Question

The Cartesian coordinates of a point are given. (a) (4, −4)(i) Find polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π.(ii) Find polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π.(b) (−1, sqrt(3))(i) Find polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π.(ii) Find polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π.

Answer 1

Answer:

$(x,y) = (r,\theta)$

i)$(4,-4) = (\pm 4\sqrt{2},\dfrac{7 \pi}{4})$

ii) $(-1, \sqrt{3}) = (\pm 2,\dfrac{2 \pi}{4})$

Thinking Process:

This can either be solved using complex variables or simply trigonometry (they are all the same in a way)

$(a,b) = a + ib = r(\cos{(\theta)} + i\sin{(\theta)}) = r e^{(i \theta)} = r\angle{\theta}$

But we'll simply go with trigonometry and good old Pythagorean theorem.

Solution:

For Cartesian coordinates are: $(x,y)$

To covert them into $r \angle{\theta}$

We can use:

• The Pythagoras theorem to find $r$

$r = \pm \sqrt{x^2 + y^2}$

• Arctangent to find $\theta$

$\theta = \tan^{-1}{\left(\dfrac{y}{x}\right)}$

These are related since:

$x = r\cos{(\theta)}$

$y = r\sin{(\theta)}$

So, let's start solving:

Part a) (4,-4)

For r:

$r = \pm \sqrt{x^2 + y^2}$

$r = \pm \sqrt{4^2 + (-4)^2}$

$r = \pm \sqrt{32}$

$r = \pm 4 \sqrt{2}$

i) for $r > 0$ we'll go with $r = + 4\sqrt{2}$

ii) for $r < 0$ we'll go with $r = - 4\sqrt{2}$

For theta:

$\theta = \tan^{-1}{\left(\dfrac{y}{x}\right)}$

$\theta = \tan^{-1}{\left(\dfrac{-4}{4}\right)}$

$\theta = \tan^{-1}{(-1)}$

$\theta = -45\texdegree \,\, or -\dfrac{\pi}{4} radians$

One thing to keep in mind is that the point (4,-4) lies in the 4th quadrant of the xy-plane, and the range given for $\theta$ is $0 \leq \theta \leq 2 \pi$ . So we have to add $2 \pi$ to our answer (we have gone around the circle and stopped at the same place as $-\frac{\pi}{4}$ radians)

$\theta=-\dfrac{\pi}{4} + 2 \pi$ radians

$\theta=\dfrac{7\pi}{4}$ radians

The Answer for part(a) is $(\pm 4\sqrt{2}, \dfrac{7 \pi}{4})$

Part b) (-1, sqrt(3))

For r:

$r = \pm \sqrt{x^2 + y^2}$

$r = \pm \sqrt{(-1)^2 + (\sqrt{3})^2}$

$r = \pm \sqrt{(-1)^2 + (\sqrt{3})^2}$

i) $r = +2$ for $r > 0$

ii) $r = -2$ for $r < 0$

For theta:

$\theta = \tan^{-1}{\left(\dfrac{y}{x}\right)}$

$\theta = \tan^{-1}{\left(\dfrac{\sqrt{3}}{-1}\right)}$

$\theta = -\dfrac{\pi}{3}$ this not in range of theta: $0 \leq \theta \leq 2 \pi$.

and the point $(-1, \sqrt{3})$ lies in the 2nd quadrant, so we can add $\pi$ radians to our answer

$\theta = -\dfrac{\pi}{3} + \pi$ in range: $0 \leq \theta \leq 2 \pi$

$\theta = \dfrac{2 \pi}{3}$

The Answer for part(b) is $(\pm 2, \dfrac{2 \pi}{3})$