Question

conversion cartesian to polar. help!! x^4 + x^2y^2 - 2x^2 y - y^3.. the answer is r= sin theta (1 + sec^2 theta)..

Answer 1

step by step process..

Answer 2

More of Satan's works.

Answer 3

hey. First plug in those identities, what do you get?

Answer 4

(rcostheta)^4 + (rcostheta)^2 (rsintheta)^2 - 2 (rcostheta)^2 (rsintheta) - (rsintheta)^3..

Answer 5

Yeah just use \(r^2\cos^2\theta +r^2\sin^2\theta = r^2\)

Answer 6

nice. Pull are of the r's to the front of their parenthesis next.

Answer 7

@AllTehMaffs ha? dont understand..

Answer 8

even the r^4?? and r^3?

Answer 9

like, factor out the r's in each grouping. ie,

r^4 (cos^4 theta) +........ -2r^3 (cos^2 theta sin theta) ..... = 0

yeah. Put em out in front of each term to make em easier to see.

Answer 10

\[r^4\cos^4\theta + r^2 (\cos^2\theta \sin^2\theta) - 2r (rcos^2\theta \sin \theta) - r^3 \sin ^3 \theta\]

Answer 11

whats the full equation

Answer 12

x^4 + x^2y^2 - 2x^2 y - y^3 @dan815

Answer 13

x^4 + x^2y^2 - 2x^2 y - y^3 = ?

Answer 14

0

Answer 15

okay

Answer 16

x=rcostheta
y=rsintheta
plug and simplify