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.

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- Apr 5th 2009, 03:17 PMxian791Rectangular 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. - Apr 5th 2009, 03:44 PMstapel
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 $\displaystyle \tan(\theta)\, =\, \frac{y}{x}\, =\, 1$, what must be the value of $\displaystyle \theta ?$

The polar coordinates $\displaystyle (8,\, \pi)$ means that r = 8 and $\displaystyle \theta\, =\, \pi$. Plug these into the formulas they gave you for x and y. (Wink) - Apr 5th 2009, 03:51 PMxian791
so to double check (3,3)=(3radical2,theta)

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

Correct? - Apr 6th 2009, 01:33 PMILoveMaths07
Recall that:

$\displaystyle x = r\ cos\ \theta$

$\displaystyle y = r\ sin\ \theta $

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

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

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

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

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

$\displaystyle \tan \theta = \frac {y}{x} = \frac {3}{3} = 1. $

So $\displaystyle \theta = \arctan 1 = \frac {\pi}{4}.$

So (3, 3) in Cartesian coordinates is ($\displaystyle 3\sqrt{2}$, $\displaystyle \frac{\pi}{4}$) 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, $\displaystyle \pi$) in polar is (-8, 0) in rectangular.

I hope that helps. :)

ILoveMaths07.