Chapter 9.2, Problem 66AYU

In Problems 63-68, graph each pair of polar equations on the same polar grid. Find the polar coordinates of the point(s) of intersection and label the point(s) on the graph.

$r=1+\cos \theta ;r=3\cos \theta$

To determine

To find: Graph the polar equations $r = 1 + \text{cos} \theta ; r = 3 \text{cos} \theta$ on the same polar grid and find the point of intersections if any.

Answer to Problem 66AYU

Solution

The graph of the polar equations $r = 1 + \text{cos} \theta ; r = 3 \text{cos} \theta$ are the equation of cardioids and an equation of circle.

From the above graph, the point of intersections are $\left(0.75, 1.3\right), \left(0, 0\right) \text{and} \left(0.75, - 1.3\right)$.

Given:

It is asked to graph the polar equations $r = 1 + \text{cos} \theta ; r = 3 \text{cos} \theta$.

Explanation of Solution

Formula to be used:

Cardioids are characterized by equations of the form

$\begin{array}{lll}r\hfill & =\hfill & a\left(1 \pm \text{cos} \theta \right)\hfill \\ r\hfill & =\hfill & a\left(1 \pm \text{sin} \theta \right)\hfill \end{array}$

where $a > 0$. The graph of a cardioid passes through the pole.

Equation of circle passing through the pole, tangent to the line $\theta = \frac{\pi }{2}$, center on the polar axis, radius $a$ is given by

$r = \pm 2a \text{cos} \theta , a > 0$

Consider $r = 1 + \text{cos} \theta$,

This polar equation can be written as

$r = 1 \left(1 + \text{cos} \theta \right)$

The above equation is an equation of cardioids passes through the pole with $a = 1$.

Consider $r = 3 \text{cos} \theta$,

This polar equation can be written as

$r = 2\left(\frac{3}{2}\right)\text{cos} \theta , 2 > 0$

The above equation is an equation of circle passing through the pole, tangent to the line $\theta = \frac{\pi }{2}$, center on the polar axis, radius $\frac{3}{2}$.

To sketch:

$r = 1 + \text{cos} \theta ; r = 3 \text{cos} \theta$

From the above graph, the point of intersections are $\left(0.75, 1.3\right), \left(0, 0\right) \text{and} \left(0.75, - 1.3\right)$.