta, psi, z), defined by $$\displaystyle{x}={a}{\cos{{h}}}\eta{\cos{\psi}},{y}={a}{\sin{{h}}}\eta{\sin{\psi}},{z}={z},\eta{G}{E}{0},{0}{L}{E}\psi{L}{E}{2}\pi,{z}{R}.{I}{n}{P}{S}#{6}$$
it was shown that this is an orthogonal coordinate system with scale factors $$\displaystyle{h}_{{1}}={h}_{{2}}={a}{\left({{\text{cosh}}^{{2}}\ }\eta-{{\cos}^{{2}}\psi}\right)}^{{{\frac{{{1}}}{{{2}}}}}}.$$
Determine the dual bases $$\displaystyle{\left({E}{1},{E}{2},{E}{3}\right)},{\left(\eta,\eta\psi,{z}\right)}.{S}{h}{o}{w}{t}\hat{:}{f}={a}\frac{{1}}{{a}}\frac{{\left({{\text{cosh}}^{{2}}{e}}{a}{t}-{{\cos}^{{s}}\psi}\right)}^{{1}}}{{2}}{\left[\frac{{f}}{\eta}{e}{1}+\frac{{f}}{\psi}{e}{2}+\frac{{f}}{{z}}{e}{3},\frac{{f}}{{w}}{h}{e}{r}{e}{\left({e}{1},{e}{2},{e}{3}\right)}\right.}$$ denotes the unit coordinate basis.
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)