Question

Given the equation sqrt(3)/xy = 2/(x^2-y^2) change the equation to polar form and simplify the answer by using fundamental identities. Where do I begin?

Answer 1

Do I change it as:\[\frac{ \sqrt{3} }{ rcos \theta rsin \theta }=\frac{ 2 }{ (rcos \theta)^{2}-(rsin \theta)^{2}}\]

Answer 2

Any luck @rorytsm?

Answer 3

@RadEn, perhaps you may know?

Answer 4

i know you can distribute the exponent on the right side of the equation and then factor out an r^2 or -r^2 and use the identity that cos^2theta+sin^2theta = 1 i'm just not 100% sure about which to factor out to get rid of the - between the two so i decided not to post

Answer 5

you can also factor an r out on the left hand side for a basic simplification, but other than that i'm not really too sure what else you can do

Answer 6

Well you had the same idea as I did. I'm not sure how to change the sqrt(3) and 2 to polar

Answer 7

pretty sure those don't change, in all the problems using spherical and cylindrical coordinates that i've worked so far this semester, we never changed any "normal" numbers over to polar coordinates, they were always just left as is.

Answer 8

unless you want to divide each side by two to get sqrt(3)/2 on the left which is equal to sin(pi/3)

Answer 9

so then you'd have (sin(pi))/3r(costhetasintheta)

Answer 10

I could make the left side:\[\frac{ \sqrt(3) }{ r(\frac{ 1 }{ 2 }\sin2 \theta )}\]correct?

Answer 11

i'm pretty sure you can

Answer 12

So then I can make the left side \[\frac{ 2 }{ r(\cos2 \theta) }\]Correct?

Answer 13

I mean right side

Answer 14

your r would have to be squared, and i really don't think there's many other options other than looking for the cos^2+sin^2 identity there unless you went with two half angle formulas (1-cos2theta)/2 and (1+cos2theta)/2 and then combined those

Answer 15

the cos2theta's would cancel leaving you with 2/2 = 1 but either way you get the same result