Chapter H.1, Problem 6E

(a)

To determine

To find:The polar coordinates of $\left(r,\theta \right)$

Answer to Problem 6E

The polar coordinates of $\left(3\sqrt{3},3\right)$ are, $\left(r,\theta \right)=\left(r,\frac{\pi }{6}\right)$ where $r>0$

  $\left(r,\theta \right)=\left(-6,\frac{7\pi }{6}\right)$ where $r<0$

Explanation of Solution

Given: The Cartesian Points $\left(3\sqrt{3},3\right)$

Calculation:

If $r>0$

  $\left(3\sqrt{3},3\right)=\left(x,y\right)$

  $\Rightarrow r=\sqrt{x^2+y^2}=\sqrt{27+9}$

  $\Rightarrow r=6$ when $r>0$

  $x=r\cos \theta$

  $y=r\sin \theta$

  $\Rightarrow 3\sqrt{3}=6\cos \theta$

  $cos\theta =\frac{\sqrt{3}}{2}>0$

  $\Rightarrow$

  $3=6\sin \theta$

  $\sin \theta =\frac{1}{2}>0$

  $\Rightarrow \theta \in I$ quadrant

  $\Rightarrow \theta =\frac{\pi }{6}\left(\because 0\le \theta <2\pi \right)$

  $\therefore \left(r,\theta \right)=\left(r,\frac{\pi }{6}\right)$ when $r>0$

If $r<0$ then

  $r=-6$

  $3\sqrt{3}=-6\cos \theta$

  $\cos \theta =\frac{-\sqrt{3}}{2}<0$

  $\Rightarrow$

  $3=-6\sin \theta$

  $\sin \theta =\frac{-1}{2}<0$

  $\Rightarrow \theta {\in }^{\left|\right||}$ Quadrant

  $\Rightarrow \theta =\pi +\frac{\pi }{6}\left(\because 0<\theta <2\pi \right)$

  $=\frac{7\pi }{\theta }$

  $\therefore \left(r,\theta \right)=\left(-6,\frac{7\pi }{6}\right)$ When $r<0$

Conclusion:

Therefore, the polar coordinates of $\left(3\sqrt{3},3\right)$ are, $\left(r,\theta \right)=\left(r,\frac{\pi }{6}\right)$ where $r>0$

  $\left(r,\theta \right)=\left(-6,\frac{7\pi }{6}\right)$ where $r<0$

(b)

To determine

To find: The polar coordinates of $\left(r,\theta \right)$

Answer to Problem 6E

If $r>0$ , $\left(r,\theta \right)=\left(\sqrt{5},\arctan \left(-2\right)\right)$ where $\tan \theta =-2;0\le \theta <2\pi$

If $r<0$ , $\left(r,\theta \right)=\left(-\sqrt{5},\arctan \left(-2\right)\right)$ where $\tan \theta =-2;0\le \theta <2\pi$

Explanation of Solution

Given: The Cartesian Points $\left(1,-2\right)$

Calculation:

  $\left(1,-2\right)=\left(x,y\right)$

If $r>0$

  $\Rightarrow r=\sqrt{x^2+y^2}$

  $=\sqrt{1+4}=\sqrt{5}$

  $1=\sqrt{5}\cos \theta$

  $\cos \theta =\frac{1}{\sqrt{5}}>0$

  $\Rightarrow$

  $-2=\sqrt{5}\sin \theta$$\sin \theta =\frac{-2}{\sqrt{5}}<0$

  $\Rightarrow \theta \in$ 4^th quadrant and $\tan \theta =\frac{\frac{-2}{\sqrt{5}}}{\frac{-1}{\sqrt{5}}}=-2$

  $\Rightarrow 0\le \theta <2\pi ;$$\tan \theta =-2$

  $\therefore \left(r,\theta \right)=\left(\sqrt{5},\arctan \left(-2\right)\right)$ where $\tan \theta =-2;0\le \theta <2\pi$

If $r<0$ then $r=-\sqrt{5}$

And $1=-\sqrt{5}\cos \theta$

  $-2=-\sqrt{5}\sin \theta$

  $\Rightarrow \cos \theta =\frac{-1}{\sqrt{5}}<0$

  $\sin \theta =\frac{2}{\sqrt{5}}>0$

  $\Rightarrow \theta \in$

  $\parallel$ quadrant ; $\tan \theta =-2$

  $\therefore \left(r,\theta \right)=\left(-\sqrt{5},\arctan \left(-2\right)\right)$ where $\tan \theta =-2;0\le \theta <2\pi$

Conclusion:

Therefore, the polar coordinates of $\left(1,-2\right)$ are,

If $r>0$ , $\left(r,\theta \right)=\left(\sqrt{5},\arctan \left(-2\right)\right)$ where $\tan \theta =-2;0\le \theta <2\pi$

If $r<0$ , $\left(r,\theta \right)=\left(-\sqrt{5},\arctan \left(-2\right)\right)$ where $\tan \theta =-2;0\le \theta <2\pi$