Question

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

Answer 1

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]}

Explanation:

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]}