How to Transform Equations to Their Polar Form

In mathematics, transforming equations to their polar form is a common and useful technique. This process allows us to analyze and solve problems involving polar coordinates more effectively. Let's explore how to transform the equations (x^3 xy^2 - 6x^2 - 2y^2 0) and (x^2 y^2 - 8x^2 - 2x^2 - 2y^2 0) into their polar forms.

Transformation of the First Equation

Consider the equation:

$$x^3 xy^2 - 6x^2 - 2y^2 0$$

First, let's simplify this equation:

$$x^3 xy^2 2y^2 - xy^2$$

$$x^3 xy^2 2 - xy^2$$

$$y^2 frac{x^2 cdot x cdot 6}{2 - x}$$

From the above, we have:

$$frac{x^6}{2 - x} geq 0 rightarrow begin{cases}x^6 geq 6 2 - x eq 0end{cases}$$

$$-6 leq x^2$$

Now, let's transform the equation into polar form. Assuming (x r cos theta) and (y r sin theta), we have:

$$r cos theta cdot sin^2 theta frac{r cos theta cdot r sin theta cdot 6}{2 - r cos theta}$$

To solve for (r), we get:

$$sin^2 theta frac{cos^2 theta cdot r sin theta cdot 6}{2 - r cos theta}$$

$$tan^2 theta cdot (2 - r cos theta) r cdot cos theta cdot 6$$

$$2 tan^2 theta - r cos theta cdot tan^2 theta r cos theta cdot 6$$

$$2 tan^2 theta - 6 r cos theta cdot (r cos theta tan^2 theta)$$

$$r frac{2 tan^2 theta - 6}{cos theta cdot (r cos theta tan^2 theta)}$$

$$r frac{2 tan^2 theta - 6}{cos theta cdot (1 tan^2 theta)}$$

$$r frac{2 tan^2 theta - 6}{cos theta cdot sec^2 theta}$$

$$r 2 cos theta tan^2 theta - 3$$

$$r 2 cos theta [sec^2 theta - 1] - 3$$

$$r 2 cos theta sec^2 theta - 4$$

$$r 2 sec theta - 8 cos theta$$

The plot for the polar form looks like this:

Transformation of the Second Equation

Now let's look at a second equation:

$$x^2 y^2 - 8x^2 - 2x^2 - 2y^2 0$$

First, simplify the equation:

$$x^2 y^2 - 8x^2 - 2x^2 - 2y^2 -9x^2 - y^2 0$$

Using the polar form (x r cos theta) and (y r sin theta), we can transform the equation into:

$$r^2 cos^2 theta r^2 sin^2 theta - 8r^2 cos^2 theta - 2r^2 sin^2 theta 0$$

$$r^2 (cos^2 theta sin^2 theta - 8 cos^2 theta - 2 sin^2 theta) 0$$

$$r^2 (1 - 8 cos^2 theta - 2 sin^2 theta) 0$$

To solve for (r), we get:

$$1 - 8 cos^2 theta - 2 sin^2 theta 0$$

$$1 - 8 cos^2 theta - 2 sin^2 theta 0$$

Since the constants don’t allow for a straightforward solution, we have:

$$r 0$$

Hence, the second equation in polar form is represented by the point (0, 0).