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]}

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