# cartesian and polar

• May 3rd 2006, 06:43 AM
simon
cartesian and polar
Hello,

I need help with this question, :confused:

a) express in polar co-cordinates the cartesian co-ordinates (2, -6)

b) express in cartesian co-ordinates the polar co-cordinates (3, 150degrees)

Thank you.
• May 3rd 2006, 08:02 AM
topsquark
Quote:

Originally Posted by simon
Hello,

I need help with this question, :confused:

a) express in polar co-cordinates the cartesian co-ordinates (2, -6)

b) express in cartesian co-ordinates the polar co-cordinates (3, 150degrees)

Thank you.

For these types of problems you need to recall that:
$\displaystyle x = r cos\theta$
$\displaystyle y = r sin\theta$

Inverting these relations gives:
$\displaystyle r = \sqrt{x^2+y^2}$
$\displaystyle \theta = tan^{-1} \left ( \frac{y}{x} \right )$

So for question a)
(2,-6) => x = 2, y = -6
$\displaystyle r = \sqrt{2^2+(-6)^2}=\sqrt{40}=2 \sqrt{10}$
$\displaystyle \theta = tan^{-1} \left ( \frac{-6}{2} \right ) = tan^{-1}(-3) \approx -71.5651^o$

$\displaystyle \theta$ is a reference angle. The point (2,-6) is in the 4th quadrant, so your angle measured from the +x axis is 360 - 71.5651 = 288.435 degrees.

Thus: (2,-6) => (2sqrt(10), 288.435 degrees).

For question b)
(3, 150 degrees) => r = 3, $\displaystyle \theta$ = 150 degrees
$\displaystyle x = 3 cos(150) \approx -2.59808$
$\displaystyle y = 3 sin(150) = 1.5$

Thus: (3, 150 degrees) => (-2.59808, 1.5).
(And note that it is, in fact, in the second quadrant as the original coordinates state.)

-Dan