Question

Find a polar equation for the curve represented by the given Cartesian equation.
x + y = 2

Answer 1

you can replace x with r cost theta, and y with r sin theta

Answer 2

so rcost + rsint = 2

But I'm confused on what to do from then

Answer 3

factor out r

Answer 4

r = 2 / ( cos t + sin t )

Answer 5

Then
r^2(cost+sint)^2 = 4

r^2 (1 + 2sintcost) = 4
?

Answer 6

that wont simplify it

Answer 7

once you have a function r = f(theta), you are done

Answer 8

ohhh I see.

Answer 9

Thank u much!

Answer 10

the other direction is harder, converting polar equation to cartesian

Answer 11

sure :)

Answer 12

here is a fun problem
find the polar equation for
the cartesian equation
y = x

Answer 13

y - x = 0

y^2 + x^2 -2xy = 0
1 -2(rcost * rsint) = 0

-2(rcost * rsint) = -1

2(rcost * rsint) = 1

(rcost * rsint) = 1/2

r(cost*sint) / 1/2
r = 1/2sintcost

r = 1/sin2t
Is that right?

Answer 14

x^2 + y^2 does not equal to 1

Answer 15

oh it's equal to r^2 right. let me try again

Answer 16

wait

Answer 17

why not just substitute y = x , right away

Answer 18

So x^2 + x^2 -2x^2 = 0
wouldn't that just give 0 = 0?

Answer 19

y = x

r sin theta = r cos theta

Answer 20

So sin theta = cos theta

sin t / cos t = 1

tan t = 1

Answer 21

looking better

Answer 22

And that makes sense because there should be no r in the function, since it's just in one direction, which is when tangent is 1 and that's at y =x.

Answer 23

right, you can keep solving

theta = arctan (1) = pi/4

Answer 24

so you have polar coordinates :
( r , pi/4) r can vary

Answer 25

I see. You made this very clear to me. Thanks again :)

Answer 26

:)