Q 46. - Calculus

Question

In Exercises 32–47 convert the equations given in polar coordinates to rectangular coordinates.

$r=\cos 4\theta$

Step-by-Step Solution

Answer

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

Step 1. Given information.

The given equation in polar coordinates is:

$r=\cos 4\theta$

Step 2. Find the equation in rectangular coordinates.

$$\begin{gathered} r=\cos 4\theta \\ r=2 {\cos }^2 2\theta -1 \left[\cos 4\theta =\left(2 {\cos }^2 2\theta -1\right)\right] \\ r=2\left(2{\cos }^2 2\theta -1 \right)-1 \left[\cos 2\theta =2 {\cos }^2\theta -1\right] \\ r=4 {\cos }^2\theta -2-1 \\ r=4 {\cos }^2\theta -3 \end{gathered}$$

Substitute $\frac{x}{r}=\cos \theta$,

$$\begin{gathered} r=4·\frac{x^2}{r^2}-3 \\ r=\frac{4x^2}{r^2}-3 \\ r=\frac{4x^2-3r^2}{r^2} \end{gathered}$$

Cross multiply,

$$\begin{gathered} r^3=4x^2-3r^2 \\ {\left(\sqrt{x^2+y^2}\right)}^3=4x^2-3\left(x^2+y^2\right) \left[r^2=x^2+y^2 \mathrm{and} r=\sqrt{x^2+y^2}\right] \\ {\left(x^2+y^2\right)}^{\frac{3}{2}}=4x^2-3x^2-3y^2 \\ {\left(x^2+y^2\right)}^{\frac{3}{2}}=x^2-3y^2 \end{gathered}$$

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