How do you graph $r = 3 θ$ $r=3theta$ ?

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How do you graph $r = 3 θ$ $r=3theta$ ?

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Answer 1 · 2017-02-09T06:27:03

Really, the graph ought to be a spiral, with $r and θ ↑$ $r and theta uarr$ from 0.

Explanation:

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

units.

The graph is a spiral, with both $r and θ \in [0, \infty)$ $r and theta in [0, oo)$ .

There are problems in using Socratic utility.

Firstly, $r = \sqrt{x^{2} + y^{2}} = 3 θ \geq 0$ $r = sqrt(x^2+y^2)=3theta>=0$ .

Upon using the Cartesian form

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

the graph obtained is inserted.

The branch in $Q_{1}$ $Q_1$ is for $r \in [0, \frac{3}{2} π)$ $r in [0, 3/2pi)$ . It is OK.

The branch in $Q_{3}$ $Q_3$ is r-negative graph, for $r = 3 a r c tan (\frac{y}{x})$ $r = 3arc tan(y/x)$ , for

$θ \in (\frac{π}{2}, π]$ $theta in (pi/2, pi]$ . Here, $r \in (- \frac{3}{2} π, 0]$ $r in (-3/2pi, 0]$ .

Really, $r ↑$ $r uarr$ with $θ$ $theta$ .

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

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

$(r, θ r a d i a n) : (0, 0) (1.18, 0.39) (2.36, 0.79) (3.53, 1.18) (4.71, 1.57)$ $(r, theta radian): (0, 0) (1.18, 0.39) (2.36, 0.79) (3.53, 1.18) (4.71, 1.57)$

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How do you graph $r = 3 θ$ $r=3theta$ ?

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