Question

polar to cartesian and cartesian to polar. can someone explain this to me? need your help for my incoming exam

Answer 1

Polar -> Cart

x=rcos(theta)
y=rsin(theta)

Cart -> Polar

x^2+y^2=r^2
theta=arctan(y/x)

Answer 2

for example 4x^2 + 9y^2 = 36.. the answer was r^2 = 36/ 4cos^2 theta + 9sin^2 theta. what is the process?

Answer 3

yes yes substituting but i always stuck up..

Answer 4

So you're trying to find tricks that will prove worthy when you come across these type of equations?

Answer 5

i don't know. ive always tried to convert that equation without seeing the real answer but i always end up nothing..

Answer 6

Cartesian from my experience are basically point values in a 3 D graph like xa +yb + zk. that pinpoints where it's at in an environment.

While polar coordinates resemble more for objects that you want to find by circular means like finding a pin point on the surface of the earth if say you were in the middle of the core where the origin of the graph should be. with r = telling you far from the earth's core is the surface point and theta being at which angle from the horizontal axis

Answer 7

yes yes but what about 4x^2 + 9y^2 = 36.. converting to polar

Answer 8

Draw the circle everytime! :)



Then you just have to remember good ol' Pythagoras and trig, since r^2 is just the hypotenuse.

r^2 = x^2 + y^2

Answer 9

In that instance just see that 4x^2 + 9y^2 = 36 fits into

(2x)^2+(3y)^2=6^2

Your r^2 would be the radius of a circle that contains the points cartesian points (X,Y), where X = 2x and Y = 3y

Answer 10

so then

\[\tan \theta = \frac{ X }{ Y } = \frac{ 2x }{ 3y }\]

Answer 11

derp, y/x

Answer 12

\[\theta = \arctan \frac{ Y }{ X } = \arctan \frac{ 3y }{ 2x }\]

and 2x = r sin theta 3y = r sin theta

Answer 13

okay then.. how come the answer is r^2 = 36/ 4cos^2 theta + 9sin^2 theta

Answer 14

is it 36 / (4cos^2theta + 9sin^2theta)? With those parenthesis there?

Answer 15

?

Answer 16

yes

Answer 17

it's because I'm dumb and like circles more than sense :P
With 4x^2 + 9y^2 = 36, like you said we're trying to find r. I thought that we could just make up a new R, but that doesn't help in this case. The only things we know for sure about the equation are our original polar definitions

x = r cos theta
y = r sin theta

So we can plug that into that equation directly

4(r cos theta)^2 + 9(r sin theta)^2 = 36

Then you can pull out the r's from those terms and get

r^2 (4cos^2 theta + 9 sin^2 theta) = 36

and that's where they got that answer.
r^2 = 36/ (4cos^2 theta + 9 sin^2 theta)

Hope I didn't confuse you before :/

Answer 18

okay i didnt pulled out r and everything went wrong. OMG thanks!!!

Answer 19

:) Welcome