# Thread: Rectangular to Polar, Polar to Rectangular

1. ## Rectangular to Polar, Polar to Rectangular

I'm having trouble solving this equation

(3,3) to polar and (8,PI) to rectangular.

I know hoe to change it when its in a equation like 4cis5theata, but i don't know how when its in coordinates.

2. Originally Posted by xian791
I'm having trouble solving this equation

(3,3) to polar and (8,PI) to rectangular.
Um... these aren't actually "equations". Were the instructions really to "solve", or were you maybe supposed to "convert the coordinates"...?

If so, then note that (3, 3) in rectangular means "x = 3 and y = 3". What then is the value of r? (Use the formula they gave you for this.) Since $\tan(\theta)\, =\, \frac{y}{x}\, =\, 1$, what must be the value of $\theta ?$

The polar coordinates $(8,\, \pi)$ means that r = 8 and $\theta\, =\, \pi$. Plug these into the formulas they gave you for x and y.

3. so to double check (3,3)=(3radical2,theta)

and (8,pi)=(-8,0)

Correct?

4. Recall that:

$x = r\ cos\ \theta$

$y = r\ sin\ \theta$

$tan\ \theta = \frac {y}{x}$

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

(3, 3) is (x, y) ---> Cartesian/Rectangular

We need (r, $\theta$) ---> Polar

$r = \sqrt {3^{2} + 3^{2}} = \sqrt {18} = \sqrt {9 \times 2} = 3\sqrt 2.$

$\tan \theta = \frac {y}{x} = \frac {3}{3} = 1.$
So $\theta = \arctan 1 = \frac {\pi}{4}.$

So (3, 3) in Cartesian coordinates is ( $3\sqrt{2}$, $\frac{\pi}{4}$) in polar coordinates.

Converting rectangular to polar is easier:

x = r cos theta = 8 cos pi = -8.
y = r sin theta = 8 sin pi = 0.

So (8, $\pi$) in polar is (-8, 0) in rectangular.

I hope that helps.

ILoveMaths07.