Question

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

$r=\sin 4\theta$

Short answer

The required equation is${\left(x^2+y^2\right)}^{\frac{5}{2}}=2xy\left(x^2-y^2\right)$.

Step-by-step solution

Step 1. Given information.

The given equation in polar coordinates is:

$r=\sin 4\theta$

Step 2. Find the equation in rectangular coordinates.

$$\begin{gathered} r=\sin 4\theta \\ r=2\sin 2\theta \cos 2\theta \left[\mathrm{Since}\sin 4\theta =2\sin 2\theta \cos 2\theta \right] \end{gathered}$$

Now substitute $\sin 2\theta =2\sin \theta \cos \theta \mathrm{and}\cos 2\theta =2{\cos }^2\theta -1$,

$r=2\sin \theta \cos \theta ·\left(2{\cos }^2\theta -1\right)$

Substitute $\frac{x}{r}=\cos \theta \mathrm{and}\frac{y}{r}=\sin \theta$,

localid="1649319886678" $$\begin{gathered} r=2·\frac{y}{r}·\frac{x}{r}\left(2·\frac{x^2}{r^2}-1\right) \\ r=2·\frac{xy}{r^2}\left(2·\frac{x^2}{r^2}-1\right) \\ r=\frac{2xy}{r^2}\left(\frac{2x^2-r^2}{r^2}\right) \end{gathered}$$

Now cross multiply,

$$\begin{gathered} r^5=2xy\left(2x^2-r^2\right) \\ {\left(\sqrt{x^2+y^2}\right)}^5=2xy\left(2x^2-\left(x^2+y^2\right)\right)\left[r^2=x^2+y^2\mathrm{and}r=\sqrt{x^2+y^2}\right] \\ {\left(x^2+y^2\right)}^{\frac{5}{2}}=2xy\left(2x^2-\left(x^2+y^2\right)\right) \\ {\left(x^2+y^2\right)}^{\frac{5}{2}}=2xy\left(x^2-y^2\right) \end{gathered}$$

Therefore, the equation in rectangular coordinates is${\left(x^2+y^2\right)}^{\frac{5}{2}}=2xy\left(x^2-y^2\right)$.