Figure 7.1.6(a) provides a slider-controlled drawing of the limaçon $r\=1\/2-\cos \left(\mathrm{\θ}\right)$. As the slider advances, the graph of the limaçon is drawn dynamically in black, while the ray from the origin to the polar point $\left(r\,\mathrm{\θ}\right)$ is drawn in red.

The graph in Figure 7.1.6(b) is most easily drawn by invoking the Plot Builder (via the Context Panel) on $1\/2-\cos \left(\mathrm{\θ}\right)$ and selecting 2-D polar plot. The appropriate polar graph will then be drawn automatically.

(Unfortunately, re-executing this worksheet via the triple exclamation icon (!!!) in the toolbar will cause the following graph to be replaced with its Cartesian version. Should that happen, simply repeat the Plot Builder option of "2-D polar plot.")

$1\/2-\cos \left(\mathrm{\θ}\right)$$\to$

Alternatively, execute either of the first two commands in Table 7.1.6(a) to obtain Figure 7.1.6(b). To obtain a graph of the limaçon on a rectangular Cartesian grid, execute the last command in the table. (Select Evaluate in the Context Panel.)

The conversion to Cartesian coordinates cannot be done naively because, unlike for the polar form of this limaçon, the right-hand side of

$\sqrt{x^2\+y^2}\=\frac{1}{2}-\frac{x}{\sqrt{x^2\+y^2}}$ 

can become negative but the left-hand side cannot. An implicit Cartesian form of the limaçon is obtained if  the square roots are eliminated by appropriate algebraic manipulations. For example, the equation

$x^2\+y^2\={\left(2x^2\+2y^2\+2x\right)}^2$ 

properly defines the limaçon implicitly in Cartesian coordinates.