Question

Let the polar coordinate of the point (x,y) be (r,$\theta$). Determine the polar coordinates for the points a)(-x,y) b)(-2x,-2y) and c) (3x,-3y).

Short answer

$$\begin{gathered} a)(r,180-\theta \left) \\ b\right)(2r,180+\theta ) \\ c)(3r,360-\theta ) \end{gathered}$$

Step-by-step solution

Define vector

A vector is a quantity that has both a direction and a magnitude, and is used to determine the relative location of two points in space.

State information used

For polar coordinates (r,$\theta )$, the Cartesian coordinates are$(x=r\cos \theta ,y=r\sin \theta )$ ,

if the angle is measured relative to positive x axis $r=\surd (x^2+y^2),\theta ={\tan }^{-1}\frac{y}{x}$.

Step 3: (a) Determine the polar coordinates for the points (-x,y)

For (-x,y)

$\begin{array}{l}{r}_1=\sqrt{\left(-x^2\right)+y^2}\\ =\sqrt{x^2+y^2}\\ =r\end{array}$

$\begin{array}{l}{\theta }_1={\tan }^{-1}\frac{y}{x}\\ =-{\tan }^{-1}\frac{y}{x}\\ =-\theta \\ =180-\theta \end{array}$

Hence the polar coordinates are (r,180-$\theta$).

Step 4: (b) Determine the polar coordinates for the points (-2x,-2y)

For (-2x, -2y)

$\begin{array}{l}{r}_2=\sqrt{{(-2x)}^2+{(-2y)}^2}\\ =\sqrt{4X^2+4y^2}\\ =2r\end{array}$

$\begin{array}{l}{\theta }_2={\tan }^{-1}\frac{-2y}{-2x}\\ ={\tan }^{-1}\frac{y}{x}\\ =\theta \\ =180+\theta \end{array}$

Hence the polar coordinates are (2r, 180+$\theta$).

Step 5: (c) Determine the polar coordinates for the points (3x, -3y)

For (3x, -3y)

$\begin{array}{l}{r}_3=\sqrt{{\left(3x\right)}^2+{(-3y)}^2}\\ =\sqrt{9x^2+9y^2}\\ =3r\end{array}$

$\begin{array}{l}{\theta }_3={\tan }^{-1}\frac{-3y}{3x}\\ =-{\tan }^{-1}\frac{y}{x}\\ =-\theta \\ =360-\theta \end{array}$

Hence the polar coordinates are (3r, 360-$\theta$).