Q. 10 - Calculus

Question

In Example 5 we converted the simple polar equation $r=\cos 2\theta$ into the messy rectangular equation $\pm \left(x^2+\right.{\left.y^2\right)}^{3/2}=x^2-y^2$. Does this rectangular equation have a simpler rectangular coordinate form? If so, what is it? If not, why not?

Step-by-Step Solution

Answer

$r=\cos 2\theta$The rectangular form of the equation is ${\left(x^2+y^2\right)}^{\frac{3}{2}}=x^2-y^2.$

Hence the explanation.

Step 1: Given information

simple polar equation $r=\cos 2\theta .$ 

Step 2: Simplification

Consider the polar equation $r=\cos 2\theta$.

The objective is to show the equation in simple rectangular form.

Take the equation $r=\cos 4\theta$

$r=2{\cos }^2\theta -1\left[\right.since\left.\cos 2\theta =2{\cos }^2\theta -1\right]$

Now substitute $\frac{x}{r}=\cos \theta$ in $r=2{\cos }^2\theta -1.$

Then,

$r=2·\frac{x^2}{r^2}-1$ By substitution

$$\begin{gathered} r=\frac{2x^2}{r^2}-1 \\ r=\frac{2x^2-r^2}{r^2} \end{gathered}$$

By doing cross multiplication,

$$\begin{gathered} r^3=2x^2-r^2 \\ {\left(\sqrt{x^2+y^2}\right)}^3=2x^2-\left(x^2+y^2\right)\left[\text{ since }{x}^2+y^2=r^2\Rightarrow r=\sqrt{x^2+y^2}\right] \\ {\left(x^2+y^2\right)}^{\frac{3}{2}}=2x^2-x^2-y^2 \\ {\left(x^2+y^2\right)}^{\frac{3}{2}}=x^2-y^2\text{ simplify } \end{gathered}$$

Thus, the equation is ${\left(x^2+y^2\right)}^{\frac{3}{2}}=x^2-y^2$.

It is not possible to give simpler rectangular form for the polar equation $r=\cos 2\theta .$

Thus, the rectangular form of the equation is ${\left(x^2+y^2\right)}^{\frac{3}{2}}=x^2-y^2$.

Hence the explanation.