How do you graph #r=3theta#?

Question

6351 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-09T06:27:03

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

I assume that #theta# is in radian and 3 in #3theta# is in distance

units.

The graph is a spiral, with both #r and theta in [0, oo)#.

There are problems in using Socratic utility.

Firstly, # r = sqrt(x^2+y^2)=3theta>=0#.

Upon using the Cartesian form

#sqrt(x^2+y^2)=arctan(y/x),#

the graph obtained is inserted.

The branch in #Q_1# is for #r in [0, 3/2pi)#. It is OK.

The branch in #Q_3# is r-negative graph, for #r = 3arc tan(y/x)#, for

#theta in (pi/2, pi]#. Here, #r in (-3/2pi, 0]#.

Really, #r uarr# with #theta#.

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

Short Table for graphing from data, with #theta = 0 ( pi/8)pi/2#:

#(r, theta radian): (0, 0) (1.18, 0.39) (2.36, 0.79) (3.53, 1.18) (4.71, 1.57) #