What is the Cartesian form of $r^{2} = 2 θ + cot (θ) - tan (θ)$ $r^2 = 2theta+cot(theta)-tan(theta)$ ?

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What is the Cartesian form of $r^{2} = 2 θ + cot (θ) - tan (θ)$ $r^2 = 2theta+cot(theta)-tan(theta)$ ?

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Answer 1 · 2018-03-03T11:02:00

The answer is going to be pretty big and so is the explanation. If you want to see only the answer, scroll to the end. Hope I was helpful!

Explanation:

In order to transform the polar coordinates $(r_{P}, θ_{P})$ $(r_P,theta_P)$ of a point $P$ $P$ to Cartesian coordinates $(x_{P}, y_{P})$ $(x_P,y_P)$ you must solve the system of equations :

$x_{P}^{2} + y_{P}^{2} = r_{P}^{2}$ $x_P^2 + y_P^2 = r_P^2$
$y_{P} = r_{P} sin (θ_{P})$ $y_P = r_P sin(theta_P)$
$x_{P} = r_{P} cos (θ_{P})$ $x_P = r_P cos(theta_P)$

Now, let $P$ $P$ be a point on the graph of of the function $r^{2} = 2 θ + cot (θ) - tan (θ)$ $r^2 = 2theta+ cot(theta)-tan(theta)$ . The point $P$ $P$ is defined by the equation
$r_{P}^{2} = 2 θ_{P} + cot (θ_{P}) - tan (θ_{P})$ $r_P^2 = 2theta_P + cot(theta_P) - tan(theta_P)$ .

We know from the first equation of the system that $r_{P}^{2} = x_{P}^{2} + y_{P}^{2}$ $color(red)(r_P^2) = x_P^2 + y_P^2$ .
From the second and third, you can get $θ_{P}$ $color(red)(theta_P)$ to be equal to $arcsin ⎛ ⎜ ⎜ ⎝ \frac{y_{P}}{\sqrt{x_{P}^{2} + y_{P}^{2}}} ⎞ ⎟ ⎟ ⎠$ $arcsin(y_P/sqrt(x_P^2 + y_P^2))$ and $arccos ⎛ ⎜ ⎜ ⎝ \frac{x_{P}}{\sqrt{x_{P}^{2} + y^{P} 2}} ⎞ ⎟ ⎟ ⎠$ $arccos(x_P/sqrt(x_P^2+y^P2))$ ,

having already substituted $r_{P}$ $r_P$ . To simplify things, let

$α_{P} = \frac{y_{P}}{\sqrt{x_{P}^{2} + y_{P}^{2}}}$ $color(blue)(alpha_P) = y_P/sqrt(x_P^2 + y_P^2)$ and $β_{P}$ $color(blue)(beta_P)$ = $\frac{x_{P}}{\sqrt{x_{P}^{2} + y_{P}^{2}}}$ $x_P/sqrt(x_P^2 + y_P^2)$ .

Now, substitute $θ_{P}$ $color(red)(theta_P)$ and $r_{P}$ $color(red)(r_P)$ into the formula that defines $P$ $P$ :

$r_{P}^{2} = 2 θ_{P} + cot (θ_{P}) - tan (θ_{P})$ $r_P^2 = 2theta_P + cot(theta_P) - tan(theta_P)$

$x_{P}^{2} + y_{P}^{2} = 2 arcsin (α_{P}) + \frac{cos (arccos (β_{P}))}{sin (arcsin (α_{P}))} + \frac{sin (arcsin (α_{P}))}{cos (arccos (β_{P}))}$ $x_P^2 + y_P^2 = 2arcsin(color(blue)(alpha_P)) + (cos(color(red)(arccos)(color(blue)(beta_P))))/(sin(color(red)(arcsin)(color(blue)(alpha_P)))) + (sin(color(red)(arcsin)(color(blue)(alpha_P))))/cos(color(red)(arccos)(color(blue)(beta_P)))$

We have substitued $θ_{P}$ $theta_P$ in such a way that it is most convienient. However, there was no convinient way to write $2 θ_{P}$ $2theta_P$ .

$x_{P}^{2} + y_{P}^{2} = 2 arcsin (α_{P}) + \frac{β_{P}}{α_{P}} + \frac{α_{P}}{β_{P}}$ $x_P^2 + y_P^2 = 2arcsin(color(blue)(alpha_P)) + (color(blue)(beta_P))/(color(blue)(alpha_P)) + (color(blue)(alpha_P))/(color(blue)(beta_P))$

$\frac{β_{P}}{α_{P}}$ $color(blue)(beta_P)/color(blue)(alpha_P)$ and it's reciprocal cancel each other pretty neatly :

$x_{P}^{2} + y_{P}^{2} = 2 arcsin ⎛ ⎜ ⎜ ⎝ \frac{y_{P}}{\sqrt{x_{P}^{2} + y_{P}^{2}}} ⎞ ⎟ ⎟ ⎠ + \frac{x_{P}}{y_{P}} + \frac{y_{P}}{x_{P}}$ $x_P^2 + y_P^2 = 2arcsin(color(blue)(y_P/sqrt(x_P^2 + y_P^2))) + color(blue)(x_P)/color(blue)(y_P) + color(blue)(y_P)/color(blue)(x_P)$ .

And finally, the equation that defines $P$ $P$ in the Cartesian coordinate system is

$x_{P}^{2} + y_{P}^{2} = 2 arcsin ⎛ ⎜ ⎜ ⎝ \frac{y_{P}}{\sqrt{x_{P}^{2} + y_{P}^{2}}} ⎞ ⎟ ⎟ ⎠ + \frac{x_{P}}{y_{P}} + \frac{y_{P}}{x_{P}}$ $x_P^2 + y_P^2 = 2arcsin(y_P/sqrt(x_P^2 + y_P^2)) + (x_P)/(y_P) + (y_P)/(x_P)$ .

Since $P$ $P$ is just a random point on the graph, we can generelise the answer to be

$x^{2} + y^{2} = 2 arcsin (\frac{y}{\sqrt{x^{2} + y^{2}}}) + \frac{x}{y} + \frac{y}{x}$ $color(blue)(x^2 + y^2 = 2arcsin(color(blue)(y/sqrt(x^2 + y^2))) + color(blue)(x)/color(blue)(y) + color(blue)(y)/color(blue)(x)$

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What is the Cartesian form of $r^{2} = 2 θ + cot (θ) - tan (θ)$ $r^2 = 2theta+cot(theta)-tan(theta)$ ?

1 Answer

Explanation:

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What is the Cartesian form of $r^{2} = 2 θ + cot (θ) - tan (θ)$ $r^2 = 2theta+cot(theta)-tan(theta)$ ?