How do you graph the equation #r = 1 + cos( theta )#?

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Answer 1 · 2016-07-15T10:39:11

Graph of #x^2+y^2 = sqrt(x^2+y^2) + x# or #r = 1 + cos(theta)#
graph{x^2+y^2=sqrt(x^2+y^2)+x [-10, 10, -5, 5]}

In case you are trying to graph the equation in rectangular form, here's a way to get it to rectangular form and graph it.

We can make use of the following formulas when trying to convert from polar to rectangular:

#x = r cos(theta)# and #y = r sin(theta)#
#r^2 = x^2+y^2#

Now we can rewrite our equation:

#r = 1 + cos(theta)#

Multiplying both sides by #r# gives us

#r^2 = r(1+cos(theta))#

#= r + rcos(theta)#

Substituting the value of #r = sqrt(x^2+y^2)# into our equation yields

#r^2 = r + rcos(theta)#

#=sqrt(x^2+y^2) + x#

So our equation becomes

#x^2+y^2 = sqrt(x^2+y^2) + x#, which is equivalent to #r = 1 + cos(theta)#.

Below are a few graphs.

Graph of #x^2+y^2 = sqrt(x^2+y^2) + x# or #r = 1 + cos(theta)#

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