Respuesta :
Answer:
[tex](x,y) = (r,\theta)\\[/tex]
i)[tex](4,-4) = (\pm 4\sqrt{2},\dfrac{7 \pi}{4})\\[/tex]
ii) [tex](-1, \sqrt{3}) = (\pm 2,\dfrac{2 \pi}{4})\\[/tex]
Thinking Process:
This can either be solved using complex variables or simply trigonometry (they are all the same in a way)
[tex](a,b) = a + ib = r(\cos{(\theta)} + i\sin{(\theta)}) = r e^{(i \theta)} = r\angle{\theta}[/tex]
But we'll simply go with trigonometry and good old Pythagorean theorem.
Solution:
For Cartesian coordinates are: [tex](x,y)[/tex]
To covert them into [tex]r \angle{\theta}[/tex]
We can use:
- The Pythagoras theorem to find [tex]r[/tex]
[tex]r = \pm \sqrt{x^2 + y^2}[/tex]
- Arctangent to find [tex]\theta[/tex]
[tex]\theta = \tan^{-1}{\left(\dfrac{y}{x}\right)}[/tex]
These are related since:
[tex] x = r\cos{(\theta)}[/tex]
[tex] y = r\sin{(\theta)}[/tex]
So, let's start solving:
Part a) (4,-4)
For r:
[tex]r = \pm \sqrt{x^2 + y^2}[/tex]
[tex]r = \pm \sqrt{4^2 + (-4)^2}[/tex]
[tex]r = \pm \sqrt{32}[/tex]
[tex]r = \pm 4 \sqrt{2}[/tex]
i) for [tex]r > 0[/tex] we'll go with [tex]r = + 4\sqrt{2} [/tex]
ii) for [tex]r < 0[/tex] we'll go with [tex]r = - 4\sqrt{2} [/tex]
For theta:
[tex]\theta = \tan^{-1}{\left(\dfrac{y}{x}\right)}[/tex]
[tex]\theta = \tan^{-1}{\left(\dfrac{-4}{4}\right)}[/tex]
[tex]\theta = \tan^{-1}{(-1)}[/tex]
[tex]\theta = -45\texdegree \,\, or -\dfrac{\pi}{4} radians[/tex]
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 [tex]\theta[/tex] is [tex]0 \leq \theta \leq 2 \pi[/tex] . So we have to add [tex]2 \pi[/tex] to our answer (we have gone around the circle and stopped at the same place as [tex]-\frac{\pi}{4}[/tex] radians)
[tex]\theta=-\dfrac{\pi}{4} + 2 \pi [/tex] radians
[tex]\theta=\dfrac{7\pi}{4} [/tex] radians
The Answer for part(a) is [tex](\pm 4\sqrt{2}, \dfrac{7 \pi}{4})[/tex]
Part b) (-1, sqrt(3))
For r:
[tex]r = \pm \sqrt{x^2 + y^2}[/tex]
[tex]r = \pm \sqrt{(-1)^2 + (\sqrt{3})^2}[/tex]
[tex]r = \pm \sqrt{(-1)^2 + (\sqrt{3})^2}[/tex]
i) [tex]r = +2 [/tex] for [tex]r > 0 [/tex]
ii) [tex]r = -2 [/tex] for [tex]r < 0 [/tex]
For theta:
[tex]\theta = \tan^{-1}{\left(\dfrac{y}{x}\right)}[/tex]
[tex]\theta = \tan^{-1}{\left(\dfrac{\sqrt{3}}{-1}\right)}[/tex]
[tex]\theta = -\dfrac{\pi}{3}[/tex] this not in range of theta: [tex]0 \leq \theta \leq 2 \pi[/tex].
and the point [tex](-1, \sqrt{3})[/tex] lies in the 2nd quadrant, so we can add [tex]\pi[/tex] radians to our answer
[tex]\theta = -\dfrac{\pi}{3} + \pi[/tex] in range: [tex]0 \leq \theta \leq 2 \pi[/tex]
[tex]\theta = \dfrac{2 \pi}{3} [/tex]
The Answer for part(b) is [tex](\pm 2, \dfrac{2 \pi}{3})[/tex]
