# How do you graph r=3theta?

Feb 9, 2017

Really, the graph ought to be a spiral, with $r \mathmr{and} \theta \uparrow$ from 0.

#### Explanation:

I assume that $\theta$ is in radian and 3 in $3 \theta$ is in distance

units.

The graph is a spiral, with both $r \mathmr{and} \theta \in \left[0 , \infty\right)$.

There are problems in using Socratic utility.

Firstly, $r = \sqrt{{x}^{2} + {y}^{2}} = 3 \theta \ge 0$.

Upon using the Cartesian form

$\sqrt{{x}^{2} + {y}^{2}} = \arctan \left(\frac{y}{x}\right) ,$

the graph obtained is inserted.

The branch in ${Q}_{1}$ is for $r \in \left[0 , \frac{3}{2} \pi\right)$. It is OK.

The branch in ${Q}_{3}$ is r-negative graph, for $r = 3 a r c \tan \left(\frac{y}{x}\right)$, for

$\theta \in \left(\frac{\pi}{2} , \pi\right]$. Here, $r \in \left(- \frac{3}{2} \pi , 0\right]$.

Really, $r \uparrow$ with $\theta$.

graph{sqrt(x^2+y^2)=3arctan(y/x)[-10, 10, -5, 5]}

Short Table for graphing from data, with $\theta = 0 \left(\frac{\pi}{8}\right) \frac{\pi}{2}$:

$\left(r , \theta r a \mathrm{di} a n\right) : \left(0 , 0\right) \left(1.18 , 0.39\right) \left(2.36 , 0.79\right) \left(3.53 , 1.18\right) \left(4.71 , 1.57\right)$