# More with Polar Curves

## Key Questions

• The easiest way is to input the equations into a graphing calculator. Look at the graph to see if both equations traced the same curve OR look at the table to see if for any given x value, the y-values of each equation are equivalent.

If the polar equations are $r = f \left(\theta\right) \mathmr{and} r = g \left(\theta\right)$ or, inversely, $\theta = {f}^{- 1} \left(r\right) \mathmr{and} \theta = {g}^{- 1} \left(r\right)$. eliminate either r or $\theta$, solve and substitute in one of the equations..

#### Explanation:

Explication:

Find the points of intersection of the cardioid

$r = a \left(1 + \cos \theta\right)$ and the circle r = a.

Eliminate r.

The equation for $\theta$ at a point of intersection is

$a = a \left(1 + \cos \theta\right)$. this is cos theta = 0 rarr theta = pi/2 and

(3pi)/2.

The common points are $\left(a , \frac{\pi}{2}\right) \mathmr{and} \left(a , \frac{3 \pi}{2}\right)$

For the graph, a = 1. Use $\left(x , y\right) = r \left(\cos \theta , \sin \theta\right)$

graph{(x^2+y^2-(x^2+y^2)^0.5-x)(x^2+y^2-1)=0[-2 4 -1.5 1.5]}

The two parabolas $1 = r \left(1 + \cos \theta\right) \mathmr{and} 1 = r \left(1 - \cos \theta\right)$

intersect at $\left(1 , \frac{\pi}{2}\right)$ and $\left(1 , 3 \frac{\pi}{2}\right)$.
graph{(x+(x^2+y^2)^0.5-1)(-x+(x^2+y^2)^0.5-1)=0[-3 3 -1.5 1.5]}

• Yes, they can; for example,

$r = \cos \theta$, which looks like:

$r = \sin \left(\theta - \frac{3 \pi}{2}\right)$, which looks like:

I hope that this was helpful.

• No. Two curves need not intersect.

Every curve can be expressed in either polar or rectangular form. Some are simpler in one form than the other, but there are not two classes (or families) of curves.

The curves ${x}^{2} + {y}^{2} = 1$ and ${x}^{2} + {y}^{2} = 9$ are concentric circles with unequal radii. They do not intersect.

In polar form, these are the curves $r = 1$ and $r = 3$. (And, of course, they still do not intersect.)