Chapter 8, Problem 19RE

(a)

To determine

The graph of the equation $r=2\sin 2\theta$ is need to drawn.

Explanation of Solution

Given:

  $r=2\sin 2\theta$

Graph:

  

Interpretation:

From the given data:

The given equation is in the polar co-ordinates is drawn as above:

  $r=2\sin 2\theta$

(b)

To determine

The given equation needs to write down in the form of rectangular -co-ordinates.

  $r=2\sin 2\theta$

Answer to Problem 19RE

The equation of the polar co-ordinates converts into a rectangular co-ordinates as $x^2+y^2\left(\sqrt{x^2+y^2}\right)=4xy$ .

Explanation of Solution

Given:

  $r=2\sin 2\theta$

Concept Used:

  $\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \\ r=\sqrt{x^2+y^2}\\ \theta ={\tan }^{-1}\left(\frac{y}{x}\right)\end{array}$

Calculation:

Know that:

  $\sin 2\theta =2\sin \theta \cos \theta$

The given equation is:

  $\begin{array}{l}r=2\sin 2\theta \\ r=2\left(2\sin \theta \cos \theta \right)\\ r=4\sin \theta \cos \theta \\ r=4\left(\frac{y}{r}\right)\left(\frac{x}{r}\right)\left[\begin{array}{l}\because x=r\cos \theta ,\cos \theta =\frac{x}{r}\\ y=r\sin \theta ,\sin \theta =\frac{y}{r}\end{array}\right]\\ r=\frac{4xy}{r^2}\\ \sqrt{x^2+y^2}=\frac{4xy}{{\left(\sqrt{x^2+y^2}\right)}^2}\\ \sqrt{x^2+y^2}=\frac{4xy}{x^2+y^2}\\ x^2+y^2\left(\sqrt{x^2+y^2}\right)=4xy\end{array}$

Conclusion:

Hence, the simplified equation is $x^2+y^2\left(\sqrt{x^2+y^2}\right)=4xy$ .